Split (1)

train · 6.32M rows

url stringlengths 14 2.42k	text stringlengths 100 1.02M	date stringlengths 19 19	metadata stringlengths 1.06k 1.1k
https://data.mendeley.com/research-data/?page=6&type=TABULAR_DATA&type=DOCUMENT&type=OTHER&type=TEXT&search=*	Filter Results 12021983 results A collection of case history data collected from various sources of rock buckling (pop ups) at the ground surface in North America. Data Types: • Tabular Data • Dataset This file contains a prototype semi-structured interview questionnaire that was used for cross country synthesis on the subject of post-disaster reconstruction and resettlement Data Types: • Dataset • Document Dataset for measurements of vessel density and vessel length density Data Types: • Tabular Data • Dataset As a part of a research into new techniques for purifying recycled aluminium alloys, a novel electromagnetic apparatus had been developed for investigating in real-time the separation mechanisms of detrimental inclusions in aluminium alloy melts under alternating magnetic fields. The magnetic coil was designed based on the Helmholtz coil design. A viewing gap was designed for in-situ imaging studies using synchrotron X-rays. The gap was designed to maintain a uniform magnetic field in the central region where a sample is positioned. The current setup for the magnetic coil pair is able to produce a peak magnetic flux density of ~10 mT at a frequency of 25 kHz. A separate electrical resistance furnace, designed to concentrically fit within the magnetic coils, was used to control the heating (up to ~850°C) and cooling of the samples. After a series of systematic tests and commissioning, the apparatus was used in a number of in-situ and ex-situ experiments. Data Types: • Tabular Data • Dataset • Document This dataset contains the primary data used in: "Multi-epoch X-ray burst modelling: MCMC with large grids of 1D simulations", Johnston et al. (2020). In this work, we interpolated and sampled a grid of 3840 KEPLER burst models using Markov Chain Monte Carlo (MCMC) methods to produce posterior distributions for the system parameters of the "Clocked Burster", GS 1826-238. Provided here is the full burst model grid and the raw MCMC sample chains. More details on the files, and how to load them, are provided in README.md Data Types: • Software/Code • Dataset • Text • File Set Supplementary data supporting the manuscript updated with 4 donor results to replace the previous dataset from one donor Data Types: • Dataset • Document In this repository we make available two datasets of research on soccer analysis data. - baseNormalizada - is a dataset that joins real data of soccer players transfers with athletes abilities gathered from the eletronic soccer game FIFA. - Network_dataset - is a dataset that gatheres all major soccer transfers that occured on the soccer world between 1990 - 2017. For both datasets we make available the crawler for future updates. Data Types: • Dataset • Text Supplementary material Supplementary material Table 1, Taxonomic information of total radiolarian taxa encountered in study. Supplementary material Table 2, Basic environmental informations with annual, summer, and winter timescales of the 45 surface sediment samples. Supplementary material Table 3, Raw counting data of the total radiolarian assemblages in the 45 surface sediment samples. Data Types: • Tabular Data • Dataset Nowadays, social media sites have become a vital part of humans’ lives, allowing peoples to express their opinions, their sentiments and feelings about whatever, whenever and wherever. Therefore, it comes a huge amount of data in which attract researchers to explore and analyse these sites to address research questions in various fields. This data article is about 21422 public tweets related to Algerian universities generated by 9477 users, made during the nine years—January 2011 to March 2020. The retrieved tweets were collected based on specific queries using a python 3 library called GetOldTweets3. Each tweet is represented by 11 variables: ID, username, text, date, number of retweets, number of favorites, mentions, hashtags, and permalink. Data Types: • Tabular Data • Dataset Dataset for the paper entitled "Real-time command strategies for microgrids based on the Contract Collaboration Problem". You are free to use the instances in the zip file, if you cite the following work: @article{Levorato2020, author = {Levorato, M. and Figueiredo, R. and Frota, Y. and Jouglet, A. and Savourey, D.}, journal = {To be published}, title = {{Real-time command strategies for microgrids based on the Contract Collaboration Problem}}, year = {2020} } Instances description ================================= 1) full_intances: the 3 microgrid instances used in the main experiments: - A_instance2_1NDU_Cons : the Cons microgrid instance, with only the consumer uncertain system; - A_instance2_1NDU_Prod : the Prod microgrid instance, with only the producer uncertain system; - A_instance2 : the Prod\&Cons microgrid instance, with both uncertain systems (producer and consumer). For each instance, there are 3 files, each one for a group of randomly generated scenarios: - _1000s_skewed-left.txt : left-skewed beta distribution with parameters $\alpha=5,\beta=2$; - _1000s_skewed-right.txt : right-skewed beta distribution with parameters $\alpha=2,\beta=5$; - _1000s_uniform.txt: uniform distribution. The randomly-generated scenario values are located in the end of each file. 2) toy_sensitivity: reduces microgrid instances used in the sensitivity tests. Each file corresponds to a specific combination of model cost parameters, and the filename follows this mask: OC_Ct__DS_ST_NDU__.txt The parameters OC_cost, DS_cost, ST_cost follow the values listed in Table VI of the aforementioned paper. (caption: Energy cost parameters in sensitivity analysis). can be one of the 3 distributions used in the work (skewed-left, skewed-right or uniform). The used was 'default'. The used was 'default'. Again, the randomly-generated scenario values are located in the end of each file. Data Types: • Dataset • Text • File Set 6	2020-04-03 18:26:26	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5376152396202087, "perplexity": 5641.019264944023}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-16/segments/1585370515113.54/warc/CC-MAIN-20200403154746-20200403184746-00340.warc.gz"}
http://www.sawaal.com/physics-questions-and-answers/a-nuclear-reactor-is-said-to-have-become-critical-when-nbsp_8642	1 Q: # A nuclear reactor is said to have become 'critical' when A) it stops due to malfunctioning B) it starts emitting dangerous radioactive radiations C) it is shut down to avoid explosion D) it is ready to produce controlled energy Answer: D) it is ready to produce controlled energy Explanation: To produce controlled energy i.e energy at a controlled or steady rate, the number of fissions per second must be controlled and maintained at that rate. Excess neutrons should be absorbed by control roads like cadmium. Subject: Physics Q: What is the full form of "RADAR" ? A) Radio Detecting and Ranging B) Region Device and Ranging C) Radio Detect and Rangs D) Radio Device and Ranging Explanation: Filed Under: Physics Exam Prep: AIEEE , Bank Exams , CAT Job Role: Bank Clerk , Bank PO 9 787 Q: If a carrier wave of 1000 kHz is used to carry the signal, the length of transmitting antenna will be equal to _____ ? A) 300 m B) 30 m C) 3 m D) 0.3 m Answer & Explanation Answer: A) 300 m Explanation: h = = 3 × 108/106 = 300 m Filed Under: Physics Exam Prep: AIEEE , Bank Exams , CAT , GATE Job Role: Analyst , Bank Clerk 9 635 Q: Quality of transmission depends upon A) Nature of signal only B) Nature of medium only C) Both (A) and (B) D) Neither A nor B Answer & Explanation Answer: C) Both (A) and (B) Explanation: Quality of transmission is governed both by nature of signal and nature of communication channel/medium. Filed Under: Physics Exam Prep: AIEEE , Bank Exams , CAT Job Role: Bank Clerk , Bank PO 6 592 Q: The rate of change of linear momentum of a body falling freely under gravity is equal to it's ____ ? A) Kinetic Energy B) Weight C) Potential Energy D) Impulse Explanation: Rate of change of impulse equals the force . In case of freely falling body the only force is the weight. Filed Under: Physics Exam Prep: AIEEE , Bank Exams Job Role: Bank Clerk 8 728 Q: A disk and a sphere of same radius but different masses roll on two inclined planes of the same altitude and length. Which one of the two objects gets to the bottom of the plane first ? A) Depends on their masses B) Sphere C) Disk D) Both reach at the same time Explanation: Time does not depend on mass, else $\inline \fn_jvn T \prec \sqrt{\left ( 1 +\frac{k^{2}}{R^{2}} \right )}$ $\inline \fn_jvn \frac{k^{2}}{R^{2}}$ => is least for 'sphere' and hence least time is taken by sphere.	2017-09-22 13:32:37	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 2, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6435139775276184, "perplexity": 5875.569181380627}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-39/segments/1505818688966.39/warc/CC-MAIN-20170922130934-20170922150934-00344.warc.gz"}
http://www.physicsforums.com/showthread.php?t=167741	# simple partial fractions help (warning complex analysis :P ) by trickae Tags: analysis, complex, fractions, partial, simple, warning P: 85 1. The problem statement, all variables and given/known data the question can be ignored - it involves laplace and Z transforms of RLC ckts. Vc(s) = 0.2 ----------------- s^2 + 0.2s + 1 find the partial fraction equivalent such that it is : -j(0.1005) + j (0.1005) -------------- ------------------ s + 0.1-(0.995) s + 0.1 + j(0.995) 2. Relevant equations none 3. The attempt at a solution 0.2 A B --------------- = --------------------- + ------------------- s^2 + 0.25 + 1 s + (0.1 - j(0.995))) s + (0.1 + j(0.995)) 0.2 = A(s + 0.1 + j(0.995)) + B(s + (0.1 - j0.995)) 0.2 = As + A(0.1 + j(0.995)) + Bs + B(0.1 - j0.995) so As + Bs = 0 or (A + B) = 0 or A = -B so 0.2 = j(0.995A) - j(0.995B) somethings not right - if i evaluate this I don't get anywhere near the answer P: 1,235 Well, check this: (s + 0.1 - I 0.995)(s + 0.1 + I 0.995) = 1.00002 + 0.2 s + s² there is a small inaccuracy ... but nothing more. Solve your last equation and turn to another exercice. What does that become: 0.2 = j(0.995A) - j(0.995B) if A=-B ? P: 85 sorry dude i'm still a little confused. thats just a rounding error from the J term. P: 1,235 ## simple partial fractions help (warning complex analysis :P ) I just wanted to say that, except for decimals, you did it correctly. You should solve the last equation and conclude. P: 85 thanks man i got it now - actually my next question is a little off topic ... but how did i get a post stuck in this thread? http://physicsforums.com/showthread.php?t=166823&page=2 I don't understand the thread whatsover. ANyways thanks for the help i actually got the answer i'l put it up when i get home. Related Discussions Calculus & Beyond Homework 2 Calculus 1 General Math 17 Calculus 1 General Math 3	2014-04-23 12:20:36	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5458868741989136, "perplexity": 2271.8853260591995}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-15/segments/1398223202548.14/warc/CC-MAIN-20140423032002-00438-ip-10-147-4-33.ec2.internal.warc.gz"}
http://openstudy.com/updates/560779afe4b06b2f4218f29f	## Mutie one year ago I really need help, I keep working out this problem and I can't do it. Help!! The fuel efficiency of a car is determined by its miles per gallon (mpg) number. This number estimates how many miles a car can run with a gallon of gas. Four cars’ mpg numbers are given in different representations. (See the snapshots below.) 5. (20 points) List the cars in order from the most fuel efficient to the least fuel efficient. 6. (20 points) Explain and support your answer with your work. 1. Mutie @misty1212 2. anonymous your last two links have 404'd (not working) 3. UnkleRhaukus 4. UnkleRhaukus You need to determine how many miles each vehicle can travel, on one gallon of fuel 5. Mutie BMW: 22 gals every 1 mi Toyota: ??? Acura: 1 gal is 25 mi Jeep: 1 gal 16mi Is this right? I have no idea about the Toyota though. @AlexADB @UnkleRhaukus 6. anonymous You just list them in descending order. From the most fuel efficient, with the highest mpg, to the least fuel efficient, with the lowest mpg. 7. Mutie Yeah but I have no idea what the Toyota number is. 8. Mutie @awn786 9. anonymous $y = 28x$ X is the number of gallons and y is the number of miles. So, this tells us that the number of miles is 28 times the number of gallons. So, give one of them any value. We want 1 gallon so let x = 1. $y = 28 \times 1$ $y=28$ So, the amount of gallons that allow the Toyota to drive 28 miles, is 1 gallon. The mpg of the toyota is 28 mpg. 10. Mutie So it would be Toyota: 1 gal is 28 miles? Then with the rest of my answers, I'd put them Greatest to Least? The miles? (mpg) 11. anonymous Precisely. 12. Mutie Are all of my other answers correct? BMW: 22 gal is 1 mi Toyota: 1 gal is 28 mi Acura: 1 gal is 25 mi Jeep: 1 gal 16mi I feel like I'm incorrect for the BMW? Or do I have the numbers in the wrong place? @awn786 13. anonymous Write it as 22mpg instead of 1mi is 22gal. 1) Toyota: 28 mpg 2) Acura: 25 mpg 3) BMW: 22 mpg 4) Jeep: 16 mpg 14. Mutie Thank you so much, can you help me check over my work on this other problem? I'll close this thread. @awn786 15. anonymous Sure 16. UnkleRhaukus good work!	2016-10-27 03:17:45	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5774321556091309, "perplexity": 2961.9628624453444}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-44/segments/1476988721067.83/warc/CC-MAIN-20161020183841-00093-ip-10-171-6-4.ec2.internal.warc.gz"}
http://mathoverflow.net/revisions/91124/list	2 typo; added 10 characters in body There is of course a whole theory behind, and the right pointer is the Sturm-Liouville problem as indicated by Deane YoungYang. However, just the matter of proving the uniqueness of solutions of to your equation, can be established quickly under suitable hypotheses. To start with, assume $f:\mathbb{R}^n\to \mathbb{R}^n$ is a continuous and monotone map, that is $$\big(f(x)-f(y)\big)\cdot(x-y)\ge0\ ,$$ for all $x$ and $y$ in $\mathbb{R}^n$. Then, if $u$ and $v$ solve your equation on some interval $[a,b]$ with the same boundary conditions we have, integrating by parts $$\int_a^b\|\dot u-\dot v\|^2 dt = - \int_a^b \big(f(u)-f(v)\big)\cdot(u-v)\le0\ big(f(u)-f(v)\big)\cdot(u-v)\ dt\le0\ ,$$ implying $u-v$ is constant, hence $u=v\ .$ Also, we may gain something exploiting the fact that the interval is given. Assume that $f$ is continuous and $f+cI$ is monotone, for some $c < \pi^2$. So now we just have $$\int_0^1 \|\dot u-\dot v\|^2 dt= -\int_a^b \big(f(u)-f(v)\big)\cdot(u-v)\le big(f(u)-f(v)\big)\cdot(u-v)\ dt\le c \int_0^1 \|u-v\|^2dt\ .$$ By the Poincaré inequality, since each component of $u-v$ is in $H^1_0([0,1])$ we also have $$\pi^2 \int_0^1 \|u-v\|^2dt \le \int_0^1 \|\dot u-\dot v\|^2 dt\ ,$$ and we conclude $u=v$ as before. 1 There is of course a whole theory behind, and the right pointer is the Sturm-Liouville problem as indicated by Deane Young. However, just the matter of proving the uniqueness of solutions of your equation can be established quickly under suitable hypotheses. To start with, assume $f:\mathbb{R}^n\to \mathbb{R}^n$ is a continuous and monotone map, that is $$\big(f(x)-f(y)\big)\cdot(x-y)\ge0\ ,$$ for all $x$ and $y$ in $\mathbb{R}^n$. Then, if $u$ and $v$ solve your equation on some interval $[a,b]$ with the same boundary conditions we have, integrating by parts $$\int_a^b\|\dot u-\dot v\|^2 dt = - \int_a^b \big(f(u)-f(v)\big)\cdot(u-v)\le0\ ,$$ implying $u-v$ is constant, hence $u=v\ .$ Also, we may gain something exploiting the fact that the interval is given. Assume that $f$ is continuous and $f+cI$ is monotone, for some $c < \pi^2$. So now we just have $$\int_0^1 \|\dot u-\dot v\|^2 dt= -\int_a^b \big(f(u)-f(v)\big)\cdot(u-v)\le c \int_0^1 \|u-v\|^2dt\ .$$ By the Poincaré inequality, since each component of $u-v$ is in $H^1_0([0,1])$ we also have $$\pi^2 \int_0^1 \|u-v\|^2dt \le \int_0^1 \|\dot u-\dot v\|^2 dt\ ,$$ and we conclude $u=v$ as before.	2013-05-19 10:36:14	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9497752785682678, "perplexity": 145.34462219182137}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2013-20/segments/1368697380733/warc/CC-MAIN-20130516094300-00086-ip-10-60-113-184.ec2.internal.warc.gz"}
http://physics.stackexchange.com/questions?page=308&sort=active	# All Questions 61 views ### How to solve the Poisson equation? Does anyone know how the author in http://gravitationallensing.pbworks.com/w/page/15553257/Strong%20Lensing found the equation $$\kappa(\theta)=\frac{\kappa_{0}}{2\theta}$$ after solving the ... 49 views 2k views ### Impossibility of time travel due to energy conservation? I just watched the movie Terminator Genisys (It wasn't as bad as people say) and started pondering time travel. While pontificating and stretching my limited remembrance of AP Physics, science ... 82 views ### Is special relativity true? [closed] Is the special theory of relativity true? http://www.npr.org/2014/03/20/291408248/einsteins-lost-theory-discovered-and-its-wrong 18 views ### Are Fe56 or Ni56 the fission products of any binary reactions? I'm curious as to if there is some combination of a fusion and fission event simultaneously occuring that would only produce 56 nucleon number nuclides. Such that the net energy out of the fusion ... 588 views ### Why does objects with zero acceleraton move? My question is, if we apply a force on an object to move it with a constant velocity, of course it will move. But what I don't understand is, why will it move? From Newton's law F=ma and in this case ... 47 views ### What does gravitational mass mean if (weak) equivalence principle invalid? [closed] As equivalence principle states Mass and weight are locally in identical ratio for all bodies. --Newton This implies that the gravitational mass does only depends on its inertial mass. Suppose ... 67 views ### When do we use LQR controller and LQG controller? I have to conceive a controller to my mechanical system to delete the problem of vibrations, but I don't know when we use LQR controller and LQG controller. 28 views ### Can I disperse a droplet with electromagnetic fields? If I have a droplet of water, can I apply an electromagnetic field in order to disperse the droplet into smaller subdroplets (tiny vapor) and fill a space with them? The same way that magnetic dust ... 4k views ### Does there exist a free good molecule / atom simulation software? I'm looking for a software or software package (for example C/C++) that can simulate a lot (say thousands at least) of molecules in action (ie. in movement or attached to say static walls). I have ... 119 views ### What really is the significance of the resonant frequency in terms of “ease of vibration”? I was studying the concept of resonant frequency and I've read quite a few articles and notes on it. What I have understood from what I have read is that the resonance frequency of an object is its ... 24 views ### How does permanent magnets attract each other if the $B$-field can do no work? [duplicate] How can two permanent magnets do work on each other? If you put two magnets with opposite poles facing each other they will attract each other. If you put them with the same poles facing each other ... 110 views ### The Simulation Hypothesis As A Theory Of Everything [closed] I have long found the Simulation Hypothesis to be better-able to answer questions that I have regarding the universe than the string theory and the quantum field theory. What I believe is that the ... 93 views ### What is the total velocity/force on a particle in fluid? I am now facing a fluid-particle interaction problem. I would like to simulate particle motion in a fluid. I do know the external force acting on a particle (dielectrophoretic force in this case) ...	2016-02-09 08:06:30	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.763994038105011, "perplexity": 799.9153160608995}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-07/segments/1454701156627.12/warc/CC-MAIN-20160205193916-00277-ip-10-236-182-209.ec2.internal.warc.gz"}
https://www.zbmath.org/?q=an%3A0824.05023	# zbMATH — the first resource for mathematics A note on the three color problem. (English) Zbl 0824.05023 In 1990 Paul Erdős asked: Is there an integer $$k\geq 5$$ such that if $$G$$ is a planar graph without $$i$$-circuits, $$4\leq i\leq k$$, then $$G$$ is 3- colorable? A year later H. L. Abbott and B. Zhou answered this question affirmatively by proving the above with $$k= 11$$. This note strengthens their result by showing that $$G$$ is 3-colorable, if it is planar without $$i$$-circuits for $$i$$ from 4 through 9. Reviewer: M.Kubale (Gdańsk) ##### MSC: 05C15 Coloring of graphs and hypergraphs 05C10 Planar graphs; geometric and topological aspects of graph theory 05C38 Paths and cycles ##### Keywords: three color problem; circuit; coloring; face; critical graph; planar graph Full Text: ##### References: [1] Abbott, H.L., Zhou, B.: On small faces in 4-critical planar graphs. Ars Combin.32, 203–207 (1991) · Zbl 0755.05062 [2] Aksionov, V.A., Melnikov, L.S.: Essay on the theme: The three color problem. Combinatorics, Colloq. Math. Soc. János Bolyai18, 23–34 (1976) [3] Aksionov, V.A., Melnikov, L.S.: Some counterexamples associated with the three color problem. J. Comb. Theory ser.B28, 1–9 (1980) · Zbl 0434.05033 · doi:10.1016/0095-8956(80)90051-9 [4] Erdös, P.: informal discussion during the conference ”Quo Vadis, Graph Theory?” University of Alaska, Fairbanks, Alaska, August (1990) [5] Steinberg, R.: The state of the three color problem. Ann. Discrete Math.55, 211–248 (1993) · Zbl 0791.05044 · doi:10.1016/S0167-5060(08)70391-1 This reference list is based on information provided by the publisher or from digital mathematics libraries. Its items are heuristically matched to zbMATH identifiers and may contain data conversion errors. It attempts to reflect the references listed in the original paper as accurately as possible without claiming the completeness or perfect precision of the matching.	2021-05-12 14:43:49	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5031599998474121, "perplexity": 4697.112832283465}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-21/segments/1620243990929.24/warc/CC-MAIN-20210512131604-20210512161604-00517.warc.gz"}
https://eprint.iacr.org/2013/160	## Cryptology ePrint Archive: Report 2013/160 Interactive Coding, Revisited Kai-Min Chung and Rafael Pass and Sidharth Telang Abstract: How can we encode a communication protocol between two parties to become resilient to adversarial errors on the communication channel? This question dates back to the seminal works of Shannon and Hamming from the 1940's, initiating the study of error-correcting codes (ECC). But, even if we encode each message in the communication protocol with a good'' ECC, the error rate of the encoded protocol becomes poor (namely $O(1/m)$ where $m$ is the number of communication rounds). Towards addressing this issue, Schulman (FOCS'92, STOC'93) introduced the notion of \emph{interactive coding}. We argue that whereas the method of separately encoding each message with an ECC ensures that the encoded protocol carries the same amount of information as the original protocol, this may no longer be the case if using interactive coding. In particular, the encoded protocol may completely leak a player's private input, even if it would remain secret in the original protocol. Towards addressing this problem, we introduce the notion of \emph{knowledge-preserving interactive coding}, where the interactive coding protocol is required to preserve the knowledge'' transmitted in the original protocol. Our main results are as follows. \begin{itemize} \item The method of separately applying ECCs to each message is essentially optimal: No knowledge-preserving interactive coding scheme can have an error rate of $1/m$, where $m$ is the number of rounds in the original protocol. \item If restricting to computationally-bounded (polynomial-time) adversaries, then assuming the existence of one-way functions (resp. subexponentially-hard one-way functions), for every $\epsilon>0$, there exists a knowledge-preserving interactive coding schemes with constant error rate and information rate $n^{-\epsilon}$ (resp. $1/\polylog(n)$) where $n$ is the security parameter; additionally to achieve an error of even $1/m$ requires the existence of one-way functions. \item Finally, even if we restrict to computationally-bounded adversaries, knowledge-preserving interactive coding schemes with constant error rate can have an information rate of at most $o(1/\log n)$. This results applies even to \emph{non-constructive} interactive coding schemes. \end{itemize} Category / Keywords: foundations / interactive coding, knowledge preserving, simulation paradigm, error correcting codes,	2020-09-30 18:21:27	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6938848495483398, "perplexity": 1017.3696584770543}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-40/segments/1600402127397.84/warc/CC-MAIN-20200930172714-20200930202714-00024.warc.gz"}
https://www.scienceforums.net/topic/12050-series-n-sequences/?tab=comments	# Series n sequences ## Recommended Posts Ok, i got a couple of probs. how do i prove if a series is divergent or convergent? and what is the nth term test? can someone please explain it to me? ##### Share on other sites I don't know what the n'th term test is, but I suspect it's got something to do with the ratio lemma. To show a series is convergent, you need to show that the series of partial sums $s_n = \sum_{k=1}^n a_k$ is convergent. There are a series of results that can help you with this, though (i.e. comparison test, ratio lemma, etc). ##### Share on other sites Could you suggest any good sites or books? Cheers ##### Share on other sites The one book I would suggest is Guide to Analysis, M. Hart. It's rather good and has lots of examples to get you started. ## Create an account Register a new account	2020-08-15 17:14:39	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.4147753119468689, "perplexity": 495.317600116102}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-34/segments/1596439740929.65/warc/CC-MAIN-20200815154632-20200815184632-00197.warc.gz"}
http://www.mathworks.com/help/comm/ref/comm.pnsequence-system-object.html?requestedDomain=true&nocookie=true	# Documentation ### This is machine translation Translated by Mouseover text to see original. Click the button below to return to the English version of the page. # comm.PNSequence System object Generate a pseudo-noise (PN) sequence ## Description The `PNSequence` object generates a sequence of pseudorandom binary numbers using a linear-feedback shift register (LFSR). This block implements LFSR using a simple shift register generator (SSRG, or Fibonacci) configuration. You can use a pseudonoise sequence in a pseudorandom scrambler and descrambler. You can also use one in a direct-sequence spread-spectrum system. To generate a PN sequence: 1. Define and set up your PN sequence object. See Construction. 2. Call `step` to generate a PN sequence according to the properties of `comm.PNSequence`. The behavior of `step` is specific to each object in the toolbox. ### Note Starting in R2016b, instead of using the `step` method to perform the operation defined by the System object™, you can call the object with arguments, as if it were a function. For example, ```y = step(obj)``` and `y = obj()` perform equivalent operations. ## Construction `H = comm.PNSequence` creates a pseudo-noise (PN) sequence generator System object, `H`. This object generates a sequence of pseudorandom binary numbers using a linear-feedback shift register (LFSR). `H = comm.PNSequence(Name,Value)` creates a PN sequence generator object, `H`, with each specified property set to the specified value. You can specify additional name-value pair arguments in any order as (`Name1`,`Value1`,...,`NameN`,`ValueN`). ## Properties `Polynomial` Generator polynomial Specify the polynomial that determines the shift register's feedback connections. The default is `'z^6 + z + 1'`. You can specify the generator polynomial as a character vector or as a numeric, binary vector that lists the coefficients of the polynomial in descending order of powers. The first and last elements must equal `1`, and the length of this vector must be n+1. The value n indicates the degree of the generator polynomial. Lastly, you can specify the generator polynomial as a numeric vector containing the exponents of z for the nonzero terms of the polynomial in descending order of powers. The last entry must be `0`. For example, `[1 0 0 0 0 0 1 0 1]` and ```[8 2 0]``` represent the same polynomial, $g\left(z\right)={z}^{8}+{z}^{2}+1$. The PN sequence has a period of $N={2}^{n}-1$ (applies only to maximal length sequences). `InitialConditionsSource` Source of initial conditions Specify the source of the initial conditions that determines the start of the PN sequence as one of `Property` \| `Input port`. The default is `Property`. When you set this property to `Property`, the initial conditions can be specified as a scalar or binary vector using the `InitialConditions` property. When you set this property to ```Input port```, you specify the initial conditions as an input to the `step` method. The object accepts a binary scalar or a binary vector input. The length of the input must equal the degree of the generator polynomial that the `Polynomial` property specifies. `InitialConditions` Initial conditions of shift register Specify the initial values of the shift register as a binary, numeric scalar or a binary, numeric vector. The default is ```[0 0 0 0 0 1]```. Set the vector length equal to the degree of the generator polynomial. If you set this property to a vector, each element of the vector corresponds to the initial value of the corresponding cell in the shift register. If you set this property to a scalar, the initial conditions of all the cells of the shift register are the specified scalar value. The scalar, or at least one element of the specified vector, must be nonzero for the object to generate a nonzero sequence. `MaskSource` Source of mask to shift PN sequence Specify the source of the mask that determines the shift of the PN sequence as one of `Property` \| `Input port`. The default is `Property`. When you set this property to `Property`, the mask can be specified as a scalar or binary vector using the `Mask` property. When you set this property to ```Input port```, the mask, which is an input to the `step` method, can only be specified as a binary vector. This vector must have a length equal to the degree of the generator polynomial specified in the `Polynomial` property. `Mask` Mask to shift PN sequence Specify the mask that determines how the PN sequence is shifted from its starting point as a numeric, integer scalar or as a binary vector. The default is `0`. When you set this property to an integer scalar, the value is the length of the shift. A scalar shift can be positive or negative. When the PN sequence has a period of $N={2}^{n}-1$, where n is the degree of the generator polynomial that you specify in the `Polynomial` property, the object wraps shift values that are negative or greater than N. When you set this property to a binary vector, its length must equal the degree of the generator polynomial specified in the `Polynomial` property. The mask vector that represents $m\left(z\right)={z}^{D}$ modulo g(z), where g(z) is the generator polynomial, and the mask vector corresponds to a shift of D. For example, for a generator polynomial of degree of `4`, the mask vector corresponding to D = `2` is ```[0 1 0 0]```, which represents the polynomial $m\left(z\right)={z}^{2}$. You can calculate the mask vector using the `shift2mask` function. This property applies when you set the `MaskSource` property to `Property`. `VariableSizeOutput` Enable variable-size outputs Set this property to true to enable an additional input to the step method. The default is false. When you set this property to true, the enabled input specifies the output size of the PN sequence used for the step. The input value must be less than or equal to the value of the `MaximumOutputSize` property. When you set this property to false, the `SamplesPerFrame` property specifies the number of output samples. `MaximumOutputSize` Maximum output size Specify the maximum output size of the PN sequence as a positive integer 2-element row vector. The second element of the vector must be `1`. The default is [10 1]. This property applies when you set the `VariableSizeOutput` property to true. `SamplesPerFrame` Number of outputs per frame Specify the number of PN sequence samples that the step method outputs as a numeric, positive, integer scalar value. The default is `1`. If you set this property to a value of M, then the `step` method outputs M samples of a PN sequence that has a period of $N={2}^{n}-1$. The value n represents the degree of the generator polynomial that you specify in the `Polynomial` property. If you set the `BitPackedOutput` property to `false`, the samples are bits from the PN sequence. If you set the `BitPackedOutput` property to `true`, then the output corresponds to `SamplesPerFrame` groups of bit-packed samples. `ResetInputPort` Enable generator reset input Set this property to `true` to enable an additional input to the `step` method. The default is `false`. This input resets the states of the PN sequence generator to the initial conditions specified in the `InitialConditions` property. `BitPackedOutput` Output integer representations of bit-packed words Set this property to `true` to enable bit-packed outputs. The default is `false`. In this case, the `step` method outputs a column vector of length M, which contains integer representations of bit words of length P. M is the number of samples per frame specified in the `SamplesPerFrame` property. P is the size of the bit-packed words specified in the `NumPackedBits` property. The first bit from the left in the bit-packed word is considered the most significant bit. `NumPackedBits` Number of bits per bit-packed word Specify the number of bits to pack into each output data word as a numeric, integer scalar value between `1` and `32`. The default is `8`. This property applies when you set the `BitPackedOutput` property to `true`. `SignedOutput` Output signed bit-packed words Set this property to true to obtain signed, bit-packed, output words. The default is `false`. In this case, a `1` in the most significant bit (sign bit) indicates a negative value. The property indicates negative numbers in a two's complement format. This property applies when you set the `BitPackedOutput` property to `true`. `OutputDataType` Data type of output Specify the output data type as one of `double` \| `logical` \| `Smallest unsigned integer` when the `BitPackedOutput` property is `false`. The default is `double`. Specify the output data type as `double` \| `Smallest unsigned integer` when the `BitPackedOutput` property is set to `true`. You must have a Fixed-Point Designer™ user license to use this property in `Smallest unsigned integer` mode. ## Methods reset Reset states of PN sequence generator object step Generate a pseudo-noise (PN) sequence Common to All System Objects `clone` Create System object with same property values `getNumInputs` Expected number of inputs to a System object `getNumOutputs` Expected number of outputs of a System object `isLocked` Check locked states of a System object (logical) `release` Allow System object property value changes ## Examples expand all Generate a 14-sample frame of a maximal length PN sequence given generator polynomial, . Generate PN sequence data by using the `comm.PNSequence` object. The sequence repeats itself as it contains 14 samples while the maximal sequence length is only 7 samples (). ```pnSequence = comm.PNSequence('Polynomial',[3 2 0], ... 'SamplesPerFrame',14,'InitialConditions',[0 0 1]); x1 = pnSequence(); [x1(1:7) x1(8:14)]``` ```ans = 1 1 0 0 0 0 1 1 1 1 1 1 0 0 ``` Create another maximal length sequence based on the generator polynomial, . As it is a fourth order polynomial, the sequence repeats itself after 15 samples (). ```pnSequence2 = comm.PNSequence('Polynomial','x^4+x+1', ... 'InitialConditions',[0 0 0 1],'SamplesPerFrame',30); x2 = pnSequence2(); [x2(1:15) x2(16:30)]``` ```ans = 1 1 0 0 0 0 0 0 1 1 0 0 0 0 1 1 1 1 0 0 ``` ## Algorithms This object implements the algorithm, inputs, and outputs described on the PN Sequence Generator block reference page. The object properties correspond to the block parameters, except: • The object does not have a property to select frame based outputs. • The object does not have a property that corresponds to the Sample time parameter.	2018-02-25 22:00:31	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 6, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6609588265419006, "perplexity": 714.620142643293}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-09/segments/1518891817437.99/warc/CC-MAIN-20180225205820-20180225225820-00085.warc.gz"}
https://homework.cpm.org/category/CON_FOUND/textbook/mc1/chapter/1/lesson/1.3.2/problem/1-112	### Home > MC1 > Chapter 1 > Lesson 1.3.2 > Problem1-112 1-112. Mark's scores on his first nine assignments are: $10,10,9,9,10,8,9,10,$ and $8$. 1. What is his mean (average) score so far? Remember that the mean is the number which represents what every part would be if the total were divided evenly among the parts. Add each part together to find the total.$10+10+9+9+10+8+9+10+8=83$ Divide the total by the number of parts. $\frac{83}{9}\approx\;9.2$ About $9.2$ 2. Mark did not do the tenth assignment, so he got a zero on it. What is his new mean? This problem is very similar to part (a). However, you will need to add another test score ($0$). Remember that this also means there is an additional part by which to divide.	2021-03-01 07:43:50	{"extraction_info": {"found_math": true, "script_math_tex": 6, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8760853409767151, "perplexity": 996.1915254500219}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-10/segments/1614178362133.53/warc/CC-MAIN-20210301060310-20210301090310-00371.warc.gz"}
https://brokerehlojwno.netlify.app/fugueroa43690fako/surds-and-indices-a-level-bym.html	## Surds and indices a level is a number with an index of 2. This is never normally said as people prefer the phrase "x to the power of 2". Likewise, surds is a heading that is not often used. 23 Feb 2020 View SolutionHelpful Tutorials. Surds - introduction & simplifying. A-Level Edexcel Core Maths C1 June 2010 Q1 : ExamSolutions - youtube Index laws are the rules for simplifying expressions involving powers of the same base Terms involving the “√ ” symbol are known as a radicals or surds. 24 Sep 2019 re 1 OCR A Level revision resources. Exam questions organised by topic, Indices and Surds 1 · Indices and Surds MS 1 · Indices and Surds 2 A surd is a number written in a form that includes a square root. Examination papers often use square numbers in questions on surds, so look out for these numbers 1 4 9 16 25 36 49 64 81 100 I.B. Standard Level, 1, Algebra and Functions, Surds, - OCR MEI AS Maths 2017, Pure Maths, Surds and Indices, Surds, -. Revise how to simplify expressions involving indices at part of National 5 maths. Simplifying expressions using the laws of indices Working with surds. ## 30 Jan 2017 Indices are used to describe the general term for in say . There are a few laws to know when manipulating expressions involving indices. Surds A-Level Maths revision section looking at Surds and how you calculate using them. C1 Algebra: Surds and Indices – Questions. 2. 1. Write a misunderstanding of the mathematical notation used in AS level mathematics is a legitimate area to is a number with an index of 2. This is never normally said as people prefer the phrase "x to the power of 2". Likewise, surds is a heading that is not often used. Questions separated by topic from Core 1 Maths A-level past papers. equations · C1 Algebra - Surds and indices · C1 Coordinate geometry - Straight lines ### 24 Sep 2019 re 1 OCR A Level revision resources. Exam questions organised by topic, Indices and Surds 1 · Indices and Surds MS 1 · Indices and Surds 2 24 Sep 2019 re 1 OCR A Level revision resources. Exam questions organised by topic, Indices and Surds 1 · Indices and Surds MS 1 · Indices and Surds 2 A surd is a number written in a form that includes a square root. Examination papers often use square numbers in questions on surds, so look out for these numbers 1 4 9 16 25 36 49 64 81 100 I.B. Standard Level, 1, Algebra and Functions, Surds, - OCR MEI AS Maths 2017, Pure Maths, Surds and Indices, Surds, -. Revise how to simplify expressions involving indices at part of National 5 maths. Simplifying expressions using the laws of indices Working with surds. In the Quiz 4 you will get Surds and Indices problems from Algebra with detailed SSC MTS, SSC LDS with shortcuts and tricks, solve surds questions fast with Aptitude » Algebra » Surds and Indices Level 1 » all » Surds and Indices Quiz 4 Review Questions. The following quiz contains 25 questions that consist of multiple choice, fill-in-the-blank, matching and pattern match types. If you wish to take Edexcel Exam Papers OCR Exam Papers · A Level Revision · A Level Revision A Level (Modular) Revision · A Level Papers KS2 Revision Resources ☰. ### Created by T. Madas. Created by T. Madas. INDICES. Exam Questions each of the following expressions, writing the final answer as a single simplified surd. In the Quiz 4 you will get Surds and Indices problems from Algebra with detailed SSC MTS, SSC LDS with shortcuts and tricks, solve surds questions fast with Aptitude » Algebra » Surds and Indices Level 1 » all » Surds and Indices Quiz 4 Review Questions. The following quiz contains 25 questions that consist of multiple choice, fill-in-the-blank, matching and pattern match types. If you wish to take Edexcel Exam Papers OCR Exam Papers · A Level Revision · A Level Revision A Level (Modular) Revision · A Level Papers KS2 Revision Resources ☰. 23 Nov 2017 mathematical ideas. At A level we are concerned only with rational powers. Notice that the rules do not naturally extend to irrational powers; ## 6 Feb 2018 Surds and Indices are some of the basics of number system theory. The questions may include simplification of equations to find the simplified Important Formulas - Surds and Indices. An integer is a whole number (positive, negative or zero). A rational number is one that can be expressed as a fraction called surds. 2. Powers and roots. We know that 2 cubed is 2 × 2 × 2, and we say that we have 2 raised to the power 3, or to the index 3. An easy way of writing Worked example 14: Surd equations. Solve for $$z$$: $$z - 4\sqrt{z} + 3 = 0$$. Factorise. \begin 3 May 2017 Surds and indices. 1. By how The level ( L cm ) of the water in the container is given by the the level of water in the container at the start is.	2022-08-19 23:05:39	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.731572687625885, "perplexity": 2341.5272456401367}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-33/segments/1659882573849.97/warc/CC-MAIN-20220819222115-20220820012115-00736.warc.gz"}
https://math.stackexchange.com/questions/2190775/how-to-prove-gcd-with-prime-factor-decomposition	# How to prove $gcd$ with prime factor decomposition? So we know that $n=\prod_{p\in\mathbb{P}} p^{v_{n}(p)}$ is formula of factor decomposition. How to show that $gcd(m,n) = \prod_{p\in\mathbb{P}} p^{min(v_{n}(p),v_{n}(p))}$ using factor decomposition? I know that for example when we have $gcd(36,360)$, we can write $36$ as $36 = 2^{p_1}3^{p_2}5^{p_3}$ which is $36 = 2^{2}3^{2}5^{0}$ and $360$ as $360 = 2^{q_1}3^{q_2}5^{q_3}$ which is $360 = 2^{3}3^{2}5^{1}$ To get $gcd$ we now need to take smallest $p$ and $q$ for every pair with same base. In this case these are 2,2,0, therefore our $gcd(36,360) = 36$ which is obvious. So, I can show that this is valid $gcd(m,n) = \prod_{p\in\mathbb{P}} p^{min(v_{n}(p),v_{n}(p))}$ through an example with numbers, but I need proof, and I don't know how to prove that. That's just because $m\|n$ iff $\nu_m(p) \leq \nu_n(p)$ for all relevant $p$, and then use the definition of greatest common divisor.	2020-08-13 09:44:58	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.87282794713974, "perplexity": 117.51318366765797}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-34/segments/1596439738964.20/warc/CC-MAIN-20200813073451-20200813103451-00086.warc.gz"}
http://interiorartistryredesign.com/c2qwqu3/bottom-margin-latex-7f3f92	# bottom margin latex The best way to correct this is using the anysize package then setting the margin size, however doing so causes the "=.5=.75" to display before my title in the pdf. You can explicitly change that. bottom page margin to above the bottom of the columns, this has no effect on the columns and they stay the same. LaTeX's margins are, by default, 1.5 inches wide on 12pt documents, 1.75 inches wide on 11pt documents, and 1.875 inches wide on 10pt documents. To create it with geometry is easy, include this one line in the preamble: The parameters passed to the command determine the layout. LaTeX documents, by default, set the page margins to be quite big. Set custom page text/line width in memoir for sidepars, Geometry does not seem to change the margins. For some reason, there is a gap of about half a cm (0.2 inches) below the bottom footnote on each page - i.e. How does one calculate effects of damage over time if one is taking a long rest? top, tmargin These two parameters represent elements 2 and 6 in the figure, combined. While the package provides several commands to adjust the page margins, text blocks or the entire page design, I find the below single line command extremely useful. You can change the page layout with intuitive parameters: Next is a list of document elements whose length can be changed. and how to fix it? In this case a4paper establishes the paper size and the total parameter determines the size of the text area. Paper size, orientation and margins are the most common page elements that must be changed depending on the type of document. This feature is not available in Word Online. The parameters have to be written in the form parameter=value, use standard LaTeX units. Finally, each margin is set to 2 in. Open an example of the geometry packa… headheight Height of the header headsep These three parameters change the length of the right margin. To subscribe to this RSS feed, copy and paste this URL into your RSS reader. It's in the Document Structure part of the Latex wikibook. These adjustments are intended to allow a maximum … In this case a4paper establishes the paper size and the totalparameter determines the size of the text area. How to split equation into a table and under square root? I recommend that you use geometry's settings only rather than mixing them with native length macros. The CSS margin properties are used to create space around elements, outside of any defined borders.. With CSS, you have full control over the margins. Wikibooks shows how but I'm not allowed to post a 2nd link. p: Put on a special page for floats only. By default, LaTeX margins are extremely generous. The following is my code: % Setup the margin \\usepackage[centering, margin={1in, 0.5in}, includeheadfoot]{ For example, let's create a document with legal paper size, landscape orientation and a 2 in margin: You can achieve the same thing in a slightly different way: As you see, the parameters are comma separated. Identify location (and painter) of old painting. ! Suppose you have to create a document in a4paper and the text shouldn't exceed 6 in width and 8 in height. Note that Overleaf uses a European LaTeX distribution, which produces documents in A4 size by default. Override internal parameters LaTeX uses for determining "good" float positions. Suppose you have to create a document in a4paper and the text shouldn't exceed 6 in width and 8 in height. the footnotes are shifted up slightly from the bottom margin of the page. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. There are properties for setting the margin for each side of an element (top, right, bottom, and left). Change the size for the Top, Bottom, Left, and Right margins. It only takes a minute to sign up. [3cm] Do not leave a space between the command and the word where you wish the footnote marker to appear, otherwise LaTeX will process that space and will leave t… Is this house-rule that has each monster/NPC roll initiative separately (even when there are multiple creatures of the same kind) game-breaking? I don't really understand why it works for every other of the three margins but not for bottom margin. This leads to good-looking documents, but there are occasions when it may be better to have smaller margins —— for instance, if you want a document to fit entirely within one page. Example of ODE not equivalent to Euler-Lagrange equation, V-brake pads make contact but don't apply pressure to wheel. Maybe a problem with the spacing between lines? What is the difference between "regresar," "volver," and "retornar"? LaTeX enforces beauty upon its documents, sometimes to the detriment of practicality. Could anybody tell me what's wrong with my code? This keeps the main body of text concise.The footnote facility is easy to use. CSS Margins. Open an example of the geometry package in Overleaf. The geometry package provides a flexible and easy interface to change page dimensions. For submitting manuscripts to journals, or proposals to funding agencies, it is important to have a document that looks good. How do politicians scrutinize bills that are thousands of pages long? ... First step, make the outer margin at the bottom of the plot large: par (oma = c (4, 1, 1, 1)) Select Layout > Margins > Custom Margins. Differences between Mage Hand, Unseen Servant and Find Familiar. Note that Overleaf uses a European LaTeX distribution, which produces documents in A4 size by default. That's not what I set up. Margins are controlled with four settings: \textwidth, the width of the text on the page; \hoffset, the horizontal … bottom, bmargin These two parameters set the distance from the bottom edge of the document to its baseline. I really don't see how this can happen. Export (png, jpg, gif, svg, pdf) and save & share with note system Requires the float package. \usepackage[margin=1in,footskip=0.25in]{geometry} gives me without problem upper, right and left margin of 1in. To set the desired values there are two ways, either you pass them as parameters to the \includepackage statement as in the example above, or use a \geometry command in the preamble. Two-sided documents differentiate the left (even) and right (odd) pages, whereas one-sided do not. Thank you very much. Because you overrode the default settings defined by the geometry package (such as \footskip) then the last settings will be applied to take effect. Elements 9 and 10 in the figure, combined. Can anyone help identify this mystery integrated circuit? Documents can be either one- or two-sided. As you can see, obviously the bottom margin is 0 and the footer is missing. Asking for help, clarification, or responding to other answers. To create it with geometryis easy, include this one line in the preamble: The parameters passed to the command determine the layout. For a complete list of predefined paper sizes, see the reference guide. Select Set As Default to set the new margin settings as the default for the current template. (3) The left margin and the top margin are measured from a default position of 1 inch. Latex can do this for you, in a manner much easier than MS Word. This is somewhat equivalent to h!. But the lower margin is 1.2in and I can't change it to be 1in even by using bottom=1in or bottom=0.8in. If you want to change them, you have several options: the \"geometry\" package, the \"fullpage\" package or changing the margins by hand. How can I set up the left/right margin of the page to 1 inch, the top/bottom one to ½ inch? (mm, cm, pt, in). Articles are by default one-sided, books are two-sided. For instance, the computers I have tested on both have old versions of the geometry and caption packages, which cause Latex to return errors. The second parameter is the orientation, its default value is portrait. To create it with geometryis easy, include this one line in the preamble: The parameters passed to the command determine the layout. Can anyone identify this biplane from a TV show? By using our site, you acknowledge that you have read and understand our Cookie Policy, Privacy Policy, and our Terms of Service. margin is conrolled by \topmargin (and \headheight and \headsep), the bottom margin is de ned implicitly by these lengths and \textheight. margins : left (inner), right (outer), top and bottom Each margin is measured from the corresponding edge of a paper. I would like the bottom margin to be 2cm, but the text does not reach the end of the body so that this "gap" adds to the bottom margin and makes it too large. By default, LaTeX uses 1.5 inches margin sizes for 12pt documents, 1.75 inches for 11pt, and 1.875 inches for 10pt—relatively large margins. To learn more, see our tips on writing great answers. Why does the EU-UK trade deal have the 7-bit ASCII table as an appendix? By clicking âPost Your Answerâ, you agree to our terms of service, privacy policy and cookie policy. set margin latex example \documentclass{article} \usepackage[left=2cm, right=5cm, top=2cm]{geometry} \begin{document} Some text … left margin is 2 cm, right margin 5 cm and the top margin is 2 cm. What's a way to safely test run untrusted JavaScript code? Let's see an example with some of the aforementioned options: Here the text area, the left margin and the top margin are set. Is it permitted to prohibit a certain individual from using software that's under the AGPL license? You set \footskip to 25pt and that is too small! online LaTeX editor with autocompletion, highlighting and 400 math symbols. Space below footnotes - not aligned with bottom margin I have an urgent problem regarding the manuscript for a book which is to be published shortly. Looking at the test … Why does the Indian PSLV rocket have tiny boosters? rev 2020.12.18.38240, The best answers are voted up and rise to the top, TeX - LaTeX Stack Exchange works best with JavaScript enabled, Start here for a quick overview of the site, Detailed answers to any questions you might have, Discuss the workings and policies of this site, Learn more about Stack Overflow the company, Learn more about hiring developers or posting ads with us. What procedures are in place to stop a U.S. Vice President from ignoring electors? H: Places the float at precisely the location in the L a T e X code. Position at the bottom of the page. Is there a word for the object of a dilettante? You can use the geometry package to specify your margins: \usepackage{geometry} \geometry{top=3cm, left=2cm, right=2cm, bottom… The following is my code: But I don't know why the result doesn't show me the correct page setup. Name of author (and anthology) of a sci-fi short story called (I think) "Gold Brick"? site design / logo © 2020 Stack Exchange Inc; user contributions licensed under cc by-sa. If you want to make the article class two-sided, use \documentclass[twoside]{article}.Many commands and variables in LaTeX take this concept into account. \reversemarginpar \marginnote { This is a margin note using the geometry package, set at 5cm vertical offset to the first line it is typeset. } Usually, it is non-essential information which can be placed at the bottom of the page. They are referred to … I am trying to format a pdf doc in Latex using texmaker. The regular LaTeX page dimensions are presented (with example values) in picture at the right of this section. Footnotes are a very useful way of providing extra information to the reader. The simplest way I found of doing it is using the changepage package available from ctan.org. LaTeX Change the page margins for a single page in a LaTeX document. The link How to avoid getting cropped marginpar? You could use dimensions like \oddsidemargin and \evensidemargin, \textheight and \textwidth, as proposed in another answer. This is what I would like the first page to do. I try to set up the left and right margins to 1 inch, top and bottom margins to 0.5 inch. How to fix the top and bottom margins in Lyx in order to conform to standard ERC proposal? Making statements based on opinion; back them up with references or personal experience. But what you really want to do is use the geometry package. However, on the second and subsequent pages the columns go all the way to the bottom page margin, and when I adjust the bottom page margin the columns adjust to fit automatically. What is Litigious Little Bow in the Welsh poem "The Wind"? Open an example of the geometry packa… Suppose you have to create a document in a4paper and the text shouldn't exceed 6 in width and 8 in height. Has Section 2 of the 14th amendment ever been enforced? I try to set up the left and right margins to 1 inch, top and bottom margins to 0.5 inch. The paper size can be set to any size you need by means of the command papersize={â¨widthâ©,â¨heightâ©}. Look at the test print and check if the bounded area falls within the print area. How to REALLY get 1-inch margins in a latex document. Otherwise, you can change the parskip value. In many cases this is too much and we need to make some adjustments. How can I negate the 1 inch TeX margins and still use TikZ? The command you need is: \footnote{text}. TeX - LaTeX Stack Exchange is a question and answer site for users of TeX, LaTeX, ConTeXt, and related typesetting systems. Podcast Episode 299: Itâs hard to get hacked worse than this, Padding left, top, right and bottom in LaTeX. The default edge to print margin notes is the left for one-sided documents, outer for double-sided documents and the closest for two-column documents. Note that Overleaf uses a European LaTeX distribution, which produces documents in A4 size by default. This margin setting will be used in each new document you create based on that template. How can I remove the indent after every section in new page? For example, left margin (inner margin) means a horizontal distance between the left (inner) edge of the paper and that of the total body. This is the standard for book margins. Therefore the left and top margins de ned in geometry are di erent from the native dimensions I'm using Latex to write my resume, however the default margins for the resume doc type are too wide. Changing \headheight pushes footer off page. The right and bottom margins are automatically computed to fit the page. Thus, \oddsidemargin=0in results in 1 inch real margin, \oddsidemargin=-0.5in results in 0.5 inch real margin, etc. and the following code snippet may help you to understand how to set up the page layout. LaTeX varies the width of its margins depending on the font size. Showing first {{hits.length}} results of {{hits_total}} for {{searchQueryText}}, {{hits.length}} results for {{searchQueryText}}, Multilingual typesetting on Overleaf using polyglossia and fontspec, Multilingual typesetting on Overleaf using babel and fontspec, a0paper, a1paper, a2paper, a3paper, a4paper, a5paper, a6paper,b0paper, b1paper, b2paper, b3paper, b4paper, b5paper, b6paper,c0paper, c1paper, c2paper, c3paper, c4paper, c5paper, c6paper,b0j, b1j, b2j, b3j, b4j, b5j, b6j,ansiapaper, ansibpaper, ansicpaper, ansidpaper, ansiepaper,letterpaper, executivepaper, legalpaper. Example of ODE not equivalent to Euler-Lagrange equation, V-brake pads make contact do. Its default value is portrait 2nd link to 2 in be seen in page margins to be even! Following native length macros should not be used in each new document you create on... What 's wrong with my code: but I do n't see how this can happen agencies! Establishes the paper size and the top and bottom margins to 1 inch this can.! Of this section precisely the location in the figure, combined question and answer for... With references or personal experience LaTeX editor with autocompletion, bottom margin latex and 400 math symbols TeX - Stack!, see our tips on writing great answers bottom edge of the page but I n't. ) the left ( even bottom margin latex and right margins text/line width in memoir for,. A very useful way of providing extra information to the reader which can be changed, obviously bottom... Is there a Word for the top, right and bottom margins Lyx! Ms Word to … online LaTeX editor with autocompletion, highlighting and math! Latex enforces beauty upon its documents, sometimes to the reader in and! Intended to allow a maximum … LaTeX documents, by default, set the new margin as. A Word for the object of a sci-fi short story called ( I think ) Gold Brick '' Episode., privacy policy and cookie policy geometryis easy, include this one line in the:. To other answers, in ) default value is portrait LaTeX units is! Body of text concise.The footnote facility is easy to use in 1 inch one is a..., include this one line in the preamble: the parameters have create. Even by using bottom=1in or bottom=0.8in ; user contributions licensed under cc by-sa doing it is the. You, in ) dimensions are presented ( with example values ) in picture at the bottom of... Been enforced to any size you need by means of the right margin volver, ., Padding left, top, bottom, and related typesetting systems tmargin two... Position of 1 inch TeX margins and still use TikZ LaTeX, ConTeXt and! I 'm using LaTeX to write my resume, however the default for the object of dilettante... What procedures are in place to stop a U.S. Vice President from ignoring electors a list predefined! Papersize= { â¨widthâ©, â¨heightâ© } and margins are automatically computed to fit the page margins to be big... Be set to any size you need by means of the three margins but not for margin. To wheel n't exceed 6 in width and 8 in height part of the geometry package in Overleaf it... Default value is portrait regular LaTeX page dimensions are presented ( with example ). Produces documents in A4 size by default geometryis easy, include this one line in the L T! Cm, pt, in a manner much easier than MS Word contact but do n't know the... Document to its baseline snippet may help you to understand how to split equation into a table and under root! Produces documents in A4 size by default in many cases this is I. Intuitive parameters: Next is a list of predefined paper sizes, see our tips writing. The Welsh poem the Wind '' separately ( even ) and right margins equation, pads. It to be 1in even by using bottom=1in or bottom=0.8in not be used in each new document you based. Thanks for contributing an answer to TeX - LaTeX Stack Exchange Inc ; user contributions licensed cc... Monster/Npc roll initiative separately ( even ) and right margins by means of the same kind ) game-breaking it in! Are by default Episode 299: Itâs hard to get hacked worse than this, Padding left, and. What 's a way to safely test run untrusted JavaScript code how I! You create based on opinion ; back them up with references or personal experience referred... Change the margins n't apply pressure to wheel open an example of the page margins and painter ) of dilettante! Question and answer site for users of TeX, LaTeX, ConTeXt, and related typesetting systems may. Identify this biplane from a default position of 1 inch there are properties for setting the margin each! Cookie policy on opinion ; back them up with references or personal experience n't apply pressure to.! Top, right and bottom margins to 0.5 inch real margin, etc even there! Me what 's wrong with my code individual from using software that 's under the AGPL?... Right margin the default for the resume doc type are too wide easy to use I... In each new document you create based on opinion ; back them up with references or personal experience non-essential which., which produces documents in A4 size by default one-sided, books are two-sided to make some adjustments geometry! You, in a manner much easier than MS Word the second parameter is difference! Overleaf uses a European LaTeX distribution, which produces documents in A4 by! From ctan.org set \footskip to 25pt and that is too much and we need to make some adjustments the is... Related typesetting systems the Welsh poem the Wind '' understand why it for. 'M using LaTeX to write my resume, however the default for the resume doc type too!, '' and retornar '' quite big, see the reference guide root... Useful way of providing extra information to the command you need is: {... Has section 2 of the 14th amendment ever been enforced to 2 in using! Important to have a document that looks good you create based on that template footnotes are very! Referred to … online LaTeX editor with autocompletion, highlighting and 400 math symbols the new margin as!: Places the float at precisely the location in the Welsh poem the Wind '' paste this into. Body of text concise.The footnote facility is easy to use permitted to a..., ConTeXt, and related typesetting systems old painting is use the geometry packa… These three change! Using LaTeX to write my resume, however the default margins for the resume type... Is Litigious Little Bow in the document to its baseline size, and... Highlighting and 400 math symbols: Places the float at precisely the location in the document to its baseline âPost! P: Put on a bottom margin latex page for floats only common page elements must! The EU-UK trade deal have the 7-bit ASCII table as an appendix Find. 'S a way to safely test run untrusted JavaScript code geometryis easy, this. ; user contributions licensed under cc by-sa a very useful way of extra. Type are too wide the reference guide you agree to our terms service... And \evensidemargin, \textheight and \textwidth, as proposed in another answer the second parameter is the difference between regresar... Is missing to any size you need is: \footnote { text } )! Depending on the type of document does the EU-UK bottom margin latex deal have the 7-bit table! The length of the three margins but not for bottom margin of the geometry package parameters. Margin for each side of an element ( top, bottom, right. Help, clarification, or responding to other answers location in the form parameter=value use... Location ( and painter ) of old painting EU-UK trade deal have 7-bit... Other answers I negate the 1 inch, top and bottom in LaTeX of damage over time one... Agencies, it is using the changepage package available from ctan.org I negate the 1.... Allow a maximum … LaTeX documents, by default one-sided, books are two-sided answer to TeX LaTeX... Bottom edge of the LaTeX wikibook be changed this RSS feed, and... Pads make contact but do n't really understand why it works for every of! The difference between regresar, '' volver, '' volver, '' volver, and. Negate the 1 inch, top and bottom in LaTeX identify location ( anthology! Is what I would like the first page to do is use the geometry packa… These three parameters the. Using software that 's under the AGPL license of service, privacy policy and cookie policy body of text footnote... Up the left/right margin of the same kind ) game-breaking select set as default set! Test run untrusted JavaScript code to any size you need by means of the text should n't exceed in. ; back them up with references or personal experience to get hacked worse than,!, clarification, or responding to other answers dimensions are presented ( example... A 2nd link, '' volver, '' and retornar '' works for every other the... On the type of document making statements based on that template one line in Welsh. With intuitive parameters: Next is a list of predefined paper sizes, see our tips on great. Print area typesetting systems adjustments are intended to allow a maximum … LaTeX documents, sometimes to the of! Margins for the resume doc type are too wide from ctan.org opinion ; back them up with references or experience. Most notable effect can be seen in page margins to allow a maximum … LaTeX documents, default. Under square root even by using bottom=1in or bottom=0.8in ERC proposal right,,! Latex editor with autocompletion, highlighting and 400 math symbols my resume, however the default the.	2021-05-08 22:31:48	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7060436010360718, "perplexity": 2582.2275677912044}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-21/segments/1620243988927.95/warc/CC-MAIN-20210508211857-20210509001857-00100.warc.gz"}
http://mathhelpforum.com/calculus/172894-when-nth-differential-non-zero.html	# Math Help - when is nth differential non zero 1. ## when is nth differential non zero >>>>>>> Question <<<<<<<<< it is easy to find it generally but i was thinking that what if in some question ... say 100th differential is 0 ... there has to be a shortcut ... some help please i even found out the general term of what differential would be after the third differential (when $2/3 x^{3}$ goes away)... its $f_{n}(x) = e^{x} + (-1)^{n-1}e^{-x} -2 sin[n\pi/2 + (-1)^{n}x]$ But i dont know how to find when $f_{n}(x)$ will be non-zero 2. Use Taylor series. In my calculations, the series for f(x) starts with $4x^7/7!$, so the answer is 7. 3. Originally Posted by cupid >>>>>>> Question <<<<<<<<< it is easy to find it generally but i was thinking that what if in some question ... say 100th differential is 0 ... there has to be a shortcut ... some help please i even found out the general term of what differential would be after the third differential (when $2/3 x^{3}$ goes away)... its $f_{n}(x) = e^{x} + (-1)^{n-1}e^{-x} -2 sin[n\pi/2 + (-1)^{n}x]$ But i dont know how to find when $f_{n}(x)$ will be non-zero Focus on that last term. $\displaystyle \left ( \frac{2}{3}x^3 \right ) ^{\prime} = 2x^2$ $( 2x ^2 ) ^{\prime} = 4x$ $( 4x ) ^{\prime} = 4$ definitely non-zero. -Dan 4. Originally Posted by topsquark Focus on that last term. $\displaystyle \left ( \frac{2}{3}x^3 \right ) ^{\prime} = 2x^2$ $( 2x ^2 ) ^{\prime} = 4x$ $( 4x ) ^{\prime} = 4$ definitely non-zero. You can't focus on the last term only because it may be canceled by other terms. 5. And that is what happens .. 1,2,3 differentials are not 0 6. I have no idea what you are talking about. You can check in WolframAlpha that $\displaystyle\frac{d^7f(x)}{dx^7}(0)=4$ and $\displaystyle\frac{d^nf(x)}{dx^n}(0)=0$ for $n=1,\dots,6$. 7. sorry i meant that they are 0 typing mistake but isnt there is any general method to do these kind of problems 8. Originally Posted by emakarov You can't focus on the last term only because it may be canceled by other terms. Good point. -Dan 9. but isnt there is any general method to do these kind of problems I don't know the answer because I don't know the exact class of problems you are referring to. For an arbitrary function f(x) given by its formula, I think the easiest way is to compute the nth derivative and evaluate it at 0. This particular problem seems to me to be designed to be solved using Taylor series. However, concerning problems that you may encounter during tests, etc., I don't know how similar to this one they will be.	2016-06-27 10:45:24	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 16, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7309879064559937, "perplexity": 387.8512524669396}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-26/segments/1466783395679.92/warc/CC-MAIN-20160624154955-00121-ip-10-164-35-72.ec2.internal.warc.gz"}
http://mail-archives.apache.org/mod_mbox/db-derby-dev/200905.mbox/%[email protected]%3E	# db-derby-dev mailing list archives ##### Site index · List index Message view Top From Kathey Marsden <[email protected]> Subject Questions about policy file and multiple FilePermission lines Date Thu, 28 May 2009 23:38:42 GMT I am working with a user that is using the network server default server.policy file and having an interesting problem. They create their database with an absolute path and sometimes they get the permission error below. When they get the failure and set java.security.debug to access:failure. They see only two or three of the file permissions getting loaded instead of the four that we have in the file for derby.jar. permission java.io.FilePermission "${derby.system.home}","read"; permission java.io.FilePermission "${derby.system.home}${/}-", "read,write,delete"; The user rebuilt derby with only the permission java.io.FilePermission "<<ALL FILES>>", "read,write,delete"; FilePermission in the server.policy file and doesn't see the issue. I actually haven't reproduced this issue on my machine with almost the same revision JVM. I see all 4 permissions listed and have no problem creating a database with an absolute path. They are using: Java(TM) SE Runtime Environment (build pwi3260sr3-20081106_07(SR3)) IBM J9 VM (build 2.4, J2RE 1.6.0 IBM J9 2.4 Windows XP x86-32 jvmwi3260-20081105_25433 (JIT enabled, AOT enabled) J9VM - 20081105_025433_lHdSMr JIT - r9_20081031_1330 GC - 20081027_AB) JCL - 20081106_01 They start their network server with an ant script. I wonder how java should handle having permission java.io.FilePermission "<<ALL FILES>>", "read"; and permission java.io.FilePermission "<<ALL FILES>>", "read,write,delete"; Should the JVM be smart enough to figure out the more liberal one and go with that? Do we need to keep all four of these or would just the one suffice? Here is the error: ERROR XBM0H: Directory D:\<snip path to database> cannot be created. at org.apache.derby.iapi.error.StandardException.newException(Unkno wn Source) at org.apache.derby.impl.services.monitor.StorageFactoryService.cre ateServiceRoot(Unknown Source) at org.apache.derby.impl.services.monitor.BaseMonitor.bootService(U nknown Source) at org.apache.derby.impl.services.monitor.BaseMonitor.createPersist entService(Unknown Source) at org.apache.derby.iapi.services.monitor.Monitor.createPersistentS ervice(Unknown Source) at org.apache.derby.impl.jdbc.EmbedConnection.createDatabase(Unknow n Source) at org.apache.derby.impl.jdbc.EmbedConnection.<init>(Unknown Source) at org.apache.derby.jdbc.Driver40.getNewEmbedConnection(Unknown Source) at org.apache.derby.jdbc.InternalDriver.connect(Unknown Source) at org.apache.derby.jdbc.AutoloadedDriver.connect(Unknown Source) at org.apache.derby.impl.drda.Database.makeConnection(Unknown Source) at org.apache.derby.impl.drda.DRDAConnThread.getConnFromDatabaseNam e(Unknown Source) at org.apache.derby.impl.drda.DRDAConnThread.verifyUserIdPassword(U nknown Source) at org.apache.derby.impl.drda.DRDAConnThread.parseSECCHK(Unknown Source) at org.apache.derby.impl.drda.DRDAConnThread.parseDRDAConnection(Un known Source) at org.apache.derby.impl.drda.DRDAConnThread.processCommands(Unknow n Source) at org.apache.derby.impl.drda.DRDAConnThread.run(Unknown Source) Caused by: java.security.AccessControlException: Access denied (java.io.FilePermission D:\<snip path to database> write) at java.security.AccessController.checkPermission(AccessController. java:108) at java.lang.SecurityManager.checkPermission(SecurityManager.java:5 32) at java.lang.SecurityManager.checkWrite(SecurityManager.java:962) at java.io.File.mkdir(File.java:1167) at java.io.File.mkdirs(File.java:1196) at org.apache.derby.impl.services.monitor.StorageFactoryService$9.r un(Unknown Source) at java.security.AccessController.doPrivileged(AccessController.jav a:251) at org.apache.derby.impl.services.monitor.StorageFactoryService.cre ateServiceRoot(Unknown Source) at org.apache.derby.impl.services.monitor.BaseMonitor.bootService(U nknown Source) Here is how I am trying to reproduce based on their description: java -Djava.security.debug="access:failure" -Dderby.system.home=C:/tmp -classpath "C:/svn/10.3/jars/sane/derbyclient.jar;C:/svn/10.3/jars/sane/derbytools.jar;C:/svn/10.3/jars/sane/derbynet.jar" org.apache.derby.drda.NetworkServerControl start -h <my machine> -p 1692 and connecting with ij with: connect 'jdbc:derby://<my machine>:1692/C:\path\to\MYDB;create=true'; but like I said I haven't been able to reproduce so far. Kathey Mime View raw message	2017-01-24 22:35:59	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5641194581985474, "perplexity": 9230.838583686344}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-04/segments/1484560285289.45/warc/CC-MAIN-20170116095125-00556-ip-10-171-10-70.ec2.internal.warc.gz"}
http://tiemice76.thesupersuper.com/post/different-designer-wholesale-brands-at-affordable-prices/	# Different Designer Wholesale Brands At Affordable Prices ## February 26, 2018 by You are needed to draw lines utilizing a compelling line. Then use the sprinkled lines to show top stitching and also for the zippers. Even so it is available drawing invisible zippers use zipper catch the attention of. The female who like reading probably will be not high education but considerable high refined person. To have an continuously reads, we can distinguish her without difficulty, especially; lousy cope with things coolly and superbly. It is said that reading people won’t verbalize casually and they conclude a effect with logical proofs. t shirts for girls don’t really mean those short and skimpy ones. Well, yes those are there all of the list of options but we also do mean the skirts that are long and free. Yes, skirts will probably be in and you should lay both on people today. Buy skirts usually are colourful and ones anyone can wear everywhere within a casual ambience. You should always remember though that your skirt is not a very transparent just one particular! That is the reason the denser and heavier fabric of the women’s fashion clothing, the thicker (and even powerful, roughly the platform) should end up being heel of one’s shoe. Conversely, the lighter the dress, the thinner the heel bone. t shirts for boys from Sean John gives a wide regarding formal and non formal shirts. Apply for storm flap shirt worth $47.99, striped jacquard woven shirt of$48.99 or ticking stripe polo worth $55.99. Black colored leather jacket or f aux leather bomber jacket with price regarding$362.99 to $144.99 could possibly be another option of winter wear for men. Then there is a variety of jacket types like royalty bubble, varsity wool jacket, royal supreme jacket or Maserati class jacket which you might go for at an economical price. boys t shirts long sleeve may be the idea that very long skirts make the legs look long. Very long, baggy skirts you could legs look shorter. men’s t shirts xxl can make you look taller when they go right down to the ankle level and if they are straight. While purchasing wholesale clothing from China, payment is mainly made through PayPal, and a secure method of payment. You will check whether a supplier is genuine, by checking the payment mode of the site. PayPal is if you want the secure methods of payment assists customers to complaint if products aren’t delivered over time. 100% cotton wear shirt from Sean John with some other style only comes in white and khaki. They are located in XL to 3XL sizes with price of$15.50 per shirt. girls t shirts 7-8 from Sean Jean can be a pack of six and prices are $12.50 per piece. Latest style T-shirts from SJ Men can also add to the wardrobes of males with only$15.50 per T-shirt. Wholesale Scarface V-neck T-shirts additionally a favorite to the newborn’s with affordable price of \$13.00.	2018-12-19 11:32:23	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.22481085360050201, "perplexity": 7655.02085953972}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-51/segments/1544376832259.90/warc/CC-MAIN-20181219110427-20181219132427-00214.warc.gz"}
http://openstudy.com/updates/4ffd093ae4b00c7a70c5ccf1	## opendiscuss The slope of the graph of f(x) = \|x\| changes abruptly when x = 0. Does this function have a derivative? If so, what is it? If not, why not? Explanation is given in the course material, but not clear...Why cant the tangent line be defined at (0,0)... one year ago one year ago • This Question is Open 1. iisthphir The function is not continuous at 0 so the derivative can't be taken. The two sided limit as x > 0 does not exist. As the function is elsewhere linear, its derivative is constant. For x < o it is -1 for x > 0 it is 1. 2. khoutir the function is continuous but hasn't a derivative because: $\lim_{x \rightarrow -0}[f(x+\Delta x)-f(x)/\Delta x] \neq \lim_{x \rightarrow +0}[f(x+\Delta x)-f(x)/\Delta x]$ for x<0 we have :$f(x+\Delta x)-f(x)/\Delta x=[-(x+\Delta x)+x]/\Delta x= -1$ for x>0 we get :$f(x+\Delta x)-f(x)/\Delta x=[(x+\Delta x)-x]/\Delta x= 1$ so the limit of the difference quotient doesn't exist and so that the derivative at x=0 3. JingleBells f(x)=\|x\| is continuous at x=0: $f(x ^{-})=\lim_{x \rightarrow 0}\left\| x \right\|=0$ $f(x ^{+})=\lim_{x \rightarrow 0}\left\| x \right\|=0$ \|dw:1342611754796:dw\| However, $f \prime(x)$ has a point of jump discontinuity at x=0: $f \prime(0^{-})=\lim_{x \rightarrow 0}-1=-1$ $f \prime(0^{+})=\lim_{x \rightarrow 0}1=1$ \|dw:1342612014321:dw\| at x=0 left hand limit $\neq$ right-hand limit as Problem Set 1 Question 1D, 3 (d) and (e) demonstrate similar cases.	2014-04-20 11:08:23	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8192628026008606, "perplexity": 475.40147755297124}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-15/segments/1397609538423.10/warc/CC-MAIN-20140416005218-00234-ip-10-147-4-33.ec2.internal.warc.gz"}
https://tex.stackexchange.com/questions/381274/in-minted-how-to-reduce-the-blank-space-between-the-line-number-and-the-code/381279	# In minted, how to reduce the blank space between the line number and the code? The following is example beamer file: \documentclass[notheorems,serif,11pt]{beamer} \usepackage{algorithm} \usepackage{float} \usepackage{minted} \renewcommand{\listingscaption}{Python code} \newminted{python}{ fontsize=\footnotesize, escapeinside=\|\|, mathescape=true, numbersep=5pt, linenos=true, gobble=2, framesep=3mm} \begin{document} \begin{frame}[fragile]{Mesh Data Strucrure} \begin{listing}[H] \begin{pythoncode} import numpy as np point = np.array( [(0.0, 0.0), (1.0, 0.0), (1.0, 1.0), (0.0, 1.0)], dtype=np.float) cell = np.array([ (1, 2, 0), (3, 0, 2)], dtype=np.int) N = point.shape[0] # the number of triangle points NC = cell.shape[0] # the number of triangle cells \end{pythoncode} \caption{The basic data structure of triangle mesh.} \end{listing} \end{frame} \end{document} Finally, I get the following result in my slide: My problem is that the blank space between the line number and code is too wide, and I want to reduce it. So how to do it? Thanks very much! • I can't give you a clear answer without an MWE, but autogobble option seems to be of help. Jul 19 '17 at 5:24 • @CarLaTex yes, you are right. I should give more information. I have updated the question. Jul 19 '17 at 5:54 • As @yudai-nkt said, if you put autogobble instead of gobble=2 it works. (To yudai-nkt: you could post an answer). Jul 19 '17 at 5:57 • Yes, I have tested it. It works! Thanks very much! @yudai-nkt Jul 19 '17 at 5:58 The autogobble option is the way to go. Here is the excerpt from the manual: Remove (gobble) all common leading whitespace from code. Essentially a version of gobble that automatically determines what should be removed. Good for code that originally is not indented, but is manually indented after being pasted into a LaTeX document. By utilizing this, \documentclass[notheorems,serif,11pt]{beamer} \usepackage{algorithm} \usepackage{float} \usepackage{minted} \renewcommand{\listingscaption}{Python code} \newminted{python}{ fontsize=\footnotesize, escapeinside=\|\|, mathescape=true, numbersep=5pt, linenos=true, autogobble, framesep=3mm} \begin{document} \begin{frame}[fragile]{Mesh Data Strucrure} \begin{listing}[H] \begin{pythoncode} import numpy as np point = np.array( [(0.0, 0.0), (1.0, 0.0), (1.0, 1.0), (0.0, 1.0)], dtype=np.float) cell = np.array([ (1, 2, 0), (3, 0, 2)], dtype=np.int) N = point.shape[0] # the number of triangle points NC = cell.shape[0] # the number of triangle cells \end{pythoncode} \caption{The basic data structure of triangle mesh.} \end{listing} \end{frame} \end{document} will produce	2021-09-20 11:17:05	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7422443628311157, "perplexity": 4026.768926665148}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-39/segments/1631780057036.89/warc/CC-MAIN-20210920101029-20210920131029-00392.warc.gz"}
http://ufldl.stanford.edu/wiki/index.php?title=Softmax_Regression&diff=prev&oldid=611	# Softmax Regression (Difference between revisions) Revision as of 23:24, 7 May 2011 (view source)Zellyn (Talk \| contribs)m (→Parameterization)← Older edit Revision as of 23:26, 7 May 2011 (view source)Zellyn (Talk \| contribs) m (→Binary Logistic Regression)Newer edit → Line 190: Line 190: &= &= - \frac{e^{ \Theta_1^T x^{(1)} } }{ 1 + e^{ \Theta_1^T x^{(i)} } } + \frac{e^{ \Theta_1^T x^{(i)} } }{ 1 + e^{ \Theta_1^T x^{(i)} } } \cdot \cdot - \frac{1}{e^{ \Theta_1^T x^{(1)} } } + \frac{1}{e^{ \Theta_1^T x^{(i)} } } \begin{bmatrix} \begin{bmatrix} e^{ \Theta_1^T x^{(i)} } \\ e^{ \Theta_1^T x^{(i)} } \\ ## Introduction Softmax regression, also known as multinomial logistic regression, is a generalisation of logistic regression to problems where there are more than 2 class labels. Recall that in logistic regression, our hypothesis was of the form: \begin{align} h_\theta(x) = \frac{1}{1+\exp(-\theta^Tx)}, \end{align} where we trained the logistic regression weights to optimize the (conditional) log-likelihood of the dataset using p(y \| x) = hθ(x). In softmax regression, we are interested in multi-class problems where each example (input image) is assigned to one of k labels. One example of a multi-class classification problem would be classifying digits on the MNIST dataset where each example has label 1 of 10 possible labels (i.e., where it is the digit 0, 1, ... or 9). To extend the logistic regression framework which only outputs a single probability value, we consider a hypothesis that outputs K values (summing to 1) that represent the predicted probability distribution. Formally, let us consider the classification problem where we have m k-dimensional inputs $x^{(1)}, x^{(2)}, \ldots, x^{(m)}$ with corresponding class labels $y^{(1)}, y^{(2)}, \ldots, y^{(m)}$, where $y^{(i)} \in \{1, 2, \ldots, n\}$, with n being the number of classes. Our hypothesis hθ(x), returns a vector of probabilities, such that \begin{align} h(x^{(i)}) = \begin{bmatrix} P(y^{(i)} = 1 \| x^{(i)}) \\ P(y^{(i)} = 2 \| x^{(i)}) \\ \vdots \\ P(y^{(i)} = n \| x^{(i)}) \end{bmatrix} = \frac{1}{ \sum_{j=1}^{n}{e^{ \theta_j^T x^{(i)} }} } \begin{bmatrix} e^{ \theta_1^T x^{(i)} } \\ e^{ \theta_2^T x^{(i)} } \\ \vdots \\ e^{ \theta_n^T x^{(i)} } \\ \end{bmatrix} \end{align} where $\theta_1, \theta_2, \ldots, \theta_n$ are each k-dimensional column vectors that constitute the parameters of our hypothesis. Notice that $\frac{1}{ \sum_{j=1}^{n}{e^{ \theta_j^T x^{(i)} }} }$ normalizes the distribution so that it sums to one. Strictly speaking, we only need n − 1 parameters for n classes, but for convenience, we use n parameters in our derivation. Now, this hypothesis defines a predicted probability distribution given some x, P(y \| x(i)) = h(x(i)). Thus to train the model, a natural choice is to maximize the (conditional) log-likelihood of the data, $l(\theta; x, y) = \sum_{i=1}^{m} \ln { P(y^{(i)} \| x^{(i)}) }$. Motivation: One reason for selecting this form of hypotheses comes from connections to linear discriminant analysis. In particular, if one assumes a generative model for the data in the form $p(x,y) = p(y) \times p(x \| y)$ and selects for p(x \| y) a member of the exponential family (which includes Gaussians, Poissons, etc.) it is possible to show that the conditional probability p(y \| x) has the same form as our chosen hypotheses h(x). For more details, see the CS 229 Lecture 2 Notes. ## Optimizing Softmax Regression Expanding the log-likelihood expression, we find that: \begin{align} \ell(\theta) &= \ln L(\theta; x, y) \\ &= \ln \prod_{i=1}^{m}{ P(y^{(i)} \| x^{(i)}) } \\ &= \sum_{i=1}^{m}{ \ln \frac{ e^{ \theta^T_{y^{(i)}} x^{(i)} } }{ \sum_{j=1}^{n}{e^{ \theta_j^T x^{(i)} }} } } \\ &= \theta^T_{y^{(i)}} x^{(i)} - \ln \sum_{j=1}^{n}{e^{ \theta_j^T x^{(i)} }} \end{align} Unfortunately, there is no closed form solution to this optimization problem (although it is concave), and we usually use an off-the-shelf optimization method (e.g., L-BFGS, stochastic gradient descent) to find the optimal parameters. Using these optimization methods require computing the gradient ($\ell(\theta)$ w.r.t. θk), which can can be derived as follows: \begin{align} \frac{\partial \ell(\theta)}{\partial \theta_k} &= \frac{\partial}{\partial \theta_k} \theta^T_{y^{(i)}} x^{(i)} - \ln \sum_{j=1}^{n}{e^{ \theta_j^T x^{(i)} }} \\ &= I_{ \{ y^{(i)} = k\} } x^{(i)} - \frac{1}{ \sum_{j=1}^{n}{e^{ \theta_j^T x^{(i)} }} } \cdot e^{ \theta_k^T x^{(i)} } \cdot x^{(i)} \qquad \text{(where } I_{ \{ y^{(i)} = k\} } \text{is 1 when } y^{(i)} = k \text{ and 0 otherwise) } \\ &= x^{(i)} ( I_{ \{ y^{(i)} = k\} } - P(y^{(i)} = k \| x^{(i)}) ) \end{align} With this, we can now find a set of parameters that maximizes $\ell(\theta)$, for instance by using L-BFGS with minFunc. ### Weight Regularization When using softmax regression in practice, it is important to use weight regularization. In particular, if there exists a linear separator that perfectly classifies all the data points, then the softmax-objective is unbounded (given any θ that separates the data perfectly, one can always scale θ to be larger and obtain a better objective value). With weight regularization, one penalizes the weights for being large and thus avoids these degenerate situations. Weight regularization is also important as it often results in models that generalize better. In particular, one can view weight regularization as placing a (Gaussian) prior on θ so as to prefer θ with smaller values. In practice, we often use a L2 weight regularization on the weights where we penalize the squared value of each element of θ. Formally, we use: \begin{align} w(\theta) = \frac{\lambda}{2} \sum_{i}{ \sum_{j}{ \theta_{ij}^2 } } \end{align} This regularization term is added together with the log-likelihood function to give a cost function, J(θ), which we want to minimize (note that we want to minimize the negative log-likelihood, which corresponds to maximizing the log-likelihood): \begin{align} J(\theta) = -\ell(\theta) + \frac{\lambda}{2} \sum_{i}{ \sum_{j}{ \theta_{ij}^2 } } \end{align} The gradients with respect to the cost function must then be adjusted to account for the weight decay term: \begin{align} \frac{\partial J(\theta)}{\partial \theta_k} &= x^{(i)} ( I_{ \{ y^{(i)} = k\} } - P(y^{(i)} = k \| x^{(i)}) ) + \lambda \theta_k \end{align} Minimizing J(θ) now performs regularized softmax regression. ## Parameterization We noted earlier that we actually only need n − 1 parameters to model n classes. To see why this is so, consider our hypothesis again: \begin{align} h(x^{(i)}) &= \frac{1}{ \sum_{j=1}^{n}{e^{ \theta_j^T x^{(i)} }} } \begin{bmatrix} e^{ \theta_1^T x^{(i)} } \\ e^{ \theta_2^T x^{(i)} } \\ \vdots \\ e^{ \theta_n^T x^{(i)} } \\ \end{bmatrix} \\ &= \frac{e^{ \theta_n^T x^{(i)} } }{ \sum_{j=1}^{n}{e^{ \theta_j^T x^{(i)} }} } \cdot \frac{1}{e^{ \theta_n^T x^{(i)} } } \begin{bmatrix} e^{ \theta_1^T x^{(i)} } \\ e^{ \theta_2^T x^{(i)} } \\ \vdots \\ e^{ \theta_n^T x^{(i)} } \\ \end{bmatrix} \\ &= \frac{1}{ \sum_{j=1}^{n}{e^{ (\theta_j^T - \theta_n^T) x^{(i)} }} } \begin{bmatrix} e^{ (\theta_1^T - \theta_n^T) x^{(i)} } \\ e^{ (\theta_2^T - \theta_n^T) x^{(i)} } \\ \vdots \\ e^{ (\theta_n^T - \theta_n^T) x^{(i)} } \\ \end{bmatrix} \\ \end{align} Letting Θj = θj − θn for $j = 1, 2 \ldots n - 1$ gives \begin{align} h(x^{(i)}) &= \frac{1}{ 1 + \sum_{j=1}^{n-1}{e^{ \Theta_j^T x^{(i)} }} } \begin{bmatrix} e^{ \Theta_1^T x^{(i)} } \\ e^{ \Theta_2^T x^{(i)} } \\ \vdots \\ 1 \\ \end{bmatrix} \\ \end{align} Showing that only n − 1 parameters are required. In practice, however, it is often easier to implement the version which is over-parametrized although both methods will lead to the same classifier. ### Binary Logistic Regression In the special case where n = 2, one can also show that softmax regression reduces to logistic regression: \begin{align} h(x^{(i)}) &= \frac{1}{ 1 + e^{ \Theta_1^T x^{(i)} } } \begin{bmatrix} e^{ \Theta_1^T x^{(i)} } \\ 1 \\ \end{bmatrix} \\ &= \frac{e^{ \Theta_1^T x^{(i)} } }{ 1 + e^{ \Theta_1^T x^{(i)} } } \cdot \frac{1}{e^{ \Theta_1^T x^{(i)} } } \begin{bmatrix} e^{ \Theta_1^T x^{(i)} } \\ 1 \\ \end{bmatrix} \\ &= \frac{1}{ e^{ -\Theta_1^T x^{(i)} } + 1 } \begin{bmatrix} 1 \\ e^{ -\Theta_1^T x^{(i)} } \\ \end{bmatrix} \\ \end{align}	2018-12-18 11:58:33	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 20, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 1.0000098943710327, "perplexity": 3002.1959747435826}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-51/segments/1544376829140.81/warc/CC-MAIN-20181218102019-20181218124019-00229.warc.gz"}
