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http://www.webassist.com/forums/post.php?pid=91500 | the error is saying that it cannot opren the Connections/db345740447.php file
send a copy of the users_LogIn.php file so i can look at the code.
it looks like it is trying to locate the connections file at:
C:\xampp\htdocs\pizzapal.co.uk\Connections\db345740447.php
Does that file exist at that location? | {"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": 0.9273410439491272, "perplexity": 1777.4429945042036}, "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-2018-13/segments/1521257647545.54/warc/CC-MAIN-20180320205242-20180320225242-00044.warc.gz"} |
https://www.sarthaks.com/1971/why-amount-collected-test-tubes-activity-double-the-amount-collected-the-other-name-this?show=1976 | # Why is the amount of gas collected in one of the test tubes in Activity 1.7 double of the amount collected in the other? Name this gas.
+1 vote
5.8k views
edited
Why is the amount of gas collected in one of the test tubes in Activity 1.7 double of the amount collected in the other? Name this gas.
by (13.6k points)
selected
During the Electrolysis of water, hydrogen and oxygen is get separated by the electricity. Water (H2O) contains two parts hydrogen and one part oxygen. Since hydrogen goes to one test tube and oxygen goes to another, the amount of gas collected in one of the test tubes is double of the amount collected in the other. | {"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": 0.8024478554725647, "perplexity": 950.4270249862134}, "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-2023-14/segments/1679296946637.95/warc/CC-MAIN-20230327025922-20230327055922-00414.warc.gz"} |
https://math.stackexchange.com/questions/1313647/removable-singularity | # removable singularity
Let $C$ be the positively oriented boundary of the square with vertices $(1,0)$, $(1,-i)$, $(-1,-i)$ and $(-1,0)$. If $$f(z)=\frac{\sin(z)}{z},$$ then clearly $f$ has a removable singularity on $z=0$. This means that $f$ is analytic on $z=0$? My real question is, since $f$ is analytic inside $C$, can I apply cauchy's thm to say that $$\int_C f(z)dz =0 ?$$ The fact that on $C$ there exist a singularity makes me think I can´t, however, since it is a removable singularity I think that I can…. What is the correct thing to do here?
• Yes $f$ is analytic and you can use the Cauchy Integral formula. – Tim Raczkowski Jun 5 '15 at 19:07
• @TimRaczkowski thanks a lot! – Leo Sera Jun 5 '15 at 19:36
$$\frac{\sin z}z=1-\frac{z^2}{3!}+\frac{z^4}{5!}-\cdots+(-1)^n\frac{z^{2n}}{(2n+1)!}+\cdots$$
Since the radius of convergence of this power series is $\infty$, this function is entire.
(Of course, the above equality has no sense for $z=0$, but RHS is defined at $z=0$ and is equal to $\sin z/z$ elsewhere). | {"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": 0.9976088404655457, "perplexity": 121.90592199167955}, "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/1614178376206.84/warc/CC-MAIN-20210307074942-20210307104942-00514.warc.gz"} |
https://www.physicsforums.com/threads/please-help-with-trig-substitution-integration.109319/ | 1. Feb 3, 2006
### silverdiesel
I am not too good with trig identities. I cant seem to figure out how to simplify these trig intergrals. I know I can use a triangle to turn the second problem into a trig integral, but once I have the trig integral, I am lost. Any help would be greatly appriciated.
$$\int\tan(x)\sec^3(x)dx$$
$$\int\frac{1}{x^2&\sqrt{16-x^2}}dx$$
Last edited: Feb 3, 2006
2. Feb 3, 2006
### Valhalla
Have you tried a u-substitution on the first one yet? Here is a hint what is the derivative of sec?
3. Feb 3, 2006
### silverdiesel
see, thats what I thought, but there seems to be an extra secant in there. The derivative of sec is sectan. So, u=sec(x), du=sec(x)tan(x)dx. That gives tan(x)sec^2(x)... right?
4. Feb 3, 2006
### Tide
HINT: Let
$$u = \cos x$$
It will save you a lot of work.
5. Feb 3, 2006
### Valhalla
$$u=sec(x)$$ $$du=sec(x) tan(x)dx$$
$$\int u^2du$$
Last edited: Feb 3, 2006
6. Feb 3, 2006
### silverdiesel
brilliant! Thanks Tide. I should have seen that.
$$\int\frac{sin(x)}{cos^4(x)}dx = \frac{1}{3cos^3(x)}$$
7. Feb 3, 2006
### silverdiesel
right, yes I can see that too Valhalla. Thanks so much. Looks like that one was much easier than I made it.
8. Feb 3, 2006
### silverdiesel
Any ideas on the second problem? Using a triangle, I have changed it to:
$$\int\frac{1}{(4sin\Theta)^2(4cos\Theta)}d\Theta$$
9. Feb 3, 2006
### silverdiesel
wait, I think got it, $$u=sin\Theta$$ | {"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": 0.9770767092704773, "perplexity": 2438.30342862617}, "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-2017-47/segments/1510934806708.81/warc/CC-MAIN-20171122233044-20171123013044-00615.warc.gz"} |
http://www.examguruadda.in/2015/09/data-interpretation.html | # Data interpretation
Study the following table and answer the following Questions.
Percentage of literate population over the total population of state for different states over the years.
Year State 2005 2006 2007 2008 2009 2010 A 65 72 69 78 79 82 B 82 81 85 85 87 88 C 78 81 84 87 89 91 D 56 65 69 71 75 77 E 89 93 94 95 95 97 F 85 88 91 93 95 96
1. Population of State C in 2005 and 2006 were in the ratio of 2:3 respectively. what was the respective ratio of their literate population?
1) 51 : 82
2) 15 : 18
3) 9 : 13
4) 26 : 27
5) None of these
2. In 2006 , all the six states had equal population of 12 lakhs. what was the average literate population of the states in that same year ?
1) 9, 88, 000
2) 10, 56, 000
3) 9, 60, 000
4) 10, 28, 000
5) None of these
3. Population of State 'D' in 2006 and 2009 was equal. Approximately, what was the percentage increase in literate population from 2006 to 2009?
1) 17
2) 15
3) 16
4) 18
5) 19
4. Population of State 'B' increased every year by 10%. If its total population in 2007 was 10 lakhs, what was its literate population in 2009?
1) 10,52,700
2) 9,57,000
3) 11,52,400
4) cannot be determined
5) None of these
5. If the average population of all States in 2007 was 12.5 lakhs what was the average literate population in that year?
1) 10.25 lakhs
2) 10.75 lakhs
3) 11 lakhs
4) Cannot be determined
5) None of these
1- 5
2- 3
3- 2
4- 1
5- 1
explanation
1. (5) let population of state C in 2005 and 2006 be 2x and 3x respectively
literate population in 2005 is= 78/100* 2x
literate population in 2006 is = 81/100*3x
required ratio= 78* 2x : 81 * 3x = 52 : 81
2. (3) Average literate population = 1200000/100 * (72+81+81+65+93+88/6)
3. (2) Required %= 75- 65/ 65* 100= 15.38=15
4. (1) Population of State B in 2009=10(1+10/100)2
= 12.1 lakhs
5.(1) Average % of literate population= 69+85+84+69+94+91/6= 82
required average literate population= 82% of 12.5 lakhs = 82*12.5/100= 10.25 lakhs
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good questions!!!!! post some more | {"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": 0.9467023611068726, "perplexity": 3363.3226277053886}, "config": {"markdown_headings": true, "markdown_code": false, "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-2016-50/segments/1480698541692.55/warc/CC-MAIN-20161202170901-00411-ip-10-31-129-80.ec2.internal.warc.gz"} |
https://repository.uantwerpen.be/link/irua/99599 | Publication
Title
Slater sum for the one-dimensional $sech^{2}$ potential in relation to the kinetic energy density
Author
Abstract
In earlier work on the one-dimensional sech(2) potential energy [I. A. Howard and N.H. March, Int. J. Quantum Chem. 91, 119 (2003)] it has been shown that both electron density rho(x) and kinetic energy t(x) are low-order polynomials in the potential V(x), for a small number of bound states. Here all attention is focused on the continuum states for the sech(2) potential with a single bound state. The tool employed is the Slater sum, which satisfies a partial differential equation. This is first solved explicitly for the bound state, and then the solution is generalized to apply to the continuum. Again, considerable simplification is exhibited for this specific choice of potential. A brief discussion is included of a central sech(2)(r) potential. (C) 2004 American Institute of Physics.
Language
English
Source (journal)
Journal of mathematical physics. - New York, N.Y.
Publication
New York, N.Y. : 2004
ISSN
0022-2488
Volume/pages
45:6(2004), p. 2411-2419
ISI
000221658500022
Full text (Publisher's DOI)
Full text (publisher's version - intranet only)
UAntwerpen
Faculty/Department Research group Publication type Subject Affiliation Publications with a UAntwerp address | {"extraction_info": {"found_math": true, "script_math_tex": 1, "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": 0.8404226303100586, "perplexity": 2107.621131587824}, "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-2017-26/segments/1498128319902.52/warc/CC-MAIN-20170622201826-20170622221826-00695.warc.gz"} |
https://scicomp.stackexchange.com/questions/33006/langevin-equation-in-4th-order-runge-kutta | Langevin equation in 4th order Runge-Kutta
I'm trying to figure out how to translate a piece of code from Velocity Verlet to Runge-Kutta, while treating the time step dependence of the thermal noise correctly.
$$ma = - \gamma v - \frac{dU}{dx} + \xi(t),$$
where $$U$$ is some interaction potential, $$\gamma$$ is damping, and $$\xi$$ is a gaussian noise term with $$\mu = 0$$ and $$\sigma = \sqrt{2\gamma m k_B T}$$.
I can use Velocity Verlet with Langevin dynamics as
$$v_{t+1} = v_t + h(a - \gamma v + \xi(t)).$$
Qualitatively, what the Langevin equation does here is that it models thermal fluctuations by adding random kicks to the acceleration while counteracting them with a constant damping term to stabilize the energy. My question then is, how does this translate to 4th order Runge-Kutta (RK4)?
In RK4 we calculate the velocity as
$$v_{t+1} = v_t + \frac{h}{6}(a_1 + 2a_2 + 2a_3 + a_4)$$
where $$a_i$$ are the partial accelerations calculated in the RK4 steps.
It is not obvious to me where to introduce the Langevin dynamics here. My best guess is that it should be applied in every separate RK-step? Meaning e.g. for $$a_1$$
$$a_1 = a_t - \gamma v_t + \xi(t).$$
Of course we would have to use the same $$\xi(t)$$ for all the $$a_i$$ during one time step for this to make sense. Meaning we generate one $$\xi(t)$$ at the start of every time step that we then use for every calculation during that time step.
Still, something is missing here... Because now the noise is not dependent on the time step, and it should be! This was not an issue in Velocity Verlet because we just multiplied the noise term with $$h$$ during every time step, but this is not the case here. It seems to me that the time step has to be included somewhere in the $$\sigma$$ term of the Langevin equation, but I can't really figure out how...
edit1: changed to a more sensible notation.
edit2: I realized, for RK4 to work, you probably have to add a timestep to the noise term as $$\sigma = \sqrt{\frac{2\gamma m k_B T}{h}}$$ for the units to come out correctly.
• You can do RK4 here, but it doesn't really help because you're missing the order 1.0 Taylor terms so it's still just going to converge like 0.5 unless you add the Milstein correction. Essentially any derivation which uses standard Newton calculus is missing terms since you have a continuous noise term. So you might as well do Euler-Maruyama / Euler-Heun, or something more advanced which utilizes the stochastic Taylor expansion like something from this list. – Chris Rackauckas Jul 5 '19 at 4:55
• Thanks, I didn't realize that, but it is a fine point. One should of course be mindful not to waste precious CPU time! – storluffarn Jul 8 '19 at 9:55
Caveat: This is not a full answer to your question on how to replace VV by RK4. Basically, I recommend another integrator method which also has potentially fourth order accuracy.
You want to integrate the above Langevin equation over time. As an analytical solution is not known, you need to do it numerically. As @Chris is correctly pointing out, this is a stochastic differential equation (SDE). Hence, your time integration consists of two parts: a deterministic part (momentum and position) and a stochastic part (affecting momentum).
One common approach is to use a splitting method: This splits the above Langevin equation (rather the associated Fokker-Planck operator) into distinct parts, each of which can be solved analytically. These analytical solutions are then used as basic building blocks from which numerical integrators of specific accuracy ("order") can be constructed.
I recommend reading the paper by Leimkuhler et al., 2012 on the splitting methods and the resulting basic building blocks -- coined "A", B" and "O" -- from section 2, especially section 2.1 on timestepping methods. (Note that the article focuses on molecular sampling and statistical physics in its terminology but this of no relevance.) I believe that this paper will give you some background on how numerical time integration schemes (for SDEs) are obtained.
So far, you have used Velocity-Verlet with Langevin. Hence, in the notation of the paper you are probably doing "BAO" (B = momentum integration, A = position integration, O = stochastic integration).
If you are only interested in obtaining a more accurate integration scheme, then you could for example use "BAOAB" (see the above paper) which has second order accuracy in the momenta and even up to fourth order in the positions and some other nice features.
The bottom line is: Simply replacing (in the deterministic integration part) VV by RK4 will not help, if you do not treat the stochastic part correctly.
• That's an interesting review paper! As noted by Chris Rackauckas in the comment section, RK4 wouldn't even be forth order in accuracy here. But between your two answers I can definitely piece together what I need, that is an applicable 4th order integrator. – storluffarn Jul 8 '19 at 9:59 | {"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": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 17, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8474252820014954, "perplexity": 441.8563139958152}, "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-2021-10/segments/1614178364932.30/warc/CC-MAIN-20210302221633-20210303011633-00129.warc.gz"} |
http://codingplayground.blogspot.com/2009/11/dbscan-clustering-algorithm.html | ## Monday, November 9, 2009
### DBSCAN clustering algorithm
DBSCAN is a well-known clustering algorithm, which is easy to implement. Quoting Wikipedia: "Basically, a point q is directly density-reachable from a point p if it is not farther away than a given distance ε (i.e., is part of its ε-neighborhood), and if p is surrounded by sufficiently many points such that one may consider p and q be part of a cluster.... For practical considerations, the time complexity is mostly governed by the number of getNeighbors queries. DBSCAN executes exactly one such query for each point, and if an indexing structure is used that executes such a neighborhood query in O(logn), an overall runtime complexity of $O(n \cdot \log n)$ is obtained."
1. DBScan does not require you to know the number of clusters in the data a priori, as opposed to k-means.
2. DBScan can find arbitrarily shaped clusters.
3. DBScan has a notion of noise.
4. DBScan requires just two parameters and is mostly insensitive to the ordering of the points in the database
1. DBScan needs to materialize the distance matrix for finding the neighbords. It has a complexity of O((n2-n)/2) since only an upper matrix is needed. Within the distance matrix the nearest neighbors can be detected by selecting a tuple with minimums functions over the rows and columns. Databases solve the neighborhood problem with indexes specifically designed for this type of application. For large scale applications, you cannot afford to materialize the distance matrix
2. Finding neighbords is an operation based on distance (generally the Euclidean distance) and the algorithm may find the curse of dimensionality problem
Here you have a DBSCAN code implemented in C++, boost and stl
1. Thanks, for sharing the code.
That would be nice to see how the DBSCAN will work if one would use the local Adaptive Metric(Mahalanobis metric) for the distance instead of Euclidean distance.
I am trying to implement it for 6D(x,y,z,vx,vy,vz)
thanks
Arman.
2. Thanks for releasing the code. However; could you specify the license under which it is released?
Thanks
3. Hi,
Thank you for sharing the codes. However, I met a problem when I try to compile the codes with visual studio 2005. I had installed boost on my computer, but it still does not work out. Can you please show me some possible solutions to this problem?
Thank you so much!
4. hi,
Thank you for sharing the codes. However, I met a problem when I tried to compile the codes in visual studio. Then, I installed boost on my computer. However, it does not work out. Could you please show me some possible solutions to this problem?
Thank you so much!
5. At lines 16 and 33 in distance.h I had to change
typedef typename VEC_T vector_type;
to
typedef VEC_T vector_type;
to get it to compile.
Dave
6. The example is clustering random numbers into 1 cluster. Actually would be good if the example was in some way illustrative of this method - on a more "shapy" data - like in the paper on DBSCAN.
7. actually I tried this code on ELKI example, it produced just one cluster, where ELKI gave several clusters with the same parameters.
8. thanks for sharing the code..where I can find real dataset for applying DBSCAN?
9. Hi
I've found a very very small bug in your code.
in clusters.cpp file
in findNeighbours function:
if ((pid != j ) && (_sim(pid, j)) > threshold)
it's supposed to be less than
if ((pid != j ) && (_sim(pid, j)) < threshold)
after changing this it works like charm, thanks for the code
10. Hi,
Thanks for sharing this code. The last commenter (M^3 Team) 's correction is wrong. Similarity, not distance is being compared! I wrote this small extra bit of code to provide the example where one can see several clusters.
void randomInit_twopopulations (Points & ps, unsigned int dims,
unsigned int num_points)
{
for (unsigned int j = 0; j < num_points; j++)
{
Point p(dims);
if (j < num_points/2)
else
for (unsigned int i = 0; i < dims; i++)
{
p(i) = (added + rand() * (2.0) / RAND_MAX);
// std::cout << p(i) << ' ';
}
ps.push_back(p);
// std::cout << std::endl;
}
} | {"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": 0.800599217414856, "perplexity": 1479.643147951549}, "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-2018-26/segments/1529267864343.37/warc/CC-MAIN-20180622030142-20180622050142-00132.warc.gz"} |
https://forum.pkp.sfu.ca/t/missing-pdf-files/68295 | # Missing pdf files
Hi;
we have a serious problem. we cannot access pdf files. They are missing.
you can see the problem here with the link
http://londonic.uk/js/index.php/ljis/article/view/33/19
(The PDF file is empty, i.e. its size is zero bytes).
OJS 3.2.1.2.
How can we solve this problem?
Did the problem begin after a software upgrade, and if so, what version were you using before the upgrade? Is this problem occurring in a particular issue of the journal, or across several issues? This additional information may help with identifying the problem with the PDFs.
Also, there are many threads on the forum describing common causes and checks for this issue – I’d suggest doing a quick search, if you haven’t already.
Best regards,
Kate
Hi Kate;
we have the same problem with the past two issues and all the articles. we haven’t upgraded the software. because it is a new journal. I have made a search but unfortunately there is not a solution.
Hi;
You posted the discussion below in the PKP Forum. A link was provided, and I understood that it was your intension to show the PDF retrieval not working. But I accessed the link and the PDF is being showed. So, did you fix the problem? Can you tell me how you did it?
I have the same problem in the same version (OJS 3.2.1.2). The upgrade was done on July 14, and the PDFs were being displayed normally. Yesterday, we realized the exact same problem you reported. We have eleven journals and all of their issues have this problem.
http://periodicos.uefs.br/index.php/sociobiology/article/view/6022/6011
Any help is wellcome!!!
================================================
Hi,
we have a serious problem. we cannot access pdf files. They are missing.
you can see the problem here with the link
http://londonic.uk/js/index.php/ljis/article/view/33/19
(The PDF file is empty, i.e. its size is zero bytes).
OJS 3.2.1.2.
How can we solve this problem?
Hi elslva;
unfortunately we cannot fix the problem. we uploaded all files again.
Hi @ensilva,
I don’t have further information about this issue so I will flag it for one of my colleagues to take a look at and weigh-in.
Kate
Yes, we also have the same problem on our site:
PDF.js v2.6.347 (build: 3be9c65f)
Message: The PDF file is empty, i.e. its size is zero bytes.
Invalid or corrupted PDF file
Message: The PDF file is empty, i.e. its size is zero bytes.
Please post How you resolved the issue | {"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": 0.8813107013702393, "perplexity": 1280.0213049595282}, "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/1669446711016.32/warc/CC-MAIN-20221205100449-20221205130449-00520.warc.gz"} |
https://brilliant.org/discussions/thread/help-pollards-p-1-correct-way-to-build-m/ | # Help: Pollard's p-1, correct way to build M
I need some help regarding one of the steps in Pollard's p-1.
What I understood: We are assuming that there exists $$p$$, a prime factor of $$n$$, such that $$p-1$$ is $$B$$-powersmooth for some $$B$$ (not $$B$$-smooth!) and that we're trying to build a number $$M$$ (to be used as an exponent) such that $$(p-1)|M$$. Most commonly this is done in two ways:
1. $$M = B!$$
2. $$M = \prod {q^a}$$, for all primes $$q <= B$$ and $$a$$ sufficiently large.
Let's focus on the second. My confusion comes from the value of a. Using the main assumption we made ($$p-1$$ is $$B$$-powersmooth) it seems clear that $$a = \lfloor log_q{B} \rfloor$$. However, some sources (including Wikipedia) use $$a = \lfloor log_q{n} \rfloor$$. Since we can assume $$n >> B$$ this works too but makes $$a$$ unnecessarily large.
At first I thought it's just an error on Wikipedia and I made a correction, but I found several revisions in page history that change this both ways ($$B$$ to $$n$$ and back), with no real discussion, which makes me unsure. Can someone explain why $$n$$ makes more sense than $$B$$ here, or confirm that $$n$$ was just an error?
Note by Nikola Jovanović
7 months, 1 week ago
MarkdownAppears as
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The confusion seems to be whether we're assuming $$p-1$$ is $$B$$-powersmooth or $$B$$-smooth, right? If it's $$B$$-powersmooth, then we can use the smaller bound, but if it's $$B$$-smooth, then we have to use the larger one.
The Wikipedia page cites the smaller bound, but the example ($$n=299,$$ $$B=5$$) appears to use the larger bound.
- 7 months ago | {"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": 0.9933263659477234, "perplexity": 1813.549092223036}, "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-2018-39/segments/1537267160233.82/warc/CC-MAIN-20180924070508-20180924090908-00040.warc.gz"} |
http://tex.stackexchange.com/questions/141319/common-bold-command-for-latin-and-greek-letters | # Common bold command for Latin and Greek letters
Following this post, Bold math: Automatic choice between \mathbf and \boldsymbol for Latin and Greek symbols?, I created the command
\newcommand{\vect}[1]{\boldsymbol{\mathbf{#1}}}
and include
\usepackage{bm}.
In a previous paper I wrote this worked great. The greek symbols were bold and the normal Latin letters were bold roman (not italic). I am writing my thesis now with a template provided by our department and using this command doesn't work (you use the command and everything shows up non-bold, non-italics roman.) Removing the \mathbf from the command so it is just
\newcommand{\vect}[1]{\boldsymbol{#1}}
makes it so it at least bolds everything, but the Latin letters are now italicized. The only possible clue I see in the log file is a warning that say "There are no bold math fonts."
Any ideas on where to look to see what would cause this?
-
Can you identify what all font packages (both math and text) does the template use? – yo' Oct 29 '13 at 17:09
It uses amsmath, amssymb, mathptmx, textcomp, and I added bm. Also, I just found that if in the original command I replace \boldsymbol with \pmb, it appears to work. – Jason Oct 29 '13 at 17:17
\pmb works, but it's completely incorrect (and very noticeably uglu). The problem here is very likely in mathptmx. – yo' Oct 29 '13 at 17:55
@tohecz you are correct--I just found these posts: tex.stackexchange.com/questions/117076/… and tex.stackexchange.com/questions/20025/… which makes it pretty clear that mathptmx is the problem. I also have to agree that \pmb is ugly. – Jason Nov 8 '13 at 18:34 | {"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": 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": 0.9278232455253601, "perplexity": 3101.2551581223893}, "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/1419447548655.55/warc/CC-MAIN-20141224185908-00046-ip-10-231-17-201.ec2.internal.warc.gz"} |
https://kr.mathworks.com/help/radar/ref/height2range.html | height2range
Convert target height to propagated range
Syntax
r = height2range(tgtht,anht,el)
r = height2range(tgtht,anht,el,Name=Value)
[r,trueSR,trueEL] = height2range(___,Method="CRPL")
Description
example
r = height2range(tgtht,anht,el) returns the propagated range to the target, r, as a function of the target height tgtht, the sensor height anht, and the local elevation angle el assuming a Curved Earth Model with a 4/3 effective Earth radius.
example
r = height2range(tgtht,anht,el,Name=Value) specifies additional inputs using name-value arguments. For example, you can specify a flat Earth model, a curved Earth model with a given radius, or a CRPL Exponential Reference Atmosphere Model with custom values.
[r,trueSR,trueEL] = height2range(___,Method="CRPL") also returns the true slant range and the true elevation angle when you specify the Earth model as "CRPL".
Examples
collapse all
Compute the range along the propagated path for a target height of 1 km, an antenna height of 10 meters, and an elevation angle of 2 degrees at the radar. Assume a curved Earth model with a 4/3 effective Earth radius.
r = height2range(1e3,10,2)
r = 2.7125e+04
Compute the range along the propagated path using the CRPL exponential reference atmosphere. Assume a target height of 1 km, an antenna height of 10 meters, and an elevation angle of 2 degrees at the radar. Additionally, compute the true slant range and the true elevation angle to the target.
[R,SRtrue,elTrue] = height2range(1e3,10,2,Method="CRPL")
R = 2.7171e+04
SRtrue = 2.7163e+04
elTrue = 1.9666
Input Arguments
collapse all
Target height in meters, specified as a nonnegative real-valued scalar or vector. If tgtht is a vector, it must have the same size as the other vector input arguments of height2range. Heights are referenced to the ground.
Data Types: double
Sensor height in meters, specified as a nonnegative real-valued scalar or vector. If anht is a vector, it must have the same size as the other vector input arguments of height2range. Heights are referenced to the ground.
Data Types: double
Local elevation angle in degrees, specified as a real-valued scalar or vector. The local elevation angle is the initial elevation angle of the ray leaving the sensor. If el is a vector, it must have the same size as the other vector input arguments of height2range.
Data Types: double
Name-Value Arguments
Specify optional pairs of arguments as Name1=Value1,...,NameN=ValueN, where Name is the argument name and Value is the corresponding value. Name-value arguments must appear after other arguments, but the order of the pairs does not matter.
Example: Method="CRPL",SurfaceRefractivity=300,RefractionExponent=0.15
Earth model used for the computation, specified as "Curved", "Flat", or "CPRL".
Data Types: char | string
Effective Earth radius in meters, specified as a positive scalar. If this argument is not specified, height2range calculates the effective Earth radius using a refractivity gradient of –39 × 10–9 N-units/meter, which results in approximately 4/3 of the real Earth radius. This argument applies only if Method is specified as "Curved".
Data Types: double
Surface refractivity in N-units, specified as a nonnegative real-valued scalar. The surface refractivity is a parameter of the CRPL Exponential Reference Atmosphere Model used by height2range. This argument applies only if Method is specified as "CRPL".
Data Types: double
Refraction exponent, specified as a nonnegative real-valued scalar. The refraction exponent is a parameter of the CRPL Exponential Reference Atmosphere Model used by height2range. This argument applies only if Method is specified as "CRPL".
Data Types: double
Output Arguments
collapse all
Propagated range between the target and the sensor in meters, returned as a real-valued scalar or row vector. If r is a vector, it has the same size as the vector input arguments of height2range.
True slant range in meters, returned as a real-valued scalar or row vector. If trueSR is a vector, it has the same size as the vector input arguments of height2range. This argument is available only if Method is specified as "CRPL".
True elevation angle in degrees, returned as a real-valued scalar or row vector. If trueEL is a vector, it has the same size as the vector input arguments of height2range. This argument is available only if Method is specified as "CRPL".
collapse all
Flat Earth Model
The flat Earth model assumes that the Earth has infinite radius and that the index of refraction of air is uniform throughout the atmosphere. The flat Earth model is applicable over short distances and is used in applications like communications, automotive radar, and synthetic aperture radar (SAR).
Given the antenna height ha and the initial elevation angle θ0, the model relates the target height hT and the slant range RT by
${h}_{T}={h}_{a}+{R}_{T}\mathrm{sin}{\theta }_{0}\text{ }⇔\text{ }{R}_{T}=\left({h}_{T}-{h}_{a}\right)\mathrm{csc}{\theta }_{0},$
so knowing one of those magnitudes enables you to compute the other. The actual range R is equal to the slant range. The true elevation angle θT is equal to the initial elevation angle.
To compute the ground range G, use
$G=\left({h}_{T}-{h}_{a}\right)\mathrm{cot}{\theta }_{0}.$
Curved Earth Model
The fact that the index of refraction of air depends on height can be treated approximately by using an effective Earth's radius larger than the actual value.
Given the effective Earth's radius R0, the antenna height ha, and the initial elevation angle θ0, the model relates the target height hT and the slant range RT by
${\left({R}_{0}+{h}_{T}\right)}^{2}={\left({R}_{0}+{h}_{a}\right)}^{2}+{R}_{T}^{2}+2{R}_{T}\left({R}_{0}+{h}_{a}\right)\mathrm{sin}{\theta }_{0},$
so knowing one of those magnitudes enables you to compute the other. In particular,
${h}_{T}=\sqrt{{\left({R}_{0}+{h}_{a}\right)}^{2}+{R}_{T}^{2}+2{R}_{T}\left({R}_{0}+{h}_{a}\right)\mathrm{sin}{\theta }_{0}}-{R}_{0}.$
The actual range R is equal to the slant range. The true elevation angle θT is equal to the initial elevation angle.
To compute the ground range G, use
$G={R}_{0}\varphi ={R}_{0}\mathrm{arcsin}\frac{{R}_{T}\mathrm{cos}{\theta }_{0}}{{R}_{0}+{h}_{T}}.$
A standard propagation model uses an effective Earth's radius that is 4/3 times the actual value. This model has two major limitations:
1. The model implies a value for the index of refraction near the Earth's surface that is valid only for certain areas and at certain times of the year. To mitigate this limitation, use an effective Earth's radius based on the near-surface refractivity value.
2. The model implies a value for the gradient of the index of refraction that is unrealistically low at heights of around 8 km. To partially mitigate this limitation, use an effective Earth's radius based on the platform altitudes.
For more information, see effearthradius.
CRPL Exponential Reference Atmosphere Model
Atmospheric refraction evidences itself as a deviation in an electromagnetic ray from a straight line due to variation in air density as a function of height. The Central Radio Propagation Laboratory (CRPL) exponential reference atmosphere model treats refraction effects by assuming that the index of refraction n(h) and the refractivity N decay exponentially with height. The model defines
$N=\left(n\left(h\right)-1\right)×{10}^{6}={N}_{\text{s}}{e}^{-{R}_{\text{exp}}h},$
where Ns is the atmospheric refractivity value (in units of 10–6) at the surface of the earth, Rexp is the decay constant, and h is the height above the surface in kilometers. Thus
$n\left(h\right)=1+\left({N}_{\text{s}}\text{\hspace{0.17em}}×{10}^{-6}\right){e}^{-{R}_{\text{exp}}h}.$
The default value of Ns is 313 N-units and can be modified using the SurfaceRefractivity name-value argument in functions that accept it. The default value of Rexp is 0.143859 km–1 and can be modified using the RefractionExponent name-value argument in functions that accept it.
CRPL Model Geometry
When the refractivity of air is incorporated into the curved Earth model, the ray paths do not follow a straight line but curve downward. (This statement assumes standard atmospheric propagation and nonnegative elevation angles.) The true elevation angle is different from the initial . The actual range , which is the distance along the curved path , is different from the slant range .
Given the Earth's radius , the antenna height , the initial elevation angle , and the height-dependent index of refraction with value at , the modified model relates the target height and the actual range by
When Method is specified as "CRPL", the integral is solved using from CRPL Exponential Reference Atmosphere Model.
To compute the ground range , use
Version History
Introduced in R2021b | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 7, "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": 0.8763687014579773, "perplexity": 1439.9614569899545}, "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-2022-49/segments/1669446710711.7/warc/CC-MAIN-20221129200438-20221129230438-00237.warc.gz"} |
https://www.khanacademy.org/math/arithmetic/fractions | # Fractions
Contents
Understanding fractions conceptually, using operations with fractions, and converting fractions.
## Understanding fractions
If you don't understand fractions, you won't be even 1/3 educated. Glasses will seem half empty rather than half full. You'll be lucky to not be duped into some type of shady real-estate scheme or putting far too many eggs in your cake batter. Good thing this tutorial is here. You'll see that fractions allow us to view the world in entirely new ways. You'll see that everything doesn't have to be a whole. You'll be able to slice and dice and then put it all back together (and if you order now, we'll throw in a spatula warmer for no extra charge).
## Visualizing equivalent fractions
Do you want 2/3 or 4/6 of this pizza? Doesn't matter because they are both the same fraction. This tutorial will help us explore this idea by really visualizing what equivalent fractions represent.
## Equivalent fractions and simplified form
There are literally infinite ways to represent any fraction (or number for that matter). Don't believe us? Let's take 1/3. 2/6, 3/9, 4/12 ... 10001/30003 are all equivalent fractions (and we could keep going)! If you know the basics of what a fraction is, this is a great tutorial for recognizing when fractions are equivalent and then simplifying them as much as possible!
## Decomposing fractions
In this tutorial, we'll see that a fraction can be broken up (or decomposed) into a bunch of other fractions. You might see the world in a completely different way after this.
You've already got 2 cups of sugar in the cupboard. Your grandmother's recipe for disgustingly-sweet-fudge-cake calls for 3 and 1/3 cups of sugar. How much sugar do you need to borrow from you robot neighbor? Adding and subtracting fractions is key. It might be a good idea to look at the equivalent fractions tutorial before tackling this one.
## Adding and subtracting fractions with unlike denominators
We've already had some good practice adding fractions with like denominators. We'll now add fractions with unlike denominators. This is a very big deal. After this tutorial, you'll be able to add, pretty much, any two (or three or four or... ) fractions!
## Adding and subtracting with unlike denominator word problems
You know what a fraction is and are now eager to apply this knowledge to real-world situations (especially ones where the denominators aren't equal)? Well, you're about to see that adding and subtracting fractions is far more powerful (and fun) then you've ever dreamed possible!
## Multiplying fractions word problems
Multiplying fractions is useful. Period. That's all we really have to say. Believe us don't believe us. You'll learn eventually. This tutorial will have you multiplying in real-world scenarios (which is almost as fun as completely artificial, fake scenarios).
## Mixed numbers and improper fractions
We can often have fractions whose numerators are not less than the denominators (like 23/4 or 3/2 or even 6/6). These top-heavy friends are called improper fractions. Since they represent a whole or more (in absolute terms), they can also be expressed as a combination of a whole number and a "proper fraction" (one where the numerator is less than the denominator) which is called a "mixed number." They are both awesome ways of representing a number and getting acquainted with both (as this tutorial does) is super useful in life!
## Mixed number multiplication and division
My recipe calls for a cup and a half of blueberries and serves 10 people. But I have 23 people coming over. How many cups of blueberries do I need? You know that mixed numbers and improper fractions are two sides of the same coin (and you can convert between the two). In this tutorial we'll learn to multiply and divide mixed numbers (mainly by converting them into improper fractions first).
## Dividing fractions
This is one exciting tutorial. In it, we will understand that fractions can represent division (and the other way around). Then we will create fractions by dividing whole numbers and then start dividing the fractions themselves. We'll see that dividing by something is the exact same thing as multiplying by that thing's reciprocal!
## Dividing fractions by fractions
In this tutorial, we'll become fraction dividing experts! In particular, we'll understand what it means to divide a fraction by another fraction. Too much fun!!!
## Number sets
The world of numbers can be split up into multiple "sets", many of which overlap with each other (integers, rational numbers, irrational numbers, etc.). This tutorial works through examples that expose you to the terminology of the various sets and how you can differentiate them. | {"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": 0.8841679692268372, "perplexity": 1054.1217994539693}, "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-2016-30/segments/1469257824226.79/warc/CC-MAIN-20160723071024-00236-ip-10-185-27-174.ec2.internal.warc.gz"} |
https://www.scienceabc.com/pure-sciences/buffer-capacity-definition-and-how-to-calculate-it.html | Buffer Capacity: Definition And Method Of Calculation
If you remember high school chemistry or took a college course like Chemistry 101, you will have conducted a titration test. Personally, the first time I let the liquid percolate at the bottom of the glass flask, patiently waiting for the solution to turn to a tinge of pink or magenta, it honestly made me feel like a scientist! However, why does this solution change color only when a certain amount of chemicals are added? To get that answer, we must understand the inherent properties of the solution.
(Image Credit: Pixabay)
Definition
Before we get into what a buffer capacity is, we should first understand buffers. A buffer is a compound that resists changes in pH when a limited amount of acid or base is added to it. The chemical composition of a buffer solution usually entails a weak acid or a weak base accompanied by its conjugate salt.
Now, Buffer Capacity can be defined as the measure of the efficiency of a buffer in resisting its change in pH. This definition does present a bit of a problem as to ‘what is the significant change?’ Sometimes, a change of 1 unit does not bring about any significant change. At other times, even a 0.1-unit change can cause a significant difference. So, to give a more clear definition, buffer capacity may be defined as the quantity of a strong acid or strong base that must be added to one liter of a solution to change it by one pH unit. The buffer capacity equation is as follows: where n is some equivalents of added strong base (per 1 L of the solution). Note that the addition of n moles of acid will change the pH by the same value, but in the opposite direction. We will derive a formula connecting buffer capacity with pH, pKa and buffer concentration.
The Calculation
Now that we have seen how the buffer equation can be written, let’s try to derive it to have a better understanding of how we arrived at the above equation. To make this derivation a bit easier, we shall make the base monoprotic (a base that will accept only one proton). We shall also assume the volume to be one, as this helps us treat concentration and the number of moles interchangeably. The charge balance of the solution we assume is demonstrated by the following equation:
[A]+[OH+]=[B+]+[H+]
[B+] denotes the presence of a strong base concentration in the solution. The [B+] is also the n present in the first buffer capacity equation. Now the total concentration of the buffer is given by the following equation:
Cbuff =[HA]+[A]
The [AH] in the above equation can be broken down into smaller constituent elements. This breaking down of a larger, more complex compound into smaller basic elements is known as the dissociation constant. The dissociation constant helps in the easier simplification of the derivation. The Ka in the below equation is the acid dissociation constant. It relates to how easily a molecule will act as an acid.
[HA]=([H+][A])/Ka
Now, the above equation can be substituted in the buffer concentrate equation, giving the following equation:
Cbuff= ([H+][A])/Ka + [A]
Now, if we were to take [A] as a common factor and LCM to simplify the above equation, we would get the following equation:
[A]= (Cbuff+Ka)/(Ka+H+)
Before moving forward, we must understand one critical definition that will serve as a prerequisite to neatly wrap this derivation up, known as the water ionization constant or the self-ionization of water. The self-ionization of water is an ionization reaction that occurs in pure water or in an aqueous solution, in which H2O loses the nucleus of one of its hydrogen atoms to become a hydroxide ion, OH.
(Photo Credit : Manuel Almagro Rivas/Wikimedia Commons)
Now, using the charge balance equation, the [A] equivalent and the water ionization constant, we can come to the following equation:
The first two terms present in the equation are not dependent on the buffer in the solution. They represent the fact that the solution of high (or low) pH is resistant to pH changes. This indicates that certain solutions with extremities in pH are resistant to changes, even without a buffer solution present.
The graph above shows the buffer capacity changes in 0.1 M of an acetic buffer. As expected, the buffer resists acid and base addition to maintain an equimolar solution (when pH=pKa). From the graph, it is obvious that the buffer capacity has reasonably high values only for pH close to the pKa value: the further from the optimal value, the lower the buffer capacity of the solution. A solution containing a conjugate base of pH 8-10 has a buffer capacity of zero, while for a higher pH, the presence of the strong base starts to play an important role. In the case of a pure acetic acid solution with a pH below 3, the pH is already low enough to be resistant to changes due to the high concentration of H+ cations.
References:
The short URL of the present article is: http://sciabc.us/LMqVS | {"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": 0.9107415676116943, "perplexity": 790.0934780257194}, "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-2020-29/segments/1593655929376.49/warc/CC-MAIN-20200711095334-20200711125334-00466.warc.gz"} |
https://www.physicsforums.com/threads/derivation-of-brst-charge.247234/ | # Derivation of BRST charge
1. Jul 28, 2008
### ismaili
I would like to derive the explicit formula of the BRST charge.
http://en.wikipedia.org/wiki/BRST_formalism
(Bottom of Wiki link, I copy the formula here)
$$Q = c^i\left(L_i - \frac{1}{2}f_{ij}{}^kb_jc_k\right)$$
where $$c$$ is the ghost field, and $$b$$ is the antighost field, $$L_i$$ is the gauge group generator.
Actually, in wiki's article, right above the formula of BRST charge, there is a Lagrangian. I tried to use Noether theorem to calculate the charge, but in vain. $$J^\mu_a\sim \frac{\partial\mathcal{L}}{\partial\psi_{,\mu}}\delta\psi_a$$, replacing the $$\psi$$ with the ghost $$c$$, I ends up with
$$J^0 = \dot{b}\,\delta c = \dot{b}\left(-\frac{1}{2}f_{ij}{}^kc^ic^j\right)$$ whose volume integral looks different as the correct answer.
In there anybody who can help me or give me some hints?
Many thanks!
Last edited: Jul 28, 2008
Can you help with the solution or looking for help too?
Draft saved Draft deleted
Similar Discussions: Derivation of BRST charge | {"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": 0.9709129333496094, "perplexity": 982.5327979226245}, "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-2016-44/segments/1476988720153.61/warc/CC-MAIN-20161020183840-00298-ip-10-171-6-4.ec2.internal.warc.gz"} |
https://wiki2.org/en/Power_(physics) | To install click the Add extension button. That's it.
The source code for the WIKI 2 extension is being checked by specialists of the Mozilla Foundation, Google, and Apple. You could also do it yourself at any point in time.
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# Power (physics)
Power
Common symbols
P
SI unitwatt (W)
In SI base unitskgm2s−3
Derivations from
other quantities
Dimension${\displaystyle {\mathsf {L}}^{2}{\mathsf {M}}{\mathsf {T}}^{-3}}$
In physics, power is the amount of energy transferred or converted per unit time. In the International System of Units, the unit of power is the watt, equal to one joule per second. In older works, power is sometimes called activity.[1][2][3] Power is a scalar quantity.
Power is related to other quantities; for example, the power involved in moving a ground vehicle is the product of the traction force on the wheels and the velocity of the vehicle. The output power of a motor is the product of the torque that the motor generates and the angular velocity of its output shaft. Likewise, the power dissipated in an electrical element of a circuit is the product of the current flowing through the element and of the voltage across the element.[4][5]
## Definition
Power is the rate with respect to time at which work is done; it is the time derivative of work:
${\displaystyle P={\frac {dW}{dt}}}$
where P is power, W is work, and t is time.
If a constant force F is applied throughout a distance x, the work done is defined as ${\displaystyle W=\mathbf {F} \cdot \mathbf {x} }$. In this case, power can be written as:
${\displaystyle P={\frac {dW}{dt}}={\frac {d}{dt}}\left(\mathbf {F} \cdot \mathbf {x} \right)=\mathbf {F} \cdot {\frac {d\mathbf {x} }{dt}}=\mathbf {F} \cdot \mathbf {v} }$
If instead the force is variable over a three-dimensional curve C, then the work is expressed in terms of the line integral:
${\displaystyle W=\int _{C}\mathbf {F} \cdot d\mathbf {r} =\int _{\Delta t}\mathbf {F} \cdot {\frac {d\mathbf {r} }{dt}}\ dt=\int _{\Delta t}\mathbf {F} \cdot \mathbf {v} \ dt}$
From the fundamental theorem of calculus, we know that ${\displaystyle P={\frac {dW}{dt}}={\frac {d}{dt}}\int _{\Delta t}\mathbf {F} \cdot \mathbf {v} \ dt=\mathbf {F} \cdot \mathbf {v} }$. Hence the formula is valid for any general situation.
## Units
The dimension of power is energy divided by time. In the International System of Units (SI), the unit of power is the watt (W), which is equal to one joule per second. Other common and traditional measures are horsepower (hp), comparing to the power of a horse; one mechanical horsepower equals about 745.7 watts. Other units of power include ergs per second (erg/s), foot-pounds per minute, dBm, a logarithmic measure relative to a reference of 1 milliwatt, calories per hour, BTU per hour (BTU/h), and tons of refrigeration.
## Average power
As a simple example, burning one kilogram of coal releases much more energy than detonating a kilogram of TNT,[6] but because the TNT reaction releases energy much more quickly, it delivers far more power than the coal. If ΔW is the amount of work performed during a period of time of duration Δt, the average power Pavg over that period is given by the formula:
${\displaystyle P_{\mathrm {avg} }={\frac {\Delta W}{\Delta t}}}$
It is the average amount of work done or energy converted per unit of time. The average power is often simply called "power" when the context makes it clear.
The instantaneous power is then the limiting value of the average power as the time interval Δt approaches zero.
${\displaystyle P=\lim _{\Delta t\rightarrow 0}P_{\mathrm {avg} }=\lim _{\Delta t\rightarrow 0}{\frac {\Delta W}{\Delta t}}={\frac {\mathrm {d} W}{\mathrm {d} t}}}$
In the case of constant power P, the amount of work performed during a period of duration t is given by:
${\displaystyle W=Pt}$
In the context of energy conversion, it is more customary to use the symbol E rather than W.
## Mechanical power
One metric horsepower is needed to lift 75 kilograms by 1 metre in 1 second.
Power in mechanical systems is the combination of forces and movement. In particular, power is the product of a force on an object and the object's velocity, or the product of a torque on a shaft and the shaft's angular velocity.
Mechanical power is also described as the time derivative of work. In mechanics, the work done by a force F on an object that travels along a curve C is given by the line integral:
${\displaystyle W_{C}=\int _{C}\mathbf {F} \cdot \mathbf {v} \,\mathrm {d} t=\int _{C}\mathbf {F} \cdot \mathrm {d} \mathbf {x} }$
where x defines the path C and v is the velocity along this path.
If the force F is derivable from a potential (conservative), then applying the gradient theorem (and remembering that force is the negative of the gradient of the potential energy) yields:
${\displaystyle W_{C}=U(A)-U(B)}$
where A and B are the beginning and end of the path along which the work was done.
The power at any point along the curve C is the time derivative:
${\displaystyle P(t)={\frac {\mathrm {d} W}{\mathrm {d} t}}=\mathbf {F} \cdot \mathbf {v} =-{\frac {\mathrm {d} U}{\mathrm {d} t}}}$
In one dimension, this can be simplified to:
${\displaystyle P(t)=F\cdot v}$
In rotational systems, power is the product of the torque τ and angular velocity ω,
${\displaystyle P(t)={\boldsymbol {\tau }}\cdot {\boldsymbol {\omega }}}$
where ω measured in radians per second. The ${\displaystyle \cdot }$ represents scalar product.
In fluid power systems such as hydraulic actuators, power is given by
${\displaystyle P(t)=pQ}$
where p is pressure in pascals, or N/m2 and Q is volumetric flow rate in m3/s in SI units.
If a mechanical system has no losses, then the input power must equal the output power. This provides a simple formula for the mechanical advantage of the system.
Let the input power to a device be a force FA acting on a point that moves with velocity vA and the output power be a force FB acts on a point that moves with velocity vB. If there are no losses in the system, then
${\displaystyle P=F_{\text{B}}v_{\text{B}}=F_{\text{A}}v_{\text{A}}}$
and the mechanical advantage of the system (output force per input force) is given by
${\displaystyle \mathrm {MA} ={\frac {F_{\text{B}}}{F_{\text{A}}}}={\frac {v_{\text{A}}}{v_{\text{B}}}}}$
The similar relationship is obtained for rotating systems, where TA and ωA are the torque and angular velocity of the input and TB and ωB are the torque and angular velocity of the output. If there are no losses in the system, then
${\displaystyle P=T_{\text{A}}\omega _{\text{A}}=T_{\text{B}}\omega _{\text{B}}}$
${\displaystyle \mathrm {MA} ={\frac {T_{\text{B}}}{T_{\text{A}}}}={\frac {\omega _{\text{A}}}{\omega _{\text{B}}}}}$
These relations are important because they define the maximum performance of a device in terms of velocity ratios determined by its physical dimensions. See for example gear ratios.
## Electrical power
Ansel Adams photograph of electrical wires of the Boulder Dam Power Units, 1941–1942
The instantaneous electrical power P delivered to a component is given by
${\displaystyle P(t)=I(t)\cdot V(t)}$
where
${\displaystyle P(t)}$ is the instantaneous power, measured in watts (joules per second)
${\displaystyle V(t)}$ is the potential difference (or voltage drop) across the component, measured in volts
${\displaystyle I(t)}$ is the current through it, measured in amperes
If the component is a resistor with time-invariant voltage to current ratio, then:
${\displaystyle P=I\cdot V=I^{2}\cdot R={\frac {V^{2}}{R}}}$
where
${\displaystyle R={\frac {V}{I}}}$
is the resistance, measured in ohms.
## Peak power and duty cycle
In a train of identical pulses, the instantaneous power is a periodic function of time. The ratio of the pulse duration to the period is equal to the ratio of the average power to the peak power. It is also called the duty cycle (see text for definitions).
In the case of a periodic signal ${\displaystyle s(t)}$ of period ${\displaystyle T}$, like a train of identical pulses, the instantaneous power ${\displaystyle p(t)=|s(t)|^{2}}$ is also a periodic function of period ${\displaystyle T}$. The peak power is simply defined by:
${\displaystyle P_{0}=\max[p(t)]}$
The peak power is not always readily measurable, however, and the measurement of the average power ${\displaystyle P_{\mathrm {avg} }}$ is more commonly performed by an instrument. If one defines the energy per pulse as:
${\displaystyle \epsilon _{\mathrm {pulse} }=\int _{0}^{T}p(t)\mathrm {d} t}$
then the average power is:
${\displaystyle P_{\mathrm {avg} }={\frac {1}{T}}\int _{0}^{T}p(t)\mathrm {d} t={\frac {\epsilon _{\mathrm {pulse} }}{T}}}$
One may define the pulse length ${\displaystyle \tau }$ such that ${\displaystyle P_{0}\tau =\epsilon _{\mathrm {pulse} }}$ so that the ratios
${\displaystyle {\frac {P_{\mathrm {avg} }}{P_{0}}}={\frac {\tau }{T}}}$
are equal. These ratios are called the duty cycle of the pulse train.
Power is related to intensity at a radius ${\displaystyle r}$; the power emitted by a source can be written as:[citation needed]
${\displaystyle P(r)=I(4\pi r^{2})}$ | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 40, "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": 0.9312617778778076, "perplexity": 538.5553078951407}, "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-2021-49/segments/1637964363376.49/warc/CC-MAIN-20211207105847-20211207135847-00113.warc.gz"} |
https://www.physicslens.com/06-motion-in-a-circle/ | # 06. Motion in a Circle
[accordions autoHeight=’true’]
[accordion title=”1. Rotational Kinematics”]
• Angular displacement $$\theta$$ is defined as the angle an object turns with respect to the centre of a circle. $$\theta=\dfrac{s}{r}$$ where s is the arc and r is the radius of the circle.
• One radian is the angular displacement when the arc length is equal to the radius of the circle.
• Angular velocity $$\omega$$ is defined as the rate of change of angular displacement. $$\omega=\dfrac{d\theta}{dt}$$
• For motion in a circle of fixed radius, $$\omega=\dfrac{d\theta}{dt}=\dfrac{d(\dfrac{s}{r})}{dt}=\dfrac{1}{r}\dfrac{ds}{dt}=\dfrac{v}{r}$$. Thus $$v=r\omega$$.
