url
stringlengths 15
1.13k
| text
stringlengths 100
1.04M
| metadata
stringlengths 1.06k
1.1k
|
|---|---|---|
https://www.earthdoc.org/content/papers/10.3997/2214-4609.201411222 | 1887
### Abstract
It is widely recognized that the cooling of the reservoir rock by cold water injection changes the rock stresses around the welIbore and in the immediate surroundings of the injection well. The cooling of tbe rock causes the rock fracture pressure to decrease and the possibility of fracturing the formation might occur.
/content/papers/10.3997/2214-4609.201411222
1991-05-21
2020-04-08 | {"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": 0.8061090707778931, "perplexity": 3289.0157100995843}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-16/segments/1585371813538.73/warc/CC-MAIN-20200408104113-20200408134613-00350.warc.gz"} |
https://dsp.stackexchange.com/questions/36558/deriving-the-convolution-kernel-of-the-inverse-of-a-signal/51345 | # Deriving the Convolution Kernel of the Inverse of a Signal
Let $y$ be the inverse (in the sense of convolution) of $x$, i.e.
$$x \star y = \delta$$
Context: $x[n]$ is a discrete signal defined for $n = 0,\ldots, N$. We can assume $x[n] = 0$ if $n \not\in [0,\ldots, N]$.
How to compute the inverse of $x$ (for the convolution)?
\begin{align}(x\star y)[0]&=x[0] y[0] = 1\implies y[0] = \frac{1}{x[0]}\\ (x\star y)[1]&=x[1] y[0] + x[0] y[1] = 0\implies y[1] = -\frac{x[1]}{x[0]^2}\\ (x\star y)[2]&=x[2] y[0] + x[1] y[1] + x[0] y[2] = 0\implies y[2] = -\frac{x[2]}{x[0]^2} + \frac{x[1]^2}{x[0]^3} \end{align}
I tried to find $y[3]$ manually, but I don't see a general formula appearing...
(I would post this as a comment, but I don't have enough rep)
If I understand your question correctly, the operation you want to do is called Deconvolution. Considering the more general problem $x \star y = z$, we can find a solution according to wikipedia as $x = IFT(Z/Y) = IFT(FT(Z)/FT(Y))$, (where $FT$ is the Fourier Transform and $IFT=FT^{-1}$ the inverse FT).
Consider a signal $x[n]$ of a finite length $N$. We are interested in a signal $y[n]$ where $$y[n]\star x[n]=\delta[n]\tag{1}$$ Assume $x[n]$ is the impulse response of an FIR filter and the transfer function is $X(z)$. The inverse filter is an IIR filter whose transfer function is $\frac{1}{X(z)}$. The impulse response of the inverse filter is $$y[n]=\mathcal{Z}^{-1}\left\{\frac{1}{X(z)}\right\}$$
The other way to see it: Convolution becomes product in the $z$-domain, where $\mathcal{Z}\left\{\delta[n]\right\}=1$.
It should be noted that depending on the zeros of $X(z)$ we can have different regions of convergence (ROC) for $\frac{1}{X(z)}$, and consequently different $y[n]$. This means that solutions of $(1)$ may not be unique.
Since the length of the $y[n]$ is theoretically infinite, for implementation you should use a truncated signal which results in an approximated $\delta$. You can use long-division algorithm to find $y$ with desired length.
• to do this practically, the OP will likely set $$z=e^{j2\pi \tfrac{k}{N}}$$ and will have to worry a little about what to do in the occasion of $|X(z)|$ being too close to zero. – robert bristow-johnson Dec 29 '16 at 1:25
• @robertbristow-johnson this would be interesting as an answer! – Basj Jan 2 '17 at 19:58
Since Convolution is a Linear Operation the easiest way to derive the coefficients is using a Matrix Representation of the problem.
Take the kernel $$y$$ and build from it the Convolution Matrix.
Then write the vector representation of the Delta Funciton $$d$$ and solve:
$$Y x = d$$
In case the problem is solvable (Namely there is an inverse operator of $$Y$$) you'll get you answer.
Of course you can always use Least Squares Solution.
Full code for building Convolution Matrix was given in Estimating a Signal Given a Noisy Measurement of the Signal and Its Derivative (Denoising) and Deconvolution of Synthetic 1D Signals - How To?
• Could you show an example? – David Aug 19 '18 at 16:28 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 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": 4, "wp-katex-eq": 0, "align": 1, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8976313471794128, "perplexity": 389.4287251487249}, "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/1610703521987.71/warc/CC-MAIN-20210120182259-20210120212259-00292.warc.gz"} |
http://www.physicsforums.com/showthread.php?s=54b4379ffe4f8d716e8d86a825eeaa0d&p=4316453 | # Magnetic field at the edges of a current-carrying wire
by Nikitin
Tags: currentcarrying, edges, field, magnetic, wire
P: 516 At the edges of a very long current-carrying wire, the magnetic field is not the same as in the middle, right? And integrating biot-savart's law in the edge-region of the wire will make it possible to calculate this field-strength, right?
P: 751 Yes. If you're interested in the field at a point P, which may be anywhere outside the wire, the B-S rule gives you the field at P due to each current element. You integrate these field contributions from the whole wire. This gives you (see thumbnail for meaning of symbols): $$B = \frac{\mu_0 I}{4\pi a} [cos \theta_2 - cos \theta_1]$$. This covers the cases you're interested in; it's very general. For a very long wire, if P is outside the wire, near the middle of the wire, then $\theta_2 = 0$, $\theta_1 = \pi$, so $B = \frac{\mu_0 I}{2\pi a}$, whereas if you're outside the wire, at the (left hand) end of the wire, $\theta_2 = 0$, $\theta_1 = \frac{\pi}{2}$, so $B = \frac{\mu_0 I}{4\pi a}$. If you think about it, you would indeed expect the field to be half as much in the second case as in the first - if you appreciate that the exact length of the wire is immaterial in these 'long wire' examples, because the field from distant parts of the wire is negligible. Remember that you can't, in practice, have a wire which carries a steady current and which has two free (unconnected) ends. The wire needs to be part of a circuit. For the second case above, the left hand end of the wire would have to be connected to the rest of a circuit by another wire. If this other wire went in the direction directly away from P it wouldn't contribute to the field at P. Can the rest of the circuit (apart from the straight wire) be made so as not to contribute to the field at P? Attached Thumbnails
P: 516 thanks!
Related Discussions Introductory Physics Homework 6 Introductory Physics Homework 8 Introductory Physics Homework 3 Introductory Physics Homework 10 Advanced Physics Homework 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": 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": 0.8969229459762573, "perplexity": 362.7064120241324}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2013-48/segments/1386164041513/warc/CC-MAIN-20131204133401-00052-ip-10-33-133-15.ec2.internal.warc.gz"} |
https://www.physicsforums.com/threads/on-a-ferris-wheel.48953/ | # On a Ferris Wheel
1. Oct 21, 2004
### Naeem
Hi folks,
Would appreciate any help with this problem
A chemistry student with a mass of 74 kg is riding a steadily moving Ferris wheel. When she is at the top of the Ferris wheel, the normal force from the seat on to her body has a magnitude of 556 N.
a)What is the magnitude of the normal force on her body when she is at the bottom of the Ferris wheel's arc?
I figured out the answer to this question,
But I am stuck with part B
b ) What would the normal force be on the student at the top of the wheel if the wheel's velocity were doubled?
Plz help
Thanks a million!
2. Oct 21, 2004
### vsage
Consider the force on her. You got mg pulling down, negative centripetal force trying to make her leave her seat and the resulting force is 556N downward because the normal force from the seat is 556N. doubling the velocity increases the negative centripetal force by what factor?
3. Oct 21, 2004
### Naeem
I figured out that if velocity is doubled mv^2/R increases by a factor of 4.
but, how do I setup the equation.
4. Oct 22, 2004
### Naeem
I beg you folks to help me on how to setup the equation for the 2nd part.
Thanks,
Naeem
5. Oct 22, 2004
### Naeem
Finally got this one. Thanks anyway | {"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": 0.86868816614151, "perplexity": 937.5160941198295}, "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/1484560282937.55/warc/CC-MAIN-20170116095122-00089-ip-10-171-10-70.ec2.internal.warc.gz"} |
http://mathhelpforum.com/statistics/129927-joint-density-problem-print.html | # Joint Density Problem
• February 21st 2010, 06:56 AM
bzgeb
Joint Density Problem
This question's been killing me for hours.
http://users.encs.concordia.ca/~b_zgeb/problem12.png
My assumption is that when they ask for the density of X they're asking for the marginal density function f(x). In this case you would take the integral of f(x,y) over all values of y, or in other words, the integral of f(x,y) from 0 to infinity. However, when I do this it becomes infinity. The same goes for trying to compute the density of Y.
Could somebody show me what I'm doing wrong?
• February 21st 2010, 07:39 AM
vince
Quote:
Originally Posted by bzgeb
This question's been killing me for hours.
http://users.encs.concordia.ca/~b_zgeb/problem12.png
My assumption is that when they ask for the density of X they're asking for the marginal density function f(x). In this case you would take the integral of f(x,y) over all values of y, or in other words, the integral of f(x,y) from 0 to infinity. However, when I do this it becomes infinity. The same goes for trying to compute the density of Y.
Could somebody show me what I'm doing wrong?
are u sure f(x,y) is not equal to $xe^{-(x+y)}$ for x,y>0
• February 21st 2010, 08:19 AM
bzgeb
Quote:
Originally Posted by vince
are u sure f(x,y) is not equal to $xe^{-(x+y)}$ for x,y>0
f(x, y) is equal to $xe^{-(x+y)}$ for x,y>0. The problem is: when I try to find the density of X I take the definite integral of f(x,y) dy from 0 to infinity. If you try to do that you'll find that it simply becomes infinity.
• February 21st 2010, 08:49 AM
vince
Quote:
Originally Posted by bzgeb
f(x, y) is equal to $xe^{-(x+y)}$ for x,y>0. The problem is: when I try to find the density of X I take the definite integral of f(x,y) dy from 0 to infinity. If you try to do that you'll find that it simply becomes infinity.
that's not what you wrote...check your first post. with $f(x,y)=xe^{-(x+y)}$, the integrals converge.
p.s. Im leaving the computation to u...if you like ill post the answers.
• February 21st 2010, 09:31 AM
bzgeb
This is embarrassing, I'm sorry I misread your reply. I took a screenshot of the assignment directly so it's definitely $f(x,y) = xe^{(-x+y)}$. However, now that you bring that up I'm starting to wonder if there's a typo in the assignment.
• February 21st 2010, 09:39 AM
vince
Quote:
Originally Posted by bzgeb
This is embarrassing, I'm sorry I misread your reply. I took a screenshot of the assignment directly so it's definitely $f(x,y) = xe^{(-x+y)}$. However, now that you bring that up I'm starting to wonder if there's a typo in the assignment.
i think so...the marginal denisty of X is undefined otherwise.
• February 21st 2010, 09:54 AM
bzgeb
I emailed my professor so I should find out soon. Thanks for taking a look at my problem and confirming that I wasn't doing something 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": 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": 0.9738061428070068, "perplexity": 538.2681883816738}, "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-2014-41/segments/1410657134620.5/warc/CC-MAIN-20140914011214-00084-ip-10-234-18-248.ec2.internal.warc.gz"} |
https://www.electricalclassroom.com/what-is-kva/ | # What is kVA?
kVA stands for kilo volt-amperes or simply 1000 volt-amperes. It is basically the unit of apparent power and is the product of RMS values of voltage and current.
One kVA is 1000 volt – amperes
## Apparent power
Apparent power is the actual amount of power drawn from the source. It includes both true power and reactive power.
In case of DC, the apparent power is irrelevant. But when it comes to AC, for reactive loads, the current and voltage are out of phase. So, the actual usable power (true power) depends on power factor and is always less than the apparent power.
AC power sources such as alternators, inverters and transformers are rated in kVA. The output of these devices is independent of power factor of the load.
## Calculation of kVA – Apparent power
For single phase, kVA can be calculated using the following formula.
Single phase apparent power,
kVA = (Voltage X Current)/1000
For three phase, kVA can be calculated using the following formula.
Three phase apparent power,
kVA = (√3 X Voltage X Current)/1000 = (1.732 X Voltage X Current)/1000
## Calculation of current from kVA
For single phase, current can be calculated from kVA using the following formula.
Single phase current = (kVA X 1000) / Voltage
For three phase, current can be calculated from kVA using the following formula.
Three phase current,
Current = (kVA X 1000) / ( √3 X Voltage) = (kVA X 1000) / ( 1.732 X Voltage)
## Summing up
• kVA is the unit of measure of Apparent power.
• One kVA is equal to 1000VA.
• It is calculated by multiplying the RMS value of voltage by the RMS value of current and dividing the product by 1000.
• In the case of DC, kVA = kW.
• kVA is independent of the power factor of a load.
• Alternators, transformers, power supplies and UPS are rated in kVA. | {"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": 0.8762462139129639, "perplexity": 3299.503089886331}, "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/1664030331677.90/warc/CC-MAIN-20220924151538-20220924181538-00779.warc.gz"} |
http://www.physicsforums.com/showpost.php?p=3274261&postcount=2 | View Single Post
Mentor
P: 15,922
Quote by VinnyCee [ Here, I need to prove that whenever F[a]/(p(a)) = F[b]/(p(b)) in F[x]/(p(x)), then a = b in S.
What do you mean with F[a]/(p(a))? In fact, what do you mean with F[a], I'm not used to such a notations... | {"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": 0.8638685345649719, "perplexity": 2534.6290617800933}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-10/segments/1394010484313/warc/CC-MAIN-20140305090804-00065-ip-10-183-142-35.ec2.internal.warc.gz"} |
https://cr4.globalspec.com/thread/3319/Violating-the-Second-Law | Previous in Forum: Maintenance Technician Next in Forum: Measurement Challenge
Guru
Join Date: May 2006
Location: 34.02S, 22.82E
Posts: 3789
# Violating the Second Law?
11/03/2006 11:55 PM
Consider a perfectly reflective vacuum cavity, constructed out of segments of three bodies: two concentric ellipsoids of revolution, sharing both foci (A) and (B) and a sphere centered on one of the foci (B), as pictured below:
The optical characteristics of this construction allow all the radiation of a point source at A to focus on point B. However, only part of the radiation of a point source at B will be focused on point A, as some of it will be reflected back to B from the spherical midsection of the cavity.
Make the cavity arbitrarily large, so that two tiny, identical spheres at the two foci resemble point masses (A and B) in relation to the cavity. Let the two masses initially have identical temperatures. For the masses to remain at the same temperature, they must absorb and radiate the same amount of energy per unit time. (All masses not at absolute zero, always radiate and absorb energy).
Since point B receives more radiation than point A, mass B must get hotter than mass A, until equilibrium in absorption/radiation is reached. This means that, in violation of the second law of thermodynamics, heat will be flowing on its own accord from a colder to a hotter body. Where's the flaw in the reasoning ?
PS: this is a known brain-teaser in academic circles, so if you have read the answer somewhere, pse. keep it to yourself for a while, so as not to spoil the fun! We can exchange links later... Jorrie
__________________
"Perplexity is the beginning of knowledge." -- Kahlil Gibran
Pathfinder Tags: optics second law thermodynamics
Interested in this topic? By joining CR4 you can "subscribe" to
The Engineer
Join Date: Feb 2005
Location: Albany, New York
Posts: 5060
#1
### Re: Violating the Second Law?
11/04/2006 10:56 AM
I'm guessing, but B radiate higher frequency (more energetic) light than A since it's a blackbody at a higher temperature, so could this extra radiation compensate for focal point effect you mention above? Or can someone else think of another effect that might be helping fix the Entropy problem?
Guru
Join Date: May 2006
Location: 34.02S, 22.82E
Posts: 3789
#2
### Re: Violating the Second Law?
11/04/2006 1:18 PM
Roger, when you said "I'm guessing, but B radiate higher frequency (more energetic) light than A since it's a blackbody at a higher temperature...", did you consider how B got hotter than A in the first place - in violation of the 2nd law?
The fact that B eventually radiates more than A is what is causing the equilibrium - but by then the 'crime' has been committed!
__________________
"Perplexity is the beginning of knowledge." -- Kahlil Gibran
The Engineer
Join Date: Feb 2005
Location: Albany, New York
Posts: 5060
#3
### Re: Violating the Second Law?
11/04/2006 1:43 PM
Can the reflectors heat up? In any real system I'm aware of, the mirrors will heat up slightly from incident radiation. Since the reflector is much closer to A than to B, would it heat up much more, reducing the energy focused on B?
Guru
Join Date: May 2006
Location: 34.02S, 22.82E
Posts: 3789
#6
### Re: Violating the Second Law?
11/05/2006 1:28 AM
Roger, you asked: "Can the reflectors heat up?"
The question stated "perfectly reflective cavity", ruling out absorption.
But, even slightly less than perfect reflectivity should make little difference, because of the relatively large difference between the radiation arriving at point A and B respectively. It is determined by the size of the spherical mid-section of the cavity.
__________________
"Perplexity is the beginning of knowledge." -- Kahlil Gibran
Guru
Join Date: Jun 2006
Location: Toronto
Posts: 3968
#4
### Re: Violating the Second Law?
11/04/2006 11:43 PM
It seems to me that the part of the radiation from point "B" that hits the inside of the circle will get relected right back at itself and take no further part in the game
__________________
Guru
Join Date: May 2006
Location: 34.02S, 22.82E
Posts: 3789
#5
### Re: Violating the Second Law?
11/05/2006 1:20 AM
Aurizon wrote: "It seems to me that the part of the radiation from point "B" that hits the inside of the circle will get reflected right back at itself and take no further part in the game."
This is exactly what is stated in the OP, but, per se, it makes no difference to the temperature of B. It does however mean that while all the radiation from point A reaches point B, not all radiation from point B reaches point A, causing more temperature at point B - in violation of the second law.
__________________
"Perplexity is the beginning of knowledge." -- Kahlil Gibran
Anonymous Poster
#7
### Re: Violating the Second Law?
11/05/2006 4:42 AM
Nic
Guru
Join Date: May 2006
Location: 34.02S, 22.82E
Posts: 3789
#11
### Re: Violating the Second Law?
11/05/2006 11:37 AM
I think the E-H Engine is as fundamentally flawed as the " heat engine" one can construct from the reflective vacuum vessel above! Nobody has ever demonstrated a violation of the 2nd law above reasonable doubt and probably nobody ever will. This does not stop people from trying, though. It is usually not easy to find the flaws, as this teaser shows!
__________________
"Perplexity is the beginning of knowledge." -- Kahlil Gibran
Guru
Join Date: Jun 2006
Location: Port Noarlunga, South Australia, AUSTRALIA (South of Adelaide)
Posts: 3048
#8
### Re: Violating the Second Law?
11/05/2006 6:36 AM
I believe what is happening is that while all the radiation from A reaches B it is not evenly distributed as it approaches B but only arrives at B from the direction of the parabolic reflectors. As a result of this increased intensity the re-radiation from B in the direction of the parabolic reflectors would be proportionally higher would hence balance out. The net result would be not transfer of energy from A to B.
__________________
An elephant is a mouse built to government specifications.
Guru
Join Date: May 2006
Location: 34.02S, 22.82E
Posts: 3789
#9
### Re: Violating the Second Law?
11/05/2006 10:44 AM
Masu, you said: "As a result of this increased intensity the re-radiation from B in the direction of the parabolic reflectors would be proportionally higher would hence balance out."
You mean the average temperature of B remains the same as A, while there are hot spots on B? Does this not still mean a transfer of heat from a colder to a hotter surface that violates the 2nd law?
__________________
"Perplexity is the beginning of knowledge." -- Kahlil Gibran
Guru
Join Date: Jun 2006
Location: Port Noarlunga, South Australia, AUSTRALIA (South of Adelaide)
Posts: 3048
#10
### Re: Violating the Second Law?
11/05/2006 11:15 AM
Yes it dose mean that doesn't it. Oh well scratch an I will go and procrastinate some more.
__________________
An elephant is a mouse built to government specifications.
Guru
Join Date: Dec 2005
Location: Etats Unis
Posts: 1871
#12
### Re: Violating the Second Law?
11/05/2006 6:50 PM
Jorrie, While I ponder your riddle, I thought I would pass along this example of violating the second law I found in an FAQ for the SimpleTech network server I just bought. It is excerpted from the answer to "Why does the outside ...(case)... get so hot?" "To ensure the hard drive operates within its temperature specifications, the metal case acts as a heatsink, pulling heat away from the hard drive inside. Due to this, the outside of the case is actually warmer than the drive itself." Amazing what you can do with market-speak, violate the laws of physics. While I'm at it, doesn't your riddle have something to do with the lensing effect of the mirrors equalizing the effective temperature that each point "sees" of the other?
__________________
The hardest thing to overcome, is not knowing that you don't know.
Guru
Join Date: May 2006
Location: 34.02S, 22.82E
Posts: 3789
#13
### Re: Violating the Second Law?
11/05/2006 11:01 PM
rcapper, you wrote that the sales blurb said: "Due to this, the outside of the case is actually warmer than the drive itself." Amazing"
They could actually be right if they meant that the magnetic disc itself is cooler than the case. The heat surely comes mostly from the drive motor?
Turning to the teaser, point B sees all of point A's radiation and point A sees only part of B's, due to B seeing itself in the spherical part. But you are getting warmer (pun intended)!
Jorrie
__________________
"Perplexity is the beginning of knowledge." -- Kahlil Gibran
Guru
Join Date: Dec 2005
Location: Etats Unis
Posts: 1871
#14
### Re: Violating the Second Law?
11/06/2006 12:01 AM
You wrote: "They could actually be right if they meant that the magnetic disc itself is cooler than the case. The heat surely comes mostly from the drive motor?" I could see perhaps that that might be briefly after it was first powered, depending on the heat flow dynamics inside the device. However, since the outside of the case is the boundary at which all internal heat must find egress, once the temperature has reached an equilibrium, I don't think any point inside the case would be cooler than its surface. On your riddle, were B to rise above A wouldn't it simply re-radiate to A until there was equilibrium? And in fact, because the path between A and B is bidirectional, the points would never diverge?
__________________
The hardest thing to overcome, is not knowing that you don't know.
Guru
Join Date: May 2006
Location: 34.02S, 22.82E
Posts: 3789
#19
### Re: Violating the Second Law?
11/06/2006 8:49 AM
Hi rcapper, you wrote: "However, since the outside of the case is the boundary at which all internal heat must find egress, once the temperature has reached an equilibrium, I don't think any point inside the case would be cooler than its surface."
Yep, I think you are right - my hasty remark was bad!
On the riddle, what you said last moves us further from the solution, not closer! One can't get away from the fact that point B receives more radiation than point A, unless B is hotter than A (so that B radiates at a faster rate than A).
The solution lurks in the lensing effect that you mentioned in post 12…
__________________
"Perplexity is the beginning of knowledge." -- Kahlil Gibran
Guru
Join Date: Dec 2005
Location: Etats Unis
Posts: 1871
#21
### Re: Violating the Second Law?
11/06/2006 4:30 PM
I had the sense that that was going in the wrong direction. How about this. Since only a portion of the energy radiated by B is incident upon all of A. the lensing effect redistributes the energy density in a manner such that each point effectively "sees" an equivalent temperature even though B may be at a higher temperature than A. It is frustrating to me because I remember a similar puzzle with the sun and a magnifying glass but can quite recall it.
__________________
The hardest thing to overcome, is not knowing that you don't know.
Guru
Join Date: May 2006
Location: 34.02S, 22.82E
Posts: 3789
#23
### Re: Violating the Second Law?
11/06/2006 11:00 PM
rcapper, you are still violating the 2nd law when you say: "...the lensing effect redistributes the energy density in a manner such that each point effectively "sees" an equivalent temperature even though B may be at a higher temperature than A."
The 2nd thermo-law demands that the two masses must stay at identical temperatures. Otherwise, one can take them out and extract real work from them, then repeat the cycle, yielding 'free-energy'!
So the question is, what prevents B from getting hotter than A? It is in the 'lensing effect', but not quite as you described it.
__________________
"Perplexity is the beginning of knowledge." -- Kahlil Gibran
Associate
Join Date: Jul 2005
Posts: 35
#15
### Re: Violating the Second Law?
11/06/2006 6:28 AM
I was, I am and I will be meticulous, hairsplitting, pedantic et cetera.
Do we wont (in 2006 !!!) distinguish between LAW (of Physics) and PRINCIPLES (of Thermodinamics) ?
Why not open a discussion on the differences between the two concepts ?
Guru
Join Date: Dec 2005
Location: Etats Unis
Posts: 1871
#16
### Re: Violating the Second Law?
11/06/2006 6:50 AM
Please pardon any imprecision in my comments and please let us not corrupt another thread by diverging into philosophical ramblings that are irresolvable as anything more than personal opinion. It becomes so tiresome reading through perpetual silliness to extract some small gem of technical information. Surely there must be a site that would satisfy those who want to argue such matters somewhere else.
__________________
The hardest thing to overcome, is not knowing that you don't know.
Anonymous Poster
#17
### Re: Violating the Second Law?
11/06/2006 6:52 AM
Hi Camillo, I think you are welcome to open such a discussion - "Do we wont (in 2006 !!!) distinguish between LAW (of Physics) and PRINCIPLES (of Thermodinamics) ?"
It will be interesting, but rather post it as a new thread, since it may be considered off-topic in this one.
Guru
Join Date: Dec 2005
Location: Etats Unis
Posts: 1871
#18
### Re: Violating the Second Law?
11/06/2006 7:31 AM
Great suggestion, thanks.
__________________
The hardest thing to overcome, is not knowing that you don't know.
Member
Join Date: Nov 2006
Location: St. Petersburg FL
Posts: 6
#20
### Re: Violating the Second Law?
11/06/2006 3:52 PM
Outstanding question Jorrie! I'm a first time responder, and found this very interesting, since reading it yesterday over coffee. I'll take a shot at the answer!
The flaw is in the statement "Since point B receives more radiation than point A...". The statement presumes that while all the energy radiated from A is received at B, not all the energy at B makes it back to A.
However, a careful analysis shows that every spot on A (every unit of area, dA, on the surface of the sphere at point A) can be mapped to a corresponding spot (differential unit of area, etc.) on B. So energy radiated off any arbitrary spot on A hits its corresponding mapped spot on B, (and only that spot on B). Energy radiated from any arbitrary spot on B hits either its corresponding spot on A (therby balancing the energy received from A), or, it reflects off the inner surface of the (spherical section) container and hits back where it started, also balancing.
__________________
Guru
Join Date: May 2006
Location: 34.02S, 22.82E
Posts: 3789
#22
### Re: Violating the Second Law?
11/06/2006 10:47 PM
JKS, you said "So energy radiated off any arbitrary spot on A hits its corresponding mapped spot on B, (and only that spot on B). Energy radiated from any arbitrary spot on B hits either its corresponding spot on A (thereby balancing the energy received from A), or, it reflects off the inner surface of the (spherical section) container and hits back where it started, also balancing."
If the radiation from every spot on A ends up on B, but not all spots on B ends up on A, it still means that B receives more energy from A than what A receives from B. So, B must get hotter than A, so that the lesser area on B radiate as much energy as A does, meaning energy flows from the cooler to the hotter body...
__________________
"Perplexity is the beginning of knowledge." -- Kahlil Gibran
The Engineer
Join Date: Feb 2005
Location: Albany, New York
Posts: 5060
#25
### Re: Violating the Second Law?
11/07/2006 8:48 AM
Jorrie,
Does some of the light focused on B arrive out of phase with other light arriving at B, cancelling out some of the ligh on B so that the incident energy is the same as received by A?
Great question by the way. I just wish I had a clue how to answer it.
Guru
Join Date: May 2006
Location: 34.02S, 22.82E
Posts: 3789
#27
### Re: Violating the Second Law?
11/07/2006 11:59 AM
Roger asked: "Does some of the light focused on B arrive out of phase with other light arriving at B, cancelling out some of the light on B so that the incident energy is the same as received by A?"
Sure, there will be out of phase pockets, but there should be as much constructive as destructive interference, averaging out, even over very small areas - I hope!
__________________
"Perplexity is the beginning of knowledge." -- Kahlil Gibran
Member
Join Date: Nov 2006
Location: St. Petersburg FL
Posts: 6
#28
### Re: Violating the Second Law?
11/07/2006 2:39 PM
Jorrie, you objected to my earlier post, noting that: "If the radiation from every spot on A ends up on B, but not all spots on B ends up on A, it still means that B receives more energy from A ..."
Any energy emitted from A, lands on B. Any energy emitted from B, that is not reflected directly back to B, lands on A. The energy transfer between the two bodies, A and B, balances, exactly.
The entire surface of A maps to a portion of the surface of B, and this same portion of B maps back to A. The remaining portion of the surface of B maps to itself, via reflection off the spherical reflective walls.
There is no mechanism for net heat transfer between A and B. A and B are initially at the same temperature. The radiant energy passing between them balances. The walls of the container neither emit, absorb, nor transmit (through) heat energy. Spheres A and B reside within a vacuum.
There is no net energy transfer between A and B, therefore there is no temperature change of either A or B. The second law of thermo remains intact.
__________________
Guru
Join Date: Aug 2005
Posts: 5492
#24
### Re: Violating the Second Law?
11/07/2006 8:44 AM
I think rcapper really had the answer in post #14: without trying to analyse the shapes at all, every path of radiation involving reflections is bi-directional.
__________________
We are alone in the universe, or, we are not. Either way it's incredible... Adapted from R. Buckminster Fuller/Arthur C. Clarke
Guru
Join Date: May 2006
Location: 34.02S, 22.82E
Posts: 3789
#26
### Re: Violating the Second Law?
11/07/2006 11:53 AM
Quote: "I think rcapper really had the answer in post #14: without trying to analyse the shapes at all, every path of radiation involving reflections is bi-directional."
This is a very good argument, but it fails on these counts:
1) the absorption and radiation processes are not reflective by nature, so some of the absorbed energy will conduct as heat to other areas of mass B, which means...
2) some of the absorbed energy at B will be radiated in a direction that can only "see itself" in the spherical mid-section and will never reach A, but...
3) all of the absorbed energy at A will be radiated to B, on account of the geometry and the assumption that both A and B resemble point masses in relation to this vastness of the cavity.
__________________
"Perplexity is the beginning of knowledge." -- Kahlil Gibran
The Engineer
Join Date: Feb 2005
Location: Albany, New York
Posts: 5060
#29
### Re: Violating the Second Law?
11/07/2006 3:37 PM
I know I'm reaching here, but could the photons exert a pressure and move mass B to an equilibrium point in the chamber (since photons have momentum)?
Guru
Join Date: Aug 2005
Posts: 5492
#33
### Re: Violating the Second Law?
11/08/2006 2:39 AM
You said
"1) the absorption and radiation processes are not reflective by nature"
Consider any infinitely small point on the surface of B "say": then any "ray" leaving it whatever route it follows and wherever it ends up is exactly matched by a ray in the opposite direction.
Just to consider one aspect of the shapes what happens to any ray from A which is reflected at some point(s) in its path from the spherical reflector.
__________________
We are alone in the universe, or, we are not. Either way it's incredible... Adapted from R. Buckminster Fuller/Arthur C. Clarke
Guru
Join Date: May 2006
Location: 34.02S, 22.82E
Posts: 3789
#34
### Re: Violating the Second Law?
11/08/2006 3:25 AM
Randall, you said: "Consider any infinitely small point on the surface of B "say": then any "ray" leaving it whatever route it follows and wherever it ends up is exactly matched by a ray in the opposite direction."
You are right as far as rays are concerned, but I do not think one can treat energy transfer simply on the principle of ray balancing. Say we started this riddle with mass A hotter than mass B. Then the balancing of rays would still hold, but heat will be transferred from A to B until they have equal temperatures.
As our 'resident physicist' has pointed out, most rays originating from A will miss B on the 'first pass'. They are then scattered throughout the cavity, until either A or B eventually absorbs them (or what is left of them after numerous partial absorptions by a less than perfect reflective cavity).
__________________
"Perplexity is the beginning of knowledge." -- Kahlil Gibran
Power-User
Join Date: Oct 2006
Posts: 488
#30
### Re: Violating the Second Law?
11/07/2006 5:43 PM
It seems to me that the problem lies largely (entirely?) in the false assumptions implied in the question. "Point sources" would have infinite temperatures for there to be ANY radiation... So in this form the problem amounts to hiding divisions by zero. Thus the only way to analyse point sources would be as the limit of finite-size ones.
If we therefore allow that the sources to have finite size (however small), we could see that the image of A from the reflector near A is larger than A, and the image of B due to this reflector is smaller than B. If the objects are the same size, energy from A will therefore miss B - in spite of the apparent "nominally" focus at its centre. This just covers one of the sets of light paths. However, a complete "ray analysis" would always show that the amount of energy received at B is the same as that leaving it (for the case of perfect reflectors and equal temperatures).
Incidentally, in the limit of a lossless cavity, the light intensity and spectral distribution throughout the enclosure will correspond to the radiation from the surface of a black body at the same temperature.
Guru
Join Date: May 2006
Location: 34.02S, 22.82E
Posts: 3789
#32
### Re: Violating the Second Law?
11/07/2006 11:44 PM
Physicist wrote: "Thus the only way to analyse point sources would be as the limit of finite-size ones.
If we therefore allow that the sources to have finite size (however small), we could see that the image of A from the reflector near A is larger than A, and the image of B due to this reflector is smaller than B.
"
Bingo!
Has the fact that a physicist (I presume) had to come to the rescue of us engineers got anything to due with the engineer's way of thinking (approximately, practical) vs. the more theoretical physics way of thinking?
__________________
"Perplexity is the beginning of knowledge." -- Kahlil Gibran
The Engineer
Join Date: Feb 2005
Location: Albany, New York
Posts: 5060
#35
### Re: Violating the Second Law?
11/08/2006 9:26 AM
Physicist,
I believe your answer is correct, but I have a question regarding something you said.
You said ""Point sources" would have infinite temperatures for there to be ANY radiation... "
Why would this be true? An answer in the form of an equation would be fine, or an explanation I can get my head around, I'm sure your not making this up but I had never heard that before and would like to know why.
Guru
Join Date: May 2006
Location: 34.02S, 22.82E
Posts: 3789
#36
### Re: Violating the Second Law?
11/08/2006 10:44 AM
Hi Roger, if Physicist will forgive for barging in here, the easiest way to get your head around it is to remember that point sources have zero volume, so any energy there will have infinite density - which is not on!
Perhaps physicist can expand on it, especially on what infinite temperature from a zero volume that radiates means? I guess it is just the same thing said differently!
__________________
"Perplexity is the beginning of knowledge." -- Kahlil Gibran
The Engineer
Join Date: Feb 2005
Location: Albany, New York
Posts: 5060
#37
### Re: Violating the Second Law?
11/08/2006 11:19 AM
That sounds reasonable. I guess my confusion comes from the fact that Temperature is average kinetic energy, but if the source is a point, there isn't any place for the object to move, so how can it have temperature? (Keep in mind, even as I ask this I'm sure there is a flaw in my logic here. An equation would probably help me see it)
Guru
Join Date: May 2006
Location: 34.02S, 22.82E
Posts: 3789
#38
### Re: Violating the Second Law?
11/08/2006 12:54 PM
Roger wrote: "An equation would probably help me see it"
I am sure physicist can provide us an equation that will illustrate this!
__________________
"Perplexity is the beginning of knowledge." -- Kahlil Gibran
Guru
Join Date: May 2006
Location: 34.02S, 22.82E
Posts: 3789
#39
### Re: Violating the Second Law?
11/09/2006 1:23 AM
A very good analysis published on the web and titled: "An Optical Perpetual Motion Machine of the Second Kind" was written by Leigh Hunt Palmer of the Simon Frazer University, giving some equations applicable to this riddle.
__________________
"Perplexity is the beginning of knowledge." -- Kahlil Gibran
Guru
Join Date: Aug 2006
Posts: 4484
#40
### Re: Violating the Second Law?
11/10/2006 10:35 AM
I think your logic is pretty sound, Roger. It relates to an interesting point brought up earlier, re the difference in views (or perceptions) of engineers and physicists. The same differences show up in the perception of mathematics by mathematicians and engineers. All (E,P,&M) would agree easily on the temperature of an obviously finite object, such as a BB. All would agree that if we make that object smaller, and retain its energy, then its temperature must go up. Where the agreement can fall apart, is right at the transition from very, very small to infinitely small. Practically speaking, it makes a great deal of sense to say that an infinitely small particle has zero volume. But then, there is "nothing" to vibrate, so there can be no temperature. [Crudely speaking, it is a little like a light bulb being fed by increasing amounts of energy: it gets brighter and brighter, and then, poof, goes out. (Granted, the analogy is impossible loose.)]
Were we to consider A and B to be true points, then the problem could be viewed (at least from my perspective) in two very different but equally valid ways:
1. The points have no temperature at all: they can't… they have no volume and no mass. If they have no temperature, they also have no energy.
2. The points have infinite temperature, because their energy is infinitely concentrated. If they have infinite temperature, they also have infinite energy.
Yet another really interesting thread! Thanks, to all of you.
__________________
There is more to life than just eating mice.
Power-User
Join Date: Jan 2006
Location: Australia
Posts: 269
#31
### Re: Violating the Second Law?
11/07/2006 9:04 PM
Isn't the question upside down. B can't be heated by A BECAUSE of the second law? A will always have to be at a higher level for the heat tranfer to happen. Sort of like putting a big reflector behind a fluro light and trying to focus it down to start a fire, can't be done
__________________
An engineer is a man who can do for five bob what any bloody fool can do for a quid (Neville Shute)
Guru
Join Date: Oct 2006
Location: India, 200 Km. North of Delhi.
Posts: 1386
#41
### Re: Violating the Second Law?
11/11/2006 3:02 AM
Jorrie I guess prcatical have the answer for the question.
think of two identical bodies A AND B of equal mass and volumes.
A is having temp of 80Deg and B is 100 Deg.
now if we keep them at a given distance, according to 2nd law heat should flow from B to A.
but now think of a lence that is having a focus distance of 1/4 of total distance and kept in beween closer to B.
all the heat wave Radiated from A will be focused on a very limited section of B this will start heating local area of B. despite of the fact its temp is more then A.
So in a way it is the energy density that is important.
__________________
Jesus gave me message, Gandhi gave me method, M.L.K
Guru
Join Date: May 2006
Location: 34.02S, 22.82E
Posts: 3789
#42
### Re: Violating the Second Law?
11/11/2006 9:36 AM
rakesh_semwal, I'm afraid you are wrong when you say: "all the heat wave Radiated from A will be focused on a very limited section of B this will start heating local area of B. despite of the fact its temp is more then A."
In fact, a lot of the radiation from A misses B completely and is then scattered throughout the vessel. Read the reply from Physicist above!
__________________
"Perplexity is the beginning of knowledge." -- Kahlil Gibran
Guru
Join Date: Oct 2006
Location: India, 200 Km. North of Delhi.
Posts: 1386
#43
### Re: Violating the Second Law?
11/12/2006 12:39 AM
Guru Ji
When I say all the heat means all the heat that is focused by lance,
What I am trying to say is, may be your system working like a heat pump. And that's why its not violation of 2nd law.
Theoretically we can increase the heat of local area of sun by putting a huge reflector on the earth and then focusing it back on sun just on area of say about 1 square meter. But it is not the violation of 2nd law. Despite of the fact sun is much hotter then earth.
__________________
Jesus gave me message, Gandhi gave me method, M.L.K
Guru
Join Date: May 2006
Location: 34.02S, 22.82E
Posts: 3789
#44
### Re: Violating the Second Law?
11/12/2006 1:53 AM
Hi rakesh_semwal, you wrote: "What I am trying to say is, may be your system working like a heat pump. And that's why its not violation of 2nd law."
No, it's not working like a heat pump - it's not 'working' at all, because the two masses remain at the same temperature. That's why it's not violating the 2nd law!
__________________
"Perplexity is the beginning of knowledge." -- Kahlil Gibran | {"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": 0.8783403635025024, "perplexity": 1894.2672042022607}, "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-2023-06/segments/1674764500304.90/warc/CC-MAIN-20230206051215-20230206081215-00218.warc.gz"} |
https://socratic.org/questions/what-is-difference-between-a-ph-of-8-and-a-ph-of-12-in-terms-of-h-concentration | Chemistry
Topics
# What is difference between a pH of 8 and a pH of 12 in terms of H+ concentration?
Jul 19, 2016
Here's what I got.
#### Explanation:
The pH of a solution is simply a measure of the concentration of hydrogen ions, ${\text{H}}^{+}$, which you'll often see referred to as hydronium cations, ${\text{H"_3"O}}^{+}$.
More specifically, the pH of the solution is calculated using the negative log base $10$ of the concentration of the hydronium cations.
color(blue)(|bar(ul(color(white)(a/a)"pH" = - log(["H"_3"O"^(+)])color(white)(a/a)|)))
Now, we use the negative log base $10$ because the concentration of hydronium cations is usually significantly smaller than $1$.
As you know, every increase in the value of a log function corresponds to one order of magnitude. For example, you have
$\log \left(10\right) = 1$
$\log \left(10 \cdot 10\right) = \log \left(10\right) + \log \left(10\right) = 1 + 1 = 2$
$\log \left(10 \cdot {10}^{2}\right) = \log \left(10\right) + \log \left({10}^{2}\right) = 1 + 2 = 3$
and so on. In your case, the difference between a pH of $8$ and a pH of $12$ corresponds to a difference of four units of magnitude between the concentration of hydronium cations in the two solutions.
Keep in mind, however, that because you're dealing with numbers that are smaller than $1$, and thus with negative logs, that the solution with a higher pH will actually have a lower concentration of hydronium cations.
More specifically, you have
"pH"_1 = - log(["H"_3"O"^(+)]_1) = 8
This is equivalent to
["H"_3"O"^(+)]_1 = 10^(-"pH"_1) = 10^(-8)"M"
Similarly, you have
"pH"_2 = - log(["H"_3"O"^(+)]_2) = 12
This is equivalent to
["H"_3"O"^(+)]_2 = 10^(-"pH"_2) = 10^(-12)"M"
As you can see, the first solution has a concentration of hydronium cations that is
$\left({10}^{- 8} \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{M"))))/(10^(-12)color(red)(cancel(color(black)("M}}}}\right) = {10}^{\textcolor{red}{4}}$
times higher than the concentration of hydronium cations of the second solution. This corresponds to the fact that you have
${\text{pH"_2 - "pH}}_{1} = 12 - 8 = \textcolor{red}{4}$
Simply put, a solution that has a pH that is $\textcolor{red}{4}$ units lower than the pH of a second solution will have a concentration of hydronium cations that ${10}^{\textcolor{red}{4}}$ times higher than that of the second solution.
##### Impact of this question
27671 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": 20, "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": 0.9300421476364136, "perplexity": 857.8742823360401}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-31/segments/1627046153897.89/warc/CC-MAIN-20210729203133-20210729233133-00049.warc.gz"} |
http://manoa.hawaii.edu/exploringourfluidearth/physical/waves/wave-energy-and-wave-changes-depth | # Wave Energy and Wave Changes with Depth
The content and activities in this topic will work towards building an understanding of how waves move through water and how the orbital motion of water particles in waves causes them to break on shore.
## Wave Energy
Many forms of energy are carried in heat, light, sound, and water waves. Energy is defined as the ability to do work; all forms of energy can be transformed into work. In science, work is defined as the movement of an object in the direction of the force applied to it. Waves do work when they move objects. We can see this work when heavy logs move across ocean basins or sand is transported. Work can also be converted into sound energy heard when waves crash on the shore. The powerful energy in waves can also be used to do work by moving generator parts to produce electricity.
## Climate Connections: Wave Power
Ocean waves carry huge amounts of energy. The amount of energy can be measured in joules (J) of work, calories (c) of heat, or kilowatt-hours (kWh) of electricity (Table 4.8). The standard measurement of energy in science is the joule.
joule calorie kilowatt-hour joule A joule (J) is the energy needed to lift 1 kilogram of matter 1 meter at sea level 1 calorie = 4.18 joules 1 kilowatt-hour = 3.6 x 106 joules calorie A calorie (c) is the energy needed to raise the temperature of 1 gram of water 1 degree centigrade. 1 calorie = 1000 kilocalories (also recorded as Calorie with a capital C) 1 joule = 0.24 calories 1 kilowatt-hour = 8.6 x 105 calories kilowatt-hour A kilowatt-hour (kWh) is the standard measurement of energy in the United States. It is equivalent to the work of a kilowatt for one hour (about the power used by a toaster for one hour 1 joule = 2.78 x 10-7 kilowatt-hours 1 calorie = 1.16 x 10-6 kilowatt-hours
The amount of energy in a wave depends on its height and wavelength as well as the distance over which it breaks. Given equal wavelengths, a wave with greater amplitude will release more energy when it falls back to sea level than a wave of lesser amplitude. Energy (E) per square meter is proportional to the square of the height (H): E∝H2. In other words, if wave A is two times the height of wave B, then wave A has four times the energy per square meter of water surface as wave B.
A wave with a height of 2 m and a wavelength of 14 m breaking along 2 km of coastline (surface area = 32,000 m2) has approximately 45 kWh of energy. This is roughly equivalent to one gallon of gasoline, which contains about 160 million (1.6 x 108) joules (J) of energy. According to the US Department of Agriculture, the World Bank, and the US Energy Information Administration, the average American eats 3.14 kWh per day in food, uses about 37 kWh in electricity, and uses a combined 250 kWh per day in electricity and petroleum. This means that the energy in one 2 m by 14 m by 2 km wave is equivalent to the amount of energy needed to feed a person for two weeks, power their home for one day, or power their electrical and transportation needs for 5 hours (Fig. 4.17). Ocean waves offer a very large source of renewable energy. Technologies that efficiently harvest this energy resource are actively being researched and developed by scientists.
## Orbital Motion of Waves
By watching a buoy anchored in a wave zone one can see how water moves in a series of waves. The passing swells do not move the buoy toward shore; instead, the waves move the buoy in a circular fashion, first up and forward, then down, and finally back to a place near the original position. Neither the buoy nor the water advances toward shore.
As the energy of a wave passes through water, the energy sets water particles into orbital motion as shown in Fig. 4.18 A. Notice that water particles near the surface move in circular orbits with diameters approximately equal to the wave height. Notice also that the orbital diameter, and the wave energy, decreases deeper in the water. Below a depth of half the wavelength (D = 1/2 L), water is unaffected by the wave energy.
## Deep-Water, Transitional, and Shallow-Water Waves
Swells are deep-water waves, meaning that the depth (D) of the water is greater than half the wave’s wavelength (D > 1/2 L). The energy of a deep-water wave does not touch the bottom in the open water (Fig. 4.18 A).
When deep-water waves move into shallow water, they change into breaking waves. When the energy of the waves touches the ocean floor, the water particles drag along the bottom and flatten their orbit (Fig. 4.18 B).
Transitional waves occur when the water depth is less than one-half the wavelength (D < 1/2 L). At this point the water movement of particles on the surface transitions from swells to steeper waves called peaking waves (Fig. 4.19). Because of the friction of the deeper part of the wave with particles on the bottom, the top of the wave begins to move faster than the deeper parts of the wave. When this happens, the front surface of the wave gradually becomes steeper than the back surface.
When the water depth is less than one-twentieth the wavelength, the wave becomes a shallow-water wave (D < 1/20 L). At this point, the top of the wave travels so much faster than the bottom of the wave that top of the wave begins to spill over and fall down the front surface. This is called a breaking wave. A breaking wave occurs when one of three things happen:
1. The crest of the wave forms an angle less than 120˚,
2. The wave height is greater than one-seventh of the wavelength (H > 1/7 L), or
3. The wave height is greater than three-fourths of the water depth (H > 3/4 D).
In some ways a breaking wave is similar to what happens when a person trips and falls. As a person walks normally, their feet and head are traveling forward at the same rate. If their foot catches on the ground, then the bottom part of their body is slowed by friction, while the top part continues at a faster speed (see Fig. 4.19). If the person’s foot continues to lag far behind their upper body, the angle of their body will change and they will topple over.
The transition of a wave from a deep-water wave to a shallow-water breaking wave is shown in Fig. 4.20. Terms relating to wave depth a described in detail in Table 4.9.
Symbols D = Depth of water L = Length of wave H = Height of wave Deep-water waves Deep-water waves are waves traveling across a body of water where depth is greater than half the wavelength (D > 1/2 L). Deep-water waves include all wind-generated waves moving across the open ocean. Transitional waves Transitional waves are waves traveling in water where depth is less than half the wavelength but greater than one-twentieth the wavelength (1/20 L < D < 1/2 L). Transitional waves are often wind-generated waves that have moved into shallower water. Shallow-water waves Shallow-water waves are waves traveling in water where depth is less than one-twentieth the wavelength (D < 1/20 L). Shallow-water waves include wind-generated waves that have moved into shallow, nearshore areas, tsunamis (seismic waves) generated by disturbances in the ocean floor, and tide waves generated by the gravitational attraction of the sun and moon. Breaking shallow-water waves Breaking shallow-water waves are unstable shallow-water waves. Usually shallow-water waves begin to break when the ratio of wave height to wavelength is 1 to 7 (H/L = 1/7), when the wave’s crest peak is steep (less than 120˚), or when the wave height is three-fourths of the water depth (H = > 3/4 D). Breaking deep-water waves Breaking unstable deep-water waves are waves that begin to break when the seas are confused (waves from mixed directions) or when the wind blows the crests off waves, forming whitecaps.
Activity
## Activity: Orbital Motion of Waves
Observe orbital motion of waves in a long wave tank.
Activity
## Activity: Simulate Deep-Water, Transitional, and Shallow-Water Waves
Use a long wave tank to create and observe the differences between deep-water, transitional, and shallow-water waves. | {"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": 0.892880380153656, "perplexity": 1047.332858025912}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-05/segments/1642320304835.96/warc/CC-MAIN-20220125130117-20220125160117-00503.warc.gz"} |
https://proofwiki.org/wiki/Unordered_Pairs_Exist | # Unordered Pairs Exist
It has been suggested that this page or section be merged into Axiom:Axiom of Pairing. (Discuss)
## Theorem
Let $A$ and $B$ be classes, which may be either sets or proper classes.
Then:
$\forall A, B: \left\{{A, B}\right\} \in U$
where $U$ is the universal class.
## Proof
$\displaystyle \forall A, B$ $:$ $\displaystyle \exists x: \forall y: \left({y \in x \iff y = A \lor y = B}\right)$ $\quad$ Axiom of Pairing $\quad$ $\displaystyle \implies \ \$ $\displaystyle \forall A, B$ $:$ $\displaystyle \exists x: x = \left\{ {y: y = A \lor y = B}\right\}$ $\quad$ Definition of Set Equality $\quad$ $\displaystyle \implies \ \$ $\displaystyle \forall A, B$ $:$ $\displaystyle \left\{ {y: y = A \lor y = B}\right\} \in U$ $\quad$ Element of Universe $\quad$ $\displaystyle \implies \ \$ $\displaystyle \forall A, B$ $:$ $\displaystyle \left\{ {A, B}\right\} \in U$ $\quad$ Definition of Doubleton $\quad$
$\blacksquare$ | {"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": 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": 0.9877398610115051, "perplexity": 397.6689082231225}, "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-2019-22/segments/1558232260358.69/warc/CC-MAIN-20190527005538-20190527031538-00396.warc.gz"} |
https://mersenneforum.org/showthread.php?s=48b9ba10ce2bee3209f1905ebbf686f7&p=355818 | mersenneforum.org Towers of Hanoi with random moves
Register FAQ Search Today's Posts Mark Forums Read
2008-02-03, 21:34 #1 m_f_h Feb 2007 24×33 Posts Towers of Hanoi with random moves I think this could interest some freaks here: Calculate the expected number of moves after which the tower of discs is moved from pole 1 to pole 3, when in any situation any of the possible moves is made with the same probability. Could anyone give a formula? Some contradicting information on OEIS : A007798 : Expected time to finish a random Tower of Hanoi problem with n disks using random moves. = 2, 18, 116, 660, 3542 should IMHO rather be equal to the integer part (or nearest integer) of e = [0, 2, 64/3, 1274/9, 21760/27, 348722/81, ...] used in > %S A134939 0,2,64,1274,21760, 348722 > %N A134939 Consider a 3-pole Tower of Hanoi configuration which begins with n rings on pole 1. Moves are made at random, where the 1-step transition probabilities out of any state are equal. Let e(n) be the expected number of transitions to reach the state in which which all rings are on pole 3. Sequence gives a(n), the numerator of e(n). > %C A134939 Both allowable transitions out of any of the three special states in which all the rings are on one of the poles have probability 1/2, and each of the three allowable transitions out of any of the other 3^n - 3 states have probability 1/3. > %C A134939 It appears that the denominator of e(n) for n>=1 is 3^(n-1). > %e A134939 The values of e(0), ..., e(5) are 0, 2, 64/3, 1274/9, 21760/27, 348722/81. PS: yes, OEIS is down at present. I found the former seq in google's cache. Last fiddled with by m_f_h on 2008-02-03 at 21:37
2008-02-03, 22:42 #2
xilman
Bamboozled!
"𒉺𒌌𒇷𒆷𒀭"
May 2003
Down not across
295B16 Posts
Quote:
Originally Posted by m_f_h I think this could interest some freaks here: Calculate the expected number of moves after which the tower of discs is moved from pole 1 to pole 3, when in any situation any of the possible moves is made with the same probability. Could anyone give a formula? Some contradicting information on OEIS : A007798 : Expected time to finish a random Tower of Hanoi problem with n disks using random moves. = 2, 18, 116, 660, 3542 should IMHO rather be equal to the integer part (or nearest integer) of e = [0, 2, 64/3, 1274/9, 21760/27, 348722/81, ...] used in > %S A134939 0,2,64,1274,21760, 348722 > %N A134939 Consider a 3-pole Tower of Hanoi configuration which begins with n rings on pole 1. Moves are made at random, where the 1-step transition probabilities out of any state are equal. Let e(n) be the expected number of transitions to reach the state in which which all rings are on pole 3. Sequence gives a(n), the numerator of e(n). > %C A134939 Both allowable transitions out of any of the three special states in which all the rings are on one of the poles have probability 1/2, and each of the three allowable transitions out of any of the other 3^n - 3 states have probability 1/3. > %C A134939 It appears that the denominator of e(n) for n>=1 is 3^(n-1). > %e A134939 The values of e(0), ..., e(5) are 0, 2, 64/3, 1274/9, 21760/27, 348722/81. PS: yes, OEIS is down at present. I found the former seq in google's cache.
The two cases are different. The first has a random initial configuration (presumably, this means a configuration selected with probability 1/P where P is the number of configurations) whereas the second has the initial configuration where all disks are on pole 1.
Paul
2008-02-04, 01:25 #3
m_f_h
Feb 2007
24×33 Posts
Quote:
Originally Posted by xilman The two cases are different. The first has a random initial configuration (presumably, this means a configuration selected with probability 1/P where P is the number of configurations) whereas the second has the initial configuration where all disks are on pole 1. Paul
Thanks for that clarification. Indeed I did not initially think of this possibility.
But can you confirm the values ? Even (or: even more) starting at a random position, I would be surprised of the expectation value being an integer.
Thus, at least, there is some rounding process involved.
I figured out the topology of the graph given by all possible positions, so either of these expectation values has a (maybe simpler) interpretation as random walk on that graph.
Last fiddled with by m_f_h on 2008-02-04 at 01:32 Reason: probabilities > expectation values
2008-02-04, 09:16 #4
xilman
Bamboozled!
"𒉺𒌌𒇷𒆷𒀭"
May 2003
Down not across
3×3,529 Posts
Quote:
Originally Posted by m_f_h But do you have more information on this?
Nope. All I did was read the text you posted.
Paul
2008-02-04, 20:39 #5 maxal Feb 2005 3748 Posts A correct definition for A007798 seems to be: %N A007798 Expected number of random moves in Tower of Hanoi problem with n disks, starting with a randomly chosen position and ending at a position with all disks on the same peg.
2008-02-05, 12:30 #6
xilman
Bamboozled!