https://math.stackexchange.com/questions/1371234/uniqueness-of-harmonic-function-with-mixed-dirichlet-neumann-condition	# Uniqueness of harmonic function with Mixed Dirichlet Neumann condition Let $u \colon \{\mbox{Im } z>0\}\subset\mathbb{C}\to \mathbb{R}$ be a positive harmonic function in the upper half plane, i.e $$\Delta u=0,\,\, \mbox{for}\,\mbox{ Im } z>0.$$ Consider now the following mixed Dirichlet-Neumann boundary condition on the boundary $\partial\{\mbox{Im } z>0\}=\{\mbox{Im }z=0\}$ $$u=0,\,\mbox{for }\{(x,0): -1<x<1\}\\ \frac{\partial u}{\partial y}=0,\,\mbox{for }\{(x,0): x<-1 \vee x>1\}$$ where obviously $z=x+iy$. For example $u(z)=\mbox{Re}(\sqrt{(z-1)(z+1)})$ is a solution of the previous problem, but it's unique upo to a constant? The question is how can i prove that this problem has an unique solution? It's possible to apply a reflection to the other half plane and then apply some Liouville type result since the function turn to be bounded from below, or a maximum principle? • It looks like you have $u(iy)=\Re(\sqrt{-y^{2}-1})=0$ for $y > 0$. If $u$ is harmonic and non-negative in the upper half-plane, then it cannot be $0$ for any $y > 0$ unless it is identically $0$. Something seems wrong. – DisintegratingByParts Jul 23 '15 at 18:49 Every non-negative harmonic function in the upper half plane can be represented as $$h(x,y) = \frac{1}{\pi}\int_{-\infty}^{\infty}\frac{yd\rho(t)}{(t-x)^{2}+y^{2}}+Ay$$ where $A$, $D$ are positive constants and $\rho(t)$ is a non-decreasing function on $\mathbb{R}$ for which $$\int_{-\infty}^{\infty}\frac{d\rho(t)}{1+t^{2}} < \infty.$$ The representation is unique if you add a normalization requirement to $\rho$ such that $\rho(t+0)=\rho(t)$. You can determine $\rho$ from limits of the function $h$ near the boundary. If $a$ and $b$ are points of continuity of $\rho$, then $$\rho(b)-\rho(a) = \lim_{y\downarrow 0}\int_{a}^{b}h(x,y)dx$$ Jumps of $\rho$ are determined by $$\frac{1}{\pi}\{\rho(t+0)-\rho(t-0)\} = \lim_{y\downarrow 0}yh(x,y)$$ And $\rho$ is differentiable at some $t$ iff the following exists $$\rho'(t) = \lim_{y\downarrow 0}h(t,y)$$ This representation theorem gives you a unique characterization of a positive harmonic function on the upper half plane.	2019-12-14 00:03:04	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9354953169822693, "perplexity": 67.56585796663632}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-51/segments/1575540569332.20/warc/CC-MAIN-20191213230200-20191214014200-00091.warc.gz"}
https://www.khanacademy.org/economics-finance-domain/microeconomics/choices-opp-cost-tutorial/marginal-utility-tutorial/v/optimal-point-on-budget-line	So let's just review what we've seen with budget lines. Let's say I'm making $20 a month. So my income is$20 per month. Let's say per month. The price of chocolate is $1 per bar. And the price of fruit is$2 per pound. And we've already done this before, but I'll just redraw a budget line. So this axis, let's say this is the quantity of chocolate. I could have picked it either way. And that is the quantity of fruit. If I spend all my money on chocolate, I could buy 20 bars of chocolate a month. So that is 20. This is 10 right over here. At these prices, if I spent all my money on fruit I could buy 10 pounds per month. So this is 10. So that's 10 pounds per month. That would be 20. And so I have a budget line that looks like this. And the equation of this budget line is going to be-- well, I could write it like this. My budget, 20, is going to be equal to the price of chocolate, which is 1, times the quantity of chocolate. So this is 1 times the quantity of chocolate, plus the price of fruit, which is 2 times the quantity of fruit. And if I want to write this explicitly in terms of my quantity of chocolate, since I put that on my vertical axis and that tends to be the more dependent axis, I can just subtract 2 times the quantity of fruit from both sides. And I can flip them. And I get my quantity of chocolate is equal to 20 minus 2 times my quantity of fruit. And I get this budget line right over there. We've also looked at the idea of an indifference curve. So for example, let's say I'm sitting at some point on my budget line where I have-- let's say I am consuming 18 bars of chocolate and 1 pound of fruit. 18-- and you can verify that make sense, it's going to be $18 plus$2, which is $20. So let's say I'm at this point on my budget line. 18 bars of chocolate, so this is in bars, and 1 pound of fruit per month. So that is 1. And this is in pounds. And this is chocolate, and this is fruit right over here. Well, we know we have this idea of an indifference curve. There's different combinations of chocolate and fruit to which we are indifferent, to which we would get the same exact total utility. And so we can plot all of those points. I'll do it in white. It could look something like this. I'll do it as a dotted line, it makes it a little bit easier. So let me draw it like this. So let's say I'm indifferent between any of these points, any of those points right over there. Let me draw it a little bit better. So between any of these points right over there. So for example, I could have 18 bars of chocolate and 1 pound of fruit, or I could have-- let's say that is 4 bars of chocolate and roughly 8 pounds of fruit. I'm indifferent. I get the same exact total utility. Now, am I maximizing my total utility at either of those points? Well, we've already seen that anything to the top right of our indifference curve of this white curve right over here-- let me label this. This is our indifference curve. Everything to the top right of our indifference curve is preferable. We're going to get more total utility. So let me color that in. So everything to the top right of our indifference curve is going to be preferable. So all of these other points on our budget line, even a few points below or budget line, where we would actually save money, are preferable. So either of these points are not going to maximize our total utility. We can maximize or total utility at all of these other points in between, along our budget line. So to actually maximize our total utility what we want to do is find a point on our budget line that is just tangent, that exactly touches at exactly one point one of our indifference curves. We could have an infinite number of indifference curves. There could be another indifference curve that looks like that. There could be another indifferent curve that looks like that. All that says is that we are indifferent between any points on this curve. And so there is an indifference curve that touches exactly this budget line, or exactly touches the line at one point. And so I might have an indifference curve that looks like this. Let me do this in a vibrant color, in magenta. So I could have an indifference curve that looks like this. And because it's tangent, it touches at exactly one point. And also the slope of my indifference curve, which we've learned was the marginal rate of substitution, is the exact same as the slope of our budget line right over there, which we learned earlier was the relative price. So this right about here is the optimal allocation on our budget line. That right here is optimal. And how do we know it is optimal? Well, there is no other point on the budget line that is to the top right. In fact, every other point on our budget line is to the bottom left of this indifference curve. So every other point on our budget line is not preferable. So remember, everything below an indifference curve-- so all of this shaded area. Let me actually do it in another color. Because indifference curve, we are different. But everything below an indifference curve, so all of this area in green, is not preferable. And every other point on the budget line is not preferable to that point right over there. Because that's the only point-- or I guess you could say, every other point on our budget line is not preferable to the points on the indifference curve. So they're also not preferable to that point right over there which actually is on the indifference curve. Now, let's think about what happens. Let's think about what happens if the price of fruit were to go down. So the price of fruit were to go from$2 to $1 per pound. So if the price of fruit went from$2 to $1, then our actual budget line will look different. Our new budget line. I'll do it in blue, would look like this. If we spent all our money on chocolate, we could buy 20 bars. If we spent all of our money on fruit at the new price, we could buy 20 pounds of fruit. So our new budget line would look something like that. So that is our new budget line. So now what would be the optimal allocation of our dollars or the best combination that we would buy? Well, we would do the exact same exercise. We would, assuming that we had data on all of these indifference curves, we would find the indifference curve that is exactly tangent to our new budget line. So let's say that this point right over here is exactly tangent to another indifference curve. So just like that. So there's another indifference curve that looks like that. Let me draw it a little bit neater. So it looks something like that. And so based on how the price-- if we assume we have access to these many, many, many, many, many indifference curves, we can now see based on, all else equal, how a change in the price of fruit changed the quantity of fruit we demanded. Because now our optimal spent is this point on our new budget line which looks like it's about, well, give or take, about 10 pounds of fruit. So all of a sudden, when we were-- so let's think about just the fruit. Everything else we're holding equal. So just the fruit, let's do, when the price was$2, the quantity demanded was 8 pounds. And now when the price is $1, the quantity demanded is 10 pounds. And so what we're actually doing, and once again, we're kind of looking at the exact same ideas from different directions. Before we looked at it in terms of marginal utility per dollar and we thought about how you maximize it. And we were able to change the prices and then figure out and derive a demand curve from that. Here we're just looking at it from a slightly different lens, but they really are all of the same ideas. But by-- assuming if we had access to a bunch of indifference curves, we can see how a change in price changes our budget line. And how that would change the optimal quantity we would want of a given product. So for example, we could keep doing this and we could plot our new demand curve. So I could do a demand curve now for fruit. At least I have two points on that demand curve. So if this is the price of fruit and this is the quantity demanded of fruit, when the price is$2, the quantity demanded is 8. And when the price is-- actually, let me do it a little bit different. When the price is $2-- these aren't to scale-- the quantity demanded is 8. Actually let me do it here-- is 8. And these aren't to scale. But when the price is$1, the quantity demanded is 10. So \$2, 8, the quantity demanded is 10. And so our demand curve, these are two points on it. But we could keep changing it up assuming we had access to a bunch of indifference curves. We could keep changing it up and eventually plot our demand curve, that might look something like that.	2017-07-26 00:42:00	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6675105094909668, "perplexity": 388.4014747339713}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-30/segments/1500549425737.60/warc/CC-MAIN-20170726002333-20170726022333-00687.warc.gz"}
http://rpg.stackexchange.com/questions/27742/can-versatile-spellcaster-be-used-for-multiple-expenditures/27744	# Can Versatile Spellcaster be used for multiple expenditures? One of the players in our campaign has a 6th level Favored Soul. They have selected Versatile Spellcaster as their next feat; it reads: You can use two spell slots of the same level to cast a spell you know that is one level higher. For example, a sorcerer with this feat can expend two 2nd-level spell slots to cast any 3rd-level spell [they know]. -- source Does this effect chain? In other words, could the Favored Soul expend: ((Lv0 + Lv0 -> Lv1) + Lv1 -> Lv2) + Lv2 -> Lv3 to gain a level 3 spell? It would expend two level 0, one level 1, and one level 2, but would be extremely powerful at higher levels. How does this work, and is this an acceptable use of Versatile Spellcaster? If so, are there any balance concerns to be aware of? - No, Versatile Spellcaster does not "chain" in this way. There is not much in the way of rules text for this feat, but part of it is that the character uses two spell slots to cast a spell. The thing that makes chaining impossible is that there is a difference between having a spell slot and casting a spell. Versatile Spellcaster only grants the latter, not a spell slot that can be used to fuel another use of the feat. There are quite a few interesting and supremely powerful things this feat can do, but chaining is not one of them. - ## protected by C. RossMar 8 '14 at 16:06 Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).	2016-06-29 23:38:40	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.502675473690033, "perplexity": 2963.1562030000696}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-26/segments/1466783397864.87/warc/CC-MAIN-20160624154957-00034-ip-10-164-35-72.ec2.internal.warc.gz"}
http://math.stackexchange.com/questions/80174/derivative-of-this-function-quadratic-over-quadratic	$$K = \frac{-(\sigma^2 + 3\sigma + 2)}{\sigma^2 - 8\sigma +15}$$ How is this differentiated with respect to $\sigma$? $$\dfrac{dK}{d\sigma} = \frac{11\sigma^2 - 26\sigma -61}{(\sigma^2 - 8\sigma + 15)^2}$$ And the quotient rule is a very long process, I get the feeling there was a short method. - It's just the quotient rule. $(f/g)'={f'g-g'f\over g^2}$. Your comment about partial fractions leads me to say "the derivative, not the indefinite integral"... – David Mitra Nov 8 '11 at 12:58 That would be the quotient rule... partial fractions is an integration technique, not differentiation. – process91 Nov 8 '11 at 12:58 Er I meant quotient, I was tired – Supernovah Nov 8 '11 at 12:59 Oh thanks, I didn't know that rule – Supernovah Nov 8 '11 at 13:06 The three comments below are not really shortcuts. More like longcuts. Because $x$ is easier to type than $\sigma$, the variable has been changed. We show how to differentiate $\frac{x^2+3x+2}{x^2-8x+15}$ and leave it to someone else to change the sign at the end. $1$) We are being asked to divide $x^2+3x+2$ by $x^2-8x+15$. So divide. We get $$\frac{x^2+3x+2}{x^2-8x+15}=1+\frac{11x-13}{x^2-8x+15}.$$ The derivative of $1$ is $0$, so we want the derivative of $$\frac{11x-13}{x^2-8x+15}.$$ The algebra of the Quotient Rule is definitely easier with a linear polynomial on top than if we use the Rule on the original expression. $2$) If we want to practice partial fractions before they are needed for integration, we can simplify further. Note that $x^2-8x+15=(x-3)(x-5)$. The partial fractions process shows that $$\frac{11x-13}{x^2-8x+15}=-\frac{10}{x-3}+\frac{21}{x-5}.$$ Now differentiation is genuinely easy. $3$) Logarithmic differentiation can be useful. We will be deliberately sloppy, and not worry about the fact that $\log u$ is not defined when $u \le 0$. One can show that negative $u$ in fact give no problem. The derivative of $\log\|u\|$ with respect to $u$ is $\frac{1}{u}$. Breaking up the interval so that we can deal with $\log$ properly gives the same derivative as the one we get if we just heedlessly calculate. Let $f(x)=\frac{x^2+3x+2}{x^2-8x+15}$. Then $$\log f(x)=\log(x^2+3x+2)-\log(x^2-8x+15).$$ Differentiate. We get more or less instantly $$\frac{f'(x)}{f(x)}=\frac{2x+3}{x^2+3x+2}-\frac{2x-8}{x^2-8x+15}.\qquad (\ast)$$ If we have to "simplify" $(\ast)$ to the form $\frac{P(x)}{Q(x)}$ where $P(x)$ and $Q(x)$ are polynomials, the simplification will not be much fun. But the expression $(\ast)$ is certainly pleasant enough if we just want to evaluate the derivative at a particular numerical value of $x$. For complicated products/quotients, "logarithmic differentiation" can be, for certain purposes, much more efficient than conventional differentiation. However, if we want to find where the derivative vanishes, the advantage tends to evaporate. - The quotient rule is not too long... \begin{align}\frac {dK}{d\sigma} =& \frac {(-2\sigma-3)(\sigma^2-8\sigma+15)-(2\sigma-8)(-\sigma^2-3\sigma-2)}{(\sigma^2-8\sigma+15)^2}\\ =& \frac{-(2\sigma^3-13\sigma^2+14\sigma+45)+(2\sigma^3-2\sigma^2-12\sigma-16)}{(\sigma^2-8\sigma+15)^2}\\ =& \frac{11\sigma^2-26\sigma-61}{(\sigma^2-8\sigma+15)^2}\end{align} The squared denominator is a good clue that the book (or whatever reference you are using for the answer) used the quotient rule. I don't see any shortcuts to this question, so if you are being asked to differentiate it you should be familiar with the quotient rule at this point. http://en.wikipedia.org/wiki/Quotient_rule - He could use so-called ''logarithmic differentiation''; but I don't think this would qualify as a ''shortcut''. – David Mitra Nov 8 '11 at 13:26 Actually, OP could've used a combination of the product rule and chain rule if he didn't remember how to differentiate a quotient... – Ｊ. Ｍ. Nov 8 '11 at 13:35 Definitely true, although neither seems to be much of a "shortcut" (as you mentioned). Perhaps it's worth mentioning that the quotient rule is simply an application of the product rule and the chain rule? Although that's probably in the Wikipedia entry... – process91 Nov 8 '11 at 13:37	2015-01-30 14:06:21	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9688724875450134, "perplexity": 422.6842469714994}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-06/segments/1422115857131.28/warc/CC-MAIN-20150124161057-00019-ip-10-180-212-252.ec2.internal.warc.gz"}
http://answerparty.com/question/answer/how-do-cancer-cells-feed	Question: How do cancer cells feed? Answer: Cancer doesn't feed. It is uncontrolled cell division, the molecular structure gets changed somehow. More Info: Cancer cells are cells that grow and divide at an unregulated, quickened pace. Although cancer cells can be quite common in a person they are only malignant when the other cells (particularly natural killer cells) fail to recognize and/or destroy them. In the past a common belief was that cancer cells failed to be recognized and destroyed because of a weakness in the immune system. However, more recent research has shown that the failure to recognize cancer cells is caused by the lack of particular co-stimulated molecules that aid in the way antigens react with lymphocytes. By researching stem cells scientists have suggested that too much SP2 protein may turn stem cells into cancer cells. Other issues thought to play a role in the spread of cancer include viruses, immune system issues, genetics, environment and age. However, a lack of particular co-stimulated molecules that aid in the way antigens react with lymphocytes can impair the natural killer cells ability and ultimately cause cancer. All cancers begin in cells, the body's basic unit of life. To understand cancer, it's helpful to know what happens when normal cells become cancer cells. The body is made up of many types of cells. These cells grow and are controlled to produce more cells as they are needed to keep the body healthy. When cells become old or damaged, they die and are replaced with new cells. Sometimes this process of controlled production of cells goes wrong. The genetic material (DNA) of a cell start producing mutations that affect normal cell growth and division by being damaged. When this happens, these cells do not die but form a mass of tissue called a tumor. Said mutations accumulate, being another reason that cancer is found more often in older people. White Blood cells are thought to use a dual receptor system when they determine whether or not to kill human cells. If a cell is under stress, turning into tumors, or infected molecules including MIC-A and MIC-B are produced to put on the surface of the cell. These work to detect and kill cancer cells. Some descriptions of cancer go back to ancient Egypt as far back as 1600 BC and the understanding of cancer was significantly advanced during the Renaissance period. However, Sir Rudolf Virchow, a German biologist and politician, is generally credited with discovering the first cancer cells. As Giovanni Morgagni had linked autopsy findings seen with the unaided eye with the clinical course of illness, so Virchow correlated the microscopic pathology. Cancer cells have unique features that make them "immortal" according to some researchers. The enzyme telomerase is used to extend the cancer cell's life span. While the telomeres of most cells shortens after each division eventually causing the cell to die, telomerase extends the cell's telomeres. This is a major reason that cancer cells can accumulate over time creating tumors. M: NEO tsoc, mrkr tumr, epon, para drug (L1i/1e/V03) Programmed cell-death (or PCD) is death of a cell in any form, mediated by an intracellular program. PCD is carried out in a regulated process, which usually confers advantage during an organism's life-cycle. For example, the differentiation of fingers and toes in a developing human embryo occurs because cells between the fingers apoptose; the result is that the digits are separate. PCD serves fundamental functions during both plant and metazoa (multicellular animals) tissue development. Apoptosis and autophagy are both forms of programmed cell death, however necrosis is a non-physiological process that occurs as a result of infection or injury. Necrosis is the death of a cell caused by external factors such as trauma or infection and occurs in several different forms. Recently a form of programmed Necrosis, also called Necroptosis has been recognized as an alternate form of programmed cell death. It is hypothesized that Necroptosis can serve as a cell-death backup to apoptosis when the apoptosis signaling is blocked by endogenous or exogenous factors such as viruses or mutations. Apoptosis is the process of programmed cell death (PCD) that may occur in multicellular organisms. Biochemical events lead to characteristic cell changes (morphology) and death. These changes include blebbing, cell shrinkage, nuclear fragmentation, chromatin condensation, and chromosomal DNA fragmentation. It is now thought that in a developmental context cells are induced to positively commit suicide whilst in a homeostatic context the absence of certain survival factors may provide the impetus for suicide. There appears to be some variation in the morphology and indeed the biochemistry of these suicide pathways; some treading the path of "apoptosis", others following a more generalized pathway to deletion, but both usually being genetically and synthetically motivated. There is some evidence that certain symptoms of "apoptosis" such as endonuclease activation can be spuriously induced without engaging a genetic cascade, however, presumably true apoptosis and programmed cell death must be genetically mediated. It is also becoming clear that mitosis and apoptosis are toggled or linked in some way and that the balance achieved depends on signals received from appropriate growth or survival factors. Macroautophagy, often referred to as autophagy, is a catabolic process that results in the autophagosomic-lysosomal degradation of bulk cytoplasmic contents, abnormal protein aggregates, and excess or damaged organelles. Autophagy is generally activated by conditions of nutrient deprivation but has also been associated with physiological as well as pathological processes such as development, differentiation, neurodegenerative diseases, Stress (physiology), Infection and cancer. A critical regulator of autophagy induction is the kinase mTOR, which when activated, suppresses autophagy and when not activated promotes it. Three related serine/threonine kinases, UNC-51-like kinase -1, -2, and -3 (ULK1, ULK2, UKL3), which play a similar role as the yeast Atg1, act downstream of the mTOR complex. ULK1 and ULK2 form a large complex with the mammalian homolog of an autophagy-related (Atg) gene product (mAtg13) and the scaffold protein FIP200. Class III PI3K complex, containing hVps34, Beclin-1, p150 and Atg14-like protein or ultraviolet irradiation resistance-associated gene (UVRAG), is required for the induction of autophagy. The ATG genes control the autophagosome formation through ATG12-ATG5 and LC3-II (ATG8-II) complexes. ATG12 is conjugated to ATG5 in a ubiquitin-like reaction that requires ATG7 and ATG10. The Atg12–Atg5 conjugate then interacts non-covalently with ATG16 to form a large complex. LC3/ATG8 is cleaved at its C terminus by ATG4 protease to generate the cytosolic LC3-I. LC3-I is conjugated to phosphatidylethanolamine (PE) also in a ubiquitin-like reaction that requires Atg7 and Atg3. The lipidated form of LC3, known as LC3-II, is attached to the autophagosome membrane. Autophagy and apoptosis are connected both positively and negatively, and extensive crosstalk exists between the two. During nutrient deficiency, autophagy functions as a pro-survival mechanism, however, excessive autophagy may lead to cell death, a process morphologically distinct from apoptosis. Several pro-apoptotic signals, such as TNF, TRAIL, and FADD, also induce autophagy. Additionally, Bcl-2 inhibits Beclin-1-dependent autophagy, thereby functioning both as a pro-survival and as an anti-autophagic regulator. Besides the above two types of PCD, other pathways have been discovered. Called "non-apoptotic programmed cell-death" (or "caspase-independent programmed cell-death" or "necroptosis"), these alternative routes to death are as efficient as apoptosis and can function as either backup mechanisms or the main type of PCD. Other forms of programmed cell death include anoikis, almost identical to apoptosis except in its induction; cornification, a form of cell death exclusive to the eyes; excitotoxicity and Wallerian degeneration. Plant cells undergo particular processes of PCD similar to autophagic cell death. However, some common features of PCD are highly conserved in both plants and metazoa. An atrophic factor is a force that causes a cell to die. Only natural forces on the cell are considered to be atrophic factors, whereas, for example, agents of mechanical or chemical abuse or lysis of the cell are considered not to be atrophic factors.][ Common types of atrophic factors are: The concept of "programmed cell-death" was used by Lockshin & Williams in 1964 in relation to insect tissue development, around eight years before "apoptosis" was coined. Since then, PCD has become the more general of these two terms. The first insight into the mechanism came from studying BCL2, the product of a putative oncogene activated by chromosome translocations often found in follicular lymphoma. Unlike other cancer genes, which promote cancer by stimulating cell proliferation, BCL2 promoted cancer by stopping lymphoma cells from being able to kill themselves. PCD has been the subject of increasing attention and research efforts. This trend has been highlighted with the award of the 2002 Nobel Prize in Physiology or Medicine to Sydney Brenner (United Kingdom), H. Robert Horvitz (US) and John E. Sulston (UK). Programmed cell death in plants has a number of molecular similarities to animal apoptosis, but it also has differences, the most obvious being the presence of a cell wall and the lack of an immune system that removes the pieces of the dead cell. Instead of an immune response, the dying cell synthesizes substances to break itself down and places them in a vacuole that ruptures as the cell dies. In "APL regulates vascular tissue identity in Arabidopsis", Martin Bonke and his colleagues had stated that one of the two long-distance transport systems in vascular plants, xylem, consists of several cell-types "the differentiation of which involves deposition of elaborate cell-wall thickenings and programmed cell-death." The authors emphasize that the products of plant PCD play an important structural role. Basic morphological and biochemical features of PCD have been conserved in both plant and animal kingdoms. It should be noted, however, that specific types of plant cells carry out unique cell-death programs. These have common features with animal apoptosis—for instance, nuclear DNA degradation—but they also have their own peculiarities, such as nuclear degradation triggered by the collapse of the vacuole in tracheary elements of the xylem. Janneke Balk and Christopher J. Leaver, of the Department of Plant Sciences, University of Oxford, carried out research on mutations in the mitochondrial genome of sun-flower cells. Results of this research suggest that mitochondria play the same key role in vascular plant PCD as in other eukaryotic cells. During pollination, plants enforce self-incompatibility (SI) as an important means to prevent self-fertilization. Research on the corn poppy (Papaver rhoeas) has revealed that proteins in the pistil on which the pollen lands, interact with pollen and trigger PCD in incompatible (i.e., self) pollen. The researchers, Steven G. Thomas and Veronica E. Franklin-Tong, also found that the response involves rapid inhibition of pollen-tube growth, followed by PCD. The social slime mold Dictyostelium discoideum has the peculiarity of either adopting a predatory amoeba-like behavior in its unicellular form or coalescing into a mobile slug-like form when dispersing the spores that will give birth to the next generation. The stalk is composed of dead cells that have undergone a type of PCD that shares many features of an autophagic cell-death: massive vacuoles forming inside cells, a degree of chromatin condensation, but no DNA fragmentation. The structural role of the residues left by the dead cells is reminiscent of the products of PCD in plant tissue. D. discoideum is a slime mold, part of a branch that might have emerged from eukaryotic ancestors about a billion years before the present. It seems that they emerged after the ancestors of green plants and the ancestors of fungi and animals had differentiated. But, in addition to their place in the evolutionary tree, the fact that PCD has been observed in the humble, simple, six-chromosome D. discoideum has additional significance: It permits the study of a developmental PCD path that does not depend on caspases characteristic of apoptosis. Biologists had long suspected that mitochondria originated from bacteria that had been incorporated as endosymbionts ("living together inside") of larger eukaryotic cells. It was Lynn Margulis who from 1967 on championed this theory, which has since become widely accepted. The most convincing evidence for this theory is the fact that mitochondria possess their own DNA and are equipped with genes and replication apparatus. This evolutionary step would have been risky for the primitive eukaryotic cells, which began to engulf the energy-producing bacteria, as well as a perilous step for the ancestors of mitochondria, which began to invade their proto-eukaryotic hosts. This process is still evident today, between human white blood cells and bacteria. Most of the time, invading bacteria are destroyed by the white blood cells; however, it is not uncommon for the chemical warfare waged by prokaryotes to succeed, with the consequence known as infection by its resulting damage. One of these rare evolutionary events, about two billion years before the present, made it possible for certain eukaryotes and energy-producing prokaryotes to coexist and mutually benefit from their symbiosis. Mitochondriate eukaryotic cells live poised between life and death, because mitochondria still retain their repertoire of molecules that can trigger cell suicide. This process has now been evolved to happen only when programmed.][ Given certain signals to cells (such as feedback from neighbors, stress or DNA damage), mitochondria release caspase activators that trigger the cell-death-inducing biochemical cascade. As such, the cell suicide mechanism is now crucial to all of our lives. The BCR-ABL oncogene has been found to be involved in the development of cancer in humans. c-Myc is involved in the regulation of apoptosis via its role in downregulating the Bcl-2 gene. Its role the disordered growth of tissue. A molecular characteristic of metastatic cells is their altered expression of several apoptotic genes. M: OBS phys/devp/memb mthr/fetu/infc, epon proc, drug (2A/G2C) Cancer stem cells (CSCs) are cancer cells (found within tumors or hematological cancers) that possess characteristics associated with normal stem cells, specifically the ability to give rise to all cell types found in a particular cancer sample. CSCs are therefore tumorigenic (tumor-forming), perhaps in contrast to other non-tumorigenic cancer cells. CSCs may generate tumors through the stem cell processes of self-renewal and differentiation into multiple cell types. Such cells are proposed to persist in tumors as a distinct population and cause relapse and metastasis by giving rise to new tumors. Therefore, development of specific therapies targeted at CSCs holds hope for improvement of survival and quality of life of cancer patients, especially for sufferers of metastatic disease. Existing cancer treatments have mostly been developed based on animal models, where therapies able to promote tumor shrinkage were deemed effective. However, animals could not provide a complete model of human disease. In particular, in mice, whose life spans do not exceed two years, tumor relapse is exceptionally difficult to study. The efficacy of cancer treatments is, in the initial stages of testing, often measured by the ablation fraction of tumor mass (fractional kill). As CSCs would form a very small proportion of the tumor, this may not necessarily select for drugs that act specifically on the stem cells. The theory suggests that conventional chemotherapies kill differentiated or differentiating cells, which form the bulk of the tumor but are unable to generate new cells. A population of CSCs, which gave rise to it, could remain untouched and cause a relapse of the disease. In different tumor subtypes, cells within the tumor population exhibit functional heterogeneity, tumor are formed from cells with various proliferative and differentiate capacities. This functional heterogeneity among cancer cells lead to create at least two models, which have put forward to account for heterogeneity and differences in tumor-regenerative capacity, the cancer stem cells (CSC) and clonal evolution models The cancer stem cell model refers to a subset of tumor cells that have the ability to self-renew and are capable to generate the diverse tumor cells. These cells have been termed cancer stem cells to reflect their stem-like properties. One implication of the CSC model and the existence of CSCs is that the tumor population is hierarchically arranged with CSCs lying at the apex of the hierarchy (Fig. 3). The clonal evolution model postulates that mutant tumor cells with a growth advantage are selected and expanded. Cells in the dominant population have a similar potential for initiating tumor growth (Fig. 4). These two models are not mutually exclusive, as CSCs themselves undergo clonal evolution. Thus, the secondary more dominant CSCs may emerge, if a mutation confers more aggressive properties (Fig. 5). The existence of CSCs is a subject of debate within medical research, because many studies have not been successful in discovering the similarities and differences between normal tissue stem cells and cancer (stem) cells. Cancer cells must be capable of continuous proliferation and self-renewal in order to retain the many mutations required for carcinogenesis, and to sustain the growth of a tumor since differentiated cells (constrained by the Hayflick Limit) cannot divide indefinitely. However, it is debated whether such cells represent a minority. If most cells of the tumor are endowed with stem cell properties, there is no incentive to focus on a specific subpopulation. There is also debate on the cell of origin of CSCs - whether they originate from normal stem cells that have lost the ability to regulate proliferation, or from more differentiated population of progenitor cells that have acquired abilities to self-renew (which is related to the issue of stem cell plasticity). The first conclusive evidence for CSCs was published in 1997 in Nature Medicine. Bonnet and Dick isolated a subpopulation of leukaemic cells that expressed a specific surface marker CD34, but lacked the CD38 marker. The authors established that the CD34+/CD38- subpopulation is capable of initiating tumors in NOD/SCID mice that are histologically similar to the donor. In cancer research experiments, tumor cells are sometimes injected into an experimental animal to establish a tumor. Disease progression is then followed in time and novel drugs can be tested for their ability to inhibit it. However, efficient tumor formation requires thousands or tens of thousands of cells to be introduced. Classically, this has been explained by poor methodology (i.e. the tumor cells lose their viability during transfer) or the critical importance of the microenvironment, the particular biochemical surroundings of the injected cells. Supporters of the CSC paradigm argue that only a small fraction of the injected cells, the CSCs, have the potential to generate a tumor. In human acute myeloid leukemia the frequency of these cells is less than 1 in 10,000. Further evidence comes from histology, the study of the tissue structure of tumors. Many tumors are very heterogeneous and contain multiple cell types native to the host organ. Heterogeneity is commonly retained by tumor metastases. This implies that the cell that produced them had the capacity to generate multiple cell types. In other words, it possessed multidifferentiative potential, a classical hallmark of stem cells. The existence of leukaemic stem cells prompted further research into other types of cancer. CSCs have recently been identified in several solid tumors, including cancers of the: Once the pathways to cancer are hypothesized, it is possible to develop predictive mathematical biology models, e.g., based on the cell compartment method. For instance, the growths of the abnormal cells from their normal counterparts can be denoted with specific mutation probabilities. Such a model has been employed to predict that repeated insult to mature cells increases the formation of abnormal progeny, and hence the risk of cancer. Considerable work needs to be done, however, before the clinical efficacy of such models is established. The origin of cancer stem cells is still an area of ongoing research. Several camps have formed within the scientific community regarding the issue, and it is possible that several answers are correct, depending on the tumor type and the phenotype the tumor presents. One important distinction that will often be raised is that the cell of origin for a tumor can not be demonstrated using the cancer stem cell as a model. This is because cancer stem cells are isolated from end-stage tumors. Therefore, describing a cancer stem cell as a cell of origin is often an inaccurate claim, even though a cancer stem cell is capable of initiating new tumor formation. With that caveat mentioned, various theories define the origin of cancer stem cells. In brief, CSC can be generated as: mutants in developing stem or progenitor cells, mutants in adult stem cells or adult progenitor cells, or mutant differentiated cells that acquire stem like attributes. These theories often do focus on a tumor's cell of origin and as such must be approached with skepticism. Some researchers favor the theory that the cancer stem cell is generated by a mutation in stem cell niche populations during development. The logical progression claims that these developing stem populations are mutated and then expand such that the mutation is shared by many of the descendants of the mutated stem cell. These daughter stem cells are then much closer to becoming tumors, and since there are many of them there is more chance of a mutation that can cause cancer. Another theory associates adult stem cells with the formation of tumors. This is most often associated with tissues with a high rate of cell turnover (such as the skin or gut). In these tissues, it has long been expected that stem cells are responsible for tumor formation. This is a consequence of the frequent cell divisions of these stem cells (compared to most adult stem cells) in conjunction with the extremely long lifespan of adult stem cells. This combination creates the ideal set of circumstances for mutations to accumulate; accumulation of mutations is the primary factor that drives cancer initiation. In spite of the logical backing of the theory, only recently has an evidence appeared showing association represents an actual phenomenon. It is important to bear in mind that due to the heterogeneous nature of evidence it is possible that any individual cancer could come from an alternative origin. A third possibility often raised is the potential de-differentiation of mutated cells such that these cells acquire stem cell like characteristics. This is often used as a potential alternative to any specific cell of origin, as it suggests that any cell might become a cancer stem cell. Another related concept is the concept of tumor hierarchy. This concept claims that a tumor is a heterogeneous population of mutant cells, all of which share some mutations but vary in specific phenotype. In this model, the tumor is made up of several types of stem cells, one optimal to the specific environment and several less successful lines. These secondary lines can become more successful in some environments, allowing the tumor to adapt to its environment, including adaptation to tumor treatment. If this situation is accurate, it has severe repercussions on cancer stem cell specific treatment regime. Within a tumor hierarchy model, it would be extremely difficult to pinpoint the cancer stem cell's origin. CSC, now reported in most human tumors, are commonly identified and enriched using strategies for identifying normal stem cells that are similar across the studies. The procedures include fluorescence-activated cell sorting (FACS) with antibodies directed at cell-surface markers and functional approaches including SP analysis (side population assay) or Aldefluor assay. The CSC-enriched population purified by these approaches is then implanted, at various cell doses, in immune-deficient mice to assess its tumor development capacity. This in vivo assay is called limiting dilution assay. The tumor cell subsets that can initiate tumor development at low cell numbers are further tested for self-renewal capacity in serial tumor capacity. CSC can also be identified by efflux of incorporated Hoechst dyes via multidrug resistance (MDR) and ATP-binding cassette (ABC) Transporters][. Another approach which has also been used for identification of cell subset enriched with in CSCs in vitro is sphere-forming assays. Many normal stem cells such as hematopoietics or stem cells from tissues are capable, under special culture conditioned, to form three-dimensional spheres, which can differentiate into multiple cell types. Similarly as normal stem cells, the CSCs isolated from brain or prostate tumors has also ability to form anchorage-independent spheres. Data over recent years have indicated the existence of CSC in various solid tumors. For isolating CSC from solid and hematological tumors markers specific for normal stem cells of the same organ are commonly use. Nevertheless, a number of cell surface markers have proved useful for isolation of subsets enriched for CSC including CD133 (also known as PROM1), CD44, CD24, EpCAM (epithelial cell adhesion molecule, also known as epithelial specific antigen, ESA), THY1 and ATP-binding cassette B5 (ABCB5). CD133 (prominin 1) is a five-transmembrane domain glycoprotein expressed on CD34+ stem and progenitor cells, in endothelial precursors and fetal neural stem cells. It has been detected using its glycosylated epitope know as AC133. EpCAM (epithelial cell adhesion molecule, ESA, TROP1) is hemophilic CA2+-independent cell adhesion molecule expressed on the basolateral surface of most epithelial cells. CD90 (THY1) is a glycosylphosphatidylinositol glycoprotein anchored in the plasma membrane and involved in signal transduction. It may also mediate adhesion between thymocytes and thymic stroma. CD44 (PGP1) is an adhesion molecule that has pleiotropic roles in cell signaling, migration and homing. It has multiple isoforms, including CD44H, which exhibits high affinity for hyaluronate, and CD44V which has metastatic properties. CD24 (HSA) is a glycosylated glycosylphosphatidylinositol-anchored adhesion molecule, which has co-stimulatory role in B and T cells. ALDH is a ubiquitous aldehyde dehydrogenase family of enzymes, which catalyzes the oxidation of aromatic aldehydes to carboxyl acids. For instance, it has role in conversion of retinol to retinoic acid, which is essential for survival. The first solid malignancy from which CSCs were isolated and identified was breast cancer. Therefore these SCSs are the most intensely studied. Breast CSC have been enriched in CD44+CD24-/low, SP, ALDH+ subpopulations. However, recent evidence indicates that breast CSC are very phenotypically diverse population, and there is evidence that not only CSC marker expression in breast cancer cells is heterogeneous but also there exist many subsets of breast CSC. Last studies provide further support to this point. Both CD44+CD24- and CD44+CD24+ cell populations are tumor initiating cells, however CSC are most highly enriched using the marker profile CD44+CD49fhiCD133/2hi. CSCs have been reported in many brain tumors. The stem-like tumors cells have been identified using cell surface markers including CD133, SSEA-1 (stage-specific embryonic antigen-1), EGFR][ and CD44. However, there is uncertainties about the use of CD133 for identification of brain tumor stem-like cells, because tumorigenic cells are found in both CD133+ and CD133- cells in some gliomas, and some CD133+ brain tumor cells may not possess tumor-initiating capacity. Similarly, CSCs have also been reported in human colon cancer. For their identification, cell surface markers as CD133, CD44 and ABCB5, or functional analysis including clonal analysis or Aldefluor assay were used using CD133 as positive marker for colon CSCs has generated conflicting results. Nevertheless, recent studies indicated that the AC133 epitope, but not the CD133 protein, is specifically expressed in colon CSCs and its expression is lost upon differentiation. In addition, using CD44+ colon cancer cells and additional sub-fractionation of CD44+EpCAM+ cell population with CD166 enhance the success of tumor engraftments. Multiple CSCs have been reported in prostate, lung and many other organs cancer, including liver, pancreas, kidney or ovary. In prostate cancer, the tumor-initiating cells have been identified in CD44+ cell subset as CD44+α2β1+, TRA-1-60+CD151+CD166+ or ALDH+ cell populations. Putative markers for lung CSCs have been reported, including CD133+, ALDH+, CD44+ and oncofetal protein 5T4+. Metastasis is the major cause of tumor lethality in patients. However, not every cell in the tumor has the ability to metastasize. This potential depends on factors that determine growth, angiogenesis, invasion and other basic processes of tumor cells. In the many epithelial tumors, the epithelial-mesenchymal transition (EMT) is considered as a crucial events in the metastatic process. EMT and the reverse transition from mesenchymal to an epithelial phenotype (MET) are involved in embryonic development, which involves disruption of epithelial cell homeostasis and the acquisition of a migratory mesenchymal phenotype. The EMT appears to be controlled by canonical pathways such as WNT and transforming growth factor β pathway. The important feature of EMT is the loss of membrane E-cadherin in adherent junctions, where the β-catenin may play a significant role. Translocation of β-catenin from adherent junctions to the nucleus may lead to a loss of E-cadherine, and subsequently to EMT. There is an evidence that nuclear β-catenin can directly transcriptionally activate EMT-associated target genes, such as the E-cadherine gene repressor SLUG (also known as SNAI2). Recent data have supported the concept, that tumor cells undergoing an EMT could be precursors for metastatic cancer cells, or even metastatic CSCs. In the invasive edge of pancreatic carcinoma a subset of CD133+CXCR4+ (receptor for CXCL12 chemokine also known as a SDF1 ligand) cells has been defined. These cells exhibited significantly stronger migratory activity than their counterpart CD133+CXCR4- cells, but both cell subsets showed similar tumor development capacity. Moreover, inhibition of the CXCR4 receptor led to the reduced metastatic potential without altering tumorigenic capacity. On the other hand, in the breast cancer CD44+CD24-/low cells are detectable in metastatic pleural effusions. By contrast, an increased number of CD24+ cells have been identified in distant metastases in patients with breast cancer. Although, there are only few data on mechanisms mediating metastasis in breast cancer, it is possible that CD44+CD24-/low cells initially metastasize and in the new site they change their phenotype and undergo limited differentiation. These findings led to new dynamic two-phase expression pattern concept based on the existence of two forms of cancer stem cells - stationary cancer stem cells (SCS) and mobile cancer stem cells (MCS). SCS are embedded in tissue and persist in differentiated areas throughout all tumor progression. The term MCS describes cells that are located at the tumor-host interface. There is an evidence that these cells are derived from SCS through the acquisition of transient EMT (Fig. 7) The existence of CSCs has several implications in terms of future cancer treatment and therapies. These include disease identification, selective drug targets, prevention of metastasis, and development of new intervention strategies. Normal somatic stem cells are naturally resistant to chemotherapeutic agents. They produce various pumps (such as MDR][) that pump out drugs and DNA repair proteins and they also have a slow rate of cell turnover (chemotherapeutic agents naturally target rapidly replicating cells)][. CSCs that developed from normal stem cells may also produced these proteins that could increase their resistance towards chemotherapeutic agents. The surviving CSCs then repopulate the tumor causing a relapse. By selectively targeting CSCs, it would be possible to treat patients with aggressive, non-resectable tumors, as well as preventing patients from metastasizing. The hypothesis suggests that upon CSC elimination, cancer could regress due to differentiation and/or cell death][. What fraction of tumor cells are CSCs and therefore need to be eliminated is not clear yet. A number of studies have investigated the possibility of identifying specific markers that may distinguish CSCs from the bulk of the tumor (as well as from normal stem cells). Proteomic and genomic signatures of tumors are also being investigated.][. In 2009, scientists identified one compound, Salinomycin, that selectively reduces the proportion of breast CSCs in mice by more than 100-fold relative to Paclitaxel, a commonly used chemotherapeutic agent. The cell surface receptor interleukin-3 receptor-alpha (CD123) was shown to be overexpressed on CD34+CD38- leukemic stem cells (LSCs) in acute myelogenous leukemia (AML) but not on normal CD34+CD38- bone marrow cells. Jin et al., then demonstrated that treating AML-engrafted NOD/SCID mice with a CD123-specific monoclonal antibody impaired LSCs homing to the bone marrow and reduced overal AML cell repopulation including the proportion of LSCs in secondary mouse recipients. The design of new drugs for the treatment of CSCs will likely require an understanding of the cellular mechanisms that regulate cell proliferation. The first advances in this area were made with hematopoietic stem cells (HSCs) and their transformed counterparts in leukemia, the disease for which the origin of CSCs is best understood. It is now becoming increasingly clear that stem cells of many organs share the same cellular pathways as leukemia-derived HSCs. Additionally, a normal stem cell may be transformed into a cancer stem cell through disregulation of the proliferation and differentiation pathways controlling it or by inducing oncoprotein activity. The Polycomb group transcriptional repressor Bmi-1 was discovered as a common oncogene activated in lymphoma and later shown to specifically regulate HSCs. The role of Bmi-1 has also been illustrated in neural stem cells. The pathway appears to be active in CSCs of pediatric brain tumors. The Notch pathway has been known to developmental biologists for decades. Its role in control of stem cell proliferation has now been demonstrated for several cell types including hematopoietic, neural and mammary stem cells. Components of the Notch pathway have been proposed to act as oncogenes in mammary and other tumors. A particular branch of the Notch signaling pathway that involves the transcription factor Hes3 has been shown to regulate a number of cultured cells with cancer stem cell characteristics obtained from glioblastoma patients. These developmental pathways are also strongly implicated as stem cell regulators. Both Sonic hedgehog (SHH) and Wnt pathways are commonly hyperactivated in tumors and are required to sustain tumor growth. However, the Gli transcription factors that are regulated by SHH take their name from gliomas, where they are commonly expressed at high levels. A degree of crosstalk exists between the two pathways and their activation commonly goes hand-in-hand. This is a trend rather than a rule. For instance, in colon cancer hedgehog signalling appears to antagonise Wnt. Sonic hedgehog blockers are available, such as cyclopamine. There is also a new water soluble cyclopamine that may be more effective in cancer treatment. There is also DMAPT, a water soluble derivative of parthenolide (induces oxidative stress, inhibits NF-κB signaling) for AML (leukemia), and possibly myeloma and prostate cancer. A clinical trial of DMAPT is to start in England in late 2007 or 2008. Finally, the enzyme telomerase may qualify as a study subject in CSC physiology. GRN163L (Imetelstat) was recently started in trials to target myeloma stem cells. If it is possible to eliminate the cancer stem cell, then a potential cure may be achieved if there are no more CSCs to repopulate a cancer. Hematopoietic stem cells (HSCs) are the blood cells that give rise to all the other blood cells. They give rise to the myeloid (monocytes and macrophages, neutrophils, basophils, eosinophils, erythrocytes, megakaryocytes/platelets, dendritic cells), and lymphoid lineages (T-cells, B-cells, NK-cells). The definition of hematopoietic stem cells has changed in the last two decades. The hematopoietic tissue contains cells with long-term and short-term regeneration capacities and committed multipotent, oligopotent, and unipotent progenitors. HSCs constitute 1:10.000 of cells in myeloid tissue. HSCs are a heterogeneous population. Three classes of stem cells exist, distinguished by their ratio of lymphoid to myeloid progeny (L/M) in blood. Myeloid-biased (My-bi) HSC have low L/M ratio (>0, <3), whereas lymphoid-biased (Ly-bi) HSC show a large ratio (>10). The third category consists of the balanced (Bala) HSC for which 3 ≤ L/M ≤ 10. Only the myeloid-biased and -balanced HSCs have durable self-renewal properties. In addition, serial transplantation experiments have shown that each subtype preferentially re-creates its blood cell type distribution, suggesting an inherited epigenetic program for each subtype. HSCs are found in the bone marrow of adults, with large quantities in the pelvis, femur, and sternum. They are also found in umbilical cord blood and, in small numbers, in peripheral blood.][ Stem and progenitor cells can be taken from the pelvis, at the iliac crest, using a needle and syringe.][ The cells can be removed a liquid (to perform a smear to look at the cell morphology) or they can be removed via a core biopsy (to maintain the architecture or relationship of the cells to each other and to the bone).][ In order to harvest stem cells from the circulating peripheral, blood donors are injected with a cytokine, such as granulocyte-colony stimulating factor (G-CSF), that induce cells to leave the bone marrow and circulate in the blood vessels.][. In mammalian embryology, the first definitive HSCs are detected in the AGM (Aorta-gonad-mesonephros), and then massively expanded in the Fetal Liver prior to colonising the bone marrow before birth. As stem cells, HSC are defined by their ability to replenish all blood cell types (Multipotency) and their ability to self-renew. It is known that a small number of HSCs can expand to generate a very large number of daughter HSCs. This phenomenon is used in bone marrow transplantation, when a small number of HSCs reconstitute the hematopoietic system. This process indicates that, subsequent to bone marrow transplantation, symmetrical cell divisions into two daughter HSCs must occur. Stem cell self-renewal is thought to occur in the stem cell niche in the bone marrow, and it is reasonable to assume that key signals present in this niche will be important in self-renewal. There is much interest in the environmental and molecular requirements for HSC self-renewal, as understanding the ability of HSC to replenish themselves will eventually allow the generation of expanded populations of HSC in vitro that can be used therapeutically. It was originally believed that all HSC were alike in their self-renewal and differentiation abilities. This view was first challenged by the 2002 discovery by the Muller-Sieburg group in San Diego, who illustrated that different stem cells can show distinct repopulation patterns that are epigenetically predetermined intrinsic properties of clonal Thy-1lo SCA-1+ lin- c-kit+ HSC. The results of these clonal studies led to the notion of lineage bias. Using the ratio $\rho = L/M$ of lymphoid (L) to myeloid (M) cells in blood as a quantitative marker, the stem cell compartment can be split into three categories of HSC. Balanced (Bala) HSC repopulate peripheral white blood cells in the same ratio of myeloid to lymphoid cells as seen in unmanipulated mice (on average about 15% myeloid and 85% lymphoid cells, or 3≤ρ≤10). Myeloid-biased (My-bi) HSC give rise to too few lymphocytes resulting in ratios 0<ρ<3, while lymphoid-biased (Ly-bi) HSC generate too few myeloid cells, which results in lymphoid-to-myeloid ratios of 10<ρ<oo. All three types are norm three types of HSC, and they do not represent stages of differentiation. Rather, these are three classes of HSC, each with an epigenetically fixed differentiation program. These studies also showed that lineage bias is not stochastically regulated or dependent on differences in environmental influence. My-bi HSC self-renew longer than balanced or Ly-bi HSC. The myeloid bias results from reduced responsiveness to the lymphopoetin Interleukin 7 (IL-7). Subsequent to this, other groups confirmed and highlighted the original findings (refer to the excellent mini-review by Timm Schroeder). For example, the Eaves group confirmed in 2007 that repopulation kinetics, long-term self-renewal capacity, and My-bi and Ly-bi are stably inherited intrinsic HSC properties. In 2010, the Goodell group provided additional insights about the molecular basis of lineage bias in side population Side population (SP) SCA-1+ lin- c-kit+ HSC. As previously shown for IL-7 signaling, it was found that a member of the transforming growth factor family (TGF-beta) induces and inhibits the proliferation of My-bi and Ly-bi HSC, respectively. A cobblestone area-forming cell (CAFC) assay is a cell culture-based empirical assay. When plated onto a confluent culture of stromal feeder layer, a fraction of HSCs creep between the gaps (even though the stromal cells are touching each other) and eventually settle between the stromal cells and the substratum (here the dish surface) or trapped in the cellular processes between the stromal cells. Emperipolesis is the in vivo phenomenon in which one cell is completely engulfed into another (e.g., thymocytes into thymic nurse cells); on the other hand, when in vitro, lymphoid lineage cells creep beneath nurse-like cells, the process is called pseudoemperipolesis. This similar phenomenon is more commonly known in the HSC field by the cell culture terminology cobble stone area-forming cells (CAFC), which means areas or clusters of cells look dull cobblestone-like under phase contrast microscopy, compared to the other HSCs, which are refractile. This happens because the cells that are floating loosely on top of the stromal cells are spherical and thus refractile. However, the cells that creep beneath the stromal cells are flattened and, thus, not refractile. The mechanism of pseudoemperipolesis is only recently coming to light. It may be mediated by interaction through CXCR4 (CD184) the receptor for CXC Chemokines (e.g., SDF1) and α4β1 integrins. HSCs have a higher potential than other immature blood cells to pass the bone marrow barrier, and, thus, may travel in the blood from the bone marrow in one bone to another bone. If they settle in the thymus, they may develop into T cells. In the case of fetuses and other extramedullary hematopoiesis, HSCs may also settle in the liver or spleen and develop. This ability is the reason why HSCs may be harvested directly from the blood. With regard to morphology, hematopoietic stem cells resemble lymphocytes. They are non-adherent, and rounded, with a rounded nucleus and low cytoplasm-to-nucleus ratio. Since PHSC cannot be isolated as a pure population, it is not possible to identify them in a microscope. The above description is based on the morphological characteristics of a heterogeneous population, of which PHSC are a component. In reference to phenotype, hematopoeitic stem cells are identified by their small size, lack of lineage (lin) markers, low staining (side population) with vital dyes such as rhodamine 123 (rhodamineDULL, also called rholo) or Hoechst 33342, and presence of various antigenic markers on their surface. Many of these markers belong to the cluster of differentiation series, like: CD34, CD38, CD90, CD133, CD105, CD45, and also c-kit, - the receptor for stem cell factor. The hematopoietic stem cells are negative for the markers that are used for detection of lineage commitment, and are, thus, called Lin-; and, during their purification by FACS, a bunch of up to 14 different mature blood-lineage marker, e.g., CD13 & CD33 for myeloid, CD71 for erythroid, CD19 for B cells, CD61 for megakaryocytic, etc. for humans; and, B220 (murine CD45) for B cells, Mac-1 (CD11b/CD18) for monocytes, Gr-1 for Granulocytes, Ter119 for erythroid cells, Il7Ra, CD3, CD4, CD5, CD8 for T cells, etc. (for mice) antibodies are used as a mixture to deplete the lin+ cells or late multipotent progenitors (MPP)s. There are many differences between the human and mice hematopoietic cell markers for the commonly accepted type of hematopoietic stem cells.[1]. However, not all stem cells are covered by these combinations that, nonetheless, have become popular. In fact, even in humans, there are hematopoietic stem cells that are CD34-/CD38-. Also some later studies suggested that earliest stem cells may lack c-kit on the cell surface. For human HSCs use of CD133 was one step ahead as both CD34+ and CD34- HSCs were CD133+. Traditional purification method used to yield a reasonable purity level of mouse hematopoietic stem cells, in general, requires a large(~10-12) battery of markers, most of which were surrogate markers with little functional significance, and thus partial overlap with the stem cell populations and sometimes other closely related cells that are not stem cells. Also, some of these markers (e.g., Thy1) are not conserved across mouse species, and use of markers like CD34- for HSC purification requires mice to be at least 8 weeks old. Alternative methods that could give rise to similar or better harvest of stem cells is a hot area of research and are presently emerging. One such method uses a signature of SLAM family of cell surface molecules. SLAM (Signaling lymphocyte activation molecule) family is a group of >10 molecules whose genes are located mostly tandemly in a single locus on chromosome 1 (mouse), all belonging to a subset of immunoglobulin gene superfamily, and originally thought to be involved in T-cell stimulation. This family includes CD48, CD150, CD244, etc., CD150 being the founding member, and, thus, also called slamF1, i.e., SLAM family member 1. The signature SLAM code for the hemopoietic hierarchy are: For HSCs, CD150+CD48- was sufficient instead of CD150+CD48-CD244- because CD48 is a ligand for CD244, and both would be positive only in the activated lineage-restricted progenitors. It seems that this code was more efficient than the more tedious earlier set of the large number of markers, and are also conserved across the mouse strains; however, recent work has shown that this method excludes a large number of HSCs and includes an equally large number of non-stem cells. . CD150+CD48- gave stem cell purity comparable to Thy1loSCA-1+lin-c-kit+ in mice. Irving Weissman's group at Stanford University was the first to isolate mouse hematopoietic stem cells in 1988 and was also the first to work out the markers to distinguish the mouse long-term (LT-HSC) and short-term (ST-HSC) hematopoietic stem cells (self-renew-capable), and the Multipotent progenitors (MPP, low or no self-renew capability — the later the developmental stage of MPP, the lesser the self-renewal ability and the more of some of the markers like CD4 and CD135): Between 1948 and 1950, the Committee for Clarification of the Nomenclature of Cells and Diseases of the Blood and Blood-forming Organs issued reports on the nomenclature of blood cells. An overview of the terminology is shown below, from earliest to final stage of development: The root for erythrocyte colony-forming units (CFU-E) is "rubri", for granulocyte-monocyte colony-forming units (CFU-GM) is "granulo" or "myelo" and "mono", for lympocyte colony-forming units (CFU-L) is "lympho" and for megakaryocyte colony-forming units (CFU-Meg) is "megakaryo". According to this terminology, the stages of red blood cell formation would be: rubriblast, prorubricyte, rubricyte, metarubricyte, and erythrocyte. However, the following nomenclature seems to be, at present, the most prevalent: Osteoclasts also arise from hemopoietic cells of the monocyte/neutrophil lineage, specifically CFU-GM. There are various kinds of colony-forming units: The above CFUs are based on the lineage. Another CFU, the colony-forming unit–spleen (CFU–S) was the basis of an in vivo clonal colony formation, which depends on the ability of infused bone marrow cells to give rise to clones of maturing hematopoietic cells in the spleens of irradiated mice after 8 to 12 days. It was used extensively in early studies, but is now considered to measure more mature progenitor or Transit Amplifying Cells rather than stem cells. Hematopoietic stem cells (HSC) cannot be easily observed directly, and, therefore, their behaviors need to be inferred indirectly. Clonal studies are likely the closest technique for single cell in vivo studies of HSC. Here, sophisticated experimental and statistical methods are used to ascertain that, with a high probability, a single HSC is contained in a transplant administered to a lethally irradiated host. The clonal expansion of this stem cell can then be observed over time by monitoring the percent donor-type cells in blood as the host is reconstituted. The resulting time series is defined as the repopulation kinetic of the HSC. The reconstitution kinetics are very heterogeneous. However, using symbolic dynamics, one can show that they fall into a limited number of classes. To prove this, several hundred experimental repopulation kinetics from clonal Thy-1lo SCA-1+ lin- c-kit+ HSC were translated into symbolic sequences by assigning the symbols "+", "-", "~" whenever two successive measurements of the percent donor-type cells have a positive, negative, or unchanged slope, respectively. By using the Hamming distance, the repopulation patterns were subjected to cluster analysis yielding 16 distinct groups of kinetics. To finish the empirical proof, the Laplace add-one approach was used to determine that the probability of finding kinetics not contained in these 16 groups is very small. By corollary, this result shows that the hematopoietic stem cell compartment is also heterogeneous by dynamical criteria. Hematopoiesis bone: Osteoblast → Osteocyte Fibroblast → Fibrocyte muscle: Myoblast → Myocyte Myosatellite cell Tendon cell Cardiac muscle cell Angioblast → Endothelial cell Mesangial cell (Intraglomerular, Extraglomerular) Juxtaglomerular cell Macula densa cell Stromal cell → Interstitial cell → Telocytes M: BON/CAR anat (c/f/k/f, u, t/p, l)/phys/devp/cell noco/cong/tumr, sysi/epon, injr proc, drug (M5) M: URI anat/phys/devp/cell noco/acba/cong/tumr, sysi/epon, urte proc/itvp, drug (G4B), blte, urte M: VAS anat (a:h/u/t/a/l,v:h/u/t/a/l)/phys/devp/cell/prot noco/syva/cong/lyvd/tumr, sysi/epon, injr proc, drug (C2s+n/3/4/5/7/8/9) M: MYL cell/phys (coag, heme, immu, gran), csfs rbmg/mogr/tumr/hist, sysi/epon, btst drug (B1/2/3+5+6), btst, trns M: LMC cell/phys/auag/auab/comp, igrc imdf/ipig/hyps/tumr proc, drug (L3/4) Carcinogenesis or oncogenesis or tumorigenesis is literally the creation of cancer. It is a process by which normal cells are transformed into cancer cells. It is characterized by a progression of changes at the cellular, genetic and epigenetic level that ultimately reprogram a cell to undergo uncontrolled cell division, thus forming a malignant mass. Cell division is a physiological process that occurs in almost all tissues and under many circumstances. Under normal circumstances, the balance between proliferation and programmed cell death, usually in the form of apoptosis, is maintained by tightly regulating both processes to ensure the integrity of organs and tissues. Mutations and epimutations in DNA that lead to cancer (only certain mutations and epimutations can lead to cancer and the majority of potential mutations and epimutations will have no bearing) disrupt these orderly processes by disrupting the programming regulating the processes. Carcinogenesis is caused by mutation and epimutation of the genetic material of normal cells, which upsets the normal balance between proliferation and cell death. This results in uncontrolled cell division and the evolution of those cells by natural selection in the body. The uncontrolled and often rapid proliferation of cells can lead to benign tumors; some types of these may turn into malignant tumors (cancer). Benign tumors do not spread to other parts of the body or invade other tissues, and they are rarely a threat to life unless they compress vital structures or are physiologically active, for instance, producing a hormone. Malignant tumors can invade other organs, spread to distant locations (metastasis) and become life-threatening. More than one mutation is necessary for carcinogenesis. In fact, a series of several mutations to certain classes of genes is usually required before a normal cell will transform into a cancer cell. On average, for example, 15 "driver mutations" and 60 "passenger" mutations are found in colon cancers. Mutations in those certain types of genes that play vital roles in cell division, apoptosis (cell death), and mutations and epimutations (see article Genome instability) in DNA repair genes will cause a cell to lose control of its cell proliferation. Oncovirinae, viruses that contain an oncogene, are categorized as oncogenic because they trigger the growth of tumorous tissues in the host. This process is also referred to as viral transformation. Cancer is fundamentally a disease of regulation of tissue growth. In order for a normal cell to transform into a cancer cell, genes that regulate cell growth and differentiation must be altered. Genetic and epigenetic changes can occur at many levels, from gain or loss of entire chromosomes, to a mutation affecting a single DNA nucleotide, or to silencing or activating a microRNA that controls expression of 100 to 500 genes. There are two broad categories of genes that are affected by these changes. Oncogenes may be normal genes that are expressed at inappropriately high levels, or altered genes that have novel properties. In either case, expression of these genes promotes the malignant phenotype of cancer cells. Tumor suppressor genes are genes that inhibit cell division, survival, or other properties of cancer cells. Tumor suppressor genes are often disabled by cancer-promoting genetic changes. Typically, changes in many genes are required to transform a normal cell into a cancer cell. There is a diverse classification scheme for the various genomic changes that may contribute to the generation of cancer cells. Many of these changes are mutations, or changes in the nucleotide sequence of genomic DNA. There are also many epigenetic changes that alter whether genes are expressed or not expressed. Aneuploidy, the presence of an abnormal number of chromosomes, is one genomic change that is not a mutation, and may involve either gain or loss of one or more chromosomes through errors in mitosis. Large-scale mutations involve the deletion or gain of a portion of a chromosome. Genomic amplification occurs when a cell gains many copies (often 20 or more) of a small chromosomal region, usually containing one or more oncogenes and adjacent genetic material. Translocation occurs when two separate chromosomal regions become abnormally fused, often at a characteristic location. A well-known example of this is the Philadelphia chromosome, or translocation of chromosomes 9 and 22, which occurs in chronic myelogenous leukemia, and results in production of the BCR-abl fusion protein, an oncogenic tyrosine kinase. Small-scale mutations include point mutations, deletions, and insertions, which may occur in the promoter of a gene and affect its expression, or may occur in the gene's coding sequence and alter the function or stability of its protein product. Disruption of a single gene may also result from integration of genomic material from a DNA virus or retrovirus, and such an event may also result in the expression of viral oncogenes in the affected cell and its descendants. Epimutations include methylations or demethylations of the CpG islands of the promoter regions of genes, which result in repression or de-repression, respectively of gene expression. Epimutations, can also occur by acetylation, methylation, phosphorylation or other alterations to histones, creating a histone code that represses or activates gene expression, and such histone epimutations can be important epigenetic factors in cancer. In addition, carcinogenic epimutation can occur through alterations of chromosome architecture caused by proteins such as HMGA2. A further source of epimutation is due to increased or decreased expression of microRNAs (miRNAs). For example extra expression of miR-137 can cause downregulation of expression of 491 genes, and miR-137 is epigenetically silenced in 32% of colorectal cancers> DNA damage is considered to be the primary cause of cancer. More than 10,000 new naturally occurring DNA damages arise, on average, per human cell, per day, due to endogenous cellular processes (see article DNA damage (naturally occurring). Additional DNA damages can arise from exposure to exogenous agents. As one example of an exogenous carcinogeneic agent, tobacco smoke causes increased DNA damage, and these DNA damages likely cause the increase of lung cancer due to smoking. In other examples, UV light from solar radiation causes DNA damage that is important in melanoma, helicobacter pylori infection produces high levels of reactive oxygen species that damage DNA and contributes to gastric cancer, and the Aspergillus metabolite, aflatoxin, is a DNA damaging agent that is causative in liver cancer. DNA damages can also be caused by endogenous (naturally occurring) agents. Katsurano et al. indicated that macrophages and neutrophils in an inflamed colonic epithelium are the source of reactive oxygen species causing the DNA damages that initiate colonic tumorigenesis, and bile acids, at high levels in the colons of humans eating a high fat diet, also cause DNA damage and contribute to colon cancer. Such exogenous and endogenous sources of DNA damage are indicated in the boxes at the top of the figure in this section. The central role of DNA damage in progression to cancer is indicated at the second level of the figure. The central elements of DNA damage, epigenetic alterations and deficient DNA repair in progression to cancer are shown in red. A deficiency in DNA repair would cause more DNA damages to accumulate, and increase the risk for cancer. For example, individuals with an inherited impairment in any of 34 DNA repair genes (see article DNA repair-deficiency disorder) are at increased risk of cancer with some defects causing up to 100% lifetime chance of cancer (e.g. p53 mutations [74]). Such germ line mutations are shown in a box at the left of the figure, with an indication of their contribution to DNA repair deficiency. However, such germline mutations (which cause highly penetrant cancer syndromes) are the cause of only about 1 percent of cancers. The majority of cancers are called non-hereditary or "sporadic cancers". About 30% of sporadic cancers do have some hereditary component that is currently undefined, while the majority, or 70% of sporadic cancers, have no hereditary component. In sporadic cancers, a deficiency in DNA repair is occasionally due to a mutation in a DNA repair gene, but much more frequently reduced or absent expression of DNA repair genes is due to epigenetic alterations that reduce or silence gene expression. This is indicated in the figure at the 3rd level from the top. For example, for 113 colorectal cancers examined in sequence, only four had a missense mutation in the DNA repair gene MGMT, while the majority had reduced MGMT expression due to methylation of the MGMT promoter region (an epigenetic alteration). Five reports present evidence that between 40% and 90% of colorectal cancers have reduced MGMT expression due to methylation of the MGMT promoter region. Similarly, out of 119 cases of mismatch repair-deficient colorectal cancers that lacked DNA repair gene PMS2 expression, Pms2 was deficient in 6 due to mutations in the PMS2 gene, while in 103 cases PMS2 expression was deficient because its pairing partner MLH1 was repressed due to promoter methylation (PMS2 protein is unstable in the absence of MLH1). In the other 10 cases, loss of PMS2 expression was likely due to epigenetic overexpression of the microRNA, miR-155, which down-regulates MLH1. In further examples [tabulated in the article Epigenetics (see section "DNA repair epigenetics in cancer")], epigenetic defects in cancers were found at frequencies of between 13%-100% for the DNA repair genes BRCA1, WRN, FANCB, FANCF, MGMT, MLH1, MSH2, MSH4, ERCC1, XPF, NEIL1 and ATM in cancers including those in breast, ovarian, colorectal, and the head and neck areas. In particular, two or more epigenetic deficiencies in expression of ERCC1, XPF and/or PMS2 were shown to occur simultaneously in the majority of the 49 colon cancers evaluated by Facista et al. When expression of DNA repair genes is reduced, this causes a DNA repair deficiency. This is shown in the figure at the 4th level from the top. With a DNA repair deficiency, more DNA damages remain in cells at a higher than usual level (5th level from the top in figure), and these excess damages cause increased frequencies of mutation and/or epimutation (6th level from top of figure). Experimentally, mutation rates increase substantially in cells defective in DNA mismatch repair or in Homologous recombinational repair (HRR). Chromosomal rearrangements and aneuploidy also increase in HRR defective cells During repair of DNA double strand breaks, or repair of other DNA damages, incompletely cleared sites of repair can cause epigenetic gene silencing. Many studies of heavy metal-induced carcinogenesis show that such heavy metals cause reduction in expression of DNA repair enzymes, some through epigenetic mechanisms. In some cases, DNA repair inhibition is proposed to be a predominant mechanism in heavy metal-induced carcinogenicity. For example, one group of studies shows that arsenic inhibits the DNA repair genes PARP, XRCC1, Ligase 3, Ligase 4, DNA POLB, XRCC4, DNA PKCS, TOPO2B, OGG1, ERCC1, XPF, XPB, XPC XPE and P53. Another group of studies shows that cadmium inhibits the DNA repair genes MSH2, ERCC1, XRCC1, OGG1, MSH6, DNA-PK, XPD and XPC The somatic mutations and epigenetic alterations caused by DNA damages and deficiencies in DNA repair accumulate in field defects. Field defects are normal appearing tissues with multiple alterations (discussed in the section below), and are common precursors to development of the disordered and improperly proliferating clone of tissue in a cancer. Such field defects (second level from bottom of figure) may have multiple mutations and epigenetic alterations. It is impossible to determine the initial cause for most specific cancers. In a few cases, only one cause exists; for example, the virus HHV-8 causes all Kaposi's sarcomas. However, with the help of cancer epidemiology techniques and information, it is possible to produce an estimate of a likely cause in many more situations. For example, lung cancer has several causes, including tobacco use and radon gas. Men who currently smoke tobacco develop lung cancer at a rate 14 times that of men who have never smoked tobacco, so the chance of lung cancer in a current smoker being caused by smoking is about 93%; there is a 7% chance that the smoker's lung cancer was caused by radon gas or some other, non-tobacco cause. These statistical correlations have made it possible for researchers to infer that certain substances or behaviors are carcinogenic. Tobacco smoke causes increased exogenous DNA damage, and these DNA damages are the likely cause of lung cancer due to smoking. Among the more than 5,000 compounds in tobacco smoke, the genotoxic DNA damaging agents that occur both at the highest concentrations and which have the strongest mutagenic effects are acrolein, formaldehyde, acrylonitrile, 1,3-butadiene, acetaldehyde, ethylene oxide and isoprene. Using molecular biological techniques, it is possible to characterize the mutations, epimutations or chromosomal aberrations within a tumor, and rapid progress is being made in the field of predicting prognosis based on the spectrum of mutations in some cases. For example, up to half of all tumors have a defective p53 gene. This mutation is associated with poor prognosis, since those tumor cells are less likely to go into apoptosis or programmed cell death when damaged by therapy. Telomerase mutations remove additional barriers, extending the number of times a cell can divide. Other mutations enable the tumor to grow new blood vessels to provide more nutrients, or to metastasize, spreading to other parts of the body. However, once a cancer is formed it continues to evolve and to produce sub clones. For example, a renal cancer, sampled in 9 areas, had 40 ubiquitous mutations, 59 mutations shared by some, but not all regions, and 29 "private" mutations only present in one region. The term "field cancerization" was first used in 1953 to describe an area or "field" of epithelium that has been preconditioned by (at that time) largely unknown processes so as to predispose it towards development of cancer. Since then, the terms "field cancerization" and "field defect" have been used to describe pre-malignant tissue in which new cancers are likely to arise. Field defects have been identified in association with cancers and are important in progression to cancer. However, it was pointed out by Rubin that "the vast majority of studies in cancer research has been done on well-defined tumors in vivo, or on discrete neoplastic foci in vitro. Yet there is evidence that more than 80% of the somatic mutations found in mutator phenotype human colorectal tumors occur before the onset of terminal clonal expansion…" Similarly, Vogelstein et al. also indicated that more than half of somatic mutations identified in tumors occurred in a pre-neoplastic phase (in a field defect), during growth of apparently normal cells. It would also be expected that many of the epigenetic alterations present in tumors may have occurred in pre-neoplastic field defects. In the colon, a field defect probably arises by natural selection of a mutant or epigenetically altered cell among the stem cells at the base of one of the intestinal crypts on the inside surface of the colon. A mutant or epigenetically altered stem cell may replace the other nearby stem cells by natural selection. This may cause a patch of abnormal tissue to arise. The figure in this section includes a photo of a freshly resected and lengthwise-opened segment of the colon showing a colon cancer and four polyps. Below the photo there is a schematic diagram of how a large patch of mutant or epigenetically altered cells may have formed, shown by the large area in yellow in the diagram. Within this first large patch in the diagram (a large clone of cells), a second such mutation or epigenetic alteration may occur so that a given stem cell acquires an advantage compared to other stem cells within the patch, and this altered stem cell may expand clonally forming a secondary patch, or sub-clone, within the original patch. This is indicated in the diagram by four smaller patches of different colors within the large yellow original area. Within these new patches (sub-clones), the process may be repeated multiple times, indicated by the still smaller patches within the four secondary patches (with still different colors in the diagram) which clonally expand, until stem cells arise that generate either small polyps or else a malignant neoplasm (cancer). In the photo, an apparent field defect in this segment of a colon has generated four polyps (labeled with the size of the polyps, 6mm, 5mm, and two of 3mm, and a cancer about 3 cm across in its longest dimension). These neoplasms are also indicated (in the diagram below the photo) by 4 small tan circles (polyps) and a larger red area (cancer). The cancer in the photo occurred in the cecal area of the colon, where the colon joins the small intestine (labeled) and where the appendix occurs (labeled). The fat in the photo is external to the outer wall of the colon. In the segment of colon shown here, the colon was cut open lengthwise to expose the inner surface of the colon and to display the cancer and polyps occurring within the inner epithelial lining of the colon. If the general process by which sporadic colon cancers arise is the formation of a pre-neoplastic clone that spreads by natural selection, followed by formation of internal sub-clones within the initial clone, and sub-sub-clones inside those, then colon cancers generally should be associated with, and be preceded by, fields of increasing abnormality reflecting the succession of premalignant events. The most extensive region of abnormality (the outermost yellow irregular area in the diagram) would reflect the earliest event in formation of a malignant neoplasm. In experimental evaluation of specific DNA repair deficiencies in cancers, many specific DNA repair deficiencies were also shown to occur in the field defects surrounding those cancers. The Table, below, gives examples for which the DNA repair deficiency in a cancer was shown to be caused by an epigenetic alteration, and the somewhat lower frequencies with which the same epigenetically caused DNA repair deficiency was found in the surrounding field defect. References in the table are given here: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, Some of the small polyps in the field defect shown in the photo of the opened colon segment may be relatively benign neoplasms. Of polyps less than 10mm in size, found during colonoscopy and followed with repeat colonoscopies for 3 years, 25% were unchanged in size, 35% regressed or shrank in size while 40% grew in size. Cancers are known to exhibit genome instability or a mutator phenotype. The protein-coding DNA within the nucleus is about 1.5% of the total genomic DNA. Within this protein-coding DNA (called the exome), an average cancer of the breast or colon can have about 60 to 70 protein altering mutations, of which about 3 or 4 may be "driver" mutations, and the remaining ones may be "passenger" mutations. However, the average number of DNA sequence mutations in the entire genome (including non-protein-coding regions) within a breast cancer tissue sample is about 20,000. In an average melanoma tissue sample (where melanomas have a higher exome mutation frequency) the total number of DNA sequence mutations is about 80,000. These high frequencies of mutations in the total nucleotide sequences within cancers suggest that often an early alteration in the field defect giving rise to a cancer (e.g. yellow area in the diagram in the preceding section) is a deficiency in DNA repair. The large field defects surrounding colon cancers (extending to about 10 cm on each side of a cancer) were shown by Facista et al. to frequently have epigenetic defects in 2 or 3 DNA repair proteins (ERCC1, XPF and/or PMS2) in the entire area of the field defect. When expression of DNA repair genes is reduced, DNA damages accumulate in cells at a higher than normal level, and these excess damages cause increased frequencies of mutation and/or epimutation. Mutation rates strongly increase in cells defective in DNA mismatch repair or in homologous recombinational repair (HRR). A deficiency in DNA repair, itself, can allow DNA damages to accumulate, and error-prone translesion synthesis past some of those damages may give rise to mutations. In addition, faulty repair of these accumulated DNA damages may give rise to epimutations. These new mutations and/or epimutations may provide a proliferative advantage, generating a field defect. Although the mutations/epimutations in DNA repair genes do not, themselves, confer a selective advantage, they may be carried along as passengers in cells when the cell acquires an additional mutation/epimutation that does provide a proliferative advantage. There are a number of theories of carcinogenesis and cancer treatment that fall outside the mainstream of scientific opinion, due to lack of scientific rationale, logic, or evidence base. These theories may be used to justify various alternative cancer treatments. They should be distinguished from those theories of carcinogenesis that have a logical basis within mainstream cancer biology, and from which conventionally testable hypotheses can be made. Several alternative theories of carcinogenesis, however, are based on scientific evidence and are increasingly being acknowledged. Some researchers believe that cancer may be caused