• Average angular velocity in one cycle. $$\omega=\dfrac{2\pi}{T}=2\pi f$$ where T is the period and f is the frequency.
[/accordion]
[accordion title=”2. Centripetal Force”]
• Centripetal acceleration $$a=\dfrac{v^2}{r} = r\omega^2$$.
• Centripetal force $$F =\dfrac{mv^2}{r} = mr\omega^2$$.
[/accordion]
[accordion title=”3. Uniform Circular Motion”]
For a body in uniform circular motion, there is a change in velocity as the direction of motion is changing. This requires an acceleration that is perpendicular to the velocity and directed towards the centre of circle. This acceleration is provided by a centripetal force. A resultant force acting on a body toward the centre of a circle provides the centripetal force.
[/accordion]
[accordion title=”4. Non-Uniform Circular Motion”]
Learn how to solve problems on circular motions of conical pendulum, cyclist, car, aircraft, swinging a pail etc.
[/accordion]
[/accordions] | {"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": 0.9900272488594055, "perplexity": 428.6165939720715}, "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/1610703506640.22/warc/CC-MAIN-20210116104719-20210116134719-00089.warc.gz"} |
http://prosseek.blogspot.com/2012/11/using-markdown-format-with-textmate.html | Sunday, November 4, 2012
Using markdown format with TextMate
You want to execute "pandoc" when "⌘-R" is clicked, what would you do? There are some ideas you need to understand.
• Scope selection
• Activation
• Command
• Environment Variables
Scope selection and Activation
Many of the keys are already assigned, so when you try to use the popular key such as "⌘-R", you need to select the scope. In other words, you need to teach TextMate that when you click "⌘-R" in markdown, it means something special.
Command
In command editor, you can give any UNIX command you want. And many of the predefined environment variables are already given.
Combining the command and variable, you can come up wit the following commands.
/usr/local/bin/pandoc "$TM_FILEPATH" -o "$TM_FILEPATH".html
/usr/bin/open "$TM_FILEPATH".html echo File: "$TM_FILEPATH" | {"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": 0.8175224661827087, "perplexity": 4432.601064661195}, "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-2017-22/segments/1495463609404.11/warc/CC-MAIN-20170528004908-20170528024908-00409.warc.gz"} |
https://www.physicsforums.com/threads/solving-6-32-a-b.722166/ | # Solving √(6 + 3√2) = √a + √b
1. Nov 11, 2013
### BMW
1. The problem statement, all variables and given/known data
Solve the equation √(6 + 3√2) = √a + √b, writing a and b in the form a + b√c.
2. Relevant equations
In the answers they say that a + b = 6, but I cannot see how they can say this.
3. The attempt at a solution
I square both sides, and that is as far as I get:
6 + 3√2 = a + 2√(ab) + b
In the answers, they say from here that a + b = 6. I am clueless as to how they can say this.
Last edited: Nov 11, 2013
2. Nov 11, 2013
### Dick
I don't think they are deducing that from the equation. They are just saying 'let's look for a solution where a+b=6 and 3√2=2√(ab)'. If you can find simple numbers a and b that satisfy that then you've got a simpler form for the radical expression.
Last edited: Nov 11, 2013
3. Nov 11, 2013
### BMW
Ah, ok. So there would be many other solutions, and they are only finding one such solution?
4. Nov 11, 2013
### Dick
Right. There are many other solutions. They are just looking for a nice simple one.
5. Nov 11, 2013
### BMW
So you would also be able to say that a + 2√(ab) = 6 and b = 3√2, and solve that way (with the risk of it being horribly complicated)?
6. Nov 11, 2013
### Dick
Yes, there are lots of ugly solutions.
Last edited: Nov 11, 2013
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http://mathhelpforum.com/math-challenge-problems/128094-three-dimensional-pythagora-s-theorem.html | 1. ## Three-dimensional Pythagoras' theorem
Suppose you have a right-angled tetrahedron, i.e. a tetrahedron with a vertex where three faces meet at right angles. Let $A$ be the area of the face opposite this vertex, and $B, C, D$ the areas of the remaining faces. Show that $A^2=B^2+C^2+D^2$.
Edit : I had first posted a more general problem but I'll put it on hold for the time being.
2. Spoiler:
Let
sides of $A:x,y,z$
sides of $B: \alpha, \gamma, y$
sides of $C: \alpha, \beta, x$
sides of $D: \beta, \gamma, z$.
Then we have
$B=\frac{1}{2}\alpha \gamma, \, C=\frac{1}{2}\alpha \beta , \, D=\frac{1}{2}\beta \gamma$
and
$x^2=\alpha^2+\beta^2$
$y^2=\alpha^2+\gamma^2$
$z^2=\beta^2+\gamma^2.$
Using Heron's formula, we have
$A^2=\frac{1}{16}(x+y+z)(y+z-x)(x+y-z)(z+x-y)$
$=\frac{1}{16}((y+z)+x)((y+z)-x)(x+(y-z))(x-(y-z))$
$=\frac{1}{16}((y+z)^2-x^2)(x^2-(y-z)^2)$
$=\frac{1}{16}(2yz-(x^2-y^2-z^2))(2yz+(x^2-y^2-z^2))$
$=\frac{1}{16}(4y^2z^2-(x^2-y^2-z^2)^2)$
$=\frac{1}{16}\left[4(\alpha^2+\gamma^2)(\beta^2+\gamma^2)-((\alpha^2+\beta^2)-(\alpha^2+\gamma^2)-(\beta^2+\gamma^2))^2\right]$
$=\frac{1}{16}\left[4(\alpha^2\beta^2+\beta^2\gamma^2+\alpha^2\gamma^2 +\gamma^4)-4\gamma^4\right]$
$=\frac{1}{4}\alpha^2\beta^2+\frac{1}{4}\beta^2\gam ma^2+\frac{1}{4}\alpha^2\gamma^2=C^2+D^2+B^2.$
3. Here's my solution
Spoiler:
Couldn't you just put your tetrahedron in the $xyz$ planes where the $xy, yz \;\text{and}\; xz$ plane are three of the four sides. Let the face A be described by the equation $ax + by + cz = 1$. If so then the area of the sides B, C and D are $\frac{1}{2} \frac{1}{ab}, \frac{1}{2} \frac{1}{ac} \; \text{and}\; \frac{1}{2} \frac{1}{bc}$. So
$
B^2+C^2+D^2 = \frac{1}{4} \left( \frac{1}{a^2b^2} + \frac{1}{a^2c^2} + \frac{1}{b^2c^2} \right)
$
The area of face $A$ is $\iint \limits_R \sqrt{1 + z_x^2 + z_y^2}dA = \sqrt{1 + \frac{a^2}{c^2} + \frac{b^2}{c^2}} \frac{1}{2ab}$
and so $A^2 = \left(1 + \frac{a^2}{c^2} + \frac{b^2}{c^2}\right) \frac{1}{4a^2b^2} = B^2 + C^2 + D^2$
4. Let
$A = A(a,0,0) , B = B(0,b,0) , C = C(0,0,c)$
together with the origin , they form a right-angled tetrahedron
we have $CA = (a,0,-c) , CB = (0,b,-c)$
and their cross product is
$CA \times CB = (a,0,-c) \times (0,b,-c)$
$= ( bc , ac , ab )$
we also know that the area of the face opposite this vertex $S$ is
the half of the modulus of the cross product so
$S^2 = \frac{(ab)^2 + (bc)^2 + (ac)^2 }{4}$
$= S^2_1 + S^2_2 + S^2_3$
where $S_1 , S_2 , S_3$ are the area of the lateral faces of the tetrahedron . | {"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": 35, "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": 0.8898197412490845, "perplexity": 383.62047479432374}, "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/1484560283475.86/warc/CC-MAIN-20170116095123-00370-ip-10-171-10-70.ec2.internal.warc.gz"} |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Atomic_Theory/Isotopes/Isotopes_II | # Isotopes II
Although all atoms of an element have the same number of protons, individual atoms may have different numbers of neutrons. These differing atoms are called isotopes. All atoms of chlorine (Cl) have 17 protons, but there are chlorine isotopes with 15 to 23 neutrons. Only two chlorine isotopes exist in significant amounts in nature: those with 18 neutrons (75.53% of all chlorine atoms found in nature), and those with 20 neutrons (24.47%). To write the symbol for an isotope, place the atomic number as a subscript and the mass number (protons plus neutrons) as a superscript to the left of the atomic symbol. The symbols for the two naturally occurring isotopes of chlorine are written as follows: $$^{35}_{17}{\rm Cl}$$ and $$^{37}_{17}{\rm Cl}$$. The subscript is somewhat unnecessary, because all atoms of chlorine have 17 protons; isotope symbols are usually written without the subscript, as in 35Cl and 37Cl. In discussing these isotopes, the terms chlorine-35 and chlorine-37 are used to differentiate between them. In general, for an atom to be stable, it must have more neutrons than protons. Nuclei with too many of either kind of fundamental particle are unstable, and break down radioactively.
Example $$\PageIndex{1}$$:
How many protons, neutrons, and electrons are there in an atom of uranium-238? Write the symbol for this isotope.
Solution
The atomic number of uranium (see periodic table) is 92, and the mass number of the isotope is given as 238. Therefore, it has 92 protons, 92 electrons, and 238 — 92 : 146 neutrons. Its symbol is $$\ce{^{238}_{92}U}$$ (or 238U).
The total mass of an atom is called its atomic weight, and is the approximate sum of the masses of its constituent protons, neutrons, and electrons. When protons, neutrons, and electrons combine to form an atom, some of their mass is converted to energy and is given off. (This is the source of energy in nuclear fusion reactions. Because the atom cannot be broken down into its fundamental particles unless the energy for the missing mass is supplied from outside it, this energy is called the binding energy of the nucleus.
Example $$\PageIndex{2}$$:
Calculate the mass that is lost when an atom of carbon-12 is formed from protons, electrons, and neutrons.
Solution
Because the atomic number of every carbon atom is 6, carbon-12 has 6 protons and therefore 6 electrons. To find the number of neutrons, we subtract the number of protons from the mass number: 12 – 6 = 6 neutrons. The data in Table 1-1 can be used to calculate the total mass of these particles:
Protons: 6 x 1.00728 amu = 6.04368 u
Neutrons: 6 x 1.00867 amu = 6.05202 u
Electrons: 6 x 0.00055 amu = 0.00330 u
Total particle mass = 12.09900 u
However, by the definition of the scale of atomic mass units, the mass of one carbon-12 atom is exactly 12 amu. Therefore, 0.0990 u of mass has disappeared in the process of building the atom from its particles.
Each isotope of an element is characterized by an atomic number (the number of protons), a mass number (the total number of protons and neutrons), and an atomic weight (mass of atom in atomic mass units). Because mass losses upon formation of an atom are small, the mass number is usually the same as the atomic weight rounded to the nearest integer (for example, the atomic weight of chlorine-37 is 36.966, which is rounded to 37). If there are several isotopes of an element in nature, then the experimentally observed atomic weight (the natural atomic weight) is the weighted average of the isotope weights. The average is weighted according to the percent abundance of the isotopes. Chlorine occurs in nature as 75.53% chlorine-35 (34.97 u) and 24.47% chlorine-37 (36.97 u), so the weighted average of the isotope weights is (07553 x 34.97 u) + (0.2447 x 36.97 u) = 35.46 u. The atomic weights found in periodic tables are all weighted averages of the isotopes occurring in nature, and these are the figures used for the remainder of this article, except when discussing one isotope specifically. In general, all isotopes of an element behave the same way chemically. Their behaviors differ with regard to mass-sensitive properties such as diffusion rates.
Example $$\PageIndex{3}$$:
Magnesium (Mg) has three significant natural isotopes: 78.70% of all magnesium atoms have an atomic weight of 23.985 u, 10.13% have an atomic weight of 24.986 u, and 11.17% have an atomic weight of 25.983 u. How many protons and neutrons are present in each of these three isotopes? How are the symbols for each isotope written? Finally, what is the weighted average of the atomic weights?
Solution
There are 12 protons in all magnesium isotopes. The isotope whose atomic weight is 23.985 u has a mass number of 24 (protons and neutrons), so 24 - 12 protons gives 12 neutrons. The symbol for this isotope is 24Mg. Similarly, the isotope whose atomic weight is 24.986 amu has a mass number of 25, 13 neutrons, and 25Mg as a symbol. The third isotope (25.983 amu) has a mass number of 26, 14 neutrons, and 26Mg as a symbol. The average atomic weight is calculated as follows:
(0.7870 x 23.985) + (0.1013 x 24.986) + (0.1117 x 25.983) = 24.31 u
Example $$\PageIndex{4}$$: Boron
Boron has two naturally occurring isotopes, 10B and 11B. In nature, 80.22% of its atoms are 11B, with atomic weight 11.009 u. From the natural atomic weight, calculate the atomic weight of the 10B isotope.
Solution
If 80.22% of all boron atoms are 11B, then 100.00 — 80.22, or 19.78%, are the unknown isotope. In the periodic table the atomic weight of boron is found to be 10.81 u. We can use W to represent the unknown atomic weight in our calculation:
$(0.8022 \times 11.009) + (0.1978 \times W) = 10.81 {\rm u} \quad {\rm (natural~atomic~weight)}$
$W=\dfrac{10.81-8.831}{0.1978}=10.01 {\rm u}$
## Contributors and Attributions
• Dickerson, Richard E. and Gray, Harry B. and Haight, Gilbert P (1979) Chemical principles. | {"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": 0.845199704170227, "perplexity": 1065.8346887416967}, "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/1614178366959.54/warc/CC-MAIN-20210303104028-20210303134028-00581.warc.gz"} |
http://mgreenbe.ucalgaryblogs.ca/2013/01/28/amat-415-matlab-activity-3-fft/ | # AMAT 415 — Matlab activity 3: Fourier series via FFT
This activity is meant to be instructive. I won’t ask you to submit your code or plots.
Let $x=(x_0,\ldots,x_{N-1})$ be an $n$-element vector. Its DFT is the vector $y=(y_0,\ldots,y_{N-1})$ defined by
$\displaystyle{y_n=\sum_{k=0}^{N-1}x_ke(-nk/N)}$,
where $e(x)=e^{2\pi i x}$. (In class yesterday, we had a factor of $1/N$ in front of the above sum. I’m dropping this because MATLAB uses different conventions.) Suppose $x_k=f(2\pi k/N)$ for some function $f$ defined on $[0,2\pi]$. Then
$\displaystyle{y_n=\frac{N}{2\pi}\sum_{k=0}^{N-1}f(2\pi k/N)e^{-in(2\pi k/N)}\frac{2\pi}{N}}$.
The sum in this expression is a “left endpoint” Riemann sum approximation to $\displaystyle{\int_0^{2\pi}f(t)e^{-int}dt}$ using a subdivision of $[0,2\pi]$ into $N$ equal subintervals of width $2\pi/N$. Thus, $y_n\approx N\hat{f}(n)$. This means that we can approximate the Fourier coefficients of a function $f(t)$ by computing the DFT of the vector $x$ with $x_k=f(2\pi k/N)$.
The Fast Fourier Transform is an algorithm for computing the DFT very efficiently. In MATLAB, invoking fft(x) gives you the DFT of the vector x. (Note that MATLAB indexes vectors starting from 1 rather than from 0. From the perspective of the mathematics of the DFT, it’s better to start indexing from 0 as above. Keep this shift in mind.) Let’s repeat some of the calculations of last week’s lab using the fft command. The FFT algorithm works best on vector of length $2^n$. So let N=2^7, say, and set t=0:2*pi/N:(2*pi-2*pi/N);. We’ll start with the step function x=t<pi;. Let y be its DFT, i.e., set y=fft(x);. Now let’s try to reconstruct x from y. Initialize the vector and, for various small values of m, perform:
s=zeros(1,length(x));
for k=1:m
s = s + y(k)/N*exp(i*(k-1)*t);
end
plot(t,real(s))
Do the graphs look right? Plot x and s on the same graph: plot(t,x,t,s). Something’s off. What’s going on?
Let’s try a different function. Let x=[t(1:N/2) t(N/2)-t(1:N/2)];. (Plot it and figure out why the command you just pasted in gives you the graph you see.) Now run through the above process with this x. Still doesn’t look right…
Wait a second. The Fourier series of a function $f$ is $\displaystyle{\sum_{n=-\infty}^\infty \hat{f}(n)e^{int}}$. We’ve been ignoring the terms with $n<0$. No wonder our pictures are wrong! Even though $y_{n}$ for $n<0$ doesn’t make sense, we recall that if $f(t)$ is real-valued, then $\hat{f}(-n)=\overline{\hat{f}(n)}$. Let’s take this into account:
s=zeros(1,length(x))+y(1)/N;
for k=2:m
s = s + y(k)/N*exp(i*(k-1)*t) + conj(y(k))/N*exp(i*(1-k)*t);
end
plot(t,real(s),t,x)
That’s more like it! Try a few more functions x like the ones you worked with last week.
Finally, I’d like to acquaint you with a nice command. Check out stem(y), stem(abs(y)) and compare them with plot(real(y)) and plot(abs(y)). Since DFTs are just N element vectors, a “plot” of them should just be N points not connected by lines. This is what stem gives you. It’s really nice for visualizing discrete signals in the frequency domain, especially when N isn’t too big so the circles it plots don’t start overlapping. What interesting symmetries do you observe in these stem plots?
This entry was posted in Uncategorized. Bookmark the permalink. | {"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": 26, "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": 0.8483632802963257, "perplexity": 781.2427673019902}, "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/1516084891706.88/warc/CC-MAIN-20180123032443-20180123052443-00450.warc.gz"} |
https://www.physicsforums.com/threads/nonconservative-force-question.186600/ | # Nonconservative Force Question- !
1. Sep 23, 2007
### perfectionist17
Nonconservative Force Question--URGENT!
A skier traveling 12 m/s reaches the foot of a steady upward 18 degree incline and glides 12.2 meters up along this slope before coming to rest. What was the average coefficient of friction?
Work nonconservative= mechanicalenergyfinal-mechanicalenergyinitial
SO
Wnc= (1/2mvf^2+mghf)- (1/2mvi^2-mghi)
i used Ff= force of friction, and 3.7 coming from sin 18= (x/12.2)- 3.7 is the height of the incline plane.
m= mass
Ff*12.2= (9.8)(3.7)m-(1/2(12.0^2))m
and got Ff(12.2)= 35.74 m
so Ff= 2.92m, which means the coefficient of friction, u= 2.92m/9.8mcos18, the masses then cancelling out.
my question is, do you have to take the x-component of weight into account at all? does it affect the force of friction and thus the coefficient of friction? or do you just ignore the x-compent of weight, mgsin18?
any help would be very much appreciated, i've been debating this all weekend- thank you in advance!
Last edited: Sep 23, 2007
2. Sep 23, 2007
### learningphysics
Looks to me like everything is correct.
3. Sep 23, 2007
### perfectionist17
thank you very much, learningphysics! you have no idea how much better that makes me feel- i have a test tomorrow in ap physics! :-) thats what i thought, i just wasn't sure and i KNOW there'll be a problem like that on the test..
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https://www.zora.uzh.ch/id/eprint/34383/ | # Relativistic redshift effects and the Galactic-center stars
Angélil, R; Saha, P (2010). Relativistic redshift effects and the Galactic-center stars. Astrophysical Journal, 711(1):157-163.
## Abstract
The high pericenter velocities (up to a few percent of light) of the S stars around the Galactic-center black hole suggest that general relativistic effects may be detectable through the time variation of the redshift during pericenter passage. Previous work has computed post-Newtonian perturbations to the stellar orbits. We study the additional redshift effects due to perturbations of the light path (what one may call "post-Minkowskian" effects), a calculation that can be elegantly formulated as a boundary-value problem. The post-Newtonian and post-Minkowskian redshift effects are comparable: both are \mathcal O(\beta ^3) and amount to a few km s-1 at pericenter for the star S2. On the other hand, the post-Minkowskian redshift contribution of spin is \mathcal O(\beta ^5) and much smaller than the \mathcal O(\beta ^4) post-Newtonian effect, which would be ~0.1 km s-1 for S2.
## Abstract
The high pericenter velocities (up to a few percent of light) of the S stars around the Galactic-center black hole suggest that general relativistic effects may be detectable through the time variation of the redshift during pericenter passage. Previous work has computed post-Newtonian perturbations to the stellar orbits. We study the additional redshift effects due to perturbations of the light path (what one may call "post-Minkowskian" effects), a calculation that can be elegantly formulated as a boundary-value problem. The post-Newtonian and post-Minkowskian redshift effects are comparable: both are \mathcal O(\beta ^3) and amount to a few km s-1 at pericenter for the star S2. On the other hand, the post-Minkowskian redshift contribution of spin is \mathcal O(\beta ^5) and much smaller than the \mathcal O(\beta ^4) post-Newtonian effect, which would be ~0.1 km s-1 for S2.
## Statistics
### Citations
Dimensions.ai Metrics
31 citations in Web of Science®
33 citations in Scopus® | {"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": 0.9210054874420166, "perplexity": 3679.7092844633466}, "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/1606141193221.49/warc/CC-MAIN-20201127131802-20201127161802-00161.warc.gz"} |
http://kile.sourceforge.net/Documentation/html/intro_mainfeat.html | Prev Introduction Next
## Kile's Main Features
### QuickStart Wizard
The QuickStart wizard built into Kile is a useful feature to quickly start creating documents in Kile. Choosing the wizard from the menubar gives you several choices for the creation of your document. You can also specify some options related to the document right away.
Class options:
• Document Class: choose the type of document you want to create: article, book, letter, report, scrartcl, scrreprt, scrbook, prosper, beamer or other custom-defined.
• Typeface Size: tell Kile what point size (pt) you want to use.
• Paper Size: choose the size or style of sheets.
• Encoding: In general it is a good idea to use your systems standard encoding. Modern systems now move more and more to UTF-8 as the standard encoding. If you can, use utf8x (which is indeed the correct spelling for LATEX documents).
• Other options: this allows you to set further options such as printing, draft, and others.
Packages
This lists some of the most common additional packages used in LATEX. Select the checkbox to include it.
Document Properties
• Author: put your name here.
• Title: add the document title here.
• Date: specify the date.
### Predefined Templates
The predefined templates in Kile are:
• Empty document: real freaks start from scratch!
• Article: sets the article format, for a document short enough not to be broken down to chapters.
• Report: sets the report format, for a middle-sized document, with for example page numbering on the page's outer edge.
• Book: sets the book format, a full-fledged flavor, so powerful that it is used to write many university textbooks.
• Letter: sets the letter format, that can automatically do those nasty indentations that nobody really remembers.
• Beamer,HA-Prosper: create nice presentations in PDF with a superior look and all LATEX power.
• Scrartcl,Scrbook,Scrreprt,Scrlttr2: the KOMA-Script document classes, especially adapted to german typography. Use them whenever you write german texts.
New users need not to worry, this list is just a brief description of features, and more attention will be paid to complete these tasks in detail later in Chapter 3, Quickstart.
### Syntax Highlighting
Kile is similar to programs that deal with source code and editing, and will automatically highlight commands, options and items that are used (and abused). Kile makes it possible to spot easily problem areas: for example, if you see major areas of text turned green, it is likely that you forgot closing a math environment somewhere; you would have noticed anyway by how crappy the output file would have looked, but highlighting really saves you time and frustration.
### Auto-Completion of Environments
The auto-completion of environments means that, when you begin a new environment by typing \begin{environment}, Kile will automatically insert a matching \end{environment} command, with a line in between them for your text. You can of course deactivate it if you want in Settings->Configure Kile...->LaTeX+Environments.
All documents are normally structured in a hierarchy of some type. LATEX allows you to break up documents into the following hierarchy (part being highest in the hierarchy, and subparagraph being lowest):
• \part
• \chapter
• \section
• \subsection
• \subsubsection
• \paragraph
• \subparagraph
When viewing a document in the Structure view, you can jump between elements by clicking on the element you would like to view.
### Inverse Search
When creating your own LATEX files, inverse search can be very helpful. Once you have created a DVI file (DeVice Independent File), you can click the middle- mouse button in the DVI viewer and Kile will jump to the corresponding line in the LATEX source code.
A DVI is a type of file containing a description of a formatted document, along with other information including character font, and is besides PDF the usual output of TEX or LATEX. A number of utilities exist to view, convert and print DVI files on various systems and devices.
### Forward Search
When using inverse search, the selection of items in the DVI file is associated with the editor, so when you click on the DVI file, the main window jumps to the corresponding section of LATEX code in the editor. Forward search is the exact opposite of this. Forward search will allow you to click on a specific section of text in the LATEX code, and jump to the associated position in the DVI viewer window.
Prev Home Next LATEX 101 Up The Toolbar | {"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": 0.9278079271316528, "perplexity": 3901.2444283683813}, "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-2015-22/segments/1432207929023.5/warc/CC-MAIN-20150521113209-00060-ip-10-180-206-219.ec2.internal.warc.gz"} |
https://georgia-james.com/find-the-domain-and-range-fx-square-root-of-3x-1/ | # Find the Domain and Range f(x) = square root of 3x-1
Set the radicand in greater than or equal to to find where the expression is defined.
Solve for .
Add to both sides of the inequality.
Divide each term by and simplify.
Divide each term in by .
Cancel the common factor of .
Cancel the common factor.
Divide by .
The domain is all values of that make the expression defined.
Interval Notation:
Set-Builder Notation:
The range of an even indexed radical starts at its starting point and extends to infinity.
Interval Notation:
Set-Builder Notation:
Determine the domain and range.
Domain:
Range:
Find the Domain and Range f(x) = square root of 3x-1 | {"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": 0.8160669803619385, "perplexity": 1450.0891503604898}, "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-2022-40/segments/1664030335059.31/warc/CC-MAIN-20220927225413-20220928015413-00720.warc.gz"} |
http://mathhelpforum.com/differential-geometry/157629-complex-integration-print.html | # Complex integration
• September 27th 2010, 04:16 PM
Danneedshelp
Complex integration
Q: Show that $\oint_{C}\frac{1}{z-z_{0}}dz$ is equal to $2\pi\\i$ for every circle $C$ centered at $z_{0}$.
First, I wrote
$\oint_{C}\frac{1}{z-z_{0}}dz=\oint_{C}\frac{1}{z-z_{0}}\frac{\bar{z-z_{0}}}{\bar{(z-z_{0})}}dz=\oint_{C}\frac{(\bar{z}-\bar{z_{0}})}{|z-z_{0}|^{2}}dz$
Now,
$\bar{z}-\bar{z_{0}}=x-iy-x_{0}+iy_{0}=(x-x_{0})-i(y-y_{0})$.
So,
$(x-x_{0})-i(y-y_{0})(dx+idy)=(x-x_{0})dx-i(y-y_{0})dx+i(x-x_{0})dy+(y-y_{0})dy$.
Thus,
$\frac{1}{|z-z_{0}|^{2}}[\oint_{C}(x-x_{0})dx+(y-y_{0})dy+i(\oint_{C}-(y-y_{0})dx+(x-x_{0})dy)]$
Here is where I am stuck. Firstly, I am assuming I will need to use the parameterization $x=cos(t)$, $y=sin(t)$, with $0\leq\\t\leq\\2\pi$.
So, I want to integrate along $C=\{(x,y):(x-x_{0})^{2}+(y-y_{0})^{2}=r\}$. I am not sure how to use the parameters and integrate along this curve.
Any help would be appreciated.
Thanks
• September 27th 2010, 05:14 PM
Ackbeet
I don't think your general method is correct. I would parametrize the circle like this:
Let $z=z_{0}+r\,e^{i\theta},$ for $r$ constant and $\theta\in[0,2\pi].$ Thus, $dz=i\,r\,e^{i\theta}\,d\theta.$ What does your integral become?
• September 28th 2010, 12:44 PM
Danneedshelp
Quote:
Originally Posted by Ackbeet
I don't think your general method is correct. I would parametrize the circle like this:
Let $z=z_{0}+r\,e^{i\theta},$ for $r$ constant and $\theta\in[0,2\pi].$ Thus, $dz=i\,r\,e^{i\theta}\,d\theta.$ What does your integral become?
Thanks.
Even so, we have not covered that material and his hint was to use greens thereom. We have just been doing integrals by seperating the real and imagenary parts into two seperate integrals, the one with the imaginary integrand has a constant i out front.
• September 28th 2010, 01:23 PM
Ackbeet
You can do Green's Theorem. So, in reviewing your work, I'd say you're good up to here:
$\displaystyle{\frac{1}{|z-z_{0}|^{2}}\left[\oint_{C}[(x-x_{0})dx+(y-y_{0})dy]+i\oint_{C}[-(y-y_{0})dx+(x-x_{0})dy)]\right]}.$
Now, what you want to do is apply Green's Theorem on each integral. Green's theorem is going to reduce your integrals quite a bit. Recall that Green's Theorem states that
$\displaystyle{\oint_{C}(L\,dx+M\,dy)=\iint_{D}\lef t(\frac{\partial M}{\partial x}-\frac{\partial L}{\partial y}\right)dx\,dy}.$
Once you have constructed your double integrals, I'd recommend converting to polar coordinates. The result pops out fairly readily.
• September 28th 2010, 04:33 PM
Danneedshelp
Quote:
Originally Posted by Ackbeet
You can do Green's Theorem. So, in reviewing your work, I'd say you're good up to here:
$\displaystyle{\frac{1}{|z-z_{0}|^{2}}\left[\oint_{C}[(x-x_{0})dx+(y-y_{0})dy]+i\oint_{C}[-(y-y_{0})dx+(x-x_{0})dy)]\right]}.$
Now, what you want to do is apply Green's Theorem on each integral. Green's theorem is going to reduce your integrals quite a bit. Recall that Green's Theorem states that
$\displaystyle{\oint_{C}(L\,dx+M\,dy)=\iint_{D}\lef t(\frac{\partial M}{\partial x}-\frac{\partial L}{\partial y}\right)dx\,dy}.$
Once you have constructed your double integrals, I'd recommend converting to polar coordinates. The result pops out fairly readily.
Ahhhh, I see. The first integral will go to zero and I will end up with the integrand 2 after taking the partails of the other chunk and applying greens thereom. So, if convert to polar and integrate over the region $D=\{(r,\theta):0 I end up with I am looking for after distributing the $\frac{1}{|z-z_{0}|^{2}}$ and the $i$.
Thanks for the help, I appreciate it. I also carried out the method you mentioned in your first post and got the answer that way.
Thanks again.
• September 29th 2010, 05:47 AM
Ackbeet
Yep, that's exactly right. You're welcome, and have a good one! | {"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": 27, "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": 0.9023738503456116, "perplexity": 466.21749011055783}, "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-2015-27/segments/1435375098924.1/warc/CC-MAIN-20150627031818-00051-ip-10-179-60-89.ec2.internal.warc.gz"} |
https://link.springer.com/article/10.1186/s42787-020-00092-6?error=cookies_not_supported&error=cookies_not_supported&code=9daf4b64-ec23-4f57-a99b-f1c3c2288cdc&code=651bad53-a5a1-4950-b0d5-6842c3197206 | # Dot product graphs and domination number
## Abstract
Let A be a commutative ring with 1≠0 and R=A×A. The unit dot product graph of R is defined to be the undirected graph UD(R) with the multiplicative group of units in R, denoted by U(R), as its vertex set. Two distinct vertices x and y are adjacent if and only if x·y=0∈A, where x·y denotes the normal dot product of x and y. In 2016, Abdulla studied this graph when $$A=\mathbb {Z}_{n}$$, $$n \in \mathbb {N}$$, n≥2. Inspired by this idea, we study this graph when A has a finite multiplicative group of units. We define the congruence unit dot product graph of R to be the undirected graph CUD(R) with the congruent classes of the relation $$\thicksim$$ defined on R as its vertices. Also, we study the domination number of the total dot product graph of the ring $$R=\mathbb {Z}_{n}\times... \times \mathbb {Z}_{n}$$, k times and k<, where all elements of the ring are vertices and adjacency of two distinct vertices is the same as in UD(R). We find an upper bound of the domination number of this graph improving that found by Abdulla.
## Introduction and preliminaries
The idea of a zero-divisor graph of a commutative ring R was introduced by Beck in [1] (1988). He considered all the elements of R to be vertices and two distinct vertices x and y are adjacent if and only if xy=0, where xy denotes the multiplication in R. He was mainly interested in colorings. Beck’s work was continued by Anderson and Naseer in [2] (1993), where they gave a counterexample of Beck’s conjecture. In 1999, Livingston and Anderson in [3] gave a modified definition of the zero-divisor graph, denoted by Γ(R), by taking the nonzero zero-divisors of the ring as vertices and adjacency of two distinct vertices remains unchanged, i.e., two distinct nonzero zero-divisors x and y are adjacent if and only if xy=0. This definition became the standard definition of the zero-divisor graph. In the same year, they continued their work on the zero-divisor graphs with Frazier and Lauve in [4]. They studied the cliques which are complete subgraphs of Γ(R) and the relationship between graph isomorphisms and ring isomorphisms.
In 2003, Redmond in [5] introduced the ideal-based zero-divisor graph ΓI(R) with vertex set {xRI | xyI for some yRI}, where I is an ideal of R and two distinct vertices x and y are adjacent if and only if xyI. This graph is considered to be a generalization of zero-divisor graphs of rings. In 2002, Mulay in [6] provided the idea of the zero-divisor graph determined by equivalence classes. Later on, Spiroff and Wickham in [7] denoted this graph by ΓE(R) and compared it with Γ(R). This graph was called the compressed graph by Anderson and LaGrange in [8] (2012). In the compressed graph, the relation on R is given by $$r\thicksim s$$ if and only if ann(r)=ann(s), where ann(r)={vR | rv=0} is the annihilator of r. This relation is an equivalence relation on R. The vertex set of the compressed graph is the set of all equivalence classes induced by $$\thicksim$$ except the classes $$[0]_{\thicksim }$$ and $$[1]_{\thicksim }$$. The equivalence class of r is $$[r]_{\thicksim }=\{a\in R\,|\,r\thicksim a\}$$ and two distinct vertices $$[r]_{\thicksim }$$ and $$[s]_{\thicksim }$$ are adjacent if and only if rs=0. There have been other ways to associate a graph to a ring R. For surveys on the topic of zero-divisor graphs, see [9, 10].
In 2015, Badawi introduced in [11] the dot product graph associated with a commutative ring R. In 2016, his student Abdulla in his master thesis [12] introduced the unit dot product graph and the equivalence dot product graph on a commutative ring with 1≠0. We are interested here primarily in these graphs.
In 2016, Anderson and Lewis introduced the congruence-based zero-divisor graph in [13], which is a generalization of the zero-divisor graphs mentioned above. The vertices of this graph are the congruence classes of the nonzero zero-divisors of R induced by a congruence relation defined on the ring R. Two distinct vertices are adjacent if and only if their product is zero. The concept of congruence relation is used in this paper.
In 2017, Chebolu and Lockridge in [14] found all cardinal numbers occurring as the cardinality of the group of all units in a commutative ring with 1≠0. This is very helpful to us as we want to graph the units of a ring R.
In the second section, we generalize a result of [12] concerning the unit dot product graph of a commutative ring R, where $$R=\mathbb {Z}_{n} \times \mathbb {Z}_{n}$$, replacing $$\mathbb {Z}_{n}$$ by a a commutative ring A such that U(A) is finite. In the third section, a congruence relation on the unit dot product graph is defined and some of its properties are characterized. In the last section, we discuss the domination number of some graphs.
We recall some definitions which are used in this paper. Let G be an undirected graph. Two vertices v1 and v2 are said to be adjacent if v1,v2 are connected by an edge of G. A finite sequence of edges from a vertex v1 of G to a vertex v2 of G is called a path of G. We say that G is connected if there is a path between any two distinct vertices and it is totally disconnected if no two vertices in G are adjacent. For two vertices x and y in G, the distance between x and y, denoted by d(x,y), is defined to be the length of a shortest path from x to y, where d(x,x)=0 and d(x,y)= if there is no such path. The diameter of G is diam(G)= sup{d(x,y)|x and y are vertices in G}. A cycle of length n, n≥3, in G is a path of the form x1x2−...−xnx1, where xixj when ij. The girth of G, denoted by gr(G), is the length of the shortest cycle in G and gr(G)= if G contains no cycle. A graph G is said to be complete if any two distinct vertices are adjacent and the complete graph with n vertices is denoted by Kn. A complete bipartite graph is a graph which may be partitioned into two disjoint nonempty vertex sets A and B such that two distinct vertices are adjacent if and only if they are in distinct vertex sets. This graph is denoted by Km,n, where |A|=m and |B|=n.
Throughout the paper, R and A denote commutative rings with 1≠0. Its set of zero-divisors is denoted by Z(R) and Z(R)=Z(R)−{0}. As usual, $$\mathbb {Z}$$, $$\mathbb {Z}_{n}$$ and GF(pn) denote the integers, integers modulo n, and finite field with pn elements, respectively, where p is a prime number and n is a positive integer. ϕ(n) is the Euler phi function of a given positive integer n, which counts the positive integers up to n that are relatively prime to n.
## Unit dot product graph of a commutative ring
The unit dot product graph of R was introduced in [12], denoted by UD(R). This graph is a subgraph of the total dot product graph, denoted by TD(R), where its vertex set is all the elements of R. Some of its properties were characterized when R=A×A and $$A=\mathbb {Z}_{n}$$. In this section, we generalize the UD(R), as A will be a commutative ring with 1≠0, whose multiplicative group of units is finite.
In the proof of Theorems 2 and 3, we use the order of the multiplicative group of units U(R) of R. In this context, the following theorem is helpful.
### Theorem 1
(Th. 8, [14]) Let λ be a cardinal number. There exists a commutative ring R with |U(R)|=λ if and only if λ is equal to
1. 1.
An odd number of the form $$\prod _{i=1}^{t} (2^{n_{i}}-1)$$ for some positive integers n1,...,nt
2. 2.
An even number
3. 3.
An infinite cardinal number
We are interested only in commutative rings R=A×A, where A is a commutative ring with 1≠0 and U(A) has a finite order. For instance, from [14], rings in the form $$R_{2m}=\frac {\mathbb {Z}[x]}{(x^{2},\, mx)}$$ are examples of such a ring. Here, U(A) has an even order equal to 2 m, where $$m \in \mathbb {N}$$. The units in these rings are in the form 1+bx and −1+bx, 0≤bm−1. If the order of U(A) is odd, then this odd number will be in the form $$\prod _{i= 1}^{t} (2^{n_{i}}-1)$$ for some positive integers n1,...,nt and the characteristic of the ring must be equal to 2.
The following two Theorems 2 and 3 characterize the graph of the rings R=R2m×R2m and R=A×A, respectively.
### Theorem 2
Let R=R2m×R2m. Then, UD(R) is the union of m disjoint K2m,2m’s.
### Proof
Since |U(R2m)|=2m, then UD(R) has exactly 4m2 vertices. Let v1=u(1,a) and v2=v(1,b) in R, for some u,v,a,bU(R2m). From [14], the units are in the form 1+ax and −1+ax, where 0≤am−1, so we have v1=u(1,1+ax) and v2=v(1,−1+bx) in R for some u,vU(R2m), 0≤a,bm−1. Hence, v1 is adjacent to v2 if and only if v1·v2=uv(ba)x=0 in R2m. This is equivalent to b=a, since uv is a unit in R2m. Thus, for each 0≤am−1, let Va={u(1,1+ax) | uU(R2m)} and Wa={u(1,−1+ax) | uU(R2m)}. For different units u and u in U(R2m), we cannot have u(1,1+ax)=u(1,1+ax) or u(1,−1+ax)=u(1,−1+ax), so |Va|=|Wa|=2m. If u(1,1+ax)=u(1,−1+ax), then u=u and u(1+ax)=u(−1+ax). So u(1+ax)=u(−1+ax) which implies that u=−u, a contradiction. Thus, VaWa=. It is clear that every two distinct vertices in Va or in Wa are not adjacent. By construction of Va and Wa, every vertex in Va is adjacent to every vertex in Wa. Thus, the vertices in VaWa form the graph K2m,2m that is a complete bipartite subgraph of TD(R). By construction, UD(R) is the union of m disjoint K2m,2m’s. □
### Example 1
When m=1, we have $$R_{2}=\frac {Z[x]}{(x^{2},\,x)}$$ which is isomorphic to $$\mathbb {Z}$$. The graph of UD(R2×R2) will be a complete bipartite graph of 4 vertices which are (1,1), (1,-1), (-1,1), and (-1,-1). Thus, its diameter = 2 and girth = 4 (Fig. 1).
The following theorem deals with the case R=A×A, where |U(A)| is odd. In this case, the unit −1 in A (from Cauchy Theorem) must have order 1. Then, Char(A)=2.
### Theorem 3
Let R=A×A. If the order of the multiplicative group U(A) is odd, then UD(R) is the union of $$\frac {m-1}{2}$$ disjoint Km,m’s and one Km.
### Proof
From [14], the order of U(A) is an odd number if and only if this odd number is of the form $$\prod _{i=1}^{t} (2^{n_{i}}-1)$$ for some positive integers n1,...,nt. Let m be the odd order of U(A), so UD(R) has exactly m2 vertices. Let v1=u(1,a) and v2=v(1,b) in R, for some u,v,a,bU(A). v1 is adjacent to v2 if and only if v1·v2=uv+uvab=0. This will occur if and only if 1+ab=0. This is equivalent to a=b−1, since uv is a unit in A and Char(R)=2. Thus, for each, a≠1∈U(A), let Va={u(1,a) | uU(A)} and Wa={u(1,a−1) | uU(A)}. For different units u and u in U(A), we cannot have u(1,a)=u(1,a−1), so |Va|=|Wa|=m. If u(1,a)=u(1,a−1), then u=u and ua=ua−1. So, u(aa−1)=0, i.e. a=a−1, which implies that a2=1 a contradiction since U(A) has an odd order. Thus, VaWa=. It is clear that every two distinct vertices in Va or in Wa are not adjacent. By construction of Va and Wa, every vertex in Va is adjacent to every vertex in Wa. Thus, the vertices in VaWa form the graph Km,m that is a complete bipartite subgraph of TD(R). By construction, there are exactly $$\frac {m-1}{2}$$ disjoint complete bipartite Km,m subgraphs of TD(R). For a=1, we have m vertices in the form of u(a,a). Since Char(R)=2, these m vertices form the graph Km, that is a complete subgraph of TD(R). Hence, UD(R) is the union of $$\frac {m-1}{2}$$ disjoint Km,m’s and one Km. □
## Congruence dot product graph of a commutative ring
In 2016, Anderson and Lewis in [13] introduced the congruence-based zero-divisor graph $$\Gamma _{\thicksim }(R)=\Gamma (R/\thicksim)$$, where $$\thicksim$$ is a multiplicative congruence relation on R and showed that $$R/\thicksim$$ is a commutative semigroup with zero. They showed that the zero-divisor graph of R, the compressed zero-divisor graph of R, and the ideal based zero-divisor graph of R are examples of the congruence-based zero-divisor graphs of R. In this paper, we are interested in the multiplicative congruence relation $$\thicksim$$ on R, which is an equivalence relation on the multiplicative monoid R with the additional property that if x,y,z,wR with $$x\thicksim y$$ and $$z\thicksim w$$, then $$xz\thicksim yw$$.
The equivalence unit dot product graph of U(R) was introduced in [12], where R=A×A and $$A=\mathbb {Z}_{n}$$. The equivalence relation $$\thicksim$$ on U(R) was defined such that $$x\thicksim y$$, where x,yU(R), if x=(c,c)y for some (c,c)∈U(R). Let EU(R) be the set of all distinct equivalence classes of U(R). If XEU(R), then ∃ aU(A) such that $$X=[(1,a)]_{\thicksim }=\{u(1,a)\,|\,u\in U(A)\}$$. Thus, the equivalence unit dot product graph of U(R) is the (undirected) graph EUD(R) with vertices EU(R). Two distinct vertices X and Y are adjacent if and only if x·y=0∈A, where x·y denotes the normal dot product of x and y.
From the definition of the congruence relation, we find that the relation defined by Abdulla is not only an equivalence relation but also a congruence relation. In fact, let $$x \thicksim y$$ and $$w \thicksim v$$. So, x=(c1,c1)y and w=(c2,c2)v for some (c1,c1),(c2,c2)∈U(R). Then, xw=(c1,c1)y(c2,c2)v=(c1,c1)(c2,c2)yv=(c,c)yv and hence $$xw \thicksim yv$$. We denote this congruence unit dot product graph by CUD(R), and its set of vertices is the set of all distinct congruence classes of U(R), denoted by CU(R).
In this section, we characterize the generalized case of the congruence unit dot product graph CUD(R), as we will apply the congruence relation on the unit dot product graph we introduced in the first section.
### Theorem 4
Let R=R2m×R2m. Then, CUD(R) is the union of m disjoint K1,1’s.
### Proof
For each aU(R2m), let Va and Wa be as in the proof of Theorem 2. Then, Va,WaCU(R). Indeed, for each aU(R2m), there exist Va and WaCU(R) each has cardinality 2m. We conclude that each K2m,2m of UD(R) is a K1,1 of CUD(R). From Theorem 2 the result follows. □
### Example 2
In Example 1, we graphed the unit dot product graph of R2×R2, and now, we graph the congruence dot product graph of the same ring. This graph will be a complete graph of 2 vertices as R2 is isomorphic to $$\mathbb {Z}$$. So, we will have only two congruence classes $$[(1,1)]_{\thicksim }=\{(1,1),(-1,-1)\}$$ and $$[(1,-1)]_{\thicksim }=\{(1,-1),(-1,1)\}$$ (Fig. 2).
### Theorem 5
Let R=A×A. If the order of U(A) is odd, then CUD(R) is the union of $$\frac {m-1}{2}$$ disjoint K1,1’s and one K1.
### Proof
For each aU(A), let Va and Wa be as in the proof of Theorem 3. Then, Va,WaCU(R). Indeed, for each aU(R) and a≠1, there exist Va and WaCU(R) each of cardinality m. For a=1, we have one congruence class V, where V={u(a,a) | uU(A)}. We conclude that each Km,m of UD(R) is a K1,1 of CUD(R), and each Km of UD(R) is a K1 of CUD(R). From Theorem 3, the result follows. □
Let $$R=\mathbb {Z}_{n}\times \mathbb {Z}_{n}$$. We make a little change on the congruence relation defined above by taking the vertices from the whole ring R not only from U(R). Define a relation on R such that $$x \thicksim y$$, where x,yR, if x=(c,c)y for some (c,c)∈U(R). It is clear that $$\thicksim$$ is an equivalence relation on R and also a congruence relation.
The congruence total dot product graph of R is defined to be the undirected graph CTD(R), and its vertices are the congruent classes of all the elements of R induced by the defined congruence relation $$\thicksim$$. Two distinct classes $$[X]_{\thicksim }$$ and $$[Y]_{\thicksim }$$ are adjacent if and only if $$x\cdot y= 0\in \mathbb {Z}_{n}$$, where x·y denotes the normal dot product of x and y. Also, the congruence zero-divisor dot product graph, denoted by CZD(R), is defined to be an undirected graph whose vertices are the congruent classes of the nonzero zero-divisor elements in R and adjacency between distinct vertices remains as defined before.
Obviously, this congruence relation is well-defined. Indeed, let x,x,y,yR be such that y=(y1,y2) and y=(y1′,y2′) and let u,uU(R) be such that u=(c1,c1) and u=(c1′,c1′), where $$y_{1}, y_{1}', y_{2}, y_{2}', c_{1}, c_ 1' \in \mathbb {Z}_{n}$$. Assume that $$x\thicksim y$$ and $$x'\thicksim y'$$. Then, x·x=0 if and only if (c1y1)(c1′y1′)+(c1y2)(c1′y2′)=0. This happens if and only if y1y1′+y2y2′=0, since c1c1′ is a unit in $$\mathbb {Z}_{n}$$.
### Theorem 6
Let $$A=\mathbb {Z}_{p}$$, where p is a prime number and R=A×A. Then, CTD(R) is disconnected and $$CZD(R)=\Gamma _{\thicksim }(R)$$ is a complete graph of 2 vertices.
### Proof
If CTD(R) was connected, then ∃ x,yR such that x is adjacent to y. x·y=0 if and only if xy=0, leads to a contradiction with (Theorem 2.1, [11]). So, $$CZD(R)=\Gamma _{\thicksim }(R)$$ is connected. Since A is a field, then all the nonzero zero-divisors in R will be in two classes only, which are $$[(a,0)]_{\thicksim }$$ and $$[(0,b)]_{\thicksim }$$, ∀ a,bU(A) and since (a,0)·(0,b)=0, so it is a complete graph of two vertices. □
If $$A=\mathbb {Z}_{p}$$ and $$R=\mathbb {Z}_{p}\times... \times \mathbb {Z}_{p}$$, k times and k<, then the diameter and girth of CZD(R) and CTD(R) are the same as the case of TD(R) and ZD(R), which was discussed before in [11]. This reduces the number of vertices but adjacency is the same in both cases.
### Example 3
If $$A=\mathbb {Z}$$, then R=$$\mathbb {Z}\times \mathbb {Z}$$. Here, the only units in the form (c,c) are (1,1) and (-1,-1) so the classes of the zero-divisors will be in the form $$[(a,0)]_{\thicksim }=\{(a,0),(-a,0)\}$$ and $$[(0,a)]_{\thicksim }=\{(0,a),(0,-a)\}$$, ∀ aU(A). For two distinct vertices (a,0)·(b,0)≠0, because ab≠0. Then, there will be an edge only between classes in the form $$[(a,0)]_{\thicksim }$$ and $$[(0,b)]_{\thicksim }$$, which means diam(CZD(R))=2 and gr(CZD(R))=4.
### Theorem 7
Let $$R=\mathbb {Z}_{n}\times \mathbb {Z}_{n}$$ for $$n\in \mathbb {N}$$ and n is not a prime number. Then, CTD(R) is a connected graph with diam(CTD(R))=3 and gr(CTD(R))=3.
### Proof
The proof is similar to that of Theorem 2.3 [11], taking into consideration that the vertices we used are in distinct classes. □
## Domination number
Let G be a graph with V as its set of vertices. We recall that a subset SV is called a dominating set of G if every vertex in V is either in S or is adjacent to a vertex in S. The domination number γ(G) of G is the minimum cardinality among the dominating sets of G. The study of the domination number started around 1960s; however, there are some domination-related problems before that date. Namely, about 100 years earlier, in 1862, De Jaenisch [15] posed the problem of finding the minimum number of queens required to cover (attack) each square of an n×n chess board. In 1892, there were three basic types of problems that chess players studied during this time reported by Rouse Ball in [16]. For more details on this topic, see [17, 18].
In this section, we find a new upper bound of the domination number of the total dot product graph of $$\mathbb {Z}_{n}\times...\times \mathbb {Z}_{n}$$, k times and k<. It is an improvement of the upper bound of the same graph given in [12].
### Theorem 8
Let n≥4 be an integer that is not prime, $$A=\mathbb {Z}_{n}$$ and R=A×A. Then, write $$n= p_{1}^{k_{1}}...p_{m}^{k_{m}}$$, where pis,1≤im, are distinct prime numbers. Let $$M=\{\frac {n}{p_{i}}\,|\,1 \leq i\leq m\}$$. Then
1. 1.