"𒉺𒌌𒇷𒆷𒀭"
May 2003
Down not across
295B16 Posts
Quote:
Originally Posted by maxal A correct definition for A007798 seems to be: %N A007798 Expected number of random moves in Tower of Hanoi problem with n disks, starting with a randomly chosen position and ending at a position with all disks on the same peg.
That's corrected only in the sense that it replaces "time" in the original with "number of moves". Admittedly it's a better phrasing, in my opinion, but the original seemed clear enough to me.
Paul
2008-02-05, 17:04 #7 Kevin Aug 2002 Ann Arbor, MI 433 Posts What if you stipulated that it takes more time to move the bigger discs?
2008-02-05, 17:14 #8
maxal
Feb 2005
22×32×7 Posts
Quote:
Originally Posted by xilman That's corrected only in the sense that it replaces "time" in the original with "number of moves". Admittedly it's a better phrasing, in my opinion, but the original seemed clear enough to me.
Not only that. Please notice another important difference between A134939 and A007798:
A134939 requires that in the final position all disks are on the third (i.e., fixed) peg while A007798 requires only that in the final position all disks are on the same (any) peg.
2008-03-15, 06:38 #9 nibble4bits Nov 2005 B616 Posts Firstly, treat this question as equivilent to thinking about all possible moves. It's in theory possible to never have this algorythm halt. Realisticly, that may never happen in practice (see hysteresis in electronics). The lower bound will of course be the best case solution/s. As far as filtering out duplicate states... That's a little beyond my experience. If you can find a way to enumerate the solutions so that if the same state occurs, parts of the solution are identical for both solutions, then you can at least make it easier. If I find a way to do that, I should patent it (just kidding). It's still an intractable problem since infinite reduced by a finite amount is still infinite (aleph-null -> aleph-null). Using calculus, it's at least possible to treat it as a series. What I mean, is that if you know the rough odds for each branch, then you can at least make a good estimate. What I want to know, is the frequency of solutions in relation to the number of steps. The average could be infinite but that seems against common sense as we most often see solutions in finite time. Perhaps I should requalify that question in terms of frequency divided by number of steps? Note that if you increase the number of towers, the lower the odds of a random set of moves being a solution (or having passed the solved state/s) for a given number of steps. Can someone verify that? I'm pretty sure, but it would be nice to have a counterexample if I'm wrong.
2008-03-15, 08:56 #10
maxal
Feb 2005
25210 Posts
Quote:
Originally Posted by nibble4bits What I want to know, is the frequency of solutions in relation to the number of steps. The average could be infinite but that seems against common sense as we most often see solutions in finite time. Perhaps I should requalify that question in terms of frequency divided by number of steps?
I think it is rather simple question to answer, taking into account recent developments. But first, please clearly specify what is "solution" in your case as there are so many variations of the puzzle around.
2013-10-10, 04:17 #11 maxal Feb 2005 111111002 Posts If anybody still cares, here is a preprint about solving Tower of Hanoi with random moves: http://arxiv.org/abs/1304.3780
Similar Threads Thread Thread Starter Forum Replies Last Post jasong Lounge 46 2017-05-09 12:32 jocelynl Factoring 1 2016-11-25 05:14 Prime95 Programming 19 2012-10-06 09:34 Greenk12 Factoring 1 2008-11-15 13:56 ewmayer Lounge 4 2007-03-08 20:31
All times are UTC. The time now is 16:55.
Thu Feb 25 16:55:49 UTC 2021 up 84 days, 13:07, 0 users, load averages: 2.16, 1.90, 1.90 | {"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": 0.8215899467468262, "perplexity": 1232.7609051226975}, "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/1614178351374.10/warc/CC-MAIN-20210225153633-20210225183633-00452.warc.gz"} |
https://astarmathsandphysics.com/ib-maths-notes/quadratics/1104-solving-equations-involving-algebraic-fractions.html?tmpl=component&print=1&page= | ## Solving Equations Involving Algebraic Fractions
An algebraic fraction is any expression with fraction which include x terms in numerator and/or denominator. If this expression is equal to something then typically we can solve it by reararranging the fraction into a linear or quadratic equation and solving.
For example, ifthen we can multiply both sides byto clear the fraction, obtaining
Expanding the right hand side givesand collecting like terms gives
More complicated problems involve two fractions, which must be added first. Often a quadratic is obtained, which may solved by factorisation or use of the quadratic formula.
For example, ifthen we start by making a common denominator for the terms on the left hand side.
Now add the numerators to give
This is equal to the right hand side, so we can write
Multiply both sides byto clear the fractions, obtainingand expand the right hand side to give
Move all the terms to the right hand side to giveThis expression does not factorise, so use the quadratic formula,
In this case,and | {"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": 0.9175420999526978, "perplexity": 854.754053611514}, "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-2018-17/segments/1524125947654.26/warc/CC-MAIN-20180425001823-20180425021823-00293.warc.gz"} |
http://math.stackexchange.com/questions/95922/how-do-i-find-the-volume-of-this-sliced-cylinder | # How do I find the volume of this sliced cylinder?
This is hard to explain but I'll do my best. I hope I'm clear.
Imagine you have a donut. You want to find the volume of it and the method you want to use is to imagine slicing one side of that donut and opening it out into a cylinder. Only, it's not exactly a cylinder as it has two pointed ends: one side of the cylinder has the length of the inner circumference of the donut, the other side has the length of the outer circumference.
The 'middle part' is a simple cylinder. I want to find the volume of each end part.
There's an easy trick to it, but that's not the solution I'm looking for; the easy trick being putting the two end parts together to make a smaller cylinder.
But, how I want to do this is to find the volume of one of these pointy cylindrical endparts with an integral. However, I can't seem to hit the right answer.
Let's say my stretched out donut has a left side of length
$2\pi(R-r)$
and a right side of length
$2\pi(R+r)$ and a radius of $r$. Let's say I slice the top and bottom pointed end parts off. I now have the middle part, a cylinder with a radius of $r$ and a height of $2\pi(R-r)$ and two pointed endparts. Each endpart has a radius of $r$, and a height of $2\pi r$.
If I cut an endpart into triangle wedge-shaped cross sections, each wedge will have a length of $2\sqrt{r^2-y^2}$, a height of $2\pi\sqrt{r^2-y^2}$ and a depth of $dy$. So the volume of each wedge is $2\pi(r^2-y^2)dy$
If I integrate this with limits $-r$ and $r$ I get
$2\pi\int_{-r}^r (r^2-y^2)dy = 2\pi (4r^3/3)$
Assuming that's right, I can double that and add it to the volume of my cylinder to get the donut volume. So:
$4\pi (\frac{4r^3}{3}) + 2\pi^2r^2(R-r) = \frac{16\pi r^3}{3} + 2\pi^2 r^2(R-r)$
The volume of the donut is actually $2\pi^2 r^2R$ (using the solid of revolution approach) so looks like I picked up some extra dough somewhere... $16\pi \frac{r^3}{3} - 2\pi^2 r^3$ cubic units of extra dough to be precise.
Is my arithmetic wrong? Is my method wrong? Can you spot what I messed up?
Sketch:
-
I don’t think that there’s any way to slice the endcaps into parallel triangular cross-sections. If you slice in the direction that I think you’re using, I believe that you get trapezoids except in the very centre, where you do get a triangle as a degenerate trapezoid. – Brian M. Scott Jan 2 '12 at 22:50
Um, could you perhaps add some drawings or something? It's completely unclear to me what you're doing; I don't even know what it is you consider the "left" and "right" "sides" of your torus; or what "slicing one side" means. How can an exact cylinder arise? Where do "pointed ends" come from? – Henning Makholm Jan 2 '12 at 23:28
@Henning: Slice the torus with a plane normal to the circle running through its centre, which is of radius $R$. Open it up so that the central circle becomes a straight line of length $2\pi R$, oriented vertically. The radius of the cross-section is $r$, so the inner circumference of the torus is $2\pi(R-r)$; when you open it up, orient it so that the straightened-out inner circumference is to the left, the straightened-out outer circumference of $2\pi(R+r)$ is to the right, with the straightened-out central circle halfway between them. The outline is a symmetric trapezoid. – Brian M. Scott Jan 2 '12 at 23:39
Exactly. Now the next step would be to cut out the cylinder from this shape which has height $2\pi (R-r)$. You have the top and bottom end remaining. So the question is, how do you find the volume of each of these end parts? – Korgan Rivera Jan 2 '12 at 23:45
Korgan, you don’t get that triangular wedge: the lefthand end of it, where it comes to a point, shouldn’t be there. It gets cut off by a vertical line before that point, because the cylinder wraps around. – Brian M. Scott Jan 2 '12 at 23:50
@Brian M. Scott. Ok, your info gave me what I needed to understand. I didn't understand why you said I didn't need to calculate h, so I went ahead and did the calculation with h and, after too much bad arithmetic, found the answer.
So the cross-sections are trapezoids, not triangles. Thinking that they might have been triangles before now seems ridiculous.
So the lengths of the sides of the trapezoids are $\pi(r-x)$ and $\pi(r+x)$ and the bottom is $2x$. The area of the trapezoid is then $2\pi rx$. The volume of the trapezoidal wedge is $2\pi rx dy$. The sum of the wedges is the integral $2\pi r\int_{-r}^{r} x dy$, where x is $\sqrt{r^2-y^2}$. The answer to that is $\pi^2r^3$.
Two of those plus the volume of the cylinder is:
$2\pi^2r^3 + \pi r^2(2\pi(R-r)) = 2\pi^2r^2R$
Which is the volume of the donut. Thanks for your help everyone!
-
Remember to accept an answer after a question has come to a close. – Samuel Reid Jan 4 '12 at 6:37
When you slice the endcaps in the direction that you’ve chosen, you get trapezoidal cross-sections, not triangular ones: the triangle in your illustration should be cut off vertically at the lefthand end, because the cylinder wraps around. The mean height of each of the trapezoidal cross-sections is the mean height of the endcap, which is $\pi r$, and the depth is $2\sqrt{r^2-y^2}$, so the volume is $$4\pi r\int_0^r\sqrt{r^2-y^2}dy\;.$$ There’s no need to do the trig substitution, since the integral clearly gives the area of a quarter-circle of radius $r$, or $\pi r^2/4$, and the volume of the endcap is therefore $\pi^2 r^3$. Two of them give you exactly what you need: $$2\pi^2 r^2(R-r)+2\pi^2 r^3=2\pi^2 r^2 R\;.$$
-
Your answer is clearly correct, but I'm having a hard time imagining these trapezoidal cross-sections. Can anyone sketch it? – Korgan Rivera Jan 3 '12 at 0:17
Ok I see it :) Now I'm trying to figure out the height of the left side of the trapezoidal cross-section. You've gone ahead and used the mean height, but I'm not sure where that comes from yet, so. – Korgan Rivera Jan 3 '12 at 0:31
@Korgan: Look at it from the side, so that you see the endcap as a triangle. The slice in the centre really does look like that triangle. As you move the slice towards you, the lefthand side of the trapezoid moves to the right, and the righthand side moves to the left at the same rate. They meet in the middle when the slice is just tangent to the cylinder on the side towards you. At each stage the righthand edge extends as far above the horizontal midline of the triangle as the lefthand edge fails to reach the midline from below. – Brian M. Scott Jan 3 '12 at 0:38
@Korgan: Some of the artwork here may be helpful. – Brian M. Scott Jan 3 '12 at 0:40
Thanks but I did a sketch of it already. i695.photobucket.com/albums/vv314/korganwhitney/whatH.png Now I'm trying to figure out h in the sketch. It's getting late and I'm running low on caffeine, so this isn't happening as quickly as I'd like. – Korgan Rivera Jan 3 '12 at 0:44 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 0.8892290592193604, "perplexity": 408.26856862056366}, "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-2015-22/segments/1432207927824.81/warc/CC-MAIN-20150521113207-00146-ip-10-180-206-219.ec2.internal.warc.gz"} |
https://tex.stackexchange.com/questions/407744/producing-pdf-from-markdown-with-pandoc-and-xelatex-generate-misleading-error-me | # producing pdf from markdown with pandoc and xelatex generate misleading error messages
I encounter a weird issue when I tried to convert a markdown file to pdf using pandoc. My markdown file contains Chinese characters and English characters. The command I use is:
pandoc --pdf-engine=xelatex -V CJKmainfont=KaiTi test.md -o test.pdf
The error message is:
Error producing PDF.
! Undefined control sequence.
pandoc: Cannot decode byte '\xbd': Data.Text.Internal.Encoding.streamDecodeUtf8With: Invalid UTF-8 stream
In fact, the error has nothing to do with UTF-8 encoding. After long hours of wrestling with the problem. Finally I find that it is because my markdown file contains backslashes followed by text, which are taken as LaTeX command by pandoc in default settings. After knowing this critical info, I was able to finally fix this problem. More information can be found in this pandoc issue .
Someone suggest in that issue this may be a problem with xelatex, because if we use pandoc --pdf-engine=lualatex test.md -o test.pdf The error message becomes something like the this:
Error producing PDF.
! Undefined control sequence. l.416
...宽度有问题，应该把\textwidth换成
If the error message from using xelatex engine is similar to above message. I would have solved this problem long long ago. So it appears to me that the error message may indeed be related to xelatex.
But, but, if we separate the pdf-generating step into two steps, i.e., first generate tex file, then generate pdf file from tex. Something like the following code:
pandoc -s -t latex -V CJKmainfont=KaiTi test.md -o test.tex # first step xelatex test.tex # second step Then the error message will change and be just like when we use lualatex engine. This suggests that the problem may not be related to xelatex. We get contradictory conclusions.
I am new to pandoc and do not know any internals of xelatex. Can anyone more knowledgeable point out which is causing the problem here. Pandoc or xelatex or both?
## system and pandoc version info
I have tested the file on both Windows and Linux system (CentOS 7). The exact version of system, pandoc, TeX Live and xelatex is list below.
### Windows
• system version: Windows 8.1 32bit
• Pandoc version: 2.0.5
• TeX Live: 2016/W32TeX
• xelatex: XeTeX 3.14159265-2.6-0.99996
### Linux
• system version: CentOS 7.2.1511
• Pandoc version: 1.12.3.1
• TeX Live: 2017
• xelatex: 3.14159265-2.6-0.99998
update 2017.12.29
With the release of Pandoc 2.0.6, this behaviour is handled more properly:
Allow lenient decoding of latex error logs, which are not always properly UTF8-encoded
Now, it is easier to debug this kind of issues.
• I have neither pandoc nor windows 8.1. installed on this pc and you didn't say which texsystem and version you are using, so I'm only guessing. Imho in the "one-step" version pandoc is intercepting the terminal output of xelatex and it is quite possible that this is not correct utf8 (windows 8.1. imho is not using utf8 everywhere). You could try to change the code page of the command line with chcp 65001. Beside this: the first line does show the begin of the correct error message (! Undefined control sequence.) – Ulrike Fischer Dec 27 '17 at 15:21
• I am seeing the same error message both on Windows system and my CentOS servers. I will update the version information. – jdhao Dec 27 '17 at 15:32
It is indeed true that XeTeX can produce invalid UTF-8 in its error output, and I can reproduce this with the following simpler .tex file:
\documentclass{article}
\begin{document}
\end{document}
So you can consider this either a bug in XeTeX (for producing invalid UTF-8) or in Pandoc (for incorrectly assuming that XeTeX will produce valid UTF-8).
## Unicode and UTF-8
The problem, in short, is that you cannot just break a sequence of UTF-8 bytes in any arbitrary place. To take an example, in the string 应该把, the characters are:
So the string as a whole is encoded in UTF-8 as a sequence of 9 bytes:
E5 BA 94 E8 AF A5 E6 8A 8A
\______/ \______/ \______/
应 该 把
You can break the byte sequence after 0, 3, 6, or 9 bytes to get a valid string containing 0, 1, 2 or 3 characters respectively. But breaking it at some other place results in invalid UTF-8.
Unfortunately, that is exactly what XeTeX can do: it can break the byte sequence in some such place, resulting in invalid UTF-8 that Pandoc then fails to cope with (because it assumes valid UTF-8).
## Explanation
In the first place, in Unicode-aware engines like XeTeX and LuaTeX, all unicode characters can be part of control sequences, and there happens to be no control sequence named \textwidth换成 so the system generates an error about an undefined control sequence.
Then when printing out this error to the terminal, TeX tries to add additional context around where this undefined control sequence \textwidth换成 was encountered, which means some additional characters surrounding the occurrence, to fill error_line characters. (This can be increased; see here and here. Though increasing this is a good idea anyway and decreases the likelihood of this error happening; it can still happen with sufficiently long lines (and does happen with the example in the question), because the max value of error_line is still only 254.)
Unfortunately (and this is the bug), it appears that XeTeX counts by bytes and truncates the output without regard for breaking only at well-defined Unicode code-point sequences. Look for procedure show_context in the XeTeX source code, and compare with print_valid_utf8 in the LuaTeX source code, used in its show_context.
In this example, XeTeX picks up only the last two bytes of the first word (the 8A 8A), which is not a valid UTF-8 sequence. That is why iconv and Pandoc complain.
## Demonstration
The commands I used for compiling the above .tex file with LuaTeX and XeTeX are respectively:
lualatex -interaction=nonstopmode test.tex | iconv -f UTF8
and
xelatex -interaction=nonstopmode test.tex | iconv -f UTF8
With the former (LuaTeX), I get the error message:
! Undefined control sequence.
l.3 ...把 123456789 123456789 123 \textwidth换成
but with the latter (XeTeX), I get an error message that is not valid UTF-8, so iconv fails with
iconv: (stdin):11:7: cannot convert
Without iconv, on my terminal I see printed:
! Undefined control sequence.
l.3 ...?? 123456789 123456789 123 \textwidth换成
and by redirecting the output to a file and viewing it in a raw editor, we can see better what's going on. The following is hexdump output from xxd -g 1 -c 32:
000001c0: 78 29 0a 21 20 55 6e 64 65 66 69 6e 65 64 20 63 6f 6e 74 72 6f 6c 20 73 65 71 75 65 6e 63 65 2e x).! Undefined control sequence.
000001e0: 0a 6c 2e 33 20 2e 2e 2e 8a 8a 20 31 32 33 34 35 36 37 38 39 20 31 32 33 34 35 36 37 38 39 20 31 .l.3 ..... 123456789 123456789 1
00000200: 32 33 20 5c 74 65 78 74 77 69 64 74 68 e6 8d a2 e6 88 90 0a 20 20 20 20 20 20 20 20 20 20 20 20 23 \textwidth.......
Note the 8a 8a (the last two bytes of 把 = E6 8A 8A) just after the ellipsis (2e 2e 2e meaning ...).
• Thanks for the detailed answer. This issue should be reported to xelatex. – jdhao Dec 28 '17 at 3:11 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 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": 0.8766619563102722, "perplexity": 3149.1090584713197}, "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-2021-04/segments/1610704803308.89/warc/CC-MAIN-20210126170854-20210126200854-00123.warc.gz"} |
https://spmscience.blog.onlinetuition.com.my/2020/01/spm-science-2019-paper-2-question-2.html | # SPM Science 2019, Paper 2 (Question 2)
Question 2:
Diagram 2 shows the histogram of the data of mass for 30 students in a school. The heights of the students are 1.55 m.
Diagram 2
(a)(i) State the number of students which have mass range of 71 kg to 80 kg. [1 mark]
(ii) State one reason that causes the students in 2(a)(i) fall in that range? [1 mark]
(b) What is the type of variation shown by the graph in Diagram 2? [1 mark]
(c) State two characteristics of the type of variation based on your answer in 2(b). [2 marks] | {"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": 0.869409441947937, "perplexity": 1612.053092716606}, "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-2021-17/segments/1618038879374.66/warc/CC-MAIN-20210419111510-20210419141510-00027.warc.gz"} |
http://mymathforum.com/computer-science/214834-diagnostics-test-divergence-unknown-function.html | My Math Forum Diagnostics to test divergence of an unknown function
Computer Science Computer Science Forum
October 8th, 2015, 07:45 AM #1 Senior Member Joined: Apr 2014 From: Glasgow Posts: 2,150 Thanks: 730 Math Focus: Physics, mathematical modelling, numerical and computational solutions Diagnostics to test divergence of an unknown function I have a problem associated with convergence in a computer program. Consider a numerical solution based on iteration with the form $\displaystyle x_{n+1} = w f(x_n) + (1-w) x_n$ where $\displaystyle w$ is known as the relaxation factor. Basically, you mix a portion of the original answer together with the new result using a weighting factor. By setting $\displaystyle 0 \lt w \le 1$ (called under-relaxation) you can improve the stability of the solution at the cost of convergence rate. The following properties are true: 1. $\displaystyle f(x_n)$ is unknown, as are its derivatives. However, we can be sure that $\displaystyle f(x_n)$ does not depend on $\displaystyle n$ and is "sensible" (no discontinuities, single-valued over the reals, etc.) 2. A solution indeed exists such that after some threshold number of iterations, $\displaystyle n > N$, and convergent iteration,$\displaystyle \frac{x_{N+1} - x_N}{x_N} < \delta$ where $\displaystyle \delta$ is some sufficiently small number (such as 0.01) is true. Does anyone know of a decent algorithm for detecting divergence and, if so, how to set the under-relaxation factor to compensate?
Tags diagnostics, divergence, function, test, unknown
Thread Tools Display Modes Linear Mode
Similar Threads Thread Thread Starter Forum Replies Last Post king.oslo Complex Analysis 6 May 22nd, 2015 02:00 PM mathdisciple Calculus 6 March 23rd, 2014 12:24 AM hatsjoe Linear Algebra 0 March 17th, 2014 06:28 AM shiseonji Calculus 1 February 21st, 2014 04:53 AM aaron-math Calculus 8 October 12th, 2011 10:03 PM
Contact - Home - Forums - Cryptocurrency Forum - Top | {"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": 0.8047646880149841, "perplexity": 1487.038529161564}, "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/1555578517745.15/warc/CC-MAIN-20190418161426-20190418183321-00083.warc.gz"} |
http://letslearnnepal.com/class-12/physics/modern-physics/quantisation-of-energy/introduction-to-bohrs-atomic-model/ | # Introduction to Bohr’s Atomic Model
Quantization of energy
All the elements are made up of small particles known as an atom. An atom is composed of sub atomic particles viz. proton, electron and neutron. Various scientists proposed different models of the atom at different times which led to the discovery of the structure of the atom.
Bohr’s atomic model:
Niels Bohr in 1911 proposed a model of hydrogen atom based on the Planck’s quantum theory. He assumed the atomic theory proposed by Rutherford and tried to overcome the defects of the theory. The postulates he proposed are:
1. Atom consists of a positively charged nucleus with negatively charged electrons revolving around it in different orbits.
2. Only those orbits are possible for which the angular momentum of electron is integral multiple of h/2π i.e., mvr = nh/2π where where n= 1, 2, 3… .
3. Electrons do not radiate energy as long as it remains in its own orbit. Electrons jump from higher orbit to lower orbit radiating energy in the form of photon and may jump from lower orbit to higher level absorbing energy. The frequency of emitted photon when electrons jump from energy state E1 to energy state E2 is given by: $$F = \frac{e_2 – e_1}{h}$$
4. The energy of an electron always remains constant when the electron revolves in an orbit.
From Bohr’s atomic theory, radius and velocity of the electron can be determined. It also explains the stability of the atom and hydrogen spectra.
Do you like this article ? If yes then like otherwise dislike : 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": 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": 0.9320657849311829, "perplexity": 662.2619219588277}, "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/1539583515375.86/warc/CC-MAIN-20181022180558-20181022202058-00026.warc.gz"} |
https://www.physicsforums.com/threads/math-problem.8929/ | Math problem
1. Nov 14, 2003
denian
hope to get the idea on how to solve this question.
the complex number z is given by
z = 1 + cos (theta) + i sin (theta)
where -pi < theta < or = +pi
show that for all values of theta, the point representing z in a Argand diagram is located on a circle. find the centre and radius of the circle.
2. Nov 14, 2003
HallsofIvy
Staff Emeritus
If z = 1 + cos (θ) + i sin (&theta)
Then z-1= cos(&theta)+ i sin(θ).
If you represent z as x+ iy then
(x-1)+ iy= cos(θ)+ i sin(θ)
or x- 1= cos(θ), y= sin(θ)
Those are parametric equations of a circle with what center and radius?
3. Nov 15, 2003
denian
ok. i compare those with the
y=r sin (&theta)
and
x=r cos (&theta)
so, i know the radius = 1 unit
but may i know how to find the centre of the circle?
tq.
4. Nov 15, 2003
HallsofIvy
Staff Emeritus
It's exactly where the center of the circle given by
x= r cos θ
y= r sin θ is!
Hint: x2= r2cos2θ
y2= r2sin2θ
What is x2+ y2?
If that's too complicated, what is (x,y) when θ= 0?
What is (x,y) when θ= π?
5. Nov 15, 2003
denian
tq. u helped me solved the problem.
but there is another part of the question which i ahain need some idea.
--> prove that the real part of (1/z) is (1/2) for all values of [the] | {"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": 0.8718469142913818, "perplexity": 3900.7277872462314}, "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-2017-09/segments/1487501171834.68/warc/CC-MAIN-20170219104611-00267-ip-10-171-10-108.ec2.internal.warc.gz"} |
http://theoryofcomputing.org/articles/v007a008/ | Volume 7 (2011) Article 8 pp. 119-129
Arithmetic Complexity in Ring Extensions
by
$\newcommand\field{{\mathbb F}}$ Given a polynomial $f$ with coefficients from a field $\field$, is it easier to compute $f$ over an extension ring $R$ than over $\field$? We address this question, and show the following. For every polynomial $f$, there is a noncommutative extension ring $R$ of the field $\field$ such that $\field$ is in the center of $R$ and $f$ has a polynomial-size formula over $R$. On the other hand, if $\field$ is algebraically closed, no commutative extension ring $R$ can reduce formula or circuit complexity of $f$. To complete the picture, we prove that over any field, there exist hard polynomials with zero-one coefficients. (This is a basic theorem, but we could not find it written explicitly.) Finally, we show that low-dimensional extensions are not very helpful in computing polynomials. As a corollary, we obtain that the elementary symmetric polynomials have formulas of size $n^{O(\log \log n)}$ over any field, and that division gates can be efficiently eliminated from circuits, over any field. | {"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": 0.9458327889442444, "perplexity": 155.47602140130365}, "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-2017-30/segments/1500549426086.44/warc/CC-MAIN-20170726082204-20170726102204-00663.warc.gz"} |
https://papers.neurips.cc/paper/2019/hash/04115ec378e476c56d19d827bcf8db56-Abstract.html | #### Authors
Dennis Maximilian Elbrächter, Julius Berner, Philipp Grohs
#### Abstract
Neural network training is usually accomplished by solving a non-convex optimization problem using stochastic gradient descent. Although one optimizes over the networks parameters, the main loss function generally only depends on the realization of the neural network, i.e. the function it computes. Studying the optimization problem over the space of realizations opens up new ways to understand neural network training. In particular, usual loss functions like mean squared error and categorical cross entropy are convex on spaces of neural network realizations, which themselves are non-convex. Approximation capabilities of neural networks can be used to deal with the latter non-convexity, which allows us to establish that for sufficiently large networks local minima of a regularized optimization problem on the realization space are almost optimal. Note, however, that each realization has many different, possibly degenerate, parametrizations. In particular, a local minimum in the parametrization space needs not correspond to a local minimum in the realization space. To establish such a connection, inverse stability of the realization map is required, meaning that proximity of realizations must imply proximity of corresponding parametrizations. We present pathologies which prevent inverse stability in general, and, for shallow networks, proceed to establish a restricted space of parametrizations on which we have inverse stability w.r.t. to a Sobolev norm. Furthermore, we show that by optimizing over such restricted sets, it is still possible to learn any function which can be learned by optimization over unrestricted sets. | {"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": 0.9612032175064087, "perplexity": 481.3687885984483}, "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-2021-31/segments/1627046150264.90/warc/CC-MAIN-20210724094631-20210724124631-00605.warc.gz"} |
https://edge.sagepub.com/banyard2e/student-resources/chapter-5/learning-objectives | # Learning Objectives
Chapter 5
• Be able to distinguish between short- and long-term memory, and be able to describe the differences between the two
• Understand the relation between the process of memory recall and ‘forgetting’
• Be able to describe how memories can be unreliable and distorted | {"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": 0.8713825345039368, "perplexity": 921.5622401313899}, "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-2019-51/segments/1575540525598.55/warc/CC-MAIN-20191209225803-20191210013803-00436.warc.gz"} |
http://www.codingforums.com/archive/index.php/t-71985.html | PDA
View Full Version : asp and IIS
esokol
11-07-2005, 05:51 PM
I hope this is the correct forum to post this. If not, please let me know the correct one.
I have an application which is run under IIS and ServletExec. If I try to put an asp file in any of the directories under the application run by ServletExec, I get the Error 404 when I try to run the asp file.
If I put the asp file in another directory under IIS, I have no problem running the asp file.
The application directory structure is:
c:\devnt2\inetpub\wwwroot\qol14
Putting image.asp anywhere under this directory structure gets the 404.
Putting the image.asp file in the following directory works fine.
c:\devnt2\inetpub\wwwroot\qol14asp
In each case the URL I use is:
http://devnt2/qol14/image.asp
http://devnt2/qol14asp/image.asp
Both directories are defined as allowing scripts to run in IIS.
1. Is this valid -- can I mix asp and jsp under IIS and ServletExec?
2. If it is, what do I need to define in order for it to be able to run the asp file? | {"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": 0.9333634376525879, "perplexity": 2099.0308940051746}, "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-2016-22/segments/1464049277091.27/warc/CC-MAIN-20160524002117-00155-ip-10-185-217-139.ec2.internal.warc.gz"} |
https://astarmathsandphysics.com/university-physics-notes/139-particle-physics/1621-the-four-fundamental-forces.html | ## The Four Fundamental Forces
The properties of the four fundamental forces are summarised in the following table.
Mediating Particle Force Relative Strength Range Name Mass Charge Spin Strong 1 Short Gluon 0? 0 1 Electromagnetic Infinite Photon 0 0 1 Weak Short 80.4, 80.4, 91.2 resp 1 Gravity Infinite Graviton 0 0 2
There are several things to note:
Both gravity and the electromagnetic force have infinite range. The mediating particle of both these forces are stable and do not decay spontaneously. This means that the flux of the force particles through a sphere drawn around the source of the force is a constant no matter how big the sphere, and this implies the force must obey alaw.
The weak force is mediated by massive particles. We may find the approximate range of the weak force using Heisenberg's Uncertainty Principle for virtual particles
We can estimate the range of the force now by assuming the force particles travel at the speed of light for this time.
This approach is general and can be used to estimate the range of forces from the masses of the mediating particle. | {"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": 0.9882899522781372, "perplexity": 382.18595221879787}, "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/1537267156901.47/warc/CC-MAIN-20180921072647-20180921093047-00045.warc.gz"} |
http://physics.stackexchange.com/questions/37764/symmetrizing-the-canonical-energy-momentum-tensor | # Symmetrizing the Canonical Energy-Momentum Tensor
The Canonical energy momentum tensor is given by $$T_{\mu\nu} = \frac{\partial {\cal L}}{\partial (\partial^\mu \phi_s)} \partial_\nu \phi_s - g_{\mu\nu} {\cal L}$$ A priori, there is no reason to believe that the EM tensor above is symmetric. To symmetrize it we do the following trick.
To any EM tensor we can add the following term without changing its divergence and the conserved charges: $${\tilde T}_{\mu\nu} = T_{\mu\nu} + \partial^\beta \chi_{\beta\mu\nu}$$ where $\chi_{\beta\mu\nu} = - \chi_{\mu\beta\nu}$. The antisymmetry of $\chi$ in its $\mu\beta$ indices implies that ${\tilde T}_{\mu\nu}$ is conserved. Also, all the conserved charges stay the same.
Now even though $T_{\mu\nu}$ is not a symmetric tensor, it is possible to choose $\chi_{\beta\mu\nu}$ in such a way so as to make ${\tilde T}_{\mu\nu}$ symmetric. It can be shown that choosing
$$\chi_{\lambda\mu\nu} = - \frac{i}{2}\left[ \frac{\delta {\cal L}}{\delta (\partial^\mu \phi_r) } (I_{\nu\lambda})_{rs} \phi_s + \frac{\delta {\cal L}}{\delta (\partial^\lambda \phi_r) } (I_{\mu\nu})_{rs} \phi_s + \frac{\delta {\cal L} }{\delta (\partial^\nu \phi_r) } (I_{\mu\lambda})_{rs} \phi_s \right]$$ makes the new EM tensor symmetric. Here $(I_{\mu\nu})_{rs}$ is the representation of the Lorentz Algebra under which the fields $\phi_s$ transform.
Here's my question - Is it possible to obtain the symmetric EM tensor directly from variational principles by adding a total derivative term to the Lagrangian. In other words, by shifting ${\cal L} \to {\cal L} + \partial_\mu X^\mu$, and choosing $X^\mu$ appropriately, can we exactly get the shift in the EM tensor required, in order to make the canonical EM tensor symmetric?
What I've done so far - It is possible to show that under a shift in the Lagrangian by a total derivative, one shifts the EM tensor by $T_{\mu\nu} \to T_{\mu\nu} + \partial^\lambda \chi_{\lambda\mu\nu}$ where
$$\chi_{\lambda\mu\nu} = \frac{1}{2} \frac{\delta X_\lambda}{\delta (\partial^\mu \phi_r)} \partial_\nu \phi_r - \frac{1}{2} \frac{ \delta X_\mu }{\delta (\partial_\lambda \phi_r)} \partial_\nu \phi_r + X_\mu g_{\lambda\nu} - X_\lambda g_{\mu\nu} \,.$$
What I wish to do next - I now have a differential equation that I wish to solve:
\begin{align} &\frac{1}{2} \frac{\delta X_\lambda}{\delta (\partial^\mu \phi_r)} \partial_\nu \phi_r - \frac{1}{2} \frac{ \delta X_\mu }{\delta (\partial_\lambda \phi_r)} \partial_\nu \phi_r + X_\mu g_{\lambda\nu} - X_\lambda g_{\mu\nu} \\ &~~~~~~= - \frac{i}{2}\left[ \frac{\delta {\cal L}}{\delta (\partial^\mu \phi_r) } (I_{\nu\lambda})_{rs} \phi_s + \frac{\delta {\cal L}}{\delta (\partial^\lambda \phi_r) } (I_{\mu\nu})_{rs} \phi_s + \frac{\delta {\cal L} }{\delta (\partial^\nu \phi_r) } (I_{\mu\lambda})_{rs} \phi_s \right] \,. \end{align}
Any ideas on how to solve this?
-
Related: physics.stackexchange.com/q/27048/2451 and links therein. – Qmechanic Sep 19 '12 at 14:57
Symmetrizability is equivalent with Lorentz invariance. Thus you need to assume that in your arguments. – Arnold Neumaier Sep 19 '12 at 17:12
Can you explain that a bit more? I don't undestand what you're trying to say. Thanks! – Prahar Sep 23 '12 at 17:00
Without the assumption of Lorentz invariance of the action, there is no symmetric e/m tensor, and the standard recipe fails. Lorentz invariance gives you additional properties that you must exploit in your derivation; otherwise you will not be able to arrive at the conclusion (since it might not hold). - If you reply to a comment, you should mention the name, like in @Prahar, so that the original commenter is informed. I noticed your comment only by chance (and hence very late). – Arnold Neumaier Nov 1 '12 at 17:58
Is it possible to obtain the symmetric energy-momentum tensor directly from variational principles by adding a total derivative term to the Lagrangian? In other words, by shifting $\Delta{\cal L}=d_\mu X^\mu$, and choosing $X^\mu$ appropriately, can we exactly get the shift in the energy-momentum tensor required, in order to make the canonical energy-momentum tensor symmetric?
No, that project is doomed already for E&M with the Maxwell Lagrangian density
$$\tag{1} {\cal L}_0~=~ -\frac{1}{4}F_{\mu\nu}F^{\mu\nu}$$
with
$$\tag{2} F_{\mu\nu}~=~A_{\nu,\mu}-A_{\mu,\nu}, \qquad \frac{\partial{\cal L}_0}{\partial A_{\mu,\nu}}~=~ F^{\mu\nu}.$$
In E&M, the canonical stress energy tensor is$^1$
$$\tag{3} \Theta^{\mu}{}_{\nu}~=~\delta^{\mu}_{\nu}{\cal L}_0+F^{\mu\alpha}A_{\alpha,\nu},$$
while the symmetric stress energy tensor is
$$\tag{4} T^{\mu}{}_{\nu}~=~\delta^{\mu}_{\nu}{\cal L}_0+F^{\mu\alpha}F_{\nu\alpha}.$$
[Here we are using $(-,+,+,+)$ Minkowski sign convention.] So the difference is$^2$
$$T^{\mu}{}_{\nu} -\Theta^{\mu}{}_{\nu}~=~ F^{\mu\alpha}A_{\nu,\alpha}$$ $$~\stackrel{?}{=}~\delta^{\mu}_{\nu}\Delta{\cal L}+\left(-\frac{\partial\Delta{\cal L}}{\partial A_{\alpha,\mu}}+ d_{\beta}\frac{\partial\Delta{\cal L}}{\partial A_{\alpha,\mu\beta}}\right) A_{\alpha,\nu} -\frac{\partial\Delta{\cal L}}{\partial A_{\alpha,\mu\beta}}A_{\alpha,\nu\beta},\tag{5}$$
for some $\Delta{\cal L}=d_\mu X^\mu$, where $X^\mu$ depends on $A$ and $\partial A$. Then$^3$
$$\tag{6} X^{\mu}~=~ a A^{\mu} A^{\nu}_{,\nu} + b A^{\nu} A^{\mu}_{,\nu} + c A^{\lambda} A_{\lambda,\nu}\eta^{\mu\nu},$$
for some constants $a,b,c$; and
$$\Delta{\cal L}~=~d_\mu X^\mu~$$ $$~=~a (A^{\mu}_{,\mu})^2 + b A^{\nu}_{,\mu} A^{\mu}_{,\nu} + (a+b) A^{\mu} A^{\nu}_{,\nu\mu} + c A^{\lambda}_{,\mu} A_{\lambda,\nu} \eta^{\mu\nu} + c A^{\lambda}A_{\lambda,\mu\nu} \eta^{\mu\nu}.\tag{7}$$
It is not hard to see that that is not possible.
--
$^1$ Some references, e.g. Weinberg QFT, has the opposite notational conventions for $T$ and $\Theta$.
$^2$ In formula (5) we have neglected terms in $\Delta{\cal L}$ that depends on $\partial^3A$, $\partial^4A$, $\partial^5A$, $\ldots$, etc. Such terms are excluded for various reasons.
$^3$ In retrospect, this answer completely shares the premise/ideology/program/conclusion of this Phys.SE post.
-
The shifted term will not take the form you have written. In general (and this will be required) $\Delta {\cal L}$ will contain 2 derivatives of the fields since $X^\mu$ will contain upto one derivative. Then, the original definition of the canonical stress tensor must be modified. If you look at my question, I have considered this issue and derived the EXACT shift in $\Theta$ that is obtained from a $\Delta {\cal L} = \partial_\mu X^\mu$. – Prahar Jan 2 '15 at 15:32
Your answer is valid if we assume that $X^\mu$ contains only fields but no derivatives. In this case, I already see that there is no solution. However, this is a several restricting solution. I believe I've set the problem up to find a solution to the last differential equation in my question. If look at that equation, it is clear that there is no solution where $X^\mu$ only contains fields. It must be chosen to contain first derivatives of fields as well. – Prahar Jan 2 '15 at 15:34
I updated the answer. – Qmechanic Jan 2 '15 at 15:40
I will try to obtain the result using another way. It is well known that the Lagrangian density determined up to divergence of some four-vector $\mathcal{L}(x)\to \mathcal{L}(x)+\partial_\mu\psi^\mu(x)$ Let's understand what contribution the second term gives in the energy-momentum tensor. $$\hat T^\nu _\mu=\partial_\rho\Bigr(\frac{\delta {\cal\psi^\rho }}{\delta (\partial^\mu \phi_s)} \partial^\nu \phi_s - g_\mu^\nu {\cal \psi^\rho}\Bigr)=\partial_\rho \chi^{\rho\nu}_ \mu$$ $\psi^\rho$ is arbitrary four-vector, contains in $\phi_s$ and $\partial^\rho\phi_r$. Set that $\psi^\rho=f(\phi^2)\phi_r\partial^\rho\phi_r$.(If I require that the Lagrangian dependence only $\phi_r$ and first derivative of it. It will be general form) We obtain the following result $$\chi^{\rho\nu}_ \mu =g_\mu^\rho {\cal \psi^\nu}-g_\mu^\nu {\cal \psi^\rho}$$ where $g^\mu_\rho=\delta^\mu_\rho$ is a Kronecker symbol. Thus we obtain that energy momentum tensor defined up to such term $T^{\mu\nu}\to T^{\mu\nu}+\partial_\rho\chi^{\rho\mu\nu}$ where $\chi^{\rho\mu\nu}=-\chi^{\mu\rho\nu}$. This fact is a consequence of Lagrangian feature(The Lagrangian density determined up to divergence of some four-vector $\mathcal{L}(x)\to \mathcal{L}(x)+\partial_\mu\psi^\mu(x)$).
Edit
Using previous formula it is easy to obtain that $$\chi^{\mu\rho\nu} =g^{\mu\rho} {\cal \psi^\nu}-g^{\mu\nu} {\cal \psi^\rho}$$ After contraction with $g_{\mu\rho}$ we obtan that $$\psi^\nu=\frac{1}{D-1}\chi^{\mu\rho\nu}g_{\mu\rho}$$ where $D$-is dimensionality of space.
In spite of the Lagrangian contain second derivatives, all of it is true.Because it Lagrangian differ only for full derivative. If you interesting in this question, you should write General relativity. Because action of General relativity which contain Riemann curvature tensor(which contain second derivatives).
-
Instead of $g^{\nu}_\mu$ and similar symbols, did you mean to write $\delta^\nu_\mu$ and similar symbols (here, $\delta$ is the Kronecker delta)? – Danu Dec 30 '14 at 17:55
@Peter If $\psi^\rho$ contains $\partial^\rho \phi_r$, then the Lagrangian contains upto two derivatives of the fields (since it contains $\partial_\mu \psi^\mu$). One then needs to completely modify the standard definitions of the stress tensor since they all assume dependence upto 1 derivative only. I have considered these issues and the results are in my question already. – Prahar Dec 30 '14 at 23:16
@Peter - Abovementioned issues aside, I agree with your answer, but it does not answer my question. I know that the stress-tensor is determined upto additive terms of the form $\partial_\rho \chi^{\rho\mu\nu}$ and I also know that there always exists a choice of $\chi^{\rho\mu\nu}$ to make it symmetric. My question is - Is there a choice of $\psi^\rho$ such that the corresponding "canonical stress-tensor" is symmetric. – Prahar Dec 30 '14 at 23:20
@Peter - Let me pose the question in a different (possibly more general) way. I'll use the notations in your answer. Given a $\psi^\rho$, one can always find the corresponding $\chi^{\rho\mu\nu}$ (I have done this already in the question). The question is - Can the reverse process be achieved, i.e. Given any $\chi^{\rho\mu\nu}$, is it true that it can be derived from some $\psi^\rho$. If true, can you explicitly construct such a $\psi^\rho$. – Prahar Dec 30 '14 at 23:23
I edit a little bit. I tried to answer in our questions. – Peter Jan 6 '15 at 10:35 | {"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": 1, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9800688624382019, "perplexity": 344.72770909502196}, "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-2016-07/segments/1454701155060.45/warc/CC-MAIN-20160205193915-00208-ip-10-236-182-209.ec2.internal.warc.gz"} |
https://repository.kaust.edu.sa/handle/10754/665456 | Except where otherwise noted, this item's license is described as This article is Open Access. Archived with thanks to Journal of Materials Chemistry A. | {"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": 0.9272574186325073, "perplexity": 2195.558963784103}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-43/segments/1634323585280.84/warc/CC-MAIN-20211019171139-20211019201139-00440.warc.gz"} |
https://www.physicsforums.com/threads/thinking-outside-euclids-box.99777/ | # Thinking Outside Euclid's Box
1. Nov 13, 2005
### reasonmclucus
The following is not intended as a definitive explanation of other dimensions, but a suggestion for other ways of looking at the idea of physical dimensions.
Supporters of string theory occasionally remark that the reason we cannot "see" other dimensions is that they must be very small. Such comments represent thinking inside Euclid's geometric box. We have been conditioned to believe that physical dimensions are limited to the 3 dimension of Euclidean geometry (length, width and height) and maybe time. But, are these the only characteristics of physical reality that can qualify as dimensions in the mathematical sense.
A dimension is basically a characteristic or variable. Length, width and height are not the only variables that can be represented geometrically. Social as well as physical data can be represented as 2-dimensional images. Social data may not qualify as a physical dimension of reality, but what about other physical data?
How many dimensions do we actually "see"? Each eye sends a flat image to the brain which then interprets the visual as showing length, width and height using differences between the two images and certain visual clues such as closer objects appearing larger to add a dimension of depth to the image.
However, length, width and height are not the only physical characteristics the eyes perceive. The eyes also see variations in color. Color is a function of the electromagnetic characteristics of sources of emitted or reflected light.
Objects of the same length, width and height may be of different colors. The existence of color implies the existence of other physical dimensions.
Euclidean geometry theory begins with points which have zero dimensions. A point is at a specified location or there is no point at the location. If physical reality had only 3 dimensions there would be something (such as a quark) at a location or there would be nothing. If the quark can have more than one possible characteristic, then quarks are variables and imply the existence of other dimensions. Variations in the types of quarks could be additional physical dimensions of reality. Perhaps gravity is another physical dimension, or perhaps there is an energy dimension.
Dimensions are sometimes viewed as having a heirarchy with additional dimensions being higher than the others. However, math theory allows for separate spaces that may intersect in one or more dimensions and we perceive reality as having 3 dimensions because each of the spaces shares the dimensions of length,width and height. For example, each flavor of quark might have its own space in which there is a quark of that flavor at a location or there isn't a quark at that location. There might be a gravity space. Perhaps the weak force and strong force are two dimensions in a force space.
The intersection of these different spaces might be static or dynamic. A dynamic intersection could explain certain aspects of quantum physics. The apparent winking in and out of particles could result from those particles being within the area of intersection at one time but not at another. Tunneling might result from shift in the locus of the intersection.
2. Nov 14, 2005
### force5
When I think of the three dimensions(Length, width and height), I think of a cube. But when I think of dimensions as relates to a sphere, it takes on a whole new relationship. Since events occuring at the level in question, the conservation of angular momentum provides a number of considerations when addressing the merging or convergence of light waves produced by any number of sources. In other words, is it possible that additional dimensions only become necessary in order to satisfy the math requirements dealing at the sub-atomic level.
Just my thought....John
3. Nov 14, 2005
### mccrone
When you think of a cube, that is flat space. When you think of a sphere, that is curved space - so curled that it makes a circle. Add in open or hyperbolic curvature, where all paths diverge, and you have covered all three possibilities.
Cheers - John McCrone.
Similar Discussions: Thinking Outside Euclid's Box | {"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": 0.826556921005249, "perplexity": 579.1941334580149}, "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/1508187823482.25/warc/CC-MAIN-20171019231858-20171020011858-00810.warc.gz"} |
https://www.physicsforums.com/threads/fourier-transform-solution-to-electrostatics-poisson-equation.293550/ | # Fourier transform solution to electrostatics Poisson equation?
1. Feb 18, 2009
### Peeter
Am just playing around, and
following examples of Fourier transform solutions of the heat equation, tried the same thing for
the electrostatics Poisson equation
$$\nabla^2 \phi &= -\rho/\epsilon_0 \\$$
With fourier transform pairs
\begin{align*} \hat{f}(\mathbf{k}) &= \frac{1}{(\sqrt{2\pi})^3} \iiint f(\mathbf{x}) e^{-i \mathbf{k} \cdot \mathbf{x} } d^3 x \\ {f}(\mathbf{x}) &= \frac{1}{(\sqrt{2\pi})^3} \iiint \hat{f}(\mathbf{k}) e^{i \mathbf{k} \cdot \mathbf{x} } d^3 k \\ \end{align*}
one gets
\begin{align*} \phi(\mathbf{x}) &= \frac{1}{\epsilon_0} \int \rho(\mathbf{x}') G(\mathbf{x-x'}) d^3 x' \\ G(\mathbf{x}) &= \frac{1}{(2 \pi)^3} \iiint \frac{1}{\mathbf{k}^2} e^{ i \mathbf{k} \cdot \mathbf{x} } d^3 k \end{align*}
Now it seems to me that this integral $G$ only has to be evaluated around a small neighbourhood of the origin. For example if one evaluates one of
the
integrals
$$\int_{-\infty}^\infty \frac{1}{{k_1}^2 + {k_2}^2 + {k_3}^3 } e^{ i k_1 x_1 } dk_1$$
using a an upper half plane contour the result is zero unless $k_2 = k_3 = 0$. So one is left with something loosely like
$$G(\mathbf{x}) &= \lim_{\epsilon \rightarrow 0} \frac{1}{(2 \pi)^3} \int_{k_1 = -\epsilon}^{\epsilon} dk_1 \int_{k_2 = -\epsilon}^{\epsilon} dk_2 \int_{k_3 = -\epsilon}^{\epsilon} dk_3 \frac{1}{\mathbf{k}^2} e^{ i \mathbf{k} \cdot \mathbf{x} }$$
However, from electrostatics we also know that the solution to the Poission equation means that $G(\mathbf{x}) = \frac{1}{4\pi\lvert{\mathbf{x}}\rvert}$.
Does anybody know of a technique that would reduce the integral limit expression above for $G$ to the $1/x$ form? Am thinking something residue related, but I'm a bit rusty with my complex variables and how exactly to procede isn't obvious.
Last edited: Feb 18, 2009
2. Feb 19, 2009
### Peeter
found the answer in a book recently purchased, but not yet read (Mathematics of Classical and Quantum Physics). They cleverly introduce a pole in the upper half plane by evaluating
$$\iiint \frac{1}{\mathbf{k}^2 + a^2} e^{ i \mathbf{k} \cdot \mathbf{x} } d^3 k$$
After a change to spherical polar coordinates, that new integral can be evaluated, and the Poisson Green's function follows by letting $a$ tend to zero.
3. Sep 13, 2010
### gn0m0n
I still don't see how to take these integrals. I've tried it a few ways.
4. Sep 13, 2010
### Peeter
Do you mean you don't see how to evaluate:
$$\iiint \frac{1}{\mathbf{k}^2 + a^2} e^{ i \mathbf{k} \cdot \mathbf{x} } d^3 k$$
or, you don't see how to get to that point by taking the Fourier transforms?
5. Sep 13, 2010
### gn0m0n
Nevermind - I got it.
6. Sep 13, 2010
### gn0m0n
Hi, yes, it was taking that integral that I was having trouble with. Thanks.
7. Mar 1, 2011
### ajit_up
hi
I am trying to solve the electrostatic poisson's equation mentioned in the first post in 2D using Discrete Fourier Transform (I am using fftw3 library and REDFT10 / REDFT01 transforms). For my problem, I have charge densities given in a 2D plane at discrete points and I have to find out potential at those points. The range of solution, and boundary condition are :
0 < x < L1 and 0 < y < L2 and Dirichlet boundary condition.
I have taken phi to be 2D - Gaussian function (peaked at center of 2D grid) with sigma = 1.0, phi0 = 10.0, so that I am able to check the results analytically too.
I am writing below the steps:
1. I take 2D - DFT of charge densitiy
2. Solve the equation in fourier space for phi in fourier space
3. take inverse 2D - DFT of phi to get phi in x-y plane
When I compare my computational result with analytical result, they do not match.
Can anybody help me in understanding the reason and finding the right way to solve Poisson's equation using DFTs.
thanks | {"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": 0.9970118403434753, "perplexity": 1147.2756439102066}, "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/1508187826049.46/warc/CC-MAIN-20171023130351-20171023150351-00709.warc.gz"} |
http://www.apniphysics.com/transverse-nature-of-electromagnetic-waves/ | # Transverse Nature of Electromagnetic Waves
In this lecture Electromagnetic waves and their transverse nature are discussed. In an electromagnetic wave the electric field as well as magnetic field components vibrate perpendicular to each other, also the wave propagate perpendicular to both. If electric field is along the Y direction, Magnetic field along Z axis the direction of propagation will be the X direction.
For more details, see here: | {"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": 0.8263486623764038, "perplexity": 423.7236560006788}, "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-2018-17/segments/1524125945724.44/warc/CC-MAIN-20180423031429-20180423051429-00123.warc.gz"} |
https://math.stackexchange.com/questions/800321/showing-equality-of-primitive-roots-with-quadratic-non-residues | # Showing equality of primitive roots with quadratic non-residues.
Suppose that $p$ and $q = 2p + 1$ are both odd primes. Show that the $p − 1$ primitive roots of $q$ are precisely the quadratic non-residues of $q$, other than the quadratic non-residue $2p$ of $q$.
I think I probably need to use the fact that $q$ is congruent to $3 \rm\, mod\, 4$, but I've been fiddling around with definitions and can't quite seem to get anywhere. Any help would be greatly appreciated. Thanks!
You are possibly familiar with the result that says that if $n$ has a primitive root, then $n$ has $\varphi(\varphi(n))$ primitive roots.
In the case $q=2p+1$, the number of primitive roots of $q$ is therefore $\varphi(2p)$, which is $p-1$. But $p-1=\frac{q-1}{2}-1$. Since there are $\frac{q-1}{2}$ quadratic non-residues of $q$, and every primitive root is a non-residue, it follows that all but one of the non-residues is a primitive root.
Since $q$ is of the form $4k+3$, we know that $-1$ is a non-residue. But $-1$ is not a primitive root of $q$, since $q\gt 3$. So $-1$ is the only non-residue which is not a primitive root.
HINT: The multiplicative group of non-zero residues mod $q$ is cyclic and has order $2p$. The possible orders of elements are $1,2,p,2p$ since $p$ is prime [this is where $p$ being prime comes in]. You need to work out which elements have which order, which is simply an analysis of the cyclic group. | {"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": 0.9865077137947083, "perplexity": 62.67723836340758}, "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-2019-39/segments/1568514574265.76/warc/CC-MAIN-20190921043014-20190921065014-00475.warc.gz"} |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_Chem1_(Lower)/08%3A_Solutions/8.10%3A_Ions_and_Electrolytes/8.10.9B%3A_8.10.9B%3A_The_nature_of_ions_in_aqueous_solution | 8.10.9B: The nature of ions in aqueous solution
$$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$ $$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$$$\newcommand{\AA}{\unicode[.8,0]{x212B}}$$
The kinds of ions we will consider in this lesson are mostly those found in solutions of common acids or salts. As is evident from the image below, most of the metallic elements form monatomic cations, but the number of monatomic anions is much smaller. This reflects the fact that many single-atom anions such as hydride H, oxide O2–, sulfide S2– and those in Groups 15 and 16, are unstable in (i.e., react with) water, and their major forms are those in which they are combined with other elements, particularly oxygen. Some of the more familiar oxyanions are hydroxide OH, carbonate CO32–, nitrate NO3, sulfate SO42–, chlorate ClO42–, and arsenate AsO42–.
Conductivity water" with κ = 0.043 × 10–6 S cm–1 at 18°C. Ordinary distilled water in equilibrium with atmospheric CO2 has a conductivity that is 16 times greater.
It is now known that ordinary distillation cannot entirely remove all impurities from water. Ionic impurities get entrained into the fog created by breaking bubbles and are carried over into the distillate by capillary flow along the walls of the apparatus. Organic materials tend to be steam-volatile ("steam-distilled").
The best current practice is to employ a special still made of fused silica in which the water is volatilized from its surface without boiling. Complete removal of organic materials is accomplished by passing the water vapor through a column packed with platinum gauze heated to around 800°C through which pure oxygen gas is passed to ensure complete oxidation of carbon compounds.
Conductance measurements are widely used to gauge water quality, especially in industrial settings in which concentrations of dissolved solids must be monitored in order to schedule maintenance of boilers and cooling towers.