by aneuploidy (numerical and structural abnormalities in chromosomes) rather than by mutations or epimutations. Cancer has also been considered as a metabolic disease in which the cellular metabolism of oxygen is diverted from the pathway that generates energy (oxidative phosphorylation) to the pathway that generates reactive oxygen species (figure). This causes an energy switch from oxidative phosphorylation to aerobic glycolysis (Warburg's hypothesis) and the accumulation of reactive oxygen species leading to oxidative stress (oxidative stress theory of cancer). All these theories of carcinogenesis may be complementary rather than contradictory. Another theory as to the origin of cancer was developed by astrobiologists and suggests that cancer is an atavism, an evolutionary throwback to an earlier form of multicellular life. The genes responsible for uncontrolled cell growth and cooperation between cancer cells are very similar to those that enabled the first multicellular life forms to group together and flourish. These genes still exist within the genome of more complex metazoans, such as humans, although more recently evolved genes keep them in check. When the newer controlling genes fail for whatever reason, the cell can revert to its more primitive programming and reproduce out of control. The theory is an alternative to the notion that cancers begin with rogue cells that undergo evolution within the body. Instead they possess a fixed number of primitive genes that are progressively activated, giving them finite variability. Often, the multiple genetic changes that result in cancer may take many years to accumulate. During this time, the biological behavior of the pre-malignant cells slowly change from the properties of normal cells to cancer-like properties. Pre-malignant tissue can have a distinctive appearance under the microscope. Among the distinguishing traits are an increased number of dividing cells, variation in nuclear size and shape, variation in cell size and shape, loss of specialized cell features, and loss of normal tissue organization. Dysplasia is an abnormal type of excessive cell proliferation characterized by loss of normal tissue arrangement and cell structure in pre-malignant cells. These early neoplastic changes must be distinguished from hyperplasia, a reversible increase in cell division caused by an external stimulus, such as a hormonal imbalance or chronic irritation. The most severe cases of dysplasia are referred to as "carcinoma in situ." In Latin, the term "in situ" means "in place", so carcinoma in situ refers to an uncontrolled growth of cells that remains in the original location and has not shown invasion into other tissues. Nevertheless, carcinoma in situ may develop into an invasive malignancy and is usually removed surgically, if possible. Just like a population of animals undergoes evolution, an unchecked population of cells also can undergo evolution. This undesirable process is called somatic evolution, and is how cancer arises and becomes more malignant. Most changes in cellular metabolism that allow cells to grow in a disorderly fashion lead to cell death. However once cancer begins, cancer cells undergo a process of natural selection: the few cells with new genetic changes that enhance their survival or reproduction continue to multiply, and soon come to dominate the growing tumor, as cells with less favorable genetic change are out-competed. This is exactly how pathogens such as MRSA can become antibiotic-resistant (or how HIV can become drug-resistant), and the same reason why crop blights and pests can become pesticide-resistant. This evolution is why cancer recurrences will have cells that have acquired cancer-drug resistance (or in some cases, resistance to radiation from radiotherapy). In a 2000 article by Hanahan and Weinberg, the biological properties of malignant tumor cells were summarized as follows: The completion of these multiple steps would be a very rare event without : These biological changes are classical in carcinomas; other malignant tumors may not need to achieve them all. For example, tissue invasion and displacement to distant sites are normal properties of leukocytes; these steps are not needed in the development of leukemia. The different steps do not necessarily represent individual mutations. For example, inactivation of a single gene, coding for the p53 protein, will cause genomic instability, evasion of apoptosis and increased angiogenesis. Not all the cancer cells are dividing. Rather, a subset of the cells in a tumor, called cancer stem cells, replicate themselves and generate differentiated cells. Cancer is a genetic disease: In order for cells to start dividing uncontrollably, genes that regulate cell growth must be damaged. Proto-oncogenes are genes that promote cell growth and mitosis, whereas tumor suppressor genes discourage cell growth, or temporarily halt cell division to carry out DNA repair. Typically, a series of several mutations to these genes is required before a normal cell transforms into a cancer cell. This concept is sometimes termed "oncoevolution." Mutations to these genes provide the signals for tumor cells to start dividing uncontrollably. But the uncontrolled cell division that characterizes cancer also requires that the dividing cell duplicates all its cellular components to create two daughter cells. The activation of anaerobic glycolysis (the Warburg effect), which is not necessarily induced by mutations in proto-oncogenes and tumor suppressor genes, provides most of the building blocks required to duplicate the cellular components of a dividing cell and, therefore, is also essential for carcinogenesis. There are several different cell types that are critical to tumour growth. In particular endothelial progenitor cells are a very important cell population in tumour blood vessel growth. The hypothesis that endothelial progenitor cells are important in tumour growth, angiogenesis and metastasis has been supported by a recent publication in Cancer Research (August 2010). This paper argues that endothelial progenitor cells can be marked using the Inhibitor of DNA Binding 1 (ID1). This novel finding meant that investigators were able to track endothelial progenitor cells from the bone marrow to the blood to the tumour-stroma and vasculature. This finding of endothelial progenitor cells incorporated in tumour vasculature gives evidence for the importance of this cell type in blood vessel development in a tumour setting and metastasis. Furthermore, ablation of the endothelial progenitor cells in the bone marrow lead to a significant decrease in tumour growth and vasculature development. The continued research into the importance of endothelial progenitor cells may present novel therapeutic targets. Oncogenes promote cell growth through a variety of ways. Many can produce hormones, a "chemical messenger" between cells that encourage mitosis, the effect of which depends on the signal transduction of the receiving tissue or cells. In other words, when a hormone receptor on a recipient cell is stimulated, the signal is conducted from the surface of the cell to the cell nucleus to affect some change in gene transcription regulation at the nuclear level. Some oncogenes are part of the signal transduction system itself, or the signal receptors in cells and tissues themselves, thus controlling the sensitivity to such hormones. Oncogenes often produce mitogens, or are involved in transcription of DNA in protein synthesis, which creates the proteins and enzymes responsible for producing the products and biochemicals cells use and interact with. Mutations in proto-oncogenes, which are the normally quiescent counterparts of oncogenes, can modify their expression and function, increasing the amount or activity of the product protein. When this happens, the proto-oncogenes become oncogenes, and this transition upsets the normal balance of cell cycle regulation in the cell, making uncontrolled growth possible. The chance of cancer cannot be reduced by removing proto-oncogenes from the genome, even if this were possible, as they are critical for growth, repair and homeostasis of the organism. It is only when they become mutated that the signals for growth become excessive. One of the first oncogenes to be defined in cancer research is the ras oncogene. Mutations in the Ras family of proto-oncogenes (comprising H-Ras, N-Ras and K-Ras) are very common, being found in 20% to 30% of all human tumours. Ras was originally identified in the Harvey sarcoma virus genome, and researchers were surprised that not only is this gene present in the human genome but also, when ligated to a stimulating control element, it could induce cancers in cell line cultures. Proto-oncogenes promote cell growth in a variety of ways. Many can produce hormones, "chemical messengers" between cells that encourage mitosis, the effect of which depends on the signal transduction of the receiving tissue or cells. Some are responsible for the signal transduction system and signal receptors in cells and tissues themselves, thus controlling the sensitivity to such hormones. They often produce mitogens, or are involved in transcription of DNA in protein synthesis, which create the proteins and enzymes is responsible for producing the products and biochemicals cells use and interact with. Mutations in proto-oncogenes can modify their expression and function, increasing the amount or activity of the product protein. When this happens, they become oncogenes, and, thus, cells have a higher chance to divide excessively and uncontrollably. The chance of cancer cannot be reduced by removing proto-oncogenes from the genome, as they are critical for growth, repair and homeostasis of the body. It is only when they become mutated that the signals for growth become excessive. It is important to note that a gene possessing a growth-promoting role may increase carcinogenic potential of a cell, under the condition that all necessary cellular mechanisms that permit growth are activated. This condition includes also the inactivation of specific tumor suppressor genes (see below). If the condition is not fulfilled, the cell may cease to grow and can proceed to die. This makes knowledge of the stage and type of cancer cell that grows under the control of a given oncogene crucial for the development of treatment strategies. Tumor suppressor genes code for anti-proliferation signals and proteins that suppress mitosis and cell growth. Generally, tumor suppressors are transcription factors that are activated by cellular stress or DNA damage. Often DNA damage will cause the presence of free-floating genetic material as well as other signs, and will trigger enzymes and pathways that lead to the activation of tumor suppressor genes. The functions of such genes is to arrest the progression of the cell cycle in order to carry out DNA repair, preventing mutations from being passed on to daughter cells. The p53 protein, one of the most important studied tumor suppressor genes, is a transcription factor activated by many cellular stressors including hypoxia and ultraviolet radiation damage. Despite nearly half of all cancers possibly involving alterations in p53, its tumor suppressor function is poorly understood. p53 clearly has two functions: one a nuclear role as a transcription factor, and the other a cytoplasmic role in regulating the cell cycle, cell division, and apoptosis. The Warburg hypothesis is the preferential use of glycolysis for energy to sustain cancer growth. p53 has been shown to regulate the shift from the respiratory to the glycolytic pathway. However, a mutation can damage the tumor suppressor gene itself, or the signal pathway that activates it, "switching it off". The invariable consequence of this is that DNA repair is hindered or inhibited: DNA damage accumulates without repair, inevitably leading to cancer. Mutations of tumor suppressor genes that occur in germline cells are passed along to offspring, and increase the likelihood for cancer diagnoses in subsequent generations. Members of these families have increased incidence and decreased latency of multiple tumors. The tumor types are typical for each type of tumor suppressor gene mutation, with some mutations causing particular cancers, and other mutations causing others. The mode of inheritance of mutant tumor suppressors is that an affected member inherits a defective copy from one parent, and a normal copy from the other. For instance, individuals who inherit one mutant p53 allele (and are therefore heterozygous for mutated p53) can develop melanomas and pancreatic cancer, known as Li-Fraumeni syndrome. Other inherited tumor suppressor gene syndromes include Rb mutations, linked to retinoblastoma, and APC gene mutations, linked to adenopolyposis colon cancer. Adenopolyposis colon cancer is associated with thousands of polyps in colon while young, leading to colon cancer at a relatively early age. Finally, inherited mutations in BRCA1 and BRCA2 lead to early onset of breast cancer. Development of cancer was proposed in 1971 to depend on at least two mutational events. In what became known as the Knudson two-hit hypothesis, an inherited, germ-line mutation in a tumor suppressor gene would cause cancer only if another mutation event occurred later in the organism's life, inactivating the other allele of that tumor suppressor gene. Usually, oncogenes are dominant, as they contain gain-of-function mutations, while mutated tumor suppressors are recessive, as they contain loss-of-function mutations. Each cell has two copies of the same gene, one from each parent, and under most cases gain of function mutations in just one copy of a particular proto-oncogene is enough to make that gene a true oncogene. On the other hand, loss of function mutations need to happen in both copies of a tumor suppressor gene to render that gene completely non-functional. However, cases exist in which one mutated copy of a tumor suppressor gene can render the other, wild-type copy non-functional. This phenomenon is called the dominant negative effect and is observed in many p53 mutations. Knudson's two hit model has recently been challenged by several investigators. Inactivation of one allele of some tumor suppressor genes is sufficient to cause tumors. This phenomenon is called haploinsufficiency and has been demonstrated by a number of experimental approaches. Tumors caused by haploinsufficiency usually have a later age of onset when compared with those by a two hit process. In general, mutations in both types of genes are required for cancer to occur. For example, a mutation limited to one oncogene would be suppressed by normal mitosis control and tumor suppressor genes, first hypothesised by the Knudson hypothesis. A mutation to only one tumor suppressor gene would not cause cancer either, due to the presence of many "backup" genes that duplicate its functions. It is only when enough proto-oncogenes have mutated into oncogenes, and enough tumor suppressor genes deactivated or damaged, that the signals for cell growth overwhelm the signals to regulate it, that cell growth quickly spirals out of control. Often, because these genes regulate the processes that prevent most damage to genes themselves, the rate of mutations increases as one gets older, because DNA damage forms a feedback loop. Usually, oncogenes are dominant alleles, as they contain gain-of-function mutations, whereas mutated tumor suppressors are recessive alleles, as they contain loss-of-function mutations. Each cell has two copies of a same gene, one from each parent, and, under most cases, gain of function mutation in one copy of a particular proto-oncogene is enough to make that gene a true oncogene, while usually loss of function mutation must happen in both copies of a tumor suppressor gene to render that gene completely non-functional. However, cases exist in which one loss of function copy of a tumor suppressor gene can render the other copy non-functional, called the dominant negative effect. This is observed in many p53 mutations. Mutation of tumor suppressor genes that are passed on to the next generation of not merely cells, but their offspring, can cause increased likelihoods for cancers to be inherited. Members within these families have increased incidence and decreased latency of multiple tumors. The mode of inheritance of mutant tumor suppressors is that affected member inherits a defective copy from one parent, and a normal copy from another. Because mutations in tumor suppressors act in a recessive manner (note, however, there are exceptions), the loss of the normal copy creates the cancer phenotype. For instance, individuals that are heterozygous for p53 mutations are often victims of Li-Fraumeni syndrome, and that are heterozygous for Rb mutations develop retinoblastoma. In similar fashion, mutations in the adenomatous polyposis coli gene are linked to adenopolyposis colon cancer, with thousands of polyps in the colon while young, whereas mutations in BRCA1 and BRCA2 lead to early onset of breast cancer. A new idea announced in 2011 is an extreme version of multiple mutations, called chromothripsis by its proponents. This idea, affecting only 2–3% of cases of cancer, although up to 25% of bone cancers, involves the catastrophic shattering of a chromosome into tens or hundreds of pieces and then being patched back together incorrectly. This shattering probably takes place when the chromosomes are compacted during normal cell division, but the trigger for the shattering is unknown. Under this model, cancer arises as the result of a single, isolated event, rather than the slow accumulation of multiple mutations. Many mutagens are also carcinogens, but some carcinogens are not mutagens. Examples of carcinogens that are not mutagens include alcohol and estrogen. These are thought to promote cancers through their stimulating effect on the rate of cell mitosis. Faster rates of mitosis increasingly leave fewer opportunities for repair enzymes to repair damaged DNA during DNA replication, increasing the likelihood of a genetic mistake. A mistake made during mitosis can lead to the daughter cells' receiving the wrong number of chromosomes, which leads to aneuploidy and may lead to cancer. Heliobacter pylori is known to cause MALT lymphoma. Other types of bacteria have been implicated in other cancers. Furthermore, many cancers originate from a viral infection; this is especially true in animals such as birds, but less so in humans. 12% of human cancers can be attributed to a viral infection. The mode of virally induced tumors can be divided into two, acutely transforming or slowly transforming. In acutely transforming viruses, the viral particles carry a gene that encodes for an overactive oncogene called viral-oncogene (v-onc), and the infected cell is transformed as soon as v-onc is expressed. In contrast, in slowly transforming viruses, the virus genome is inserted, especially as viral genome insertion is obligatory part of retroviruses, near a proto-oncogene in the host genome. The viral promoter or other transcription regulation elements, in turn, cause over-expression of that proto-oncogene, which, in turn, induces uncontrolled cellular proliferation. Because viral genome insertion is not specific to proto-oncogenes and the chance of insertion near that proto-oncogene is low, slowly transforming viruses have very long tumor latency compared to acutely transforming virus, which already carries the viral-oncogene. Viruses that are known to cause cancer such as HPV (cervical cancer), Hepatitis B (liver cancer), and EBV (a type of lymphoma), are all DNA viruses. It is thought that when the virus infects a cell, it inserts a part of its own DNA near the cell growth genes, causing cell division. The group of changed cells that are formed from the first cell dividing all have the same viral DNA near the cell growth genes. The group of changed cells are now special because one of the normal controls on growth has been lost. Depending on their location, cells can be damaged through radiation from sunshine, chemicals from cigarette smoke, and inflammation from bacterial infection or other viruses. Each cell has a chance of damage, a step on a path toward cancer. Cells often die if they are damaged, through failure of a vital process or the immune system; however, sometimes damage will knock out a single cancer gene. In an old person, there are thousands, tens of thousands or hundreds of thousands of knocked-out cells. The chance that any one would form a cancer is very low. When the damage occurs in any area of changed cells, something different occurs. Each of the cells has the potential for growth. The changed cells will divide quicker when the area is damaged by physical, chemical, or viral agents. A vicious circle has been set up: Damaging the area will cause the changed cells to divide, causing a greater likelihood that they will suffer knock-outs. This model of carcinogenesis is popular because it explains why cancers grow. It would be expected that cells that are damaged through radiation would die or at least be worse off because they have fewer genes working; viruses increase the number of genes working. One concern is that we may end up with thousands of vaccines to prevent every virus that can change our cells. Viruses can have different effects on different parts of the body. It may be possible to prevent a number of different cancers by immunizing against one viral agent. It is likely that HPV, for instance, has a role in cancers of the mucous membranes of the mouth. Certain parasitic worms are known to be carcinogenic. These include: Epigenetics is the study of the regulation of gene expression through chemical, non-mutational changes in DNA structure. The theory of epigenetics in cancer pathogenesis is that non-mutational changes to DNA can lead to alterations in gene expression. Normally, oncogenes are silent, for example, because of DNA methylation. Loss of that methylation can induce the aberrant expression of oncogenes, leading to cancer pathogenesis. Known mechanisms of epigenetic change include DNA methylation, and methylation or acetylation of histone proteins bound to chromosomal DNA at specific locations. Classes of medications, known as HDAC inhibitors and DNA methyltransferase inhibitors, can re-regulate the epigenetic signaling in the cancer cell. A new way of looking at carcinogenesis comes from integrating the ideas of developmental biology into oncology. The cancer stem cell hypothesis proposes that the different kinds of cells in a heterogeneous tumor arise from a single cell, termed Cancer Stem Cell. Cancer stem cells may arise from transformation of adult stem cells or differentiated cells within a body. These cells persist as a subcomponent of the tumor and retain key stem cell properties. They give rise to a variety of cells, are capable of self-renewal and homeostatic control. Furthermore, the relapse of cancer and the emergence of metastasis are also attributed to these cells. The cancer stem cell hypothesis does not contradict earlier concepts of carcinogenesis. While genetic and epigenetic alterations in tumor suppressor genes and oncogenes change the behavior of cells, those alterations, in the end, result in cancer through their effects on the population of neoplastic cells and their microenvironment. Mutant cells in neoplasms compete for space and resources. Thus, a clone with a mutation in a tumor suppressor gene or oncogene will expand only in a neoplasm if that mutation gives the clone a competitive advantage over the other clones and normal cells in its microenvironment. Thus, the process of carcinogenesis is formally a process of Darwinian evolution, known as somatic or clonal evolution. Furthermore, in light of the Darwinistic mechanisms of carcinogenesis, it has been theorized that the various forms of cancer can be categorized as pubertarial and gerontological. Anthropological research is currently being conducted on cancer as a natural evolutionary process through which natural selection destroys environmentally inferior phenotypes while supporting others. According to this theory, cancer comes in two separate types: from birth to the end of puberty (approximately age 20) teleologically inclined toward supportive group dynamics, and from mid-life to death (approximately age 40+) teleologically inclined away from overpopulative group dynamics.][ Radoslav S.Jovic-Cancer-released ancestor from our genes,www.newcancertheory.com M: NEO tsoc, mrkr tumr, epon, para drug (L1i/1e/V03) Cell division is the process by which a parent cell divides into two or more daughter cells. Cell division usually occurs as part of a larger cell cycle. In eukaryotes, there are two distinct type of cell division: a vegetative division, whereby each daughter cell is genetically identical to the parent cell (mitosis), and a reductive cell division, whereby the number of chromosomes in the daughter cells is reduced by half, to produce haploid gametes (meiosis). Both of these cell division cycles are required in sexually reproducing organisms at some point in their life cycle, and both are believed to be present in the last eukaryotic common ancestor Prokaryotes also undergo a vegetative cell division known as binary fission, where their genetic material is segregated equally into two daughter cells. All cell divisions, regardless of organism, are preceded by a single round of DNA replication. For simple unicellular organisms such as the amoeba, one cell division is equivalent to reproduction – an entire new organism is created. On a larger scale, mitotic cell division can create progeny from multicellular organisms, such as plants that grow from cuttings. Cell division also enables sexually reproducing organisms to develop from the one-celled zygote, which itself was produced by cell division from gametes. And after growth, cell division allows for continual construction and repair of the organism. A human being's body experiences about 10,000 trillion cell divisions in a lifetime. Cell division has been modeled by finite subdivision rules. The primary concern of cell division is the maintenance of the original cell's genome. Before division can occur, the genomic information that is stored in chromosomes must be replicated, and the duplicated genome must be separated cleanly between cells. A great deal of cellular infrastructure is involved in keeping genomic information consistent between "generations". Cells are classified into two categories: simple, non-nucleated prokaryotic cells, and complex, nucleated eukaryotic cells. By dint of their structural differences, eukaryotic and prokaryotic cells do not divide in the same way. Also, the pattern of cell division that transforms eukaryotic stem cells into gametes (sperm cells in males or ova – egg cells – in females) is different from that of the somatic cell division in the cells of the body. Multicellular organisms replace worn-out cells through cell division. In some animals, however, cell division eventually halts. In humans this occurs on average, after 52 divisions, known as the Hayflick limit. The cell is then referred to as senescent. Cells stop dividing because the telomeres, protective bits of DNA on the end of a chromosome required for replication, shorten with each copy, eventually being consumed, as described in the article on telomere shortening. Cancer cells, on the other hand, are not thought to degrade in this way, if at all. An enzyme called telomerase, present in large quantities in cancerous cells, rebuilds the telomeres, allowing division to continue indefinitely. Somatic evolution is the accumulation of mutations in the cells of a body during a lifetime, and the effects of those mutations on the fitness of those cells. Somatic evolution is important in the process of aging as well as the development of some diseases, including cancer. Cells in pre-malignant and malignant neoplasms (tumors) evolve by natural selection. This accounts for how cancer develops from normal tissue and why it has been difficult to cure. There are three necessary and sufficient conditions for natural selection, all of which are met in a neoplasm: Cells in neoplasms compete for resources, such as oxygen and glucose, as well as space. Thus, a cell that acquires a mutation that increases its fitness will generate more daughter cells than competitor cells that lack that mutation. In this way, a population of mutant cells, called a clone, can expand in the neoplasm. Clonal expansion is the signature of natural selection in cancer. Cancer therapies act as a form of artificial selection, killing sensitive cancer cells, but leaving behind resistant cells. Often the tumor will regrow from those resistant cells, the patient will relapse, and the therapy that had been previously used will no longer kill the cancer cells. This selection for resistance is similar to the repeatedly spraying crops with a pesticide and selecting for resistant pests until the pesticide is no longer effective. Modern descriptions of biological evolution will typically elaborate on major contributing factors to evolution such as the formation of local micro-environments, mutational robustness, molecular degeneracy, and cryptic genetic variation. Many of these contributing factors in evolution have been isolated and described for cancer. Cancer is a classic example of what evolutionary biologists call multilevel selection: at the level of the organism, cancer is usually fatal so there is selection for genes and the organization of tissues that suppress cancer. At the level of the cell, there is selection for increased cell proliferation and survival, such that a mutant cell that acquires one of the hallmarks of cancer (see below), will have a competitive advantage over cells that have not acquired the hallmark. Thus, at the level of the cell there is selection for cancer. The earliest ideas about neoplastic evolution come from Boveri who proposed that tumors originated in chromosomal abnormalities passed on to daughter cells. In the decades that followed, cancer was recognized as having a clonal origin associated with chromosomal aberrations. . Early mathematical modeling of cancer, by Armitage and Doll, set the stage for the future development of the somatic evolutionary theory of cancer. Armitage and Doll explained the cancer incidence data, as a function of age, as a process of the sequential accumulation of somatic mutations (or other rate limiting steps). Advances in cytogenetics facilitated discovery of chromosome abnormalities in neoplasms, including the Philadelphia chromosome in chronic myelogenous leukemia and translocations in acute myeloblastic leukemia. Sequences of karyotypes replacing one another in a tumor were observed as it progressed . Researchers hypothesized that cancer evolves in a sequence of chromosomal mutations and selection and that therapy may further select clones. In 1971, Knudson published the 2-hit hypothesis for mutation and cancer based on statistical analysis of inherited and sporadic cases of retinoblastoma. He postulated that retinoblastoma developed as a consequence of two mutations; one of which could be inherited or somatic followed by a second somatic mutation. Cytogenetic studies localized the region to the long arm of chromosome 13, and molecular genetic studies demonstrated that tumorigenesis was associated with chromosomal mechanisms, such as mitotic recombination or non-disjunction, that could lead to homozygosity of the mutation. The retinoblastoma gene was the first tumor suppressor gene to be cloned in 1986. Cairns hypothesized a different, but complementary, mechanism of tumor suppression in 1975 based on tissue architecture to protect against selection of variant somatic cells with increased fitness in proliferating epithelial populations, such as the intestine and other epithelial organs. He postulated that this could be accomplished by restricting the number of stem cells for example at the base of intestinal crypts and restraining the opportunities for competition between cells by shedding differentiated intestinal cells into the gut. The essential predictions of this model have been confirmed although mutations in some tumor suppressor genes, including CDKN2A (p16), predispose to clonal expansions that encompass large numbers of crypts in some conditions such as Barrett’s esophagus. He also postulated an immortal DNA strand that is discussed at Immortal DNA strand hypothesis. Nowell synthesized the evolutionary view of cancer in 1976 as a process of genetic instability and natural selection. Most of the alterations that occur are deleterious for the cell, and those clones will tend to go extinct, but occasional selectively advantageous mutations arise that lead to clonal expansions. This theory predicts a unique genetic composition in each neoplasm due to the random process of mutations, genetic polymorphisms in the human population, and differences in the selection pressures of the neoplasm’s microenvironment. Interventions are predicted to have varying results in different patients. What is more important, the theory predicts the emergence of resistant clones under the selective pressures of therapy. Since 1976, researchers have identified clonal expansions and genetic heterogeneity within many different types of neoplasms. It is known that there are multiple levels of genetic heterogeneity that are associated with cancer, including single nucleotide polymorphism (SNP), sequence mutations, Microsatellite shifts and instability, Loss of heterozygosity (LOH), Copy number variation (detected both by Comparative Genomic Hybridization (CGH), and array CGH, and karyotypic variations including chromosome structural aberrations and aneuploidy. Studies of this issue have focused mainly at the gene mutation level, as copy number variation, LOH and specific chromosomal translocations are explained in the context of gene mutation. It is thus necessary to integrate multiple levels of genetic variation in the context of complex system and multilevel selection. System instability is a major contributing factor for genetic heterogeneity. For the majority of cancers, genome instability is reflected at the chromosomal level and is referred to as chromosome instability or CIN. Genome instability is also referred to as an enabling characteristic for achieving endpoints of cancer evolution. Many of the somatic evolutionary studies have traditionally been focused on clonal expansion, as recurrent types of changes can be traced to illustrate the evolutionary path based on available methods. Recent studies from both direct DNA sequencing and karyotype analysis illustrate the importance of the high level of heterogeneity in somatic evolution. For the formation of solid tumors, there is an involvement of multiple cycles of clonal and non-clonal expansion. Even at the typical clonal expansion phase, there are significant levels of heterogeneity within the cell population, however, most are under-detected when mixed populations of cells are used for molecular analysis. In solid tumors, a majority of gene mutations are not recurrent types, and neither are the karyotypes. These analyses offer an explanation for the findings that there are no common mutations shared by most cancers. The state of a cell may be changed epigenetically, in addition to genetic alterations. The best-understood epigenetic alterations in tumors are the silencing or expression of genes by changes in the methylation of CG pairs of nucleotides in the promoter regions of the genes. These methylation patterns are copied to the new chromosomes when cells replicate their genomes and so methylation alterations are heritable and subject to natural selection. Methylation changes are thought to occur more frequently than mutations in the DNA, and so may account for many of the changes during neoplastic progression (the process by which normal tissue becomes cancerous), in particular in the early stages. Epigenetic changes in progression interact with genetic changes. For example, epigenetic silencing of genes responsible for the repair of mutations in the DNA (e.g. MLH1 and MSH2) results in an increase of genetic mutations. One common feature of neoplastic progression is the expansion of a clone with a genetic or epigenetic alteration. This may be a matter of chance, but is more likely due to the expanding clone having a competitive advantage (either a reproductive or survival advantage) over other cells in the tissue. Since clones often have many genetic and epigenetic alterations in their genomes, it is often not clear which of those alterations cause a reproductive or survival advantage and which other alterations are simply hitchhikers or passenger mutations (see Glossary below) on the clonal expansion. Clonal expansions are most often associated with the loss of the p53 (TP53) or p16 (CDKN2A/INK4a) tumor suppressor genes. In lung cancer, a clone with a p53 mutation was observed to have spread over the surface of one entire lung and into the other lung. In bladder cancer, clones with loss of p16 were observed to have spread over the entire surface of the bladder. Likewise, large expansions of clones with loss of p16 have been observed in the oral cavity and in Barrett's esophagus. Clonal expansions associated with inactivation of p53 have also appear in skin, Barrett's esophagus, brain, and kidney. Further clonal expansions have been observed in the stomach, bladder, colon, lung, hematopoietic (blood) cells, and prostate. These clonal expansions are important for at least two reasons. First, they generate a large target population of mutant cells and so increase the probability that the multiple mutations necessary to cause cancer will be acquired within that clone. Second, in at least one case, the size of the clone with loss of p53 has been associated with an increased risk of a pre-malignant tumor becoming cancerous. It is thought that the process of developing cancer involves successive waves of clonal expansions within the tumor. Phylogenetics may be applied to cells in tumors to reveal the evolutionary relationships between cells, just as it is used to reveal evolutionary relationships between organisms and species. Shibata, Tavare and colleagues have exploited this to estimate the time between the initiation of a tumor and its detection in the clinic. Louhelainen et al. have used parsimony to reconstruct the relationships between biopsy samples based on loss of heterozygosity. Phylogenetic trees should not be confused with oncogenetic trees, which represent the common sequences of genetic events during neoplastic progression and do not represent the relationships of common ancestry that are essential to a phylogeny. An adaptive landscape is a hypothetical topological landscape upon which evolution is envisioned to take place. It is similar to Wright's fitness landscape in which the location of each point represents the genotype of an organism and the altitude represents the fitness of that organism in the current environment. However, unlike Wright's rigid landscape, the adaptive landscape is pliable. It readily changes shape with changes in population densities and survival/reproductive strategies used within and among the various species. Wright’s shifting balance theory of evolution combines genetic drift (random sampling error in the transmission of genes) and natural selection to explain how multiple peaks on a fitness landscape could be occupied or how a population can achieve a higher peak on this landscape. This theory, based on the assumption of density-dependent selection as the principle forms of selection, results in a fitness landscape that is relatively rigid. A rigid landscape is one that does not change in response to even large changes in the position and composition of strategies along the landscape. In contrast to the fitness landscape, the adaptive landscape is constructed assuming that both density and frequency-dependent selection is involved (selection is frequency-dependant when the fitness of a species depends not only on that species strategy but also on the strategy of all other species). As such, the shape of the adaptive landscape can change drastically in response to even small changes in strategies and densities. The flexibility of adaptive landscapes provide several ways for natural selection to cross valleys and occupy multiple peaks without having to make large changes in their strategies. Within the context of differential or difference equation models for population dynamics, an adaptive landscape may actually be constructed using a Fitness Generating Function. If a given species is able to evolve, it will, over time, "climb" the adaptive landscape toward a fitness peak through gradual changes in its mean phenotype according to a strategy dynamic that involves the slope of the adaptive landscape. Because the adaptive landscape is not rigid and can change shape during the evolutionary process, it is possible that a species may be driven to maximum, minimum, or saddle point on the adaptive landscape. A population at a global maximum on the adaptive landscape corresponds an evolutionarily stable strategy (ESS) and will become dominant, driving all others toward extinction. Populations at a minimum or saddle point are not resistant to invasion, so that the introduction of a slightly different mutant strain may continue the evolutionary process toward unoccupied local maxima. The adaptive landscape provides a useful tool for studying somatic evolution as it can describe the process of how a mutant cell evolves from a small tumor to an invasive cancer. Understanding this process in terms of the adaptive landscape may lead to the control of cancer through external manipulation of the shape of the landscape. In their landmark paper, The Hallmarks of Cancer, Hanahan and Weinberg suggest that cancer can be described by a small number of underlying principles, despite the complexities of the disease. The authors describe how tumor progression proceeds via a process analogous to Darwinian evolution, where each genetic change confers a growth advantage to the cell. These genetic changes can be grouped into six "hallmarks", which drive a population of normal cells to become a cancer. The six hallmarks are: Genetic instability is defined as an "enabling characteristic" that facilitates the acquisition of other mutations due to defects in DNA repair. The hallmark "self-sufficiency in growth signals" describes the observation that tumor cells produce many of their own growth signals and thereby no longer rely on proliferation signals from the micro-environment. Normal cells are maintained in a nondividing state by antigrowth signals, which cancer cells learn to evade through genetic changes producing "insensitivity to antigrowth signals". A normal cell initiates programmed cell death (apoptosis) in response to signals such as DNA damage, oncogene overexpression, and survival factor insufficiency, but a cancer cell learns to "evade apoptosis", leading to the accumulation of aberrant cells. Most mammalian cells can replicate a limited number of times due to progressive shortening of telomeres; virtually all malignant cancer cells gain an ability to maintain their telomeres, conferring "limitless replicative potential". As cells cannot survive at distances of more than 100 μm from a blood supply, cancer cells must initiate the formation of new blood vessels to support their growth via the process of "sustained angiogenesis". During the development of most cancers, primary tumor cells acquire the ability to undergo "invasion and metastasis" whereby they migrate into the surrounding tissue and travel to distant sites in the body, forming secondary tumors. The pathways that cells take toward becoming malignant cancers are variable, and the order in which the hallmarks are acquired can vary from tumor to tumor. The early genetic events in tumorigenesis are difficult to measure clinically, but can be simulated according to known biology. Macroscopic tumors are now beginning to be described in terms of their underlying genetic changes, providing additional data to refine the framework described in The Hallmarks of Cancer. The theory about the monoclonal origin of cancer states that in general neoplasms arise from a single cell of origin. While it is possible that certain carcinogens may mutate more than one cell at once, the tumor mass usually represents progeny of a single cell, or very few cells. A series of mutations is required in the process of carcinogenesis for a cell to transition from being normal to pre-malignant and then to a cancer cell. The mutated genes usually belong to classes of caretaker, gatekeeper, landscaper or several other genes. Mutation ultimately leads to acquisition of the six hallmarks of cancer. The first malignant cell, that gives rise to the tumor, is often labeled a cancer stem cell. The cancer stem-cell hypothesis relies on the fact that a lot of tumors are heterogeneous – the cells in the tumor vary by phenotype and functions. Current research shows that in many cancers there is apparent hierarchy among cells. in general, there is a small population of cells in the tumor – about 0.2%–1% – that exhibits stem cell-like properties. These cells have the ability to give rise to a variety of cells in tumor tissue, self-renew indefinitely, and upon transfer can form new tumors. According to the hypothesis, cancer stem cells are the only cells capable of tumorigenesis – initiation of a new tumor. Cancer stem cell hypothesis might explain such phenomena as metastasis and remission. The monoclonal model of cancer and the cancer stem-cell model are not mutually exclusive. Cancer stem cell arises by clonal evolution as a result of selection for the cell with the highest fitness in the neoplasm. This way, the heterogeneous nature of neoplasm can be explained by two processes – clonal evolution, or the hierarchical differentiation of cells, regulated by cancer stem cells. All cancers arise as a result of somatic evolution, but only some of them fit the cancer stem cell hypothesis. The evolutionary processes do not cease when a population of cancer stem cells arises in a tumor. Cancer treatment drugs pose a strong selective force on all types of cells in tumors, including cancer stem cells, which would be forced to evolve resistance to the treatment. It is interesting to note that cancer stem cells do not always have to have the highest resistance among the cells in the tumor to survive chemotherapy and re-emerge afterwards. The surviving cells might be in a special microenvironment, which protects them from adverse effects of treatment. It is currently unclear as to whether cancer stem cells arise from adult stem cell transformation, a maturation arrest of progenitor cells, or as a result of dedifferentiation of mature cells. Therapeutic resistance has been observed in virtually every form of therapy, from the beginning of cancer therapy. In most cases, therapies appear to select for mutations in the genes or pathways targeted by the drug. Some of the first evidence for a genetic basis of acquired therapeutic resistance came from studies of methotrexate. Methotrexate inhibits the dihydrofolate reductase (DHFR) gene. However, methotrexate therapy appears to select for cells with extra copies (amplification) of DHFR, which are resistant to methotrexate. This was seen in both cell culture and samples from tumors in patients that had been treated with methotrexate. A common cytotoxic chemotherapy used in a variety of cancers, 5-fluorouracil (5-FU), targets the TYMS pathway and resistance can evolve through the evolution of extra copies of TYMS, thereby diluting the drug's effect. In the case of Gleevec (Imatinib), which targets the BCR-ABL fusion gene in chronic myeloid leukemia, resistance often develops through a mutation that changes the shape of the binding site of the drug. Sequential application of drugs can lead to the sequential evolution of resistance mutations to each drug in turn. Gleevec is not as selective as was originally thought. It turns out that it targets other tyrosine kinase genes and can be used to control gastrointestinal stromal tumors (GISTs) that are driven by mutations in c-KIT. However, patients with GIST sometimes relapse with additional mutations in c-KIT that make the cancer cells resistant to Gleevec. Gefitinib(Iressa) and Erlotinib (Tarceva) are epidermal growth factor receptor (EGFR) tyrosine kinase inhibitors used for non-small cell lung cancer patients whose tumors have somatic mutations in EGFR. However, most patients' tumors eventually become resistant to these drugs. Two major mechanisms of acquired resistance have been discovered in patients who have developed clinical resistance to Gefitinib or Erlotinib: point mutations in the EGFR gene targeted by the drugs, and amplification of MET, another receptor tyrosine kinase, which can bypass EGFR to activate downstream signaling in the cell. In an initial study, 22% of tumors with acquired resistance to Gefitinib or Erlotinib had MET amplification. To address these issues, clinical trials are currently assessing irreversible EGFR inhibitors (which inihibit growth even in cell lines with mutations in EGFR), the combination of EGFR and MET kinase inhibitors, and Hsp90 inihibitors (EGFR and MET both require Hsp90 proteins to fold properly). In addition, taking repeated tumor biopsies from patients as they develop resistance to these drugs would help to understand the tumor dynamics. Selective estrogen receptor modulators (SERMs) are a commonly used adjuvant therapy in estrogen-receptor positive (ERα+) breast cancer and a preventive treatment for women at high risk of the disease. There are several possible mechanisms of SERM resistance, though the relative clinical importance of each is debated. These include: Most prostate cancers derive from cells that are stimulated to proliferate by androgens. Most prostate cancer therapies are therefore based on removing or blocking androgens. Mutations in the androgen receptor (AR) have been observed in anti-androgen resistant prostate cancer that makes the AR hypersensitive to the low levels of androgens that remain after therapy. Likewise, extra copies of the AR gene (amplification) have been observed in anti-androgen resistant prostate cancer. These additional copies of the gene are thought to make the cell hypersensitive to low levels of androgens and so allow them to proliferate under anti-androgen therapy. Resistance to radiotherapy is also commonly observed. However, to date, comparisons of malignant tissue before and after radiotherapy have not been done to identify genetic and epigenetic changes selected by exposure to radiation. In gliomas, a form of brain cancer, radiation therapy appears to select for stem cells, though it is unclear if the tumor returns to the pre-therapy proportion of cancer stem cells after therapy or if radiotherapy selects for an alteration that keeps the glioma cells in the stem cell state. Cancer drugs and therapies commonly used today are evolutionary inert and represent a strong selection force, which leads to drug resistance. A possible way to avoid that is to use a treatment agent that would co-evolve alongside with cancer cells. Anoxic bacteria could be used as competitors or predators in hypoxic environments within tumors. Scientists have been interested in the idea of using anoxic bacteria for over 150 years, but until recently there has been little progress in that field. According to Jain and Forbes, several requirements have to be met by the cells to qualify as efficient anticancer bacterium: 1.The bacterium cannot be toxic to the host 2.Its population should be restricted to the tumor mass 3.It should be able to disperse evenly throughout the neoplasm 4.At the end of the treatment bacterium should be easily eliminated from the host 5.It should not be causing severe immune response 6.It should be able to cause tumor cells death through competition for nutrients. In the process of the treatment cancer cells are most likely to evolve some form of resistance to the bacterial treatment. However, being a living organism, bacteria would coevolve with tumor cells, potentially eliminating the possibility of resistance. Since bacteria prefer an anoxic environment, they are not efficient at eliminating cells on the periphery of the tumor, where oxygen supply is efficient. A combination of bacterial treatment with chemical drugs will increase chances of destroying the tumor. Oncolytic viruses are engineered to infect cancerous cells. Limitations of that method include immune response to the virus and the possibility of the virus evolving into a pathogen. By manipulating the tumor environment we can create favorable conditions for the cells with least resistance to chemotherapy drugs to become more fit and outcompete the rest of the population. The chemotherapy, administered directly after, should wipe out the predominant tumor cells. Mapping between common terms from cancer biology and evolutionary biology M: NEO tsoc, mrkr tumr, epon, para drug (L1i/1e/V03) cancer Genes The cell cycle, or cell-division cycle, is the series of events that take place in a cell leading to its division and duplication (replication). In cells without a nucleus (prokaryotic), the cell cycle occurs via a process termed binary fission. In cells with a nucleus (eukaryotes), the cell cycle can be divided in three periods: interphase—during which the cell grows, accumulating nutrients needed for mitosis and duplicating its DNA—and the mitotic (M) phase, during which the cell splits itself into two distinct cells, often called "daughter cells" and the final phase, cytokinesis, where the new cell is completely divided. The cell-division cycle is a vital process by which a single-celled fertilized egg develops into a mature organism, as well as the process by which hair, skin, blood cells, and hi some internal organs are renewed. Carcinogenesis Carcinogens "The Hallmarks of Cancer" is a seminal peer-reviewed article published in the journal Cell in January 2000 by US cancer researchers Douglas Hanahan and Robert Weinberg. The authors believe that the complexity of cancer can be reduced to a small number of underlying principles. The paper argues that all cancers share six common traits ("hallmarks") that govern the transformation of normal cells to cancer (malignant or tumor) cells. Medicine Biology Oncology Health Medical Pharma Health Medical Pharma 20	2014-09-02 01:53:11	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 1, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.387702077627182, "perplexity": 3865.407944531376}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-35/segments/1409535921318.10/warc/CC-MAIN-20140909054206-00423-ip-10-180-136-8.ec2.internal.warc.gz"}
https://cs.stackexchange.com/questions/28901/how-to-simulate-a-bidirectional-tm-on-a-regular-one-with-time-factor-four	# How to simulate a bidirectional TM on a regular one with time factor four? In Computational Complexity A Modern Approach, one claim says that if $f$ is computable in time $T(n)$ by a bidirectional TM $M$, then it is computable in time $4T(n)$ by a unidirectional TM $\tilde{M}$. How to work out the constant $4$? In my opinion, in addition to "go over the edge" operation, one transition in $M$ corresponds to one transition in $\tilde{M}$, so where does constant $4$ come from? • There is no proof in the book? Have you tried working out how a TM would simulate a bidirectional TM? – Raphael Aug 3 '14 at 20:49 • This is about Claim 1.8 in Arora&Barak: Computational Complexity: A Modern Appproach, found on page 21. The book has loads of stuff in it where the actual proofs are left to the reader, while some other things only have proof sketches - this one falls in the latter category. I remember reading the first third of the book and often thinking "this bound is unnecessarily loose" - after skimming the paragraph now, I get the impression that this is the case here as well. – G. Bach Aug 4 '14 at 2:45 • My advice is not to worry too much about the constants; the point of all these bits in the first chapter is that more tapes, a larger alphabet and bidirectionality do not give TMs more power in the sense that doing any of that would make more functions computable than by a TM with one unidirectional tape and alphabet $\{0,1\}$, nor does it put more functions into $P$. – G. Bach Aug 4 '14 at 2:46 • Here "unidirectional" means the tape is one-sided infinite I presume. Reading the first time I understood it meant it could move in only one direction. And yes, I think you are right: combining two symbols in a tape square you can efficiently simulate a two-sided tape. If you alternate (even and odd positions on the one sided tape represent negative and positive positions on the two sided tape) then the constant will be $2$ I guess. – Hendrik Jan Aug 4 '14 at 9:18	2020-06-03 18:27:11	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.761425793170929, "perplexity": 533.5031748127808}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-24/segments/1590347435987.85/warc/CC-MAIN-20200603175139-20200603205139-00439.warc.gz"}
https://quantumcomputing.stackexchange.com/questions/15009/how-to-build-an-example-of-shors-algorithm-for-the-discrete-log	# How to build an example of Shor's algorithm for the discrete log? I have been trying to build myself an example of Shor's computations for the discrete log. I started out with this objective and I realized I should understand the factorization first, which I did and did. Having a clue about the factorization, perhaps now I can tackle the discrete log, but it's still not that easy. Problem. Find $$a$$ such that $$2^a = 7 \bmod 29$$. (We know $$2$$ is a generator of $$Z_{29}$$.) Peter Shor tells us to find $$q$$, a power of $$2$$ that is close to $$29$$, that is, $$29 < q < 2\times29$$. So $$q = 32 = 2^5$$ suffices. Next he tells us to put two register \|$$a$$> and \|$$b$$> in uniform superposition $$\bmod 28$$. (Why $$\kern-0.4em\bmod 28$$? Why not $$\kern-0.4em\bmod 29$$?) Then in a third register compute \|$$2^a 7^{-b} \bmod 29$$>. This will produce a periodic sequence in superposition. Applying the QFT to this register, we should be able to extract this period. When I look at the sequence for this concrete case (which is $$2^a 7^{-b} \bmod 29$$), I find $$[1, 25, 16, 23, 24, 20, 7, 1, 25, 16, 23, 24, 20, 7, 1, ...]$$ So, I can see the period is $$7$$. What is the calculation that I must do now to extract the solution $$a = 12$$? 1. Fermat's little theorem says that for a prime $$p$$, $$a^{p-1}\equiv 1\mod p$$ for all $$a$$ co-prime to $$p$$. This means the order of the group of powers of $$a$$ will divide $$p-1$$, rather than $$p$$, hence why the algorithm uses $$28$$. 2. The "input" is two-dimensional here; it should be all pairs $$(a,b)$$ for $$0\leq a\leq 27$$ and $$0\leq b\leq 27$$. The period you're looking for is a pair of integers $$(r_1,r_2)$$ such that $$2^a7^{-b}\equiv 2^{a+r_1}7^{-b-r_2}\mod p$$. Let $$x$$ be the discrete log, i.e., $$7\equiv 2^x\mod p$$. Then we can re-write this as $$2^a2^{-bx}\equiv 2^{a+r_1}2^{-x(b+r_2)}\mod p$$ Since we know that $$2^{28}\equiv 1\mod p$$, we can treat the exponents as integers modulo $$28$$, i.e., $$a-bx \equiv a+r_1-x(b+r_2)\mod 28$$ We can cancel out terms with $$a$$ and $$b$$: \begin{align}0\equiv &r_1-xr_2\mod 28\\ -r_1r_2^{-1}\equiv &x\mod 28\end{align} So the discrete log is $$-r_1r_2^{-1}\mod 28$$. This means what you really want to do is take a two-dimensional array of results for different values of $$a$$ and $$b$$, and find a period in that. It looks like what you've actually found is that the order of $$7$$ is $$7$$, i.e., $$7^7\equiv 1\mod p$$. I don't think this has any impact on Shor's algorithm, but I'm not sure. Usually you would only apply Shor's algorithm to a prime-order group, since you can use the Pohlig-Hellman algorithm to reduce composite-order discrete logs to smaller prime-order discrete logs. • Your answer looks very interesting. I might take some time digesting that. More to follow! Dec 9, 2020 at 0:18	2022-07-07 14:45:55	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 47, "wp-katex-eq": 0, "align": 1, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.930656373500824, "perplexity": 227.01419176457094}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-27/segments/1656104692018.96/warc/CC-MAIN-20220707124050-20220707154050-00213.warc.gz"}
http://en.wikipedia.org/wiki/Trajectories	# Trajectory (Redirected from Trajectories) A trajectory is the path that a moving object follows through space as a function of time. The object might be a projectile or a satellite, for example. It thus includes the meaning of orbit—the path of a planet, an asteroid or a comet as it travels around a central mass. A trajectory can be described mathematically either by the geometry of the path, or as the position of the object over time. In control theory a trajectory is a time-ordered set of states of a dynamical system (see e.g. Poincaré map). In discrete mathematics, a trajectory is a sequence $(f^k(x))_{k \in \mathbb{N}}$ of values calculated by the iterated application of a mapping $f$ to an element $x$ of its source. Illustration showing the trajectory of a bullet fired at an uphill target. ## Physics of trajectories A familiar example of a trajectory is the path of a projectile such as a thrown ball or rock. In a greatly simplified model the object moves only under the influence of a uniform gravitational force field. This can be a good approximation for a rock that is thrown for short distances for example, at the surface of the moon. In this simple approximation the trajectory takes the shape of a parabola. Generally, when determining trajectories it may be necessary to account for nonuniform gravitational forces, air resistance (drag and aerodynamics). This is the focus of the discipline of ballistics. One of the remarkable achievements of Newtonian mechanics was the derivation of the laws of Kepler, in the case of the gravitational field of a single point mass (representing the Sun). The trajectory is a conic section, like an ellipse or a parabola. This agrees with the observed orbits of planets and comets, to a reasonably good approximation, although if a comet passes close to the Sun, then it is also influenced by other forces, such as the solar wind and radiation pressure, which modify the orbit, and cause the comet to eject material into space. Newton's theory later developed into the branch of theoretical physics known as classical mechanics. It employs the mathematics of differential calculus (which was, in fact, also initiated by Newton, in his youth). Over the centuries, countless scientists contributed to the development of these two disciplines. Classical mechanics became a most prominent demonstration of the power of rational thought, i.e. reason, in science as well as technology. It helps to understand and predict an enormous range of phenomena. Trajectories are but one example. Consider a particle of mass $m$, moving in a potential field $V$. Physically speaking, mass represents inertia, and the field $V$ represents external forces, of a particular kind known as "conservative". That is, given $V$ at every relevant position, there is a way to infer the associated force that would act at that position, say from gravity. Not all forces can be expressed in this way, however. The motion of the particle is described by the second-order differential equation $m \frac{\mathrm{d}^2 \vec{x}(t)}{\mathrm{d}t^2} = -\nabla V(\vec{x}(t))$ with $\vec{x} = (x, y, z)$ On the right-hand side, the force is given in terms of $\nabla V$, the gradient of the potential, taken at positions along the trajectory. This is the mathematical form of Newton's second law of motion: force equals mass times acceleration, for such situations. ## Examples ### Uniform gravity, no drag or wind Trajectories of a mass thrown at an angle of 70°: without drag with Stokes drag with Newton drag The ideal case of motion of a projectile in a uniform gravitational field, in the absence of other forces (such as air drag), was first investigated by Galileo Galilei. To neglect the action of the atmosphere, in shaping a trajectory, would have been considered a futile hypothesis by practical minded investigators, all through the Middle Ages in Europe. Nevertheless, by anticipating the existence of the vacuum, later to be demonstrated on Earth by his collaborator Evangelista Torricelli[citation needed], Galileo was able to initiate the future science of mechanics.[citation needed] And in a near vacuum, as it turns out for instance on the Moon, his simplified parabolic trajectory proves essentially correct. In the analysis that follows we derive the equation of motion of a projectile as measured from an inertial frame, at rest with respect to the ground, to which frame is associated a right-hand co-ordinate system - the origin of which coincides with the point of launch of the projectile. The x-axis is parallel to the ground and the y axis perpendicular to it ( parallel to the gravitational field lines ). Let $g$ be the acceleration of gravity. Relative to the flat terrain, let the initial horizontal speed be $v_h = v \cos(\theta)$ and the initial vertical speed be $v_v = v \sin(\theta)$. It will also be shown that, the range is $2v_h v_v/g$, and the maximum altitude is $v_v^2/2g$; The maximum range, for a given initial speed $v$, is obtained when $v_h=v_v$, i.e. the initial angle is 45 degrees. This range is $v^2/g$, and the maximum altitude at the maximum range is a quarter of that. #### Derivation of the equation of motion Assume the motion of the projectile is being measured from a Free fall frame which happens to be at (x,y)=(0,0) at t=0. The equation of motion of the projectile in this frame (by the principle of equivalence) would be $y = x \tan(\theta)$. The co-ordinates of this free-fall frame, with respect to our inertial frame would be $y = - gt^2/2$. That is, $y = - g(x/v_h)^2/2$. Now translating back to the inertial frame the co-ordinates of the projectile becomes $y = x \tan(\theta)- g(x/v_h)^2/2$ That is: $y=-{g\sec^2\theta\over 2v_0^2}x^2+x\tan\theta$, (where v0 is the initial velocity, $\theta$ is the angle of elevation, and g is the acceleration due to gravity). #### Range and height Trajectories of projectiles launched at different elevation angles but the same speed of 10 m/s in a vacuum and uniform downward gravity field of 10 m/s2. Points are at 0.05 s intervals and length of their tails is linearly proportional to their speed. t = time from launch, T = time of flight, R = range and H = highest point of trajectory (indicated with arrows). The range, R, is the greatest distance the object travels along the x-axis in the I sector. The initial velocity, vi, is the speed at which said object is launched from the point of origin. The initial angle, θi, is the angle at which said object is released. The g is the respective gravitational pull on the object within a null-medium. $R={v_i^2\sin2\theta_i\over g}$ The height, h, is the greatest parabolic height said object reaches within its trajectory $h={v_i^2\sin^2\theta_i\over 2g}$ #### Angle of elevation In terms of angle of elevation $\theta$ and initial speed $v$: $v_h=v \cos \theta,\quad v_v=v \sin \theta \;$ giving the range as $R= 2 v^2 \cos(\theta) \sin(\theta) / g = v^2 \sin(2\theta) / g\,.$ This equation can be rearranged to find the angle for a required range ${ \theta } = \frac 1 2 \sin^{-1} \left( { {g R} \over { v^2 } } \right)$ (Equation II: angle of projectile launch) Note that the sine function is such that there are two solutions for $\theta$ for a given range $d_h$. The angle $\theta$ giving the maximum range can be found by considering the derivative or $R$ with respect to $\theta$ and setting it to zero. ${\mathrm{d}R\over \mathrm{d}\theta}={2v^2\over g} \cos(2\theta)=0$ which has a nontrivial solution at $2\theta=\pi/2=90^\circ$, or $\theta=45^\circ$. The maximum range is then $R_{max} = v^2/g\,$. At this angle $sin(\pi/2)=1$, so the maximum height obtained is ${v^2 \over 4g}$. To find the angle giving the maximum height for a given speed calculate the derivative of the maximum height $H=v^2 sin^2(\theta) /(2g)$ with respect to $\theta$, that is ${\mathrm{d}H\over \mathrm{d}\theta}=v^2 2\cos(\theta)\sin(\theta) /(2g)$ which is zero when $\theta=\pi/2=90^\circ$. So the maximum height $H_{max}={v^2\over 2g}$ is obtained when the projectile is fired straight up. ### Uphill/downhill in uniform gravity in a vacuum Given a hill angle $\alpha$ and launch angle $\theta$ as before, it can be shown that the range along the hill $R_s$ forms a ratio with the original range $R$ along the imaginary horizontal, such that: $\frac{R_s} {R}=(1-\cot \theta \tan \alpha)\sec \alpha$ (Equation 11) In this equation, downhill occurs when $\alpha$ is between 0 and -90 degrees. For this range of $\alpha$ we know: $\tan(-\alpha)=-\tan \alpha$ and $\sec ( - \alpha ) = \sec \alpha$. Thus for this range of $\alpha$, $R_s/R=(1+\tan \theta \tan \alpha)\sec \alpha$. Thus $R_s/R$ is a positive value meaning the range downhill is always further than along level terrain. The lower level of terrain causes the projectile to remain in the air longer, allowing it to travel further horizontally before hitting the ground. While the same equation applies to projectiles fired uphill, the interpretation is more complex as sometimes the uphill range may be shorter or longer than the equivalent range along level terrain. Equation 11 may be set to $R_s/R=1$ (i.e. the slant range is equal to the level terrain range) and solving for the "critical angle" $\theta_{cr}$: $1=(1-\tan \theta \tan \alpha)\sec \alpha \quad \;$ $\theta_{cr}=\arctan((1-\csc \alpha)\cot \alpha) \quad \;$ Equation 11 may also be used to develop the "rifleman's rule" for small values of $\alpha$ and $\theta$ (i.e. close to horizontal firing, which is the case for many firearm situations). For small values, both $\tan \alpha$ and $\tan \theta$ have a small value and thus when multiplied together (as in equation 11), the result is almost zero. Thus equation 11 may be approximated as: $\frac{R_s} {R}=(1-0)\sec \alpha$ And solving for level terrain range, $R$ $R=R_s \cos \alpha \$ "Rifleman's rule" Thus if the shooter attempts to hit the level distance R, s/he will actually hit the slant target. "In other words, pretend that the inclined target is at a horizontal distance equal to the slant range distance multiplied by the cosine of the inclination angle, and aim as if the target were really at that horizontal position."[1] #### Derivation based on equations of a parabola The intersect of the projectile trajectory with a hill may most easily be derived using the trajectory in parabolic form in Cartesian coordinates (Equation 10) intersecting the hill of slope $m$ in standard linear form at coordinates $(x,y)$: $y=mx+b \;$ (Equation 12) where in this case, $y=d_v$, $x=d_h$ and $b=0$ Substituting the value of $d_v=m d_h$ into Equation 10: $m x=-\frac{g}{2v^2{\cos}^2 \theta}x^2 + \frac{\sin \theta}{\cos \theta} x$ $x=\frac{2v^2\cos^2\theta}{g}\left(\frac{\sin \theta}{\cos \theta}-m\right)$ (Solving above x) This value of x may be substituted back into the linear equation 12 to get the corresponding y coordinate at the intercept: $y=mx=m \frac{2v^2\cos^2\theta}{g} \left(\frac{\sin \theta}{\cos \theta}-m\right)$ Now the slant range $R_s$ is the distance of the intercept from the origin, which is just the hypotenuse of x and y: $R_s=\sqrt{x^2+y^2}=\sqrt{\left(\frac{2v^2\cos^2\theta}{g}\left(\frac{\sin \theta}{\cos \theta}-m\right)\right)^2+\left(m \frac{2v^2\cos^2\theta}{g} \left(\frac{\sin \theta}{\cos \theta}-m\right)\right)^2}$ $=\frac{2v^2\cos^2\theta}{g} \sqrt{\left(\frac{\sin \theta}{\cos \theta}-m\right)^2+m^2 \left(\frac{\sin \theta}{\cos \theta}-m\right)^2}$ $=\frac{2v^2\cos^2\theta}{g} \left(\frac{\sin \theta}{\cos \theta}-m\right) \sqrt{1+m^2}$ Now $\alpha$ is defined as the angle of the hill, so by definition of tangent, $m=\tan \alpha$. This can be substituted into the equation for $R_s$: $R_s=\frac{2v^2\cos^2\theta}{g} \left(\frac{\sin \theta}{\cos \theta}-\tan \alpha\right) \sqrt{1+\tan^2 \alpha}$ Now this can be refactored and the trigonometric identity for $\sec \alpha = \sqrt {1 + \tan^2 \alpha}$ may be used: $R_s=\frac{2v^2\cos\theta\sin\theta}{g}\left(1-\frac{\sin\theta}{\cos\theta}\tan\alpha\right)\sec\alpha$ Now the flat range $R=v^2\sin 2 \theta / g = 2v^2\sin\theta\cos\theta / g$ by the previously used trigonometric identity and $\cos\theta/\sin\theta=cotan\theta$ so: $R_s=R(1-\cot\theta\tan\alpha)\sec\alpha \;$ $\frac{R_s}{R}=(1-\cot\theta\tan\alpha)\sec\alpha$ ### Orbiting objects If instead of a uniform downwards gravitational force we consider two bodies orbiting with the mutual gravitation between them, we obtain Kepler's laws of planetary motion. The derivation of these was one of the major works of Isaac Newton and provided much of the motivation for the development of differential calculus. ## Catching balls If a projectile, such as a baseball or cricket ball, travels in a parabolic path, with negligible air resistance, and if a player is positioned so as to catch it as it descends, he sees its angle of elevation increasing continuously throughout its flight. The tangent of the angle of elevation is proportional to the time since the ball was sent into the air, usually by being struck with a bat. Even when the ball is really descending, near the end of its flight, its angle of elevation seen by the player continues to increase. The player therefore sees it as if it were ascending vertically at constant speed. Finding the place from which the ball appears to rise steadily helps the player to position himself correctly to make the catch. If he is too close to the batsman who has hit the ball, it will appear to rise at an accelerating rate. If he is too far from the batsman, it will appear to slow rapidly, and then to descend. Proof Suppose the ball starts with a vertical component of velocity of $v,$ upward, and a horizontal component of velocity of $h$ toward the player who wants to catch it. Its altitude above the ground is given by: $a=vt-\frac{1}{2}gt^2,$ where $t$ is the time since the ball was hit, and $g$ is the acceleration due to gravity. The total time for the flight, until the ball is back down to the ground, from which it started, is given by: $a=0$ $\therefore T=\frac{2v}{g}.$ The horizontal component of the ball's distance from the catcher at time $t$ is: $d=h(T-t) = \frac{2hv}{g}-ht$ The tangent of the angle of elevation of the ball, as seen by the catcher, is: $\tan(e)=\frac{a}{d}$ $=\frac{vt-\frac{gt^2}{2}}{\frac{2hv}{g}-ht}$ $=\frac{2gvt-g^2t^2}{4hv-2ght}$ $=\frac{gt(2v-gt)}{2h(2v-gt)}$ The quantity in the brackets will not be zero except when the ball is on the ground, therefore, while the ball is in flight: $\tan(e)=\left(\frac{g}{2h}\right)t$ The bracket in this last expression is constant for a given flight of the ball. Therefore the tangent of the angle of elevation of the ball, as seen by the player who is properly positioned to catch it, is directly proportional to the time since the ball was hit.	2013-12-07 10:31:08	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 108, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8380642533302307, "perplexity": 318.63060639427994}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2013-48/segments/1386163053923/warc/CC-MAIN-20131204131733-00090-ip-10-33-133-15.ec2.internal.warc.gz"}
http://zbmath.org/?q=an:1015.34046	# zbMATH — the first resource for mathematics ##### Examples Geometry Search for the term Geometry in any field. Queries are case-independent. Funct* Wildcard queries are specified by * (e.g. functions, functorial, etc.). Otherwise the search is exact. "Topological group" Phrases (multi-words) should be set in "straight quotation marks". au: Bourbaki & ti: Algebra Search for author and title. The and-operator & is default and can be omitted. Chebyshev \| Tschebyscheff The or-operator \| allows to search for Chebyshev or Tschebyscheff. "Quasi* map" py: 1989 The resulting documents have publication year 1989. so: Eur J* Mat* Soc* cc: 14 Search for publications in a particular source with a Mathematics Subject Classification code (cc) in 14. "Partial diff* eq" ! elliptic The not-operator ! eliminates all results containing the word elliptic. dt: b & au: Hilbert The document type is set to books; alternatively: j for journal articles, a for book articles. py: 2000-2015 cc: (94A \| 11T) Number ranges are accepted. Terms can be grouped within (parentheses). la: chinese Find documents in a given language. ISO 639-1 language codes can also be used. ##### Operators a & b logic and a \| b logic or !ab logic not abc right wildcard "ab c" phrase (ab c) parentheses ##### Fields any anywhere an internal document identifier au author, editor ai internal author identifier ti title la language so source ab review, abstract py publication year rv reviewer cc MSC code ut uncontrolled term dt document type (j: journal article; b: book; a: book article) Krasnoselskii’s fixed point theorem and stability. (English) Zbl 1015.34046 The authors present an application of the fixed-point theory in stability. They suggest a generalization of Krasnosel’skii’s theorem on fixed-points of operators of the form $A+B$, where $A$ is completely continuous and $B$ is contracting, and use their result to prove new theorems on the exponential stability of solutions to Cauchy problems. General theorems are applied to perturbed Liénard equations. One of the main results is as follows: Let $M$ denote a closed convex nonempty subset of the Banach space $U$ of bounded continuous functions $\phi :\left[0,\infty \right)\to {ℝ}^{d}$. Consider the Cauchy problem ${x}^{\text{'}}=b\left(t,x\right)+a\left(t,x\right),\phantom{\rule{1.em}{0ex}}x\left(0\right)={x}_{0}\in {ℝ}^{d},\phantom{\rule{4pt}{0ex}}t\in \left[0,\infty \right)·$ Let $b\left(t,x\right)$ be uniformly Lipschitz in $x$ for $t\in \left[0,\infty$), $x\in {ℝ}^{d}$. Let the operator $A$ defined by $y↦{\int }_{0}^{t}a\left(s,y\left(s\right)\right)ds$ be continuous on $M$ and the image $A\left(M\right)$ of the set $M$ be compact. Let the operator $B$ defined by $y↦{\int }_{0}^{t}b\left(s,y\left(s\right)\right)ds$ be contracting on $U$ with the constant $\alpha <1$. Then if for each $y\in M$ a unique solution $x$ to ${x}^{\text{'}}=b\left(t,x\right)+a\left(t,y\right)$, $x\left(0\right)={x}_{0}$, is in $M$, then a solution to the Cauchy problem above is also in $M$. ##### MSC: 34D30 Structural stability of ODE and analogous concepts 34G20 Nonlinear ODE in abstract spaces ##### Keywords: exponential stability; solutions; Liénard equations	2014-04-24 23:18:02	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 26, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8988066911697388, "perplexity": 3570.1001894371198}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-15/segments/1398223206770.7/warc/CC-MAIN-20140423032006-00376-ip-10-147-4-33.ec2.internal.warc.gz"}
https://www.gradesaver.com/textbooks/math/algebra/elementary-algebra/chapter-1-some-basic-concepts-of-arithmetic-and-algebra-1-3-integers-addition-and-subtraction-problem-set-1-3-page-20/80	# Chapter 1 - Some Basic Concepts of Arithmetic and Algebra - 1.3 - Integers: Addition and Subtraction - Problem Set 1.3 - Page 20: 80 $-11$ #### Work Step by Step Written horizontally, this problem reads $8-19$ In order to subtract $19$ from $8$, we find $19-8$, and put a negative out front because the number being subtracted ($19$) is larger. Therefore, $8-19=-(19-8)=-11$ After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.	2018-09-20 12:18:57	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.49007362127304077, "perplexity": 696.0741867729747}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-39/segments/1537267156471.4/warc/CC-MAIN-20180920120835-20180920141235-00249.warc.gz"}
http://mathhelpforum.com/math-topics/3543-question-sets-need-confirmation.html	# Math Help - Question on Sets - Need confirmation 1. ## Question on Sets - Need confirmation Hi, In the attached document, i have mentioned one question on sets and its corresponding solution. Would like to confirm whether the solution mentioned is correct or not. If not, pld provide me the correct answer. Thanks for the help. Best Regards, Lalit Chugh[B] 2. Originally Posted by lalitchugh Hi, In the attached document, i have mentioned one question on sets and its corresponding solution. Would like to confirm whether the solution mentioned is correct or not. If not, pld provide me the correct answer. Thanks for the help. Best Regards, Lalit Chugh[B] Your diagram should have 8,9,10,..20, in U but outside of A, B and C RonL 3. ## Question on sets - Need confirmation Hi, Will you be able to advice how i can dram my questions in maths format easily. currently, i am using word which is not having all the math related signs. Someone advised me to install LAtex but not sure how and from where to install it for windows. Also if any other software is available, pls let me know. Best Regards, Lalit Chugh 4. Originally Posted by lalitchugh Hi, Will you be able to advice how i can dram my questions in maths format easily. currently, i am using word which is not having all the math related signs. Someone advised me to install LAtex but not sure how and from where to install it for windows. Also if any other software is available, pls let me know. Best Regards, Lalit Chugh This site has a TeX system which alows you to write mathematical notation. If you click on the typeset mathematics in a post a window will pop up with the TeX that generated it. There is also a tutorial here RonL Hi,	2014-10-20 23:16:59	{"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8701783418655396, "perplexity": 1497.514819466444}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-42/segments/1413507443451.12/warc/CC-MAIN-20141017005723-00288-ip-10-16-133-185.ec2.internal.warc.gz"}
https://zbmath.org/?q=an:06804592	# zbMATH — the first resource for mathematics Disjunctive answer set solvers via templates. (English) Zbl 1379.68046 Summary: Answer set programming is a declarative programming paradigm oriented towards difficult combinatorial search problems. A fundamental task in answer set programming is to compute stable models, i.e., solutions of logic programs. Answer set solvers are the programs that perform this task. The problem of deciding whether a disjunctive program has a stable model is $$\Sigma^{\mathrm P}_2$$-complete. The high complexity of reasoning within disjunctive logic programming is responsible for few solvers capable of dealing with such programs, namely dlv, gnt, cmodels, clasp and wasp. In this paper, we show that transition systems introduced by Nieuwenhuis, Oliveras, and Tinelli to model and analyze satisfiability solvers can be adapted for disjunctive answer set solvers. Transition systems give a unifying perspective and bring clarity in the description and comparison of solvers. They can be effectively used for analyzing, comparing and proving correctness of search algorithms as well as inspiring new ideas in the design of disjunctive answer set solvers. In this light, we introduce a general template, which accounts for major techniques implemented in disjunctive solvers. We then illustrate how this general template captures solvers dlv, gnt, and cmodels. We also show how this framework provides a convenient tool for designing new solving algorithms by means of combinations of techniques employed in different solvers. ##### MSC: 68N17 Logic programming 68T20 Problem solving in the context of artificial intelligence (heuristics, search strategies, etc.)	2021-06-22 07:37:44	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6033570170402527, "perplexity": 1709.0094075013358}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-25/segments/1623488512243.88/warc/CC-MAIN-20210622063335-20210622093335-00115.warc.gz"}
http://tex.stackexchange.com/questions/149524/the-equals-sign-is-missing	# The equals sign is missing When I use the fonts fourier and iwona at the same time, the equals sign = does not appear. For example, the source \documentclass{article} \usepackage{fourier} \usepackage[math]{iwona} \begin{document} $A=B$ \end{document} produces A B Question: How can I fix this problem? () I want to use iwona and to replace fonts for \mathbb with ones of fourier. Remarks: 1. If I use the package fourier first, then = appears as usual. But all of the fonts are replaced by fourier fonts, so this is not the solution which I want. 2. The following error could relate to the problem. LaTeX Font Warning: Font shape OT1/iwonam/m/n' undefined (Font) usingOT1/cmr/m/n' instead on input line 5. LaTeX Font Warning: Some font shapes were not available, defaults substituted. 3. If I use only iwona then the equals sign appears. 4. If there is a way to satisfy (), please let me know. - You can load only the \mathbb alphabet from fourier: \DeclareFontFamily{U}{futm}{} \DeclareFontShape{U}{futm}{m}{n}{ <-> fourier-bb % changed from .92 to 1 }{} \DeclareMathAlphabet{\mathbb}{U}{futm}{m}{n} Note that I've also changed the magnification from .92 to 1 to match iwona sizes. MWE: \documentclass{article} \usepackage[math]{iwona} \DeclareFontFamily{U}{futm}{} \DeclareFontShape{U}{futm}{m}{n}{ <-> fourier-bb % changed from .92 to 1 }{} \DeclareMathAlphabet{\mathbb}{U}{futm}{m}{n} \begin{document} $A(x)=B(x),\quad x\in\mathbb{R}$ \end{document} Output: - s*[1] doesn't harm, but removing it is more efficient. – egreg Dec 11 '13 at 17:29 @egreg Edited, thanks. – karlkoeller Dec 11 '13 at 18:04	2014-12-28 01:02:06	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.843867301940918, "perplexity": 3265.261016023305}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-52/segments/1419447554429.44/warc/CC-MAIN-20141224185914-00096-ip-10-231-17-201.ec2.internal.warc.gz"}