If n is even, then $$D=\{(0,b)\,|\,b\in M\} \cup \{(d,0)\,|\,d\in M\} \cup \{(\frac {n}{2},\frac {n}{2})\}$$ is a minimal dominating set of TD(R), and thus, γ(TD(R))≤2m+1.
2. 2.
If n is odd, then D={(0,b) | bM}∪{(d,0) | dM}∪{(1,c) | cU(A))} is a minimal dominating set of TD(R), and thus, γ(TD(R))≤2m+ϕ(n).
### Proof
1. 1.
Let n be even and x=(x1,x2) a vertex in TD(R). We consider two cases:
1. (a)
Assume that x is a unit. Since $$(x_{1},x_{2})\cdot (\frac {n}{2},\frac {n}{2})=\frac {n}{2}(x_{1}+x_{2})= nc'= 0$$ (because x1+x2 is an even number), then x is adjacent to a vertex in D,
2. (b)
Assume that x2 is a zero-divisor of A, i.e., pi|x2 in A for some pi,1≤im. Then, $$v=(0,\frac {n}{p_{i}})\in D$$ is adjacent to x in TD(R) (the same case is true if x1 is a zero-divisor of A).
This shows that D is a dominating set of TD(R). We show that it is minimal. We have to find when $$(a,p_{i})\cdot (\frac {n}{2},\frac {n}{2})= 0,\ a\in A$$. It is clear that if both a and pi are even or odd together, then $$(a,p_{i})\cdot (\frac {n}{2},\frac {n}{2})=0$$. But if a is even and pi is odd or the opposite, we have $$(a,p_{i})\cdot (\frac {n}{2},\frac {n}{2})\neq 0$$. Now, when we remove the vertex $$(0,\frac {n}{p_{i}})$$, the vertex (a,pi) is not adjacent to any other vertex in D, where a and pi are different. The same argument holds if we remove the vertex $$(\frac {n}{p_{i}},0)$$. Moreover, when we remove the vertex $$(\frac {n}{2},\frac {n}{2})$$, the vertex u=(u1,u2) which is a unit will not be adjacent to any other vertex in D. Thus, D is a minimal dominating set and therefore γ(CTD(R))≤2m+1.
2. 2.
Let n be odd, ϕ(n)=r and x=(x1,x2) a vertex in TD(R). We consider two cases:
1. (a)
Assume that x is a unit. From Theorem 3.3 parts 2 & 3 [12], UD(R) is either a union of $$\frac {r}{2}$$ disjoint Kr, r’s or a union of $$\frac {r}{2}-2^{m-1}$$ disjoint Kr, r and 2m disjoint Kr’s. In both graphs, every unit (1,c) is adjacent to r units in the form u(1,c) for all uU(A),
2. (b)
Assume that x2 is a zero-divisor of A, i.e., pi|x2 in A for some pi,1≤im. Then, $$v=(0,\frac {n}{p_{i}})\in D$$ is adjacent to x in TD(R) (the same case takes place if x1 is a zero-divisor of A).
This shows that D is a dominating set of TD(R). In order to show that it is minimal, let us first remove the vertex $$v=(0,\frac {n}{p_{i}})$$ from D for some i, 1≤im. We have (a,pi)·(1,c)=a+cpi=0 if and only if a=−cpi. So, when we remove $$(0,\frac {n}{p_{i}})$$, we will find a vertex (a,pi) for some aA which is not adjacent to any other vertices in D (as an example, take a=1). Thus, v cannot be removed from D. The same argument is true if we remove $$(\frac {n}{p_{i}},0)$$. If we remove the unit (1,c), we will have distinct r units that are not adjacent to any other vertex in D. Thus, D is a minimal dominating set, and then, γ(TD(R))≤2m+ϕ(n).
### Example 4
Let $$R=\mathbb {Z}_{4}\times \mathbb {Z}_{4}$$. As a result of part 1 in the previous theorem, γ(TD(R))≤2×1+1=3. The following figure shows that the dominating set of R is {(0,2),(2,0),(2,2)} (Fig. 3).
We note that the upper bound of the domination number of the congruence total dot product graph of $$\mathbb {Z}_{n}\times \mathbb {Z}_{n}$$ is the same as the previous result of the total dot product graph, taking into consideration that the vertices we used are in distinct classes.
### Example 5
Let $$R=\mathbb {Z}_{4}\times \mathbb {Z}_{4}$$ as in example 4. The vertices of CTD(R) are the congruence classes $$[(0,1)]_{\thicksim }=\{(0,1),(0,3)\}$$, $$[(0,2)]_{\thicksim }=\{(0,2)\}$$, $$[(1,0)]_{\thicksim }=\{(1,0),(3,0)\}$$, $$[(1,1)]_{\thicksim }=\{(1,1),(3,3)\}$$, $$[(1,2)]_{\thicksim }=\{(1,2),(3,2)\}$$, $$[(1,3)]_{\thicksim }=\{(1,3),(3,1)\}$$, $$[(2,0)]_{\thicksim }=\{(2,0)\}$$, $$[(2,1)]_{\thicksim }=\{(2,1),(2,3)\}$$, $$[(2,2)]_{\thicksim }=\{(2,2)\}$$. The following graph shows that γ(CTD(R))=3 and its dominating set is $$\{ [(0,2)]_{\thicksim }, [(2,0)]_{\thicksim }, [(2,2)]_{\thicksim }\}$$ (Fig. 4).
The following corollary is a generalization of Theorem 8 when $$R=\mathbb {Z}_{n}\times...\times \mathbb {Z}_{n}$$, k times, k< and n is even.
### Corollary 1
Let n≥4 be an even integer, $$A=\mathbb {Z}_{n}$$ and R=A×...×A, k times and k<. Then, write $$n= p_{1}^{k_{1}}...p_{m}^{k_{m}}$$, where pis,1≤im, are distinct prime numbers. Let $$M=\{\frac {n}{p_{i}}\,|\,1 \leq i\leq m\}$$. Then,$$D=\{(0,...,0,b)\,|\,b\in M\} \cup \{(d,0,...,0)\,|\,d\in M\} \cup \{(\frac {n}{2},0,...,0,\frac {n}{2})\}$$ is a minimal dominating set of TD(R), and thus, γ(TD(R))≤2m+1.
### Proof
Let x=(x1,...,xk) be a vertex in TD(R). We consider two cases:
1. 1.
Assume that x is a unit, i.e., each coordinate is an odd number. Then, $$(x_{1},...,x_{k})\cdot (\frac {n}{2},0,...,0,\frac {n}{2})=0$$,
2. 2.
Assume that xi is a zero-divisor of A, 1≤in. If i=1 or n, then x is adjacent to $$(0,...,0,\frac {n}{p_{i}})$$ or $$(\frac {n}{p_{i}},0,...,0)$$ and both of them are in D. But if i≠1 or in such that x1 and xn are units, then it is adjacent to $$(\frac {n}{2},0,...,0,\frac {n}{2})\in D$$.
This shows that D is a dominating set of TD(R). We show that it is minimal. Since $$(a,0,0,p_{i})\cdot (\frac {n}{2},0,...,0,\frac {n}{2})=\frac {n}{2} (a+p_{i})$$ for some aA, which is equal to zero if and only if a and pi are odd or even together. By removing $$(0,...,0,\frac {n}{p_{i}})$$ from D, we will find a vertex (a,0,0,pi) in TD(R) that is not adjacent to any other vertex in D. The same argument works if we remove $$(\frac {n}{p_{i}},0,...,0)$$ from D. Also, by removing $$(\frac {n}{2},0,...,0,\frac {n}{2})$$, we will find a vertex in TD(R) where the first and the last coordinates are units that is not adjacent to any other vertex in D. Thus, D is a minimal dominating set, and then, γ(TD(R))≤2m+1. □
Again here, we note that the upper bound of the domination number of the congruence total dot product graph of $$\mathbb {Z}_{n}\times...\times \mathbb {Z}_{n}$$ is the same as the previous result of the total dot product graph, taking into account that the vertices we used are in distinct classes.
### Theorem 9
For the unit dot product graph of R:
1. 1.
If R=R2m×R2m, then D={(1,a) | aU(R2m)} and γ(UD(R))=2m.
2. 2.
If R=A×A, where A is a commutative ring with 1≠0 and |U(A)| is an odd number, say m, then D={(1,c) | cU(A)} and γ(UD(R))=m.
### Proof
1. 1.
Let x=(c1,c2) be a vertex in UD(R) and assume that x is not in D. Let $$c=c_{1}c_{2}^{-1} \in U(R_{2m})$$. Then, (c1,c2) is adjacent to (1,c) in D. Assume that (1,c) is removed from D for some c in U(R2m). Then, (−c,1) is not adjacent to any other vertex in D. Hence, D is a minimal dominating set, and thus, γ(UD(R))=2m.
2. 2.
Holds by the same idea of the proof of part 1.
The following theorem is a direct consequence of Theorem 6 in this paper. In that theorem, we find that CTD(R) is a complete graph of 2 vertices, and as a result its dominating set contains only one vertex of these 2 vertices.
### Theorem 10
Let p be a prime number, n≥1, m=pn−1, A=GF(pn) and R=A×A. Then γ(CZD(R))=1.
## Conclusion
In our future work, we are looking forward to working on one of the following open questions:
1. 1.
In the first section, we studied the case when R=R2m×R2m. We are interested in defining the unit dot product graph in the general case when R=R2m×...×R2m, n times and n<. For n odd, the simplest case is R=R2m×...×R2m, (i.e., m=1). By straight forward calculations, the unit dot product graph will be isolated vertices. But the case when m≥2 and n is odd is still an open question. For even n, the case is more complicated.
2. 2.
Define the unit dot product graph on a commutative ring R=A×A relaxing the conditions on the ring A and characterizing the case when |U(A)| is infinite.
3. 3.
Determine the domination number for all of the previous cases using the results of this paper.
## Availability of data and materials
Data sharing is not applicable to this article as no datasets were generated or analyzed during the current study.
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Not applicable
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Correspondence to Dina Saleh.
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Saleh, D., Megahed, N. Dot product graphs and domination number. J Egypt Math Soc 28, 31 (2020). https://doi.org/10.1186/s42787-020-00092-6
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### Keywords
• Unit dot product graph
• Congruence dot product graph
• Congruence total dot product graph
• Domination number
### AMS Subject Classification
• Primary 05C25; secondary 13A99
• 05C69 | {"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": 0.9217565655708313, "perplexity": 762.5648048131465}, "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/1606141686635.62/warc/CC-MAIN-20201202021743-20201202051743-00238.warc.gz"} |
http://mathhelpforum.com/calculus/23409-analysis-print.html | # analysis
• November 24th 2007, 01:39 PM
eigenvector11
analysis
Let S be a nonempty set of real numbers that is bounded above, and let Beta be the least upper bound of S. Prove that for every Epsilon greater than 0, there exists an element x such that x is greater than Beta minus Epsilon.
Any ideas or help would be appreciated. Thanks.
• November 24th 2007, 01:53 PM
ThePerfectHacker
Quote:
Originally Posted by eigenvector11
Let S be a nonempty set of real numbers that is bounded above, and let Beta be the least upper bound of S. Prove that for every Epsilon greater than 0, there exists an element x such that x is greater than Beta minus Epsilon.
Any ideas or help would be appreciated. Thanks.
If $\beta$ is a least upper bound then $\beta - \epsilon$ cannot be an upper bound on the set if $\epsilon > 0$. Why?
• November 24th 2007, 02:02 PM
Plato
The statement that $\lambda$ is not an upper bound of S means that $\left( {\exists x \in S} \right)\left[ {\lambda < x} \right]$. That is some number in S is strictly greater that $\lambda$.
The statement that $\beta$ is the least upper bound of S means that $\beta$ is an upper bound of S no number less than $\beta$ is an upper bound.
If $\varepsilon > 0$ then $\beta - \varepsilon < \beta$.
Now you put those three facts together to form a proof of the proposition. | {"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": 11, "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": 0.9922861456871033, "perplexity": 113.63311284851943}, "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-2015-14/segments/1427131299496.98/warc/CC-MAIN-20150323172139-00196-ip-10-168-14-71.ec2.internal.warc.gz"} |
http://nzjm.math.auckland.ac.nz/index.php/On_Extension_and_Structure_of_Generalized_Derivations | # On Extension and Structure of Generalized Derivations
### New Zealand Journal of Mathematics
Vol. 42, (2012), Pages 83-90
Carlos Peña
UNCPBA,
NUCOMPA
Abstract We are concerned with the extension problem and the nature of ge-neralized derivations of an underlying associative algebra $\mathcal{A}$ on a field k to the corresponding double centralizer algebra and to the unitized algebra. This investigation will be made in the frame of some general associative algebras. We shall also seek on the continuity of generalized derivations on normed algebras.
Keywords Semiprime ring, separating space of a linear map.
Classification (MSC2000) Primary: 16W25, Secondary: 47B47 | {"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": 0.9362326264381409, "perplexity": 1120.1560034069814}, "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/1505818687766.41/warc/CC-MAIN-20170921115822-20170921135822-00200.warc.gz"} |
https://planetmath.org/ExampleOfFalseImpliesTrue | # example of false implies true
A common mistake in writing proofs is to overlook that false can imply true. Consider the following example:
###### Claim 1.
$2=5$.
###### False proof..
$2=5\Rightarrow 0\cdot 2=0\cdot 5\Rightarrow 0=0$. Clearly $0=0$ so $2=5$. ∎
Of course this is wrong and $2\neq 5$. The mistake is that we start from a false premise, that of $2=5$, and then use completely correct methods (multiply an equation on both sides by the same number) to arrive a true statement, that $0=0$. That process is logical: false can imply true. However, the mistake in the “proof” is to think that only true can imply true, as then we are lead to believe that $0=0$ makes $2=5$.
This line of argument is absurd in the example just given, but it is often hard to spot and many times a good looking proof is hiding such a mistake. For example:
###### Claim 2.
$\displaystyle\sum_{k=1}^{n}k=\frac{n(n+1)}{2}$ for all integers $n\geq 1$.
###### Incorrect proof..
We proceed by induction. In the base case $n=1$:
$\displaystyle\sum_{k=1}^{1}k$ $\displaystyle=\frac{1(1+1)}{2}$ $\displaystyle 1$ $\displaystyle=1.$
Therefore the base case is true.
Induction hypothesis: suppose that for some $n\geq 1$, $\displaystyle\sum_{k=1}^{n}k=\frac{n(n+1)}{2}$. Now consider the case $n+1$.
$\displaystyle\sum_{k=1}^{n+1}k$ $\displaystyle=\frac{(n+1)((n+1)+1)}{2}$ $\displaystyle(n+1)+\sum_{k=1}^{n}k$ $\displaystyle=\frac{(n+1)(n+2)}{2}$ $\displaystyle(n+1)+\frac{n(n+1)}{2}$ $\displaystyle=\frac{n^{2}+3n+2}{2}$ $\displaystyle\frac{2n+2}{2}+\frac{n^{2}+n}{2}$ $\displaystyle=\frac{n^{2}+3n+2}{2}$ $\displaystyle\frac{n^{2}+3n+2}{2}$ $\displaystyle=\frac{n^{2}+3n+2}{2}.$
So by induction the theorem is true for all $n\geq 1$. ∎
Notice both the base case and the inductive step assume what is to be proved and then using only logical implications arrive at a place where something is obviously true, such as $1=1$ or $\frac{n^{2}+3n+2}{2}=\frac{n^{2}+3n+2}{2}$. What makes these both incorrect is that:
We may arrive at truth by starting with either truth or fiction!
That is the way we define implication (at least in logic).
Some common signs of this mistake include:
1. 1.
The proof begins by assuming what is to be proved.
The exception to this is a proof by contradiction. Then the proof will end with something false. Implications do not allow this: true must imply only true. Hence it is accurate to then conclude that the assumption at the start was false.
Because of the possible confusion, it is almost always necessary to start a proof by contradiction with a phrase of the form ”Suppose not” or ”We proceed by contradiction”.
2. 2.
The proof ends with something that is always true, such as $1=1$ or $0=0$ or $x=x$.
3. 3.
The proof cannot be read aloud.
This is an often overlooked tool in writing good proofs. Try to insert words like “implies” or “ from this it follows that ” for symbols like $\rightarrow$ or $\Rightarrow$ or (even just new lines). If it sounds silly when read aloud in this way the chances are high that something is wrong. For example, the base case in the inductive proof above could be read as:
The sum from $k=1$ to $1$ of $k$ equals $\frac{1(1+1)}{2}$ implies that $1=1$.
Think: isn’t $1=1$ obvious? Why would that need to be implied by some ugly formula? Ah, it would not need to be. Perhaps the implication is not meaningful.
Correcting the mistake
For many problems the mistaken proof is nevertheless a good way to begin thinking about a correct proof. This is because by assuming what is to be proved one has something to begin with. By simplifying the problem down to something recognizably true, for example $1=1$, it then seems certain that the equation is correct. To prove the result it is often a matter of only reversing the scratch work. To see this we use the ingredients of the incorrect proof of $\sum_{k=1}^{n}k=\frac{n(n+1)}{2}$ in the following correct proof. Notice also that we have included many more words and punctuation to the proof which makes it both readable and also helps us to understand the logic of the proof more easily.
###### Proof.
We proceed by induction. In the base case $n=1$ and so on the left hand side of our equation we find that $\sum_{k=1}^{n}k=1$. On the right hand side we have $\frac{1(1+1)}{2}=1$. As the left hand side equals the right hand side, the base case is settled.
Now suppose for induction that $\sum_{k=1}^{n}k=\frac{n(n+1)}{2}$ for some integer $n\geq 1$. We then consider the case of the equation for $n+1$. On the left hand side we have: $\sum_{k=1}^{n+1}k=(n+1)+\sum_{k=1}^{n}k=n+1+\frac{n(n+1)}{2}$ with that last step applying the induction hypothesis. From here we use basic algebra to see that $n+1+\frac{n(n+1)}{2}=\frac{(n+1)(n+2)}{2}=\frac{(n+1)((n+1)+1)}{2}$. Hence, by induction the equation holds for all $n\geq 1$. ∎
Title example of false implies true ExampleOfFalseImpliesTrue 2013-03-22 18:49:52 2013-03-22 18:49:52 Algeboy (12884) Algeboy (12884) 9 Algeboy (12884) Example msc 03B05 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 54, "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": 0.9242063760757446, "perplexity": 266.95394873478017}, "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/1585370495413.19/warc/CC-MAIN-20200329171027-20200329201027-00170.warc.gz"} |
http://math.stackexchange.com/questions/24797/curvature-and-intersection-of-convex-functions | # curvature and intersection of convex functions
Take two weakly convex, weakly increasing non-negative functions, $g(x)$ and $h(x)$, domain of $x$ is $[a,d]$, $g(x)=0$ for $x \in [a,b]$, $h(x)=0$ for $x \in [a,c]$, and $g(d)=h(d)$. So $g$ and $h$ start at zero, stay at zero over some (possibly different) domain(s), and intersect again at $d$. Can we make a general statement on the relative convexities and slopes of $g$ and $h$ that achieves a weak ranking of $g$ and $h$ over all $x$ in $[a,d]$? i.e. $g(x)\geq h(x)$?
Clearly if $b<c$ and $g'(x)\leq h'(x)$ for $x \in [c,d]$ that will do it, but i'm more interested in situations where the first derivatives can't necessarily be ranked in this way (and, preferably, where we don't have to invoke a condition on $b$ or $c$). Intuitively it seems like there must be some condition on the scaled second derivative? like on $g''(x)/g'(x)$ vs $h''(x)/h'(x)$?
-
For future reference, please use the LaTeX capabilities of this website for displayed mathematics. Also please double check your question after you posted it: a stray, unescaped less than sign (<) can "chop off" the tail end of your post. – Willie Wong Mar 3 '11 at 11:41
A pointwise condition on the scaled second derivative is actually not really weaker than a condition on the first derivative. If you have a comparison result $g''/g' \leq h''/h'$, note that $f''/f' = (\log f')'$, you can integrate and get $g' \leq C h'$ using monotonicity of $\log$. In any case, your question feels a bit open-ended, it would help if you give some motivation or some more detailed restrictions on what properties of $g$ and $h$ you are allowed to use. – Willie Wong Mar 3 '11 at 11:52
Also, the $b<c$ condition is unnecessary: if $b > c$ and $g'(x) \leq h'(x)$ for $x \in [c,d]$, then you get a contradiction to the assumption that $h(y) \neq 0$ for some $y\in (c,b]$ and $g(d) = h(d)$. So you can't really object to that condition on the basis of it depending on ranking of $b$ versus $c$. – Willie Wong Mar 3 '11 at 11:57
Thanks very much. Sorry, I can't really give a detailed motivation without being very long-winded. I take your point that the scaled 2nd derivative is equivalent to a condition on the first. But the latter is not constructive since $C$ depends on $g$ and $h$. Is it true that the condition on the scaled second derivative achieves a ranking of g and h? If not, does it help that $g'(x), h'(x)\in[0,1]\forall x$? – user7728 Mar 7 '11 at 10:47 | {"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": 0.9521886110305786, "perplexity": 193.40019568825207}, "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-2014-10/segments/1394010638293/warc/CC-MAIN-20140305091038-00051-ip-10-183-142-35.ec2.internal.warc.gz"} |
https://tex.stackexchange.com/questions/491488/how-to-compact-text-up-arrow-in-tikzcd | # how to compact text up arrow in tikzcd
How to compact the size of text written upper an arrow in tikz-cd?
\documentclass{article}
\usepackage{amsmath}
\usepackage{tikz-cd}
\usetikzlibrary{arrows, matrix}
\tikzcdset{every label/.append style = {font = \tiny}}
\begin{document}
$\begin{tikzcd} 0\rar & \mathrm{Hom}_R(L, E) \arrow{r}{\mathrm{Hom}(\beta, \mathrm{id}_E)} & \mathrm{Hom}_R(N, E) \arrow{r}{\mathrm{Hom}(\alpha, \mathrm{id}_E)} & \mathrm{Hom}_R(M, E) \end{tikzcd}$
\end{document}
You can scale it to any size
\documentclass{article}
\usepackage{amsmath}
\usepackage{tikz-cd}
\usetikzlibrary{arrows, matrix}
\tikzcdset{every label/.append style = {scale=0.5,yshift=0.2ex}}
\begin{document}
$\begin{tikzcd} 0\rar & \mathrm{Hom}_R(L, E) \arrow{r}{\mathrm{Hom}(\beta, \mathrm{id}_E)} & \mathrm{Hom}_R(N, E) \arrow{r}{\mathrm{Hom}(\alpha, \mathrm{id}_E)} & \mathrm{Hom}_R(M, E) \end{tikzcd}$
\end{document}
but I personally would rather increase the distance between the columns as suggested by Joule V.
• thank you all. How ever i mean how to reduce the distance between the letters in for all text upper the arrow. May 18, 2019 at 14:40
• @H.Gorbanzad Do you really think that there is substantial space between the letters? You could do \tikzcdset{every label/.append style = {xscale=0.5,yshift=0.2ex}} but this doesn't look good IMHO.
– user121799
May 18, 2019 at 14:43
• Dear friend i really mean to reduce the space between the letters. May 18, 2019 at 14:55
• Is it possible? May 18, 2019 at 15:07
• @H.Gorbanzad How about \documentclass{article} \usepackage{amsmath} \usepackage{tikz-cd} \usetikzlibrary{arrows, matrix} \tikzcdset{every label/.append style = {scale=0.6,yshift=0.2ex}} \begin{document} $\begin{tikzcd} 0\rar & \mathrm{Hom}_R(L, E) \arrow{r}{\mathrm{Hom}\!(\!\beta\!, \mathrm{id}\!_E\!)} & \mathrm{Hom}_R(N, E) \arrow{r}{\mathrm{Hom}(\alpha, \mathrm{id}_E)} & \mathrm{Hom}_R(M, E) \end{tikzcd}$ \end{document} ?
– user121799
May 18, 2019 at 15:14
You just have to enlarge the column width.
\documentclass{article}
\usepackage{amsmath}
\usepackage{tikz-cd}
\usetikzlibrary{arrows, matrix}
\tikzcdset{every label/.append style = {font = \tiny}}
\begin{document}
$\begin{tikzcd}[column sep=1.5cm] 0\rar & \mathrm{Hom}_R(L, E) \arrow{r}{\mathrm{Hom}(\beta, \mathrm{id}_E)} & \mathrm{Hom}_R(N, E) \arrow{r}{\mathrm{Hom}(\alpha, \mathrm{id}_E)} & \mathrm{Hom}_R(M, E) \end{tikzcd}$
\end{document}
I wouldn't use tikzcd for this. Anyway, by reducing the font size and enlarging the column separation you get some result. Below I also show how I'd do it.
\documentclass{article}
\usepackage{amsmath}
\usepackage{tikz-cd}
\DeclareMathOperator{\Hom}{Hom}
\newcommand{\id}{\mathrm{id}}
\begin{document}
$\begin{tikzcd}[column sep=large] 0\arrow[r] & \Hom_R(L, E) \arrow[r,"{\scriptscriptstyle\Hom(\beta, \id_E)}"] & \Hom_R(N, E) \arrow[r,"{\scriptscriptstyle\Hom(\alpha, \id_E)}"] & \Hom_R(M, E) \end{tikzcd}$
$0\longrightarrow \Hom_R(L, E) \xrightarrow{\Hom(\beta, \id_E)} \Hom_R(N, E) \xrightarrow{\Hom(\alpha, \id_E)} \Hom_R(M, E)$
\end{document} | {"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": 0.9340142011642456, "perplexity": 2950.2875380927435}, "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/1656103850139.45/warc/CC-MAIN-20220630153307-20220630183307-00533.warc.gz"} |
http://mathoverflow.net/feeds/question/17132 | Bounds on $\sum {n ((\alpha n))}$ where $((x))$ is the sawtooth function - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-18T21:34:19Z http://mathoverflow.net/feeds/question/17132 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/17132/bounds-on-sum-n-alpha-n-where-x-is-the-sawtooth-function Bounds on $\sum {n ((\alpha n))}$ where $((x))$ is the sawtooth function Deinst 2010-03-04T20:35:36Z 2010-03-07T18:32:10Z <p>Let the sawtooth function $((x)) = x-\lfloor x \rfloor - 1/2$ if x is not an integer, 0 if it is. </p> <p>For an arbitrary irrational $\alpha$ are there good bounds for $\sum_{i=0}^n i((i \alpha))$?</p> <p>Experiment seems to indicate something of the the form $O(n^{1+\epsilon})$, and that would be great, but anything better than $O(n^2)$ would be nice. Explicitly computable constants depending on $\alpha$ would be great as well. </p> http://mathoverflow.net/questions/17132/bounds-on-sum-n-alpha-n-where-x-is-the-sawtooth-function/17141#17141 Answer by Gerry Myerson for Bounds on $\sum {n ((\alpha n))}$ where $((x))$ is the sawtooth function Gerry Myerson 2010-03-04T22:26:59Z 2010-03-04T22:26:59Z <p>For $\alpha$ irrational, the sequence $i\alpha$ is uniformly distributed modulo 1. A theorem of Weyl says that if $u_n$ is uniformly distributed in $[0,1)$ and $f$ is Riemann-integrable then $\lim_{N\to\infty}(1/N)\sum_1^Nf(u_n)=\int_0^1f(x)dx$. There's a quantitative version of this theorem that says the difference between the sum and the integral is bounded by the discrepancy of the sequence times the variation of the function. The discrepancy of the sequence $i\alpha$ (modulo 1) is well-understood. Maybe this gives an approach. </p> <p>A reference is Kuipers and Niederreiter, Uniform Distribution of Sequences. </p> http://mathoverflow.net/questions/17132/bounds-on-sum-n-alpha-n-where-x-is-the-sawtooth-function/17149#17149 Answer by maks for Bounds on $\sum {n ((\alpha n))}$ where $((x))$ is the sawtooth function maks 2010-03-05T00:46:59Z 2010-03-05T00:46:59Z <p>Here is how to obtain the $o(N^2)$ bound. If $\alpha$ is irrational then by Weyl's bound (as the previous poster remarked) we have $\sum_{n \leq N} ((n\cdot\alpha)) = o(N)$. Now note that that $$\sum_{i \leq N} i\cdot ((i\cdot\alpha)) = N\sum_{n \leq N} ((n\cdot\alpha)) - \int_{1}^{N} \sum_{n \leq t} ((n \cdot \alpha)) \text{d}t$$ For irrational $\alpha$ by Weyl's bound both terms are $o(N^2)$. You can certainly improve on the $o(N^2)$ but if you want a very good improvement you may need to know more about $\alpha$.</p> http://mathoverflow.net/questions/17132/bounds-on-sum-n-alpha-n-where-x-is-the-sawtooth-function/17176#17176 Answer by Douglas Zare for Bounds on $\sum {n ((\alpha n))}$ where $((x))$ is the sawtooth function Douglas Zare 2010-03-05T12:43:00Z 2010-03-07T18:32:10Z <p>Here is an approach which may give some better estimates for particular values of $\alpha$:</p> <p>$$\sum_{i= 1}^N i((i\alpha)) = \sum_{i=1}^N \sum_{j=i}^N ((j \alpha)) = \sum_{i=1}^N \sum_{k=0}^{N-i}((N\alpha -k\alpha))$$.</p> <p>So, if you can estimate </p> <p>$$\sup_{\substack{x\in (0,1) \\ M \le N}} \bigg|\sum_{k=0}^{M} ((x-k\alpha))\bigg|,$$ </p> <p>then you can crudely multiply by $N$ to get an estimate for $|\sum i((i\alpha))|$.</p> <p>Specifically, my guess is that for quadratic irrationals $\alpha$, there is an upper bound for </p> <p>$$\bigg|\sum_{k=0}^M ((x-k\alpha))\bigg|$$</p> <p>which is $O (\log M)$, which would give you a bound of $O(N \log N)$, and more generally that there is a bound in terms of the coefficients of the simple continued fraction for $\alpha$, so that if those are bounded, then you still get $O(N \log N)$. </p> <p>For the particular value $\phi = (\sqrt5 + 1)/2$, $\sum_{i=0}^{M} ((i \phi))$ has logarithmic growth $c + (5\sqrt5 - 11)/4 \log_\phi M$ (achieved at indices in the sequences <a href="http://www.research.att.com/~njas/sequences/A064831" rel="nofollow">A064831</a> (+) and <a href="http://www.research.att.com/~njas/sequences/A059840" rel="nofollow">A059840</a> (-)), which suggests that $\sup \sum_{i=0}^M ((x-i\phi))$ also has logarithmic growth, which would give an $N \log N$ bound for the sum. </p> <hr> <p>In the opposite direction, for all $\alpha \not\in \frac 12\mathbb Z$, $$\limsup \bigg(\log_N \bigg|\sum_{i=0}^N i((i\alpha))\bigg|\bigg) \ge 1$$ since there are terms proportional to $N$.</p> <p>The sum can be greater than $N^{2-\epsilon}$ infinitely often by choosing $\alpha$ so that it is extremely well approximated by infinitely many rational numbers. When $\alpha$ is very closely approximated by $p/q$, then for $N$ a small multiple of $q$ (where "small" is relative to how well $p/q$ approximates $\alpha$), about $1/q$ of the terms can be moved past integers with a small perturbation of $\alpha$ to $\alpha'$, which causes a jump of about $N^2/q$ in the sum. So, either the sum for $\alpha$ or $\alpha'$ is large. We can choose a sequence $p_n/q_n$ which converges to an $\alpha$ which produces large sums infinitely often, so that for these $\alpha,$ $$\limsup \bigg(\log_N \bigg|\sum_{i=0}^N i((i\alpha))\bigg|\bigg) = 2$$.</p> | {"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": 0.9818082451820374, "perplexity": 295.6108558144651}, "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-2013-20/segments/1368707188217/warc/CC-MAIN-20130516122628-00068-ip-10-60-113-184.ec2.internal.warc.gz"} |
http://encoretheatercompany.com/hrwmd8oo/0532f8-latex-table-of-contents | 30
Dec
This command can be used to edit files for table of contents, list of figures, or list of tables. I just ran into the following problem: For a technical document I’m creating with LaTeX, there are a lot of sections that have a repeated/consistent format. Imports the bibtex data file sample.bib, this file is the one that includes information about each referenced book, article, etc. \documentclass[11pt]{article} \usepackage{blindtext} \usepackage[toc]{multitoc} \renewcommand*{\multicolumntoc}{2} \setlength{\columnseprule}{1pt} \begin{document} … LaTeX processes all source code sequentially, so when it first encounters the \tableofcontents command, it doesn't yet know anything about the chapters, sections etc. A clear, concise, and well formatted TOC is the first indicator of a good research paper. Let's start with a minimal working example, by simply importing the hyperref package all cross-referenced elementsbecome hyperlinked. LaTeX has all the facilities for these classic requirements and this tutorial will guide you through them. To make the table of contents LaTeX stores the information in an auxiliary file named root-file.toc (see Splitting the input). Unnumbered sections in the Table of Contents The way the relevant parts of sectioning commands work is exemplified by the way the \chapter command uses the counter secnumdepth (described in Appendix C of the LaTeX manual):. The tocloft package implements a number of commands to customize the table of contents, list of figures and list of tables. Chapters, sections and subsections are included in the table of contents. Intro to Beamer Overlaying Concepts Sparkle References About … If you use a KOMA class: use the document option liststotoc; Else: use the package tocbibind from CTAN:/, i.e. This page may contain information about the author, institution, event, logo, and so on. Thus every input file must contain the commandsThe area between \documentclass{...} and \begin{document} is called the preamble. Make use of \href{Url} command for adding a hyperlink without its description. A short introduction to table of contents when creating presentations using the LaTeX beamer package. To manually add entries, for example when you want an unnumbered section, use the command \addcontentsline as shown in the example. \usepackage{tocbibind}; If you use \usepackage[nottoc]{tocbibind} instead, the toc will not show up in the toc. Open an example of the hyperref package in Overleaf \cite{einstein} 1. LaTeX question: Is there a way I can control the depth of the table of contents (\tableofcontents) using LaTeX? The commands \listoffigures and \listoftables are self explanatory, the first one generates the list of figures and the second one the list of tables. They can be modified to fit a specific style. This article explains how? Imports the package biblatex. The Table of Contents (TOC) is an organized listing of the chapters and major sections of your document. You put the command right where you want the table of contents to go; LaTeX does the rest for you. I am using document class "report" and package {tocloft}. Tables of Contents A table of contents typically sits at the beginning of a large manuscript and lists all the chapters and sections within along with a page number. For example, this LaTeX file test.tex See the bibliography filesection for more information. Section or chapter numbering is restarted and the representation of the counter switches to alphabetic. It is actually taking the size of contents as they are present in report but I want to alter the font size of table of contents … Usually, it has to be the last package to be imported, but there might be some exceptions to this rule. The first page is the titlepage, and the second one contains sample content. The first statement in the document declares this is a Beamer slideshow: \documentclass{beamer} The first command after the preamble, \frame{\titlepage}, generates the title page. Table of Contents. If you want a page break after the table of contents, write a \newpage command after the \tableofcontents command, as above. However, there is one di erence be- See the titl… Appendixes LaTeX provides an exceedingly simple mechanism for appendixes: the command \appendix switches the document from generating sections (in article class) or chapters (in report or book classes) to producing appendixes. Include the hyperref package in the document preamble. It normally contains commands that affect the entire document.After the preamble, the text of your document is enclosed between two commands which identify the beginning and end of the actual document:You would put your text where the dots are. A minimal working example of a simple beamerpresentation is presented below. LaTeX uses the commands \section, \subsection and \subsubsection to define sections in your document; The sections will have successive numbers and appear in the table of contents; Paragraphs are not numbered and thus don't appear in the table of contents; Next Lesson: 03 Packages In this example there are two more relevant commands: \thispagestyle{empty} Removes the page numbering. 4 posts • Page 1 of 1 §How can I get Bibliography, Lists of Figures/Tables and Appendix to appear in the table of contents (TOC)? The multitoc package provides an easy and clean way to save space when producing a table of contents. Summary. Generally, default color is OK. However, if you wish to change the default settings of hyperlink color, you have to use \hypersetup command in your document preamble. § How do I add the index to the TOC? It produces a heading, but it does not automatically start a new page. This is particularly neat for short documents, but might also improve the readability in longer content lists. For example: Code: [Expand ... See the LaTeX manual or LaTeX Companion for explanation. Note: For the table of contents to work properly you must compile the document twice or use latexmk -pdf Open an example in Overleaf After compilation, a two-page PDF file will be produced. The lines in the table of contents become links to the corresponding pages in the document by simply adding in the preamble of the document the line One must be careful when importing hyperref. header and footer lines, page formats, page numbers). \pagenumbering{arabic} Re-start the page numbering with arabic style. Creating Table of Contents. Auxiliary Files ). One of these commands provided by the package adds whitespace before entries. All you have to do is include a chapter before the table of contents section is start it in the same way as Chapter 1 but before where the table of contents appears in the document. Add a title on the top of the Table of Contents. Changing LaTeX Hyperlink Color. This command inserts a reference within the document, in this case, that … It allows for two or more columns. So the first time the document is LaTeXed the necessary information is written to the table of contents ( .toc) file (see §2.4. When LaTeX processes an input file, it expects it to follow a certain structure. To add a line to the table of contents, use "toc" for file, and "chapter" for section_unit, and place the command in your document immediately after the section heading. When working with large documents with tens (or hundreds) of pages, it’s useful to be able to scroll directly to the section you’re interested in by clicking the section in the table of contents. Readers will immediately be able to see how your manuscript is organized and then skip down to sections that are most relevant to them. A minimal working example of the biblatexpackage is shown below: There are four bibliography-related commands in this example: \usepackage{biblatex} 1. because LaTeX matters. \tableofcontents The ToC, LoF, and LoT are printed at the point in the document where these \listoffigures \listoftables commands are called, as per normal LATEX. LaTeX Table of Contents. The tocloft package provides means of specifying the typography of the Table of Contents (ToC), the List of Figures (LoF) and the List of Tables (LoT). Similarly, list of figures and list of tables can be compressed. \begin{table}[] command initiate the start of table, \centering align the table into center and caption is used for table heading as shown below: \addbibresource{sample.bib} 1. Sections, subsections and chapters are included in the table of contents. Title the Table of Contents. A table of contents is produced with the \tableofcontents command. LaTeX forum ⇒ Page Layout ⇒ Table of contents only on odd pages Information and discussion about page layout specific issues (e.g. This code is written in LaTex (Generate Table in LaTeX) for creating an empty table without containing any contents in the cells. LaTeX table of contents with clickable links. LaTeX offers features to generate a table of contents, changing its title, list of figures and tables, captions. I am having a problem in changing font size of sections, chapters, subsections explicitly on Table of Contents generated by "\tableofcontents". It can be done using a few simple commands. Now just look at the code given in the following table. Usually the title is “Table … Table of Contents 1 Intro to Beamer About Beamer Basic Structure 2 Overlaying Concepts Speci cations Examples: Lists, Graphics, Tables 3 Adding that Sparkle Sections Themes Prathik Naidu and Adam Pahlavan Fun with Beamer 1 / 30. Summary. 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https://socratic.org/questions/how-do-you-find-the-next-two-terms-of-the-geometric-sequence-1-3-5-6-25-12 | Precalculus
Topics
# How do you find the next two terms of the geometric sequence 1/3, 5/6, 25/12,...?
Oct 23, 2016
The next two terms are found by multiplying by the common ratio $r = \frac{5}{2}$ which results in $\frac{125}{24}$ and $\frac{625}{48}$.
#### Explanation:
A geometric sequence is a sequence of numbers generated by mutiplying the previous term by the common ratio or $r$.
Find the common ration $r$ by dividing any term by the previous term.
In this example, $r = \frac{5}{6} \div \frac{1}{3} = \frac{5}{2}$.
The second term is generated by multiplying the first term by $\frac{5}{2}$
or $\frac{1}{3} \cdot \frac{5}{2} = \frac{5}{6}$
The third term is generated by multiplying the second term by $\frac{5}{2}$ or $\frac{5}{6} \cdot \frac{5}{2} = \frac{25}{12}$
Similarly, the fourth term is $\frac{25}{12} \cdot \frac{5}{2} = \frac{125}{24}$
The fifth term is $\frac{125}{24} \cdot \frac{5}{2} = \frac{625}{48}$
##### Impact of this question
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https://www.physicsforums.com/threads/find-dy-dx-if-f-x-y-0.660922/ | # Find dy/dx if f(x,y) = 0
1. Dec 26, 2012
### unscientific
1. The problem statement, all variables and given/known data
Given f(x,y) = 0, find an expression for dy/dx.
2. Relevant equations
3. The attempt at a solution
f(x,y) = 0
df = (∂f/∂x)dx + (∂f/∂y)dy = 0
df/dx = (∂f/∂x) + (∂f/∂y)(dy/dx) = 0
dy/dx = -(∂f/∂x)/(∂f/∂y)
2. Dec 26, 2012
### Dick
Looks ok to me. Do you have a question about that?
3. Dec 26, 2012
### unscientific
Hmmm, so I guess i'm right?
My only question is
f(x,y) = 0,
df = 0?
4. Dec 26, 2012
### Dick
Well, yes. Why would you think not?
5. Dec 26, 2012
### unscientific
That makes absolute sense. If a function is zero for all x,y then the infinitesimal change df would be 0 for all x,y.
6. Dec 26, 2012
### Dick
In this problem you shouldn't think of f as identically 0. Here's an example to think about. Suppose y=x (defining a curve). Take f(x,y)=x^2-y^2. Then f(x,y)=0 along the line y=x, but f(x,y) is not identically 0. But df=0 along the line y=x.
7. Dec 26, 2012
### unscientific
So, what does f(x,y) = 0 actually mean?
8. Dec 26, 2012
### Dick
You think of y as being some function of x. That's why they can ask you to find an expression for dy/dx. Then saying f(x,y)=0 means f is zero along the curve defined by y(x).
9. Dec 26, 2012
### unscientific
That makes sense, thanks! But still it doesn't explain why df = 0..
10. Dec 26, 2012
### haruspex
You're given that f is always zero, does not change as x and y change. So df=0.
11. Dec 27, 2012
### epenguin
Congratulations, you have worked out what is a fairly fundamental and often needed little formula. Which surprises many students as with a superficial look of the formula they can imagine it is going to be the other way up and the minus sign surprises.
If you draw a little picture it should become less mysterious and surprising.
Also, as mentioned, it holds for f constant - the constant doesn't have to be 0.
12. Dec 27, 2012
### algebrat
This reply doesn't really stand alone, I was trying to build on what others had previously said.
If f is a nice function (for instance, a polynomial in x and y, or some other equation not doing anything crazy. Look up space filling curve.), the solution set is one-dimensional. So it is a curve in the plane.
In other words, the equation f(x,y)=0 has a set of solution points (x,y) that is a curve. Along some sections of that curve, we might be able to write the curve explicitly, as y=y(x), or g(x), whatever. Notice that along the curve, f=0.
We can use f to look at other parts of the ambient plane. f may take many values; in fact, for each value of c, we expect the equation f(x,y)=c to give various curves in the plane, a solution set for each value c. Notice that once we pick a curve as specified, f is constant, along that curve. In other words, df=0. (We could say something like, "because d(c)=0"; i.e., the differential of a constant is zero.)
The idea of an equation, or relation, determining a 1-d subset of the plane, ie curve, generalizes and is a very useful and frequently used concept in math and science, and you should totally try to absorb the idea into your brain, and keep an eye out for when it comes up.
Similar Discussions: Find dy/dx if f(x,y) = 0 | {"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": 0.8955989480018616, "perplexity": 1170.4095560261412}, "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-2017-43/segments/1508187824068.35/warc/CC-MAIN-20171020101632-20171020121632-00553.warc.gz"} |
https://arxiv.org/abs/1202.3203 | cond-mat.str-el
(what is this?)
# Title: A study on correlation effects in two dimensional topological insulators
Abstract: We investigate correlation effects in two dimensional topological insulators (TI). In the first part, we discuss finite size effects for interacting systems of different sizes in a ribbon geometry. For large systems, there are two pairs of well separated massless modes on both edges. For these systems, we analyze the finite size effects using a standard bosonization approach. For small systems, where the edge states are massive Dirac fermions, we use the inhomogeneous dynamical mean field theory (DMFT) combined with iterative perturbation theory as an impurity solver to study interaction effects. We show that the finite size gap in the edge states is renormalized for weak interactions, which is consistent with a Fermi-liquid picture for small size TIs. In the second part, we investigate phase transitions in finite size TIs at zero temperature focusing on the effects of possible inter-edge Umklapp scattering for the edge states within the inhomogeneous DMFT using the numerical renormalization group. We show that correlation effects are effectively stronger near the edge sites because the coordination number is smaller than in the bulk. Therefore, the localization of the edge states around the edge sites, which is a fundamental property in TIs, is weakened for strong coupling strengths. However, we find no signs for "edge Mott insulating states" and the system stays in the topological insulating state, which is adiabatically connected to the non-interacting state, for all interaction strengths smaller than the critical value. Increasing the interaction further, a nearly homogeneous Mott insulating state is stabilized.
Comments: 20 pages Subjects: Strongly Correlated Electrons (cond-mat.str-el) Journal reference: Physical Review B 85, 165138 (2012) DOI: 10.1103/PhysRevB.85.165138 Cite as: arXiv:1202.3203 [cond-mat.str-el] (or arXiv:1202.3203v1 [cond-mat.str-el] for this version) | {"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": 0.8568015694618225, "perplexity": 860.7220001305548}, "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-2018-34/segments/1534221218122.85/warc/CC-MAIN-20180821112537-20180821132537-00539.warc.gz"} |
https://physics.stackexchange.com/questions/327851/scalar-field-in-a-curved-space-time | # Scalar field in a curved space-time
We know that action of a scalar field in a curved space-time in Jordan frame is given by:
$$\int \left( \frac{1}{2} M^2 R -\frac{1}{2} \partial_\mu \phi \partial^\mu \phi –V(\phi) -\frac{1}{2} \xi \phi^2 R \right) \sqrt{-g}\ d^4x$$
1. Why does this action break the equivalence principle?
2. It is said in the literature that when coupling to gravity is minimal $$\xi = 0$$, then M is the Planck scale.
Firstly Planck mass is given by $m_p = \sqrt{\frac{\hbar c}{G}}$. I cannot see any $\hbar$ in this equation, so where is the Planck scale coming from?
Secondly why does M correspond to Planck mass only when $\xi = 0$ ?
Any ideas are appreciated.
• Which literature? – Qmechanic Apr 20 '17 at 13:40
• @Qmechanic for instance please look at this one link when they discuss about equation (1). – Ramtin Apr 21 '17 at 13:05
• So, in natural units $\hbar=c=1$, the reduced Planck mass is defined by $m_p=(8\pi G)^{-1/2}$. Since in the standard Einstein-Hilbert action, $R$ should be multiplied by $1/16\pi G=m_p^2/2$, it is clear why $M$ should be identified with $m_p$ in the limit $\xi\to 0$. There is no need for $\hbar$. I hope that clarified it! – Bob Knighton Apr 22 '17 at 12:19
If you have a matter field $\psi$ which is coupled minimally to the curved background, and a field $\phi$ which is coupled non-minimally, objects made of $\psi$ will move on different free-fall trajectories than $\phi$. This is a direct violation of the weak equivalence principle.
But the fact that the field $\phi$ is coupled non-minimally also means that you can observe its behaviour and essentially measure the local value of $R$. This is a violation of the Einstein equivalence principle even if you do not have a reference field $\psi$.
Now for the question of $M$: The reason why $M$ is the Planck mass is directly dependent on the context you are considering! The paper you link talks about $\phi$ being the Higgs field undergoing electroweak symmetry breaking which drives the inflation. Thus, $\phi$ reaches an essentially constant value $\phi_0$ in the post-inflationary era.
That is, the gravitational part of your action reduces effectively to $$S_\mathrm{grav} = \int \sqrt{-g} \frac{1}{2}(M^2 - \xi \phi_0^2)R \mathrm{d}^4 x$$ in the post-inflationary era.
However, we measure experimentally right here and right now in the post-inflationary era that the gravitational term in the action is, at least phenomenologically, $R/(4\pi G_\mathrm{N})$ where $G_\mathrm{N}$ is Newton's gravitational constant. In Planck units this term is written as $M^2_\mathrm{p} R/2$. This means that if we want to fit the respective term in the action you give in the postinflationary era, we must fulfill $$M^2_\mathrm{p} = M^2 - \xi \phi_0^2$$ This means that if we impose the phenomenological constrain, $M$ will never be equal to $M_\mathrm{p}$ for a non-minimally coupled field with symmetry breaking. The magnitude of $\phi_0$ depends on the details of the symmetry breaking and then imposes the range of $\xi$ which do not take $M$ too far from $M_\mathrm{p}$. However, if $\xi=0$, the phenomenological constraint gives immediately $M=M_\mathrm{p}$. | {"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": 0.9377986192703247, "perplexity": 161.3598413621044}, "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/1575540534443.68/warc/CC-MAIN-20191212000437-20191212024437-00499.warc.gz"} |
https://clifford.readthedocs.io/en/latest/resources.html | # Other resources for Geometric Algebra¶
If you think Geometric Algebra looks interesting and want to learn more, check out these websites and textbooks | {"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": 0.860429048538208, "perplexity": 970.8402811270073}, "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-2020-45/segments/1603107889574.66/warc/CC-MAIN-20201025154704-20201025184704-00634.warc.gz"} |
http://mathhelpforum.com/new-users/207991-probability-question-please-help-print.html | # Probability question please help
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• Nov 19th 2012, 12:49 PM
Nasc0
Probability question please help
if i guess at five choice questions. each question has 5 anwsers to pick from. whats the probability that i l get three correct? tell me how u did it also thanks.
• Nov 19th 2012, 01:19 PM
skeeter
Re: Probability question please help
probability problems go in the statistics forum
• Nov 19th 2012, 01:57 PM
Plato
Re: Probability question please help
Quote:
Originally Posted by Nasc0
if i guess at five choice questions. each question has 5 anwsers to pick from. whats the probability that i l get three correct? tell me how u did it also thanks.
If $p$ is the probability of success and $N$ is the number of trials then the probability of exactly $k$ successes, where $0\le k\le N$, is $\binom{N}{k}p^k(1-p)^{N-k}$. | {"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": 5, "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": 0.9483499526977539, "perplexity": 1851.5639500259808}, "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/1476988720356.2/warc/CC-MAIN-20161020183840-00477-ip-10-171-6-4.ec2.internal.warc.gz"} |
https://math.stackexchange.com/questions/3864236/the-dual-of-ellp-textbfr-is-ellq-textbfr-where-textbfr-is | # The dual of $\ell^p(\textbf{r})$ is $\ell^q(\textbf{r})$, where $\textbf{r}$ is a weighted vector
I'm trying to show that the dual space of $$\ell^p$$ is $$\ell^q$$ with the typical conditions, only that we will include a weight to our space. The proof that I want to imitate is the Kreyszig, but I have two problems. So I star with
For each $$k\in\mathbb{N}$$, we consider the canonical sequence in $$\ell^p(\textbf{r})$$ defined by $$e_k = \left(\delta_{kj}\right)_{j\in\mathbb{N}}$$, where $$\delta_{kj}$$ is the known Kronecker delta, that is, $$\delta_{kj} = 1$$ if $$k=j$$ y $$0$$ otherwise for $$k, j \in \mathbb{N}$$. Then, for each $$\textbf{x}=\left(x_k\right) \in \ell^p(\textbf{r})$$ it is true that $$\lim_{N\to\infty} \left\|\textbf{x}- \displaystyle\sum^{N}_{k=1} x_k e_k \right\|_r^p = \lim_{N\to\infty}\displaystyle\sum^{\infty}_{k=N+1} \left|x_k\right|^p r_k = 0$$ and then all $$\textbf{x} \in \ell^p(\textbf{r})$$ has a single representation of the form $$\textbf{x} = \displaystyle\sum^{\infty}_{k=1} x_k e_k.$$ which means that $$\left\{e_k\right\}_{k\in\mathbb{N}}$$ is a Schauder basis for $$\ell^p(\textbf{r})$$.