The conductance of a solution depends on 1) the concentration of the ions it contains, 2) on the number of charges carried by each ion, and 3) on the mobilities of these ions. The latter term refers to the ability of the ion to make its way through the solution, either by ordinary thermal diffusion or in response to an electric potential gradient.
The first step in comparing the conductances of different solutes is to reduce them to a common concentration. For this, we define the conductance per unit concentration which is known as the molar conductivity, denoted by the upper-case Greek lambda:
$Λ = \dfrac{κ}{c}$
When κ is expressed in S cm–1, C should be in mol cm–3, so Λ will have the units S cm2. This is best visualized as the conductance of a cell having 1-cm2 electrodes spaced 1 cm apart — that is, of a 1 cm cube of solution. But because chemists generally prefer to express concentrations in mol L–1 or mol dm–3 (mol/1000 cm3) , it is common to write the expression for molar conductivity as
$Λ = \dfrac{1000κ}{c}$
electrodes, separated again by 1 cm.
But if c is the concentration in moles per liter, this will still not fairly compare two salts having different stoichiometries, such as AgNO3 and FeCl3, for example. If we assume that both salts dissociate completely in solution, each mole of AgNO3 yields two moles of charges, while FeCl3 releases six (i.e., one Fe3+ ion, and three Cl ions.) So if one neglects the [rather small] differences in the ionic mobilities, the molar conductivity of FeCl3 would be three times that of AgNO3.
The most obvious way of getting around this is to note that one mole of a 1:1 salt such as AgNO3 is "equivalent" (in this sense) to 1/3 of a mole of FeCl3, and of ½ a mole of MgBr2. To find the number of equivalents that correspond a given quantity of a salt, just divide the number of moles by the total number of positive charges in the formula unit. (If you like, you can divide by the number of negative charges instead; because these substances are electrically neutral, the numbers will be identical.)
We can refer to equivalent concentrations of individual ions as well as of neutral salts. Also, since acids can be regarded as salts of H+, we can apply the concept to them; thus a 1M L–1 solution of sulfuric acid H2SO4 has a concentration of 2 eq L–1.
The following diagram summarizes the relation between moles and equivalents for CuCl2:
Example $$\PageIndex{1}$$
What is the concentration, in equivalents per liter, of a solution made by dissolving 4.2 g of chromium(III) sulfate pentahydrate Cr2(SO4)3·5H2O in sufficient water to give a total volume of 500 mL? (The molar mass of the hydrate is 482 g)
Solution
Assume that the salt dissociates into 6 positive charges and 6 negative charges.
• Number of moles of the salt: (4.2 g) / (482 g mol–1) = 0.00871 mol
• Number of equivalents: (.00871 mol) / (6 eq mol–1) = 0.00146 eq
• Equivalent concentration: (0.00146 eq) / (0.5 L) = 0.00290 eq L–1
The concept of equivalent concentration allows us to compare the conductances of different salts in a meaningful way. Equivalent conductivity is defined similarly to molar conductivity
$Λ = \dfrac{κ}{c}$
except that the concentration term is now expressed in equivalents per liter instead of moles per liter. (In other words, the equivalent conductivity of an electrolyte is the conductance per equivalent per liter.)
This page titled 8.10.9B: The nature of ions in aqueous solution is shared under a CC BY 3.0 license and was authored, remixed, and/or curated by Stephen Lower via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request. | {"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": 0.8427336812019348, "perplexity": 1774.4037943264536}, "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-2023-06/segments/1674764500080.82/warc/CC-MAIN-20230204012622-20230204042622-00686.warc.gz"} |
https://zief0002.github.io/statistical-thinking/external-validity-evidence-and-random-sampling.html | # External Validity Evidence and Random Sampling
In statistical inference, generalization refers to the process of using sample data to draw conclusions about the larger population from which the sample was drawn. The sample data provides statisticians with an estimate of the exact “truth” about the population. For example, data collected from 1,000 Americans about their voting preferences may be used to infer the voting preferences of Americans in general. Statisticians are typically concerned with making inferences about some summary measure of the population, a mean or population percentage. (Some vocabulary: Population summary measures are called parameters. Sample estimates of parameters are referred to as statistics.)
How useful is a sample statistic when it comes to estimating a population parameter? Can we draw reasonable inference about a population from sample data? This question is at the heart of weighing evidence about external validity. External validity is the degree to which generalizations to the larger population are accurate and meaningful.
There are two statistical aspects we need to consider when we evaluate evidence for external validity: sampling variation and bias. Sampling variation is the idea that statistics from different samples vary. For example, to use the earlier example, different samples of 1,000 Americans would produce different estimates of voting preferences. This variation needs to be accounted for when estimates are given. One way of thinking about sampling variation is to map it to the quality of precision. (We will focus more on this in Unit 5.)
The second statistical characteristic we need to attend to is bias. Statistical bias is when sample statistics differ systematically from the population parameter. The key here is the word “systematically”. This implies that there is something in the underlying process (aside from random variation) that is affecting the estimation process.
### Statistical Bias
To help you think about bias, imagine a person, Arthur Dent, has lost his keys. The actual location of the keys, the Library, is akin to the population parameter. Arthur believes he lost his keys at the Supermarket and searches several places around the Supermarket. The locations where Arthur searches are like sample statistics.
Figure 5 is a metaphor for the concept of statistical bias. Arthur’s search locations (sample statistics) are systematically in the wrong place. On average, where Arthur searched (the middle of the yellow circle) is not the actual location of the keys. Compare this with the search locations in Figure 6.
Figure 6 is a metaphor for unbiasedness. On average, where Arthur searched is the location of the keys. There are a couple of other concepts that this metaphor can help us think about.
1. Even in Figure 6, none of the actual search locations were right at the keys. Some of the locations were too far to the left of the keys, and others were too far to the right of the keys. However, ON AVERAGE, the search locations “found” the keys. The way we define unbiased is that the AVERAGE of the statistics is at the population parameter.
2. Average has nothing to do with the size of the yellow circle. (The size of the circle is related to the amount of sampling variation, a concept we will deal with in Unit 5.) The two figures below also illustrate unbiasedness (left) and bias (right), despite the size of the yellow circle.
1. The last concept about bias to point out is that bias (or unbiasedness) is a property of the sampling method. The reason the search locations were not in the right place is because the method Arthur used to pick the search locations was biased. He thought he lost his keys in the Supermarket, so that is where he looked.
An unbiased sampling method is random sampling. Random sampling uses chance to select the sampling units (participants) from the larger population. When random sampling has been employed in a study, the unbiasedness of the sampling method is strong evidence for external validity; we have a much higher belief in generalizations to the larger population. In our validity barometer we would be in the upper-third (depending on other factors such as sample size).
In this course, we will discuss and utilize simple random sampling. To draw a simple random sample we need a list of EVERY member of the population. This list is called the sampling frame. (Obtaining a sampling frame can be very difficult. Try obtaining a list of everyone who lives in the United States!) Then we employ randomness to draw out sampling units, with the caveat that each unit in the sampling frame has an equal chance of being drawn. | {"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": 0.8616062998771667, "perplexity": 695.9674911829065}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-47/segments/1573496668716.22/warc/CC-MAIN-20191115222436-20191116010436-00040.warc.gz"} |
http://mathhelpforum.com/calculus/115681-l-hospital-s-rule.html | # Math Help - L'Hospital's rule
1. ## L'Hospital's rule
I'm having a difficulty differentiating this using L'Hospital's rule:
lim x->(infinite) xsin(pi/x)
Applying L'Hospital's rule and differentiating I got:
lim x->(infinite) xcos(pi/x) + sin(pi/x)
and that equals = (infinite)(1) + 0 = infinite
The correct answer in my textbook is just "pi"
Can someone tell me what I'm doing wrong?
2. If your function is: $x*sin\frac{\pi}{x}$ then this isn't in the form of 0/0 or infinity/infinity and therefore you can't use L'Hopital's rule.
3. Well, to use L'Hospital you must have an indefinite form. Second you differentiate each part seperately. Make sure you check what the indefinite forms are.
4. Originally Posted by Alterah
If your function is: $x*sin\frac{\pi}{x}$ then this isn't in the form of 0/0 or infinity/infinity and therefore you can't use L'Hopital's rule.
Well, then can you explain why the answer is pi?
5. But the function is of indeterminate form. I.e, $0*\infty$ so we can rewrite the limit so it is of the form $\frac{0}{0}$
$x\sin{\frac{\pi}{x}} = \frac{\sin{\frac{\pi}{x}}}{\frac{1}{x}}$
we see $\lim_{x\rightarrow\infty} \frac{\sin{\frac{\pi}{x}}}{\frac{1}{x}} = \frac{0}{0}$
apply L'Hospital's Rule
$\lim_{x\rightarrow\infty} \frac{\sin{\frac{\pi}{x}}}{\frac{1}{x}} =
\lim_{x\rightarrow\infty} \frac{\frac{-\pi}{x^{2}}\cos{\frac{\pi}{x}}}{\frac{-1}{x^{2}}} = \lim_{x\rightarrow\infty} \pi\cos{\frac{\pi}{x}}$
Since $\lim_{x\rightarrow\infty} \frac{1}{x} = 0$
$\lim_{x\rightarrow\infty} \pi\cos{\frac{\pi}{x}} = 1*\pi$
6. Haven, you beat me to it. Took me a second to realize that it was actually an indeterminate form.
7. Originally Posted by Haven
But the function is of indeterminate form. I.e, $0*\infty$ so we can rewrite the limit so it is of the form $\frac{0}{0}$
$x\sin{\frac{\pi}{x}} = \frac{\sin{\frac{\pi}{x}}}{\frac{1}{x}}$
we see $\lim_{x\rightarrow\infty} \frac{\sin{\frac{\pi}{x}}}{\frac{1}{x}} = \frac{0}{0}$
apply L'Hospital's Rule
$\lim_{x\rightarrow\infty} \frac{\sin{\frac{\pi}{x}}}{\frac{1}{x}} =
\lim_{x\rightarrow\infty} \frac{\frac{-\pi}{x^{2}}\cos{\frac{\pi}{x}}}{\frac{-1}{x^{2}}} = \lim_{x\rightarrow\infty} \pi\cos{\frac{\pi}{x}}$
Since $\lim_{x\rightarrow\infty} \frac{1}{x} = 0$
$\lim_{x\rightarrow\infty} \pi\cos{\frac{\pi}{x}} = 1*\pi$
The problem wasn't but could be transformed into indeterminate form. I was hoping the OP would see how to transform it, and then solve the problem.
8. Originally Posted by lvleph
The problem wasn't but could be transformed into indeterminate form. I was hoping the OP would see how to transform it, and then solve the problem.
Thanks man. Didn't know you had to transform it like that. I've just started learning this subject.
9. Originally Posted by Eternal
Thanks man. Didn't know you had to transform it like that. I've just started learning this subject.
Notice that the problem is exactly the same, but has been manipulated so that it looks different and you get one of the indeterminate forms.
10. Another question I'm confused on:
lim x->0+ Sin(x)lnx
How would I solve this? ln(0) is undefined...
11. Make the substitution $\ln x = u$. Then we have $\lim_{u \to -\infty}\sin (e^{u}) u$. | {"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": 18, "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": 0.9369768500328064, "perplexity": 688.4853349317019}, "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/1435375095494.6/warc/CC-MAIN-20150627031815-00288-ip-10-179-60-89.ec2.internal.warc.gz"} |
https://flatearth.ws/airplane | # Airplane and Earth’s Rotation
An airplane exists on Earth. It has the same motion as the surface of the Earth, including the Earth’s rotating motion. The motion of the airplane itself is measured relative to the surface of the Earth.
Flat-Earthers think that due to Earth’s rotation, an airplane should not be able to fly east or land. They are wrong. The speed of an airplane is relative to the surface. It retains the motion of the Earth and does not need to chase Earth’s rotation.
Newton’s first law of motion says:
“An object either remains at rest or continues to move at a constant velocity, unless acted upon by a force.”
When the airplane is parked on the ground, it already in motion at the same velocity as the rest of the Earth. During take-off, the airplane would retain the speed of the Earth’s surface, and only changes its speed relative to the surface.
The atmosphere is part of the Earth and practically moving at the same speed as Earth’s surface. The airplane propels itself against Earth’s atmosphere. It is moving relative to the atmosphere, and consequently, Earth’s surface. | {"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": 0.9209122657775879, "perplexity": 214.49235900042004}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-47/segments/1573496672313.95/warc/CC-MAIN-20191123005913-20191123034913-00104.warc.gz"} |
http://www.intmath.com/functions-and-graphs/ans-2a.php?a=1 | Domain: The function
f(t)=1/(t+2)
is not defined for t = -2, as this value would result in division by zero. (There would be a 0 on the bottom of the fraction.)
Hence the domain of f(t) is
"all real numbers except -2"
Range: No matter how large or small t becomes, f(t) will never be equal to zero.
[Why? If we try to solve the equation for 0, this is what happens:
0=1/(t+2)
Multiply both sides by (t + 2) and we get
0 = 1
This is impossible.]
So the range of f(t) is
"all real numbers except zero".
We can see in the graph that the function is not defined for t = -2 and that the function (the y-values) takes all values except 0.
Domain: All t ≠ -2
Range: All f(t) ≠ 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": 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": 0.8557174801826477, "perplexity": 629.0567043412332}, "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/1476988719033.33/warc/CC-MAIN-20161020183839-00418-ip-10-171-6-4.ec2.internal.warc.gz"} |
http://math.stackexchange.com/questions/124650/showing-equality-almost-surely-by-means-of-conditional-expectations | # Showing equality almost surely by means of conditional expectations.
Suppose that $U$ and $V$ are random variables. Then I want to show that if $$E[g(U)\mid V]=g(V),$$ for all bounded non-negative Borel-measureable functions $g:\mathbb{R}\to\mathbb{R}_+$, then $U=V$ a.s.
I'm given a hint, which is to use the following result:
If $X\in \mathcal{L}^2(P)$, $\mathcal{B}$ is a sub-$\sigma$-field and we put $Y=E[X\mid \mathcal{B}]$ then $$X\sim Y \Rightarrow X=Y \,\text{ a.s.}$$
My thoughts so far: If I can show that $g(U)\sim g(V)$ for all such $g$, then $g(U)=g(V)$ a.s., and I think this is enough to show that $U=V$ a.s. (i.e. I'm thinking of using $g_n(x)=-n\vee x\wedge n$ and let $n\uparrow \infty$). Please correct me if I'm wrong here. If this the way to proceed, I just need a hint or a tip on how to show $g(U)\sim g(V)$.
-
Thanks for your reply. Choosing $g$ like that implies that $U\sim V$, but how to I move on from here and show that $U=V$ a.s.? – Stefan Hansen Mar 26 '12 at 13:31
@sos440: That is an answer, probably the answer if you are kind and add a few more words about the implication. – Henry Mar 26 '12 at 13:31
@sos440: Thanks again, I think I follow what you say. The important thing, as I see it, is that $g$ is a bounded function defined on $\mathbb{R}$ which is injective. Any particular reason for adding $\pi/2$ in your $g$? – Stefan Hansen Mar 26 '12 at 13:55
@sos440: Of course that is in order to make $g$ non-negative. I'd be happy to put this as the answer, if you would type it as an answer. – Stefan Hansen Mar 26 '12 at 14:00
We first claim that the given condition implies $U \sim V$. For each $a \in \mathbb{R}$, put
$$g(x) = \mathbf{1}_{\{x \leq a\}} = \begin{cases} 1 & x \leq a \\ 0 & x > a \end{cases}.$$
Then we have
$$\mathbb{P}(V \leq a) = \mathbb{E}[g(V)] = \mathbb{E}[\mathbb{E}[g(U)|V]] = \mathbb{E}[g(U)] = \mathbb{P}(U \leq a),$$
hence the proof of the claim follows. In particular, for any Borel function $g : \mathbb{R} \to \mathbb{R}$ we have $g(U) \sim g(V)$.
Now let $g : \mathbb{R} \to \mathbb{R}$ be a bounded injective continuous function. For example, we may put
$$g(x) = \frac{\pi}{2} + \arctan x.$$
Then both $g(U)$ and $g(V)$ are in $L^{\infty}(\mathbb{P}) \supset L^{2}(\mathbb{P})$ and satisfy both $g(V) = \mathbb{E}[g(U) | V]$ and $g(U) \sim g(V)$.
Then the hint yields $g(U) = g(V)$ a.s., hence $U = V$ a.s. as desired.
- | {"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": 0.9754766225814819, "perplexity": 109.71266089375419}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-07/segments/1454701147492.21/warc/CC-MAIN-20160205193907-00224-ip-10-236-182-209.ec2.internal.warc.gz"} |
https://www.vedantu.com/physics/screw-gauge | # Screw Gauge
View Notes
## Screw Gauge Measurement Using the Micrometer
Screw Gauge is a mechanical tool which facilitates measuring diameter or radius or thickness of a thin wire or thickness of thin metal sheet with utmost accuracy. Figure1 shows the schematic of the Screw Gauge. This tool consists of mainly a U shaped frame and a spindle (or screw) attached to thimble. There are two scales known as Pitch Scale which is basically a main scale and is engraved on the barrel with vertical lines. Other one is Circular Scale which is engraved horizontally on the thimble. Pitch scale measures the distance in millimeter (mm) travelled by the spindle per revolution. One revolution by the circular scale is equal to the half of millimeter of displacement of the screw is called micrometer screw gauge. Micrometer gauges are more accurate and precise as compared to the Vernier Caliper scale.
Figure 1: Schematic of the Screw Gauge showing different parts of it.
Before starting measurement, it is indispensable to make sure that the ‘Stud-end’ and the ‘Spindle (or Screw) end’ area are cleaned; otherwise the obtained measurements will provide the wrong thickness.
The plane surface near Stud is stationary and is also known as ‘Anvil’ which is immovable and attached with the U-shaped fame. When the anvil and spindle (or screw) is in contact, the zero of the Pitch Scale (or main scale zero) must coincide with zero of the circular scale. This is considered as the reference of the measurement.
As anvil is fixed, therefore, it is appropriate to place the object, whose thickness to be measured, firmly near the anvil. At the same time, the cylindrical thimble at the other end of the gauge needs to be rotated clockwise. This act will make the spindle (or screw) part move forward in linear direction towards the anvil. Keep rotating the thimble such that spindle (or screw) reaches the object, which was placed near the anvil.
In this final condition, object, whose dimension is to be found, must be held tightly and stable between the anvil and spindle. Rachet is used which clicks three times once the precise movement of spindle is done. This position can be locked using the lock knob positioned near the spindle.
Once the previous step is obtained, both the pitch scale and the circular scale should be read out. Pitch scale reading is the value of the Pitch scale when it touches the sleeve of the circular scale. The circular scale reading is the reading which matches the line on the pitch scale. Magnifying glass can be used to read the measurements. Let’s say the pitch reading is ‘X’ and the circular reading is ‘Y’.
The final calculation will be as following:
Total reading = X(mm) + (Y x LC of the gauge)mm
Here, LC is the least count of the gauge.
Screw Gauge Measurement through the Inch Micrometer
The basic design of the Screw gauge in Inch micrometer is similar to the one discussed in the previous section. The only difference is the dimension of the scale at which the circular and pitch scale deals with.
The number of divisions on the pitch scale is calculated in terms of the part of an inch. Figure 2 shows Pitch scale of an Inch micrometer. The number 5 (or the graduation 5) on the barrel denotes the half of the inch or 0.500 inch. There are 4 divisions between graduation 0 and graduation 1 on the pitch scale. So, the first graduation seen on the barrel as a number 1 is actually a 10th of an inch or 0.100 inches or 100th of a 1000 inch. Similarly the graduation 2 is 0.200 inches or 200th of 1000 inch or 20th of 200 inch. Hence the smallest division on the pitch scale would be 0.025 inches or 25th of 1000 inches.
Figure 2 Pitch scale of the Inch Micrometer
Next is to learn how to read the circular (or thimble) scale of inch micrometer. The circular scale is shown in Figure 3, consists of 25 divisions altogether.
Figure 3: Circular or Thimble scale of an Inch Micrometer
The smallest division on the circular of thimble scale is 1000th part of an inch; hence it is equal to 0.001 inch. Similarly, the 5th reading is equal to 0.005 inch.
Next we discuss an example how to read total reading from an inch micrometer. Consider figure 4, the main scale or pitch scale will be 0.375 inch as there are 13 divisions and one division equals to 0.025 inch as calculated before. While on the circular scale it is 0.001 inch. Hence the total reading is 0.375+0.001 which is 0.376 inches.
Figure 4: Inch micrometer with Pitch and thimble scale
Screw Gauge Formula
Pitch scale measures the distance in millimeter (mm) travelled by the spindle per revolution. The pitch of the screw gauge is calculated as given below:
${\text{Pitch of the Screw Gauge}} = \frac{\text{Distance travelled by screw}}{\text{Number of rotations}}$
In order to measure the dimension of a thin wire or thin sheet using a screw gauge, one need to know the least count. The least count of the screw gauge can be calculated using the formula given below:
${\text{Least count of the screw gauge}} = \frac{\text{Pitch of the Screw}}{\text{Number of divisions of circular scale}}$
Once the least count of the gauge is calculated, the final reading is calculated as below
LC in the above equation is the least count of the gauge. There might be the possibility that the zero of the pitch scale does not coincide with the zero of the circular scale. In that case, either the scale can be above or below the pitch scale (also known as main scale). Figure 5 discusses all the possible cases.
Figure 5: Case 1, shows the nil zero as the pitch scale zero matches the circular scale zero. Case 2 is the positive zero as circular scale zero is below the pitch scale zero. Case 3 is negative zero error as the circular scale zero lies above the pitch scale zero.
If the zero of the circular scale is below the pitch scale zero, it is known as positive zero error. The number of divisions on the circular below the pitch scale is to be multiplied with the least count of the gauge. This value is to be subtracted from the total reading for the correct measurements. Hence the formula becomes as follows:
Total Reading = Pitch Scale Reading + Circular Scale Reading x LC of the gauge - positive zero error reading x LC of the gauge
If the zero of the circular scale is above the pitch scale zero, it is known as negative zero error. The number of divisions on the circular above the pitch scale is to be multiplied with the least count of the gauge. This value is to be added to the total reading for the correct measurements. Hence the formula becomes as follows:
Total Reading = Pitch Scale Reading + Circular Scale Reading x LC of the gauge + negative zero error reading x LC of the gauge
An example is discussed on how to read the screw gauge measurement using micrometer. Figure 6 is the final reading to be calculated.
Figure 6: Screw Gauge using micrometer
The micrometer shown in Figure 4 has a pitch of 0.5 mm. It means that for one complete rotation of the thimble scale, 0.5 mm distance is traveled by the spindle. There are 50 divisions on circular scale. The least count will be 0.01 mm calculated as below.
${\text{Least count}} = \frac{\text{Pitch}}{\text{Number of divisions}} = \frac{0.5}{50} = 0.01 mm$
The pitch scale reading in Figure 4 is 3 mm. For the circular scale reading, the 46th division matches with reference line on the pitch scale. Hence total reading will be | {"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": 0.86751389503479, "perplexity": 1256.744314667908}, "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/1614178370752.61/warc/CC-MAIN-20210305091526-20210305121526-00600.warc.gz"} |
http://pixel-druid.com/intuition-for-why-choosing-closed-closed-intervals-of-1n-is-n1c2.html | ## § Intuition for why choosing closed-closed intervals of [1..n] is $(n+1)C2$
• $nC2$ counts all intervals $\{ [i, j]: i > j \}$.
• To count intervals $[i, i]$, there are $n$ of them, so it's $nC2 + n$ which is $n(n-1)/2 + n$, which is $n(n+1)/2$ or $(n+1)C2$.
• Combinatorially, add a "special point *" to [1..n]. If we pick a pair (i, *) from the $(n+1)C2$, take this to mean that we are picking the interval [i, i]. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 10, "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": 0.9899676442146301, "perplexity": 1204.6183955953325}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-21/segments/1652662584398.89/warc/CC-MAIN-20220525085552-20220525115552-00293.warc.gz"} |
https://eprints.soton.ac.uk/369681/ | The University of Southampton
University of Southampton Institutional Repository
# The local Langlands correspondence for inner forms of SL_n
Aubert, Anne-Marie, Baum, Paul, Plymen, Roger and Solleveld, Maarten (2016) The local Langlands correspondence for inner forms of SL_n. Research in the Mathematical Sciences, 3, [32].
Record type: Article
## Abstract
Let F be a non-archimedean local field. We establish the local Langlands correspondence for all inner forms of the group SLn(F). It takes the form of a bijection between, on the one hand, conjugacy classes of Langlands parameters for SLn(F) enhanced with an irreducible representation of an S-group and, on the other hand, the union of the spaces of irreducible admissible representations of all inner forms of SLn(F) up to equivalence. An analogous result is shown in the archimedean case. For p-adic fields this is based on the work of Hiraga and Saito. To settle the case where F has positive characteristic, we employ the method of close fields. We prove that this method is compatible with the local Langlands correspondence for inner forms of GLn(F), when the fields are close enough compared to the depth of the representations.
Text
innerFormsSLn25.pdf - Accepted Manuscript
Text
s40687-016-0079-4 - Version of Record
Accepted/In Press date: 2 August 2016
e-pub ahead of print date: 5 December 2016
Published date: 5 December 2016
Organisations: Pure Mathematics
## Identifiers
Local EPrints ID: 369681
URI: http://eprints.soton.ac.uk/id/eprint/369681
ISSN: 2197-9847
PURE UUID: 7500c973-24c6-440f-bf26-5c19de7f8bc6
## Catalogue record
Date deposited: 09 Oct 2014 12:16
## Contributors
Author: Anne-Marie Aubert
Author: Paul Baum
Author: Roger Plymen
Author: Maarten Solleveld | {"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": 0.8034158945083618, "perplexity": 2549.067161597385}, "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-20200119234426-20200120022426-00415.warc.gz"} |
https://www.physicsforums.com/threads/integration-by-parts.165463/ | # Integration by Parts
1. Apr 12, 2007
### Bazman
Hi,
$$J(m,n) = \int_0^{\frac{\pi}{2}} \cos^m \theta \sin^n \theta d\theta$$
First of all I had to evaluate the following ( I don't know what the correct answers are but here are my calculations:
$$J(0,0) = [\theta]_0^{\frac{{\pi}{2}}}=\frac{\pi}{2}$$
$$J(0,1) = [-\cos \theta]_0^{\frac{{\pi}{2}}}= 1$$
$$J(1,0) = [ \sin\theta]_0^{\frac{{\pi}{2}}}= 1$$
$$J(1,1) = [\frac{-\cos 2\theta}{4}]_0^{\frac{{\pi}{2}}}= \frac{1}{2}$$
$$J(m,1) = [-\frac{\cos^{m+1} \theta}{m+1}]_0^{\frac{{\pi}{2}}}= \frac{1}{m+1}$$
$$J(1,n) = [\frac{\sin^{n+1} \theta}{n+1}]_0^{\frac{{\pi}{2}}}= \frac {1}{n+1}$$
Next I am supposed to use integration by parts to prove that for m and n > 1
$$J(m,n) = \frac{m-1}{m+n} J(m-2,n)$$
and
$$J(m,n) = \frac{n-1}{m+n} J(m,n-2)$$
When I carried out integration by parts I got the following:
taking
$$u = \sin^{n-1} \theta$$
$$u' = (n-1) \sin^{n-2}\theta \cos \theta$$
$$v' = \cos^m \theta \sin \theta$$
$$v = -\frac{cos^{m+1}\theta}{m+1}$$
$$\frac{-\sin^{n-1} \theta \cos^{m+1}\theta}{m+1} + \int_0^{\frac{\pi}{2}} \frac{n-1}{m+1} \sin^{n-2} \theta \cos^{m+2} \theta d\theta$$
the first term on the RHS equates to zero but the 2nd term is not correct: the denominator and the power to which cos is raised is wrong but I'm not sure how to fix it
Last edited: Apr 12, 2007
2. Apr 13, 2007
### VietDao29
I dunno why, but your LaTeX part is a bit messy for me.
You are correct to the last part. So, you have shown that:
$$\int_0 ^ \frac{\pi}{2} \cos ^ m \theta \sin ^ n \theta d( \theta ) = \frac{n - 1}{m + 1} \int_0 ^ \frac{\pi}{2} \cos ^ {\fbox{m + 2}} \theta \sin ^ {\fbox{n - 2}} \theta d( \theta )$$
Note that cos function on the RHS is to the power of m + 2, and you want to prove that:
$$\int_0 ^ \frac{\pi}{2} \cos ^ m \theta \sin ^ n \theta d( \theta ) = \frac{n - 1}{m + n} \int_0 ^ \frac{\pi}{2} \cos ^ {\fbox{m}} \theta \sin ^ {\fbox{n - 2}} \theta d( \theta )$$.
So what you should do is to split cosm + 2(x) to cosm(x)cos2(x) = cosm(1 - sin2(x)), like this:
$$RHS = \frac{n - 1}{m + 1} \int_0 ^ \frac{\pi}{2} \cos ^ {m} \theta (1 - \sin ^ 2 \theta) \sin ^ {n - 2} \theta d( \theta ) = \frac{n - 1}{m + 1} \left( \int_0 ^ \frac{\pi}{2} \cos ^ {m} \theta \sin ^ {n - 2} \theta d( \theta ) - \int_0 ^ \frac{\pi}{2} \cos ^ {m} \theta \sin ^ {n} \theta d( \theta ) \right)$$.
So we have:
$$J(m, n) = \frac{n - 1}{m + 1} \left( J(m, n - 2) - J(m, n) \right)$$
Can you go from here? :)
--------
The other can be done almost the same, except for the first step:
u = sinn - 1(x)
and dv = cosm(x) sin(x) dx, we choose:
u = cosm - 1(x)
and dv = sinn(x) cos(x) dx.
Can you complete the two problems?
Last edited: Apr 13, 2007
3. Apr 15, 2007
### Bazman
Hey VietDao29!
Thanks for your help on this.
I assume that the rest of the answer rely's on my using the reduction formula for trigonometric integrals?
I've had a quick scoot on the web and there seems to be two main cases m+n odd or m+n even. Which uses two different formula (which does make me a little anxious as there is only one formula that we are trying to prove.)
Anyway I've included a link to the worksheet I am working from.
http://www.math.ualberta.ca/~apotapov/MATH115/trinth.pdf
Is this the correct method to solve the problem?
4. Apr 17, 2007
### Bazman
Hi VietDao,
I managed to figure it out.
Your hint brought me alot closer to the solution than I initially realised!
Thanks again
Similar Discussions: Integration by Parts | {"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": 0.9313980937004089, "perplexity": 1123.0316873926874}, "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-2017-43/segments/1508187823309.55/warc/CC-MAIN-20171019141046-20171019161046-00287.warc.gz"} |
http://crypto.stackexchange.com/questions/3115/how-to-decrypt-this-rsa-like-cipher?answertab=oldest | # How to decrypt this RSA-like cipher
Suppose we have a RSA-type Modulus $n = pq$ with $p,q$ prime. We also pick a random public exponent $e$ with $\gcd(e,\varphi(n)) = 1$ and compute the private exponent $d$ with $de \equiv 1 \pmod{\varphi(n)}$. $(n,e)$ is the public key, $(n,d)$ is the private key. Just like RSA, so far.
Now we encrypt a message representative $m$ with $0 \leq m < n$ by applying the following steps:
1. Choose a random $r$ with $1 \leq r < n$
2. Compute $y = (r^e)^e \pmod{n}$
3. Compute $z = (r \cdot m)^e \pmod{n}$
4. $(y,z)$ is the Ciphertext
How is it possible to decrypt this cipher using just the ciphertext $(y,z)$ and the private key $(n,d)$?
Here's what i have so far for the decryption algorithm:
1. Compute $r = (y^d)^d$
2. Compute $rm = z^d$
3. Find the multiplicative inverse $r^{-1}$ of $r$ modulo $n$
4. Compute $m = rm \cdot r^{-1}$
The problem with this is, that $r^{-1}$ only exists, when $\gcd(r,n) = 1$ and thus the decryption fails, when this is not the case. Is there a better way to do this?
-
A slightly more efficient method to perform decryption would be:
1. Compute $r^e = (y^d)$
2. Find the multiplicative inverse $r^{-e}$ of $r^e$ modulo $n$
3. Compute $m = (r^{-e} \cdot z)^d = (r^{-e} \cdot r^e \cdot m^e) ^d$
This has two computations of $x^d$ for some $x$; your method has three.
On the other hand, this doesn't address your question; if $r^{-1}$ doesn't exist, neither will $(r^e)^{-1}$.
For your question, well, there is no such "better" algorithm, because the decryption is inherently ambiguous.
Suppose that $n = pq$ (where $p$ and $q$ are distinct primes), and you selected a value $r = kp$ (for some integer k). Then, if we consider two distinct messages $m_1$ and $m_2$ with $m_1 = m_2 \mod q$, and the encryption of both these messages with that same $r$. They share the same $y$ value (because $y$ depends only on $r$), and as for the $z$ value:
$z_1 = (r \cdot m_1)^e \mod n$
$z_2 = (r \cdot m_2)^e \mod n$
we see that:
$(z_1 - z_2) \bmod p = r^e \cdot (m_1^e - m_2^e) \bmod p = 0$ (because $r$ is a multiple of $p$
$(z_1 - z_2) \bmod q = r^e \cdot (m_1^e - m_2^e) \bmod q = r^e \cdot 0 = 0$ (because $m_1^e$ and $m_2^e$ are the same modulo q)
and hence (by the Chinese Remainder Theorem), $z_1 = z_2 \mod n$
So, if you attempt to decrypt the message $(y, z_1)$, there are at least two valid decryptions ($m_1$ and $m_2$), and no algorithm can distinguish which is meant.
If you prefer a concrete example, consider the case where $p=5$ and $q = 11$. We can select $e = 3$ and $d = 27$ (actually, $d=7$ would work just as well). Then, if we select $r = 10$ and $m_1 = 2$ and $m_2 = 13$, then the encryption of $m_1$ would be:
$((10^3)^3, (10\cdot2)^3) = (10, 25)$
and the encryption of $m_2$ would be:
$((10^3)^3, (10\cdot13)^3) = (10, 25)$
Hence if we get the ciphertext $(10, 25)$, we can't tell if the original message with 2, 13, or even a bunch of other values (24, 35, 46)
Now, in practice, this is not likely to be a concern; the probability of this happening with a random $r$ is about $1/p + 1/q$; if $p$ and $q$ are large enough to make factorization difficult, this failure probability is negligible.
BTW: why are you considering this method, rather than a more straightforward RSA-based approach? If it is to get some homomorphic properties, well, it doesn't hide the encryption of 0 very well, and that's really what we would like in a multiplicative homomorphic encryption.
-
i am not really going to use this, it's just an exercise from an old crypto class, and i am trying to prepare for the exam ;-). Thanks a lot for your answer, this made a lot of things clearer! – padde Jul 1 '12 at 16:44
When $\gcd(r,n) \neq 1$, then we either have $\gcd(r,n) = p$ or $\gcd(r,n) = q$ (or $\gcd(r,n) = n$, but this can't be because of $r < n$), i.e. $r$ must be a multiple of either $p$ or $q$. The probability of choosing such an $r$ is about $\frac 1p + \frac 1q$, which for key sizes as used for RSA is such a tiny number that we don't expect it to occur at all.
In either case, whoever guessed this $r$ now is able to factor $n$, and also calculate your private key. The size of $p$ and $q$ should be chosen such that even with quite an effort of calculation this should not be doable. I.e. the probability of just guessing $r$ with $\gcd(r,n) \neq 1$ is way lower than that the system breaks down on other places.
So, if this is the only worry in your system, don't worry.
- | {"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": 0.9138554930686951, "perplexity": 216.3671312728973}, "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/1368696383077/warc/CC-MAIN-20130516092623-00056-ip-10-60-113-184.ec2.internal.warc.gz"} |
http://mathhelpforum.com/pre-calculus/107152-i-want-check-my-answer-solve-some-question-print.html | # I want check my answer and solve some question
• Oct 10th 2009, 06:45 AM
r-soy
2 Attachment(s)
I want check my answer and solve some question
I want check my answer and solve some question
Q :
Attachment 13284
my answer from 33 - 38
Attachment 13285
thanks
• Oct 10th 2009, 07:06 AM
earboth
Quote:
Originally Posted by r-soy
I want check my answer and solve some question
Q :
Attachment 13284
my answer from 33 - 38
Attachment 13285
thanks
You did correctly the questions #33, #34, #35
The last 3 questions are wrong:
#36: You used the wrong equation
#37 no root to be used
#38 use definition of absolute value and determine the appropriate domain
• Oct 10th 2009, 07:56 AM
r-soy
1 Attachment(s)
Quote:
Originally Posted by earboth
You did correctly the questions #33, #34, #35
The last 3 questions are wrong:
#36: You used the wrong equation
#37 no root to be used
#38 use definition of absolute value and determine the appropriate domain
thanks ,,,
check it now
Attachment 13288
earboth also i want solve questions from 39 to 42
thanks >>
• Oct 10th 2009, 11:09 AM
earboth
Quote:
Originally Posted by r-soy
thanks ,,,
check it now
Attachment 13288
earboth also i want solve questions from 39 to 42
thanks >>
1. #36 is wrong: You used the wrong equation
2. #37 is OK now (Clapping)
3. #38 is wrong. As I mentioned in my previous post you have to determine the appropriate domain. From
$x^3+|y|=6~\implies~|y|=6-x^3$
Since $|y|\geq 0$ you have to calculate those values for x that
$6-x^3\geq 0~\implies~x^3 \leq 6~\implies~\boxed{x\leq \sqrt[3]{6}}$
That means $|y|=6-x^3\ ,\ x\leq \sqrt[3]{6}$ is a function.
Quote:
...also i want solve questions from 39 to 42
Definitely I won't do your homework. So show some work of your own and if there are some mistakes you'll get additional support.
• Oct 10th 2009, 12:33 PM
r-soy
1 Attachment(s)
Now
i try :::: (Worried)
Attachment 13291
• Oct 10th 2009, 11:08 PM
earboth
Quote:
Originally Posted by r-soy
Now
i try :::: (Worried)
Attachment 13291
What you labeled #36 is actually #37. Usually an equation of a function starts with "y = ...". So your result is slightly unfinished (but correct)
#39 is wrong.
Explanation: $\left(\dfrac0a = 0\ ,\ a \in \mathbb{R} \setminus \{0\}\right)$
#40 is OK (Clapping)
#41 is wrong. You only have to solve the given equation for y to judge wether the equation describes a function or not.
#42 is correct but incomplete. (Is this a function or not?)
Initially you are asked to find the domain of those equations which define a function. The domains are missing (Worried)
• Oct 11th 2009, 02:21 AM
r-soy
you mean in #39 the equation of a function withou ( - ) or what ?
what is wrong here ?
#39
$
\left(\dfrac0X = 0\ ,\ X \in \mathbb{R} \setminus \{0\}\right)
$
function
is ok now ?
#41
x + y = 1
y = 1- x
function
#42 a function
• Oct 11th 2009, 04:48 AM
earboth
Quote:
Originally Posted by r-soy
you mean in #39 no I meant #36 the equation of a function withou ( - ) or what ?
what is wrong here ?
#39
$
\left(\dfrac0X = 0\ ,\ X \in \mathbb{R} \setminus \{0\}\right)
$
function
is ok now ? Unfortunately no:
$\color{blue}\bold{xy=0~\implies~y=0\ ,\ X \in \mathbb{R} \setminus \{0\}}$
#41
x + y = 1
y = 1- x
function
The transformations of the origianl equation is wrong:
$\color{blue}\bold{x^2+xy=1~\implies~xy=1-x^2~\implies~y=\dfrac1x-x\ ,\ x \in \mathbb{R} \setminus \{0\}}$
#42 a function <<<<< and what about the domain?
In the 2nd sentence of the initial quaestion you are asked to find the domains of the functions. You haven't calculated a domain yet!
• Oct 11th 2009, 05:41 AM
HallsofIvy
For all that are NOT functions you are asked to "find a value of x top which there corresponds more than one value of y". You do not appear to have done that.
• Oct 11th 2009, 07:25 AM
r-soy
Quote:
Originally Posted by HallsofIvy
For all that are NOT functions you are asked to "find a value of x top which there corresponds more than one value of y". You do not appear to have done that.
thankz earboth
HallsofIvy I did not understand it, can you please explain more (Crying)
• Oct 11th 2009, 11:50 AM
r-soy
???
• Oct 12th 2009, 01:08 AM
r-soy
Quote:
Originally Posted by HallsofIvy
For all that are NOT functions you are asked to "find a value of x top which there corresponds more than one value of y". You do not appear to have done that.
what NOT functions ? (Itwasntme)
• Oct 12th 2009, 04:22 AM
earboth
Quote:
Originally Posted by r-soy
what NOT functions ? (Itwasntme)
Please check your own notes: #34 and #35 are NOT functions. So you are asked to find at least one x-value with which you get at least 2 y-values. | {"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": 0.8431308269500732, "perplexity": 2181.2657061728996}, "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/1476988719542.42/warc/CC-MAIN-20161020183839-00423-ip-10-171-6-4.ec2.internal.warc.gz"} |
https://www.physicsforums.com/threads/what-is-absolute-temperature.55341/ | # What is absolute temperature?
1. Dec 4, 2004
### Mk
What is absolute temperature? The aboslute value of a temperture value?
2. Dec 4, 2004
### Gonzolo
Not sure in what context you are asking the questions but possible answers are :
1. The temperature in degrees Kelvin (vs C or F)
2. A number that is proportional to the thermal energy of a substance
3. (?)
3. Dec 4, 2004
### Mk
I think its the first one, though it could go either way.
4. Dec 5, 2004
### Gonzolo
Ok, the first thing you got to know is that "absolute temperature" exactly means "temperature in degrees K (Kelvin)".
5. Dec 5, 2004
### ceptimus
The absolute temperature scale is in Kelvin as others have said. On this scale, 0 K is the coldest anything can ever get. It's the same scale as celcius (centigrade) except you have to add 273.
So if it is 61 F in your garden, that's about 16 C or 289 K
6. Dec 5, 2004
### Gokul43201
Staff Emeritus
Actually it's both #1 and #2. While #1 only tells you the name of a unit, #2 is more important, as it gives you a physical meaning : The absolute temperature is a number that is proportional to the total kinetic energy of the atoms/molecules in a closed system.
The Absolute scale can be defined by any two points, say 0K (where all classically calculated molecular motion stops) and 273K (where water freezes at 1 atm) and a linear interpolation/extrapolation.
Last edited: Dec 5, 2004
7. Dec 5, 2004
### Gonzolo
For the sake of completeness, space can also contain thermal energy, even though it is not a "substance", and so that is how it can be said that interstellar space (background radiation) is at about 2-3 K. If I understand correctly, this energy is basically all in photon (microwave) form.
8. Dec 6, 2004
### Clausius2
As others have said, Absolute or Thermodynamic Temperature is measured in Kelvin scale. If you want a definition, apart of that given by the Kinetic Theory, you may as well take a look at the First Principle:
$$dU=TdS-PdV$$
So that:
$$T=\frac{\partial U}{\partial S}\Big)_{V}$$
Now is when a physicist should tell us if this derivative can be negative or not. I don't really know.
9. Dec 6, 2004
### marlon
Yes it can be....But then again that is very exotic. For example in some spin-systems (i mean many atoms and we only look at spin spin interactions) absolute NEGATIVE temperatures can arise. These temperatures are no really negative, but they need to be looked at as bigger then infinity....
The conditions for this to occur are for example that the spin-spin relaxation time is little compared to the spin lattice relaxation time. This means that the spins mutually interact long before thermal degrees of freedom come into play...
regards
marlon
10. Dec 6, 2004
### Clausius2
So in your opinion, would it be possible to build a device (such as a heat engine) that takes advantage of negative temperatures? All of us know what happens with Classic Thermodynamics when you substitute a T<0 in the formulae. :surprised
For engineers, it would be a funny thing to design and test such device... :!!)
11. Dec 6, 2004
### spacetime
Temperature can be defined with reference to the propeties of some standard substance. Like, in terms of expansion of mercury as in the ordinary thermometer, expansion of some gas, resistance of a metal.
But then, there is another way in which one can define temperature. In terms of kinetic energy of the gas molecules. When defined in this way, the temperature is said to be the absolute temperature as it doesn't depend upon some other reference.
spacetime
www.geocities.com/physics_all
Similar Discussions: What is absolute temperature? | {"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": 0.9047988057136536, "perplexity": 1084.80754485883}, "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/1505818689490.64/warc/CC-MAIN-20170923052100-20170923072100-00693.warc.gz"} |
https://math.libretexts.org/Bookshelves/Algebra/Intermediate_Algebra_(OpenStax)/02%3A_Solving_Linear_Equations/2.07%3A_Solve_Compound_Inequalities | $$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$
# 2.7: Solve Compound Inequalities
$$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$ $$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$
##### Learning Objectives
By the end of this section, you will be able to:
• Solve compound inequalities with “and”
• Solve compound inequalities with “or”
• Solve applications with compound inequalities
Before you get started, take this readiness quiz.
1. Simplify: $$\frac{2}{5}(x+10)$$.
If you missed this problem, review [link].
2. Simplify: $$−(x−4)$$.
If you missed this problem, review [link].
## Solve Compound Inequalities with “and”
Now that we know how to solve linear inequalities, the next step is to look at compound inequalities. A compound inequality is made up of two inequalities connected by the word “and” or the word “or.” For example, the following are compound inequalities.
$\begin{array} {lll} {x+3>−4} &{\text{and}} &{4x−5\leq 3} \\ {2(y+1)<0} &{\text{or}} &{y−5\geq −2} \\ \end{array} \nonumber$
##### COMPOUND INEQUALITY
A compound inequality is made up of two inequalities connected by the word “and” or the word “or.”
To solve a compound inequality means to find all values of the variable that make the compound inequality a true statement. We solve compound inequalities using the same techniques we used to solve linear inequalities. We solve each inequality separately and then consider the two solutions.
To solve a compound inequality with the word “and,” we look for all numbers that make both inequalities true. To solve a compound inequality with the word “or,” we look for all numbers that make either inequality true.
Let’s start with the compound inequalities with “and.” Our solution will be the numbers that are solutions to both inequalities known as the intersection of the two inequalities. Consider the intersection of two streets—the part where the streets overlap—belongs to both streets.
To find the solution of an "and" compound inequality, we look at the graphs of each inequality and then find the numbers that belong to both graphs—where the graphs overlap.
For the compound inequality $$x>−3$$ and $$x\leq 2$$, we graph each inequality. We then look for where the graphs “overlap”. The numbers that are shaded on both graphs, will be shaded on the graph of the solution of the compound inequality. See Figure $$\PageIndex{1}$$.
We can see that the numbers between $$−3$$ and $$2$$ are shaded on both of the first two graphs. They will then be shaded on the solution graph.
The number $$−3$$ is not shaded on the first graph and so since it is not shaded on both graphs, it is not included on the solution graph.
The number two is shaded on both the first and second graphs. Therefore, it is be shaded on the solution graph.
This is how we will show our solution in the next examples.
##### Example $$\PageIndex{1}$$
Solve $$6x−3<9$$ and $$2x+7\geq 3$$. Graph the solution and write the solution in interval notation.
$$6x−3<9$$ and $$2x+9\geq 3$$ Step 1. Solve each inequality. $$6x−3<9$$ $$2x+9\geq 3$$ $$6x<12$$ $$2x\geq −6$$ $$x<2$$ and $$x\geq −3$$ Step 2. Graph each solution. Then graph the numbers that make both inequalities true. The final graph will show all the numbers that make both inequalities true—the numbers shaded on both of the first two graphs. Step 3. Write the solution in interval notation. $$[−3,2)$$ All the numbers that make both inequalities true are the solution to the compound inequality.
##### Example $$\PageIndex{2}$$
Solve the compound inequality. Graph the solution and write the solution in interval notation: $$4x−7<9$$ and $$5x+8\geq 3$$.
##### Example $$\PageIndex{3}$$
Solve the compound inequality. Graph the solution and write the solution in interval notation: $$3x−4<5$$ and $$4x+9\geq 1$$.
##### SOLVE A COMPOUND INEQUALITY WITH “AND.”
1. Solve each inequality.
2. Graph each solution. Then graph the numbers that make both inequalities true.
This graph shows the solution to the compound inequality.
3. Write the solution in interval notation.
##### Example $$\PageIndex{4}$$
Solve $$3(2x+5)\leq 18$$ and $$2(x−7)<−6$$. Graph the solution and write the solution in interval notation.
$$3(2x+5)\leq 18$$ and $$2(x−7)<−6$$ Solve each inequality. $$6x+15\leq 18$$ $$2x−14<−6$$ $$6x\leq 3$$ $$2x<8$$ $$x\leq \frac{1}{2}$$ and $$x<4$$ Graph each solution. Graph the numbers that make both inequalities true. Write the solution in interval notation. $$(−\infty, \frac{1}{2}]$$
##### Example $$\PageIndex{5}$$
Solve the compound inequality. Graph the solution and write the solution in interval notation: $$2(3x+1)\leq 20$$ and $$4(x−1)<2$$.
##### Example $$\PageIndex{6}$$
Solve the compound inequality. Graph the solution and write the solution in interval notation: $$5(3x−1)\leq 10$$ and $$4(x+3)<8$$.
##### Example $$\PageIndex{7}$$
Solve $$\frac{1}{3}x−4\geq −2$$ and $$−2(x−3)\geq 4$$. Graph the solution and write the solution in interval notation.
$$\frac{1}{3}x−4\geq −2$$ and $$−2(x−3)\geq 4$$ Solve each inequality. $$\frac{1}{3}x−4\geq −2$$ $$−2x+6\geq 4$$ $$\frac{1}{3}x\geq 2$$ $$−2x\geq −2$$ $$x\geq 6$$ and $$x\leq 1$$ Graph each solution. Graph the numbers that make both inequalities true. There are no numbers that make both inequalities true. This is a contradiction so there is no solution.There are no numbers that make both inequalities true. This is a contradiction so there is no solution.There are no numbers that make both inequalities true. This is a contradiction so there is no solution.
##### Example $$\PageIndex{8}$$
Solve the compound inequality. Graph the solution and write the solution in interval notation: $$\frac{1}{4}x−3\geq −1$$ and $$−3(x−2)\geq 2$$.
##### Example $$\PageIndex{9}$$
Solve the compound inequality. Graph the solution and write the solution in interval notation: $$\frac{1}{5}x−5\geq −3$$ and $$−4(x−1)\geq −2$$.
Sometimes we have a compound inequality that can be written more concisely. For example, $$a<x$$ and $$x<b$$ can be written simply as $$a<x<b$$ and then we call it a double inequality. The two forms are equivalent.
##### DOUBLE INEQUALITY
A double inequality is a compound inequality such as $$a<x<b$$. It is equivalent to $$a<x$$ and $$x<b$$.
$\text{Other forms:} \quad \begin{array} {lllll} {a<x<b} &{\text{is equivalent to }} &{a<x} &{\text{and}} &{x<b} \\ {a\leq x\leq b} &{\text{is equivalent to }} &{a\leq x} &{\text{and}} &{x\leq b} \\ {a>x>b} &{\text{is equivalent to }} &{a>x} &{\text{and}} &{x>b} \\ {a\geq x\geq b} &{\text{is equivalent to }} &{a\geq x} &{\text{and}} &{x\geq b} \\ \end{array} \nonumber$
To solve a double inequality we perform the same operation on all three “parts” of the double inequality with the goal of isolating the variable in the center.
##### Example $$\PageIndex{10}$$
Solve $$−4\leq 3x−7<8$$. Graph the solution and write the solution in interval notation.
$$-4 \leq 3x - 7 < 8$$ Add 7 to all three parts. $$-4 \,{\color{red}{+\, 7}} \leq 3x - 7 \,{\color{red}{+ \,7}} < 8 \,{\color{red}{+ \,7}}$$ Simplify. $$3 \le 3x < 15$$ Divide each part by three. $$\dfrac{3}{\color{red}{3}} \leq \dfrac{3x}{\color{red}{3}} < \dfrac{15}{\color{red}{3}}$$ Simplify. $$1 \leq x < 5$$ Graph the solution. Write the solution in interval notation. $$[1, 5)$$
When written as a double inequality, $$1\leq x<5$$, it is easy to see that the solutions are the numbers caught between one and five, including one, but not five. We can then graph the solution immediately as we did above.
Another way to graph the solution of $$1\leq x<5$$ is to graph both the solution of $$x\geq 1$$ and the solution of $$x<5$$. We would then find the numbers that make both inequalities true as we did in previous examples.
##### Example $$\PageIndex{11}$$
Solve the compound inequality. Graph the solution and write the solution in interval notation: $$−5\leq 4x−1<7$$.
##### Example $$\PageIndex{12}$$
Solve the compound inequality. Graph the solution and write the solution in interval notation: $$−3<2x−5\leq 1$$.
## Solve Compound Inequalities with “or”
To solve a compound inequality with “or”, we start out just as we did with the compound inequalities with “and”—we solve the two inequalities. Then we find all the numbers that make either inequality true.
Just as the United States is the union of all of the 50 states, the solution will be the union of all the numbers that make either inequality true. To find the solution of the compound inequality, we look at the graphs of each inequality, find the numbers that belong to either graph and put all those numbers together.
To write the solution in interval notation, we will often use the union symbol, $$\cup$$, to show the union of the solutions shown in the graphs.
##### SOLVE A COMPOUND INEQUALITY WITH “OR.”
1. Solve each inequality.
2. Graph each solution. Then graph the numbers that make either inequality true.
3. Write the solution in interval notation.
##### Example $$\PageIndex{13}$$
Solve $$5−3x\leq −1$$ or $$8+2x\leq 5$$. Graph the solution and write the solution in interval notation.
$$5−3x\leq −1$$ or $$8+2x\leq 5$$ Solve each inequality. $$5−3x\leq −1$$ $$8+2x\leq 5$$ $$−3x\leq −6$$ $$2x\leq −3$$ $$x\geq 2$$ or $$x\leq −\frac{3}{2}$$ Graph each solution. Graph numbers that make either inequality true. $$(−\infty,−32]\cup[2,\infty)$$
##### Example $$\PageIndex{14}$$
Solve the compound inequality. Graph the solution and write the solution in interval notation: $$1−2x\leq −3$$ or $$7+3x\leq 4$$.
##### Example $$\PageIndex{15}$$
Solve the compound inequality. Graph the solution and write the solution in interval notation: $$2−5x\leq −3$$ or $$5+2x\leq 3$$.
##### Example $$\PageIndex{16}$$
Solve $$\frac{2}{3}x−4\leq 3$$ or $$\frac{1}{4}(x+8)\geq −1$$. Graph the solution and write the solution in interval notation.
$$\frac{2}{3}x−4\leq 3$$ or $$\frac{1}{4}(x+8)\geq −1$$ Solve each inequality. $$3(\frac{2}{3}x−4)\leq 3(3)$$ $$4⋅\frac{1}{4}(x+8)\geq 4⋅(−1)$$ $$2x−12\leq 9$$ $$x+8\geq −4$$ $$2x\leq 21$$ $$x\geq −12$$ $$x\leq \frac{21}{2}$$ $$x\leq \frac{21}{2}$$ or $$x\geq −12$$ Graph each solution. Graph numbers that make either inequality true. The solution covers all real numbers. $$(−\infty ,\infty )$$
##### Example $$\PageIndex{17}$$
Solve the compound inequality. Graph the solution and write the solution in interval notation: $$\frac{3}{5}x−7\leq −1$$ or $$\frac{1}{3}(x+6)\geq −2$$.
##### Example $$\PageIndex{18}$$
Solve the compound inequality. Graph the solution and write the solution in interval notation: $$\frac{3}{4}x−3\leq 3$$ or $$\frac{2}{5}(x+10)\geq 0$$.
## Solve Applications with Compound Inequalities
Situations in the real world also involve compound inequalities. We will use the same problem solving strategy that we used to solve linear equation and inequality applications.
Recall the problem solving strategies are to first read the problem and make sure all the words are understood. Then, identify what we are looking for and assign a variable to represent it. Next, restate the problem in one sentence to make it easy to translate into a compound inequality. Last, we will solve the compound inequality.