https://zoldterv.hu/4mvslr/aab2c6-hopfield-network-algorithm	(1949). e k . 2. i In Section 2, we applied Hopfield networks to clustering, feature selection and network inference on a small example dataset. 1 m {\displaystyle w_{ij}^{\nu }=w_{ij}^{\nu -1}+{\frac {1}{n}}\epsilon _{i}^{\nu }\epsilon _{j}^{\nu }-{\frac {1}{n}}\epsilon _{i}^{\nu }h_{ji}^{\nu }-{\frac {1}{n}}\epsilon _{j}^{\nu }h_{ij}^{\nu }}. Weights should be symmetrical, i.e. μ V Model − The model or architecture can be build up by adding electrical components such as amplifiers which can map the input voltage to the output voltage over a sigmoid activation function. i k The net can be used to recover from a distorted input to the trained state that is most similar to that input. ( s j ( {\displaystyle C_{1}(k)} is a set of McCulloch–Pitts neurons and It is an energy-based auto-associative memory, recurrent, and biologically inspired network. f Step 6 − Calculate the net input of the network as follows −, $$y_{ini}\:=\:x_{i}\:+\:\displaystyle\sum\limits_{j}y_{j}w_{ji}$$, Step 7 − Apply the activation as follows over the net input to calculate the output −. The arrangement of the nodes in a binary tree greatly improves both learning complexity and retrieval time. + [15] The weight matrix of an attractor neural network[clarification needed] is said to follow the Storkey learning rule if it obeys: w s j Introduction to the theory of neural computation. Application Hopfield and Tank used the following parameter values in their solution of the problem: A = B = 500, C = 200, D = 500, N = 15, = 50. As a result, the weights of the network remain fixed, showing that the model is able to switch from a learning stage to a recall stage. N = ∑ Introduction What is Hopfield network? i ϵ The Hopfield nets are mainly used as associative memories and for solving optimization problems. For further details, see the recent paper. 2 + 2 2 j Notice that every pair of units i and j in a Hopfield network has a connection that is described by the connectivity weight In associative memory for the Hopfield network, there are two types of operations: auto-association and hetero-association. See Chapter 17 Section 2 for an introduction to Hopfield networks.. Python classes. IEEE, vol. n Lawrence Erlbaum, 2002. s {\displaystyle \epsilon _{i}^{\mu }} Biological Cybernetics 55, pp:141-146, (1985). [12] Since then, the Hopfield network has been widely used for optimization. Abstract: In this paper a Hopfield neural network (HNN) based parallel algorithm is presented for predicting the secondary structure of ribonucleic acids (RNA). [6] At a certain time, the state of the neural net is described by a vector j C When this operated in discrete line fashion it is called discrete Hopfield network and its architecture as a single-layer feedback network can be called as recurrent. Consider the connection weight Firstly, the network is initialized to specified states, then each neuron is evolved into a steady state or fixed point according to certain rules. 1 A Hopfield network is one of the simplest and oldest types of neural network. Algorithm 30. ∑ Blog post on the same. otherwise. Updating one unit (node in the graph simulating the artificial neuron) in the Hopfield network is performed using the following rule: s 2 1 ( In this arrangement, the neurons transmit signals back and forth to each other in a closed-feedback loop, … Introduction What is Hopfield network? An energy function is defined as a function that is bonded and non-increasing function of the state of the system. If ϵ ≥ In this way, Hopfield networks have the ability to "remember" states stored in the interaction matrix, because if a new state The Hopfield network is commonly used for auto-association and optimization tasks. C + ( i h ∑ in Facebook’s facial recognition algorithm, the input is pixels and the output is the name of the person). Hopfield networks were invented in 1982 by J.J. Hopfield, and by then a number of different neural network models have been put together giving way better performance and robustness in comparison.To my knowledge, they are mostly introduced and mentioned in textbooks when approaching Boltzmann Machines and Deep Belief Networks, since they are built upon Hopfield’s work. Overall input to neu… Hopfield Networks with Retina. Bruck shed light on the behavior of a neuron in the discrete Hopfield network when proving its convergence in his paper in 1990. j k Step 2 − Perform steps 3-9, if the activations of the network is not consolidated. The Hopfield network explained here works in the same way. This allows the net to serve as a content addressable memory system, that is to say, the network will converge to a "remembered" state if it is given only part of the state. 0 The Hebbian rule is both local and incremental. n "On the Working Principle of the Hopfield Neural Networks and its Equivalence to the GADIA in Optimization", IEEE Transactions on Neural Networks and Learning Systems, pp.1-11, 2019. = s matlab computational-neuroscience schizophrenia point-attractor energy-landscapes signal-to-noise hopfield-neural-network N 1 wij = wji The ou… {\displaystyle U(k)=\sum _{i=1}^{N}\sum _{j=1}^{N}w_{ij}(s_{i}(k)-s_{j}(k))^{2}+2\sum _{j=1}^{N}{\theta _{j}}s_{j}(k)}, The continuous-time Hopfield network always minimizes an upper bound to the following weighted cut [10], V J.J. Hopfield, and D.W. h μ ϵ Figure 2: Hopfield network reconstructing degraded images from noisy (top) or partial (bottom) cues. Hopfield network is a form of recurrent artificial network that was invented by Dr. john Hopfield in 1982. Hopfield neural network was invented by Dr. John J. Hopfield in 1982. is a function that links pairs of units to a real value, the connectivity weight. n Recurrent neural networks were based on David Rumelhart's work in 1986. Step 1 − Initialize the weights, which are obtained from training algorithm by using Hebbian principle. ϵ 5. Example 2. = μ 1 J Therefore, it is evident that many mistakes will occur if one tries to store a large number of vectors. 3 This type of network is mostly used for the auto-association and optimization tasks. w Hopfield Network model of associative memory¶. s This is called associative memory because it recovers memories on the basis of similarity. This page was last edited on 14 January 2021, at 13:26. Here, we focus on the clustering aspect and study the performance of Hopfield networks in comparison with a selection of other clustering algorithms on a larger suite of datasets. j Book chapters. "The basins of attraction of a new Hopfield learning rule." k w j 2. Updates in the Hopfield network can be performed in two different ways: The weight between two units has a powerful impact upon the values of the neurons. ϵ ∈ 1579–1585, Oct. 1990. {\displaystyle w_{ii}=0} ν Hopfield network. ϵ Updating a node in a Hopfield network is very much like updating a perceptron. . It is a customizable matrix of weights that can be used to recognize a patter. Hopfield networks conjointly give a model for understanding human memory. As part of its machine learning module, Retina provides a full implementation of a general Hopfield Network along with classes for visualizing its training and action on data. 8 i Similarly, they will diverge if the weight is negative. The original Hopfield net [1982] used model neurons with two values of activity, that can be taken as 0 and 1. i ν ( μ Application Hopfield and Tank used the following parameter values in their solution of the problem: A = B = 500, C = 200, D = 500, N = 15, = 50. = θ It is calculated by converging iterative process. Furthermore, both types of operations are possible to store within a single memory matrix, but only if that given representation matrix is not one or the other of the operations, but rather the combination (auto-associative and hetero-associative) of the two. n Memory vectors can be slightly used, and this would spark the retrieval of the most similar vector in the network. i So in a few words, Hopfield recurrent artificial neural network shown in Fig 1 is not an exception and is a customizable matrix of weights which is used to find the local minimum (recognize a pattern). j Algorithm 30. Example 1. The output of each neuron should be the input of other neurons but not the input of self. i The HNN here is used to find the near-maximum independent set of an adjacent graph made of RNA base pairs and then compute the stable secondary structure of RNA. Hopfield network is a special kind of neural network whose response is different from other neural networks. ± 1 ν j ) Memory vectors can be slightly used, and this would spark the retrieval of the most similar vector in the network. c 1 s ( ≠ Strength of synaptic connection from neuron to neuron is 3. It is important to note that Hopfield's network model utilizes the same learning rule as Hebb's (1949) learning rule, which basically tried to show that learning occurs as a result of the strengthening of the weights by when activity is occurring. If you are updating node 3 of a Hopfield network, then you can think of that as the perceptron, and the values of all the other nodes as input values, and the weights from those nodes to node 3 as the weights. It is capable of storing information, optimizing calculations and so on. j Discrete Hopfield Network is a type of algorithms which is called - Autoassociative memories Don’t be scared of the word Autoassociative. {\displaystyle k} i s Introduction to the theory of neural computation. When such a network recognizes, for example, digits, we present a list of correctly rendered digits to the network. k The Bumptree Network An even newer algorithm is the Bumptree Network which combines the advantages of a binary tree with an advanced classification method using hyper ellipsoids in the pattern space instead of lines, planes or curves. θ Summary Hopfield networks are mainly used to solve problems of pattern identification problems (or recognition) and optimization. ∑ ) j The connections in a Hopfield net typically have the following restrictions: The constraint that weights are symmetric guarantees that the energy function decreases monotonically while following the activation rules. Discrete Hopfield Network is a type of algorithms which is called - Autoassociative memories Don’t be scared of the word Autoassociative. Thus, a great variety of ,optimization problems can be solving by the modified ,Hopfield network in association with the genetic ,algorithm, verifying that the network equilibrium ,points, correspondents to values ,v, that minimize the ,energy function ,E,conf, given in (5), and minimize the ,optimization term ,E,op, of the problem, all of them ,belong to the same solutions valid subspace. ∑ {\displaystyle h_{ij}^{\nu }=\sum _{k=1~:~i\neq k\neq j}^{n}w_{ik}^{\nu -1}\epsilon _{k}^{\nu }} ± Hopfield Network. They are recurrent or fully interconnected neural networks. , J. Bruck, “On the convergence properties of the Hopfield model,” Proc. represents bit i from pattern Direct input (e.g. i → This would, in turn, have a positive effect on the weight During training of discrete Hopfield network, weights will be updated. h {\displaystyle V^{s'}} {\displaystyle f:V^{2}\rightarrow \mathbb {R} } I will briefly explore its continuous version as a mean to understand Boltzmann Machines. For example, when using 3 patterns The output of each neuron should be the input of other neurons but not the input of self. ± ϵ (DOI: 10.1109/TNNLS.2020.2980237). ) ∑ In this article, we will go through in depth along with an implementation. Bruck shows[9] that neuron j changes its state if and only if it further decreases the following biased pseudo-cut. ) When the network is presented with an input, i.e. Following are some important points to keep in mind about discrete Hopfield network − 1. w Section 3-Provides a basic comparison of various TSP Algorithms. This learning rule is local, since the synapses take into account only neurons at their sides. In 1993, a neural history compressor system solved a “Very Deep Learning” task that required more than 1000 subsequent layers in an RNN unfolded in time. It consists of a single layer which contains one or more fully connected recurrent neurons. {\displaystyle w_{ij}=(2V_{i}^{s}-1)(2V_{j}^{s}-1)}, but A learning system that was not incremental would generally be trained only once, with a huge batch of training data. f − ν − Figure 2 shows the results of a Hopfield network which was trained on the Chipmunk and Bugs Bunny images on the left hand side and then presented with either a noisy cue (top) or a partial cue (bottom). ( ( ∑ ) . $$y_{i}\:=\begin{cases}1 & if\:y_{ini}\:>\:\theta_{i}\\y_{i} & if\:y_{ini}\:=\:\theta_{i}\\0 & if\:y_{ini}\: Step 8 − Broadcast this output yi to all other units. ⁡ ) Here λ is gain parameter and gri input conductance. Hopfield networks can be used to retrieve binary patterns when given a corrupted binary string by repeatedly updating the network until it reaches a stable state. Hopfield networks can be used as associative memories for information storage and retrieval, and to solve combinatorial optimization problems. Hopfield and Tank claimed a high rate of success in finding valid tours; they found 16 from 20 starting configurations. 0 = {\displaystyle w_{ij}} V A Hopfield network is a kind of typical feedback neural network that can be regarded as a nonlinear dynamic system. 1 = j In this Python exercise we focus on visualization and simulation to develop our intuition about Hopfield … Associative memory … ( For the Hopfield network, we found that, in the retrieval phase favored when the network wants to memory one of stored patterns, all the reconstruction algorithms fail to extract interactions within a desired accuracy, … During the retrieval process, no learning occurs. i j 1 Furthermore, it was shown that the recall accuracy between vectors and nodes was 0.138 (approximately 138 vectors can be recalled from storage for every 1000 nodes) (Hertz et al., 1991). i θ Each neuron has a binary value of either +1 or -1 (not +1 or 0!) ≠ 1 where is a form of local field [13] at neuron i. 3 . − The Hopfield model accounts for associative memorythrough the incorporation of memory vectors. between neurons have units that usually take on values of 1 or -1, and this convention will be used throughout this article. V w Rizzuto and Kahana (2001) were able to show that the neural network model can account for repetition on recall accuracy by incorporating a probabilistic-learning algorithm. Step 4 − Make initial activation of the network equal to the external input vector X as follows −,$$y_{i}\:=\:x_{i}\:\:\:for\:i\:=\:1\:to\:n. 7. N The rule makes use of more information from the patterns and weights than the generalized Hebbian rule, due to the effect of the local field. Artificial Neural Networks – ICANN'97 (1997): Hertz, John A., Anders S. Krogh, and Richard G. Palmer. j j N [Show full abstract] using the modified Hopfield neural network with two updating modes : the algorithm with a sequential updates and the algorithm with … > t 1 Hopfield networks can be analyzed mathematically. This type of network is mostly used for the auto-association and optimization tasks. Neural Networks 12.6 (1999): Hebb, Donald Olding. Rather, the same neurons are used both to enter input and to read off output. ) [7] A network with asymmetric weights may exhibit some periodic or chaotic behaviour; however, Hopfield found that this behavior is confined to relatively small parts of the phase space and does not impair the network's ability to act as a content-addressable associative memory system. ≅ V Section 4-Contains details of the case study on TSP algorithm using Hopfield neural network and Simulated Annealing. 2. N i t the units only take on two different values for their states and the value is determined by whether or not the units' input exceeds their threshold This will only change the state of the input pattern not the state of the actualnetwork. 4. = 8 i This model consists of neurons with one inverting and one non-inverting output. {\displaystyle \mu } It would be excitatory, if the output of the neuron is same as the input, otherwise inhibitory. {\displaystyle \mu _{1},\mu _{2},\mu _{3}} Similarly, other arcs have the weights on them. f Hopfield networks - a special kind of RNN - were discovered by John Hopfield in 1982. 2 {\displaystyle w_{ij}>0} = = When the network is presented with an input, i.e. HOPFIELD NETWORK ALGORITHM PROBLEM STATEMENT Construct a Hopfield net with two neurons and generate its phase portrait. . k The idea of using the Hopfield network in optimization problems is straightforward: If a constrained/unconstrained cost function can be written in the form of the Hopfield energy function E, then there exists a Hopfield network whose equilibrium points represent solutions to the constrained/unconstrained optimization problem. log w n Note that this energy function belongs to a general class of models in physics under the name of Ising models; these in turn are a special case of Markov networks, since the associated probability measure, the Gibbs measure, has the Markov property. k I will briefly explore its continuous version as a mean to understand Boltzmann Machines. Hopfield Network model of associative memory¶. , . {\displaystyle V(t)=\sum _{i=1}^{N}\sum _{j=1}^{N}w_{ij}({f(s_{i}(t))}-f(s_{j}(t))^{2}+2\sum _{j=1}^{N}{\theta _{j}}{f(s_{j}(t))}}. : Discrete Hopfield nets describe relationships between binary (firing or not-firing) neurons Hopfield Network is a recurrent neural network with bipolar threshold neurons. ) ν i k Although sometimes obscured by inappropriate interpretations, the relevant algorithms … Cambridge university press, 1992, Rolls, Edmund T. Cerebral cortex: principles of operation. The Hopfield network, a point attractor network, is modified here to investigate the behavior of the resting state challenged with varying degrees of noise. 1 If the weights of the neural network were trained correctly we would hope for the stable states to correspond to memories. The energy level of a pattern is the result of removing these products and resulting from negative 2. {\displaystyle f(.)} k 1 ) The network has symmetrical weights with no self-connections i.e., w ij = w ji and w ii = 0. {\displaystyle s_{i}\leftarrow \left\{{\begin{array}{ll}+1&{\mbox{if }}\sum _{j}{w_{ij}s_{j}}\geq \theta _{i},\\-1&{\mbox{otherwise.}}\end{array}}\right.}. The entire network contributes to the change in the activation of any single node. ϵ s 1 1 ⁡ = j , , In hierarchical neural nets, the network has a directional flow of information (e.g. + ∑ i Save / Trainstores / trains the curre… In Section 2, we applied Hopfield networks to clustering, feature selection and network inference on a small example dataset. Patterns that the network uses for training (called retrieval states) become attractors of the system. ∑ Vol. The learning algorithm “stores” a given pattern in the network … put in a state, the networks nodes will start to update and converge to a state which is a previously stored pattern. ∈ Modern neural networks is just playing with matrices. j i Although including the optimization constraints into the synaptic weights in the best possible way is a challenging task, indeed many various difficult optimization problems with constraints in different disciplines have been converted to the Hopfield energy function: Associative memory systems, Analog-to-Digital conversion, job-shop scheduling problem, quadratic assignment and other related NP-complete problems, channel allocation problem in wireless networks, mobile ad-hoc network routing problem, image restoration, system identification, combinatorial optimization, etc, just to name a few. Algorithm. i Hopfield nets function content-addressable memory systems with binary threshold nodes. 1 Exploiting the reducibility property and the capability of Hopfield Networks to provide approximate solutions in polynomial time we propose a Hopfield Network based approximation engine to solve these NP complete problems. ϵ Storkey, Amos. [8] He found that this type of network was also able to store and reproduce memorized states. The Hopfield network calculates the product of the values of each possible node pair and the weights between them. Although performances of these network reconstruction algorithms on the simulated network of spiking neurons are extensively studied recently, the analysis of Hopfield networks is lacking so far. Do not have self-loops ( Figure 6.3 ) this rule has a greater capacity than a network... A form of recurrent artificial neural network Hebbian principle vector in the network the! ( 1985 ) in contrast to perceptron training, the negation -x is also spurious. 2 ] Hopfield networks and Gintaras v. Reklaitis Dept memories for information storage and retrieval.! N { \displaystyle 1,2,... i, j,... ( e.g 1985! Parameter and gri input conductance model, ” Proc training data pattern is the predecessor of Restricted Machine! Neuron outputs x i units in Hopfield nets are binary threshold nodes improves both learning complexity retrieval! A learning system that was invented by Dr. John Hopfield ) are a family of recurrent artificial neural networks introduced. Net with two neurons i and j local, since the human brain always! Input pattern not the input and to read off output at 13:26 of information ( e.g 1982 by John in... To one of the most similar vector in the energy level of any single node the above energy function defined. 1992, Rolls, Edmund T. Cerebral cortex: principles of hopfield network algorithm, Anders S. Krogh, and Richard Palmer... Simulated Annealing, perform steps 3-9, if the output of the actualnetwork =.... Strength of synaptic connection from neuron to neuron is 3 Hopfield networks conjointly give model. Step 3 − for each unit Yi, perform steps 3-9, if the weights which. Huge batch of training data his paper in 1990 training ( called retrieval states on TSP algorithm using neural! Are obtained from training algorithm by using Hebbian principle associative memories and for solving optimization problems. their! For an introduction to Hopfield networks and nonlinear optimization 355 generalized Hopfield to. Setting the values of 0 and 1 by Hopfield are known as Hopfield networks and Gintaras v. Reklaitis Dept they! And nonlinear optimization 355 generalized Hopfield networks.. Python classes input pattern not the of. The simplest and oldest types of neurons with one inverting and one non-inverting output patterns that the neurons never! J are different weights between them incorporation of memory vectors can be used to store a number... A repetitious fashion x, the networks nodes will start to update and converge to spurious (... Distinguish between different types of neural networks to clustering, feature selection network! & Palmer, R.G Amos Storkey in 1997 and is both local incremental... … introduction What is Hopfield network, whenever the state of the retrieval of the model! Layer which contains one or more fully connected recurrent neurons image encryption algorithm based on learning... Network uses for training and applying the structure sync, hopfield network algorithm to link '' occurs in a binary greatly. Neuroscience Deep learning Generic Machine learning Machine learning algorithms hopfield network algorithm neural networks (. Networks Python 2 Comments not have self-loops ( Figure 6.3 ), pp:141-146, 1985... Nodes will start to update and converge to spurious patterns is also a local minimum in the discrete Hopfield,. 1997 and is both local and incremental 2 − perform steps 3-9, if a state, network... Going to Y2, Yi and Yn have the weights between them of 0 and 1 } between. Reason that human learning is incremental 355 generalized Hopfield networks and nonlinear 355... Gain parameter and gri input conductance update and converge to a state is a stored... Content-Addressable memory systems with binary threshold nodes give a model in the function! Of a single or more fully connect neurons … Hopfield network trained using the rule. Defined as a nonlinear dynamic system neural network is mostly used for auto-association optimization. The simplest and oldest types of neurons with one inverting and one non-inverting output changes, the has... And w1n respectively network that was invented by Dr. John Hopfield in 1982 of operation, Edmund T. Cerebral:... The basins of attraction of a pattern is the result of removing these products and from... Learning rule is local, since the synapses take into account only neurons at sides! Rule was introduced by Amos Storkey in 1997 and is commonly used for the Hopfield nets are mainly used associative! Vectors are associated in storage 3-9, if hopfield network algorithm output of each possible node and! Into account only neurons at their sides weights on them, perform steps 3-9, if the output of neuron... The binary input vectors in comparison with discrete Hopfield network is one of the pattern! A given pattern in the network so in a repetitious fashion Klawonn Moewes! Universally agreed [ 13 ], literature suggests that the neurons popularized by John Hopfield Tank! Network explained here works in the 1970s, Hopfield networks.. Python.... Will decrease Hopfield net rules that can be used to recognize a.... Their sides ( MLP ) or content addressable memory 1982 conforming to network. For pattern classification if a state is a local minimum state is a form of recurrent artificial network can... John Hopfield in 1982 ij = w ji and w ii = 0 as! Feature selection and network inference on a small example dataset of attractor neural networks were popularised by John Hopfield they. Distorted pattern distorted input to the change in the activation of any given pattern in the discrete Hopfield network symmetrical! Developed a model in the 1970s, Hopfield networks are one of the )! Going into Hopfield network, weights will be updated after the scientist John Hopfield and Tank claimed a rate! Store and reproduce memorized states so in a binary tree greatly improves both learning complexity and retrieval time 17 2!: Hopfield network is commonly used for optimization digits, we will find out due!, a color image encryption algorithm based on Hebbian learning algorithm “ stores ” a given pattern array... Can occur hand and the weights bruck hopfield network algorithm “ on the basis of.... And j to +1, accordingly by to right-clickto -1 are able be. Networks also provide a model in the year 1982 conforming to the artificial Computational! And so on in a state which is called - Autoassociative memories Don ’ t scared... Hopfield chaotic neural network with the buttons below: 1 of discrete Hopfield network each input x... So on bruck shows [ 9 ] that neuron j changes its state if and only if it further the. Feedback neural network and perceptron network will converge to a state, networks... Operations: auto-association and optimization tasks Cerebral cortex: principles of operation optimization algorithm model is shown to one. Network inference on a small example dataset typical feedback neural network and.. It uses … introduction What is Hopfield network is a form of recurrent artificial that. We focus on visualization and simulation to develop our intuition about Hopfield … Hopfield network is customizable... Algorithm “ stores ” a given pattern in the Hopfield network is a type of algorithms is much! Is a customizable matrix of weights that can be transfered to the artificial Intelligence field in 1990 Intelligence Computational Deep! Relevant algorithms … algorithm 30 our intuition about Hopfield … Hopfield network is a recurrent neural network perceptron... A time it has just one layer of neurons with one inverting and one output! Input vector x, perform steps 6-9 network that can be used to recover from a distorted pattern that in! In 1997 and is commonly used for the Hopfield network trained using Hebbian! Rules that can be regarded as a nonlinear dynamic system it implements so! Figure 2: Hopfield network is a form of recurrent artificial network was. Be transfered to the desired start pattern serve as content-addressable ( associative '' ) memory systems with threshold. Neurons is fully connected, although neurons do not have self-loops ( 6.3... Output is the result of removing these products and resulting from negative 2 eventually lead convergence. Boltzmann Machines network application in solving the classical traveling-salesman problem in 1985 9 ] hopfield network algorithm neuron j its! Replaced by more efficient models, they will diverge if the weights hopfield network algorithm. Activations of the nodes in a Hopfield network is commonly used for auto-association and optimization problems such travelling... Purdue university... specific problem at hand and the neuron is same as the start configuration the. And gri input conductance were discovered by John Hopfield in 1982 by Hopfield... The year 1982 conforming to the artificial Intelligence field states to correspond to.. Is 4 world of attractor neural networks – ICANN'97 ( 1997 ): Hebb, Donald.. Used, and the weights conjointly give a model in the network proving its convergence in his paper in.! ( 1997 ): Hebb, Donald Olding store a large number of states. Perform steps 4-8 changes, the negation -x is also a spurious pattern is possible in same. To note that, in contrast to perceptron training, the input, i.e repeated would... Present a list of correctly rendered digits to the trained state that is most similar vector in the has! Pixels and the implemented optimization algorithm recovers memories on the basis of similarity of training data by adjusting the.! ( input, i.e updates would eventually lead to convergence to one of the neural network a continuous.! W i j { \displaystyle 1,2,... n } rule has a greater capacity than corresponding. Memory, recurrent, and this would spark the retrieval of the word Autoassociative ii = 0 memories... Network application in associative memory for the stable states to correspond to memories patterns ) called associative and... Information storage and retrieval time because it recovers memories on the fact that only unit... Tantri Syalindri Ichlasariflow Alkaline Water Walmart, Www Spring-green Com Customer Care, Linear Pair Theorem Proof, Bring To Knees Synonym, Tds Rate Reduction Notification, Cs Lewis Space Trilogy, Req Meaning Mobile Legend, Authors Like Lynn Austin,	2021-05-06 10:54:15	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7177562117576599, "perplexity": 1575.9804275524225}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-21/segments/1620243988753.91/warc/CC-MAIN-20210506083716-20210506113716-00433.warc.gz"}
https://taoofmac.com/space/blog/2009/08/10/2205	# Shortening and Expanding URLs with Python Opinions on shortened URLs are a dime a dozen these days, but the basic facts are: 1. They’re awfully convenient for passing around (and this was true even before Twitter came about) 2. They are, by nature, short-lived (either the services or the URLs) 3. You should never rely on their being around later on So basically you have absolutely no excuse to not be able to handle them. I decided to mess around with the concept a few weeks back to see how simple I could make it all work, and came up with a couple of useful Python classes that I can share with the world: ### Creating short URLs The trouble with creating short URLs is that there are entirely too many shortening services, and far too many variations on APIs – in fact, nearly all of them suffer from “not invented here” syndrome and try to “enhance” their APIs to give you a lot of stuff that you basically don’t (ever) need, and wrap their results in JSON or XML Me, I refuse to put up with that kind of crap. So I poked around a bit, found the simplest services to work against and created the following class, which will try all its known services in turn until it gives you a working URL: import urllib, urllib2, urlparse, httplib class URLShortener: services = { 'api.bit.ly': "http://api.bit.ly/shorten?version=2.0.1&%s&format=text&longUrl=" % BITLY_AUTH, 'api.tr.im': '/api/trim_simple?url=', 'tinyurl.com': '/api-create.php?url=', 'is.gd': '/api.php?longurl=' } def query(self, url): for shortener in self.services.keys(): c = httplib.HTTPConnection(shortener) c.request("GET", self.services[shortener] + urllib.quote(url)) r = c.getresponse() if ("Error" not in shorturl) and ("http://" + urlparse.urlparse(shortener)[1] in shorturl): return shorturl else: continue raise IOError Yes, the error handling is naïve – any network exceptions and stuff ought to be caught upstream from this – but it works fine so far. ### Expanding short URLs This is the really fun bit, because it is not immediately obvious whether or not a short URL will actually be immediately useful – there are plenty of times when you’ll actually be redirected to something else, and while fooling around with the Google Reader API (something I’ll eventually write about alter), I found that also applied (in spades) to Feedburner links and whatnot. So I decided to build some smarts into the process and have it not only ping some known hosts twice, but also turn it into a link checker of sorts, and learning which hosts were actually redirecting to other places: import urllib, urllib2, urlparse, httplib class URLExpander: # known shortening services shorteners = ['tr.im','is.gd','tinyurl.com','bit.ly','snipurl.com','cli.gs', # learned hosts learned = [] def resolve(self, url, components): """ Try to resolve a single URL """ c = httplib.HTTPConnection(components.netloc) c.request("GET", components.path) r = c.getresponse() if l == None: return url # it might be impossible to resolve, so best leave it as is else: return l def query(self, url, recurse = True): """ Resolve a URL """ components = urlparse.urlparse(url) # Check weird shortening services first if (components.netloc in self.twofers) and recurse: return self.query(self.resolve(url, components), False) # Check known shortening services first if components.netloc in self.shorteners: return self.resolve(url, components) # If we haven't seen this host before, ping it, just in case if components.netloc not in self.learned: ping = self.resolve(url, components) if ping != url: self.shorteners.append(components.netloc) self.learned.append(components.netloc) return ping # The original URL was OK return url This one’s a bit more convoluted but has turned out to be very useful indeed, and you can simply pickle the whole object to preserve its learned hosts.	2019-01-17 02:57:31	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.3229130208492279, "perplexity": 5225.077041533661}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-04/segments/1547583658681.7/warc/CC-MAIN-20190117020806-20190117042806-00424.warc.gz"}
https://study.com/academy/answer/on-a-map-2-cm-represents-68-miles-how-many-miles-away-are-two-cities-that-are-7-cm-apart-on-the-map.html	# On a map, 2 cm represents 68 miles. How many miles away are two cities that are 7 cm apart on the... ## Question: On a map, {eq}2 \textrm{ cm} {/eq} represents {eq}68 \textrm{ miles} {/eq}. How many miles away are two cities that are {eq}7 \textrm{ cm} {/eq} apart on the map? ## Proportions and Variation: In any numerical ratio, this denotes a relationship between two numbers {eq}x {/eq} and {eq}y {/eq}. On the other hand, a proportion shows the similarity between two ratios. When two variables are dependent, variations in the magnitude of one variable will have a proportional effect on the other. When there is an increase or decrease of a variable {eq}x {/eq} with respect to another {eq}y {/eq}, for a ratio or constant K, variations are present. In the case that we have a direct variation, it happens that when one variable increases the other increases, which can also be written as: {eq}\frac{{{y_1}}}{{{x_1}}} = \frac{{{y_2}}}{{{x_2}}} {/eq}. {eq}\eqalign{ & {\text{In this specific case }}{\text{,we have two proportional values }}\,x\,\left( {centimeters} \right){\text{ }} \cr & {\text{and }}y\,\left( {miles} \right){\text{ that have a variation in directly proportional form}}{\text{. }} \cr & {\text{So we have:}} \cr & \,\,\,\,{x_1} = 2\,cm \cr & \,\,\,\,{y_1} = 68\,miles \cr & \,\,\,\,{x_2} = 7\,cm \cr & \,\,\,\,{y_2} = ?\,\,miles \cr & {\text{Since}}{\text{, }}x{\text{ and }}y{\text{ vary directly}}{\text{, then}}{\text{, when }}x{\text{ increases it also }} \cr & {\text{increases }}y{\text{. For this reason}}{\text{, it must be satisfied that:}} \cr & \,\,\,\,\frac{{{y_2}}}{{{x_2}}} = \frac{{{y_1}}}{{{x_1}}} \cr & {\text{So if we do cross - multiplying:}} \cr & \,\,\,\,{y_2} \cdot {x_1} = {y_1} \cdot {x_2} \cr & {\text{Now}}{\text{, solving for }}\,{y_2}{\text{:}} \cr & \,\,\,\,{y_2} = \frac{{{y_1} \cdot {x_2}}}{{{x_1}}} \cr & {\text{So}}{\text{, substituting the given values:}} \cr & \,\,\,\,{y_2} = \frac{{68 \times 7}}{2} = 238\,miles \cr & {\text{Therefore}}{\text{, }}\boxed{7{\text{ }}cm}{\text{ represent 238 miles}}{\text{.}} \cr} {/eq}	2020-04-08 16:46:40	{"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9974043369293213, "perplexity": 7964.236269002437}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-16/segments/1585371818008.97/warc/CC-MAIN-20200408135412-20200408165912-00117.warc.gz"}
http://www.annstclair.com/article/coordination-number-of-copper-04bab3	Caribbean Weather January Celsius, Biomechanical Preparation In Primary Teeth, Whole Roasted Cauliflower With Sauce, Russian Sayings Funny, Suave Daily Clarifying Conditioner Curly Girl Method, " /> Caribbean Weather January Celsius, Biomechanical Preparation In Primary Teeth, Whole Roasted Cauliflower With Sauce, Russian Sayings Funny, Suave Daily Clarifying Conditioner Curly Girl Method, " /> coordination number of copper What is the coordination number of copper? The following five rules are used for naming complexes: When the complex is either a cation or a neutral molecule, the name of the central metal atom is spelled exactly like the name of the element and is followed by a Roman numeral in parentheses to indicate its … Furthermore, there is no difference in metal coordination number on going from oxy-hemocyanin to met-hemocyanin, as deduced from the fitting of the first-shell Identify compound X in the following sequence of reactions: Identify a molecule which does not exist. Related Videos. tions of copper ions are occupied by imidazole nitrogens and no bridging ligand in addition to the p-$:q2 peroxide is present in the oxy-hemocyanin. 4 and 2. The term was originally defined in 1893 by Swiss chemist Alfred Werner (1866–1919). Analysis of structural and solution data shows a specific stabilisation of copper (II) in ternary complexes containing one 2,2′:6′,2″-terpyridine and one 2,2′-bipyridine ligand. In chemistry and crystallography, the coordination number describes the number of neighbor atoms with respect to a central atom. D. 3 and 1. Which of the following is non-ionizable. here Cobalt ,atomic no. COORDINATION NUMBER 4: SQUARE PLANAR This is common with d8 complexes that contain large central metal atoms like Pd and … A copper complex of this ligand is shown below but the charge (and thus oxidation state) is not specified. The nickel complex exhibits the usual, very symmetrical, octahedral geometry. This is not a general phenomenon relating to five-coordinate copper (II), but rather a specific consequence of the two ligand donor sets. Aluminum, tin, and lead, for example, form complexes such as the AlF 6 3-, SnCl 4 2-and PbI 4 2-ions. Among these, Cu + more commonly gives tetrahedral complexes but can be coaxed to give linear ones. chemistry. The oxidation number and coordination number, of copper in [C u (N H 3 ) 4 ] 2 + are: A. Primary valency. Click here for solution to problem 17. of ligands that are surrounding a centrl metal ion. The copper industry provides conductors of heat, electricity and fluids to growing economies with expanding infrastructures. Coordination number means no. Discuss this situation in the context of electronic structure and preferred coordination geometry. Although coordination complexes are particularly important in the chemistry of the transition metals, some main group elements also form complexes. Soc. Coordination Number 2. Typically the chemistry of complexes is dominated by interactions between s and p molecular orbitals of the ligan… CHEM1902 Coordination Chemistry The total number of points of attachment to the central element is termed the coordination numberand this can vary from 2 to as many as 16, but is usually 6. Metal centers of electrons required to get octect configuraion or pseudo inert gas configuration. Chemistry. 2004;Rubino and Franz 2012). Ligand, Coordination Number, Coordination Sphere & Oxidation Number Ligand . Thus the metal atom has coordination number 8 in the coordination complexes [Mo(CN)8]4- and [Sr(H2O)8]2+; 7 in the complex That would extend copper (II)’s coordination count range from 2 to 10 — always given beneficial circumstances for a certian coordination number. The interaction between a metal atom and the ligands can be thought of as Lewis acid-base reaction. B –1. Coordination Number 7 Capped trigonal prism (C2v) Pentagonal Bipyramid (D5h) C. 4.5. Answered by Ramandeep \| 27th Feb, 2018, 10:15: AM. In this experiment a solution of the six-coordinate complex ion [Cu(H 2O) 6]2+(aq) will be converted to the four-coordinate complex … The coordination number of copper in the complex [Cu(en)2(H2O)2] is _. O2 O none of these 06 4 . Given that the ionic product of$Ni(OH)_2$is$2 \times 10^{-15}$. Which of the following set of molecules will have zero dipole moment ? C. Effective atomic number. This video explains about types of ligands and coordination number. View Lab Report - Copper Coordination Compound from CHEM 162 at Seattle Central College. is 6 because 6 NH3 are surronuding the Co. Oxidation no. Coordination number s generally range between 2 and 12, with 4 (tetracoordinate) and 6 (hexacoordinate) being the most common. 2. eg: [Co(NH3)6] here coordination no. Herein, we report efficient single copper atom catalysts that consist of dense atomic Cu sites dispersed on a three-dimensional carbon matrix with highly enhanced mesoporous structures and improved active site accessibility (Cu-SA/NC(meso)). Class 12 Class 11 Class 10 Class 9 … For example, both [Cu(NH 3) 2] + and [CuCl 2]-are copper(I) complexes which don't disproportionate. This coordination number is rare outside of d 10 complexes of the group 11 metals and mercury, specifically, Cu +, Ag +, Au +, and Hg 2 +. 2+ What is a chelate? NCERT RD Sharma Cengage KC Sinha. A +3. Publication Date (Print): October 1, 1987. Copper exhibits a rich coordination chemistry with complexes known in oxidation states ranging from 0 to +4, although the +2 (cupric) and the +1 (cuprous) oxidation states are by far the most common, with +2 … What is the oxidation number of the copper in the copper-EDTA complex? The molecules or ions that are attached to the metal in a complex ion are called ligands. NCERT DC Pandey Sunil Batra HC Verma Pradeep Errorless. six. The chlorine-containing complex is formed if copper(I) oxide is dissolved in concentrated hydrochloric acid. Determining Empirical Formula of Copper (2) Coordination Compound by Gravimetric, Volumetric, and Copper is widely used for its heat and electricity conduction capacity. I hope this helps you. When excess of KCN is added, a new coordination entity, [Cu(CN)4]2– is formed due to following reaction: CuSO4 (aq) + 4KCN → K2[Cu(CN)4] + K2SO4(aq) Cu(H2O)42+ + 4CN(from aq. What is the coordination number of the copper in the copper-EDTA complex? Forming copper(I) complexes (other than the one with water as a ligand) also stabalises the copper(I) oxidation state. Which of the following complex is optically inactive, The compound$\ce{[Pt(NH3)2Cl2]}$can exhibit. Coordination number is the term proposed by Werner to denote the total number of bonds from the ligands to the metal atom. Maths. The coordination number of the complex is not known but is expected to be 4 or 6. The nomenclature of the complexes is patterned after a system suggested by Alfred Werner, a Swiss chemist and Nobel laureate, whose outstanding work more than 100 years ago laid the foundation for a clearer understanding of these compounds. The total number of points of attachment to the central element is termed the coordination number and this can vary from 2 to as many as 16, but is usually 6. The arrangement of these components, however, is fascinating. The coordination sphere consists of the central metal ion or atom plus its attached ligands. In [NiCl 4] 2–, the number of unpaired electron is. B. Biological copper is coordinated predominantly by just three ligand types: the side chains of histidine, cysteine, and methionine, with of course some exceptions. Copper can lose 1 electron (4s1) leaving a full 3d shell or lose an extra electron from the d shell (3d9). The ion/molecule/atom surrounding the central ion/molecule/atom is called a ligand.This number is determined somewhat differently for molecules than for crystals. The oxidation number of cobalt in K[Co(CO) 4] is. As expected, they are distorted by the presence of Al tetrahedra and by the tendency of Cu 2+ to maximize its coordination number to 4 or 5. Am. Co-ordination number is straightforward - it's the number of ligands attached to the central metal atom/ion. Most complexes have a coordination number of 6, and in almost all of these complexes, the ligands are arranged around the metal center in octahedral geometry. The coordination number of the central metal ion or atom is the number of donor atoms bonded to it. In this experiment, we will study reactions of two octahedral complexes: [Ni(H 26 O) ]2+ and [Cu(H 26 O) ]2+. The answer is: The coordination number of copper in [Cu(NH3)4]2+ is 4. Examples of various coordination numbers are shown in planar geometry in Figure 2. Answer. The coordination number for the silver ion in [Ag(NH 3 ) 2 ] + is two ( Figure 19.14 ). A chelate is a ligand that forms a complex with ions through multiple coordinate-covalent bonds. The complex compounds which result from the coordination of carbon monoxide are known as (a) Electronic 1 and 3. In [Fe(CN)6]3- the co-ordination number is 6 because there are six cyanide ions. Coordination number. The coordination number of the central metal ion or atom is the number of donor atoms bonded to it. Brackets in a formula enclose the coordination sphere; species outside the brackets are not part of the coordination sphere. In chemistry, crystallography, and materials science, the coordination number, also called ligancy, of a central atom in a molecule or crystal is the number of atoms, molecules or ions bonded to it. (a) Would conductivity measurements provide information about the coordination number? So tl;dr: There is no simple explanation for the choice of exactly that coordination number. B. Which of the following complexes exists as pair of enantiomers? Reaction between acetone and methyl magnesium chloride followed by hydrolysis will give : Identify the correct statements from the following: Question 19. Since copper is surrounded by four ammonia group hence its coordination number is 4. D [Co(NH3)6]Cl2. The highest known coordination number of a coordination polymer is 14, though coordination numbers are most often between 2 and 10. Download PDF's. Answer By Toppr. Coordination numbers are normally between two and nine. Expert Answer: The coordination number of Cu 2+ is either 4 or 6. (a) How many joules did the water absorb? The atomic radiusis: Find out the solubility of$Ni(OH)_2$in 0.1 M NaOH. Lung Shan Kau; Darlene J. Spira-Solomon ; James E. Penner-Hahn; Keith O. Hodgson; Edward I. Solomon; Cite This: J. 4. The number of ligands bound to the transition metal ion is called the coordination number. B. Which kind of isomerism is exhibited by octahedral$[Co(NH_3)_4Br_2]Cl$? In coordination chemistry, the coordination number is the number of ligands attached to the central ion (more specifically, the number of donor atoms). Copper can lose 1 electron (4s1) leaving a full 3d shell or lose an extra electron from the d shell (3d9). Aqueous solution of copper sulphate contains Cu2+ ions in form of complex entity, [Cu(H2O)4]2+ and H2O ligand is a weak ligand. In a particular isomer of$\left[Co\left(NH_{3}\right)_{4}Cl_{2}\right]^{0},$the$Cl-Co-Cl$angle is$90^\circ$, the isomer is known as, Compounds,$ {{[PtC{{l}_{2}}N{{H}_{3}})}_{4}}]B{{r}_{2}} $and$ [PtB{{r}_{2}}{{(N{{H}_{3}})}_{4}}]C{{l}_{2}} $, shows the following type of isomerism. The temperature of the resulting mixture became 28.0°C. A [Co(NH3)4 Cl2]Cl. These include four coordinate, square-planar complexes and five- and six-coordinate derivatives of the sp3d2 hybridized octahedral structure. Which one of the following is heteroleptic complex? The coordination number of copper in cuprammonium sulphate is . Coordination Numbers and Geometry Lecture 2. NCERT NCERT Exemplar NCERT Fingertips Errorless Vol-1 Errorless Vol-2. Required fields are marked *, What Is The Coordination Number Of Copper Incunh342plus. Chem. To support copper producers, Fives develops a wide range of thermal systems with an emphasis on improved fuel and melter efficiency. B [Co(NH3)3Cl3] C [Co(NH3)5 Cl]Cl2 . The tridimensional shape of the Tetramino-Copper(II) Ion seems Planar Picture which displaces Ammonia Molecules at Vertex of a Square. The coordination number of an atom in a molecule is the number of atoms bonded to the atom. On electrolysis of dil.sulphuric acid using Platinum (Pt) electrode, the product obtained at anode will be: An element has a body centered cubic (bcc) structure with a cell edge of 288 pm. Copper(I) complexes. The 3d9 configuration is stabilised by the coordination sphere. A 37.0 g mass of a metal was heated to 100°C and then plunged into 74 g of water at 24.0°C. Coordination number, the number of atoms, ions, or molecules that a central atom or ion holds as its nearest neighbours in a complex or coordination compound or in a crystal. Application to the type 3 site in Rhus vernicifera laccase and its reaction with oxygen. Correct increasing order for the wavelength of absorption in the visible region for the complexes of$Co^{3+}$is: When$Ag^+$reacts with excess of sodium-thiosulphate then he obtained species having charge and geometry respectively : In Wolff‐Kishner reduction, the carbonyl group of aldehydes and ketones is converted into. Books. Copper is a chemical element with the symbol Cu (from Latin: cuprum) and atomic number 29. For the copper(II) ion in [CuCl 4 ] 2− , the coordination number is four, whereas for the cobalt(II) ion in [Co(H 2 O) 6 ] 2+ the coordination number is six. NCERT P Bahadur IIT-JEE Previous Year Narendra Awasthi MS Chauhan. C +1. Access detailed answers to various other Science and Maths questions at BYJU'S. (b) In using . In Figure 1 the 1D structure is 2-coordinated, the planar is 4-coordinated, and the 3D is 6-coordinated. Option C is correct. 1987 109 21 6433-6442. The above examples are tetrafluoroborate, permanganate, nickel tetracarbonyl, and tetrakis (pyridine)copper (I), respectively. Coordination Chemistry MCQs for practice. X-ray absorption edge determination of the oxidation state and coordination number of copper. x=+2. The 3d9 configuration is stabilised by the coordination sphere. Get more help from Chegg. The charge on the complex ion is also the same as the charge on the copper ion, 2+ because the ligands have no charge. Biology. Why Is Boron Trifluoride Written In Two Ways In The Lewis Dot Diagram Structure, Why Is It Incorrect To Write A Chemical Formula For Petrol, Why Is The Water Cycle Important To All Life On The Earth, Why Would Fluorine Have A Positive Oxidation State In Hof Even Though It Is More Electronegative, What Is The Molar Mass Of Helium In Kg Mol, What Is The Percentage Composition Of Ammonium Nitrate Nh 4no 3, What Is The Structure And Function Of The Heart Of A Fish, What Is The Unit For The Index Of Refraction, CBSE Previous Year Question Papers Class 10, CBSE Previous Year Question Papers Class 12, NCERT Solutions Class 11 Business Studies, NCERT Solutions Class 12 Business Studies, NCERT Solutions Class 12 Accountancy Part 1, NCERT Solutions Class 12 Accountancy Part 2, NCERT Solutions For Class 6 Social Science, NCERT Solutions for Class 7 Social Science, NCERT Solutions for Class 8 Social Science, NCERT Solutions For Class 9 Social Science, NCERT Solutions For Class 9 Maths Chapter 1, NCERT Solutions For Class 9 Maths Chapter 2, NCERT Solutions For Class 9 Maths Chapter 3, NCERT Solutions For Class 9 Maths Chapter 4, NCERT Solutions For Class 9 Maths Chapter 5, NCERT Solutions For Class 9 Maths Chapter 6, NCERT Solutions For Class 9 Maths Chapter 7, NCERT Solutions For Class 9 Maths Chapter 8, NCERT Solutions For Class 9 Maths Chapter 9, NCERT Solutions For Class 9 Maths Chapter 10, NCERT Solutions For Class 9 Maths Chapter 11, NCERT Solutions For Class 9 Maths Chapter 12, NCERT Solutions For Class 9 Maths Chapter 13, NCERT Solutions For Class 9 Maths Chapter 14, NCERT Solutions For Class 9 Maths Chapter 15, NCERT Solutions for Class 9 Science Chapter 1, NCERT Solutions for Class 9 Science Chapter 2, NCERT Solutions for Class 9 Science Chapter 3, NCERT Solutions for Class 9 Science Chapter 4, NCERT Solutions for Class 9 Science Chapter 5, NCERT Solutions for Class 9 Science Chapter 6, NCERT Solutions for Class 9 Science Chapter 7, NCERT Solutions for Class 9 Science Chapter 8, NCERT Solutions for Class 9 Science Chapter 9, NCERT Solutions for Class 9 Science Chapter 10, NCERT Solutions for Class 9 Science Chapter 12, NCERT Solutions for Class 9 Science Chapter 11, NCERT Solutions for Class 9 Science Chapter 13, NCERT Solutions for Class 9 Science Chapter 14, NCERT Solutions for Class 9 Science Chapter 15, NCERT Solutions for Class 10 Social Science, NCERT Solutions for Class 10 Maths Chapter 1, NCERT Solutions for Class 10 Maths Chapter 2, NCERT Solutions for Class 10 Maths Chapter 3, NCERT Solutions for Class 10 Maths Chapter 4, NCERT Solutions for Class 10 Maths Chapter 5, NCERT Solutions for Class 10 Maths Chapter 6, NCERT Solutions for Class 10 Maths Chapter 7, NCERT Solutions for Class 10 Maths Chapter 8, NCERT Solutions for Class 10 Maths Chapter 9, NCERT Solutions for Class 10 Maths Chapter 10, NCERT Solutions for Class 10 Maths Chapter 11, NCERT Solutions for Class 10 Maths Chapter 12, NCERT Solutions for Class 10 Maths Chapter 13, NCERT Solutions for Class 10 Maths Chapter 14, NCERT Solutions for Class 10 Maths Chapter 15, NCERT Solutions for Class 10 Science Chapter 1, NCERT Solutions for Class 10 Science Chapter 2, NCERT Solutions for Class 10 Science Chapter 3, NCERT Solutions for Class 10 Science Chapter 4, NCERT Solutions for Class 10 Science Chapter 5, NCERT Solutions for Class 10 Science Chapter 6, NCERT Solutions for Class 10 Science Chapter 7, NCERT Solutions for Class 10 Science Chapter 8, NCERT Solutions for Class 10 Science Chapter 9, NCERT Solutions for Class 10 Science Chapter 10, NCERT Solutions for Class 10 Science Chapter 11, NCERT Solutions for Class 10 Science Chapter 12, NCERT Solutions for Class 10 Science Chapter 13, NCERT Solutions for Class 10 Science Chapter 14, NCERT Solutions for Class 10 Science Chapter 15, NCERT Solutions for Class 10 Science Chapter 16, What Is The Working Principle Of Electromagnet, What Is The Partial Charges Present On An Atom, Eddy Currents Are Produced Due To Change In, Induced E M F Is Given By It Will Be Maximum When, What Criteria Do We Use To Decide Something Is Alive, Give Definition And Example Of Translational Motion, In Which Of The Following Process Heat Is Released, Maize Seed Has Fill In The Blank Cotyledons, Give Three Examples Exhibiting Inertia In Our Daily Life. The number of ligands is equal to the coordination number for copper, 4. In accordance with the predictions of the Kepert model these give linear complexes. Access detailed answers to various other Science and Maths questions at BYJU'S. 18. The number of bonds depends on the size, charge, and electron configuration of the metal ion and the ligands. (a)$CO_2(g)\$ is used as refrigerant for ice-cream and frozen food. D. Atomic number. The coordination number of copper in [Cu(NH3)4]2+ is 4, Your email address will not be published. Co+3. Question 17. Coordination number = 4. The coordination numbers and geometries of the copper complexes vary with oxidation state (Blumberger et al. Physics. The copper(II) ion readily forms coordination complexes with a variety of coordination numbers and geometries. means no. Check Answer and Solution for above question from The answer is: The coordination number of copper in [Cu(NH3)4]2+ is 4. D –3. C. 2 and 4. The coordination number of copper in cupram- monium sulphate is. Now, oxidation number of copper is given as: x+0=+2 i.e. A. eg:1. Asked by Saurav \| 27th Feb, 2018, 09:57: AM. So in [CuCl4]2- the co-ordination number of the copper is 4 as there are four chloride ion ligands. Which of the following compounds show optical isomerism? KCET 1990: The coordination number of copper in cupram- monium sulphate is (A) 2 (B) 3 (C) 4 (D) 6. Question 18. Hence, oxidation number of copper in given complex is +2. The Coordination Number express how many NH3 molecules are bound to "Cu++" ion : C. N. is 4. For the spherically symmetric d 10 Cu + … Two geometries are possible for this coordination number. Your email address will not be published. Comments on this entry are closed.	2021-01-27 04:32:17	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.3602483570575714, "perplexity": 5111.284665052756}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-04/segments/1610704820894.84/warc/CC-MAIN-20210127024104-20210127054104-00114.warc.gz"}
http://math.stackexchange.com/questions/252243/property-of-derivative-of-dirac-delta-function-in-mathbbrn	# Property of derivative of Dirac delta function in $\mathbb{R}^n$ With reference to Property of Dirac delta function in $\mathbb{R}^n$, is there a similar formula for $\langle g^\delta', f \rangle$ (or even $\langle g^\delta^{(n)}, f \rangle$)? By similar I mean a representation by an integral over $g^{-1}(0)$. - I found Wagner's paper in Appl. Anal. 89 (2010), no. 8, 1183–1199. Maybe I should read that first. – kal Dec 6 '12 at 13:22	2016-02-14 02:18:28	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7877677083015442, "perplexity": 230.05044727109683}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-07/segments/1454701168076.20/warc/CC-MAIN-20160205193928-00286-ip-10-236-182-209.ec2.internal.warc.gz"}
https://dmoj.ca/problem/tsoc16c1p4	## Alex and Animal Rights View as PDF Points: 7 (partial) Time limit: 0.6s Memory limit: 256M Authors: Problem types Allowed languages Ada, Assembly, Awk, Brain****, C, C#, C++, COBOL, CommonLisp, D, Dart, F#, Forth, Fortran, Go, Groovy, Haskell, Intercal, Java, JS, Kotlin, Lisp, Lua, Nim, ObjC, OCaml, Octave, Pascal, Perl, PHP, Pike, Prolog, Python, Racket, Ruby, Rust, Scala, Scheme, Sed, Swift, TCL, Text, Turing, VB, Zig Alex is an animal rights activist out to protect monkeys from the evil corporations that use them to test shampoo. He is planning a daring breakout of a group of monkeys contained in a room in a secret underground lab. The monkeys are contained in a system of cages that border each other, all without doors. Some cages have monkeys, and some are empty. A cage is a set of adjacent squares that is enclosed by wall squares. Alex wants to know how many cages have monkeys in them so that he knows how many holes he'll need to drill in the ceiling to rappel down from. On the floor plan, a wall is denoted by a #, an empty space is denoted by a . and a monkey is denoted by an M. It is guaranteed that the edges of the room will consist entirely of walls. #### Input Specification Two space-separated integers and , denoting the height and width of the room, respectively. lines of length each, denoting the layout of the floor. #### Output Specification The number of cages that contain monkeys. #### Sample Input 9 10 ########## ##M#.M.#.# #MM####..# #M#.#.#..# ##..#.#.M# #..##..#M# ##M####### #.#MM#.M.# ########## #### Sample Output 6 • commented on Dec. 13, 2017, 4:42 p.m. This comment is hidden due to too much negative feedback. Click here to view it. • commented on Jan. 7, 2016, 1:04 p.m. edited Is only: ### #.# ### considered a room, or # #. # # as well? • commented on Jan. 11, 2016, 4:17 p.m. If the diagonals are also considered (8 directions in total), then the answer to the sample input would be 3. • commented on Jan. 7, 2016, 4:52 p.m. Not sure what you mean as the second test case you've given wouldn't be valid. A room is a strongly connected component of room nodes (one node per character) when there exist edges connecting each node with its left, upper, lower, or right neighbour if said neighbour exists. • commented on Jan. 7, 2016, 4:53 p.m. This comment is hidden due to too much negative feedback. Click here to view it. • commented on Jan. 7, 2016, 7:36 p.m. I didn't write this problem. I was just trying to help.	2020-09-20 04:57:04	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.18951255083084106, "perplexity": 5813.699385455466}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-40/segments/1600400193391.9/warc/CC-MAIN-20200920031425-20200920061425-00241.warc.gz"}
http://cvgmt.sns.it/paper/2983/	# Non-Existence of Theta-Shaped Self-Similarly Shrinking Networks Moving by Curvature created by root on 05 Apr 2016 modified on 16 Jun 2018 [BibTeX] Published Paper Inserted: 5 apr 2016 Last Updated: 16 jun 2018 Journal: Comm. PDE Volume: 43 Number: 3 Pages: 403-427 Year: 2018 Abstract: We prove that there are no networks homeomorphic to the Greek theta'' letter (a double cell) embedded in the plane with two triple junctions with angles of 120 degrees, such that under the motion by curvature they are self-similarly shrinking. This fact completes the classification of the self-similarly shrinking networks in the plane with at most two triple junctions.	2021-03-07 15:28:17	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.3092314898967743, "perplexity": 7251.721636116224}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-10/segments/1614178377821.94/warc/CC-MAIN-20210307135518-20210307165518-00320.warc.gz"}
https://en.wikibooks.org/wiki/Linear_Algebra/Rangespace_and_Nullspace/Solutions	# Linear Algebra/Rangespace and Nullspace/Solutions ## Solutions This exercise is recommended for all readers. Problem 1 Let ${\displaystyle h:{\mathcal {P}}_{3}\to {\mathcal {P}}_{4}}$ be given by ${\displaystyle p(x)\mapsto x\cdot p(x)}$. Which of these are in the nullspace? Which are in the rangespace? 1. ${\displaystyle x^{3}}$ 2. ${\displaystyle 0}$ 3. ${\displaystyle 7}$ 4. ${\displaystyle 12x-0.5x^{3}}$ 5. ${\displaystyle 1+3x^{2}-x^{3}}$ First, to answer whether a polynomial is in the nullspace, we have to consider it as a member of the domain ${\displaystyle {\mathcal {P}}_{3}}$. To answer whether it is in the rangespace, we consider it as a member of the codomain ${\displaystyle {\mathcal {P}}_{4}}$. That is, for ${\displaystyle p(x)=x^{4}}$, the question of whether it is in the rangespace is sensible but the question of whether it is in the nullspace is not because it is not even in the domain. 1. The polynomial ${\displaystyle x^{3}\in {\mathcal {P}}_{3}}$ is not in the nullspace because ${\displaystyle h(x^{3})=x^{4}}$ is not the zero polynomial in ${\displaystyle {\mathcal {P}}_{4}}$. The polynomial ${\displaystyle x^{3}\in {\mathcal {P}}_{4}}$ is in the rangespace because ${\displaystyle x^{2}\in {\mathcal {P}}_{3}}$ is mapped by ${\displaystyle h}$ to ${\displaystyle x^{3}}$. 2. The answer to both questions is, "Yes, because ${\displaystyle h(0)=0}$." The polynomial ${\displaystyle 0\in {\mathcal {P}}_{3}}$ is in the nullspace because it is mapped by ${\displaystyle h}$ to the zero polynomial in ${\displaystyle {\mathcal {P}}_{4}}$. The polynomial ${\displaystyle 0\in {\mathcal {P}}_{4}}$ is in the rangespace because it is the image, under ${\displaystyle h}$, of ${\displaystyle 0\in {\mathcal {P}}_{3}}$. 3. The polynomial ${\displaystyle 7\in {\mathcal {P}}_{3}}$ is not in the nullspace because ${\displaystyle h(7)=7x}$ is not the zero polynomial in ${\displaystyle {\mathcal {P}}_{4}}$. The polynomial ${\displaystyle x^{3}\in {\mathcal {P}}_{4}}$ is not in the rangespace because there is no member of the domain that when multiplied by ${\displaystyle x}$ gives the constant polynomial ${\displaystyle p(x)=7}$. 4. The polynomial ${\displaystyle 12x-0.5x^{3}\in {\mathcal {P}}_{3}}$ is not in the nullspace because ${\displaystyle h(12x-0.5x^{3})=12x^{2}-0.5x^{4}}$. The polynomial ${\displaystyle 12x-0.5x^{3}\in {\mathcal {P}}_{4}}$ is in the rangespace because it is the image of ${\displaystyle 12-0.5x^{2}}$. 5. The polynomial ${\displaystyle 1+3x^{2}-x^{3}\in {\mathcal {P}}_{3}}$ is not in the nullspace because ${\displaystyle h(1+3x^{2}-x^{3})=x+3x^{3}-x^{4}}$. The polynomial ${\displaystyle 1+3x^{2}-x^{3}\in {\mathcal {P}}_{4}}$ is not in the rangespace because of the constant term. This exercise is recommended for all readers. Problem 2 Find the nullspace, nullity, rangespace, and rank of each map. 1. ${\displaystyle h:\mathbb {R} ^{2}\to {\mathcal {P}}_{3}}$ given by ${\displaystyle {\begin{pmatrix}a\\b\end{pmatrix}}\mapsto a+ax+ax^{2}}$ 2. ${\displaystyle h:{\mathcal {M}}_{2\!\times \!2}\to \mathbb {R} }$ given by ${\displaystyle {\begin{pmatrix}a&b\\c&d\end{pmatrix}}\mapsto a+d}$ 3. ${\displaystyle h:{\mathcal {M}}_{2\!\times \!2}\to {\mathcal {P}}_{2}}$ given by ${\displaystyle {\begin{pmatrix}a&b\\c&d\end{pmatrix}}\mapsto a+b+c+dx^{2}}$ 4. the zero map ${\displaystyle Z:\mathbb {R} ^{3}\to \mathbb {R} ^{4}}$ 1. The nullspace is ${\displaystyle {\mathcal {N}}(h)=\{{\begin{pmatrix}a\\b\end{pmatrix}}\in \mathbb {R} ^{2}\,{\big \|}\,a+ax+ax^{2}+0x^{3}=0+0x+0x^{2}+0x^{3}\}=\{{\begin{pmatrix}0\\b\end{pmatrix}}\,{\big \|}\,b\in \mathbb {R} \}}$ while the rangespace is ${\displaystyle {\mathcal {R}}(h)=\{a+ax+ax^{2}\in {\mathcal {P}}_{3}\,{\big \|}\,a,b\in \mathbb {R} \}=\{a\cdot (1+x+x^{2})\,{\big \|}\,a\in \mathbb {R} \}}$ and so the nullity is one and the rank is one. 2. The nullspace is this. ${\displaystyle {\mathcal {N}}(h)=\{{\begin{pmatrix}a&b\\c&d\end{pmatrix}}\,{\big \|}\,a+d=0\}=\{{\begin{pmatrix}-d&b\\c&d\end{pmatrix}}\,{\big \|}\,b,c,d\in \mathbb {R} \}}$ The rangespace ${\displaystyle {\mathcal {R}}(h)\{a+d\,{\big \|}\,a,b,c,d\in \mathbb {R} \}}$ is all of ${\displaystyle \mathbb {R} }$ (we can get any real number by taking ${\displaystyle d}$ to be ${\displaystyle 0}$ and taking ${\displaystyle a}$ to be the desired number). Thus, the nullity is three and the rank is one. 3. The nullspace is ${\displaystyle {\mathcal {N}}(h)=\{{\begin{pmatrix}a&b\\c&d\end{pmatrix}}\,{\big \|}\,a+b+c=0{\text{ and }}d=0\}=\{{\begin{pmatrix}-b-c&b\\c&0\end{pmatrix}}\,{\big \|}\,b,c\in \mathbb {R} \}}$ while the rangespace is ${\displaystyle {\mathcal {R}}(h)=\{r+sx^{2}\,{\big \|}\,r,s\in \mathbb {R} \}}$. Thus, the nullity is two and the rank is two. 4. The nullspace is all of ${\displaystyle \mathbb {R} ^{3}}$ so the nullity is three. The rangespace is the trivial subspace of ${\displaystyle \mathbb {R} ^{4}}$ so the rank is zero. This exercise is recommended for all readers. Problem 3 Find the nullity of each map. 1. ${\displaystyle h:\mathbb {R} ^{5}\to \mathbb {R} ^{8}}$ of rank five 2. ${\displaystyle h:{\mathcal {P}}_{3}\to {\mathcal {P}}_{3}}$ of rank one 3. ${\displaystyle h:\mathbb {R} ^{6}\to \mathbb {R} ^{3}}$, an onto map 4. ${\displaystyle h:{\mathcal {M}}_{3\!\times \!3}\to {\mathcal {M}}_{3\!\times \!3}}$, onto For each, use the result that the rank plus the nullity equals the dimension of the domain. 1. ${\displaystyle 0}$ 2. ${\displaystyle 3}$ 3. ${\displaystyle 3}$ 4. ${\displaystyle 0}$ This exercise is recommended for all readers. Problem 4 What is the nullspace of the differentiation transformation ${\displaystyle d/dx:{\mathcal {P}}_{n}\to {\mathcal {P}}_{n}}$? What is the nullspace of the second derivative, as a transformation of ${\displaystyle {\mathcal {P}}_{n}}$? The ${\displaystyle k}$-th derivative? Because ${\displaystyle {\frac {d}{dx}}\,(a_{0}+a_{1}x+\dots +a_{n}x^{n})=a_{1}+2a_{2}x+3a_{3}x^{2}+\dots +na_{n}x^{n-1}}$ we have this. ${\displaystyle {\begin{array}{rl}{\mathcal {N}}({\frac {d}{dx}})&=\{a_{0}+\dots +a_{n}x^{n}\,{\big \|}\,a_{1}+2a_{2}x+\dots +na_{n}x^{n-1}=0+0x+\dots +0x^{n-1}\}\\&=\{a_{0}+\dots +a_{n}x^{n}\,{\big \|}\,a_{1}=0,{\text{ and }}a_{2}=0,\ldots ,a_{n}=0\}\\&=\{a_{0}+0x+0x^{2}+\dots +0x^{n}\,{\big \|}\,a_{0}\in \mathbb {R} \}\end{array}}}$ In the same way, ${\displaystyle {\mathcal {N}}({\frac {d^{k}}{dx^{k}}})=\{a_{0}+a_{1}x+\dots +a_{n}x^{n}\,{\big \|}\,a_{0},\dots ,a_{k-1}\in \mathbb {R} \}}$ for ${\displaystyle k\leq n}$. Problem 5 Example 2.7 restates the first condition in the definition of homomorphism as "the shadow of a sum is the sum of the shadows". Restate the second condition in the same style. The shadow of a scalar multiple is the scalar multiple of the shadow. Problem 6 For the homomorphism ${\displaystyle h:{\mathcal {P}}_{3}\to {\mathcal {P}}_{3}}$ given by ${\displaystyle h(a_{0}+a_{1}x+a_{2}x^{2}+a_{3}x^{3})=a_{0}+(a_{0}+a_{1})x+(a_{2}+a_{3})x^{3}}$ find these. 1. ${\displaystyle {\mathcal {N}}(h)}$ 2. ${\displaystyle h^{-1}(2-x^{3})}$ 3. ${\displaystyle h^{-1}(1+x^{2})}$ 1. Setting ${\displaystyle a_{0}+(a_{0}+a_{1})x+(a_{2}+a_{3})x^{3}=0+0x+0x^{2}+0x^{3}}$ gives ${\displaystyle a_{0}=0}$ and ${\displaystyle a_{0}+a_{1}=0}$ and ${\displaystyle a_{2}+a_{3}=0}$, so the nullspace is ${\displaystyle \{-a_{3}x^{2}+a_{3}x^{3}\,{\big \|}\,a_{3}\in \mathbb {R} \}}$. 2. Setting ${\displaystyle a_{0}+(a_{0}+a_{1})x+(a_{2}+a_{3})x^{3}=2+0x+0x^{2}-x^{3}}$ gives that ${\displaystyle a_{0}=2}$, and ${\displaystyle a_{1}=-2}$, and ${\displaystyle a_{2}+a_{3}=-1}$. Taking ${\displaystyle a_{3}}$ as a parameter, and renaming it ${\displaystyle a_{3}=a}$ gives this set description ${\displaystyle \{2-2x+(-1-a)x^{2}+ax^{3}\,{\big \|}\,a\in \mathbb {R} \}=\{(2-2x-x^{2})+a\cdot (-x^{2}+x^{3})\,{\big \|}\,a\in \mathbb {R} \}}$. 3. This set is empty because the range of ${\displaystyle h}$ includes only those polynomials with a ${\displaystyle 0x^{2}}$ term. This exercise is recommended for all readers. Problem 7 For the map ${\displaystyle f:\mathbb {R} ^{2}\to \mathbb {R} }$ given by ${\displaystyle f({\begin{pmatrix}x\\y\end{pmatrix}})=2x+y}$ sketch these inverse image sets: ${\displaystyle f^{-1}(-3)}$, ${\displaystyle f^{-1}(0)}$, and ${\displaystyle f^{-1}(1)}$. All inverse images are lines with slope ${\displaystyle -2}$. This exercise is recommended for all readers. Problem 8 Each of these transformations of ${\displaystyle {\mathcal {P}}_{3}}$ is nonsingular. Find the inverse function of each. 1. ${\displaystyle a_{0}+a_{1}x+a_{2}x^{2}+a_{3}x^{3}\mapsto a_{0}+a_{1}x+2a_{2}x^{2}+3a_{3}x^{3}}$ 2. ${\displaystyle a_{0}+a_{1}x+a_{2}x^{2}+a_{3}x^{3}\mapsto a_{0}+a_{2}x+a_{1}x^{2}+a_{3}x^{3}}$ 3. ${\displaystyle a_{0}+a_{1}x+a_{2}x^{2}+a_{3}x^{3}\mapsto a_{1}+a_{2}x+a_{3}x^{2}+a_{0}x^{3}}$ 4. ${\displaystyle a_{0}+a_{1}x+a_{2}x^{2}+a_{3}x^{3}\mapsto a_{0}+(a_{0}+a_{1})x+(a_{0}+a_{1}+a_{2})x^{2}+(a_{0}+a_{1}+a_{2}+a_{3})x^{3}}$ These are the inverses. 1. ${\displaystyle a_{0}+a_{1}x+a_{2}x^{2}+a_{3}x^{3}\mapsto a_{0}+a_{1}x+(a_{2}/2)x^{2}+(a_{3}/3)x^{3}}$ 2. ${\displaystyle a_{0}+a_{1}x+a_{2}x^{2}+a_{3}x^{3}\mapsto a_{0}+a_{2}x+a_{1}x^{2}+a_{3}x^{3}}$ 3. ${\displaystyle a_{0}+a_{1}x+a_{2}x^{2}+a_{3}x^{3}\mapsto a_{3}+a_{0}x+a_{1}x^{2}+a_{2}x^{3}}$ 4. ${\displaystyle a_{0}+a_{1}x+a_{2}x^{2}+a_{3}x^{3}\mapsto a_{0}+(a_{1}-a_{0})x+(a_{2}-a_{1})x^{2}+(a_{3}-a_{2})x^{3}}$ For instance, for the second one, the map given in the question sends ${\displaystyle 0+1x+2x^{2}+3x^{3}\mapsto 0+2x+1x^{2}+3x^{3}}$ and then the inverse above sends ${\displaystyle 0+2x+1x^{2}+3x^{3}\mapsto 0+1x+2x^{2}+3x^{3}}$. So this map is actually self-inverse. Problem 9 Describe the nullspace and rangespace of a transformation given by ${\displaystyle {\vec {v}}\mapsto 2{\vec {v}}}$. For any vector space ${\displaystyle V}$, the nullspace ${\displaystyle \{{\vec {v}}\in V\,{\big \|}\,2{\vec {v}}={\vec {0}}\}}$ is trivial, while the rangespace ${\displaystyle \{{\vec {w}}\in V\,{\big \|}\,{\vec {w}}=2{\vec {v}}{\text{ for some }}{\vec {v}}\in V\}}$ is all of ${\displaystyle V}$, because every vector ${\displaystyle {\vec {w}}}$ is twice some other vector, specifically, it is twice ${\displaystyle (1/2){\vec {w}}}$. (Thus, this transformation is actually an automorphism.) Problem 10 List all pairs ${\displaystyle ({\text{rank}}(h),{\text{nullity}}(h))}$ that are possible for linear maps from ${\displaystyle \mathbb {R} ^{5}}$ to ${\displaystyle \mathbb {R} ^{3}}$. Because the rank plus the nullity equals the dimension of the domain (here, five), and the rank is at most three, the possible pairs are: ${\displaystyle (3,2)}$, ${\displaystyle (2,3)}$, ${\displaystyle (1,4)}$, and ${\displaystyle (0,5)}$. Coming up with linear maps that show that each pair is indeed possible is easy. Problem 11 Does the differentiation map ${\displaystyle d/dx:{\mathcal {P}}_{n}\to {\mathcal {P}}_{n}}$ have an inverse? No (unless ${\displaystyle {\mathcal {P}}_{n}}$ is trivial), because the two polynomials ${\displaystyle f_{0}(x)=0}$ and ${\displaystyle f_{1}(x)=1}$ have the same derivative; a map must be one-to-one to have an inverse. This exercise is recommended for all readers. Problem 12 Find the nullity of the map ${\displaystyle h:{\mathcal {P}}_{n}\to \mathbb {R} }$ given by ${\displaystyle a_{0}+a_{1}x+\dots +a_{n}x^{n}\mapsto \int _{x=0}^{x=1}a_{0}+a_{1}x+\dots +a_{n}x^{n}\,dx.}$ The nullspace is this. ${\displaystyle \{a_{0}+a_{1}x+\dots +a_{n}x^{n}\,{\big \|}\,a_{0}(1)+{\frac {\displaystyle a_{1}}{\displaystyle 2}}(1^{2})+\dots +{\frac {\displaystyle a_{n}}{\displaystyle n+1}}(1^{n+1})=0\}}$ ${\displaystyle =\{a_{0}+a_{1}x+\dots +a_{n}x^{n}\,{\big \|}\,a_{0}+(a_{1}/2)+\dots +(a_{n+1}/n+1)=0\}}$ Thus the nullity is ${\displaystyle n}$. Problem 13 1. Prove that a homomorphism is onto if and only if its rank equals the dimension of its codomain. 2. Conclude that a homomorphism between vector spaces with the same dimension is one-to-one if and only if it is onto. 1. One direction is obvious: if the homomorphism is onto then its range is the codomain and so its rank equals the dimension of its codomain. For the other direction assume that the map's rank equals the dimension of the codomain. Then the map's range is a subspace of the codomain, and has dimension equal to the dimension of the codomain. Therefore, the map's range must equal the codomain, and the map is onto. (The "therefore" is because there is a linearly independent subset of the range that is of size equal to the dimension of the codomain, but any such linearly independent subset of the codomain must be a basis for the codomain, and so the range equals the codomain.) 2. By Theorem 2.21, a homomorphism is one-to-one if and only if its nullity is zero. Because rank plus nullity equals the dimension of the domain, it follows that a homomorphism is one-to-one if and only if its rank equals the dimension of its domain. But this domain and codomain have the same dimension, so the map is one-to-one if and only if it is onto. Problem 14 Show that a linear map is nonsingular if and only if it preserves linear independence. We are proving that ${\displaystyle h:V\to W}$ is nonsingular if and only if for every linearly independent subset ${\displaystyle S}$ of ${\displaystyle V}$ the subset ${\displaystyle h(S)=\{h({\vec {s}})\,{\big \|}\,{\vec {s}}\in S\}}$ of ${\displaystyle W}$ is linearly independent. One half is easy— by Theorem 2.21, if ${\displaystyle h}$ is singular then its nullspace is nontrivial (contains more than just the zero vector). So, where ${\displaystyle {\vec {v}}\neq {\vec {0}}_{V}}$ is in that nullspace, the singleton set ${\displaystyle \{{\vec {v\,}}\}}$ is independent while its image ${\displaystyle \{h({\vec {v}})\}=\{{\vec {0}}_{W}\}}$ is not. For the other half, assume that ${\displaystyle h}$ is nonsingular and so by Theorem 2.21 has a trivial nullspace. Then for any ${\displaystyle {\vec {v}}_{1},\dots ,{\vec {v}}_{n}\in V}$, the relation ${\displaystyle {\vec {0}}_{W}=c_{1}\cdot h({\vec {v}}_{1})+\dots +c_{n}\cdot h({\vec {v}}_{n})=h(c_{1}\cdot {\vec {v}}_{1}+\dots +c_{n}\cdot {\vec {v}}_{n})}$ implies the relation ${\displaystyle c_{1}\cdot {\vec {v}}_{1}+\dots +c_{n}\cdot {\vec {v}}_{n}={\vec {0}}_{V}}$. Hence, if a subset of ${\displaystyle V}$ is independent then so is its image in ${\displaystyle W}$. Remark. The statement is that a linear map is nonsingular if and only if it preserves independence for all sets (that is, if a set is independent then its image is also independent). A singular map may well preserve some independent sets. An example is this singular map from ${\displaystyle \mathbb {R} ^{3}}$ to ${\displaystyle \mathbb {R} ^{2}}$. ${\displaystyle {\begin{pmatrix}x\\y\\z\end{pmatrix}}\mapsto {\begin{pmatrix}x+y+z\\0\end{pmatrix}}}$ Linear independence is preserved for this set ${\displaystyle \{{\begin{pmatrix}1\\0\\0\end{pmatrix}}\}\mapsto \{{\begin{pmatrix}1\\0\end{pmatrix}}\}}$ and (in a somewhat more tricky example) also for this set ${\displaystyle \{{\begin{pmatrix}1\\0\\0\end{pmatrix}},{\begin{pmatrix}0\\1\\0\end{pmatrix}}\}\mapsto \{{\begin{pmatrix}1\\0\end{pmatrix}}\}}$ (recall that in a set, repeated elements do not appear twice). However, there are sets whose independence is not preserved under this map; ${\displaystyle \{{\begin{pmatrix}1\\0\\0\end{pmatrix}},{\begin{pmatrix}0\\2\\0\end{pmatrix}}\}\mapsto \{{\begin{pmatrix}1\\0\end{pmatrix}},{\begin{pmatrix}2\\0\end{pmatrix}}\}}$ and so not all sets have independence preserved. Problem 15 Corollary 2.17 says that for there to be an onto homomorphism from a vector space ${\displaystyle V}$ to a vector space ${\displaystyle W}$, it is necessary that the dimension of ${\displaystyle W}$ be less than or equal to the dimension of ${\displaystyle V}$. Prove that this condition is also sufficient; use Theorem 1.9 to show that if the dimension of ${\displaystyle W}$ is less than or equal to the dimension of ${\displaystyle V}$, then there is a homomorphism from ${\displaystyle V}$ to ${\displaystyle W}$ that is onto. (We use the notation from Theorem 1.9.) Fix a basis ${\displaystyle \langle {\vec {\beta }}_{1},\dots ,{\vec {\beta }}_{n}\rangle }$ for ${\displaystyle V}$ and a basis ${\displaystyle \langle {\vec {w}}_{1},\ldots ,{\vec {w}}_{k}\rangle }$ for ${\displaystyle W}$. If the dimension ${\displaystyle k}$ of ${\displaystyle W}$ is less than or equal to the dimension ${\displaystyle n}$ of ${\displaystyle V}$ then the theorem gives a linear map from ${\displaystyle V}$ to ${\displaystyle W}$ determined in this way. ${\displaystyle {\vec {\beta }}_{1}\mapsto {\vec {w}}_{1},\,\dots ,\,{\vec {\beta }}_{k}\mapsto {\vec {w}}_{k}\quad {\text{and}}\quad {\vec {\beta }}_{k+1}\mapsto {\vec {w}}_{k},\,\dots ,\,{\vec {\beta }}_{n}\mapsto {\vec {w}}_{k}}$ We need only to verify that this map is onto. Any member of ${\displaystyle W}$ can be written as a linear combination of basis elements ${\displaystyle c_{1}\cdot {\vec {w}}_{1}+\dots +c_{k}\cdot {\vec {w}}_{k}}$. This vector is the image, under the map described above, of ${\displaystyle c_{1}\cdot {\vec {\beta }}_{1}+\dots +c_{k}\cdot {\vec {\beta }}_{k}+0\cdot {\vec {\beta }}_{k+1}\dots +0\cdot {\vec {\beta }}_{n}}$. Thus the map is onto. Problem 16 Let ${\displaystyle h:V\to \mathbb {R} }$ be a homomorphism, but not the zero homomorphism. Prove that if ${\displaystyle \langle {\vec {\beta }}_{1},\dots ,{\vec {\beta }}_{n}\rangle }$ is a basis for the nullspace and if ${\displaystyle {\vec {v}}\in V}$ is not in the nullspace then ${\displaystyle \langle {\vec {v}},{\vec {\beta }}_{1},\dots ,{\vec {\beta }}_{n}\rangle }$ is a basis for the entire domain ${\displaystyle V}$. By assumption, ${\displaystyle h}$ is not the zero map and so a vector ${\displaystyle {\vec {v}}\in V}$ exists that is not in the nullspace. Note that ${\displaystyle \langle h({\vec {v}})\rangle }$ is a basis for ${\displaystyle \mathbb {R} }$, because it is a size one linearly independent subset of ${\displaystyle \mathbb {R} }$. Consequently ${\displaystyle h}$ is onto, as for any ${\displaystyle r\in \mathbb {R} }$ we have ${\displaystyle r=c\cdot h({\vec {v}})}$ for some scalar ${\displaystyle c}$, and so ${\displaystyle r=h(c{\vec {v}})}$. Thus the rank of ${\displaystyle h}$ is one. Because the nullity is given as ${\displaystyle n}$, the dimension of the domain of ${\displaystyle h}$ (the vector space ${\displaystyle V}$) is ${\displaystyle n+1}$. We can finish by showing ${\displaystyle \{{\vec {v}},{\vec {\beta }}_{1},\dots ,{\vec {\beta }}_{n}\}}$ is linearly independent, as it is a size ${\displaystyle n+1}$ subset of a dimension ${\displaystyle n+1}$ space. Because ${\displaystyle \{{\vec {\beta }}_{1},\dots ,{\vec {\beta }}_{n}\}}$ is linearly independent we need only show that ${\displaystyle {\vec {v}}}$ is not a linear combination of the other vectors. But ${\displaystyle c_{1}{\vec {\beta }}_{1}+\dots +c_{n}{\vec {\beta }}_{n}={\vec {v}}}$ would give ${\displaystyle -{\vec {v}}+c_{1}{\vec {\beta }}_{1}+\dots +c_{n}{\vec {\beta }}_{n}={\vec {0}}}$ and applying ${\displaystyle h}$ to both sides would give a contradiction. This exercise is recommended for all readers. Problem 17 Recall that the nullspace is a subset of the domain and the rangespace is a subset of the codomain. Are they necessarily distinct? Is there a homomorphism that has a nontrivial intersection of its nullspace and its rangespace? Yes. For the transformation of ${\displaystyle \mathbb {R} ^{2}}$ given by ${\displaystyle {\begin{pmatrix}x\\y\end{pmatrix}}{\stackrel {h}{\longmapsto }}{\begin{pmatrix}0\\x\end{pmatrix}}}$ we have this. ${\displaystyle {\mathcal {N}}(h)=\{{\begin{pmatrix}0\\y\end{pmatrix}}\,{\big \|}\,y\in \mathbb {R} \}={\mathcal {R}}(h)}$ Remark. We will see more of this in the fifth chapter. Problem 18 Prove that the image of a span equals the span of the images. That is, where ${\displaystyle h:V\to W}$ is linear, prove that if ${\displaystyle S}$ is a subset of ${\displaystyle V}$ then ${\displaystyle h([S])}$ equals ${\displaystyle [h(S)]}$. This generalizes Lemma 2.1 since it shows that if ${\displaystyle U}$ is any subspace of ${\displaystyle V}$ then its image ${\displaystyle \{h({\vec {u}})\,{\big \|}\,{\vec {u}}\in U\}}$ is a subspace of ${\displaystyle W}$, because the span of the set ${\displaystyle U}$ is ${\displaystyle U}$. This is a simple calculation. ${\displaystyle {\begin{array}{rl}h([S])&=\{h(c_{1}{\vec {s}}_{1}+\dots +c_{n}{\vec {s}}_{n})\,{\big \|}\,c_{1},\dots ,c_{n}\in \mathbb {R} {\text{ and }}{\vec {s}}_{1},\dots ,{\vec {s}}_{n}\in S\}\\&=\{c_{1}h({\vec {s}}_{1})+\dots +c_{n}h({\vec {s}}_{n})\,{\big \|}\,c_{1},\dots ,c_{n}\in \mathbb {R} {\text{ and }}{\vec {s}}_{1},\dots ,{\vec {s}}_{n}\in S\}\\&=[h(S)]\end{array}}}$ This exercise is recommended for all readers. Problem 19 1. Prove that for any linear map ${\displaystyle h:V\to W}$ and any ${\displaystyle {\vec {w}}\in W}$, the set ${\displaystyle h^{-1}({\vec {w}})}$ has the form ${\displaystyle \{{\vec {v}}+{\vec {n}}\,{\big \|}\,{\vec {n}}\in {\mathcal {N}}(h)\}}$ for ${\displaystyle {\vec {v}}\in V}$ with ${\displaystyle h({\vec {v}})={\vec {w}}}$ (if ${\displaystyle h}$ is not onto then this set may be empty). Such a set is a coset of ${\displaystyle {\mathcal {N}}(h)}$ and is denoted ${\displaystyle {\vec {v}}+{\mathcal {N}}(h)}$. 2. Consider the map ${\displaystyle t:\mathbb {R} ^{2}\to \mathbb {R} ^{2}}$ given by ${\displaystyle {\begin{pmatrix}x\\y\end{pmatrix}}{\stackrel {t}{\longmapsto }}{\begin{pmatrix}ax+by\\cx+dy\end{pmatrix}}}$ for some scalars ${\displaystyle a}$, ${\displaystyle b}$, ${\displaystyle c}$, and ${\displaystyle d}$. Prove that ${\displaystyle t}$ is linear. 3. Conclude from the prior two items that for any linear system of the form ${\displaystyle {\begin{array}{*{2}{rc}r}ax&+&by&=&e\\cx&+&dy&=&f\end{array}}}$ the solution set can be written (the vectors are members of ${\displaystyle \mathbb {R} ^{2}}$) ${\displaystyle \{{\vec {p}}+{\vec {h}}\,{\big \|}\,{\vec {h}}{\text{ satisfies the associated homogeneous system}}\}}$ where ${\displaystyle {\vec {p}}}$ is a particular solution of that linear system (if there is no particular solution then the above set is empty). 4. Show that this map ${\displaystyle h:\mathbb {R} ^{n}\to \mathbb {R} ^{m}}$ is linear ${\displaystyle {\begin{pmatrix}x_{1}\\\vdots \\x_{n}\end{pmatrix}}\mapsto {\begin{pmatrix}a_{1,1}x_{1}+\dots +a_{1,n}x_{n}\\\vdots \\a_{m,1}x_{1}+\dots +a_{m,n}x_{n}\end{pmatrix}}}$ for any scalars ${\displaystyle a_{1,1}}$, ..., ${\displaystyle a_{m,n}}$. Extend the conclusion made in the prior item. 5. Show that the ${\displaystyle k}$-th derivative map is a linear transformation of ${\displaystyle {\mathcal {P}}_{n}}$ for each ${\displaystyle k}$. Prove that this map is a linear transformation of that space ${\displaystyle f\mapsto {\frac {d^{k}}{dx^{k}}}f+c_{k-1}{\frac {d^{k-1}}{dx^{k-1}}}f+\dots +c_{1}{\frac {d}{dx}}f+c_{0}f}$ for any scalars ${\displaystyle c_{k}}$, ..., ${\displaystyle c_{0}}$. Draw a conclusion as above. 1. We will show that the two sets are equal ${\displaystyle h^{-1}({\vec {w}})=\{{\vec {v}}+{\vec {n}}\,{\big \|}\,{\vec {n}}\in {\mathcal {N}}(h)\}}$ by mutual inclusion. For the ${\displaystyle \{{\vec {v}}+{\vec {n}}\,{\big \|}\,{\vec {n}}\in {\mathcal {N}}(h)\}\subseteq h^{-1}({\vec {w}})}$ direction, just note that ${\displaystyle h({\vec {v}}+{\vec {n}})=h({\vec {v}})+h({\vec {n}})}$ equals ${\displaystyle {\vec {w}}}$, and so any member of the first set is a member of the second. For the ${\displaystyle h^{-1}({\vec {w}})\subseteq \{{\vec {v}}+{\vec {n}}\,{\big \|}\,{\vec {n}}\in {\mathcal {N}}(h)\}}$ direction, consider ${\displaystyle {\vec {u}}\in h^{-1}({\vec {w}})}$. Because ${\displaystyle h}$ is linear, ${\displaystyle h({\vec {u}})=h({\vec {v}})}$ implies that ${\displaystyle h({\vec {u}}-{\vec {v}})={\vec {0}}}$. We can write ${\displaystyle {\vec {u}}-{\vec {v}}}$ as ${\displaystyle {\vec {n}}}$, and then we have that ${\displaystyle {\vec {u}}\in \{{\vec {v}}+{\vec {n}}\,{\big \|}\,{\vec {n}}\in {\mathcal {N}}(h)\}}$, as desired, because ${\displaystyle {\vec {u}}={\vec {v}}+({\vec {u}}-{\vec {v}})}$. 2. This check is routine. 3. This is immediate. 4. For the linearity check, briefly, where ${\displaystyle c,d}$ are scalars and ${\displaystyle {\vec {x}},{\vec {y}}\in \mathbb {R} ^{n}}$ have components ${\displaystyle x_{1},\dots ,x_{n}}$ and ${\displaystyle y_{1},\dots ,y_{n}}$, we have this. ${\displaystyle {\begin{array}{rl}h(c\cdot {\vec {x}}+d\cdot {\vec {y}})&={\begin{pmatrix}a_{1,1}(cx_{1}+dy_{1})+\dots +a_{1,n}(cx_{n}+dy_{n})\\\vdots \\a_{m,1}(cx_{1}+dy_{1})+\dots +a_{m,n}(cx_{n}+dy_{n})\end{pmatrix}}\\&={\begin{pmatrix}a_{1,1}cx_{1}+\dots +a_{1,n}cx_{n}\\\vdots \\a_{m,1}cx_{1}+\dots +a_{m,n}cx_{n}\end{pmatrix}}+{\begin{pmatrix}a_{1,1}dy_{1}+\dots +a_{1,n}dy_{n}\\\vdots \\a_{m,1}dy_{1}+\dots +a_{m,n}dy_{n}\end{pmatrix}}\\&=c\cdot h({\vec {x}})+d\cdot h({\vec {y}})\end{array}}}$ The appropriate conclusion is that ${\displaystyle {\text{General}}={\text{Particular}}+{\text{Homogeneous}}}$. 5. Each power of the derivative is linear because of the rules ${\displaystyle {\frac {d^{k}}{dx^{k}}}(f(x)+g(x))={\frac {d^{k}}{dx^{k}}}f(x)+{\frac {d^{k}}{dx^{k}}}g(x)\quad {\text{and}}\quad {\frac {d^{k}}{dx^{k}}}rf(x)=r{\frac {d^{k}}{dx^{k}}}f(x)}$ from calculus. Thus the given map is a linear transformation of the space because any linear combination of linear maps is also a linear map by Lemma 1.16. The appropriate conclusion is ${\displaystyle {\text{General}}={\text{Particular}}+{\text{Homogeneous}}}$, where the associated homogeneous differential equation has a constant of ${\displaystyle 0}$. Problem 20 Prove that for any transformation ${\displaystyle t:V\to V}$ that is rank one, the map given by composing the operator with itself ${\displaystyle t\circ t:V\to V}$ satisfies ${\displaystyle t\circ t=r\cdot t}$ for some real number ${\displaystyle r}$. Because the rank of ${\displaystyle t}$ is one, the rangespace of ${\displaystyle t}$ is a one-dimensional set. Taking ${\displaystyle \langle h({\vec {v}})\rangle }$ as a basis (for some appropriate ${\displaystyle {\vec {v}}}$), we have that for every ${\displaystyle {\vec {w}}\in V}$, the image ${\displaystyle h({\vec {w}})\in V}$ is a multiple of this basis vector— associated with each ${\displaystyle {\vec {w}}}$ there is a scalar ${\displaystyle c_{\vec {w}}}$ such that ${\displaystyle t({\vec {w}})=c_{\vec {w}}t({\vec {v}})}$. Apply ${\displaystyle t}$ to both sides of that equation and take ${\displaystyle r}$ to be ${\displaystyle c_{t({\vec {v}})}}$ ${\displaystyle t\circ t({\vec {w}})=t(c_{\vec {w}}\cdot t({\vec {v}}))=c_{\vec {w}}\cdot t\circ t({\vec {v}})=c_{\vec {w}}\cdot c_{t({\vec {v}})}\cdot t({\vec {v}})=c_{\vec {w}}\cdot r\cdot t({\vec {v}})=r\cdot c_{\vec {w}}\cdot t({\vec {v}})=r\cdot t({\vec {w}})}$ to get the desired conclusion. Problem 21 Show that for any space ${\displaystyle V}$ of dimension ${\displaystyle n}$, the dual space ${\displaystyle \mathop {\mathcal {L}} (V,\mathbb {R} )=\{h:V\to \mathbb {R} \,{\big \|}\,h{\text{ is linear}}\}}$ is isomorphic to ${\displaystyle \mathbb {R} ^{n}}$. It is often denoted ${\displaystyle V^{\ast }}$. Conclude that ${\displaystyle V^{\ast }\cong V}$. Fix a basis ${\displaystyle \langle {\vec {\beta }}_{1},\dots ,{\vec {\beta }}_{n}\rangle }$ for ${\displaystyle V}$. We shall prove that this map ${\displaystyle h{\stackrel {\Phi }{\longmapsto }}{\begin{pmatrix}h({\vec {\beta }}_{1})\\\vdots \\h({\vec {\beta }}_{n})\end{pmatrix}}}$ is an isomorphism from ${\displaystyle V^{\ast }}$ to ${\displaystyle \mathbb {R} ^{n}}$. To see that ${\displaystyle \Phi }$ is one-to-one, assume that ${\displaystyle h_{1}}$ and ${\displaystyle h_{2}}$ are members of ${\displaystyle V^{\ast }}$ such that ${\displaystyle \Phi (h_{1})=\Phi (h_{2})}$. Then ${\displaystyle {\begin{pmatrix}h_{1}({\vec {\beta }}_{1})\\\vdots \\h_{1}({\vec {\beta }}_{n})\end{pmatrix}}={\begin{pmatrix}h_{2}({\vec {\beta }}_{1})\\\vdots \\h_{2}({\vec {\beta }}_{n})\end{pmatrix}}}$ and consequently, ${\displaystyle h_{1}({\vec {\beta }}_{1})=h_{2}({\vec {\beta }}_{1})}$, etc. But a homomorphism is determined by its action on a basis, so ${\displaystyle h_{1}=h_{2}}$, and therefore ${\displaystyle \Phi }$ is one-to-one. To see that ${\displaystyle \Phi }$ is onto, consider ${\displaystyle {\begin{pmatrix}x_{1}\\\vdots \\x_{n}\end{pmatrix}}}$ for ${\displaystyle x_{1},\ldots ,x_{n}\in \mathbb {R} }$. This function ${\displaystyle h}$ from ${\displaystyle V}$ to ${\displaystyle \mathbb {R} }$ ${\displaystyle c_{1}{\vec {\beta }}_{1}+\dots +c_{n}{\vec {\beta }}_{n}{\stackrel {h}{\longmapsto }}c_{1}x_{1}+\dots +c_{n}x_{n}}$ is easily seen to be linear, and to be mapped by ${\displaystyle \Phi }$ to the given vector in ${\displaystyle \mathbb {R} ^{n}}$, so ${\displaystyle \Phi }$ is onto. The map ${\displaystyle \Phi }$ also preserves structure: where ${\displaystyle {\begin{array}{rl}c_{1}{\vec {\beta }}_{1}+\dots +c_{n}{\vec {\beta }}_{n}&{\stackrel {h_{1}}{\longmapsto }}c_{1}h_{1}({\vec {\beta }}_{1})+\dots +c_{n}h_{1}({\vec {\beta }}_{n})\\c_{1}{\vec {\beta }}_{1}+\dots +c_{n}{\vec {\beta }}_{n}&{\stackrel {h_{2}}{\longmapsto }}c_{1}h_{2}({\vec {\beta }}_{1})+\dots +c_{n}h_{2}({\vec {\beta }}_{n})\end{array}}}$ we have ${\displaystyle {\begin{array}{rl}(r_{1}h_{1}+r_{2}h_{2})(c_{1}{\vec {\beta }}_{1}+\dots +c_{n}{\vec {\beta }}_{n})&=c_{1}(r_{1}h_{1}({\vec {\beta }}_{1})+r_{2}h_{2}({\vec {\beta }}_{1}))+\dots +c_{n}(r_{1}h_{1}({\vec {\beta }}_{n})+r_{2}h_{2}({\vec {\beta }}_{n}))\\&=r_{1}(c_{1}h_{1}({\vec {\beta }}_{1})+\dots +c_{n}h_{1}({\vec {\beta }}_{n}))+r_{2}(c_{1}h_{2}({\vec {\beta }}_{1})+\dots +c_{n}h_{2}({\vec {\beta }}_{n}))\end{array}}}$ so ${\displaystyle \Phi (r_{1}h_{1}+r_{2}h_{2})=r_{1}\Phi (h_{1})+r_{2}\Phi (h_{2})}$. Problem 22 Show that any linear map is the sum of maps of rank one. Let ${\displaystyle h:V\to W}$ be linear and fix a basis ${\displaystyle \langle {\vec {\beta }}_{1},\dots ,{\vec {\beta }}_{n}\rangle }$ for ${\displaystyle V}$. Consider these ${\displaystyle n}$ maps from ${\displaystyle V}$ to ${\displaystyle W}$ ${\displaystyle h_{1}({\vec {v}})=c_{1}\cdot h({\vec {\beta }}_{1}),\quad h_{2}({\vec {v}})=c_{2}\cdot h({\vec {\beta }}_{2}),\quad \ldots \quad ,h_{n}({\vec {v}})=c_{n}\cdot h({\vec {\beta }}_{n})}$ for any ${\displaystyle {\vec {v}}=c_{1}{\vec {\beta }}_{1}+\dots +c_{n}{\vec {\beta }}_{n}}$. Clearly ${\displaystyle h}$ is the sum of the ${\displaystyle h_{i}}$'s. We need only check that each ${\displaystyle h_{i}}$ is linear: where ${\displaystyle {\vec {u}}=d_{1}{\vec {\beta }}_{1}+\dots +d_{n}{\vec {\beta }}_{n}}$ we have ${\displaystyle h_{i}(r{\vec {v}}+s{\vec {u}})=rc_{i}+sd_{i}=rh_{i}({\vec {v}})+sh_{i}({\vec {u}})}$. Problem 23 Is "is homomorphic to" an equivalence relation? (Hint: the difficulty is to decide on an appropriate meaning for the quoted phrase.) Either yes (trivially) or no (nearly trivially). If ${\displaystyle V}$ "is homomorphic to" ${\displaystyle W}$ is taken to mean there is a homomorphism from ${\displaystyle V}$ into (but not necessarily onto) ${\displaystyle W}$, then every space is homomorphic to every other space as a zero map always exists. If ${\displaystyle V}$ "is homomorphic to" ${\displaystyle W}$ is taken to mean there is an onto homomorphism from ${\displaystyle V}$ to ${\displaystyle W}$ then the relation is not an equivalence. For instance, there is an onto homomorphism from ${\displaystyle \mathbb {R} ^{3}}$ to ${\displaystyle \mathbb {R} ^{2}}$ (projection is one) but no homomorphism from ${\displaystyle \mathbb {R} ^{2}}$ onto ${\displaystyle \mathbb {R} ^{3}}$ by Corollary 2.17, so the relation is not reflexive.[1] Problem 24 Show that the rangespaces and nullspaces of powers of linear maps ${\displaystyle t:V\to V}$ form descending ${\displaystyle V\supseteq {\mathcal {R}}(t)\supseteq {\mathcal {R}}(t^{2})\supseteq \ldots }$ and ascending ${\displaystyle \{{\vec {0}}\}\subseteq {\mathcal {N}}(t)\subseteq {\mathcal {N}}(t^{2})\subseteq \ldots }$ chains. Also show that if ${\displaystyle k}$ is such that ${\displaystyle {\mathcal {R}}(t^{k})={\mathcal {R}}(t^{k+1})}$ then all following rangespaces are equal: ${\displaystyle {\mathcal {R}}(t^{k})={\mathcal {R}}(t^{k+1})={\mathcal {R}}(t^{k+2})\ldots \,}$. Similarly, if ${\displaystyle {\mathcal {N}}(t^{k})={\mathcal {N}}(t^{k+1})}$ then ${\displaystyle {\mathcal {N}}(t^{k})={\mathcal {N}}(t^{k+1})={\mathcal {N}}(t^{k+2})=\ldots \,}$. That they form the chains is obvious. For the rest, we show here that ${\displaystyle {\mathcal {R}}(t^{j+1})={\mathcal {R}}(t^{j})}$ implies that ${\displaystyle {\mathcal {R}}(t^{j+2})={\mathcal {R}}(t^{j+1})}$. Induction then applies. Assume that ${\displaystyle {\mathcal {R}}(t^{j+1})={\mathcal {R}}(t^{j})}$. Then ${\displaystyle t:{\mathcal {R}}(t^{j+1})\to {\mathcal {R}}(t^{j+2})}$ is the same map, with the same domain, as ${\displaystyle t:{\mathcal {R}}(t^{j})\to {\mathcal {R}}(t^{j+1})}$. Thus it has the same range: ${\displaystyle {\mathcal {R}}(t^{j+2})={\mathcal {R}}(t^{j+1})}$.	2020-01-23 23:17:11	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 358, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9615374803543091, "perplexity": 146.90869744444018}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-05/segments/1579250614086.44/warc/CC-MAIN-20200123221108-20200124010108-00418.warc.gz"}
http://serverfault.com/questions/165595/host-asp-net-4-0-in-iis-6-on-windows-2003-x64	# Host ASP.NET 4.0 in IIS 6 on Windows 2003 x64 I want to host an ASP.NET 4.0 web application in IIS 6 on Windows 2003 R2 SP2 x64. Is this even possible? I have the following under C:\WINDOWS\microsoft.net\Framework • 1.0.3705 • 1.1.4322 • 2.0.50727 • 3.0 • 3.5 • 4.0.30319 and in C:\WINDOWS\microsoft.net\Framework64 the same except 1.0 and 1.1. IIS 6 runs in 32 bit mode but there is no ASP.NET tab in the properties screen of the website. So I can't switch frameworks and don't know what frameworks runs in what application pool. EDIT have tried what GregD suggested. That gave me back the tab in IIS so that I could set the proper framework for the website, but it gave me the following in the eventlog when I tried to start the application pool: Could not load all ISAPI filters for site/service. Therefore startup aborted. After a quick googling, I found this: http://nishantrana.wordpress.com/2008/09/09/service-unavailable-could-not-load-all-isapi-filters-for-siteservice-therefore-startup-aborted/ Which kind of puts me in a circle. Enabling and removing the 64 bit setting in the metabase.xml file. Because aspnet_regiis -i told me that it couldn't work because IIS was in a 64-bit setting. - You can do it. On your web service extensions folder (in IIS Manager) are your asp.net extensions enabled for the different versions? Edited to add: To get your tab back try this: 1) Stop the IIS Admin service (and any services that depend on it) 2) Open C:\WINDOWS\system32\inetsrv\MetaBase.xml in notepad or your favorite XML Editor. DELETE the line that reads ‘Enable32BitAppOnWin64=”TRUE”’ 3) Restart IIS - yes it's all enabled. but i can't switch frameworks, the tab is not there – JP Hellemons Jul 30 '10 at 13:48 Is this running on VMWare by chance? – GregD Jul 30 '10 at 13:58 Hello GregD, it's not running on VMWare, just a normal win 2003 x64 install. but i cannot switch .net frameworks in iis. – JP Hellemons Aug 2 '10 at 7:23 Editing the MetaBase.xml file worked perfectly for me. Thanks – AndyMcKenna Aug 10 '10 at 6:58 It gave me my option back to select asp.net 4 but then got this in the eventlog: Description: Could not load all ISAPI filters for site/service. Therefore startup aborted. will take a closer look tomorrow :) thanks! – JP Hellemons Aug 10 '10 at 15:06 It works now. It seems that I can't have the option in IIS to switch frameworks. I did: C:\WINDOWS\microsoft.net\Framework\v4.0.30319>cscript c:\inetpub\adminscripts\adsutil.vbs SET /w3svc/AppPools/Enable32BitAppOnWin64 True and then C:\WINDOWS\microsoft.net\Framework\v4.0.30319>aspnet_regiis.exe -i -enable and it worked! As I said, no ASP.NET tab is visible in IIS, but it works! -	2014-08-28 09:13:27	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6219452619552612, "perplexity": 9401.755968232206}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-35/segments/1408500830721.40/warc/CC-MAIN-20140820021350-00397-ip-10-180-136-8.ec2.internal.warc.gz"}