Consider some $$f \in \left(\ell^p(\textbf{r})\right)'$$, the dual space of $$\ell^p(\textbf{r})$$, and define the sequence $$\textbf{y}=\left(y_k\right)$$ by $$$$\label{def-yk} y_k = f\left(e_k\right). \hspace{10cm} (1)$$$$ Since $$f$$ is linear and continuous, for any $$\textbf{x}=\left(x_k\right) \in \ell^p(\textbf{r})$$ it is true that $$\begin{eqnarray*} f\left(\textbf{x}\right) &= & f\left( \displaystyle\sum^{\infty}_{k=1} x_k e_k\right)\\ &=& f\left(\lim_{N\to\infty} \displaystyle\sum^{N}_{k=1} x_k e_k\right)\\ &=& \lim_{N\to\infty} f\left( \displaystyle\sum^{N}_{k=1} x_k e_k\right)\hspace{1cm}\text{(by continuity)}\\ &=& \lim_{N\to\infty} \displaystyle\sum^{N}_{k=1} x_k f\left(e_k\right)\hspace{1.3cm}\text{(by linearity)}\\ &=& \lim_{N\to\infty} \displaystyle\sum^{N}_{k=1} x_k y_k = \displaystyle\sum^{\infty}_{k=1} x_k y_k \end{eqnarray*}$$ and the formula $$f\left(\textbf{x}\right) =\displaystyle\sum^{\infty}_{k=1} x_k y_k$$ holds; so now we have to show that the sequence $$\textbf{y}=\left(y_k\right)$$ defined in (1) is in $$\ell^q(\textbf{r})$$.
Indeed, for each $$n\in\mathbb{N}$$ the sequence $$\textbf{x}_n = (\xi^{(n)}_k)$$ is considered with ( THE FIRST PROBLEM IS ADDING WEIGHT TO THIS SUCCESSION, WHICH SEEMS NATURAL, BUT LATER IT AFFECTS ME PROOF (or so I think)) $$$$\label{d2} \xi^{(n)}_k = \begin{cases} \frac{|y_k|^q}{y_k}, & \mbox{si } k \le n \hspace{2mm} \mbox{y } y_k \neq 0 \\ 0, & \mbox{si } k > n \hspace{2mm} \mbox{o } y_k = 0. \end{cases}$$$$ Then $$\textbf{x}_n\in\ell^p\left(\textbf{r}\right)$$ since it has a finite amount of non-null elements; so by the formula $$f\left(\textbf{x}\right) =\displaystyle\sum^{\infty}_{k=1} x_k y_k$$ it is allowed to write $$f(\textbf{x}_n) = \displaystyle\sum^{\infty}_{k=1} \xi^{(n)}_k y_k = \displaystyle\sum^{n}_{k=1} |y_k|^q.$$ Now using the definition of $$\xi^{(n)}_k$$ and the fact that $$(q - 1)p = q$$, \begin{aligned} \left|f(\textbf{x}_n)\right| &\le \left\| f \right\| \left\| \textbf{x}_n \right\|_r\\ & = \left\| f \right\| \left( \displaystyle\sum^{n}_{k=1} |\xi^{(n)}_k|^p r_k \right)^{1/p}\\ & = \left\| f \right\| \left( \displaystyle\sum^{n}_{k=1} |y_k|^{(q-1)p} r_k \right)^{1/p}\\ & = \left\| f \right\| \left( \displaystyle\sum^{n}_{k=1} |y_k|^q r_k\right)^{1/p} \end{aligned} and when joining the ends you have to $$\left|f(\textbf{x}_n)\right| = \displaystyle\sum^{n}_{k=1} |y_k|^q \le \left\| f \right\| \left( \displaystyle\sum^{n}_{k=1} |y_k|^q r_k \right)^{1/p}.$$ (HERE IS THE OTHER PROBLEM, BECAUSE ONE PART HAS THE WEIGHTED VECTOR, BUT AND THE OTHER DOES NOT)
Thanks for the help
The functional needs three terms, $$f\left(\textbf{x}\right) = \sum^{\infty}_{k=1} x_k y_k r_k$$ then you must define the sequence $$(y_k)$$ as follows $$y_k = \frac{f\left(e_k\right)}{r_k}$$ thus, you no longer need to multiply by the weighted vector in the sequence $$\xi^{(n)}$$.
Finally, with that substitution, you should easily arrive at what $$f(\textbf{x}_n) = \displaystyle\sum^{\infty}_{k=1} \xi^{(n)}_k y_k r_k = \displaystyle\sum^{n}_{k=1} |y_k|^q r_k.$$ Achieving the expected result $$\left|f(\textbf{x}_n)\right| = \displaystyle\sum^{n}_{k=1} |y_k|^q r_k \le \left\| f \right\| \left( \displaystyle\sum^{n}_{k=1} |y_k|^q r_k \right)^{1/p}.$$ although you still haven't shown that $$\ell^q$$ is its dual space, already the proof is natural | {"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": 43, "wp-katex-eq": 0, "align": 0, "equation": 2, "x-ck12": 0, "texerror": 0, "math_score": 0.9979895949363708, "perplexity": 294.0721333401477}, "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/1637964362918.89/warc/CC-MAIN-20211203182358-20211203212358-00077.warc.gz"} |
http://physics.stackexchange.com/questions/76282/magnetic-force-at-the-edge | # Magnetic force at the edge?
When sliding a magnet over a ferromagnetic surface there is no magnetic force at all, however,there is a magnetic force only at the edge, when the magnet is being pulled of I assume the magnetic field is trying to resist. However, a fact it noticeable, the sliding/lateral magnetic force at the edge is a lot less, by a factor of half - ten times less.
What could be the reasons that detains the magnitude of the lateral force? Pelase do explain in detail.
Thank you.
- | {"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": 0.947475790977478, "perplexity": 489.5000631089144}, "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-2015-48/segments/1448398447913.86/warc/CC-MAIN-20151124205407-00354-ip-10-71-132-137.ec2.internal.warc.gz"} |
https://philosophy.stackexchange.com/questions/70575/a-non-circular-definition-of-not/72764#72764 | # A non-circular definition of "not"
The notion of "not" is used throughout all languages as far as I know. For instance, consider the sentence,
"Trees are not blue."
There are various ways of expressing this notion without using the word "not", but they still mean the same. For instance,
"No tree is blue."
"All parts of trees have colors that are different than blue."
"The negation of the statement, 'some trees are blue', is true."
I'm curious if there is a definition of "not" which is not circular, that does not use any synonyms (as were used above) to define it.
• Would a "set theory" explanation work? Inside vs outside a set, with "not" being outside? "Me and not me. Sometimes you all go away, but I'm always right here." Mar 2 '20 at 20:14
• A popular way to "define" the meaning of all logical connectives is to give introduction and elimination rules for them, as in natural deduction. This is called Gentzen semantics, see logical harmony. The introduction rule adds ¬ to A when any B can be derived from A, and the elimination rule removes it when the same happens with ¬A in place of A. Mar 2 '20 at 20:48
• @Conifold I didn't see your comment when I wrote my answer! To the OP, Conifold's comment (besides introducing very useful terms) is very similar to my answer but not quite identical: the former introduces the further level of explicit syntax and concrete manipulations of strings. The underlying idea is the same, though: we understand logical operations by their interactions with deduction, and can try to look for characterizations in terms of what does (as opposed to does not) occur in the relevant context. Mar 2 '20 at 21:21
• Further elaborating on Conifold's comment, it's worth noting that there are serious subtleties with the Gentzen approach to logical connectives. Consider an imagined logical connective @ ("tonk") defined by the introduction rule "From a infer a@b" and the elimination rule "From a@b infer b." Obviously as soon as our syntax admits this connective everything collapses; so what criteria do we use to determine "meaningfulness" of proposed intro/elim definitions? This is exactly what logical harmony is meant to address. Mar 2 '20 at 21:29
• Maybe useful Jaakko Hintikka, Negation in Logic and in Natural Language (2002) as well as Dov Gabbay & Heinrich Wansing (editors), What is Negation? (Springer, 1999) Mar 3 '20 at 7:21
There's a "purely algebraic" approach, albeit one of dubious success: namely, we take the view that negation is characterized by explosion (avoiding non-contradiction since the latter is negative).
We work in the context of abstract deductive systems. There are many different notions of such floating around in the literature; what I'll mean in this answer is a rather simple one:
An a.s.d. is a pair (X, >) where X is some set of things called "sentences" and > is a relation between sets of sentences and individual sentences called "entailment."
(If this site supported mathjax, I'd replace ">" with the symbol with LaTeX code "$\vdash$" - this is standard usage in algebraic logic.)
A priori a.s.d.s could be extremely boring or pathological; generally we consider them in the presence of additional assumptions, e.g. compactness (if A>p then there is some finite subset B of A such that B>p), recursive enumerability (elements of X are canonically coded by natural numbers and > is compact and when restricted to finite sets of hypotheses is r.e.), presence of conjunction (for all p, q in X there is some r in X such that for all s in X we have {p,q}>s iff {r}>s), or so forth.
The key point I care about here is having negation:
An a.s.d. (X,>) has negations if for each p in X there is some q in X such that the following hold:
• For all r in X we have {p,q}>r.
• If q' is in X and for all r in X we have {p,q}>r, then we have {q'}>q.
Such a q is said to be a negation of p. (Note that negations need not be unique, merely "unique up to equivalence"!)
Note that there is no negative information here. In particular, ">" isn't assumed to be asymmetric at all (and it isn't in most natural examples, e.g. when (X,>) is just classical propositional logic we have {p&p}>p and {p}> p&p).
To repeat the remark at the beginning, the above reflects the following perspective on negation:
## Negation is characterized by explosion.
This is the "positive" flipside to non-contradiction, which is the other main way to think about negation.
Of course, this is extremely objectionable to some people: e.g. paraconsistent logic does not accept the principle of explosion (note that constructivism/intuitionism does - it's the excluded middle they reject, which is different). Since paraconsistent logic - whether I hold with it or not (I don't) - is something which makes sense and is interesting, it's clear to me that the approach above is not good. However, it's also not nothing, and it introduces algebraic logic as a topic of interest.
The other big potential dissatisfaction with the approach above is that it steps away from natural language. But that doesn't mean it's irrelevant: thinking of formal/algebraic logic as a "laboratory" for analyzing natural language, we can still talk about informal deductions and more-or-less follow the same theme as above. So I'd argue that that's ultimately not as serious an issue.
• It says in the answer, "If q' is any other element of X with the above property". What above property are you referring to? Mar 2 '20 at 22:06
• @CraigFeinstein The property of the previous bulletpoint - I've edited to clarify (and remove a possible worry about negativity in the word "other"). Mar 2 '20 at 23:42
• I’ve seen this in natural deduction interpretations a bit, and it makes sense in that context. Interestingly, for non-classical logics, there is a variation using Sequent Calculus frameworks with multiple conclusions; you can “left introduce” a negation by removing a conclusion from an entailment sequence. This does something pragmatically similar, and explosion can be seen as a specific case of a more general phenomenon! Mar 3 '20 at 0:36
• I'm not 100% sure this answer is correct, but I am 100% sure that if there is a correct answer, this is it. Mar 3 '20 at 14:47
According to the research of the Natural semantic metalanguage project, the problem of circular definitions is solved through the identification of semantic primes, the basic blocks of meaning which are shared by all languages and cannot be meaningfully subdivided. Any attempt to define a prime will end up circular or more convoluted than the word itself.
NOT is one of these primes. Proponents of NSM say that empirical research from languages all around the world show that NOT is a fundamental and irreducible concept in all languages. Being a prime also means that NOT doesn't really need to be defined - it's a base level concept which everyone will already understand. Any attempt to give a definition will be more complex than the word itself. The meaning of "not" is "not"!
• This doesnt answer the question. Jul 10 at 13:21
I'm curious if there is a definition of "not" which is not circular, that does not use any synonyms…to define it.
I am not certain that this problem has a solution. The central dilemma is that at some point the definition must exclude some category of things, and describing exclusion requires some form of the concept “not”. As curiousdannii discussed, perhaps there are some notions which are semantically prime, and are simply a part of what languages are.
The best that I can offer is this: Can the dilemma be avoided by crafting a definition whose content is both necessary and sufficient to completely define a set? Only affirmative statements would be used to create the definition. But anything whose qualities were not identical to those in the definition would, by implication, not be the same thing.
So, for example, assume that when Bertrand Russell proved that 1+1=2, he did so affirmatively. He began from a set of assumptions and followed them until he reached 1+1=2. When he did so, he did not need to add that 1+1 does not equal 5 or 12. That conclusion follows, but it does not need to be spoken to conclude that 1+1=2. The erroneous conclusions simply do not follow from the premises.
The precision of the premises enables the implication that anything outside the premises is superfluous or contradictory.
Negation is a (unary) logical operator, and can be considered either syntactically or semantically.
(1) From a syntactical point of view , negation is a function which associates any propositional input "phi" to the output " ~ phi". No circularity here, since the symbol" ~ " ( at the purely syntactic level) has no meaning.
All this amounts to saying that, among our formation rules, we have the following one :
If "phi" is a well-formed-formula , then "~ phi" is also a wff.
(2) From a semantical point of views, no circularity either.
Semantically , negation is a function from the set {0,1} to the set [0,1} , let's call it function N.
This function is defined in the following way :
N(1)= 0
N(0) =1.
Saying that Q is the negation of P means that
the image of the value of Q by function N is always identical to the value of proposition P, and reciprocally.
"Always" meaning " for all possible value assignements, namely 0 or 1 for each atomic sentence of the language".
The words " no" or " not " nowhere appear in this defnition.
Now, we are free to say that "1" means "true", and that "0" means "false". We are also free to say that "false" means "not true", and that "true" means " not false". There is indeed some circularity here, but it does not affect the definition of the function. It only affects the interpretation you give to this function, and the use you make of it.
• What about non-binary things? "Not tall" doesn't equal "short". Apr 23 '20 at 3:01
We can define "not" in terms of the numbers one and two:
Consider the sentence "A tree is a tree." There is one noun in the sentence, "tree".
Now consider the sentence "A tree is not grass." There are two nouns in the sentence, "tree" and "grass".
The word "is not" is used when there are two. The word "is" is used when there is one.
Is this a circular definition?
• How do you determine inequality of nouns? This is just a linguistic repackaging of defining "not" in terms of "not equal to," which seems circular to me. (This is why an "explosion" rule is useful: it's a purely positive characterization.) May 13 '20 at 15:34
• It is circular in the sense that one can argue that it is linguistic repackaging. However, if one regards numbers as more fundamental than logic, it seems like a better definition than other circular definitions. May 13 '20 at 17:02
• I regard numbers as more fundamental than logic for the following reason: Binary logical operations like "and", "or", "if then" always need two statements, and two is a number. May 13 '20 at 17:55
Here is a three-line algorithm for determining whether X = Y or X != Y.
1) Define the variable XisNotY=true.
2) If X = Y then let XisNotY=false.
3) End
At the end of the algorithm, XisNotY will be true if and only if X != Y. Note that the algorithm never uses the word "not" or anything like it, so it is a noncircular definition of "not".
• A criticism of this answer that I just realized is "how does one define an "if then" statement?" Can one define it without using the notion of "not"? May 15 '20 at 15:24
You can define negation by paraphrasing the law of non-contradiction:
not can be defined as a function such that given a subject A and a predicate B, A can never be both B and not(B).
• But what does "never" mean? Mar 2 '20 at 20:57 | {"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": 0.8382439017295837, "perplexity": 821.7984713282552}, "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-2021-39/segments/1631780058467.95/warc/CC-MAIN-20210927181724-20210927211724-00396.warc.gz"} |
http://tex.stackexchange.com/users/18/jos%c3%a9-figueroa-ofarrill?tab=activity&sort=all&page=5 | José Figueroa-O'Farrill
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Aug6 accepted Is there an incompatibility between eulervm and fourier packages? Aug6 comment What professions use TeX/LaTeX besides CS? Well, Hull all but obliterated their Maths department a few years back, so somehow I'm not surprised... Aug6 comment Which command should I use for displayed equations? Emacsen (with AUCTeX) autocomplete many environments, so you can be lazy and still use equation* :) Aug5 comment Include section number in list number I am not sure you can have a \section inside an enumeration, but why would you ever want to do that? Aug5 comment Include section number in list number Yes, this is a better answer than mine, because it does not require doing this at every level. I had not read the previous question. Aug5 comment Include section number in list number Yep, I just checked and it works. Aug5 answered Include section number in list number Aug4 comment How can I recognize documents prepared in (La)TeX? I think you'll find Computer Modern fonts being used less and less, though. Proper ligatures certainly. Also an overall feel of aesthetic balance of the page which I'm unable to define more precisely. Aug4 comment What is the difference between TeX and LaTeX? "LaTeX is all about content" is correct, but that's only with LaTeX2e, because in LaTeX 2.09 the distinction between formatting and content was not clear cut. In fact, it was only when LaTeX2e was introduced that I stopped using TeX in favour of LaTeX. Aug4 comment What is the difference between TeX and LaTeX? I didn't vote this down, but I don't agree that TeX and LaTeX are different languages. LaTeX (as of today) is still a set of TeX macros. Aug4 comment Installing bibexport on Mac OS X @Jukka: indeed. Aug4 comment Installing bibexport on Mac OS X I take it as given that one installs the GNU textools in any *nix system! Aug4 answered Installing bibexport on Mac OS X Aug3 awarded Beta Aug3 comment Is there an automatic process to create index creation? What do you mean by "suggested to be included in the index"? I would expect it to be the author of the content who should know what is suitable for the index. Or are machines already getting so smart? :) Aug2 comment Is it already possible to use the STIX fonts? I'm running a pretest TL2010 on my Mac and have just installed the XITS fonts. As noted above \mathscr works fine, but \mathcal does not. The fonts look great, though :) Aug2 revised Cyrillic in (La)TeX changed distributive (adjective) to distribution (noun), which was the intention of the OP Aug2 comment Cyrillic in (La)TeX Does "(la)tex" preclude the use of XeLaTeX? Because unless there is something I'm missing, in the same way that I can typeset in Japanese with XeLaTeX, you should be able to typeset in Cyrillic, provided you have the fonts. Aug2 comment Is there a way to get two pages in one with LaTeX? I think it's fair to say that we are living in a transitional period where typography is still very much linked to paper and printing and, in my opinion, the end of the transition cannot arrive too quickly. This is why I make a clear distinction, as articulated by Caramdir, between the typsetting system (in this context, *TeX) and the different ways of rendering the output of that system. This does not contradict what anyone else is saying, but it helps perhaps explain my original commment. Aug2 comment Is there a way to get two pages in one with LaTeX? I must echo Caramdir's question. It's usually the print driver which handles that, assuming that you are doing this for printing, and even if you don't want to print, you can always "print to file". | {"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": 0.9131605625152588, "perplexity": 2202.071889560832}, "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-2015-32/segments/1438042988051.33/warc/CC-MAIN-20150728002308-00260-ip-10-236-191-2.ec2.internal.warc.gz"} |
https://www.gradesaver.com/textbooks/math/algebra/algebra-1-common-core-15th-edition/chapter-9-quadratic-functions-and-equations-chapter-review-page-606/55 | ## Algebra 1: Common Core (15th Edition)
The solutions of the system are $(-8,3)$ and $(12,123)$.
$y=6x+51$ $y=x^2+2x-45$ Substitute $6x+51$ for y $6x+51=x^2+2x-45$ $x^2-4x-96=0$ $(x-12)(x+8)=0$ $x+8=0$ or $x-12=0$ $x=-8$ or $x=12$ Find corresponding y-values. Use either original equation $y=6x+51$ $y=6(-8)+51$ or $y=6(12)+51$ $y=3$ or $y=123$ The solutions of the system are $(-8,3)$ and $(12,123)$. | {"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": 0.8496958017349243, "perplexity": 203.37113285132492}, "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-2020-40/segments/1600400279782.77/warc/CC-MAIN-20200927121105-20200927151105-00647.warc.gz"} |
http://mathhelpforum.com/trigonometry/7363-identities.html | 1. ## Identities
Ok, I'm having a bad of a time understanding identities for my pre-calc class...all of trig is hard for me to grasp since this is the first time learning any of it (didn't take trig in high school), so basically my question is with subtraction and addition identities. Say I have a problem that says find the exact value of sin 15 degrees and I break it up with sin (x + y) = sinxsiny + cosxcosy (I think that's the right formula?) so it's sin (45 D + 30 D) = sin45sin30 + cos45cos30...ok that's easy enough but I need to get the exact values for each part to find the final solution and I know for instance that sin45 is square root 2/2 (I think anyways) but how would I find that out without already knowing? If it's relevant, I'm using a 84 SE, thanks.
2. Originally Posted by premedtim
I know for instance that sin45 is square root 2/2 (I think anyways) but how would I find that out without already knowing? If it's relevant, I'm using a 84 SE, thanks.
You should know that 45 degrees is the angle in an equilateral right triangle,
and that the sine of an angle in a right triangle is the side opposite the angle
divided by the hypotenuse (that’s the longest side).
In the case of an equilateral right triangle we apply Pythagoras's theorem
to get the length of the longest side as sqrt(2)s, where s is the length
of one of the other sides.
So sin(45) = cos(45) = s/(sqrt(2)s)=1/sqrt(2)=sqrt(2)/2.
You should also know that if you take an equilateral triangle and cut it into
two congruent right triangles you have two 30, 60, 90 degree angle triangles
with sides 1/2, 1, sqrt(3)/2 times the side of the original equilateral triangle.
Which should be enough to allow you to deduce that sin(30)=1/2, and
that cos(30)=sqrt(3)/2.
Also you should be using:
sin(15) = sin(45-30) = sin(45)cos(30) - cos(45)sin(30)
RonL
3. Hello, premedtim!
You're expected to know the trig values for 30°, 45°, and 60°.
I'll add diagrams to what Captain Black said.
For the 45° angle, we have "half a square".
Code:
+ - - - - - *
: * *
: x * *
: * * 1
: * *
: * 45° *
* * * * * * *
1
If the sides of the square are of length 1,
. . we use Pythagorus to find the length of the diagonal.
Hence:. $x^2\,=\,1^2+1^2\quad\Rightarrow\quad x = \sqrt{2}$
For a 45° angle, the sides of the right triangle are: . $\{1,\,1,\,\sqrt{2}\}$
If you memorize that, you can write all the trig values for 45°.
. . $\begin{array}{cccc}\sin45^o \:= \\ \cos45^o\:= \\ \tan45^o\:= \\ \vdots \end{array}
\begin{array}{cccc}\frac{1}{\sqrt{2}}\:= \\ \frac{1}{\sqrt{2}}\:= \\ \frac{1}{1} \:\:=\\ \vdots\end{array}
\begin{array}{cccc}\frac{\sqrt{2}}{2} \\ \frac{\sqrt{2}}{2} \\ 1 \\ \vdots\end{array}$
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
For the 30° and 60° angles, we have an equilateral triangle.
Draw an altitude, bisecting the vertex angle and the base.
Code:
*
*:*
* : *
* : *
1 *30*: *
* : *
* : _ *
* :½√3 *
* : *
* 60* : *
* * * * * * * * * * *
½
We have a 30-60 right triangle with $base \,= \,\frac{1}{2},\:hyp\,=\,1$
Pythagorus says that: $height\,=\,\frac{\sqrt{3}}{2}$
The sides are: . $\frac{1}{2},\:\frac{\sqrt{3}}{2},\:1$
Multiply by 2: . $\underbrace{1}_\uparrow\:\underbrace{\sqrt{3}}_\up arrow \:\underbrace{2}_\uparrow$
. . . . . . . . .
opp 30° . opp 60° .hyp
If you memorize that, you can recreate these triangles any time.
Code:
*
2 * *
* * 1
* 30° *
* * * * * * * * *
√3
*
**
* *
2 * * _
* *√3
* *
* *
* 60° *
* * * * *
1
4. Ahh ok, thanks guys...so outside 30, 45 and 60 degree angles, it's just repeats of those for say something like 180 degrees? | {"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": 7, "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": 0.9474412202835083, "perplexity": 771.2929535203125}, "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-2017-13/segments/1490218189802.18/warc/CC-MAIN-20170322212949-00460-ip-10-233-31-227.ec2.internal.warc.gz"} |
https://brilliant.org/problems/find-that-factorial/ | # Find that Factorial!
Note that 40585 = 4! + 0! + 5! + 8! + 5! (i.e. the number can be expressed as the sum of the factorial of each of its digits).
There is only one three digit number that can be expressed this way. Find the sum of the digits of that number.
(P.S: I would really appreciate it if somebody could share their complete solutions to this problem.)
× | {"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": 0.9654778242111206, "perplexity": 107.87135435809522}, "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-2020-29/segments/1593655887319.41/warc/CC-MAIN-20200705090648-20200705120648-00029.warc.gz"} |
https://cs.stackexchange.com/questions/70112/decidability-of-checking-whether-all-words-accepted-by-a-tm-have-an-even-number | Decidability of checking whether all words accepted by a TM have an even number of 1s
Let $\Sigma = \{0,1\}$, and consider the language $$L = \{\langle M \rangle \mid M \text{ is a TM and } L(M) \text{ is a subset of } Σ^* \text{ and all strings in this set have an even number of 1s}\}.$$
I think $L$ is undecidable. But I am stuck on proving it. I think it is equivalent to prove $L = \{\langle M \rangle \mid M \text{ is a TM and } L(M) \text{ is a specific set with an even number of 1s}\}$ is undecidable. How can I do some reduction for it?
First, it can be shown easily by Rice's theorem.
Alternatively, you can show a reduction from $\overline{HP}$ as follows:
$$f(\left< M \right>,x)=\left<M_x\right>$$
$M_x$ on input $w$:
1. run $M$ on $x$ for $|w|$ steps.
2. if M stopped then accept.
3. if the number of 1's in $w$ is even then accept, otherwise reject.
I'll let you verify the correctness. | {"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": 0.9713952541351318, "perplexity": 305.51188260072234}, "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/1563195526508.29/warc/CC-MAIN-20190720111631-20190720133631-00503.warc.gz"} |
https://forum.allaboutcircuits.com/threads/how-to-design-a-peak-current-limiter-using-opamps.158592/ | # How to design a peak current limiter using opamps?
#### Xavier Pacheco Paulino
Joined Oct 21, 2015
727
Hi,
I'm simulating a non-inverting buck-boost converter using LTSpice. I'm getting the expected output voltages at each operating point, but the inductor saturates in the transition from buck mode to boost mode. See simulation files and pictures. You can see in Fig1 how blue waveform, which is the inductor current, goes too high at that point indicating that it has saturated. How can I eliminate this behavior? I'm using average current mode control, and I think that's why I have issues with the peak current, as this method does not actually handle it. So I thought of designing another opamp current limiter circuit. What are your thoughts? If you need more details, let me know. I took this exercise from Power Electronics online course on Coursera.
#### Attachments
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• 14.6 KB Views: 2 | {"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": 0.8550949096679688, "perplexity": 2434.4653701065604}, "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-2020-16/segments/1585371700247.99/warc/CC-MAIN-20200407085717-20200407120217-00130.warc.gz"} |
http://math.stackexchange.com/questions/258911/i-need-help-with-proofs-pertaining-to-countability | # I need help with proofs pertaining to countability [duplicate]
Let $A$ and $B$ be countable sets.
(a) Show that $A \times B$ is countable. Hint: Show that there is a bijection from $A\times B$ onto a subset of $\Bbb Z \times\Bbb Z$:
(b) Use induction on $n$ to show that $A_1 \times A_2 \times \ldots \times A_n$ is countable if $A_1, A_2,\ldots, A_n$ are countable.
-
## marked as duplicate by Asaf Karagila, Amr, JSchlather, TMM, QiL Dec 14 '12 at 22:39
Have you tried searching the site? – Asaf Karagila Dec 14 '12 at 21:04
Was the hint not helpful enough? – Hagen von Eitzen Dec 14 '12 at 21:08
Note, though that the proofs in the answers to the question that @Amr mentions are very different from the one suggested in your hint. They are direct proofs; yours makes use of the result that $\Bbb Z\times\Bbb Z$ is countable, which presumably you’ve already proved. – Brian M. Scott Dec 14 '12 at 21:19
Note that the question found by @Asaf covers only (b). – Brian M. Scott Dec 14 '12 at 21:21
@Brian: Thanks, luckily Amr's duplicate covers (a). – Asaf Karagila Dec 14 '12 at 21:26
A small additional nudge for (a): if $f:W\to Y$ and $g:X\to Z$ are injections, the map
$$W\times X\to Y\times Z:\langle w,x\rangle\mapsto\big\langle f(w),g(x)\big\rangle$$ | {"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": 0.8590720295906067, "perplexity": 835.9183447914746}, "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/1429246654264.98/warc/CC-MAIN-20150417045734-00254-ip-10-235-10-82.ec2.internal.warc.gz"} |
https://www.computer.org/csdl/trans/td/2013/08/ttd2013081546-abs.html | Issue No. 08 - Aug. (2013 vol. 24)
ISSN: 1045-9219
pp: 1546-1555
Yiling Yang , Nanjing University, Nanjing
Yu Huang , Nanjing University, Nanjing
Jiannong Cao , The Hong Kong Polytechnic University, Hong Kong
Xiaoxing Ma , Nanjing University, Nanjing
Jian Lu , Nanjing University, Nanjing
ABSTRACT
Formal specification and runtime detection of contextual properties is one of the primary approaches to enabling context awareness in pervasive computing environments. Due to the intrinsic dynamism of the pervasive computing environment, dynamic properties, which delineate concerns of context-aware applications on the temporal evolution of the environment state, are of great importance. However, detection of dynamic properties is challenging, mainly due to the intrinsic asynchrony among computing entities in the pervasive computing environment. Moreover, the detection must be conducted at runtime in pervasive computing scenarios, which makes existing schemes do not work. To address these challenges, we propose the property detection for asynchronous context (PDAC) framework, which consists of three essential parts: 1) Logical time is employed to model the temporal evolution of environment state as a lattice. The active surface of the lattice is introduced as the key notion to model the runtime evolution of the environment state; 2) Specification of dynamic properties is viewed as a formal language defined over the trace of environment state evolution; and 3) The SurfMaint algorithm is proposed to achieve runtime maintenance of the active surface of the lattice, which further enables runtime detection of dynamic properties. A case study is conducted to demonstrate how the PDAC framework enables context awareness in asynchronous pervasive computing scenarios. The SurfMaint algorithm is implemented and evaluated over MIPA--the open-source context-aware middleware we developed. Performance measurements show the accuracy and cost-effectiveness of SurfMaint, even when faced with dynamic changes in the asynchronous pervasive computing environment.
INDEX TERMS
Lattices, Runtime, Labeling, Pervasive computing, Context, Heuristic algorithms, Formal languages, predicate detection, Lattices, Runtime, Labeling, Pervasive computing, Context, Heuristic algorithms, Formal languages, lattice, Dynamic property, context awareness, asynchrony
CITATION
J. Lu, X. Ma, Y. Huang, J. Cao and Y. Yang, "Formal Specification and Runtime Detection of Dynamic Properties in Asynchronous Pervasive Computing Environments," in IEEE Transactions on Parallel & Distributed Systems, vol. 24, no. , pp. 1546-1555, 2013.
doi:10.1109/TPDS.2012.259 | {"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": 0.868256151676178, "perplexity": 3884.0554109302025}, "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-2018-26/segments/1529267866358.52/warc/CC-MAIN-20180624044127-20180624064127-00050.warc.gz"} |
https://www.physicsforums.com/threads/understanding-dual-basis-scale-factors.822257/ | # Understanding dual basis & scale factors
1. Jul 7, 2015
### Incand
I'm confused by the following passage in our book (translated).
An alternative too choosing the normed tangent vectors $\vec e_i = \frac{1}{h_1}\frac{\partial \vec r}{\partial u_i}$ with scale factors $h_i = \left| \frac{\partial \vec r }{\partial u_i} \right|$ is to choose the normal vector $\nabla u_i$.
If the system is ortogonal the vectors $\nabla u_i$ and $\frac{\partial \vec r}{\partial u_i}$ point in the same direction so we can write $\vec e_i = h_i \nabla u_i$.
$\{ u_i \}_{i=1}^3$ is supposed to be curvilinear coordinates with a transformation describing the position $\vec r = \vec r (u_1,u_2,u_3)$.
What does it mean to take $\nabla u_i$? am I supposed to express $u_i$ in cartesian coordinates and then take the gradient? And why multiply with the scale factors instead of dividing?
Another question I'm wondering about is if it's always true that the jacobian $J$ is equal too $h_1h_2h_3$.
2. Jul 7, 2015
### Orodruin
Staff Emeritus
That would be one way of doing it yes. The coordinates $u_i$ are functions and therefore you can take the gradient of them.
Because this is how you would obtain normalised base vectors. The norm of the dual basis is the reciprocal of the norm of the tangent vector basis.
It is only true in orthogonal coordinate systems (which are the only ones for which it makes sense to deal with the scale factors rather than the full metric tensor).
Similar Discussions: Understanding dual basis & scale factors | {"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": 0.9439123272895813, "perplexity": 264.66398458521076}, "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-2017-34/segments/1502886109470.15/warc/CC-MAIN-20170821172333-20170821192333-00319.warc.gz"} |
http://www.techyv.com/questions/how-manage-paragraph-and-line-spacing/ | ## How to manage the Paragraph and Line spacing?
Asked By 220 points N/A Posted on -
Hello friends,
As a new user, I want to try type writing. It has lots of options to learn typing. File create, save file, font change, size maximum or minimum and a lot of functions. Actually I can already learn this thing, but not nicely arranged. Need to manage line space matchup others. Then my paragraph will be right to able print. Please help.
Thanks.
SHARE
Answered By 0 points N/A #127628
## How to manage the Paragraph and Line spacing?
Hello Mr. Red,
I understand how confusing it might be for a new user. When I first started working with Microsoft Word, I got overwhelmed with all the controls, too. But once you figure everything out, it will be so easy.
Here are some tips I can give you:
Microsoft Word 2010.
To edit the spacing between paragraphs, click on Page Layout, then you will see the options there for the spacing. Set it to "1" if you want single spacing.
Click on the Margins icon to set your margins. The default is 1 inch on all sides.
You can actually drag your mouse to the icons and a "tip" will be shown. Just let your mouse cursor hover and the tip will show.
You can also click on the small box with the error (see encircled object on screenshot below), to open the paragraph settings.
Once you open the Paragraph dialog box, click on the dropdown where it says spacing, then choose the desired spacing (single, 1.5 or double).
To change the justification or alignment of the paragraphs, just click on the Home button. In the center you will see there "paragraph" and slightly above it, the justification of the paragraphs. Just choose whether you want it centered, or justified (even on both sides), justified to the left (only the left side is even), justified to the right.
Word 2003
Word 2003 is easier to use. Just click on Format, then go to Paragraph, and you will see the same options for the paragraph settings.
Most of the shortcuts are also shown on the menu bar.
You can also try these keyboard shortcuts:
When copying text, highlight the text by holding the left click button. Hold the CTRL key on your keyboard then press C to copy.
To paste, just click on where you want to paste the text then press CTRL and V.
Hope this will make your typing easier!
Answered By 0 points N/A #127629
## How to manage the Paragraph and Line spacing?
How to manage the Paragraph and Line spacing?
Dear Mr. Red thanks for your question about How to manage the paragraph and Line spacing.
Here we can solve how to:
In Microsoft Office open word document paragraph spacing means a white area in between two paragraphs and after paragraphs. Same as line spacing means also a white area in between two lines.
We can manage or adjust Paragraph and line spacing, in open word documents first we place the insertion point onto the Format menu option then Scroll down Select Paragraph sub menu, see picture:1
Now we click paragraph sub menu, then Paragraph dialog box window appear. See picture: 2
Here we select/clicks Indents and Spacing tab from the paragraph dialog box.
Look down spacing select box then specify amount of space before and type the number 6pt/12pt/18pt what's needed, same doing on space after and type the number 6pt/12pt/18pt what's needed and then select line spacing from line spacing box (Single, 1.5 lines, Double) etc. what's needed.
These ways we can solved our Paragraph and Line spacing problem and You also.
Thanks. | {"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": 0.9076985716819763, "perplexity": 3002.9266595316153}, "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-2017-34/segments/1502886124662.41/warc/CC-MAIN-20170823225412-20170824005412-00600.warc.gz"} |
https://forum.allaboutcircuits.com/threads/3-db-point-of-non-inv-op-amp-with-capacitor-in-parallel-with-feedback-resistor.69639/ | # -3 dB point of non-inv op-amp with capacitor in parallel with feedback resistor
Discussion in 'Homework Help' started by Mango Symphony, May 3, 2012.
1. ### Mango Symphony Thread Starter New Member
Mar 1, 2011
7
0
Hi
For a non-inverting op-amp, it is easy to show that the gain is given by 1 + RF/R1, where RF is the feedback resistor between the op-amp output and the inverting input, and R1 is the resistor between the inverting input and ground.
With a capacitor added across RF, is easy to see intuitively that as frequency increases, the reactance of the capacitor decreases; therefore the impedance of the cap and RF in parallel becomes dominated by the reactance of the cap. Thus the expression 1 + RF/R1 tends towards 1, because the RF/R1 part tends to 0. So it is clear that a low-pass filter has been implemented - the higher the frequency, the more the gain reduces from 1 + RF/R1 towards 1, where it will stay.
However, using complex arithmetic, I am unsure how to predict where the - 3 dB point will occur, or to be absolutely clear, the frequency where the gain has fallen from the midband gain (which is given by 1 + RF/R1) to 0.707, or 1/√2 of that value.
I assumed initially, that the cap and RF will form the time constant and thus determine where the gain has fallen 3 dB from the midband value. But after deriving the transfer function of the above circuit, in operator j, I get the following:
(1) H(jω) = 1 + (RF/R1 x 1/√(1+jωxRFxC))
I then turned the whole expression on the RHS from rectangular to polar form in order to get an expression for the magnitude of the overall transfer function.
My plan then was to set this equal to 1/√2, and solve for ω.
However, when I did, it seemed a real pickle as I ended up with ω to the power of 4 and all sorts of mess!
It would have made sense to me if the number crunching had returned the following:
(2) H(jω) = (1 + RF/R1) x 1/√(1+jωxRFxC))
NB note the difference between this and (1); in (1) the '1' at the start is outside the brackets.
It would have made sense because... here we have 2 transfer functions, the first is real and constant and is the expression for the midband gain; the second is that of a LPF, whose value will become 0.707 when ωRC = 1, and everything will be nice and simple, and the reciprocal of RC will indeed define the radian - 3 dB point.
Am I overlooking some complex number issue here?
If anyone one can spot where I am going wrong on this I would be most grateful. The derivation does not appear in any of my text books or on the internet!
Thanks
Mango
2. ### t_n_k AAC Fanatic!
Mar 6, 2009
5,448
789
It's a good deal of algebraic manipulation to come up with a result but I end up with
$\omega_{(-3dB)}=\frac{A_{DC}}{R_FC_F\sqrt{(A_{DC}^2 -2)}}$
Where
$A_{DC}=1+\frac{R_F}{R_1}$
3. ### Mango Symphony Thread Starter New Member
Mar 1, 2011
7
0
Did you follow a similar procedure to mine, where you convert the a + jb expression to polar to find the magnitude, or some other method?
Here is my working if anyone is interested, with R1 substituted for RF, and R2 for R1 (sorry for the confusion!).
I feel like I am missing a point here on how to go about the exercise...
File size:
115.1 KB
Views:
27
4. ### t_n_k AAC Fanatic!
Mar 6, 2009
5,448
789
I checked the first four lines and the gain looks fine. I don't have the patience to look much further but the approach looks much the same as I would take.
What it boils down to is finding the frequency at which the gain has fallen to the DC gain divided by √2.
You will obtain a quadratic equation in ω^2 as the unknown variable - even though it looks like a fourth order equation. You solve for the real root with ω^2 as the unknown then take the square root of the positive real root to give the actual -3dB angular frequency.
My approach was identical with the exception I got rid of a lot of the clutter by deducing that the gain is given by
$A_{v}=\frac{A_{DC}\omega_{F}+j \omega}{ \omega_{F}+j\omega}$
Where for your component tags I have the composite tags
$A_{DC}=1+\frac{R_1}{R_2}$
$\omega_{F}=\frac{1}{R_1C_1}$
I then worked through the process using the composite tags. This made the algebraic burden more bearable & less prone to error for me.
Last edited: May 4, 2012
5. ### Jony130 AAC Fanatic!
Feb 17, 2009
4,480
1,264
But do we really need to do all those calculations to find -3 dB point?
This circuit is a simply first order circuit. So for first order circuit the -3 dB point isn't always equal
$Fc = \frac{1}{2 \pi C R}$
So for the circuit under discussion the transfer function is equal to
$\frac{Vout}{Vin} = 1 + \frac{Rf||Xc1}{R1} = 1 + \frac{\frac{Rf}{1 + j \omega Rf C1} }{R1} = \frac{ R1 + Rf + j\omega Rf C1}{R1(1 + j \omega Rf C1)} = 1+\frac{Rf}{R1} * \frac{1+ j\omega \frac{Rf*R1}{Rf+R1}C1}{1+ j\omega Rf C1}$
So we have one Pole at
$F_p = \frac{1}{2 \pi C1 Rf}$
And one Zero at
$F_Z = \frac{1}{ 2 \pi \frac{Rf*R1}{Rf+R1}C1}$
File size:
14.3 KB
Views:
133
6. ### t_n_k AAC Fanatic!
Mar 6, 2009
5,448
789
The -3dB point in this case depends on the DC Gain. At DC gains of less than 5 there is a modest difference in the typical 1/(RC) value for a first order case. At higher gains the difference is negligible.
In the equation in post #2 the factor of interest would be
$\frac{A_{DC}}{\sqrt{(A_{DC}^2-2)}}$
At a DC gain of 10 the the factor is 1.01, or close enough to unity.
Similarly
If DC gain = 5 the factor is 1.04
If DC gain = 1.5 the factor is 3.0
So with the feedback R=100k and C=10nF
The nominal frequency 1/(2∏RC)=159Hz
At Adc=10 the actual -3dB frequency is 161Hz.
At Adc=5 the actual -3dB frequency is 165Hz.
At Adc=1.5 the actual -3dB frequency is 477Hz.
Which is probably all fairly academic, albeit with some vague interest.
Last edited: May 4, 2012
7. ### The Electrician AAC Fanatic!
Oct 9, 2007
2,576
461
I notice that t_n_k says in post #2:
"It's a good deal of algebraic manipulation to come up with a result but I end up with"
And again in post #4:
"I then worked through the process using the composite tags. This made the algebraic burden more bearable & less prone to error for me."
Nowadays when I see a problem like this I say to myself, "It's good for a beginning student to go through all the algebra by hand in the beginning, but for myself, let's see in how few steps I can get a modern symbolic algebra program to produce the answer".
Herewith the Mathematica result in an attachment:
File size:
16.8 KB
Views:
37
8. ### t_n_k AAC Fanatic!
Mar 6, 2009
5,448
789
Thanks Electrician - maybe I should try Mathematica.
On the other hand the doctors & chiropractors keep telling me that an older person such as myself needs lots of mental gymnastics to reduce the likelihood of early dementia.
9. ### Mango Symphony Thread Starter New Member
Mar 1, 2011
7
0
Thank you very much for all your help with this.
t_n_k - could you briefly describe or show how you arrive at your expression
$image=http://forum.allaboutcircuits.com/mimetex.cgi?A_{v}=\frac{A_{DC}\omega_{F}+j%20\omega}{%20\omega_{F}+j\omega}&hash=7dee1c832c062750460266d6c0f7ab68$
Thanks again,
Mango
10. ### t_n_k AAC Fanatic!
Mar 6, 2009
5,448
789
$Z_F=\frac{R_F\frac{1}{j\omega C_F}}{R_F+\frac{1}{j\omega C_F}}=\frac{R_F}{1+j\omega C_F R_F}$
$A_v=\frac{Z_F}{R_1}+1=\frac{\frac{R_F}{1+j\omega C_F R_F}}{R_1}+1=\frac{R_F+R_1(1+j\omega C_F R_F)}{R_1(1+j\omega C_F R_F)}$
or
$A_v=\frac{R_F+R_1+j\omega C_FR_1R_F}{R_1C_FR_F(\frac{1}{C_FR_F}+j\omega)} = \frac{R_1C_FR_F$$( \frac{R_1+R_F}{R_1C_FR_F})+j\omega$$}{R_1C_FR_F(\frac{1}{C_FR_F}+j\omega)}=\frac{ ( \frac{R_1+R_F}{R_1C_FR_F})+j\omega }{(\frac{1}{C_FR_F}+j\omega)}$
or
$A_v=\frac{\frac{1}{R_FC_F}$$\frac{R_1+R_F}{R_1}$$+j\omega }{$$\frac{1}{C_FR_F}+j\omega$$}=\frac{\omega_F A_{DC}+j\omega}{\omega_F+j\omega}=\frac{A_{DC} \omega_F +j\omega}{\omega_F+j\omega}$
where
$\omega_F=\frac{1}{R_FC_F}$
and
$A_{DC}=\frac{R_F}{R_1}+1=\frac{R_1+R_F}{R_1}$
11. ### Mango Symphony Thread Starter New Member
Mar 1, 2011
7
0
Thanks t_n_k
I have to say that is some very nifty algebraic manipulation...
Is there a term/description to this type of parameter juggling, where you are expressing the complex transfer function in terms of the time constant formed by Rf and Cf, and the mid-band/DC gain, in order to get a final expression that lends itself to actually finding out something useful? I couldn't see this wood for my trees when I ended up with the quadratic...
I.e. is this a known standard procedure to try out when dealing with transfer functions or were you just being nifty and clever?
Mango
12. ### t_n_k AAC Fanatic!
Mar 6, 2009
5,448
789
Nothing particularly "nifty" in my methods. I try to see patterns in formulae which might lead to simplifications of this type. Algebra can be a drudgery sometimes. In the end it probably comes down to years of 'playing' with these ideas.
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17,442 | {"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": 17, "/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": 0.804470419883728, "perplexity": 1432.8111907490375}, "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-2018-43/segments/1539583512268.20/warc/CC-MAIN-20181019020142-20181019041642-00316.warc.gz"} |
http://mathoverflow.net/questions/47601/lebesgue-measure-on-frechet-space | # Lebesgue measure on Frechet space?
It is well known that there are no Lebesgue measures on infinite-dimensional Banach spaces (see e.g. http://en.wikipedia.org/wiki/There_is_no_infinite-dimensional_Lebesgue_measure). However, I couldn't find anything about Lebesgue measures on infinite-dimensional Frechet spaces. The question seems very natural in the context of creating a mathematically rigorous definition of the path integral of quantum field theory.
So: 1. Can an infinite-dimensional Frechet space have a measure which is locally finite, strictly positive and translation-invariant? 2. Can a separable infinite-dimensional Frechet space have a measure which is locally finite, non-zero and translation-invariant?
The motivation for the formulations of 1+2 are that the analogous statements for Banach spaces are false. 2 is more important for me since all interesting examples as separable, as far as I can tell.
If such measures exist, I'd be glad to get some references on whatever is known about them.
Thx!
-
No. The proof on the Wikipedia page you cite can be adapted to the Frechet setting. – Ryan Budney Nov 28 '10 at 18:51
If you're interested in fresh approaches to defining the path integral, I recommend Kevin Costello's in-progress book on the topic: math.northwestern.edu/~costello/renormalization – j.c. Nov 28 '10 at 21:41
In an appendix to Weil's book on integration on topological groups he proves a theorem to the effect that a topological group with a measure that resembles Haar measure (there are precise conditions, which I don't recall right now) has to be locally compact. There is a theorem of Riesz that any locally compact Hausdorff topological real vector space is finite-dimensional. (Well, Riesz gave the proof only for normed real vector spaces, but in Weil's Basic Number Theory I think he gives a proof for top. vector spaces.) – KConrad Nov 29 '10 at 5:01
The same argument works for any infinite dimensional separable TVS $X$. A non-negative and non-zero measure $\mu$ on $X$ vanishes on no non-empty open set, because countably many translates of the latter cover $X$. On the other hand, any non-empty open set of $X$ does contains infinitely many disjoint translates of some non-empty open set, thus it has infinite $\mu$ measure. – Pietro Majer May 31 '11 at 20:39 | {"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": 0.9804384112358093, "perplexity": 286.8557916559991}, "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-2015-11/segments/1424936462710.27/warc/CC-MAIN-20150226074102-00272-ip-10-28-5-156.ec2.internal.warc.gz"} |
http://math.stackexchange.com/questions/103071/negative-binomial-distribution | # Negative Binomial Distribution
Why is the negative binomial distribution defined as $$P(X=x|r,p)= \binom{x-1}{r-1}p^{r}(1-p)^{x-r}$$
Basically this is the probability that $x$ Bernoulli trials are needed for $r$ successes. So we need $r-1$ successes in the first $x-1$ trials. Then success on the $r^{th}$ trial happens with probability $p$. Why can't we write it as the following:
$$P(X = x|r,p) = \binom{x}{r}p^{r} (1-p)^{x-r}$$
This means that you have $r$ successes in the first $x$ trials.
-
You wrote that $x$ trials are needed for $r$ successes. That means in particular that $x-1$ trials were not enough. So we hit our goal of $r$ successes at the $x$-th trial.
Suppose for example that we are tossing a fair coin until we get the first head. What is the probability that $2$ tosses are needed for $1$ success? Here $x=2$ and $r=1$. Two tosses are needed precisely if we get TH. This has probability $1/4$. By way of contrast, the probability of exactly one head in two tosses is $1/2$.
-
@David Mitra: Thanks for spotting the missing not! Fixed. – André Nicolas Jan 27 '12 at 20:34
One reason the negative binomial distribution is written that way is that $$\sum_{x=r}^\infty \binom{x-1}{r-1}p^{r}(1-p)^{x-r} =1$$ while $\sum_{x=r}^\infty \binom{x}{r}p^{r}(1-p)^{x-r} \gt 1$ and so is not a probability distribution.
The cause of this is non-mutually exclusive events: if you take a sequence of trials, then that sequence can include both 4 trials and 2 successes, and (if the fifth trial is then a failure) 5 trials and 2 successes; you only get mutually exclusive events if you stop when you first have 2 successes.