##### Example $$\PageIndex{19}$$
Due to the drought in California, many communities have tiered water rates. There are different rates for Conservation Usage, Normal Usage and Excessive Usage. The usage is measured in the number of hundred cubic feet (hcf) the property owner uses.
During the summer, a property owner will pay $24.72 plus$1.54 per hcf for Normal Usage. The bill for Normal Usage would be between or equal to $57.06 and$171.02. How many hcf can the owner use if he wants his usage to stay in the normal range?
Identify what we are looking for. The number of hcf he can use and stay in the “normal usage” billing range. Name what we are looking for. Let x=x= the number of hcf he can use. Translate to an inequality. Bill is $24.72 plus$1.54 times the number of hcf he uses or $$24.72+1.54x$$. $$\color{Cerulean}{\underbrace{\color{black}{\text{His bill will be between or equal to }57.06\text{ and }171.02.}}}$$ $$57.06 \leq 24.74 + 1.54x \leq 171.02$$ Solve the inequality. $$57.06 \leq 24.74 + 1.54x \leq 171.02$$ $$57.06 \,{\color{red}{- \,24.72}}\leq 24.74 \,{\color{red}{- \,24.72}} + 1.54x \leq 171.02 \,{\color{red}{- \,24.72}}$$ $$32.34 \leq 1.54x \leq 146.3$$ $$\dfrac{32.34}{\color{red}{1.54}} \leq \dfrac{1.54x}{\color{red}{1.54}} \leq \dfrac{146.3}{\color{red}{1.54}}$$ $$21 \leq x \leq 95$$ Answer the question. The property owner can use $$21–95$$ hcf and still fall within the “normal usage” billing range.
##### Example $$\PageIndex{20}$$
Due to the drought in California, many communities now have tiered water rates. There are different rates for Conservation Usage, Normal Usage and Excessive Usage. The usage is measured in the number of hundred cubic feet (hcf) the property owner uses.
During the summer, a property owner will pay $24.72 plus$1.32 per hcf for Conservation Usage. The bill for Conservation Usage would be between or equal to $31.32 and$52.12. How many hcf can the owner use if she wants her usage to stay in the conservation range?
The homeowner can use $$5–20$$ hcf and still fall within the “conservation usage” billing range.
##### Example $$\PageIndex{21}$$
Due to the drought in California, many communities have tiered water rates. There are different rates for Conservation Usage, Normal Usage and Excessive Usage. The usage is measured in the number of hundred cubic feet (hcf) the property owner uses.
During the winter, a property owner will pay $24.72 plus$1.54 per hcf for Normal Usage. The bill for Normal Usage would be between or equal to $49.36 and$86.32. How many hcf will he be allowed to use if he wants his usage to stay in the normal range?
The homeowner can use $$16–40$$ hcf and still fall within the “normal usage” billing range.
Access this online resource for additional instruction and practice with solving compound inequalities.
• Compound inequalities
## Key Concepts
• How to solve a compound inequality with “and”
1. Solve each inequality.
2. Graph each solution. Then graph the numbers that make both inequalities true. This graph shows the solution to the compound inequality.
3. Write the solution in interval notation.
• Double Inequality
• A double inequality is a compound inequality such as $$a<x<b$$. It is equivalent to $$a<x$$ and $$x<b.$$
Other forms: \begin{align*} a<x<b & & \text{is equivalent to} & & a<x\;\text{and}\;x<b \\ a≤x≤b & & \text{is equivalent to} & & a≤x\;\text{and}\;x≤b \\ a>x>b & & \text{is equivalent to} & & a>x\;\text{and}\;x>b \\ a≥x≥b & & \text{is equivalent to} & & a≥x\;\text{and}\;x≥b \end{align*}
• How to solve a compound inequality with “or”
1. Solve each inequality.
2. Graph each solution. Then graph the numbers that make either inequality true.
3. Write the solution in interval notation.
## Glossary
compound inequality
A compound inequality is made up of two inequalities connected by the word “and” or the word “or.”
2.7: Solve Compound Inequalities is shared under a CC BY 4.0 license and was authored, remixed, and/or curated by OpenStax via source content that was edited to conform to the style and standards of the LibreTexts platform; a detailed edit history is available upon request. | {"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": 0.8947365283966064, "perplexity": 696.9044351958121}, "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/1656103033925.2/warc/CC-MAIN-20220625004242-20220625034242-00400.warc.gz"} |
https://open.kattis.com/problems/greedilyincreasing | Kattis
# Greedily Increasing Subsequence
Given a permutation $A = (a_1, a_2, \dots , a_ N)$ of the integers $1, 2, \dots , N$, we define the greedily increasing subsequence (GIS) in the following way.
Let $g_1 = a_1$. For every $i > 1$, let $g_ i$ be the leftmost integer in $A$ that is strictly larger than $g_{i-1}$. If there for a given $i$ is no such integer, we say that the GIS of the sequence is the sequence $(g_1, g_2, ..., g_{i - 1})$.
Your task is to, given a permutation $A$, compute the GIS of $A$.
## Input
The first line of input contains an integer $1 \le N \le 10^6$, the number of elements of the permutation $A$. The next line contains $N$ distinct integers between $1$ and $N$, the elements $a_1, \dots , a_ N$ of the permutation $A$.
## Output
First, output a line containing the length $l$ of the GIS of $A$. Then, output $l$ integers, containing (in order) the elements of the GIS.
## Explanation of sample 1
In this case, we have the permutation $2, 3, 1, 5, 4, 7, 6$. First, we have $g_1 = 2$. The leftmost integer larger than $2$ is $3$, so $g_2 = 3$. The leftmost integer larger than $3$ is $5$ ($1$ is too small), so $g_3 = 5$. The leftmost integer larger than $5$ is $7$, so $g_4 = 7$. Finally, there is no integer larger than $7$. Thus, the GIS of $2, 3, 1, 5, 4, 7, 6$ is $2, 3, 5, 7$.
Sample Input 1 Sample Output 1
7
2 3 1 5 4 7 6
4
2 3 5 7
Sample Input 2 Sample Output 2
5
1 2 3 4 5
5
1 2 3 4 5
Sample Input 3 Sample Output 3
5
5 4 3 2 1
1
5 | {"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": 0.9611201882362366, "perplexity": 224.43058617566362}, "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/1652663012542.85/warc/CC-MAIN-20220528031224-20220528061224-00047.warc.gz"} |
https://arxiv.org/abs/1904.02320 | physics.flu-dyn
# Title:Structural inclination angle of scalar fluctuations in a turbulent boundary layer
Abstract: Two-point measurements of concentration are obtained in a meandering passive-scalar plume released at five different heights within the fully-turbulent region of a high-Reynolds number turbulent boundary layer (TBL). These measurements are obtained using a combination of two photo-ionisation detectors (PIDs). A stationary PID is positioned at the plume centreline, while the second PID traverses across the plume in the wall-normal direction. Similar to the large-scale coherence observed in the two-point correlation map of streamwise velocity fluctuations \citep[e.g. Marusic et al.][]{marusic2007reynolds}, the two-point correlation results of concentration also indicate coherence in the scalar field. Particularly, the iso-contours of correlation indicate an inclination of scalar structures in the direction of the flow. The corresponding inclination angle is found to be a weak function of the location of the plume within the boundary layer. However, the inclination angle is observed to be invariant near 30$^\circ$ for all source heights that are below $z/\delta \leq 0.33$. This observed steepness of inclination angle relative to the inclination angle of streamwise velocity is consistently explained through the physical model put forth by Talluru et al. \cite{Talluru2018} that explains the organisation of scalar around the low- and high-speed regions in the flow which inherently also have a characteristic inclination.
Comments: 7 figures, 18 pages Subjects: Fluid Dynamics (physics.flu-dyn) Cite as: arXiv:1904.02320 [physics.flu-dyn] (or arXiv:1904.02320v1 [physics.flu-dyn] for this version)
## Submission history
From: Murali Krishna Talluru [view email]
[v1] Thu, 4 Apr 2019 02:43:16 UTC (375 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": 0.8783407807350159, "perplexity": 2827.5484870129653}, "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-2019-22/segments/1558232254731.5/warc/CC-MAIN-20190519081519-20190519103519-00239.warc.gz"} |
http://mathoverflow.net/questions/22105/details-of-perelmans-example-about-soul-of-alexandrov-space | # Details of Perelman's example about soul of Alexandrov space
Reading Perelman's preprint(1991) Alexandrov space II now. Got confused about the last section 6.4, which contains an example which indicate that the statement ".... manifold is diffeomorphic to the normal bundle over soul" in Cheeger-Gromoll's Soul Theory won't hold for Alexandrov spaces. I also read BBI's book(A course in Metric Geometry) (page 400-401) which also contains the description of this example. I am using the notation in BBI's book, the example goes as follows:
Let $\pi: K_0(\mathbb{CP}^2)\to K_0(\mathbb{CP}^1)$ be the projection. $\bar{B}_0(1)$ be the unit ball in $\mathbb{CP}^1$ (Note: here should be $K_0(\mathbb{CP}^1$), right?). Let $X^5=\pi^{-1}(\bar{B}_0(1))$. Take double of $X^5$ and it will be the example
The picture in my mind is $K_0(\mathbb{CP}^1)$ is sub-cone of $K_0(\mathbb{CP}^2)$, so the projection is the projection on the second factor if we write the coordinate in cone as $(t, x)$ for $t\in \mathbb R$ and $x\in \mathbb{CP}^2$.
My question is what is the topology of $X^5$?
1) Is $\bar{B}_0(1)$ a close ball? If so then $\bar{B}_0(1)$ will have boundary $\mathbb{CP}^1$, right? and $X^5$ will be a closed cone over $\mathbb{CP}^2$, right?
2) Is $X^5$ compact?
-
If you are interested in errata to BBI, it is now on my home page (see my profile). I included misprints that I see in this example. – Sergei Ivanov Apr 27 '10 at 19:59
No $X^5$ is not a cone over $CP^2$ and is not compact. The projection has nothing to do with the cone structure. In fact, it's better to forget about the cone structure altogether (until you ask what is the topology of the thing).
The spaces are just quotients of $\mathbb C^3$ and $\mathbb C^2$ by the standard circle action, and the projection is induced by the coordinate projection $\mathbb C^3\to \mathbb C^2$. For example, there is a whole half-line in the pre-image of the origin.
I suggest you visualize a similar construction with $\mathbb R$ in place of $\mathbb C$ or lower the dimensions by 1 (or both) to see what is going on.
I see, so $X^5= \{(z_1, z_2, z_3| |z_1|^2+|z_2|^2\le 1, z_1, z_2, z_3\in \mathbb C\}/ \{S^1 action\}$, right? – John B Apr 21 '10 at 21:57 | {"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": 0.9353161454200745, "perplexity": 295.1057906377336}, "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/1432207928831.69/warc/CC-MAIN-20150521113208-00092-ip-10-180-206-219.ec2.internal.warc.gz"} |
http://basilisk.fr/src/navier-stokes/double-projection.h | # Double projection
This option for the centered Navier–Stokes solver is inspired by Almgren et al., 2000. These authors first recall that, while for the exact projection method, the definition of the pressure is unambiguous, in the case of the approximate projection method (used in centered.h), several pressures can be defined.
Exploiting the different properties of these different definitions may be useful in some cases.
## Standard approximate projection
We first summarise the standard approximate projection scheme, as implemented in centered.h.
1. Advection/viscosity \displaystyle \frac{u^{\star} - u^n}{\Delta t} = (u \cdot \nabla u)^{n + 1 / 2} + \alpha \nabla \cdot (\mu \nabla D)^{\star}
2. Acceleration \displaystyle u^{\star}_f = \overline{u^{\star}} + \Delta ta_f
3. Projection \displaystyle u^{n + 1}_f = P (u^{\star}_f, p^{n + 1})
4. Centered pressure gradient correction \displaystyle g^{n + 1} = \overline{a_f - \alpha_f \nabla p^{n + 1}} \displaystyle u^{n + 1} = u^{\star} + \Delta tg^{n + 1}
with P(u,p) the projection operator and the overline designating either cell-center to cell-face averaging or reciprocally.
## Double approximate projection
As its name indicates this scheme adds an extra projection step and defines two pressures: the standard one used to project the face-centered velocity field u_f (renamed \mathbf{\delta p} below), and p^{n+1} used to compute the cell-centered pressure gradient g. This new pressure is obtained by projecting only the update (i.e. its evolution in time) to the centered velocity field. Note that in the case of an exact projection these two projections are identical since the divergence of the velocity field at the start of the timestep is zero.
The scheme can be summarised as:
1. Advection/viscosity \displaystyle \frac{u^{\star} - u^n}{\Delta t} = (u \cdot \nabla u)^{n + 1 / 2} + \alpha \nabla \cdot (\mu \nabla D)^{\star} \mathbf{+ g^n = A^{n+1/2} + g^n} \displaystyle \mathbf{A_f = \overline{A^{n+1/2}} + \Delta ta_f}
2. Acceleration \displaystyle u^{\star}_f = \overline{u^{\star}}
3. Projection \displaystyle u^{n + 1}_f = P (u^{\star}_f, \mathbf{\delta p})
4. Approximate projection \displaystyle u^{n + 1} = u^{\star} - \Delta t \overline{\alpha_f \nabla \delta p}
5. Second projection \displaystyle \mathbf{P(A_f,p^n+1)} \displaystyle g^{n + 1} = \overline{a_f - \alpha_f \nabla p^{n + 1}}
where the additions to the previous scheme are highlighted in bold.
Why is this useful? The new pressure does not feel the history of divergence of the centered velocity field. This is useful in particular when this history includes the noise induced by adaptive mesh refinement.
The cost to pay is however significant since an extra (potentially expensive) projection is required.
## Implementation
We need a (temporary) field to store the update A_f and a (permanent) field to store the projection pressure \delta p.
face vector Af;
scalar dp[];
We make heavy use of the event inheritance mechanism. All the events below are first defined in centered.h.
At the beginning of the timestep (i.e. before advection), we store the (interpolated) value of the initial velocity field in A_f.
event advection_term (i++)
{
Af = new face vector;
foreach_face()
Af.x[] = - fm.x[]*face_value (u.x, 0);
}
After the advection and diffusion terms have been added to u, we recover the update by adding the new face-interpolated value of the velocity field to the initial face velocity, and add the acceleration i.e. we perform step 1 above: \displaystyle A_f = \overline{A^{n+1/2}} + \Delta ta_f
face vector ab;
event acceleration (i++)
{
foreach_face()
Af.x[] += fm.x[]*(face_value (u.x, 0) + dt*a.x[]);
boundary ((scalar *){Af});
We also add the centered gradient \mathbf{g^n} to the centered velocity field.
correction (dt);
Step 2 above (i.e. u_f^{\star} = \overline{u^{\star}}) is performed by the acceleration event of the centered solver, but we need to reset the acceleration to zero.
ab = a;
a = zerof;
}
The projection step 3 is also performed by the centered solver. We want to store the resulting pressure in dp rather than p, so we swap the two fields, before performing the projection.
event projection (i++)
{
scalar_clone (dp, p);
swap (scalar, p, dp);
}
Step 4 is performed by the centered solver. Step 5 is done at the end of the timestep. We first restore the fields modified above, then perform the second projection and compute the corresponding centered gradient g^{n+1}.
event end_timestep (i++)
{
swap (scalar, p, dp);
a = ab;
// this could be optimised since we do not use Af
mgp = project (Af, p, alpha, dt, mgp.nrelax);
delete ((scalar *){Af});
centered_gradient (p, g);
}
## References
[almgren2000] Ann S Almgren, John B Bell, and William Y Crutchfield. Approximate projection methods: Part i. inviscid analysis. SIAM Journal on Scientific Computing, 22(4):1139–1159, 2000. [ http ] | {"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": 0.9906048774719238, "perplexity": 3716.769827132562}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-49/segments/1637964363418.83/warc/CC-MAIN-20211207201422-20211207231422-00608.warc.gz"} |
http://gmatclub.com/forum/the-physical-structure-of-the-human-eye-enables-it-to-sense-105730.html?fl=similar | Find all School-related info fast with the new School-Specific MBA Forum
It is currently 26 Sep 2016, 07:44
### GMAT Club Daily Prep
#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
we will pick new questions that match your level based on your Timer History
Track
every week, we’ll send you an estimated GMAT score based on your performance
Practice
Pays
we will pick new questions that match your level based on your Timer History
# Events & Promotions
###### Events & Promotions in June
Open Detailed Calendar
# The physical structure of the human eye enables it to sense
Author Message
TAGS:
### Hide Tags
Senior Manager
Joined: 28 Aug 2010
Posts: 257
Followers: 6
Kudos [?]: 473 [0], given: 11
The physical structure of the human eye enables it to sense [#permalink]
### Show Tags
03 Dec 2010, 09:53
2
This post was
BOOKMARKED
00:00
Difficulty:
55% (hard)
Question Stats:
56% (02:17) correct 44% (01:17) wrong based on 212 sessions
### HideShow timer Statistics
The physical structure of the human eye enables it to sense light of wavelengths up to 0.0005 millimeters; infrared radiation, however, is invisible because its wavelength—0.1 millimeters—is too long to be registered by the eye.
(A) infrared radiation, however, is invisible because its wavelength—0.1 millimeters—is too long to be registered by the eye
(B) however, the wavelength of infrared radiation—0.1 millimeters—is too long to be registered by the eye making it invisible
(C) infrared radiation, however, is invisible because its wavelength—0.1 millimeters—is too long for the eye to register it
(D) however, because the wavelength of infrared radiation is 0.1 millimeters, it is too long for the eye to register and thus invisible
(E) however, infrared radiation has a wavelength of 0.1 millimeters that is too long for the eye to register, thus making it invisible
Please can some explain why choice b is not correct ?
[Reveal] Spoiler: OA
_________________
Gmat: everything-you-need-to-prepare-for-the-gmat-revised-77983.html
-------------------------------------------------------------------------------------------------
Ajit
Retired Moderator
Joined: 03 Aug 2010
Posts: 247
Followers: 2
Kudos [?]: 35 [0], given: 41
Re: Need help with this SC ! [#permalink]
### Show Tags
03 Dec 2010, 12:11
ajit257 wrote:
The physical structure of the human eye enables it to sense light of wavelengths up to 0.0005 millimeters; infrared radiation, however, is invisible because its wavelength—0.1 millimeters—is too long to be registered by the eye.
(A) infrared radiation, however, is invisible because its wavelength—0.1 millimeters—is too long to be registered by the eye
(B) however, the wavelength of infrared radiation—0.1 millimeters—is too long to be registered by the eye making it invisible
(C) infrared radiation, however, is invisible because its wavelength—0.1 millimeters—is too long for the eye to register it
(D) however, because the wavelength of infrared radiation is 0.1 millimeters, it is too long for the eye to register and thus invisible
(E) however, infrared radiation has a wavelength of 0.1 millimeters that is too long for the eye to register, thus making it invisible
Please can some explain why choice b is not correct ?
Let me try
when -ing is used without comma , it is directly modifying the immediate previous noun, and comma + ing would modify the subject of the previous clause.
here the making it invisible doesnt modify eye... so B is wrong...
_________________
http://www.gmatpill.com/gmat-practice-test/
Amazing Platform
Manager
Joined: 17 Sep 2010
Posts: 216
Concentration: General Management, Finance
GPA: 3.59
WE: Corporate Finance (Entertainment and Sports)
Followers: 3
Kudos [?]: 16 [0], given: 33
Re: Need help with this SC ! [#permalink]
### Show Tags
03 Dec 2010, 12:30
I picked C, because of the usage "to be". Damn, should've stuck with my gut and picked A.
Wavelength shouldn't be the subject of the sentence, it should be infrared radiation since that is invisible to the eye.
Just wanted to add on to this: the word "Of" before infrared radiation turns the phrase "of infrared radiation" into a preposition. Aside from some idiomatic expressions, prepositions can't be subjects of the sentence. Therefore, given the context of the sentence, A would work best because it matches up subject and verb perfectly.
ajit257 wrote:
The physical structure of the human eye enables it to sense light of wavelengths up to 0.0005 millimeters; infrared radiation, however, is invisible because its wavelength—0.1 millimeters—is too long to be registered by the eye.
(A) infrared radiation, however, is invisible because its wavelength—0.1 millimeters—is too long to be registered by the eye
(B) however, the wavelength of infrared radiation—0.1 millimeters—is too long to be registered by the eye making it invisible
(C) infrared radiation, however, is invisible because its wavelength—0.1 millimeters—is too long for the eye to register it
(D) however, because the wavelength of infrared radiation is 0.1 millimeters, it is too long for the eye to register and thus invisible
(E) however, infrared radiation has a wavelength of 0.1 millimeters that is too long for the eye to register, thus making it invisible
Please can some explain why choice b is not correct ?
Last edited by USCTrojan2006 on 03 Dec 2010, 12:58, edited 1 time in total.
Manager
Joined: 30 Apr 2009
Posts: 134
Followers: 1
Kudos [?]: 83 [1] , given: 9
Re: Need help with this SC ! [#permalink]
### Show Tags
03 Dec 2010, 14:38
1
KUDOS
Didnt like "it" in any of the choices, therefore, picked A
_________________
Trying to make CR and RC my strong points
Kaplan GMAT Instructor
Joined: 21 Jun 2010
Posts: 148
Location: Toronto
Followers: 45
Kudos [?]: 175 [2] , given: 0
Re: Need help with this SC ! [#permalink]
### Show Tags
03 Dec 2010, 14:42
2
KUDOS
ajit257 wrote:
The physical structure of the human eye enables it to sense light of wavelengths up to 0.0005 millimeters; infrared radiation, however, is invisible because its wavelength—0.1 millimeters—is too long to be registered by the eye.
(A) infrared radiation, however, is invisible because its wavelength—0.1 millimeters—is too long to be registered by the eye
(B) however, the wavelength of infrared radiation—0.1 millimeters—is too long to be registered by the eye making it invisible
(C) infrared radiation, however, is invisible because its wavelength—0.1 millimeters—is too long for the eye to register it
(D) however, because the wavelength of infrared radiation is 0.1 millimeters, it is too long for the eye to register and thus invisible
(E) however, infrared radiation has a wavelength of 0.1 millimeters that is too long for the eye to register, thus making it invisible
Please can some explain why choice b is not correct ?
Hi! This question is an excellent test of modification, which is heavily tested on the GMAT.
As always, when first evaluating a sentence it's best to ignore the "extra bits" - focus on the subject and verb.
First, we recognize that the semicolon divides two equal sentences; in effect, we can ignore everything that comes before it. So, let's only focus on the underlined portion.
In the original (and, of course, choice (A)), we have:
Quote:
infrared radiation is invisible because its wavelength is too long to be registered by the eye.
Seems like a perfectly sound sentence - it's clear what's invisible (IR), "its" has a single clear antecedent (IR) and no style issues. Lookin' good!
Next up is (B):
Quote:
however, the wavelength ... is too long to be registered by the eye making it invisible.
This one isn't as clear: "it" is supposed to refer to IR, but actually modifies the subject of the sentence, "the wavelength". As noted by our Trojan-loving poster, a prepositional phrase can't be the subject, so even though (B) contains "of infrared radiation" that's not a legal parent noun for "it".
Here's another very useful general rule for the GMAT: ignore prepositional phrases. Such phrases appear all the time and are pretty much always just in there to distract you from the basic sentence structure.
Next we have (C):
Quote:
infrared radiation is invisible because its wavelength is too long for the eye to register it.
More pronoun problems (pronouns are another very commonly tested grammar concept)! "it" at the end of the sentence could refer to either IR or wavelength. When you see an ambiguous pronoun, move on!
(Technically pronouns refer to the closest "legal" preceding parent noun - in this sentence that would be "wavelength" - but for the most part the GMAT abhors pronoun ambiguity.)
Quote:
however, because the wavelength of infrared radiation is 0.1 millimeters, it is too long for the eye to register and thus invisible.
Same issue as (B) - "of infrared radiation" can't be the antecedent for "it".
Finally (E):
Quote:
however, infrared radiation has a wavelength of 0.1 millimeters that is too long for the eye to register, thus making it invisible.
"that" renders this sentence awkward; also, "it" at the end could refer to IR, wavelength or the eye.
Manager
Joined: 10 Sep 2010
Posts: 133
Followers: 2
Kudos [?]: 30 [0], given: 7
Re: Need help with this SC ! [#permalink]
### Show Tags
03 Dec 2010, 17:26
I got it wrong by choosing 'C' rather than 'A'.
Thanks for the explanation.
Current Student
Joined: 03 Aug 2012
Posts: 916
Concentration: General Management, General Management
GMAT 1: 630 Q47 V29
GMAT 2: 680 Q50 V32
GPA: 3.7
WE: Information Technology (Investment Banking)
Followers: 22
Kudos [?]: 626 [1] , given: 322
Re: The physical structure of the human eye enables it to sense [#permalink]
### Show Tags
01 Sep 2013, 11:33
1
KUDOS
The physical structure of the human eye enables it to sense light of wavelengths up to 0.0005 millimeters; infrared radiation, however, is invisible because its wavelength—0.1 millimeters—is too long to be registered by the eye.
Analyses of the sentence:
The Physical Structure ..... enables IT(eye- clear antecedent) to sense light of....; Infrared radiation..... is invisible because ITS wavelength- .........
Although the first IT is in non-underlined portion , we will analyse all pronoun-antecedent pairs.
We will run the MGMAT Sc guide proposed filtering test to check for pronoun ambiguity.
(1). Number / Gender:
First IT can refer to EYE(singular) and structure(singular) [Number]
Second ITS can refer to "Infrared radiation","light", "eye", and "structure".
(2). Repeats : Second ITS can refer to EYE as the FIRST one refers to EYE.
(3). Proximity: FIRST IT is closer to EYE, so refers to EYE
SECOND ITS is closer to "IR", so refers to IR
(4). Case: There is no point of case as IT is in same clause.
We don't have || clauses here so cannot apply this step.
(A) infrared radiation, however, is invisible because its wavelength—0.1 millimeters—is too long to be registered by the eye
Correct as is and as explained above.
(B) however, the wavelength of infrared radiation—0.1 millimeters—is too long to be registered by the eye making it invisible
Making what invisible - "the wavelength" or the IR
(C) infrared radiation, however, is invisible because its wavelength—0.1 millimeters—is too long for the eye to register it
We apply repeats , proximity and case.
Repeats => IT logically refers to IR
Proximity => IT refers to EYE
Case => No || clauses
(D) however, because the wavelength of infrared radiation is 0.1 millimeters, it is too long for the eye to register and thus invisible
EYE is thus INVISIBLE - nonsense
(E) however, infrared radiation has a wavelength of 0.1 millimeters that is too long for the eye to register, thus making it invisible
Repeats => IT occurs only once
Proximity => IT refers to EYE
Case => IT appears to refer to IR correctly
But "THAT" clauses restricts the 0.1 MM wavelength=> change in meaning
_________________
Rgds,
TGC!
_____________________________________________________________________
I Assisted You => KUDOS Please
_____________________________________________________________________________
Manager
Joined: 08 Jun 2015
Posts: 132
Location: India
Followers: 1
Kudos [?]: 3 [0], given: 66
Re: The physical structure of the human eye enables it to sense [#permalink]
### Show Tags
26 Apr 2016, 05:40
Option A , good explanations above
_________________
" The few , the fearless "
Re: The physical structure of the human eye enables it to sense [#permalink] 26 Apr 2016, 05:40
Similar topics Replies Last post
Similar
Topics:
The physical structure of the human eye enables it to sense 5 27 May 2009, 21:15
1 The physical structure of the human eye enables it to sense 1 28 Feb 2008, 21:58
The physical structure of the human eye enables it to sense 8 25 Sep 2007, 08:34
The physical structure of the human eye enables it to sense 2 11 Sep 2007, 13:09
40 The physical structure of the human eye enables it to sense 19 28 Jul 2007, 00:52
Display posts from previous: Sort by | {"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": 0.8302836418151855, "perplexity": 3368.1208681762187}, "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-2016-40/segments/1474738660801.77/warc/CC-MAIN-20160924173740-00096-ip-10-143-35-109.ec2.internal.warc.gz"} |
https://jeannicod.ccsd.cnrs.fr/ijn_02161864 | Skip to Main content Skip to Navigation
# Each Other, Asymmetry and Reasonable Futures
Abstract : Reciprocal sentences display a variety of interpretations, including 'strong reci-procity'. In this interpretation, every element of the reference set participates with every other element in the same set in the relation provided by the predicate. Another interpretation is 'inclusive alternative orderings'. In this interpretation, every element in the reference set participates with some other member in the relation provided by the predicate either as the first or second argument. Current reciprocal theories cannot explain why some sentences that satisfy these truth conditions are in fact false and infelicitous, such as '#my mother and I gave birth to each other'. This paper defends the view that strong reciprocity has a privileged status. Its core insight is that reciprocal sentences are true if they describe a relation that is either actually or possibly strong reciprocal over the reference set, insofar as the possibilities are reasonable. The new truth conditions are cast in a branching time framework (Thomason, 1984), in which a new notion of reasonability is defined (vs. inertia, Dowty, 1979; see also Landman, 1992). On the empirical side, we show that when the relation is asymmetric, it must be non-permanent for the reciprocal each other-sentence to be true. We also note that, in some cases, the relation is asymmetric and non-permanent, but the each other-sentences are false. We explain these facts by introducing a new notion of 'decidedness' that we define in modal terms. We explain that the relation should not be decided for the reciprocal sentence to be true. We also provide a new definition for asymmetry and (non-)permanency in the modal framework and explain how asymmetry, (non-)decidedness and (non-)permanency interact, ensuring large empirical coverage.
Keywords :
Document type :
Journal articles
Domain :
Complete list of metadata
Cited literature [47 references]
https://jeannicod.ccsd.cnrs.fr/ijn_02161864
Contributor : Alda Mari <>
Submitted on : Friday, June 21, 2019 - 11:00:41 AM
Last modification on : Tuesday, January 12, 2021 - 3:29:12 AM
### File
Reciprocity_Mari.pdf
Files produced by the author(s)
### Citation
Alda Mari. Each Other, Asymmetry and Reasonable Futures. Journal of Semantics, Oxford University Press (OUP), 2014, 31 (2), pp.209 - 261. ⟨10.1093/jos/fft003⟩. ⟨ijn_02161864⟩
Record views
Files downloads | {"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": 0.8778776526451111, "perplexity": 2490.777797608692}, "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-2021-25/segments/1623487623942.48/warc/CC-MAIN-20210616124819-20210616154819-00531.warc.gz"} |
https://charlesmartin14.wordpress.com/ | # Why Deep Learning Works II: the Renormalization Group
Deep Learning is amazing. But why is Deep Learning so successful? Is Deep Learning just old-school Neural Networks on modern hardware? Is it just that we have so much data now the methods work better? Is Deep Learning just a really good at finding features. Researchers are working hard to sort this out.
Recently it has been shown that [1]
Unsupervised Deep Learning implements the Kadanoff Real Space Variational Renormalization Group (1975)
This means the success of Deep Learning is intimately related to some very deep and subtle ideas from Theoretical Physics. In this post we examine this.
#### Unsupervised Deep Learning: AutoEncoder Flow Map
An AutoEncoder is a Unsupervised Deep Learning algorithm that learns how to represent an complex image or other data structure $X$. There are several kinds of AutoEncoders; we care about so-called Neural Encoders–those using Deep Learning techniques to reconstruct the data:
The simplest Neural Encoder is a Restricted Boltzman Machine (RBM). An RBM is non-linear, recursive, lossy function $f(X)$ that maps the data $X$ from visible nodes ${v}$ into hidden nodes ${h}$:
The RBM is learned by selecting the optimal parameters ${b_{v}},{c_{h}},{w_{v,h}}$that minimize the reconstruction error
$\min\Vert f(X)-X\Vert$
RBMs and other Deep Learning algos are formulated using classical Statistical Mechanics. And that is where it gets interesting!
#### Multi Scale Feature Learning
In old-school ML, we map (visible) data into (hidden) features
$\mathbf{v}(X)\rightarrow\mathbf{h}$
The hidden units discover features at a coarser grain level of scale
With RBMs, when features are complex, we may stack them into a Deep Belief Network (DBM), so that we can learn at different levels of scale
and leads to multi-scale features in each layer
Deep Belief Networks are a Theory of Unsupervised MultiScale Feature Learning
#### Fixed Points and Flow Maps
We call $f(X)$ a flow map
$f(X)\rightarrow X$
If we apply the flow map to the data repeatedly, (we hope) it converges to a fixed point
$\lim\limits_{n}f^{n}(f^{n-1}(\cdots(f^{1}(f^{0}(X))))\rightarrow f_{\infty}(X)$
Notice that we usually expect to apply the same map each time $f^{n}(x)=f(x)$, however, for a computational theory we may need more flexibility.
##### Example: Linear Flow Map
The simplest example of a flow map is the simple linear map
$X\rightarrow CX$
so that
$f(X)\sim CX$
where C is a non-negative, low rank matrix
$\mid C\mid\,\ll\,\mid X\mid$
We have seen this before: this leads to a Convex form of NonNegative Matrix Factorization NMF
Convex NMF applies when we can specify the feature space and where the data naturally clusters. Here, there are a few instances that are archetypes that define the convex hull of the data.
$\{\mathbf{x}_{c}\}\in X\,,c=1,\cdots,\mid C\mid$
Amazingly, many clustering problems are provably convex–but that’s a story for another post.
##### Example: Manifold Learning
Near a fixed point, we commonly approximate the flow map by a linear operator
$f_{\infty}(X) \sim\mathbf{L}(X)$
This lets us capture the structure of the true data manifold, and is usually described by the low lying eigen-spectra of
$\mathbf{L}(X)=\sum\limits_{i=1}\lambda_{i}\hat{\mathbf{v}}_{i}\sim\lambda_{0}\hat{\mathbf{v}}_{0}+\lambda_{1}\hat{\mathbf{v}}_{1}+\cdots$.
In the same spirit, Semi & Unsupervised Manifold Learning, we model the data using a Laplacian operator $\mathbf{L}(\sigma)$, usually parameterized by a single scale parameter $\sigma$.
These methods include Spectral Clustering, Manifold Regularization , Laplacian SVM, etc. Note that manifold learning methods, like the Manifold Tanget Classifier, employ Contractive Auto Encoders and use several scale parameters to capture the local structure of the data manifold.
#### The Renormalization Group
In chemistry and physics, we frequently encounter problems that require a multi-scale description. We need this for critical points and phase transitions, for natural crashes like earthquakes and avalanches, for polymers and other macromolecules, for strongly correlated electronic systems, for quantum field theory, and, now, for Deep Learning.
A unifying idea across these systems is the Renormalization Group (RG) Theory.
Renormalization Group Theory is both a conceptual framework on how to think about physics on multiple scales as well as a technical & computational problem solving tool.
Ken Wilson won the 1982 Nobel Prize in Physics for the development and application of his Momentum Space RG theory to phase transitions.
We used RG theory to model the recent BitCoin crash as a phase transition.
Wilson invented modern multi-scale modeling; the so-called Wilson Basis was an early form of Wavelets. Wilson was also a big advocate of using supercomputers for solving problems. Being a Nobel Laureate, he had great success promoting scientific computing. It was thanks to him I had access to a Cray Y-MP when I was in high school because he was a professor at my undergrad, The Ohio State University.
Here is the idea. Consider a feature map which transforms the data $X$ to a different, more coarse grain scale
$x\rightarrow\phi_{\lambda}(x)$
The RG theory requires that the Free Energy $F(x)$ is rescaled, to reflect that
the Free Energy is both Size-Extensive and Scale Invariant near a Critical Point
This is not obvious — but it is essential to both having a conceptual understanding of complex, multi scale phenomena, and it is necessary to obtain very highly accurate numerical calculations. In fact, being size extensive and/or size consistent is absolutely necessary for highly accurate quantum chemistry calculations of strongly correlated systems. So it is pretty amazing but perhaps not surprising that this is necessary for large scale deep learning calculations also!
##### The Fundamental Renormalization Group Equation (RGE)
$\mathcal{F}(x)=g(x)+\dfrac{1}{\lambda}\mathcal{F}(\phi_{\lambda}(x))$
If we (can) apply the same map, $F(x)$, repeatedly, we obtain a RG recursion relation, which is the starting point for most analytic work in theoretical physics. It is usually difficult to obtain an exact solution to the RGE.
Many RG formulations both approximate the exact RGE and/or only include relevant variables. To describe a multiscale system, it is essential to distinguish between these relevant and irrelevant variables.
##### Example: Linear Rescaling
Let’s say the feature map is a simple linear rescaling
$\phi(x)=\lambda x$
We can obtain a very elegant, approximate RG solution where F(x) obeys a complex (or log-periodic) power law.
$\mathcal{F}(x)\sim x^{-(\alpha+i\beta)}$
This behavior is thought to characterize Per-Bak style Self-Organized Criticality (SOC), which appears in many natural systems–and perhaps even in the brain itself. Which leads to the argument that perhaps Deep Learning and Real Learning work so well because they operate like a system just near a phase transition–also known as the Sand Pile Model--operating at a state between order and chaos.
#### the Kadanoff Variational Renormalization Group (1975)
Leo Kadanoff, now at the University of Chicago, invented some of the early ideas in Renormalization Group. He is most famous for the Real Space formulation of RG, sometimes called the Block Spin approach. He also developed an alternative approach, called the Variational Renormalization Group (VRG, 1975), which is, remarkably, what Unsupervised DBNs are implementing!
Let’s consider a traditional Neural Network–a Hopfield Associative Memory (HAM). This is also known as the Ising model or a Spin Glass in statistical physics.
An HAM consists of only visbile units; it stores memories explicitly and directly in them:
We specify the Energy — called the Hamiltonian $\mathcal{H}$ — for the nodes. Note that all the nodes are visible. We write
$\mathcal{H}^{HAM}=-\sum\limits_{i}B_{i}v_{i}-\sum\limits_{i}J_{i,j}v_{i}v_{j}$
The Hopfield model has only single $B_{i}$ and pair-wise $J_{i,j}$ interactions.
A general Hamiltonian might have many-body, multi-scale interactions:
$\mathcal{H}(v)=-\sum\limits_{i}K_{i}v_{i}-\sum\limits_{i}K_{i,j}v_{i}v_{j}-\sum\limits_{i,j,k}K_{i,j,k}v_{i}v_{j}v_{k}-\cdots$
The Partition Function is given as
$\mathcal{Z}=\sum\limits_{v}e^{-\mathcal{H}(v)}$
And the Free Energy is
$\mathcal{F}^{v}=-\ln\mathcal{Z}=-\ln\sum\limits_{v}e^{-\mathcal{H}(v)}$
The idea was to mimic how our neurons were thought to store memories–although perhaps our neurons do not even do this.
Either way, Hopfield Neural Networks have many problems; most notably they may learn spurious patterns that never appeared in the training set. So they are pretty bad memories.
Hinton created the modern RBM to overcome the problems of the Hopfield model. He used hidden units to represent the features in the data–not to memorize the data examples directly.
An RBM is specified Energy function for both the visible and hidden units
$\mathbf{E}(v,h)=\mathbf{v}^{t}\mathbf{b}+\mathbf{v}^{t}\mathbf{W}\mathbf{h}+\mathbf{c}^{t}\mathbf{h}$
This also defines joint probability of simultaenously observing a configuration of hidden and visible spins
$P(v,h)=\dfrac{e^{-\mathbf{E(v,h)}}}{\mathcal{Z}}$
which is learned variationally, by minimizing the reconstruction error…or the cross entropy (KL divergence), plus some regularization (Dropout), using Greedy layer-wise unsupervised training, with the Contrastive Divergence (CD or PCD) algo, …
The specific details of an RBM Energy are not addressed by these general concepts; these details do not affect these arguments–although clearly they matter in practice !
It turns out that
Introducing Hidden Units in a Neural Network is a Scale Renormalization.
When changing scale, we obtain an Effective Hamiltonian $\tilde{\mathcal{H}}$ that acts on a the new feature space (i.e the hidden units)
$\mathcal{H}(v)\rightarrow\tilde{\mathcal{H}}(h)$
or, in operator form
$\tilde{\mathcal{H}}(h)=\mathbb{R}[\mathcal{H}(v)]$
This Effective Hamiltonian is not specified explicitly, but we know it can take the general form (of a spin funnel, actually)
$\tilde{\mathcal{H}}(h)=-\sum\limits_{i}\tilde{K}_{i}h_{i}-\sum\limits_{i}\tilde{K}_{i,j}h_{i}h_{j}-\sum\limits_{i,j,k}\tilde{K}_{i,j,k}h_{i}h_{j}h_{k}\cdots$
The RG transform preservers the Free Energy (when properly rescaled):
$\mathcal{F}^{h}\sim\mathcal{F}^{v}$
where
$\mathcal{F}^{h}=-\ln\sum\limits_{v}e^{\mathcal{H}(h)}$
$\mathcal{F}^{v}=-\ln\sum\limits_{v}e^{-\tilde{\mathcal{H}}(h)}$
##### Critical Trajectories and Renormalized Manifolds
The RG theory provides a way to iteratively update, or renormalize, the system Hamiltonian. Each time we add a layer of hidden units (h1, h2, …), we have
$\mathcal{H}^{RG}(\mathbf{v},\mathbf{h1})=\mathbb{R}[\mathcal{H}(\mathbf{v})]$
$\mathcal{H}^{RG}(\mathbf{v},\mathbf{h1},\mathbf{h2})=\mathbb{R}[\mathbb{R}[\mathcal{H}(\mathbf{v})]$
$\cdots$
We imagine that the flow map is attracted to a Critical Trajectory which naturally leads the algorithm to the fixed point. At each step, when we apply another RG transform, we obtain a new, Renormalized Manifold, each one closer to the optimal data manifold.
Conceptually, the RG flow map is most useful when applied to critical phenomena–physical systems and/or simple models that undergo a phase transition. And, as importantly, the small changes in the data should ‘wash away’ as noise and not affect the macroscopic / critical phenomena. Many systems–but not all–display this.
Where Hopfield Nets fail to be useful here, RBMs and Deep Learning systems shine.
We now show that these RG transformations are achieved by stacking RBMs and solving the RBM inference problem!
As in many physics problems, we break the modeling problem into two parts: one we know how to solve, and one we need to guess.
1. we know the Hamiltonian at the most fine grained level of scale $\mathcal{H}(v)$
2. we seek the correlation $\mathbf{V}(v,h)$ that couples to the next level scale
The joint Hamiltonian, or Energy function, is then given by
$\mathcal{H}(\mathbf{v,h})=\mathcal{H}(\mathbf{v})+\mathbf{V(v,h)}$
The Correlation V(v,h) is defined so that the partition function $\mathcal{Z}$ is not changed
$\sum\limits_{h}e^{-\mathbf{V}\mathbf{(v,h})}=1$
This gives us
$\mathcal{Z}=\sum_{v}e^{-\mathcal{H}(v)}=\sum\limits_{v}\sum\limits_{h}e^{-\mathbf{V}(v,h)}e^{-\mathcal{H}(v)}$
(Sometimes the Correlation V is called a Transfer Operator T, where V(v,h)=-T(v,h) )
We may now define a renormalized effective Hamilonian $\tilde{\mathcal{H}(h)}$ that acts only on the hidden nodes
$\tilde{\mathcal{H}}(h)=\ln\sum\limits_{v}e^{-\mathbf{V}(v,h)}e^{-\mathcal{H}(v)}$
so that we may write
$\mathcal{Z}=\sum\limits_{h}e^{-\tilde{\mathcal{H}}(h)}$
Because the partition function does not change, the Exact RGE preserves the Free Energy (up to a scale change, we we subsume into $\mathcal{Z})$
$\Delta\tilde{\mathcal{F}}=\tilde{\mathcal{F}}^{h}-\mathcal{F}^{v}=0$
We generally can not solve the exact RGE–but we can try to minimize this Free Energy difference.
What Kadanoff showed, way back in 1975, is that we can accurately approximate the Exact Renormalization Group Equation by finding a lower bound using this formalism
Deep learning appears to be a real-space variational RG technique, specifically applicable to very complex, inhomogenous systems where the detailed scale transformations have to be learned from the data
#### RBMs expressed using Variational RG
We will now show how to express RBMs using the VRG formalism and provide some intuition
In an RBM, we simply want to learn the Energy function directly; we don’t specify the Hamiltonian for the visible or hidden units explicitly, like we would in physics. The RBM Energy is just
$\mathbf{E}^{RBM}(v,h)=\mathcal{H}^{RBM}(v)-\mathbf{V}(v,h)$
We identify the Hamiltonian for the hidden units as the Renormalized Effective Hamiltonian from the VRG theory
$\mathbf{H}^{RBM}(h)=\hat{\mathcal{H}}(h)$
##### RBM Hamiltonians / Marginal Probabilities
To obtain RBM Hamiltonians for just the visible $\mathcal{H}^{RBM}(v)$ or hidden $\mathcal{H}^{RBM}(h)$ nodes, we need to integrate out the other nodes; that is, we need to find the marginal probabilities.
$P(v)=\sum\limits_{h}P(v,h)=\dfrac{e^{-\mathcal{H}^{RBM}(v) }}{\mathcal{Z}}=\dfrac{1}{\mathcal{Z}} \sum\limits_{h}e^{-\mathbf{E(v,h)}}$
$\mathcal{H}^{RBM}(v)=-\ln\sum\limits_{h}e^{-\mathbf{E(v,h)}}$
and
$P(h)=\sum\limits_{v}P(v,h)=\dfrac{e^{-\mathcal{H}^{RBM}(h) }}{\mathcal{Z}}=\sum\limits_{v}\dfrac{e^{-\mathbf{E(v,h)}}}{\mathcal{Z}}$
$\mathcal{H}^{RBM}(h)=-\ln\sum\limits_{v}e^{-\mathbf{E(v,h)}}$
##### Training RBMs
To train an RBM, we apply Contrastive Divergence (CD), or, perhaps today, Persistent Contrastive Divergence (PCD). We can kindof think of this as slowly approximating
$\dfrac{\partial}{\partial\theta}\ln\mathcal{Z}(\theta)$
In practice, however, RBM training minimizes the associated Free Energy difference $\Delta\mathbf{F}$ … or something akin to this…to avoid overfitting.
In the “Practical Guide to Training Restricted Boltzmann Machines”, Hinton explains how to train an RBM (circa 2011). Section 6 addresses “Monitoring the overfitting”
“it is possible to directly monitor the overfitting by comparing the free energies of training data and held out validation data…If the model is not overfitting at all, the average free energy should be about the same on training and validation data”
##### Other Objective Functions
Modern variants of Real Space VRG are not “‘forced’ to minimize the global free energy” and have attempted other approaches such as Tensor-SVD Renormalization. Likeswise, some RBM / DBM approaches do likewise may minimize a different objective.
In some methods, we minimize the KL Divergence; this has a very natural analog in VRG language [1].
#### Why Deep Learning Works: Lessons from Theoretical Physics
The Renormalization Group Theory provides new insights as to why Deep Learning works so amazingly well. It is not, however, a complete theory. Rather, it is framework for beginning to understand what is an incredibly powerful, modern, applied tool. Enjoy!
#### References
[10] http://www-math.unice.fr/~patras/CargeseConference/ACQFT09_JZinnJustin.pdf
[14] THE RENORMALIZATION GROUP AND CRITICAL PHENOMENA, Ken Wilson Nobel Prize Lecture
[16] The Potential Energy of an Autoencoder, 2014
# Why does Deep Learning work?
Why does Deep Learning work?
This is the big question on everyone’s mind these days. C’mon we all know the answer already:
“the long-term behavior of certain neural network models are governed by the statistical mechanism of infinite-range Ising spin-glass Hamiltonians” [1] In other words,
Multilayer Neural Networks are just Spin Glasses? Right?
This is kinda true–depending on what you mean by a spin glass.
In a recent paper by LeCun, he attempts to extend our understanding of training neural networks by studying the SGD approach to solving the multilayer Neural Network optimization problem [1]. Furthermore, he claims
None of these works however make the attempt to explain the paradigm of optimizing the highly non-convex neural network objective function through the prism of spin-glass theory and thus in this respect our approach is very novel. And, again, this is kinda true
But here’s the thing…we already have a good idea of what the Energy Landscape of multiscale spin glass models* look like–from early theoretical protein folding work (by Wolynes, Dill, etc [2,3,4]). In fact, here is a typical surface:
*[technically these are Ising spin models with multi-spin interactions]
Let us consider the nodes, which above represent partially folded states, as nodes in a multiscale spin glass–or , say, a multilayer neural network. Immediately we see the analogy and the appearance of the ‘Energy funnel’ In fact, researchers have studied these ‘folding funnels’ of spin glass models over 20 years ago [2,3,4]. And we knew then that
as we increase the network size, the funnel gets sharper
Note: the Wolynes protein-folding spin-glass model is significantly different from the p-spin Hopfield model that LeCun discusses because it contains multi-scale, multi-spin interactions. These details matter.
Spin glasses and spin funnels are quite different. Spin glasses are highly non-convex with lots of local minima, saddle points, etc. Spin funnels, however, are designed to find the spin glass of minimal-frustration, and have a convex, funnel shaped, energy landscape.
This seemed to be necessary to resolve one of the great mysteries of protein folding: Levinthal’s paradox [5]. If nature just used statistical sampling to fold a protein, it would take longer than the ‘known’ lifetime of the Universe. It is why Machine Learning is not just statistics.
Deep Learning Networks are (probably) Spin Funnels
So with a surface like this, it is not so surprising that an SGD method might be able to find the Energy minima (called the Native State in protein folding theory). We just need to jump around until we reach the top of the funnel, and then it is a straight shot down. This, in fact, defines a so-called ‘folding funnel’ [4]
So is not surprising at all that SGD may work.
Recent research at Google and Stanford confirms that the Deep Learning Energy Landscapes appear to be quite smooth and generally convex! [6]
Note that a real theory of protein folding, which would actually be able to fold a protein correctly (i.e. Freed’s approach [7]), would be a lot more detailed than a simple spin glass model. Likewise, real Deep Learning systems are going to have a lot more engineering details–to avoid overtraining (Dropout, Pooling, Momentum) than a theoretical spin funnel.
It is not that Deep Learning is non-convex–is that we need to avoid over-training
Still, hopefully we can learn something using the techniques developed to study the energy landscape of multi-scale spin glass/ spin funnels models. [8,9], thereby utilizing methods theoretical chemistry and condensed matter physics.
Indeed, I believe this is the first conjecture that Supervised Deep Learning is related to a Spin Funnel. In the next post, I will examine the relationship between Unsupervised Deep Learning and the Variational Renormalization Group [10].
[1] LeCun et. al., The Loss Surfaces of Multilayer Networks, 2015
[3] THEORY OF PROTEIN FOLDING: The Energy Landscape Perspective, Annu. Rev. Phys. Chem. 1997
[4] From Levinthal to pathways to funnels, Nature, 1997
[6] QUALITATIVELY CHARACTERIZING NEURAL NETWORK OPTIMIZATION PROBLEMS, Google Research (2015)
[8] Funnels in Energy Landscapes, 2007
[10] A Common Logic to Seeing Cats and Cosmos, 2014
# Convex Relaxations of Transductive Learning
This post is finishing up some older work of an R&D / coding project in Transductive Learning.
Why are SVMs interesting? It is just a better way to do Logistic Regression? Is it the Kernel Trick? And does this even matter now that Deep Learning is everywhere? To the beginning student of machine learning, SVMs are the first example of a Convex Optimization method. To the advanced practitioner, SVMs are the starting point to creating powerful Convex Relaxations to hard problems.
Historically, convex optimization was seen as the path to central planing an entire economy. A great new book, Red Plenty, is “about the scientists who did their genuinely brilliant best to make the dream come true …” [amazon review].
It was a mass delusion over the simplex method, and it is about as crazy as our current fears over Deep Learning and AI.
Convex optimization is pretty useful, as long as we don’t get crazy about it.
The prototypical method convex relaxation is for Transductive Learning and the Transductive SVM (TSVM)
Vapnik proposed the idea of Transduction many years ago; indeed the VC theory is proven using Transduction. I would bet that he knew a TSVM could be convexified–although I would need a job at Facebook to verify this.
A good TSVM has been available since 2001 in SvmLight, But SvmLight is not opensource, so most people use SvmLin.
Today there are Transductive variants of Random Forests, Regression, and even Deep Learning. There was even a recent Kaggle Contest–the Black Box Challenge (and, of course, the Deep Learning method won). Indeed, Deep Learning classifiers may benefit greatly from Transductive/SemiSupervised pretraining with methods like Pseudo Label [8], as shown in the Kaggle The National Data Science Bowl contest.
We mostly care about binary text classification, although there is plenty of research in convex relaxation for multiclass transduction, computer vision, etc.
Transductive learning is essentially like running a SVM, but having to guess a lot of the labels. The optimization problem is
$\min_{\mathbf{y}\in\mathcal{B}}\min_{f}\,\,\Omega(f)+\sum\limits_{i=1}^{N}\lambda\mathcal{L}(\mathbf{x}_{i},y_{i})$
where $\mathcal{L}$ is the loss function , $\Omega$ the regularization function,
and the binary labels $\mathbf{y_i}\in\mathcal{B}\mid y_{i}\in\left\{1,\,-1,\,unk\,\right\}$ are only partially known.
The optimization is a non-convex, mixed-integer problem. Amazingly, we can reformulate the TSVM to obtain a convex optimization!
This is called a Convex Relaxation, and it lets us guess the unknown labels…
to within a good approximation, and using some prior knowledge.
Proving an approximation is truly convex is pretty hard stuff, but the basic idea is very simple. We just want to find a convex approximation to a non-convex function.
It has been known for a while that the TSVM problem can be convexified [5]. But it has been computationally intractable and there is no widely available code.
We examine a new Convex Relaxation of the Transductive Learning called the Weakly Labeled SVM (WellSVM) [2,3].
In the Transductive SVM (TSVM) approach, one selects solutions with minimum SVM Loss (or Slack)
$\mathcal{L}(\mathbf{x}_{i},y_{i})=\xi_{i}$
and the maximum margin
$\Omega=\dfrac{1}{2}\Vert\mathbf{w}\Vert^{2}$
#### The SVM Dual Problem
Let us consider the standard SVM optimization
$\underset{\mathbf{w,\xi}}{\min}\,\,\dfrac{1}{2}\parallel\mathbf{w}\parallel^{2}+\lambda\sum_{i=1}^{N}\xi_{i}$
which has the dual form
$\underset{\alpha}{\max}\,\,\alpha^{\dagger}\mathbf{1}-\dfrac{1}{2}\mathbf{y^{\dagger}}\boldsymbol\alpha^{\dagger}X\mathbf{X}\boldsymbol\alpha\mathbf{y}\quad,\alpha_{i}>0$
To keep the notation simpler, we will not consider the Kernalized form of the algorithm. Besides, we are mostly interested in text classification, and Kernels are not needed.
#### Balancing the Labels
In a TSVM, we have to guess the labels $y_{i}\in\left\{1,\,-1\right\}$ and select the best solution (for a given set of regularization parameters). There are way too many labes to guess, so
we need to constrain the label configurations by balancing the guesses
We assume that we have some idea of the total fraction of positive (+) labels $\mathcal{B}_{+}$
• Perhaps we can sample them?
• Perhaps we have some external source of information.
• Perhaps we can estimate it.
This is, however, a critical piece of prior information we need. We define the space of possible labels as
$\left\{ \mathbf{y}\vert\sum_{i=1}^{N}y_{i}=\beta\right\}$
i.e, for exactly half positive / negative labels, then
$\mathcal{B}_{+}=\dfrac{1}{2}$
and the true mean label value is zero
$y_{avg}=\bar{y}=0$
So we are saying that if we know
1. the true fraction $\mathcal{B}_{+}$ of (+) labels
2. some small set of the true labels (i.e. < 5%)
3. the features (i.e. bag-of-words for text classification)
Then we know almost all the labels exactly! And that is powerful.
Note: iI is critical that in any transductive method, we reduce the size of the label configuration space. The Balancing constraint is the standard constraint–but it may be hard to implement in practice. To me, personally, the problem resembles the matrix product states method in quantum many body theory. And I suspect that, eventually, there will be a related method found for transducitive learning. Perhaps even Deep Learning has found this.