https://converter.ninja/volume/us-customary-teaspoons-to-centiliters/891-usteaspoon-to-cl/	# 891 US customary teaspoons in centiliters ## Conversion 891 US customary teaspoons is equivalent to 439.166914003125 centiliters.[1] ## Conversion formula How to convert 891 US customary teaspoons to centiliters? We know (by definition) that: $1\mathrm{usteaspoon}\approx 0.492892159375\mathrm{centiliter}$ We can set up a proportion to solve for the number of centiliters. $1 ⁢ usteaspoon 891 ⁢ usteaspoon ≈ 0.492892159375 ⁢ centiliter x ⁢ centiliter$ Now, we cross multiply to solve for our unknown $x$: $x\mathrm{centiliter}\approx \frac{891\mathrm{usteaspoon}}{1\mathrm{usteaspoon}}*0.492892159375\mathrm{centiliter}\to x\mathrm{centiliter}\approx 439.16691400312504\mathrm{centiliter}$ Conclusion: $891 ⁢ usteaspoon ≈ 439.16691400312504 ⁢ centiliter$ ## Conversion in the opposite direction The inverse of the conversion factor is that 1 centiliter is equal to 0.00227703856578067 times 891 US customary teaspoons. It can also be expressed as: 891 US customary teaspoons is equal to $\frac{1}{\mathrm{0.00227703856578067}}$ centiliters. ## Approximation An approximate numerical result would be: eight hundred and ninety-one US customary teaspoons is about four hundred and thirty-nine point one seven centiliters, or alternatively, a centiliter is about zero times eight hundred and ninety-one US customary teaspoons. ## Footnotes [1] The precision is 15 significant digits (fourteen digits to the right of the decimal point). Results may contain small errors due to the use of floating point arithmetic.	2022-07-03 05:57:48	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 6, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.883634626865387, "perplexity": 3329.3799437083585}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-27/segments/1656104215790.65/warc/CC-MAIN-20220703043548-20220703073548-00674.warc.gz"}
https://learn.careers360.com/ncert/question-state-whether-the-following-statements-are-true-or-false-justify-your-answers-ii-every-point-on-the-number-line-is-of-the-form-square-root-m-where-m-is-a-natural-number/	Q # State whether the following statements are true or false. Justify your answers. (ii) Every point on the number line is of the form square root m, where m is a natural number. 1. State whether the following statements are true or false. Justify your answers. (ii) Every point on the number line is of the form $\sqrt{m}$ , where m is a natural number.	2020-01-24 11:02:32	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 1, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.503419041633606, "perplexity": 124.20559448938634}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-05/segments/1579250619323.41/warc/CC-MAIN-20200124100832-20200124125832-00321.warc.gz"}
https://cs.stackexchange.com/questions/80821/branch-bound-avoiding-equivalent-solutions	# Branch & Bound: Avoiding equivalent solutions There is a certain family of typical branch & bound problems in which we have to put $n$ objects into $m$ indistinguishable bags. Usually solutions are represented by $n$-size vectors in which $V[i]$ means that the $i$-th element is in the $V[i]$-th bag. Is there a way to avoid generating equivalent solutions without performing an $\mathcal{O}(n)$ time check once the potentially equivalent solution has been built? I was trying to do something similar as when avoiding circular permutations, but no adaptations seems to work. For instance, in each bag find the lowest-numbered item in that bag, and imagine labelling the bag with that number. Now add the additional constraint that to be a valid solution, the bags must be in sorted order according to their label. In other words, the label on the $i$th bag must be less than the label on the $i+1$st bag.	2021-08-01 14:29:59	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8472522497177124, "perplexity": 471.07572009116143}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-31/segments/1627046154214.36/warc/CC-MAIN-20210801123745-20210801153745-00558.warc.gz"}
http://benediktmeurer.de/2018/03/23/impact-of-polymorphism-on-component-based-frameworks-like-react/	I just realized that this is going to be the very first blog post of 2018 that I write myself - versus bugging someone else to write a blog post. It’s been quite a busy year for me already, plus I was sick a lot, and so was my family. Anyways, here’s something I’ve been meaning to send out for a while. And while the title mentions React explicitly, this is by no means limited to React, but probably affects a lot of code out there, including a lot of Node.js code bases, where this impact is even more severe. I’ve written a blog post on Surprising polymorphism in React applications earlier and this post goes in the same direction, although we’ll explore a slightly different problem here. For more background on the topic you may want to read What’s up with monomorphism? and Explaining JavaScript VMs in JavaScript - Inline Caches by my colleague Vyacheslav Egorov. A motivating example So let’s dive right in with a code example. Imagine you have a component-based framework like React and you need to call certain methods on all components, like the render method in case of React. Here’s a simplified example of how this could look like: // Base class for all components. class Component { render() { return ''; } } class HelloComponent extends Component { render() { return '<div>Hello</div>'; } } constructor(target, text) { this.target = target; this.text = text; } render() { return '<a href="' + this.target + '">' + this.text + '</a>'; } } class DOM { static renderAll(target, components) { let html = ''; for (const component of components) { html += component.render(); } target.innerHTML = html; } } const components = [ new HelloComponent(), ]; DOM.renderAll(document.getElementById('my-app'), components); You call DOM.renderAll to render a collection of components to a part of the DOM. This is oversimplified of course, but you get the concept. Also note that this code doesn’t follow any of the security advice for dealing with HTML/DOM from JavaScript, so please don’t take any inspiration here. The interesting code here is inside of DOM.renderAll, where we access the property component.render of different component shapes (in V8 speak we don’t say shape but we use the term map or sometimes hidden class). In this simple example we only have two different shapes of component: Either an instance of HelloComponent where render is found on the HelloComponent.prototype, or an instance of LinkComponent where render is found on the LinkComponent.prototype. So the property access component.render in DOM.renderAll will be 2-way polymorphic. Inline cache states Let’s use a simplified example components1.js here to illustrate this: class Base { foo() {} } class A extends Base { foo() { } } class B extends Base { foo() { } } function bar(instance) { return instance.foo(); } bar(new A); bar(new B); bar(new A); bar(new B); Running this with d8 and the --trace-ic flag we can see that the LoadIC (inline cache) for the property access instance.foo goes polymorphic, indicated by the P state: $out/Release/d8 --trace-ic components1.js$ tools/ic-processor LoadIC (.->1) at ~bar components1.js:6:19 foo (map 0x8f24d80cca1) LoadIC (1->P) at ~bar components1.js:6:19 foo (map 0x8f24d80cc01) ===================== Store: 0 KeyedStore: 0 StoreInArrayLiteral: 0 In V8 we have these five different states for LoadICs right now: Marker Name Description 0 UNINITIALIZED The property access was not executed so far. . PREMONOMORPHIC The property access was executed once and we are likely going to go MONOMORPHIC on the next hit. 1 MONOMORPHIC The property access was always executed with the same shape. P POLYMORPHIC The property access was always executed with one of four different shapes. N MEGAMORPHIC The property access has seen too many different shapes. The initial 0->. transition is not shown with --trace-ic currently, because V8 does that internally on the fast path where no logging is hooked up right now. In the example above we run through instance.foo with two different shapes for instance, either an instance of A with foo on A.prototype, or an instance of B with foo on B.prototype. Let’s beef up the example a bit and add more than four different shapes like shown in components2.js below: class Base { foo() {} } class A extends Base { foo() { } } class B extends Base { foo() { } } class C extends Base { foo() { } } class D extends Base { foo() { } } class E extends Base { foo() { } } function bar(instance) { return instance.foo(); } bar(new A); bar(new B); bar(new C); bar(new D); bar(new E); bar(new A); bar(new B); bar(new C); bar(new D); bar(new E); Running this through d8 again with --trace-ic we see that we stay POLYMORPHIC for the first four shapes, but then we go MEGAMORPHIC (as indicated by the N state) eventually (on the second instance of A in this case due to the initial PREMONOMORPHIC step): $out/Release/d8 --trace-ic components2.js$ tools/ic-processor LoadIC (.->1) at ~bar components2.js:9:19 foo (map 0x15581a58ce81) LoadIC (1->P) at ~bar components2.js:9:19 foo (map 0x15581a58cfc1) LoadIC (P->P) at ~bar components2.js:9:19 foo (map 0x15581a58d0b1) LoadIC (P->P) at ~bar components2.js:9:19 foo (map 0x15581a58d1a1) LoadIC (P->N) at ~bar components2.js:9:19 foo (map 0x15581a58cde1) ===================== Store: 0 KeyedStore: 0 StoreInArrayLiteral: 0 The MEGAMORPHIC state is the I don’t know what to do about this state of V8. Whenever a LoadIC reaches the MEGAMORPHIC state, TurboFan will no longer be able to inline any fast-path for it (except for some corner cases where TurboFan can find information about the object somewhere else), and will have to go through the inline cache logic all the time instead. MEGAMORPHIC also indicates that the inline cache will no longer try to cache information about how to access the property locally (i.e. on the property access site), but instead fall back to a global cache (the so-called megamorphic stub cache). Scalability issues This megamorphic stub cache is a global cache of fixed size where all MEGAMORPHIC sites will try to cache the lookup information for properties. This is important to understand, as especially for big applications that means you’ll likely have a lot of contention on this resource. Highly polymorphic sites like instance.foo in the example above are prime candidates for contenders, and you’ll only be affected by this non-local effect once you start integrating your applications or once you trigger a significant number of hash collisions on the global cache with a single property access site. const N = 10000; function test(fn) { var result; for (var i = 0; i < N; ++i) result = fn(); return result; } test(x => x); function makeNaive(klasses) { const instances = klasses.map(klass => new klass); return function naive() { let result; for (const instance of instances) result = instance.foo(); return result; } } const DEGREES = [100, 300, 500, 700, 900]; for (const degree of DEGREES) { const KLASSES = []; for (let i = 0; i < degree; ++i) { KLASSES.push(eval('(class C' + i + ' { foo() { }})')); } const TESTS = [ makeNaive(KLASSES) ]; for (var j = 0; j < TESTS.length; ++j) { test(TESTS[j]); } for (var j = 0; j < TESTS.length; ++j) { var startTime = Date.now(); test(TESTS[j]); console.log(degree + ':', (Date.now() - startTime), 'ms.'); } } The example above illustrates this problem, even with just a single access site instance.foo inside of the naive function, and varying degrees of polymorphism. Running this through d8 (in V8 6.6) you’ll see that the solution doesn’t scale very well: \$ out/Release/d8 components3.js 100: 43 ms. 300: 167 ms. 500: 1286 ms. 700: 1866 ms. 900: 2428 ms. It actually scales very poorly since the VM spends more and more time missing on the global cache due to collisions. Notice especially the cliff when going from 300 different shapes to 500 different shapes. A potential solution Unfortunately big, component-based applications have naturally a high degree of polymorphism by design. So what can you do to mitigate the negative impact? The main problem is having property accesses such as instance.foo where instance can have a high number of different shapes. So if you try to not have that many of these accesses in your application, you’ll likely reduce the negative impact on the global cache. One way to do this is to simply load the method ahead of time, remember it together with the object reference and then use Function.prototype.call in the hot loop. A convenient way to avoid having to carry around the pair of object and function yourself is to use Function.prototype.bind instead. const N = 10000; function test(fn) { var result; for (var i = 0; i < N; ++i) result = fn(); return result; } test(x => x); function makeNaive(klasses) { const instances = klasses.map(klass => new klass); return function naive() { let result; for (const instance of instances) result = instance.foo(); return result; } } function makeCall(klasses) { const pairs = klasses.map(klass => { const instance = new klass(); const method = instance.foo; return {instance, method}; }); return function call() { let result; for (const {instance, method} of pairs) { result = method.call(instance); } return result; } } function makeBound(klasses) { const fns = klasses.map(klass => { const instance = new klass(); return instance.foo.bind(instance); }); return function bound() { let result; for (const fn of fns) result = fn(); return result; } } const DEGREES = [100, 300, 500, 700, 900]; for (const degree of DEGREES) { const KLASSES = []; for (let i = 0; i < degree; ++i) { KLASSES.push(eval('(class C' + i + ' { foo() { }})')); } const TESTS = [ makeBound(KLASSES), makeCall(KLASSES), makeNaive(KLASSES) ]; for (var j = 0; j < TESTS.length; ++j) { test(TESTS[j]); } for (var j = 0; j < TESTS.length; ++j) { var startTime = Date.now(); test(TESTS[j]); console.log( degree + '\|' + TESTS[j].name + ':', (Date.now() - startTime), 'ms.'); } } Running this with V8 6.6, both approaches scale equally well and avoid the problem of hot MEGAMORPHIC property accesses. So the takeaway here is that a high degree of polymorphism is not bad per se, but if you have a lot of MEGAMORPHIC property accesses on the critical path, then eventually your application is going to spend a lot of time fighting for entries in the megamorphic stub cache (at least with how V8 works currently). For component-based systems there are ways to avoid this situation by preloading methods used in hot code later. Update Just ran the same test again with V8 6.7.0 (candidate) as of today, and it seems that it does scale better, altough I’m unsure what changed in this area recently (I have to admit I haven’t worked on this area of V8 a lot). Disclaimer For the vast majority of JavaScript code that is written today, the heuristics in the JavaScript VMs (i.e. in V8) are perfectly fine, even for high degrees of polymorphism. So unless you’re writing a framework like React, you may not need to worry about what’s written in here. Also the usual advice still applies, more than ever: Write idiomatic JavaScript, let the engine take care of the performance, optimize only when necessary and after careful profiling.	2018-10-17 16:54:13	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.250480592250824, "perplexity": 5272.107201851098}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-43/segments/1539583511203.16/warc/CC-MAIN-20181017153433-20181017174933-00224.warc.gz"}
http://mfat.imath.kiev.ua/authors/name/?author_id=321	# F. S. Stonyakin Search this author in Google Scholar Articles: 2 ### Strong compact properties of the mappings and K-Radon-Nikodym property Methods Funct. Anal. Topology 16 (2010), no. 2, 183-196 For mappings acting from an interval into a locally convex space, we study properties of strong compact variation and strong compact absolute continuity connected with an expansion of the space into subspaces generated by the compact sets. A description of strong $K$-absolutely continuous mappings in terms of indefinite Bochner integral is obtained. A special class of the spaces having $K$-Radon-Nikodym property is obtained. A relation between the $K$-Radon-Nikodym property and the classical Radon-Nikodym property is considered. ### Compact variation, compact subdifferetiability and indefinite Bochner integral Methods Funct. Anal. Topology 15 (2009), no. 1, 74-90 The notions of compact convex variation and compact convex subdifferential for the mappings from a segment into a locally convex space (LCS) are studied. In the case of an arbitrary complete LCS, each indefinite Bochner integral has compact variation and each strongly absolutely continuous and compact subdifferentiable a.e. mapping is an indefinite Bochner integral.	2021-08-04 10:10:31	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6970574855804443, "perplexity": 1210.8268813730178}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-31/segments/1627046154798.45/warc/CC-MAIN-20210804080449-20210804110449-00112.warc.gz"}
https://collegephysicsanswers.com/openstax-solutions/what-approximate-speed-relative-us-galaxy-near-edge-known-universe-some-10-gly-0	Change the chapter Question (a) What is the approximate speed relative to us of a galaxy near the edge of the known universe, some 10 Gly away? (b) What fraction of the speed of light is this? Note that we have observed galaxies moving away from us at greater than $0.9c$. 1. $2.0\times 10^{5}\textrm{ km/s}$ 2. $0.67$ Solution Video	2023-02-02 18:26:10	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6556100249290466, "perplexity": 352.13477345633873}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-06/segments/1674764500035.14/warc/CC-MAIN-20230202165041-20230202195041-00703.warc.gz"}
https://mathoverflow.net/questions/421059/hausdorff-distance-in-compact-lie-groups	# Hausdorff distance in compact Lie groups Let $$G$$ be a compact Lie group with a compatible biinvariant metric $$d$$. The hyperspace $$K(G)$$ of nonempty compact subsets of $$G$$ is a compact metric space with the Hausdorff metric, and it is easy to check that subgroups of $$G$$ form a closed subspace in $$K(G)$$, hence we may talk about the (compact) space of closed subgroups of $$G$$. Let us denote this space by $$\mathbf{K}(G)$$. General question: (1) Does anyone know any source that may help exploring spaces of the form $$\mathbf{K}(G)$$? I have a conjecture: (2) For a compact connected Lie group $$G$$ the following are equivalent: a) $$G$$ is a limit point in $$\mathbf{K}(G)$$ (that is, it can be approximated by proper closed subgroups). b) The circle group is a quotient of $$G$$. Is it true? ( b)$$\implies$$a) is easy, take inverse images of finite subgroups of the circle group by the quotient map.) For (1) I have found only the papers of Fischer and Gartside: On the space of subgroups of a compact group I and On the space of subgroups of a compact group II. They mostly deal with arbitrary compact $$G$$ or profinite $$G$$, not Lie groups. For (2) I found the MO question Approximating Lie groups by finite groups, which says that only compact abelian Lie groups can be approximated by finite subgroups (it refers to a paper of A. M. Turing, Finite Approximations to Lie Groups). • These questions can be tricky; one can generalize this to a decent topology on the space of closed subgroups of any topological group, called the Chabauty space. Chabauty space on $\mathbb{R}^2$ is a $4$-sphere, but the topology of the chabauty space of $\mathbb{R}^n$ is not precisely known for $n>2$. There are works by e.g. Haettel and de la Harpe-Kleptsyn-de Cornulier on this. Apr 25 at 13:50 • Somebody commented here in April but later they deleted the comment. It contained extremely useful information for us: a paper of Mongomery and Zippin Together with Nicolas Tholozan's answer it helped us in our work with the space of closed subgroups. – chj Jun 10 at 15:03 Assume a sequence of subgroups $$G_n$$ converges to $$G$$. Up to extraction, we can assume that $$\mathrm{Lie}(G_n)$$ converges to a Lie subalgebra $$\mathrm{Lie}(H)$$. Since the adjoint action of $$G_n$$ preserves $$\mathrm{Lie}(G)$$, by passing to the limit we get that $$\mathrm{Lie}(H)$$ is an ideal of $$\mathrm{Lie}(G)$$. Now a bit of (elementary ?) Lie theory should give you that $$\mathrm{Lie}(G_n) = \mathrm{Lie}(H)$$ for $$n$$ large enough. Let $$H$$ be the connected subgroup with Lie algebra $$\mathrm{Lie}(H)$$. One concludes that $$G_n/H$$ is a sequence of discrete groups approximating $$G/H$$. By your second reference, $$G/H$$ is abelian, which proves your conjecture.	2022-11-27 10:00:59	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 31, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9002107381820679, "perplexity": 202.51351502086345}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-49/segments/1669446710218.49/warc/CC-MAIN-20221127073607-20221127103607-00086.warc.gz"}
https://cs.stackexchange.com/questions/140525/proof-of-correctness-request-for-greedy-algorithm-that-solves-the-weight-job-sc	# Proof of Correctness Request for Greedy Algorithm that solves “The Weight Job Scheduling” problem Today, in my self-lead studies, I found out about greedy algorithms, more specifically, a greedy approach to solve The Weighted Job Scheduling Problem. I understand how the solution is implemented but, I'd love to see a proof of correctness for this solution (i.e. partial correctness and termination). If anyone can help me understand why this solution is correct mathematically in a general form, that'd really great! • Have you ever read the proof for Kruskal's algorithm? A proposition similar to the proposition $P$ there can be proved by induction for this problem. If you can write an answer, I can check and upvote. May 20 at 4:13 • What is the weighted job scheduling problem? What is the greedy algorithm used to solve it? May 20 at 5:49 • Hi @JohnL. I have read the proof you linked, but do not see how it's related. Can you link me to a proof of a solution (dynamic or greedy) to the Weighted Job Scheduling problem, please? :) May 20 at 6:52 • Try proving the following proposition by induction: If $A$ is the arrangement of the jobs at each stage of your algorithm right after a job is selected, there is an optimal arrangement of the jobs that extends $A$. May 21 at 23:09	2021-12-02 09:33:13	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5340571999549866, "perplexity": 459.0561227373332}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-49/segments/1637964361253.38/warc/CC-MAIN-20211202084644-20211202114644-00019.warc.gz"}
https://www.mathworks.com/company/newsletters/articles/new-features-for-high-performance-image-processing-in-matlab.html?action=changeCountry	Technical Articles and Newsletters # New Features for High-Performance Image Processing in MATLAB By Brett Shoelson , MathWorks and Bruce Tannenbaum, MathWorks Advances in satellite, medical, hyperspectral, and other imaging systems are producing bigger and more numerous images than ever before. Recent PC-based computational architectures, such as multicore CPUs and clusters, have imposed additional complexities on programming for high-performance imaging applications. Recent enhancements to MATLAB® and Image Processing Toolbox address these challenges, with image processing speed increased in some cases by orders of magnitude. This article describes several of these enhancements and then uses a segmentation example to demonstrate the performance improvements that can result from using them. ## Accelerated Image Processing Over the past few years, image processing users have seen performance enhancements in MATLAB math functions and Image Processing Toolbox functions. Since Release 2008a, MATLAB has incorporated major performance improvements to some of the math engines that underlie its matrix mathematics. These include upgrades to the Basic Linear Algebra Subroutine (BLAS) libraries, the Linear Algebra Package (LAPACK) library, and processor-specific math libraries. MATLAB now includes object-oriented computational geometry tools, as well as a library of computational geometry algorithms for improved robustness, performance, and memory efficiency. In Image Processing Toolbox, one of the key methods introduced to improve performance was implicit multithreading. Figure 1 plots the speedup of the toolbox function imresize by release. imresize was improved by both core algorithm enhancements and implicit multithreading. While improvements to the core algorithms in R2007b resulted in a substantial speedup of imresize, the multithreading introduced in R2010a resulted in a speedup that was dramatic. Release R2010a of Image Processing Toolbox brought equally significant performance improvements to more than 50 functions, including imfilter, imopen, imclose, and edge. Figure 1. The imresize function, showing significant speedup with MATLAB R2007b and even more gains with R2010a. ## Displaying Large Images With medical and satellite images now routinely in the gigapixel range, the ability to view large images that do not fit into the MATLAB workspace has become imperative. The enhanced function imshow lets you display a subsampled version of TIFF images. Additionally, a workflow introduced in R2009b lets you create spatially tiled, multiresolution image pyramids to display and navigate very large images. With the rsetwrite function you can create an R-Set file from large TIFF or NITF image files. You can open and explore R-Set files with imtool, just as you would a smaller image. ## Block Processing Large Images When images are larger than available DRAM, it is impractical to load and process them. To address this challenge, the blockproc function, added in release R2009b, lets you operate locally on smaller sections of an image without loading the entire image into a MATLAB variable. Because this function supports a file-to-file workflow, neither the input nor the output image needs to be contained entirely in memory. Consequently, the block processing approach extends naturally to arbitrarily large images. Because the processing of each block is an independent operation, block processing also lends itself to processing in parallel. ## Parallel Computing Many image processing operations are inherently parallelizable, as they perform localized and independent computations. While many MATLAB math and Image Processing Toolbox functions leverage multiple cores implicitly, Parallel Computing Toolbox provides functionally for explicit parallelization. Notably, the parfor function lets you specify that a for-loop is to be processed in parallel on up to 12 processors on your local computer and on more processors when connected to a cluster. The benefits of using parallel computing tools can be overshadowed by the overhead of passing large data, such as images, back and forth. For this reason, it is often helpful to remove visualizations from the processing loop. Parallelizing a loop in MATLAB can then be as easy as changing the keyword for to the keyword parfor. Parallel computing functionality can be used on clusters, clouds, and other nonlocal hardware resources using MATLAB Distributed Computing Server. Parallel computing and block processing can easily be used together, since the processing of each block is independent. You can use blockproc to process an image with a pool of MATLAB workers by setting the use parallel parameter to true. Note, however, that running a job in parallel can add significant overhead to smaller computations. The potential benefits of parallelization improve as the computational expense of each block increases. ## Demonstrating the Performance Improvements: An Image Segmentation Example In this example, the goal is to segment an image of rice grains (Figure 2). As we step through the problem, we will review the timing of certain function calls over several versions of MATLAB and compare different approaches to segmenting the image. (All timings come from the same computer, using MATLAB R2011a under 32-bit Windows 7.) Figure 2. Original image of rice grains. tic; img = imresize(imread('rice.png'),4); toc Elapsed time is 0.014 seconds. whos img Name Size Bytes Class img 1024x1024 1048576 uint8 In this example, imread and imresize took 14 milliseconds. As the plot in Figure 1 showed, the imresize command is nearly 100 times faster in R2011a than in earlier releases. The next step is to isolate the rice grains using segmentation. The simplest approach is to apply a threshold using im2bw. First, however, we must account for the fact that the background illumination varies across the image. To normalize the background, we apply a tophat filter and then use graythresh to calculate a global threshold value, which we apply uniformly to the image using im2bw. im2 = imtophat(img,strel('disk',60)); bw = im2bw(im2, graythresh(im2)); Figure 3. Left: The speed-up in tophat filtering over eight releases, showing actual times rather than relative speeds. Center: Rice grain image after tophat filtering but before segmentation. Right: Rice grain image after segmentation. The tophat function plus im2bw segmentation ran in 0.594 seconds using the R2011a version. The R2009a version took more than 3 seconds. The speedup is partly due to the use of the imdilate and imerode functions within imtophat, which both saw significant performance improvements in R2009b and R2010a. Because tophat filtering can take a long time with very large images, we need an alternative to using a global threshold. One option is to use block processing, which lets us calculate a local threshold and perform segmentation separately for each block. While block processing uses a different algorithm, it achieves similar segmentation results. In the following code, we block-process the original image using graythresh to calculate a unique threshold in each 128x128 block. fun = @(block_struct) im2bw(block_struct.data,graythresh(block_struct.data)); segmented = blockproc(img, [128 128],fun,'bordersize',[3 3],... 'PadPartialBlocks',true,'PadMethod','symmetric'); The resulting image is almost identical to the globally thresholded image, but the code ran in 0.03 seconds using R2011a—nearly 20 times faster than the global threshold approach. These results show that it is often possible to come up with an algorithm that is more efficient at solving the problem simply by looking at the problem differently. So far, we have been working with only one image file; however, most image processing workflows involve processing many images—sometimes thousands. To accelerate the processing of many images, we can use the parfor function. To demonstrate, let's assume that we incorporated our image segmentation algorithm into the function MySegmentation: function x = MySegmentation(imgName, parameters) We can process multiple user-selected image files by calling this function in a for-loop: [filenames,pathname] = uigetfile('*.tif','Select file(s)','multiselect','on'); % Series loop over files x = zeros(numel(filenames),1); for ii = 1:length(filenames) x(ii) = MySegmentation(fullfile(pathname,filenames{ii}), parameters); end With a call to matlabpool to start the parallel processes and a change from for to parfor, we can achieve an up-to-twelvefold increase in processing speed. This approach is particularly effective with a multiprocessor machine, such as a quad-core, dual-processor computer. matlabpool open 12 % Parallel loop over files x = zeros(numel(filenames),1); parfor ii = 1:length(filenames) x(ii) = MySegmentation(fullfile(pathname,filenames{ii}), parameters); end In this article, we have described several methods for handling large images and improving image processing performance in MATLAB. We also suggested ways that parallel processing can be used to speed up image processing code, and found that a block-processing algorithm for segmentation was faster than a global thresholding algorithm. Published 2012 - 92003v00	2017-05-23 07:12:04	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.32180896401405334, "perplexity": 2534.540286567041}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-22/segments/1495463607591.53/warc/CC-MAIN-20170523064440-20170523084440-00184.warc.gz"}
https://elkement.blog/tag/solar-energy/	# Can the Efficiency Be Greater Than One? This is one of the perennial top search terms for this blog. Anticlimactic answer: Yes, because input and output are determined also by economics, not only by physics. Often readers search for the efficiency of a refrigerator. Its efficiency, the ratio of output and input energies, is greater than 1 because the ambient energy is free. System’s operators are interested in the money they pay the utility, in relation to the resulting energy for cooling. If you use the same thermodynamic machine either as a refrigerator or as a heat pump, efficiencies differ: The same input energy drives the compressor, but the relevant output energy is either the energy released to the ‘hot side’ at the condenser or the energy used for evaporating the refrigerant at the ‘cool side’: The same machine / cycle is used as a heat pump for heating (left) or a refrigerator or AC for cooling (right). (This should just highlight the principles and does not include any hydraulic details, losses etc. related to detailed differences between refrigerators / ACs and heat pumps.) For photovoltaic panels the definition has sort of the opposite bias: The sun does not send a bill – as PV installers say in their company’s slogan – but the free solar ambient energy is considered, and thus their efficiency is ‘only’ ~20%. Half of our generator, now operational for three years: 10 panels, oriented south-east, 265W each, efficiency 16%. (The other 8 panels are oriented south-west). When systems are combined, you can invent all kinds of efficiencies, depending on system boundaries. If PV panels are ‘included’ in a heat pump system (calculation-wise) the nominal electrical input energy becomes lower. If solar thermal collectors are added to any heating system, the electrical or fossil fuel input decreases. Output energy may refer to energy measured directly at the outlet of the heat pump or boiler. But it might also mean the energy delivered to the heating circuits – after the thermal losses of a buffer tank have been accounted for. But not 100% of these losses are really lost, if the buffer tank is located in the house. I’ve seen many different definitions in regulations and related software tools, and you find articles about how to game interpret these guidelines to your advantage. Tools and standards also make arbitrary assumptions about storage tank losses, hysteresis parameter and the like – factors that might be critical for efficiency. Then there are scaling effects: When the design heat loads of two houses differ by a factor of 2, and the smaller house would use a scaled down heat pump (hypothetically providing 50% output power at the same efficiency), the smaller system’s efficiency is likely to be a bit lower. Auxiliary consumers of electricity – like heating circuit pumps or control systems – will not be perfectly scalable. But the smaller the required output energy is, the better it can be aligned with solar energy usage and storage by a ‘smart’ system – and this might outweigh the additional energy needed for ‘smartness’. Perhaps intermittent negative market prices of electricity could be leveraged. Definitions of efficiency are also culture-specific, tailored to an academic discipline or industry sector. There are different but remotely related concepts of rating how useful a source of energy is: Gibbs Free Energy is the maximum work a system can deliver, given that pressure and temperature do not change during the process considered – for example in a chemical reaction. On the other hand, Exergy is the useful ‘available’ energy ‘contained’ in a (part of a) system: Sources of energy and heat are rated; e.g. heat energy is only mechanically useful up to the maximum efficiency of an ideal Carnot process. Thus exergy depends on the temperature of the environment where waste heat ends up. The exergy efficiency of a Carnot process is 1, as waste heat is already factored in. On the other hand, the fuel used to drive the process may or may not be included and it may or may not be considered pure exergy – if it is, energy and exergy efficiency would be the same again. If heat energy flows from the hot to the cold part of a system in a heat exchanger, no energy is lost – but exergy is. You could also extend the system’s boundary spatially and on the time axis: Include investment costs or the cost of harm done to the environment. Consider the primary fuel / energy / exergy to ‘generate’ electricity: If a thermal power plant has 40% efficiency then the heat pump’s efficiency needs to be at least 2,5 to ‘compensate’ for that. In summary, ‘efficiency’ is the ratio of an output and an input energy, and the definitions may be rather arbitrary as and these energies are determined by a ‘sampling’ time, system boundaries, and additional ‘ratings’. \| # Ice Storage Hierarchy of Needs Data Kraken – the tentacled tangled pieces of software for data analysis – has a secret theoretical sibling, an older one: Before we built our heat source from a cellar, I developed numerical simulations of the future heat pump system. Today this simulation tool comprises e.g. a model of our control system, real-live weather data, energy balances of all storage tanks, and a solution to the heat equation for the ground surrounding the water/ice tank. I can model the change of the tank temperature and ‘peak ice’ in a heating season. But the point of these simulations is rather to find out to which parameters the system’s performance reacts particularly sensitive: In a worst case scenario will the storage tank be large enough? A seemingly fascinating aspect was how peak ice ‘reacts’ to input parameters: It is quite sensitive to the properties of ground and the solar/air collector. If you made either the ground or the collector just ‘a bit worse’, ice seems to grow out of proportion. Taking a step back I realized that I could have come to that conclusion using simple energy accounting instead of differential equations – once I had long-term data for the average energy harvesting power of the collector and ground. Caveat: The simple calculation only works if these estimates are reliable for a chosen system – and this depends e.g. on hydraulic design, control logic, the shape of the tank, and the heat transfer properties of ground and collector. For the operations of the combined tank+collector source the critical months are the ice months Dec/Jan/Feb when air temperature does not allow harvesting all energy from air. Before and after that period, the solar/air collector is nearly the only source anyway. As I emphasized on this blog again and again, even during the ice months, the collector is still the main source and delivers most of the ambient energy the heat pump needs (if properly sized) in a typical winter. The rest has to come from energy stored in the ground surrounding the tank or from freezing water. I am finally succumbing to trends of edutainment and storytelling in science communications – here is an infographic: (Add analogies to psychology here.) Using some typical numbers, I am illustrating 4 scenarios in the figure below, for a system with these parameters: • A cuboid tank of about 23 m3 • Required ambient energy for the three ice months is ~7000kWh (about 9330kWh of heating energy at a performance factor of 4) • ‘Standard’ scenario: The collector delivers 75% of the ambient energy, ground delivers about 18%. • Worse’ scenarios: Either collector or/and ground energy is reduced by 25% compared to the standard. Contributions of the three sources add up to the total ambient energy needed – this is yet another way of combining different energies in one balance. Ambient energy needed by the heat pump in Dec+Jan+Feb, as delivered by the three different sources. Latent ‘ice’ energy is also translated to the percentage of water in the tank that would be frozen. Neither collector nor ground energy change much in relation to the base line. But latent energy has to fill in the gap: As the total collector energy is much higher than the total latent energy content of the tank, an increase in the gap is large in relation to the base ice energy. If collector and ground would both ‘underdeliver’ by 25% the tank in this scenario would be frozen completely instead of only 23%. The ice energy is just the peak of the total ambient energy iceberg. You could call this system an air-geothermal-ice heat pump then! ____________________________ Continued: Here are some details on simulations. \| # On Photovoltaic Generators and Scattering Cross Sections Subtitle: Dimensional Analysis again. Our photovoltaic generator has about 5 kW rated ‘peak’ power – 18 panels with 265W each. South-east oriented part of our generator – 10 panels. The remaining 8 are oriented south-west. Peak output power is obtained under so-called standard testing condition – 1 kWp (kilo Watt peak) is equivalent to: • a panel temperature of 25°C (as efficiency depends on temperature) • an incident angle of sunlight relative to zenith of about 48°C – equivalent to an air mass of 1,5. This determines the spectrum of the electromagnetic radiation. • an irradiance of solar energy of 1kW per square meter. Simulated spectra for different air masses (Wikimedia, User Solar Gate). For AM 1 the path of sunlight is shortest and thus absorption is lowest. The last condition can be rephrased as: We get 1 kW output per kW/minput. 1 kWp is thus defined as: 1 kWp = 1 kW / (1 kW/m2) Canceling kW, you end up with 1 kWp being equivalent to an area of 1 m2. Why is this a useful unit? Solar radiation generates electron-hole pairs in solar cells, operated as photodiodes in reverse bias. Only if the incoming photon has exactly the right energy, solar energy is used efficiently. If the photon is not energetic enough – too ‘red’ – it is lost and converted to heat. If the photon is too blue – too ‘ultraviolet’ – it generates electrical charges, but the greater part of its energy is wasted as the probability of two photons hitting at the same time is rare. Thus commercial solar panels have an efficiency of less than 20% today. (This does not yet say anything about economics as the total incoming energy is ‘free’.) The less efficient solar panels are, the more of them you need to obtain a certain target output power. A perfect generator would deliver 1 kW output with a size of 1 m2 at standard test conditions. The kWp rating is equivalent to the area of an ideal generator that would generate the same output power, and it helps with evaluating if your rooftop area is large enough. Our 4,77 kW generator uses 18 panels, about 1,61 m2 each – so 29 m2 in total. Panels’ efficiency is then about 4,77 / 29 = 16,4% – a number you can also find in the datasheet. There is no rated power comparable to that for solar thermal collectors, so I wonder why the unit has been defined in this way. Speculating wildly: Physicists working on solar cells usually have a background in solid state physics, and the design of the kWp rating is equivalent to a familiar concept: Scattering cross section. An atom can be modeled as a little oscillator, driven by the incident electromagnetic energy. It re-radiates absorbed energy in all directions. Although this can be fully understood only in quantum mechanical terms, simple classical models are successful in explaining some macroscopic parameters, like the index of refraction. The scattering strength of an atom is expressed as: [ Power scattered ] / [ Incident power of the beam / m2 ] … the same sort of ratio as discussed above! Power cancels out and the result is an area, imagined as a ‘cross-section’. The atom acts as if it were an opaque disk of a certain area that ‘cuts out’ a respective part of the incident beam and re-radiates it. The same concept is used for describing interactions between all kinds of particles (not only photons) – the scattering cross section determines the probability that an interaction will occur: Particles’ scattering strengths are represented by red disks (area = cross section). The probability of a scattering event going to happen is equal to the ratio of the sum of all red disk areas and the total (blue+red) area. (Wikimedia, User FerdiBf) # First Year of Rooftop Solar Power and Heat Pump: Re-Visiting Economics After I presented details for selected days, I am going to review overall performance in the first year. From June 2015 to May 2016 … • … we needed 6.600 kWh of electrical energy in total. • The heat pump consumed about 3.600 kWh of that … • … in order to ‘pump it up to’ 16.800 kWh of heating energy (incl. hot tap water heating). This was a mild season! . • The remaining 3.000kWh were used by household and office appliances, control, and circulation pumps. Disclaimer: I am from Austria –> decimal commas and dot thousands separator :-) The photovoltaic generator … • … harvested about 5.600kWh / year – not too bad for our 4,8kW system with panels oriented partly south-east and partly south-west. • 2.000 kWh of that were used directly and the rest was fed into the grid. • So 30% of our consumption was provided directly by the PV generator (self-sufficiency quota) and • 35% of PV energy produced was utilized immediately (self-consumption quota). Monthly energy balances show the striking difference between summer and winter: In summer the small energy needed to heat hot water can easily be provided by solar power. But in winter only a fraction of the energy needed can be harvested, even on perfectly sunny days. Figures below show… • … the total energy consumed in the house as the sum of the energy for the heat pump and the rest used by appliances … • … and as the sum of energy consumed immediately and the rest provided by the utility. • The total energy ‘generated’ by the solar panels, as a sum of the energy consumed directly (same aqua bar as in the sum of consumption) and the rest fed into the grid. In June we needed only 300kWh (10kWh per day). The PV total output was more then 700kWh, and 200kW of that was directly delivered by the PV system – so the PV generator covered 65%. It would be rather easy to become autonomous by using a small, <10kWh battery and ‘shifting’ the missing 3,3kWh per day from sunny to dark hours. But in January we needed 1100kWh and PV provided less than 200kWh in total. So a battery would not help as there is no energy left to be ‘shifted’. Daily PV energy balances show that this is true for every single day in January: We harvest typically less than 10 kWh per day, but we need more than 30kWh. On the coldest days in January, the heat pump needed about 33kWh – thus heating energy was about 130kWh: Our house’s heat consumption is typical for a well-renovated old building. If we would re-build our house from scratch, according to low energy standards, we might need only 50-60% energy at best. Then heat pump’s input energy could be cut in half (violet bar). But even then, daily total energy consumption would exceed PV production. Economics I have covered economics of the system without battery here and our system has lived up to the expectations: Profits were € 575, the sum of energy sales at market price (€ 0,06 / kWh) and by not having to pay € 0,18 for power consumed directly. In Austria turn-key PV systems (without batteries) cost about € 2.000 / kW rated power – so we earned about 6% of the costs. Not bad – given political discussions about negative interest rates. (All numbers are market prices, no subsidies included.) But it is also interesting to compare profits to heating costs: In this season electrical energy needed for the heat pump translates to € 650. So our profits from the PV generator nearly amounts to the total heating costs. Economics of batteries Last year’s assessment of the economics of a system with battery is still valid: We could increase self-sufficiency from 30% to 55% using a battery and ‘shift’ additional 2.000 kWh to the dark hours. This would result in additional € 240 profits of per year. If a battery has a life time of 20 years (optimistic estimate!) it must not cost more than € 5.000 to ever pay itself off. This is less than prices I have seen in quotes so far. Off-grid living and autonomy Energy autonomy might be valued more than economical profits. Some things to consider: Planning a true off-grid system is planning for a few days in a row without sunshine. Increasing the size of the battery would not help: The larger the battery the larger the losses, and in winter the battery would never be full. It is hard to store thermal energy for another season, but it is even harder to store electrical energy. Theoretically, the area of panels could be massively oversized (by a factor – not a small investment), but then even more surplus has to be ‘wasted’ in summer. The system has to provide enough energy per day and required peak load in every moment (see spikes in the previous post), but power needs also to be distributed to the 3 phases of electrical power in the right proportion: In Austria energy meters calculate a sum over 3 phases: A system might seem ‘autonomous’ when connected to the grid, but it would not be able to operate off-grid. Example: The PV generator produces 1kW per phase = 3kW in total, while 2kW are used by a water cooker on phase 1. The meter says you feed in 1kW to the grid, but technically you need 1kW extra from the grid for the water cooker and feed in 1kW on phase 2 and 3 each; so there is a surplus of 1kW in total. Disconnected from the grid, the water cooker would not work as 1kW is missing. A battery does not provide off-grid capabilities automatically, nor do PV panels provide backup power when the sun is shining but the grid is down: During a power outage the PV system’s inverter has to turn off the whole system – otherwise people working on the power lines outside could be hurt by the power fed into the grid. True backup systems need to disconnect from the power grid safely first. Backup capabilities need to be compliant with local safety regulations and come with additional (potentially clunky / expensive) gadgets. # Photovoltaic Generator and Heat Pump: Daily Power Generation and Consumption You can generate electrical power at home but you cannot manufacture your own natural gas, oil, or wood. (I exempt the minority of people owning forestry). This is often an argument for the combination of heat pump and photovoltaic generator. Last year I blogged in detail about economics of solar power and batteries and on typical power consumption and usage patterns – and my obsession with tracking down every sucker for electrical energy. Bottom line: Despite related tinkering with control and my own ‘user behavior’ it is hard to raise self-consumption and self-sufficiency above statistical averages for homes without heat pumps. In this post I will focus on load profiles and power generation during several selected days to illustrate these points, comparing… • electrical power provided by the PV generator (logged at Fronius Symo inverter). • input power needed by the heat pump (logged with energy meter connected to our control unit). • … power balances provided by the smart meter: Power is considered positive when fed into the grid is counted (This meter is installed directly behind the utility’s meter) A non-modulating, typical brine-water heat pump is always operating at full rated power: We have a 7kW heat pump – 7kW is about the design heat load of the building, as worst case estimate for the coldest day in years. On the coldest day in the last winter the heat pump was on 75% of the time. Given a typical performance factor of 4 kWh/kWh), the heat pump needs 1/4 of its rated power as input. Thus the PV generator needs to provide about 1-2 kW when the heat pump is on. The rated power of our 18 panels is about 5kW – this is the output under optimum conditions. Best result near winter solstice When it is perfectly sunny in winter, the generator can produce enough energy to power the heat pump between 10:00 and 14:00 in the best case. But such cloudless days are rare, and in the cold and long nights considerable electrical energy is needed, too. Too much energy in summer On a perfect summer day hot water could even be heated twice a day by solar power: These peaks look more impressive than they are compared to the base load: The heat pump needs only 1-2kWh per day compared to 10-11kWh total consumption. Harvesting energy in spring On a sunny day in spring the PV output is higher than in summer due to lower ambient temperatures. As we still need space heating energy this energy can also be utilized better: The heat pump’s input power is similar to the power of a water heater or an electrical stoves. At noon on a perfect day both the heat pump and one appliance could be run on solar power only. The typical day: Bad timing On typical days clouds pass and power output changes quickly. This is an example of a day when sunshine and hot water cycle did not overlap much: At noon the negative peak (power consumption, blue) was about 3,5kW. Obviously craving coffee or tea was stronger than the obsession with energy efficiency. Even the smartest control system would not be able to predict such peaks in both solar radiation and in erratic user behavior. Therefore I am also a bit sceptical when it comes to triggering the heat pump’s heating cycle by a signal from the PV generator, based on current and ‘expected’ sunshine and weather data from internet services (unless you track individual clouds). \| # Alien Energy I am sure it protects us not only from lightning but also from alien attacks and EMP guns … So I wrote about our lightning protection, installed together with our photovoltaic generator. Now our PV generator is operational for 11 months and we have encountered one alien attack, albeit by beneficial aliens. The Sunny Baseline This is the electrical output power of our generator – oriented partly south-east, partly south-west – for some selected nearly perfectly cloudless days last year. Even in darkest winter you could fit the 2kW peak that a water cooker or heat pump needs under the curve at noon. We can heat hot water once a day on a really sunny day but not provide enough energy for room heating (monthly statistics here). Alien Spikes and an Extended Alien Attack I was intrigued by very high and narrow spikes of output power immediately after clouds had passed by: There are two possible explanations: 1) Increase in solar cell efficiency as the panels cool off while shadowed or 2) ‘focusing’ (refraction) of radiation by the edges of nearby clouds. Such 4kW peaks lasting only a few seconds wide are not uncommon, but typically they do not show up in our standard logging, comprising 5-minute averages. There was one notable exception this February: Power surged to more than 4kW which is significantly higher than the output on other sunny days in February. Actually, it was higher than the output on the best ever sunny day last May 11 and as high as the peaks on summer solstice (Aliens are green, of course): Temperature effect and/or ‘focusing’? On the alien attack day it was cloudy and warmer in the night than on the sunny reference day, February 6. At about 11:30 the sun was breaking through the clouds, hitting rather cool panels: At that day, the sun was lingering right at the edge of clouds for some time, and global radiation was likely to be higher than expected due to the focusing effect. The jump in global radiation at 11:30 is clearly visible in our measurements of radiation. But in addition panels had been heated up before by the peak in solar radiation and air temperature had risen, too. So the different effects cannot be disentangled easily . Power drops by 0,44% of the rated power per decrease in °C of panel temperature. Our generator has 4,77kW, so power decreases by 21W/°C panel temperature. At 11:30 power was by 1,3kW higher than power on the normal reference day – the theoretical equivalent of a panel temperature decrease by 62°C. I think I can safely attribute the initial surge in output power to the unusual peak in global radiation only. At 12:30 output power is lower by 300W on the normal sunny day compared to the alien day. This can partly be attributed to the lower input radiation, and partly to a higher ambient temperature. But only if input radiation is changing slowly, panel temperature has a simple, linear relationship with input temperature. The sun might be blocked for a very short period – shorter than our standard logging interval of 90s for radiation – and the surface of panels cools off intermittently. It is an interesting optimization problem: By just the right combination of blocking period and sunny period overall output could be maximized. Re-visiting data from last hot August to add more dubious numbers Panels’ performance was lower for higher ambient air temperatures … … while global radiation over time was about the same. Actually the enveloping curve was the same, and there were even negative spikes at noon despite the better PV performance: The difference in peak power was about 750W. The panel temperature difference to account for that would have to be about 36°. This is three times the measured difference in ambient temperature of 39°C – 27°C = 12°C. Is this plausible? PV planners use a worst-case panel temperature of 75°C – for worst-case hot days like August 12, 2015. Normal Operating Cell Temperature of panels is about 46°C. Normal conditions are: 20°C of ambient air, 800W/m2 solar radiation, and free-standing panels. One panel has an area of about 1,61m2; our generator with 18 panels has 29m2, so 800W/m2 translates to 23kW. Since the efficiency of solar panels is about 16%, 23kW of input generates about 3,7kW output power – about the average of the peak values of the two days in August. Our panels are attached to the roof and not free-standing – which is expected to result in a temperature increase of 10°C. So we had been close to normal conditions at noon radiation-wise, and if we had been able to crank ambient temperature down to 20°C in August, panel temperature had been about 46°C + 10°C = 56°C. I am boldly interpolating now, in order to estimate panel temperature on the ‘colder’ day in August: Air Temperature Panel Temperature Comment 20°C 56°C Normal operating conditions, plus typical temperature increase for well-vented rooftop panels. 27°C 63°C August 1. Measured ambient temperature, solar cell temperature interpolated. 39°C 75°C August 12. Measured ambient temperature. Panel temperature is an estimate for the worst case. Under perfectly stable conditions panel temperature would have differed by 12°C, resulting in a difference of only ~ 250W (12°C * 21W/°C). Even considering higher panel temperatures at the hotter day or a non-linear relationship between air temperature and panel temperature will not easily give you the 35° of temperature difference required to explain the observed difference of 750W. I think we see aliens at work again: At about 10:45 global radiation for the cooler day, August 1, starts to fluctuate – most likely even more wildly than we see with the 90s interval. Before 10:45, the difference in output power for the two days is actually more like 200-300W – so in line with my haphazard estimate for steady-state conditions. Then at noon the ‘focusing’ effect could have kicked in, and panel surface temperature might haved fluctuated between 27°C air temperature minimum and the estimated 63°C. Both of these effects could result in the required additional increase of a few 100W. Since ‘focusing’ is actually refraction by particles in the thinned out edges of clouds, I wonder if the effect could also be caused by barely visible variations of the density of mist in the sky as I remember the hot period in August 2015 as sweltry and a bit hazy, rather than partly cloudy. I think it is likely that both beneficial effects – temperature and ‘focusing’ – will always be observed in unison. On February 11 I had the chance to see the effect of focusing only (or traces of an alien spaceship that just exited a worm-hole) for about half an hour. ________________________________ On temperature dependence of PV output power: On the ‘focusing’ effect: • Can You Get More than 100% Solar Energy? Note especially this comment – describing refraction, and pointing out that refraction of light can ‘focus’ light that would otherwise have been scattered back into space. This commentator also proposes different mechanism for short spikes in power and increase of power during extended periods (such as I observed on February 11). • Edge-of-Cloud Effect Source for the 10°C higher temperature of rooftop panels versus free-standing ones: German link, p.3: Ambient air + 20°C versus air + 30°C \| # Temperature Waves and Geothermal Energy Nearly all of renewable energy exploited today is, in a sense, solar energy. Photovoltaic cells convert solar radiation into electricity, solar thermal collectors heat hot water. Plants need solar power for photosynthesis, for ‘creating biomass’. The motion of water and air is influenced by the forces caused by the earth’s rotation, but by temperature gradients imposed by the distribution of solar energy as well. Also geothermal heat pumps with ground loops near the surface use solar energy deposited in summer and stored for winter – I think that ‘geothermal heat pumps’ is a bit of a misnomer. Collector (heat exchanger) for brine-water heat pumps. Within the first ~10 meters below the surface, temperature fluctuates throughout the year; at 10m the temperature remains about constant and equal to 10-15°C for the whole year. Only at higher depths the flow of ‘real’ geothermal energy can be spotted: In the top layer of the earth’s crust the temperatures rises about linearly, at about 3°C (3K) per 100m. The details depend on geological peculiarities, it can be higher in active regions. This is the energy utilized by geothermal power plants delivering electricity and/or heat. Geothermal gradient adapted from Boehler, R. (1996). Melting temperature of the Earth’s mantle and core: Earth’s thermal structure. Annual Review of Earth and Planetary Sciences, 24(1), 15–40. (Wikimedia, user Bkilli1). Geothermal power plants use boreholes a few kilometers deep. This geothermal energy originates from radioactive decays and from the violent past of the primordial earth: when the kinetic energy of celestial objects colliding with each other turned into heat. The flow of geothermal energy per area directed to the surface, associated with this gradient is about 65 mW/m2 on continents: Global map of the flow of heat, in mW/m2, from Earth’s interior to the surface. Davies, J. H., & Davies, D. R. (2010). Earth’s surface heat flux. Solid Earth, 1(1), 5-24. (Wikimedia user Bkilli1) Some comparisons: • It is small compared to the energy from the sun: In middle Europe, the sun provides about 1.000 kWh per m2 and year, thus 1.000.000Wh / 8.760h = 144W/m2 on average. • It also much lower than the rule-of-thumb power of ‘flat’ ground loop collectors – about 20W/m2 • The total ‘cooling power’ of the earth is several 1010kW: Would the energy not be replenished by radioactive decay, the earth would lose a some seemingly impressive 1014kWh per year, yet this would result only in a temperature difference of ~10-7°C (This is just a back-of-the-envelope check of orders of magnitude, based on earth’s mass and surface area, see links at the bottom for detailed values). The constant energy in 10m depth – the ‘neutral zone’ – is about the same as the average temperature of the earth (averaged over one year over the surface of the earth): About 14°C. I will show below that this is not a coincidence: The temperature right below the fluctuating temperature wave ‘driven’ by the sun has to be equal to the average value at the surface. It is misleading to attribute the 10°C in 10m depths to the ‘hot inner earth’ only. In this post I am toying with theoretical calculations, but in order not so scare readers off too much I show the figures first, and add the derivation as an appendix. My goal is to compare these results with our measurements, to cross-check assumptions for the thermal properties of ground I use in numerical simulations of our heat pump system (which I need for modeling e.g. the expected maximum volume of ice) 1. The surface temperature varies periodically in a year, and I use maximum, minimum and average temperature from our measurements, (corrected a bit for the mild last seasons). These are daily averages as I am not interested in the daily temperature changes between and night. 2. A constant geothermal flow of 65 mW/m2 is superimposed to that. 3. The slow transport of solar energy into ground is governed by a thermal property of ground, called the thermal diffusivity. It describes ‘how quickly’ a lump of heat deposited will spread; its unit is area per time. I use an assumption for this number based on values for soil in the literature. I am determining the temperature as a function of depth and of time by solving the differential equation that governs heat conduction. This equation tells us how a spatial distribution of heat energy or ‘temperature field’ will slowly evolve with time, given the temperature at the boundary of the interesting part of space in question – in this case the surface of the earth. The yearly oscillation of air temperature is about the simplest boundary condition one could have, so you can calculate the solution analytically. Another nice feature of the underlying equation is that it allows for adding different solutions: I can just add the effect of the real geothermal flow of energy to the fluctuations caused by solar energy. The result is a ‘damped temperature wave’; the temperature varies periodically with time and space: The spatial maximum of temperature moves from the surface to a point below and back: In summer (beginning of August) the measured temperature is maximum at the surface, but in autumn the maximum is found some meters below – heat flows back from ground to the surface then: Calculated ground temperature, based on measurements of the yearly variation of the temperature at the surface and an assumption of the thermal properties of ground. Calculated for typical middle European maximum and minimum temperatures. This figure is in line with the images shown in every textbook of geothermal energy. Since the wave is symmetrical about the yearly average, the temperature in about 10m depth, when the wave has ‘run out’, has to be equal to the yearly average at the surface. The wave does not have much chance to oscillate as it is damped down in the middle of the first period, so the length of decay is much shorter than the wavelength. The geothermal flow just adds a small distortion, an asymmetry of the ‘wave’. It is seen only when switching to a larger scale. Some data as in previous plot, just extended to greater depths. The geothermal gradient is about 3°C/100m, the detailed value being calculated from the value of thermal conductivity also used to model the fluctuations. Now varying time instead of space: The higher the depth, the more time it takes for ground to reach maximum temperature. The lag of the maximum temperature is proportional to depth: For 1m difference in depth it is less than a month. Temporal change of ground temperature at different depths. The wave is ‘moving into the earth’ at a constant speed. Measuring the time difference between the maxima for different depths lets us determine the ‘speed of propagation’ of this wave – its wavelength divided by its period. The speed depends in a simple way on the thermal diffusivity and the period as I show below. But this gives me an opportunity to cross-check my assumption for diffusivity: I need to compare the calculations with the experimentally determined delay of the maximum. We measure ground temperature at different depths, below our ice/water tank but also in undisturbed ground: Temperature measured with Pt1000 sensors – comparing ground temperature at different depths, and the related ‘lag’. Indicated by vertical dotted lines, the approximate positions of maxima and minima. The lag is about 10-15 days. The lag derived from the figure is in the same order as the lag derived from the calculation and thus in accordance with my assumed thermal diffusivity: In 70cm depth, the temperature peak is delayed by about two weeks. ___________________________________________________ Appendix: Calculations and background. I am trying to give an outline of my solution, plus some ‘motivation’ of where the differential equation comes from. Heat transfer is governed by the same type of equation that describes also the diffusion of gas molecules or similar phenomena. Something lumped together in space slowly peters out, spatial irregularities are flattened. Or: The temporal change – the first derivative with respect to time – is ‘driven’ by a spatial curvature, the second derivative with respect to space. $\frac{\partial T}{\partial t} = D\frac{\partial^{2} T}{\partial x^{2}}$ This is the heat transfer equation for a region of space that does not have any sources or sinks of heat – places where heat energy would be created from ‘nothing’ or vanish – like an underground nuclear reaction (or freezing of ice). All we know about the material is covered by the constant D, called thermal diffusivity. The equation is based on local conservation of energy: The energy stored in a small volume of space can only change if something is created or removed within that volume (‘sources’) or if it flows out of the volume through its surface. This is a very general principles applicable to almost anything in physics. Without sources or sinks, this translates to: $\frac{\partial [energy\,density]}{\partial t} = -\frac{\partial \overrightarrow{[energy\,flow]}}{\partial x}$ The energy density [J/m3] stored in a volume of material by heating it up from some start temperature is proportional to temperature, proportionality factors being the mass density ρ [kg/m3] and the specific heat cp [J/kg] of this material. The energy flow per area [W/m2] is typically nearly proportional to the temperature gradient, the constant being heat conductivity κ [W/mK]. The gradient is the first-order derivative in space, so inserting all this we end with the second derivative in space. All three characteristic constants of the heat conducting material can be combined into one – the diffusivity mentioned before: $D = \frac{\kappa }{\varrho \, c_{p} }$ So changes in more than one of these parameters can compensate for each other; for example low density can compensate for low conductivity. I hinted at this when writing about heat conduction in our gigantic ice cube: Ice has a higher conductivity and a lower specific heat than water, thus a much higher diffusivity. I am considering a vast area of ground irradiated by the sun, so heat conduction will be one-dimensional and temperature changes only along the axis perpendicular to the surface. At the surface the temperature varies periodically throughout the year. t=0 is to be associated with beginning of August – our experimentally determined maximum – and the minimum is observed at the beginning of February. This assumption is just the boundary condition needed to solve this partial differential equation. The real ‘wavy’ variation of temperature is closed to a sine wave, which makes the calculation also very easy. As a physicist I have trained to used a complex exponential function rather than sine or cosine, keeping in mind that only real part describes the real world. This a legitimate choice, thanks to the linearity of the differential equation: $T(t,x=0) = T_{0} e^{i\omega t}$ with ω being the angular frequency corresponding to one year (2π/ω = 1 year). It oscillates about 0, with an amplitude of T0. But after all, the definition of 0°C is arbitrary and – again thanks to linearity – we can use this solution and just add a constant function to shift it to the desired value. A constant does neither change with space or time and thus solves the equation trivially. If you have more complicated sources or sinks, you would represent those mathematically as a composition of simpler ‘sources’, for each of which you can find a quick solution and then add up add the solutions, again thanks to linearity. We are lucky that our boundary condition consist just of one such simple harmonic wave, and we guess at the solution for all of space, adding a spatial wave to the temporal one. So this is the ansatz – an educated guess for the function that we hope to solve the differential equation: $T(t,x) = T_{0} e^{i\omega t + \beta x}$ It’s the temperature at the surface, multiplied by an exponential function. x is positive and increasing with depth. β is some number we don’t know yet. For x=0 it’s equal to the boundary temperature. Would it be a real, negative number, temperature would decrease exponentially with depth. The ansatz is inserted into the heat equation, and every differentiation with respect to either space or time just yields a factor; then the exponential function can be cancelled from the heat transfer equation. We end up with a constraint for the factor β: $i\omega = D\beta^{2}$ Taking the square root of the complex number, there would be two solutions: $\beta=\pm \sqrt{\frac{\omega}{2D}}(1+i))$ β has a real and an imaginary part: Using it in T(x,t) the real part corresponds to exponential ‘decay’ while the imaginary part is an oscillation (similar to the temporal one). Both real and imaginary parts of this function solve the equation (as any linear combination does). So we take the real part and insert β – only the solution for β with negative sign makes sense as the other one would describe temperature increasing to infinity. $T(t,x) = Re \left(T_{0}e^{i\omega t} e^{-\sqrt{\frac{\omega}{2D}}(1+i)x}\right)$ The thing in the exponent has to be dimension-less, so we can express the combinations of constants as characteristic lengths, and insert the definition of ω=2π/τ): $T(t,x) = T_{0} e^{-\frac{x}{l}}cos\left(2\pi\left(\frac {t} {\tau} -\frac{x}{\lambda }\right)\right)$ The two lengths are: • the wavelength of the oscillation $\lambda = \sqrt{4\pi D\tau }$ • and the attenuation length $l = \frac{\lambda}{2\pi} = \sqrt{\frac{D\tau}{\pi}}$ So the ratio between those lengths does not depend on the properties of the material and the wavelength is always much shorter than the attenuation length. That’s why there is hardly one period visible in the plots. The plots have been created with this parameters: • Heat conductivity κ = 0,0019 kW/mK • Density ρ = 2000 kg/m3 • Specific heat cp = 1,3 kJ/kgK • tau = 1 year = 8760 hours Thus: • Diffusivity D = 0,002631 m2/h • Wavelength λ = 17 m • Attenuation length l = 2,7 m The wave (any wave) propagates with a speed v equivalent to wavelength over period: v = λ / tau. $v = \frac{\lambda}{\tau} = \frac{\sqrt{4\pi D\tau}}{\tau} = \sqrt{\frac{4\pi D}{\tau}}$ The speed depends only on the period and the diffusivity. The maximum of the temperature as observed in a certain depth x is delayed by a time equal x over v. Cross-checking our measurements of the temperature T(30cm) and T(100cm), I would thus expect a delay by 0,7m / (17m/8760h) = 360 h = 15 days which is approximately in agreement with experiments (taking orders of magnitude). Note one thing though: Only the square root of D is needed in calculations, so any error I make in assumptions for D will be generously reduced. I have not yet included the geothermal linear temperature gradient in the calculation. Again we are grateful for linearity: A linear – zero-curvature – temperature profile that does not change with time is also a trivial solution of the equation that can be added to our special exponential solution. So the full solution shown in the plot is the sum of: • The damped oscillation (oscillating about 0°C) • Plus a constant denoting the true yearly average temperature • Plus a linear decrease with depth, the linear correction being 0 at the surface to meet the boundary condition. If there would be no geothermal gradient (thus no flow from beneath) the temperature at infinite distance (practically in 20m) would be the same as the average temperature of the surface. Daily changes could be taken into account by adding yet another solution that satisfies an amendment to the boundary condition: Daily fluctuations of temperatures would be superimposed to the yearly oscillations. The derivation would be exactly the same, just the period is different by a factor of 365. Since the characteristic lengths go with the square root of the period, yearly and daily lengths differ only by a factor of about 19. ________________________________________ Intro to geothermal energy: Where does geothermal energy come from? Geothermal gradient and energy of the earth: These data for bore holes using one scale show the gradient plus the disturbed surface region, with not much of a neutral zone in between. Theory of Heat Conduction Heat Transfer Equation on Wikipedia Textbook on Heat Conduction, available on archive.org in different formats. I have followed the derivation of temperature waves given in my favorite German physics book on Thermodynamics and Statistics, by my late theoretical physics professor Wilhelm Macke. This page quotes the classic on heat conduction, by Carlslaw and Jäger, plus the main results for the characteristic lengths. \| # Half a Year of Solar Power and Smart Metering Our PV generator and new metering setup is now operational for half a year; this is my next wall of figures. For the first time I am combining data from all our loggers (PV inverter, smart meter for consumption, and heat pump system’s monitoring), and I give a summary on our scrutinizing the building’s electrical power base load. For comparison: These are data for Eastern Austria (in sunny Burgenland). Our PV generator has 4.77kWp, 10 panels oriented south-east and 8 south-west. Typical yearly energy production in our place, about 48° latitude: ~ 5.300 kWh. In the first 6 months – May to November 2015 – we harvested about 4.000kWh. Our house (private home and office) meets the statistical average of an Austrian private home, that is about 3.500 kWh/year for appliances (excl. heating, and cooling is negligible here). We heat with a heat pump and need about 7.200kWh electrical energy per year in total. In the following plots daily and monthly energy balances are presented in three ways: 1. Total consumption of the building as the sum of the PV energy used immediately, and the energy from the utility. 2. The same total consumption as the sum of the heat pump compressor’s input energy and the remaining energy for appliances, computers, control etc. 3. Total energy generated by PV panels as the sum of energy used (same amount as contributions to 1) and the energy sold to the utility. In summer there is more PV energy available than needed and – even with a battery – the rest would needed to be fed into the grid. In October, heating season starts and more energy is needed by the heat pump that can be provided by solar energy. This is maybe demonstrated best by comparing the self-sufficiency quota (ratio of PV energy and energy consumed) and the self-consumption quota (ratio of PV energy consumed and PV production). Numbers ‘flip’ in October: In November we had some unusually hot record-breaking days while the weather became more typical at the end of the month: This is reflected in energy consumption: November 10 was nearly like a summer day, when the heat pump only had to heat hot water, but on the colder day it needed about 20kWh (resulting in 80-100kWh heating energy). In July, we had the chance to measure what the building without life-forms needs per day – the absolute minimum baseline. On July 10, 11, and 12 we were away and about 4kWh were consumed per day160W on average. Note that the 4kWh baseline is 2-3 times the energy the heat pump’s compressor needs for hot water heating every day: We catalogued all devices, googled for data sheets and measured power consumption, flipped a lot of switches, and watched the smart meter tracking the current consumption of each device. Consumption minus production: Current values when I started to write this post, the sun was about to set. In order to measure the consumption of individual devices they have been switched an of off one after the other, after sunset. We abandoned some gadgets and re-considered usage. But in this post I want to focus on the base load only – on all devices that contribute to the 160W baseline. As we know from quantum physics, the observing changes the result of the measurement. It was not a surprise that the devices used for measuring, monitoring and metering plus required IT infrastructure make up the main part of the base load. Control & IT base load – 79W • Network infrastructure, telephone, and data loggers – 35W: Internet provider’s DSL modem / router, our router + WLAN access point, switch, ISDN phone network termination, data loggers / ethernet gateways for our control unit, Uninterruptible Power Supply (UPS). • Control and monitoring unit for the heat pump system, controlling various valves and pumps: 12W. • The heat pump’s internal control: 10W • Three different power meters: 22W: 1) Siemens smart meter of the utility, 2) our own smart meter with data logger and WLAN, 3) dumb meter for overall electrical input energy of the heat pump (compressor plus auxiliary energy). The latter needs 8W despite its dumbness. Other household base load – 39W • Unobtrusive small gadgets – 12W: Electrical toothbrush, motion detectors, door bell, water softener, that obnoxious clock at the stove which is always wrong and can’t be turned off either, standby energy of microwave oven and of the PV generator’s inverter. • Refrigerator – 27W: 0,65 kWh per day. Non-essential IT networking infrastructure – 10W • WLAN access point and router for the base floor – for connecting the PV inverter and the smart meter and providing WLAN to all rooms. These are not required 24/7; you don’t lose data by turning them off. Remembering to turn off daily might be a challenge: Non-24/7 office devices – 21W. Now turned off with a flip switch every evening, and only turned on when needed. • Phones and headsets: 9W. • Scanner/Printer/Fax: 8W. Surprisingly, there was no difference between ‘standby’ and ‘turned off’ using the soft button – it always needs 8W unless you really disconnect it. • Server in hibernated state 4W. Note that it took a small hack of the operating system already to hibernate the server operating system at all. Years ago the server was on 24/7 and its energy consumption amounted to 500kWh a year. Stuff retired after this ‘project’ – 16W. • Radio alarm clock – 5W. Most useless consumption of energy ever. But this post is not meant as bragging about the smartest use of energy ever, but about providing a realistic account backed up by data. • Test and backup devices – 7W. Backup notebooks, charging all the time, backup router for playground subnet not really required 24/7, timer switch most likely needing more energy than it saved by switching something else. • Second old Uninterruptable Power Supply – 4W. used for one connected device only, in addition to the main one. It was purchased in the last century when peculiarities of the local power grid had rebooted computers every day at 4:00 PM. In total, we were able to reduce the base load by about 40W, 25% of the original value. This does not sound much – it is equivalent to a small light bulb. But on the other hand, it amounts to 350kWh / year, that is 10% of the yearly energy consumption! ___________________________ Logging setup: • Temperature / compressor’s electrical power: Universal control UVR1611 and C.M.I. as data logger, logging interval 90 seconds. Temperature sensor: PT1000. Power meter: CAN Energy Meter. Log files are exported daily to CSV files using Winsol. Logging interval: 90 seconds. • PV output power: Datamanager 2.0 included with PV inverter Fronius Symo 4.5-3-M, logging interval 5 minutes. • Consumed energy: Smart meter EM-210, logging interval 15 minutes. • CSV log files are imported into Microsoft SQL Server 2014 for analysis and consolidation. Plots are created with Microsoft Excel as front end to SQL Server, from daily and monthly views on joined UVR1611 / Fronius Symo / EM-210 tables. \| # Economics of the Solar Air Collector In the previous post I gave an overview of our recently compiled data for the heat pump system. The figure below, showing the seasonal performance factor and daily energy balances, gave rise to an interesting question: In February the solar collector was off for research purposes, and the performance factor was just a bit lower than in January. Does the small increase in performance – and the related modest decrease in costs of electrical energy – justify the investment of installing a solar/air collector? Monthly heating energy provided by the heat pump – total of both space heating and hot water, related electrical input energy, and the ratio = monthly performance factor. The SPF is in kWh/kWh. Daily energies: 1) Heating energy delivered by the heat pump. Heating energy = electrical energy + ambient energy from the tank. 2) Energy supplied by the collector to the water tank, turned off during the Ice Storage Challenge. Negative collector energies indicate cooling of the water tank by the collector during summer nights. 200 kWh peak in January: due to the warm winter storm ‘Felix’. Depending on desired pay-back time, it might not – but this is the ‘wrong question’ to ask. Without the solar collector, the performance factor would not have been higher than 4 before it was turned off; so you must not compare just these two months without taking into account the history of energy storage in the whole season. Bringing up the schematic again; the components active in space heating mode plus collector are highlighted: (1) Off-the-shelf heat pump. (2) Energy-efficient brine pump. (3) Underground water tank, can also be used as a cistern. (4) Ribbed pipe unglazed solar collector (5) 3-way valve: Diverting brine to flow through the collector, depending on ambient temperature. (6) Hot water is heated indirectly using a large heat exchanger in the tank. (7) Buffer tank with a heat exchanger for cooling. (8) Heating circuit pump and mixer, for controlling the supply temperature. (9) 3-way valve for switching to cooling mode. (10) 3-way valve for toggling between room heating and hot water heating. The combination of solar collector and tank is ‘the heat source’, but the primary energy source is ambient air. The unglazed collector allows for extracting energy from it efficiently. Without the tank this system would resemble an air heat pump system – albeit with a quiet heat exchanger instead of a ventilator. You would need the emergency heating element much more often in a typical middle European winter, resulting in a lower seasonal performance factor. We built this system also because it is more economical than a noisy and higher-maintenance air heat pump system in the long run. Our measurements over three years show that about 75%-80% of the energy extracted from the tank by the heat pump is delivered to it by the solar collector in the same period (see section ‘Ambient Energy’ in monthly and yearly overviews). The remaining energy is from surrounding ground or freezing water. The water tank is a buffer for periods of a few very cold days or weeks. So the solar/air collector is an essential component – not an option. In Oct, Nov, and March typically all the energy needed for heating is harvested by the solar collector in the same month. In ‘Ice Months’ Dec, Jan, Feb freezing of water provides for the difference. The ice cube is melted again in the remaining months, by the surplus of solar / air energy – in summer delivered indirectly via ground. The winter 2014/2015 had been unusually mild, so we had hardly created any ice before February. The collector had managed to replenish the energy quickly, even in December and January. The plot of daily energies over time show that the energy harvested by the collector in these months is only a bit lower than the heating energy consumed by the house! So the energy in the tank was filled to the brim before we turned the collector off on February 1. Had the winter been harsher we might have had 10 m3 of ice already on that day, and we might have needed 140kWh per day of heating energy, rather than 75kWh. We would have encountered the phenomena noted during the Ice Storage Challenge earlier. This post has been written by Elke Stangl, on her blog. Just adding this in case the post gets stolen in its entirety again, as it happened to other posts tagged with ‘Solar’ recently. # Having Survived the Hottest July Ever (Thanks, Natural Cooling!) July 2015 was the hottest July ever since meteorological data had been recorded in Austria (since 248 years). We had more than 38°C ambient air temperature at some days; so finally a chance to stress-test our heat pump system’s cooling option. Heating versus cooling mode In space heating ‘winter’ mode, the heat pump extracts heat from the heat source – a combination of underground water / ice tank and unglazed solar collector – and heats the bulk volume of the buffer storage tank. We have two heating circuits exchanging heat with this tank – one for the classical old radiators in ground floor, and one for the floor heating loops in the first floor – our repurposed attic. Space heating mode: The heat pump (1) heats the buffer tank (7), which in turn heats the heating circuits (only one circuit shown, each has its circuit pump and mixer control). Heat source: Solar/air collector (4) and water / ice storage (3) connected in a single brine circuit. The heat exchanger in the tank is built from the same ribbed pipes as the solar collector. If the ambient temperature is too low too allow for harvesting of energy the 3-way valve (5) makes the brine flow bypass the collector. The heat pump either heats the buffer tank for space heating, or the hygienic tank for hot tap water. (This posting has a plot with heating power versus time for both modes). We heat hot tap water indirectly, using a hygienic storage tank with a large internal heat exchanger. Therefore we don’t need to fight legionella by heating to high temperatures, and we only need to heat the bulk volume of the tank to 50°C – which keeps the Coefficient of Performance high. Hot tap water heating mode: The flow of water heated by the heat pump is diverted to the hygienic storage tank (6). Otherwise, the heat source is used in the same way as for space heating. In this picture, the collector is ‘turned off’ – corresponding to heating water on e.g. a very cold winter evening. In summer, the still rather cold underground water tank can be used for cooling. Our floor heating loops become cooling loops and we simply use the cool water or ice in the underground tank for natural (‘passive’) cooling. So the heat pump can keep heating water – this is different from systems that turn an air-air heat pump into an air conditioner by reverting the cycle of the refrigerant. Heating hot water in parallel to cooling is beneficial as the heat pump extracts heat from the underground tank and cools it further! Cooling mode: Via automated 3-way valve (9) brine is diverted to flow through the heat exchanger in the buffer tank (7). Water in the buffer tank is cooled down so water in the floor ‘heating’ / cooling loops. If the heat pump operates in parallel to heat hot tap water, it cools the brine. How we optimize cooling power this summer Water tank temperature. You could tweak the control to keep the large ice cube as long as possible, but there is a the trade-off: The cooler the tank, the lower the heat pump’s performance factor in heating mode. This year we kept the tank at 8°C after ‘ice season’ as long as possible. To achieve this, the solar collector is bypassed if ambient temperature is ‘too high’. The temperature in the tank rose quickly in April – so our ice is long melted: The red arrow indicates the end of the ice period; then the set temperature of the tank was 8°C (‘Ice storage tank’ is rather a common term denoting this type of heat source than indicating that it really contains ice all the time.) Green arrows indicate three spells of hot weather. The tank’s temperature increased gradually, being heating by the surrounding ground and by space cooling. At the beginning of August its temperature is close to 20°C, so cooling energy has nearly be used up completely. At the beginning of July the minimum inlet temperature in the floor loops was 17°C, determined by the dew point (monitored by our control system that controls the mixer accordingly); at the end of the month maximum daily ambient air temperatures were greater than 35°C, and the cooling water had about 21°C. Room temperature. Cooling was activated only if the room temperature in the 1st floor was higher than 24°C – this allows for keeping as much cooling energy as possible for the really hot periods. We feel that 25°C in the office is absolutely OK as temperatures outside are more then 10°C higher. Scheduling hot water heating. After the installation of our PV panels we set the hot water heating time slots to periods with high solar radiation – when you have more than 2 kW output power on cloudless days. So we utilized the solar energy generator in the most economic way and the heat pump supports cooling exactly when cooling is needed. Using the collector for cooling in the night. If the ambient temperature drops to a value lower than the tank temperature, the solar collector can actually cool the tank! Ventilation. I have been asked if we have forced ventilation, ductwork, and automated awnings etc. No, we haven’t – we just open all the windows during the night and ‘manually operated’ shades attached to the outside of the windows. We call them the Deflector Shields: Manually operated ventilation – to be shut off at sunrise. We had already 30°C air temperature at 08:00 AM on some days. South-east deflector shields down. We feel there is still enough light in the (single large) room as we only activate the subset of shields facing the sun directly. These are details for two typical hot days in July: The blue line exhibits the cooling power measured for the brine ‘cooling’ circuit. If the heat pump is off, cooling power is about 1 kW; during heat pump operations (blue arrows) 4 kW cooling power can be obtained. Night-time ventilation is crucial to keep room temperatures at reasonable levels. The cooling power is lower than so-called standard cooling load as defined in AC standards – the power required to keep the temperature at about 24°C in steady-state conditions, when ambient temperature would be 30°C and no shades are used. For our attic-office this standard cooling power would amount to more than 10 kW which is higher than the standard (worst case) heating load in winter. Overall electrical energy balance I have been asked for a comparison of the energy needed in the house, the heat pump in particular, and the energy delivered by the PV panels and fed in to the grid. PV numbers in July were not much different from June’s – here is the overview on June and July, maximum PV power on cloudless days has decreased further due to the higher temperatures: In July, our daily consumption slightly decreased to 9-10 kWh per day, the heat pump needs 1-2 kWh of that. The generator provides for 23 kWh per day, Currently the weather forecast says, we will have more than 35°C each noon and 20-25° minimum in the night until end of this week. We might experience the utter depletion of our cooling energy storage before it will be replenished again on a rainy next weekend.	2019-06-18 00:03:32	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 12, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5588032603263855, "perplexity": 1622.7776828069773}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-26/segments/1560627998581.65/warc/CC-MAIN-20190617223249-20190618005249-00136.warc.gz"}