An alternative approach would be to look at $Y$ the number of failures before the $r$th success with $\Pr(Y=y) = \binom{y+r-1}{r-1}p^{r}(1-p)^{y}$ noting $$\sum_{y=0}^\infty \binom{y+r-1}{r-1}p^{r}(1-p)^{y} =1.$$
You can of course look at the binomial distribution, taking the sum over $r$ (with $x$ fixed) using your formulation and getting $$\sum_{r=0}^x \binom{x}{r}p^{r}(1-p)^{x-r} = 1.$$
- | {"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": 0.9468426704406738, "perplexity": 218.72734624951744}, "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-2015-32/segments/1438042987402.78/warc/CC-MAIN-20150728002307-00067-ip-10-236-191-2.ec2.internal.warc.gz"} |
http://mathhelpforum.com/discrete-math/38587-counting-question-print.html | # Counting question?
• May 17th 2008, 04:38 AM
sakhr98
Counting question?
Greetings,
I have the following question in couting. It might be straight forward but I could not find a formula for the answer.
Let N and m be positive integers.
How many combinations are there of the form:
n1, n2, ..., nN such that
n1 >=0, n2>=0, ..., nN >= 0, and
n1 + n2 + n3 + ... + nN = m.
Example: N = 3, m = 4:
All combinations are given by
1) 4, 0, 0,
2) 3, 1, 0,
3) 3, 0, 1,
4) 2, 2, 0,
5) 2, 1, 1,
6) 2, 0, 1,
7) 1, 3, 0,
8) 1, 2, 1,
9) 1, 1, 2,
10) 1, 0, 3,
11) 0, 4, 0,
12) 0, 3, 1,
13) 0, 2, 2,
14) 0, 1, 3,
15) 0, 0, 4
Therefore, the number of combinations for this example is 15.
Best regards.
• May 17th 2008, 04:46 AM
Moo
Hello,
Quote:
Originally Posted by sakhr98
Greetings,
I have the following question in couting. It might be straight forward but I could not find a formula for the answer.
Let N and m be positive integers.
How many combinations are there of the form:
n1, n2, ..., nN such that
n1 >=0, n2>=0, ..., nN >= 0, and
n1 + n2 + n3 + ... + nN = m.
Example: N = 3, m = 4:
All combinations are given by
1) 4, 0, 0,
2) 3, 1, 0,
3) 3, 0, 1,
4) 2, 2, 0,
5) 2, 1, 1,
6) 2, 0, 1,
7) 1, 3, 0,
8) 1, 2, 1,
9) 1, 1, 2,
10) 1, 0, 3,
11) 0, 4, 0,
12) 0, 3, 1,
13) 0, 2, 2,
14) 0, 1, 3,
15) 0, 0, 4
Therefore, the number of combinations for this example is 15.
Best regards.
If I've understood correctly your question, the formula is :
${m \choose N+m-1}={N-1 \choose N+m-1}$
I can try to explain..
• May 17th 2008, 05:19 AM
sakhr98
Thank you for the prompt reply.
I am not sure I get your answer. For the example I put, N = 3 and m = 4, substituting in your formula results in C(2, 6) = C(4, 6).
But what is the definition of C(2,6)?
Best regards, Ashraf.
• May 17th 2008, 05:21 AM
Moo
The definition ?
$C(p,n)={p \choose n}=\frac{n!}{(n-p)!p!}$
Is it what you want ?
• May 17th 2008, 05:34 AM
sakhr98
I apologize for my hastiness. Your answer is correct except for the typo where the answer is C(N+m-1, m) = C(N+m-1, N-1).
Thank you very much.
• May 17th 2008, 05:39 AM
Moo
Quote:
Originally Posted by sakhr98
I apologize for my hastiness. Your answer is correct except for the typo where the answer is C(N+m-1, m) = C(N+m-1, N-1).
Thank you very much.
Yeah, I have real difficulties with this formula...
You're welcome
• May 18th 2008, 06:47 PM
awkward
Quote:
Originally Posted by Moo
Hello,
If I've understood correctly your question, the formula is :
${m \choose N+m-1}={N-1 \choose N+m-1}$
I can try to explain..
Ahem.
Moo, I believe the usual convention on the notation for combinations is that $\binom{n}{m}$ is the number of combinations of n objects taken m at a time.
$\binom{N+m-1}{m} = \binom{N+m-1}{N-1}$. | {"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": 5, "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": 0.8037901520729065, "perplexity": 1675.311157099745}, "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-2017-51/segments/1512948568283.66/warc/CC-MAIN-20171215095015-20171215115015-00031.warc.gz"} |
http://en.wikipedia.org/wiki/Coherence_length | # Coherence length
Jump to: navigation, search
In physics, coherence length is the propagation distance over which a coherent wave (e.g. an electromagnetic wave) maintains a specified degree of coherence. Wave interference is strong when the paths taken by all of the interfering waves differ by less than the coherence length. A wave with a longer coherence length is closer to a perfect sinusoidal wave. Coherence length is important in holography and telecommunications engineering.
This article focuses on the coherence of classical electromagnetic fields. In quantum mechanics, there is a mathematically analogous concept of the quantum coherence length of a wave function.
## Formulas
In radio-band systems, the coherence length is approximated by
$L={c \over n\, \Delta f},$
where c is the speed of light in a vacuum, n is the refractive index of the medium, and $\Delta f$ is the bandwidth of the source.
In optical communications, the coherence length $L$ is given by
$L={2 \ln(2) \over \pi n} {\lambda^2 \over \Delta\lambda},$
where $\lambda$ is the central wavelength of the source, $n$ is the refractive index of the medium, and $\Delta\lambda$ is the spectral width of the source. If the source has a Gaussian spectrum with FWHM spectral width $\Delta\lambda$, then a path offset of ±$L$ will reduce the fringe visibility to 50%.
Coherence length is usually applied to the optical regime.
The expression above is a frequently used approximation. Due to ambiguities in the definition of spectral width of a source, however, the following definition of coherence length has been suggested:
The coherence length can be measured using a Michelson interferometer and is the optical path length difference of a self-interfering laser beam which corresponds to a $1/e=37%$ fringe visibility,[1] where the fringe visibility is defined as
$V = {I_\max - I_\min \over I_\max + I_\min} ,\,$
where $I$ is the fringe intensity.
In long-distance transmission systems, the coherence length may be reduced by propagation factors such as dispersion, scattering, and diffraction.
## Lasers
Multimode helium–neon lasers have a typical coherence length of 20 cm, while the coherence length of singlemode ones can exceed 100 m. Semiconductor lasers reach some 100 m. Singlemode fiber lasers with linewidths of a few kHz can have coherence lengths exceeding 100 km. Similar coherence lengths can be reached with optical frequency combs due to the narrow linewidth of each tooth. Non-zero visibility is present only for short intervals of pulses repeated after cavity length distances up to this long coherence length.
## References
1. ^ Ackermann, Gerhard K. (2007). Holography: A Practical Approach. Wiley-VCH. ISBN 3-527-40663-8. | {"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": 0.9249675273895264, "perplexity": 634.1656984095887}, "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-2014-35/segments/1408500835872.63/warc/CC-MAIN-20140820021355-00220-ip-10-180-136-8.ec2.internal.warc.gz"} |
https://terrytao.wordpress.com/2020/01/25/equidistribution-of-syracuse-random-variables-and-density-of-collatz-preimages/ | Define the Collatz map ${\mathrm{Col}: {\bf N}+1 \rightarrow {\bf N}+1}$ on the natural numbers ${{\bf N}+1 = \{1,2,\dots\}}$ by setting ${\mathrm{Col}(N)}$ to equal ${3N+1}$ when ${N}$ is odd and ${N/2}$ when ${N}$ is even, and let ${\mathrm{Col}^{\bf N}(N) := \{ N, \mathrm{Col}(N), \mathrm{Col}^2(N), \dots \}}$ denote the forward Collatz orbit of ${N}$. The notorious Collatz conjecture asserts that ${1 \in \mathrm{Col}^{\bf N}(N)}$ for all ${N \in {\bf N}+1}$. Equivalently, if we define the backwards Collatz orbit ${(\mathrm{Col}^{\bf N})^*(N) := \{ M \in {\bf N}+1: N \in \mathrm{Col}^{\bf N}(M) \}}$ to be all the natural numbers ${M}$ that encounter ${N}$ in their forward Collatz orbit, then the Collatz conjecture asserts that ${(\mathrm{Col}^{\bf N})^*(1) = {\bf N}+1}$. As a partial result towards this latter statement, Krasikov and Lagarias in 2003 established the bound
$\displaystyle \# \{ N \leq x: N \in (\mathrm{Col}^{\bf N})^*(1) \} \gg x^\gamma \ \ \ \ \ (1)$
for all ${x \geq 1}$ and ${\gamma = 0.84}$. (This improved upon previous values of ${\gamma = 0.81}$ obtained by Applegate and Lagarias in 1995, ${\gamma = 0.65}$ by Applegate and Lagarias in 1995 by a different method, ${\gamma=0.48}$ by Wirsching in 1993, ${\gamma=0.43}$ by Krasikov in 1989, ${\gamma=0.3}$ by Sander in 1990, and some ${\gamma>0}$ by Crandall in 1978.) This is still the largest value of ${\gamma}$ for which (1) has been established. Of course, the Collatz conjecture would imply that we can take ${\gamma}$ equal to ${1}$, which is the assertion that a positive density set of natural numbers obeys the Collatz conjecture. This is not yet established, although the results in my previous paper do at least imply that a positive density set of natural numbers iterates to an (explicitly computable) bounded set, so in principle the ${\gamma=1}$ case of (1) could now be verified by an (enormous) finite computation in which one verifies that every number in this explicit bounded set iterates to ${1}$. In this post I would like to record a possible alternate route to this problem that depends on the distribution of a certain family of random variables that appeared in my previous paper, that I called Syracuse random variables.
Definition 1 (Syracuse random variables) For any natural number ${n}$, a Syracuse random variable ${\mathbf{Syrac}({\bf Z}/3^n{\bf Z})}$ on the cyclic group ${{\bf Z}/3^n{\bf Z}}$ is defined as a random variable of the form
$\displaystyle \mathbf{Syrac}({\bf Z}/3^n{\bf Z}) = \sum_{m=1}^n 3^{n-m} 2^{-{\mathbf a}_m-\dots-{\mathbf a}_n} \ \ \ \ \ (2)$
where ${\mathbf{a}_1,\dots,\mathbf{a_n}}$ are independent copies of a geometric random variable ${\mathbf{Geom}(2)}$ on the natural numbers with mean ${2}$, thus
$\displaystyle \mathop{\bf P}( \mathbf{a}_1=a_1,\dots,\mathbf{a}_n=a_n) = 2^{-a_1-\dots-a_n}$
} for ${a_1,\dots,a_n \in {\bf N}+1}$. In (2) the arithmetic is performed in the ring ${{\bf Z}/3^n{\bf Z}}$.
Thus for instance
$\displaystyle \mathrm{Syrac}({\bf Z}/3{\bf Z}) = 2^{-\mathbf{a}_1} \hbox{ mod } 3$
$\displaystyle \mathrm{Syrac}({\bf Z}/3^2{\bf Z}) = 2^{-\mathbf{a}_1-\mathbf{a}_2} + 3 \times 2^{-\mathbf{a}_2} \hbox{ mod } 3^2$
$\displaystyle \mathrm{Syrac}({\bf Z}/3^3{\bf Z}) = 2^{-\mathbf{a}_1-\mathbf{a}_2-\mathbf{a}_3} + 3 \times 2^{-\mathbf{a}_2-\mathbf{a}_3} + 3^2 \times 2^{-\mathbf{a}_3} \hbox{ mod } 3^3$
and so forth. After reversing the labeling of the ${\mathbf{a}_1,\dots,\mathbf{a}_n}$, one could also view ${\mathrm{Syrac}({\bf Z}/3^n{\bf Z})}$ as the mod ${3^n}$ reduction of a ${3}$-adic random variable
$\displaystyle \mathbf{Syrac}({\bf Z}_3) = \sum_{m=1}^\infty 3^{m-1} 2^{-{\mathbf a}_1-\dots-{\mathbf a}_m}.$
The probability density function ${b \mapsto \mathbf{P}( \mathbf{Syrac}({\bf Z}/3^n{\bf Z}) = b )}$ of the Syracuse random variable can be explicitly computed by a recursive formula (see Lemma 1.12 of my previous paper). For instance, when ${n=1}$, ${\mathbf{P}( \mathbf{Syrac}({\bf Z}/3{\bf Z}) = b )}$ is equal to ${0,1/3,2/3}$ for ${x=b,1,2 \hbox{ mod } 3}$ respectively, while when ${n=2}$, ${\mathbf{P}( \mathbf{Syrac}({\bf Z}/3^2{\bf Z}) = b )}$ is equal to
$\displaystyle 0, \frac{8}{63}, \frac{16}{63}, 0, \frac{11}{63}, \frac{4}{63}, 0, \frac{2}{63}, \frac{22}{63}$
when ${b=0,\dots,8 \hbox{ mod } 9}$ respectively.
The relationship of these random variables to the Collatz problem can be explained as follows. Let ${2{\bf N}+1 = \{1,3,5,\dots\}}$ denote the odd natural numbers, and define the Syracuse map ${\mathrm{Syr}: 2{\bf N}+1 \rightarrow 2{\bf N}+1}$ by
$\displaystyle \mathrm{Syr}(N) := \frac{3n+1}{2^{\nu_2(3N+1)}}$
where the ${2}$valuation ${\nu_2(3n+1) \in {\bf N}}$ is the number of times ${2}$ divides ${3N+1}$. We can define the forward orbit ${\mathrm{Syr}^{\bf N}(n)}$ and backward orbit ${(\mathrm{Syr}^{\bf N})^*(N)}$ of the Syracuse map as before. It is not difficult to then see that the Collatz conjecture is equivalent to the assertion ${(\mathrm{Syr}^{\bf N})^*(1) = 2{\bf N}+1}$, and that the assertion (1) for a given ${\gamma}$ is equivalent to the assertion
$\displaystyle \# \{ N \leq x: N \in (\mathrm{Syr}^{\bf N})^*(1) \} \gg x^\gamma \ \ \ \ \ (3)$
for all ${x \geq 1}$, where ${N}$ is now understood to range over odd natural numbers. A brief calculation then shows that for any odd natural number ${N}$ and natural number ${n}$, one has
$\displaystyle \mathrm{Syr}^n(N) = 3^n 2^{-a_1-\dots-a_n} N + \sum_{m=1}^n 3^{n-m} 2^{-a_m-\dots-a_n}$
where the natural numbers ${a_1,\dots,a_n}$ are defined by the formula
$\displaystyle a_i := \nu_2( 3 \mathrm{Syr}^{i-1}(N) + 1 ),$
so in particular
$\displaystyle \mathrm{Syr}^n(N) = \sum_{m=1}^n 3^{n-m} 2^{-a_m-\dots-a_n} \hbox{ mod } 3^n.$
Heuristically, one expects the ${2}$-valuation ${a = \nu_2(N)}$ of a typical odd number ${N}$ to be approximately distributed according to the geometric distribution ${\mathbf{Geom}(2)}$, so one therefore expects the residue class ${\mathrm{Syr}^n(N) \hbox{ mod } 3^n}$ to be distributed approximately according to the random variable ${\mathbf{Syrac}({\bf Z}/3^n{\bf Z})}$.
The Syracuse random variables ${\mathbf{Syrac}({\bf Z}/3^n{\bf Z})}$ will always avoid multiples of three (this reflects the fact that ${\mathrm{Syr}(N)}$ is never a multiple of three), but attains any non-multiple of three in ${{\bf Z}/3^n{\bf Z}}$ with positive probability. For any natural number ${n}$, set
$\displaystyle c_n := \inf_{b \in {\bf Z}/3^n{\bf Z}: 3 \nmid b} \mathbf{P}( \mathbf{Syrac}({\bf Z}/3^n{\bf Z}) = b ).$
Equivalently, ${c_n}$ is the greatest quantity for which we have the inequality
$\displaystyle \sum_{(a_1,\dots,a_n) \in S_{n,N}} 2^{-a_1-\dots-a_m} \geq c_n \ \ \ \ \ (4)$
for all integers ${N}$ not divisible by three, where ${S_{n,N} \subset ({\bf N}+1)^n}$ is the set of all tuples ${(a_1,\dots,a_n)}$ for which
$\displaystyle N = \sum_{m=1}^n 3^{m-1} 2^{-a_1-\dots-a_m} \hbox{ mod } 3^n.$
Thus for instance ${c_0=1}$, ${c_1 = 1/3}$, and ${c_2 = 2/63}$. On the other hand, since all the probabilities ${\mathbf{P}( \mathbf{Syrac}({\bf Z}/3^n{\bf Z}) = b)}$ sum to ${1}$ as ${b \in {\bf Z}/3^n{\bf Z}}$ ranges over the non-multiples of ${3}$, we have the trivial upper bound
$\displaystyle c_n \leq \frac{3}{2} 3^{-n}.$
There is also an easy submultiplicativity result:
Lemma 2 For any natural numbers ${n_1,n_2}$, we have
$\displaystyle c_{n_1+n_2-1} \geq c_{n_1} c_{n_2}.$
Proof: Let ${N}$ be an integer not divisible by ${3}$, then by (4) we have
$\displaystyle \sum_{(a_1,\dots,a_{n_1}) \in S_{n_1,N}} 2^{-a_1-\dots-a_{n_1}} \geq c_{n_1}.$
If we let ${S'_{n_1,N}}$ denote the set of tuples ${(a_1,\dots,a_{n_1-1})}$ that can be formed from the tuples in ${S_{n_1,N}}$ by deleting the final component ${a_{n_1}}$ from each tuple, then we have
$\displaystyle \sum_{(a_1,\dots,a_{n_1-1}) \in S'_{n_1,N}} 2^{-a_1-\dots-a_{n_1-1}} \geq c_{n_1}. \ \ \ \ \ (5)$
Next, observe that if ${(a_1,\dots,a_{n_1-1}) \in S'_{n_1,N}}$, then
$\displaystyle N = \sum_{m=1}^{n_1-1} 3^{m-1} 2^{-a_1-\dots-a_m} + 3^{n_1-1} 2^{-a_1-\dots-a_{n_1-1}} M$
with ${M = M_{N,n_1,a_1,\dots,a_{n_1-1}}}$ an integer not divisible by three. By definition of ${S_{n_2,M}}$ and a relabeling, we then have
$\displaystyle M = \sum_{m=1}^{n_2} 3^{m-1} 2^{-a_{n_1}-\dots-a_{m+n_1-1}} \hbox{ mod } 3^{n_2}$
for all ${(a_{n_1},\dots,a_{n_1+n_2-1}) \in S_{n_2,M}}$. For such tuples we then have
$\displaystyle N = \sum_{m=1}^{n_1+n_2-1} 3^{m-1} 2^{-a_1-\dots-a_{n_1+n_2-1}} \hbox{ mod } 3^{n_1+n_2-1}$
so that ${(a_1,\dots,a_{n_1+n_2-1}) \in S_{n_1+n_2-1,N}}$. Since
$\displaystyle \sum_{(a_{n_1},\dots,a_{n_1+n_2-1}) \in S_{n_2,M}} 2^{-a_{n_1}-\dots-a_{n_1+n_2-1}} \geq c_{n_2}$
for each ${M}$, the claim follows. $\Box$
From this lemma we see that ${c_n = 3^{-\beta n + o(n)}}$ for some absolute constant ${\beta \geq 1}$. Heuristically, we expect the Syracuse random variables to be somewhat approximately equidistributed amongst the multiples of ${{\bf Z}/3^n{\bf Z}}$ (in Proposition 1.4 of my previous paper I prove a fine scale mixing result that supports this heuristic). As a consequence it is natural to conjecture that ${\beta=1}$. I cannot prove this, but I can show that this conjecture would imply that we can take the exponent ${\gamma}$ in (1), (3) arbitrarily close to one:
Proposition 3 Suppose that ${\beta=1}$ (that is to say, ${c_n = 3^{-n+o(n)}}$ as ${n \rightarrow \infty}$). Then
$\displaystyle \# \{ N \leq x: N \in (\mathrm{Syr}^{\bf N})^*(1) \} \gg x^{1-o(1)}$
as ${x \rightarrow \infty}$, or equivalently
$\displaystyle \# \{ N \leq x: N \in (\mathrm{Col}^{\bf N})^*(1) \} \gg x^{1-o(1)}$
as ${x \rightarrow \infty}$. In other words, (1), (3) hold for all ${\gamma < 1}$.
I prove this proposition below the fold. A variant of the argument shows that for any value of ${\beta}$, (1), (3) holds whenever ${\gamma < f(\beta)}$, where ${f: [0,1] \rightarrow [0,1]}$ is an explicitly computable function with ${f(\beta) \rightarrow 1}$ as ${\beta \rightarrow 1}$. In principle, one could then improve the Krasikov-Lagarias result ${\gamma = 0.84}$ by getting a sufficiently good upper bound on ${\beta}$, which is in principle achievable numerically (note for instance that Lemma 2 implies the bound ${c_n \leq 3^{-\beta(n-1)}}$ for any ${n}$, since ${c_{kn-k+1} \geq c_n^k}$ for any ${k}$).
— 1. Proof of proposition —
Assume ${\beta=1}$. Let ${\varepsilon>0}$ be sufficiently small, and let ${n_0}$ be sufficiently large depending on ${\varepsilon}$. We first establish the following proposition, that shows that elements in a certain residue class have a lot of Syracuse preimages:
Proposition 4 There exists a residue class of ${{\bf Z}/3^{n_0}{\bf Z}}$ with the property that for all integers ${N}$ in this class, and all non-negative integers ${j}$, there exist natural numbers ${n_j, L_j}$ with
$\displaystyle (2-\varepsilon^2) n_j \leq L_j \leq (2+\varepsilon^2) n_j$
and
$\displaystyle (4/3)^{(1-\varepsilon^2) (1+\varepsilon)^j n_0} \leq 3^{-n_j} 2^{L_j} \leq (4/3)^{(1+\varepsilon^2) (1+\varepsilon)^j n_0}$
and at least ${3^{-n_j - \varepsilon^4 n_j} 2^{L_j}}$ tuples
$\displaystyle (a_1,\dots,a_{n_j-1}) \in S'_{n_j,N}$
$\displaystyle a_1+\dots+a_{n_j-1} = L_j \ \ \ \ \ (6)$
and
$\displaystyle a_1+\dots+a_i - \frac{\log 3}{\log 2} i \geq - \varepsilon^5 n_0 \ \ \ \ \ (7)$
for all ${1 \leq i \leq n_j-1}$.
Proof: We begin with the base case ${j=0}$. By (4) and the hypothesis ${\beta=1}$, we see that
$\displaystyle \sum_{(a_1,\dots,a_{n_0-1}) \in S'_{n_0,N}} 2^{-a_1-\dots-a_{n_0-1}} \gg 3^{-(1+\varepsilon^6) n_0}$
for all integers ${N}$ not divisible by ${3}$. Let ${S''_{n_0,N}}$ denote the tuples ${(a_1,\dots,a_{n_0-1})}$ in ${S'_{n_0,N}}$ that obey the additional regularity hypotheses
$\displaystyle |a_1 + \dots + a_i - 2i| \leq - \varepsilon^5 n_0 \ \ \ \ \ (8)$
for all ${1 \leq i \leq n_0-1}$,note that this implies in particular the ${j=0}$ case of (7). From the Chernoff inequality (noting that the geometric random variable ${\mathrm{Geom}(2)}$ has mean ${2}$) and the union bound we have
$\displaystyle \sum_{b \in {\bf Z}/3^{n_0}{\bf Z}: 3 \not | b} \sum_{(a_1,\dots,a_{n_0-1}) \in S'_{n_0,b} \backslash S''_{n_0,b}} 2^{-a_1-\dots-a_{n_0-1}} \ll 3^{-c \varepsilon^5 n_0}$
for an absolute constant ${c>0}$ (where we use the periodicity of ${S'_{n_0,N}, S''_{n_0,N}}$ in ${N}$ to define ${S'_{n_0,b}, S''_{n_0,b}}$ for ${b \in {\bf Z}/3^{n_0}{\bf Z}}$ by abuse of notation). Hence by the pigeonhole principle we can find a residue class ${b}$ not divisible by ${3}$ such that
$\displaystyle \sum_{(a_1,\dots,a_{n_0-1}) \in S'_{n_0,b} \backslash S''_{n_0,b}} 2^{-a_1-\dots-a_{n_0-1}} \ll 3^{-(1+c \varepsilon^5) n_0}$
and hence by the triangle inequality we have
$\displaystyle \sum_{(a_1,\dots,a_{n_0-1}) \in S''_{n_0,N}} 2^{-a_1-\dots-a_{n_0-1}} \gg 3^{-(1+\varepsilon^6) n_0}$
for all ${N}$ in this residue class.
Henceforth ${N}$ is assumed to be an element of this residue class. For ${(a_1,\dots,a_{n_0-1}) \in S''_{n_0,N}}$, we see from (8)
$\displaystyle a_1 + \dots + a_{n_0-1} = (2+O(\varepsilon^5)) n_0,$
hence by the pigeonhole principle there exists ${L_0 = (2+O(\varepsilon^5)) n_0}$ (so in particular ${3^{-n_0} 2^{L_0} = (4/3)^{(1+O(\varepsilon^5))n_0}}$) such that
$\displaystyle \sum_{(a_1,\dots,a_{n_0-1}) \in S''_{n_0,N}: a_1+\dots+a_{n_0-1} = L_0} 2^{-L_0} \gg 3^{-(1+\varepsilon^6) n_0}$
so the number of summands here is at least ${\gg 2^{L_0} 3^{-(1+\varepsilon^6) n_0}}$. This establishes the base case ${j=0}$.
Now suppose inductively that ${j \geq 1}$, and that the claim has already been proven for ${j-1}$. By induction hypothesis, there exists natural numbers ${n_{j-1}, L_{j-1}}$ with
$\displaystyle (2-\varepsilon^2) n_{j-1} \leq L_{j-1} \leq (2+\varepsilon^2) n_{j-1}$
and
$\displaystyle (4/3)^{(1-\varepsilon^2) (1+\varepsilon)^{j-1} n_0} \leq 3^{-n_{j-1}} 2^{L_{j-1}} \leq (4/3)^{(1+\varepsilon^2) (1+\varepsilon)^{j-1} n_0} \ \ \ \ \ (9)$
(which in particular imply that ${n_{j-1} = (1+O(\varepsilon^2)) (1+\varepsilon)^{j-1} n_0}$) and at least ${3^{-n_{j-1} - \varepsilon^4 n_{j-1}} 2^{L_{j-1}}}$ tuples
$\displaystyle (a_1,\dots,a_{n_{j-1}-1}) \in S'_{n_{j-1},N} \ \ \ \ \ (10)$
$\displaystyle a_1+\dots+a_{n_{j-1}-1} = L_{j-1} \ \ \ \ \ (11)$
and (7) for all ${1 \leq i \leq n_{j-1}-1}$.
Let ${n_{j}}$ be an integer such that
$\displaystyle 3^{-n_{j}} 2^{L_{j-1} + 2(n_{j}-n_{j-1})} \asymp (4/3)^{(1+\varepsilon)^j n_0} N. \ \ \ \ \ (12)$
One easily checks that
$\displaystyle n_{j} = (1+\varepsilon+O(\varepsilon^2)) n_{j-1} = (1+O(\varepsilon^2)) (1+\varepsilon)^{j-1} n_0.$
For each tuple (10), we may write (as in the proof of Lemma 2)
$\displaystyle N = \sum_{m=1}^{n_{j-1}-1} 3^{m-1} 2^{-a_1-\dots-a_m} + 3^{n_{j-1}-1} 2^{-L_{j-1}} M_{\vec a}$
for some integers ${M_{\vec a}}$. We claim that these integers lie in distinct residue classes modulo ${3^k}$ where
$\displaystyle k :=\lfloor \frac{\log 2}{\log 3} L_{j-1} - n_{j-1} + \varepsilon^4 n_{j-1} \rfloor.$
Indeed, suppose that ${M_{\vec a} = M_{\vec b} \hbox{ mod } 3^k}$ for two tuples ${\vec a = (a_1,\dots,a_{n_{j-1}-1})}$, ${\vec b = (b_1,\dots,b_{n_{j-1}-1})}$ of the above form. Then
$\displaystyle \sum_{m=1}^{n_{j-1}-1} 3^{m-1} 2^{-a_1-\dots-a_m} = \sum_{m=1}^{n_{j-1}-1} 3^{m-1} 2^{-b_1-\dots-b_m} \hbox{ mod } 3^{n_{j-1}-1+k}$
(where we now invert ${2}$ in the ring ${{\bf Z}/3^{n_{j-1}-1+k}{\bf Z}}$), or equivalently
$\displaystyle \sum_{m=1}^{n_{j-1}-1} 3^{m-1} 2^{a_{m+1}+\dots+a_{n_{j-1}-1}} = \sum_{m=1}^{n_{j-1}-1} 3^{m-1} 2^{b_{m+1}+\dots+b_{n_{j-1}-1}} \hbox{ mod } 3^{n_{j-1}-1+k}.$
By (11), (7), all the summands on the left-hand side are natural numbers of size ${O( 2^{L_{j-1}} 3^{O(\varepsilon^5 n_{j-1})})}$, hence the sum also has this size; similarly for the right-hand side. From the estimates of ${n_{j-1}, n_{j}}$, we thus see that both sides are natural numbers between ${1}$ and ${3^{n_{j-1}-1+k}}$, by hypothesis on ${k}$. Thus we may remove the modular constraint and conclude that
$\displaystyle \sum_{m=1}^{n_{j-1}-1} 3^{m-1} 2^{a_{m+1}+\dots+a_{n_{j-1}-1}} = \sum_{m=1}^{n_{j-1}-1} 3^{m-1} 2^{b_{m+1}+\dots+b_{n_{j-1}-1}}$
and then a routine induction (see Lemma 6.2 of my paper) shows that ${(a_1,\dots,a_{n_{j-1}-1}) = (b_1,\dots,b_{n_{j-1}-1})}$. This establishes the claim.
As a corollary, we see that every residue class modulo ${3^{n_j-n_{j-1}}}$ contains
$\displaystyle O( 3^{k - (n_j-n_{j-1})} ) = O( 2^{L_{j-1}} 3^{-n_j + \varepsilon^4 n_{j-1}} )$
of the ${M_{\vec a}}$ at most. Since there were at least ${3^{-n_{j-1} - \varepsilon^4 n_{j-1}} 2^{L_{j-1}}}$ tuples ${\vec a}$ to begin with, we may therefore forbid up to ${O(3^{n_j-n_{j-1} - 3 \varepsilon^4 n_{j-1}})}$ residue classes modulo ${3^{n_j-n_{j-1}}}$, and still have ${\gg 3^{-n_{j-1} - \varepsilon^4 n_{j-1}} 2^{L_{j-1}}}$ surviving tuples ${\vec a}$ with the property that ${M_{\vec a}}$ avoids all the forbidden classes.
Let ${\vec a}$ be one of the tuples (10). By the hypothesis ${\beta = 1}$, we have
$\displaystyle \sum_{(a_{n_{j-1}},\dots,a_{n_j-1}) \in S'_{n_j-n_{j-1},M_{\vec a}}} 2^{-a_{n_{j-1}}-\dots-a_{n_j-1}} \gg 3^{-(1+\varepsilon^6) (n_j-n_{j-1})}.$
Let ${S'''_{n_j-n_{j-1},M}}$ denote the set of tuples ${(a_{n_{j-1}},\dots,a_{n_j-1}) \in S'_{n_j-n_{j-1},M}}$ with the additional property
$\displaystyle |a_{n_{j-1}} + \dots + a_i - 2(i-n_{j-1}+1)| \leq - \varepsilon^3 (n_j - n_{j-1})$
for all ${n_{j-1} \leq i \leq n_j - 1}$, then by the Chernoff bound we have
$\displaystyle \sum_{b \in {\bf Z}/3^{n_j-n_{j-1}}{\bf Z}} \sum_{(a_{n_{j-1}},\dots,a_{n_j-1}) \in S'_{n_j-n_{j-1},b} \backslash S'''_{n_j-n_{j-1},b}} 2^{-a_{n_{j-1}}-\dots-a_{n_j-1}}$
$\displaystyle \ll 3^{-c\varepsilon^3 (n_j-n_{j-1})}$
for some absolute constant ${c>0}$. Thus, by the Markov inequality, by forbidding up to ${O(3^{n_j-n_{j-1} - 3 \varepsilon^4 n_{j-1}})}$ classes, we may ensure that
$\displaystyle \sum_{(a_{n_{j-1}},\dots,a_{n_j-1}) \in S'_{n_j-n_{j-1},M_{\vec a}} \backslash S'''_{n_j-n_{j-1},M_{\vec a}}} 2^{-a_{n_{j-1}}-\dots-a_{n_j-1}} \ll 3^{-(1+\varepsilon^5) (n_j-n_{j-1})}$
and hence
$\displaystyle \sum_{(a_{n_{j-1}},\dots,a_{n_j-1}) \in S'''_{n_j-n_{j-1},M_{\vec a}}} 2^{-a_{n_{j-1}}-\dots-a_{n_j-1}} \gg 3^{-(1+\varepsilon^6) (n_j-n_{j-1})}.$
We thus have
$\displaystyle \sum_{a_1,\dots,a_{n_j-1}} 2^{-a_{n_{j-1}}-\dots-a_{n_j-1}} \gg 3^{-n_{j-1} - \varepsilon^4 n_{j-1}} 2^{L_{j-1}} 3^{-(1+\varepsilon^6) (n_j-n_{j-1})}$
where ${(a_1,\dots,a_{n_j-1})}$ run over all tuples with ${\vec a = (a_1,\dots,a_{n_{j-1}-1})}$ being one of the previously surviving tuples, and ${(a_{n_{j-1}},\dots,a_{n_j-1}) \in S'''_{n_j-n_{j-1},M_{\vec a}}}$. By (11) we may rearrange this a little as
$\displaystyle \sum_{a_1,\dots,a_{n_j-1}} 2^{-a_1-\dots-a_{n_j-1}} \gg 3^{-n_{j} - \varepsilon^4 n_{j-1}-\varepsilon^6 (n_j-n_{j-1})}.$
By construction, we have
$\displaystyle a_1 + \dots + a_{n_j-1} = L_{j-1} + (2 + O(\varepsilon^3)) (n_j - n_{j-1})$
for any tuple in the above sum, hence by the pigeonhole principle we may find an integer
$\displaystyle L_j = L_{j-1} + (2 + O(\varepsilon^3)) (n_j - n_{j-1}) \ \ \ \ \ (13)$
for which
$\displaystyle \sum_{a_1,\dots,a_{n_j-1}: a_1+\dots+a_{n_j-1}=L_j} 2^{-a_1-\dots-a_{n_j-1}} \geq 3^{-n_{j} - \varepsilon^4 n_j}.$
In particular the number of summands is at least ${3^{-n_{j} - \varepsilon^4 n_j} 2^{L_j}}$. Also observe from (13), (12) that
$\displaystyle 3^{-n_j} 2^{L_j} = 3^{-n_{j} + O( \varepsilon^3 (n_j - n_{j-1})} 2^{L_{j-1} + 2(n_j - n_{j-1})}$
$\displaystyle = (4/3)^{(1+\varepsilon)^j n_0} 3^{( \varepsilon^3 (n_j - n_{j-1})}$
so in particular
$\displaystyle (4/3)^{(1-\varepsilon^2) (1+\varepsilon)^j n_0} \leq 3^{-n_j} 2^{L_j} \leq (4/3)^{(1+\varepsilon^2) (1+\varepsilon)^j n_0}.$
It is a routine matter to verify that all tuples in this sum lie in ${S'_{n_j,N}}$ and obeys the requirements (6), (7), closing the induction hypothesis. $\Box$
Corollary 5 For all ${N}$ in the residue class from the previous proposition, and all ${j \geq 0}$, we have
$\displaystyle \{ M \in (\mathrm{Syr}^{\bf N})^*(N): M \leq 3 (4/3)^{(1+\varepsilon^2) (1+\varepsilon)^j n_0} N \}$
$\displaystyle \gg (4/3)^{(1-\varepsilon) (1+\varepsilon)^j n_0}.$
In particular, we have
$\displaystyle \{ M \in (\mathrm{Syr}^{\bf N})^*(N): M \leq x \} \gg_{\varepsilon,n_0,N} x^{1-\varepsilon}$
as ${x \rightarrow \infty}$.
Proof: For every tuple ${(a_1,\dots,a_{n_j-1})}$ in the previous proposition, we have
$\displaystyle N = \sum_{m=1}^{n_{j}-1} 3^{m-1} 2^{-a_1-\dots-a_m} + 3^{n_{j}-1} 2^{-L_{j}} M$
for some integer ${M}$. As before, all these integers ${M}$ are distinct, and have magnitude
$\displaystyle M \leq 3^{-n_j+1} 2^{L_j} N \leq \leq 3 (4/3)^{(1+\varepsilon^2) (1+\varepsilon)^j n_0} N.$
From construction we also have ${\mathrm{Syr}^{n_j}(M) = N}$, so that ${M \in (\mathrm{Syr}^{\bf N})^*(N)}$. The number of tuples is at least
$\displaystyle 3^{-n_j - \varepsilon^4 n_j} 2^{L_j}$
which can be computed from the properties of ${n_j,L_j}$ to be of size at least ${(4/3)^{(1-\varepsilon) (1+\varepsilon)^j n_0}}$. This gives the first claim, and the second claim follows by taking ${j}$ to be the first integer for which ${3 (4/3)^{(1+\varepsilon^2) (1+\varepsilon)^j n_0} N \geq x}$. $\Box$
To conclude the proof of Proposition 3, it thus suffices to show that
Lemma 6 Every residue class ${b \hbox{ mod } 3^{n_0}}$ has a non-trivial intersection with ${(\mathrm{Syr}^{\bf N})^*(1)}$.
Indeed, if we let ${b \hbox{ mod } 3^{n_0}}$ be the residue class from the preceding propositions, and use this lemma to produce an element ${N}$ of ${(\mathrm{Syr}^{\bf N})^*(1)}$ that lies in this class, then from the inclusion ${(\mathrm{Syr}^{\bf N})^*(N) \subset (\mathrm{Syr}^{\bf N})^*(1)}$ we obtain (3) with ${\gamma = 1-O(\varepsilon)}$, and then on sending ${\varepsilon}$ to zero we obtain the claim.
Proof: An easy induction (based on first establishing that ${2^{2 \times 3^n} = 1 + 3^{n+1} \hbox{ mod } 3^{n+2}}$ for all natural numbers ${n}$) shows that the powers of two modulo ${3^{n_0+1}}$ occupy every residue class not divisible by ${3}$. From this we can locate an integer ${N}$ in ${b \hbox{ mod } 3^{n_0}}$ of the form ${N = \frac{2^n-1}{3}}$. Since ${\mathrm{Syr}(N)=1}$, the claim follows. $\Box$
We remark that the same argument in fact shows (assuming ${\beta=1}$ of course) that
$\displaystyle \# \{ N \leq x: N \in (\mathrm{Col}^{\bf N})^*(N_0) \} \gg_{N_0} x^{1-o(1)}$
in the limit ${x \rightarrow \infty}$ for any natural number ${N_0}$ not divisible by three. | {"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": 541, "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": 0.9839063882827759, "perplexity": 122.19063812703153}, "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-20200219092153-20200219122153-00302.warc.gz"} |
https://arxiv.org/abs/1711.04980 | hep-lat
(what is this?)
# Title:Hadronic vacuum polarization contribution to the anomalous magnetic moments of leptons from first principles
Abstract: We compute the leading, strong-interaction contribution to the anomalous magnetic moment of the electron, muon and tau using lattice quantum chromodynamics (QCD) simulations. Calculations include the effects of $u$, $d$, $s$ and $c$ quarks and are performed directly at the physical values of the quark masses and in volumes of linear extent larger than $6\,\mathrm{fm}$. All connected and disconnected Wick contractions are calculated. Continuum limits are carried out using six lattice spacings. We obtain $a_e^\mathrm{LO-HVP}=189.3(2.6)(5.6)\times 10^{-14}$, $a_\mu^\mathrm{LO-HVP}=711.1(7.5)(17.4)\times 10^{-10}$ and $a_\tau^\mathrm{LO-HVP}=341.0(0.8)(3.2)\times 10^{-8}$, where the first error is statistical and the second is systematic.
Comments: 17 pages, 8 figures (in 13 PDF files), RevTeX 4.1. Minor changes to results and to text. References updated. Matches version published in Physical Review Letters Subjects: High Energy Physics - Lattice (hep-lat); High Energy Physics - Phenomenology (hep-ph) Journal reference: Phys. Rev. Lett. 121, 022002 (2018) DOI: 10.1103/PhysRevLett.121.022002 Cite as: arXiv:1711.04980 [hep-lat] (or arXiv:1711.04980v2 [hep-lat] for this version)
## Submission history
From: Laurent Lellouch [view email]
[v1] Tue, 14 Nov 2017 07:20:22 UTC (461 KB)
[v2] Wed, 18 Jul 2018 10:35:52 UTC (464 KB) | {"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": 0.8334584832191467, "perplexity": 2247.9106502526247}, "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-2019-18/segments/1555578534596.13/warc/CC-MAIN-20190422035654-20190422061532-00023.warc.gz"} |
https://documen.tv/question/find-the-value-of-and-the-measure-of-each-angle-21529399-11/ | ## find the value of x and the measure of each angle
Question
find the value of x and the measure of each angle
in progress 0
4 weeks 2021-08-18T10:27:43+00:00 1 Answers 1 views 0 | {"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": 0.8303301334381104, "perplexity": 1062.3208365633275}, "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/1631780057524.58/warc/CC-MAIN-20210924110455-20210924140455-00292.warc.gz"} |
http://math.stackexchange.com/questions/272546/find-the-density-function-of-the-random-variable-z-xy | # Find the density function of the random variable $Z=X+Y$
Let $X$ and $Y$ be independent and uniformly distributed random variable on the intervals $[0,3]$ and $[0,1]$, respectively. Find the density function of the random variable $Z=X+Y$.
I find $f(x,y)=1/3$ and the range of $Z$ to be $[0,4]$, but I cannot find the density function of $Z$.
-
Find $P\{Z \leq \alpha\}$ for $\alpha \in [0,1]$, for $\alpha \in (1,3)$, and $\alpha \in [3,4]$. Then differentiate with respect to $\alpha$ to find the density function. Look, for example, at this question and its answers for ideas on how to proceed. – Dilip Sarwate Jan 8 '13 at 2:11
In general, to find the pdf, you should find the cdf first, and then differentiate the function.
$F[z] = P(Z \leq z) = \begin{cases} \frac {1}{6} z^2 & 0\leq z \leq 1 \\ \frac {1}{6} + \frac {1}{3} (x-1) & 1 \leq z \leq 3 \\ 1- \frac {1}{6} (4-z)^2 & 3 \leq z \leq 4 \\ \end{cases}$
This gives that
$f(z) = \begin{cases} \frac {1}{3} z & 0 \leq z \leq 1 \\ \frac {1}{3} & 1 \leq z \leq 3 \\ \frac {1}{3} (4-z) & 3 \leq z \leq 4 \\\ \end{cases}$
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The "hard" part of this problem is determining the CDF and perhaps you could provide some hints to the OP as to how to go about doing that. – Dilip Sarwate Jan 8 '13 at 12:44
Another way, if you are confortable with convolutions, is to recall that the sum of independent variables has a density that is the convolution of the components. In this case, we must convolve a unit rectangle with a rectangle in $[0,3]$, which is very similar to this image (except for some axis shifting and scaling) (in our case we'd have $f_X$ in red, $f_Y$ in blue, and the result in green), and which results in the function given in Calvin's answer.
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When one realizes that the result of the convolution is a trapezoid whose base is of length $4$ and the "top" (the side parallel to the base) is of length $2$, very little additional calculation is required. The area of the trapezoid is $$\frac{1}{2}\times(4+2)\times(\text{distance between parallel sides})$$ and since the area must be $1$, we have that $f_Z(z)$ increases linearly from $0$ to $\frac{1}{3}$ as $z$ increases from $0$ to $1$, has constant value $\frac{1}{3}$ for $z\in[1,3]$ and decreases linearly to $0$ as $z$ increases from $3$ to $4$; additional details as in Calvin's answer. – Dilip Sarwate Jan 8 '13 at 12:55 | {"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": 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": 0.9204870462417603, "perplexity": 115.59403765523321}, "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/1429246633972.52/warc/CC-MAIN-20150417045713-00291-ip-10-235-10-82.ec2.internal.warc.gz"} |
http://mathhelpforum.com/algebra/115732-quadratic-problem-frustrum-cone.html | # Math Help - Quadratic Problem/Frustrum of a cone
1. ## Quadratic Problem/Frustrum of a cone
Hi Everyone,
Can someone have a look at the attachment and let me know where I am going wrong?
Thanks....Mark
Attached Thumbnails
2. Hello Mark
Originally Posted by manich44
Hi Everyone,
Can someone have a look at the attachment and let me know where I am going wrong?
Thanks....Mark
I can't see anything wrong at all.
Obviously, you haven't shown us the complete question, but if you want the value of $r$ that gives the volume of the frustum to be $3.46 \text{ m}^3$, it looks OK to me. The positive root of your quadratic $\approx 0.856$, and this value of $r$ gives the volume to be $3.48$ - the slight error being due to the approximation $\frac{3V}{\pi h} \approx 3.2$, where $3.178$ would have been better.
Is this not the correct answer, then?
3. ## RE Complete question
Sorry the complete question is....change the size of the upper radius "r" (without any changes to "h" height, and "R" lower Radius) to make the weight of this concrete pile to be at least 8300kg (density of used concrete = 2400kg/m3)
Calculate the new upper Radius r
r=55cm R=1.2m h=104cm
Thanks....
4. Originally Posted by manich44
Sorry the complete question is....change the size of the upper radius "r" (without any changes to "h" height, and "R" lower Radius) to make the weight of this concrete pile to be at least 8300kg (density of used concrete = 2400kg/m3)
Calculate the new upper Radius r
r=55cm R=1.2m h=104cm
Thanks....
OK, I got that much. And that gives the volume $3.46 \text{ m} ^3$ as you said. So where is the problem? Do you have a different answer for $r$? If so, I suspect the answer you have may be wrong! | {"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": 9, "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": 0.9703100919723511, "perplexity": 854.3100842050534}, "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/1408500959239.73/warc/CC-MAIN-20140820021559-00222-ip-10-180-136-8.ec2.internal.warc.gz"} |
https://www.physicsforums.com/threads/derivation-of-lagranges-eqs.873617/ | # Homework Help: Derivation of Lagrange's eqs
Tags:
1. May 28, 2016
### lampCable
1. The problem statement, all variables and given/known data
So I'm deriving Lagrange's equations using Hamilton's principle which states that the motion of a dynamical system follows the path, consistent with any constraints, that minimise the time integral over the lagrangian $L = T-U$, where $T$ is the kinetic energy and $U$ is the potential energy.
We define the lagrangian as $L = L(q_j,\dot{q}_j,t)$. Now I want to let $q_j = q_j^{(0)}+\delta q_j$, where $\delta q_j$ is the variation of $q_j$. We also define
$$S=\int_{t_1}^{t_2}L(q_j,\dot{q}_j,t)dt$$
So according to Hamilton's principle we would now like to minimise $S$. At extremum we have $\delta S = 0$, i.e. the variation of S is zero.
Now, my problem:
My first experience with calculus of variations was to find Euler's equation. We considered then the functional
$$J = \int_{x1}^{x2}f(y,y',x)dx$$
and our goal was to find the function $y$ that minimise S. To do this we let $y(x,\alpha) = y(x)+\alpha\eta(x)$, and set $\frac{\partial J}{\partial\alpha}|_{\alpha=0}=0$, where $\eta$ is some arbitrary function. This would give us an equation in $y(x)$ since we evaluate at $\alpha=0$.
So, returning to the derivation of Lagrange's equations. I set to find $q_j$ that minimise $S$ in similar fashion as we did for $J$. But this time I do not have any $\alpha$ that I can put to zero. Should I instead take $\delta S|_{\delta q_j = 0}=0$? For unless I have understood it wrong, it is actually $q_j^{(0)}$ that we want to find, right? I mean it seems strange to find $q_j$ to be some curve $q_j^{(0)}$ added by some arbitrary variations.
At the same time, I am not sure whether evaluating $\delta q_j$ at makes sense. What increase my doubts is that neither in my text book (Marion and Thornton) nor at any lecture has it been emphasized that we evaluate with $\delta q_j = 0$.
Anyhow, I am thankful for answers.
2. May 29, 2016
### ShayanJ
Remember the definition of a derivative ? $\displaystyle f'(x)=\lim_{h\to 0} \frac{f(x+h)-f(x)}{h}$. And to find an extremum of the function, you just set $f'(x)=0$ .
Its the same here, the only difference being that here we're talking about a functional, a function in the space of functions defined on a given interval with given boundary conditions. There is just one difference though. What we do here to derive the EL equations is, after doing the calculations and arriving at the following equation:
$\delta S=\int_{t_1}^{t_2} \left[ \frac{\partial L}{\partial q_j}-\frac{d}{dt} \frac{\partial L}{\partial \dot{q_j}} \right]\delta q_j dt=0$
We reason that this equation should be true for any non-zero $\delta q_j$ so we should have $\frac{\partial L}{\partial q_j}-\frac{d}{dt} \frac{\partial L}{\partial \dot{q_j}} =0$. The point is, although $\delta q_j$ is supposed to be very small, its not exactly zero. But there should be somewhere that we actually use the fact that $\delta q_j$ is very small, and its actually in the derivation of the above equation were we expand the Lagrangian in a Taylor series and treat $\delta q_j$ as the small displacement from $q_j^{(0)}$.
Anyway, I think you still can use the $y(x,\alpha)=y(x)+\alpha \eta(x)$ approach here. But even here, its actually $\alpha\to 0$ not $\alpha=0$. | {"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": 2, "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": 0.9837729930877686, "perplexity": 154.26237747606055}, "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-2018-22/segments/1526794864657.58/warc/CC-MAIN-20180522092655-20180522112655-00491.warc.gz"} |
http://stats.stackexchange.com/questions/34446/stationary-arma-model-as-infinite-ar-or-ma-process | # Stationary ARMA model as infinite AR or MA process
How can a stationary, invertible ARMA(1,1) process be represented as either an infinite order AR or infinite order MA process?
- | {"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": 0.9973587989807129, "perplexity": 3547.3142237622505}, "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-2014-10/segments/1394011183468/warc/CC-MAIN-20140305091943-00082-ip-10-183-142-35.ec2.internal.warc.gz"} |
https://math.stackexchange.com/questions/2498364/weingarten-surfaces-and-its-caustic-surfaces | # Weingarten surfaces and its caustic surfaces
This is an exercises in A Course in Differential Geometry, which is 3.9.3 on page 68.
Needed definitions are in the image:
I have trouble with (iii) in $\Rightarrow$ direction.