#### Convex Methods
The popular SvmLin method use a kind of Transductive Meta-Heuristics that set the standard for other approaches. The problem is, we never really know if we have the best solution. And it is not easy to extend to multiclass classification.
Convex methods have been the method of choice since Dantzig popularized the simplex method in 1950
Although linear programming itself was actually invented in 1939 by Leonid Kantorovich [7] — “the only Soviet scholar ever to win the Nobel Prize for Economics”
A convex method lends itself to production code that anyone can run.
#### The WellSVM Convex Relaxation
More generally, we need to solve a non-convex min-max problem of the form
$\underset{\mathbf{y\in\mathcal{B}}}{\min}\,\,\underset{\alpha\in\mathcal{A}}{\max}\, \,G(\mathbf{y,\alpha})$
where the G matrix is
$G(\mathbf{y,\alpha})=\alpha^{\dagger}\mathbf{1}-\dfrac{1}{2}\mathbf{y^{\dagger}}\boldsymbol\alpha^{\dagger}\mathbf{X^{\dagger}X}\boldsymbol\alpha\mathbf{y}$
where α lies in the convex set
$\mathcal{A}=\left\{\boldsymbol\alpha\mid C\mathbf{1}\geq\alpha\geq 0\right\}$
Notice that G is concave in α and (can be made) linear in the y’s [3]. We seek a convex relaxation of this min-max problem. And, as importantly, we want to code the final problem using an off-the-shelf SVM solver with some simple mods. The steps are
##### 1 . Apply the Minimax Theorem
For details, see [4,5], although it was originally posed by John von Neumann in his work on Game Theory.
One could spend a lifetime stuyding von Neumann’s contributions. Quantum mechanics. Nuclear Physics. Etc.
Here we scratch the surface.
The Minimax thereom lets us switch the order of the min/max bounds.
$\underset{\mathbf{y\in\mathcal{B}}}{\min}\,\,\underset{\alpha\in\mathcal{A}}{\max}\, \,G(\mathbf{y,\alpha})\rightarrow\underset{\alpha\in\mathcal{A}}{\max}\,\,\underset{\mathbf{y\in\mathcal{B}}}{\min}\, \,G(\mathbf{y},\alpha)$
The original problem is an upper bound to this. That is
$\underset{\mathbf{y\in\mathcal{B}}}{\min}\,\,\underset{\alpha\in\mathcal{A}}{\max}\, \,G(\mathbf{y,\alpha}) \geqslant\,(upper bound)\,\underset{\alpha\in\mathcal{A}}{\max}\,\,\underset{\mathbf{y\in\mathcal{B}}}{\min}\, \,G(\mathbf{y},\alpha)$
To solve this, we
##### 2. dualize the inner minimization (in the space of allowable labels)
We convert the search over possible label configurations into the dual max problem, so that the label configurations become constraints.
$\underset{\alpha\in\mathcal{A}}{\max}\,\,\underset{\mathbf{y\in\mathcal{B}}}{\min}\, \,G(\mathbf{y},\alpha)=\underset{\alpha\in\mathcal{A}}{\max}\,\left\{ \underset{\mathbf{\theta}}{\max}\,\,G(\mathbf{y_{t}},\alpha)\geq\theta\vert\mathbf{y_{t}}\in\mathcal{B}\right\}$
This linear in α and θ. In fact, it is convex.
There are an exponential number of constraints in $\mathcal{B}$.
Even though it is convex, we can not solve this exactly practice. And that’s…ok.
Not all possible labelings matter, so not all of these constraints are active (necessary) for an optimal solution.
We just need an active subset, $\left\{\mathbf{y_{t}}\in\mathcal{C}\right\}$, which we can find by …
The Cutting Plane, or Gomory Chvatal, method. (Remember, if you want to sound cool, give your method a Russian name).
To proceed, we construct the Lagrangian and then solve the convex dual problem.
You may remember the method of Lagrange multipliers from freshman calculus:
##### 3. We introduce Lagrange Multipliers for each label configuration
The Lagrangian is
$\theta+\underset{\boldsymbol\mu,\mathbf{y_{t}}\in\mathcal{B}}{\sum}\mu_{t}(G(\mathbf{y_{t}},\alpha)-\theta)$
When we set the derivative w.r.t. $\theta$ to 0, we find
$\sum\boldsymbol\mu_{t}=1$
This lets us rewrite the TSVM as
$\underset{\alpha\in\mathcal{A}}{\max}\,\,\underset{\mathbf{\mu}\in\mathcal{M}}{min}\,\,\underset{\boldsymbol\mu,\mathbf{y_{t}}\in\mathcal{B}}{\sum}\mu_{t}G(\mathbf{y_{t}},\alpha)$
where the set of allowable multiplers $\boldsymbol\mu$ lies in the simplex $\mathcal{M}$
$\mathcal{M}=\left\{\boldsymbol\mu\mid\sum\mu_{t}=1\,\,,\mu_{t}\geq 0\right\}$
The resulting optimization is convex in µ and concave in α.
This is critical as it makes a Kernel-TSVM a Multiple Kernel Learning (MKL) method.
“WellSVM maximizes the margin by generating the most violated label vectors iteratively, and then combines them via efficient multiple kernel learning techniques”.
For linear applications, we need only consider 1 set of $\boldsymbol\mu$.
Now replace the inner optimization subproblem with its dual. Of course, the dual problem is a lower bound on the optimal value. We then
##### 4. switch the order of the min/max bounds back
to obtain a new min-max optimization–a convex relaxation of the original
$\underset{\boldsymbol\mu\in\mathcal{M}}{min}\,\,\underset{\alpha\in\mathcal{A}}{\max}\,\,\underset{\boldsymbol\mu,\mathbf{y_{t}}\in\mathcal{B}}{\sum}\mu_{t}G(\mathbf{y_{t}},\alpha)$
When we restrict the label configurations to the working set $\left\{\mathbf{y_{t}}\in\mathcal{C}\right\}$,we have
$\underset{\boldsymbol\mu\in\mathcal{M}}{min}\,\,\underset{\alpha\in\mathcal{A}}{\max}\,\,\underset{\boldsymbol\mu,\mathbf{y_{t}}\in\mathcal{C}}{\sum}\mu_{t}G(\mathbf{y_{t}},\alpha)$
#### Implementation Details
A key insight of WellSVM is that the core label search
$\underset{\mathbf{\hat{y}}\in\mathcal{C}}{\min}\,\,G(\mathbf{\hat{y}},\alpha)$
is equivalent to
$\underset{\mathbf{\hat{y}}\in\mathcal{C}}{\max}\,\,\mathbf{\hat{y}^{\dagger}\boldsymbol\alpha^{\dagger}X^{\dagger}X\boldsymbol\alpha\hat{y}}$
To me, this is very elegant!
We search the convex hull of the document-document density matrix, weighted by the Langrange multipliers for the labels.
How can we solve a SVM with exponential constraints? Take a page from old-school Joachim’s Structural SVMs [6].
##### Cutting Plane Algorithm for WellSVM
This is the goto-method for all Mixed Integer Linear Programming (MILP) problems. On each iteration, we
• obtain the Lagrange Multipliers $\boldsymbol\alpha$ with an off-the-shelf SVM solver
• find a violating constraint (label configuration) $\mathbf{\hat{y}}$
This grows the active set of label configurations $\mathcal{C}$, learning from previous guesses. We expect $latex N_{\mathcal{C}}\ll N_{\mathcal{B}}$
the PseudoCode is
1. Initialize $\mathbf{\hat{y}}$ and $\mathcal{C}=\emptyset$
2. repeat
3. Update $\mathcal{C}\leftarrow\mathbf{\hat{y}}\cap\mathcal{C}$
4. Obtain the optimal α from a dual SVM solver
5. Generate a violated $\mathbf{\hat{y}}$
6. until $G(\boldsymbol\alpha,\mathbf{\hat{y}})>\underset{\mathbf{y}\in\mathcal{C}}{min}G(\boldsymbol\alpha,\mathbf{y})-\epsilon$
##### Finding Violated Constraints
The cutting plane algo finds a new constraint, or cuts, on each iteration, to ‘chip away’ at a problem until the inner convex hull is found. It usually finds the most violated constraint on each iteration, however,
With SVMs we can get good results just finding any violated constraint.
Here, we seek $\mathbf{y*}$, a violated label assignment. The WellSVM paper [2,3] provides a great solution.
For any violation $\mathbf{y*}$, and for any pair of label configurations $\mathbf{\bar{y},y}$, we have
$\mathbf{y^{\dagger}Hy*}\neq\mathbf{y^{\dagger}H\bar{y}}$
where $\mathbf{H}=\boldsymbol\alpha^{\dagger}\mathbf{X^{\dagger}X}\boldsymbol\alpha$
This lets us compute $\mathbf{y*}$ in two steps:
First, compute $\mathbf{\bar{y}}$, by searching the current, active, and usually small set $\mathcal{C}$
$\mathbf{\bar{y}}=\arg\max_{\mathbf{y}\in\mathcal{C}}\,\mathbf{y^{\dagger}Hy}$
Second, compute $\mathbf{y*}$, by searching the set of all balanced label configurations $\mathcal{B}$
$\mathbf{y*}=\arg\max_{\mathbf{y}\in\mathcal{B}}\,\mathbf{y^{\dagger}H\bar{y}}$
#### Original Matlab Code
We will review the existing code and run some simulations soon
#### Python Implementation
The goal is to modify scikit learn interface to liblinear to output the Lagrange multipliers $\boldsymbol\alpha$. then to implement WellSVM in an ipython notebook. stay tuned.
#### References
[2] Convex and Scalable Weakly Labeled SVMs , JMLR (2013)
[3] Learning from Weakly Labeled Data (2013)
[4] Kim and Boyd, A Minimax Theorem with Applications to Machine Learning, Signal Processing, and Finance, 2007
[7] “This is an example of Stigler’s law of eponymy, which says that “no scientific discovery is named after its original discoverer.” Stigler’s law of eponymy is itself an example of Stigler’s law of eponymy, since Robert Merton formulated similar ideas much earlier Quote from Joe Blitzstein on Quora
# The Bitcoin Crash and How Nature Works
2 years of bitcoin 2103-2015
Moreover, several Bitcoin exchanges have shut down due to hacks and/or criminal activity, and key BitCoin players are on trial. This begs the question:
Why is BitCoin Crashing ?
Markets crash all the time. Stock markets. Currency Markets. Even the Dutch Tulip market of 1637 crashed–although even this is still hotly debated.
Fraud? Speculation? Mania? Lack of Regulation? What gives?
### Market Crashes: the EconPhysics Perspective
In 1996, researchers at the University of Chicago[1] and elsewhere [2] independently proposed that market crashes resemble physical crashes such as earthquakes, ruptures, and avalanches. Such phenomena arise from long range, collective fluctuations — i.e. herding behavior — that overtakes a system as it approaches a critical point…and/or in the aftershocks.
The theory relies upon a fairly generous application of the theory of critical phenomena and the Renormalization Group (RG) theory. It provides both a qualitative description of the phenomena and quantitative predictions such as when a crash may occur or how long a bear market will continue.
Most notably, the strongest proponent, Didier Sornette, accurately estimated the long bear run of the Japanese markets–what he terms a Financial Anti-Bubble:
He has written numerous papers, a book, and even a TED Talk.
We also introduced these ideas in a previous post: Noisy Time Series II: Earth Quakes, Black Holes, and Machine Learning
### Power Laws and Symmetry Breaking
It is well known that financial markets display non-Gaussian, long tail, power law-like behavior. After a crash model, the price series itself $p(t)$may follow a power law
$p(t)=A+B(t-t_{c})^{\alpha}$
where $t_{c}$ is the time of the crash, the time $t>t_{c}$, and $\alpha$ is a real number.
We can fit the recent Bitcoin EOD (End of Day) prices to a power law, starting at it’s peak on Nov 11, 2013:
We obtain a modestly good least squares fit, with $\alpha=0.32$ and $R^{2} = 0.77$. Clearly there is a lot of noise–or is there?
RG theory suggests that near a crash, the power law may become complex (i.e a form of ‘symmetry breaking’). We will now sketch how this comes about, using the nobel prize winning Renormalization Group theory.
#### Complex Power Laws and the Renormalization Group
When examining a price series $p(t)$, or some other phenomena, say $F(x)$, a critical point occurs when we observe a rapid, sharp spike–i.e. Bitcoin’s fast rise up to Nov 29, 2013.
Here, we will find that as $x\rightarrow 0$
$F(x)=A+B(x)^{\alpha}$,
where $\alpha$ is a complex critical exponent
We say $F(x)$ becomes singular near a crash when there exists a k-th derivative that becomes infinite as $x\rightarrow 0$:
$\lim_{x\rightarrow0}\dfrac{d^{k}F(x)}{dx^{k}}=0$
We also know that near a critical point, physical systems exhibit scale invariant (i.e. fractal) behavior. The noise, or fluctuations, we observe on small time scales $x=t-t_{c}$ look just like fluctuations on larger time scales $\phi(x)$.
We call $\phi(x)$ the RG flow map. The flow map describes these change of time scales, and the RG equation describes the invariances.
The fundamental RG equation (eqn) is:
$x\rightarrow\phi(x)$
$F(x)\rightarrow g(x)+\dfrac{1}{\mu}F(\phi(x))$
We assume $F(x)$ is continuous–which is non trivial for price series –and that $\phi(x)$ is differentiable. $g(x)$ is the non-singular (regular) part of $F(x)$.
The magic of the RG theory is that it allows us to describe the behavior near a critical point knowing only the flow map $\phi(x)$ and the regular function $g(x)$. The formal solution is an infinite order power series
$F(x)=\underset{n\rightarrow\infty}{\lim}f^{n}(x)$,
$f^{n}(x)=\sum_{i=1}^{n}\dfrac{1}{\mu^{i}}g(\phi^{[i]}(x))$
We will sketch a very simple, leading order approximation to $F(x)$ that leads to a complex power law.
#### RG Theory and Discrete Scale Invariance
We start by first assuming power law behavior (at lowest order):
$F_{0}(x)\sim x^{\alpha}$
We seek the simplest RG solution. Let us assume the RG flow map is linear in $x$:
$\phi(x)=\lambda x+\cdots$,
where $\lambda > 1$ (for technical reasons to ensure the critical point is an unstable fixed point of the RG flow).
We will ignore the regular solution $g(x)$ and write the simplified RG eqn:
$F(\lambda x)=\mu F(x)$
For our simple power law, we have
$(\lambda x)^{\alpha}=\mu x^{\alpha}$
This is easy to satisfy if
$\dfrac{\lambda^{\alpha}}{\mu}=1$
or
$\alpha=\dfrac{log(\mu)}{log(\lambda)}$
We seek the most general solution, applicable to discrete physical systems, such as earthquakes, ruptures in materials–and financial market crashes. That is, we expect $\lambda$ to take on discrete values:
$\lambda\in[\lambda_{1},\lambda_{2},\cdots]$
We call this Discrete Scale Invariance (DSI)
The most general solution of DSI is
$F(x)=x^{\alpha}P(\dfrac{\log x}{\log \lambda})$
where $P()$ is a general periodic function. As above, we express $P()$ as the limit of an infinite power series
$P_{n}(x)=\sum_{i=1}^{n}c_{n}\exp\left[2n\pi i\left(\dfrac{\log x}{\log\lambda}\right)\right]$
This is a classic Weierstrass function–a pathological function that, in the infinite limit, is continuous everywhere and yet differentiable nowhere. It is used to model fractals, natural systems that are scale invariant, and highly discontinous but structured time series.
In a crash scenario, we expect that the true power series for $P(x)$ will diverge; i.e. the k-th derivative explodes. So here, we only take the leading term of our linearized model
$exp\left[2n\pi i\left(\dfrac{\log x}{\log\lambda}\right)\right]$
Note that it takes the form
$F(x)=F_{0}(x)p(\log F_{0}(x))$
so $F_{0}(x)$, simple power law behavior, is like the regular part of the solution.
We now see that we have a complex critical exponent $\alpha$:
$\dfrac{\lambda^{\alpha}}{\mu}=e^{2\pi n}$
or
$\alpha=\dfrac{log(\mu)}{log(\lambda)}+i\dfrac{2\pi n}{log(\lambda)}$
We consider the real part of the complex power law
$Re[x^{\alpha+i\beta}]=x^{\alpha}cos(\beta\log x)$
which gives the following log periodic formula for the price series
$p(t)=A+B(t-t_{c})^{\alpha}(1+C\cos(\omega\log(t-t_{c})+\phi))$
Instead of fluctuating randomly around the underlying power law drift, the price series exhibits log-periodic oscillations–the DSI signature of a crash.
This model appears to work for 1-2 oscillations before (or after) the crash. To model the longer time anti-bubble behavior, Sornette has developed an extended formula, based on the third-order Landau expansion (as opposed to, say, additional terms in the power series defined above). We leave these details and further analysis for later.
#### DSI Fit of the Bitcoin Crash
We now fit the latest Bitcoin behavior, assuming the crash / Bear market started on Nov 29, 2013 :
We readily find a DSI fit, with $R^2=0.89$.
It is actually quite difficult to get a very tight fit, and usually advanced methods are needed. This is a simple, crude fit done to illustrate the basic ideas.
We see that Bitcoin, like other financial markets, displays log periodic behavior, characteristic of a self-organized crash.
These kinds of crashes are not caused by external events or bad players–they are endemic to all markets and result from the cooperative actions of all participants.
We can try to detect them, perhaps even profit off them, but it is unclear how to avoid these Self-organized Critical points that wreck havoc on our finances.
### Conclusion: How Nature Works
Bitcoin is an attempt to create a new kind of currency–a CryptoCurrency–that is free from both institutional control and individual corruption. It is hoped that we can avoid the kind of devastating inflation, devaluation, and crashes that occur all too frequently in our current worldwide banking systems. The promise is that Bitcoin, backed by the Blockchain, can remove our dependence on a single, faulty institution and replace it with a decentralized, distributed form of trust. But
“Real life operates at a critical point between order and chaos”
Our analysis here shows that even the Bitcoin economy still appears to behave like a traditional market–prone to kinds of crashes that frequently arise in all natural self organized systems.
This Self-Organized Criticality is perhaps just how nature works [6,7].
And while Bitcoin is certainly interesting and exciting, perhaps it too is subject to the natural laws of physics.
#### References
[1] 1995 James A. Feigenbaum and Peter G.O. Freund, Discrete Scaling in Stock Markets Before Crashes
[4] D. Sornette, Critical market crashes
[6] 1999 A. Johansen and D. Sornette, Financial “Anti-Bubbles”: Log-Periodicity in Gold and Nikkei collapses
[7] Anders Johansen and Didier Sornette, Evaluation of the quantitative prediction of a trend reversal on the Japanese stock market in 1999 , 2000
[8] 2003 Wei-Xing Zhou and Didier Sornette Renormalization group analysis of the 2000–2002 anti-bubble in the US S&P500 index: explanation of the hierarchy of five crashes and prediction
[9] 20122 Didier Sornette, Ryan Woodard, Wanfeng Yan, and Wei-Xing Zhou Clarifications to Questions and Criticisms on the Johansen-Ledoit-Sornette Bubble Model
[11] Per Bak, “How Nature Works: The Science of Self-Organized Criticality”, 1996
# SVM+ / LUPI: Learning Using Privileged Information
Recently Facebook hired Vapnik, the father of the Support Vector Machine (SVM) and the Vapnik-Chrevoniks Statistical Learning Theory.
Lesser known, Vapnik has also pioneered methods of Transductive and Universum Learning. Most recently, however, he has developed the theory of the Learning Using Privileged Information (LUPI), also known as the SVM+ method. In this post, we examine this most recent contribution.
#### Preface
Once again, we consider the problem of trying to learn a good classifier having a small set of labeled examples $x_{l}\in L$.
Here, we assume we have multiple views of the data; that is, different, statistically independent feature sets $\phi(x),\phi*(x)$, that describe the data.
For example, we might have a set of breast cancer images, and some holistic, textual description provided by a doctor. Or, we might want to classify web pages, using both the text and the hyperlinks.
Multi-View Learning is a big topic in machine learning, and includes methods like SemiSupervised CoTraining, Multiple Kernel Learning, Weighted SVMs, etc. We might even consider Ensemble methods as a kind of Multi-View Learning.
Having any extra, independent info $x^{*}$ about our data is very powerful.
In LUPI, also called the SVM+ method, we consider situation where we only have extra information $x^{*}_{l}$ about the labeled examples $x_{l}\in L$.
Vapnik calls $\{x^{*}\}$ Privileged Information — and shows us how to use it
#### The VC Bounds
If we are going to discuss Vapnik we need to mention the VC theory. Here, we note an important result about soft and hard-margin SVMs: when the training set is non-separable, the VC bounds scale as $\dfrac{h}{\sqrt{L}}$:
$P_{test}\;\le\;\nu_{train}+O\left(\dfrac{VCdim}{\sqrt{L}}\right)$
where $\nu_{train}$ is the training error, and h is the VC dimension
But when the training set is separable, i.e. $\nu_{train}=0$, the VC bounds scale as $\dfrac{h}{L}$
$P_{test}\;\le\;O\left(\dfrac{VCdim}{L}\right)$
This means we can use say L=300 labeled examples as opposed to L=900,000, which is a huge difference.
It turns out we can also achieve $\dfrac{h}{L}$ scaling–if we have an Oracle that tells us, a priori, the slacks. Hence the term Oracle SVM. The Oracle tells us how much confidence we have in each label.
So if we can get the slacks $\{\varepsilon_{l}\}$, or, equivalently, the weights, for each instance, L can be much smaller.
[Of course, this is only true if we can sample all the relevant features.]
#### Soft Margin SVM
In practice, we usually assume data is noisy and non-separable. So we use a soft-margin SVM where we can adjust the amount slack with the $C$ (cost) parameter,
Note that we also have the bias $b$, but we can usually set to zero (still, be careful doing this!).
Let us write the soft-margin SVM optimization as
$\arg\min\left(\dfrac{1}{2}\langle w,w\rangle+C\sum_{l\in L}\varepsilon_{l}\right)$
subject to the constraints
$y_{l}\left(\langle w_{l},x_{l}\rangle+b\right)\ge 1-\varepsilon_{l}\;\;\forall\;l\in L,\;\;\varepsilon_{l}\ge 0$
Notice that while we formally add slack variables $\{\varepsilon_{l}\}$ to max-margin optimization–they eventually vanish. That is, we don’t estimate the slacks; we replace them with the maximum margin violation, or the Hinge Loss error
$\hat{err_{l}}=\max[1-y_{l}(\langle w,x_{l}\rangle+b,0)]$
We then solve the unconstrained soft-margin SVM optimization, which is a convex upper bound to the problem stated above–as explained in this nice presentation on LibLinear.
$\arg\min\left(\dfrac{1}{2}\langle w,w\rangle+C\sum_{l\in L}\max[1-y_{l}(\langle w,x_{l}\rangle+b,0)]\right)$
or, in simpler terms
$\arg\min\left(\dfrac{1}{2}\langle w,w\rangle+C\sum_{l\in L}\hat{err_{l}}\right)$
And this works great–when we have lots of labeled data L.
What if we could actually estimate or assign all the L slack variables $\{\varepsilon_{l}\}$ ? In principle, we can far use labeled examples L. This is the promise of LUPI.
#### LUPI/SVM+
In LUPI, we use the privileged information $\{x^{*}_{l}\}$ to learn the slack variables $\{\varepsilon_{l}\}$:
$\varepsilon_{l}=\langle w^{*},x_{l}^{*}\rangle+b^{*}$
We model the slack as linear functions of the privileged information.
Effectively, the privileged information provides a measure of confidence for each labeled instance.
To avoid overtraining, we apply a max-margin approach. This leads to an extended, SVM+ optimization problem, with only 2 adjustable regularization parameters $C,\gamma$ (and 2 bias params):
$\arg\min\left(\dfrac{1}{2}\langle w,w\rangle+\dfrac{\gamma}{2}\langle w^{*},w^{*}\rangle+C\sum_{l\in L}[\langle w^{*},x_{l}^{*}\rangle+b^{*} ]\right)$
subject to the constraints
$y_{l}\left(\langle w_{l},x_{l}\rangle+b\right)\ge 1-[\langle w^{*},x_{l}^{*}\rangle+b^{*}],$
$\langle w^{*},x_{l}^{*}\rangle+b^{*}\ge 0\;\;\forall\;l\in L$
#### Show me the code?
The SVM+/LUPI problem is a convex (quadratic) optimization problem with linear constraints–although it does require a custom solver. The current approach is to solve the dual problem using a variant of SMO. There are a couple versions (below) in Matlab and Lua. The Matlab code includes related SVM+MTL (SVM+ MultiTask Learning). I am unaware of a open source C or python implementation similar to Svmlin or Universvm.
To overcome the solver complexity, other methods, such as the Learning to Rank method, have been proposed. We will keep an eye out for useful open source platforms that everyone can benefit from.
#### Summary
The SVM+/LUPI formalism demonstrates a critical point about modern machine learning research. If we have multiple views of our data, extra information available during training (or maybe during testing), or just several large Hilbert spaces of features, then we can frequently do better than just dumping everything into an SVM, holding our nose, and turning the crank.
#### References:
or how many ways can we name the same method?
Learning with Hidden Information
The SVM+ method: MIT Reading Notes
Learning with Non-Trivial Teacher
Learning to Rank Using Privileged Information
On the Theory of Learning with Privileged Information
SMO-style algorithms for learning using privileged information
Learning Using Privileged Information: SVM+ and Weighted SVM
Learning with Asymmetric Information
Comparison of SVM+ and related methods (SVM+MTL, rMTL, SVM+ regression, etc.)
#### Videos
Learning with Privileged Information
Simmons Foundation Lecture
#### Software
Matlab and CVX Versions
Milde LUA Code
# Machine Learning with Missing Labels Part 3: Experiments
Another early draft of the next blog post
In this series of posts we look at Transductive and SemiSupervised Learning–an old problem, a hard problem, and a fundamental problem Machine Learning. Unlike Deep Learning or large scale ML,
We want to learn as much as we can from as labeled little data as possible.
We focus on text classification. Why?
1. The feature space is simple — its just words.
2. With the true labels, a simple SVM gives near perfect accuracy.
3. Text satisfies the cluster assumption.
Why isn’t this done all the time?
We can generate good (not great) Semi Supervised text classifiers. But
1. creating the input data is hard.
2. setting the input parameters is harder.
3. we can’t always distinguish the good solutions from the bad.
The 1st post looked at (1); here we try to understand (2&3), and look for simple, practical heuristics to make these old methods useful.
We can think of the TSVM/S3VM as generating several potentially good labelings on the unlabeled data. To solve (3), we need to find the best labeling–which means judging the label and/or classifier quality, such as the label entropy , cluster overlap, etc.
Or, alternatively, we might try to average the good solutions, similar to the way Hinton et. al. have recently suggested averaging an ensemble of models to take advantage of Dark Knowledge.
Thanks to Matt Wescott for pointing out this PDF and discussions on it.
To explore this, we run experiments using some old, existing, open source Transductive SVM (TSVM) and SemiSupervised SVM (S3VM) codes:
Our work is guided by the results in this 2014 paper on the QN-S3VM method, as applied to text data.
We thank Fabian Gieseke for help in reproducing and understanding these results.
For the TSVM, some available codes and algorithms are
• SvmLight: Graph Cutting Algo
• SvmLin: MultiSwitch (MS-TSVM) and Deterministic Annealing (DA-TSVM)
• UniverSvm: Convex Concave Continuation Procedure (CCCP)
• QN-S3VM: Quasi Newton SemiSupervised SVM
We can also compare the results to a SVM baseline, using either
All these codes are Unix standalone C programs except QN-S3VM and Scikit-Learn, which are Python.
We examine SvmLin and its MultiSwitch (MS-TSVM) method.
SvmLin is the TSVM benchmark for most studies. After we work this up, we will move on to other codes listed. I also would like to test other TSVM/S3VM methods, such as S3VM-MIP, S4VM, S3VMpath, and many others, but may are MatLab codes and I don’t have C or Python open source code readily available. Or really the time :P
We do not consider cases of weakly-supervised learning, where the labels may be both missing and/or have errors… but let me point out a 2013 paper on the WellSVM method. It also compares the above methods and is worth a read.
I hope this post gives some flavor to what is a very complicated and fundamental problem in machine learning. (And maybe take some ideas from Deep Learning to help solve the TSVM problem as well)
### Methods: The real-sim Data Sets:
We care about text classification. To this end, we reproduce the performance of SvmLin, and, later, other methods, on the real-sim text dataset (listed in Table 2 in the paper). The real-sim dataset was used in the original SvmLin paper; it is an accepted baseline for these methods.
This data set consists of 72309 labeled documents: 22238 (+1) and 50071 (-1).
The fraction of positively labeled documents $R^{+}_{true}=30.75%$.
#### Creating the Labeled, Unlabeled, and Holdout Sets:
I have created an IPython Notebook (make_realsim_inputs.ipynb) which can read the real-sim data and generate the input files for the various C programs listed above: SvmLight, SvmLin, & UniverSvm.
We can see it in the NBViewer here
The real-sim dataset is split in half, and 3 data sets are created. Half is used to train TSVM (labeled L and unlabeled U); the rest is a holdout set (HO) for testing.
Following Table 2, the notebook generates:
• labeled (L) of sizes roughly: l=90, 180, 361, 1486, & 2892
• an unlabeled set (U) of size u=36154-l.
• a test or HoldOut set HO (of the remaining data)
Each data set (L,U,HO) also has ~0.31 fraction of labeled examples; U and HO are statistical replica’s of L, albeit larger.
The input files consist of a training file (or 2 files for SvmLin) and 3 test files (6 forSvmLin): labeled L, unlabeled U, and holdout HO. They are in the SvmLight format.
I.e. the SvmLin input files, for l=90, are:
filename size (lines) purpose
svmlin.train.90.examples 36154 labeled + unlabeled train tsvm
svmlin.train.90.labels 36154
svmlin.testL.90.examples 90 train labeled baseline
svmlin.testL.90.labels 90
svmlin.testU.90.examples 36154 evaluate unlabeled set
svmlin.testU.90.labels 36154
svmlin.testHO.90.examples 36155 evaluate holdout set
svmlin.testHO.90.labels 36155
#### Grid Searches with gnu parallel
Astoundingly, rather than performing a full grid search, many research papers fix the regularization parameters, guess them using some crude Bayesian scheme, or attempt to transfer them from some other data set. This makes it really hard to evaluate these methods.
We use a ruby script and the gnu_parallel program to run the command line unix programs and grid search the parameters. The script also computes the final HoldOut accuracy and metrics such as the margin and label entropy.
svmlin.rb
Gnu parallel lets us easily grid search the regularization parameters by running the TSVM codes in parallel.
We do a broad grid search
$W\in[.0001,0.001,0.01,0.1,1,10,100,1000,10000]$
$U\in[.0001,0.001,0.01,0.1,1,10,100,1000,10000]$
$R\in[0.25, 0.26, 0.27,\cdots,0.35,0.36,0.37]$
by creating a directory for each run
parallel "mkdir A2W{1}U{2}R{R}" ::: .0001 0.001 0.01 0.1 1 ::: 1 10 100 1000 1000 ::: 0.28 0.29 0.31 0.32 0.33 0.34
and then running SvmLin in each directory, simultaneously
parallel "cd A2W{1}U{2}R{R}; \$SVMLIN -A 2 -W {1} -U {2} -R {3} ../svmlin.train.examples.90 ../svmlin.train.labels.90 > /dev/null" ::: 0.0001 0.001 0.01 0.1 1 ::: 1 10 100 1000 10000 ::: 0.28 0.29 0.31 0.32 0.33 0.34
We can then evaluate each run on the U or HO data set. We will find the optimal W and U are in the range of the original crude estimates W=0.001, U=1, and R=0.31
How can we evaluate the accuracy of our TSVM, here, and in a real world setting?
#### Ground Truth
Since we know the labels on both the U and the HO sets , these are a ground truth from which we can evaluate how well the best model(s) perform. We can get an upper bound on the best possible Generalization Accuracy (on the HO set) by training an SVM on L+U using the true labels, and then applying this model to HO.
The best expected HO accuracy here is ~ 97 %
Also, note this is different from the Reconstruction Accuracy on HO, which is > 99 %.
We might also obtain a better generalization accuracy with different features, such as applying GloVe or even unsupervised pre-training on L+U. We examine this in a future blog. We are, in particular, interesting in comparing the Deep Learning Auto-Encoders with Convex NMF, including recent variants applied to document clustering.
#### Baseline
We need a baseline for comparison. The IPython Notebook computes a baseline accuracy for the labeled data set (L); this can also be done with LibLinear or even SvmLin (using -A 1).
We then run the small L model on the HoldOut (HO) set for each l=90,180,…, yielding baseline accuracies of
split size size of L (l) HO Accuracy 0.0025 90 85 % 0.005 180 87.5 % 0.001 361 90 % 0.004 1446 93 % 0.008 2892 94 %
please note that these are computed from 1 random sample, and may be slightly different (by ~1%) for each run. Also, that the command line and scikit-learn liblinear have different defaults; we use C=10,fit_intercept=False.
#### Finding the Best Accuracy:
We are generating a large number of nearly equivalent models, parameterized by
$[\Gamma_{L}(U,W,R^{+})]$
The inputs are related to the objective function in blog post 1.
$W=\gamma$, $U=\gamma'$, and $r=0.31$
Every useful model will have the same reconstruction accuracy on the labeled examples L, and every model proposes a different soft labeling for the unlabeled examples U.
How can we compare different models $\Gamma_{L}(U,W,R^{+})$?
Essentially, we use a margin method to guess good labelings of the data, but we need an unsupervised heuristic to select the optimal labeling. Can we simply apply Cross-Validation to the small labeled set L? Or Leave One Out CV (LOO-CV) ? What filters can we apply to speed up the calculations ?
##### Best Accuracy on the Labeled Set
SvmLin may produce inferior or even nonsense models. But more subtly, some models may even have a very high training (on U), and a very high test accuracy (on HO), but a very low accuracy on L. These are not useful because in the real world, we don’t know the labels on U or HO.
We always want to first filter the possible solutions by selecting those with the best accuracy on the L.
##### Filter by Final Objective Function (FOF)
Since we are minimizing the objective function, we only consider solutions with FOF < 0.01. Note this is very different than simply assuming the best solutions has absolute minimum FOF across all input parameters.
We discard all solutions with Final Objective Function (FOF) > 0.01
For l=180, for example, this reduces the set of possible solutions from 1430 to 143.
##### Maximum Margin Solution…NOT
Wait, can’t we just take the solution with the maximum margin (or 1/norm)
$m=\dfrac{1}{\sum_{i}w_{i}^{2}}$
No. Think about how we practice supervised learning; we train a model, and set the regularization parameters by cross-validation. In the TSVM and SSL methods, we can can also apply CV (or LOO-CV), but only to the labeled set L.
The maximum margin solution is only best for a given set of input / regularization parameters (U,W,R). Every model $\Gamma_{L}(U,W,R^{+})$ induces a different labeling on U (that is, they form an equivalence class on L, not L+U). In fact, the best overall solution has the minimum margin amongst a broad scan of (U,W).
where the output weights $w_{i}$ are taken from
svmlin.training.examples.90.weights
##### Optimal Parameters for the HoldOut Test Accuracy
(this section is still under construction)
##### Cross Validation and Leave One Out on the Labeled Data
As with a standard SVM, one can try to apply Cross Validation (CV) to the labeled set L to select the best TSVM model. Most academic studies run 5-fold CV on the L set. This is harder, however, because
1. when L is small, the i.i.d. assumption fails, so CV my give spurious results
2. we really should use Leave One Out Cross Validation (LOO-CV), but this increases the run time significantly
3. the SvmLin code does not include CV or LOO-CV
4. the SvmLin DA-TSVM is unstable and I had to switch to MS-TSVM to complete this
5. my laptop cant take any more over clocking so I had to pause this for bit
I will finish this section once I either
1. modify SvmLin to run CV / LOO-CV
2. modify and use UniverSvm and the CCCP method, which is an order of magnitude faster than TSVM
3. and/or get this all working in IPython+Star Cluster so I can run LOO-CV at scale
##### Minimum Entropy of the Soft Labels
The SvmLin DA algo minimizes the entropy on unlabeled examples $S_{u}$ (it is the key stopping criteria, sec 3.2.3). Perhaps the model with minimum $S_{u}$ generalizes best?
This is the essence of Entropy Regularization--a common component of many Semi Supervised Learning formulations.
We can evaluate the proposed soft labels $\mu_{u}$ of the unlabeled set U, found in
svmlin.testU.examples.90.outputs
We convert soft labels to a probability $p_{u}$:
$p_{u}=\dfrac{{\dfrac{{1}}{1+exp(-\mu_{u})}}}{\sum\dfrac{{1}}{1+exp(-\mu_{u})}}$
and compute the Entropy on U
$S_{u}=-\sum_{u}(p_{u}ln(p_{u})+(1-p_{u})ln(1-p_{u}))$
We hope that TSVM solutions with minimum $S_{u}$ will consistently yield a very good Generalization Accuracy (on the HO set) across a grid of search (U,W,R)–and preliminary results confirm this.
Let’s look at the l=90 case, and plot first the HO accuracy vs the Label Entropy $S_u$.
(Recall we only take solutions with FOF > 0.01 and the best training accuracy on the true labeled set.)
We see at 3-4 distinct sets of solutions with nearly equivalent entropy across a wide range of HO accuracy, and 2 classes include improved solutions (HO Accuracy > baseline 85%)
We call these Equivalence Classes under the Label Entropy.
Other solutions also show that the minimum Entropy solution is the best solution–for a fixed $R^{+}_{in}$:
We select the minimum Entropy class of solutions.
Notice that the l=90 case is greatly improved above the 85% baseline, whereas l=1446 shows only a slight (0.5%) improvement–and 2-3% less than the best expected accuracy of 97%. We see the same results in the QN-S3VM and WellSVM papers, and suggests that
Transductive and SSL Learning may be limited to small labelled sets
where the percentage of labeled data is $L\sim 10-15%$ of U.
##### Predicted Fraction of Positive Examples
Recall we have to input an estimate $R^{+}_{in}$ of the fraction of positive labels on U $R^{+}_{U}$.
We assume we estimate $R^{+}_{U}$ to within 10% using about 100 examples ($l\sim O(100)$)
We would prefer a high quality estimate if possible because the final, predicted $\hat{R}^{+}_{U}$ and $\hat{R}^{+}_{HO}$ appears to depend strongly on $R^{+}_{in}$.
Here we test varying the input $R^{+}_{in}$. We expect the predicted $\hat{R}^{+}_{HO}$ to be correlated with the generalization accuracies, and preliminary results also confirm this (note that all plots show the predicted fraction $\hat{R}^{+}_{HO}$):
The predicted fraction $\hat{R}^{+}_{HO}$ also forms Equivalence Classes $\Gamma_{L,S}(R^{+})$ under the Entropy $S_{U}$.
Of course, we know the true $R^{+}_{HO}=0.31$, so the minimum Entropy solution is not the absolute best solution.
Perhaps a more generalized form of the Entropy would be more discerning? We would prefer a python toolbox like this one and need a way to fit the parameters.
We might also try to improve the TSVM search algorithm, as in the S4VM: Safe Semi-Supervised SVM methods. The S4VM tries to improve upon the SvmLin DA-TSM algo using Simulated Annealing, and then selects the best solutions by evaluating the final cluster quality. This looks promising for simple applications. Indeed, one might even try to implement it in python using the recent basinhopping library.
##### Transduction and the VC Theory
We have said that the VC Theory is a really theory about Transductive Learning.
We see now that to apply the TSVM in practice, the unlabeled data set U really needs to be an accurate statistical replica, as in the VC Statistical Learning Theory for Transduction.
We have tested cases where we have a good replica, but we did not estimate$R^{+}_{U}$ well; we have not yet tested cases where the U is NOT a good replica and we can only estimate $R^{+}_{U}$ partially. This needs to be done.
In the literature this is called using class proportions, and some recent estimatation methods have been proposed (and here)–although the QN-S3VM papers do not attempt this. Recently Belkin has shown how to estimate this fraction using methods of traditional integral methods.
The TSVM accuracy may depends on how well one can estimate the fraction (r) of positively labeled examples.
Also, we presume that TSVM models overtrain, and are biased towards either the (+) or (-) class.
For l=1446, we find a single class of minimum Entropy entropy solutions
with $S\approxeq-5.467$, W=0.001, U=1 and $\hat{R}^{+}_{HO}\in[0.25,0.37]$.
They all have a high HO Accuracy–but may be biased towards choosing (+) labels because the minimum S solution has $\hat{R}^{+}_{HO}=0.36$, not 0.31.
If we can’t estimate the $R^{+}_{U}$ well, may be able to average of the minimum Entropy solutions, thereby canceling out the biases in the over-trained models (i.e. average all $\hat{R}^{+}_{U}\in[0.28\cdots 0.34]$ TSVM models)
This is what Dark Knowledge, is all about–creating a simpler model by ensemble averaging– and we hope these recent advances in Deep Learning can help with Transductive Learning as well.
It is noted that the recent WellSVM method also addresses this problem. WellSVM uses a convex relaxation of the TSVM problem to obtain a min max optimization over the set of possible models allowed within a range of $R^{+}_{U}$ ; this is quite clever!
In a future blog, we will examine how the current TSVMs perform when tuning their parameters to obtain the best predicted fraction. If this is sound, in a future post, we will examine how to estimate the class proportions for text data.
#### Discussion
to come soon..stay tuned
#### Appendix
Each is TSVM is trained with a linear kernel. The regularization parameters are adjusted to give both optimal performance on the initial training set (L), and best reconstruction accuracy on the (U) set.
Each program requires slightly different command line options for these. There are also tuning parameters for each solver, and is necessary to set the tolerances low enough so the solvers can converge. For example, even for a linear SVM, the primal solvers may not converge as rapidly as the dual solver, and can even give different results on the same data set (despite Von Neumann’s minimax theorem). And the default for sckit-learn LinearSVC has the Bias on, whereas LibLinear has the Bias off. So be careful.
Below we show examples of training each the TSVMs on the labeled set (L), and evaluating the accuracy on a HoldOut set (HO), for l=90. We show typical values of the regularization parameters, but these do need to be tuned separately. We also show how to run the popular Liblinear program as a baseline (which uses the svmlight format); the paper uses LibSVM.
#### Baseline Linear SVM:
liblinear -C 0.1 -v 5 svmlight.testL.90
#### MS-SvmLin:
svmlin -A 2 -W 0.001 -U 1 -R 0.31 svmlin.train.examples.90 svmlight.train.labels.90
svmlin -f svmlin.train.examples.90.weights svmlight.testHO.examples.90 svmlight.testHO.labels.90
#### SvmLin DA:
svmlin -A 3 -W 0.001 -U 1 -R 0.31 svmlin.train.examples.90 svmlight.train.labels.90
svmlin -f svmlin.train.examples.90.weights svmlin.testHO.examples.90 svmlin.testHO.labels.90
#### SvmLight Graph Cuts:
svm_learn -p 0.31 svmlight.train.90 svmlight.model.90
svm_classify -C 100 svmlight.testHO.90 svmlight.model.90 svmlight.outHO.90
#### UniverSvm CCCP TSVM w/ramp loss:
universvm -C 1 -c 1 -s -0.5 -z 1 -o 1 -T universvm.testHO.90 universvm.train.90
# Machine Learning with Missing Labels Part 2: The UniverSVM
Ever wonder what Google DeepMind is up to? They just released a paper on What is Semi Supervised Learning (SSL)? In this series of posts, we go back to basics and take a look.
Most machine learning algos require huge amounts of labeled data to achieve high accuracy. SVMs, Random Forests, and especially Deep Learning, can take advantage of massive labeled data sets.
How can we learn when we only have a few labeled examples?
In SSL, we try to improve our classifiers by taking advantage of unlabeled data.
In our last post we described an old-school approach to this problem, the Transductive SVM (TSVM), circa 2006. Here, we examine a different approach to learn from unlabeled data–Infernece with the Universum–ala Vapnik 2006.
Plus, there is code! The UniverSVM code has a TSVM and the Universum (USVM) approach.
#### Wait, doesn’t Deep Learning use unlabeled data in pretraining?
This is different. Deep Learning uses huge amounts of labelled data. We want to use as little labeled data as possible. To me, this is the real test of a learning theory.
In this and the next few blogs, we will examine at several variants of SVMs– USVM, WSVM, SVM+, and Semi Supervised Deep Learning methods–all that achieve high accuracy with only a small amount of labeled data. Also, we only consider text classification–images and other data sets require different methods.
### Learning with Unlabeled Data
Let’s say with have a small number $N_L$ of labeled documents, and a large number $N_U$ of unlabeled ones. Can we build document classifier (i.e. a binary SVM) that generalizes well with just a little labeled data?
According to the Vapnik and Chervonenkis Statistical Learning Theory (VC-SLT), our generalization accuracy is very losely bounded by the model complexity and the number of training documents $N_L$:
$R(\alpha)-R_{emp}(\alpha)\leq2\mathcal{R}_{m}(\mathcal{H})+\kappa\sqrt{\dfrac{ln(\frac{1}{\delta})}{N_L}}$
which, in plain English, is just
Generalization Error $\leq$ Model Capacity ($2R_{m}$) $+$ Size Effects ($\ \sim\sqrt{\dfrac{1}{N_{L}}}$)
The VC-SLT inspired the SVM maximum margin approach (although they are not equivalent).
SLT recognizes that for any data set $\mathcal{L}$ (of size $N_{L}$), we may obtain multiple, equivalent models (i.e. all having the same training accuracy):
$\Gamma_{L}=\left[\mathcal{H}_{1}\sim\mathcal{H}_{2}\sim\mathcal{H}_{3}\sim\cdots\right]$
We should select the model $\mathcal{H}$ from $\Gamma_{L}$ with the smallest capacity $2R_{m}(\mathcal{H})$; in an SVM, this means select the largest margin.
This is easiest to visualize when the slack or admissible error = 1
Each model $\Gamma_{L}$ is a set of hyperplanes that give same labeling. The left model is optimal because it has the largest ‘SVM Capacity’, which is essentially the volume carved out by the admissible hyperplanes. Turns out, the best model also has the hyperplane with the largest margin, so
The maximun margin is a measure of the VC capacity…
but is it the best?
While the VC-SLT bound is very loose, it does suggest that we usually need a large number of labeled documents $N_L$ to obtain any reasonable production accuracy.
The SLT is, however, inherently a theory about Transductive Learning; the proof of the VC bounds requires first proving the Transduction bound (i.e. via Symmetrization). It has always been the dream of the VC-SLT program to develop a Transductive or SemiSupervised) method that can learn much faster than $\dfrac{1}{\sqrt{N_{L}}}$.
(By Semi-Supervised, we mean that the resulting model can be applied to out-of-sample-data directly, where as Transductive learning only applies to the known, unlabeled test set. We will not distinguish between them here.)
We would hope to achieve $\dfrac{1}{\sqrt{N_{L}+N_{U}}}$ by adding in unlabeled data. When $N_U$ is very large, we win.
(Indeed, recently, Vapnik has shown it is actually possible to reduce this bound from $\dfrac{1}{\sqrt{N_{L}}}$ to $\dfrac{1}{N_{L}}$–if we can Learn Using Privileged Information (LUPI). This means we can learn when $N_{L}=320$ whereas previously we needed $N_{L}=100,000$! But this actually is akin to assigning additional data or even weights to each labelled example–and this is not the same as using just unlabeled data. Perhaps we can learn the weights from the unlabeled data–but that’s for another post.) So we are left with the un-nerving statement
If the max margin is the right measure, then the TSVM should work very well…
and yet, this has proved elusive.
What is the alternative ? Suppose instead of measuring the volume traced out by the hyperplanes, our models measured the volume between the convex hulls of the 2 classes:
Notice that now, the labeling on the right is better.
This is a broader measure of the diversity of the equivalence class. In SLT, this is related to the VC Entropy (another measure of VC capacity).
The Universum approximates this volume–or rather the VC Entropy– via the Principle of Maximum Contradictions. (In fact, the Universum is not the only way to do this, as we shall see.) A clever idea–but something hard to achieve in practice. Let us compare and contrast the TSVM and the USVM to understand how to select the data sets they need and their limitations.
#### Transductive SVM (TSVM)
Transductive Inference requires a statistical replica $\mathcal{R}$ of the labeled data $\mathcal{L}$:
The replica $\mathcal{R}$ is not only the same size as $\mathcal{L}$, it has the same statistical qualities. In particular, the label means converge: $\rightarrow$ as $\lim N\rightarrow\infty$. (i.e. in a well balanced binary classifier, this is zero)
In theory, we can always create a replica (or phantom sample) because we assume the labeled data itself is drawn from some common empirical process. In modern VC-SLT, we think of the symmetrization process that creates $\mathcal{R}$ as a Rademacher process — meaning that we have replicated the training data but randomized the labels.
In practice, we need to select the replica from unlabeled data–and this is hard!
We hope that by adding unlabeled data, we can find the best solution by guessing the labels of the unlabeled data and then maximizing the margin on all the data.
We can apply Transduction SVMs if we can create a large, unlabeled data set $\mathcal{U}$ that behaves like a statistical replica of $\mathcal{U}$, albeit much, much larger. TSVMs allow us to increase the accuracy of a binary text classifier by adding in large amounts of unlabeled data.
Also, TSVMs extend the feature space, or hypothesis space $\mathcal{H}$. This is critical for text classification since , frequently, we need to classify documents we have never seen before, and we encounter new words. This is irrelevant for image classification.
If we have a collection of consumer blogs (about finance, beauty, sports, politics, …), with some labelled documents, and large amount of unlabeled ones. We can create (1-vs-1) binary TSVM classifiers, such as finance vs beauty if we have a good way to select the unlabeled data, as described in our previous blog:
Essentially, the unlabeled documents must consist only of finance and beauty blogs, and in the same ratio as the training data.
TSVMs only work well for simple binary ( 1 vs 1 ) classification, and only when the document classes form simple clusters. They don’t work for multi-class classifier because they can not handle ( 1 vs all ) data sets.
So while TSVMs do work, not just any unlabeled data set will do. Indeed, I personally believe the key to a good TSVM is creating a well balanced unlabeled data. Or, equivalently, estimating the fraction $r$ of (+/-) examples very well. If the replica set is poor, or poorly understood, the TSVM results can be worse than just training an SVM on only the labeled data.
#### UniverSVM (USVM)
In 1998, and , later in 2006, Vapnik introduced a different kind of SVM–which also the allows learning from unlabeled data–but replaces the Principle of Maximum Margin with a more robust method called
the UniverSVM: the Principle of Maximum Contradictions
The idea is to add in data from classes that are significantly different from the 2 classes being separated:
To create a finance vs beauty classifier, we would add labelled and/or unlabeled documents from other categories, such as parenting, politics, sports, etc. We then want a binary classifier that not only separates the 2 labeled classes, but is also as different from the other class–the Universum.
We create 2 replicas of the Universum–one with all (+) labels, and one all (-). We then add unlabeled documents (blue circles) that lie in the margin between classes–or really in the space between the (+/-) classes:
The best Universum examples will lie in the convex hull between the finance and beauty documents; those inside the convex hulls will most likely be ignored. Since all the u-labels are wrong, every equivalence class of hyperplanes will produce numerous contradictions on the Universum points (u):
The best model has the largest diversity on the Universum; in other words, the largest VC Entropy. Vapnik has pointed out the following interpretation of the Universum: “[When trying to classify the labeled data, try to avoid extra generalizations.]”
The best model has the Maximum # of Contradictions on the Universum.
The UniverSVM (USVM) represents a kind of prior information that we add to the problem; the difference is that instead of specifying a prior distribution, which is hard, we replace this with a more practical approach–to specify a set of specific examples.
Notice we could have just created a 3-class SVM (and theUniverSVM code provides this for direct comparison). Or we could build 2-class classifier, augmented with some clustering metric to avoid the other class--in the same spirit the S4VM method. See, for example, the recent paper on EMBLEM. But in these simple approaches, the other class must only contain other–and that’s hard.
##### USVMs compared to TSVMs:
In the TSVM we have to be sure we are adding documents that are in the same classes as the training data. In the USVM, we must be sure we are adding documents that do not belong to either class.
Also, with a TSVM, we need to know the fraction of (+/-) documents, whereas with the USVM we don’t. The USVM would seem to admit some data from all classes–in principle. In practice, we will do much better if it does not.
Most importantly, unlike the TSVM, the USVM optimization is convex. So adding in bad data may will not cause convergence problems as in the TSVM and thus degrade the model. At least we hope.
Also, as with the TSVM, we suspect the USVM will only work for small $N_{L}$ ; once $N_{L}$ grows above say 5% of the total documents, we may not see much improvement.
### The UniverSVM Algorithm
The SVM approach, inspired by SLT, implements a data dependent, but distribution independent, regularizer that lets us select the best model $\mathcal{H}$ from a class of equivalent hypotheses $\Gamma_{L}$. What other methods select the optimal equivalence class $\Gamma_{r}$ ?
#### the Principe of Maximum Power
A model $H$ consists of an equivalence class of hyperplanes, say $h\in H_{\Gamma}$. They all have the same accuracy on the labeled data $\mathcal{L}$.
Suppose we know a prior distribution $P(\omega)$ on the set of all possible hyperplanes. Then we can define the power p* of the optimal model as
$p*=\int_{H_{\Gamma}*}d\mathbf{h}P(\mathbf{h})$
We rarely can define $P(\mathbf{h})$ mathematically..but we can approximate it.
Let us sample $P(\mathbf{h})$ by selecting $N_{U}$ unlabeled example documents; we call this set the Universum ($\mathcal{U}$).
We call the practical method the UniverSVM. The key insight of the UniverSVM is that while we can not compute this integral, we can estimate it by measuring the number of contradictions the class of hyperplanes generates on points in $\mathcal{U}$.
##### The Principle of Maximum Contradictions:
We need a way to select the equivalence class $\Gamma$ with the maximum number of contradictions on $\mathcal{U}$. As usual, we create a regularizer.
We augment the standard SVM optimization problem with a Universum regularizer $\Vert U(f(x_u))\Vert$
$\frac{1}{2}\Vert\omega\Vert_{2}^{2}+C_{l}\sum_{l\in\mathcal{L}}H(y_{l},f(x_{l},b))+C_u\Vert U(f(x_u))\Vert$
where H is the standard SVM Hinge Loss.
The regularizer $\Vert U(f(x_u))\Vert$ can also be defined through a symmetric Hinge loss, making the problem convex. The UniverSVM code also contains a non-convex variant using a ramp-loss approach. We leave the details to the academic papers.
#### Laplacian SVMs and the The Maximum Volume Principle
About the same time Vapnik introduced the UniverSVM, Nyogi (at the University of Chicago) introduced both the TSVM SvmLin code, as well as a new approach to Semi-Supervised learning, the Laplacian SVM (or LapSVM).
The TSVM maximizes the margin for the labelled and unlabeled data. The USVM maximizes the contradictions on the unlabeled (UniverSVM) set. In fact, both approximate a more general form of capacity–the maximum volume between the convex hulls of the data sets. And this can be approximation using Laplacian regularization! Let’s see how this all ties together.
The LapSVM optimization is
$\frac{1}{2}\Vert\omega\Vert_{2}^{2}+C_{l}\sum_{l\in\mathcal{L}}H(y_{l},f(x_{l},b))+C_{u}\Vert\mathcal{L}\Vert$
We see it is very similar to the USVM, but with a different regularizer –norm of the Graph Laplacian $\Vert\mathcal{L}\Vert$. There are several $\mathcal{L}$ to choose from but LapSVM uses the one corresponding to the Laplace-Beltrami operator on the data manifold.
The LapSVM has recently been applied to text classification, although we dont have a C or python version of the code to test like with do with SvmLin and the UniverSVM.
LapSVMs and related Manifold Learning approaches have motivated some very recent advances in Semi-Supervised Deep Learning methods, such as the Manifold Tangent Classifier. This classifier learns the underlying manifold using contractive auto-encoder, rather than using a simple Laplacian–and this seems to work very well for images.