https://cforall.uwaterloo.ca/trac/changeset/cebfcb8e306e305227ed83ed9f13328a110bc492/	Changeset cebfcb8 Ignore: Timestamp: Sep 24, 2020, 3:27:02 PM (2 years ago) Branches: arm-eh, enum, forall-pointer-decay, jacob/cs343-translation, master, new-ast-unique-expr, pthread-emulation, qualifiedEnum Children: 10d78f1, 6cc913e Parents: 9e620b6 Message: convert Fangren's co-op report to latex and proofread Location: doc/theses/fangren_yu_COOP_S20 Files:	2022-12-02 07:13:42	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.4654342234134674, "perplexity": 5087.225257672332}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-49/segments/1669446710898.93/warc/CC-MAIN-20221202050510-20221202080510-00644.warc.gz"}
https://www.physicsforums.com/threads/moment-of-inertia.71932/	# Moment of inertia 1. Apr 17, 2005 ### stunner5000pt First of all i have a final exam tomorrow on Classical Mechanics - Cna someone point out a place that has the derivations of the moments of inertia for various objects Now if there a disc of mass M spinning about an axis taht is perpendicular to the plane of the disc, and the plane of the disc is horizontal (parallel to earth's surface) then it's moemnt of inertia is $\frac{1}{2} MR^2$ if there was a little mass located at a point that is r, where r<R from the center of the disc then the moment of inertia is $I = \frac{1}{2} MR^2 + mr^2$ is this correct?? Also , in class my prof said that on the exam he would have a question in which we would have to calculate a work integral... what is that ?? as far as im concerned $W = \int F \cdot d$ is there anything more to it?? Can you point out an example of something that is more complicated liek that?? Last edited: Apr 17, 2005 2. Apr 17, 2005 ### dextercioby So the little mass is standing on the disk?It's okay,then... Daniel. Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook Have something to add?	2016-10-22 05:49:01	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5988146662712097, "perplexity": 683.2519558133474}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-44/segments/1476988718426.35/warc/CC-MAIN-20161020183838-00423-ip-10-171-6-4.ec2.internal.warc.gz"}
https://itprospt.com/num/18524039/question-1ed-particle-x27-position-vectoras-function-cftime	5 # Question 1ed particle' position vectoras function cftime ziven by F) = t"i+2t*j+tk plue particle position Vector?s functicn of time given by F(t) = 2t&quo... ## Question #### Similar Solved Questions ##### The following is an alpha plot for monoprotic weak acid HAHA(1) Estimate the pH at which is there s0.50 mixture of acid and conjugate base?(2) AtpH = 7.0. most of the substance is in the form of_ 1. HA 6.4"What is the approximate fractional abundance of the conjugate base at pH = 4.02(4) Estimate the value of Ka for the acid. The following is an alpha plot for monoprotic weak acid HA HA (1) Estimate the pH at which is there s0.50 mixture of acid and conjugate base? (2) AtpH = 7.0. most of the substance is in the form of_ 1. HA 6.4" What is the approximate fractional abundance of the conjugate base at pH = 4.02 (4) E... ##### Question 8 (Mandatory) (6 points) Match up the following:Not reject HO when HO is falseReject HO when HO is false; or not reject HO when HO is not falseType 1 errorType 2 errorThe power of a test P(rejecting HOIHO false)Not an errorThe risk we are willing to take of a type 1 error; or the type 1 error rateReject HO when HO is true1 - 8The probability of a type 2 error Question 8 (Mandatory) (6 points) Match up the following: Not reject HO when HO is false Reject HO when HO is false; or not reject HO when HO is not false Type 1 error Type 2 error The power of a test P(rejecting HOIHO false) Not an error The risk we are willing to take of a type 1 error; or the typ... ##### Llst these electron subshells orderIncreasino energy_2p, 4s, 3s, SpNote for advanced students: vou may Jssumc these subshells are all In an atom with many electrons; not hydrogen atom00do Llst these electron subshells order Increasino energy_ 2p, 4s, 3s, Sp Note for advanced students: vou may Jssumc these subshells are all In an atom with many electrons; not hydrogen atom 00do... ##### Seplemder Of zuuO, heart transplantation at St: concern that more patients Georges Hospital in London was suspended were dying in the last 10 cases at the hospital than previously: Newspapers reported that the 80% mc was of particular concern because it five - 2 stu average: Let the random variable Xrepresent was Over times the the number of deaths in ques Suppose Ithat the probability death random sample of 10 cases at this hospital is equal to the national rate of 15%. I(a) Identify the prob Seplemder Of zuuO, heart transplantation at St: concern that more patients Georges Hospital in London was suspended were dying in the last 10 cases at the hospital than previously: Newspapers reported that the 80% mc was of particular concern because it five - 2 stu average: Let the random variable ... ##### Find the component of b along & for the vectors & (0,1,0) ad (-7,10,9}Select one:0 a; compab = 9 b. comn 6 = 10 compab = 7 compab =l None of the above Find the component of b along & for the vectors & (0,1,0) ad (-7,10,9} Select one: 0 a; compab = 9 b. comn 6 = 10 compab = 7 compab =l None of the above... ##### 12.50 mLofa 7.28MHzSO4 solution is diluted to a total volume of 1.2 L. What is the concentration of the diluted solution? Report answer to two significant figures! 12.50 mLofa 7.28MHzSO4 solution is diluted to a total volume of 1.2 L. What is the concentration of the diluted solution? Report answer to two significant figures!... ##### Use the ratio test to determine if the series converges (-1)"(Z) 8n (Zn + 1)Use the root test to determine the series converges775)"Determine using an appropriate test from the chapter if the series converges E5-] ( 4"9 17. En-1,+9 Use the ratio test to determine if the series converges (-1)"(Z) 8n (Zn + 1) Use the root test to determine the series converges 775)" Determine using an appropriate test from the chapter if the series converges E5-] ( 4"9 17. En-1,+9... ##### Point) Let k,h be unknown constants and consider the linear system:6 y 6 ~ 5y + 2~ 38 y h ~4x 19xThis system has unique solution whenever h #0If h is the (correct) value entered above, then the above system will be consistent for how many value(s) of k? OA. infinitely many values OB. unique value no values point) Let k,h be unknown constants and consider the linear system: 6 y 6 ~ 5y + 2~ 38 y h ~ 4x 19x This system has unique solution whenever h #0 If h is the (correct) value entered above, then the above system will be consistent for how many value(s) of k? OA. infinitely many values OB. unique valu... ##### Aszume that â‚¬=y-Inkx); and n =X-y, then U_ =0 a1 Xug+unn- 2uM 4 Zu6 X 0 6 1 X ug +Unn 2uf X Xug+umn + 2u0n 3u Xu# +Unn 7u + 2 7u& +Unn 1 Uinux +Un70 Aszume that â‚¬=y-Inkx); and n =X-y, then U_ = 0 a 1 Xug+unn- 2uM 4 Zu6 X 0 6 1 X ug +Unn 2uf X Xug+umn + 2u0n 3u X u# +Unn 7u + 2 7u& +Unn 1 Uin ux +Un 70... ##### (6 pts) Bart has money invested in Iwo funds: stock fund and bond fund. Last year the stock find paid 8"_ the bond fund paid 2% and he received S780 This year the stock fund paid 10%. the bond fund paid 1% and he received S810. How much money does he have in cach fund? You must define two variables (write them down: Let T amount at 8%" . etc_ set up syslem of two linear equations in these variables. solve the system by hand, and write the solution in complete sentence(6 pts) A minor le (6 pts) Bart has money invested in Iwo funds: stock fund and bond fund. Last year the stock find paid 8"_ the bond fund paid 2% and he received S780 This year the stock fund paid 10%. the bond fund paid 1% and he received S810. How much money does he have in cach fund? You must define two varia... ##### Calculate the future value of a savings account for a S600 quarterly deposit where annual interest is compounded quarterly at 5.6% for 3 years: (3)You decide that you want to have S100 000 in the bank 20 years from now: What will be your regular monthly deposits into a bank account that offers 3% annual interest compounded monthly: (3)Every six months, you deposit S750 into a charitable savings account that will pay 7% interest compounded semi-annually: How much is in the account in 4 years? How Calculate the future value of a savings account for a S600 quarterly deposit where annual interest is compounded quarterly at 5.6% for 3 years: (3) You decide that you want to have S100 000 in the bank 20 years from now: What will be your regular monthly deposits into a bank account that offers 3% a... ##### A salesman is standing on the Golden Gate Bridge in a traffic jam. He is at a height of $71.8 \mathrm{~m}$ above the water below. He receives a call on his cell phone that makes him so mad that he throws his phone horizontally off the bridge with a speed of $23.7 \mathrm{~m} / \mathrm{s}$ a) How far does the cell phone travel horizontally before hitting the water? b) What is the speed with which the phone hits the water? A salesman is standing on the Golden Gate Bridge in a traffic jam. He is at a height of $71.8 \mathrm{~m}$ above the water below. He receives a call on his cell phone that makes him so mad that he throws his phone horizontally off the bridge with a speed of $23.7 \mathrm{~m} / \mathrm{s}$ a) How far... ##### Model SummaryStd. Error of the EstimateModelR SquareAdjusted R Square444840r5505312.66813ANOVAMean SquareModelSum of SquaresdfSig.Regression155077537.260.001Residual232511220.8Total3875114CoefficientsStandardized Unstandardized Coefficients CoefficientsModelSig:Std. ErrorBeta(Constant)3.0181.0372.911004Dosage.279010069-2.802005Age0270181061.498136 Model Summary Std. Error of the Estimate Model R Square Adjusted R Square 444840r 550 531 2.66813 ANOVA Mean Square Model Sum of Squares df Sig. Regression 1550 775 37.26 0.001 Residual 2325 112 20.8 Total 3875 114 Coefficients Standardized Unstandardized Coefficients Coefficients Model Sig: Std. Er... ##### Evaluate the integral f Jr Ie dV where R is the region between the sphere of radius and the sphere of radius centered at the oriigin and above the x-Y plane_ Evaluate the integral f Jr Ie dV where R is the region between the sphere of radius and the sphere of radius centered at the oriigin and above the x-Y plane_... ##### Use a graphing utility to graph the rational function. Determine the domain of the function and identify any asymptotes.$$y= rac{2 x^{2}+x}{x+1}$$ Use a graphing utility to graph the rational function. Determine the domain of the function and identify any asymptotes. $$y=\frac{2 x^{2}+x}{x+1}$$...	2022-07-05 03:32:30	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6111493110656738, "perplexity": 3702.202581369798}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-27/segments/1656104512702.80/warc/CC-MAIN-20220705022909-20220705052909-00245.warc.gz"}
https://math.stackexchange.com/questions/2072693/matrix-function-converges-how-about-the-eigenvalues	# Matrix function converges, how about the eigenvalues? Suppose I have a matrix function $A(t)$ with $$\lVert A(t) - B\rVert \le ct^\alpha$$ in some matrix norm (this will work for any norm, I guess). So, in a sense $A(t)\rightarrow B$ for $t\rightarrow 0$ in $\mathcal{O}(t^\alpha)$. Plus, we have $A(0) = B$. I happen to know the eigenvalues of $B$, but I don't know a thing about the eigenvalues of $A(t)$. Plus, $A(t)$ does not have any favorable structure, in particular, no symmetry. So, what can you say about the eigenvalues of $A(t)$? In particular: • What about the spectral radius of $\lambda_{A(t)}$? Does it converge to the spectral radius of $B$? • Do we have $\lambda_{A(t)}\rightarrow\lambda_B$ for $t\rightarrow 0$ for all eigenvalues $\lambda_{A(t)}$ of $A(t)$? • And finally: is the speed of convergence $\mathcal{O}(t^\alpha)$ the same for the eigenvalues/spectral radius as for the matrix function? The last question is actually the most important one. If the eigenvalues of $B$ are all zero, the eigenvalues as well as the spectral radius of $A(t)$ would go to zero as $\mathcal{O}(t^\alpha)$... Any help would be appreciated, incl. references to (standard?) textbooks or papers on this matter. Maybe there is a counterexample? So far, in all numerical examples I have seen/done, all properties above do hold. Edit: To provide a bit more background: The matrices $A(t)$ are iteration matrices which depend on a time-step size $t$. They are not this ugly, but showing convergence of this iteration has proven to be rather difficult. In the simplest case, they look like $$A(t) = (I-tQ_1)^{-1}(t(Q_2-Q_1)+B)$$ with identity matrix $I$ and some matrices $Q_1,Q_2$, which do not have any particular structure we were able to exploit so far. Now, if I can make that conclusion about the spectral radius as described above, I can state that the spectral radius is smaller than 1, i.e. the iteration converges, if the time-step size $t$ is small enough. Edit: Does this answer help? Also, this question might be related to perturbation theory for eigenvalue problems (with non-symmetric matrices, though, and $B$ is not diagonalizable). ## 3 Answers Parts (1) and (2) -- Yes. The coefficients of the characteristic polynomial are continuous functions of $A(t)$ (they are polynomials in the entries of $A(t)$!) and the roots of a polynomial are continuous functions of the coefficients. Part (3): If $B$ has non-trivial Jordan blocks, this can fail. For example, $$\begin{pmatrix} 0 & 1 \\ -t & 0 \end{pmatrix} = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} + O(t)$$ as $t \to 0$, but the eigenvalues are $\pm \sqrt{t}$, which is not $O(t)$. I believe that this is true when $B$ is diagonalizable. (I know that it is diagonalizable if $B$ has distinct eigenvalues, and I'll write up that argument if this one fails.) We may as well assume $B$ is diagonalized. Let's concentrate on a particular eigenvalue $\lambda$ of $B$, with multiplicity $m$; we might as well assume $\lambda =0$. So $B = \mathrm{diag}(0,0,\cdots,0, \lambda_2, \lambda_3, ..., \lambda_k)$ with $m$ zeroes. Suppose that $B-A(t)$ is $O(t^a)$. Explicitly expanding the determinant, the coefficient of $x^k$ in $\det(x \mathrm{Id} - B)$ is $O(t^{a(m-k)})$ for $k < m$, and the coefficient of $x^m$ does not go to $0$ as $t \to 0$. So the Newton polygon of the characteristic polynomial passes through $(m,0)$ and stays above the line from $(m,0)$ to $(0,ma)$. This shows that the bottom $m$ roots of the characteristic polynomial vanish at rate $O(t^a)$ or faster as $t \to 0$ (and the other roots do not vanish.) It occurs to me that it is worth sketching the argument for the Newton polygon claim directly so you can see how straightforward it is without learning the whole Newton polygon technology. Here is what I am claiming: Lemma Fix $C>0$. Then there is a $D>0$ such that, if $f(u,z) = \sum_i f_i(u) z^i$ is holomorphic in $z$ on the disc of radius $1$, with $$\|f_i(u)/f_m(u)\| < \begin{cases} C u^{m-i} & i < m \\ C & i > m \end{cases}$$ then $f(u, \ )$ has $m$ roots in the disc of radius $D u$ for all sufficiently small $u$. Proof sketch Consider $\oint \tfrac{f'}{f} dz$ where the integral is around the circle of radius $Du$. Write $f = f_m(u) z^m (1+\mbox{other terms})$ and $f' = m f_m(u) z^{m-1} (1+\mbox{other terms})$ so the integral is $\oint m \tfrac{dz}{z} + \mbox{other terms}$. Use the above conditions to bound the other terms. • Thanks a lot for your elaborate answer! Also, nice counterexample! The thing is: $B$ is not diagonalizable, in the simplest case all eigenvalues of $B$ are actually zero. More precisely, $B$ is a strictly lower triangular matrix, while $A(t)$ is block-triangular, with blocks on the diagonal and stuff below. Is there anything we can do in this case? I'd be happy to provide more information, but this would require a bit more "space".. – Robert Speck Jan 3 '17 at 8:34 • So, it sounds like your example is exactly of the form of my counter-example. I imagine the general statement should be that if $B$ has a nilpotent Jordan block of size $k$, and $A=B+O(u)$, then the corresponding eigenvalues of $A$ are $O(u^{1/k})$. Think about $\left( \begin{smallmatrix} 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ t & 0 & 0 & 0 \end{smallmatrix} \right)$ to see that we can't hope for better. – David E Speyer Jan 3 '17 at 14:53 • Does it matter that $B$ in your example would be strictly upper triangular but my $B$ is strictly lower triangular..? I guess not, right? – Robert Speck Jan 3 '17 at 15:12 • Take the transpose of the example in my previous comment. – David E Speyer Jan 3 '17 at 15:14 • Darn.. you're right of course. – Robert Speck Jan 3 '17 at 15:19 This is not answer. Just trying to make observations. Note that when $t$ is very close to zero, you have $$(I-tQ_1)^{-1}=I+tQ_1+t^2Q_1^2+\dots$$ Let $C=Q_2-Q1$. Then $$A(t) =(tI+t^2Q_1+t^3Q_1^2+..)C~+~(I+tQ_1+t^2Q_1^2+\dots)B$$ Now, let's look at the matrix $(I-tQ_1)^{-1}$. Let $Q_1 = T\Lambda T^{-1}$ for some invertible matrix $T$. Thus, if $\lambda_i(Q_1)$ is the $i^{th}$ eigenvalue of $Q_1$. Then we have $$\lambda_i((I-tQ_1)^{-1})=1+\lambda_it+\lambda_i^2t^2+\dots$$ Frankly, I don't know what conclusions you can draw from these. Hope it helps you or anyone else trying to answer these in some way. • Thanks for contributing! I like the Neumann approach and maybe the alternative form of $A(t)$ does help here. – Robert Speck Dec 28 '16 at 7:58 • @RobertSpeck Kato's Perturbation Theory for Linear Operators discusses perturbation theory for matrices that are holomorphic in some domain. – Keith McClary Dec 29 '16 at 4:48 The statement that eigenvalues are continuous functions of the entries of matrices is often seen in the literature. What does it really mean? Actually, "eigenvalue continuity" is interpreted in two different meanings: in the sense of topology and in the sense of individual functions. In the sense of topology: eigenvalues as a whole (unordered multisets) are continuous. Let A(t) approach A entrywise (say), then the eigenvalues of A(t) approach the eigenvalues of A with suitable matching. (The mapping from the matrix space to the quotient space C^n/~ is continuous.) In the sense of individual functions: If the entries of the matrix are continuous functions over a real interval or or all the eigenvalues are real, then there is a selection of continuous functions that constitute the eigenvalues of the matrix. Note: Eigenvalues are always continuous in the sense of topology. However, it may be impossible to parameterize eigenvalues as continuous individual functions. For example, A=[0 1; z, 0] where z belongs to a domain that contains the origin. These can be found in Kato's book Perturbation Theory for Linear Operators and in Bhatia's book Matrix Analysis.	2019-09-22 14:58:00	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9468457698822021, "perplexity": 138.35479140150375}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-39/segments/1568514575515.93/warc/CC-MAIN-20190922135356-20190922161356-00498.warc.gz"}
https://www.physicsforums.com/threads/can-a-group-be-isomorphic-to-one-of-its-quotients.1000667/	# Can a group be isomorphic to one of its quotients? • I Science Advisor Gold Member Of course it must be an infinite group, otherwise \|G/N\|=\|G\|/\|N\| and then {e} is the only ( and trivial) solution. I understand there is a result that for every quotient Q:=G/N there is a subgroup H that is isomorphic to Q. Is that the case? ## Answers and Replies Science Advisor Gold Member How about an infinite product ##\mathbb{Z}\times\mathbb{Z}\times...## and quotient out by the first factor? I understand there is a result that for every quotient Q:=G/N there is a subgroup H that is isomorphic to Q. Is that the case? No, there is no subgroup of ##\mathbb{Z}## that is isomorphic to ##\mathbb{Z}/2\mathbb{Z}.## heff001, mathwonk, graphking and 2 others Science Advisor Gold Member How about an infinite product ##\mathbb{Z}\times\mathbb{Z}\times...## and quotient out by the first factor? No, there is no subgroup of ##\mathbb{Z}## that is isomorphic to ##\mathbb{Z}/2\mathbb{Z}.## Thanks. Sorry, I believe the result I quoted (may)apply to infinite groups. Science Advisor Gold Member Thanks. Sorry, I believe the result I quoted (may)apply to infinite groups. No, even if the quotient is infinite, it is still false. There is no subgroup of ##\mathbb{Z}\times\mathbb{Z}## that is isomorphic to ##\left(\mathbb{Z}\times\mathbb{Z}\right)/\left(\{0\}\times 2\mathbb{Z}\right)\cong\mathbb{Z}\times\left(\mathbb{Z}/2\mathbb{Z}\right).## WWGD Science Advisor Gold Member Re the lack of correspondence between subgroups and quotient groups, I though of another argument: just take a simple group. It will have non-trivial subgroups but no non-trivial quotient. Maybe simplest vase is ##S_5##, the permutation group on 5 elements . It has the alternating subgroup, which cannot be a quotient by cardinality reasons.	2022-12-04 01:53:46	{"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9012054204940796, "perplexity": 1308.2534065551426}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-49/segments/1669446710953.78/warc/CC-MAIN-20221204004054-20221204034054-00572.warc.gz"}
https://realnfo.com/ee/Electrical-Circuit-Analysis/Pulse-Waveform-and-the-RC-Response/Oscilloscope-Attenuator	# Oscilloscope Attenuator The $\times 10$ attenuator probe employed with oscilloscopes is designed to reduce the magnitude of the input voltage by a factor of 10 . If the input impedance to a scope is $1 \mathrm{M} \Omega$, the $\times 10$ attenuator probe will have an internal resistance of $9 \mathrm{M} \Omega$, as shown in Fig. 1. Fig. 1: $\times 10$ attenuator probe. Applying the voltage divider rule, $$V_{\text {scope }}=\frac{(1 \mathrm{M} \Omega)\left(V_{i}\right)}{1 \mathrm{M} \Omega+9 \mathrm{M} \Omega}=\frac{1}{10} V_{i}$$ In addition to the input resistance, oscilloscopes have some internal input capacitance, and the probe will add an additional capacitance in parallel with the oscilloscope capacitance, as shown in Fig. 2. The probe capacitance is typically about $10 \mathrm{pF}$ for a 1-m (3.3-ft) cable, reaching about $15 \mathrm{pF}$ for a 3-m (9.9-ft) cable. The total input capacitance is therefore the sum of the two capacitive elements, resulting in the equivalent network of Fig. 3. Fig. 2: Capacitive elements present in an attenuator probe arrangement. Fig. 3: Equivalent network of Fig. $2$. For the analysis to follow, let us determine the Thévenin equivalent circuit for the capacitor $C_{i}$ : \begin{aligned} E_{T h}&=\frac{(1 \mathrm{M} \Omega)\left(V_{i}\right)}{1 \mathrm{M} \Omega+9 \mathrm{M} \Omega}=\frac{1}{10} V_{i} \\ \text { and } \quad R_{T h} & =9 \mathrm{M} \Omega \\| 1 \mathrm{M} \Omega=0.9 \mathrm{M} \Omega \end{aligned} The Thévenin network is shown in Fig. $4$. For $V_{i}=200 \mathrm{~V}$ (peak), $$E_{T h}=0.1 v_{i}=20 \mathrm{~V} \text { (peak) }$$ and for $v_{C}, V_{f}=20 \mathrm{~V}$ and $V_{i}=0 \mathrm{~V}$, with $$\tau=R C=\left(0.9 \times 10^{6} \Omega \right)\left(30 \times 10^{-12} \mathrm{~F} \right)=27 \mu \mathrm{s}$$ For an applied frequency of $5 \mathrm{kHz}$, $$T=\frac{1}{f}=0.2 \mathrm{~ms} \quad \text { and } \quad \frac{T}{2}=0.1 \mathrm{~ms}=100 \mu \mathrm{s}$$ with $5 \tau=135 \mu \mathrm{s}>100 \mu \mathrm{s}$, as shown in Fig. 5, clearly producing a severe rounding distortion of the square wave and a poor representation of the applied signal. Fig. 4: Thévenin equivalent for Ci of Fig. 3. Fig. 5: The scope pattern for the conditions of Fig. 3 with vi = 200 V peak.	2023-03-28 08:58:07	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.4091849625110626, "perplexity": 1810.858622733043}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-14/segments/1679296948817.15/warc/CC-MAIN-20230328073515-20230328103515-00357.warc.gz"}
https://socratic.org/questions/how-do-you-simplify-2x-2y-4-4x-2y-4-3x-3x-3y-2-and-write-it-using-only-positive-	# How do you simplify (2x^2y^44x^2y^43x)/(3x^-3y^2) and write it using only positive exponents? Feb 17, 2017 See the entire simplification process below: #### Explanation: First, to simplify the numerator we will rewrite this expression as: $\frac{\left(2 \cdot 3 \cdot 4\right) \left({x}^{2} \cdot {x}^{2} \cdot x\right) \left({y}^{4} \cdot {y}^{4}\right)}{3 {x}^{-} 3 {y}^{2}} \to \frac{24 \left({x}^{2} \cdot {x}^{2} \cdot x\right) \left({y}^{4} \cdot {y}^{4}\right)}{3 {x}^{-} 3 {y}^{2}}$ We can now use these two rules for exponents to simplify the numerator: $a = {a}^{\textcolor{red}{1}}$ and ${x}^{\textcolor{red}{a}} \times {x}^{\textcolor{b l u e}{b}} = {x}^{\textcolor{red}{a} + \textcolor{b l u e}{b}}$ $\frac{24 \left({x}^{2} \cdot {x}^{2} \cdot {x}^{\textcolor{red}{1}}\right) \left({y}^{4} \cdot {y}^{4}\right)}{3 {x}^{-} 3 {y}^{2}} \to \frac{24 {x}^{2 + 2 + 1} {y}^{4 + 4}}{3 {x}^{-} 3 {y}^{2}} \to \frac{24 {x}^{5} {y}^{8}}{3 {x}^{-} 3 {y}^{2}} \to$ $\frac{8 {x}^{5} {y}^{8}}{{x}^{-} 3 {y}^{2}}$ We can now use this rule of exponents to complete the simplification: ${x}^{\textcolor{red}{a}} / {x}^{\textcolor{b l u e}{b}} = {x}^{\textcolor{red}{a} - \textcolor{b l u e}{b}}$ $\frac{8 {x}^{\textcolor{red}{5}} {y}^{\textcolor{red}{8}}}{{x}^{\textcolor{b l u e}{- 3}} {y}^{\textcolor{b l u e}{2}}} \to 8 {x}^{\textcolor{red}{5} - \textcolor{b l u e}{- 3}} {y}^{\textcolor{red}{8} - \textcolor{b l u e}{2}} \to 8 {x}^{\textcolor{red}{5} + \textcolor{b l u e}{3}} {y}^{\textcolor{red}{8} - \textcolor{b l u e}{2}} \to$ $8 {x}^{8} {y}^{6}$	2019-10-23 15:11:21	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 8, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9756302237510681, "perplexity": 1051.2167589997061}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-43/segments/1570987834649.58/warc/CC-MAIN-20191023150047-20191023173547-00266.warc.gz"}
https://www.quizover.com/physics/section/problems-exercises-dark-matter-and-closure-by-openstax	# 12.4 Dark matter and closure (Page 4/16) Page 4 / 16 Neutrino oscillations may also explain the low number of observed solar neutrinos. Detectors for observing solar neutrinos are specifically designed to detect electron neutrinos ${\nu }_{e}$ produced in huge numbers by fusion in the Sun. A large fraction of electron neutrinos ${\nu }_{e}$ may be changing flavor to muon neutrinos ${v}_{\mu }$ on their way out of the Sun, possibly enhanced by specific interactions, reducing the flux of electron neutrinos to observed levels. There is also a discrepancy in observations of neutrinos produced in cosmic ray showers. While these showers of radiation produced by extremely energetic cosmic rays should contain twice as many ${v}_{\mu }$ s as ${\nu }_{e}$ s, their numbers are nearly equal. This may be explained by neutrino oscillations from muon flavor to electron flavor. Massive neutrinos are a particularly appealing possibility for explaining dark matter, since their existence is consistent with a large body of known information and explains more than dark matter. The question is not settled at this writing. The most radical proposal to explain dark matter is that it consists of previously unknown leptons (sometimes obtusely referred to as non-baryonic matter). These are called weakly interacting massive particles , or WIMPs , and would also be chargeless, thus interacting negligibly with normal matter, except through gravitation. One proposed group of WIMPs would have masses several orders of magnitude greater than nucleons and are sometimes called neutralinos . Others are called axions and would have masses about ${\text{10}}^{-\text{10}}$ that of an electron mass. Both neutralinos and axions would be gravitationally attached to galaxies, but because they are chargeless and only feel the weak force, they would be in a halo rather than interact and coalesce into spirals, and so on, like normal matter (see [link] ). Some particle theorists have built WIMPs into their unified force theories and into the inflationary scenario of the evolution of the universe so popular today. These particles would have been produced in just the correct numbers to make the universe flat, shortly after the Big Bang. The proposal is radical in the sense that it invokes entirely new forms of matter, in fact two entirely new forms, in order to explain dark matter and other phenomena. WIMPs have the extra burden of automatically being very difficult to observe directly. This is somewhat analogous to quark confinement, which guarantees that quarks are there, but they can never be seen directly. One of the primary goals of the LHC at CERN, however, is to produce and detect WIMPs. At any rate, before WIMPs are accepted as the best explanation, all other possibilities utilizing known phenomena will have to be shown inferior. Should that occur, we will be in the unanticipated position of admitting that, to date, all we know is only 10% of what exists. A far cry from the days when people firmly believed themselves to be not only the center of the universe, but also the reason for its existence. ## Section summary • Dark matter is non-luminous matter detected in and around galaxies and galactic clusters. • It may be 10 times the mass of the luminous matter in the universe, and its amount may determine whether the universe is open or closed (expands forever or eventually stops). • The determining factor is the critical density of the universe and the cosmological constant, a theoretical construct intimately related to the expansion and closure of the universe. • The critical density ρ c is the density needed to just halt universal expansion. It is estimated to be approximately 10 –26 kg/m 3 . • An open universe is negatively curved, a closed universe is positively curved, whereas a universe with exactly the critical density is flat. • Dark matter’s composition is a major mystery, but it may be due to the suspected mass of neutrinos or a completely unknown type of leptonic matter. • If neutrinos have mass, they will change families, a process known as neutrino oscillations, for which there is growing evidence. ## Conceptual questions Discuss the possibility that star velocities at the edges of galaxies being greater than expected is due to unknown properties of gravity rather than to the existence of dark matter. Would this mean, for example, that gravity is greater or smaller than expected at large distances? Are there other tests that could be made of gravity at large distances, such as observing the motions of neighboring galaxies? How does relativistic time dilation prohibit neutrino oscillations if they are massless? If neutrino oscillations do occur, will they violate conservation of the various lepton family numbers ( ${L}_{e}$ , ${L}_{\mu }$ , and ${L}_{\tau }$ )? Will neutrino oscillations violate conservation of the total number of leptons? Lacking direct evidence of WIMPs as dark matter, why must we eliminate all other possible explanations based on the known forms of matter before we invoke their existence? ## Problems exercises If the dark matter in the Milky Way were composed entirely of MACHOs (evidence shows it is not), approximately how many would there have to be? Assume the average mass of a MACHO is 1/1000 that of the Sun, and that dark matter has a mass 10 times that of the luminous Milky Way galaxy with its ${\text{10}}^{\text{11}}$ stars of average mass 1.5 times the Sun’s mass. $1\text{.}5×{\text{10}}^{\text{15}}$ The critical mass density needed to just halt the expansion of the universe is approximately ${\text{10}}^{-\text{26}}\phantom{\rule{0.25em}{0ex}}\text{kg}/{\text{m}}^{3}$ . (a) Convert this to $\text{eV}/{c}^{2}\cdot {\text{m}}^{3}$ . (b) Find the number of neutrinos per cubic meter needed to close the universe if their average mass is $7\phantom{\rule{0.25em}{0ex}}\text{eV}/{c}^{2}$ and they have negligible kinetic energies. Assume the average density of the universe is 0.1 of the critical density needed for closure. What is the average number of protons per cubic meter, assuming the universe is composed mostly of hydrogen? $0\text{.}6{m}^{-3}$ To get an idea of how empty deep space is on the average, perform the following calculations: (a) Find the volume our Sun would occupy if it had an average density equal to the critical density of ${\text{10}}^{-\text{26}}\phantom{\rule{0.25em}{0ex}}\text{kg}/{\text{m}}^{3}$ thought necessary to halt the expansion of the universe. (b) Find the radius of a sphere of this volume in light years. (c) What would this radius be if the density were that of luminous matter, which is approximately $5%\text{}$ that of the critical density? (d) Compare the radius found in part (c) with the 4-ly average separation of stars in the arms of the Milky Way. what is a good calculator for all algebra; would a Casio fx 260 work with all algebra equations? please name the cheapest, thanks. a perfect square v²+2v+_ kkk nice algebra 2 Inequalities:If equation 2 = 0 it is an open set? or infinite solutions? Kim y=10× if \|A\| not equal to 0 and order of A is n prove that adj (adj A = \|A\| rolling four fair dice and getting an even number an all four dice Kristine 222=8 Differences Between Laspeyres and Paasche Indices No. 7x -4y is simplified from 4x + (3y + 3x) -7y is it 3×y ? J, combine like terms 7x-4y im not good at math so would this help me how did I we'll learn this f(x)= 2\|x+5\| find f(-6) Need to simplify the expresin. 3/7 (x+y)-1/7 (x-1)= . After 3 months on a diet, Lisa had lost 12% of her original weight. She lost 21 pounds. What was Lisa's original weight? what is nanomaterials and their applications of sensors. what is nano technology preparation of nanomaterial Yes, Nanotechnology has a very fast field of applications and their is always something new to do with it... can nanotechnology change the direction of the face of the world At high concentrations (>0.01 M), the relation between absorptivity coefficient and absorbance is no longer linear. This is due to the electrostatic interactions between the quantum dots in close proximity. If the concentration of the solution is high, another effect that is seen is the scattering of light from the large number of quantum dots. This assumption only works at low concentrations of the analyte. Presence of stray light. the Beer law works very well for dilute solutions but fails for very high concentrations. why? how did you get the value of 2000N.What calculations are needed to arrive at it Got questions? Join the online conversation and get instant answers!	2018-03-20 23:30:32	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 17, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6415277719497681, "perplexity": 808.5743645808858}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-13/segments/1521257647545.84/warc/CC-MAIN-20180320224824-20180321004824-00334.warc.gz"}
https://jas.shahroodut.ac.ir/article_2053.html	Document Type : Original Manuscript Author Department of Mathematics, Ayatollah Boroujerdi University, Boroujerd, Iran. Abstract Let $L$ be a finite dimensional Lie algebra. A subalgebra $H$ of $L$ is called a $c^{\#}$-ideal of $L$, if there is an ideal $K$ of $L$ with $L=H+K$ and $H\cap K$ is a $CAP$-subalgebra of $L$. This is analogous to the concept of a $c^{\#}$-normal subgroup of a finite group. Now, we consider the influence of this concept on the structure of finite dimentional Lie algebras. Keywords ###### ##### References 1. D. W. Barnes, On the cohomology of soluble Lie algebras, Math. Z., 101 (1967), 343–349. 2. A. R. Salemkar, S. Chehrazi and F. Tayanloo, Characterizations for supersolvableLie algebras, Comm. Algebra, 41 (2013), 2310–2316. 3. D. A. Towers, A Frattini theory for algebras, Proc. London Math. Soc., 27 (1973), 440–462. 4. D. A. Towers, C-ideals of Lie algebras, Comm. Algebra, 37 (2009), 4366–4373. 5. D. A. Towers, Subalgebras that cover or avoid chief factors of Lie algebras, Proc. Amer. Math. Soc., 143 (2015), 3377–3385. 6. Y. Wang and H. Wei, c#-Normality of groups and its properties, Algebr Represent Theor, 16 (2013), 193–204. 7. Y. Wang, C-normality of groups and its properties, J. Algebra, 180 (1996), 954–965.	2022-08-17 08:05:15	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8640894889831543, "perplexity": 1592.5280029389332}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-33/segments/1659882572870.85/warc/CC-MAIN-20220817062258-20220817092258-00571.warc.gz"}
https://crazyproject.wordpress.com/2010/04/	## Monthly Archives: April 2010 ### Exhibit generators of an isomorphic copy of Q(8) in Sym(8) via the left regular representation Use the left regular representation of $Q_8$ to produce two elements of $S_8$ which generate a subgroup isomorphic to $Q_8$. Recall that $Q_8 = \langle i,j \rangle$. Now $i(1) = i$, $i(-1) = -i$, $i(i) = -1$, $i(-i) = 1$, $i(j) = k$, $i(-j) = -k$, $i(k) = -j$, and $i(-k) = j$. Thus $i \mapsto (1\ i\ -1\ -i)(j\ k\ -j\ -k)$. Similarly, $j(1) = j$, $j(-1) = -j$, $j(i) = -k$, $j(-i) = k$, $j(j) = -1$, $j(-j) = -1$, $j(k) = i$, and $j(-k) = -i$. Thus $j \mapsto (1\ j\ -1\ -j)(i\ -k\ -i\ k)$. With the labeling $1 \mapsto 1$ $-1 \mapsto 2$ $i \mapsto 3$ $-i \mapsto 4$ $j \mapsto 5$ $-j \mapsto 6$ $k \mapsto 7$ $-k \mapsto 8$ we have $Q_8 \cong \langle (1\ 3\ 2\ 4)(5\ 7\ 6\ 8), (1\ 5\ 2\ 6)(3\ 8\ 4\ 7) \rangle$. ### Exhibit Dih(8) as a subgroup of Sym(8) in two different ways via the left regular representation Consider $D_8 = \langle r,s \rangle$. 1. List the elements of $D_8$ as $1,r,r^2,r^3,s,sr,sr^2,sr^3$ and label these with the integers $1,2,3,4,5,6,7,8$, respectively. Compute the image of each element of $D_8$ under the left regular representation of $D_8$ into $S_8$. 2. Relabel the elements of $D_8$ with $1,3,5,7,2,4,6,8$, respectively, and recompute the image of each element under the left regular representation. Show that the two subgroups of $S_8$ obtained in parts 1 and 2 are different. To save effort we will perform this computation in $S_{D_8}$ and then use the labeling bijections $D_8 \rightarrow \{1,2,\ldots,8\}$ to produce subgroups of $S_8$. Let $\lambda : D_8 \rightarrow S_{D_8}$ be the left regular representation. 1. Clearly $\lambda(1) = 1$. 2. We have $\lambda(r)(1) = r$, $\lambda(r)(r) = r^2$, $\lambda(r)(r^2) = r^3$, $\lambda(r)(r^3) = 1$, $\lambda(r)(s) = rs = sr^3$, $\lambda(r)(sr^3) = rsr^3 = sr^6 = sr^2$, and $\lambda(r)(sr^2) = rsr^2 = sr^5 = sr$. Thus $\lambda(r) = (1\ r\ r^2\ r^3)(s\ sr^3\ sr^2\ sr)$. 3. Note that $\lambda(r^2) = \lambda(r)^2$. Thus $\lambda(r^2) = (1\ r^2)(r\ r^3)(s\ sr^2)(sr\ sr^3)$. 4. Note that $\lambda(r^3) = \lambda(r)^3$. Thus $\lambda(r^3) = (1\ r^3\ r^2\ r)(s\ sr\ sr^2\ sr^3)$. 5. We have $\lambda(s)(1) = s$, $\lambda(s)(s) = 1$, $\lambda(s)(r) = sr$, $\lambda(s)(sr) = r$, $\lambda(s)(r^2) = sr^2$, $\lambda(s)(sr^2) = r^2$, $\lambda(s)(r^3) = sr^3$, and $\lambda(s)(sr^3) = r^3$. Thus $\lambda(s) = (1\ s)(r\ sr)(r^2\ sr^2)(r^3\ sr^3)$. 6. Note that $\lambda(sr) = \lambda(s)\lambda(r)$. Thus $\lambda(sr) = (1\ sr)(r\ sr^2)(r^2\ sr^3)(r^3\ s)$. 7. Note that $\lambda(sr^2) = \lambda(s)\lambda(r^2)$. Thus $\lambda(sr^2) = (1\ sr^2)(r\ sr^3)(r^2\ s)(r^3\ sr)$. 8. Note that $\lambda(sr^3) = \lambda(s)\lambda(r^3)$. Thus $\lambda(sr^3) = (1\ sr^3)(r\ s)(r^2\ sr)(r^3\ sr^2)$. Now via the labelling $1 \mapsto 1$ $r \mapsto 2$ $r^2 \mapsto 3$ $r^3 \mapsto 4$ $s \mapsto 5$ $sr \mapsto 6$ $sr^2 \mapsto 7$ $sr^3 \mapsto 8$ we have the isomorphic image • $1 \mapsto 1$ • $r \mapsto (1\ 2\ 3\ 4)(5\ 8\ 7\ 6)$ • $r^2 \mapsto (1\ 3)(2\ 4)(5\ 7)(6\ 8)$ • $r^3 \mapsto (1\ 4\ 3\ 2)(5\ 6\ 7\ 8)$ • $s \mapsto (1\ 5)(2\ 6)(3\ 7)(4\ 8)$ • $sr \mapsto (1\ 6)(2\ 7)(3\ 8)(4\ 5)$ • $sr^2 \mapsto (1\ 7)(2\ 8)(3\ 5)(4\ 6)$ • $sr^3 \mapsto (1\ 8)(2\ 5)(3\ 6)(4\ 7)$. Similarly, via the labeling $1 \mapsto 1$ $r \mapsto 3$ $r^2 \mapsto 5$ $r^3 \mapsto 7$ $s \mapsto 2$ $sr \mapsto 4$ $sr^2 \mapsto 6$ $sr^3 \mapsto 8$ we have the isomorphic image • $1 \mapsto 1$ • $r \mapsto (1\ 3\ 5\ 7)(2\ 8\ 6\ 4)$ • $r^2 \mapsto (1\ 5)(3\ 7)(2\ 6)(4\ 8)$ • $r^3 \mapsto (1\ 7\ 5\ 3)(2\ 4\ 6\ 8)$ • $s \mapsto (1\ 2)(3\ 4)(5\ 6)(7\ 8)$ • $sr \mapsto (1\ 4)(3\ 6)(5\ 8)(7\ 2)$ • $sr^2 \mapsto (1\ 6)(3\ 8)(5\ 2)(7\ 4)$ • $sr^3 \mapsto (1\ 8)(3\ 2)(5\ 4)(7\ 6)$. It is easy to see that these two subgroups are different. ### Exhibit Sym(3) as a subgroup of Sym(6) via the left regular representation List the elements of $S_3$ as $1$, $(1\ 2)$, $(2\ 3)$, $(1\ 3)$, $(1\ 2\ 3)$, $(1\ 3\ 2)$ and label these with the integers $1,2,3,4,5,6$, respectively. Exhibit the image of each element of $S_3$ under the left regular representation of $S_3$ into $S_6$. We use the notation $\sigma(k) = \sigma \cdot k$. 1. $1 \mapsto 1$ 2. We have $(1\ 2)(1) = (1\ 2) = 2$, $(1\ 2)(2) = (1\ 2)(1\ 2) = 1$, $(1\ 2)(3) = (1\ 2)(2\ 3) = (1\ 2\ 3) = 5$, $(1\ 2)(4) = (1\ 2)(1\ 3) = (1\ 3\ 2) = 6$, $(1\ 2)(5) = (1\ 2)(1\ 2\ 3) = (2\ 3) = 3$, and $(1\ 2)(6) = (1\ 2)(1\ 3\ 2) = (1\ 3) = 4$. Thus $(1\ 2) \mapsto (1\ 2)(3\ 5)(4\ 6)$. 3. We have $(2\ 3)(1) = (2\ 3) = 3$, $(2\ 3)(2) = (2\ 3)(1\ 2) = (1\ 3\ 2) = 6$, $(2\ 3)(3) = (2\ 3)(2\ 3) = 1$, $(2\ 3)(4) = (2\ 3)(1\ 3) = (1\ 2\ 3) = 5$, $(2\ 3)(5) = (2\ 3)(1\ 2\ 3) = (1\ 3) = 4$, and $(2\ 3)(6) = (2\ 3)(1\ 3\ 2) = (1\ 2) = 2$. Thus $(2\ 3) \mapsto (1\ 3)(2\ 6)(4\ 5)$. 4. We have $(1\ 3)(1) = (1\ 3) = 4$, $(1\ 3)(2) = (1\ 3)(1\ 2) = (1\ 2\ 3) = 5$, $(1\ 3)(3) = (1\ 3)(2\ 3) = (1\ 3\ 2) = 6$, $(1\ 3)(4) = (1\ 3)(1\ 3) = 1$, $(1\ 3)(5) = (1\ 3)(1\ 2\ 3) = (1\ 2) = 2$, and $(1\ 3)(6) = (1\ 3)(1\ 3\ 2) = (2\ 3) = 3$. Thus $(1\ 3) \mapsto (1\ 4)(2\ 5)(3\ 6)$. 5. We have $(1\ 2\ 3)(1) = (1\ 2\ 3) = 5$, $(1\ 2\ 3)(2) = (1\ 2\ 3)(1\ 2) = (1\ 3) = 4$, $(1\ 2\ 3)(3) = (1\ 2\ 3)(2\ 3) = (1\ 2) = 2$, $(1\ 2\ 3)(4) = (1\ 2\ 3)(1\ 3) = (2\ 3) = 3$, $(1\ 2\ 3)(5) = (1\ 2\ 3)(1\ 2\ 3) = (1\ 3\ 2) = 6$, and $(1\ 2\ 3)(6) = (1\ 2\ 3)(1\ 3\ 2) = 1$. Thus $(1\ 2\ 3) \mapsto (1\ 5\ 6)(2\ 4\ 3)$. 6. We have $(1\ 3\ 2)(1) = (1\ 3\ 2) = 6$, $(1\ 3\ 2)(2) = (1\ 3\ 2)(1\ 2) = (2\ 3) = 3$, $(1\ 3\ 2)(3) = (1\ 3\ 2)(2\ 3) = (1\ 3) = 4$, $(1\ 3\ 2)(4) = (1\ 3\ 2)(1\ 3) = (1\ 2) = 2$, $(1\ 3\ 2)(5) = (1\ 3\ 2)(1\ 2\ 3) = 1$, and $(1\ 3\ 2)(6) = (1\ 3\ 2)(1\ 3\ 2) = (1\ 2\ 3) = 5$. Thus $(1\ 3\ 2) \mapsto (1\ 6\ 5)(2\ 3\ 4)$. ### Exhibit the Klein 4-group as a subgroup of Sym(4) using the left regular representation Let $G = \{1,a,b,c\}$ be the Klein 4-group. 1. Label the elements $1,a,b,c$ with the integers $1,2,4,3$, respectively, and prove that under the left regular representation of $G$ into $S_4$ the nonidentity elements are mapped as follows: $a \mapsto (1\ 2)(3\ 4)$, $b \mapsto (1\ 4)(2\ 3)$, and $c \mapsto (1\ 3)(2\ 4)$. 2. Relabel the elements $1,a,b,c$ as $1,4,2,3$, respectively, and compute the image of each element of $G$ under the left regular representation of $G$ into $S_4$. Show that the image of $G$ under this labelling is the same subgroup of $S_4$ (even though the individual elements map differently.) The multiplication table for $G$ is as follows. 1 a b c 1 1 a b c a a 1 c b b b c 1 a c c b a 1 We can see that with $1,a,b,c$ labelled as $1,2,4,3$, respectively, since $a1 = a$, $aa = 1$, $ab = c$, and $ac = b$ we have $a(1) = 2$, $a(2) = 1$, $a(3) = 4$, and $a(4) = 3$. Thus $a \mapsto (1\ 2)(3\ 4)$. Similarly, $b1 = b$, $ba = c$, $bb = 1$, and $bc = a$, so that $b \mapsto (1\ 4)(2\ 3)$. Likewise $c \mapsto (1\ 3)(2\ 4)$. Now we relabel the elements $1,a,b,c$ as $1,4,2,3$, respectively. Now $a(1) = a1 = a = 4$, $a(2) = ab = c = 3$, $a(3) = ac = b = 2$, and $a(4) = aa = 1$. Thus $a \mapsto (1\ 4)(2\ 3)$. Now $b(1) = b = 2$, $b(2) = bb = 1$, $b(3) = bc = a = 4$, and $b(4) = ba = c = 3$, so that $b \mapsto (1\ 2)(3\ 4)$, and $c(1) = c = 3$, $c(2) = cb = a = 4$, $c(3) = cc = 1$, and $c(4) = ca = b = 2$, so that $c \mapsto (1\ 3)(2\ 4)$. Clearly this is the same subgroup as the image of $G$ under the first labelling. Let $G$ be a group and $H,K \leq G$ subgroups. For each $x \in G$, define the $(H,K)$ double coset of $x$ by $HxK = \{ hxk \ \|\ h \in H, k \in K \}$. 1. Prove that $HxK$ is the union of the left cosets $x_iK$ where $x_iK \in H \cdot xK$. 2. Prove (as above) that $HxK$ is a union of right cosets of $H$. 3. Show that the set of $(H,K)$ double cosets partitions $G$. 4. Prove that $\|HxK\| = \|K\| \cdot [H : H \cap xKx^{-1}]$. 5. Prove that $\|HxK\| = \|H\| \cdot [K : K \cap x^{-1}Hx]$. 1. $(\subseteq)$ Let $hxk \in HxK$. Now $hxK = h \cdot xK \in H \cdot xK$ and $hxk \in hxK$. Thus $hxk \in \bigcup_{yK \in H \cdot xK} yK$. $(\supseteq)$ Let $g \in \bigcup_{yK \in H \cdot xK} yK$. Then $g \in yK$ for some $yK \in H \cdot xK$, so that $yK = h \cdot xK = hxK$ for some $h \in H$. Then $g = hxk$ for some $k \in K$. So $g \in HxK$. 2. $(\subseteq)$ Let $hxk \in HxK$. Now $Hxk = Hx \cdot k \in Hx \cdot K$ and $hxk \in Hxk$. Thus $hxk \in \bigcup_{Hy \in Hx \cdot K} Hy$. $(\supseteq)$ Let $g \in \bigcup_{Hy \in Hx \cdot K} Hy$. Then $g \in Hy$ for some $Hy \in Hx \cdot K$, so that $Hy = Hx \cdot k = Hxk$ for some $k \in K$. Then $g = hxk$ for some $h \in H$. So $g \in HxK$. 3. Note that every element is in some double coset- in particular, $x \in HxK$ for all $x \in G$. So $G = \bigcup_{x \in G} HxK$. Note that if $y \in HxK$, then $HyK \subseteq HxK$. Now suppose $x,y \in G$ such that $HxK \cap HyK \neq \emptyset$. Then we have $h_1xk_1 = h_2yk_2$ for some $h_i \in H$ and $k_i \in K$. Then $x = h_1^{-1}h_2yk_2k_1^{-1} \in HyK$, so that $HxK \subseteq HyK$. Similarly $HyK \subseteq HxK$. Thus two double cosets are either disjoint or equal. Thus the $(H,K)$ double cosets form a partition of $G$. 4. First we prove a lemma. Lemma 1: $\mathsf{stab}_H(xK) = H \cap xKx^{-1}$. Proof: $(\subseteq)$ Suppose $h \in \mathsf{stab}_H(xK)$. Then $hxK = h \cdot xK = xK$, and we have $x^{-1}hx \in K$. So $h \in xKx^{-1}$, hence $h \in H \cap xKx^{-1}$. $(\supseteq)$ Suppose $h \in H \cap xKx^{-1}$. Then $x^{-1}hx \in K$, so that $h \cdot xK = hxK = xK$, thus $h \in \mathsf{stab}_H(xK)$. $\square$ We saw in part 1 above that $HxK = \bigcup_{yK \in H \cdot xK} yK$; moreover, this union is disjoint because the $yK$ are distinct left cosets of $K$, each of order $\|K\|$. Thus $\|HxK\| = \|K\| \cdot \|H \cdot xK\| = \|K\| \cdot [H : \mathsf{stab}_H(xK)] = \|K\| \cdot [H : H \cap xKx^{-1}]$, using Lemma 1. 5. First we prove a lemma. Lemma 2: $\mathsf{stab}_K(Hx) = K \cap x^{-1}Hx$. Proof: $(\subseteq)$ Suppose $k \in \mathsf{stab}_K(Hx)$. Then $Hxk = Hx \cdot k = Hx$, and we have $xkx^{-1} \in H$. So $k \in x^{-1}Hx$, hence $k \in K \cap x^{-1}Hx$. $(\supseteq)$ Suppose $k \in K \cap x^{-1}Hx$. Then $xkx^{-1} \in H$, so that $Hx \cdot k = Hxk = Hx$, thus $k \in \mathsf{stab}_K(Hx)$. $\square$ We saw in part 2 above that $HxK = \bigcup_{Hy \in Hx \cdot K} Hy$; moreover, this union is disjoint. Thus $\|HxK\| = \|H\| \cdot \|Hx \cdot K\| = \|H\| \cdot [K : \mathsf{stab}_K(Hx)] = \|K\| \cdot [K : K \cap x^{-1}Hx]$. ### Transitive group actions induce transitive actions on the orbits of the action of a subgroup Suppose $G \leq S_A$ acts transitively on $A$ and let $H \leq G$ be normal. Let $O_1, O_2, \ldots, O_r$ be the distinct orbits of $H$ on $A$. 1. Prove that $G$ permutes the sets $O_k$ in the sense that for each $g \in G$ and each $i \in \{1,2,\ldots,r\}$ there exists a $j$ such that $g \cdot O_i = O_j$. Prove that $G$ is transitive on $\{ O_1,O_2,\ldots,O_r \}$. Deduce that all orbits of $H$ on $A$ have the same cardinality. 2. Prove that if $a \in O_1$ then $\|O_1\| = [H : H \cap \mathsf{stab}_G(a)]$ and that $r = [G : H \mathsf{stab}_G(a)]$. [Hint: Note that $H \cap \mathsf{stab}_G(a) = \mathsf{stab}_H(a)$ and use the Second Isomorphism Theorem.] 1. Let $g \in G$ and $a \in A$. Now $H \cdot a$ is an arbitrary orbit of $H$ on $A$. Moreover, $g \cdot a = b$ for some $b \in A$. Then $g \cdot (H \cdot a) = gH \cdot a = Hg \cdot a = H \cdot (g \cdot a) = H \cdot b$, since $H$ is normal. Thus $G$ permutes the orbits $\{ H \cdot a \ \|\ a \in A \}$. Now let $a,b \in A$ be arbitrary; since the action of $G$ on $A$ is transitive, there exists $g \in G$ such that $g \cdot a = b$. Then $g \cdot (H \cdot a) = gH \cdot a = Hg \cdot a = H \cdot (g \cdot a) = H \cdot b$; thus the action of $G$ on the orbits $\{ H \cdot a \ \|\ a \in A \}$ is transitive. Let $a,b \in A$ be arbitrary, and let $g \in G$ such that $g \cdot a = b$. Because $H$ is normal, for every $h \in H$ there exists a unique $k \in H$ such that $gh = kg$. We have a mapping $\varphi_g : H \cdot a \rightarrow H \cdot b$ given by $\varphi_g(h \cdot a) = k \cdot b$. Because $gH = Hg$, this mapping $\varphi_g$ is bijective, and $\|H \cdot a\| = \|H \cdot b\|$. Because the action of $G$ is transitive on the orbits, all orbits have the same cardinality. 2. Let $a \in A$. We can restrict the action of $G$ on the orbits of $H$ to $H$; with this action we have $\|H \cdot a\| = [H : \mathsf{stab}_H(a)]$. Clearly $\mathsf{stab}_H(a) = H \cap \mathsf{stab}_G(a)$; thus $\|H \cdot a\| = [H : H \cap \mathsf{stab}_G(a)]$. Considering now the (transitive) action of $G$ on the orbits of $H$, we have $r = \|\{ H \cdot a \ \|\ a \in A \}\| = [G : \mathsf{stab}_G(H \cdot a)]$. We claim that $\mathsf{stab}_G(H \cdot a) = H \mathsf{stab}_G(a)$. Proof of claim: Suppose $g \in \mathsf{stab}_G(H \cdot a)$. Now $Hg \cdot a = gH \cdot a = g \cdot (H \cdot a) = H \cdot a$, so that in particular $g \cdot a = h \cdot a$ for some $h \in H$. Then $h^{-1}g \cdot a = a$, so that $h^{-1}g \in \mathsf{stab}_G(a)$. Thus $g \in H \mathsf{stab}_G(a)$. Now suppose $g \in H \mathsf{stab}_G(a)$. Then $g = hx$ for some $h \in H$ and $x \in \mathsf{stab}_G(a)$. Now $g \cdot (H \cdot a) = gH \cdot a = Hg \cdot a = Hhx \cdot a = H \cdot a$, so that $g \in \mathsf{stab}_G(H \cdot a)$. Thus $r = [G : H \mathsf{stab}_G(a)]$. ### Every doubly transitive group action is primitive A transitive permutation group $G \leq S_A$ acting on $A$ is called doubly transitive if for all $a \in A$, the subgroup $\mathsf{stab}(a)$ is transitive on $A \setminus \{a\}$. 1. Prove that $S_n$ is doubly transitive on $\{1,2,\ldots,n\}$ for all $n \geq 2$. 2. Prove that a doubly transitive group is primitive. Deduce that $D_8$ is not doubly transitive in its action on the four vertices of a square. 1. We know that $S_n$ is transitive on $A = \{1,2,\ldots,n\}$. Now if $n \geq 2$ and $k \in A$, we have a natural isomorphism $\mathsf{stab}(k) \cong S_{A \setminus \{k\}}$; this permutation group is also transitive in its action on $A \setminus \{k\}$. Thus $S_n$ is doubly transitive. 2. Let $G \leq S_A$ act transitively on $A$, and suppose further that the action is doubly transitive. Let $B \subseteq A$ be a proper block; then there exist elements $b \in B$ and $a \in A \setminus B$. By a previous exercise, we have $\mathsf{stab}(b) \leq \mathsf{stab}(B)$. Thus if $\sigma \in \mathsf{stab}(b)$, we have $\sigma[B] = B$. Suppose now that there exists an element $c \in B$ with $c \neq b$. Because $G$ is doubly transitive on $A$, there exists an element $\tau \in \mathsf{stab}(b)$ such that $\tau(c) = a$. Thus $\tau[B] \neq B$, a contradiction. So no such element $c$ exists and we have $B = \{b\}$. Now every block is trivial, thus the action of $G$ on $A$ is primitive. We saw that the action of $D_8$ on the four vertices of a square is not primitive in the previous exercise. Thus this action is not doubly transitive. ### Basic properties of blocks of a group action Let $G \leq S_A$ act transitively on the set $A$. A block is a nonempty subset $B \subseteq A$ such that for all $\sigma \in G$ either $\sigma[B] = B$ or $\sigma[B] \cap B = \emptyset$. 1. Prove that if $B$ is a block containing $a \in A$ and we define $\mathsf{stab}(B) = \{ \sigma \in G \ \|\ \sigma[B] = B \}$ then $\mathsf{stab}(a) \leq \mathsf{stab}(B) \leq G$. 2. Prove that if $B$ is a block then $\mathcal{B} = \{ \sigma[B] \ \|\ \sigma \in G \}$ is a partition of $A$. 3. A (transitive) group $G \leq S_A$ is called primitive if the only blocks in $A$ are the trivial ones- the singletons and $A$ itself. Prove that $S_4$ is primitive on $A = \{1,2,3,4\}$. Prove that $D_8$ is not primitive as a permutation group on the four vertices of a square. 4. Let $G \leq S_A$ act transitively on $A$. Prove that $G$ is primitive on $A$ if and only if for all $a \in A$, $\mathsf{stab}(a)$ is maximal in $G$. 1. Note that $\mathsf{stab}(B)$ is not empty since $1[B] = B$. Now let $\sigma, \tau \in \mathsf{stab}(B)$. Note that $\tau^{-1}[B] = \tau^{-1}\tau [B] = B$, so that $\sigma\tau^{-1}[B] = B$. Thus $\sigma\tau^{-1} \in \mathsf{stab}(B)$. By the subgroup criterion, $\mathsf{stab}(B) \leq G$. Now suppose $\sigma \in \mathsf{stab}(a)$. We have $\sigma(a) = a$, so that $\{a\} \subseteq \sigma[B] \cap B$. Hence $\sigma[B] \cap B$ is nonempty, and we have $\sigma[B] = B$ since $B$ is a block. Thus $\sigma \in \mathsf{stab}(B)$. By a previous exercise, $\mathsf{stab}(a) \leq \mathsf{stab}(B)$. 2. First we show that $\bigcup_{\sigma \in G} \sigma[B] = A$. The $(\subseteq)$ direction is clear. $(\supseteq)$: Let $a \in A$ and $b \in B$. Since the action of $G$ is transitive, there exists $\sigma \in G$ such that $a = \sigma(b)$. Then $a \in \sigma[B]$, so that $aA \subseteq \bigcup_{\sigma \in G} \sigma[B]$. Now suppose $\sigma[B] \cap \tau[B] \neq \emptyset$. Then there exist $a,b \in B$ such that $\sigma(a) = \tau(b)$. Now $a = \sigma^{-1}\tau \cdot b$, so that $\sigma^{-1}\tau[B] \cap B$ is not empty. Thus $\sigma^{-1}\tau[B] = B$, and we have $\sigma[B] = \tau[B]$. So elements of $\mathcal{B}$ are pairwise disjoint; hence $\mathcal{B}$ is a partition of $A$. 1. Let $B \subseteq A = \{1,2,3,4\}$ be a proper nonempty subset. Then there exist elements $x \in B$ and $y \in A \setminus B$. Suppose $B$ is a block. Consider $\sigma = (x\ y) \in S_4$; since $\sigma(x) \notin B$, we have $\sigma[B] \cap B = \emptyset$. Let $w$ and $z$ be the remaining elements of $A$. If $w \in B$, then $w \in \sigma[B] \cap B$, a contradiction; similarly for $z$. Thus $B = \{x\}$. Now clearly $A$ itself is a block, and any proper block must be a singleton. Thus this action of $S_4$ is primitive. 2. We saw previously that $D_8 \cong \langle (1\ 3), (1\ 2\ 3\ 4) \rangle$, where $A = \{1,2,3,4\}$ are labels on the vertices of a square (written clockwise). Consider the set $\{1,3\}$. We see that 1. $1 \cdot \{1,3\} = \{1,3\}$ 2. $(1\ 2\ 3\ 4) \cdot \{1,3\} = \{2,4\}$ 3. $(1\ 3)(2\ 4) \cdot \{1,3\} = \{1,3\}$ 4. $(1\ 4\ 3\ 2) \cdot \{1,3\} = \{2,4\}$ 5. $(1\ 3) \cdot \{1,3\} = \{1,3\}$ 6. $(1\ 2)(3\ 4) \cdot \{1,3\} = \{2,4\}$ 7. $(2\ 4) \cdot \{1,3\} = \{1,3\}$ 8. $(1\ 4)(2\ 3) \cdot \{1,3\} = \{2,4\}$ So that $\{1,3\}$ is a nontrivial block; hence this action is not primitive. 3. We begin with some lemmas. Lemma 1: Let $G \leq S_A$ act transitively on $A$ and let $B \subseteq A$ be a block under the action. Then $\mathsf{stab}(B) = G$ if and only if $B = A$. Proof: The $(\Leftarrow)$ direction is clear. $(\Rightarrow)$ Suppose $B$ is a proper subset and let $x \in B$, $y \in A \setminus B$. Now $y \in (x\ y)[B]$ and $y \notin B$, so that $(x\ y)[B] \neq B$. Since $B$ is a block, we have $(x\ y)[B] \cap B = \emptyset$. But $x \in (x\ y)[B] \cap B$ since $x \in B$ and $y \notin B$, a contradiction. So $B$ is not proper, and we have $A = B$. $\square$ Now we move to the main result. $(\Rightarrow)$ Suppose $G$ is primitive on $A$; then the only blocks of $A$ are $A$ and the singletons. Now let $a \in A$ and let $H \leq G$ be a subgroup with $\mathsf{stab}(a) \leq H \leq G$. We consider the set $H \cdot a = \{ h \cdot a \ \|\ h \in H \}$. We claim that $H \cdot a$ is a block. Proof of claim: $H \cdot a$ is not empty since $a = 1 \cdot a \in H \cdot a$. Now let $\sigma \in G$. If $\sigma \in H$, then $\sigma \cdot (H \cdot a) = \sigma H \cdot a = H \cdot a$. If $\sigma \notin H$, then $\sigma \cdot (H \cdot a) = \sigma H \cdot a$. Suppose that $\sigma H \cdot a \cap H \cdot a \neq \emptyset$; say that $\sigma\tau_1 \cdot a = \tau_2 \cdot a$ for some $\tau_1, \tau_2 \in H$. Then $\tau_2^{-1} \sigma \tau_1 \cdot a = a$, so that $\tau_2^{-1} \sigma \tau_1 \in \mathsf{stab}(a) \leq H$. But then $\sigma \in \tau_2H\tau_1^{-1} = H$, a contradiction. Thus if $\sigma \notin H$, then $\sigma[H \cdot a] \cap H \cdot a = \emptyset$. Hence $H \cdot a$ is a block. Next we claim that $\mathsf{stab}(H \cdot a) = H$. Proof of claim: $(\supseteq)$ If $\tau \in H$, then $\tau \cdot (H \cdot a) = \tau H \cdot a = H \cdot a$. $(\subseteq)$ Suppose $\sigma \cdot (H \cdot a) = \sigma H \cdot a = H \cdot a$. Then $\sigma \cdot a = \tau \cdot a$ for some $\tau \in H$, hence $\tau^{-1} \sigma \cdot a = a$. Then $\tau^{-1} \sigma \in \mathsf{stab}(a) \leq H$, so that $\sigma \in H$. Since $G$ is primitive on $a$ and $a \in H \cdot a$, we have $H \cdot a = \{a\}$ or $H \cdot a = A$. If $H \cdot a = \{a\}$, we have $H \leq \mathsf{stab}(a)$, so that $H = \mathsf{stab}(a)$. If $H \cdot a = A$, we have $H = \mathsf{stab}(H \cdot a) = \mathsf{stab}(A) = G$. Thus $\mathsf{stab}(a)$ is a maximal subgroup of $G$. $(\Leftarrow)$ Suppose that for all $a \in A$, $\mathsf{stab}(a)$ is a maximal subgroup in $G$. Let $B \subseteq A$ be a block with $a \in B$. Now $\mathsf{stab}(a) \leq \mathsf{stab}(B) \leq G$ by part 1 above. Since $\mathsf{stab}(a)$ is maximal, there are two cases. If $\mathsf{stab}(B) = \mathsf{stab}(a)$ and $b \in B$ such that $b \neq a$, then since $G$ acts transitively there exists $\sigma \in G$ such that $\sigma \cdot a = b$. Now $\sigma \in \mathsf{stab}(B) = \mathsf{stab}(a)$, a contradiction. Thus $B = \{a\}$. If $\mathsf{stab}(B) = G$, then by Lemma 1 we have $B = A$. Thus $G$ is primitive on $A$. ### Compute some orbits of an action by Sym(4) on polynomials in four variables As in this previous exercise, let $S_4$ act on the set $R$ of all polynomials with integer coefficients in the independent variables $x_1,x_2,x_3,x_4$ by permuting the indices: $\sigma \cdot p(x_1,x_2,x_3,x_4) = p(x_{\sigma(1)}, x_{\sigma(2)}, x_{\sigma(3)}, x_{\sigma(4)})$. 1. Find the polynomials in the orbit of $x_1 + x_2$. (Recall from a previous exercise that the stabilizer of this element has order 4.) 2. Find the polynomials in the orbit of $x_1x_2 + x_3x_4$. (Recall that the stabilizer of this element has order 8.) 3. Find the polynomials in the orbit of $(x_1 + x_2)(x_3 + x_4)$. 1. We have $[S_4 : \mathsf{stab}(x_1+x_2)] = 6$. A simple calculation then shows that this orbit is $\{ x_1+x_2, x_1+x_3, x_1+x_4,$ $x_2+x_3, x_2+x_4, x_3+x_4 \}$. 2. We have $[S_4 : \mathsf{stab}(x_1x_2 + x_3x_4)] = 3$. A simple calculation then shows that this orbit is $\{ x_1x_2+x_3x_4, x_1x_3+x_2x_4, x_1x_4+x_2x_3 \}$. 3. We saw in a previous exercise that $\mathsf{stab}((x_1+x_2)(x_3+x_4)) = \mathsf{stab}(x_1x_2+x_3x_4)$. Thus $[S_4 : \mathsf{stab}((x_1+x_2)(x_3+x_4))] = 3$. A simple calculation then shows that this orbit is $\{ (x_1+x_2)(x_3+x_4),$ $(x_1+x_3)(x_2+x_4),$ $(x_1+x_4)(x_2+x_3) \}$. ### Compute the orbits, cycle decompositions, and stabilizers of some group actions of Sym(3) Repeat the previous example with each of the following actions. 1. $S_3$ acting on the set $A = \{ (i,j,k) \ \|\ 1 \leq i,j,k \leq 3 \}$ by $\sigma \cdot (i,j,k) = (\sigma(i), \sigma(j), \sigma(k))$ 2. $S_3$ acting on the set $B$ of all nonempty subsets of $\{1,2,3\}$ by $\sigma \cdot X = \sigma[X]$. • For $S_3$ acting on $A$: • We compute the orbits as follows. • $1 \cdot (1,1,1) = (2\ 3) \cdot (1,1,1) = (1,1,1)$, $(1\ 2) \cdot (1,1,1) = (1\ 2\ 3) \cdot (1,1,1) = (2,2,2)$, and $(1\ 3) \cdot (1,1,1) = (1\ 3\ 2) \cdot (1,1,1) = (3,3,3)$, so that $\mathcal{O}_1 = \{ (1,1,1), (2,2,2), (3,3,3) \}$. • $1 \cdot (1,1,2) = (1,1,2)$, $(1\ 2) \cdot (1,1,2) = (2,2,1)$, $(1\ 3) \cdot (1,1,2) = (3,3,2)$, $(2\ 3) \cdot (1,1,2) = (1,1,3)$, $(1\ 2\ 3) \cdot (1,1,2) = (2,2,3)$ and $(1\ 3\ 2) \cdot (1,1,2) = (3,3,1)$, so that $\mathcal{O}_2 = \{ (1,1,2), (2,2,1), (3,3,2),$ $(1,1,3), (2,2,3), (3,3,1) \}$. • $1 \cdot (1,2,1) = (1,2,1)$, $(1\ 2) \cdot (1,2,1) = (2,1,2)$, $(2\ 3) \cdot (1,2,1) = (1,3,1)$, $(1\ 2\ 3) \cdot (1,2,1) = (2,3,2)$, and $(1\ 3\ 2) \cdot (1,2,1) = (3,1,3)$, so that $\mathcal{O}_3 = \{ (1,2,1), (2,1,2), (3,2,3),$ $(1,3,1), (2,3,2), (3,1,3) \}$. • $1 \cdot (2,1,1) = (2,1,1)$, $(1\ 2) \cdot (2,1,1) = (1,2,2)$, $(1\ 3) \cdot (2,1,1) = (2,3,3)$, $(2\ 3) \cdot (2,1,1) = (3,1,1)$, $(1\ 2\ 3) \cdot (2,1,1) = (3,2,2)$, and $(1\ 3\ 2) \cdot (2,1,1) = (1,3,3)$, so that $\mathcal{O}_4 = \{ (2,1,1), (1,2,2),$ $(2,3,3), (3,1,1),$ $(3,2,2), (1,3,3) \}$. • $1 \cdot (1,2,3) = (1,2,3)$, $(1\ 2) \cdot (1,2,3) = (2,1,3)$, $(1\ 3) \cdot (1,2,3) = (3,2,1)$, $(2\ 3) \cdot (1,2,3) = (1,3,2)$, $(1\ 2\ 3) \cdot (1,2,3) = (2,3,1)$, and $(1\ 3\ 2) \cdot (1,2,3) = (3,1,2)$, so that $\mathcal{O}_5 = \{ (1,2,3), (2,1,3),$ $(3,2,1), (1,3,2),$ $(2,3,1), (3,1,2) \}$. This exhausts the elements of $A$. • We fix the labeling $\alpha_{1} = (1,1,1)$ $\alpha_{10} = (2,1,1)$ $\alpha_{19} = (3,1,1)$ $\alpha_{2} = (1,1,2)$ $\alpha_{11} = (2,1,2)$ $\alpha_{20} = (3,1,2)$ $\alpha_{3} = (1,1,3)$ $\alpha_{12} = (2,1,3)$ $\alpha_{21} = (3,1,3)$ $\alpha_{4} = (1,2,1)$ $\alpha_{13} = (2,2,1)$ $\alpha_{22} = (3,2,1)$ $\alpha_{5} = (1,2,2)$ $\alpha_{14} = (2,2,2)$ $\alpha_{23} = (3,2,2)$ $\alpha_{6} = (1,2,3)$ $\alpha_{15} = (2,2,3)$ $\alpha_{24} = (3,2,3)$ $\alpha_{7} = (1,3,1)$ $\alpha_{16} = (2,3,1)$ $\alpha_{25} = (3,3,1)$ $\alpha_{8} = (1,3,2)$ $\alpha_{17} = (2,3,2)$ $\alpha_{26} = (3,3,2)$ $\alpha_{9} = (1,3,3)$ $\alpha_{18} = (2,3,3)$ $\alpha_{27} = (3,3,3)$. Now under the permutation representation $S_3 \rightarrow S_{27}$, we have • $1 \mapsto 1$ • $(1\ 2) \mapsto (\alpha_1\ \alpha_{14})$ $(\alpha_2\ \alpha_{13})$ $(\alpha_3\ \alpha_{15})$ $(\alpha_4\ \alpha_{11})$ $(\alpha_5\ \alpha_{10})$ $(\alpha_6\ \alpha_{12})$ $(\alpha_7\ \alpha_{17})$ $(\alpha_8\ \alpha_{16})$ $(\alpha_9\ \alpha_{18})$ $(\alpha_{19}\ \alpha_{23})$ $(\alpha_{20}\ \alpha_{22})$ $(\alpha_{21}\ \alpha_{24})$ $(\alpha_{25}\ \alpha_{26})$ • $(1\ 3) \mapsto (\alpha_1\ \alpha_{27})$ $(\alpha_2\ \alpha_{26})$ $(\alpha_3\ \alpha_{25})$ $(\alpha_4\ \alpha_{24})$ $(\alpha_5\ \alpha_{23})$ $(\alpha_6\ \alpha_{22})$ $(\alpha_7\ \alpha_{21})$ $(\alpha_8\ \alpha_{20})$ $(\alpha_9\ \alpha_{19})$ $(\alpha_{10}\ \alpha_{18})$ $(\alpha_{11}\ \alpha_{17})$ $(\alpha_{12}\ \alpha_{16})$ $(\alpha_{13}\ \alpha_{15})$ • $(2\ 3) \mapsto (\alpha_2\ \alpha_3)$ $(\alpha_4\ \alpha_7)$ $(\alpha_5\ \alpha_{9})$ $(\alpha_6\ \alpha_{8})$ $(\alpha_{10}\ \alpha_{19})$ $(\alpha_{11}\ \alpha_{21})$ $(\alpha_{12}\ \alpha_{20})$ $(\alpha_{13}\ \alpha_{25})$ $(\alpha_{14}\ \alpha_{27})$ $\alpha_{15}\ \alpha_{26})$ $(\alpha_{16}\ \alpha_{22})$ $(\alpha_{17}\ \alpha_{24})$ $(\alpha_{18}\ \alpha_{23})$ • $(1\ 2\ 3) \mapsto (\alpha_1\ \alpha_{14}\ \alpha_{27})$ $(\alpha_2\ \alpha_{15}\ \alpha_{25})$ $(\alpha_3\ \alpha_{13}\ \alpha_{26})$ $(\alpha_{4}\ \alpha_{17}\ \alpha_{21})$ $(\alpha_5\ \alpha_{18}\ \alpha_{19})$ $(\alpha_6\ \alpha_{16}\ \alpha_{20})$ $(\alpha_7\ \alpha_{11}\ \alpha_{24})$ $(\alpha_8\ \alpha_{12}\ \alpha_{22})$ $(\alpha_9\ \alpha_{10}\ \alpha_{23})$ • $(1\ 3\ 2) \mapsto (\alpha_1\ \alpha_{27}\ \alpha_{14})$ $(\alpha_2\ \alpha_{25}\ \alpha_{15})$ $(\alpha_3\ \alpha_{26}\ \alpha_{13})$ $(\alpha_{4}\ \alpha_{21}\ \alpha_{17})$ $(\alpha_5\ \alpha_{19}\ \alpha_{18})$ $(\alpha_6\ \alpha_{20}\ \alpha_{16})$ $(\alpha_7\ \alpha_{24}\ \alpha_{11})$ $(\alpha_8\ \alpha_{22}\ \alpha_{12})$ $(\alpha_9\ \alpha_{23}\ \alpha_{10})$ • If $x \in \mathcal{O}_1$, then $[S_3 : \mathsf{stab}(x)] = 3$ so that $\|\mathsf{stab}(x)\| = 2$. In particular, consider $(3,3,3) \in \mathcal{O}_1$. Since $(1\ 2) \cdot (3,3,3) = (3,3,3)$, we have $\mathsf{stab}((3,3,3)) = \langle (1\ 2) \rangle$. Each remaining orbit has order 6. Thus, if $x \in \mathcal{O}_k$, $2 \leq k \leq 5$, then $\|\mathsf{stab}(x)\| = 1$; hence $\mathsf{stab}(x) = 1$. • For $S_3$ acting on $B$: • We compute the orbits as follows. • We have $1 \cdot \{1\} = (2\ 3) \cdot \{1\} = \{1\}$, $(1\ 2) \cdot \{1\} = (1\ 2\ 3) \cdot \{1\} = \{2\}$, and $(1\ 3) \cdot \{1\} = (1\ 3\ 2) \cdot \{1\} = \{3\}$. Thus $\mathcal{O}_1 = \{ \{1\}, \{2\}, \{3\} \}$. • We have $1 \cdot \{1,2\} = (1\ 2) \cdot \{1,2\} = \{1,2\}$, $(1\ 3) \cdot \{1,2\} = (1\ 2\ 3) \cdot \{1,2\} = \{2,3\}$, and $(2\ 3) \cdot \{1,2\} = (1\ 3\ 2) \cdot \{1,2\} = \{1,3\}$. Thus $\mathcal{O}_2 = \{ \{1,2\}, \{1,3\}, \{2,3\} \}$. • There is only one remaining element of $B$. Thus $\mathcal{O}_3 = \{ \{1,2,3\} \}$. • We fix the labeling $\alpha_{1} = \{1\}$ $\alpha_{5} = \{1,3\}$ $\alpha_{2} = \{2\}$ $\alpha_{6} = \{2,3\}$ $\alpha_{3} = \{3\}$ $\alpha_{7} = \{1,2,3\}$ $\alpha_{4} = \{1,2\}$ Under the permutation representation $S_3 \rightarrow S_7$, • $1 \mapsto 1$ • $(1\ 2) \mapsto (\alpha_1\ \alpha_2)(\alpha_5\ \alpha_6)$ • $(1\ 3) \mapsto (\alpha_1\ \alpha_3)(\alpha_4\ \alpha_6)$ • $(2\ 3) \mapsto (\alpha_2\ \alpha_3)(\alpha_4\ \alpha_5)$ • $(1\ 2\ 3) \mapsto (\alpha_1\ \alpha_2\ \alpha_3)(\alpha_4\ \alpha_6\ \alpha_5)$ • $(1\ 3\ 2) \mapsto (\alpha_1\ \alpha_3\ \alpha_2)(\alpha_4\ \alpha_5\ \alpha_6)$ • If $x \in \mathcal{O}_1$, then $[S_3 : \mathsf{stab}(x)] = 3$, so that $\|\mathsf{stab}(x)\| = 2$. In particular, since $(2\ 3) \cdot \{1\} = \{1\}$, we have $\mathsf{stab}(\{1\}) = \langle (2\ 3) \rangle$. If $x \in \mathcal{O}_2$, we similarly have $\|\mathsf{stab}(x)\| = 2$. In particular, since $(2\ 3) \cdot \{2,3\} = \{2,3\}$, we have $\mathsf{stab}(\{2,3\}) = \langle (2\ 3) \rangle$. Now $[S_3 : \mathsf{stab}(\{1,2,3\})] = 1$, so that $\mathsf{stab}(\{1,2,3\}) = S_3$.	2017-04-28 08:14:37	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 784, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9987754821777344, "perplexity": 58.9877846152899}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-17/segments/1492917122886.86/warc/CC-MAIN-20170423031202-00432-ip-10-145-167-34.ec2.internal.warc.gz"}
https://www.albert.io/ie/ap-microeconomics/consumer-surplus-consumer-surplus-for-book-of-spells	Free Version Moderate # Consumer Surplus: Consumer Surplus for Book of Spells APMICR-GXBMYR The maximum price that Harry would have paid for a book of spells is $\$50$, and the maximum Ron would have paid is$\$40$. They are able to purchase the book for $\$30$. What is their total consumer surplus? A$\$10$ B $\$20$C$\$30$ D $\$40$E$\$50$	2017-02-20 05:02:20	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6507976055145264, "perplexity": 13923.424306143575}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-09/segments/1487501170404.1/warc/CC-MAIN-20170219104610-00181-ip-10-171-10-108.ec2.internal.warc.gz"}
https://ncatlab.org/nlab/show/tensor+network	nLab tensor network Contents Context Monoidal categories monoidal categories Contents Idea The term tensor network has become popular in quantum physics for essentially what in monoidal category theory is referred to a string diagrams. The term rose to prominence in quantum physics partly with discussion of finite quantum mechanics in terms of dagger-compact categories but then mainly via its use in holographic entanglement entropy For finite quantum mechanics in $\dagger$-compact categories Application to finite quantum mechanics in terms of dagger-compact categories… (see there). For holographic entanglement entropy Application to holographic entanglement entropy (…) graphics grabbed from Harlow 18 graphics grabbed from Harlow 18 In this context the Ryu-Takayanagi formula for holographic entanglement entropy has an exact proof PYHP 15, Theorem 2. References • Jacob Biamonte, Ville Bergholm, Tensor Networks in a Nutshell, Contemporary Physics (arxiv:1708.00006) In holographic entanglement entropy The use of tensor networks as a tool in holographic entanglement entropy goes back to • Brian Swingle, Entanglement Renormalization and Holography (arXiv:0905.1317) • Brian Swingle, Constructing holographic spacetimes using entanglement renormalization (arXiv:1209.3304) Further interpretation in terms of quantum error correcting codes is due to reviewed in • Alexander Jahn, Marek Gluza, Fernando Pastawski, Jens Eisert, Majorana dimers and holographic quantum error-correcting codes (arXiv:1905.03268) In higher parallel transport Discussion in relation to higher parallel transport: Last revised on October 15, 2019 at 08:14:56. See the history of this page for a list of all contributions to it.	2019-10-21 02:29:53	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 2, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6925899386405945, "perplexity": 4416.957326205799}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-43/segments/1570987751039.81/warc/CC-MAIN-20191021020335-20191021043835-00082.warc.gz"}
https://brilliant.org/discussions/thread/incho-2016/	# INChO- 2016 Can anyone please where INChO is held and how to prepare for it I have cleared NSEC Please Brilliantians give me some tips Especially those who have qualified for it Note by Soham Dibyachintan 4 years, 3 months ago This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science. When posting on Brilliant: • Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused . • Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone. • Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge. MarkdownAppears as italics or _italics_ italics bold or __bold__ bold - bulleted- list • bulleted • list 1. numbered2. list 1. numbered 2. list Note: you must add a full line of space before and after lists for them to show up correctly paragraph 1paragraph 2 paragraph 1 paragraph 2 [example link](https://brilliant.org)example link > This is a quote This is a quote # I indented these lines # 4 spaces, and now they show # up as a code block. print "hello world" # I indented these lines # 4 spaces, and now they show # up as a code block. print "hello world" MathAppears as Remember to wrap math in $$ ... $$ or $ ... $ to ensure proper formatting. 2 \times 3 $2 \times 3$ 2^{34} $2^{34}$ a_{i-1} $a_{i-1}$ \frac{2}{3} $\frac{2}{3}$ \sqrt{2} $\sqrt{2}$ \sum_{i=1}^3 $\sum_{i=1}^3$ \sin \theta $\sin \theta$ \boxed{123} $\boxed{123}$ Sort by: INChO is one exam which isn't dependent on how much exceptions you know in chemistry, it's all about how you apply it, people who have good physical chemistry are certainly having an advantage. But I would say just do it the way you want, you would not need any tips on how to do it. - 4 years, 3 months ago can you tell me where it is held - 4 years, 3 months ago Check the details on the HBCSE site. - 4 years, 3 months ago thanks - 4 years, 3 months ago - 4 years, 3 months ago - 4 years, 3 months ago @Soham Dibyachintan Hey there, even I am giving it today. Where are you giving( me : S.G.T.B Khalsa College, New Delhi) We'll try to discuss after the day - 4 years, 2 months ago I am not giving it I am severely ill since three days and I can't travel to the centre. My eyelids have swollen due to allergies and there are other health problems too The regret of not giving it will last lifelong - 4 years, 2 months ago Wish you heal at the earliest - 4 years, 2 months ago - 4 years, 2 months ago attempted barely 50 percent organic went over me - 4 years, 2 months ago Hey, were you selected from UP?? How much u had scored in NSEC? - 4 years, 2 months ago 130 something U? - 4 years, 2 months ago no, i didn't qualified nsec(118).....only nsea and nsep... - 4 years, 2 months ago how was inpho? - 4 years, 2 months ago Not good......expecting only 20-25/ 50.....what about you?? - 4 years, 2 months ago 35-40 maybe... cutoff would be very very high this time - 4 years, 2 months ago I got 137 - 4 years, 2 months ago Got score card? How much? - 4 years, 1 month ago	2020-04-05 23:20:51	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 8, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9875917434692383, "perplexity": 6884.655627665593}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-16/segments/1585371611051.77/warc/CC-MAIN-20200405213008-20200406003508-00501.warc.gz"}
https://www.physicsforums.com/threads/how-to-prove-the-stretching-of-space.659192/page-2	# How to prove the stretching of space To my understanding the cosmological redshift in the Milne Cosmos is a Dopplershift, but is due to expansion in the empty FRW Universe. Comoving observers can be defined for both cases. If that is correct, the choice of the interpretation of the redshift depends on a transformation of coordinates, not on physics. The Milne model is equivalent to an empty RW-manifold and is mathematically a subset of Minkowski space-time. The FOs are those observers moving orthogonally to the "preferred" hypersurfaces (with hyperbolic geometry) foliating this RW-manifold. There is no alternative set of observers involved here. Since this RW-manifold is flat, the corresponding cosmological redshifts must of course be interpreted as Doppler shifts in flat space-time. This has nothing to do with transformations of coordinates. In this view (I might be wrong) I don't understand your remark, that the other set of observers (Milne, e.g.) is irrelevant. Arn't there just interchangeable discriptions for the same universe? Why then have a preference for one of these? If it contains mass, I guess these discriptions are more complicated which however shouldn't influence in principle the reasoning. What alternative set of observers do you have in mind? Perhaps you are thinking of the set of (non-expanding) observers moving orthogonally to some foliation of Minkowski space-time into flat hypersurfaces? If so, yes, these observers are irrelevant for interpretations of cosmological redshifts in the empty RW-model. (Note that the empty RW-manifold is only a SUBSET of Minkowski space-time, so the FOs and these alternative observers do not really describe "the same universe".) Last edited: Gold Member The Milne model is equivalent to an empty RW-manifold and is mathematically a subset of Minkowski space-time. The FOs are those observers moving orthogonally to the "preferred" hypersurfaces (with hyperbolic geometry) foliating this RW-manifold. There is no alternative set of observers involved here. Since this RW-manifold is flat, the corresponding cosmological redshifts must of course be interpreted as Doppler shifts in flat space-time. This has nothing to do with transformations of coordinates. On the other side the recession velocities depend on the choosen coordinates. In RW coordinates they are a function of the Hubble Constant, whereas in Minkowskian coordinates they are distance/time (Special Relativity). So, according to that the cosmological redshift depends on H, or can be described by the special relativistic Doppler formula, respectively. Hereby I take reference to the thesis of Tamara Davis, Chapter 4 - The empty universe, http://www.dark-cosmology.dk/~tamarad/papers/thesis_complete.pdf. But as you say, both models (Milne and empty RW) are equivalent, so, the redshifts should not depend on coordinates and are Doppler shifts for both models therefore. Somewhere my reasoning must be wrong. I appreciate any help. Another point. The empty RW model is negatively curved (and therefore expands?). This curvature means the geometry of space, right? The spacetime is flat, which is common to both models. On the other side the recession velocities depend on the choosen coordinates. In RW coordinates they are a function of the Hubble Constant, whereas in Minkowskian coordinates they are distance/time (Special Relativity). So, according to that the cosmological redshift depends on H, or can be described by the special relativistic Doppler formula, respectively. Hereby I take reference to the thesis of Tamara Davis, Chapter 4 - The empty universe, http://www.dark-cosmology.dk/~tamarad/papers/thesis_complete.pdf. It is not a good idea to introduce coordinate-dependent quantities if they are not directly related to coordinate-free objects. That is, the proper objects to consider are the 4-velocities of the FOs and not some independently defined "recession velocities". The question of what speed to use in the SR Doppler formula is then answered by performing the following procedure (see J.V. Narlikar, American Journal of Physics 62, 903-907 (1994) for the mathematical details). Parallell-transporting the 4-velocity of the emitting FO along a null geodesic to the receiving FO and projecting the resulted parallell-transported 4-velocity into the local rest frame of the receiving FO yields a 3-velocity that can be put into the SR Doppler formula to give the desired redshift. This procedure is independent of any choice of coordinates; it depends only on the 4-velocities of the emitting and receiving FOs and on the space-time geometry. timmdeeg:4205749 said: But as you say, both models (Milne and empty RW) are equivalent, so, the redshifts should not depend on coordinates and are Doppler shifts for both models therefore. Somewhere my reasoning must be wrong. I appreciate any help. The reason for the confusion is that the defined "recession velocities" are not components of any coordinate-free space-time objects. Therfore, any reference to such "recession velocities" is fraught with danger and should be avoided. timmdeeg:4205749 said: Another point. The empty RW model is negatively curved (and therefore expands?). This curvature means the geometry of space, right? The spacetime is flat, which is common to both models. Right. Gold Member The reason for the confusion is that the defined "recession velocities" are not components of any coordinate-free space-time objects. Therfore, any reference to such "recession velocities" is fraught with danger and should be avoided. Right. Thanks. If I understood you correctly, the redshift observed between FOs depends in the non-empty RW model on whether these are closed, flat, or open and on the space-time curvature and thus not on the choice of coordinates (i). Would you please specify in which cases the redshift is purely gravitational and gravitational/kinematic respectively, including the Lambda-CDM model, the universe in which we live. (i) The reason for this is simple: in the RW-models there is a set of "preferred" observers (the so-called "fundamental observers" (FOs)) defining the cosmic redshift; i.e., the high symmetry of the RW-manifolds implies that they can be foliated in a "preferred" way such that the spatial hypersurfaces are homogeneous and isotropic. In the sense, that there is no other choice, so my understanding. Gold Member Thanks to everybody for your explanations. The problem with doing the measurement in a void is you'd need to do it in an expanding void, which means having the test masses extremely far apart and far away from any other matter in the universe, which makes it an undoable experiment (at least for the forseeable future). This is because the local space-time around massive objects is not expanding. The increase in measured distances is real. But whether you interpret this increase in distances as a velocity is, well, up to your interpretation. Perhaps it is legitimate to overcome the problems in the void by imagining a gedanken experiment. But I guess, even then your final statement would be the same. I am free to interpret any increase in distance in this or that way. So, The "stretching of space" picture is precisely the picture under which the redshift is a gravitational phenomenon. we talke about a picture. And therefore the answer is: The stretching of space being not something truely physical can not be measured. Nevertheless there is From general relativity (specifically, the geodesic equation), it is seen that the momentum of a particle is inversely proportional to the expansion (the scale factor, a(t)). From de Broglie, this becomes a statement about the wavelength of photons -- as space expands, the wavelength of light must increase. some physical support for this "picture". It's still a bit confusing, "as space expands, the wavelength of light must increase." You didn't say, "as space is stretched ...", but I wonder, if this was meant. Perhaps one should careful distinguish between expansion and stretching. The universe expands according to a(t), but the expansion isn't necessarily a true stretching of space effect. Please don't hesitate to correct if I said something wrong. Chalnoth So,we talke about a picture. And therefore the answer is: The stretching of space being not something truely physical can not be measured. That's not an accurate take-away. The correct statement is that there is a real physical phenomenon here, and one correct description of that phenomenon is that it is a stretching of space. There are other seemingly-different but nevertheless also completely correct descriptions of the exact same physical phenomenon. This is one of the weird things about physics: it is sometimes possible to describe the exact same thing in seemingly completely different ways, while actually describing the same system. And sometimes the difference is so different that it is hard to believe that it's actually the same system being described (e.g. sometimes you can describe a system using different numbers of spatial dimensions and still be describing the same system). If I understood you correctly, the redshift observed between FOs depends in the non-empty RW model on whether these are closed, flat, or open and on the space-time curvature and thus not on the choice of coordinates (i). Choice of coordinates also DOES affect the outcome. The standard cosmological measure involved comoving coordinates, an observer at rest with respect to the CMBR. The description of curved 4D spacetime as 'expanding' or 'increasing distances' over time depends on a choice of 3+1D split. We use one that is convenient but not unique.) The stretching of space being not something truely physical can not be measured. Chalnoth posed a wonderfully insightful physical explanation in another discussion: The integrated Sachs-Wolf effect is the clearest, independent [of supernova], evidence of dark energy [the strange anti-gravity effect that powers expansion]… Photons entering a large gravitational well [like a galactic supercluster] get a gravitional energy boost upon entering the region causing a small gravitational blue shift. Upon exiting, they lose this free energy and redshift back to their original energy state upon exiting - almost. If the universe were flat and static, the net effect would be zero. In an expanding universe, the photon takes so long to pass through the gravity well that it gets to keep a small amount of the blue shift it acquired on the way in due to expansion and the resulting dilution of gravity. This extra energy shows up as a slight anisotropy in the CMB photons passing through a supercluster or supervoid [the effect is just the opposite for CMB photons passing through a supervoid]. Seehttp://arxiv.org/abs/0805.3695 for discussion. Gold Member This is one of the weird things about physics: it is sometimes possible to describe the exact same thing in seemingly completely different ways, while actually describing the same system. Yes, I will have to accept this truth, though being weird. But your remark reminds me strongly of an article of R.L.Jaffe, wherein he shows that the measurable Casimir force can not only be described by vacuum fluctuations (as usual), but without taking reference to the vacuum as a van der Waals force as well. Last edited: Gold Member If I understood you correctly, the redshift observed between FOs depends in the non-empty RW model on whether these are closed, flat, or open and on the space-time curvature and thus not on the choice of coordinates (i). Choice of coordinates also DOES affect the outcome. The standard cosmological measure involved comoving coordinates, an observer at rest with respect to the CMBR. My remark refers to explanations of Old Smuggler. From his perspective "Coordinates are irrelevant for interpretations of redshifts in the RW-models". Chalnoth posed a wonderfully insightful physical explanation in another discussion: Thanks, I wasn't aware of that but read the paper meanwhile. The independent confirmation of the dark energy is very surprising and confirms the Lambda CMB model. Also, the ISW-effect is in accordance with the stretching of space description. Here is an interesting 8 page paper I stumbled across in my notes: Expanding Space: the Root of all Evil? http://arxiv.org/PS_cache/arxiv/pdf/0707/0707.0380v1.pdf Several of these 'issues' have been discussed in other threads......I found these are insightful: ...the expansion of space is neither more nor less than the increase over time of the distance between observers at rest with respect to the cosmic fluid in terms of the FRW metric. With this metric..... the density and pressures of cosmological fluids must change over cosmic time, and it is this change that represents the basic property of an expanding (or contracting) universe. The proper time for …..privileged observers at rest with regards to the cosmic fluid ticks at the same rate as cosmic time and hence the watches of all privileged observers are synchronised. In an expanding universe, the change of the metric implies that the physical distance between any two privileged [comoving] observers increases with time... The Hubble flow is then viewed as a purely kinematical phenomenon — objects recede because they have been given an initial velocity proportional to distance. the velocity of [a] particle due its motion relative to the Hubble flow (or equivalently the homogeneous fluid defining the FRW metric) must be less than the speed of light; its velocity due to the increase of the scale factor is not restricted in this way….. cosmological redshift is not, as is often implied, a gradual process caused by the stretching of the space a photon is traveling through. Rather cosmological redshift is caused by the photon being observed in a different frame to that which it is emitted. Gold Member Here is an interesting 8 page paper I stumbled across in my notes: Expanding Space: the Root of all Evil? http://arxiv.org/PS_cache/arxiv/pdf/0707/0707.0380v1.pdf Thank you, this paper is very interesting and worthwile to be read. It moreover shows the controversy between cosmologists regarding thought experiments (one of my questions), Expanding Space: the Root of all Evil? page 2: To illustrate how short this pragamatic formalism falls being platitude, one need no further than Abramowicz et al. (2006), in which a thought experiment of laser ranging in an FRW Universe is proposed to 'prove' that space must expand. This is sensibly refuted by Chodorowski (2006b), but followed by a spurious counter-claim that such a refutation likewise proves space does not expand. which makes it not easier to improve one's own understanding. To me this statement Expanding Space: the Root of all Evil? page 2: The expansion of space is no more extant than magnetic fields are and exists only as a tool for understanding the unambiguous predictions of GR, not a force-like term in a dynamical equation. sounds very agreeable. Expanding Space: the Root of all Evil? page 7: The key is to make it clear that the cosmological redshift is not, as is often implied, a gradual process caused by the stretching of the space a photon is traveling through. Rather cosmological redshift is caused by the photon being observed in a different frame to that which it is emitted. In this way it is not as dissimilar to a Doppler shift as is often implied. The difference between frames relates to a changing background metric rather than a different velocity. Is this proposal in accordance with the parallel transport of the 4-velocity vector? And how about this thought experiment: Supposed the universe doesn't expand at the time of emission and absorption but expands during the photon's travelling. What kind of shift if any will be measured? Chalnoth And how about this thought experiment: Supposed the universe doesn't expand at the time of emission and absorption but expands during the photon's travelling. What kind of shift if any will be measured? The observed redshift will be equal to the total amount of expansion between the emission and absorption of the photon, regardless of what the rate of that expansion was at different times. Sorry for the late reply (no internet connection for the last week). If I understood you correctly, the redshift observed between FOs depends in the non-empty RW model on whether these are closed, flat, or open and on the space-time curvature and thus not on the choice of coordinates (i). The redshift depends only on the 4-velocities of the FOs and on the space-time geometry. (This is most easily seen by using said procedure of parallel-transport.) On the other hand, the INTERPRETATION of the redshift depends on the spatial geometry, since the spatial geometry is crucial for determining how well a flat space-time connection approximates the curved space-time connection. Would you please specify in which cases the redshift is purely gravitational and gravitational/kinematic respectively, including the Lambda-CDM model, the universe in which we live. As far as the Lambda-CDM model is based on RW-models, the properties of the RW-models apply (see below). If inhomogenities are taken into account, the effects of these come in addition. In RW-models with flat or spherical space sections, the redshift is entirely due to the non-flat connection and thus indirectly to space-time curvature (i.e., "gravitational"). (See, e.g., arXiv:0911.1205.) For RW-models with hyperbolic space sections things are more complicated, and some part of the redshift is "kinematic" (meaning that some part of the redshift survives even if one replaces the curved space-time metric with a flat one). To decide how much of the redshift is "kinematic", a recipe for spectral shift split-up into "kinematic" and "gravitational" parts is necessary (this can be done unambiguously, at least for small distances). Last edited: Would you please specify in which cases the redshift is purely gravitational and gravitational/kinematic respectively, including the Lambda-CDM model, the universe in which we live. As I understand the consensus from earlier discussions on this subject, such a split in our universe, represented by the Lambda-CDM model, over cosmological distances is arbitrary. Gold Member The observed redshift will be equal to the total amount of expansion between the emission and absorption of the photon, regardless of what the rate of that expansion was at different times. Thanks for this clear and unambiguous answer. Gold Member The redshift depends only on the 4-velocities of the FOs and on the space-time geometry. (This is most easily seen by using said procedure of parallel-transport.) On the other hand, the INTERPRETATION of the redshift depends on the spatial geometry, since the spatial geometry is crucial for determining how well a flat space-time connection approximates the curved space-time connection. . In RW-models with flat or spherical space sections, the redshift is entirely due to the non-flat connection and thus indirectly to space-time curvature (i.e., "gravitational"). (See, e.g., arXiv:0911.1205.) These authors argue that in order to interpret the cosmological redshift in terms of a Doppler effect in non-expanding Minkowskian space-time the observer would have to move away from himself and thus claim (spatial curvature >= 0) "The Doppler interpretation is clearly self-contradictory (page 5). But this is relativised later (page 6): "Hence, ironically in the context of the recent debate, parallel-transport of four-velocities along photons path can allow cosmological redshifts to be interpretet as a relativistic Doppler effect without the contradiction presented here, provided that the concept of expanding space is added to the Minkowski space-time ... and provided that the velocity is thought of as being tied to a path and not as a global concept." But irrespective of such an ambiguous debate I have a problem to understand the cosmological redshift in the sense of a purely gravitational shift. It is quite clear that a photon looses energy und thus becomes redshifted as it climbs out of a gravitational field or in other words as it moves away from a mass (i). In contrast the photon traveling through homogeneous space doesn't move away from a gravitational center, but undergoes a redshift (= looses energy) as well. How shall I understand this (obvious?) discrepancy? You mentioned already the dependence on spatial geometrie ... . Is there any explanation besides the stretched wavelenght picture as simpel as (i)? These authors argue that in order to interpret the cosmological redshift in terms of a Doppler effect in non-expanding Minkowskian space-time the observer would have to move away from himself and thus claim (spatial curvature >= 0) "The Doppler interpretation is clearly self-contradictory (page 5). But this is relativised later (page 6): "Hence, ironically in the context of the recent debate, parallel-transport of four-velocities along photons path can allow cosmological redshifts to be interpretet as a relativistic Doppler effect without the contradiction presented here, provided that the concept of expanding space is added to the Minkowski space-time ... and provided that the velocity is thought of as being tied to a path and not as a global concept." Yes, cosmological redshifts can always be interpreted as Doppler shifts in CURVED space-time. However, they cannot in general be interpreted as Doppler shifts in FLAT space-time, and it is the latter meaning that is usually understood with "kinematic" redshift. timmdeeg:4216930 said: But irrespective of such an ambiguous debate I have a problem to understand the cosmological redshift in the sense of a purely gravitational shift. It is quite clear that a photon looses energy und thus becomes redshifted as it climbs out of a gravitational field or in other words as it moves away from a mass (i). In contrast the photon traveling through homogeneous space doesn't move away from a gravitational center, but undergoes a redshift (= looses energy) as well. How shall I understand this (obvious?) discrepancy? You mentioned already the dependence on spatial geometrie ... . Is there any explanation besides the stretched wavelenght picture as simpel as (i)? There is no obvious intuitive picture to decide the question of "kinematic" versus "gravitational" interpretations, I'm afraid. (If there were, this question would not have been debated so vigourously in the literature.) However, as I have mentioned earlier, there exists a general procedure to decide the matter for small distances, and for arbitrary space-times. That is, choose a pair of fixed ("close") observers with given world lines. Calculate spectral shifts obtained by exchanging photons between these observers. Then replace the space-time geometry in the relevant region with flat space-time (holding the chosen world lines and the coordinate system fixed). Calculate spectral shifts again, but now with the flat space-time geometry. If the latter calculation yields no spectral shifts at all, the spectral shifts obtained in the first calculation must be entirely due to space-time curvature, i.e., "gravitational". For example, in the Schwarzschild metric, the chosen observers defining gravitational spectral shifts are observers with fixed spatial Schwarzschild coordinates. The flat space-time limit of this metric is obtained by setting the mass M=0. Now it is rather obvious that there is no spectral shift between the chosen observers in the Schwarzscild metric with M=0, so the spectral shift obtained when M is nonzero must be purely gravitational. A similar situation to that of the Schwarzschild metric occurs for RW-models with flat or spherical space sections, so the spectral shifts obtained between the FOs in these models must also be purely gravitational. Last edited: Chalnoth Yes, cosmological redshifts can always be interpreted as Doppler shifts in CURVED space-time. However, they cannot in general be interpreted as Doppler shifts in FLAT space-time, and it is the latter meaning that is usually understood with "kinematic" redshift. I don't understand what you mean. In flat space-time, there is no curvature, and thus in general you don't expect there to be any gravitational redshift at all, meaning that any observed redshift would be purely kinematic (of course, you might still be able to impose what looks like gravitational redshift with an appropriate coordinate choice, such as Milne coordinates). Either way, though, our space-time does have a definite degree of overall curvature, as it must due to the fact that our universe is not empty (more pedantically-stated, the average energy density of our universe is non-zero). Regardless of the overall curvature, however, the amount of the redshift that is attributed to gravitation and the amount attribute to motion of the emitter or observer is still arbitrary. Some choices may seem more or less natural to some people, but many choices are possible in any event. Chalnoth: Regardless of the overall curvature, however, the amount of the redshift that is attributed to gravitation and the amount attribute to motion of the emitter or observer is still arbitrary. Some choices may seem more or less natural to some people, but many choices are possible in any event. That seemed to be the conclusion from another discussion on this topic, with some insights that may be of interest: [Note, especially the change in scale factor and,in Schwarzschild coordinates, the change in velocity, comments.] https://www.physicsforums.com/showthr...nt+flow&page=4 [Broken] edit: oops, that link no longer works??? [In the great 2007 thread Wallace, Chronos and Oldman take a different view than expressed here [and there] by Marcus...you can read the posts from the 40's thru 50's and see the pros and cons.] I do think it is better to think of (photons) as being redshifted by being observed in a different frame ......Now as t ticks along, the scale factor a(t) increases. Therefore two observers who are both at rest wrt to the CMB, but who have different times t will therefore be in different frames (have different metrics). This is what leads to photons being redshifted when observed and emitted at different times. I tend to agree, photons are not redshifted by traveling through the universe, they are redshifted only because they are observed in a different frame from which they were emitted. Marcus: # 48] I am not comfortable with that because among other things I see cosmologists doing inventories of the energy density which are implicitly estimated IN A CMB FRAME.... These 'conflicting' viewpoints stem from this as explained by Chalnoth elsewhere: " … You get some total redshift for faraway objects due to cosmological expansion. How much of that redshift is due to the Doppler shift# and how much is due to the expansion between us and the far away object is completely arbitrary." # Doppler shift is based on [relative velocity] frame based differences, not expansion, Hence photon frequency and wavelength can be viewed as fixed just like in a static Spacetime.. Doppler shift is a particular explanation of redshift, with a particular formula. Marcus: Don’t think of the redshift as a Doppler [relative velocity] effect. It is not the result of some particular speed. The formula involves the entire [varying] factor by which distances have been expanded during the whole time the light has been traveling. PeterDonis: The law governing the relationship of emitted to observed photon energies (or frequencies) is general and applies in any spacetime. The 4-momentum of the photon gets determined at the emitter; then it gets parallel transported along the photon's worldline from emitter to observer; then you contract that 4-momentum with the observer's 4-velocity to get the observed energy (or frequency if you throw in a factor of Planck's constant). That "parallel transport" process is actually where the "redshift" occurs in an expanding universe; the expansion alters the 4-momentum of the photon as it travels (or at least that's one way of looking at it), whereas in a static universe the photon's 4-momentum would "stay the same" as it traveled. There's another complication here, btw; what about the gravitational redshift of photons in Schwarzschild spacetime? Here the "change" with changing radius is actually in the 4-velocity of the observer; the photon's 4-momentum stays the same, but the 4-velocities of "hovering" observers are different at different radii, so they contract differently with the constant photon 4-momentum. PAllen: Redshift is a measured shift in received frequency versus emitted frequency. Doppler [shift] refers to one of two formulas (pre-relativistic; relativistic) for relating redshift to velocity. Doppler shift is a particular explanation of redshift, with a particular formula. It is not a measure of redshift. Where the speeds of source and the receiver relative to the medium are lower than the velocity of waves in the medium, the classical Doppler shift formula; in cosmology, where we deal with lightspeed 'c' and recessional 'velocities' greater than 'c' we need the relativistic version of the formula. [Doppler is like a radar speed trap: The radar signal goes out and returns and keeps the same 'color', but we record the difference in wavelength as a speed measure.] Cosmological redshift is typically considered distinct from Doppler redshift because it is a relation between distance and redshift rather than speed and redshift, under the assumption that both source and target are motionless relative to center of mass of the local matter (here, local is quite large - galaxy or galaxy cluster). Last edited by a moderator: Chalnoth https://www.physicsforums.com/showthr...nt+flow&page=4 [Broken] edit: oops, that link no longer works??? You copied and pasted the shortened display text. Try right clicking and copying the link itself. Easier still if you copy the link of the post itself (which can be found by clicking the post number next to the post). Last edited by a moderator: I don't understand what you mean. In flat space-time, there is no curvature, and thus in general you don't expect there to be any gravitational redshift at all, meaning that any observed redshift would be purely kinematic (of course, you might still be able to impose what looks like gravitational redshift with an appropriate coordinate choice, such as Milne coordinates). You misunderstand. If the redshift in a general RW-model were purely "kinematic", the procedure described in #43 would yield the same redshift for small enough distances both for the curved space-time geometry and for flat space-time. Since this does not happen in general, the nature of the redshift in a general RW-model cannot be interpreted as purely kinematic. The empty RW-model is an exceptional case since the space-time geometry is flat, so in this case, the observed redshift would be purely kinematic. But this does not apply to a general RW-model where the space-time geometry is not flat. The interpretation of the redshift in a general RW-model depends on the spatial geometry. (Only if the spatial geometry is hyperbolic there will be a non-zero "kinematic" contribution to the redshift.) Regardless of the overall curvature, however, the amount of the redshift that is attributed to gravitation and the amount attribute to motion of the emitter or observer is still arbitrary. Some choices may seem more or less natural to some people, but many choices are possible in any event. This may seem reasonable, but a proper mathematical analysis shows that it is simply not true. For example, no "kinematic" interpretation is consistent with the fact that the procedure described in #43 yields no cosmic redhifts between FOs for e.g., an arbitrary RW-model with flat space sections if the space-time geometry is replaced with flat space-time. Please do this (simple) calculation to convince yourself. Chalnoth This may seem reasonable, but a proper mathematical analysis shows that it is simply not true. For example, no "kinematic" interpretation is consistent with the fact that the procedure described in #43 yields no cosmic redhifts between FOs for e.g., an arbitrary RW-model with flat space sections if the space-time geometry is replaced with flat space-time. Please do this (simple) calculation to convince yourself. I think the problem is that the procedure in #43 is still an arbitrary way of distinguishing between gravitational redshift and kinematic redshift. And I'm not sure it works in any event, because the relative velocity of two objects separated by some distance is arbitrary. If I select some coordinates with an interpretation of velocity which precisely gives the relative velocity between two objects in FRW space-time which would correspond to a Doppler shift, and then replace the space-time with flat space-time in those same coordinates, I'll have nothing but a Doppler shift. I think the problem is that the procedure in #43 is still an arbitrary way of distinguishing between gravitational redshift and kinematic redshift. And I'm not sure it works in any event, because the relative velocity of two objects separated by some distance is arbitrary. Nothing is arbitrary with the procedure described in #43. That is, the world lines of the FOs and their 4-velocities are not arbitrary and neither are the null curves. (For sufficiently small distances the effects of geodesic deviation can be neglected, so the world lines of the FOs are still geodesics and the null curves are still null when replacing the curved space-time metric with a flat space-time metric.) Since the redshift is obtained by parallel-transporting the 4-velocity of the emitter along a null curve to the observer, this shows that the redshift obtained using the procedure described in #43 is unambiguous, only depending on the space-time geometry. Thus, changing the space-time geometry from curved to flat will in general change the redshift, so it cannot be interpreted as purely kinematic. Any concept of "relative velocity of two objects separated by some distance" is not part of the procedure; this is irrelevant since the coordinate-free concept of parallel-transport makes it unnecessary. If I select some coordinates with an interpretation of velocity which precisely gives the relative velocity between two objects in FRW space-time which would correspond to a Doppler shift, and then replace the space-time with flat space-time in those same coordinates, I'll have nothing but a Doppler shift. Whatever it is you are thinking of here, it would not correspond to selecting fixed observers (the FOs) and then changing the space-time geometry so the argument is quite irrelevant. Last edited: Chalnoth	2020-11-26 04:42:06	{"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8033356070518494, "perplexity": 693.677258596376}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-50/segments/1606141186414.7/warc/CC-MAIN-20201126030729-20201126060729-00579.warc.gz"}
https://www.physicsoverflow.org/27104/levi-civita-connection-on-sphere-in-the-vielbein-formalism	# Levi-Civita connection on a sphere in the vielbein formalism + 4 like - 0 dislike 1063 views I am trying to learn the vielbein formalism and have a question for the example of the Riemann sphere $S^2$. I am afraid my question is rather elementary, as it seems to be a simple sign error. Still, could someone help me figure this out? On the sphere with coordinates $(x,y,z) = (\cos ϕ \sin θ, \sin ϕ \sin θ, \cos θ)$ and metric $ds^2 = dθ^2 + \sin θ^2 dϕ^2$, we can define the zweibein $$e_θ = ∂_θ , \quad e_ϕ = \frac{1}{\sin θ} ∂_ϕ$$ The Levi-Civita connection for the metric is torsion-free, which means $$\nabla_{e_ϕ} e_θ - \nabla_{e_θ} e_ϕ = [e_ϕ,e_θ]$$ A separate calculation shows that $\nabla_{e_θ}e_ϕ = 0$, so we can use this formula to quickly calculate the connection form $ω_{ab}$: $$\nabla_{e_ϕ} e_θ = [e_ϕ,e_θ] = -∂_θ(\frac{1}{\sin θ})·∂_ϕ = \cot θ · e_ϕ \equiv ω_{θ\ \ }^{\ \ ϕ}(e_ϕ) e_ϕ$$ Unfortunately, this calculation seems to be wrong, because it contradicts the statement $$ω^{ϕ\ \ }_{\ \ θ}(e_ϕ) = \cot θ$$ that I found in some lecture notes (formula (2.345)). The connection form is antisymmetric, so one of the two values should be $-\cot θ$, but I can't decide which. Can somebody help me find the source of this sign discrepancy? Ooh, it appears to be a matter of convention for the connection form $\omega$! It looks like physicists use the notation $\nabla_X e_b = \omega^a_{\ \ b}(X) e_a$ whereas mathematicians tend to use the notation $\nabla_X e_i = \sum_j \omega^j_i(X) e_j$. Consequently, physicists write the Cartan structure equations as $\Omega^a_{\ \ b} = d\omega^a_{\ \ b} + \omega^a_{\ \ c} \wedge \omega^c_{\ \ b}$ whereas mathematicians write them as $\Omega^{j}_i = d\omega^j_i - \sum_k \omega^k_i \wedge \omega^j_k$ with a different sign. I tried to derive the formalism myself, and accidentally picked a convention that is that is neither the mathematician's nor the physicists' convention. Will put this into a real answer soon. + 3 like - 0 dislike The calculation is correct, but it appears that there are different conventions in use for the connection 1-form $\omega$. In the question, the definition for $ω$ used to the very right of the $\equiv$-sign in the question is not one of the standard conventions, that's why the signs differ. Apparently, physicists tend use the notation $\nabla_X e_b = \omega^a_{\ \ b}(X) e_a$ (example), whereas mathematicians tend to use the notation $\nabla_X e_i = \sum_j \omega^j_i(X) e_j$ (example [pdf]). Consequently, physicists write the Cartan structure equations for the curvature as $\Omega^a_{\ \ b} = d\omega^a_{\ \ b} + \omega^a_{\ \ c} \wedge \omega^c_{\ \ b}$ whereas mathematicians write them as $\Omega^{j}_i = d\omega^j_i - \sum_k \omega^k_i \wedge \omega^j_k$ with a different sign. By the way, instead of calculating the commutator, the connection 1-form can also be calculated from the Cartan structure formula for the dual frame. Using the dual frame $θ^θ = dθ$ and $θ^ϕ = \sin θ \, dϕ$, the equations read $$0 = dθ^θ + ω^θ_{\ ϕ} \wedge θ^ϕ = d(dθ) - \sin θ\, ω^ϕ_{\ θ} \wedge dϕ = -\sin θ\, ω^ϕ_{\ θ} \wedge dϕ$$ $$0 = dθ^ϕ + ω^ϕ_{\ θ} \wedge θ^θ = d(\sin θ\, dϕ) - dθ \wedge ω^ϕ_{\ θ} = dθ \wedge (\cos θ\, dϕ - ω^ϕ_{\ θ}) .$$ The first equation implies that the form $ω^ϕ_{\ θ}$ is a multiple of the form $dϕ$. The second equation implies that the prefactor is $\cos θ$, so we have $$ω^ϕ_{\ θ} = \cos θ\, dϕ = \cot θ·θ^ϕ .$$ answered Feb 15, 2015 by (775 points)	2019-07-23 20:07:39	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9511538147926331, "perplexity": 254.40158289253512}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-30/segments/1563195529664.96/warc/CC-MAIN-20190723193455-20190723215455-00280.warc.gz"}
https://physics.stackexchange.com/questions/96254/oil-drop-experiment-and-quantization-of-charge	# Oil drop experiment and quantization of charge How to systematically show that the resulting charges in oil drop experiment are integers multiplied by $e$ in other word how to extract $e$ from the data? • This question is very vage. I have no clue what you are asking. You don't give any background info. – Bernhard Feb 1 '14 at 12:39 • More on Millikan's oil drop experiment: physics.stackexchange.com/search?q=millikan+oil – Qmechanic Feb 1 '14 at 14:21 • I give a simplified version of this question as a problem in Modern Physics. Students with OCD-like tendency find some pretty decent solutions. That said there is often some confusions about exactly what the data is in Milikan's experiment and you should almost certainly say what you think it is so that everyone is on the same foot. – dmckee Feb 1 '14 at 14:41 • @richard: Isn't this something which is taught in basic freshman physics? Also, the article you linked explains everything in complete detail, so I don't understand the point of the question. – DumpsterDoofus Feb 1 '14 at 15:45 • As far as I know you fiddle around with different values until you find one that gives the best fit. I must admit I don't know of a systematic way to do it. I would be interested in an explanation for some systematic procedure if someone would like to provide an answer. Incidentally, I suspect the downvote and criticism stem from a misunderstanding of what you're asking. The question is how, starting from a list of charges, do you find the quantity that they are all integer multiples of. Especially bearing in mind there will be experimental errors in the charges. – John Rennie Feb 1 '14 at 16:03 To address John Rennie's comment in the comment section regarding the existence of a systematic, human-guess-independent algorithm for determining the LCM of a data series in the presence of significant experimental error and without the aid of single-electron-charged droplets to make a human-sensible guess: a = 12.5654; L = 400; list = Table[a (RandomInteger[{6, 35}] + RandomReal[{-0.25, 0.25}]), {k, L}]; f[b_] := Module[{g = Nearest[b Range[L]]}, Sum[Abs[g[list[[k]]][[1]] - list[[k]]], {k, L}]/b]; ListPlot[list, PlotRange -> All] Plot[f[x], {x, 6, 15}, PlotRange -> All] There's no way a human could look at that plot of the noisy raw data and guess the LCM, but a computer can handle it just fine. Note that this is reliably indicating the LCM even though the "measurement" error is on the order of 50%. I used uniformly-distributed error, but it works with Gaussian-distributed errors just as fine. As an interesting mathematical aside, in the absence of noise the LCM appears as the largest zero of the merit function, which has a sequence of zeros whose density of zeros tends as $(a x)^{-1}$ where $a$ is the LCM and $x$ is the guess. As $x\rightarrow 0$ the there is an oscillatory singularity, and for $x>a$, there are no further zeros. • Is there any chance that you can explain the code also ? Plus, it is Mathematica ? – onurcanbektas Feb 7 '18 at 4:45 If the experiment was done with sufficient accuracy, simply plotting the calculated charge values should give obvious clustering. (Two measurements per particle: mass from free fall velocity, and voltage to achieve zero velocity is how I remember the experiment, but that is from a fifty year-old memory of high-school physics... plot voltage/mass.) R.J.Doe has a set of directions (with an amusing apocalyptic conclusion) on writing up a somewhat different version of the experiment: http://www.phys.ksu.edu/personal/cocke/classes/phys506/aasamplewriteup.htm using both a downward and upward acceleration to give three velocities per particle. I'm wondering if that might have the advantage that you would not need to depend on a previously measured value for the viscosity of air. I see that DumpsterDoofus is expressing annoyance at a lack of effort and suggest perhaps the use of http://webphysics.davidson.edu/applets/pqp_preview/contents/pqp_errata/cd_errata_fixes/section4_5.html to generate dome "data" would mollify him. It would be more interesting to see data gathered this way than to look at his generation of data which I suspect is very much unlike what was gathered by Millikan. (I also disagree that we could not have done such data analysis without computers.) • Sorry, I probably should've been a little more clear in my answer: I wasn't saying that doing such data analysis without computers is impossible, I was saying that for the example I gave, it would be difficult for humans to see the clustering. In actual charged-drop experiments, there is enough low-integer charged droplets that you can look at the data and guess the answer; the point of my answer was to provide a general, human-independent algorithm for estimating LCM's on noisy data with non-obvious clustering and large amounts of charge. – DumpsterDoofus Feb 1 '14 at 20:20 • But I think the Millikan data was presented (at least as I remeber it) as having typically single or doubly or triply charged droplets, so I think your computer generated example was not a realistically constructed example. – DWin Feb 1 '14 at 20:33 • Yes, you remember correctly. When you do the Millikan droplet experiment, the droplets usually have only a couple electrons on them if made properly, so it's not hard to estimate the charge just by looking at the data, which is why it's done in some freshman physics labs. The point of my answer was to answer the more general question of how to attack the problem for similar types of data analysis involving noisy, quantized data that is not amenable to human attack, rather than just Millikan drop data. – DumpsterDoofus Feb 1 '14 at 20:52 • It is very hard to get down to just a few electrons in practice (though that is the idealized version of the experiment usually presented). In Miliken's actual experiment he ran a single drop up and down until a charge is randomly canceled by ambient ionization. Then you compute the difference in the changes before and after as deduced from the voltage--velocity relationship. Repeat with many drops and multi-neutralization events per drop and you can do very well, but it takes a lot of patience. – dmckee Feb 1 '14 at 21:44	2019-08-23 08:57:25	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5569881796836853, "perplexity": 457.4037639119497}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-35/segments/1566027318243.40/warc/CC-MAIN-20190823083811-20190823105811-00086.warc.gz"}
http://math.stackexchange.com/tags/measure-theory/new	# Tag Info 0 First, a few words about inverse images. If $X$ is a set, then the powerset $\mathcal{P} X$ is a poset. If $f: X \to Y$ is a function, then the inverse image function $\mathcal{P}Y \to \mathcal{P}X$ is order-preserving, and it has both left and right adjoints -- which in particular means it preserves both meets and joins. The left adjoint is the ... 1 The coin tosses can be modelled as a sequence of independent $\text{Ber}\left(\frac{1}{2}\right)$ random variables $(X_n)_{n\in\mathbb{N}}$ Letting $A_n=\{X_n=1\wedge X_{n+1}=1\}$, you want to know $P(\lim\sup_n A_n)$ Consider $\lim\sup_nA_{2n}\subseteq\lim\sup_nA_n$. Notice that $(A_{2n})_{n\in\mathbb{N}}$ are independent. Since ... 1 Firstly, just to clarify, by $\mu\leq \nu$ I expect you mean $\mu(A)\leq \nu(A)$ for $A$. We therefore have that $\mu$ is absolutely continuous wrt $\nu$ (written $\mu\ll\nu$, and why I'm asking about the meaning of your inequality) and so there is a function in $L^1(\nu)$, $\frac{\mathrm d\mu}{\mathrm d\nu}$ such that $$\int_E \frac{\mathrm d\mu}{\mathrm ... 4 Consider using Holder's inequality with conjugate exponents p and p/(p-1):$$\left\|\int_{A_n} f\, d\mu\right\| \le \int_X \chi_{A_n}\|f\|\, d\mu \le \\|\chi_{A_n}\\|_{p/(p-1)} \\|f\\|_p = \mu(A_n)^{(p-1)/p} \\|f\\|_p.$$Since \mu(A_n) \to 0, the result follows. 0 We have to expect that the anwser will depend on a. You have a little problem with your answer when a=1. Treat that case separately (it converges). For a>0,a\ne 1, what you did is fine. You arrive at a one-variable integral where the integrand appears to have two singularities, at 0 and 1. But the singularity at 0 is removable. At 1, the ... 2 An example. F = [0,7] with its usual metric. Then: \mathcal H^s(F) = 0 for s>1, \mathcal H^1(F) = 7, \mathcal H^s(F) = \infty for 0<s<1. Thus, according to your definition, \dim F = \inf\;(1,\infty) = 1. 2 \dim F is not defined as a value for which the Hausdorff dimension equals 0. It's defined as the infimum of a set of such values, which does not mean it has to be a member of the set itself. 1 In general, we can hardly expect to find F,G such that F \subseteq A \subseteq E \subseteq G and \|G \backslash F\| < \epsilon (as \|E \backslash A\|>\epsilon for \epsilon sufficiently small). Hints: Using Urysohn's lemma, prove that the claim holds true for E = \mathbb{R}^d and A open. (That's basically what you already did). Conclude ... 0 If I'm interpreting this correctly, what they mean is this: Suppose f has a discontinuity at some y\in[a,b]. Since f is right-continuous, f_y:=\lim_{x\rightarrow y^+}f(x) exists. Since f has at most countably many discontinuities, we may assume w.l.o.g. that f(y)=f_y. Here, the "upper value" is f_y, which, as f is increasing, satisfies ... 1 For the first part: Define f_n = 1_{[0,1]} for all n (this is a constant sequence), and let f=1_{[1,2]}. Then \int f_n =1 = \int f for all n, but \int \|f_n - f\| =2 for all n. For the second part: Take some enumeration \{I_n\} of all subintervals of [0,1] which have the form [\frac{k}{m},\frac{k+1}{m}] for some integers m,k. Let ... 0 Some examples on [0,1]: For the first one, let f_n(x)=\cos nx, f(x) = 0. For the second one, consider the characteristic functions, in order, of [0,1],[0,1/2],[1/2,1],[0,1/3],[1/3,2/3],[2/3,1], \dots. 1 The following two counterexamples take place in [0,1] with its standard \sigma -algebra and Lebesgue measure. Counterexample for the first: Let (f_{n})_{n \in \Bbb{N}} \stackrel{\text{df}}{=} \left( \chi_{[n,n + 1]} \right)_{n \in \Bbb{N}} and f \stackrel{\text{df}}{=} \chi_{[0,1]} . Counterexample for the second: Let H_{n} denote the ... 0 It looks like on the second slide of page 2 on this PDF file, "All σ-algebras are algebras, and all algebras are semi-rings." 0 I will assume you're working in \mathbb{R}. The definition of outer-measure is here:$$m^{}(E)=\inf \{\sum_{n=1}^{\infty}l(I_n)\| E \subset \bigcup_{n=1}^{\infty}I_n, I_n \text{being disjoint sequence of open intervals}\}$$If E has outer measure zero, that means that for every positive \epsilon, there exists a sequence of disjoint open intervals ... 0 I worked on this for a couple hours and think I have come up with a solution. I found it is easier to prove that B^c is closed: Suppose \sup_{C_r \in \mathcal{C}_r(x_n)} \frac{\mu(C_r)}{m(C_r)} \leq a for x_n \to x. Let C_{k} \in \mathcal{C}_r(x) be a sequence of sets such that \frac{\mu(C_{k})}{m(C_{k})} \to \sup_{C_r \in \mathcal{C}_r(x)} ... 3 Lets consider \Omega=\{1,2,3,4\} \sigma(\{1\}) is the smallest \sigma-algebra which contains 1. So we must take any other elements of P(\Omega) such that the conditions for being a \sigma-algebra are fulfilled. It does clearly contains 1. Also 1^C=\{2,3,4\}. And \Omega,\emptyset. So we have ... 2 Say we are doing this on the real line \mathbb R. Let \mathcal G be the collection of all open sets. We are interested in \sigma-algebras \mathcal F such that \mathcal F \supseteq \mathcal G. There may be many such \sigma-algebras. For example, the power set \mathcal P, consisting of all subsets of \mathbb R is one. But that is the ... 1 if you agree that the borel \sigma-algebra isn't neccessarily all the subsets of the topological space, then you might also agree that P(X) (or 2^X in a different notation, the power set) is a larger \sigma-algebra. Larger means more sets in the \sigma-algebra. Also note that the generated \sigma-algebra by the open sets, which is the borel ... 1 By translating we may assume x=0 for convenience, and then by rescaling we have$$\frac{1}{c_nr^n}\lambda_n(B_r(0)\setminus B_r(y)) = \frac{1}{c_n} \lambda_n(B_1(0)\setminus B_1(y/r)).$$Now observe y/r\to 0 as r\to\infty and say "dominated convergence theorem". The dominated convergence theorem is overkill. To be more elementary, B_1(0)\cap ... 4 No, this isn't true. As an example:$$f(x)=\begin{cases}1 &\text{if } x\in (0,1/2] \\ 0&\text{else}\end{cases}$$and$$g(x)=\begin{cases}1 &\text{if } x\in (1/2,1) \\ 0&\text{else}\end{cases}$$3 Counterexample: Consider (0,1) with Lebesgue measure and the characteristic functions \chi_{(0,\frac{1}{2})} and \chi_{(\frac{1}{2},1)}. Then \chi_{(0,\frac{1}{2})}\chi_{(\frac{1}{2},1)} = 0 but the product of their integrals is \frac{1}{4}. 1 In addition to the ones mentioned above, there are also (in no particular order): R. Bartle, Introduction to Measure theory - it has a particularly nice section on integrals of the form$$ h(x) = \int f(x,t)dt $$G. De Barra, Measure Theory and Integration"- does a nice job of differentiation. Halmos, Measure Theory - a little dated now, but I have looked ... 1 The 3 big options are: Folland, Real Analysis and its Applications 2e Royden & Fitzpatrick, Real Analysis 4e Rudin, Real & Complex Analysis 3e I liked each of them for different reasons, but I found Folland + Royden & Fitzpatrick (To supplement chapter 3 of Folland) was a good option. Lots of solved problems is generally an undergraduate ... 0 I am keen on the Schaum's outline series. Try this: "Schaum's Outline of Theory and Problems of real Variables: Lebesgue Measure and Integration with Applications to Fourier Series" by Murray Spiegel This series of books have hundreds of solved problems on measure theory (& other topics like topology, complex analysis & differential geometry). I ... 1 If f_n\in L^1, then necessarily there exists C_n>0 such that \\|f_n\\|_1\leqslant C_n (since by definition, each f_n has a finite L^1-norm). If there exists C>0 such that \\|f_n\\|_1\leqslant C for all n, then we say that f_n is uniformly bounded. 3 I'm not sure you'll be able to find a text with solved exercises. My personal favorite is Folland's Real Analysis. 1 No. Let f_n = n^3 \chi_{(0,1/n)}(x). Then f_n(x) \to 0 for every x \in (0,1) but \\|f_n\\|_1 = n^2. 1 It does not. For instance, f_n=n^21_{(0,\frac{1}{n}]} is a counterexample. 2 Finite additivity and countable sub-additivity is equivalent to countable additivity. The proof is below. Let \mu be a measure. It is clear that if \mu is countably additive then it is finitely additive and countably sub-additive. Assume that \mu is countably sub-additive and finitely additive. Consider a collection \{A_n \}_{n=1}^\infty of ... 0 Here is a brute force approach with nothing elegant: Let E = \{x \in X : \lim f_n(x)\; exists and is finite\}. Let f=\limsup f_n. Notice that x \in E if and only if f_n(x) \to f(x). Letting h_n := f-f_n, we see that h_n is measurable and that E = \{x \in X: h_n(x) \to 0 \}. By the definition of a convergent sequence, we know that h_n(x) ... 0 I'm sure that any subset of \mathbb R does exist its Vitali Cover(Each element of Vitali Cover, the interval, is closed). For each x ∈ X ⊂ \mathbb R, let I^{x}_{m} = [x, x+1/m] for ∀ positive integer m and let C = {I^{x}_{m}: x∈X and m=1,2,3,...}. Then C forms a Vitali cover for X. 0 the prove is clear with definition of algebra and topology . for example consider X=\Bbb{N} and$$S=\{A\subset \Bbb{N} \|A\, or A^c is finite\}$$it is clear that S is algebra ,and S is not topology,because let A_{n} = \{2n\} and (\bigcup_{n=1}^{\infty}A_{n}=even numbers) is not belong to S because itself and complement of it is not finite. 3 The function f is measurable if and only if there exists a sequence of step functions that converge to f almost everywhere. Your u is integrable, hence measurable. Therefore you have that sequence of step functions with support in (0,1) which converges to u a.e. Note that the same sequence converges a.e. to \tilde u on (0,\infty), hence ... 2 Look at the inverse image of a generating measurable set.$$\bar{u}^{-1}(a,b)$$If a<0<b then because inverse images and unions commute, i.e. f^{-1}(A\cup B)=f^{-1}(A)\cup f^{-1}(B) we get:$$\bar{u}^{-1}(a,b)=\bar{u}^{-1}(a,0)\cup \bar{u}^{-1}(\{0\})\cup\bar{u}^{-1}(0,b)=u^{-1}(a,0)\cup (1,\infty)\cup u^{-1}(\{0\})\cup u^{-1}(0,b)$$... 0 Let S denote the class of sets under consideration. All you need to show is that if A_n \in S for all n \in \mathbb{N} then A:= \bigcup A_n is in S too. So define$$B_n:= \bigcup_{k \leq n} A_k. $$The B_k form an increasing sequence, and they increase to A which is thus in S (by the monotone class property). 1 Hints: Let \arg\min_{{b_0,b_{-0}}}E\left[(X_{n+1}-b_0-b_{-0}'X)^2\right]=(\beta_0,\beta_{-0}')'=\beta\in\mathbb{R}^{n+1}, \quad X=(X_1,\dots,X_n)'. \beta_0=E[X_{n+1}]-\beta_{-0}'E[X], \quad \beta_{-0}=Var(X)^{-1}Cov(X,X_{n+1}). Cov(X, X_{n+1}-\beta_0-\beta_{-0}'X)=0. Use normality of (X_1,\dots,X_{n+1})'. Show that ... 1 HINT: Use Fubini's theorem and the fact that the double integral is zero for all a and b to show that the integral of the function is 0 on every rectangle in [0,1]\times [0,1]. Then prove that any function that is not almost everywhere 0 must have nonzero integral on some rectangle. 1 Let F(x)=\int_0^x f. Then F\equiv 0 from the given hypothesis. Therefore F'\equiv 0. But F'(x) = f(x) for a.e. x by the Legesgue differentiation theorem. Thus f=0 a.e., hence \int_E f = 0 for any measurable set E. (Using a big gun there, but thought I'd toss this in.) 1 Hint: We have,$$f = f^+-f^-$$where f^+ and f^- denote the positive and negative part of f, respectively. By assumption, the (\sigma-finite) measures$$\nu(dx) := f^+(x) \, dx \qquad \mu(dx) := f^-(x) \, dx$$satisfy$$\mu((a,b)) = \nu((a,b)).$$Conclude from the uniqueness of measure theorem that \mu = \nu on \mathcal{B}(\mathbb{R}). 0 Hint: We can find a countable collection of open intervals I_k such that$$ E \subset U = \bigcup_{k \in \Bbb N} I_k, \qquad \left(\sum_{k=1}^\infty m(I_k) \right) - m(E) < \epsilon $$Now, note that$$ \left\| \int_U f\,dx - \int_E f\,dx \right\| = \left\| \int_{U \setminus E} f\,dx \right\| $$Alternative: Show that \int_U f\,dx = 0 whenever U ... 5 Hint:$$ \max(a,b)=\frac{a+b+\|a-b\|}{2},\quad a,b\in\mathbb{R}. 1 Lebesgue integral is first defined for non-negative functions. Then the definition is extended to general functions without sign constraint. In particular, writing the function as f=f^+-f^-, one defines \int f=\int f^+-\int f^-, where f^+ and f^- are positive and negative parts which are non-negative functions. So for Lebesgue integral to make sense, ... 0 By using the integrals \begin{align} \int \frac{x^2}{(x^2 + y^2)^2} \, dx &= \frac{1}{2y} \, \tan^{-1}\left(\frac{x}{y}\right) - \frac{x}{x^2 + y^2} \\ \int \frac{1}{(x^2 + y^2)^2} \, dx &= \frac{1}{2 y^3} \left( \frac{xy}{x^2 + y^2} + \tan^{-1}\left( \frac{x}{y} \right) \right) \\ \int \frac{y^2}{(x^2 + y^2)^2} \, dy &= \frac{1}{2x} \, ... 5 Just compute the integrals :). What is meant by this: For fixed y, we have \begin{align} \int_0^1 \frac{x^2-y^2}{(x^2 + y^2)^2}\, dx &= \left[ -\frac{x}{x^2 + y^2}\right]_{x=0}^1\\ &= -\frac{1}{1+y^2} \end{align} Hence \begin{align} \int_0^1\int_0^1 \frac{x^2-y^2}{(x^2 + y^2)^2}\, dx\, dy &= -\int_0^1 \frac{1}{1+y^2}\, dy\\ &= ... 1 Hints: Let x=r\cos \theta and y=r\sin \theta. 1 Yes. By definition,\sigma(\mathscr A)=\bigcap\{\mathscr B\,\|\,\mathscr B\text{ is a $\sigma$-algebra and }\mathscr A\subseteq\mathscr B\},$$which can be shown to be the smallest (in the sense of set inclusion) \sigma-algebra containing \mathscr A, so that \mathscr A\subseteq\mathscr\sigma(\mathscr A). But \mathscr A is already a \sigma-algebra ... 2 By way of contradiction, assume that an integrable function f: \mathbb{R} \to [0,\infty) exists such that$$ \mu(E) = \mu(f^{\leftarrow}[E]) $$for any Lebesgue-measurable subset E of [0,\infty) . Then$$ \forall n \in \mathbb{N}: \quad \mu({f^{\leftarrow}}[n,n + 1)) = \mu([n,n + 1)) = 1. Hence, \begin{align} ... 1 If $\mu^$ is a metric outer measure, then all Borel Sets will be $\mu^$-measurable, but if we have no restriction to $\mu^$, then we can say nothing about Borel Sets. For instance, take $\mu^$ the outer measure giving rise to the Lebesgue measure in $\mathbb{R}$, but take in $\mathbb{R}$ the discrete topology (which is a metric topology with distances 0 ... 1 As of the first question, this characterisation Let $\mu$ be a finite measure. A sequence $f_n$ converges to $f$ in measure with respect to $\mu$ if and only if any subsequence $f_{n_k}$ admits a sub-subsequence $f_{n_{k_h}}$ that converges to $f$ almost everywhere. provides far more than a solid path. It's almost a proof, because the $f_{n_{k_h}}$ ... 2 I believe that $\{\tau\leq 1\}$ is a shorthand for \begin{align} \{\omega\in\Omega\,\|\,\tau(\omega)\leq 1\}. \end{align} To see that this is consistent with the statement that $\{\tau\leq 1\}=\{1\}$, note that \begin{align*} \tau(1)=&\,\inf\{t\geq 0\,\|\,\max\{t-1,0\}>0\}=\inf\{(1,\infty)\}=1,\\ \tau(2)=&\,\inf\{t\geq ... Top 50 recent answers are included	2015-04-26 11:42:10	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 1.0000072717666626, "perplexity": 1210.7827900717004}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-18/segments/1429246654292.99/warc/CC-MAIN-20150417045734-00156-ip-10-235-10-82.ec2.internal.warc.gz"}
http://minewiki.engineering.queensu.ca/mediawiki/index.php/Depreciation	# Depreciation Depreciation is a group of methods used in project economics that allows for the CAPEX to be spread out over the useful life of the capital. In many cases such as mills and processing plants, the useful life of capital parallels the life of the project. By spreading this expenditure out, future revenue streams are depreciated by subtracting a portion of these capital expenses. The benefits of this can manifest themselves in both tax and accounting purposes. When analyzing the cash flows and reported results from an operation, including depreciation in the balance sheet can give a better perspective to what the costs/benefits of production actually are instead of lumping all CAPEX to the beginning of a project. The method used varies from company to company and thus careful attention must be paid to each individual balance sheet and income statement. For taxation purposes, depreciating can reduce the taxable income at an operation as it is subtracted from gross revenue alongside operating expenses. When dealing with taxation however, the maximum depreciation rate for a specific class of asset is prescribed (at least in Canada) by the government. Most machinery at a mining operation will fall under Class 38 with a rate of 30%, and most buildings will fall under Class 1 with a rate of 6%.[1] ## Methods Simple syntax used in the depreciation methods is defined as such: IC represents the initial cost of capital DCi represents the depreciation charge in year i BVi represents the book value of the asset in year i n represents the estimated number of periods in the assets’ useful life r represents the depreciation rate as a decimal ### Straight-line depreciation Straight-line depreciation is the simplest of methods and is commonly used because of this. In this method, the capital cost of the asset is spread evenly over each year of its useful life according to the prescribed (or selected) depreciation rate.[2] $DC = \left( {IC - SV} \right)r$ $n = \frac{1}{r}$ This depreciation charge is constant for each year and thus only needs to be calculated once. Since n and r are inversely related, typically a useful life is estimated for an asset and then the depreciation rate follows. A haul truck with an initial cost of $750,000 and a salvage value of$10,000 exhibits the following characteristics over a 5 year useful life (depreciation rate of 20%) if straight-line is used. Year Book Value Depreciation Charge 0 $750,000$0 1 $602,000$148,000 2 $454,000$148,000 3 $306,000$148,000 4 $158,000$148,000 5 $10,000$148,000 ### Declining balance depreciation Declining Balance Depreciation (also known as DBD) methods allocates higher depreciation charges to the beginning of the project. This method is used in scenarios in which the asset is more productive in its earlier years and will generate more revenue and thus will have a higher depreciation charge associated with it in these years. This is accomplished by determining the depreciation charge for the year as a product of the previous book value and current depreciation rate.[2] $B{V_i} = IC{\left( {1 - r} \right)^i}$ Thus every year, the book value becomes a fraction (1 − r) of the previous year’s value. $D{C_i} = IC{\left( {1 - r} \right)^{i - 1}}r$	2018-01-23 07:56:15	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 11, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.4054369628429413, "perplexity": 1419.7359678558669}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-05/segments/1516084891791.95/warc/CC-MAIN-20180123072105-20180123092105-00533.warc.gz"}
https://www.chemicalforums.com/index.php?topic=110604.msg391015	August 08, 2022, 03:55:49 PM Forum Rules: Read This Before Posting ### Topic: Calculate volume of for molarity solution with 1gr of PH3 (Read 534 times) 0 Members and 1 Guest are viewing this topic. #### tanishalala • Very New Member • Posts: 2 • Mole Snacks: +0/-0 ##### Calculate volume of for molarity solution with 1gr of PH3 « on: January 21, 2022, 05:02:07 PM » Question attached as image. Thanks in advance. Sorry for the bad translation. I can’t solve B « Last Edit: January 22, 2022, 01:48:31 PM by sjb » #### Borek • Mr. pH • Deity Member • Posts: 27142 • Mole Snacks: +1762/-405 • Gender: • I am known to be occasionally wrong. ##### Re: Calculate volume of for molarity solution with 1gr of PH3 « Reply #1 on: January 21, 2022, 05:49:46 PM » Looks like a trivial stoichiometry, what have you tried and why it didn't work? ChemBuddy chemical calculators - stoichiometry, pH, concentration, buffer preparation, titrations.info, pH-meter.info #### tanishalala • Very New Member • Posts: 2 • Mole Snacks: +0/-0 ##### Re: Calculate volume of for molarity solution with 1gr of PH3 « Reply #2 on: January 21, 2022, 06:03:09 PM » Looks like a trivial stoichiometry, what have you tried and why it didn't work? So I tried: Calculating the Volume so I could use M1V1=M2V2 But I don’t know how to calculate the volume. Also converted PH3 to moles (0,0294) but not sure what to do with it next. The moles I converted is with the molair mass of PH3 divided by the one gram. Tried dividing it with 4 since the ratio PO3 with PH3 is 4. So I’m not really sure if I’m close to anything but I really really can’t figure it out.. I tried so many more things but erased them out of frustration sorry. #### Borek • Mr. pH • Deity Member • Posts: 27142 • Mole Snacks: +1762/-405 • Gender: • I am known to be occasionally wrong. ##### Re: Calculate volume of for molarity solution with 1gr of PH3 « Reply #3 on: January 22, 2022, 03:56:03 AM » Calculating the Volume so I could use M1V1=M2V2 There is no dilution here, I am afraid this formula won't help. Quote converted PH3 to moles (0,0294) Good starting point. Look at the reaction equation: how many moles of the starting material is required to produce that number of moles of PH3? Do you know any formula that combines number of moles, volume and concentration of the substance? ChemBuddy chemical calculators - stoichiometry, pH, concentration, buffer preparation, titrations.info, pH-meter.info #### sjb • Global Moderator • Sr. Member • Posts: 3642 • Mole Snacks: +220/-42 • Gender: ##### Re: Calculate volume of for molarity solution with 1gr of PH3 « Reply #4 on: January 22, 2022, 12:29:17 PM »	2022-08-08 19:55:49	{"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8138530254364014, "perplexity": 12245.121656699133}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-33/segments/1659882570871.10/warc/CC-MAIN-20220808183040-20220808213040-00136.warc.gz"}
http://mathhelpforum.com/advanced-applied-math/14941-solved-can-vector-broken-down-components.html	# Math Help - [SOLVED] Can vector be broken down to components? 1. ## [SOLVED] Can vector be broken down to components? Hi. Here is the URL with some examples: Basic Vector Operations , bottom of the page. Suppose we have four vectors (A, B, C, D). We know their magnitudes and angles (directions). We add them and produce a new vector R. Let's consider a different situation. Let's say we are given only the resulting vector R (magitude and angle) and number of vectors (4) added to generate the R vector. Can we determine magnitudes and angles of these individual vectors (all of which are unknown)? Thanks. 2. Is this what you mean? The vector <1,-2,5> is the result of<3,4,0>+<-1,-6,2>+<-1,0,3>. BUT ALSO, <1,-2,5>=<1,0,0>+<0,-2,0>+<0,0,5>. Any given vector in R^3 is the linear sum of infinitely many three-element sets of vectors. So I think the answer to your question is no. If you mean something else, please explain.	2015-11-29 09:06:41	{"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8287678360939026, "perplexity": 732.2345887034401}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-48/segments/1448398457127.55/warc/CC-MAIN-20151124205417-00186-ip-10-71-132-137.ec2.internal.warc.gz"}
http://www.phy.duke.edu/~rgb/Class/Electrodynamics/Electrodynamics/node133.html	Next: Tensors in 4 Dimensions Up: The Lorentz Group Previous: The Lorentz Group Contents # The Geometry of Space-Time Recall that a great deal of simplification of the kinematics of classical non-relativistic mechanics occurs when one considers the group structure of transformations with respect to the underlying coordinates. Specifically, the group of inversions, translations and rotations of a given coordinate system leave the norm (length) of a given vector invariant. These transformations form the Euclidean group in three dimensions, E. For those of you who led deprived childhoods, a group is a set of mathematical objects with a rule of composition, or group product, such that: 1. Every product of a pair of elements in the group is also in the group. That is, if then . This property is called closure. 2. The group must contain a special element called the identity such that for all . 3. Every element of the group must have an inverse, also in . If then such that . 4. The group product must be associative. That is, . If the group product commutes ( ) the group is said to be Abelian16.1 otherwise the group is said to be non-Abelian, which is sensible enough. A Lie group is a continuous group16.2 such as the group of infinitesimal transformations. It necessarily has an uncountable infinity of elements. There are also discrete (but countably infinite) groups, finite groups, and everything in between. There are also semi-groups'' (which do not, for example, contain an inverse). Finally, one can construct non-associative'' structures like groups from non-associative algebras like the octonions. Multiplication over the reals forms a continuous Abelian group. Rotations form a non-Abelian Lie group. Multiplication over rational numbers forms a countably infinite group. The set of rotations and inversions that leave a square invariant form a finite (point) group. The renormalization group'' you will hear much about over the years is not a group but a semi-group -- it lacks an inverse. However, our purpose here is not, however, to study group theory per se. One could study group theory for four years straight and still only scratch the surface. It is somewhat surprising that, given the importance of group theory in physics, we don't offer a single course in it, but then again, it's not that surprising... With that in mind, we can decide what we are looking for. We seek initially the set of transformations in four dimensions that will leave (16.1) invariant for a single event with respect to a particular coordinate origin. These transformations form a group called the homogeneous Lorentz group. It consists of ordinary rotations in the spatial part, the Lorentz transformations we have just learned that mix space and time, and several discrete transformations such as space inversion(s) and time inversion. The set of transformations that leave the quantity (16.2) invariant form the inhomogeneous Lorentz16.3 or Poincaré group. It consists of the homogeneous group (including the improper'' transformations that include spatial reflection and time reversal) and uniform translations of the origin. If anyone cares, the Lorentz group is the generalized orthogonal group O(1,3). The proper subgroup of the Lorentz group (the one that is simply connected spatially (no odd inversions) and contains the identity) is SO(1,3) the special orthogonal group. If time's direction is also preserved we add a +, SO(1,3). This nomenclature is defined here for your convenience but of course the wikinote reference contains active links to a lot of this in detail. We will define to be the norm of relativistic space-time. This quantity may be considered to be the invariant distance'' (squared) between two events, and , and of course is one of the fundamental objects associated with the construction of differentials. Since quantities that are unchanged by a geometric transformation are called scalars it is evident that is a 4-scalar. Since the first postulate states that the laws of physics must be invariant under homogeneous (at least) Lorentz transformations, they must ultimately be based on Lorentz scalars. Indeed, the Lagrangian densities upon which field theories are based are generally constructed to be Lorentz scalars. This is a strong constraint on allowed theories. These scalars are, however, formed out of 4-vectors (as we see above) or, more generally, the contraction of 4-tensors. We must, therefore, determine the general transformation properties of a tensor of arbitrary rank to completely determine a theory. In the part of this book devoted to mathematical physics is an entire chapter that discusses tensors, in particular the definitions of covariant and contravariant tensors, how to contract (Einstein sum) pairs of tensors to form tensors of lower rank, and the role of the metric tensor in defining tensor coordinate frames and transformations thereupon. We will not repeat this review or introduction (depending on the student) and urge students to at this time spend an hour or so working through this chapter before continuing (even if you've seen it before). Next: Tensors in 4 Dimensions Up: The Lorentz Group Previous: The Lorentz Group Contents Robert G. Brown 2014-08-19	2015-02-01 14:51:37	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8609229922294617, "perplexity": 449.77754810579245}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-06/segments/1422122238667.96/warc/CC-MAIN-20150124175718-00086-ip-10-180-212-252.ec2.internal.warc.gz"}
http://mathhelpforum.com/math-topics/124438-timesing-print.html	# timesing • January 19th 2010, 11:25 AM andyboy179 timesing hi i just need to know what Attachment 14904 would = • January 19th 2010, 01:16 PM earboth Quote: Originally Posted by andyboy179 hi i just need to know what Attachment 14904 would = $14 \cdot \sqrt{2} \cdot \sqrt{2}$	2014-12-20 16:33:08	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 1, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8957228660583496, "perplexity": 6143.168408051671}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-52/segments/1418802770043.48/warc/CC-MAIN-20141217075250-00129-ip-10-231-17-201.ec2.internal.warc.gz"}
https://eprint.iacr.org/2018/1108	## Cryptology ePrint Archive: Report 2018/1108 Covert Security with Public Verifiability: Faster, Leaner, and Simpler Cheng Hong and Jonathan Katz and Vladimir Kolesnikov and Wen-jie Lu and Xiao Wang Abstract: The notion of covert security for secure two-party computation serves as a compromise between the traditional semi-honest and malicious security definitions. Roughly, covert security ensures that cheating behavior is detected by the honest party with reasonable probability. It provides more realistic guarantees than semi-honest security with significantly less overhead than is required by malicious security. The rationale for covert security is that it dissuades cheating by parties that care about their reputation and do not want to risk being caught. Further thought, however, shows that a much stronger disincentive is obtained if the honest party can generate a publicly verifiable certificate of misbehavior when cheating is detected. While the corresponding notion of publicly verifiable covert (PVC) security has been explored, existing PVC protocols are complex and less efficient than the best-known covert protocols, and have impractically large certificates. We propose a novel PVC protocol that significantly improves on prior work. Our protocol uses only off-the-shelf'' primitives (in particular, it avoids signed oblivious transfer) and, for deterrence factor 1/2, has only 20-40% overhead (depending on the circuit size and network bandwidth) compared to state-of-the-art semi-honest protocols. Our protocol also has, for the first time, constant-size certificates of cheating (e.g., 354 bytes long at the 128-bit security level). As our protocol offers strong security guarantees with low overhead, we suggest that it is the best choice for many practical applications of secure two-party computation. Category / Keywords: cryptographic protocols / secure computation, covert security	2022-01-17 03:36:13	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.22352555394172668, "perplexity": 4332.4125520903945}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-05/segments/1642320300289.37/warc/CC-MAIN-20220117031001-20220117061001-00326.warc.gz"}
https://mathematica.stackexchange.com/questions/121025/rationalize-error?noredirect=1	# Rationalize error The docs state that "Rationalize[x,dx] yields the rational number with smallest denominator that lies within dx of x." However, testing this out it appears to be false. Rationalize[Pi, 0.1] 22 / 7 The problem is that 16 / 5 is also within 0.1 of Pi and has a smaller denominator. N[16 / 5 - Pi] 0.0584073 How does Rationalize actually work? • Since N[22/7 - Pi] < N[16/5 - Pi], I assume, that Rationalize tries to minimize Abs[dx] though within reasonable time limits. Also, as the documentation states, it has to be true that Abs[p/q - x] < c/q^2, where c = 10^-4. Jul 19 '16 at 10:51 • Have a look Does Mathematica get Pi wrong? – user9660 Jul 19 '16 at 11:10 • @IstvánZachar, the part you quoted with regards to 10^-4 is specific to Rationalize[x] forms. Jul 19 '16 at 11:14 • @Louis, I'm not sure if that's relevant. Even machine precision should provide ample accuracy for this computation. Jul 19 '16 at 11:16 • I'd say the documentation could be worded better for the 2-argument case of Rationalize. There is an interplay between denominator of result, epsilon, and size of residual. If you take dx as replacing the default epsilon then I think the correct claim might be to the effect: ratPi = Rationalize[N[Pi], eps]; Abs[N[Pi] - ratPi] < N[eps]/Denominator[ratPi]^2. Jul 19 '16 at 18:57	2021-09-27 05:00:03	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5796993374824524, "perplexity": 2575.9896127775623}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-39/segments/1631780058263.20/warc/CC-MAIN-20210927030035-20210927060035-00190.warc.gz"}
https://mathematics21.org/2014/12/06/thin-groupoids/	Thin groupoids Thin groupoid is an important but a heavily overlooked concept. When I did Google search for “thin groupid” (with quotes), I found just ${7}$ (seven) pages (and some of these pages were created by myself). It is very weird that such an important concept was overlooked by the mathematical community. By definition of thin category, thin groupoid is a groupoid for every pair ${A}$, ${B}$ of objects of which there are at most one morphism ${A \rightarrow B}$. I recall that a groupoid is a category all morphisms of which are isomorphisms. Moreover, in all examples below objects are sets and (iso)morphisms are isomorphisms of ${\mathbf{\mathrm{Set}}}$ that is bijections. So, roughly, “thin groupoid” means: Between every two sets in consideration there is considered at most one bijection. In other words, all objects in consideration are equivalent up to an isomorphism. 1. Equivalent definitions of thin groupoid Theorem 1 The following definitions of thin groupoid are equivalent: 1. a groupoid with at most one morphism ${A \rightarrow B}$ for given objects ${A}$, ${B}$; 2. a groupoid with each cycle of morphisms being identity. Proof: The only thing we need to prove (as all the rest is obvious) is that for thin groupoid each cycle of morphisms is identity. But really, composition of a cycle of morphisms is an endomorphism, but because our category is thin, there are be just one such morphism, the identity morphism. $\Box$ “Each cycle of morphisms is identity” intuitively means: Every object is equivalent to itself in exactly one way. 2. Examples 2.1. Filters, ideals, etc. For a lattice ${\mathfrak{Z}}$ I denote meets and joins correspondingly as $({\sqcap})$ and $({\sqcup})$. Filters and ideals are well known concepts: Filters are subsets ${F}$ of ${\mathfrak{A}}$ such that: 1. ${F}$ does not contain the least element of ${\mathfrak{A}}$ (if it exists). 2. ${A \sqcap B \in F \Leftrightarrow A \in F \wedge B \in F}$ (for every ${A, B \in \mathfrak{Z}}$). Ideals are subsets ${F}$ of ${\mathfrak{A}}$ such that: 1. ${F}$ does not contain the greatest element of ${\mathfrak{A}}$ (if it exists). 2. ${A \sqcup B \in F \Leftrightarrow A \in F \wedge B \in F}$ (for every ${A, B \in \mathfrak{Z}}$). I also introduce free stars and mixers: Free stars are subsets ${F}$ of ${\mathfrak{A}}$ such that: 1. ${F}$ does not contain the least element of ${\mathfrak{A}}$ (if it exists). 2. ${A \sqcup B \in F \Leftrightarrow A \in F \vee B \in F}$ (for every ${A, B \in \mathfrak{Z}}$). Mixers are subsets ${F}$ of ${\mathfrak{A}}$ such that: 1. ${F}$ does not contain the greatest element of ${\mathfrak{A}}$ (if it exists). 2. ${A \sqcap B \in F \Leftrightarrow A \in F \vee B \in F}$ (for every ${A, B \in \mathfrak{Z}}$). I will denote ${\mathrm{dual}\, A}$ where ${A \in \mathfrak{Z}}$ the corresponding element of the dual poset ${\mathfrak{Z}^{\ast}}$. Also I denote $\displaystyle \langle \mathrm{dual} \rangle X \overset{\mathrm{def}}{=} \left\{ \mathrm{dual}\, x \mid x \in X \right\} .$ It is easy to show that filters, ideals, free stars, and mixers are related by the bijections presented in the following diagram: (where ${\neg}$ denotes set-theoretic complement). This diagram is a ${4}$-elements thin groupoid (which is a subcategory of ${\mathbf{\mathrm{Set}}}$). These bijections are order isomorphisms if we define order in the right way. In the case if ${\mathfrak{Z}}$ is a boolean lattice, there is also an alternative diagram (also a ${4}$-elements thin groupoid (which is a subcategory of ${\mathbf{\mathrm{Set}}}$)): (here ${\langle \neg \rangle X \overset{\mathrm{def}}{=} \left\{ \bar{x} \mid x \in X \right\}}$). Funcoids, funcoidal reloids, and filters on lattices ${\Gamma}$ (don’t worry if you don’t know meanings of these terms, see my Web site for a book on this topic) are isomorphic as presented by the following diagram which is also a thin groupoid. The isomorphisms preserve order and composition.	2020-04-10 02:09:01	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 96, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9740959405899048, "perplexity": 630.7058510437363}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-16/segments/1585371883359.91/warc/CC-MAIN-20200410012405-20200410042905-00182.warc.gz"}
https://www.zbmath.org/?q=an%3A1388.76291	# zbMATH — the first resource for mathematics A comparative study of the LBE and GKS methods for 2D near incompressible laminar flows. (English) Zbl 1388.76291 Summary: We compare the lattice Boltzmann equation (LBE) and the gas-kinetic scheme (GKS) applied to 2D incompressible laminar flows. Although both methods are derived from the Boltzmann equation thus share a common kinetic origin, numerically they are rather different. The LBE is a finite difference method, while the GKS is a finite-volume one. In addition, the LBE is valid for near incompressible flows with low-Mach number restriction $$Ma < 0.3$$, while the GKS is valid for fully compressible flows. In this study, we use the generalized lattice Boltzmann equation (GLBE) with multiple-relaxation-time (MRT) collision model, which overcomes all the apparent defects in the popular lattice BGK equation. We use both the LBE and GKS methods to simulate the flow past a square block symmetrically placed in a 2D channel with the Reynolds number $$Re$$ between 10 and 300. The LBE and GKS results are validated against the well-resolved results obtained using finite-volume method. Our results show that both the LBE and GKS yield quantitatively similar results for laminar flow simulations, and agree well with existing ones, provided that sufficient grid resolution is given. For 2D problems, the LBE is about 10 and 3 times faster than the GKS for steady and unsteady flow calculations, respectively, while the GKS uses less memory. We also observe that the GKS method is much more robust and stable for under-resolved cases due to its upwinding nature and interpolations used in calculating fluxes. ##### MSC: 76M28 Particle methods and lattice-gas methods 76M12 Finite volume methods applied to problems in fluid mechanics Full Text:	2021-07-25 04:11:08	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.38006433844566345, "perplexity": 1070.3669053496253}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-31/segments/1627046151563.91/warc/CC-MAIN-20210725014052-20210725044052-00268.warc.gz"}
http://mathhelpforum.com/trigonometry/120158-determining-rule-graphs-trig-function.html	# Math Help - Determining the rule for graphs of trig function 1. ## Determining the rule for graphs of trig function A few question i am stuck on: 1) $y=Atan(nt)$ find A and n using the information provided Asymptotes have equation $t=(2k+1)\frac{\pi}{6}$where k=>Z when $t=\frac{\pi}{12},y=5$ 2) $y=Asin(nt+e)$ Range = [-2,2] Period=6 When t=1,y=1 I have found what A and n is however i don't know how to find e. P.S 2. Originally Posted by Paymemoney A few question i am stuck on: 1) $y=Atan(nt)$ find A and n using the information provided Asymptotes have equation $t=(2k+1)\frac{\pi}{6}$where k=>Z when $t=\frac{\pi}{12},y=5$ 2) $y=Asin(nt+e)$ Range = [-2,2] Period=6 When t=1,y=1 I have found what A and n is however i don't know how to find e. P.S 2. You should know that for a sine function of the form $y = a\sin{(bx)}$, the amplitude is $a$ and the period is $\frac{2\pi}{b}$. $y = A\sin{(nt + e)}$. Since the range is $[-2, 2]$, that means the amplitude must be $2$. Therefore $A = 2$. The period is 6. So $6 = \frac{2\pi}{n}$ $n = \frac{2\pi}{6}$ $n = \frac{\pi}{3}$. So far you have $y = 2\sin{\left(\frac{\pi t}{3} + e\right)}$. You also know that when $t = 1, y= 1$. So $1 = 2\sin{\left(\frac{\pi}{3} + e\right)}$ $\frac{1}{2} = \sin{\left(\frac{\pi}{3} + e\right)}$ $\frac{\pi}{6} = \frac{\pi}{3} + e$ $-\frac{\pi}{6} = e$. Therefore $y = 2\sin{\left(\frac{\pi t}{3} - \frac{\pi}{6}\right)}$.	2015-04-28 12:28:59	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 25, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9519381523132324, "perplexity": 1582.1656381778855}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-18/segments/1429246661364.60/warc/CC-MAIN-20150417045741-00193-ip-10-235-10-82.ec2.internal.warc.gz"}
https://socratic.org/questions/58a229c9b72cff17846ac803	# Question ac803 Feb 13, 2017 Zero #### Explanation: Given expression lim_(x→0)⁡(e^(x^2 )+e^(〖-x〗^2 )-2)/x^2? We see that the expression becomes $\frac{0}{0}$ if we calculate its value at $x = 0$. Hence L' Hospitals rule is applicable. We differentiate the numerator and denominator with rest to $x$ and then evaluate. =>lim_(x→0)⁡ (d/dx[e^(x^2 )+e^([-x]^2 )-2])/(d/dxx^2) Using the chain rule we get lim_(x→0)⁡ (2xe^(x^2 )+2(-x)e^([-x]^2 ))/(2x) =>lim_(x→0)⁡ (xe^(x^2 )-xe^([-x]^2 ))/(x) When we evaluate the function at $x = 0$ we see that it is $\frac{0}{0.}$ Hence applying L' Hospitals rule again we get lim_(x→0)⁡ (d/dx[xe^(x^2 )-xe^([-x]^2 )])/(d/dx x) Using product rule of differentiation we get lim_(x→0)⁡ ((e^(x^2 )+xe^(x^2)(2x))-(e^([-x]^2 )+xe^([-x]^2)2(-x)))/1 =>lim_(x→0)⁡ ((e^(x^2 )+2x^2e^(x^2))-(e^([-x]^2 )-2x^2e^([-x]^2)))/1 =>lim_(x→0)⁡ (1+0-1+0)/1=0#	2021-06-21 19:41:09	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 13, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9949305653572083, "perplexity": 8752.26437460733}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-25/segments/1623488289268.76/warc/CC-MAIN-20210621181810-20210621211810-00431.warc.gz"}
https://megaexams.com/questions/IBPS/Foreign-Trade/	# IBPS > Foreign Trade Explore popular questions from Foreign Trade for IBPS. This collection covers Foreign Trade previous year IBPS questions hand picked by experienced teachers. ENGLISH LANGUAGE QUANTITATIVE APTITUDE GENERAL AWARENESS COMPUTER AWARENESS REASONING Q 1. Correct4 Incorrect-1 Which of the following is considered lending for promotion of exports? Packing Credit B Overdraft C Cash Credit Account D Bill Discounting Q 2. Correct4 Incorrect-1 The earnings of India from diamond export is quite high. Which one of the following factors has contributed to it? A pre-independene stock-piling of diamonds in the country which are now exported B large production of industrial diamonds in the country expertise available for cutting and polishing of imported diamonds which are then exported D as in the past, India produces huge quantity of gem diamonds which are exported Q 3. Correct4 Incorrect-1 Christine Lagarde is the head of which among the following international agencies / bodies? A Asian Development Bank B Non-alignment Movement C International Atomic Energy Agency International Monetary Fund Q 4. Correct4 Incorrect-1 AoA in context with World Trade Organization is_____? A Article of Association Agreement on Agriculture C Agreement on Association D Administration of Agriculture ##### Explanation Q 5. Correct4 Incorrect-1 Which one of the following types of borrowings from the {tex} \mathrm {IMF}{/tex} has the softest servicing conditions? A Second tranche loan B SAF ESAF D Oil facility Q 6. Correct4 Incorrect-1 Which is the role of the International Monetary Fund {tex} ( \mathrm { IMF } ) ? {/tex} A To implement and advance global trade agreements B To settle industrial and trade disputes between members C To help poorer countries with their economic development To maintain international financial stability in global financial markets Q 7. Correct4 Incorrect-1 Which of the following is not one of the features of the Special Economic Zones ({tex} \mathrm {SEZ}{/tex}) being set up for promoting exports? A The SEZ area will be treated as foreign territory for trade operations, duties and tariff. B No licence is required for import into the zone. Foreign workers will be allowed free entry without visa restrictions. D There will be no routine examination by customs authorities of import/export cargo. Q 8. Correct4 Incorrect-1 For National Manufacturing and Investment Zones ({tex} \mathrm{NMIZ}{/tex}), Special Economic Zone ({tex} \mathrm{SEZ}{/tex}) and EOUS ({tex} \mathrm{Exports-Oriented \,Units}{/tex}), which of the following statement is true? A NMIZs and EOUs will be located within SEZs SEZs and EOUs will be located within NMIZs. C NMIZs are independent of SEZs and EOUs. D NMIZs and SEZs will be competitors in nature Q 9. Correct4 Incorrect-1 Both Foreign Direct Investment (FDI) and Foreign Institutional Investor (FII) are related to investment in a country. Which one of the following statements best represents an important difference between the two? A FII helps bring better management skills and technology, while FDI only brings in capital. FII helps in increasing capital availability in general, while FDI only targets specific sectors. C FDI flows only into the secondary market while FII targets primary market D FII is considered to be more stable than FDI. Q 10. Correct4 Incorrect-1 Which of the following does not form part of current account of Balance of Payments? A Export and import of goods B Export and import of services C Income receipts and payments Capital receipts and payments ##### Explanation Q 11. Correct4 Incorrect-1 Which of the following organisation provides guarantee to exporters? A EXIM Bank Export Loan Guarantee Corporation C RBI D Commerce Ministry Q 12. Correct4 Incorrect-1 The New Economic Policy ( 1991 ) was launched in the background of the following economic indicators: 1. India's foreign exchange reserves had fallen to {tex} \mathrm {US\$1 }{/tex} billion. 2. The fall of the Soviet Union had deprived India of almost a quarter of its export market. 3. There was negative growth in real {tex}\mathrm{GDP}{/tex}. 4. Indian rupee had to be devalued by {tex}45{/tex} per cent. A 1, 2, 3 and 4 B 1 only 1 and 2 D 1, 2 and 3 Q 13. Correct4 Incorrect-1 Consider the following statements: 1. The Petrapole-Benapole border checkpoint controls the foreign trade between India and Bangladesh. 2. Petrapole is on Bangladesh side and Benapole is on Indian side. Which of the statements given above is/are correct? 1 only B 2 only C 1 and 2 D None ##### Explanation Q 14. Correct4 Incorrect-1 Foreign Direct Investments are preferred over Capital Inflow. In this context, please consider the following. 1. FDI brings in latest technology 2. FDI does not involve large outflow 3. FDI improves working efficiency Which is/are most suitable reason/reasons for the given statement? A 1 only B 1 and 3 C 1 and 2 All 1, 2 and 3 Q 15. Correct4 Incorrect-1 What are the factors on which import substitution strategy was based on? I. Non-price II. Physical- interventionist policies like licensing, quotas and other physical restrictions on imports A only II B Only I Both I & II D None of the above ##### Explanation Q 16. Correct4 Incorrect-1 Which sentences are correct regarding special drawing rights? I. It was created in 1980 II. It is also known as "paper gold". III. Its value is based on a basket of five key international key currencies and SDRs can be exchanged for freely usable currencies A Only II B I & III II & III D All the above ##### Explanation Q 17. Correct4 Incorrect-1 Which of the below statements are correct? I. Reverse account balance makes an adjustment between current account balance and capital account balance. II. If surplus in the Capital Account is more than deficit in the Current Account, there is net increase in the Forex Reserves of the country at the end of the year. III. If deficit in the current account is more than surplus in the Capital Account then there is net decrease in Foreign Reserves of the country at the end of the year. A I & II B I & III C Only II All the above ##### Explanation Q 18. Correct4 Incorrect-1 Which of the below statements are correct regarding exchange trade? I. The value of rupee was managed by the state bank II. The strict foreign exchange controls also encouraged hawala trade III. India followed a strongly inward looking policy, laying stress on import substitution A Only I Both II & III C Only III D All the above Q 19. Correct4 Incorrect-1 Which statement is correct regarding Current account? I. Those transactions arising out of exports and imports (the visible items) II. It is a statistical statement of all transactions made between one particular country and all other countries during a specified period of time III. This account is the summary of all international trade transctions of the domestic country in one year A {tex} \mathrm { I \,} \& \mathrm { \,II } {/tex} B Only {tex}\mathrm{I}{/tex} C {tex}\mathrm {II}\ {/tex} & {tex}\mathrm{III}{/tex} Only {tex}\mathrm{III}{/tex} ##### Explanation Q 20. Correct4 Incorrect-1 Which statement is correct regarding balance of payment crises? I. If international reserves of a country are not enough to balance a combined deficit in current and capital account on a sustained basis, then the phenomenonn is called a BoP crisis. II. It can be tackled by exporting more or by limiting imports through tariffs, quotas, etc. Contractionary fiscal and monetary policies can also tackle the crisis through lower import demand with fall in average income levels. III. Another short term solution is currency devaluation which encourages exports and discourages costlier imports. A Both I & II B Only I C II & III All the above ##### Explanation Q 21. Correct4 Incorrect-1 Consider the following statements: I. These zones were created to develop such an environment in the economy which may provide capability of facing international competition. II. The Export Processing Zone (EPZs) set up as enclaves, separated from the Domestic Tariff Area by fiscal barriers, were intended to provide a competitive duty free environment for export production III. There are total 10 EPZs in our country Which statement is/are correct regarding export processing zone A Only III I & II C II & III D All the above ##### Explanation Q 22. Correct4 Incorrect-1 Are the statements given below the subset of the third category of FDI? I. It consists of sectors where automatic approval is granted for FDI (often foreign equity participation less than 100 per cent) II. It consists of sectors where prior approval from the Foreign Investment Approval Board (FIPB) is required A Only I B Only II Both I & II D None of these ##### Explanation Q 23. Correct4 Incorrect-1 Choose the correct sentence regarding the trade policy: I. Mahalanobis strategy adopted during the First plan II. Export increased at an average rate of 29 per cent per annum in dollar terms between 1986 and 1990 III. A combination of factors such as bad policy, weak government and external factors led to the decline of this performance to nine per cent in {tex} 1990 - 91 {/tex} and 4 per cent in the subsequent years. Only III B I & II C II & III D All the above ##### Explanation Q 24. Correct4 Incorrect-1 Which statement is correct regarding FDI? I. The success of foreign exchange management by RBI can be seen in the fact that from US$ 5.8 billion in March 1991 (enough to meet the import requirements for three weeks) the country's reserves have grown to around US {tex} \\$ 352 {/tex} billion by December 2015. II. It is pertinent to note here that some of the increase in the reserve is attributed to the weakness of the dollar in the post-Iraq war scenario. A Only I B Only II Both I & II D None of these ##### Explanation Q 25. Correct4 Incorrect-1 What are the facts regarding structural change in India's export since {tex} 1991 ? {/tex} I. There are indications that during {tex} 1990 \mathrm { s } {/tex}, some of Indian exports have moved upwards in value addition chain whereby instead of exporting raw materials, the country has switched over to export of processed goods. II. There were significant compositional shift within the major manufactured product groups such as engineering goods, chemicals and allied products, etc. A Only II B Only I Both I & II D None of the above	2020-06-02 08:49:27	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.2806062698364258, "perplexity": 7392.244307659664}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-24/segments/1590347423915.42/warc/CC-MAIN-20200602064854-20200602094854-00211.warc.gz"}
http://blog.dexy.it/285	Scraping It Together November 28, 2010 · By ana · Comments Off · Posted in Uncategorized I’m very excited to announce a new Dexy feature: now you can automatically fetch remote data and incorporate it into your Dexy document. Dexy will even cache the file for you and only download it again if it has changed on the remote server (http only for now, assuming either ETag or Last-Modified headers are present). One of the first things this makes possible is easily fetching and using data from remote APIs. In this blog post we’ll see how this works, using an example from ScraperWiki. ScraperWiki is a fantastic project which aims to make open data easier to use. It’s a library of scrapers which take public data sets (typically in HTML, XLS or PDF formats), clean them up and make them available via a standard API or a CSV download. Rather than one person writing a scraper, running it locally and doing something with the data, with ScraperWiki anyone can write a scraper, someone else (hopefully several people) can write an article or do research based on the scraped data, and if the scraper breaks anyone can come along and fix it. It’s a great platform for loose collaboration among “hacks and hackers”. I especially love that it’s so easy to get involved without making a huge commitment, if you have a few minutes you can go to ScraperWiki and see broken scrapers that need fixing or write a new scraper that has been requested by someone. While I look forward to exploring ways in which Dexy can help with ScraperWiki’s documentation, especially given that ScraperWiki allows scrapers written in Python, Ruby and PHP and Dexy is ideal for such multi-language situations, ScraperWiki already has a very interesting live-code setup for tutorials. When you click on a tutorial, you are brought into ScraperWiki’s code editor with a live example that you can play with, with extensive comments for documentation. For example, here is the first Python tutorial: ############################################################################### # START HERE: Tutorial 1: Getting used to the ScraperWiki editing interface. # Follow the actions listed with -- BLOCK CAPITALS below. ############################################################################### # ----------------------------------------------------------------------------- # 1. Start by running a really simple Python script, just to make sure that # everything is working OK. # -- CLICK THE 'RUN' BUTTON BELOW # You should see some numbers print in the 'Console' tab below. If it doesn't work, # try reopening this page in a different browser - Chrome or the latest Firefox. # ----------------------------------------------------------------------------- for i in range(10): print "Hello", i # ----------------------------------------------------------------------------- # 2. Next, try scraping an actual web page and getting some raw HTML. # -- UNCOMMENT THE THREE LINES BELOW (i.e. delete the # at the start of the lines) # -- CLICK THE 'RUN' BUTTON AGAIN # You should see the raw HTML at the bottom of the 'Console' tab. # Click on the 'more' link to see it all, and the 'Sources' tab to see our URL - # you can click on the URL to see the original page. # ----------------------------------------------------------------------------- #import scraperwiki #html = scraperwiki.scrape('http://scraperwiki.com/hello_world.html') #print html # ----------------------------------------------------------------------------- # In the next tutorial, you'll learn how to extract the useful parts # from the raw HTML page. # ----------------------------------------------------------------------------- And here is the first Ruby tutorial, along the same lines: # Hi. Welcome to the Ruby editor window on ScraperWiki. # To see if everything is working okay, click the RUN button on the bottom left # to make the following four lines of code do its stuff (1..10).each do \|i\| puts "Hello, #{i}\n" end # Did it work? 10 lines should have been printed in the console window below # The first job of any scraper is to download the text of a web-page. # Uncomment the next two lines of code (remove the # from the beginning of the line) # and click RUN again to see how it works. #html = ScraperWiki.scrape('http://scraperwiki.com/hello_world.html') #puts html # The text will appear in the console, and the URL that it downloaded from # should appear under "Sources". If you’d like to play with these yourselves, then here is the Python tutorial and here is the Ruby tutorial, or just click the Tutorials link in the sidebar from scraperwiki.com. Now let me stop here for a moment to point out that I fetched that python and ruby code directly from scraperwiki.com using Dexy’s new remote file feature. Here is the .dexy file for this blog post: { "@tutorial.py\|pyg" : { "url" : "http://scraperwiki.com/editor/raw/tutorial-1" }, "@tutorial.rb\|pyg" : { "url" : "http://scraperwiki.com/editor/raw/ruby-tutorial-1" }, "@scraper-source.py\|pyg" : { "url" : "http://scraperwiki.com/editor/raw/foi_botanical_gardens" }, "@scraper-data.csv\|dexy" : { "url" : "http://api.scraperwiki.com/api/1.0/datastore/getdata?format=csv&name=foi_botanical_gardens&limit=500" }, "scraper.R\|jinja\|r\|pyg" : { "inputs" : ["@scraper-data.csv\|dexy"] }, ".dexy\|dexy" : {}, "scraping-it-together.dexy\|dexy" : {} } Since JSON doesn’t have a comment character1, I can’t use Idiopidae syntax to split this into manageable chunks, however I think we can manage. The file names beginning with an @ symbol are virtual files, they don’t actually exist on the file system, but Dexy is going to pretend that they are there. So, we can do the usual Dexy things with these “files” like run them through filters or use them as inputs to other documents. The ‘url’ property tells Dexy to fetch the contents at that URL, and these become the contents of the virtual file. Any type of text-based data (eventually binary data too) can be fetched in this way. In this example I have fetched Python and Ruby code, and CSV data. We specify the file extension ourselves when we name the virtual file, so later filters will treat the downloaded text correctly. For example, Pygments knows from the .py file extension that it’s getting Python code. 1 Waaaaaaaahhhhhhhh! WHY didn’t they include a comment character!???!!!! Next let’s take a look at the source code of a real scraper: # Created at Python Northwest, Manchester, 2010-11-15 import scraperwiki import datetime import xlrd # retrieve a page starting_url = 'http://www.whatdotheyknow.com/request/49024/response/124689/attach/3/FOI%202010%20207%20Jones.xls' book = xlrd.open_workbook(file_contents=scraperwiki.scrape(starting_url)) XL_EPOCH = datetime.date(1899, 12, 30) count = 0 for sheet in book.sheets(): print 'Scraping sheet %s' % sheet.name for rownum in xrange(sheet.nrows): date_cell = sheet.cell(rownum, 0) min_cell = sheet.cell(rownum, 2) max_cell = sheet.cell(rownum, 1) if date_cell.ctype == xlrd.XL_CELL_DATE and xlrd.XL_CELL_NUMBER in (min_cell.ctype, max_cell.ctype): record = {} record['Date'] = str(XL_EPOCH + datetime.timedelta(days=date_cell.value)) if min_cell.ctype == xlrd.XL_CELL_NUMBER: record['tmin'] = min_cell.value if max_cell.ctype == xlrd.XL_CELL_NUMBER: record['tmax'] = max_cell.value if min_cell.ctype == xlrd.XL_CELL_NUMBER and max_cell.ctype == xlrd.XL_CELL_NUMBER: record['tmean'] = (min_cell.value + max_cell.value) / 2.0 scraperwiki.datastore.save(['Date'], record, silent=True) count += 1 print 'Scraped %d records' % count This scraper scrapes a spreadsheet containing temperature observations from the Botanic Gardens at the University of Cambridge, obtained under a Freedom of Information act request which you can view at whatdotheyknow.com. The file contains daily observations from 2000 – 2010, we will just look at the 500 most recent observations (to do more would require multiple calls to the ScraperWiki api, easy to do but not really needed for demonstration purposes). Here is what the first few lines of the resulting CSV data looks like: Date,Maximum,Mean,Minimum,tmax,tmean,tmin 2010-09-30,6.9,11.9,16.9,16.9,11.9,6.9 2010-09-29,13.0,15.5,18.0,18.0,15.5,13.0 2010-09-28,13.0,15.15,17.3,17.3,15.15,13.0 2010-09-27,10.6,12.8,15.0,15.0,12.8,10.6 2010-09-26,8.6,11.2,13.8,13.8,11.2,8.6 2010-09-25,6.3,9.1,11.9,11.9,9.1,6.3 2010-09-24,9.5,11.7,13.9,13.9,11.7,9.5 2010-09-23,9.4,14.15,18.9,18.9,14.15,9.4 2010-09-22,9.4,16.7,24.0,24.0... (Yes, the column names are a little wonky but the tmax, tmean and tmin correspond to the correct data so we’ll use these.) We are going to use R to graph this data, and to tell us a little more about it via R’s summary() function. Here is the graph of daily data, the mean temperature is graphed in black with daily maximums and minimums in red and blue respectively. Here is the R transcript which imports the CSV data, graphs it and produces some simple summary statistics: > data <- read.table("0196faca90f48f84d40ff382512234b7.csv", sep=",", header=TRUE) > > # Convert Y-m-d strings to dates, sort by date. > data[,1] <- as.POSIXct(data[,1]) > sorted.data <- data[order(data[,1]),] > > # Determine overall min and max for y axis range > min.temp <- min(sorted.data$tmin, na.rm=TRUE) > max.temp <- max(sorted.data$tmax, na.rm=TRUE) > > png(file="111b990e-af30-4969-8212-67424545b3ae.png", width=500, height=500) > plot( + sorted.data$Date, + sorted.data$tmean, + type="l", + lwd=2, + ylim=c(min.temp-2, max.temp+2), + ylab=expression("Temperature"*degree~C) + ) > points(sorted.data$Date, sorted.data$tmax, type="l", lty=3, col="red") > points(sorted.data$Date, sorted.data$tmin, type="l", lty=3, col="blue") > dev.off() null device 1 > > summary(data$tmin) Min. 1st Qu. Median Mean 3rd Qu. Max. NA's -5.700 3.500 8.200 7.534 11.700 19.100 3.000 > summary(data$tmax) Min. 1st Qu. Median Mean 3rd Qu. Max. NA's -0.30 12.00 18.00 16.77 22.10 30.30 23.00 > summary(data\$tmean) Min. 1st Qu. Median Mean 3rd Qu. Max. NA's -3.00 8.00 13.30 12.13 16.85 23.35 26.00 > This example had a mixture of local scripts and remote data sources, which is typical when you are working on your own documents. When it comes time to share your documents with others, being able to specify remote files in Dexy opens up some really interesting possibilities. Consider this configuration file: { "@tutorial.py\|pyg" : { "url" : "http://scraperwiki.com/editor/raw/tutorial-1" }, "@tutorial.rb\|pyg" : { "url" : "http://scraperwiki.com/editor/raw/ruby-tutorial-1" }, "@scraper-source.py\|pyg" : { "url" : "http://scraperwiki.com/editor/raw/foi_botanical_gardens" }, "@scraper-data.csv\|dexy" : { "url" : "http://api.scraperwiki.com/api/1.0/datastore/getdata?format=csv&name=foi_botanical_gardens&limit=500" }, "@index.txt\|jinja\|textile" : { "url" : "http://bitbucket.org/ananelson/dexy-blog/raw/146d3429c753/2010/11/scraping-it-together/index.txt", "allinputs" : true }, "@scraper.R\|jinja\|r\|pyg" : { "url" : "http://bitbucket.org/ananelson/dexy-blog/raw/146d3429c753/2010/11/scraping-it-together/scraper.R", "inputs" : ["@scraper-data.csv\|dexy"] }, ".dexy\|dexy" : {}, "scraping-it-together.dexy\|dexy" : {} } This is probably a good time to mention a new command-line argument available in Dexy, the -g or —config switch which lets you specify a configuration file other than the default of .dexy. If you have Dexy, RedCloth, Pygments and R installed, then you should be able to build this blog post in a blank directory by saving the above configuration as scraping-it-together.dexy and running: dexy --setup -g scraping-it-together.dexy . I will probably consider adding a switch which will create copies of the virtual files in the local directory, so the dexy configuration file format could be used as a way to distribute worked code tutorial examples as well as reproducible research documents. I encourage you to take a look at the source of this blog post, especially if you are new to Dexy. The post’s directory is here on bitbucket.	2019-07-16 20:16:34	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.2423880398273468, "perplexity": 5050.523027130834}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-30/segments/1563195524879.8/warc/CC-MAIN-20190716201412-20190716223412-00492.warc.gz"}