I think it is sufficient to show that the Gauss curvature is less than zero so I compute the curvature of $b_1$,
$b_{1,1}=f_{,1}+\frac{n_{,1}}{\kappa_1}-\frac{\kappa_{1,1}}{\kappa_1^2}n=f_{,1}-\frac{\kappa_1}{\kappa_1}f_{,1}-\frac{\kappa_{1,1}}{\kappa_1^2}n=-\frac{\kappa_{1,1}}{\kappa_1^2}n$
$b_{1,2}=(1-\frac{\kappa_2}{\kappa_1})f_{,2}-\frac{\kappa_{1,2}}{\kappa_1^2}n ,$
and the unit norm of $b_1$ is $n_{b_1}=\frac{f_{,1}}{|f_{,1}|}$, so
$n_{b_1,1}=\frac{f_{,11}}{|f_{,1}|}+\partial_1(\frac{1}{|f_{,1}|})f_{,1}$
$n_{b_1,2}=\frac{f_{,12}}{|f_{,1}|}+\partial_2(\frac{1}{|f_{,1}|})f_{,1} .$
Then, I compute the entries of the second fundamental form
$-L = n_{b_1,1} \cdot b_{1,1}=-\frac{\kappa_{1,1}}{\kappa_1^2|f_{,1}|}(n \cdot f_{,11})$
$-M= n_{b_1,1} \cdot b_{1,2} = (1-\frac{\kappa_2}{\kappa_1})\frac{1}{|f_{,1}|}(f_{,11}\cdot f_{,2})-\frac{\kappa_{1,2}}{\kappa_1^2|f_{,1}|}(n\cdot f_{,11})$
$-M =n_{b_1,2} \cdot b_{1,1}=-\frac{\kappa_{1,1}}{\kappa_1^2|f_{,1}|}(n \cdot f_{,12})$
$-N = n_{b_1,1} \cdot b_{1,2} = (1-\frac{\kappa_2}{\kappa_1})\frac{1}{|f_{,1}|}(f_{,12}\cdot f_{,2})-\frac{\kappa_{1,2}}{\kappa_1^2|f_{,1}|}(n\cdot f_{,12}) .$
I notice that $-M=n_{b_1,2} \cdot b_{1,1}=0$ since
$n \cdot f_{,12}=-n_{,1} \cdot f_{,2}=\kappa_1f_{,1}\cdot f_{,2}=0$
and $(n \cdot f_{,1}) = 0$, so
$n \cdot f_{,11} = (n \cdot f_{,1})_{,1} - n_{,1} \cdot f_{,1} = \kappa_{1} |f_{,1}|^{2} .$
Thus,
$LN-M^2=LN=-\frac{\kappa_{1,1}}{\kappa_1}(1-\frac{\kappa_2}{\kappa_1})(f_{,2}\cdot f_{,12}) .$
I expect that determinent of 2nd fundamental form is less than zero, but I don't know how to use the condition of being a Weingarten surface or the equality in (i). | {"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": 0.9818768501281738, "perplexity": 300.95064604114935}, "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-2019-26/segments/1560627998513.14/warc/CC-MAIN-20190617163111-20190617185111-00310.warc.gz"} |
http://www.dancummins.tv/forum/merely-reluctant-submit-commands-someone-who-had-been-one-40531 | # merely reluctant to submit to the commands of someone who had been one
• Started by
(14 Oct '13)
one. Those who think there is only one because it admits of degrees
<a href=http://almorranas2013.tumblr.com>almorranas2013.tumblr.com</a>
rules. In the end it was found necessary to leave all to the conscience
other hand are strictly bound by the precise terms of their commission,
if he is to accomplish any of the acts that correspond to his virtue,
only at a particular time and for a short period, though not without
nothing to him is great. Nor is he mindful of wrongs; for it is not
true of the good man too that he does many acts for the sake of his
This will be even more apparent from the
Category:
0
on October 14, 2013
one. Those who think there is only one because it admits of degrees
<a href=http://almorranas2013.tumblr.com>almorranas2013.tumblr.com</a>
rules. In the end it was found necessary to leave all to the conscience
other hand are strictly bound by the precise terms of their commission,
if he is to accomplish any of the acts that correspond to his virtue,
only at a particular time and for a short period, though not without
nothing to him is great. Nor is he mindful of wrongs; for it is not
true of the good man too that he does many acts for the sake of his
This will be even more apparent from the fact that activities are
knowledge. Discussion of his accuracy in using his sources must however lie
I have said that the magistrate has a public power of commanding, to
says the Germans only allowed one wife. Sometimes they even lived
It is a principle of all sound definition that one should pay no regard to
advocate, in the hope that the protector, being bound not only by his
corruption of soul." Thomas answered and said , "What have we to say in the face
this reason it is most ill-advised and dangerous to condemn a prince without
Mowgli in sleep: Ummmm...
all nations in their turn separated the callings of arms, and of justice
I conclude then that the subject is never justified in any circumstances in
citizens. The consequences have been most unfortunate. During the last
necessary, but carries with it some risks. Anyone who wished to abolish all
form of a revocable grant, the recipient certainly is, and should be
eight armed Spaniards were hardly sufficient to hold their own against
Forum category: | {"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": 0.8490992188453674, "perplexity": 2677.370485093887}, "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/1408500830834.3/warc/CC-MAIN-20140820021350-00207-ip-10-180-136-8.ec2.internal.warc.gz"} |
https://www.physicsforums.com/threads/problem-understanding-theorm-of-cross-product.712447/ | # Problem Understanding theorm of Cross Product
1. Sep 24, 2013
### PsychonautQQ
1. The problem statement, all variables and given/known data
My textbook says:
The length of the cross product a x b is equal to the area of the parallelogram determined by a and b.
How can a length equal and area? They have different units?
2. Sep 24, 2013
### Staff: Mentor
They are numerically equal. That doesn't say anything about units, just that the numbers are the same.
3. Sep 24, 2013
### Dick
If a and b have dimension length, then axb properly has dimension (length)^2. So does the area of a parallelogram.
Last edited: Sep 24, 2013
4. Sep 24, 2013
### Staff: Mentor
You're going to have to convince me, Dick. a X b is a vector, and its magnitude is a length. I don't see how you can get (length)2 out of a vector.
5. Sep 24, 2013
### Dick
I'll try. The components of axb are products of components of a and b. If the components of a and b have dimension length, that means the components of axb have dimension (length)^2. Not all vectors have dimension length. An example from physics is that angular momentum is defined by l=rxp. r has dimensions of length (m), p has the dimensions of momentum (kg*m/s). l has dimensions of angular momentum (kg*m^2/s).
6. Sep 24, 2013
### Office_Shredder
Staff Emeritus
Dick, my mind is blown right now.
7. Sep 24, 2013
### Dick
I'm kind of surprised this is of mind-blowing proportions. It's just consistently tracking units. In math, you largely treat vectors as dimensionless, so the issue never comes up. If your vectors are dimensionful and you want a length vector perpendicular to a and b, you should use axb/(length units). Which you usually do anyway, just by ignoring the (length)^2 aspect of axb. It should really be satisfying that |axb| correctly has the dimensions of a parallelogram area. Don't get me talking about differential forms and duality. It's why there is a 'cross product' only in three dimensions. The dimension funnyness reflects that.
Last edited: Sep 24, 2013
8. Sep 25, 2013
### D H
Staff Emeritus
I never use "length" as a synonym for the magnitude of a physical vector. What is the length of a velocity vector? That doesn't quite make sense. Velocity has dimensionality of length per time, not length. The magnitude of that velocity vector? That makes sense, and it even has a name: Speed.
Draft saved Draft deleted
Similar Discussions: Problem Understanding theorm of Cross Product | {"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": 0.811682403087616, "perplexity": 1443.575523228719}, "config": {"markdown_headings": true, "markdown_code": false, "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-2018-09/segments/1518891814872.58/warc/CC-MAIN-20180223233832-20180224013832-00507.warc.gz"} |
http://www.research.lancs.ac.uk/portal/en/publications/search-for-anomalous-boldmathwtb-couplings-in-single-top-quark-production(68dc47b8-7dc4-49c6-860a-57fd5b386f30).html | Home > Research > Publications & Outputs > Search for anomalous $\boldmath{Wtb}$ couplings...
Electronic data
• PhysRevLett
Rights statement: © 2008 The American Physical Society
Final published version, 242 KB, PDF-document
Search for anomalous $\boldmath{Wtb}$ couplings in single top quark production
Research output: Contribution to journalJournal article
Published
Article number 221801 25/11/2008 Physical Review Letters 22 101 7 Published English
Abstract
In 0.9 fb$^{-1}$ of $p \bar p$ collisions, D0 has observed an excess of events with an isolated lepton, missing transve rse momentum, and two to four jets. This excess is consistent with single top quark production. We examine these data to study the Lorentz structure of the $Wtb$ coupling. The standard model predicts a left-handed vector coupling at the $Wt b$ vertex. The most general lowest dimension, CP-conserving Lagrangian admits right-handed vector and left- or right-h anded tensor couplings as well. We find that the data prefer the left-handed vector coupling and set upper limits on the anomalous couplings. These are the first direct constraints on a general $Wtb$ interaction and the first direct limits on left- and right-handed tensor couplings.
Bibliographic note
© 2008 The American Physical Society 7 pages, 4 figures | {"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": 0.8479074239730835, "perplexity": 3509.816053149678}, "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-2018-26/segments/1529267864848.47/warc/CC-MAIN-20180623000334-20180623020334-00340.warc.gz"} |
https://www.physicsforums.com/threads/a-kronecker-delta-problem.642019/ | # Homework Help: A Kronecker Delta problem
1. Oct 7, 2012
### bossman007
1. The problem statement, all variables and given/known data
[PLAIN]http://postimage.org/image/s7m1kohst/ [Broken][/PLAIN]
2. Relevant equations
The Kronecker Delta = 1 ; if i=j
The Kronecker Delta = 0 ; i (not equal) j
3. The attempt at a solution
I have no idea what to do from here, or even if I did this first step right?
[PLAIN]http://postimage.org/image/t3d3u9qg7/ [Broken][/PLAIN]
Last edited by a moderator: May 6, 2017
2. Oct 7, 2012
### bossman007
Still no luck on my own :(
3. Oct 7, 2012
### Zondrina
Okay so expanding your sum :
$$\sum_{i=1}^{3}a_iδ_{ij} = a_1δ_{1j} + a_2δ_{2j} + a_3δ_{3j}$$
Note that the only term which survives is the term $a_jδ_{jj}$ where i=j, but $δ_{jj} = 1$ as per the Delta Kronecker. So $a_jδ_{jj} = a_j$
Are you sure that's the question? I feel as if you're missing something. Some important intervals are not mentioned here.
4. Oct 7, 2012
### bossman007
Thats the entire problem...im still confused from ur response, but I feel u gave me a good nudge in the right direction. thx
Last edited: Oct 7, 2012
5. Oct 7, 2012
### bossman007
That's the entire problem...hmmm
6. Oct 7, 2012
### bossman007
I solved it for good, I understand it now. many thanks. The Kronecker Delta has the role of a substitution operator, basically replacing a repeated indice | {"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": 0.8798186779022217, "perplexity": 2291.1190253335258}, "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-2018-30/segments/1531676592001.81/warc/CC-MAIN-20180720232914-20180721012914-00274.warc.gz"} |
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NEW!!! Tough and tricky exponents and roots questions
Author Message
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Math Expert
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NEW!!! Tough and tricky exponents and roots questions [#permalink] 12 Jan 2012, 02:03
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Exponents and roots problems are very common on the GMAT. So, it's extremely important to know how to manipulate them, how to factor out, take roots, multiply, divide, etc. Below are 11 problems to test your skills. Please post your thought process/solutions along with the answers.
I'll post OA's with detailed solutions tomorrow. Good luck.
1. What is the value of \sqrt{25+10\sqrt{6}}+\sqrt{25-10\sqrt{6}}?
A. 2\sqrt{5}
B. \sqrt{55}
C. 2\sqrt{15}
D. 50
E. 60
Solution: tough-and-tricky-exponents-and-roots-questions-125956-40.html#p1029216
2. What is the units digit of (17^3)^4-1973^{3^2}?
A. 0
B. 2
C. 4
D. 6
E. 8
Solution: tough-and-tricky-exponents-and-roots-questions-125956-40.html#p1029219
3. If 5^{10x}=4,900 and 2^{\sqrt{y}}=25 what is the value of \frac{(5^{(x-1)})^5}{4^{-\sqrt{y}}?
A. 14/5
B. 5
C. 28/5
D. 13
E. 14
Solution: tough-and-tricky-exponents-and-roots-questions-125956-40.html#p1029221
4. What is the value of 5+4*5+4*5^2+4*5^3+4*5^4+4*5^5?
A. 5^6
B. 5^7
C. 5^8
D. 5^9
E. 5^10
Solution: tough-and-tricky-exponents-and-roots-questions-125956-40.html#p1029222
5. If x=23^2*25^4*27^6*29^8 and is a multiple of 26^n, where n is a non-negative integer, then what is the value of n^{26}-26^n?
A. -26
B. -25
C. -1
D. 0
E. 1
Solution: tough-and-tricky-exponents-and-roots-questions-125956-40.html#p1029223
6. If x=\sqrt[5]{-37} then which of the following must be true?
A. \sqrt{-x}>2
B. x>-2
C. x^2<4
D. x^3<-8
E. x^4>32
Solution: tough-and-tricky-exponents-and-roots-questions-125956-40.html#p1029224
7. If x=\sqrt{10}+\sqrt[3]{9}+\sqrt[4]{8}+\sqrt[5]{7}+\sqrt[6]{6}+\sqrt[7]{5}+\sqrt[8]{4}+\sqrt[9]{3}+\sqrt[10]{2}, then which of the following must be true:
A. x<6
B. 6<x<8
C. 8<x<10
D. 10<x<12
E. x>12
Solution: tough-and-tricky-exponents-and-roots-questions-125956-40.html#p1029227
8. If x is a positive number and equals to \sqrt{6+{\sqrt{6+\sqrt{6+\sqrt{6+...}}}}}, where the given expression extends to an infinite number of roots, then what is the value of x?
A. \sqrt{6}
B. 3
C. 1+\sqrt{6}
D. 2\sqrt{3}
E. 6
Solution: tough-and-tricky-exponents-and-roots-questions-125956-40.html#p1029228
9. If x is a positive integer then the value of \frac{22^{22x}-22^{2x}}{11^{11x}-11^x} is closest to which of the following?
A. 2^{11x}
B. 11^{11x}
C. 22^{11x}
D. 2^{22x}*11^{11x}
E. 2^{22x}*11^{22x}
Solution: tough-and-tricky-exponents-and-roots-questions-125956-40.html#p1029229
10. Given that 5x=125-3y+z and \sqrt{5x}-5-\sqrt{z-3y}=0, then what is the value of \sqrt{\frac{45(z-3y)}{x}}?
A. 5
B. 10
C. 15
D. 20
E. Can not be determined
Solution: tough-and-tricky-exponents-and-roots-questions-125956-40.html#p1029231
11. If x>0, x^2=2^{64} and x^x=2^y then what is the value of y?
A. 2
B. 2^(11)
C. 2^(32)
D. 2^(37)
E. 2^(64)
Solution: tough-and-tricky-exponents-and-roots-questions-125956-40.html#p1029232
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Re: NEW!!! Tough and tricky exponents and roots questions [#permalink] 25 Jun 2012, 08:59
Expert's post
kuttingchai wrote:
Bunuel wrote:
2. What is the units digit of (17^3)^4-1973^{3^2}?
A. 0
B. 2
C. 4
D. 6
E. 8
Must know for the GMAT:
I. The units digit of (abc)^n is the same as that of c^n, which means that the units digit of (17^3)^4 is that same as that of (7^3)^4 and the units digit of 1973^{3^2} is that same as that of 3^{3^2}.
II. If exponentiation is indicated by stacked symbols, the rule is to work from the top down, thus:
a^m^n=a^{(m^n)} and not (a^m)^n, which on the other hand equals to a^{mn}.
So:
(a^m)^n=a^{mn};
a^m^n=a^{(m^n)}.
Thus, (7^3)^4=7^{(3*4)}=7^{12} and 3^{3^2}=3^{(3^2)}=3^9.
III. The units digit of integers in positive integer power repeats in specific pattern (cyclicity): The units digit of 7 and 3 in positive integer power repeats in patterns of 4:
1. 7^1=7 (last digit is 7)
2. 7^2=9 (last digit is 9)
3. 7^3=3 (last digit is 3)
4. 7^4=1 (last digit is 1)
5. 7^5=7 (last digit is 7 again!)
...
1. 3^1=3 (last digit is 3)
2. 3^2=9 (last digit is 9)
3. 3^3=27 (last digit is 7)
4. 3^4=81 (last digit is 1)
5. 3^5=243 (last digit is 3 again!)
...
Thus th units digit of 7^{12} will be 1 (4th in pattern, as 12 is a multiple of cyclicty number 4) and the units digit of 3^9 will be 3 (first in pattern, as 9=4*2+1).
So, we have that the units digit of (17^3)^4=17^{12} is 1 and the units digit of 1973^3^2=1973^9 is 3. Also notice that the second number is much larger then the first one, thus their difference will be negative, something like 11-13=-2, which gives the final answer that the units digit of (17^3)^4-1973^{3^2} is 2.
Hey Bunuel,
I understood the logic behind finding the unit places, but how can u determine if the reminder is 2 or 8
in this example
(17^12) - (1973^9) = unit place is 2 (agreed- i actually did the calculation using calculator )
because : (17^12) unit place = 1
(1973^9) unit place = 3
but if we have just - (7^12) - (3^9) or (3^9) - (7^12) we have unit place as "8"
3^9 = 19683 - unit place will still be 2
7^12 = 13841287201 - unit place will still be 1
so my question is how will u determine if the answer is 2/8? clearly we have 2 different answers??
am i missing anything??
The units digit of 17^{12}-1973^{9} is 2 and not 8 since 1973^{9} is much larger number than 17^{12}, thus their difference will be negative, something like 11-13=-2.
If we had something like 21-13 (if the first number were greater than the second one), then the units digit of their difference would be 8.
Hope i's clear.
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Re: NEW!!! Tough and tricky exponents and roots questions [#permalink] 29 Jun 2012, 10:34
Bunuel wrote:
SOLUTIONS:
1. What is the value of \sqrt{25+10\sqrt{6}}+\sqrt{25-10\sqrt{6}}?
A. 2\sqrt{5}
B. \sqrt{55}
C. 2\sqrt{15}
D. 50
E. 60
Square the given expression to get rid of the roots, though don't forget to un-square the value you get at the end to balance this operation and obtain the right answer:
Must know fro the GMAT: (x+y)^2=x^2+2xy+y^2 (while (x-y)^2=x^2-2xy+y^2).
So we get: (\sqrt{25+10\sqrt{6}}+\sqrt{25-10\sqrt{6}})^2=(\sqrt{25+10\sqrt{6}})^2+2(\sqrt{25+10\sqrt{6}})(\sqrt{25-10\sqrt{6}})+(\sqrt{25-10\sqrt{6}})^2=
=(25+10\sqrt{6})+2(\sqrt{25+10\sqrt{6}})(\sqrt{25-10\sqrt{6}})+(25-10\sqrt{6}).
Note that sum of the first and the third terms simplifies to (25+10\sqrt{6})+(25-10\sqrt{6})=50, so we have 50+2(\sqrt{25+10\sqrt{6}})(\sqrt{25-10\sqrt{6}}) --> 50+2(\sqrt{25+10\sqrt{6}})(\sqrt{25-10\sqrt{6}})=50+2\sqrt{(25+10\sqrt{6})(25-10\sqrt{6})}.
Also must know for the GMAT: (x+y)(x-y)=x^2-y^2, thus 50+2\sqrt{(25+10\sqrt{6})(25-10\sqrt{6})}=50+2\sqrt{25^2-(10\sqrt{6})^2)}=50+2\sqrt{625-600}=50+2\sqrt{25}=60.
Recall that we should un-square this value to get the right the answer: \sqrt{60}=2\sqrt{15}.
Another way to do it, using the same formulas:
\sqrt{25+10\sqrt{6}}+\sqrt{25-10\sqrt{6}}=\sqrt{5(5+2\sqrt{6})}+\sqrt{5(5-2\sqrt{6})}=\sqrt{5}\sqrt{5+2\sqrt{6}}+\sqrt{5}\sqrt{5-2\sqrt{6}}=
=\sqrt{5}(\sqrt{5+2\sqrt{6}}+\sqrt{5-2\sqrt{6}})=\sqrt{5}(\sqrt{(\sqrt{3}+\sqrt{2})^2}+\sqrt{(\sqrt{3}-\sqrt{2})^2})=\sqrt{5}(\sqrt{3}+\sqrt{2}+\sqrt{3}-\sqrt{2})=\sqrt{5}(2\sqrt{3})=2\sqrt{15}
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Last edited by EvaJager on 15 Oct 2012, 00:35, edited 1 time in total.
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Re: NEW!!! Tough and tricky exponents and roots questions [#permalink] 25 Jul 2012, 12:53
Bunuel wrote:
SOLUTIONS:
1. What is the value of \sqrt{25+10\sqrt{6}}+\sqrt{25-10\sqrt{6}}?
A. 2\sqrt{5}
B. \sqrt{55}
C. 2\sqrt{15}
D. 50
E. 60
Square the given expression to get rid of the roots, though don't forget to un-square the value you get at the end to balance this operation and obtain the right answer:
Must know fro the GMAT: (x+y)^2=x^2+2xy+y^2 (while (x-y)^2=x^2-2xy+y^2).
So we get: (\sqrt{25+10\sqrt{6}}+\sqrt{25-10\sqrt{6}})^2=(\sqrt{25+10\sqrt{6}})^2+2(\sqrt{25+10\sqrt{6}})(\sqrt{25-10\sqrt{6}})+(\sqrt{25-10\sqrt{6}})^2=
=(25+10\sqrt{6})+2(\sqrt{25+10\sqrt{6}})(\sqrt{25-10\sqrt{6}})+(25-10\sqrt{6}).
Note that sum of the first and the third terms simplifies to (25+10\sqrt{6})+(25-10\sqrt{6})=50, so we have 50+2(\sqrt{25+10\sqrt{6}})(\sqrt{25-10\sqrt{6}}) --> 50+2(\sqrt{25+10\sqrt{6}})(\sqrt{25-10\sqrt{6}})=50+2\sqrt{(25+10\sqrt{6})(25-10\sqrt{6})}.
Also must know for the GMAT: (x+y)(x-y)=x^2-y^2, thus 50+2\sqrt{(25+10\sqrt{6})(25-10\sqrt{6})}=50+2\sqrt{25^2-(10\sqrt{6})^2)}=50+2\sqrt{625-600}=50+2\sqrt{25}=60.
Recall that we should un-square this value to get the right the answer: \sqrt{60}=2\sqrt{15}.
I don't understand why we should un-square the value at the end?
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Re: NEW!!! Tough and tricky exponents and roots questions [#permalink] 26 Jul 2012, 05:10
Expert's post
Stiv wrote:
Bunuel wrote:
SOLUTIONS:
1. What is the value of \sqrt{25+10\sqrt{6}}+\sqrt{25-10\sqrt{6}}?
A. 2\sqrt{5}
B. \sqrt{55}
C. 2\sqrt{15}
D. 50
E. 60
Square the given expression to get rid of the roots, though don't forget to un-square the value you get at the end to balance this operation and obtain the right answer:
Must know fro the GMAT: (x+y)^2=x^2+2xy+y^2 (while (x-y)^2=x^2-2xy+y^2).
So we get: (\sqrt{25+10\sqrt{6}}+\sqrt{25-10\sqrt{6}})^2=(\sqrt{25+10\sqrt{6}})^2+2(\sqrt{25+10\sqrt{6}})(\sqrt{25-10\sqrt{6}})+(\sqrt{25-10\sqrt{6}})^2=
=(25+10\sqrt{6})+2(\sqrt{25+10\sqrt{6}})(\sqrt{25-10\sqrt{6}})+(25-10\sqrt{6}).
Note that sum of the first and the third terms simplifies to (25+10\sqrt{6})+(25-10\sqrt{6})=50, so we have 50+2(\sqrt{25+10\sqrt{6}})(\sqrt{25-10\sqrt{6}}) --> 50+2(\sqrt{25+10\sqrt{6}})(\sqrt{25-10\sqrt{6}})=50+2\sqrt{(25+10\sqrt{6})(25-10\sqrt{6})}.
Also must know for the GMAT: (x+y)(x-y)=x^2-y^2, thus 50+2\sqrt{(25+10\sqrt{6})(25-10\sqrt{6})}=50+2\sqrt{25^2-(10\sqrt{6})^2)}=50+2\sqrt{625-600}=50+2\sqrt{25}=60.
Recall that we should un-square this value to get the right the answer: \sqrt{60}=2\sqrt{15}.
I don't understand why we should un-square the value at the end?
Please read the solution: "Square the given expression to get rid of the roots, though don't forget to un-square the value you get at the end to balance this operation and obtain the right answer."
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Re: NEW!!! Tough and tricky exponents and roots questions [#permalink] 27 Aug 2012, 15:58
These questions are absolutely insane, do people really solve these on the GMAT under 2 or 3 minutes? I just recently took a GMATprep practice test and got a 690 (a measly 44Q) and nowhere did I see any problems of this caliber. Is it truly what people deal with to get to 46 or 47 Quant?
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Re: NEW!!! Tough and tricky exponents and roots questions [#permalink] 01 Sep 2012, 02:29
Bunuel wrote:
11. If x>0, x^2=2^{64} and x^x=2^y then what is the value of y?
A. 2
B. 2^(11)
C. 2^(32)
D. 2^(37)
E. 2^(64)
x^2=2^(64) --> x=\sqrt{2^{64}}=2^{\frac{64}{2}}=2^{32} (note that x=-\sqrt{2^{64}} is not a valid solution as given that x>0).
Second step: x^x=(2^{32})^{(2^{32})}=2^{32*2^{32}}=2^{2^{5}*2^{32}}=2^{2^{37}}=2^y --> y=2^{37}.
OR second step: x^x=(2^{32})^x=2^{32x}=2^y --> y=32x --> since x=2^{32} then y=32x=32*2^{32}=2^5*2^{32}=2^{37}.
Bunuel I have little confusion about the last problem
given x^2= 2^{64}so we can write 2^{64} as (2^{32})^{2} = x^2 now we have x = 2^{32}
now second part x^x = {2^{32}}^ {x} (i)
or does it mean
x^x= 2^{{32}{x}} ( ii)
in (i) we have 32^xand in (ii) we have 32*x obviously they cannot mean the same thing {{2^{32}}^2}^{32} \neq{2^{32.2}}^{32} ,as {2^{32}}^x \neq 2^{32.x} (x>1)
in the solution you have taken {2^{32}}^{x} = 2^{{32}{x}} so I am confused
so we are saying {{2^{32}}^2}^{32} = {2^{32.2}}^{32}..how ?
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Re: NEW!!! Tough and tricky exponents and roots questions [#permalink] 03 Sep 2012, 00:57
stne wrote:
Bunuel wrote:
11. If x>0, x^2=2^{64} and x^x=2^y then what is the value of y?
A. 2
B. 2^(11)
C. 2^(32)
D. 2^(37)
E. 2^(64)
x^2=2^(64) --> x=\sqrt{2^{64}}=2^{\frac{64}{2}}=2^{32} (note that x=-\sqrt{2^{64}} is not a valid solution as given that x>0).
Second step: x^x=(2^{32})^{(2^{32})}=2^{32*2^{32}}=2^{2^{5}*2^{32}}=2^{2^{37}}=2^y --> y=2^{37}.
OR second step: x^x=(2^{32})^x=2^{32x}=2^y --> y=32x --> since x=2^{32} then y=32x=32*2^{32}=2^5*2^{32}=2^{37}.
Bunuel I have little confusion about the last problem
given x^2= 2^{64}so we can write 2^{64} as (2^{32})^{2} = x^2 now we have x = 2^{32}
now second part x^x = {2^{32}}^ {x} (i)
or does it mean
x^x= 2^{{32}{x}} ( ii)
in (i) we have 32^xand in (ii) we have 32*x obviously they cannot mean the same thing {{2^{32}}^2}^{32} \neq{2^{32.2}}^{32} ,as {2^{32}}^x \neq 2^{32.x} (x>1)
in the solution you have taken {2^{32}}^{x} = 2^{{32}{x}} so I am confused
so we are saying {{2^{32}}^2}^{32} = {2^{32.2}}^{32}..how ?
Bunuel ( preferably ) or any one else , can you please provide solution to the above query , This question was given by you
seems something is odd with the solution as pointed in my post above, unless I am missing something .
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Re: NEW!!! Tough and tricky exponents and roots questions [#permalink] 03 Sep 2012, 01:07
Expert's post
stne wrote:
Bunuel wrote:
11. If x>0, x^2=2^{64} and x^x=2^y then what is the value of y?
A. 2
B. 2^(11)
C. 2^(32)
D. 2^(37)
E. 2^(64)
x^2=2^(64) --> x=\sqrt{2^{64}}=2^{\frac{64}{2}}=2^{32} (note that x=-\sqrt{2^{64}} is not a valid solution as given that x>0).
Second step: x^x=(2^{32})^{(2^{32})}=2^{32*2^{32}}=2^{2^{5}*2^{32}}=2^{2^{37}}=2^y --> y=2^{37}.
OR second step: x^x=(2^{32})^x=2^{32x}=2^y --> y=32x --> since x=2^{32} then y=32x=32*2^{32}=2^5*2^{32}=2^{37}.
Bunuel I have little confusion about the last problem
given x^2= 2^{64}so we can write 2^{64} as (2^{32})^{2} = x^2 now we have x = 2^{32}
now second part x^x = {2^{32}}^ {x} (i)
or does it mean
x^x= 2^{{32}{x}} ( ii)
in (i) we have 32^xand in (ii) we have 32*x obviously they cannot mean the same thing {{2^{32}}^2}^{32} \neq{2^{32.2}}^{32} ,as {2^{32}}^x \neq 2^{32.x} (x>1)
in the solution you have taken {2^{32}}^{x} = 2^{{32}{x}} so I am confused
so we are saying {{2^{32}}^2}^{32} = {2^{32.2}}^{32}..how ?
(a^m)^n=a^{mn}, so x^x=(2^{32})^{(2^{32})}=2^{32*2^{32}}=2^{2^{5}*2^{32}}=2^{2^{37}}.
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Re: NEW!!! Tough and tricky exponents and roots questions [#permalink] 14 Oct 2012, 23:54
Bunuel wrote:
8. If x is a positive number and equals to \sqrt{6+{\sqrt{6+\sqrt{6+\sqrt{6+...}}}}}, where the given expression extends to an infinite number of roots, then what is the value of x?
A. \sqrt{6}
B. 3
C. 1+\sqrt{6}
D. 2\sqrt{3}
E. 6
Given: x>0 and x=\sqrt{6+{\sqrt{6+\sqrt{6+\sqrt{6+...}}}}} --> x=\sqrt{6+({\sqrt{6+\sqrt{6+\sqrt{6+...})}}}}, as the expression under the square root extends infinitely then expression in brackets would equal to x itself and we can safely replace it with x and rewrite the given expression as x=\sqrt{6+x}. Square both sides: x^2=6+x --> (x+2)(x-3)=0 --> x=-2 or x=3, but since x>0 then: x=3.
Hi Bunuel - All sols are crystal clear except this one logic mentioned here. I'm not able to understand the below part. Could you please elaborate?
x=\sqrt{6+({\sqrt{6+\sqrt{6+\sqrt{6+...})}}}}, as the expression under the square root extends infinitely then expression in brackets would equal to x itself and we can safely replace it with x and rewrite the given expression as x=\sqrt{6+x}.
Cheers
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Re: NEW!!! Tough and tricky exponents and roots questions [#permalink] 15 Oct 2012, 02:09
Expert's post
Jp27 wrote:
Bunuel wrote:
8. If x is a positive number and equals to \sqrt{6+{\sqrt{6+\sqrt{6+\sqrt{6+...}}}}}, where the given expression extends to an infinite number of roots, then what is the value of x?
A. \sqrt{6}
B. 3
C. 1+\sqrt{6}
D. 2\sqrt{3}
E. 6
Given: x>0 and x=\sqrt{6+{\sqrt{6+\sqrt{6+\sqrt{6+...}}}}} --> x=\sqrt{6+({\sqrt{6+\sqrt{6+\sqrt{6+...})}}}}, as the expression under the square root extends infinitely then expression in brackets would equal to x itself and we can safely replace it with x and rewrite the given expression as x=\sqrt{6+x}. Square both sides: x^2=6+x --> (x+2)(x-3)=0 --> x=-2 or x=3, but since x>0 then: x=3.
Hi Bunuel - All sols are crystal clear except this one logic mentioned here. I'm not able to understand the below part. Could you please elaborate?
x=\sqrt{6+({\sqrt{6+\sqrt{6+\sqrt{6+...})}}}}, as the expression under the square root extends infinitely then expression in brackets would equal to x itself and we can safely replace it with x and rewrite the given expression as x=\sqrt{6+x}.
Cheers
Given: x=\sqrt{6+({\sqrt{6+\sqrt{6+\sqrt{6+...})}}}}. Consier the expression in brackets: {(\sqrt{6+\sqrt{6+\sqrt{6+...}}). It's the same as the right hand side of the initial expression, thus it also equals to x. When replaced we'll have: x=\sqrt{6+x}.
Hope it's clear.
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Re: NEW!!! Tough and tricky exponents and roots questions [#permalink] 03 Nov 2012, 05:21
Bunuel wrote:
4. What is the value of 5+4*5+4*5^2+4*5^3+4*5^4+4*5^5?
A. 5^6
B. 5^7
C. 5^8
D. 5^9
E. 5^10
This question can be solved in several ways:
Traditional approach: 5+4*5+4*5^2+4*5^3+4*5^4+4*5^5=5+4(5+5^2+5^3+5^4+5^5) Note that we have the sum of geometric progression in brackets with first term equal to 5 and common ratio also equal to 5. The sum of the first n terms of geometric progression is given by: sum=\frac{b*(r^{n}-1)}{r-1}, where b is the first term, n # of terms and r is a common ratio \neq{1}.
So in our case: 5+4(5+5^2+5^3+5^4+5^5)=5+4(\frac{5(5^5-1)}{5-1})=5^6.
30 sec approach based on answer choices:
We have the sum of 6 terms. Now, if all terms were equal to the largest term 4*5^5 we would have: sum=6*(4*5^5)=24*5^5\approx{5^2*5^5}\approx{5^7}, so the actual sum must be less than 5^7, thus the answer must be A: 5^6.
I am not familiar with GP, could you please let me know how we have consider "r". Using other examples???
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Re: NEW!!! Tough and tricky exponents and roots questions [#permalink] 03 Nov 2012, 05:30
Expert's post
breakit wrote:
Bunuel wrote:
4. What is the value of 5+4*5+4*5^2+4*5^3+4*5^4+4*5^5?
A. 5^6
B. 5^7
C. 5^8
D. 5^9
E. 5^10
This question can be solved in several ways:
Traditional approach: 5+4*5+4*5^2+4*5^3+4*5^4+4*5^5=5+4(5+5^2+5^3+5^4+5^5) Note that we have the sum of geometric progression in brackets with first term equal to 5 and common ratio also equal to 5. The sum of the first n terms of geometric progression is given by: sum=\frac{b*(r^{n}-1)}{r-1}, where b is the first term, n # of terms and r is a common ratio \neq{1}.
So in our case: 5+4(5+5^2+5^3+5^4+5^5)=5+4(\frac{5(5^5-1)}{5-1})=5^6.
30 sec approach based on answer choices:
We have the sum of 6 terms. Now, if all terms were equal to the largest term 4*5^5 we would have: sum=6*(4*5^5)=24*5^5\approx{5^2*5^5}\approx{5^7}, so the actual sum must be less than 5^7, thus the answer must be A: 5^6.
I am not familiar with GP, could you please let me know how we have consider "r". Using other examples???
Check here: sequences-progressions-101891.html
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Re: NEW!!! Tough and tricky exponents and roots questions [#permalink] 03 Nov 2012, 05:55
Bunuel wrote:
breakit wrote:
Bunuel wrote:
4. What is the value of 5+4*5+4*5^2+4*5^3+4*5^4+4*5^5?
A. 5^6
B. 5^7
C. 5^8
D. 5^9
E. 5^10
This question can be solved in several ways:
Traditional approach: 5+4*5+4*5^2+4*5^3+4*5^4+4*5^5=5+4(5+5^2+5^3+5^4+5^5) Note that we have the sum of geometric progression in brackets with first term equal to 5 and common ratio also equal to 5. The sum of the first n terms of geometric progression is given by: sum=\frac{b*(r^{n}-1)}{r-1}, where b is the first term, n # of terms and r is a common ratio \neq{1}.
So in our case: 5+4(5+5^2+5^3+5^4+5^5)=5+4(\frac{5(5^5-1)}{5-1})=5^6.
30 sec approach based on answer choices:
We have the sum of 6 terms. Now, if all terms were equal to the largest term 4*5^5 we would have: sum=6*(4*5^5)=24*5^5\approx{5^2*5^5}\approx{5^7}, so the actual sum must be less than 5^7, thus the answer must be A: 5^6.
I am not familiar with GP, could you please let me know how we have consider "r". Using other examples???
Check here: sequences-progressions-101891.html
Thanks a lot ... I suppose this is not available in Gmatclub math book
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Re: NEW!!! Tough and tricky exponents and roots questions [#permalink] 03 Nov 2012, 05:58
Expert's post
breakit wrote:
Thanks a lot ... I suppose this is not available in Gmatclub math book
Actually it is, check Sequences and Progressions here: gmat-math-book-87417.html
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Re: NEW!!! Tough and tricky exponents and roots questions [#permalink] 13 Jan 2013, 23:38
Thanks Bunuel for these interesting problems.
With respect to problem number 2
"So, we have that the units digit of is 1 and the units digit of is 3. Also notice that the second number is much larger then the first one, thus their difference will be negative, something like 11-13=-2, which gives the final answer that the units digit of is 2."
I don't understand the above statement. If the last digit is 1 for the smaller number and the last digit is 3 for the larger number.
And we are trying to find different between smaller number - larger number
eg 757571 - 58299374483. We borrow one from the previous digit and make 1 as 11 and then subtract 3 from 11. So shouldn't the last digit be 8.
I am unable to think any other way. Please let me know where I am going wrong.
Thanks
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Re: NEW!!! Tough and tricky exponents and roots questions [#permalink] 14 Jan 2013, 00:20
Expert's post
gmatrant wrote:
Thanks Bunuel for these interesting problems.
With respect to problem number 2
"So, we have that the units digit of is 1 and the units digit of is 3. Also notice that the second number is much larger then the first one, thus their difference will be negative, something like 11-13=-2, which gives the final answer that the units digit of is 2."
I don't understand the above statement. If the last digit is 1 for the smaller number and the last digit is 3 for the larger number.
And we are trying to find different between smaller number - larger number
eg 757571 - 58299374483. We borrow one from the previous digit and make 1 as 11 and then subtract 3 from 11. So shouldn't the last digit be 8.
I am unable to think any other way. Please let me know where I am going wrong.
Thanks
gmatrant
First of all 757,571 - 58,299,374,483 = -58,298,616,912
You could consider easier cases: 11-13=-2, 11-23=-12, 21-83=-62, ...
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Re: NEW!!! Tough and tricky exponents and roots questions [#permalink] 21 Jan 2013, 09:03
This is as stated correctly a tricky Question. But if we do a little arranging before we start to solve the problem then we can eliminate the confusion. From the question we can easily determine that 1973^9 is greater than 17^12. So lets rewrite the question as -(1973^9 - 17^12). And now no body will be confusing this and the answer is 2 clearly.
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Re: NEW!!! Tough and tricky exponents and roots questions [#permalink] 23 Mar 2013, 00:25
2
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Bunuel wrote:
6. If x=\sqrt[5]{-37} then which of the following must be true?
A. \sqrt{-x}>2
B. x>-2
C. x^2<4
D. x^3<-8
E. x^4>32
Must know for the GMAT: Even roots from negative number is undefined on the GMAT (as GMAT is dealing only with Real Numbers): \sqrt[{even}]{negative}=undefined, for example \sqrt{-25}=undefined.
Odd roots have the same sign as the base of the root. For example, \sqrt[3]{125} =5 and \sqrt[3]{-64} =-4.
Back to the original question:
As -2^5=-32 then x must be a little bit less than -2 --> x=\sqrt[5]{-37}\approx{-2.1}<-2. Thus x^3\approx{(-2.1)^3}\approx{-8.something}<-8, so option D must be true.
As for the other options:
A. \sqrt{-x}=\sqrt{-(-2.1)}=\sqrt{2.1}<2, \sqrt{-x}>2 is not true.
B. x\approx{-2.1}<-2, thus x>-2 is also not true.
C. x^2\approx{(-2.1)}^2=4.something>4, thus x^2<4 is also not true.
E. x^4\approx{(-2.1)}^4\approx17, (2^4=16, so anyway -2.1^4 can not be more than 32) thus x^4>32 is also not true.
Please correct the GC test Q : M26-05. Error highlighted in red
If x=\sqrt[5]{-37} then which of the following must be true?
A. \sqrt{-x}>2
B. x>-2
C. x^2<4
D. x^3<8
E. x^4>32
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Re: NEW!!! Tough and tricky exponents and roots questions [#permalink] 23 Mar 2013, 04:21
Expert's post
greatps24 wrote:
Bunuel wrote:
6. If x=\sqrt[5]{-37} then which of the following must be true?
A. \sqrt{-x}>2
B. x>-2
C. x^2<4
D. x^3<-8
E. x^4>32
Must know for the GMAT: Even roots from negative number is undefined on the GMAT (as GMAT is dealing only with Real Numbers): \sqrt[{even}]{negative}=undefined, for example \sqrt{-25}=undefined.
Odd roots have the same sign as the base of the root. For example, \sqrt[3]{125} =5 and \sqrt[3]{-64} =-4.
Back to the original question:
As -2^5=-32 then x must be a little bit less than -2 --> x=\sqrt[5]{-37}\approx{-2.1}<-2. Thus x^3\approx{(-2.1)^3}\approx{-8.something}<-8, so option D must be true.
As for the other options:
A. \sqrt{-x}=\sqrt{-(-2.1)}=\sqrt{2.1}<2, \sqrt{-x}>2 is not true.
B. x\approx{-2.1}<-2, thus x>-2 is also not true.
C. x^2\approx{(-2.1)}^2=4.something>4, thus x^2<4 is also not true.
E. x^4\approx{(-2.1)}^4\approx17, (2^4=16, so anyway -2.1^4 can not be more than 32) thus x^4>32 is also not true.
Please correct the GC test Q : M26-05. Error highlighted in red
If x=\sqrt[5]{-37} then which of the following must be true?
A. \sqrt{-x}>2
B. x>-2
C. x^2<4
D. x^3<8
E. x^4>32
Edited. Thank you.
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Re: NEW!!! Tough and tricky exponents and roots questions [#permalink] 30 Mar 2013, 10:53
Great questions and great explanations! Thanks a lot this is really great for practicing. However, with a few of these problems I am running in to the same problem that I have with some of the questions in the gmatclub test's. For example the question below. Even if I assume it only takes me only seconds to read the question 5 seconds to figure out how to handle it: I find it very hard to finish in 2 minutes (litterarly even by copying your answer it almost takes me 2 minutes...) Does this mean I just need to learn to go quicker or are these questions kind of long indeed ? Thanks a lot in advance for your time
Bunuel wrote:
Stiv wrote:
Bunuel wrote:
SOLUTIONS:
1. What is the value of \sqrt{25+10\sqrt{6}}+\sqrt{25-10\sqrt{6}}?
A. 2\sqrt{5}
B. \sqrt{55}
C. 2\sqrt{15}
D. 50
E. 60
Square the given expression to get rid of the roots, though don't forget to un-square the value you get at the end to balance this operation and obtain the right answer:
Must know fro the GMAT: (x+y)^2=x^2+2xy+y^2 (while (x-y)^2=x^2-2xy+y^2).
So we get: (\sqrt{25+10\sqrt{6}}+\sqrt{25-10\sqrt{6}})^2=(\sqrt{25+10\sqrt{6}})^2+2(\sqrt{25+10\sqrt{6}})(\sqrt{25-10\sqrt{6}})+(\sqrt{25-10\sqrt{6}})^2=
=(25+10\sqrt{6})+2(\sqrt{25+10\sqrt{6}})(\sqrt{25-10\sqrt{6}})+(25-10\sqrt{6}).
Note that sum of the first and the third terms simplifies to (25+10\sqrt{6})+(25-10\sqrt{6})=50, so we have 50+2(\sqrt{25+10\sqrt{6}})(\sqrt{25-10\sqrt{6}}) --> 50+2(\sqrt{25+10\sqrt{6}})(\sqrt{25-10\sqrt{6}})=50+2\sqrt{(25+10\sqrt{6})(25-10\sqrt{6})}.
Also must know for the GMAT: (x+y)(x-y)=x^2-y^2, thus 50+2\sqrt{(25+10\sqrt{6})(25-10\sqrt{6})}=50+2\sqrt{25^2-(10\sqrt{6})^2)}=50+2\sqrt{625-600}=50+2\sqrt{25}=60.
Recall that we should un-square this value to get the right the answer: \sqrt{60}=2\sqrt{15}.
I don't understand why we should un-square the value at the end?
Please read the solution: "Square the given expression to get rid of the roots, though don't forget to un-square the value you get at the end to balance this operation and obtain the right answer."
Re: NEW!!! Tough and tricky exponents and roots questions [#permalink] 30 Mar 2013, 10:53
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https://www.simonsfoundation.org/people/olga-zhaxybayeva/ | # Olga Zhaxybayeva, Ph.D.
Dartmouth College
Olga Zhaxybayeva’s work focuses on how horizontal gene transfer influences (or influenced) the evolution of bacteria and archaea. Her work developing and implementing statistical techniques for monitoring the evolution of all of the genes in a bacterial genome showed that horizontal gene transfer has affected the evolution of much of the genomes of cyanobacteria, thermophilic bacteria and halophilic archaea. This and related work established horizontal gene transfer as an important driver of microbial evolution. Her research plan involves studying gene transfer agents (virus-like particles produced by some bacteria and archaea) to explore the possibility that horizontal gene transfer can provide an evolutionary force favoring cooperation and the emergence of complexity. | {"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": 0.9172384142875671, "perplexity": 3313.740914246615}, "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/1664030337595.1/warc/CC-MAIN-20221005073953-20221005103953-00349.warc.gz"} |
https://www.physicsforums.com/threads/plotting-vector-fields-in-matlab-or-maple.279103/ | # Plotting vector fields in MATLAB or Maple
1. Dec 11, 2008
### JefeNorte
1. The problem statement, all variables and given/known data
The original problem was x'=(-2 1; 1 -2)*x and I needed to find two linearly independent solutions.
2. Relevant equations
3. The attempt at a solution
I found that x1=(1;1)e^(-t) and x2=(1;-1)e^(-3t). Now I am trying to plot a vector field of this. Is there an easy way to do this using MATLAB or Maple or do i need to do each point by hand? I would like to avoid unnecessary headaches if possible.
Thanks
Can you offer guidance or do you also need help?
Similar Discussions: Plotting vector fields in MATLAB or Maple | {"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": 0.8971399664878845, "perplexity": 539.8090926039446}, "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-2017-17/segments/1492917123635.74/warc/CC-MAIN-20170423031203-00298-ip-10-145-167-34.ec2.internal.warc.gz"} |
https://proofwiki.org/wiki/Intermediate_Value_Theorem/Corollary | # Intermediate Value Theorem/Corollary
## Theorem
Let $I$ be a real interval.
Let $f: I \to \R$ be a real function which is continuous on $\left({a \,.\,.\, b}\right)$.
Let $a, b \in I$ such that $\left({a \,.\,.\, b}\right)$ is an open interval.
Let $0 \in \R$ lie between $f \left({a}\right)$ and $f \left({b}\right)$.
That is, either:
$f \left({a}\right) < 0 < f \left({b}\right)$
or:
$f \left({b}\right) < 0 < f \left({a}\right)$
Then $f$ has a root in $\left({a \,.\,.\, b}\right)$.
## Proof
Follows directly from the Intermediate Value Theorem and from the definition of root.
$\blacksquare$ | {"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": 0.9678229689598083, "perplexity": 183.27946964583953}, "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-2020-29/segments/1593657151197.83/warc/CC-MAIN-20200714181325-20200714211325-00538.warc.gz"} |
https://acems.org.au/file/social-network | # social-network
A graphical representation of a social network constructed from discussion on Twitter. | {"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": 0.9757862687110901, "perplexity": 2663.2330416518544}, "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-2019-43/segments/1570986649841.6/warc/CC-MAIN-20191014074313-20191014101313-00277.warc.gz"} |
http://mathhelpforum.com/algebra/103028-solved-i-cant-get-one-print.html | # [SOLVED] I can't get this one....
• Sep 18th 2009, 04:21 PM
cnotez
[SOLVED] I can't get this one....
i thought the answer to this question would be 700, but on the answer sheet it says 500. I don't get how it was done. Please help, thanks.
If two planes leave the same airport at 1:00 PM, how many miles apart will they be at 3:00 PM if one travels directly north at 150 mph and the other travels directly west at 200 mph?
• Sep 18th 2009, 04:23 PM
galactus
In 2 hours, plane A travels 300 miles.
In 2 hours, plane B travels 400 miles
It is just a 3-4-5 triangle. Use Pythagoras.
• Sep 18th 2009, 04:23 PM
artvandalay11
2 hours have past, so the planes move 300 miles and 400 miles perpindicular to eachother, so how far apart they are is a pythagorean theorem question
We have to find the hypotneuse of this triangle, so let d be how far apart they are
$d^2=300^2+400^2$
you will see d is 500
• Sep 18th 2009, 04:49 PM
cnotez
i'm still not seeing where 500 is the answer. 25000 divided by 50 would give me 500 but where is the 50 coming from?
• Sep 18th 2009, 04:52 PM
cnotez
never mind i got it. thanks
• Sep 18th 2009, 04:54 PM
skeeter
Quote:
Originally Posted by cnotez
i'm still not seeing where 500 is the answer. 25000 divided by 50 would give me 500 but where is the 50 coming from?
50 is not part of the problem, and there is no division involved.
$d = \sqrt{300^2 + 400^2}
$
calculate it. | {"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": 0.9241306185722351, "perplexity": 1249.8706569078054}, "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-2017-13/segments/1490218188891.62/warc/CC-MAIN-20170322212948-00399-ip-10-233-31-227.ec2.internal.warc.gz"} |
https://chem.libretexts.org/Bookshelves/Ancillary_Materials/Worksheets/Worksheets%3A_General_Chemistry/Worksheets%3A_General_Chemistry_(Traditional)/Ions_in_Solution_(Worksheet) | # Ions in Solution (Worksheet)
Name: ______________________________
Section: _____________________________
Student ID#:__________________________
Work in groups on these problems. You should try to answer the questions without referring to your textbook. If you get stuck, try asking another group for help.
“Begin at the beginning…and go on till you come to the end: then stop."
Lewis Carroll
Early ideas of atoms and compounds, developed primarily through the reactions of solids and gases, did not include the concept of charge. Atoms and molecules were seen as neutral particles. However, as the study of chemistry progressed to include solutions, new models were needed because the old models could not explain electrical conductivity. Studies of the electrical conductivity of solutions, and other properties of solutions such as freezing point depression and osmotic pressure, showed an interesting dichotomy. Solutions of compounds like sugar did not increase the electrical conductivity of water, yet they had lower freezing points than pure water. Solutions of compounds such as sodium chloride greatly affected the electrical conductivity of water, and they also caused the freezing point of the solution to be reduced twice as much as was observed in sugar water solutions.