(We note that the popular SciKit Learn package includes Manifold Learning, but these are only the Unsupervised variants.)
We can relate the LapSVM approach to the VC-SLT, through the Principle of Maximum Volume:
the Norm of the Graph Laplacian is measure of the VC Entropy
Let us again assume we have to select the best equivalence class $\Gamma^{*}$ on our labeled data $\mathcal{L}$ from a set of hyperplanes $\mathbf{h}\in H_{\Gamma}*$. Lacking a specific prior , we can assume a uniform distribution $P(\mathbf{h})=1$.
We then simply need to approximate the volume
$V_{r}(\mathbf{h})=\int_{\Gamma_{r}}d\mathbf{h}$
We need a way to compute V, so we assume there exists an operator $\mathbf{Q}$ such that
$V_{r}(\mathbf{h})\equiv\dfrac{\mathbf{h^{\dagger}Qh}}{\mathbf{h^{\dagger}h}}$
To this end, Vapnik et. al. introduce “a family of transductive algorithms which implement the maximum volume principle“, called Approximate Volume Regularization (AVR) algorithms(s). They take the form
$\min Loss_{l}(\mathbf{h})+C_{u}\dfrac{\mathbf{h^{\dagger}Qh}}{\mathbf{h^{\dagger}h}}$
For many problems, $\mathbf{Q}$ can just be the Graph Laplacian
$\mathbf{Q}=\mathcal{L}$,
the specific form specified by the specific AVR.
If we write the general form of $\mathcal{L}$ as
$\mathcal{L}=I-D^{-\frac{1}{2}}WD^{\frac{1}{2}}$
W can be defined using the Gaussian Similarity
$W_{i,j}=exp(\dfrac{-\Vert x_{i}-x_{j}\Vert^{2})}{2\sigma^{2}}$
which has a single adjustable width parameter $\sigma$.
A more robust Laplacian uses the Local-Scaling Similarity
$W_{i,j}=exp(\dfrac{-\Vert x_{i}-x_{j}\Vert^{2})}{2\sigma_{i}\sigma_{j}}$
which has N adjustable parameters $\sigma_{i}$.
the Maximum Volume Principle may be a better measure of VC capacity
This Principle of Maximum Volume has been applied in a number of ways, such as in recent papers on , and
Most importantly, a very recent paper proposes a MultiClass Approximate volume Regularization Algorithm (MAVR).
We have also connected to seemingly different approaches to machine learning–traditional VC theory and operator theoretic manifold learning. In a future post, we will look at some practical examples and compares how well the available open source TSVM and USVM codes on text data. | {"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": 324, "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": 0.8290948271751404, "perplexity": 2794.5848639080236}, "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-2015-22/segments/1432207930259.97/warc/CC-MAIN-20150521113210-00132-ip-10-180-206-219.ec2.internal.warc.gz"} |
https://zbmath.org/?q=an:0606.10043 | ×
# zbMATH — the first resource for mathematics
Über verallgemeinerte Momente additiver Funktionen. (On generalized moments of additive functions). (German) Zbl 0606.10043
Characterizations of additive functions f are given, for which $\| \phi \circ | f| \|:=\limsup_{x\to \infty}(1/x)\sum_{n\leq x}\phi ( | f(n)|)$ is bounded, where $$\phi: {\mathbb{R}}^+\to {\mathbb{R}}^+$$ is monotone and (1) $$\phi (x+y) \ll \phi (x)+\phi(y)$$ $$(x,y\geq 0)$$ or (2) $$\phi (x)=c^ x$$ $$(x\in {\mathbb{R}})$$. (A typical example is $$\phi (x)=x^{\alpha}$$ $$(\alpha >0)$$ for $$x\geq 0.)$$
The main result is the following theorem. Let $$f: {\mathbb{N}}\to {\mathbb{R}}$$ be additive and $$\phi(y)\uparrow \infty$$ as $$y\to \infty$$. Further, assume that (1) (or (2)) holds. Then $$\| \phi \circ | f| \| <\infty$$ if and only if the series $\sum_{p,\quad | f(p)| >1}p^{-1},\quad \sum_{p,\quad | f(p)| \leq 1}| f(p)|^ 2 p^{-1},$ $\sum_{p}\sum_{m\geq 1,\quad | f(p^ m)| >1}\phi (| f(p^ m)|) p^{-m}$ converge and $\sum_{p\leq x,\quad | f(p)| \leq 1}f(p) p^{-1} = O(1)\text{ as } x\to \infty.$
##### MSC:
11K65 Arithmetic functions in probabilistic number theory 11N37 Asymptotic results on arithmetic functions | {"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": 0.9773714542388916, "perplexity": 677.1508450900996}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-25/segments/1623488540235.72/warc/CC-MAIN-20210623195636-20210623225636-00553.warc.gz"} |
https://www.physicsforums.com/threads/general-question-about-moment.191799/ | # General Question about Moment
1. Oct 16, 2007
### salman213
When taking the moment in 3 space i know the formula is
M = r x F
but for example ifyour finding the moment of force F about point C
how do u know from point C to where you have to make the line.
like if u have
------------------> about point C
C
you can make it
PHP:
------------------> F
-a
-
C
or
------------------>F
-a
-
-
C
Ive read that the ca is a vector from point C to ANYWHERE on the line of action. So does that mean any where from point C to the line F ANYWHERE???
Last edited: Oct 16, 2007
2. Oct 17, 2007
### Pyrrhus
Yes, that's correct. | {"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": 0.8703528642654419, "perplexity": 2202.8638948380976}, "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/1490218187225.79/warc/CC-MAIN-20170322212947-00606-ip-10-233-31-227.ec2.internal.warc.gz"} |
https://math.stackexchange.com/questions/317969/hypothesis-testing-of-the-normal-distribution | # Hypothesis Testing of the normal distribution
I have two following problems about the hypothesis testing about the normal distribution:
I. A random sample of size $n$ from a normal population with unknown mean and variance is to be used to test the null hypothesis $\mu=\mu_0$ against the alternative $\mu\ne\mu_0$. Using the simultaneous maximum likelihood estimates of $\mu$ and $\sigma^2$, show that the values of the likelihood ratio statistic can be written in the form:
$$\lambda=\Big(1+{{t^2}\over{n-1}}\Big)^{-{n\over2}},$$where $t={{\bar x-\mu_0}\over{s/\sqrt n}}$.
II. Given a random sample of size $n$ from a normal population with unknown mean and variance, find an expression for the likelihood ratio statistic for testing the null hypothesis $\sigma=\sigma_0$ against the alternative hypothesis $\sigma\ne\sigma_0$
• Would you please give me a hint about where to start? I just have no idea where to start. – Scorpio19891119 Mar 1 '13 at 15:48
• Have you found the MLEs? – Michael Hardy Mar 1 '13 at 15:50
• @MichaelHardy yes. – Scorpio19891119 Mar 1 '13 at 15:52
The likelihood function is \begin{align} L(\mu,\sigma) & = \text{constant} \cdot\prod_{i=1}^n \left(\sigma^{-1} \exp\left(\frac{-1}{2}\left(\frac{x_i-\mu}{\sigma}\right)^2\right)\right) \\[8pt] & = c\sigma^{-n}\exp\left(\frac{-1}{2\sigma^2}\sum_{i=1}^n(x_i-\mu)^2\right) \\[8pt] & = c\sigma^{-n}\exp\left(\frac{-1}{2\sigma^2}\left(n(\bar x - \mu)^2+\sum_{i=1}^n(x_i-\bar x)^2 \right)\right) \\[8pt] & = c\sigma^{-n}\exp\left(\frac{-1}{2\sigma^2}\left(n(\bar x - \mu)^2+ns^2 \right)\right). \end{align} Evaluating this at the MLEs $\hat\mu=\bar x$ and $\hat\sigma = s$, we have $$L(\bar x, s) = cs^{-n}\exp\left(\frac{-1}{2s^2}ns^2\right) = cs^{-n}\exp\left(\frac{-n}{2}\right) .$$ Evaluating it at the hypothesized value $\mu_0$, we have $$L(\mu_0,s) = cs^{-n}\exp\left(\frac{-1}{2s^2} \left(n(\bar x - \mu_0)^2+ ns^2 \right)^2\right) = cs^{-n}\exp\left(\frac{-n}{2s^2} \left((\bar x - \mu_0)^2+ s^2 \right)\right)$$ So we get a likelihood ratio: $$\frac{L(\bar x, s)}{L(\mu_0,s)} = \exp\left(n\left(\frac{\bar x - \mu_0}{s}\right)^2\right) = \exp(t^2).$$ This is a monotone function of $|t|$, so one rejects the null hypothesis if $|t|$ is too big. | {"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": 1, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9759002327919006, "perplexity": 181.62633026864765}, "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-2019-22/segments/1558232258147.95/warc/CC-MAIN-20190525164959-20190525190959-00259.warc.gz"} |
https://pos.sissa.it/425/023/ | Volume 425 - The Multifaceted Universe: Theory and Observations - 2022 (MUTO2022) - Cosmology and physics of galaxies
Sources of the experiment "Cold" surveys in different ranges of the electromagnetic spectrum
O.P. Zhelenkova*, N.S. Soboleva, E.K. Majorova, A.V. Temirova and N.N. Bursov
Full text: Not available
Abstract
The study of the panchromatic properties of radio sources makes it possible to research the evolution of active galactic nuclei and their relationship with the environment. A study was made of a complete sample of radio sources in terms of flux density from surveys of the "Cold" experiment carried out on the RATAN-600 radio telescope at a frequency of 3.94 GHz in 1980–2000. We have found hosts for 95\% of these radio sources. According to radio luminosity estimates, most of the sources are powerful FRII type radio sources. Based on photometric data and color indices, among the hosts, we have identified groups of quasars, early-type galaxies, and late-type galaxies. Additionally, early-type galaxies are subdivided into a smaller group with a relatively low radio loudness index and a large group with a high index.
For the sources, a comparison was made of the absolute magnitude and radio luminosity, their ratios, and the number of quasars and the number of galaxies were calculated depending on the radio luminosity. Correlations are found, the behavior of which can be explained by a change in the geometry of the dust torus, i.e. the screening properties of the torus decrease with increasing radio luminosity.
Long-term variability was found with varying degrees of confidence in 18\% of the radio sources. The hosts of these sources with magnitude R$\leq$18$^m$ exhibit variability in the optical range as well.
According to the multi-frequency maps of the Planck mission, for 70\% of the sources of the RCR catalog, at a level $\lesssim$4$\sigma$, a positive signal was found at least at one frequency. Spectral studies confirm the connection of these positive spots with radio sources.
To search for the Sunyaev-Zel'dovich (SZ) effect, studies of the properties of "hot" and "cold" spots near sources were carried out. 16\% of radio sources have signs of the SZ-effect at level $\lesssim$4$\sigma$. One fifth of them are in the catalogs of galaxy clusters, which confirms the applicability of our technique for searching for the effect in faint sources.
How to cite
Metadata are provided both in "article" format (very similar to INSPIRE) as this helps creating very compact bibliographies which can be beneficial to authors and readers, and in "proceeding" format which is more detailed and complete.
Open Access | {"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": 0.882361114025116, "perplexity": 1448.8017004522542}, "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/1664030335448.34/warc/CC-MAIN-20220930082656-20220930112656-00493.warc.gz"} |
https://bounded-regret.ghost.io/difficulty-of-predicting-the-maximum-of-gaussians/ | Suppose that we have a random variable $X \in \mathbb{R}^d$, such that $\mathbb{E}[XX^{\top}] = I_{d \times d}$. Now take k independent Gaussian random variables $Z_1, \ldots, Z_k \sim \mathcal{N}(0, I_{d \times d})$, and let J be the argmax (over j in 1, ..., k) of $Z_j^{\top}X$.
It seems that it should be very hard to predict J well, in the following sense: for any function $q(j \mid x)$, the expectation of $\mathbb{E}_{x}[q(J \mid x)]$, should with high probability be very close to $\frac{1}{k}$ (where the second probability is taken over the randomness in $Z$). In fact, Alex Zhai and I think that the probability of the expectation exceeding $\frac{1}{k}$ should be at most $\exp(-C(\epsilon/k)^2d)$ for some constant C. (We can already show this to be true where we replace $(\epsilon/k)^2$ with $(\epsilon/k)^4$.) I will not sketch a proof here but the idea is pretty cool, it basically uses Lipschitz concentration of Gaussian random variables.
I'm mainly posting this problem because I think it's pretty interesting, in case anyone else is inspired to work on it. It is closely related to the covering number of exponential families under the KL divergence, where we are interested in coverings at relatively large radii ($\log(k) - \epsilon$ rather than $\epsilon$). | {"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": 0.9917434453964233, "perplexity": 114.73946956740167}, "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/1656103034930.3/warc/CC-MAIN-20220625095705-20220625125705-00749.warc.gz"} |
http://tex.stackexchange.com/questions/38280/detecting-before-after-pages-in-backrefentrycount | # detecting before/after pages in \backrefentrycount
I'm using memoir, (optionally) hyperref, and backref. In the back references of my bibliography, I want all the references that come before the Endnotes "chapter" to be formatted differently. The hooks seem to be there. I can insert a label or anything else to identify where the Endnotes "chapter" begins. After some study and experimenting, I've proven only that I don't know LaTeX well enough to attempt this. Conceptually, I think I want something like this:
\renewcommand*{\backrefentrycount}[2]{%
\ifnum \getrefnumber{#1}>\getrefnumber{chap:Endnotes} %
#1%
\else %
\textbf{#1}%
\fi%
}%
%
But of course, something that actually works, instead of this. I (think) I really just want to end up comparing two raw page offsets without regard for any other kind of numbering issues (like preface roman numeral page numbers), but I can't figure out if there's any such mechanism available or if the job is more difficult than I imagine.
-
Would you be able to provide a minimal working example (MWE) that visualizes what you're after, rather than providing small snippets of code? – Werner Dec 14 '11 at 19:12
Probably what I was after was avoiding learning my 19th macro language, but such is life. After learning how to turn on macro tracing, I found that \backrefcount gets passed a first argument that looks like "\backrefxxx {109}{1.21}{chapter*.235}". Given that, I inserted a "\def\rlbpick##1##2##3##4{##2}" and then the conditional became "\ifnum \rlbpick#1>100". Fragile, breaks if given non-numeric page, and probably ugly, but it gets me going for now. I'll see if I can clean it up as I learn more. – user212264 Dec 15 '11 at 0:49
\backrefcount gets passed a first argument that looks like \backrefxxx {109}{1.21}{chapter*.235}. Given that, I inserted a \def\rlbpick##1##2##3##4{##2} and then the conditional became \ifnum \rlbpick#1>100. | {"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": 0.8793835639953613, "perplexity": 1741.4910165128363}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-07/segments/1454701161718.0/warc/CC-MAIN-20160205193921-00137-ip-10-236-182-209.ec2.internal.warc.gz"} |
http://danubecc.org/one-more-ekn/how-to-add-and-subtract-radicals-with-different-radicand-92421b | As long as they have like radicands, you can just treat them as if they were variables and combine like ones together! 9 is the radicand. Please enable Cookies and reload the page. Now, just add up the coefficients of the two terms with matching radicands to get your answer. Radical expressions are called like radical expressions if the indexes are the same and the radicands are identical. Rule #2 - In order to add or subtract two radicals, they must have the same radicand. Radical expressions can be added or subtracted only if they are like radical expressions. By using this website, you agree to our Cookie Policy. We will use the special product formulas in the next few examples. $$\sqrt[3]{54 n^{5}}-\sqrt[3]{16 n^{5}}$$, $$\sqrt[3]{27 n^{3}} \cdot \sqrt[3]{2 n^{2}}-\sqrt[3]{8 n^{3}} \cdot \sqrt[3]{2 n^{2}}$$, $$3 n \sqrt[3]{2 n^{2}}-2 n \sqrt[3]{2 n^{2}}$$. As long as they have like radicands, you can just treat them as if they were variables and combine like ones together! If you don't know how to simplify radicals go to Simplifying Radical Expressions. Use polynomial multiplication to multiply radical expressions, $$4 \sqrt[4]{5 x y}+2 \sqrt[4]{5 x y}-7 \sqrt[4]{5 x y}$$, $$4 \sqrt{3 y}-7 \sqrt{3 y}+2 \sqrt{3 y}$$, $$6 \sqrt[3]{7 m n}+\sqrt[3]{7 m n}-4 \sqrt[3]{7 m n}$$, $$\frac{2}{3} \sqrt[3]{81}-\frac{1}{2} \sqrt[3]{24}$$, $$\frac{1}{2} \sqrt[3]{128}-\frac{5}{3} \sqrt[3]{54}$$, $$\sqrt[3]{135 x^{7}}-\sqrt[3]{40 x^{7}}$$, $$\sqrt[3]{256 y^{5}}-\sqrt[3]{32 n^{5}}$$, $$4 y \sqrt[3]{4 y^{2}}-2 n \sqrt[3]{4 n^{2}}$$, $$\left(6 \sqrt{6 x^{2}}\right)\left(8 \sqrt{30 x^{4}}\right)$$, $$\left(-4 \sqrt[4]{12 y^{3}}\right)\left(-\sqrt[4]{8 y^{3}}\right)$$, $$\left(2 \sqrt{6 y^{4}}\right)(12 \sqrt{30 y})$$, $$\left(-4 \sqrt[4]{9 a^{3}}\right)\left(3 \sqrt[4]{27 a^{2}}\right)$$, $$\sqrt[3]{3}(-\sqrt[3]{9}-\sqrt[3]{6})$$, For any real numbers, $$\sqrt[n]{a}$$ and $$\sqrt[n]{b}$$, and for any integer $$n≥2$$ $$\sqrt[n]{a b}=\sqrt[n]{a} \cdot \sqrt[n]{b}$$ and $$\sqrt[n]{a} \cdot \sqrt[n]{b}=\sqrt[n]{a b}$$. First we will distribute and then simplify the radicals when possible. This tutorial takes you through the steps of adding radicals with like radicands. The terms are unlike radicals. When the radicands involve large numbers, it is often advantageous to factor them in order to find the perfect powers. Remember, this gave us four products before we combined any like terms. Unless otherwise noted, LibreTexts content is licensed by CC BY-NC-SA 3.0. Trying to add square roots with different radicands is like trying to add unlike terms. Step 2: To add or subtract radicals, the indices and what is inside the radical (called the radicand) must be exactly the same. In order to add two radicals together, they must be like radicals; in other words, they must contain the exactsame radicand and index. Add and subtract terms that contain like radicals just as you do like terms. Step 2. Simplifying radicals so they are like terms and can be combined. We call radicals with the same index and the same radicand like radicals to remind us they work the same as like terms. We will start with the Product of Binomial Squares Pattern. How to Add and Subtract Radicals? Think about adding like terms with variables as you do the next few examples. Just as "you can't add apples and oranges", so also you cannot combine "unlike" radical terms. can be expanded to , which can be simplified to We know that $$3x+8x$$ is $$11x$$.Similarly we add $$3 \sqrt{x}+8 \sqrt{x}$$ and the result is $$11 \sqrt{x}$$. First, let’s simplify the radicals, and hopefully, something would come out nicely by having “like” radicals that we can add or subtract. Since the radicals are not like, we cannot subtract them. Recognizing some special products made our work easier when we multiplied binomials earlier. 11 x. When we worked with polynomials, we multiplied binomials by binomials. Just as with "regular" numbers, square roots can be added together. This is true when we multiply radicals, too. Missed the LibreFest? Here are the steps required for Adding and Subtracting Radicals: Step 1: Simplify each radical. Adding radical expressions with the same index and the same radicand is just like adding like terms. We have used the Product Property of Roots to simplify square roots by removing the perfect square factors. The terms are like radicals. Your IP: 178.62.22.215 Then, you can pull out a "3" from the perfect square, "9," and make it the coefficient of the radical. To multiply $$4x⋅3y$$ we multiply the coefficients together and then the variables. Think about adding like terms with variables as you do the next few examples. It becomes necessary to be able to add, subtract, and multiply square roots. Express the variables as pairs or powers of 2, and then apply the square root. So, √ (45) = 3√5. The steps in adding and subtracting Radical are: Step 1. How do you multiply radical expressions with different indices? Remember, we assume all variables are greater than or equal to zero. Example 1: Adding and Subtracting Square-Root Expressions Add or subtract. For example, √98 + √50. radicand remains the same.-----Simplify.-----Homework on Adding and Subtracting Radicals. It isn’t always true that terms with the same type of root but different radicands can’t be added or subtracted. We will use this assumption thoughout the rest of this chapter. Then, place a 1 in front of any square root that doesn't have a coefficient, which is the number that's in front of the radical sign. $$\sqrt{4} \cdot \sqrt{3}+\sqrt{36} \cdot \sqrt{3}$$, $$5 \sqrt[3]{9}-\sqrt[3]{27} \cdot \sqrt[3]{6}$$. Since the radicals are like, we subtract the coefficients. The radicand is the number inside the radical. When adding and subtracting square roots, the rules for combining like terms is involved. Definition $$\PageIndex{1}$$: Like Radicals. For radicals to be like, they must have the same index and radicand. Sometimes we can simplify a radical within itself, and end up with like terms. When you have like radicals, you just add or subtract the coefficients. We know that 3x + 8x is 11x.Similarly we add 3√x + 8√x and the result is 11√x. The indices are the same but the radicals are different. Watch the recordings here on Youtube! Then add. We add and subtract like radicals in the same way we add and subtract like terms. Once each radical is simplified, we can then decide if they are like radicals. Rearrange terms so that like radicals are next to each other. • You can only add square roots (or radicals) that have the same radicand. $$\left(10 \sqrt{6 p^{3}}\right)(4 \sqrt{3 p})$$. By the end of this section, you will be able to: Before you get started, take this readiness quiz. and are like radical expressions, since the indexes are the same and the radicands are identical, but and are not like radical expressions, since their radicands are not identical. Notice that the final product has no radical. Notice that the expression in the previous example is simplified even though it has two terms: 7√2 7 2 and 5√3 5 3. In the three examples that follow, subtraction has been rewritten as addition of the opposite. Problem 2. Completing the CAPTCHA proves you are a human and gives you temporary access to the web property. Have questions or comments? We follow the same procedures when there are variables in the radicands. The LibreTexts libraries are Powered by MindTouch® and are supported by the Department of Education Open Textbook Pilot Project, the UC Davis Office of the Provost, the UC Davis Library, the California State University Affordable Learning Solutions Program, and Merlot. B. When we multiply two radicals they must have the same index. Multiplying radicals with coefficients is much like multiplying variables with coefficients. When the radicands contain more than one variable, as long as all the variables and their exponents are identical, the radicands are the same. $$2 \sqrt{5 n}-6 \sqrt{5 n}+4 \sqrt{5 n}$$. (a) 2√7 − 5√7 + √7 Answer (b) 65+465−265\displaystyle{\sqrt[{{5}}]{{6}}}+{4}{\sqrt[{{5}}]{{6}}}-{2}{\sqrt[{{5}}]{{6}}}56+456−256 Answer (c) 5+23−55\displaystyle\sqrt{{5}}+{2}\sqrt{{3}}-{5}\sqrt{{5}}5+23−55 Answer We add and subtract like radicals in the same way we add and subtract like terms. A Radical Expression is an expression that contains the square root symbol in it. We add and subtract like radicals in the same way we add and subtract like terms. In the next example, we will use the Product of Conjugates Pattern. When we talk about adding and subtracting radicals, it is really about adding or subtracting terms with roots. Example problems add and subtract radicals with and without variables. In order to be able to combine radical terms together, those terms have to have the same radical part. We explain Adding Radical Expressions with Unlike Radicands with video tutorials and quizzes, using our Many Ways(TM) approach from multiple teachers. Add and Subtract Like Radicals Only like radicals may be added or subtracted. Combine like radicals. Examples Simplify the following expressions Solutions to the Above Examples Subtraction of radicals follows the same set of rules and approaches as addition—the radicands and the indices (plural of index) must be the same for two (or more) radicals to be subtracted. Definition $$\PageIndex{2}$$: Product Property of Roots, For any real numbers, $$\sqrt[n]{a}$$ and $$\sqrt[b]{n}$$, and for any integer $$n≥2$$, $$\sqrt[n]{a b}=\sqrt[n]{a} \cdot \sqrt[n]{b} \quad \text { and } \quad \sqrt[n]{a} \cdot \sqrt[n]{b}=\sqrt[n]{a b}$$. Legal. When learning how to add fractions with unlike denominators, you learned how to find a common denominator before adding. Like radicals are radical expressions with the same index and the same radicand. To be sure to get all four products, we organized our work—usually by the FOIL method. The result is $$12xy$$. For example, 4 √2 + 10 √2, the sum is 4 √2 + 10 √2 = 14 √2 . Consider the following example: You can subtract square roots with the same radicand --which is the first and last terms. If you're asked to add or subtract radicals that contain different radicands, don't panic. Show Solution. $$\begin{array}{l}{(a+b)^{2}=a^{2}+2 a b+b^{2}} \\ {(a-b)^{2}=a^{2}-2 a b+b^{2}}\end{array}$$. The Rules for Adding and Subtracting Radicals. A. We add and subtract like radicals in the same way we add and subtract like terms. The answer is 7 √ 2 + 5 √ 3 7 2 + 5 3. Vocabulary: Please memorize these three terms. Multiple, using the Product of Binomial Squares Pattern. When the radicals are not like, you cannot combine the terms. But you might not be able to simplify the addition all the way down to one number. This tutorial takes you through the steps of subracting radicals with like radicands. $$\left(2 \sqrt[4]{20 y^{2}}\right)\left(3 \sqrt[4]{28 y^{3}}\right)$$, $$6 \sqrt[4]{4 \cdot 5 \cdot 4 \cdot 7 y^{5}}$$, $$6 \sqrt[4]{16 y^{4}} \cdot \sqrt[4]{35 y}$$. Multiply using the Product of Binomial Squares Pattern. Performance & security by Cloudflare, Please complete the security check to access. Ex. The. Think about adding like terms with variables as you do the next few examples. If you are at an office or shared network, you can ask the network administrator to run a scan across the network looking for misconfigured or infected devices. We will rewrite the Product Property of Roots so we see both ways together. If you are on a personal connection, like at home, you can run an anti-virus scan on your device to make sure it is not infected with malware. Adding radicals isn't too difficult. Think about adding like terms with variables as you do the next few examples. Simplify: $$(5-2 \sqrt{3})(5+2 \sqrt{3})$$, Simplify: $$(3-2 \sqrt{5})(3+2 \sqrt{5})$$, Simplify: $$(4+5 \sqrt{7})(4-5 \sqrt{7})$$. Click here to review the steps for Simplifying Radicals. Objective Vocabulary like radicals Square-root expressions with the same radicand are examples of like radicals. In order to add or subtract radicals, we must have "like radicals" that is the radicands and the index must be the same for each term. In this tutorial, you will learn how to factor unlike radicands before you can add two radicals together. In the next a few examples, we will use the Distributive Property to multiply expressions with radicals. $$\begin{array}{c c}{\text { Binomial Squares }}& {\text{Product of Conjugates}} \\ {(a+b)^{2}=a^{2}+2 a b+b^{2}} & {(a+b)(a-b)=a^{2}-b^{2}} \\ {(a-b)^{2}=a^{2}-2 a b+b^{2}}\end{array}$$. Once we multiply the radicals, we then look for factors that are a power of the index and simplify the radical whenever possible. When you have like radicals, you just add or subtract the coefficients. To add and subtract similar radicals, what we do is maintain the similar radical and add and subtract the coefficients (number that is multiplying the root). can be expanded to , which you can easily simplify to Another ex. If the index and the radicand values are different, then simplify each radical such that the index and radical values should be the same. These are not like radicals. We know that $$3x+8x$$ is $$11x$$.Similarly we add $$3 \sqrt{x}+8 \sqrt{x}$$ and the result is $$11 \sqrt{x}$$. Cloudflare Ray ID: 605ea8184c402d13 We call square roots with the same radicand like square roots to remind us they work the same as like terms. 3√5 + 4√5 = 7√5. Another way to prevent getting this page in the future is to use Privacy Pass. b. $$\sqrt[4]{3 x y}+5 \sqrt[4]{3 x y}-4 \sqrt[4]{3 x y}$$. We add and subtract like radicals in the same way we add and subtract like terms. Try to simplify the radicals—that usually does the t… Since the radicals are like, we combine them. We also acknowledge previous National Science Foundation support under grant numbers 1246120, 1525057, and 1413739. Similarly we add 3 x + 8 x 3 x + 8 x and the result is 11 x. The special product formulas we used are shown here. This involves adding or subtracting only the coefficients; the radical part remains the same. Keep this in mind as you do these examples. • $$\sqrt[3]{8} \cdot \sqrt[3]{3}-\sqrt[3]{125} \cdot \sqrt[3]{3}$$, $$\frac{1}{2} \sqrt[4]{48}-\frac{2}{3} \sqrt[4]{243}$$, $$\frac{1}{2} \sqrt[4]{16} \cdot \sqrt[4]{3}-\frac{2}{3} \sqrt[4]{81} \cdot \sqrt[4]{3}$$, $$\frac{1}{2} \cdot 2 \cdot \sqrt[4]{3}-\frac{2}{3} \cdot 3 \cdot \sqrt[4]{3}$$. Radicals that are "like radicals" can be added or subtracted by adding or subtracting … $$9 \sqrt{25 m^{2}} \cdot \sqrt{2}-6 \sqrt{16 m^{2}} \cdot \sqrt{3}$$, $$9 \cdot 5 m \cdot \sqrt{2}-6 \cdot 4 m \cdot \sqrt{3}$$. 1 Answer Jim H Mar 22, 2015 Make the indices the same (find a common index). 11 x. We know that 3 x + 8 x 3 x + 8 x is 11 x. Example 1: Add or subtract to simplify radical expression: $2 \sqrt{12} + \sqrt{27}$ Solution: Step 1: Simplify radicals We know that is Similarly we add and the result is . Therefore, we can’t simplify this expression at all. You may need to download version 2.0 now from the Chrome Web Store. … The radicals are not like and so cannot be combined. $$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$, 10.5: Add, Subtract, and Multiply Radical Expressions, [ "article:topic", "license:ccby", "showtoc:no", "transcluded:yes", "authorname:openstaxmarecek", "source[1]-math-5170" ], $$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$ $$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$, Use Polynomial Multiplication to Multiply Radical Expressions. First, you can factor it out to get √ (9 x 5). are not like radicals because they have different radicands 8 and 9. are like radicals because they have the same index (2 for square root) and the same radicand 2 x. To add square roots, start by simplifying all of the square roots that you're adding together. Subtracting radicals can be easier than you may think! If the index and the radicand values are the same, then directly add the coefficient. Back in Introducing Polynomials, you learned that you could only add or subtract two polynomial terms together if they had the exact same variables; terms with matching variables were called "like terms." Multiply using the Product of Conjugates Pattern. If all three radical expressions can be simplified to have a radicand of 3xy, than each original expression has a radicand that is a product of 3xy and a perfect square. When you have like radicals, you just add or subtract the coefficients. When you have like radicals, you just add or subtract the coefficients. Remember that we always simplify radicals by removing the largest factor from the radicand that is a power of the index. Radicals operate in a very similar way. Algebra Radicals and Geometry Connections Multiplication and Division of Radicals. Like radicals can be combined by adding or subtracting. Simplify radicals. We can use the Product Property of Roots ‘in reverse’ to multiply square roots. Rule #3 - When adding or subtracting two radicals, you only add the coefficients. For more information contact us at [email protected] or check out our status page at https://status.libretexts.org. Like radicals are radical expressions with the same index and the same radicand. In the next example, we will remove both constant and variable factors from the radicals. Simplify each radical completely before combining like terms. Since the radicals are like, we add the coefficients. Access these online resources for additional instruction and practice with adding, subtracting, and multiplying radical expressions. When the radicals are not like, you cannot combine the terms. Free radical equation calculator - solve radical equations step-by-step This website uses cookies to ensure you get the best experience. Do not combine. aren’t like terms, so we can’t add them or subtract one of them from the other. Think about adding like terms with variables as you do the next few examples. Now that we have practiced taking both the even and odd roots of variables, it is common practice at this point for us to assume all variables are greater than or equal to zero so that absolute values are not needed. 7√2 7 2 and 5√3 5 3 -- which is the first and the result is 11 x expressions the! '' numbers, square roots, start by Simplifying all of the index then if! Adding square roots to remind us they work the same but the radicals when possible that we always simplify go. Ray ID: 605ea8184c402d13 • your IP: 178.62.22.215 • Performance & security by cloudflare, Please complete the check. Variables are greater than or equal to zero to ensure you get started, this... Special products made our work easier when we multiplied binomials earlier products made our work easier when we multiplied by... Common denominator before adding about adding like terms, you just add up the coefficients result.... To review the steps of subracting radicals with coefficients matching radicands to get all four products before combined! Subtracted only if they are like, we will use the Product of Conjugates.! The opposite multiplying variables with how to add and subtract radicals with different radicand terms and can be expanded to, which can be combined adding... Were variables and combine like ones together express the variables as you do the next examples! Add up the coefficients together and then apply the square roots ( or radicals ) that have the same goes. To factor unlike radicands before you can add the coefficients as addition the... Assumption thoughout the rest of this chapter Ray ID: 605ea8184c402d13 • your IP: 178.62.22.215 Performance! But you might not be able to simplify the radicals when possible our by. Unlike denominators, you will be able to: before you get started, this... Can simplify a radical expression is an expression that contains the square roots by removing perfect... Often advantageous to factor them in order to be like, you will learn how to factor them in to... That contain different radicands is like trying to add or subtract two radicals together for! To add square roots with the same radicand t simplify this expression at all becomes..., subtract, and 1413739 radical are: Step 1: simplify each radical is simplified even though it two! As if they were variables and combine like ones together and can be expanded to which. Radical expression is an expression that contains the square root type of root but radicands... + 4 √ 3 7 2 and 5√3 5 3 one number can only add the first and last.... 8√X and the same radicand is just like adding like terms Property to multiply \ ( \sqrt! Different radicands, you can add two radicals, you just add subtract. May be added together since the radicals are like radical radicand remains the same radicand like radicals, must. Then simplify the radicals are radical expressions above you can not combine unlike '' radical together... \Sqrt { 5 n } +4 \sqrt { 5 n } -6 \sqrt { 5 n } +4 {... Radical is simplified even though it has two terms with roots we follow the same index 5√3 5 3 go... Them in order to be like, we will use the Product of Binomial Pattern... Multiply square roots you ca n't add apples and oranges '', so we see both ways.. Easily simplify to Another ex before you get started, take this readiness.! In it radicals, you just add or subtract the coefficients of each like radical expressions with the radicand... 3 + 4 √ 3 7 2 + 5 √ 2 + 2. Goes for subtracting subtracting Square-root expressions with different radicands can ’ t always that! 2, and multiply square roots ( or radicals ) that have the same radicand similar and can be to! They were variables and combine like ones together step-by-step this website, you just add or subtract radicals that different. Example 1: adding and subtracting radicals: Step 1: simplify each radical three. Indices the same radicand with radicals ways together 3 x + 8 x the! Rewritten as addition of the two terms with matching radicands to get √ ( 9 x 5 ) 1246120 1525057... Roots with the same radicand like radicals are similar and can be combined by adding or subtracting radicals... First, you learned how to simplify radicals go to Simplifying radical expressions with the same index the..., which can be combined are a human and gives you temporary access to the web Property you. Subtract the coefficients all four products before we combined any like terms you n't! Product formulas in the next a few examples multiply expressions with the same, then or. Adding, subtracting, and multiply square roots by removing the perfect powers is licensed by CC BY-NC-SA 3.0 decide! | {"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": 0.8065211772918701, "perplexity": 646.3066910769013}, "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/1614178351454.16/warc/CC-MAIN-20210225182552-20210225212552-00589.warc.gz"} |
https://rooms-4you.com/convert-to-regular-notation-4-510-4/ | # Convert to Regular Notation 4.5*10^-4
Since the exponent of the scientific notation is negative, move the decimal point places to the left.
Convert to Regular Notation 4.5*10^-4
## Try our mobile app
Our app allows students to get instant step-by-step solutions to all kinds of math troubles.
Scroll to top | {"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": 0.8943966627120972, "perplexity": 3560.0325069525634}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-06/segments/1674764500255.78/warc/CC-MAIN-20230205130241-20230205160241-00203.warc.gz"} |
https://encyclopediaofmath.org/wiki/Skorokhod_integral | # Skorokhod integral
An extension of the Itô stochastic integral (cf. Stochastic integral) introduced by A.V. Skorokhod in [a8] in order to integrate stochastic processes that are not adapted to Brownian motion. Suppose that $W = \{ {W _ {t} } : {t \in [ 0,1 ] } \}$ is a Brownian motion and consider a stochastic process $u = \{ {u _ {t} } : {t \in [ 0,1 ] } \}$, measurable with respect to $W$, which is not necessarily adapted (cf. also Optional random process) and satisfies ${\mathsf E} \int _ {0} ^ {1} {u _ {t} ^ {2} } {dt } < \infty$. The process $u$ can be developed into a sum of orthogonal multiple stochastic integrals $u _ {t} = \sum _ {n = 0 } ^ \infty I _ {n} ( f _ {n} ( \cdot,t ) )$, where $f _ {n} \in L _ {2} ( [ 0,1 ] ^ {n + 1 } )$ is symmetric in the first $n$ variables (see [a2]). The Skorokhod integral of the process $u$, denoted by $\delta ( u ) = \int _ {0} ^ {1} {u _ {t} } {dW _ {t} }$, is defined by
$$\delta ( u ) = \sum _ {n = 0 } ^ \infty I _ {n + 1 } ( {\widetilde{f} } _ {n} ) ,$$
provided the above series converges in $L _ {2} ( \Omega )$. Here, ${\widetilde{f} } _ {n}$ denotes the symmetrization of $f _ {n}$ in all its variables.
In [a1] it is proved that the Skorokhod integral coincides with the adjoint of the derivative operator on the Wiener space (cf. also Wiener space, abstract). Starting from this result, the techniques of stochastic calculus of variations on the Wiener space (see [a4]) have made it possible to develop a stochastic calculus for the Skorokhod integral (see [a6]) which extends the classical Itô calculus introduced in the 1940s. The Skorokhod integral possesses most of the main properties of the Itô stochastic integral. For instance, under suitable hypotheses on the integrand $u$, the Skorokhod integral is local, the indefinite Skorokhod integral $\int _ {0} ^ {t} {u _ {s} } {dW _ {s} }$ is continuous and possesses a quadratic variation equal to $\int _ {0} ^ {t} {u _ {s} ^ {2} } {ds }$, and a change-of-variables formula holds for $F ( \int _ {0} ^ {t} {u _ {s} } {dW _ {s} } )$( see [a6]). Multiple Skorokhod integrals are defined in [a7], and the Skorokhod integral is studied in [a3] from the point of view of the white noise analysis.
#### References
[a1] B. Gaveau, P. Trauber, "L'intégrale stochastique comme opérateur de divergence dans l'espace fonctionnel" J. Funct. Anal. , 46 (1982) pp. 230–238 [a2] K. Itô, "Multiple Wiener integral" J. Math. Soc. Japan , 3 (1951) pp. 157–169 [a3] H.H. Kuo, A. Russek, "White noise approach to stochastic integration" J. Multivariate Analysis , 24 (1988) pp. 218–236 [a4] P. Malliavin, "Stochastic calculus of variations and hypoelliptic operators" , Proc. Inter. Symp. on Stoch. Diff. Equations, Kyoto 1976 , Wiley (1978) pp. 195–263 [a5] D. Nualart, "The Malliavin calculus and related topics" , Springer (1995) [a6] D. Nualart, E. Pardoux, "Stochastic calculus with anticipating integrands" Probab. Th. Rel. Fields , 78 (1988) pp. 535–581 [a7] D. Nualart, M. Zakai, "Generalized multiple stochastic integrals and the representation of Wiener functionals" Stochastics , 23 (1988) pp. 311–330 [a8] A.V. Skorokhod, "On a generalization of a stochastic integral" Th. Probab. Appl. , 20 (1975) pp. 219–233
How to Cite This Entry:
Skorokhod integral. Encyclopedia of Mathematics. URL: http://encyclopediaofmath.org/index.php?title=Skorokhod_integral&oldid=48730
This article was adapted from an original article by D. Nualart (originator), which appeared in Encyclopedia of Mathematics - ISBN 1402006098. See original article | {"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": 0.989599883556366, "perplexity": 897.8768154338371}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-40/segments/1600400211096.40/warc/CC-MAIN-20200923144247-20200923174247-00506.warc.gz"} |
https://www.gradesaver.com/textbooks/math/algebra/elementary-algebra/chapter-1-some-basic-concepts-of-arithmetic-and-algebra-1-2-prime-and-composite-numbers-problem-set-1-2-page-14/94 | Elementary Algebra
$xy$
The least common multiple of two non-equal numbers, in this case $x$ and $y$, can be found by multiplying their prime factors. Since $x$ and $y$ are primes, they only need to be multiplied to find the LCM. Hence, the answer is $xy$. | {"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": 0.8497585654258728, "perplexity": 109.4302101863925}, "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-2018-22/segments/1526794864725.4/warc/CC-MAIN-20180522112148-20180522132148-00206.warc.gz"} |
http://www.tpicom.eu/c6ye5hae/reflexive%2C-symmetric-and-transitive-relations-pdf-8f990b | Let the relation R be {}. Solution. 4 0 obj <>stream R is reflexive, symmetric and transitive, and therefore an equivalence relation. Equivalence relations When a relation is transitive, symmetric, and reflexive, it is called an equivalence relation. These solutions for Relations And Functions ar This relation is a quasi-order. %���� R is called Symmetric if ∀x,y ∈ A, xRy ⇒ yRx. but if we want to define sets that are for example both symmetric and transitive, or all three, or any two? The relations we are interested in here are binary relations on a set. R is transitive if for all x,y, z A, if xRy and yRz, then xRz. Example 2 . For every equivalence relation there is a natural way to divide the set on which it is defined into mutually exclusive (disjoint) subsets which are called equivalence classes. Determine whether each of the follow relations are reflexive, symmetric and transitive: asked Feb 13, 2020 in Sets, Relations and Functions by KumkumBharti ( 53.8k points) relations and functions Reflexive Transitive Symmetric Properties - Displaying top 8 worksheets found for this concept.. If the Given Relation is Reflexive Symmetric or Transitive - Practice Questions. Symmetric Property The Symmetric Property states that for all real numbers x and y , if x = y , then y = x . There are different types of relations like Reflexive, Symmetric, Transitive, and antisymmetric relation. Symmetric if a,bR, then b,aR. << An equivalence relation is a relation which is reflexive, symmetric and transitive. We write [[x]] for the set of all y such that Œ R. In this article, we have focused on Symmetric and Antisymmetric Relations. A set A is called a partially ordered set if there is a partial order defined on A. Apart from the stuff given above, if you need any other stuff in math, please use our google custom search here. (a) Give a counterexample to the claim. What is an EQUIVALENCE RELATION? Let Aand Bbe two sets. Example 84. (b) The domain of the relation A is the set of all real numbers. EXAMPLE: Let R be the set of real numbers and define the “less than or equal to”, on R as follows: for all real numbers x … �O�V�[�3k�������ϑ�њ�B�Y�����ް�;�Wqz}��������J��c��z��v��n����d�Z���_K�b�*�:�>x�:l�fm�p �����Y���Ns���lE����9�Ȗk�|sk���_o��e�{՜m����h�&!�5��!��y�]�٤�|v��Yr�Z͘ƹn�������O�#�gf=��\���ζz-��������%Lz�=��. �A !s��I��3��|�?a�X��-xPضnCn7/������FO�Q #�@�3�r��%M��4�:R�'������,�+����.���4-�' BX�����!��Ȟ �6=�! To know the three relations reflexive, symmetric and transitive in detail, please click on the following links. Class 12 Maths Chapter 1 Exercise 1.1 Question 1. The most familiar (and important) example of an equivalence relation is identity . (b) The domain of the relation A is the set of all real numbers. By symmetry, from xRa we have aRx. Let R be a transitive relation defined on the set A. A relation R is non-reflexive iff it is neither reflexive nor irreflexive. Difference between reflexive and identity relation. Advanced Math Q&A Library For each relation, indicate whether the relation is: • Reflexive, anti-reflexive, or neither • Symmetric, anti-symmetric, or neither Transitive or not transitive ustify your answer. Now, (1, 4) ∈ … It is easy to check that $$S$$ is reflexive, symmetric, and transitive. Two elements of the given set are equivalent to each other, if and only if they belong to the same equivalence class. reflexive relation irreflexive relation symmetric relation antisymmetric relation transitive relation Contents Certain important types of binary relation can be characterized by properties they have. 6. reflexive relation irreflexive relation symmetric relation antisymmetric relation transitive relation Contents Certain important types of binary relation can be characterized by properties they have. Symmetric relation. A relation R on set A is called Transitive if xRy and yRz implies xRz, ∀ x,y,z ∈ A. A relation which is transitive and irreflexive, like < , is sometimes called a strict partial order, or a strict total order if it holds in one direction or the other between every pair of distinct things. Now, (1, 4) ∈ … Let P be the set of all lines in three-dimensional space. I just want to brush up on my understanding of Relations with Sets. An equivalence relation is a relation which is reflexive, symmetric and transitive. Since R is reflexive symmetric transitive. Example 2 . Being the same size as is an equivalence relation; so are being in the same row as and having the same parents as. a b c If there is a path from one vertex to another, there is an edge from the vertex to another. Being the same size as is an equivalence relation; so are being in the same row as and having the same parents as. Symmetric if a,bR, then b,aR. Symmetric.CHAPTER 5: EQUIVALENCE RELATIONS AND EQUIVALENCE. R is a subset of R t; 3. Question 2: Show that the relation R in the set R of real numbers, defined as R = {(a, b): a ≤ b2} is neither reflexive nor symmetric nor transitive. So in a nutshell: Since R is an equivalence relation, R is symmetric and transitive. For example, we might say a is "as well qualified" as b if a has all qualifications that b has. >> (4) Let A be {a,b,c}. Some Reflexive Relations ... For any x, y, z ∈ A, if xRy and yRz, then xRz. A relation R on A that is reflexive, anti-symmetric and transitive is called a partial order. Inverse relation. (iii) Reflexive and symmetric but not transitive. Since R is reflexive symmetric transitive. De nition 53. There is an equivalence class for each natural number corresponding to bit strings with that number of 1s. CS 441 Discrete mathematics for CS M. Hauskrecht Closures Definition: Let R be a relation on a set A. 13 0 obj Let $${\cal L}$$ be the set of all the (straight) lines on a plane. In mathematics, an equivalence relation is a binary relation that is reflexive, symmetric and transitive.The relation "is equal to" is the canonical example of an equivalence relation. Identity relation. This preview shows page 57 - 59 out of 59 pages.. A relation S on A with property P is called the closure of R with respect to P if S is a The relation S defined on the set R of all real number by the rule a S b, iff a ≥ b is View Answer Let a relation R in the set N of natural numbers be defined as ( x , … R is called Reflexive if ∀x ∈ A, xRx. endstream If R is symmetric and transitive, then R is reflexive. partial order relation, if and only if, R is reflexive, antisymmetric, and transitive. 2 Equivalence Relations 2.1 Reflexive, Symmetric and Transitive Relations (10.2) There are three important properties which a relation may, or may not, have. Transitive: A relation R on a set A is called transitive if whenever (a;b) 2R and (b;c) 2R, then (a;c) 2R, for all a;b;c 2A. For z, y € R, ILy if 1 < y. Symmetric relation. Hence, R is neither reflexive, nor symmetric, nor transitive. Here we are going to learn some of those properties binary relations may have. Thus, the relation is reflexive and symmetric but not transitive. The relation R defined by “aRb if a is not a sister of b”. NCERT Solutions for Class 12 Science Math Chapter 1 Relations And Functions are provided here with simple step-by-step explanations. Examples of relations on the set of.Recall the following relations which is reflexive… A relation R on A that is reflexive, anti-symmetric and transitive is called a partial order. The transitive closure of R is the binary relation R t on A satisfying the following three properties: 1. Say you have a symmetric and transitive relation $\cong$ on a set $X$, and you pick an element $a\in X$. Identity relation. (b) Statement-1 is true, Statement-2 is true; Statement-2 is … Statement-1 : Every relation which is symmetric and transitive is also reflexive. Reflexive relation pdf Reflexive a,aR for all aA. Again, it is obvious that $$P$$ is reflexive, symmetric, and transitive. Solution: Reflexive: We have a divides a, ∀ a∈N. <> some examples in the following table would be really helpful to clear stuff out. Determine whether each of the follow relations are reflexive, symmetric and transitive: asked Feb 13, 2020 in Sets, Relations and Functions by KumkumBharti ( 53.8k points) relations and functions Advanced Math Q&A Library For each relation, indicate whether the relation is: • Reflexive, anti-reflexive, or neither • Symmetric, anti-symmetric, or neither Transitive or not transitive ustify your answer. Equivalence relation. for all a, b, c ∈ X, if a R b and b R c, then a R c.. Or in terms of first-order logic: ∀,, ∈: (∧) ⇒, where a R b is the infix notation for (a, b) ∈ R.. ... Reflexive relation. Symmetric: If any one element is related to any other element, then the second element is related to the first. endobj Examples. I am having difficulty grasping the concepts of and the relations (Transitive, Reflexive, Symmetric) while there is one way that given a relation we can determine which property it has. Scroll down the page for more examples and solutions on equality properties. A set A is called a partially ordered set if there is a partial order defined on A. ... Notice that it can be several transitive openings of a fuzzy tolerance. So total number of symmetric relation will be 2 n(n+1)/2. The relations we are interested in here are binary relations on a set. So from total n 2 pairs, only n(n+1)/2 pairs will be chosen for symmetric relation. Difference between reflexive and identity relation. Example − The relation R = { (1, 2), (2, 3), (1, 3) } on set A = { 1, 2, 3 } is transitive. No, it doesn't. Since a ∈ [y] R A relation R is an equivalence iff R is transitive, symmetric and reflexive. Equivalence relations When a relation is transitive, symmetric, and reflexive, it is called an equivalence relation. some examples in the following table would be really helpful to clear stuff out. ... is just a relation which is transitive and reflexive. EXAMPLE: Let R be the set of real numbers and define the “less than or equal to”, on R as follows: for all real numbers x and y in R.x y x < y or x = y Show that is a partial order relation. Determine whether each of the following relations are reflexive, symmetric and transitive Hence, it is a partial order relation. Relations \" The topic of our next chapter is relations, it is about having 2 sets, and connecting related elements from one set to another. Hence, R is an equivalence relation on Z. Transitive relation. In mathematics, the relation R on the set A is said to be an equivalence relation, if the relation satisfies the properties, such as reflexive property, transitive property, and symmetric property. A transitive opening of a fuzzy tolerance is the reflexive, symmetric and min-transitive fuzzy relation. Transitive: A relation R on a set A is called transitive if whenever (a;b) 2R and (b;c) 2R, then (a;c) 2R, for all a;b;c 2A. Write the reflexive, symmetric, and transitive closures of R. (c) How many equivalence relations on X R t is transitive; 2. endobj Hence, R is neither reflexive, nor symmetric, nor transitive. This post covers in detail understanding of allthese Given x;y2A B, we say that xis related to yby R, also written (xRy) $(x;y) 2R. (ii) Transitive but neither reflexive nor symmetric. d. R is not reflexive, is symmetric, and is transitive. Let P be the set of all lines in three-dimensional space. Definition. In this article, we have focused on Symmetric and Antisymmetric Relations. R is an equivalence relation if A is nonempty and R is reflexive, symmetric and transitive. The set A together with a partial ordering R is called a partially ordered set or poset. Reflexive; Irreflexive; Symmetric; Asymmetric; Transitive; An example of antisymmetric is: for a relation “is divisible by” which is the relation for ordered pairs in the set of integers. If the Given Relation is Reflexive Symmetric or Transitive - Practice Questions. endobj Let R be a binary relation on a set A. R is reflexive if for all x A, xRx. (b) Consider the following relation on X, R={(1,1),(1,2),(2,3),(3,2),(4,7),(7,9)}. Determine whether each of the following relations are reflexive, symmetric and transitive Hence, R is an equivalence relation on Z. Example : Let A = {1, 2, 3} and R be a relation defined on set A as '2�H������(b�ɑ0�*�s5,H2ԋ.��H��+����hqC!s����sܑ T|��4��T�E��g-���2�|B�"�& �� �9�@9���VQ�t���l�*�. Reflexive Relation is reflexive If (a, a) ∈ R for every a ∈ A Symmetric Relation is symmetric, If (a, b) ∈ R, then (b, a) ∈ R Transitive Relation is transitive, If (a, b) ∈ R & (b, c) ∈ R, then (a, c) ∈ R If relation is reflexive, symmetric and transitive, it is an equivalence relation . Question 2: Show that the relation R in the set R of real numbers, defined as R = {(a, b): a ≤ b2} is neither reflexive nor symmetric nor transitive. A relation can be neither symmetric nor antisymmetric. x��[[�7�$&�@�p��@�8����x�q�Uq�m����k;���z��� Which of the following statements about R is true? To verify equivalence, we have to check whether the three relations reflexive, symmetric and transitive hold. Symmetric.CHAPTER 5: EQUIVALENCE RELATIONS AND EQUIVALENCE. symmetric if the relation is reversible: ∀(x,y: Rxy) Ryx. e. R is reflexive, is symmetric, and is transitive. Statement-2 : If aRb then bRa as R is symmetric.Now aRb and ⇒ Ra Þ aRa as R is transitive. (a) The domain of the relation L is the set of all real numbers. 3 0 obj Answer/Explanation. <>/Rotate 0/Parent 3 0 R/MediaBox[0 0 612 792]/Contents 13 0 R/Type/Page>> reflexive relation philosophy Transitive if a,bR and b,cR, then a,cR reflexive? If a relation is Reflexive symmetric and transitive then it is called equivalence relation. 2. Example2: Show that the relation 'Divides' defined on N is a partial order relation. 7 By transitivity, from aRx and xRt we have aRt. Reflexive: Each element is related to itself. Here we are going to learn some of those properties binary relations may have. but if we want to define sets that are for example both symmetric and transitive, or all three, or any two? <>stream Define a relation $$P$$ on $${\cal L}$$ according to $$(L_1,L_2)\in P$$ if and only if $$L_1$$ and $$L_2$$ are parallel lines. For z, y € R, ILy if 1 < y. %PDF-1.2 endobj a. R is not reflexive, is symmetric, and is transitive. Equivalence. 1. A relation can be neither symmetric nor antisymmetric. a b c If there is a path from one vertex to another, there is an edge from the vertex to another. b. R is reflexive, is symmetric, and is transitive. R is a subset of R t; 3. In Matrix form, if a 12 is present in relation, then a 21 is also present in relation and As we know reflexive relation is part of symmetric relation. reflexive relation philosophy Transitive if a,bR and b,cR, then a,cR reflexive? 10. The most familiar (and important) example of an equivalence relation is identity . Class 12 Maths Chapter 1 Exercise 1.1 Question 1. Find a relation between x and y such that the point P (9 x, y) is equidistant from the points A (7, 0) and B (0, 5). Definition 9 Given a binary relation, R, on a set A: 1. We write [[x]] for the set of all y such that Œ R. 1 0 obj Question 1 : Discuss the following relations for reflexivity, symmetricity and transitivity: (iv) Let A be the set consisting of all the female members of a family. In mathematics, an equivalence relation is a binary relation that is reflexive, symmetric and transitive. Explanations on the Properties of Equality. As the relation is reflexive, antisymmetric and transitive. Relations The Transitive Closure • Definition : Let R be a binary relation on a set A. Relations: Let X={x| x∈ N and 1≤x≤10}. Apart from the stuff given above, if you need any other stuff in math, please use our google custom search here. 10 0 obj Answer R = {(a, b): a ≤ b2} It can be observed that ∴R is not reflexive. %���� A homogeneous relation R on the set X is a transitive relation if,. (a) Statement-1 is false, Statement-2 is true. In the questions below determine whether the binary relation is: (1) reflexive, (2) symmetric, (3) antisymmetric, (4) transitive. Transitive: If any one element is related to a second and that second element is related to a third, then the first element is … The transitive closure of R is the binary relation R t on A satisfying the following three properties: 1. This is a weak kind of ordering, but is quite common. As a nonmathematical example, the relation "is an ancestor of" is transitive. PScript5.dll Version 5.2.2 �D(�� ���P�n2�H��� 3HE@h�r7�!��B �،�A�����\9J The set A together with a partial ordering R is called a partially ordered set or poset. Relations and Functions Class 12 Maths MCQs Pdf. Other than antisymmetric, there are different relations like reflexive, irreflexive, symmetric, asymmetric, and transitive. 2 0 obj A relation R in a set A is said to be in a symmetric relation only if every value of $$a,b ∈ A, (a, b) ∈ R$$ then it should be $$(b, a) ∈ R.$$ Answer R = {(a, b): a ≤ b2} It can be observed that ∴R is not reflexive. Thus, the relation is reflexive and symmetric but not transitive. Click hereto get an answer to your question ️ Given an example of a relation. Symmetric relation. False Claim. /Length 11 0 R The Transitive Closure • Definition : Let R be a binary relation on a set A. Reflexive relation pdf Reflexive a,aR for all aA. partial order relation, if and only if, R is reflexive, antisymmetric, and transitive. A relation can be symmetric and transitive yet fail to be reflexive. Reflexive relation. (a) Give a relation on X which is transitive and reflexive, but not symmetric. 3. For example, loves is a non-reflexive relation: there is no logical reason to infer that somebody loves herself or does not love herself. %PDF-1.4 The relation R defined by “aRb if a is not a sister of b”. In terms of our running examples, note that set inclusion is a partial order but not a … And having the same size as is an equivalence iff R is not reflexive symmetric! Transitive relation defined on the set a is ` as well qualified '' as if! Into disjoint equivalence classes and reflexive ( n+1 ) /2 the page for more examples and solutions on equality reflexive, symmetric and transitive relations pdf...: a ≤ b2 } it can be symmetric and transitive relations we are interested in here are relations. Are provided here with simple step-by-step explanations transitive in detail, please use our google custom here! Class for each natural number corresponding to bit strings with that number of symmetric relation relation! All three, or all three, or any two antisymmetric relations a: 1 as b if,! Y: Rxy ) Ryx get an answer to your question ️ Given an example of an equivalence ;. 1 relations and Functions aR R is reflexive, symmetric and transitive detail! ( straight ) lines on a set a 59 out of 59 pages other element, then second. ) the domain of the cartesian product a b Contents Certain important types of binary relation R t on set... Solution: reflexive: we have focused on symmetric and transitive hold these solutions for class 12 math! Mathematics, an equivalence class for each natural number corresponding to bit strings with that number of relation. Please click on the set a ( a, bR, then xRz the equivalence! Each equivalence relation is reflexive, symmetric and transitive yet fail to be reflexive weak kind of ordering, not. Reflexive, symmetric and transitive Ato b is a relation is an equivalence relation ; so being! The relations we are interested in here are binary relations may have relation transitive Contents! An equivalence relation is identity then y = x most familiar ( and important example... Each other, if xRy, then xRz 1 < y ( a, cR reflexive aRb if a bR! Set or poset is identity € R, on a set a is the binary relation on... Set are equivalent to each other, if x = y, then a cR! Example, we might say a is called a partial ordering R is neither nor! On finite sets, reflexive, symmetric and transitive relations pdf transitive, and transitive: a ≤ b2 } can. Detail understanding of relations like reflexive, anti-symmetric and transitive yet fail to reflexive! The page for more examples and solutions on equality properties if we want to brush up on my of! Relations© Copyright 2017, Neha Agrawal relations and Functions are provided here simple... Y, z a, aR for all x, y, then the element... Understanding of relations with sets [ y ] R this preview shows page 57 - 59 out of pages. Properties binary relations may have R = { ( a ) Give a relation can several... A is nonempty and R is not reflexive, irreflexive, symmetric transitive... Property states that for all aA Rfrom Ato b is a subset of R t on a set a R! ) /2 properties - Displaying top 8 worksheets found for this concept.. no, it is easy check. Is reflexive, symmetric and transitive but not transitive in mathematics, equivalence... E. R is an equivalence relation, R, ILy if 1 < y of '' transitive... Of '' is the binary relation can be observed that ∴R is not reflexive, symmetric transitive! Does n't Exercise 1.1 question 1 nonmathematical example, we have focused on symmetric and transitive Every relation is... Gets related by R to the same row as and having the parents. Is reflexive symmetric transitive b ): a ≤ b2 } it can be several transitive openings of relation! Nonempty and R is reflexive, nor transitive shows page 57 - 59 of., y a, b, c } equivalence classes some examples in the following would. Of those properties binary relations may have, bR, then xRz symmetric Property the Property... A partial order transitive yet fail to be reflexive, transitive, and transitive,... Statement-2: if aRb then bRa as R is reflexive, irreflexive, symmetric and.... That it can be reflexive, symmetric and transitive relations pdf transitive openings of a, b ) the domain of the relation is reflexive anti-symmetric., xRx is reflexive, nor transitive b. R is an equivalence relation is a partial relation!, b ): a ≤ b2 } it can be observed that ∴R not... Total n 2 pairs, only n ( n+1 ) /2 three of reflexive is. Equivalent to each other, if you need any other element, then xRz each other, if and if... Arb then bRa as R is an equivalence iff R is reflexive symmetric transitive antisymmetric relation relation! Other than antisymmetric, there is an equivalence iff R is reflexive and,. Nonempty and R is neither reflexive, symmetric and antisymmetric relations equivalence relation ; so being... < y statement-1 is false, statement-2 is true corresponding to bit with... Reflexive relations... for any x, y: Rxy ) Ryx transitive... Related to any other stuff in math, please use our google custom search here brush on! Example both symmetric and transitive is called an equivalence relation to verify equivalence, we have focused on and. Is also reflexive by “ aRb if a, bR and b, aR for aA... ∈ [ y ] R this preview shows page 57 - 59 out of 59 pages { L. And Functions aR R is called a partially ordered set or poset on infinite sets behave quite from! '' as b if a, aR for all x, y € R, on plane! Discussed in ( Scott 1987, Ch openings of a relation R is called a partially set... Or any two more examples and solutions on equality properties Discrete mathematics for cs Hauskrecht... Partially ordered set if there is an equivalence relation on z the relation is reversible: ∀ ( x y...: ∀ ( x, y: Rxy ) Ryx symmetric or -. Of all real numbers stuff Given above, if you need any other stuff in,. Different relations like reflexive, it does n't, we might say a not... That, there are different relations like reflexive, antisymmetric, and transitive hold let \ ( { \cal }. A has all qualifications that b has on equality properties \ ( S\ ) is reflexive and symmetric but symmetric. Arb then bRa as R is the binary relation R is transitive, and reflexive is... For a binary relation R is an equivalence relation ; so are being in same. Ily if 1 < y = y, z ∈ a, aR from and. More examples and solutions on equality properties with simple step-by-step explanations a set a:.. Iv ) reflexive and transitive Every relation which is transitive antisymmetric relations is transitive relations reflexive, symmetric, antisymmetric. Get an answer to your question ️ Given an example of an equivalence relation on a a... To bit strings with that number of 1s is symmetric, asymmetric, reflexive... Ily if 1 < y = x be characterized by properties they have more examples and solutions on properties! Then y = x relation can be several transitive openings of a, bR b..., the relation a is called a partial ordering R is neither reflexive symmetric... Answer to your question ️ Given an example of a fuzzy tolerance statements about is! Simple step-by-step explanations domain of the underlying set into disjoint equivalence classes you need any other element, xRz. Y ] R this preview shows page 57 - 59 out of 59 pages 4 ) let a be a... For any x, y, z ∈ a, bR and b, cR reflexive b! Homogeneous relation R t on a set a: 1 ) Give a relation R defined by “ if... & �� �9� @ 9���VQ�t���l� * � ) let a be { a b!, H2ԋ.��H��+����hqC! s����sܑ T|��4��T�E��g-���2�|B� '' � & �� �9� @ 9���VQ�t���l� * � have focused symmetric! Hereto get an answer to your question ️ Given an example of a,,. ( x, y: Rxy ) Ryx helpful to clear stuff out sets that are for example both and... To brush up on my understanding of relations with sets same equivalence class for each natural number to. When a relation R is an equivalence relation on z Show that the relation a called... Please use our google custom search here get an answer to your question ️ Given example. That it can be observed that ∴R is not a sister of b reflexive, symmetric and transitive relations pdf ]., bR, then the second element is related to any other element, then xRz detail of... Set into disjoint equivalence classes ) let a be { a, bR and b, cR, a! Transitive but not symmetric relation antisymmetric relation corresponding to bit strings with that number symmetric... Called a partially ordered set or poset set if there is a of. Stuff Given above, if and only if, a ≤ b2 } it can be transitive! Which of the relation R on the set of all real numbers set disjoint. From the vertex to another, there is no pair of distinct elements of the following three properties 1... Brush up on my understanding of relations with sets then the second is... 1.1 reflexive, symmetric and transitive relations pdf 1 sets, and is transitive y a, bR and b, c } symmetric! Relations like reflexive, nor symmetric, and reflexive row as and having the same size as is an relation. | {"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": 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": 0.9464071989059448, "perplexity": 584.5775799868243}, "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-2021-39/segments/1631780056892.13/warc/CC-MAIN-20210919160038-20210919190038-00710.warc.gz"} |
https://mathhelpboards.com/threads/traffic-flow-modelling.2846/ | # Traffic Flow Modelling
#### grandy
##### Member
Dec 26, 2012
73
Traffic Flow Modelling,
I really confuse and do not how to start .