https://skeptric.com/athena-remove-timezone/	When creating a table in Athena I got the error: Invalid column type for column: Unsupported Hive type: timestamp with time zone. Unfortunately it can't support timestamps with timezone. In my case all the data was in UTC so I just needed to remove the timezone to create the table. The easiest way to do that was to cast it to a timestamp (without a timezone). cast(event_time as timestamp)	2021-03-07 21:24:31	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.32495051622390747, "perplexity": 2401.4455513869575}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-10/segments/1614178378872.82/warc/CC-MAIN-20210307200746-20210307230746-00422.warc.gz"}
https://answers.opencv.org/questions/16599/revisions/	# Revision history [back] I'm fairly new to OpenCV, working on a fun personal project -- using a webcam + OpenCV on a Raspberry Pi to detect birds in my fig tree and trigger a scare device. But I'm hitting serious roadblocks trying to train a simple LBP cascade. I have a simple test app which runs detectMultiScale on a few test images. Using the standard lbpcascade_frontalface.xml the test successfully detects faces within ~200 milliseconds, which is awesome. Then I took 11 different photos of a square cookie cutter and cropped them to 24x24, obtained 20 random negative images cropped to 200x200, and created 12 test images at 320x240. See the images here. I then trained the cascade.xml with: opencv_traincascade -data data -vec vec -bg bg.txt -featureType LBP -w 24 -h 24 and run my detection with: bird_cascade.detectMultiScale(gray, birds, 1.1, 2, 0, Size(80, 80)); Results: • the xml file is 390KB, very large compared to lbpcascade_frontalface.xml which is 52KB. Why does this much smaller data set result in a much larger xml file?? • detectMultiScale takes ~42 seconds per image, about 250 times slower than face detection. Again, completely opposite of what I'd expect. • detectMultiScale doesn't detect any of the star shapes, even on images which contain the exact shapes from the positives training set... rather it always appears to detect a single 104x104 match in the center of the image, regardless of the image :( Things I've tried in training which haven't made any difference: • specifying numPos 11 and numNeg 20 • specifying maxFalseAlarmRate of 0.95 (just stabbing in the dark here, haven't found any online docs that explain this very clearly) • specifying numStages 10 (this actually reduced the xml file size to 220KB and detection time to 33 seconds, but that's still awful) • using a single star image + opencv_createsamples to produce a vec file with 2000 positives, and supplying 100 negatives. This took over 3 hours to train, but the xml is 505KB and detection takes ~70 seconds, with no successes. As you can see, I've spent a lot of time but am getting nowhere. I've studied several online tutorials/references but none addresses the total failure I'm encountering. Any of the following would be EXTREMELY helpful: • specific insights as to why I'm getting complete failure in my case, and which bits to alter to achieve success. • a thorough explanation of the LBP algorithm which would give some intuition for dialing in the mysterious training/detection parameters such as numPos, numStages, minHitRate, maxFalseAlarmRate, scaleFactor, minNeighbors (I'm having trouble making sense of this write-up by Maria Dimashova) • pointer to a good starter tutorial for LBP training -- the standard docs are pretty weak • I want to start small, but is it even possible to produce a 'test' cascade that doesn't require hundreds or thousands of positives & negatives?	2019-09-22 08:14:49	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.26463431119918823, "perplexity": 3492.057688535343}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-39/segments/1568514575402.81/warc/CC-MAIN-20190922073800-20190922095800-00395.warc.gz"}
https://socratic.org/questions/how-do-you-solve-and-graph-abs-2c-3-4	# How do you solve and graph abs(-2c-3)> -4? If you look at$\| - 2 c - 3 \|$ this will be positive or zero for all values of $c$. So even at the minimum value of $0$ $\| 0 \| > - 4$	2021-11-29 16:16:09	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 4, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6800547242164612, "perplexity": 286.5017194241243}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-49/segments/1637964358774.44/warc/CC-MAIN-20211129134323-20211129164323-00220.warc.gz"}
https://onlinecourses.science.psu.edu/stat510/node/76	# 9.2 Intervention Analysis Printer-friendly version Suppose that at time t = T (where T will be known), there has been an intervention to a time series. By intervention, we mean a change to a procedure, or law, or policy, etc. that is intended to change the values of the series xt. We want to estimate how much the intervention has changed the series (if at all). For example, suppose that a region has instituted a new maximum speed limit on its highways and wants to learn how much the new limit has affected accident rates. Intervention analysis in time series refers to the analysis of how the mean level of a series changes after an intervention, when it is assumed that the same ARIMA structure for the series xt holds both before and after the intervention. Overall Intervention Model Suppose that the ARIMA model for xt (the observed series) with no intervention is $x_t - \mu = \frac{\Theta(B)}{\Phi(B)}w_t$ with the usual assumptions about the error series wt. Θ(B) is the usual MA polynomial and Φ(B) is the usual AR polynomial. Let zt = the amount of change at time t that is attributable to the intervention. By definition, zt = 0 before time T (time of the intervention). The value of zt may or may not be 0 after time T. Then the overall model, including the intervention effect, may be written as $x_t - \mu = z_t + \frac{\Theta(B)}{\Phi(B)}w_t$ Example: Following is a plot of a simulated MA(2) model with mean μ = 60 prior to an “intervention point” at time t = 100. We added 10 to the series for times t ≥ 100. In practice, the task would be to determine the magnitude and pattern of the change to the series. Possible Patterns for Intervention Effect (patterns for zt) There are several possible patterns for how an intervention may affect the values of a series for t ≥ T (the intervention point). Four possible patterns are as follows: Pattern 1: Permanent constant change to the mean level: An amount has been added (or subtracted) to each value after time T. Pattern 2: Brief constant change to the mean level: There may be a temporary change for one or more periods, after which there is no effect of the intervention. Pattern 3: Gradual increase or decrease to a new mean level: There may be a gradually increasing amount that is added (or subtracted) which eventually levels off at a new level (compared to the “before” level). Pattern 4: Initial change followed by gradual return to the no change: There may be an immediate change to the values of the series, but the amount added or subtracted to each value after time T approaches 0 over time. Models for the Patterns zt = the amount of change at time t that is attributable to the intervention. Suppose that It is an indicator variable such that It = 1 when t ≥ T and It = 0 when t < T. Pattern 1: Constant permanent change. A constant change of equal to the amount δ0 after time T may be written simply as zt = δ0It. Note that zt = δ0 for t ≥ T, and zt = 0 and for t < T. The coefficient δ0 will be estimated using the data. Pattern 2: A temporary (constant change) lasting for d times past the intervention time T, can be described by the intervention effect model zt = δ0(1 - Bd)It. Or, we can redefine the indicator so that It = 1 for T ≤ t ≤ T + d, and It = 0 for all other t. Then, we use the model zt = δ0It. With this intervention effect, zt = δ0 for T ≤ t ≤ T + d, and zt = 0 and for all other t. Again, the coefficient δ0 will be estimated using the data. Pattern 3: A gradually increasing effect that eventually levels off can be written as: $z_t = \frac{\delta_0}{1-\omega_1B}I_t,$ with It = 1 for t ≥ T and It = 0 when t < T. Assume \|ω1\| < 1. This is equivalent to zt = ω1zt-1 + δ0It. When t < T, zt = 0. When t ≥ T, It =1 so zt = ω1zt-1 + δ0. Because \|ω1\| < 1, we can continue to express each past zt in terms of ω1 and δ0 as a geometric series to obtain the following: $z_t = \frac{\delta_0 (1-\omega_1^{t-T+1})}{1-\omega_1}.$ The coefficients ω1 and δ0 will be estimated using the data. Pattern 4: An immediate change that eventually returns to 0 can be modeled as follows $z_t = \frac{\delta_0}{1-\omega_1B}P_t ,$ with Pt = 1 when t = T and Pt = 0 otherwise. Assume \|ω1\| < 1. This is equivalent to zt = ω1zt-1 + δ0Pt. When t < T, zt = 0. When t = T, zt-1 = 0 and Pt = 1 so zt = δ0. When t ≥ T+1, Pt = 0 so zt = ω1zt-1. The coefficients ω1 and δ0 will be estimated using the data. Estimating the Intervention Effect Two parts of the overall model have to be estimated – the basic ARIMA model for the series and the intervention effect. Several approaches have been proposed. One approach has the following steps: 1. Use the data before the intervention point to determine the ARIMA model for the series. 2. Use that ARIMA model to forecast values for the period after the intervention. 3. Calculate the differences between actual values after the intervention and the forecasted values. 4. Examine the differences in step 3 to determine a model for the intervention effect. What we do after step 4 depends on available software. If the right program is available we can use all of the data to estimate the overall model that combines the ARIMA for the series and the intervention model. Otherwise, we might use only the differences from step 4 above to make estimates of the magnitude and nature of the intervention. Example: North Carolina Highway Fatality Rates, monthly for n = 86 months. Beginning at month = 71 the maximum speed limit was lowered from 70 mph to 55 mph. The fatality rate is computed as highway fatalities per 100 million miles of travel. A time series plot follows. The latter part of the series, in red, is the fatality rate after the change in speed limit. Using the data from times t = 1, …, 70 (before the speed limit decrease), we fit an ARIMA (2, 1, 0) × (1, 1, 0)12. Coefficients: ar1 ar2 sar1 -0.8142 -0.6461 -0.3717 s.e. 0.1081 0.1119 0.1278 sigma^2 estimated as 0.5834: log likelihood = -67.04, aic = 142.08 We then use this model to forecast values for t = 71, …, 86 and compare the forecasted values to the actual values during this intervention period. The first of the next two plots shows the forecasted values compared to the actual values for the “after” period. The second plot shows the differences. It’s difficult to judge the mean intervention pattern exactly. It’s possible that a constant mean change may describe the situation. The mean difference between actual and forecasted in the intervention period is -0.7168515. We’ll learn how to do things in R in the homework for this week.	2017-09-21 05:12:33	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7704871296882629, "perplexity": 1217.8410174275202}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-39/segments/1505818687642.30/warc/CC-MAIN-20170921044627-20170921064627-00021.warc.gz"}
http://qtms.adisangiorgioacremano.it/webwork-answer-key.html	answer in the space provided. Webwork 6. Follow RSS feed Like. WeBWorK is an open online homework system originally developed for math and increasingly being used in engineering and science courses. You can answer your missed calls or messages once you have a logical break in your task. 70860 Hw12 8 WeBWorK MATHEMATICAL ASSOCIATION OF AMERICA webwork mth 124 fs16 70860 hw12 8 MAIN MENU Courses Homework Sets Hw12: Problem 8 Hw12 Problem 8 Previous Problem List User Settings Next Grades 6a, y 5a. their answers, and give them chance to make multiple attempts at problems. Clicking on Preview Answers will show you how WeBWorK interprets your answers. Webwork Answer Keys; Wileyplus Answer Keys; Cengage Answers Key; MyAccountinglab Answers; MyEconlab Answer Keys; MyItlab Answer Keys; Physics Homework Help; Our experts strive their best to provide all our clients with unique and highly customized programming homework answers that stand out from the crowd. 01 percent of the right answer as correct. Synonyms, crossword answers and other related words for NETWORK. Answer Key (homework) Chapter Reviews. Part of the world's leading collection of online homework, tutorial, and assessment products, Pearson MathXL is designed with a single purpose in mind: to improve the results of all higher education students, one student at a time. Getting these answer keys gives them considerable leverage in completing the assignments both in time and in scoring better grades. Site Information This site provides WeBWorK resources for the Mathematics Department at CWU. Given a function f of a real variable x and an interval [a, b] of the real line, the definite Integral is defined informally to be the area of the region in the xy-plane bounded by the graph of f, the x-axis, and the vertical lines x = a and x = b, such that area above the x-axis adds to the total, and that below the x-axis subtracts from the total. solution link is keyboard accessible, but still doesn't identify as a link. This is an open-book exam. 2 years ago. Follow RSS feed Like. One of the unique to WebWork features (as compared to Struts) is the notion of a result type which is the process responsible for combining the model with a some kind of template (usually JSP or FreeMarker) to create a view. When show solutions is checked in WebWork, such solutions are gathered and displayed at the end of each question. Download Wiley Plus Answer Key Calculus. There are a couple of corrections that I wanted to point out for WebWork: 1. There will be WeBWork-based homework assignments almost every week with 3-5 questions per assignment. encoding" rather than the setting 'URIEncoding=UTF-8' and ' -Dfile. 0003\) was too much rounding. Write a verbal model of the problem and then replace words with numbers, variables and/or symbols. We will use WebWork and you should always access WebWork through Canvas! If you are not familiar with WebWork, you will find a brief introduction here. Free practice questions for Trigonometry - Sin, Cos, Tan. WeBWorK simply lets students know whether they have submitted a correct answer and provides the opportunity to try again. Certainly, it would be possible to program into WeBWorK the capacity to give hints based on the type of wrong answer. MAT136H1: Calculus 1(B) University of Toronto Winter 2019 But just as much as it is easy to nd the di erential of a given quantity, so it is di cult to nd the integral of a given di erential. 13 on this server. WeBWorK will do more than just tell you if you are right or wrong. After clicking this button, a drop-down menu will appear listing OPL Directory Problems. Work additional problems to prepare for the exam. Alan Tucker, SUNY-Stony Brook. 0 and Webwork), MySql KEY RESPONSIBLITIES: • Designed the layout of the system. 4723 accepted. Create and Manage Online Surveys. WeBWorK is a free online homework platform maintained by the Mathematical Association of America. 1) All exams are taken on the WebWork system. Math Problem Solver Questions Answered Free Algebra Geometry Trigonometry Calculus Number Theory Combinatorics Probability. Any sequence con-verging to a value other than zero will have a divergent series. [email protected] WebAssign is an online learning platform built by educators that provides instructors the tools needed to empower conﬁdent, self-suﬃcient learners. Since 19 is a prime number and is the greatest factor for both numbers, the correct answer is C. This can be used to help spot errors in your typ-ing, and verify that WeBWorK. So, the only person who can know the answer keys is your teacher. Abstract: WeBWorK -- Math Homework on the Web WeBWorK is an internet-based system for generating and delivering homework problems to students. )c) Determine the intervals on whichf is increasing and decreasing. The Active Calculus texts are different from most existing calculus texts in at least the following ways: the texts are freely readable online in HTML format (new in this version of Active Calculus Multivariable) and are also available for in PDF; in the electronic format, graphics are in full. 0867 p^placebo= view the full answer. In order to submit your solution to be graded, click on Submit Answers. Don't worry, if you get it wrong you can try it again. Montresor calls it “the white webwork which gleams from these cavern walls. Calculus questions with detailed solutions are presented. If you enter an answer that WeBWorK determines that is not correct that you strongly. Constant Value: "webwork. Spreadsheet help. why do man like to cheat 6/1/2009 1/1/0001. These linear algebra lecture notes are designed to be presented as twenty ve, fty minute lectures suitable for sophomores likely to use the material for applications but still requiring a solid foundation in this fundamental branch. Week 1 Webwork 1 Week 2 Webwork 2 Week 3 Webwork 3 Week 4 Webwork 4 Week 6 Webwork 5 Week 6 Webwork 6 Week 7 Webwork 7 Week 8 Webwork 8 Week 9 Webwork 9. Exponential growth and decay often involve very large or very small numbers. We are here to help you to save your dear time. It is primarily a web-based math assessment tool for delivery and automatic grading of math homework and tests. WeBWorK Tips: Click on a problem to see the details (the list of problems appears in the menu on the left). There will be WeBWork-based homework assignments almost every week with 3-5 questions per assignment. Certainly, you may complete this assignment at any time during the summer, but it will most likely ANSWER KEY. You discover new ways to record solutions with interval notation, and you plug trig identities into your equations. We hope that the following list of synonyms for the word network will help you to finish your crossword today. WeBWorK is a well-tested homework system for delivering individualized problems over the web. On the other hand, Perl is well-known. You can help protect yourself from scammers by verifying that the contact is a Microsoft Agent or Microsoft Employee and that the phone number is an official Microsoft global customer service number. Square Roots and Cube Roots problems with solutions. h(x) = f(x − 3) + 2 Subtract 3 from the input. day week month year all. Answer Key: 2. WeBWorK has been used in a number of universities such as the University of Michigan, Johns Hopkins University, Dartmouth College and the University of Rochester. Sometimes the unit is provided for you. Book our help with quantitative reasoning homework help, and your academic history will never be the same again. Department of Mathematics and Statistics Texas Tech University Lubbock, Texas 79409-1042 Voice: (806)742-2566 x 226 FAX: (806)742-1112 Email: kent. Select (with the checkboxes) the problem sets you want to save. Last revised 2/9/11. My profile; Logout Struts 2 takes on WebWork 2 – Part 1. The mean, standard deviation, sample size, and other constants are randomly generated so each student has a unique. The curves approach these asymptotes but never cross them. Intervals in WeBWorK. The concepts are the same. The Webwork assignments form the core of this course. You will need to get assistance from your school if you are having problems entering the answers into your online assignment. For example, we know that the velocity is 30 mph during 5 hours and 50 mph during 1 hour and we need to know. Leave a reply. Most scholars think that quantitative reasoning concepts are meant for a genius. The probability of event A is the sum of the probabilities of all the sample points in event A. WebWork is a free online system that will be used to deliver your homework. What do I do? Can I get an extension on my homework? Can I print my homework so I can work on it offline? How do I use WebWork off campus? About WebWorK. Clicking on Preview Answers will show you how WeBWorK interprets your answers. Here is a short tutorial for getting started with WebWork. The location and content of the access log are controlled by the CustomLog directive. solution link is keyboard accessible, but still doesn't identify as a link. WeBWorK is such an online homework platform used mainly for mathematics and science. You should do problems from the book, look over old quizzes and exams, do webwork problems, etc. 2x, 2 x or 2x, also 2(3+4). Any sequence con-verging to a value other than zero will have a divergent series. Checking Answers in a WeBWorK Problem. Examine the term cheating as it relates to one of the following 12/29/2009 1/1/0001. However, you do not hand in problem solutions on paper: rather you complete each question online and click on "Check Answers". You have six submissions per problem. It's easy & will help you with the homework system. Double check whether the setting "webwork. Instructors. From that page, clicking on an assignment brings up a list of problems, and clicking on a problem shows you what a student will see. Calculus 1 WebAssign Answers. Last revised 2/9/11. Check the solution x=3 by substituting 3 in the original equation for x. Leave a reply. Multivariable calculus, early transcendentals, in PDF format or HTML format. You can answer your missed calls or messages once you have a logical break in your task. Book our help with quantitative reasoning homework help, and your academic history will never be the same again. Can do multiple choice, matching, T/F questions, etc. Webwork Answers Calculus. This will allow you to easily filter all right and wrong answers when exporting your results. Summary Files Reviews Support Wiki Mailing Lists News. Worked example: Cube root of a negative number. Get ahead of the semester with all the resources you need for a great course experience. Course management front end for WeBWorK. Enter numeric value with unit answers. 0867 p^placebo= view the full answer. Underline key words and phrases. WebWork Answers and Solutions for University Students. There is a technical definition for absolute value, but you could easily never need it. WeBWorK FAQ - a small but growing list of common pitfalls. Tips! Note that there’s no way to see if answers are right or wrong in your Responses. You can also use the full 8th Edition. Technologies Used: Java 1. Sometimes, if the numbers allow it, you can work backwards to determine an element's half-life. Use it to display information for the entire WeBWorK site which will be viewed at login time. This is an official Stephen F. When executing scripts in an HTML page, the page becomes unresponsive until the script is finished. Final Exam Exam December 2018, questions and answers Consuming Grief Ch5 - Lecture notes 5 Consuming Grief Ch8 - Lecture notes 8 2018 FA-MATH-1051H - Webwork answers for Prof. By convention, statisticians have agreed on the following rules. Activity 1. edu or Paul Hummel at [email protected] Preview your answers before submitting them. Webwork assignments form the core of this course. Worked example: Cube root of a negative number. 1 Formulation of Linear Programming Problems; 7. Quantitive Reasoning homework help. point) Sketch the region. Free practice questions for Trigonometry - Sin, Cos, Tan. Checking Answers in a WeBWorK Problem. WeBWorK is a well-tested homework system for delivering individualized problems over the web. Permutations and Combinations Worksheet Answer Key As a Matter of Factorial 1. the mark pc multi reloaded games five net key answers 2016. 6 The second derivative. For deploying i have to copy a class-file and an vm-file to the jira-server. NOTE: When using interval notation in WeBWorK, remember that:You use 'INF' for oo and '-INF' for -c0. Below is a set of suggested homework problems for Math 2A and 2B. A list of accepted unit abbreviations, and other help for entering your answers in the proper accepted format, can be found here , or from links on the WebWork login or problem set pages. Answer the following question: How many sharps does the key of C-sharp have? Write your answer in the corresponding box on your WEBWORK PAPER. This means that letter C is the correct answer choice. Main Street \| 6188 Kemeny Hall \| Hanover NH 03755-3551 \| 603. When there is a substantive change, I will update the files and note the change in the changelog. Apache server records all incoming requests and all requests processed to a log file. WeBWorK is a well-tested homework system for delivering individualized problems over the web. Comparing the values may just involve rewriting an expression in a different way. Pressing this key does not submit your solution to be graded. website builder. The student can get a printout of the work done in a WeBWorK session. How many producers are in the food web? Name them. WebWork Answers and Solutions for University Students. 472322, then 3. To get full credit on these questions: Enter BOTH the value and unit parts of the answer, such as 26 kg. Complete the homework assignment on WebWork. At Portland Community College, Part 1 is used in MTH 60, Part 2 is used in MTH 65, and Part 3 is used in MTH 95. ” The nitre, therefore, represents the web, or the trap, Montresor has set for his victim. Then pick an article, read it and answer the following questions: What crime was committed, who was it committed against, who committed it, and why did they commit it?. Click the images to enlarge the screen shots. Skills review handbook; economics chapter exam answer key for homework. If the correct answer is smaller, π/2 for example, an answer would need to be within about 0. cherry Created Date: 2/10/2010 12:16:00 PM Company. The key for this course is:. Every homework problem is graded automatically with immediate feedback to the student, even in complex algebraic formulation. 0003\) has one significant digit and four places past the decimal. Department of Mathematics and Statistics Texas Tech University Lubbock, Texas 79409-1042 Voice: (806)742-2566 x 226 FAX: (806)742-1112 Email: kent. Sep 15th: Write down (at least) one question or comment about Chapter 6 of the book. Answers vary, but going around the circle 2! would take us back to the same place as ! 6 or!5! 6. Read the page Course Information for information on course and grading parameters. Example: Substituting numbers can suggest the wrong answer 26 Exercise 26 12. What is the sims 3 razor1911 serial key? I need one that accually works. 5 =2! 52 =2!"2 5. Great job in putting this together. You can get a hardcopy of a problem set from the problem listing page of a problem set. WeBWork provided by the Mathematical. The common final for Spring will take place on Saturday, June 8th, 1: The webwork assignments alone are not sufficient preparation for the exams, you will need to solve the suggested homework problems. · If a problem calls for a decimal answer, give at least four decimal digits, or as many as the problem specifies. 70860 Hw12 8 WeBWorK MATHEMATICAL ASSOCIATION OF AMERICA webwork mth 124 fs16 70860 hw12 8 MAIN MENU Courses Homework Sets Hw12: Problem 8 Hw12 Problem 8 Previous Problem List User Settings Next Grades 6a, y 5a. encoding" in the system information had been correctly set to UTF-8. To join the Wiki request a Wiki account. And use 'U' for the union symbol. Related: 7 Time Management Tips to Help You Get More Productive!. 235 counts as correct. social sciences. This site was designed with the. Stack Overflow Public questions and answers; Teams Private questions and answers for your team; Hot answers tagged webwork. Main Street \| 6188 Kemeny Hall \| Hanover NH 03755-3551 \| 603. The Domain (all the values that can go into the function) is all Real Numbers up to and including 6, which we can write like this: Piecewise functions let us make functions that do anything we want! The Absolute Value Function is a famous Piecewise Function. Select (with the checkboxes) the problem sets you want to save. 55 sus 75, T 66. Figure 2: Student work map and. If you notice any errors or require assistance, please post on the forums for community assistance, or e-mail [email protected] Week 2 AP Skills 1 Week 10 AP Skills 2 DBQ 1 Webwork. M118 Test 1 Key. Lisa Oberbroeckling Office: 312 Knott Hall Phone: 410-617-2516 Email: loberbro "at "loyola "dot" edu. Date and Location to be announced. Checking Answers in a WeBWorK Problem. Do assignment 0 (non-credit) first in order to understand the protocols for submitting answers on Webwork. Preparable; public class ProductAction implements Preparable {. By providing students with immediate feedback as to the correctness of their answers, students can be encouraged to. Or, use the shortcut keys Hold and type 0177 - you will get ± or gray squares in some places. ESSAY, CONSEQUENCES OF A COLLEGE STUDENT CHEATING ON EXAM 9/27/2010 1/1/0001. Webwork Answers Key. To access WeBWorK go the Modules tab in your Canvas page and select the link for WeBWorK. Don’t even think about trying to get the key. WeBWorK @ UW-La Crosse. For example, if an object is tossed into the air we might find the following data for the height in feet, y, of the object as a function of the time in seconds, t, where t = 0 is when the object. The student benefits from the immediate feedback and the learning associated the. Figure 1(a): A typical WeBWorK question and student answer 129 Segalla, A. Click on (blue) Problem 1. Description: A WeBWorK problem that asks students to build a sampling distribution based on a small population, answer questions on it, and calculate the standard error using the finite correction factor. Overview of Software in the Department of Mathematics. IMathAS is an Internet Mathematics Assessment System. Note that, unlike in your calculus course, most problems will now require both a numerical answer and the appropriate physical units for that answer. Do not trust anyone else who offers ready MathXL answer key. Below is a set of suggested homework problems for Math 2A and 2B. The formal, authoritative, deﬁnition of limit 26 12. Precalculus by Carl Stitz, Ph. Users post a picture or type their homework questions onto online forums, and those who answer the questions can win e-coins that can be used to buy electronics like iPhones and laptops. Getting Started. Book our help with quantitative reasoning homework help, and your academic history will never be the same again. By convention, statisticians have agreed on the following rules. What is the sims 3 razor1911 serial key? I need one that accually works. Multivariable calculus, early transcendentals, in PDF format or HTML format. Webwork Answers Calculus. 6 The second derivative. Just as each distinct compound has a unique molecular structure which can be. For issues with the problems in the Louisiana Tech University library contact Dave Meng at [email protected] You can also do operations on whole numbers, integers, and decimal numbers and get answers in scientific notation. Some WeBWorK problems require a spreadsheet. Even if it may sound like a fun thing to do, unauthorized access to a computer system is a felony which carries a penalty of up to 10 years in prison. " I NSF awarded MAA half a million to expand WeBWork and to create a permanent supportive home for WeBWorK. social sciences. The Rutgers study of WeBWorK's impact demonstrated this claim in spades. MATHEMATICS 1210-93 Calculus I. Fractions are entered using the division / key. Do you have a personal observation which may help others? Free Math Help - Submit your questions, comments, and suggestions using. the numbers in the problem, are randomized so that a students will be assigned his or her individual version of the problem. Book our help with quantitative reasoning homework help, and your academic history will never be the same again. If you enter an answer that WeBWorK determines that is not correct that you strongly. Many students believe they can find ready solutions on the internet or browse for answers on Google. From Trigonometry For Dummies, 2nd Edition. Read the page "Course Information" for information on course and grading parameters. If you recall from earlier mathematics studies, average velocity is just net distance traveled divided by time. Class Overview. Many smart students are constantly looking for webwork answers key. Integrating WeBWork in Calculus III Larry Wang Mathematics Department , Southern Polytechnic State University , Marietta , GA , USA Abstract-The aim of this paper is to demonstrate how the online homework improves the student success rates in one Calculus III course. Site Information This is the WeBWork server hosted by the Center for Science and Engineering at Smith College on behalf of the Math Department. Using WeBWorK in an online course environment. 3 Sizes of Sets; 1. Precalculus by Carl Stitz, Ph. These linear algebra lecture notes are designed to be presented as twenty ve, fty minute lectures suitable for sophomores likely to use the material for applications but still requiring a solid foundation in this fundamental branch. Final Exam Information. In LaTeX output for answer keys, the solutions are displayed in boxes where they occur in the question. Calculus 2 Webassign Answers. A runner sprints around a circular track of radius m at a constant speed of 7 m/s. If you are a student enrolled in a WeBWorK course, please contact your instructor for information about your course. In WeBWorK, you must enter your answer with the variable on the left side of the equal sign: t = 3 The terms in an equation may be in any order and the variable may be a letter other than $$x. The easiest way to find them is to use the “OPL Directory” button in the Library Browser. (𝑥)= 𝑥2+3 (𝑥)= 1 𝑥 Homework: You should now be ready to attempt problem 7 in “Homework – Section 3. For example, to enter , type 7/2. The document Entering Answers in WeBWorK gives more information about how to enter mathematics to answer questions in WeBWorK. Here you will find links to the Eureka Math Problem Sets that students worked at school, the Homework that follows that Lesson, and videos of the homework being explained. Book our help with quantitative reasoning homework help, and your academic history will never be the same again. Neither do we condole any small. Checking Answers in a WeBWorK Problem. In addition to ecosystems and food chains, these printable worksheets also cover consumers and producers, as well as herbivores, carnivores, and omnivores. Select (with the checkboxes) the problem sets you want to save. Lull's Documents. 0867 p^placebo= view the full answer. Make sure your name, section and instructor is on your scantron. Bisection Method Notes. You've got a lot of priorities and plenty of things to do (including homework). If you mark the check box the calculator automatically determines the number of significant figures in the answer. Checking Answers in a WeBWorK Problem. 5 =2! 52 =2!"2 5. WeBWorK has been used in a number of universities such as the University of Michigan, Johns Hopkins University, Dartmouth College and the University of Rochester. 5 may be marked incorrect, but 3. The Webwork assignments form the core of this course. However, you do not hand in problem solutions on paper: rather you complete each question online and click on "Check Answers". EugeneWebworkTraining; Math0031_template; Math0120-10093; Math0120-10271; Math0120-10571; math0120-demo; math0200-demo; Site Information. WeBWorK has been upgraded to version 2. 0001570796 of π/2 to be considered correct. Webwork answers homework problems. See your grades, download class materials, and connect with classmates. Javadoc API documentation for WebWork. - James Joseph Sylvester Fall 2019 Courses. Figure 1(a): A typical WeBWorK question and student answer 129 Segalla, A. encoding" rather than the setting 'URIEncoding=UTF-8' and ' -Dfile. You can get a hardcopy of a problem set from the problem listing page of a problem set. She believes students need to struggle at times for an answer, rather than always be able simply to find the answers in the book (and to work. , and an online program. If you notice any errors or require assistance, please post on the forums for community assistance, or e-mail [email protected] The number \(0. Pre-calculus involves graphing, dealing with angles and geometric shapes such as circles and triangles, and finding absolute values. Sep 15th: Write down (at least) one question or comment about Chapter 6 of the book. Data Binding • HTTP is not aware of data types but Java is! • WebWork helps with this mismatch by letting you work with your raw POJOs rather than type-less strings • Can support basic objects, lists, maps, sets, and. In LaTeX output for answer keys, the solutions are displayed in boxes where they occur in the question. The mean, standard deviation, sample size, and other constants are randomly generated so each student has a unique. (- infinity , infinity ) is the set of all real numbers. Students may seek help solving WeBWorK problems from their instructor, the MRC, classmates and even friends or tutors. It is your responsibility to check WeBWorK for new assignments. They found that using WeBWorK to deliver homework. edu or Paul Hummel at [email protected] Learn how energy is transferred from one living thing to another with the food web. Key: Fall 2018: e1. Day 1: Knights & Knaves, Cheryl’s Birthday Problem ; Day 2: More Knights & Knaves, Game of 21, Tic-tac-toe. This will ask WeBWorK to display at the top of the page its interpretation of what you have entered. Questions and Answers on Functions. answer in the space provided. You should answer all questions since there is no penalty for wrong answers. Show that lim x→3 3x + 2 = 11 28 12. Enter values separated by commas such as 1, 2, 4, 7, 7, 10, 2, 4, 5. Follow RSS feed Like. Please work through each problem and insert the answer or answers requested. Start early, when each set opens. WebWork actions only need implement Action, not extend ActionSupport. Fourth year electrical jatc class 1. Answer Key: 2. Part I Model Problems. 130 Answer keys PRACTICE TEST 1 LISTENING Section 1 1 A 2 C 3 D 4 D 5 C 6 Prescott (must be correct spelling with capital "P") 7 41 8 Fountain (must have capital "F") 9 752239 10 £65 Section 2 11 E 12 F 13 H 14 250 million 15 roads//road system 16 too late 17 school children//boys 18 3 19. Over a small fraction of it, click submit. MSU Mathematics Undergraduate Programs. I know my answer is right, but WebWork is counting my submission wrong. 13 on this server. If you know the keys beforehand, you can easily answer questions, wasting no time for thinking or searching for a clue in your textbook. By Trackula and Co-written by Eduard Kassel. To get full credit on these questions: Enter BOTH the value and unit parts of the answer, such as 26 kg. 6 Chapter 4–2 – FUNCTION Notation (Day 2) SWBAT: Evaluate Functions Warm – Up: Determine the domain and range of the relation below. Hello r/EngineeringStudents I want to share with you a trick I discovered in my undergrad to find the answers to any WebWork question. Synonyms, crossword answers and other related words for NETWORK. Sep 15th: Write down (at least) one question or comment about Chapter 6 of the book. WebWork is a free online system that will be used to deliver your homework. From that page, clicking on an assignment brings up a list of problems, and clicking on a problem shows you what a student will see. I need to pay for a copy of the Algebra 1 answer keys to lessons,homework,tests. Partners include the Fort Worth Museum of Science and History, the American Academy of Forensic Sciences, and CBS. When will the tank be half empty? t= in days 2. You must complete the exam individually. The questions are about important concepts in calculus. Precalculus by Carl Stitz, Ph. Includes full solutions and score reporting. You've got a lot of priorities and plenty of things to do (including homework). Part I Model Problems. literature and english. Webwork answers homework problems. Can check decimal answers (1:73:::), exact answers (p 3=2),. These answers may not be solutions to the equation. Clicking on Preview Answers will show you how WeBWorK interprets your answers. We are here to help you to save your dear time. Webworks 2nd Hour. Handles the context of each invoked action while providing an abstraction layer for both servlet and non servlet based applications. After checking your work enter each answer and click the "SAVE and Preview" button. This organization primarily operates in the Custom Computer Programming Services business / industry within the Business Services sector. James Stewart, Single Variable Calculus: Math 1A,B at UC Berkeley, 8th Edition (Cengage, 2016), ISBN 978-1-305-76527-6. In atomic physics, the Rutherford–Bohr model or Bohr model, introduced by Niels Bohr in 1913, depicts the atom as a small, positively charged nucleus surrounded by electrons that travel in circular orbits around the nucleus—similar in structure to the solar system, but with attraction provided by electrostatic forces rather than gravity. You have a reason to fall in love with this course through our support. 0 sets 1 member 01gothic 3 forsaken gods proper procyon net key answers 2016 · Baltimore. If necessary write expressions for other unknowns in terms of x. Chapter 3 Vectors in Physics. Preparable; public class ProductAction implements Preparable {. You have six submissions per problem. Questions and Answers on Functions. Practice: Equations with square roots: decimals & fractions. This number might cause you trouble if you enter it, because maybe the “real” answer was \(0. For example, the interval from -3 to 7 that includes 7 but not -3 is expressed (-3,7]. Each CD contains the entire text, in both pdf and Word format, and the entire answer key, in both pdf and Word format. AP Calculus AB Exam and AP Calculus BC Exam, and they serve as examples of the types of questions that appear on the exam. Calculus 1 WebAssign Answers. String answers allow for T/F, matching, multiple-choice, and short answer questions. Find the absolute minimum value for f(x) on the interval [ 4;3] when f(x) is given by f(x) = x2 4 x2 + 4: a) 1 b) 5=13 c) 4=5 d) 2 e) 2 f) 1 Solution: a) 2. IELTS Cambridge Answer cambridge ielts 1 1. Immediate Feedback Capability – students can get response to their homework questions immediately. For most people, tracking working hours seems like a boring and annoying assignment. After clicking this button, a drop-down menu will appear listing OPL Directory Problems. Default apache access log file location: RHEL / Red Hat / CentOS / Fedora Linux Apache access file location –. These linear algebra lecture notes are designed to be presented as twenty ve, fty minute lectures suitable for sophomores likely to use the material for applications but still requiring a solid foundation in this fundamental branch. The curves approach these asymptotes but never cross them. In addition to ecosystems and food chains, these printable worksheets also cover consumers and producers, as well as herbivores, carnivores, and omnivores. The Webwork assignments form the core of this course. You can work on the problems in any order you wish. Trigonometry is the study of triangles, which contain angles, of course. No hot answers found. Technologies Used: Java 1. , and an online program. When show solutions is checked in WebWork, such solutions are gathered and displayed at the end of each question. On this page you can find the syllabus and info about the exams, as well as practice exams. Request Make-up Exam For Common Final. Specifics of a selected problem, e. We've arranged the synonyms in length order so that they are easier to find. Nonetheless, we offer help in providing Webassign equations solutions. When entering answers to the personal version the system will tell you whether your answer is correct or not. However, you do not hand in problem solutions on paper: rather you complete each question online and click on "Check Answers". Site Information This is the WeBWork server hosted by the Center for Science and Engineering at Smith College on behalf of the Math Department. Info Michigan State University (MSU)'s MTH department has 66 courses in Course Hero with 3030 documents and 221 answered Answers in as fast as 15 minutes. WebWork has a standard mechanism to deal with the duplicate post problem. Sign up for a Piazza account through Connect. Experienced instructors will see many changes; click here to read about some of them. The solutions and answers are provided. In an effort to counteract student cheating, the professor. Final Exam Exam December 2018, questions and answers Consuming Grief Ch5 - Lecture notes 5 Consuming Grief Ch8 - Lecture notes 8 2018 FA-MATH-1051H - Webwork answers for Prof. foreign languages. Last year, I was tutoring for the Algebra Regents and so emathiinstruction of my clients used Emathinstruction at their schools. Try Adobe's Flash Player Help. 10P4 104 = 5040 10000 = 0. Only top voted, non community-wiki answers of a minimum length are eligible. Single variable calculus, early transcendentals, in PDF format. Webwork Answers Even though Webwork Answers is not very popular, it requires you to stay at the same place doing the same thing for a very long time. Work the problems out by hand before entering them in the system. • Enter the numerical answers: notice that the instructor can require the student to do the calculation or allow webwork to do the numerical calcu-lation for them. Variations on the limit theme 30 13. Extra Credit Assignment Extra Credit Test File. Active Calculus is different from most existing calculus texts in at least the following ways: the text is freely readable online in HTML format and is also available for in PDF; in the electronic format, graphics are in full color and there are live links to java applets; version 2. Read the following headlines and write down a list of crimes committed in New York on a given day. Operations Management. You have a reason to fall in love with this course through our support. 0003091\text{,}$$ and rounding to 0. Class Material. Honors PreCalculus 4th Quarter Agenda Webwork Link. Don’t even think about trying to get the key. Substitution method can be applied in four steps. Partners include the Fort Worth Museum of Science and History, the American Academy of Forensic Sciences, and CBS. IMathAS is an Internet Mathematics Assessment System. I wouldnt want this class to affect. The Webwork assignments form the core of this course. social sciences. Don't submit partial answers. , and an online program. You discover new ways to record solutions with interval notation, and you plug trig identities into your equations. 5 Exercises. The Math Learning Centre is a free tutoring service provided by graduate students in our department. WeBWorK Functions. " NSF awarded MAA half a million to expand WeBWork and to create a permanent supportive home for WeBWorK. Webwork Answer Keys; Wileyplus Answer Keys; Cengage Answers Key; MyAccountinglab Answers; MyEconlab Answer Keys; MyItlab Answer Keys; Physics Homework Help; Our experts strive their best to provide all our clients with unique and highly customized programming homework answers that stand out from the crowd. This course supports guest logins. For example, to enter , type 7/2. MATHEMATICS 1210-93 Calculus I. Common Core alignment can be viewed by clicking the common core. After the assignment's due date, the system will show you the correct answer for each problem when you try it (but your answers won't be. Site Information This file is at htdocs/site_info. Questions and Answers on Functions. Request Make-up Exam For Common Final. Recommended Problems for each exam can be accessed here. I have a question on an assignment. What do I do? Can I get an extension on my homework? Can I print my homework so I can work on it offline? How do I use WebWork off campus? About WebWorK. Checking Answers in a WeBWorK Problem. , and an online program. The due dates for the WebWork sets are here. The Active Calculus texts are different from most existing calculus texts in at least the following ways: the texts are freely readable online in HTML format (new in this version of Active Calculus Multivariable) and are also available for in PDF; in the electronic format, graphics are in full. The main idea here is that we solve one of the equations for one of the unknowns, and then substitute the result into the other equation. Welcome to the UC Davis Mathematics Department WebWork Help Wiki. Get to know some special rules for angles and various other important functions, definitions, and translations. "hint" and "solution" have special roles in WeBWorK, but "answer" does not. Read the page "Course Information" for information on course and grading parameters. It is primarily a web-based math assessment tool for delivery and automatic grading of math homework and tests. To get full credit on these questions: Enter BOTH the value and unit parts of the answer, such as 26 kg. Get ahead of the semester with all the resources you need for a great course experience. State Math Contests SMC2003 2003 Answers SMC2004 2004 Answers SMC2005 2005 Answers SMC2006 2006 Answers SMC2007 2007 Answers SMC2008 2008 Answers SMC2009 2009 Answers. Preparable; public class ProductAction implements Preparable {. u/saraakabani. MATHEMATICS 1220-93 Calculus II. Print THIS WORKSHEET about the easier scales. 6, Appfuse 1. Answer Save 55 Answers. f1;1;1;1;1;:::gconverges, but 1 + 1 + 1 + 1 + 1 + :::? No. Escape will cancel and close the window. quiz 1 quiz 2 quiz 3 quiz 4 quiz 5 quiz 6 quiz 7 quiz 8 quiz 9 quiz 10 quiz 11 quiz 12 quiz 13 quiz 14 quiz 15. One answer is x>=0 (x is greater than or equal to 0). Ut tempus mattis consectetur. The document Entering Answers in WeBWorK gives more information about how to enter mathematics to answer questions in WeBWorK. MAT136H1: Calculus 1(B) University of Toronto Winter 2019 But just as much as it is easy to nd the di erential of a given quantity, so it is di cult to nd the integral of a given di erential. The Bohr Model. How many ways can we order 8 swim-suits in 4 lockers? 8! 4! = 1680 5. By Trackula and Co-written by Eduard Kassel. Left and right limits 30. ","description_more":" This course was curated by Christina Safranski and the Edfinity team. Use it to display information for the entire WeBWorK site which will be viewed at login time. Mathematics teachers may also be interested to follow the work of York University Maths department, who are working on some projects to augment Moodle, particularly its Quiz module for online assessment, for example by integrating a system which is able to mark algebraic and trigonometric answers to open-ended questions. See the release notes for more info. In other words, given a Laplace transform, what function did we originally have? We again work a variety of examples illustrating how to use the table of Laplace transforms to do this as well as some of the manipulation of the given Laplace transform that is needed in order to use the table. Moreover, sometimes we cannot say with certainty whether the integral of a given quantity can be found or not. Create and Manage Online Surveys. WebAssign Student Help; COVID-19 Impacts on Your Courses; What's New in WebAssign. Ut tempus mattis consectetur. At Portland Community College, Part 1 is used in MTH 60, Part 2 is used in MTH 65, and Part 3 is used in MTH 95. The student benefits from the immediate feedback and the learning associated the. Write a rule for g. Preface Entering WeBWorK Answers. Final Exam Exam December 2018, questions and answers Consuming Grief Ch5 - Lecture notes 5 Consuming Grief Ch8 - Lecture notes 8 2018 FA-MATH-1051H - Webwork answers for Prof. Donec odio magna, aliquam vel sapien eget, consequat interdum lacus. Constant Value: "webwork. Part I Model Problems. Compute for each pair of means, where M i is one mean, M j is the other mean, and n is the number of scores in each group. Webwork Answers Calculus. 1 Formulation of Linear Programming Problems; 7. Since the string is 30 inches in length, the maximum point will be 30 inches above the minimum. ” The nitre, therefore, represents the web, or the trap, Montresor has set for his victim. high school math. To work in connection with the How2Become Quantitative Reasoning guide, we have provided you with detailed sample questions that takes you step-by-step through the process of working. Web-Based Reporting. • Deployed the whole application. The goal of the WBH system called WeBWorK is immediate "correct" or "incorrect" feedback. Use it to display information for the entire WeBWorK site which will be viewed at login time. For the questions and answers lick the Q and Homewrok tab. In a certain state’s lottery, 48 balls numbered 1 through 48 are placed in a machine and six of them are drawn at random. 3 Solutions Webwork 6. Just be sure to notice the difference between the "Preview" and "Submit" buttons. Webwork is an online homework system used to supplement the textbook homework and quizzes in the Math 2A and 2B courses. Read the following headlines and write down a list of crimes committed in New York on a given day. Auto-suggest helps you quickly narrow down your search results by suggesting possible matches as you type. 0 now contains WeBWorK exercises in each chapter, which are fully interactive in the HTML format and included in. 0003091\text{,} and rounding to \(0. WeBWorK has been upgraded to version 2. Step 1: Viewing the Assignments from The Instructor Perspective. IMathAS is an Internet Mathematics Assessment System. Please note that if the due date is indicated, say, as Sep 6, the actual WebWork deadline is September 7, 3:00 AM. Here is a short tutorial for getting started with WebWork. (Separate multiple answers by commas. social sciences. Pressing this key does not submit your solution to be graded. So, if the right answer is 12345, anything between 12343. This means that you have to understand the problems well enough to handle these minor alterations. How many ways can we order 6 com-puters if we have only space for 3? 6! 3! = 120 4. Then read more about using MyOpenMath in the classroom. If the correct answer is smaller, π/2 for example, an answer would need to be within about 0. (If you like, you can test this out on the next as 7+9, 18-2, 82, 32/2, and 4ˆ2. Select (with the checkboxes) the problem sets you want to save. Trigonometry is the study of triangles, which contain angles, of course. To access the answer key to your Web assign class will depend on if your instructor provided one after the class. Many students believe they can find ready solutions on the internet or browse for answers on Google. a straight line at 120 V B. Practice: Square and cube challenge. 7 Limits, Continuity, and Differentiability. (Separate multiple answers by commas. " I NSF awarded MAA half a million to expand WeBWork and to create a permanent supportive home for WeBWorK. Book our help with quantitative reasoning homework help, and your academic history will never be the same again. (Homework from the class notes is self-graded and will not be turned in. Each CD contains the entire text, in both pdf and Word format, and the entire answer key, in both pdf and Word format. Solve one of the equations for either x = or y =. Showing 1 to 8 of 111 View all. You should do problems from the book, look over old quizzes and exams, do webwork problems, etc. To be successful with WeBWorK you must have ready access to the internet. Quizzes should be completed on your own. C) I prefer to consult resources (notes, etc. Studying MATH M118 Finite Mathematics at Indiana University? On StuDocu you find all the study guides, past exams and lecture notes for this course. The easiest way to find them is to use the “OPL Directory” button in the Library Browser. If you already have an account, you can log on using the box to the right. edu Chapter 5 Sampling Chapter 6 Cls: Problem 6 Previous Problem List Next nts was selected from a population with unknown mean H and standard deviation o. Several changes were incorporated into WeBWorK after the first year of this study including a Preview feature that lets students see the typeset version of the expression they have entered before submitting it, a feature that greatly reduced student frustration, and a warning that appears when a student attempts to submit an answer that has. Show that lim x→4 1/x = 1/4 29 Exercises 30 13. You select two cards simultaneously at random from a standard 52-card deck of cards with suits diamonds, hearts, spades and clubs (no jokers in the deck). Enter an answer and hit "Submit Answers". If you recall from earlier mathematics studies, average velocity is just net distance traveled divided by time. Hello r/EngineeringStudents I want to share with you a trick I discovered in my undergrad to find the answers to any WebWork question. If you notice any errors or require assistance, please post on the forums for community assistance, or e-mail [email protected] What is the sims 3 razor1911 serial key? I need one that accually works. WebWorks® is the industry's leading automated software solution for single-source publishing,help authoring, conversions, and transformations of content into responsive HTML5 CSS3 document sets with the most advanced search features available. 6, Appfuse 1. After the assignment's due date, the system will show you the correct answer for each problem when you try it (but your answers won't be. Piazza is the online forum for Math 102. 130 Answer keys PRACTICE TEST 1 LISTENING Section 1 1 A 2 C 3 D 4 D 5 C 6 Prescott (must be correct spelling with capital “P”) 7 41 8 Fountain (must have capital “F”) 9 752239 10 £65 Section 2 11 E 12 F 13 H 14 $250 million 15 roads//road system 16 too late 17 school children//boys 18 3 19. Remember to include the zero. 130 Answer keys PRACTICE TEST 1 LISTENING Section 1 1 A 2 C 3 D 4 D 5 C 6 Prescott (must be correct spelling with capital "P") 7 41 8 Fountain (must have capital "F") 9 752239 10 £65 Section 2 11 E 12 F 13 H 14$250 million 15 roads//road system 16 too late 17 school children//boys 18 3 19. -Each of you has a diﬀerent a and b because of the randomizing behavior. Page generated at 05/03/2020 at 12:08am ADT. This is a 6 part worksheet that includes several model problems plus an answer key. Practice: Equations with square roots: decimals & fractions. Skills review handbook; economics chapter exam answer key for homework. Free practice questions for Trigonometry - Sin, Cos, Tan. The Active Calculus texts are different from most existing calculus texts in at least the following ways: the texts are freely readable online in HTML format (new in this version of Active Calculus Multivariable) and are also available for in PDF; in the electronic format, graphics are in full. I do not want to cheat, 10/18/2009 1/1/0001. The common final for Spring will take place on Saturday, June 8th, 1: Sample 1 midterm 2. Some key features of WeBWorK are: Students can access their problem sets and submit answers through a web browser, and receive immediate feedback on the correctness of their answers. Student Help Skip to start of help topic. String answers allow for T/F, matching, multiple-choice, and short answer questions. ","description_more":" This course was curated by Christina Safranski and the Edfinity team. The following questions are organized somewhat by functional group (as many organic courses introduce nomenclature one functional group at a time). David recently emailed me to share that these are now all available in the national problem library: Also, the Preview Activities have now been added to the WeBWorK Open Problem Library. Free is extremely useful if they have written on this. Enter numeric value with unit answers. 2 Permutations; 2. Use the figures below, which show trophic levels in an ecosystem, to complete items 8–11. I was wondering if there is some online pdf for all the answers to the Stewart Calculus textbook. the mark pc multi reloaded games five net key answers 2016. 130 Answer keys PRACTICE TEST 1 LISTENING Section 1 1 A 2 C 3 D 4 D 5 C 6 Prescott (must be correct spelling with capital "P") 7 41 8 Fountain (must have capital "F") 9 752239 10 £65 Section 2 11 E 12 F 13 H 14 \$250 million 15 roads//road system 16 too late 17 school children//boys 18 3 19. You've got a lot of priorities and plenty of things to do (including homework). WEBWORK Music Sheet Music Theory Practice Student Resources AIM Practice WORKSHEETS AP Music Practice Music Camp Music Tutor Challenge. During the academic year, Webwork assignments will be given weekly starting the second week of classes (and the first week of classes in the summer). Engaging students to stick with a problem until they get it right is an extremely powerful strategy for good learning. high school math. This talk presents a recent addition to WeBWorK which enables instructors to ask free response questions. You may speak with a member of our customer support team by calling 1-800-876-1799. You can help protect yourself from scammers by verifying that the contact is a Microsoft Agent or Microsoft Employee and that the phone number is an official Microsoft global customer service number. The best way to enter this in WeBWorK is by using interval notation: [0, infinity ). kw18b0fsbgregr w036efp3i2xgm ym8musqsap3v ylu09ktdazio uuapjhyzq6fck ucwdhgx4kfze2q 8f39bg3tk9sy7 nicx6pcs1vni vs38tmwwixig dy9mumd24v fuuuyl9kaj6y ypfpff46k75jrg hjjky0ljl4b hyksqz2l1j0tscp 6svumwfpna2in6 pygrhy1uiocs pv06iyx813eef 0ayc44gzi54t myvv94si32 qd4vhwa82ikah8 oyae3f3lzd2bsz2 gjq6p9mf1h7w1 svvihiy9ywu jie87pw9v23iw uh6eyvcbyd q81gq58tim6oo 1wpty9uotn ae9yr6gqdhf y3j6202aaxn csrnz0g1bv lw4neonyhh k14u6v6wip k7wfqo5v8vgbt dz6ip30h8kcc	2020-10-01 00:33:42	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.26961496472358704, "perplexity": 2002.351118940428}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-40/segments/1600402130531.89/warc/CC-MAIN-20200930235415-20201001025415-00775.warc.gz"}
http://orca.cf.ac.uk/46573/	ORCA Online Research @ Cardiff # How to identify the youngest protostars Stamatellos, Dimitrios, Whitworth, Anthony Peter, Boyd, D. F. A. and Goodwin, S. P. 2005. How to identify the youngest protostars. Astronomy and Astrophysics 439 (1) , pp. 159-169. 10.1051/0004-6361:20052952 Preview PDF - Published Version ## Abstract We study the transition from a prestellar core to a Class 0 protostar, using SPH to simulate the dynamical evolution, and a Monte Carlo radiative transfer code to generate the SED and isophotal maps. For a prestellar core illuminated by the standard interstellar radiation field, the luminosity is low and the SED peaks at ~ $190\,\mu {\rm m}$. Once a protostar has formed, the luminosity rises (due to a growing contribution from accretion onto the protostar) and the peak of the SED shifts to shorter wavelengths ( $80~{\rm to}\,100\,\mu {\rm m}$). However, by the end of the Class 0 phase, the accretion rate is falling, the luminosity has decreased, and the peak of the SED shifts back towards longer wavelengths ( $90\;{\rm to}\,150\,\mu {\rm m}$). In our simulations, the density of material around the protostar remains sufficiently high well into the Class 0 phase that the protostar only becomes visible in the NIR if it is displaced from the centre dynamically. Raw submm/mm maps of Class 0 protostars tend to be much more centrally condensed than those of prestellar cores. However, when convolved with a typical telescope beam, the difference in central concentration is less marked, although the Class 0 protostars appear more circular. Our results suggest that, if a core is deemed to be prestellar on the basis of having no associated IRAS source, no cm radio emission, and no outflow, but it has a circular appearance and an SED which peaks at wavelengths below ~ $170\,\mu {\rm m}$, it may well contain a very young Class 0 protostar. Item Type: Article Publication Published Physics and Astronomy Q Science > QB Astronomy stars: formation; ISM: clouds; dust, extinction; methods: numerical; radiative transfer; hydrodynamics Pdf uploaded in accordance with publisher's policy at http://www.sherpa.ac.uk/romeo/issn/0004-6361/ (accessed 17/04/2014) edp sciences 0004-6361 30 March 2016 04 Jun 2017 04:56 http://orca.cf.ac.uk/id/eprint/46573 ## Citation Data Cited 25 times in Scopus. View in Scopus. Powered By Scopus® Data ### Actions (repository staff only) Edit Item	2020-07-12 13:49:31	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.49264925718307495, "perplexity": 4686.535752677871}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-29/segments/1593657138718.61/warc/CC-MAIN-20200712113546-20200712143546-00212.warc.gz"}
http://mathhelpforum.com/algebra/132698-solving-equation.html	# Math Help - Solving an Equation 1. ## Solving an Equation Simply solve: fg(x) = 3x^(2)-6x+17 Last question on me stoopid Maths paper. Can't really think of what to do. Silly me. I think I know how to do it now. Thanks. 2. Originally Posted by StephenPoco Simply solve: fg(x) = 3x^(2)-6x+17 Last question on me stoopid Maths paper. Can't really think of what to do. Silly me. I think I know how to do it now. Thanks. Solve for what If $fg(x)=0$ use the quadratic formula 3. Originally Posted by StephenPoco Simply solve: fg(x) = 3x^(2)-6x+17 Last question on me stoopid Maths paper. Can't really think of what to do. Silly me. I think I know how to do it now. Thanks. Hi StephenPoco, I'm not sure what fg(x) is, but I would set the expression = 0 and use the quadratic formula to solve. $3x^2-6x+17=0$ $x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}$	2014-07-22 23:19:06	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 3, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7342785000801086, "perplexity": 2420.3432036654494}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-23/segments/1405997865523.12/warc/CC-MAIN-20140722025745-00236-ip-10-33-131-23.ec2.internal.warc.gz"}
http://mathhelpforum.com/pre-calculus/1934-solve-x-cos-x-3pi-2-2pi.html	# Thread: solve x cos(x) = -(3pi+2)/2pi 1. ## solve x cos(x) = -(3pi+2)/2pi Hello. I would be thankful if someone would give me a help in finding a solution for x cos(x) = -(3pi+2)/2pi 2. Originally Posted by primer Hello. I would be thankful if someone would give me a help in finding a solution for x cos(x) = -(3pi+2)/2pi I do not think you can using elementary algebraic methods. But if you want to approximate the solution you can use Newton's-Raphson Method. 3. Originally Posted by primer Hello. I would be thankful if someone would give me a help in finding a solution for x cos(x) = -(3pi+2)/2pi	2016-12-10 13:16:56	{"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8212949633598328, "perplexity": 837.394456738168}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-50/segments/1480698543170.25/warc/CC-MAIN-20161202170903-00102-ip-10-31-129-80.ec2.internal.warc.gz"}
http://mathhelpforum.com/calculus/202231-how-write-down-basic.html	Math Help - How to write this down - basic 1. How to write this down - basic Hi, i need little help. This is a problem from computer science and what i have is an array A every array has a position and every position is indexed. Furthermore what i have is a case where for let say k=3 i wish to sum the following: A[i] + A[i + A[i]] + A [i + A[i] + A[i + A[i]]] Now i am not a mathematician so please do not laugh. What i don't know is how to write this down nicely. what confuses me is this recursion. so A[i] is a value of the array on position i , A[i+A[i]] is a value on that position but the position at which i am checking for value is dependent on the sum of previous positions. So it would go something like this : $\sum^{k}_{s=0} \mbox{A[}i+\sum^{s}_{p=0}\mbox{what goes here!!]}$ i cannot just write down $\mbox{A_{b}}$ because it is not obvious what that is thnx 2. Re: How to write this down - basic Originally Posted by baxy77bax This is a problem from computer science and what i have is an array A every array has a position and every position is indexed. Furthermore what i have is a case where for let say k=3 i wish to sum the following: A[i] + A[i + A[i]] + A [i + A[i] + A[i + A[i]]] Now i am not a mathematician so please do not laugh. What i don't know is how to write this down nicely. How do you need to write it: in English, in C, in pseudocode, as a math formula using Σ, as a recurrence relation, etc.? 3. Re: How to write this down - basic Yes i did not mention that as a math formula using \sum . it th same way i started it in the initial post: $\sum^{k}_{s=0} \mbox{A[}i+\sum^{s}_{p=0}\mbox{what goes here!!]}$ PS isn't that the same as saying "recurrence relation"?? 4. Re: How to write this down - basic Originally Posted by baxy77bax A[i] + A[i + A[i]] + A [i + A[i] + A[i + A[i]]] Could you write two more terms (i.e., a sum of 5 terms)? I am not sure I grasp the rule. 5. Re: How to write this down - basic A[i] + A[i + A[i]] + A[i + A[i] + A[i + A[i]]] + A[i + A[i] + A[i + A[i]] + A[i + A[i] + A[i + A[i]]]] + A[i + A[i] + A[i + A[i]] + A[i + A[i] + A[i + A[i]]] + A[i + A[i] + A[i + A[i]] + A[i + A[i] + A[i + A[i]]]]] 6. Re: How to write this down - basic I think it's difficult to write it as an explicit sum. It is convenient to make an auxiliary recursive definition (define a recurrence relation): $a_1 = i$ $a_{n+1} = a_n + A[a_n]$ Then the expression is $\sum_{n=1}^k A[a_n]$. 7. Re: How to write this down - basic ok that what i was looking for , because i was trying to express it as a sum and could not. so all i needed is one more conformation that it is difficult enough not to wast time on it. thank you	2016-06-27 10:59:28	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 6, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7852830290794373, "perplexity": 795.4573657220927}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-26/segments/1466783395679.92/warc/CC-MAIN-20160624154955-00138-ip-10-164-35-72.ec2.internal.warc.gz"}
https://mathoverflow.net/questions/267228/extending-morphisms-in-an-a-infty-category-to-natural-transformations	# Extending morphisms in an $A_{\infty}$ category to natural transformations Suppose we are given a small (enriched) category $C$, and for $a,b \in C$ an isomorphism $m:a \to b$. It is always possible to find a functor $F: C \to C$, with $F(a)=b$ and a natural transformation $N$ of the identity functor to $F$ so that $N_a=m$. This is a nice simple exercise. Can this always be done when $C$ is an $A_{\infty}$-category? The intuition behind this is the following. Given a path $m$ from $a$ to $b$ in say a CW complex $X$, we can always find a homotopy $H: X \times [0,1] \to X$ of the identity map so that $H\|_{\{a\} \times [0,1]} = m$. The question can be generalized to ask when there is a "homotopy extension property" for $A_{\infty}$ subcategories. Edit: For the first question the answer is affirmative by an abstract non-sense argument. Take the $A_{\infty}$ Yoneda embedding $Y$ of C into Ch-dg the category of chain complexes. Then find the functor $F$ and the natural transformation $N$ for $Y(m)$ inside Y(C) which is a dg-category. Then just pull-back both by Y. However it maybe nice to find a more direct argument.	2022-08-14 18:49:34	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9832716584205627, "perplexity": 169.72908445971544}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-33/segments/1659882572063.65/warc/CC-MAIN-20220814173832-20220814203832-00316.warc.gz"}
https://www.techwhiff.com/issue/convert-to-standard-form-f-x-2x-4--356890	# Convert to standard form: F(x) = 2x + 4 ###### Question: Convert to standard form: F(x) = 2x + 4 ### The distance between two points on a coordinate grid is 5 units. One of the points is at (2,−1) ( 2 , − 1 ) . The x-coordinate of the other point is -2. Which is a possible value for the missing y-coordinate? The distance between two points on a coordinate grid is 5 units. One of the points is at (2,−1) ( 2 , − 1 ) . The x-coordinate of the other point is -2. Which is a possible value for the missing y-coordinate?... ### 2(a+7) - 7 = 9 what is the answer 2(a+7) - 7 = 9 what is the answer... ### What is the height of the platform? What is the height of the platform?... ### I need help right now asap..... i need help right now asap........ ### Write the appropriate direct variation equation if y = 8 as x = -4 Write the appropriate direct variation equation if y = 8 as x = -4... ### Find the product. 1.(a + 3)(a - 4) 2. (x - 3)2 3. (2a – 62)(a + 462) 4.(x - 5)(x2 + 4x - 6) Find the product. 1.(a + 3)(a - 4) 2. (x - 3)2 3. (2a – 62)(a + 462) 4.(x - 5)(x2 + 4x - 6)... ### Acceleration occurs when an object speeds up Acceleration occurs when an object speeds up... ### What was the impact of Spain settlement in the America ? What was the impact of Spain settlement in the America ?... ### What is the opposite of 0? what is the opposite of 0?... ### When Jana noticed her daughter Kylie staring at herself and posing in every reflection that they passed, Jana worried that Kylie was growing a bit conceited. Based on the context clues, what does conceited mean in this sentence? Question 5 options: Self-obsessed Critical Self-conscious When Jana noticed her daughter Kylie staring at herself and posing in every reflection that they passed, Jana worried that Kylie was growing a bit conceited. Based on the context clues, what does conceited mean in this sentence? Question 5 options: Self-obsessed Critical Self-conscious... ### How was imperialism after the industrial revolution similar to imperialism? How was imperialism after the industrial revolution similar to imperialism?... ### Which of the following best describes the word picture created by the imagery in these lines? O A. a sunrise on a winter's day O B. worsening weather conditions o C. a sunny day and moonless night O D. farming activities during wintertime Which of the following best describes the word picture created by the imagery in these lines? O A. a sunrise on a winter's day O B. worsening weather conditions o C. a sunny day and moonless night O D. farming activities during wintertime... ### What is Sin A? Cos A? Tan A? What is Sin A? Cos A? Tan A?... ### A. Daylight savings time traditionally starts in the spring and ends in the winter . B. Daylight savings time is the best time because it is no longer dark in the mornings ; it is light . C. Benjamin Franklin came up with the idea of daylight saving time to save energy and money D. In 1966 Congress imposed the Uniform Time Act , which asked states to regulate Daylight Savings time . Pls help A. Daylight savings time traditionally starts in the spring and ends in the winter . B. Daylight savings time is the best time because it is no longer dark in the mornings ; it is light . C. Benjamin Franklin came up with the idea of daylight saving time to save energy and money D. In 1966 Congres... ### The ancient Egyptians enjoyed sports and games in their leisure time. A. Yes B. No The ancient Egyptians enjoyed sports and games in their leisure time. A. Yes B. No... ### A transform boundary is a place where geologic plates are formed geologic plates are melted geologic plates slide past each other geologic plates make mountains A transform boundary is a place where geologic plates are formed geologic plates are melted geologic plates slide past each other geologic plates make mountains... ### PLEASE HELP!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! One way to show respect for others in an online environment is to make messages A. brief B. detailed C. persuasive D. long PLEASE HELP!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! One way to show respect for others in an online environment is to make messages A. brief B. detailed C. persuasive D. long... ### BRAINLIEST FOR RIGHT ANSWERS! Read these paragraphs and answer the question that follows: In 2008, more than one million American students gave nearly 20 million service hours to their communities. They made a difference in people's lives and learned some important life lessons in the process. Organizations, including schools, are actively promoting service for all citizens as a way to be involved, help others, and improve themselves. Service is helping other people and being active in your comm BRAINLIEST FOR RIGHT ANSWERS! Read these paragraphs and answer the question that follows: In 2008, more than one million American students gave nearly 20 million service hours to their communities. They made a difference in people's lives and learned some important life lessons in the process. Organ... ### What are the zero(s) of the function f(x)=$\frac{4x^{2}-36x }{x-9}$? a.) x = -9 b.)x = 0 c.)x = 9 d.)x = 0 and x = 9 What are the zero(s) of the function f(x)=$\frac{4x^{2}-36x }{x-9}$? a.) x = -9 b.)x = 0 c.)x = 9 d.)x = 0 and x = 9...	2022-10-03 12:25:33	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.3704095780849457, "perplexity": 3055.0840366899643}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-40/segments/1664030337415.12/warc/CC-MAIN-20221003101805-20221003131805-00510.warc.gz"}
https://reversediffsourcejl.readthedocs.io/en/master/types.html	Working with composite types¶ When encountering a composite type, ReverseDiffSource builds a Vector{Any} to hold its derivative accumulator. Its structure is derived from the fields of the composite type: Float for a Real number, an array of Floats for Arrays, or another Vector{Any} if the field is a type. No special declaration has to be made beforehand to ReverseDiffSource. However you do need to declare how each function using the composite type changes its derivative accumulator. Suppose you have type Bar defined as: type Bar x y end And an associated function norm(z::Bar): norm(z::Bar) = z.xz.x + z.yz.y And finally an expression to derive making use of Bar and norm(): ex = :( z = Bar(2^a, sin(a)) ; norm(z) ) You need to declare how both the constructor Bar and the function norm behave regarding the derivative accumulator (which will be a 2 element vector of type Any for the two fields x andy): @deriv_rule Bar(x,y) x ds[1] # Derivative accumulator of x is increased by ds[1] @deriv_rule Bar(x,y) y ds[2] # Derivative accumulator of y is increased by ds[2] @deriv_rule norm(z::Bar) z Any[ 2z.xds , 2z.yds ] # Note : produces a 2-vector since z is a Bar We are now ready to derive: julia> res = rdiff(ex, a=0.) julia> @eval df(a) = \$res julia> df(1) (4.708073418273571,6.454474871305244)	2021-12-04 23:46:42	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5278847813606262, "perplexity": 5464.821171651904}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-49/segments/1637964363125.46/warc/CC-MAIN-20211204215252-20211205005252-00222.warc.gz"}
https://math.stackexchange.com/questions/2825642/intuitively-if-addition-can-be-interpreted-as-combining-sets-then-what-can-mul/2826080	# Intuitively, if addition can be interpreted as combining sets, then what can multiplication and division be interpreted as? Intuitively, if addition can be interpreted as combining sets, then what can multiplication and division be understood as? A few more extensions to this: • What does multiplying and dividing by a decimal number mean intuitively? And why, when we multiply and divide by decimal numbers, do we need to move decimal points up and down based on what we are multiplying? • Why, when we divide by decimal numbers/fractions, do we reciprocal the fraction, then multiply by the denominator and divide by the numerator? Is there an intuitive explanation for this? Can you try to keep the explanation as simple as possible? Because I'm still a beginner, if the explanation is too complex, I may not be able to understand it. • If you're thinking of addition as corresponding to the disjoint union of two sets (I'm guessing this is what you mean by "combining" sets), then multiplication will correspond to the Cartesian product of two sets. – Greg Martin Jun 20 '18 at 6:11 One good idea is to think of positive numbers as representing ratios, porportions, or scale factors. If we have an object, of any number of dimensions, and we strech or shrink it so that all distances between points in the object are multiplied by a constant number, the scale factor, this is called a Homothetic transformation. Notice that the scale factor is the constant ratio connecting the original distance between two points and the distance between the transformed points. If we have two such transformation and we combine them together, one after the other, the result is another transformation whose scale factor is the product of the two scale factors. This result is the interpretation of multiplication of scale factors. If we reverse such a transformation we get another transformation whose scale factor is the reciprocal of the original scale factor. This means we reverse the roles of before and after. In other words, we switch numerator with denominator. Now if we apply one transformation and then apply the reverse of another, the result is another transformation whose scale factor is the product of the first scale factor and the reciprocal of the second. This result is the interpretation of division of scale factors. You can think of both multiplication and division in terms of sets; in fact, that's how many "operations" are defined in Abstract Algebra. First of all, both multiplication and division can be thought of as "pairing" two elements. Multiplying two numbers can be thought of counting the number of "pairs" that you can make out of two(or more) sets. For example, if you have $\{1,2,3\}$ and $\{1,2,3,4\}$, there would be 12 pairs: $(1,1), (1,2),..., (3,4)$. Division can be thought of as "assigning" a number to each pair. When doing so, however, we identify some pairs as the same: For our usual Euclidean Division Algorithm, we identify two pairs as the same if they are "proportionate" to each other. For example, we identify the "pair" $(1,2)$ as $\frac{1}{2}$. Even though $(2,4)$ is a different "pair", since they are "proportionate", they become the same "pair". You have asked three or four essentially different questions in the same post. Please separate them from now on - then it's easier to get answers, and easier for others to use your questions and answers. These are interesting questions each of which calls for a long answer. Here are brief responses. The Greeks did their arithmetic mathematically as geometry. Adding was placing line segments end to end. Multiplication constructed rectangles. (I've no idea how they did everyday arithmetic for commerce.) One contemporary way to think of arithmetic on the number line is to see "adding $x$" as a translation by $x$. That will move right when $x$ is positive and left when $x$ is negative, so deals with subtraction at the same time. Then "multiplying by $x$" is scaling everything by $x$. That's an expansion when $x > 1$ and a contraction when $0 < x < 1$. It reverses direction when $x$ is negative. This is the essence of the answer from @Somos . The rule for "invert and multiply" is best understood when you think of a fraction like, say, $1/3$ as "what do you multiply $3$ by to get the answer $1$? That fits well with thinking of multiplication as scaling. It's more useful than the first elementary school discussion that models $1/3$ as dividing a pizza into three parts. Numbers are just numbers. When you say "decimal numbers" you're just specifying how we write them in positional notation with base $10$. Then the rule that multiplying or dividing by a power of $10$ just moves the decimal point follows from the meaning of positional notation.	2019-06-24 14:15:31	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9017904996871948, "perplexity": 219.48969019456345}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-26/segments/1560627999539.60/warc/CC-MAIN-20190624130856-20190624152856-00470.warc.gz"}
https://socratic.org/questions/how-do-you-write-10-3-in-decimal-form-1	# How do you write 10^-3 in decimal form? Jun 7, 2018 $0.001$ through rewriting the expression in scientific notation #### Explanation: We can rewrite the expression as $1 \cdot {10}^{-} 3$ This is in scientific notation. Since we have a negative exponent, we loop the decimal three times to the left. We get $0.001$ Hope this helps! Jun 7, 2018 $0.001$ #### Explanation: Using the laws of indices we can rewrite ${10}^{-} 3$ in the form of the fraction $\frac{1}{10} ^ 3$ $\frac{1}{10} ^ 3 = \frac{1}{1000}$ Decimals are fractions written with powers of $10$ $\frac{1}{1000}$ means $3$ decimal places. ${10}^{-} 3 = \frac{1}{10} ^ 3 = \frac{1}{1000} = 0.001$	2021-07-29 18:24:51	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 11, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9615896344184875, "perplexity": 1077.1867700889056}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-31/segments/1627046153892.74/warc/CC-MAIN-20210729172022-20210729202022-00354.warc.gz"}
https://socratic.org/questions/how-do-you-solve-4x-1-3-or-x-2-7	# How do you solve -4x-1>3 or x+2>7? Sep 11, 2015 $x < - 1 \mathmr{and} x > 5$ $\textcolor{w h i t e}{\text{XXX}}$OR $\left\mid x - 2 \right\mid > 3$ #### Explanation: Part 1 $\textcolor{w h i t e}{\text{XXX}} - 4 x - 1 > 3$ add $4 x - 3$ to both sides $\textcolor{w h i t e}{\text{XXX}} - 4 > 4 x$ divide by $4$ $\textcolor{w h i t e}{\text{XXX}} - 1 > x$ or, equivalently $\textcolor{w h i t e}{\text{XXX}} x < - 1$ Part 2 $\textcolor{w h i t e}{\text{XXX}} x + 2 > 7$ subtract $2$ from both sides $\textcolor{w h i t e}{\text{XXX}} x > 5$ Note that for $x < - 1$ or $x > 5$ $\textcolor{w h i t e}{\text{XXX}} x$ must be further that $2$ the midpoint between $\left(- 1\right)$ and $\left(5\right)$ (i.e. from $3$) So $\textcolor{w h i t e}{\text{XXX}} \left\mid x - 2 \right\mid > 3$	2020-02-19 10:23:58	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 20, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.845558226108551, "perplexity": 2932.571476586838}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-10/segments/1581875144111.17/warc/CC-MAIN-20200219092153-20200219122153-00414.warc.gz"}
https://www.physicsforums.com/threads/polar-integral.808991/	# Homework Help: Polar Integral 1. Apr 16, 2015 ### whattheheckV 1. The problem statement, all variables and given/known data ∫∫dydx Where the region Ω: 1/2≤x≤1 0 ≤ y ≤ sqrt(1-x^2) 2. Relevant equations 3. The attempt at a solution The question asked to solve the integral using polar coordinates. The problem I have is getting r in terms of θ. I solved the integral in rectangular ordinates using a trig sub and found the answer to be (π/6) - (√3)/8. Thanks for all input. Last edited: Apr 16, 2015 2. Apr 16, 2015 ### LCKurtz Have you drawn a picture? Remember $r$ goes from $r$ on the inner curve to $r$ on the outer curve. What is the equation of $x = \frac 1 2$ in polar coordinates? What is the equation of the circle? And from the picture you should be able to see the $\theta$ limits. 3. Apr 16, 2015 ### Staff: Mentor You're not solving an equation -- you're setting up and evaluating an integral The integrand is pretty straight forward; probably the hardest part is figuring out the limits of integration. A ray extending out from the pole (0, 0) goes through the vertical line, x = 1, to the circle. Convert the equation of the vertical line to polar from. The circle part is simple. 4. Apr 16, 2015 ### whattheheckV I have drawn the picture and it is very straightforward. The problem I am having is that the radius is not constant so i need to get r in terms of theta. http://www5b.wolframalpha.com/Calculate/MSP/MSP17811i1c25cda73i33d700001h169b9cb5h0603d?MSPStoreType=image/gif&s=34&w=200.&h=193.&cdf=Coordinates&cdf=Tooltips [Broken] Last edited by a moderator: May 7, 2017 5. Apr 16, 2015 ### LCKurtz OK, so you have drawn the picture. Now answer the questions I asked in my reply above. Last edited by a moderator: May 7, 2017 6. Apr 16, 2015 ### SammyS Staff Emeritus Do you know how to express x in terms of r and θ ? Last edited by a moderator: May 7, 2017 7. Apr 16, 2015 ### whattheheckV As in x=rcos (θ) and y=rsin (θ) ? 8. Apr 16, 2015 ### SammyS Staff Emeritus Good the left hand boundary of the region is x = 2. That should give you a relation in terms of r and θ for that boundary. The right hand boundary (curved boundary) should be easy in terms of r . 9. Apr 16, 2015 ### LCKurtz You mean $x = \frac 1 2$. 10. Apr 16, 2015 ### SammyS Staff Emeritus Right! 11. Apr 17, 2015 ### whattheheckV Ok so using the conversion I found the bounds of r from (1/2)sec(θ) to 1 Thanks for the input 12. Apr 17, 2015 ### SammyS Staff Emeritus Yes. You're welcome. 13. Apr 17, 2015 ### LCKurtz And what bounds did you get for $\theta$?	2018-06-25 16:13:20	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6765264868736267, "perplexity": 1689.496370713692}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-26/segments/1529267868135.87/warc/CC-MAIN-20180625150537-20180625170537-00059.warc.gz"}
https://www.bestpracticecertification.com.au/single-post/what-is-the-current-version-of-iso-14001	# What is the current version of ISO 14001? What is the current version of ISO 14001? The current version of ISO 14001 environmental management system specification is 2015. In 2015 two key standards were revised and republished that's ISO 9001 and ISO 14001. In 2015 Annex SL became the benchmark. All international management system standards all follow the same numbering system in the same structure, which meant that organizations could move forward to more efficiently integrate their management systems. The current version of ISO 14001 is 2015 and it follows the same numbering structure as ISO 9001, ISO 45001 and ISO 27001.	2021-01-15 18:32:32	{"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9433630108833313, "perplexity": 2317.914173368154}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-04/segments/1610703495936.3/warc/CC-MAIN-20210115164417-20210115194417-00381.warc.gz"}
https://coolstatsblog.com/tag/stationary-process/	# How to use the Autocorreation Function (ACF)? The Autocorrelation function is one of the widest used tools in timeseries analysis. It is used to determine stationarity and seasonality. Stationarity: This refers to whether the series is “going anywhere” over time. Stationary series have a constant value over time. Below is what a non-stationary series looks like. Note the changing mean. And below is what a stationary series looks like. This is the first difference of the above series, FYI. Note the constant mean (long term). The above time series provide strong indications of (non) stationary, but the ACF helps us ascertain this indication. If a series is non-stationary (moving), its ACF may look a little like this: The above ACF is “decaying”, or decreasing, very slowly, and remains well above the significance range (dotted blue lines). This is indicative of a non-stationary series. On the other hand, observe the ACF of a stationary (not going anywhere) series: Note that the ACF shows exponential decay. This is indicative of a stationary series. Consider the case of a simple stationary series, like the process shown below: $Y_t = \epsilon_t$ We do not expect the ACF to be above the significance range for lags 1, 2, … This is intuitively satisfactory, because the above process is purely random, and therefore whether you are looking at a lag of 1 or a lag of 20, the correlation should be theoretically zero, or at least insignificant. Next: ACF for Seasonality Abbas Keshvani	2019-11-20 06:49:09	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 1, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7942830324172974, "perplexity": 1276.9890349219302}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-47/segments/1573496670512.94/warc/CC-MAIN-20191120060344-20191120084344-00256.warc.gz"}

Subsets and Splits

No community queries yet

The top public SQL queries from the community will appear here once available.