A new model that explained these observations was based on the concept that charged particles, which were called ions, formed in solutions. If compounds like sodium chloride broke apart into charged particles when in solution, the ions could carry electrical current. Substances such as sugar must not break into ions in solution because they did not conduct electricity. These studies of the characteristics of solutions led to a more complete and accurate understanding of chemistry at the particulate level.
## Solvent and Solute
When a solid dissolves in a liquid to form a solution, the solid is called the solute, and the liquid is called the solvent. This is the only case that we will consider in this workshop. Note, these few terms are insufficient to describe solutions in general. They will be expanded upon in the workshop Solutions.
Mass of potassium chloride crystals
## Solubility
Solubility is a measure of how much solute can dissolve in a given amount of solvent. A wide variety of units of concentration can be used for this purpose. When describing how much of a given solute dissolves in water, the most common, "universal" solvent, the semi-quantitative terms: soluble, slightly soluble, and insoluble, can be used. These terms are applied quite loosely, and there tends to be substantial variation in the ranges of these categories and other descriptive terms. A more detailed treatment will be provided in a later workshop, Solutions. A table and a set of a few generalized rules follow that summarize the solubility characteristics of a number of compounds.
## Simple Solubility Rules for Ionic Salts in Water
1. Most nitrate (NO3) salts are soluble.
2. Most sodium, potassium, and ammonium (Na+, K+, NH4+) salts are soluble.
3. Most chloride (Cl) salts are soluble. Notable exceptions are AgCl, PbCl2, and Hg2Cl2.
4. Most sulfate (SO42–) salts are soluble. Combinations with sulfate ion which form insoluble compounds are: SrSO4, BaSO4, Hg2SO4, Ag2SO4, and PbSO4.
5. Most hydroxide (OH) salts are only slightly soluble. The important soluble hydroxides are NaOH, KOH, and Ca(OH)2.
6. Most sulfide, carbonate, and phosphate (S2, CO32–, PO43–) salts are only slightly soluble.
## Electrolytes and Nonelectrolytes
An electrolyte is a compound whose aqueous solution contains ions. When NaCl dissolves in water, the compound dissociates into Na+ and Cl ions. A good test to determine whether or not a compound is an electrolyte is to measure the ability of its water solution to conduct an electrical current. Consider a battery which has both a positive and a negative pole. If the poles are immersed in a solution via conductive metal electrodes, such as copper wires, the positively charged sodium ions in the solution will move toward the negative pole and the negatively charged chloride ions will move toward the positive pole. Such a solution has a high conductivity.
In contrast, if a neutral molecule such as sugar is in solution, it will not move toward either pole and the solution will be a non-conductor.
## Strong and Weak Electrolytes
Electrolytes can be further classified as strong or weak electrolytes. Strong electrolytes are compounds like NaCl, which are nearly 100% dissociated in solution. This means that nearly every sodium chloride formula unit exists as sodium ions surrounded by water molecules and chloride ions surrounded by water molecules. We can represent this by the following equation:
NaCl(s) --> Na+(aq) + Cl(aq)
## The Proton in Chemistry
Acids and bases form a special and very important class of electrolytes. Some acids, such as hydrochloric acid, HCl, almost completely dissociate in aqueous solution. They are strong electrolytes. These acids are similar to sodium chloride in that they exist as ions when in solution. Other acids such as acetic acid, CH3COOH, dissociate only slightly when dissolved in water. These are classified as weak electrolytes. For example, at a certain concentration and temperature, only four of every one hundred acetic acid molecules will ionize in solution. We represent the ionization of weak electrolytes in solution with double arrows as shown below:
CH3COOH (aq) --0> CH3COO(aq) + H+(aq)
In the case of a strong acid such as HCl, a single arrow is used in the reaction equation. This indicates that essentially all the HCl molecules dissociate.
HCl(aq) --> H+(aq) + Cl(aq)
Acids can be defined as substances that release hydrogen ions, H+(aq), in solution. The concentration of H+(aq) in solution is an important factor in a great number of chemical processes, including many of biological interest. Common acids that you may be familiar with include hydrochloric acid (sometimes called muriatic acid), which is used to control the acidity of swimming pools, sulfuric acid, found in automobile batteries, and phosphoric acid, which is widely used in colas for flavoring.
Now we will consider the chemical “opposite” of acids, which are compounds known as bases. A base is a compound that produces hydroxide ions, OH–(aq), in solution. As with acids, bases can be classified as either weak or strong. An example of a strong base is sodium hydroxide:
NaOH(s) --> Na+(aq) + OH(aq)
Ammonia is a common weak base:
NH3(aq) + H2O (l) --> NH4+(aq) + OH(aq)
When acids and bases react with each other, they form an ionic salt and water in what is called a neutralization reaction. Examples include:
NaOH(aq) + HCl(aq) --> Na+(aq) + Cl(aq) + H2O(l)
Ba(OH)2(s) + H2SO4(aq) --> BaSO4(s) + 2 H2O(l)
The salt may be soluble in water, as is sodium chloride, or it may precipitate out as a solid, as does barium sulfate.
## Molarity: Chemistry's Most Often Used Concentration Unit
Chemists for ease of application use a specialized mole-based system of concentration units to express the amount of solute in a solution. This system allows chemists to easily extend stoichiometric calculations to reactions that occur in solution.
Molarity is defined as the number of moles of solute per liter of solution and is given the symbol M:
M = (moles of solute)/(volume of solution in L )
When chemists prepare solutions, they usually refer to the molarity of the compound dissolved in the solution, whether or not it exists as ions. For example, if sufficient water is added to dissolve 1.0 mole of NaCl and bring the total solution volume to 1.0 L, then the solution is called a 1.0 M NaCl solution. We know that there are no “NaCl” particles in the solution, but rather Na+(aq) and Cl(aq) particles.
## Chemical Analysis by Titration
Reactions in solution are useful for determining the amount of a particular chemical species present in a given aqueous sample. For example, swimming pool water is often analyzed for its acid content. One way to determine the amount of acid in a solution is to titrate the solution with a base of known concentration. To analyze a solution that contains the acid HCl, for example, you can add a known concentration of the base NaOH in small amounts until all of the acid is neutralized. By measuring the volume of NaOH solution needed to neutralize the HCl, the moles of NaOH added can be determined. Since the reaction of HCl and NaOH occurs in a 1:1 ratio, at the equivalence point of the titration, the moles of NaOH added must be equal to the moles of HCl in solution. The reaction equation is
HCl (aq) + NaOH (aq) --> Na+(aq) + Cl(aq) + H2O (l)
In every titration, we need a way to determine the point at which the reaction is complete. In the case of our example titration of NaOH into HCl, the net ionic equation is
H+(aq) + OH(aq) --> H2O(l)
The equivalence point is reached when all of the H+(aq) ions in the HCl solution have reacted. The consumption of H+ (aq) can be detected by employing a chemical dye known as an indicator. Indicators change color when the hydrogen ion concentration of a solution changes substantially. The color change signals the endpoint of the titration. A pH meter can also be used to measure the H+(aq) in solution and signal the endpoint.
As we noted earlier, we can apply stoichiometric calculations to reactions that occur in solution. The macroscopic–particulate conversion is made with the molarity concentration unit. Molarity allows us to convert from moles to liters and vice versa. Continuing to consider the titration of sodium hydroxide solution into hydrochloric acid:
Assume that we want to know the concentration of a 25.0 mL (0.0250 L) sample of HCl. From the balanced chemical equation we know that 1 mole of HCl reacts with 1 mole of NaOH. If the titration required 17.9 mL (0.0179 L) of 0.122 M NaOH solution, the concentration of the HCl solution is calculated as follows:
0.0179 L 0.122 mol NaOH L 1 mol HCl1 mol NaOH = 0.00218 mol HCl
0.00218 mol HCl 0.0250 L = 0.0872 mol HClL = 0.0872 M HCl | {"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": 0.8855990767478943, "perplexity": 1975.3350000156029}, "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-2020-24/segments/1590347401004.26/warc/CC-MAIN-20200528232803-20200529022803-00305.warc.gz"} |
http://latexref.xyz/Font-sizes.html | Next: , Previous: , Up: Fonts [Contents][Index]
### 4.3 Font sizes
The following standard type size commands are supported by LaTeX. The table shows the command name and the corresponding actual font size used (in points) with the ‘10pt’, ‘11pt’, and ‘12pt’ document size options, respectively (see Document class options).
Command10pt11pt12pt
\tiny566
\scriptsize788
\footnotesize8910
\small91010.95
\normalsize (default)1010.9512
\large121214.4
\Large14.414.417.28
\LARGE17.2817.2820.74
\huge20.7420.7424.88
\Huge24.8824.8824.88
The commands are listed here in declaration (not environment) form, since that is how they are typically used. For example.
\begin{quotation} \small
The Tao that can be named is not the eternal Tao.
\end{quotation}
Here, the scope of the \small lasts until the end of the quotation environment. It would also end at the next type style command or the end of the current group, so you could enclose it in curly braces {\small This text is typeset in the small font.}.
An environment form of each of these commands is also defined; for instance, \begin{tiny}...\end{tiny}. However, in practice this form can easily lead to unwanted spaces at the beginning and/or end of the environment without careful consideration, so it’s generally less error-prone to stick to the declaration form.
(Aside: Technically, due to the way LaTeX defines \begin and \end, nearly every command that does not take an argument technically has an environment form. But in almost all cases, it would only cause confusion to use it. The reason for mentioning the environment form of the font size declarations specifically is that this particular use is not rare.) | {"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": 0.9985612034797668, "perplexity": 1807.9293266212599}, "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/1631780057158.19/warc/CC-MAIN-20210921041059-20210921071059-00195.warc.gz"} |
https://highnoongmt.wordpress.com/2013/06/18/formalized_evaluation_in_machine_learning/ | # Formalized evaluation in machine learning?
I am wondering if experimental design, or evaluation in machine learning,
has ever been formalized? I show below an example of what I wish to have.
In short, one thing I want is a nice way to define an evaluation that addresses some claim or hypothesis, which shows where weak points in validity arise.
Also, it would be a nice way to explore the possibilities of other evaluation.
An evaluation of some machine learning systems is completely specified by three things:
an experimental design ($$\mathcal{E}$$),
test data ($$\mathcal{U}$$), and a relevant figure of merit
($$\mathcal{F}$$).
In more formal terms,
define a unit $$u \in \Omega$$ a member of the universal set of units,
and thus the test data $$\mathcal{U} := \{u_n : n \in \mathcal{N}\}$$
as an indexed set of units.
Define the figure of merit $$\mathcal{F}$$ as a set of
functions $$f \in \mathcal{F}$$ where
the range of $$f$$ is $$\mathcal{R}(f)$$.
For instance, if $$f$$ is a function that produces a confusion table,
then $$\mathcal{R}(f) = {N}_0^{T\times T}$$.
Now, we can see $$\mathcal{E}$$ as a map of $$\mathcal{U}$$ by $$\mathcal{X}$$
into the set of ranges of each member of $$\mathcal{F}$$:
$$\mathcal{E}: \mathcal{U} \times \mathcal{X} \to \{\mathcal{R}(f) : f \in \mathcal{F}\}.$$
As an example, consider the classification system to be the map
$$\mathcal{X} : \Omega \to [0,1]^{T}$$,
and define the function $$\mathcal{L} : \Omega \to [0,1]^{T}$$,
which produces the “ground truth” of a unit.
The experimental design Classify
is thus defined as
$$\mathcal{E}_{\textrm{Cl}}(\mathcal{U},\mathcal{X}) := \bigl \{f\{(\mathcal{L}(u_n),\mathcal{X}(u_n)) : u_n \in \mathcal{U}\} : f \in \mathcal{F} \bigr \}.$$
A relevant $$f$$ produces a confusion table.
Now consider a system that retrieves a set of $$M$$
units from $$\mathcal{U}$$ based on observing
a $$u \in \Omega$$, i.e., $$\mathcal{X} : \Omega \to \mathcal{U}^M$$.
The experimental design Retrieve is defined by
$$\mathcal{E}_{\textrm{Re}} := \bigl \{f\{(\mathcal{L}(u),\mathcal{L}(u’)) : u’ \in \mathcal{X}(u) \} : f \in \mathcal{F} \bigr \}.$$
A relevant $$f$$ is precision at $$M$$ for each class.
Another experimental design is Generalize,
which is defined by crossing the experimental design
Classify with several datasets, i.e.,
\begin{align} \mathcal{E}_{\textrm{Ge}} & := \bigl \{\mathcal{E}_{\textrm{Cl}} \times \{\mathcal{U}_1,\mathcal{U}_2, \ldots\} \bigr\} \\ & := \Bigl \{f \bigl \{ \{(\mathcal{L}(u_n),\mathcal{X}(u_n)) : u_n \in \mathcal{U}\} : \mathcal{U} \in \{\mathcal{U}_1, \mathcal{U}_2, \ldots \} \bigr \} : f \in \mathcal{F} \Bigr \}. \end{align}
A relevant $$f$$ is classification accuracy.
Now, consider a system $$\mathcal{X} : [0,1]^T \to \Omega$$, i.e., it maps a label
to a $$u \in \Omega$$.
The experimental design Compose is defined by
$$\mathcal{E}_{\textrm{Co}} := \bigl \{f\{(l,\mathcal{L}(\mathcal{X}(l))) : l \in \mathcal{L} \} : f \in \mathcal{F} \bigr \}$$
where $$\mathcal{L}$$ is a set of labels.
In this case, $$\mathcal{L}$$ is an expert or other system labeling the output of $$\mathcal{X}$$.
A relevant $$f$$ is a confusion table. | {"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": 2, "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": 0.9942235350608826, "perplexity": 1138.8710028721248}, "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-2017-43/segments/1508187822488.34/warc/CC-MAIN-20171017200905-20171017220905-00231.warc.gz"} |
https://www.esaral.com/q/find-the-direction-cosines-of-a-line-which-makes-equal-angles-with-the-coordinate-axes-93769 | Find the direction cosines of a line which makes equal angles with the coordinate axes.
Question:
Find the direction cosines of a line which makes equal angles with the coordinate axes.
Solution:
Let the direction cosines of the line make an angle α with each of the coordinate axes.
∴ l = cos αm = cos αn = cos α
$l^{2}+m^{2}+n^{2}=1$
$\Rightarrow \cos ^{2} \alpha+\cos ^{2} \alpha+\cos ^{2} \alpha=1$
$\Rightarrow 3 \cos ^{2} \alpha=1$
$\Rightarrow \cos ^{2} \alpha=\frac{1}{3}$
$\Rightarrow \cos \alpha=\pm \frac{1}{\sqrt{3}}$
Thus, the direction cosines of the line, which is equally inclined to the coordinate axes, are $\pm \frac{1}{\sqrt{3}}, \pm \frac{1}{\sqrt{3}}$, and $\pm \frac{1}{\sqrt{3}}$. | {"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": 0.88185715675354, "perplexity": 163.68201515885008}, "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-2023-14/segments/1679296950247.65/warc/CC-MAIN-20230401191131-20230401221131-00295.warc.gz"} |
https://socratic.org/questions/56f1664c7c01495d90f38149 | Chemistry
Topics
# Question #38149
The formula must be written ${\left(N {H}_{4}\right)}_{2} S {O}_{4}$ because ammonium sulfate is an ionic compound.
#### Explanation:
Ionic compounds combine positively charged ions (cations) with negatively charged ions (anions).
The resulting compound will be electrically neutral - that means the compound has an overall charge of zero.
The ammonium ion ($N {H}_{4}^{+}$) has a charge of +1, so you need to have two of them to balance the charge of the sulfate ion ($S {O}_{4}^{-} 2$) which has a charge of -2.
This video discusses how to determine the chemical formula for some additional examples of ternary ionic compounds. The video discusses how to write formulas for; lithium hydroxide, magnesium nitrate and aluminum sulfate.
Hope this helps!
##### Impact of this question
1392 views around the world | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 3, "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": 0.8081099987030029, "perplexity": 2451.614480769954}, "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/1579251671078.88/warc/CC-MAIN-20200125071430-20200125100430-00353.warc.gz"} |
https://www.ann-geophys.net/32/449/2014/ | Journal topic
Ann. Geophys., 32, 449–463, 2014
https://doi.org/10.5194/angeo-32-449-2014
Special issue: Dynamical processes in space plasmas II
Ann. Geophys., 32, 449–463, 2014
https://doi.org/10.5194/angeo-32-449-2014
Regular paper 25 Apr 2014
Regular paper | 25 Apr 2014
# Excitation of planetary electromagnetic waves in the inhomogeneous ionosphere
Yu. Rapoport1, Yu. Selivanov2, V. Ivchenko1, V. Grimalsky3, E. Tkachenko1, A. Rozhnoi4, and V. Fedun5 Yu. Rapoport et al.
• 1Kyiv National Taras Shevchenko University, 2, Glushkova Av., Build. 1, 03680 Kyiv, Ukraine
• 2Space Research Institute NASU-NSAU, 40, Glushkov Av., 03680, Kiev 187, Ukraine
• 3University of Morelos, Av. Universidad, 1001, Z.P. 62210, Cuernavaca, Morelos, Mexico
• 4Institute of Physics of the Earth, Russian Academy of Sciences, 10 B. Gruzinskaya, Moscow, 123995, Russia
• 5Space Systems Laboratory, Department of Automatic Control and Systems Engineering, University of Sheffield, Sheffield, S1 3JD, UK
Abstract. In this paper we develop a new method for the analysis of excitation and propagation of planetary electromagnetic waves (PEMW) in the ionosphere of the Earth. The nonlinear system of equations for PEMW, valid for any height, from D to F regions, including intermediate altitudes between D and E and between E and F regions, is derived. In particular, we have found the system of nonlinear one-fluid MHD equations in the β-plane approximation valid for the ionospheric F region (Aburjania et al., 2003a, 2005). The series expansion in a "small" (relative to the local geomagnetic field) non-stationary magnetic field has been applied only at the last step of the derivation of the equations. The small mechanical vertical displacement of the media is taken into account. We have shown that obtained equations can be reduced to the well-known system with Larichev–Reznik vortex solution in the equatorial region (see e.g. Aburjania et al., 2002). The excitation of planetary electromagnetic waves by different initial perturbations has been investigated numerically. Some means for the PEMW detection and data processing are discussed. | {"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": 0.8072004318237305, "perplexity": 3635.8217635199485}, "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/1585370505730.14/warc/CC-MAIN-20200401100029-20200401130029-00344.warc.gz"} |
https://philarchive.org/rec/MCKSAA-10 | # Structure and applied mathematics
Synthese 200 (5):1-31 (2022)
# Abstract
‘Mapping accounts’ of applied mathematics hold that the application of mathematics in physical science is best understood in terms of ‘mappings’ between mathematical structures and physical structures. In this paper, I suggest that mapping accounts rely on the assumption that the mathematics relevant to any application of mathematics in empirical science can be captured in an appropriate mathematical structure. If we are interested in assessing the plausibility of mapping accounts, we must ask ourselves: how plausible is this assumption as a working hypothesis about applied mathematics? In order to do so, we examine the role played by mathematics in the multiscalar modelling of sea ice melting behaviour and examine whether we can indeed capture the mathematics involved in the kind of mathematical structure employed by the mapping account. Along the way, we note that the cases of applied mathematics that appear in discussions of mapping accounts almost exclusively involve the employment of a single clearly circumscribed mathematical field or domain. While the core assumption of mapping accounts may appear plausible in such situations, we ultimately suggest that the mapping account is not able to handle the important added complexities involved in our sea ice case study. In particular, the notion of mathematical structure around which such accounts are framed does not seem to be able to capture the way in which some applications of mathematics require that very different pieces of mathematics be related to one another on the basis of both mathematical and empirical information.
# Author's Profile
Travis McKenna
University of Pittsburgh | {"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": 0.8689032196998596, "perplexity": 503.5478456681487}, "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/1669446710968.29/warc/CC-MAIN-20221204072040-20221204102040-00269.warc.gz"} |
http://support.sas.com/documentation/cdl/en/statug/66859/HTML/default/statug_genmod_details22.htm | ### Predicted Values of the Mean
Subsections:
#### Predicted Values
A predicted value, or fitted value, of the mean corresponding to the vector of covariates is given by
where g is the link function, regardless of whether corresponds to an observation or not. That is, the response variable can be missing and the predicted value is still computed for valid . In the case where does not correspond to a valid observation, is not checked for estimability. You should check the estimability of in this case in order to ensure the uniqueness of the predicted value of the mean. If there is an offset, it is included in the predicted value computation.
#### Confidence Intervals on Predicted Values
Approximate confidence intervals for predicted values of the mean can be computed as follows. The variance of the linear predictor is estimated by
where is the estimated covariance of . The robust estimate of the covariance is used for in the case of models fit with GEEs.
Approximate confidence intervals are computed as
where is the percentile of the standard normal distribution and g is the link function. If either endpoint in the argument is outside the valid range of arguments for the inverse link function, the corresponding confidence interval endpoint is set to missing. | {"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": 0.9242327809333801, "perplexity": 326.4754225066257}, "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-2017-43/segments/1508187827662.87/warc/CC-MAIN-20171023235958-20171024015958-00357.warc.gz"} |
http://mathhelpforum.com/calculus/11278-multivarible-extremum-point.html | # Math Help - multivarible extremum point
1. ## multivarible extremum point
Let f(x,y)=((x^(2)y^(2))/(x^(2)+y^(2))), classify the behavior of f near the critical point (0,0).
2. Originally Posted by bobby77
Let f(x,y)=((x^(2)y^(2))/(x^(2)+y^(2))), classify the behavior of f near the critical point (0,0).
I don't know how you are supposed to approach this on your course,
but the following works:
Along the $x \,$ and $y\,$ axes the function is a constant equal to zero. Along
any other ray through the origin we may put $y=\lambda x\,$, when:
$f(x,\lambda x)=\frac{\lambda^2 x^2}{1+\lambda^2}\,$
which has a quadratic like mininum at $x=0\,$.
A surface plot shows what this looks like (see attachment)
(this assumes that we give the function a value of $0\,$ at $x=y=0\,$)
RonL
Attached Thumbnails
3. Originally Posted by CaptainBlank
I don't know how you are supposed to approach this on your course,
but the following works:
I think he wants it done through the second partials test.
4. Originally Posted by ThePerfectHacker
I think he wants it done through the second partials test.
In this case that will be inconclustve (I think, I haven't checked)
RonL
5. Originally Posted by CaptainBlack
In this case that will be inconclustve (I think, I haven't checked)
RonL
I agree with that, because by different views (planes) the point is both a maximum and minimum. Hence it is neither.
6. Originally Posted by ThePerfectHacker | {"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": 7, "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": 0.9315493106842041, "perplexity": 1332.1875042277966}, "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/1429246652631.96/warc/CC-MAIN-20150417045732-00090-ip-10-235-10-82.ec2.internal.warc.gz"} |
https://www.esaral.com/q/at-room-temperature-27-0-c-the-resistance-of-a-heating-element-is-100-38618/ | At room temperature (27.0 °C) the resistance of a heating element is 100 Ω.
Question:
At room temperature $\left(27.0^{\circ} \mathrm{C}\right)$ the resistance of a heating element is $100 \Omega$. What is the temperature of the element if the resistance is found to be $117 \Omega$, given that the temperature coefficient of the material of the resistor is $1.70 \times 10^{-4}{ }^{\circ} \mathrm{C}^{-1}$
Solution:
Room temperature, T = 27°C
Resistance of the heating element at TR = 100 Ω
Let T1 is the increased temperature of the filament.
Resistance of the heating element at T1R1 = 117 Ω
Temperature co-efficient of the material of the filament,
$\alpha=1.70 \times 10^{-4} \mathrm{C}^{-1}$
$\alpha$ is given by the relation,
$\alpha=\frac{R_{1}-R}{R\left(T_{1}-T\right)}$
$T_{1}-T=\frac{R_{1}-R}{R \alpha}$
$T_{1}-27=\frac{117-100}{100\left(1.7 \times 10^{-4}\right)}$
$T_{1}-27=1000$
$T_{1}=1027^{\circ} \mathrm{C}$
Therefore, at $1027^{\circ} \mathrm{C}$, the resistance of the element is $117 \Omega$. | {"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": 0.8942722082138062, "perplexity": 382.5366052257227}, "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/1669446711278.74/warc/CC-MAIN-20221208050236-20221208080236-00347.warc.gz"} |
https://cris.bgu.ac.il/en/publications/on-the-chain-pair-simplification-problem-3 | # On the chain pair simplification problem
Chenglin Fan, Omrit Filtser, Matthew J. Katz, Tim Wylie, Binhai Zhu
Research output: Chapter in Book/Report/Conference proceedingConference contributionpeer-review
2 Scopus citations
## Abstract
The problem of efficiently computing and visualizing the structural resemblance between a pair of protein backbones in 3D has led Bereg et al. [4] to pose the Chain Pair Simplification problem (CPS). In this problem, given two polygonal chains A and B of lengths m and n, respectively, one needs to simplify them simultaneously, such that each of the resulting simplified chains, Aʹ and Bʹ, is of length at most k and the discrete Fréchet distance between Aʹ and Bʹ is at most δ, where k and δ are given parameters. In this paper we study the complexity of CPS under the discrete Fréchet distance (CPS-3F), i.e., where the quality of the simplifications is also measured by the discrete Fréchet distance. Since CPS-3F was posed in 2008, its complexity has remained open. In this paper, we prove that CPS-3F is actually polynomially solvable, by presenting an O(m2n2 min{m, n}) time algorithm for the corresponding minimization problem. On the other hand, we prove that if the vertices of the chains have integral weights then the problem is weakly NP-complete.
Original language English Algorithms and Data Structures - 14th International Symposium, WADS 2015, Proceedings Frank Dehne, Jorg-Rudiger Sack, Ulrike Stege Springer Verlag 351-362 12 9783319218397 https://doi.org/10.1007/978-3-319-21840-3_29 Published - 1 Jan 2015 14th International Symposium on Algorithms and Data Structures, WADS 2015 - Victoria, CanadaDuration: 5 Aug 2015 → 7 Aug 2015
### Publication series
Name Lecture Notes in Computer Science (including subseries Lecture Notes in Artificial Intelligence and Lecture Notes in Bioinformatics) 9214 0302-9743 1611-3349
### Conference
Conference 14th International Symposium on Algorithms and Data Structures, WADS 2015 Canada Victoria 5/08/15 → 7/08/15
## Fingerprint
Dive into the research topics of 'On the chain pair simplification problem'. Together they form a unique fingerprint. | {"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": 0.909866452217102, "perplexity": 2711.1606032005743}, "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-2022-21/segments/1652662562410.53/warc/CC-MAIN-20220524014636-20220524044636-00517.warc.gz"} |
https://en.m.wikipedia.org/wiki/Automatic_differentiation | # Automatic differentiation
In mathematics and computer algebra, automatic differentiation (AD), also called algorithmic differentiation or computational differentiation,[1][2] is a set of techniques to numerically evaluate the derivative of a function specified by a computer program. AD exploits the fact that every computer program, no matter how complicated, executes a sequence of elementary arithmetic operations (addition, subtraction, multiplication, division, etc.) and elementary functions (exp, log, sin, cos, etc.). By applying the chain rule repeatedly to these operations, derivatives of arbitrary order can be computed automatically, accurately to working precision, and using at most a small constant factor more arithmetic operations than the original program.
Automatic differentiation is not:
Figure 1: How automatic differentiation relates to symbolic differentiation
These classical methods run into problems: symbolic differentiation leads to inefficient code (unless done carefully) and faces the difficulty of converting a computer program into a single expression, while numerical differentiation can introduce round-off errors in the discretization process and cancellation. Both classical methods have problems with calculating higher derivatives, where the complexity and errors increase. Finally, both classical methods are slow at computing the partial derivatives of a function with respect to many inputs, as is needed for gradient-based optimization algorithms. Automatic differentiation solves all of these problems, at the expense of introducing more software dependencies[citation needed].
## The chain rule, forward and reverse accumulationEdit
Fundamental to AD is the decomposition of differentials provided by the chain rule. For the simple composition ${\displaystyle y=f(g(h(x)))=f(g(h(w_{0})))=f(g(w_{1}))=f(w_{2})=w_{3}}$ the chain rule gives
${\displaystyle {\frac {dy}{dx}}={\frac {dy}{dw_{2}}}{\frac {dw_{2}}{dw_{1}}}{\frac {dw_{1}}{dx}}}$
Usually, two distinct modes of AD are presented, forward accumulation (or forward mode) and reverse accumulation (or reverse mode). Forward accumulation specifies that one traverses the chain rule from inside to outside (that is, first compute ${\displaystyle dw_{1}/dx}$ and then ${\displaystyle dw_{2}/dx}$ and at last ${\displaystyle dy/dx}$ ), while reverse accumulation has the traversal from outside to inside (first compute ${\displaystyle dy/dw_{2}}$ and then ${\displaystyle dy/dw_{1}}$ and at last ${\displaystyle dy/dx}$ ). More succinctly,
1. forward accumulation computes the recursive relation: ${\displaystyle {\frac {dw_{i}}{dx}}={\frac {dw_{i}}{dw_{i-1}}}{\frac {dw_{i-1}}{dx}}}$ with ${\displaystyle w_{3}=y}$ , and,
2. reverse accumulation computes the recursive relation: ${\displaystyle {\frac {dy}{dw_{i}}}={\frac {dy}{dw_{i+1}}}{\frac {dw_{i+1}}{dw_{i}}}}$ with ${\displaystyle w_{0}=x}$ .
Generally, both forward and reverse accumulation are specific manifestations of applying the operator of program composition, with the appropriate one of the two mappings ${\displaystyle (w,y)}$ being fixed.
### Forward accumulationEdit
Figure 2: Example of forward accumulation with computational graph
In forward accumulation AD, one first fixes the independent variable to which differentiation is performed and computes the derivative of each sub-expression recursively. In a pen-and-paper calculation, one can do so by repeatedly substituting the derivative of the inner functions in the chain rule:
${\displaystyle {\frac {\partial y}{\partial x}}={\frac {\partial y}{\partial w_{n-1}}}{\frac {\partial w_{n-1}}{\partial x}}={\frac {\partial y}{\partial w_{n-1}}}\left({\frac {\partial w_{n-1}}{\partial w_{n-2}}}{\frac {\partial w_{n-2}}{\partial x}}\right)={\frac {\partial y}{\partial w_{n-1}}}\left({\frac {\partial w_{n-1}}{\partial w_{n-2}}}\left({\frac {\partial w_{n-2}}{\partial w_{n-3}}}{\frac {\partial w_{n-3}}{\partial x}}\right)\right)=\cdots }$
This can be generalized to multiple variables as a matrix product of Jacobians.
Compared to reverse accumulation, forward accumulation is very natural and easy to implement as the flow of derivative information coincides with the order of evaluation. One simply augments each variable w with its derivative (stored as a numerical value, not a symbolic expression),
${\displaystyle {\dot {w}}={\frac {\partial w}{\partial x}}}$
as denoted by the dot. The derivatives are then computed in sync with the evaluation steps and combined with other derivatives via the chain rule.
As an example, consider the function:
{\displaystyle {\begin{aligned}z&=f(x_{1},x_{2})\\&=x_{1}x_{2}+\sin x_{1}\\&=w_{1}w_{2}+\sin w_{1}\\&=w_{3}+w_{4}\\&=w_{5}\end{aligned}}}
For clarity, the individual sub-expressions have been labeled with the variables wi.
The choice of the independent variable to which differentiation is performed affects the seed values 1 and 2. Suppose one is interested in the derivative of this function with respect to x1. In this case, the seed values should be set to:
{\displaystyle {\begin{aligned}{\dot {w}}_{1}={\frac {\partial x_{1}}{\partial x_{1}}}=1\\{\dot {w}}_{2}={\frac {\partial x_{2}}{\partial x_{1}}}=0\end{aligned}}}
With the seed values set, one may then propagate the values using the chain rule as shown in the table below. Figure 2 shows a pictorial depiction of this process as a computational graph.
${\displaystyle {\begin{array}{l|l}{\text{Operations to compute value}}&{\text{Operations to compute derivative}}\\\hline w_{1}=x_{1}&{\dot {w}}_{1}=1{\text{ (seed)}}\\w_{2}=x_{2}&{\dot {w}}_{2}=0{\text{ (seed)}}\\w_{3}=w_{1}\cdot w_{2}&{\dot {w}}_{3}=w_{2}\cdot {\dot {w}}_{1}+w_{1}\cdot {\dot {w}}_{2}\\w_{4}=\sin w_{1}&{\dot {w}}_{4}=\cos w_{1}\cdot {\dot {w}}_{1}\\w_{5}=w_{3}+w_{4}&{\dot {w}}_{5}={\dot {w}}_{3}+{\dot {w}}_{4}\end{array}}}$
To compute the gradient of this example function, which requires the derivatives of f with respect to not only x1 but also x2, one must perform an additional sweep over the computational graph using the seed values ${\displaystyle {\dot {w}}_{1}=0;{\dot {w}}_{2}=1}$ .
The computational complexity of one sweep of forward accumulation is proportional to the complexity of the original code.
Forward accumulation is more efficient than reverse accumulation for functions f : ℝn → ℝm with mn as only n sweeps are necessary, compared to m sweeps for reverse accumulation.
### Reverse accumulationEdit
Figure 3: Example of reverse accumulation with computational graph
In reverse accumulation AD, one first fixes the dependent variable to be differentiated and computes the derivative with respect to each sub-expression recursively. In a pen-and-paper calculation, one can perform the equivalent by repeatedly substituting the derivative of the outer functions in the chain rule:
${\displaystyle {\frac {\partial y}{\partial x}}={\frac {\partial y}{\partial w_{1}}}{\frac {\partial w_{1}}{\partial x}}=\left({\frac {\partial y}{\partial w_{2}}}{\frac {\partial w_{2}}{\partial w_{1}}}\right){\frac {\partial w_{1}}{\partial x}}=\left(\left({\frac {\partial y}{\partial w_{3}}}{\frac {\partial w_{3}}{\partial w_{2}}}\right){\frac {\partial w_{2}}{\partial w_{1}}}\right){\frac {\partial w_{1}}{\partial x}}=\cdots }$
In reverse accumulation, the quantity of interest is the adjoint, denoted with a bar (); it is a derivative of a chosen dependent variable with respect to a subexpression w:
${\displaystyle {\bar {w}}={\frac {\partial y}{\partial w}}}$
Reverse accumulation traverses the chain rule from outside to inside, or in the case of the computational graph in Figure 3, from top to bottom. The example function is scalar-valued, and thus there is only one seed for the derivative computation, and only one sweep of the computational graph is needed in order to calculate the (two-component) gradient. This is only half the work when compared to forward accumulation, but reverse accumulation requires the storage of the intermediate variables wi as well as the instructions that produced them in a data structure known as a Wengert list (or "tape"),[3][4] which may represent a significant memory issue if the computational graph is large. This can be mitigated to some extent by storing only a subset of the intermediate variables and then reconstructing the necessary work variables by repeating the evaluations, a technique known as checkpointing.
The operations to compute the derivative using reverse accumulation are shown in the table below (note the reversed order):
${\displaystyle {\begin{array}{l}{\text{Operations to compute derivative}}\\\hline {\bar {w}}_{5}=1{\text{ (seed)}}\\{\bar {w}}_{4}={\bar {w}}_{5}\\{\bar {w}}_{3}={\bar {w}}_{5}\\{\bar {w}}_{2}={\bar {w}}_{3}\cdot w_{1}\\{\bar {w}}_{1}={\bar {w}}_{3}\cdot w_{2}+{\bar {w}}_{4}\cdot \cos w_{1}\end{array}}}$
The data flow graph of a computation can be manipulated to calculate the gradient of its original calculation. This is done by adding an adjoint node for each primal node, connected by adjoint edges which parallel the primal edges but flow in the opposite direction. The nodes in the adjoint graph represent multiplication by the derivatives of the functions calculated by the nodes in the primal. For instance, addition in the primal causes fanout in the adjoint; fanout in the primal causes addition in the adjoint; a unary function y = f(x) in the primal causes = ȳ f′(x) in the adjoint; etc.
Reverse accumulation is more efficient than forward accumulation for functions f : ℝn → ℝm with mn as only m sweeps are necessary, compared to n sweeps for forward accumulation.
Reverse mode AD was first published in 1970 by Seppo Linnainmaa in his master thesis.[5][6][7]
Backpropagation of errors in multilayer perceptrons, a technique used in machine learning, is a special case of reverse mode AD.
### Beyond forward and reverse accumulationEdit
Forward and reverse accumulation are just two (extreme) ways of traversing the chain rule. The problem of computing a full Jacobian of f : ℝn → ℝm with a minimum number of arithmetic operations is known as the optimal Jacobian accumulation (OJA) problem, which is NP-complete.[8] Central to this proof is the idea that there may exist algebraic dependencies between the local partials that label the edges of the graph. In particular, two or more edge labels may be recognized as equal. The complexity of the problem is still open if it is assumed that all edge labels are unique and algebraically independent.
## Automatic differentiation using dual numbersEdit
Forward mode automatic differentiation is accomplished by augmenting the algebra of real numbers and obtaining a new arithmetic. An additional component is added to every number which will represent the derivative of a function at the number, and all arithmetic operators are extended for the augmented algebra. The augmented algebra is the algebra of dual numbers. This approach was generalized by the theory of operational calculus on programming spaces (see Analytic programming space), through tensor algebra of the dual space.
Replace every number ${\displaystyle \,x}$ with the number ${\displaystyle x+x'\varepsilon }$ , where ${\displaystyle x'}$ is a real number, but ${\displaystyle \varepsilon }$ is an abstract number with the property ${\displaystyle \varepsilon ^{2}=0}$ (an infinitesimal; see Smooth infinitesimal analysis). Using only this, we get for the regular arithmetic
{\displaystyle {\begin{aligned}(x+x'\varepsilon )+(y+y'\varepsilon )&=x+y+(x'+y')\varepsilon \\(x+x'\varepsilon )\cdot (y+y'\varepsilon )&=xy+xy'\varepsilon +yx'\varepsilon +x'y'\varepsilon ^{2}=xy+(xy'+yx')\varepsilon \end{aligned}}}
and likewise for subtraction and division.
Now, we may calculate polynomials in this augmented arithmetic. If ${\displaystyle P(x)=p_{0}+p_{1}x+p_{2}x^{2}+\cdots +p_{n}x^{n}}$ , then
{\displaystyle {\begin{aligned}P(x+x'\varepsilon )&=p_{0}+p_{1}(x+x'\varepsilon )+\cdots +p_{n}(x+x'\varepsilon )^{n}\\&=p_{0}+p_{1}x+\cdots +p_{n}x^{n}+p_{1}x'\varepsilon +2p_{2}xx'\varepsilon +\cdots +np_{n}x^{n-1}x'\varepsilon \\&=P(x)+P^{(1)}(x)x'\varepsilon \end{aligned}}}
where ${\displaystyle P^{(1)}}$ denotes the derivative of ${\displaystyle P}$ with respect to its first argument, and ${\displaystyle x'}$ , called a seed, can be chosen arbitrarily.
The new arithmetic consists of ordered pairs, elements written ${\displaystyle \langle x,x'\rangle }$ , with ordinary arithmetics on the first component, and first order differentiation arithmetic on the second component, as described above. Extending the above results on polynomials to analytic functions we obtain a list of the basic arithmetic and some standard functions for the new arithmetic:
{\displaystyle {\begin{aligned}\left\langle u,u'\right\rangle +\left\langle v,v'\right\rangle &=\left\langle u+v,u'+v'\right\rangle \\\left\langle u,u'\right\rangle -\left\langle v,v'\right\rangle &=\left\langle u-v,u'-v'\right\rangle \\\left\langle u,u'\right\rangle *\left\langle v,v'\right\rangle &=\left\langle uv,u'v+uv'\right\rangle \\\left\langle u,u'\right\rangle /\left\langle v,v'\right\rangle &=\left\langle {\frac {u}{v}},{\frac {u'v-uv'}{v^{2}}}\right\rangle \quad (v\neq 0)\\\sin \left\langle u,u'\right\rangle &=\left\langle \sin(u),u'\cos(u)\right\rangle \\\cos \left\langle u,u'\right\rangle &=\left\langle \cos(u),-u'\sin(u)\right\rangle \\\exp \left\langle u,u'\right\rangle &=\left\langle \exp u,u'\exp u\right\rangle \\\log \left\langle u,u'\right\rangle &=\left\langle \log(u),u'/u\right\rangle \quad (u>0)\\\left\langle u,u'\right\rangle ^{k}&=\left\langle u^{k},ku^{k-1}u'\right\rangle \quad (u\neq 0)\\\left|\left\langle u,u'\right\rangle \right|&=\left\langle \left|u\right|,u'{\mbox{sign}}u\right\rangle \quad (u\neq 0)\end{aligned}}}
and in general for the primitive function ${\displaystyle g}$ ,
${\displaystyle g(\langle u,u'\rangle ,\langle v,v'\rangle )=\langle g(u,v),g_{u}(u,v)u'+g_{v}(u,v)v'\rangle }$
where ${\displaystyle g_{u}}$ and ${\displaystyle g_{v}}$ are the derivatives of ${\displaystyle g}$ with respect to its first and second arguments, respectively.
When a binary basic arithmetic operation is applied to mixed arguments—the pair ${\displaystyle \langle u,u'\rangle }$ and the real number ${\displaystyle c}$ —the real number is first lifted to ${\displaystyle \langle c,0\rangle }$ . The derivative of a function ${\displaystyle f:\mathbb {R} \rightarrow \mathbb {R} }$ at the point ${\displaystyle x_{0}}$ is now found by calculating ${\displaystyle f(\langle x_{0},1\rangle )}$ using the above arithmetic, which gives ${\displaystyle \langle f(x_{0}),f'(x_{0})\rangle }$ as the result.
### Vector arguments and functionsEdit
Multivariate functions can be handled with the same efficiency and mechanisms as univariate functions by adopting a directional derivative operator. That is, if it is sufficient to compute ${\displaystyle y'=\nabla f(x)\cdot x'}$ , the directional derivative ${\displaystyle y'\in \mathbb {R} ^{m}}$ of ${\displaystyle f:\mathbb {R} ^{n}\rightarrow \mathbb {R} ^{m}}$ at ${\displaystyle x\in \mathbb {R} ^{n}}$ in the direction ${\displaystyle x'\in \mathbb {R} ^{n}}$ , this may be calculated as ${\displaystyle (\langle y_{1},y'_{1}\rangle ,\ldots ,\langle y_{m},y'_{m}\rangle )=f(\langle x_{1},x'_{1}\rangle ,\ldots ,\langle x_{n},x'_{n}\rangle )}$ using the same arithmetic as above. If all the elements of ${\displaystyle \nabla f}$ are desired, then ${\displaystyle n}$ function evaluations are required. Note that in many optimization applications, the directional derivative is indeed sufficient.
### High order and many variablesEdit
The above arithmetic can be generalized to calculate second order and higher derivatives of multivariate functions. However, the arithmetic rules quickly grow very complicated: complexity will be quadratic in the highest derivative degree. Instead, truncated Taylor polynomial algebra can be used. The resulting arithmetic, defined on generalized dual numbers, allows to efficiently compute using functions as if they were a new data type. Once the Taylor polynomial of a function is known, the derivatives are easily extracted. A rigorous, general formulation is achieved through the tensor series expansion using operational calculus on programming spaces.
## Operational calculus on programming spacesEdit
Operational calculus on programming spaces [9] generalizes concepts of automatic differentiation and provides deep learning with a formal calculus. This formulation using tensor algebra is a generalization of the dual numbers approach.
### Differentiable programming spaceEdit
A differentiable programming space ${\displaystyle {\mathcal {P}}_{0}}$ is any subspace of ${\displaystyle {\mathcal {F}}_{0}:{\mathcal {V}}\to {\mathcal {V}}}$ such that
${\displaystyle \partial {\mathcal {P}}_{0}\subset {\mathcal {P}}_{0}\otimes T({\mathcal {V}}^{*}),}$
where ${\displaystyle T({\mathcal {V}}^{*})}$ is the tensor algebra of the dual space ${\displaystyle {\mathcal {V}}^{*}}$ . When all elements of ${\displaystyle {\mathcal {P}}_{0}}$ are analytic, we call ${\displaystyle {\mathcal {P}}_{0}}$ an analytic programming space.
Theorem. Any differentiable programming space ${\displaystyle {\mathcal {P}}_{0}}$ is an infinitely differentiable programming space, meaning that
${\displaystyle \partial ^{k}{\mathcal {P}}_{0}\subset {\mathcal {P}}_{0}\otimes T({\mathcal {V}}^{*})}$
for any ${\displaystyle k\in \mathbb {N} .}$ If all elements of ${\displaystyle {\mathcal {P}}_{0}}$ are analytic, than so are the elements of ${\displaystyle {\mathcal {P}}_{n}}$ .
Definition. Let ${\displaystyle {\mathcal {P}}_{0}}$ be a differentiable programming space. The space ${\displaystyle {\mathcal {P}}_{n}<{\mathcal {F}}_{n}:{\mathcal {V}}\to {\mathcal {V}}\otimes T({\mathcal {V}}^{*})}$ spanned by ${\displaystyle {\mathcal {D}}^{n}{\mathcal {P}}_{0}}$ over ${\displaystyle K}$ , where ${\displaystyle {\mathcal {D}}^{n}=\{\partial ^{k}|0\leq k\leq n\},}$ is called a differentiable programming space of order ${\displaystyle n}$ .
Corollary. A differentiable programming space of order ${\displaystyle n,{\mathcal {P}}_{n}:{\mathcal {V}}\to {\mathcal {V}}\otimes T({\mathcal {V}}^{*}),}$ can be embedded into the tensor product of the function space ${\displaystyle {\mathcal {P}}_{0}:{\mathcal {V}}\to {\mathcal {V}}}$ and the subspace ${\displaystyle T_{n}({\mathcal {V}}^{*})}$ of the tensor algebra of the dual of the virtual space ${\displaystyle {\mathcal {P}}}$ . By taking the limit as ${\displaystyle n\to \infty }$ , we consider
${\displaystyle {\mathcal {P}}_{\infty }<{\mathcal {P}}_{0}\otimes {\mathcal {T}}({\mathcal {V}}^{*}),}$
where
${\displaystyle {\mathcal {T}}({\mathcal {V}}^{*})=\prod _{k=0}^{\infty }({\mathcal {V}}^{*})^{\otimes k}}$
is the tensor series algebra, the algebra of the infinite formal tensor series, which is a completion of the tensor algebra ${\displaystyle T({\mathcal {V}}^{*})}$ in suitable topology.
Proofs can be found in.[9]
This means that we can represent calculation of derivatives of the map ${\displaystyle P:{\mathcal {V}}\to {\mathcal {V}}}$ , with only one mapping ${\displaystyle \tau }$ . We define the operator ${\displaystyle \tau _{n}}$ as a direct sum of operators
${\displaystyle \tau _{n}=1+\partial +\partial ^{2}+\cdots +\partial ^{n}}$
The image ${\displaystyle \tau _{k}P(\mathbf {x} )}$ is a multitensor of order ${\displaystyle k}$ , which is a direct sum of the maps value and all derivatives of order ${\displaystyle n\leq k}$ , all evaluated at the point ${\displaystyle \mathbf {x} }$
${\displaystyle \tau _{k}P(\mathbf {x} )=P(\mathbf {x} )+\partial _{\mathbf {x} }P(\mathbf {x} )+\partial _{\mathbf {x} }^{2}P(\mathbf {x} )+\cdots +\partial _{\mathbf {x} }^{k}P(\mathbf {x} ).}$
The operator ${\displaystyle \tau _{n}}$ satisfies the recursive relation.
${\displaystyle \tau _{k+1}=1+\partial \tau _{k},}$
that can be used to recursively construct programming spaces of arbitrary order. Only explicit knowledge of ${\displaystyle \tau :{\mathcal {P}}_{0}\to {\mathcal {P}}_{1}}$ is required for the construction of ${\displaystyle {\mathcal {P}}_{n}}$ from ${\displaystyle {\mathcal {P}}_{1}}$ , which is evident from the above theorem.
### Virtual tensor machineEdit
The paper [9] proposed an abstract virtual machine capable of constructing and implementing the theory. Such a machine provides a framework for analytic study of algorithmic procedures through algebraic means.
Claim. The tuple ${\displaystyle ({\mathcal {V}},{\mathcal {P}}_{0})}$ and the belonging tensor series algebra are sufficient conditions for the existence and construction of infinitely differentiable programming spaces ${\displaystyle {\mathcal {P}}_{\infty }}$ , through linear combinations of elements of ${\displaystyle {\mathcal {P}}_{0}\otimes {\mathcal {T}}({\mathcal {V}}^{*})}$ .
This claim allows a simple definition of such a machine.
Definition (Virtual tensor machine). The tuple ${\displaystyle M=\langle {\mathcal {V}},{\mathcal {P}}_{0}\rangle }$ is a virtual tensor machine, where
• ${\displaystyle {\mathcal {V}}}$ is a finite dimensional vector space
• ${\displaystyle {\mathcal {V}}\otimes {\mathcal {T}}({\mathcal {V}}^{*})}$ is the virtual memory space
• ${\displaystyle {\mathcal {P}}_{0}}$ is an analytic programming space over ${\displaystyle {\mathcal {V}}}$
### Tensor series expansionEdit
Expansion into a series offers valuable insights into programs through methods of analysis.
There exists a space spanned by the set ${\displaystyle {\mathcal {D}}^{n}=\{\partial ^{k}|0\leq k\leq n\}}$ over a field ${\displaystyle K}$ . Thus, the expression
${\displaystyle e^{h\partial }=\sum _{n=0}^{\infty }{\frac {(h\partial )^{n}}{n!}}}$
is well defined. The operator ${\displaystyle e^{h\partial }}$ is a mapping between function spaces
${\displaystyle e^{h\partial }:{\mathcal {P}}\to {\mathcal {P}}_{\infty }.}$
It also defines a map
${\displaystyle e^{h\partial }:{\mathcal {P}}\times {\mathcal {V}}\to {\mathcal {V}}\otimes {\mathcal {T}}({\mathcal {V}}^{*}),}$
by taking the image of the map ${\displaystyle e^{h\partial }(P)}$ at a certain point ${\displaystyle \mathbf {v} \in {\mathcal {V}}}$ .
We may construct a map from the space of programs, to the space of polynomials. Note that the space of multivariate polynomials ${\displaystyle {\mathcal {V}}\to K}$ is isomorphic to symmetric algebra ${\displaystyle S({\mathcal {V}}^{*})}$ , which is in turn a quotient of tensor algebra ${\displaystyle T({\mathcal {V}}^{*})}$ . To any element of ${\displaystyle {\mathcal {V}}\otimes T({\mathcal {V}}^{*})}$ one can attach corresponding element of ${\displaystyle {\mathcal {V}}\otimes S({\mathcal {V}}^{i*})}$ namely a polynomial map ${\displaystyle {\mathcal {V}}\to {\mathcal {V}}}$ . Thus, we consider the completion of the symmetric algebra ${\displaystyle S({\mathcal {V}}^{*})}$ as the Formal power series ${\displaystyle {\mathcal {S}}({\mathcal {V}}^{*})}$ , which is in turn isomorphic to a quotient of tensor series algebra ${\displaystyle {\mathcal {T}}({\mathcal {V}}^{*})}$ , arriving at
${\displaystyle e^{h\partial }:{\mathcal {P}}\times {\mathcal {V}}\to {\mathcal {V}}\otimes {\mathcal {S}}({\mathcal {V}}^{i*})}$
For any element ${\displaystyle \mathbf {v} _{0}\in {\mathcal {V}}}$ , the expression ${\displaystyle e^{h\partial }(\cdot ,\mathbf {v} _{0})}$ is a map ${\displaystyle {\mathcal {P}}\to {\mathcal {V}}\otimes {\mathcal {S}}({\mathcal {V}}^{*})}$ , mapping a program to a Formal power series. We can express the correspondence between multi-tensors in ${\displaystyle {\mathcal {V}}\otimes T({\mathcal {V}}^{*})}$ and polynomial maps ${\displaystyle {\mathcal {V}}\to {\mathcal {V}}}$ given by multiple contractions for all possible indices.