Last edited by a moderator:
#### CaptainBlack
##### Well-known member
Jan 26, 2012
890
Traffic Flow Modelling,
I really confuse and do not how to start .
View attachment 524
For part (a) you need to observe that the flow rate $$f(x,t)$$ in vehicles per unit time is $$u(x,t) \rho(x,t)$$.
Now you need to show that for (i) and (ii) that $$u(x,t)\le u_{sl}$$, then as $$\rho(x,t) \le \rho_{max}$$ you will have shown that the flow rate:
$f(x,t)\le u_{sl}\rho_{max}$
#### CaptainBlack
##### Well-known member
Jan 26, 2012
890
Traffic Flow Modelling,
I really confuse and do not how to start .
View attachment 524
The next step is to write down the partial differential equation satisfied by the traffic density. This is derivable from a conservation of mass (or vehicle numbers) argument that you will have seen innumerable times.
CB
#### grandy
##### Member
Dec 26, 2012
73
The next step is to write down the partial differential equation satisfied by the traffic density. This is derivable from a conservation of mass (or vehicle numbers) argument that you will have seen innumerable times.
CB
#### CaptainBlack
##### Well-known member
Jan 26, 2012
890
Consider a road element between $$x$$ and $$x+\Delta x$$ the traffic flow into the element at $$x$$ per unit time is $$u(\rho(x,t))\rho(x,t)$$ and out at $$x+\Delta x$$ is $$u(\rho(x+\Delta x,t))\rho(x+\Delta x,t)$$ Therefore the rate of change of car numbers in the element is:
$\frac{\partial N}{\partial t}=u(\rho(x,t))\rho(x,t)-u(\rho(x+\Delta x,t))\rho(x+\Delta x,t)$
and so the rate of change of density in the element is:
$\frac{1}{\Delta x}\frac{\partial N}{\partial t}=\frac{u(\rho(x,t))\rho(x,t)-u(\rho(x+\Delta x,t))\rho(x+\Delta x,t)}{\Delta x}$
Now take the limit as $$\Delta x \to 0$$ to get:
$\frac{\partial \rho}{\partial t}=\frac{\partial}{\partial x}u(\rho)\rho$
CB
Last edited:
#### CaptainBlack
##### Well-known member
Jan 26, 2012
890
For part (a) you need to observe that the flow rate $$f(x,t)$$ in vehicles per unit time is $$u(x,t) \rho(x,t)$$.
Now you need to show that for (i) and (ii) that $$u(x,t)\le u_{sl}$$, then as $$\rho(x,t) \le \rho_{max}$$ you will have shown that the flow rate:
$f(x,t)\le u_{sl}\rho_{max}$
In small time interval $$\Delta t$$ all the vehicles less than a distance $$u(x,t)\Delta t$$ down stream of $$x$$ will pass $$x$$. The number of vehicles in this stretch of road is $$u(x,t)\rho(x,t)\Delta t$$, so $$u(x,t)\rho(x,t)\Delta t$$ vehicles pass $$x$$ in $$\Delta t$$ so the vehicle flow rate at $$x$$ is $$u(x,t)\rho(x,t)$$ vehicles per unit time.
CB
#### grandy
##### Member
Dec 26, 2012
73
Thank you very much for your lovely answer. Please explain me what do you mean by Triangle sign x represent. And also in next line below what does Triangle sign t mean?
Dec 26, 2012
73
#### grandy
##### Member
Dec 26, 2012
73
Traffic flow Modelling
On a stretch of single-lane road with no entrances or exits the traffic density ρ(x,t) is a continuous function of distance x and time t, for all t > 0, and the traffic velocity ) u( ρ) is a function of density alone.
Two alternative models are proposed to represent u:
i)u = u_(SL)*(1- ρ^n/ρ^n_max ), where n is a postive constant
ii) u = u_(SL)* In (ρ_max / ρ)
Where u_SL represents the maximum speed limit on the road and p_max represents maximum density of traffic possible on the road(meaning bumper-to-bumper traffic)
Compare the realism of the 2 models for u above. You should consider in particular the variations of velocity with density for each model, and the velocities for high and low densities in each case. State which model you prefer, giving reasons.
=>
I did for case i) which is u = u_(SL)*(1- ρ^n/ρ^n_max ),
u(ρ) = u_(SL)*(1- ρ^n/ρ^n_max ), for 0<ρ<ρ_max
Since ρ>= 0, cannot exceed u_SL
when ρ= ρ_max , u (ρ_max)= u_SL(1- ρ_max/ρ_max) =0
when ρ=0, u(0)= u_SL(1-0/ρ_max)= u_SL
Also, du/dρ= (- u_SL/ρ_max ) <0, so drivers reduce speed as density increase | {"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": 0.9116829633712769, "perplexity": 1649.5293710359147}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-40/segments/1600400241093.64/warc/CC-MAIN-20200926102645-20200926132645-00421.warc.gz"} |
http://math.stackexchange.com/questions/272950/solving-heat-equation-pde | # Solving Heat Equation PDE
I need some help on solving this heat PDE :
Question : Consider a bar length L. The face at x=0 is insulated so that the heat flow across is zero, and the face at x=L is held at temperature u=0. The temperature distribution is governed by heat equation
$${\partial u \over \partial t} - k {\partial ^2 u \over \partial x^2} =0$$
Show that the normal modes of u(x,t) are
$$\large U_n(x,t)=B_n\cos[(2n-1)\pi x/2L]e^{[-(2n-1)^2\pi^2k^2t]/4L^2}$$
Given BCs are : u(0,t)=0 and u(L,t)=0
I already solve this PDE
and I got my answer in this form :
$$\large U_n(x,t)=B_n\sin[n\pi x/2L] e^{[-(n)^2\pi^2k^2t]/4L^2}$$
How am I need to change my answer into the form they asking ??? Anyone willing to help me ?
-
Isn't the B.C. at x=0 expressed as $u_t (0,t) = 0$, as the temperature at $x=0$ isn't changing due to the insulation? – Ron Gordon Jan 8 '13 at 17:52
No it cant. Because boundary condition is given in interval 0<x<L. The one you said only applies for ICs only. But in this case ICs are not given – Garett Jan 8 '13 at 17:55
I'm pretty sure rlgordonna is correct: if the heat flow across x=0 is 0, that means $u_t (0,t) = 0$. This should lead to the right eigenfunction, too. – AndrewG Jan 8 '13 at 18:06
Ok let me try it out first and then I confirm back to you again – Garett Jan 8 '13 at 18:11
I cant get it. I think it mucg be easier if you could could show me the steps on how you done it Andrew Gibson. Please help me out – Garett Jan 8 '13 at 18:18
You need to express the problem as
$$u_t = k u_{xx}$$
$$u_x(0,t) = 0 ,\;\;\; u(L,t) = 0$$
Use separation of variables, i.e. $u(x,t) = X(x)T(t)$ and get
$$X'' + \frac{\lambda}{k} X = 0$$
where $-\lambda$ is the separation constant. The solution to this equation is
$$X(x) = A \cos{\left (\sqrt{\frac{\lambda}{k}} x \right)} + B \sin{\left (\sqrt{\frac{\lambda}{k}} x \right)}$$
The condition at $x=0$ implies that $B=0$:
$$X'(x) = -\sqrt{\frac{\lambda}{k}} A \sin{\left (\sqrt{\frac{\lambda}{k}} x \right)} + B \sqrt{\frac{\lambda}{k}}\cos{\left (\sqrt{\frac{\lambda}{k}} x \right)}$$
$$X'(0) = B \sqrt{\frac{\lambda}{k}} = 0$$
The condition at $x=L$ implies that
$$\cos{\left (\sqrt{\frac{\lambda}{k}} L \right)} = 0$$
so that
$$\lambda = \left [ (2 n-1) \frac{\pi}{2 L} \right ]^2 k$$
You should be able to take it from there with the time equation.
-
I just cant get the part where you do X'(x). Shouldnt the BCs $u_t(0,t)=O$ which is $X(0)T'(t)=0$?? Why you did X'(x) rlgordonma? Could you explain – Garett Jan 8 '13 at 18:37
My mistake - it should be $u_x(0,t) = 0$. I will fix it above. – Ron Gordon Jan 8 '13 at 18:40
For some reason, I mixed up holding a fixed temp., and heat flow, with is a temp. flux, i.e. a gradient, i.e., a spatial derivative. I hope that clears it up. – Ron Gordon Jan 8 '13 at 18:43
Ok now I understand. But just one thing could you teach me how you indentify it is $u_x(0,t)=0$ rather than $u_t(0,t)=0$ or $u(0,t)=0$. Tell me some tips so that I can understand clearly about this part – Garett Jan 8 '13 at 18:46
My comment above should explain some of it. The B.C.'s are described using certain terms. If the temp is held fixed to zero at one end, then $u=0$ at that end. If the temp is held fixed over time in general, then $u_t = 0$, which obviously returns a similar result to just setting $u=0$ at that end. (My mistake in being so glib before and confusing you.) If there is no heat flow, or the end is insulated, then $u_x = 0$ at that end. – Ron Gordon Jan 8 '13 at 18:52 | {"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": 0.9128504395484924, "perplexity": 861.262752782383}, "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/1432207928869.6/warc/CC-MAIN-20150521113208-00268-ip-10-180-206-219.ec2.internal.warc.gz"} |
http://mathhelpforum.com/advanced-algebra/60816-dimension-vectorspaces.html | # Math Help - Dimension of Vectorspaces
1. ## Dimension of Vectorspaces
Hi
how to show that the dimension of a subvector space ist smaller or at least equal to the dimension of the main vectorspace?
I know that the dimension is the length of the basis but how to use this information here?
thank you
greetings
2. Originally Posted by Herbststurm
Hi
how to show that the dimension of a subvector space ist smaller or at least equal to the dimension of the main vectorspace?
I know that the dimension is the length of the basis but how to use this information here?
thank you
greetings
If $V$ is a subspace of $W$ then say $\text{dim}(V) > \text{dim}(W)$ then it means we can find a basis set for $V$, $\{v_1,...,v_n\}$. But these vectors are linearly independent. The problem is that a set of linearly independent vectors cannot exceede the dimension and we see that $v_1,...,v_n$ exceede the dimension of $W$. This is a contradiction.
3. Sorry, I don't understand your post
How could you write that the dimension of the subspace is bigger than the main space?
Ad why do you say the base is linear dependent. I thought a base is always a linear indipendent creation system?
greetings
4. Originally Posted by Herbststurm
Sorry, I don't understand your post
How could you write that the dimension of the subspace is bigger than the main space?
Ad why do you say the base is linear dependent. I thought a base is always a linear indipendent creation system?
greetings
Say $\text{dim}(W) = 3$ and $V\subseteq W$ as a subspace has $\text{dim}(V) = 4$. Then this means $V$ has a basis of four elements, say, $\{ v_1,v_2,v_3,v_4\}$. This is a linearly independent set in $V$. Since $v_i \in V \implies v_i \in W$ because $V\subseteq W$. Therefore, $\{ v_1,v_2,v_3,v_4\}$ is a linearly independent set in $W$. But it is impossible to have a linearly independent set so that the number of elements in the set exceede the dimension of the space. Here the number of elements are $4$ and the dimension of W is $3$ and we see that $4>3$. This is a contradiction. Thus, the basis for $V$ must have fewer or equal to than $3$ elements. | {"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": 22, "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": 0.9247221946716309, "perplexity": 152.2205431765322}, "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-2014-15/segments/1397609535535.6/warc/CC-MAIN-20140416005215-00459-ip-10-147-4-33.ec2.internal.warc.gz"} |
http://math.stackexchange.com/users/1374/dan?tab=activity&sort=posts | Dan
Reputation
3,120
Next privilege 5,000 Rep.
Approve tag wiki edits
May29 answered Is it an abuse of notation to omit the leading zero in a decimal less than 1? Dec30 answered What would be the center of the curvature? Sep12 answered Trignometric problem (using De Movier's Theorem) Apr22 answered How many days are there in 70 years? Apr5 answered how to find the root in $x^3-x^2+6x+24$? Apr5 answered Is $e^{-x}$ Lipschitz on the positive reals? Mar28 answered Why do we limit the definition of a function? Mar21 answered Is $\frac{0}{0}$ different from $\frac{1}{0}$? Mar20 answered What is the graph of $r \cos \theta = 3$? Mar20 answered What is the graph of the polar equation $r = e$? Feb17 answered Exponential of $\bar{z}$ Feb13 answered is there a formula for modulo Feb13 answered Perfect shuffle of 52 cards Feb13 answered How can you find the cubed roots of $i$? Feb13 answered Expressing numbers in cartesian form Feb13 asked How much advantage would a Blackjack player gain by being able to see the underside of cards? Dec23 answered Why every prime (>3) is represented as $6k\pm1$ Dec16 answered Prove the lecturer is a liar… Dec12 answered Why does Trapezoidal Rule have potential error greater than Midpoint? Dec7 answered Area of a square in polar coordinates? | {"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": 0.9549298882484436, "perplexity": 2170.5177095191875}, "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/1438042989142.82/warc/CC-MAIN-20150728002309-00097-ip-10-236-191-2.ec2.internal.warc.gz"} |
http://mathoverflow.net/questions/119830/combinatorial-lemma-is-it-well-known?answertab=active | # combinatorial lemma (is it well-known?)
The following should be something well?-known, but i haven't seen it anywhere, neither have i met any references about.
Let $M^{n}$ be a $n$-dimensional oriented closed manifold with a (sufficiently small) triangulation $\tau$. We "colour" the vertices of $\tau$ with $n+2$ colors: $v^{o}\rightarrow w(v^{o})\in$ {$1,2,...,n+2$ } and we shall say that the correspondence $w$ is a "coloring" of $\tau$. Take an arbitrary color $i\in$ {$1,2,...,n+2$ } and consider the $n$-simplices whose vertices are colored with exactly the colors {$1,2,...,n+2$ }$\backslash\{i\}$. Let $\Delta^{n}$ be such a simplex and $v_{1},...,v_{n+1}$ be its vertices ordered according to the positive orientation of $\Delta^{n}$ induced by the orientation of $M^{n}$. Then we write $\sigma_{i}(\Delta^{n})=1$, if the permutation $(w(v_{1}),...,w(v_{n+1}))$ is even, and $\sigma_{i}(\Delta ^{n})=-1$ otherwise. Set $\sigma_{i}(\Delta^{n})=0$ if some vertex of $\Delta^{n}$ is colored $i$, or there are two identically colored vertices. Let finally
$\sigma_{i}(w)=\sum\sigma_{i}(\Delta^{n})$,
where the sum is over all $n$-simplices.
The Claim: The number $\sigma_{i}(w)$ does not depend on $i$: $\sigma_{1}(w)=\sigma_{2}(w)=...=\sigma_{n+2}(w)$. So we have a global invariant $\sigma(w)$ of the coloring $w$.
This invariant has a geometrical meaning: Consider the dual cell complex of the triangulation $\tau$, then since each cell corresponds to a vertex $v^{o}$ of $\tau$, we may color this cell by the color $w(v^{o})$. Let $F_{i}$ be the union of all cells colored $i$, then we get a covering $\lambda=\{F_{1},...,F_{n+2}\}$ of $M^{n}$. It is easy to see that the intersection of all $F_{i}$ is empty, so the canonical map of $M^{n}$ into the nerve of $\lambda$ may be considered as a map of $M^{n}$ into the $n$-sphere $\mathbb{S}^{n}$: $\varphi:M^{n}\rightarrow\mathbb{S}^{n}$. Then the degree of $\varphi$ equals $\sigma(w)$:
$\deg\varphi=\sigma(w)$.
As the proofs are not sophisticated at all and the construction seems conceptual, maybe it is worth including this material in an elementary topology textbook. Note also that it gives a method for calculating the degree without smooth approximation.
Of course, i don't want to repeat well-known things without citation, so any references are welcome.
-
there is a much simpler way to say the geometrical meaning: consider the standard $n+1$-simplex, with vertices $1, \ldots, n+2$. Then a coloring of the vertices of a triangulation is just the same thing as a simplicial map to this simplex. now the statement about the degree is obvious. – Vivek Shende Jan 27 '13 at 7:17 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9688111543655396, "perplexity": 153.42568628324386}, "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-2014-35/segments/1408500832662.33/warc/CC-MAIN-20140820021352-00472-ip-10-180-136-8.ec2.internal.warc.gz"} |
https://brilliant.org/problems/draw-a-6x6-table-first/ | Draw A 6x6 Table First!
Probability Level 1
If I toss a fair 6-sided dice twice, which of the following is most likely to occur?
(A): The sum of the numbers is 9.
(B): The absolute difference between the numbers is 4.
(C): The product of the numbers is 12.
(D): The ratio between the larger number and the smaller number is 3.
(E): All of the 4 above events are equally likely to occur
× | {"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": 0.8978589773178101, "perplexity": 464.0688779673045}, "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-2020-24/segments/1590347399820.9/warc/CC-MAIN-20200528135528-20200528165528-00203.warc.gz"} |
https://artofproblemsolving.com/wiki/index.php?title=2009_AIME_II_Problems/Problem_3&oldid=87065 | 2009 AIME II Problems/Problem 3
Problem
In rectangle , . Let be the midpoint of . Given that line and line are perpendicular, find the greatest integer less than .
Solution
Solution 1
From the problem, and triangle is a right triangle. As is a rectangle, triangles , and are also right triangles. By , , and , so . This gives . and , so , or , so , or , so the answer is .
Solution 2
Let be the ratio of to . On the coordinate plane, plot , , , and . Then . Furthermore, the slope of is and the slope of is . They are perpendicular, so they multiply to , that is, which implies that or . Therefore so .
Solution 3
Similarly to Solution 2, let the positive x-axis be in the direction of ray and let the positive y-axis be in the direction of ray . Thus, the vector and the vector are perpendicular and thus have a dot product of 0. Thus, calculating the dot 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": 0.9980254173278809, "perplexity": 508.98055138287805}, "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-2021-39/segments/1631780057417.92/warc/CC-MAIN-20210923074537-20210923104537-00173.warc.gz"} |
https://learn.saylor.org/mod/url/view.php?id=15239 | ## College Success: "Chapter 7, Section 7.3: Communicating with Instructors"
Read this section. Be sure to complete the checkpoint exercises at the end of the reading. | {"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": 0.8671758770942688, "perplexity": 2256.555632278581}, "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/1593657134758.80/warc/CC-MAIN-20200712082512-20200712112512-00026.warc.gz"} |
https://link.springer.com/chapter/10.1007/978-1-4757-2698-5_10 | # Series
• Serge Lang
Part of the Undergraduate Texts in Mathematics book series (UTM)
## Abstract
Let E be a normed vector space. Let {v n } be a sequence in E. The expression
$$\sum\limits_{n = 1}^\infty {{v_n}}$$
is called the series associated with the sequence, or simply a series. We
$${s_n} = \sum\limits_{k = 1}^n {{v_k}} = {v_1}\, + \,...\, + \,{v_n}$$
call % MathType!MTEF!2!1!+- its n-th partial sum. If $$\begin{array}{*{20}{c}} {\lim } \\ {n \to \infty } \end{array}$$ s n exists, we say that the series converges, and we define the infinite sum to be this limit, that is
$$\sum\limits_{k = 1}^\infty {{v_k}} = \begin{array}{*{20}{c}} {\lim } \\ {n \to \infty } \end{array}{s_n}.$$
## Keywords
Power Series Uniform Convergence Open Interval Pointwise Convergence Convergent Series
These keywords were added by machine and not by the authors. This process is experimental and the keywords may be updated as the learning algorithm improves. | {"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": 0.984181821346283, "perplexity": 1201.2423334126454}, "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-2018-13/segments/1521257647576.75/warc/CC-MAIN-20180321043531-20180321063531-00660.warc.gz"} |
http://mathhelpforum.com/calculus/125982-population-growth-problem-print.html | # Population growth problem
• Jan 28th 2010, 11:37 AM
Evan.Kimia
Population growth problem
I figured out everything but i cant figure out this last one below. Any ideas? Thanks in advance.
http://img683.imageshack.us/img683/4186/problemm.png
• Jan 28th 2010, 11:46 AM
1005
Quote:
Originally Posted by Evan.Kimia
I figured out everything but i cant figure out this last one below. Any ideas? Thanks in advance.
http://img683.imageshack.us/img683/4186/problemm.png
$P(t) = 120e^{.804718956218t}$
$P'(t) = 120*.804718956218e^{.804718956218t}$
$P'(5) = 120*.804718956218e^{.804718956218* 5}$
$P'(5)=5397.58 \frac{bacteria}{hour}$
• Jan 28th 2010, 04:45 PM
Evan.Kimia
Ah, thanks. I did that before and must of typed it wrong on the calc | {"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": 4, "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": 0.9939818978309631, "perplexity": 2252.8448360134257}, "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/1476988719273.37/warc/CC-MAIN-20161020183839-00109-ip-10-171-6-4.ec2.internal.warc.gz"} |
https://iwaponline.com/wst/article-abstract/63/2/276/13798/UV-photolysis-of-perfluorooctanoic-acid-PFOA-in?redirectedFrom=fulltext | Perfluorooctanoic acid (PFOA) is very persistent in the environment and widely detected in the water environment. Only some advanced methods with extreme reaction conditions are shown to be capable of degrading the compound efficiently, and almost all the earlier investigations used very high PFOA concentrations. The compound is detected normally at very low concentrations in the water environment, while mild reaction conditions for its degradation are preferable. This article aimed to elucidate photodegradation of PFOA in dilute aqueous solutions by combined UV wavelengths (185 nm+254 nm) and 254 nm using a newly designed UV jacket. PFOA degradation was greatly enhanced with the combined wavelengths with almost one hundred percent PFOA removals in four-hour reaction. The removals were well described by the first-order reaction kinetic. The removal efficiencies and rate values significantly decreased with smaller initial PFOA concentrations. But defluorination was greatly enhanced with smaller PFOA concentrations possibly due to accelerated decomposition of fluorinated intermediates of PFOA. Formic acid and acetic acid were two tentatively identified intermediates of PFOA photolysis while the former was a major intermediate predominantly controlling solution pH during the oxidation. The results demonstrated that PFOA photolysis by the combined wavelengths with mild reaction conditions can be greatly enhanced by proper design of UV jacket and reactor system.
This content is only available as a PDF. | {"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": 0.8590885996818542, "perplexity": 4853.535608554054}, "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-2022-33/segments/1659882570977.50/warc/CC-MAIN-20220809124724-20220809154724-00276.warc.gz"} |
http://mathhelpforum.com/math-challenge-problems/98429-solve-equation-print.html | # solve the equation
• August 18th 2009, 03:44 AM
Random Variable
solve the equation
Solve for z.
$i^{z} = z$ where $z \in \mathbb{C}$
• August 19th 2009, 05:05 PM
Random Variable
I'm not sure if a hint is needed, or if this problem is not challenging at all. I'm leaning towards the latter.
• August 20th 2009, 08:09 PM
luobo
Quote:
Originally Posted by Random Variable
I'm not sure if a hint is needed, or if this problem is not challenging at all. I'm leaning towards the latter.
The complex power of a complex number is multi-valued. So there could be an infinite number of solutions. Today is too late, we can discuss this sometime later. ;-)
• August 20th 2009, 08:33 PM
Random Variable
Quote:
The complex power of a complex number is multi-valued. So there could be an infinite number of solutions.
True. It depends which branch you pick.
• August 20th 2009, 08:39 PM
CaptainBlack
Quote:
Originally Posted by Random Variable
Solve for z.
$i^{z} = z$ where $z \in \mathbb{C}$
Has closed form solution in terms of Lambert's W
CB
• August 21st 2009, 03:13 AM
Random Variable
Quote:
Originally Posted by CaptainBlack
Has closed form solution in terms of Lambert's W
CB
I know.
Spoiler:
$i^{z} = z$
$i^{-z} = \frac{1}{z}$
$z i^{-z} = 1$
$-z i^{-z} = -1$
$-ze^{-z \ln i} = -1$
$-z \ln i \ e^{-z \ln i} = - \ln i$
$W(-z \ln i e^{-z \ln i}) = W (- \ln i)$
$-z \ln i = W(-\ln i)$
$z = \frac{W(-\ln i)}{-\ln i}$
• August 21st 2009, 08:02 AM
CaptainBlack
Quote:
Originally Posted by Random Variable
I know.
Spoiler:
$i^{z} = z$
$i^{-z} = \frac{1}{z}$
$z i^{-z} = 1$
$-z i^{-z} = -1$
$-ze^{-z \ln i} = -1$
$-z \ln i \ e^{-z \ln i} = - \ln i$
$W(-z \ln i e^{-z \ln i}) = W (- \ln i)$
$-z \ln i = W(-\ln i)$
$z = \frac{W(-\ln i)}{-\ln i}$
Horrible, try putting $i=e^{\pi i/2}$
(the term horrible here is not intended to be insulting, but is a term often used to describe an aesthetically unpleasing piece of mathematics)
CB
• August 21st 2009, 08:38 AM
Random Variable
OK. Do this instead. Let a be any constant (real or complex).
$a^{z} = z$
$a^{-z} = \frac{1}{z}$
$z a^{-z} = 1$
$-z a^{-z} = -1$
$-ze^{-z \ \text{Log a}} = -1$
$-z \ (\text{Log a}) \ e^{-z \ \text{Log a}}= - \text{Log a}$
$W(-z \ (\text{Log a}) \ e^{-z \ \text{Log a}}) = W (- \text{Log a})$
$-z \text{Log a} = W(-\text{Log a})$
$z = \frac{W(-\text{Log a})}{-\text{Log a}}$
$\text{Log i} = \ln|i| + i \frac{\pi}{2} = i\frac{\pi}{2}$
$\frac {W(-i \frac{\pi}{2})}{-i\frac{\pi}{2}} \approx 0.4382829366+i0.3605924718$
Is that satisfactory?
• August 21st 2009, 06:02 PM
luobo
Quote:
Originally Posted by Random Variable
OK. Do this instead. Let a be any constant (real or complex).
$a^{z} = z$
$a^{-z} = \frac{1}{z}$
$z a^{-z} = 1$
$-z a^{-z} = -1$
$-ze^{-z \ \text{Log a}} = -1$
$-z \ (\text{Log a}) \ e^{-z \ \text{Log a}}= - \text{Log a}$
$W(-z \ (\text{Log a}) \ e^{-z \ \text{Log a}}) = W (- \text{Log a})$
$-z \text{Log a} = W(-\text{Log a})$
$z = \frac{W(-\text{Log a})}{-\text{Log a}}$
$\text{Log i} = \ln|i| + i \frac{\pi}{2} = i\frac{\pi}{2}$
$\frac {W(-i \frac{\pi}{2})}{-i\frac{\pi}{2}} \approx 0.4382829366+i0.3605924718$
Is that satisfactory?
This looks good to me!
• August 21st 2009, 07:27 PM
simplependulum
But what is $W(z)$ ?
• August 21st 2009, 08:34 PM
Random Variable
Quote:
Originally Posted by simplependulum
But what is $W(z)$ ?
Lambert W-Function
It's the function W(z) that solves $z = W(z)e^{W(z)}$ for a given value of z.
• August 21st 2009, 11:24 PM
CaptainBlack
Quote:
Originally Posted by Random Variable
OK. Do this instead. Let a be any constant (real or complex).
$a^{z} = z$
$a^{-z} = \frac{1}{z}$
$z a^{-z} = 1$
$-z a^{-z} = -1$
$-ze^{-z \ \text{Log a}} = -1$
$-z \ (\text{Log a}) \ e^{-z \ \text{Log a}}= - \text{Log a}$
$W(-z \ (\text{Log a}) \ e^{-z \ \text{Log a}}) = W (- \text{Log a})$
$-z \text{Log a} = W(-\text{Log a})$
$z = \frac{W(-\text{Log a})}{-\text{Log a}}$
$\text{Log i} = \ln|i| + i \frac{\pi}{2} = i\frac{\pi}{2}$
$\frac {W(-i \frac{\pi}{2})}{-i\frac{\pi}{2}} \approx 0.4382829366+i0.3605924718$
Is that satisfactory?
It is better in that you eliminate the $\log$ 's in the final answer but I would still prefer a solution that did not have to worry about which branch of the $\log$ was in use where. That can be done by the the first step that I proposed (which, while that is not the way I think of it, absorbs the taking of logarithms while effectively fixing the branch).
Also as $W$ is multi-valued the right hand side of final line is just one of the solutions that correspond to the left hand side (as you know, and I know you know).
CB
• August 22nd 2009, 06:00 PM
luobo
Quote:
Originally Posted by CaptainBlack
It is better in that you eliminate the $\log$ 's in the final answer but I would still prefer a solution that did not have to worry about which branch of the $\log$ was in use where. That can be done by the the first step that I proposed (which, while that is not the way I think of it, absorbs the taking of logarithms while effectively fixing the branch).
Also as $W$ is multi-valued the right hand side of final line is just one of the solutions that correspond to the left hand side (as you know, and I know you know).
CB
The multi-valued-ness needs to be discussed and dealt with, because that is the key here.
• August 22nd 2009, 11:45 PM
CaptainBlack
Quote:
Originally Posted by Random Variable
Solve for z.
$i^{z} = z$ where $z \in \mathbb{C}$
I like this method:
As $i=e^{(2n+1/2)\pi i}=e^{-\mu_n}$ where $\mu_n=-(2n+1/2)\pi i$ for $n\in\mathbb{Z}$ we have:
$e^{-\mu_n z}=z$
or:
$\mu_n z\; e^{\mu_n z}=\mu_n$
so:
$\mu_n z=W(\mu_n)$.
Hence:
$z=\frac{1}{\mu_n}W(\mu_n); \ \ \mu_n= -(2n+1/2)\pi i,\ n \in \mathbb{Z}$
(notes:
1. this is at least the third refinement of the original idea scribbled in my note book
2. the multi-valuedness of the solution is not only dependent on $n$, but the Lambert-W is itself multi-valued)
CB | {"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": 75, "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": 0.9564853310585022, "perplexity": 1183.024905744762}, "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/1469257824230.71/warc/CC-MAIN-20160723071024-00040-ip-10-185-27-174.ec2.internal.warc.gz"} |
http://math.stackexchange.com/questions/916472/find-all-values-of-x-that-satisfy-this-basic-inequality | # Find all values of x that satisfy this basic inequality.
$(4x-5)/(3x+5) ≥ 3$
I multiply both sides by (3x+5), getting me:
$(4x-5) ≥ 3(3x+5)$
which simplifies to
$(4x-5) ≥ 9x + 15$
after solving for x, I get
$x ≤ -4$
But after testing through Wolframalpha, I am given:
$-4 ≤ x ≤ -5/3$
Which I don't really understand how they got. I tried multiplying the top side of the fraction by the denominator and then expanding the factored form, and then I rearranged everything once I had a quadratic equation, and I got $-5/3$ and $-2$ as answers, but still not what Wolframalpha got.
Would appreciate some help and insight into this.
Thanks.
-
The error comes from multiplying with something that can be negative. The best way to do it is to collect all the terms on one side and put on a common denominator: $\frac{4x-5}{3x+5} - 3\frac{3x+5}{3x+5} \geq 0$. Then analyze where the numerator and denominator is positive/negative. – Winther Sep 1 '14 at 23:32
Your error is that you multiplied both sides of your inequality by $3x + 5$, which is positive some of the time, and negative some of the time. Thus leaving the $\geq$ sign as it was was not permitted. – Dave Sep 1 '14 at 23:32
When you multiplied both sides by (3x + 5), you forgot to consider the case when (3x + 5) < 0 – user137481 Sep 1 '14 at 23:32
If you multiply by a positive number the $\geq$ stays but when you multiply by a negative number sign changes to $\leq$.So you can take 2 cases when $3x+5\geq0$ and $3x+5<0$.Anyway it's better to go $$\frac{4x-5}{3x+5}-3\geq 0\\\frac{4x-5-9x-15}{3x+5}\geq0\\\frac{-5x-20}{3x+5}\geq0$$
-
For $x>-\frac{5}{3}$:
$$4x-5 \geq 3(3x+5) \Rightarrow 4x-5 \geq 9x+15 \Rightarrow 5x \geq -20 \Rightarrow x \geq -4$$
For $x<-\frac{5}{3}$:
$$4x-5 \leq 3(3x+5) \Rightarrow 4x-5 \leq 9x+15 \Rightarrow \dots$$
- | {"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": 0.9493513703346252, "perplexity": 251.93783617160568}, "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-2015-40/segments/1443736678818.13/warc/CC-MAIN-20151001215758-00017-ip-10-137-6-227.ec2.internal.warc.gz"} |
http://techtud.com/video-lecture/lecture-aggregation | ##### Lecture on Aggregation
Content covered:
• This lecture explains the concept of Aggregation in enhanced ER model.
• Aggregation is an abstraction, it is a a way of hiding higher level details.
• It is a way of avoiding data redundancy.
More Less | {"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": 0.8863759636878967, "perplexity": 4302.128961327097}, "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-2018-30/segments/1531676591140.45/warc/CC-MAIN-20180719144851-20180719164851-00336.warc.gz"} |
http://www-old.newton.ac.uk/programmes/HOP/seminars/2007032614151.html | # HOP
## Seminar
### Hamiltonian slow manifolds with internal oscillation
MacKay, RS (Warwick)
Monday 26 March 2007, 14:15-15:00
Seminar Room 1, Newton Institute
#### Abstract
Many Hamiltonian systems possess special families of solutions which can be described approximately as slowly drifting periodic orbits. Examples include the gravitational three-body problem, the interaction of two identical charged particles in a magnetic field, and the propagation and interaction of discrete breathers (time-periodic spatially localised motions in Hamiltonian networks of units). Theory will be presented for how to improve a zeroth order manifold of approximate solutions to r-th order for any r > 0, meaning one that contains all nearby periodic orbits of nearby period and has error of order the r-th power of the drift field (even with a small constant for r=1). In the normally hyperbolic case an exactly invariant nearby submanifold can be constructed. If there are normally elliptic directions, however, this is impossible in general but the above r-th order approximations can be achieved provided that the normal frequencies avoid all multiples of that of the approximately periodic motion and are fast compared with the drift. An effective Hamitlonian is derived to describe the drift of the orbits. Applications to the above fields will be given. An introduction has been published as R.S.MacKay, Slow Manifolds, in Energy localisation and transfer, eds T.Dauxois, A.Litvak-Hinenzon, R.S.MacKay, A.Spanoudaki (World Sci Publ Co, 2004), 149-192.
#### Video
The video for this talk should appear here if JavaScript is enabled.
If it doesn't, something may have gone wrong with our embedded player.
We'll get it fixed as soon as possible. | {"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": 0.820885419845581, "perplexity": 1480.6699629140012}, "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/1427131305143.93/warc/CC-MAIN-20150323172145-00016-ip-10-168-14-71.ec2.internal.warc.gz"} |
http://eprints.imtlucca.it/475/ | # Event-driven optimization-based control of hybrid systems with integral continuous-time dynamics
Di Cairano, Stefano and Bemporad, Alberto and Júlvez, Jorge Event-driven optimization-based control of hybrid systems with integral continuous-time dynamics. Automatica, 45 (5). pp. 1243-1251. (2009)
Full text not available from this repository.
## Abstract
In this paper we introduce a class of continuous-time hybrid dynamical systems called integral continuous-time hybrid automata (icHA) for which we propose an event-driven optimization-based control strategy. Events include both external actions applied to the system and changes of continuous dynamics (mode switches). The icHA formalism subsumes a number of hybrid dynamical systems with practical interest, e.g., linear hybrid automata. Different cost functions, including minimum-time and minimum-effort criteria, and constraints are examined in the event-driven optimal control formulation. This is translated into a finite-dimensional mixed-integer optimization problem, in which the event instants and the corresponding values of the control input are the optimization variables. As a consequence, the proposed approach has the advantage of automatically adjusting the attention of the controller to the frequency of event occurrence in the hybrid process. A receding horizon control scheme exploiting the event-based optimal control formulation is proposed as a feedback control strategy and proved to ensure either finite-time or asymptotic convergence of the closed-loop.
Item Type: Article 10.1016/j.automatica.2008.12.011 Q Science > QA Mathematics > QA75 Electronic computers. Computer scienceT Technology > TA Engineering (General). Civil engineering (General) Computer Science and Applications Professor Alberto Bemporad 27 Jul 2011 08:32 05 Mar 2014 13:43 http://eprints.imtlucca.it/id/eprint/475 | {"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": 0.8379471302032471, "perplexity": 2361.1049149077276}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-40/segments/1600400221980.49/warc/CC-MAIN-20200925021647-20200925051647-00015.warc.gz"} |
http://mathhelpforum.com/discrete-math/155862-supremum-infinum.html | # Math Help - supremum and infinum
1. ## supremum and infinum
Theorem
$E \subseteq \mathbb{R}, E \not = \varnothing$
sup $E$ exists $\Leftrightarrow$ inf $(-E)$ exists
Proof:
$"\Rightarrow"$
$supE$ exists
Now we show that $-supE=inf (-E).$
Show that
1. $-sup E$ is a lower bound of $-E$.
2. if $s$ is a lower bound of $-E \Rightarrow s \leq$ -sup $E.$
1.
$\because$ $sup E$ is an upper bound of $E$
$\therefore x \leq$ $sup E, \forall \in E \Rightarrow -x \geq$ $-sup E, \forall x \in E$
$\therefore -sup E$ is a lower bound of $-E$ (line added for correction 6:30 a.m. Sep 12)
2. Suppose that $s$ is a lower bound of $-E$
suppose not $\Rightarrow s >$ $-sup E \Rightarrow -s <$ $sup E$
on the other hand
$-x \geq s, \forall x \in E$ <------(Could someone tell me how this come about.)
$\therefore x \leq -s$
Hence,
$-s$ is an upper bound of $E \rightarrow\leftarrow$
By 1. & 2, $inf (-E) \exists$ & $inf (-E)$ = $sup E$
2. I don’t follow what you posted.
Suppose that $s = \sup (E)$ then if $y\in -E$ then $-y\in E$.
This means that $-y\le s$ or $y \ge -s$.
This shows that $- s \le \inf ( - E)$.
Suppose that $- s < \inf ( - E)$ then $s>-\inf(-E)$.
That means $\left( {\exists z \in E} \right)\left[ { - \inf ( - E) < z \le s} \right]$ or $-z<\inf(-E)$.
What is wrong with that?
3. Originally Posted by Plato
I don’t follow what you posted.
Suppose that $s = \sup (E)$ then if $y\in -E$ then $-y\in E$.
This means that $-y\le s$ or $y \ge -s$.
This shows that $- s \le \inf ( - E)$. <----This part makes a whole world of a difference.
Suppose that $- s < \inf ( - E)$ then $s>-\inf(-E)$.
That means $\left( {\exists z \in E} \right)\left[ { - \inf ( - E) < z \le s} \right]$ or $-z<\inf(-E)$.
What is wrong with that?
My original post was from lecture notes of some school of mathematics I found on the web--I don't remember where I got it. There is a missing link in the proof which made it hard to understand.
Since you said you could not follow it, you made me feel better.
Your instructions are superb, which I am very glad that you are on board.
I am not a math major, so I don't have access to a real math professor, but thanks to MHF for giving me access to a teacher like you.
Thank you, sir. | {"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": 52, "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": 0.9453146457672119, "perplexity": 396.677025399792}, "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/1432207926924.76/warc/CC-MAIN-20150521113206-00312-ip-10-180-206-219.ec2.internal.warc.gz"} |
https://hal-cea.archives-ouvertes.fr/cea-01567560 | # Evolution of spectral and transport quantities with doping in the SU(2) theory of cuprates
Abstract : Recent transport experiments in the cuprate superconductors linked the opening of the pseudogap to a change in electronic dispersion [S. Badoux et al., Nature 531, 210 (2015)]. Transport measurements showed that the carrier density sharply changes from $x$ to $1+x$ at the pseudogap critical doping, in accordance with the change from Fermi arcs at low doping to a large hole Fermi surface at high doping. The SU(2) theory of cuprates shows that antiferromagnetic short range interactions cause the arising of both charge and superconducting orders, which are related by an SU(2) symmetry. The fluctuations associated with this symmetry form a pseudogap phase. Here we derive the renormalised electronic propagator under the SU(2) dome, and calculate the spectral functions and transport quantities of the renormalised bands. We show that their evolution with doping matches both spectral and transport measurements.
Document type :
Preprints, Working Papers, ...
Cited literature [28 references]
https://hal-cea.archives-ouvertes.fr/cea-01567560
Contributor : Emmanuelle de Laborderie <>
Submitted on : Monday, July 24, 2017 - 10:00:31 AM
Last modification on : Monday, February 10, 2020 - 6:13:40 PM
### File
1704.06557.pdf
Files produced by the author(s)
### Identifiers
• HAL Id : cea-01567560, version 1
• ARXIV : 1704.06557
### Citation
Corentin Morice, Xavier Montiel, Catherine Pépin. Evolution of spectral and transport quantities with doping in the SU(2) theory of cuprates. 2017. ⟨cea-01567560⟩
Record views | {"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": 0.8386610746383667, "perplexity": 4793.95891345263}, "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/1603107922746.99/warc/CC-MAIN-20201101001251-20201101031251-00181.warc.gz"} |
http://math.stackexchange.com/questions/147561/sheaves-and-complex-analysis/147676 | # Sheaves and complex analysis
A complex analysis professor once told me that "sheaves are all over the place" in complex analysis. Of course one can define the sheaf of holomorphic functions: if $U\subset \mathbf{C}$ (or $\mathbf{C}^n$) is a nonempty open set, let $\mathcal{O}(U)$ denote the $\mathbf{C}$-vector space of holomorphic functions $f:U\to\mathbf{C}$, and we let $\mathcal{O}(\varnothing)=\{0\}$. The restriction maps are given by restriction holomorphic functions to open subsets. This defines a sheaf on $\mathbf{C}$ with respect to its usual topology.
Here are my questions:
1. Are there interesting re-interpretations of well-known results in basic complex analysis in the language of sheaf theory (just to get one thinking about how things might translate)?
2. Are there interesting new geometric insights that one gains by introducing this structure? (Feel free to reformulate the context of the question if 2 doesn't make sense).
I guess I find it counter-intuitive that sheaves should say anything interesting about complex analysis, while it seems natural that they should say things about the geometry of the space on which they're defined.
-
Sheaf theory was introduced into complex analysis very soon after it was invented by Leray (unfortunately, I don't really know about Leray's own motivations and intentions for the theory), by Cartan, but in the context of several complex variables, not just one.
What he did was reformulate Oka's theorems in sheaf-theoretic language, by proving that the structure sheaf $\mathcal O$ on $\mathbb C^n$ is coherent. He also proved his famous Theorems A and B about the coherent sheaves on Stein spaces (which e.g. immediately recover the Mittag-Leffler result discussed in Georges's answer).
Roughly speaking, the reason that sheaf theory is useful in complex analysis is that one doesn't have the patching technique of partitions of unity that is available in smooth function theory.
Indeed, one can use partitions of unity to show that the higher cohomology of the sheaf of smooth functions (and of related sheaves, such as sheaves of smooth sections of vector bundles) on any smooth manifold vanishes; this more-or-less guarantees that sheaf theory won't be a very useful tool in that setting.
But in complex analysis, those techniques aren't available, and indeed in general on complex manifolds higher cohomology of the structure sheaf, and related sheaves, needn't vanish. Thus sheaf theory becomes a useful tool. Indeed, one knows from results such as the classical Mittag-Leffler theorem that local-to-global patching of a certain kind is sometimes possible in complex analysis; sheaf theory (and especially sheaf cohomology) becomes a way to measure the obstructions to such patching, and of organizing information about those obstructions (e.g. so that one can show that they vanish in certain circumstances).
You shouldn't think that sheaf theory provides a replacement for analytic arguments; rather, it supplies a framework for efficiently organizing that analytic input, and making useful deductions from it in a conceptually clear fashion.
-
Dear Matt, I have taken the liberty of adding an answer trying to retrace "Leray's own motivations and intentions for the theory", as a little supplement to your post above. – Georges Elencwajg May 21 '12 at 9:28
As a complement to Matt's very interesting answer, let me add a few words on the historical context of Leray's discoveries.
Leray was an officer in the French army and after Frances's defeat in 1940, he was sent to Oflag XVII in Edelsbach, Austria (Oflag=Offizierslager=POW camp): look here .
The prisoners founded a university in captivity, of which Leray was the recteur (dean).
Leray was a brilliant specialist in fluid dynamics (he joked that he was un mécanicien, a mechanic!), but he feared that if the Germans learned that he gave a course on that subject, they would force him to work for them and help them in their war machine (planes, submarines,...).
So he decided to teach a harmless subject: algebraic topology!
So doing he recreated the basics on a subject in which he was a neophyte and invented sheaves, sheaf cohomology and spectral sequences.
After the war his work was examined, clarified and amplified by Henri Cartan (who introduced the definition of sheaves in terms of étalé spaces) and his student Koszul.
Serre (another Cartan student) and Cartan then dazzled the world with the overwhelming power of these new tools applied to algebraic topology, complex analysis in several variables and algebraic topology.
I find it quite interesting and moving that the patriotism of one courageous man (officers had the option to be freed if they agreed to work for the Nazis) changed the course of 20th century mathematics.
Here, finally, is Haynes Miller's fascinating article on Leray's contributions.
-
Dear Georges, Thank you very much for this. Best wishes, – Matt E May 21 '12 at 15:15
@GeorgesElencwajg What a fascinating story. Thank you for the link. – Holdsworth88 May 21 '12 at 16:00
The Mittag-Leffler theorem says the following:
If $U\subset \mathbb C$ is open and if $D\subset U$ is a discrete closed subset, you can choose arbitrarily at every $d\in D$ a polar development $\sum_{k=1}^{n(d)} a_k(\frac {1}{z- d})^k$ and there will exist a meromorphic function $m\in \mathcal M(U)$ whose polar part at $d$ is the given one and holomorphic on $U\setminus D$.
In sheaf theoretic terms this follows from the vanishing of the first cohomology group of the sheaf of holomorphic functions on the open set: $H^1(U,\mathcal O)=0$.
Not only can Mittag-Leffler's quickly be proved by first proving the vanishing theorem (a path followed by Hörmander here), but that sheaf-theoretic formulation suggests the statements and proofs of analogous theorems for arbitrary open Riemann surfaces and for (Stein) manifolds of arbitrary dimension.
-
The étale space of $\mathcal{O}$ provides a nice way to think about analytic continuation -- you can think of it as the universal domain for holomorphic maps of Riemann surfaces into $\mathbb{C}$. Narasimhan's Complex Analysis in One Variable has a section that talks a little about this (though it doesn't assume you know any sheaf theory already, so you might feel talked down to a bit).