Theorem. For a program ${\displaystyle P\in {\mathcal {P}}}$ the expansion into an infinite tensor series at the point ${\displaystyle \mathbf {v} _{0}\in {\mathcal {V}}}$ is expressed by multiple contractions
${\displaystyle P(\mathbf {v} _{0}+h\mathbf {v} )=\left((e^{h\partial }P)(\mathbf {v} _{0})\right)\cdot \mathbf {v} =\sum _{n=0}^{\infty }{\frac {h^{n}}{n!}}\partial ^{n}P(\mathbf {v} _{0})\cdot (\mathbf {v} ^{\otimes n})}$
Proof can be found in.[9] Evaluated at ${\displaystyle h=1}$ , the operator is a generalization of the Shift operator widely used in physics. For a specific ${\displaystyle v_{0}\in {\mathcal {V}}}$ it is here on denoted by
${\displaystyle e^{\partial }\vert _{v_{0}}:{\mathcal {P}}\to {\mathcal {V}}\otimes {\mathcal {T}}({\mathcal {V}}^{*}).}$
When the choice of ${\displaystyle v_{0}\in {\mathcal {V}}}$ is arbitrary, we omit it from expressions for brevity. Following this work, a similar approach was taken by others [10].
### Operator of program compositionEdit
Theory offers a generalization of both forward and reverse mode of automatic differentiation to arbitrary order, under a single invariant operator in the theory. This condenses complex notions to simple expressions allowing meaningful manipulations before being applied to a particular programming space.
Theorem. Composition of maps ${\displaystyle {\mathcal {P}}}$ is expressed as
${\displaystyle e^{h\partial }(f\circ g)=\exp(\partial _{f}e^{h\partial _{g}})(g,f)}$
where ${\displaystyle \exp(\partial _{f}e^{h\partial _{g}}):{\mathcal {P}}\times {\mathcal {P}}\to {\mathcal {P}}_{\infty }}$ is an operator on pairs of maps ${\displaystyle (g,f)}$ , where ${\displaystyle \partial _{g}}$ is applied to ${\displaystyle g}$ and ${\displaystyle \partial _{f}}$ to ${\displaystyle f}$ .
Proof can be found in.[9]
Both forward and reverse mode (generalized to arbitrary order) are obtainable using this operator, by fixing the appropriate one of the two maps. This generalizes both concepts under a single operator in the theory. For example, by considering projections of the operator onto the space spanned by ${\displaystyle {\mathcal {D}}=\{1,\partial \}}$ , and fixing the second map ${\displaystyle g}$ , we retrieve the basic first order forward mode of automatic differentiation, or reverse mode, by fixing ${\displaystyle f}$ .
Thus the operator alleviates the need for explicit implementation of the higher order chain rule (see Faà di Bruno's formula), as it is encoded in the structure of the operator itself, which can be efficiently implemented by manipulating its generating map (see [9]).
### Order reduction for nested applicationsEdit
It is useful to be able to use the ${\displaystyle k}$ -th derivative of a program ${\displaystyle P\in {\mathcal {P}}}$ as part of a different differentiable program ${\displaystyle P_{1}}$ . As such, we must be able to treat the derivative itself as a differentiable program ${\displaystyle P^{\prime k}\in {\mathcal {P}}}$ , while only coding the original program ${\displaystyle P}$ .
Theorem. There exists a reduction of order map ${\displaystyle \phi :{\mathcal {P}}_{n}\to {\mathcal {P}}_{n-1}}$ satisfying
${\displaystyle \forall P_{1}\in {\mathcal {P}}_{0}\quad \exists P_{2}\in {\mathcal {P}}_{0}\qquad \phi ^{k}\circ e_{n}^{\partial }(P_{1})=e_{n-k}^{\partial }(P_{2})}$
for each ${\displaystyle n\geq 1,}$ where ${\displaystyle e_{n}^{\partial }}$ is the projection of the operator ${\displaystyle e^{\partial }}$ onto the set ${\displaystyle {\mathcal {D}}^{n}=\{\partial ^{k}|0\leq k\leq n\}}$ .
By the above Theorem, ${\displaystyle n}$ -differentiable ${\displaystyle k}$ -th derivatives of a program ${\displaystyle P\in {\mathcal {P}}_{0}}$ can be extracted by
${\displaystyle ^{n}P^{k\prime }=\phi ^{k}\circ e_{n+k}^{\partial }(P)\in {\mathcal {P}}_{n}.}$
Thus, we gained the ability of writing a differentiable program acting on derivatives of another program, stressed as crucial by other authors.[11]
## ImplementationEdit
Forward-mode AD is implemented by a nonstandard interpretation of the program in which real numbers are replaced by dual numbers, constants are lifted to dual numbers with a zero epsilon coefficient, and the numeric primitives are lifted to operate on dual numbers. This nonstandard interpretation is generally implemented using one of two strategies: source code transformation or operator overloading.
### Source code transformation (SCT)Edit
Figure 4: Example of how source code transformation could work
The source code for a function is replaced by an automatically generated source code that includes statements for calculating the derivatives interleaved with the original instructions.
Source code transformation can be implemented for all programming languages, and it is also easier for the compiler to do compile time optimizations. However, the implementation of the AD tool itself is more difficult.
Operator overloading is a possibility for source code written in a language supporting it. Objects for real numbers and elementary mathematical operations must be overloaded to cater for the augmented arithmetic depicted above. This requires no change in the form or sequence of operations in the original source code for the function to be differentiated, but often requires changes in basic data types for numbers and vectors to support overloading and often also involves the insertion of special flagging operations.
Operator overloading for forward accumulation is easy to implement, and also possible for reverse accumulation. However, current compilers lag behind in optimizing the code when compared to forward accumulation.
Operator overloading, for both forward and reverse accumulation, can be well-suited to applications where the objects are vectors of real numbers rather than scalars. This is because the tape then comprises vector operations; this can facilitate computationally efficient implementations where each vector operation performs many scalar operations. Vector adjoint algorithmic differentiation (vector AAD) techniques may be used, for example, to differentiate values calculated by Monte-Carlo simulation.
## ReferencesEdit
1. ^ Neidinger, Richard D. (2010). "Introduction to Automatic Differentiation and MATLAB Object-Oriented Programming" (PDF). SIAM Review. 52 (3): 545–563. doi:10.1137/080743627.
2. ^ http://www.ec-securehost.com/SIAM/SE24.html
3. ^ R.E. Wengert (1964). "A simple automatic derivative evaluation program". Comm. ACM. 7: 463–464. doi:10.1145/355586.364791.
4. ^ Bartholomew-Biggs, Michael; Brown, Steven; Christianson, Bruce; Dixon, Laurence (2000). "Automatic differentiation of algorithms" (PDF). Journal of Computational and Applied Mathematics. 124 (1-2): 171–190. doi:10.1016/S0377-0427(00)00422-2.
5. ^ Linnainmaa, Seppo (1970). The representation of the cumulative rounding error of an algorithm as a Taylor expansion of the local rounding errors. Master's Thesis (in Finnish), Univ. Helsinki, 6-7.
6. ^ Linnainmaa, Seppo (1976). Taylor expansion of the accumulated rounding error. BIT Numerical Mathematics, 16(2), 146-160.
7. ^ Griewank, Andreas (2012). Who Invented the Reverse Mode of Differentiation?. Optimization Stories, Documenta Matematica, Extra Volume ISMP (2012), 389-400.
8. ^ Naumann, Uwe (April 2008). "Optimal Jacobian accumulation is NP-complete". Mathematical Programming. 112 (2): 427–441. doi:10.1007/s10107-006-0042-z.
9. Sajovic, Žiga; Vuk, Martin (2016). "Operational calculus on programming spaces". arXiv: [math.FA].
10. ^ izzo, Dario; Biscani, Francesci (2016). "Differentiable Genetic Programming". arXiv: [math.FA].
11. ^ Pearlmutter, Barak A.; Siskind, Jeffrey M (May 2008). "PPutting the Automatic Back into AD: Part II". ECE Technical Reports. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 159, "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": 0.94989013671875, "perplexity": 1260.6426431318212}, "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-2018-22/segments/1526794865702.43/warc/CC-MAIN-20180523180641-20180523200641-00234.warc.gz"} |
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[/B]1 ) Dry air at normal temperature and pressure has a mass density of 0.0012 g/cm3 with a mass fraction of oxygen of 0.23. What is the atom density of 18O?
2) In the centre of the core of a 1000 MWe BWR, the observed fission rate is 1.7×1012 cm-3 s-1 and the observed temperature of the fuel is 800 oC. What will be the fission rate at the same location if the temperature is raised to 1000 oC? (Hint, Σf is a non- 1/v cross-section.)
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Astronuc
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[/B]1 ) Dry air at normal temperature and pressure has a mass density of 0.0012 g/cm3 with a mass fraction of oxygen of 0.23. What is the atom density of 18O?
Please show work. Determine the mole fraction of oxygen in air. Then apply the isotopic fraction of 18O.
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2K | {"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": 0.9400287866592407, "perplexity": 4151.577240293418}, "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/1620243989012.26/warc/CC-MAIN-20210509183309-20210509213309-00319.warc.gz"} |
https://physics.stackexchange.com/questions/80432/are-the-second-and-third-generations-of-matter-required-or-optional | # Are the second and third generations of matter required or optional?
If the second and third matter families (so-called generations: muon, muon neutrino, tau, tau neutrino, ...) didn't exist, would that affect how the universe runs? Are they optional or required?
This question can be split into two parts: How do three generations affect the way the universe runs within the standard model, and how do they affect beyond standard model physics?
It is likely that the three generations are important in beyond standard model physics and without them particle physics would be very different. However we do not have any definite idea about how that works so let's concentrate on standard model physics.
It is true that almost all physics that affects the way the universe runs depends on only the first generation of matter particles, i.e. the up and down quarks, electron and electron neutrino, plus the gauge bosons. Although we can now easily observe particles from the second generation i.e muons, strange particles and charm, these have almost no significance in chemistry or even main processes of nuclear physics. Hence their effect on how the universe runs is very small. The third generation has even less influence.
However there are some exceptions and possible exceptions.
(1) Muons are an important component of cosmic rays and it is now thought that cosmic rays can play an important part in weather and climate. There have been studies to observe correlations between muon flux and upper atmosphere temperature. Even if this is observed it does not necessarily mean that muons have a causal affect on weather but it is possible. Atmospheric muons could have other important influences such as gene mutation, but this is speculative as far as I know.
(2) CP violation can occur in the standard model at observable rates only through the CKM and PMNS mixing matrices for three generations. With only one or two generations CP violation in the standard model can only happen via unobservable non-perturbative effects. While CP violation does not have any affect on the way the universe runs now, it is thought to be essential in big bang cosmology in order to create an imbalance between matter and anti-matter. Without it all matter would have annihilated with antimatter as the universe cooled. However, it is not known if other CP violating processes from beyond standard model physics that do not depend on the three generations are more important.
(3) In calculations of particle masses and decay rates the existence of second generation particles as virtual particles has some effect. You could argue that the masses of protons and neutron might be slightly different etc if there were no muons etc.
(4) The three generations of neutrino have observable effects on decay rates of weak gauge bosons and also on cosmological parameters measured in the cosmic microwave background. Whether this affects the running of the universe in some significant way is another question.
(5) The top quark mass is an important factor affecting the stability of the vacuum and the allowable mass range of the Higgs boson for stability. You could therefore argue that the top quark and perhaps some of the other heavy flavour particles are needed to stop the vacuum decaying.
In several of these possibilities you could make the point that without the second and third generation the parameters of the standard model could be adjusted to produce the same effect, so it is not certain that the second and third generations have any real necessity in the running of the universe.
• to elaborate on 5, if there were no higher generations then the Higgs would couple with the heaviest quark the d, which is 4.8 Mev, but the u is only 1/2 that and even the electron is 10%. I think this would mean that higher order contributions to cross sections and decays would be different by large factors. For example electromagnetic interactions which hold the atoms together would be affected, etc. I cannot easily envision a universe similar to ours under these conditions. – anna v Oct 12 '13 at 11:10
• @annav The SM is still a renormalizable theory with only one generation, and below the kaon mass ($\sim 493\ \mathrm{MeV}$) you have an effective theory of light leptons and pions in both cases. So by adjusting the parameters you should be able to match the all the physics below that scale up to corrections of the order $E/m_K$ at the worst, right? In particular low energy EM wouldn't necessarily be affected. Can't remember ever hearing atomic physicists talk about having to include kaon loops... I don't know how protons and neutrons would be affected though. Possibly nontrivial stuff there.:) – Michael Brown Oct 17 '13 at 3:10
• Re the Higgs, that's an interesting question. Without the top around you might not be able to get away with a simple scalar Higgs, but might need a strongly coupled Higgs sector, something like technicolour? Unless you fiddle with the Higgs mass as well... – Michael Brown Oct 17 '13 at 3:13
I guess protons need strange quarks. eg.
http://arxiv.org/abs/1203.4051
The a particle later to be known as the electron neutrino was proposed to exist by Wolfgang Pauli because physicists already knew then that it must exist--otherwise the known laws of conservation of energy, momentum and angular momentum would be violated. They were studying beta decay of neutrons: $n^0 → p^+ + e^−$, and found that while the charges conserved, the energy and momentums did not. A hypothetical particle was proposed to fill in these gaps (this is why scientific laws are so useful in that they have predictive quality). It was subsequently shown that just one particle which filled in all the gaps of beta decay existed--the electron neutrino (Cowan, Reines, 1956).
With the discovery of the electron neutrino, physicists were puzzled by how few were reaching earth from the sun's fusion reactions. From known values of solar radiation and knowledge of fusion mechanisms including beta decay, we were detecting way too few electron neutrinos--a big hole in our understanding (based on the Standard Solar Model) later known as the solar neutrino problem. It was again, another implied violation of conservation laws. A year later Fermi's assistant proposed that neutrinos change into different flavours (muon, tau neutrinos) just like quarks were known to do. It was just a matter of time that experimental physicists found the next generations of the neutrino.
As you may know this year's Nobel in physics was awarded to the people who hypothesized the Higgs boson, which had to exist in order to explain the masses of the W and Z bosons and in a way which did not conflict with gauge theory which was by then too successful to abandon.
To answer your question, yes there is a theoretical need for three generations of leptons to exist mostly due to the fact that the foundations of physics would collapse and everything else would follow. Instead we assume that everything does not collapse, that everything works, and we just have to discover what those things are that make everything work the way it works. | {"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": 0.9202283024787903, "perplexity": 449.1279777820967}, "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/1579250595787.7/warc/CC-MAIN-20200119234426-20200120022426-00342.warc.gz"} |
https://www.zbmath.org/?q=an%3A1183.68613 | ×
# zbMATH — the first resource for mathematics
Axiomatic foundations for ranking systems. (English) Zbl 1183.68613
Summary: Reasoning about agent preferences on a set of alternatives, and the aggregation of such preferences into some social ranking is a fundamental issue in reasoning about multi-agent systems. When the set of agents and the set of alternatives coincide, we get the ranking systems setting. A famous type of ranking systems are page ranking systems in the context of search engines. In this paper we present an extensive axiomatic study of ranking systems. In particular, we consider two fundamental axioms: Transitivity, and Ranked Independence of Irrelevant Alternatives. Surprisingly, we find that there is no general social ranking rule that satisfies both requirements. Furthermore, we show that our impossibility result holds under various restrictions on the class of ranking problems considered. However, when transitivity is weakened, an interesting possibility result is obtained. In addition, we show a complete axiomatization of approval voting using ranked IIA.
##### MSC:
68T35 Theory of languages and software systems (knowledge-based systems, expert systems, etc.) for artificial intelligence 68T05 Learning and adaptive systems in artificial intelligence
##### Keywords:
multi-agent systems
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https://www.lmfdb.org/ModularForm/GL2/Q/holomorphic/1216/3/e/m/ | # Properties
Label 1216.3.e.m Level $1216$ Weight $3$ Character orbit 1216.e Analytic conductor $33.134$ Analytic rank $0$ Dimension $8$ CM no Inner twists $2$
# Related objects
## Newspace parameters
Level: $$N$$ $$=$$ $$1216 = 2^{6} \cdot 19$$ Weight: $$k$$ $$=$$ $$3$$ Character orbit: $$[\chi]$$ $$=$$ 1216.e (of order $$2$$, degree $$1$$, not minimal)
## Newform invariants
Self dual: no Analytic conductor: $$33.1336001462$$ Analytic rank: $$0$$ Dimension: $$8$$ Coefficient field: $$\mathbb{Q}[x]/(x^{8} + \cdots)$$ Defining polynomial: $$x^{8} + 34 x^{6} + 345 x^{4} + 1064 x^{2} + 256$$ Coefficient ring: $$\Z[a_1, \ldots, a_{19}]$$ Coefficient ring index: $$2^{5}$$ Twist minimal: no (minimal twist has level 152) Sato-Tate group: $\mathrm{SU}(2)[C_{2}]$
## $q$-expansion
Coefficients of the $$q$$-expansion are expressed in terms of a basis $$1,\beta_1,\ldots,\beta_{7}$$ for the coefficient ring described below. We also show the integral $$q$$-expansion of the trace form.
$$f(q)$$ $$=$$ $$q + \beta_{1} q^{3} + ( 2 + \beta_{4} ) q^{5} + ( -1 + \beta_{3} ) q^{7} + ( -\beta_{4} + \beta_{6} ) q^{9} +O(q^{10})$$ $$q + \beta_{1} q^{3} + ( 2 + \beta_{4} ) q^{5} + ( -1 + \beta_{3} ) q^{7} + ( -\beta_{4} + \beta_{6} ) q^{9} + ( 4 + \beta_{4} - 2 \beta_{6} ) q^{11} + ( 2 \beta_{1} - \beta_{7} ) q^{13} + ( 4 \beta_{1} - \beta_{2} ) q^{15} + ( -3 - 2 \beta_{4} + \beta_{6} ) q^{17} + ( -2 + \beta_{2} + \beta_{4} + \beta_{5} + \beta_{6} + \beta_{7} ) q^{19} + ( -2 \beta_{1} - \beta_{2} - 6 \beta_{5} - \beta_{7} ) q^{21} + ( 1 + \beta_{3} + \beta_{4} + 2 \beta_{6} ) q^{23} + ( 5 + 2 \beta_{3} + \beta_{4} - 4 \beta_{6} ) q^{25} + ( 5 \beta_{1} + \beta_{2} + 2 \beta_{5} ) q^{27} + ( 6 \beta_{1} - \beta_{2} + 4 \beta_{5} - \beta_{7} ) q^{29} + ( -4 \beta_{1} + \beta_{2} - 2 \beta_{7} ) q^{31} + ( 10 \beta_{1} - \beta_{2} - 4 \beta_{5} ) q^{33} + ( 6 + 4 \beta_{3} + \beta_{4} - 2 \beta_{6} ) q^{35} + ( -2 \beta_{1} - 3 \beta_{2} - 6 \beta_{5} ) q^{37} + ( -13 - 3 \beta_{3} + \beta_{4} + 2 \beta_{6} ) q^{39} + ( -8 \beta_{1} + \beta_{2} - 4 \beta_{5} - 2 \beta_{7} ) q^{41} + ( -10 + 4 \beta_{3} - 5 \beta_{4} ) q^{43} + ( -22 - 2 \beta_{3} - 3 \beta_{4} + 6 \beta_{6} ) q^{45} + ( 4 + 6 \beta_{3} + 3 \beta_{4} - 8 \beta_{6} ) q^{47} + ( 6 - 8 \beta_{3} + 6 \beta_{4} + \beta_{6} ) q^{49} + ( -9 \beta_{1} + 2 \beta_{2} + 2 \beta_{5} ) q^{51} + ( -8 \beta_{1} + \beta_{2} + 4 \beta_{5} - \beta_{7} ) q^{53} + ( 26 + 2 \beta_{3} + 11 \beta_{4} - 8 \beta_{6} ) q^{55} + ( 1 - 2 \beta_{1} - \beta_{2} + 6 \beta_{3} + 5 \beta_{4} + 2 \beta_{5} - 3 \beta_{6} ) q^{57} + ( -13 \beta_{1} - 3 \beta_{2} - 20 \beta_{5} ) q^{59} + ( 18 + 6 \beta_{3} + 5 \beta_{4} + 6 \beta_{6} ) q^{61} + ( -2 - 2 \beta_{3} - 3 \beta_{4} + 6 \beta_{6} ) q^{63} + ( 2 \beta_{1} - 4 \beta_{2} + 4 \beta_{5} - 2 \beta_{7} ) q^{65} + ( -9 \beta_{1} + 16 \beta_{5} + 2 \beta_{7} ) q^{67} + ( -2 \beta_{1} - 2 \beta_{2} - 2 \beta_{5} - \beta_{7} ) q^{69} + ( -16 \beta_{1} - 16 \beta_{5} + 2 \beta_{7} ) q^{71} + ( -21 - 6 \beta_{3} + 8 \beta_{4} + 13 \beta_{6} ) q^{73} + ( 13 \beta_{1} - 3 \beta_{2} - 20 \beta_{5} - 2 \beta_{7} ) q^{75} + ( -8 + 6 \beta_{3} + 5 \beta_{4} - 10 \beta_{6} ) q^{77} + ( 14 \beta_{1} + \beta_{2} - 16 \beta_{5} ) q^{79} + ( -37 + 4 \beta_{3} - 6 \beta_{4} + 10 \beta_{6} ) q^{81} + ( 12 - 10 \beta_{3} + 14 \beta_{4} + 8 \beta_{6} ) q^{83} + ( -54 - 4 \beta_{3} - 5 \beta_{4} + 10 \beta_{6} ) q^{85} + ( -45 - \beta_{3} - 11 \beta_{4} + 4 \beta_{6} ) q^{87} + ( -12 \beta_{1} - 5 \beta_{2} - 4 \beta_{7} ) q^{89} + ( 19 \beta_{1} - \beta_{2} + 18 \beta_{5} - 2 \beta_{7} ) q^{91} + ( 50 - 4 \beta_{3} + 18 \beta_{4} - 6 \beta_{6} ) q^{93} + ( 26 - 16 \beta_{1} + 2 \beta_{2} + 2 \beta_{3} - 7 \beta_{4} + 18 \beta_{5} - 2 \beta_{6} + 4 \beta_{7} ) q^{95} + ( -26 \beta_{1} + 20 \beta_{5} - 4 \beta_{7} ) q^{97} + ( -66 - 6 \beta_{3} - 9 \beta_{4} - 2 \beta_{6} ) q^{99} +O(q^{100})$$ $$\operatorname{Tr}(f)(q)$$ $$=$$ $$8q + 14q^{5} - 6q^{7} + 4q^{9} + O(q^{10})$$ $$8q + 14q^{5} - 6q^{7} + 4q^{9} + 26q^{11} - 18q^{17} - 16q^{19} + 12q^{23} + 34q^{25} + 50q^{35} - 108q^{39} - 62q^{43} - 162q^{45} + 22q^{47} + 22q^{49} + 174q^{55} + 4q^{57} + 158q^{61} - 2q^{63} - 170q^{73} - 82q^{77} - 256q^{81} + 64q^{83} - 410q^{85} - 332q^{87} + 344q^{93} + 222q^{95} - 526q^{99} + O(q^{100})$$
Basis of coefficient ring in terms of a root $$\nu$$ of $$x^{8} + 34 x^{6} + 345 x^{4} + 1064 x^{2} + 256$$:
$$\beta_{0}$$ $$=$$ $$1$$ $$\beta_{1}$$ $$=$$ $$\nu$$ $$\beta_{2}$$ $$=$$ $$($$$$-\nu^{7} - 30 \nu^{5} - 205 \nu^{3} + 16 \nu$$$$)/40$$ $$\beta_{3}$$ $$=$$ $$($$$$-\nu^{6} - 20 \nu^{4} - 35 \nu^{2} + 216$$$$)/40$$ $$\beta_{4}$$ $$=$$ $$($$$$\nu^{6} + 30 \nu^{4} + 205 \nu^{2} + 64$$$$)/40$$ $$\beta_{5}$$ $$=$$ $$($$$$\nu^{7} + 30 \nu^{5} + 245 \nu^{3} + 504 \nu$$$$)/80$$ $$\beta_{6}$$ $$=$$ $$($$$$\nu^{6} + 30 \nu^{4} + 245 \nu^{2} + 424$$$$)/40$$ $$\beta_{7}$$ $$=$$ $$($$$$-\nu^{7} - 40 \nu^{5} - 495 \nu^{3} - 1784 \nu$$$$)/40$$
$$1$$ $$=$$ $$\beta_0$$ $$\nu$$ $$=$$ $$\beta_{1}$$ $$\nu^{2}$$ $$=$$ $$\beta_{6} - \beta_{4} - 9$$ $$\nu^{3}$$ $$=$$ $$2 \beta_{5} + \beta_{2} - 13 \beta_{1}$$ $$\nu^{4}$$ $$=$$ $$-17 \beta_{6} + 21 \beta_{4} + 4 \beta_{3} + 125$$ $$\nu^{5}$$ $$=$$ $$-4 \beta_{7} - 58 \beta_{5} - 25 \beta_{2} + 197 \beta_{1}$$ $$\nu^{6}$$ $$=$$ $$305 \beta_{6} - 385 \beta_{4} - 120 \beta_{3} - 1969$$ $$\nu^{7}$$ $$=$$ $$120 \beta_{7} + 1330 \beta_{5} + 505 \beta_{2} - 3229 \beta_{1}$$
## Character values
We give the values of $$\chi$$ on generators for $$\left(\mathbb{Z}/1216\mathbb{Z}\right)^\times$$.
$$n$$ $$191$$ $$705$$ $$837$$ $$\chi(n)$$ $$1$$ $$-1$$ $$1$$
## Embeddings
For each embedding $$\iota_m$$ of the coefficient field, the values $$\iota_m(a_n)$$ are shown below.
For more information on an embedded modular form you can click on its label.
Label $$\iota_m(\nu)$$ $$a_{2}$$ $$a_{3}$$ $$a_{4}$$ $$a_{5}$$ $$a_{6}$$ $$a_{7}$$ $$a_{8}$$ $$a_{9}$$ $$a_{10}$$
1025.1
− 4.27138i − 3.20945i − 2.27869i − 0.512197i 0.512197i 2.27869i 3.20945i 4.27138i
0 4.27138i 0 7.92033 0 5.75693 0 −9.24469 0
1025.2 0 3.20945i 0 3.06310 0 −12.3151 0 −1.30054 0
1025.3 0 2.27869i 0 −6.29008 0 −1.03740 0 3.80758 0
1025.4 0 0.512197i 0 2.30665 0 4.59559 0 8.73765 0
1025.5 0 0.512197i 0 2.30665 0 4.59559 0 8.73765 0
1025.6 0 2.27869i 0 −6.29008 0 −1.03740 0 3.80758 0
1025.7 0 3.20945i 0 3.06310 0 −12.3151 0 −1.30054 0
1025.8 0 4.27138i 0 7.92033 0 5.75693 0 −9.24469 0
$$n$$: e.g. 2-40 or 990-1000 Embeddings: e.g. 1-3 or 1025.8 Significant digits: Format: Complex embeddings Normalized embeddings Satake parameters Satake angles
## Inner twists
Char Parity Ord Mult Type
1.a even 1 1 trivial
19.b odd 2 1 inner
## Twists
By twisting character orbit
Char Parity Ord Mult Type Twist Min Dim
1.a even 1 1 trivial 1216.3.e.m 8
4.b odd 2 1 1216.3.e.n 8
8.b even 2 1 152.3.e.b 8
8.d odd 2 1 304.3.e.g 8
19.b odd 2 1 inner 1216.3.e.m 8
24.f even 2 1 2736.3.o.p 8
24.h odd 2 1 1368.3.o.b 8
76.d even 2 1 1216.3.e.n 8
152.b even 2 1 304.3.e.g 8
152.g odd 2 1 152.3.e.b 8
456.l odd 2 1 2736.3.o.p 8
456.p even 2 1 1368.3.o.b 8
By twisted newform orbit
Twist Min Dim Char Parity Ord Mult Type
152.3.e.b 8 8.b even 2 1
152.3.e.b 8 152.g odd 2 1
304.3.e.g 8 8.d odd 2 1
304.3.e.g 8 152.b even 2 1
1216.3.e.m 8 1.a even 1 1 trivial
1216.3.e.m 8 19.b odd 2 1 inner
1216.3.e.n 8 4.b odd 2 1
1216.3.e.n 8 76.d even 2 1
1368.3.o.b 8 24.h odd 2 1
1368.3.o.b 8 456.p even 2 1
2736.3.o.p 8 24.f even 2 1
2736.3.o.p 8 456.l odd 2 1
## Hecke kernels
This newform subspace can be constructed as the intersection of the kernels of the following linear operators acting on $$S_{3}^{\mathrm{new}}(1216, [\chi])$$:
$$T_{3}^{8} + 34 T_{3}^{6} + 345 T_{3}^{4} + 1064 T_{3}^{2} + 256$$ $$T_{5}^{4} - 7 T_{5}^{3} - 34 T_{5}^{2} + 256 T_{5} - 352$$ $$T_{7}^{4} + 3 T_{7}^{3} - 99 T_{7}^{2} + 221 T_{7} + 338$$
## Hecke characteristic polynomials
$p$ $F_p(T)$
$2$ $$T^{8}$$
$3$ $$256 + 1064 T^{2} + 345 T^{4} + 34 T^{6} + T^{8}$$
$5$ $$( -352 + 256 T - 34 T^{2} - 7 T^{3} + T^{4} )^{2}$$
$7$ $$( 338 + 221 T - 99 T^{2} + 3 T^{3} + T^{4} )^{2}$$
$11$ $$( -5912 + 2260 T - 174 T^{2} - 13 T^{3} + T^{4} )^{2}$$
$13$ $$246866944 + 11290176 T^{2} + 161977 T^{4} + 778 T^{6} + T^{8}$$
$17$ $$( 4814 - 253 T - 195 T^{2} + 9 T^{3} + T^{4} )^{2}$$
$19$ $$16983563041 + 752734096 T + 133448704 T^{2} + 6059024 T^{3} + 479902 T^{4} + 16784 T^{5} + 1024 T^{6} + 16 T^{7} + T^{8}$$
$23$ $$( 27836 - 504 T - 499 T^{2} - 6 T^{3} + T^{4} )^{2}$$
$29$ $$36014930176 + 474986408 T^{2} + 1927129 T^{4} + 2498 T^{6} + T^{8}$$
$31$ $$2723080830976 + 8628732032 T^{2} + 10137808 T^{4} + 5232 T^{6} + T^{8}$$
$37$ $$5146582257664 + 20060072576 T^{2} + 22018512 T^{4} + 8272 T^{6} + T^{8}$$
$41$ $$5382400000000 + 16332880000 T^{2} + 17109456 T^{4} + 7216 T^{6} + T^{8}$$
$43$ $$( 258656 - 7264 T - 1982 T^{2} + 31 T^{3} + T^{4} )^{2}$$
$47$ $$( 2613512 - 84772 T - 6302 T^{2} - 11 T^{3} + T^{4} )^{2}$$
$53$ $$13467138304 + 2071871016 T^{2} + 6302521 T^{4} + 5410 T^{6} + T^{8}$$
$59$ $$247119406081024 + 303682748480 T^{2} + 126378201 T^{4} + 20362 T^{6} + T^{8}$$
$61$ $$( 17033600 + 298720 T - 7306 T^{2} - 79 T^{3} + T^{4} )^{2}$$
$67$ $$273018790144 + 85706332456 T^{2} + 72017257 T^{4} + 15538 T^{6} + T^{8}$$
$71$ $$1596725395456 + 25682937856 T^{2} + 53362320 T^{4} + 14696 T^{6} + T^{8}$$
$73$ $$( 28567486 - 288325 T - 12111 T^{2} + 85 T^{3} + T^{4} )^{2}$$
$79$ $$787748552704 + 110072696960 T^{2} + 103889616 T^{4} + 19312 T^{6} + T^{8}$$
$83$ $$( -454016 + 644096 T - 19300 T^{2} - 32 T^{3} + T^{4} )^{2}$$
$89$ $$682745962430464 + 992323117056 T^{2} + 291139072 T^{4} + 29768 T^{6} + T^{8}$$
$97$ $$19669519455748096 + 7988311560704 T^{2} + 1099281040 T^{4} + 58184 T^{6} + T^{8}$$ | {"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": 0.9851225018501282, "perplexity": 4400.251594558218}, "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/1620243988774.96/warc/CC-MAIN-20210507025943-20210507055943-00252.warc.gz"} |
http://www.playquiz2win.com/mathematics/ratio-and-proportion-formula.html | You are here: Home >> Mathematics >> Ratio and Proportion Formula
### Formulas of Problems on Ratio and Proportion - Aptitude Questions and Answers.
TIPS FOR SOLVING QUESTIONS RELATED TO RATIO AND PROPORTION:
Ratio: The ratio of two quantities of the same kind and in the same unit is a comparison by division of the measure of two quantities.
It determines how many times one quantity is greater or lesser than the other quantity.
Thus, the ratio of two quantities 'a' and 'b' in the same units, is the fraction \begin{aligned} \frac{a}{b} \end{aligned} and is generally expressed as a : b.
Here, 'a' is called the antecedent and 'b' is called the consequent.
Proportion: The equality of two ratios is called proportion.
For example, 5 : 7 = 10 : 14 i.e. 5 : 7 :: 10 : 14.
Thus, If a : b = c : d, then a : b :: c : d and a, b, c, d are said to be in proportion. Here first and fourth terms i.e. a and d are called extremes, while second and third terms i.e. b and c are called mean terms.
1. If four quantities are in proportion, then
Product of means = Product of extremes
Thus, if a, b, c and d are in proportion i.e. a : b :: c : d, then
b x c = a x d
2. The mean proportion between any two numbers is equal to the square root of their product.
For example, if a : x :: x : b, then
x2 = ab
x = ab
3. If a : b and c : d are two ratios, and
\begin{aligned}
\frac{a}{b} > \frac{c}{d}
\end{aligned}
\begin{aligned}
\frac{a}{b} < \frac{c}{d}
\end{aligned}
\begin{aligned}
\frac{a}{b} = \frac{c}{d}
\end{aligned}
4. Duplicate ratio of a : b = a2 : b2
5. Sub-duplicate ratio of a : b = a: b
4. Triplicate ratio of a : b = 3 : b3
5. Sub-triplicate ratio of a : b = \begin{aligned}
({a}^\frac{1}{3}:{b}^\frac{1}{3})
\end{aligned}
6. If a, b, c and d are four quantities such that a : b :: c : d, then
\begin{aligned} \frac{a+b}{a-b} = \frac{c+d}{c-d}
\end{aligned}
7. Important formula-
\begin{aligned} \frac{a}{b} = \frac{c}{d} = \frac{e}{f} = \frac{a + c + e}{b + d + f}
\end{aligned}
8. If a quantity 'k' has to be divided in the ratio of a : b : c, then the proportional parts are
\begin{aligned} \frac{ka}{a + b + c}, \frac{kb}{a + b + c}, \frac{kc}{a + b + c}
\end{aligned} | {"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": 0.9578729271888733, "perplexity": 892.2480139520014}, "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-2017-30/segments/1500549425381.3/warc/CC-MAIN-20170725202416-20170725222416-00068.warc.gz"} |
http://mathoverflow.net/questions/117238/how-to-calculate-zeroth-crystalline-cohomology?sort=newest | # How to calculate zeroth crystalline cohomology
I am just learning crystalline cohomology, so I understand the basic set-ups. But I can't really do any calculations.
For example, let's choose the base $S=W(k)/p^n$, and let $X$ be an affine scheme over $S$, so it is represented by a $W(k)/p^n$-algebra $A$. Let's just consider the structural sheaf $\mathcal O_{X/S}$ and the sheaf of PD ideal $\mathcal J_{X/S}$ . Now, what is the zero-th cohomology:
1. $H^0_{cris}((X/S) ,\mathcal O_{X/S})$
2. $H^0_{cris}((X/S) ,\mathcal J_{X/S})$
3. $H^0_{cris}((X/S) ,\mathcal J_{X/S}^{[i]})$
I believe that there must be some books explaining this basic example, but I still can't find the reference..
Thanks!
-
## 1 Answer
In this generality, this is unfortunately not so easy because it requires to compute universal PD-envelopes. The case where $X/S$ is embedded in a smooth scheme $Y/S$ is explained in the book by Pierre Berthelot, Cohomologie cristalline des schémas de caractéristique $p>0$, Lecture notes in mathematics 407, 1974 (see chapter V).
You may also look at a paper by Jean-Marc Fontaine (« Cohomologie de de Rham, cohomologie cristalline et représentations $p$-adiques », in Algebraic geometry Kyoto-Tokyo, Lecture notes in mathematics 1016, 1983, p. 86-108). There, he proves that the 0th crystalline cohomology of $O_{\bar K}$ ($K$ local field) is the ring $B_{\rm{cris}}$ he had defined earlier.
-
Thanks! Do you know any other simple example besides the B_cris one? What if, say, X is the affine space $A^n$ or a quotient of $A^n$ by some "nice" ideals? – natura Dec 26 '12 at 15:20 | {"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": 0.9718313217163086, "perplexity": 150.83392482227526}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "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-2014-49/segments/1416931010792.55/warc/CC-MAIN-20141125155650-00092-ip-10-235-23-156.ec2.internal.warc.gz"} |
http://mathhelpforum.com/calculus/183923-taylor-series-any-x-function-x-any-x.html | # Thread: Taylor Series for any (x) = Function (x) for any (x) ?
1. ## Taylor Series for any (x) = Function (x) for any (x) ?
When a Taylor Series is generated from a functions n derivatives at a single
point, then is that series for any value of x equal to the original function for
any value of x.
For example graph the original function (x) from x = 0 to x = 10.
Now plug into the Taylor expansion for x, values from 0 to 10 and graph each point.
Are the two plots approximate or equal ?
2. ## Re: Taylor Series for any (x) = Function (x) for any (x) ?
Originally Posted by morrobay
When a Taylor Series is generated from a functions n derivatives at a single
point, then is that series for any value of x equal to the original function for
any value of x.
For example graph the original function (x) from x = 0 to x = 10.
Now plug into the Taylor expansion for x, values from 0 to 10 and graph each point.
Are the two plots approximate or equal ?
A function f(*) is said to be entire if it is analytic in the whole complex plane. In that case the Taylor expansion of f(*) around any x has radious of convergence infinity. Examples of entire functions are polynomials, circular functions sin and cos, exponential function, hyperbolic functions sinh and cosh, ...
Kind regards
$\chi$ $\sigma$
3. ## Re: Taylor Series for any (x) = Function (x) for any (x) ?
Im asking a more basic question : Not asking for problem to be worked but :
Suppose F(x) = 6x^4 + 3x^3 - 4x^2 + 2x
Now generate a Taylor Expansion ( n derivatives at a point )
Graph the original function for x = 0 to 10
Now plug into this Taylor Expansions (x) , values from 0 to 10 and graph.
Are these two graphs geometrically equal ?
4. ## Re: Taylor Series for any (x) = Function (x) for any (x) ?
Originally Posted by morrobay
Im asking a more basic question : Not asking for problem to be worked but :
Suppose F(x) = 6x^4 + 3x^3 - 4x^2 + 2x
Now generate a Taylor Expansion ( n derivatives at a point )
Graph the original function for x = 0 to 10
Now plug into this Taylor Expansions (x) , values from 0 to 10 and graph.
Are these two graphs geometrically equal ?
Your $f(x)= 6\ x^{4}+ 3\ x^{3} -4\ x^{2} + 2\ x$ is a polynomial and that means that is 'well defined' for all the real or complex x, not only for $x \in [0,10]$. The fact that $f(*)$ is entire guarantees You that the Taylor expansion of $f(*)$ around any value of $x$ allows You to compute $f(x)$ for any other $x$. In case of a polynomial the Taylor expansion around $x=0$ is the polynomial itself and the fastest way to compute $f(x)$ is, of course, to use the 'original expression'...
Kind regards
$\chi$ $\sigma$
5. ## Re: Taylor Series for any (x) = Function (x) for any (x) ?
the answer is a somewhat qualified "yes, they are equal". not every Taylor series for a function x converges everywhere, and even if the Taylor series converges, it may not converge to f. but within the radius of convergence (the term radius is used because the natural setting for studying Taylor series is in the complex plane, where open intervals are replaced by open disks) the LIMIT of the series at a point x equals the value f(x). there are things that happen "off the real line" that account for some of the unexpected ill behavior of some Taylor expansions.
however, the point is that the n-th partial sum of a Taylor series is in fact a polynomial function (the n-th approximating polynomial of "best fit" of degree n). if we only take a finite partial sum (for some finite value of n, say 6 perhaps), there will be some discrepancy between f and the Taylor series sum for f (unless, of course, f is already a polynomial of degree ≤ n).
on this page: Taylor series - Wikipedia, the free encyclopedia you can see a picture of how well various Taylor series partial sums aproximate sin(x) (for n = 1,3,5,7,9,11 and 13). as you can see from the picture, even at a 13-th degree polynomial approximation (about 0, the point at which we are evaluating the derivatives) is only decent in the range (-5,5). if one desires the approximating partial sum to be within a certain range of accuracy (say, so many decimal places) one finds a bound for the n-th remainder term, so as to know how many terms one has to compute.
6. ## Re: Taylor Series for any (x) = Function (x) for any (x) ?
A function is said to be "analytic" at x= a if and only if its Taylor's series at x= a exists, and the value of the function is equal to the value of the Taylor's series in some neighborhood of a. An example of a smooth function that is NOT analytic is $f(x)= e^{-1/x}$ if x is not 0, 0 if x= 0. That function is infinitely differentiable for all x and every derivative is a rational function times $e^{-1/x}$ if x is not 0, 0, if x= 0. That is, the Taylor's series at x= 0 is identically 0. But clearly $e^{-1/x}\ne 0]$ for x not equal to 0. So while the Taylor's series exists and trivially converges for all x, the value of the function is not equal to the value of the Taylor's series for any x except x= 0. | {"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": 0.9094449877738953, "perplexity": 354.8065294500159}, "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-2013-20/segments/1368706153698/warc/CC-MAIN-20130516120913-00051-ip-10-60-113-184.ec2.internal.warc.gz"} |
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Topic: Re: Circular arrangements of n objects taken r at a time allowing (fwd)
Replies: 0
Brian Alspach Posts: 3 Registered: 12/6/04
Re: Circular arrangements of n objects taken r at a time allowing (fwd)
Posted: Nov 2, 1999 11:36 AM
>From: Andrew Wayne Graff <[email protected]> (by way of r!chard
> tchen)
>Subject: Circular arrangements of n objects taken r at a time allowing
This is a problem which is easiest done using Polya's enumeration
theorem. My favorite exposition on Polya's theorem is what de Bruijn
wrote as Chapter 5 in the book "Applied Combinatorics" edited by
appears after the problem statement below.
>I would like to see a formula for the total number of distinct
>circular combinations of n distinct objects taken r at a time
>allowing repetition. In other words, if you have r positions
>around in a circle, and n objects which may be placed in any
>position, then any element of the set of n objects may appear in
>any location, but when rotated, no one arrangement may be
>identical.
>
>ex.:
>n=1,r=1 CCR(1,1)=1 (CCR, meaning circular combinations with
>repetition)
>n=2,r=1 CCR(2,1)=2
>n=3,r=1 CCR(3,1)=3
>...
>n=1,r=2 CCR(1,2)=1
>n=2,r=2 CCR(2,2)=3
>n=3,r=2 CCR(3,2)=6
>...
>n=1,r=3 CCR(1,3)=1
>n=2,r=3 CCR(2,3)=4
>n=3,r=3 CCR(3,3)=11
>n=4,r=3 CCR(4,3)=24
>...
>n=1,r=4 CCR(1,4)=1
>n=2,r=4 CCR(2,4)=6
>n=3,r=4 CCR(3,4)=24
>n=4,r=4 CCR(4,4)=70
>...
>...
>
>I have a general formula for these calculations, however, I am
>not satisfied with how complicated it is. There is a particular
>formula for each value of r, none of which are the same.
>
>The one for r=1 is
>
>f(x)=n
>
>for r=2,
>
> 2
> n +n
>f(x)=----
> 2
>
>for r=3,
>
> 3
> n +2n
>f(x)=-----
> 3
>
>for r=4, however,
>
> 4 2
> n +n +2n
>f(x)=--------
> 4
>
>, ...
>
>As it turns out for all prime values of r, including r=1,
>
> r
> n +(r-1)n
>f(x)=---------
> r
>
>my general formula is something like the following,
>
> p
> ___ g(a )
> \ i
>f(x)= > ----
> /__ a
> i=1 i
>
>where
> q
> ___
> x \
>g(x)=n - > g(b )
> /__ j
> j=1
>
>g(1)=n
>
>{a ,a ,...,a }={positive factors of r}
> 1 2 p
>
>{b ,b ,...,b }={positive factors of x less than x}
> 1 2 j
>
>This is the best generalization I could come up with. Please
>post a simpler solution, if one exists.
>
>Andrew Wayne Graff
What one does is calculate the cycle index for the permutation
group generated by an $r$-cycle. To obtain the cycle index, one
forms a monomial for each permutation in the group; the monomial
is $x_1^{e_l} + x_2^{e_2} + \cdots + x_r^{e_r}$ for a permutation
whose disjoint cycle decomposition has $e_1$ fixed points, $e_2$
transpositions, $e_3$ 3-cycles, ..., $e_r$ r-cycles. Note that
$e_1 + 2e_2 + \cdots + re_r = r$ for a permutation acting on $r$
elements. To get the cycle index, sum all the monomials over the
elements of the group and divide by the order of the group.
In the present case, for the group generated by an $r$-cycle,
one obtains
$\frac{1}{r}\sum_{d|r}\phi(d)x_d^{r/d},$
where $\phi$ denotes the Euler totient function and the sum is taken
over all divisors of $r$.
The first few cycle indexes are given by $x_1$ for $r = 1$,
$1/2(x_1^2 + x_2)$ for $r = 2$, $1/3(x_1^3 + 2x_3)$ for $r = 3$,
and $1/4(x_1^4 + x_2^2 + 2x_4)$ for $r = 4$.
To obtain the answer to the enumeration problem posed above,
simply substitute $n$ for each variable in the cycle index.
One can do more. For example, suppose you allow yourself the
same symmetries as you allow an $n$-gon, that is, you allow the
dihedral flips. Then you do the same thing except you now use the
dihedral group rather than the cyclic group.
Also, suppose you would like to know how many patterns there are
using fixed amounts of each of the symbols. Then you assign a
different weight to each of the objects - say $n = 3$; give one
object weight $b$, one object weight $c$ and one object weight $d$ -
and substitute the sum of the weights to the i-th powers for $x_i$ in
the cycle index and look at the coeficients of various terms in the
evaluated expression. For example, in the preceding case of $n = 3$,
the coefficient of $b^2c^2d$ would be the number of arrangements using
two objects of weight $b$, two objects of weight $c$, and one object of
weight $d$ when $r = 5$.
Brian Alspach | {"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": 2, "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": 0.808528482913971, "perplexity": 2013.4912021171401}, "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-2017-13/segments/1490218203515.32/warc/CC-MAIN-20170322213003-00629-ip-10-233-31-227.ec2.internal.warc.gz"} |
http://mathhelpforum.com/calculus/99015-finding-limit-print.html | # Finding the limit
• Aug 23rd 2009, 01:10 PM
Finding the limit
How do I find the limit of (x^2)(cosh(1/x) when x is approching zero, I know its is ∞ but how do I show the steps?
• Aug 23rd 2009, 01:21 PM
Chris L T521
Quote:
How do I find the limit of (x^2)(cosh(1/x) when x is approching zero, I know its is ∞ but how do I show the steps?
Let $u=\frac{1}{x}$. Thus, $x\to0^+\implies u\to\infty$
Therefore, $\lim_{x\to0^+}x^2\cosh\left(\tfrac{1}{x}\right)=\l im_{u\to\infty}\frac{\cosh\left(u\right)}{u^2}=\tf rac{1}{2}\lim_{u\to\infty}\frac{\sinh u}{u}=\tfrac{1}{2}\lim_{u\to\infty}\cosh u=\infty$
Let $u=\frac{1}{x}$. This time, $x\to0^-\implies u\to-\infty$
Therefore , $\lim_{x\to0^-}x^2\cosh\left(\tfrac{1}{x}\right)=\lim_{u\to-\infty}\frac{\cosh\left(u\right)}{u^2}=\tfrac{1}{2 }\lim_{u\to-\infty}\frac{\sinh u}{u}=\tfrac{1}{2}\lim_{u\to-\infty}\cosh u=\infty$
Therefore, $\lim_{x\to 0}x^2\cosh\left(\tfrac{1}{x}\right)=\infty$ | {"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": 7, "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": 0.9814394116401672, "perplexity": 2633.709658744385}, "config": {"markdown_headings": true, "markdown_code": false, "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-2016-50/segments/1480698542246.21/warc/CC-MAIN-20161202170902-00107-ip-10-31-129-80.ec2.internal.warc.gz"} |
http://unimodular.net/blog/?m=200411 | # Monthly Archives: November 2004
## Finally …
Finally my maths blog is up and running, all thanks to that web-savvy someone.
It took me almost a week to prove $\int_0^{\pi} \log (1+4\cos^2(u)) du = 2\pi \log{\phi}$ where $\phi = (\sqrt{5}+1)/2$. A time consuming but fruitful exercise. I had tried all the integration tricks I know, resorted to computer packages Maple and … Continue reading | {"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": 0.8960374593734741, "perplexity": 2860.082185462868}, "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-2018-05/segments/1516084889660.55/warc/CC-MAIN-20180120142458-20180120162458-00185.warc.gz"} |
https://www.gradesaver.com/textbooks/math/algebra/elementary-algebra/chapter-11-additional-topics-11-5-quadratic-equations-complex-solutions-problem-set-11-5-page-496/3 | ## Elementary Algebra
{$2-i,2+i$}
We know that if $x^{2}=a$, then $x=\pm \sqrt{a}$. Thus, we obtain: Step 1: $(x-2)^{2}=-1$ Step 2: $x-2=\pm \sqrt {-1}$ Step 3: $x=2\pm \sqrt {-1}$ Step 4: $x=2\pm i$ [as $i=\sqrt {-1}$] Step 5: $x=2+i$ or $x=2-i$ Therefore, the solution set is {$2-i,2+i$}. | {"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": 0.830062985420227, "perplexity": 181.00494077518914}, "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/1669446711417.46/warc/CC-MAIN-20221209144722-20221209174722-00192.warc.gz"} |
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