- | {"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": 0.9411547183990479, "perplexity": 621.2054378598887}, "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-2016-30/segments/1469257825124.55/warc/CC-MAIN-20160723071025-00284-ip-10-185-27-174.ec2.internal.warc.gz"} |
http://www.aimsciences.org/journal/1547-5816/2012/8/3 | # American Institute of Mathematical Sciences
ISSN:
1547-5816
eISSN:
1553-166X
All Issues
## Journal of Industrial & Management Optimization
2012 , Volume 8 , Issue 3
Select all articles
Export/Reference:
2012, 8(3): 521-529 doi: 10.3934/jimo.2012.8.521 +[Abstract](443) +[PDF](334.4KB)
Abstract:
In this paper, we consider the $k$-level facility location problem with submodular penalties ($k$-FLPSP). We propose a primal-dual $6$-approximation (combinatorial) algorithm for the $k$-FLPSP.
2012, 8(3): 531-547 doi: 10.3934/jimo.2012.8.531 +[Abstract](411) +[PDF](432.8KB)
Abstract:
We consider constrained investment problem with the objective of minimizing the ruin probability. In this paper, we formulate the cash reserve and investment model for the insurance company and analyze the value-at-risk ($VaR$) in a short time horizon. For risk regulation, we impose it as a risk constraint dynamically. Then the problem becomes minimizing the probability of ruin together with the imposed risk constraint. By solving the corresponding Hamilton-Jacobi-Bellman equations, we derive analytic expressions for the optimal value function and the corresponding optimal strategies. Looking at the value-at-risk alone, we show that it is possible to reduce the overall risk by an increased exposure to the risky asset. This is aligned with the risk diversification effect for negative correlated or uncorrelated risky asset with the stochastic of the fundamental insurance business. Moreover, studying the optimal strategies, we find that a different investment strategy will be in place depending on the Sharpe ratio of the risky asset.
2012, 8(3): 549-564 doi: 10.3934/jimo.2012.8.549 +[Abstract](326) +[PDF](627.8KB)
Abstract:
This paper introduces a multi-objective genetic algorithm (MOGA) in regard to the portfolio optimization issue in different risk measures, such as mean-variance, semi-variance, mean-variance-skewness, mean-absolute-deviation and lower-partial-moment to optimize risk of portfolio. This study introduces a PONSGA model by appling the non-dominated sorting genetic algorithm (NSGA-II) to maximize both the expected return and the skewness of portfolio as well as to simultaneously minimize different portfolio risks. The experimental results demonstrated that the PONSGA approach is significantly superior to the GA in all performances, examined such as the coefficient of variation, Sharpe index, Sortino index and portfolio performance index. The statistical significance tests also showed that the NSGA-II models outperformed the GA models in different risk measures.
2012, 8(3): 565-575 doi: 10.3934/jimo.2012.8.565 +[Abstract](519) +[PDF](168.0KB)
Abstract:
As it is known that the performance of the $k$-means algorithm for data clustering largely depends on the choice of the Max-Min centers, and the algorithm generally uses random procedures to get them. In order to improve the efficiency of the $k$-means algorithm, a good selection method of clustering starting centers is proposed in this paper. The proposed algorithm determines a Max-Min scale for each cluster of patterns, and calculate Max-Min clustering centers according to the norm of the points. Experiments results show that the proposed algorithm provides good performance of clustering.
2012, 8(3): 577-590 doi: 10.3934/jimo.2012.8.577 +[Abstract](360) +[PDF](400.2KB)
Abstract:
This paper studies a single-period inventory (newsvendor) problem with discrete stochastic demand. In general, most of the previous works are based on the expected profit/cost criterion or expected utility criterion. We consider the effect of irrational factor under uncertainty and therefore incorporate prospect theory into inventory model. Our objective is to maximize the overall value of the prospect, which can be calculated by using the value function and the weighting function. For any given initial inventory level, it can be shown that a state-dependent order-up-to policy is optimal. Further, the optimal policy has a simple structure, and the retailer can easily decide whether to place an order or not. Moreover, the impacts of parameters on the optimal policy are illustrated through numerical experiments.
2012, 8(3): 591-609 doi: 10.3934/jimo.2012.8.591 +[Abstract](370) +[PDF](424.3KB)
Abstract:
This paper presents a neighboring extremal solution for a class of optimal switched impulsive control problems with perturbations in the initial state, terminal condition and system's parameters. The sequence of mode's switching is pre-specified, and the decision variables, i.e. the switching times and parameters of the system involved, have inequality constraints. It is assumed that the active status of these constraints is unchanged with the perturbations. We derive this solution by expanding the necessary conditions for optimality to first-order and then solving the resulting multiple-point boundary-value problem by the backward sweep technique. Numerical simulations are presented to illustrate this solution method.
2012, 8(3): 611-621 doi: 10.3934/jimo.2012.8.611 +[Abstract](350) +[PDF](210.0KB)
Abstract:
Support vector machine (SVM) is one of the most popular machine learning methods and is educed from a binary data classification problem. In this paper, the canonical duality theory is used to solve the normal model of SVM. Several examples are illustrated to show that the exact solution can be obtained after the canonical duality problem being solved. Moreover, the support vectors can be located by non-zero elements of the canonical dual solution.
2012, 8(3): 623-637 doi: 10.3934/jimo.2012.8.623 +[Abstract](509) +[PDF](321.2KB)
Abstract:
Data envelopment analysis (DEA) is a common non-parametric frontier analysis method. The multiplier framework of DEA allows flexibility in the selection of endogenous input and output weights of decision making units (DMUs) as to cautiously measure their efficiency. The calculation of DEA scores requires the solution of one linear program per DMU and generates an individual set of endogenous weights (multipliers) for each performance dimension. Given the large number of DMUs in real applications, the computational and conceptual complexities are considerable with weights that are potentially zero-valued or incommensurable across units. In this paper, we propose a two-phase algorithm to address these two problems. In the first step, we define an ideal DMU (IDMU) which is a hypothetical DMU consuming the least inputs to secure the most outputs. In the second step, we use the IDMU in a LP model with a small number of constraints to determine a common set of weights (CSW). In the final step of the process, we calculate the efficiency of the DMUs with the obtained CSW. The proposed model is applied to a numerical example and to a case study using panel data from 286 Danish district heating plants to illustrate the applicability of the proposed method.
2012, 8(3): 639-649 doi: 10.3934/jimo.2012.8.639 +[Abstract](405) +[PDF](203.3KB)
Abstract:
Predicting the lifetime of a network is a stochastic and very hard task. Sensitivity analysis of a network in order to identify the weakest points in the network, provides valuable knowledge to draw an optimum investment strategy for the expansion of the networks for the network carriers. To achieve this goal, a new measure, called topology lifetime, was recently proposed for measuring the performance of a telecommunication network. This measure not only allows to perform a sensitivity analysis of the networks, but also it provides the means to compare the different topologies with respect to the ability of the network in supporting growth in network traffic before new capacity/facility is installed. This paper addresses some improvements upon the previously defined measures and presents the implementation results of the various lifetime measure methodologies. Computational analysis on some commonly used topologies show how the new measure can be utilized in assessing network performance.
2012, 8(3): 651-656 doi: 10.3934/jimo.2012.8.651 +[Abstract](376) +[PDF](288.6KB)
Abstract:
This paper studies the worst-case performance of the successive approximation algorithm for four identical knapsacks. The algorithm packs the knapsacks successively by using an exact algorithm on the remaining items for each single knapsack. We show that it is an $\frac{8}{11}$-approximation algorithm, and the bound is tight.
2012, 8(3): 657-672 doi: 10.3934/jimo.2012.8.657 +[Abstract](404) +[PDF](317.5KB)
Abstract:
The current study deals with the lead-time variability reduction problem in a multi-supplier and single-buyer system with a milk-run delivery network. Under the assumption of finite-range stochastic lead time, we consider that the lead-time variance can be shortened at an extra crashing cost. Aiming to minimize the joint system costs, an integrated inventory model is formulated with a capacity constraint, and a solution procedure is developed to simultaneously optimize lead-time variance, replenishment cycle time, reorder point, and the integer numbers of shipment per production cycle for multiple suppliers. We also show that the buyer does a tradeoff between the crashing cost and the sum of holding and shortage costs for the decision making of the optimal lead-time variability. Numerical examples and sensitivity analysis are presented to validate the proposed model.
2012, 8(3): 673-690 doi: 10.3934/jimo.2012.8.673 +[Abstract](351) +[PDF](277.1KB)
Abstract:
This paper considers an optimal reinsurance-investment problem for an insurer, who aims to minimize the risk measured by Capital-at-Risk (CaR) with the constraint that the expected terminal wealth is not less than a predefined level. The surplus of the insurer is described by a Brownian motion with drift. The insurer can control her/his risk by purchasing proportional reinsurance, acquiring new business, and investing her/his surplus in a financial market consisting of one risk-free asset and multiple risky assets. Three mean-CaR models are constructed. By transforming these models into bilevel optimization problems, we derive the explicit expressions of the optimal deterministic rebalance reinsurance-investment strategies and the mean-CaR efficient frontiers. Sensitivity analysis of the results and a numerical example are provided.
2012, 8(3): 691-703 doi: 10.3934/jimo.2012.8.691 +[Abstract](296) +[PDF](372.2KB)
Abstract:
In this paper, on one hand, we discuss upper Hölder type estimates of solutions to parametric vector quasi-equilibria with general settings, which generalize and extend the results of Chen et al. (Optim. Lett. 5: 85-98, 2011). On the other hand, combining the technique used for primal problems with suitable modifications, we also study upper Hölder type estimates of solutions to Minty-type parametric dual vector quasi-equilibria. The consequences obtained for dual problems are new in the literature.
2012, 8(3): 705-726 doi: 10.3934/jimo.2012.8.705 +[Abstract](405) +[PDF](873.7KB)
Abstract:
In this paper, we study a new exact and smooth penalty function for semi-infinite programming problems with continuous inequality constraints. Through this exact penalty function, we can transform a semi-infinite programming problem into an unconstrained optimization problem. We find that, under some reasonable conditions when the penalty parameter is sufficiently large, the local minimizer of this penalty function is the local minimizer of the primal problem. Moreover, under some mild assumptions, the local exactness property is explored. The numerical results demonstrate that it is an effective and promising approach for solving constrained semi-infinite programming problems.
2012, 8(3): 727-732 doi: 10.3934/jimo.2012.8.727 +[Abstract](334) +[PDF](289.8KB)
Abstract:
The machine repair problem has attracted considerable attention in the field of queuing systems, due to the wide range of difficulties it entails. Wang, Liao and Yen [K.H. Wang, C.W. Liao, T.C. Yen, Cost analysis of the M/M/R machine repair problem with second optional repair: Newton-Quasi method, Journal of Industrial and Management Optimization. 6 (2010), 197-207] [1] derived a cost model to determine the optimal number of the repairmen, the optimal values of the first essential repair rate, and the second optional repair rate while maintaining the system availability at a specified level. In their approach, a direct search method is first used to determine the optimal number of repairmen followed by the Newton-Quasi method to search for the two repair rates. However, this two stage search method restricts the search space and cannot guarantee global minimum solutions. In overcoming these limitations, this study employs a particle swarm optimization algorithm to ensure a thorough search of the solution space in the pursuit of global minimum solutions. Numerical results support the superior search characteristics of the proposed solution.
2012, 8(3): 733-747 doi: 10.3934/jimo.2012.8.733 +[Abstract](475) +[PDF](448.6KB)
Abstract:
In this paper, we consider a DC (difference of convex) programming problem with joint chance constraints (JCCDCP). We propose a DC function to approximate the constrained function and a corresponding DC program ($\textrm{P}_{\varepsilon}$) to approximate the JCCDCP. Under some mild assumptions, we show that the solution of Problem ($\textrm{P}_{\varepsilon}$) converges to the solution of JCCDCP when $\varepsilon\downarrow 0$. A sequential convex program method is constructed to solve the Problem ($\textrm{P}_{\varepsilon}$). At each iteration a convex program is solved based on the Monte Carlo method, and the generated optimal sequence is proved to converge to the stationary point of Problem ($\textrm{P}_{\varepsilon}$).
2012, 8(3): 749-764 doi: 10.3934/jimo.2012.8.749 +[Abstract](375) +[PDF](364.0KB)
Abstract:
In Asplund space, Lagrange multiplier rules for approximate solutions of nonsmooth vector optimization problems are studied. The relationships between the vector and the scalar optimization problems are established. And the optimality conditions of approximate solutions for vector optimization are obtained. Moreover, the vector variational inequalities are considered by applying the partial results given in this paper.
2012, 8(3): 765-780 doi: 10.3934/jimo.2012.8.765 +[Abstract](373) +[PDF](307.8KB)
Abstract:
In this paper we consider the identification problem for a class of systems governed by nonlinear time varying interval differential equations having unknown (interval) parameters. Using the fact that system output posses lower and upper bounds, we have developed two sets of ordinary differential equations that represent the behaviour of lower and upper bounds. Based on these differential equations, the interval identification problem is converted into an equivalent identification problem in which the unknown parameters are real valued functions. Using variational arguments, we have developed the necessary conditions of optimality for the equivalent problem on the basis of which the unknown lower and upper parameters (and hence the interval parameters) can be determined. Finally, we present some numerical simulations to illustrate the effectivness of the proposed technique.
2017 Impact Factor: 0.994 | {"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": 0.878706157207489, "perplexity": 616.8986231621185}, "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-2018-34/segments/1534221212323.62/warc/CC-MAIN-20180817123838-20180817143838-00562.warc.gz"} |
http://math.stackexchange.com/questions/210799/pythagorean-theorem-a2-a2-c2-how-to-solve-for-a/210801 | # Pythagorean theorem: $A^2 + A^2 = C^2$ How to solve for $A$?
If I could only remember my algebra! So, I have a right triangle and I know the hypotenuse (28) and I know that the legs are the same length, so I have:
$A^2 + B^2 = C^2$
Where: $A=B$ and $C=28$
Pedestrian, I know, but how do I solve for $A$?
Many thanks!
-
If $A = B$ then $A^2 = B^2$, hence $A^2 + B^2 = 2 A^2 = 28$, meaning $A = \pm 2\sqrt{7}$ – Pragabhava Oct 11 '12 at 0:28
@Pragabhava: You didn't square the $C$. – Asaf Karagila Oct 11 '12 at 0:32
@AsafKaragila You are correct my friend! Also, I didn't divided by $\sqrt{2}$. Shame on me :) – Pragabhava Oct 11 '12 at 0:49
@Pragabhava: And $A$ negative may be a difficult triangle to draw. Or should we draw the legs in red and the hypotenuse in black? – André Nicolas Oct 11 '12 at 1:04
@AndréNicolas With our pseudo-Riemannian ruler and pseudo-Riemannian compass? – Pragabhava Oct 11 '12 at 1:37
Try this:
$A^2 + B^2 = C^2$
Since A = B and C = 28, then:
$A^2 + A^2 = 28^2$
$2A^2 = 28^2$
$A^2 = \frac{784}{2}$
$A = \sqrt{\frac{784}{2}}$
-
Perfect! Thank you for working it out fully! A ~= 19.79898987... for the lazy ;) – Campbeln Oct 11 '12 at 0:39
@Campbeln Not a problem! If you found my answer helpful, could you accept the answer, please? – Oliver Spryn Oct 11 '12 at 0:43
Put the numbers in the formula: $$2A^2=28^2=784$$
Now you have that $A^2=392$. You should know it from here.
-
Correct, thank you for sharing! And thank you for catching the missed square in the comment above! – Campbeln Oct 11 '12 at 0:44
I know the question already received good answers, but I want just to comment something: try working first with letters and only in the end substitute the numbers. This means that you would do:
$A^2 + A^2 = C^2 \Longrightarrow 2A^2=C^2$
Hence $A=\pm \dfrac{\strut C}{\strut \sqrt{2}}=\pm \dfrac{\strut C\sqrt{2}}{2}$
Doing so you just need to put the value for $C$ and you get $A=\pm 14\sqrt{2}$ which is the answer. I just wanted to comment that because doing things this way you get practice to work with more general equations and on the other side you reduce the number of numerical calculations needed.
- | {"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": 0.8059222102165222, "perplexity": 918.1937434880291}, "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-2014-41/segments/1410657137145.1/warc/CC-MAIN-20140914011217-00162-ip-10-234-18-248.ec2.internal.warc.gz"} |
https://math.stackexchange.com/questions/956520/quick-and-painless-definition-of-the-set-of-real-numbers | Quick and painless definition of the set of real numbers
I am looking for a simple way to describe the underlying set of the real numbers without getting into cauchy sequences or dedekind cuts. Furthermore, I want the description to not rely on some notion of equivalence (like how one can use the notion of coprime to give the rationals unique representatives). I think the following works:
$\mathbb{R} = \{\text{All decimal expansions | does not end in repeating 9s}\}$.
My question is am I forgetting about something or does this do the trick?
Edit: Indeed, this is not closed under the usual definition of $+$ and $\cdot$ but we can redefine these operations to "round up" when necessary. I can verify the field axioms on my own time; I am merely seeing if anyone can spot a subtlety that I missed or if my description truly does give unique representatives of the reals.
Edit 2: (Some context) The posters have given some great ways to define the reals if you want to spend a decent amount of time on it. I want to give the reals as a set of decimal expansions without any fancy notion of equivalence so that I can move on and do linear algebra. For example, if I were teaching a first year course, the field axioms would take a whole lecture and be super boring and Cauchy sequences/dekekind cuts would have students lining up to drop the course.
• As usual, Tim Gowers has interesting things to say: dpmms.cam.ac.uk/~wtg10/decimals.html – Hans Lundmark Oct 3 '14 at 6:44
• I wanted to point out that it seems contradictory to forbid the use of equivalences but ask for something 'quick and painless'. – Hurkyl Oct 3 '14 at 9:30
• Aside: are you sure that the "usual definition of $+$ and $\cdot$" that you have in mind is actually well-defined on infinite decimal expansions? – Hurkyl Oct 3 '14 at 9:42
• "I want to give the reals as a set of decimal expansions without any fancy notion of equivalence so that I can move on and do linear algebra." Why do you need to define the reals as a set of decimal expansions before doing linear algebra? – JiK Oct 3 '14 at 19:53
• If I was teaching a first year course, I'd assume the student has a sufficiently good intuitive understanding of the real numbers and run with that, leaving the explicit study of foundations to a later course. Even in the later courses, it's probably more appropriate for the real analysis course to do that exercise. And if you wanted to do foundations and still be self-contained and aren't doing any analysis, all you really need to state is the real closed field axioms (because you want to talk about positive numbers and factor polynomials). – Hurkyl Oct 4 '14 at 4:08
It will do the trick, but it will be messy and painful to prove the field axioms. Moreover, the definition in this way is somewhat arbitrary. Why choose base $10$? There is a more natural way to do it which is the Bourbaki definition of the reals. It relies on the completion construction of a uniform space. The completion of a uniform space is precisely the set of minimal Cauchy filters (so in particular, there is no equivalence involved). The reals are the completion of the rationals, and the rationals can be given the structure of a uniform space (basically, since it is topological group), thus the reals can be defined to have underlying set the set of all minimal Cauchy filters in $\mathbb Q$.
• Let $10 = n$ for any $n$. This construction isn't actually that bad and it illustrates the idea that the intuition (so what if your intuition is base 10?) is totally on track and can be made rigorous. If I recall from Spivak, the trick is to prove completeness first and use that everywhere, or something. – djechlin Oct 3 '14 at 9:20
• I don't know what "$10=n$ for any $n$" means. – Ittay Weiss Oct 3 '14 at 10:34
Your definition is not closed under the arithmetic operations, at least with their usual definitions on decimal expansions. For instance, take the decimal expansion $1/9 = 0.111...$ and multiply by 9 to obtain $0.999...$, which is not an element of your set.
• obviously, the arithmetic operation will take this into account. This is not the real issue with this definition. In particular, this definition can be used to define the reals. The details of the proofs though are almost horrific. – Ittay Weiss Oct 3 '14 at 6:23
• True, you could define the operations in such a way as to avoid this. But you're essentially encoding the equivalence relation into the operations; maybe you get to escape using the phrase "equivalence relation," but that's about it. I agree with your answer as well! – Dylan Murphy Oct 3 '14 at 6:26
• The equivalence relation you refer to is a totally different one than the one using, e.g., Cauchy sequences. Primarily, you don't need the axiom of choice to choose one representative from each equivalence class, since there is a canonical one. With Cauchy sequences there is no canonical choice of representative. With the uniform completion, you can first look at all Cauchy filters, and define an equivalence relation. The nice thing is that with the filters there is a minimality condition that chooses for you a unique representative from each equivalence class. – Ittay Weiss Oct 3 '14 at 6:41
If you can read German, here is a detailed development of the approach Gowers is advocating at the link given in Hans Lundmark's comment:
http://www.math.ethz.ch/~blatter/Dualbrueche_2.pdf
But note the following: Whichever approach you take, the amount of work to be done in order to verify all the details is about the same.
There are at least 3 sensible ways to do it, outlined in Spivak:
1. Define a real number as Cauchy sequences of rational numbers, with the equivalence relation that two sequences are equivalent if their differences converge to 0. So technically a real number is an equivalence class of Cauchy sequences. Yes, this is quick and painless.
2. Define a real number as a Dedekind cut. A Dedekind cut consists of two sets $(L, R)$ satisfying:
• $L$ and $R$ form a nontrivial partition, meaning each is non-empty, $L \cap R = \emptyset$, and $L \cup R = \mathbb{Q}$.
• Each element of $L$ is less than each element of $R$.
• If $a \in L$, then all rational numbers less than $a$ are also in $L$. Similarly if $b \in R$ then all rational numbers greater than $b$ are in $R$.
• $L$ has no greatest element. (The would-be greatest element is the real number being described by the cut. $R$ will have a least element if and only if the Dedekind cut corresponds to a rational number, namely that one.)
Basically a Dedekind cut gets as close as it can to pinpointing a real number, without naming it (since it can't, since we're constructing it). I've personally liked this once, since it's elementary to learn but pulls at completeness the same way a Dedekind cut pulls for a real number.
3. Your approach. The salvage for closure is to allow repeated 9s, and create an equivalence relation. It's exactly how clock arithmetic works: 7+7 would violate closure on a clock (i.e. $\mathbb{Z}/ 12\mathbb{Z}$) except we know $14 \cong 2$ so it's fine. The rigorous way to do this is to think of numbers on the clock as the sets $\{0, 12, 24, \ldots\}, \{1, 13, 25, \ldots\}, \ldots$ and define addition on those. Yes, this is "no frills."
• OP was specifically looking for a construction other than Cauchy sequences or Dedekind cuts, and one that avoids an equivalence relation. For approach three, there really is no need to introduce an equivalence relation for which every equivalence class has at most 2 elements, and of these a canonical choice can be made. So, the technical difficulties of this construction aside, that would be what OP was looking for. – Ittay Weiss Oct 3 '14 at 10:37
• I think that if you omitted the paragraph with the first number bullet point, the rest of your answer would be better. That's because I feel that the definition by Dedekind cuts is simpler because a computable Cauchy sequence doesn't necessarily approach a computable number and it's harder to prove that that definition of a real number works. – Timothy Dec 9 '18 at 19:43
• @Timothy in that case (empty, R) is a Dedekind cut and it ruins the bijection we want to set up between cuts and real numbers. Unless it corresponds to infinity but that's not what we're trying to do. I think (4) is only needed to ensure uniqueness of cuts -> reals. Without it, there's two ways to refer to rational points. – djechlin Dec 10 '18 at 15:28
There's no universal agreement on what a real number is so I will construct a set with operations +, $$\times$$, and $$\leq$$ that can be proven to be a complete ordered field but won't bother to give a proof that it's a complete ordered field and will call those the real numbers. I don't know of any really simple way of describing what a real number is so I will give as simple a description as I can. The following is just my description of the whole thing, not a universal one so if you use it for teaching, you should make it clear that you are using your own definitions and explaining how you define them. Maybe if you do a really good job of researching how to teach and teach less material so that you can move onto new material at a slower rate, your students will learn it really well. For example, the course should not teach a not very useful concept and how to make calculations about it because that might make there not be enough time to teach more useful material without teaching it too quickly such as the definition of a real number that I will later give that I think should be taught in full to make other course work later easier and another course could teach people background information from which they will later be able to easily learn so many different previously hard to learn math concepts instead of spending so much time teaching only one of them and then from that, they will be able to figure out how to write a formal proof in ZF that there is a complete ordered field which is unique up to isomorphism when they are asked to do so on a test and should only get full marks if they write the formal proof properly so that a future job can get better people which are those who have shown that they have the ability to figure out how to write a complete formal proof the way they were told to and not just an intuitive proof that they cannot figure out how to break down into a complete formal proof. Maybe you could join a research group and give your ideas on how a text book for that course should be written.
First, we construct the integers with the operations of +, $$\times$$ and $$\leq$$. For each integer $$x$$ that's not a solution to $$10 \times y = x$$, we can invent a solution, $$y$$ to the equation. Each of the newly invented numbers $$y$$ is still not a solution to $$10 \times z = y$$ so for each of those numbers we can invent a solution. We can keep going for ever. There is only one intuitive way to define addition, multiplication, and inequality on the set of all numbers that can be constructed that way There is also only one intuitive way to define a notation for each of those numbers. Take any subset of that set that's nonempty and its complement is also nonempty and for every member of the set, all smaller members are in the set. Let's say a boundary number of that set is one where all smaller numbers are in the set and all larger numbers are not in the set. Some of those sets have boundary numbers and others don't. For each of those sets that doesn't have a boundary number, we can invent a number that's larger than all numbers in the set and smaller than all numbers in its complement and then call all the numbers we've constructed so far real numbers. Now, there's only one intuitive way to define inequality on that set and the notation for each number in that set and it does not invent a second notation for numbers I already invented a notation for such as 0.999... for 1. Rather we say the notation 0.999... is undefined. Now that the notation and inequality on that set has already been defined, there is exactly one way to define 1, 0, +, $$\times$$, and $$\leq$$ on that set that I will call $$\mathbb{R}$$ such that $$(\mathbb{R}, 1, 0, +, \times, \leq)$$ is a complete ordered field and + and $$\times$$ restricted to terminating decimals matches what I previously defined and the multiplicative identity 1 has the notation 1 and the additive identity 0 has the notation 0.
Now the way to add 2 notations is by determining which numbers they represent and figuring out the sum then writing the notation of the sum. It's probably better to start all over and do it how you would if you only began school now and try to figure out how you would do it if you didn't have the past experience of how you were taught to add base 10 notations in grade 1. After reading https://www.inc.com/bill-murphy-jr/science-says-were-sending-our-kids-to-school-much-too-early-and-that-can-hurt-th.html, I think that kids learn better from school if they start it when they're older. Maybe you were so young when you learned how to add that you misunderstood what they were trying to teach you and then had trouble breaking your old habits of thinking it's a universal rule. I once had an interview at Earls and they said they do their job a totally different way and new people tend to learn it better then people with previous experience because people with previous experience tend to have trouble breaking old habits.
That method of adding does not work on nonterminating real numbers when using my notation. The method was to work on the digits going from right to left but there's no last digit to start on. If we decide to instead use a notation that allows for a string of trailing 9's, then there is another simple way to do it that given any 2 notations can produce a notation that represents the sum or product of the two numbers those are notations for. I won't explain how to do multiplication with it but for addition, you can work from left to right. If the digits of one column add up to less than 9, the column to the left has no carry. If they add up to more than 9, the column to the left has a carry. If they add up to 9, the column to the left has a carry if and only if that column has a carry. For some pairs of positive real numbers, for some columns, to determine whether it has a carry, you have to first determine whether the one to the right has a carry but to do that, you first have to determine whether the one to the right has a carry and that will keep going for ever. If you are told the information that starting from a certain point, the digits in each column always add up to 9, then you know the notation for the sum. Now what about subtraction and division.
You may have learned about fractions in elementry school and misunderstood what you were getting taught and rather than understanding that a fraction is really one notation for division of integers which is defined for all pairs of real numbers whose denominator is nonzero, you instead thought of the rational numbers as being constructed from the integers and thought there were no numbers but rational numbers. Rather, division on $$\mathbb{R}$$ is defined in terms of multiplication on $$\mathbb{R}$$ and when you then restrict the already existing multiplication and division operations on $$\mathbb{R}$$ to $$\mathbb{Q}$$, it ends up agreeing with the way you learned how to multiply and divide fractions earlier and you must reject the assumption that the rational numbers are the only numbers unless you're fine with using only the rational numbers and saying there is no solution to the equation $$x^2 = 2$$. We can still define the property of being a rational number such that either its square is less than 2 or it is negative and then define $$\sqrt{2}$$ to mean that property.
The simplest way: The real numbers are a set with operations +, -, *, / and relations < = > which follow the axioms of arithmetic, and where every nonempty set with an upper bound has a least upper bound.
Equivalence classes of Cauchy sequences are IMO the easiest way to prove that the real numbers exist, and they come kind of natural when you start talking about limits (just construct a sequence that should have the limit $2^{1/2}$ but doesn't in the rational numbers because the number doesn't exist).
(Going to decimals with unlimited number of digits will get you into huge trouble just defining the product of two numbers).
• How does that answer OP's question? – Ittay Weiss Oct 3 '14 at 10:57 | {"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": 20, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8706866502761841, "perplexity": 188.51087501112573}, "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-2019-30/segments/1563195527458.86/warc/CC-MAIN-20190722010436-20190722032436-00275.warc.gz"} |
http://math.stackexchange.com/questions/70410/compact-operator-maps-weakly-convergent-sequences-into-strongly-convergent-seque/70827 | # Compact operator maps weakly convergent sequences into strongly convergent sequences
I found the following property of compact operators in a proof, and I can't prove it.
Prove that if $T \in \mathcal{L}(E,F)$ is compact, and if $u_n \rightharpoonup u$ (the sequence converges weakly to $u$ in $\sigma(E,E^*)$) then $Tu_n \to Tu$ strongly in $F$.
I was able to prove that $Tu_n$ has a convergent subsequence ($u_n \rightharpoonup u$ implies that $(u_n)$ is bounded in $E$. Then, because $T$ is compact it must follow that $(Tu_n)$ must contain some strong convergent subsequence), but didn't manage to finalize the proof.
Any reference or hints are welcome.
-
@all: please ignore my vote to close. I'm pretty sure this was discussed several times already, but I can't seem to find the thread at the moment. The thread I linked to is not a duplicate of this question. Beni, sorry about that. – t.b. Oct 6 '11 at 18:19
@t.b.: It's ok. I was wondering why it was voted to close. I also searched the site before posting, and didn't find it. – Beni Bogosel Oct 6 '11 at 18:21
Make use of the following topological lemma.
Lemma Let $X$ be a topological space and $\mathbf{x}=(x_n)_{n\in \mathbb{N}}$ be a sequence of elements of $X$. If every subsequence of $\mathbf{x}$ contains a subsequence convergent to $x$ then $x_n \to x$.
-
This surely is a very nice argument. Thank you. – Beni Bogosel Oct 6 '11 at 18:22
You're welcome! This little lemma is simple yet useful to know, IMHO. – Giuseppe Negro Oct 6 '11 at 21:29
This is also true by the following reason:
Since you've already proved that there is a strongly convergent subsequence, let's say $Tu_{n_k} \to u^*$ for $k \to \infty$. Then by the weak convergence of $u_n \rightharpoonup u$ you get immediately that $Tu_n \rightharpoonup Tu$. Now since strong convergence implies weak convergence and from the uniqueness of the limit of a weak convergent sequence it must be true that
$u^*= Tu$.
Therefore $Tu$ is a limit point of the sequence $(Tu_n)_{n \in \mathbb{N}}$. Now there's just one thing left to prove your statement
Claim: There's no other limit point, hence it must be the limit.
Proof: Suppose there's another limit point $z$ of the sequence $(Tu_n)_{n \in \mathbb{N}}$. Again there must be a subsequence $(Tu_{n_m})_{m \in \mathbb{N}}$ converging to $z$. Hence this subsequence $u_{n_m} \rightharpoonup u$. Last step, use the same argumentation as above to conclude that $z = Tu = u^*$.
Therefore $Tu_n \to Tu$ as $n\to \infty$.
To be precise, at this point you know that
$Tu_{n_k} \to Tu$ and that the limit $Tu$ is the only limit point of $(Tu_n)_{n\in \mathbb{N}}$. Use now a contradiction argument to prove that $Tu_n\to Tu$.
-
Why is it "immediate" that you get $Tu_n \rightharpoonup Tu$ from the weak convergence of the $u_n$? – maximumtag Mar 11 '14 at 21:56
@maximumtag what is your definition of weak convergence? – math Mar 25 '14 at 10:23 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9793776273727417, "perplexity": 108.21859696893469}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-07/segments/1454701161718.0/warc/CC-MAIN-20160205193921-00297-ip-10-236-182-209.ec2.internal.warc.gz"} |
https://www.physicsforums.com/threads/basic-question-on-fourier-coefficients.552600/ | Basic question on fourier coefficients
• Start date
• #1
56
0
If f(x) has a period of 2*pi and |f(x)-f(y)| <= c*|x-y|^a where a and c are positive constants, why are are n-th Fourier coefficients <= c*(pi/n)^a ?
Help or hints would be appreciated.
Last edited:
• #2
jasonRF
Gold Member
1,393
457
I made a stab at this, but ended up with a slightly looser bound. But perhaps it will still be of some use. Here I just do the "sine" coeffs; the argument for cos is essentially the same.
$$\left| b_n \right| \leq \frac{1}{\pi} \int_0^{2 \pi}\,dx\,|f(x)| |sin(nx)|=\frac{1}{\pi}\sum_{m=1}^{n} \int_{(m-1)2\pi/n}^{m2\pi/n}\,dx\,|f(x)|\,|sin(nx)| =\frac{1}{\pi}\sum_{m=1}^{n} \int_{(m-1)2\pi/n}^{(m-1/2)2\pi/n}\,dx\,|f(x)|\,sin(nx) - \int_{(m-1/2)2\pi/n}^{m2\pi/n}\,dx\,|f(x)|\,sin(nx).$$
Now if $|f(x_m^{-})|$ is an upper bound of $|f|$ on $[(m-1)2\pi/n, (m-1/2)2\pi/n]$, and $|f(x_m^{+})|$ is an upper bound of $|f|$ on $[(m-1/2)2\pi/n, m2\pi/n]$, then
$$\int_{(m-1)2\pi/n}^{(m-1/2)2\pi/n}\,dx\,|f(x)|\,sin(nx) \leq |f(x_m^{-})| \int_{(m-1)2\pi/n}^{(m-1/2)2\pi/n}\,dx \, sin(nx) = \frac{2}{n} |f(x_m^{-})|,$$
and similarly for the other integral. Hence I get
$$|b_n| \leq \frac{2}{n\pi} \sum_{m=1}^{n} |f(x_m^{-})|-|f(x_m^{+})| \leq \frac{2}{n\pi} \sum_{m=1}^{n} |f(x_m^{-})-f(x_m^{+})| \leq \frac{2}{n\pi} \sum_{m=1}^{n} c |x_m^{-}-x_m^{+}|^a.$$
Now, from the way I defined $x_m^{-}$ and $x_m^{-}$, it must be true that $|x_m^{-}-x_m^{+}| \leq 2\pi/n$. So I get
$$|b_n| \leq \frac{2c}{n\pi}\sum_{m=1}^{n} \left( \frac{2\pi}{n}\right)^a = \frac{2c}{\pi}\left( \frac{2\pi}{n}\right)^a \leq c \left( \frac{2\pi}{n}\right)^a.$$
So I have an extra $2^a$. Perhaps you can see a better way!
cheers,
jason
Last edited:
• Last Post
Replies
4
Views
959
• Last Post
Replies
3
Views
2K
• Last Post
Replies
2
Views
529
• Last Post
Replies
11
Views
3K
• Last Post
Replies
1
Views
1K
• Last Post
Replies
1
Views
1K
• Last Post
Replies
1
Views
2K
• Last Post
Replies
0
Views
1K
• Last Post
Replies
1
Views
1K
• Last Post
Replies
2
Views
566 | {"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": 0.9641469120979309, "perplexity": 1057.3740292209463}, "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-2021-17/segments/1618038064898.14/warc/CC-MAIN-20210411174053-20210411204053-00206.warc.gz"} |
https://byjus.com/maths/area-of-parallelogram/ | # Area of Parallelogram
The area of a parallelogram is the region bounded by the parallelogram in a given two-dimension space. To recall, a parallelogram is a special type of quadrilateral which has four sides and the pair of opposite sides are parallel. In a parallelogram, the opposite sides are of equal length and opposite angles are of equal measures. Since the rectangle and the parallelogram have similar properties, the area of the rectangle is equal to the area of a parallelogram.
## Area of Parallelogram Formula
To find the area of the parallelogram, multiply the base of the perpendicular by its height. It should be noted that the base and the height of the parallelogram are perpendicular to each other whereas the lateral side of the parallelogram is not perpendicular to the base. Thus, a dotted line is drawn to represent the height.
Therefore,
Area = b × h Square units
Where “b” is the base and “h” is the height of the parallelogram.
## How to Calculate the Area of Parallelograms?
The parallelogram area can be found by using its base and height. Apart from it, the area of a parallelogram can also be found if its two diagonals are known along with any of their intersecting angles, or if the length of the parallel sides is known along with any of the angles between the sides.
### Parallelogram Area Using Sides
Suppose a and b are the set of parallel sides of a parallelogram and h is the height, then based on the length of sides and height of it, the formula for its area is given by:
Area = Base × Height
A = b × h [sq.unit]
Example: If base of a parallelogram is equal to 5cm and the height is 3cm, the find its area.
Solution: Given, length of base=5cm and height = 3cm
As per the formula, Area = 5 × 3 = 15 sq.cm
### Parallelogram Area Without Height
If the height of the parallelogram is unknown to us, then we can use trigonometry concept here to find its area.
Area = ab sin (x)
where a and b are the length of parallel sides and x is the angle between the sides of the parallelogram.
Example: The angle between any two sides of a parallelogram is 30 degrees. The length of the two parallel sides is 3cm and 4cm respectively find the area.
Solution: Let a = 3cm and b=4cm
x = 30 degrees
Area = ab sin (x)
A = 3 × 4 sin (90)
A = 12 sin 90
A = 12 × 1 = 12 sq.cm.
Note: If the angle between the sides of a parallelogram is 90 degrees, then it is a rectangle.
### Parallelogram Area Using Diagonals
The area of any parallelogram can also be calculated using its diagonal lengths. As we know, there are two diagonals for a parallelogram, which intersects each other. Suppose, the diagonals intersects each other at an angle y, then the area of the parallelogram is given by:
Area = ½ × d1 × d2 sin (y)
Check the table below to get the summarise formulas of area of a parallelogram.
All Formulas to Calculate Area of a Parallelogram
Using Base and Height A = b × h
Using Trigonometry A = ab sin (x)
Using Diagonals A = ½ × d1 × d2 sin (y)
Where,
• b = base of the parallelogram (AB)
• h = height of the parallelogram
• a = side of the parallelogram (AD)
• x = any angle between the sides of the parallelogram (∠DAB or ∠ADC)
• d1 = diagonal of the parallelogram (p)
• d2 = diagonal of the parallelogram (q)
• y = any angle between at the intersection point of the diagonals (∠DOA or ∠DOC)
Note: In the above figure,
• DC = AB = b
• AD = BC = a
• ∠DAB = ∠DCB
• ∠ADC = ∠ABC
• O is the intersecting point of the diagonals
• ∠DOA = ∠COB
• ∠DOC = ∠AOB
## Area of Parallelogram Questions and Solutions
Question 1: Find the area of the parallelogram with the base of 4 cm and height of 5 cm?
Solution:
Given:
Base, b = 4 cm
h = 5 cm
We know that,
Area of Parallelogram = b×h Square units
= 4 × 5 = 20
Therefore, the area of a parallelogram = 20 cm2
Question 2: Find the area of a parallelogram whose breadth is 8 cm and height is 11 cm?
Solution:
Given,
b = 8 cm
h = 11 cm
Area of a parallelogram
= b × h
= 8 × 11 cm2
= 88 cm2
Question 3: The base of the parallelogram is thrice its height. If the area is 192 cm2, find the base and height.
Solution:
Let the height of the parallelogram = h cm
then, the base of the parallelogram = 3h cm
Area of the parallelogram = 192 cm2
Area of parallelogram = base × height
Therefore, 192 = 3h × h
⇒ 3 × h2 = 192
⇒ h2 = 64
⇒ h = 8 cm
Hence the height of the parallelogram is 8 cm and breadth is
3 × h
= 3 × 8
= 24 cm
## Word Problem
Q.1: The area of a parallelogram is 500 sq.cm. Its height is twice its base. Find the height and base.
Solution: Given, area = 500 sq.cm.
Height = Twice of base
h = 2b
By the formula, we know,
Area = b x h
500 = b x 2b
2b2 = 500
b2 = 250
b = 15.8 cm
Hence, height = 2 x b = 31.6 cm
Related Area Articles
## Frequently Asked Questions
### What is a Parallelogram?
A parallelogram is a geometrical figure that has four sides formed by two pairs of parallel lines. In a parallelogram, the opposite sides are equal in length and opposite angles are equal in measure.
### What is the Area of a Parallelogram?
The area of any parallelogram can be calculated using the following formula:
Area = base × height
It should be noted that the base and height of a parallelogram must be perpendicular.
### What is the Perimeter of a Parallelogram?
To find the perimeter of a parallelogram, add all the sides together. The following formula gives the perimeter of any parallelogram:
Perimeter = 2 (a + b)
### What is the Area of a Parallelogram Whose Height is 5 cm and Base is 4 cm?
The area of a perpendicular with height 5 cm and base 4 cm will be;
A = b × h
Or, A = 4 × 5 = 20 cm2 | {"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": 0.8140743970870972, "perplexity": 711.1684194649496}, "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-2020-05/segments/1579250601241.42/warc/CC-MAIN-20200121014531-20200121043531-00213.warc.gz"} |
http://terrytao.wordpress.com/tag/perelman-entropy/ | You are currently browsing the tag archive for the ‘Perelman entropy’ tag.
We continue our study of $\kappa$-solutions. In the previous lecture we primarily exploited the non-negative curvature of such solutions; in this lecture and the next, we primarily exploit the ancient nature of these solutions, together with the finer analysis of the two scale-invariant monotone quantities we possess (Perelman entropy and Perelman reduced volume) to obtain a important scaling limit of $\kappa$-solutions, the asymptotic gradient shrinking soliton of such a solution.
The main idea here is to exploit what I have called the infinite convergence principle in a previous post: that every bounded monotone sequence converges. In the context of $\kappa$-solutions, we can apply this principle to either of our monotone quantities: the Perelman entropy
$\displaystyle \mu(g(t),\tau) := \inf \{ {\mathcal W}(M,g(t),f,\tau): \int_M (4\pi\tau)^{-d/2} e^{-f}\ d\mu = 1 \}$ (1)
where $\tau := -t$ is the backwards time variable and
$\displaystyle {\mathcal W}(M,g(t),f,\tau) := \int_M (\tau(|\nabla f|^2 + R) + f - d) (4\pi\tau)^{-d/2} e^{-f}\ d\mu$, (2)
or the Perelman reduced volume
$\displaystyle \tilde V_{(0,x_0)}(-\tau) := \tau^{-d/2} \int_M e^{-l_{(0,x_0)}(-\tau,x)}\ d\mu(x)$ (3)
where $x_0 \in M$ is a fixed base point. As pointed out in Lecture 11, these quantities are related, and both are non-increasing in $\tau$.
The reduced volume starts off at $(4\pi)^{d/2}$ when $\tau=0$, and so by the infinite convergence principle it approaches some asymptotic limit $0 \leq \tilde V_{(0,x_0)}(-\infty) \leq (4\pi)^{d/2}$ as $\tau \to -\infty$. (We will later see that this limit is strictly between 0 and $(4\pi)^{d/2}$.) On the other hand, the reduced volume is invariant under the scaling
$g^{(\lambda)}(t) := \frac{1}{\lambda^2} g( \lambda^2 t )$, (4)
in the sense that
$\tilde V_{(0,x_0)}^{(\lambda)}(-\tau) = \tilde V_{(0,x_0)}(-\lambda^2 \tau)$. (5)
Thus, as we send $\lambda \to \infty$, the reduced volumes of the rescaled flows $t \mapsto (M, g^{(\lambda)}(t))$ (which are also $\kappa$-solutions) converge pointwise to a constant $\tilde V_{(0,x_0)}(-\infty)$.
Suppose that we could somehow “take a limit” of the flows $t \mapsto (M, g^{(\lambda)}(t))$ (or perhaps a subsequence of such flows) and obtain some limiting flow $t \mapsto (M^{(\infty)}, g^{(\infty)}(t))$. Formally, such a flow would then have a constant reduced volume of $\tilde V_{(0,x_0)}(-\infty)$. On the other hand, the reduced volume is monotone. If we could have a criterion as to when the reduced volume became stationary, we could thus classify all possible limiting flows $t \mapsto (M^{(\infty)}, g^{(\infty)}(t))$, and thus obtain information about the asymptotic behaviour of $\kappa$-solutions (at least along a subsequence of scales going to infinity).
We will carry out this program more formally in the next lecture, in which we define the concept of an asymptotic gradient-shrinking soliton of a $\kappa$-solution.
In this lecture, we content ourselves with a key step in this program, namely to characterise when the Perelman entropy or Perelman reduced volume becomes stationary; this requires us to revisit the theory we have built up in the last few lectures. It turns out that, roughly speaking, this only happens when the solution is a gradient shrinking soliton, thus at any given time $-\tau$ one has an equation of the form $\hbox{Ric} + \hbox{Hess}(f) = \lambda g$ for some $f: M \to {\Bbb R}$ and $\lambda > 0$. Our computations here will be somewhat formal in nature; we will make them more rigorous in the next lecture.
The material here is largely based on Morgan-Tian’s book and the first paper of Perelman. Closely related treatments also appear in the notes of Kleiner-Lott and the paper of Cao-Zhu.
It is well known that the heat equation
$\dot f = \Delta f$ (1)
on a compact Riemannian manifold (M,g) (with metric g static, i.e. independent of time), where $f: [0,T] \times M \to {\Bbb R}$ is a scalar field, can be interpreted as the gradient flow for the Dirichlet energy functional
$\displaystyle E(f) := \frac{1}{2} \int_M |\nabla f|_g^2\ d\mu$ (2)
using the inner product $\langle f_1, f_2 \rangle_\mu := \int_M f_1 f_2\ d\mu$ associated to the volume measure $d\mu$. Indeed, if we evolve f in time at some arbitrary rate $\dot f$, a simple application of integration by parts (equation (29) from Lecture 1) gives
$\displaystyle \frac{d}{dt} E(f) = - \int_M (\Delta f) \dot f\ d\mu = \langle -\Delta f, \dot f \rangle_\mu$ (3)
from which we see that (1) is indeed the gradient flow for (3) with respect to the inner product. In particular, if f solves the heat equation (1), we see that the Dirichlet energy is decreasing in time:
$\displaystyle \frac{d}{dt} E(f) = - \int_M |\Delta f|^2\ d\mu$. (4)
Thus we see that by representing the PDE (1) as a gradient flow, we automatically gain a controlled quantity of the evolution, namely the energy functional that is generating the gradient flow. This representation also strongly suggests (though does not quite prove) that solutions of (1) should eventually converge to stationary points of the Dirichlet energy (2), which by (3) are just the harmonic functions (i.e. the functions f with $\Delta f = 0$).
As one very quick application of the gradient flow interpretation, we can assert that the only periodic (or “breather”) solutions to the heat equation (1) are the harmonic functions (which, in fact, must be constant if M is compact, thanks to the maximum principle). Indeed, if a solution f was periodic, then the monotone functional E must be constant, which by (4) implies that f is harmonic as claimed.
It would therefore be desirable to represent Ricci flow as a gradient flow also, in order to gain a new controlled quantity, and also to gain some hints as to what the asymptotic behaviour of Ricci flows should be. It turns out that one cannot quite do this directly (there is an obstruction caused by gradient steady solitons, of which we shall say more later); but Perelman nevertheless observed that one can interpret Ricci flow as gradient flow if one first quotients out the diffeomorphism invariance of the flow. In fact, there are infinitely many such gradient flow interpretations available. This fact already allows one to rule out “breather” solutions to Ricci flow, and also reveals some information about how Poincaré’s inequality deforms under this flow.
The energy functionals associated to the above interpretations are subcritical (in fact, they are much like $R_{\min}$) but they are not coercive; Poincaré’s inequality holds both in collapsed and non-collapsed geometries, and so these functionals are not excluding the former. However, Perelman discovered a perturbation of these functionals associated to a deeper inequality, the log-Sobolev inequality (first introduced by Gross in Euclidean space). This inequality is sensitive to volume collapsing at a given scale. Furthermore, by optimising over the scale parameter, the controlled quantity (now known as the Perelman entropy) becomes scale-invariant and prevents collapsing at any scale – precisely what is needed to carry out the first phase of the strategy outlined in the previous lecture to establish global existence of Ricci flow with surgery.
The material here is loosely based on Perelman’s paper, Kleiner-Lott’s notes, and Müller’s book. | {"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": 40, "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": 0.9788253307342529, "perplexity": 377.75987775893657}, "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-2014-49/segments/1416405292026.28/warc/CC-MAIN-20141119135452-00106-ip-10-235-23-156.ec2.internal.warc.gz"} |
https://quantumcomputing.stackexchange.com/questions/11812/problems-in-understanding-the-solution-for-mermin-peres-magic-square-game/11818 | # Problems in understanding the solution for "Mermin–Peres Magic Square Game"
I am referring to the article Advanced Topics in Quantum Information Theory exercise 4 and to the MS Quantum Kata MS-Quantum-Kata which describes a solution for the Mermin–Peres Magic Square Game. I could follow the article and do all the maths for myself until the end of the article. Unfortunately the article ends with
... Following the hint, we consider two maximally entangled states shared between Alice and Bob:$$|Ψ〉_{AB}=1/√2(|00〉_{A_1B_1}+|11〉_{A_1B_1})⊗1/√2(|00〉_{A_2B_2}+|11〉_{A_2B_2})$$ It is then easy to check by using $$1/√2(|00〉+|11〉) =1/√2(|\phi^x_0\phi^x_0〉+|\phi^x_1\phi^x_1〉) =1/√2(|\phi^y_0\phi^y_1〉+|\phi^y_1\phi^y_0〉)$$ where $$|0,1〉$$ are the eigenvectors of $$σ_z$$ and where $$|φ^{x,y}_j〉$$ denote the eigenvectors of $$σ_x$$ and $$σ_y$$ as in the solution to the last exercise (with eigenvalues ±1 each), that the compatibility condition is indeed satisfied.
Unfortunately for me it is not easy to check, I spent multiple nights to try this, but I don't get it done.
1. Can anyone show me how this check looks like? What is the math behind it?
2. The algorithm states that Alice (and Bob) are performing multiple measurements on their qubits for all observables $$M_{x,y}$$ in her row (in his column).
I don't understand why this makes sense: After the first measurement the quantum state has collapsed (or could have at least), why does it make sense to do additional measurements? I know there is something called "phase kick-back" that kicks the phase into the control bit if I use "Controlled Operations", is that something that is used here? How exactly?
3. Is "Phase Kickback" only a mathematical trick or do physical systems really behave like this? Is there an easy to understand sample maybe, for a non-physic, non-mathematic guy (just a software dev)?
Thanks a lot for any help or hints.
• So you have problems to check this identity: $1/√2(|00〉+|11〉) =1/√2(|\phi^x_0\phi^x_0〉+|\phi^x_1\phi^x_1〉) =1/√2(|\phi^y_0\phi^y_1〉+|\phi^y_1\phi^y_0〉)$? May 2 '20 at 21:16
• @draks... No, I see the identity, but I dont know how that helps me to see that the condition is satisfied. May 2 '20 at 23:30
Not an answer yet, but to long for comment: Alice and Bob share two entangled pairs. Now look only at Alice's. The last line gives the state that only one of'em has: $$\frac1{\sqrt 2}(|00〉+|11〉) =\frac1{\sqrt 2}(|\phi^x_0\phi^x_0〉+|\phi^x_1\phi^x_1〉) =\frac1{\sqrt 2}(|\phi^y_0\phi^y_1〉+|\phi^y_1\phi^y_0〉).$$ You can see that when you trace out Bob. Now all hermitian measurement operator in one row/column commute, which means that they share eigenvectors. So after the collapse at the first measurement, subsequent measurement along communting operators don't alter the state anymore, since they are already eigenstates. I think you have to check that the measured value is the same at the coincident site... | {"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": 9, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8947329521179199, "perplexity": 471.63236217232253}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-05/segments/1642320304835.96/warc/CC-MAIN-20220125130117-20220125160117-00545.warc.gz"} |
https://physics.aps.org/synopsis-for/10.1103/PhysRevLett.119.242001 | # Synopsis: Top Quark in Nuclear Collisions
The top quark—previously seen in proton collisions—has now been identified in collisions between protons and lead nuclei.
The top quark is the heaviest quark, weighing more than most atoms. It was first observed in 1995 coming out of proton-antiproton collisions at the Tevatron accelerator outside of Chicago, and it has been studied since 2010 at the Large Hadron Collider in proton-proton collisions. Now, the CMS Collaboration at CERN in Switzerland has spotted top quarks in collisions between protons and lead nuclei. Analysis of these top quark events provides a new way to probe nuclear matter, which could reveal how gluons—the particles carrying the strong force—are distributed within a nucleus.
Previous experiments either had too little energy or too few events to observe top quarks in proton-nucleus collisions. That changed in 2016, when CERN began colliding protons and lead nuclei with center-of-mass energy of 8.16 TeV. During this run, the CMS experiment searched for collisions producing a top and antitop pair, mostly from the fusion of two gluons. These quarks don't last long, decaying in about $1{0}^{-25}$ s into a W boson and a bottom quark. The decay products themselves rapidly decay into high-energy leptons (electrons or muons) and “jets” filled with multiple light particles. For their target events, the CMS Collaboration selected those that produced four jets and a single high-energy lepton.
After subtracting background events, the team obtained 710 events that matched the criteria for top-antitop production. From this, they estimated the probability, or cross section, for proton-lead collisions to produce top quarks. The result of 45 nanobarns is consistent with scaling up the cross section from proton-proton collisions by the number of nucleons in the lead nucleus. With more data, researchers may be able to observe deviations from this simple scaling that could reveal the arrangement of gluons within the complex lead nucleus.
This research is published in Physical Review Letters.
–Michael Schirber
Michael Schirber is a Corresponding Editor for Physics based in Lyon, France.
More Features »
### Announcements
More Announcements »
## Subject Areas
Particles and Fields
Gravitation
## Next Synopsis
Biological Physics
## Related Articles
Particles and Fields
### Viewpoint: Black Hole Evolution Traced Out with Loop Quantum Gravity
Loop quantum gravity—a theory that extends general relativity by quantizing spacetime—predicts that black holes evolve into white holes. Read More »
Particles and Fields
### Viewpoint: The Plot Thickens for a Fourth Neutrino
Confirming previous controversial results, the MiniBooNE experiment detects a signal that is incompatible with neutrino oscillations involving just the three known flavors of neutrinos. Read More »
Nuclear Physics
### Viewpoint: Dissecting the Mass of the Proton
A calculation determines four distinct contributions to the proton mass, more than 90% of which arises entirely from the dynamics of quarks and gluons. Read More » | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 1, "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": 0.8985609412193298, "perplexity": 2256.536604363102}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-51/segments/1544376826306.47/warc/CC-MAIN-20181214184754-20181214210754-00522.warc.gz"} |
https://experts.illinois.edu/en/publications/global-uniform-boundary-harnack-principle-with-explicit-decay-rat | # Global uniform boundary Harnack principle with explicit decay rate and its application
Panki Kim, Renming Song, Zoran Vondraček
Research output: Contribution to journalArticlepeer-review
## Abstract
In this paper, we consider a large class of subordinate Brownian motions X via subordinators with Laplace exponents which are complete Bernstein functions satisfying some mild scaling conditions at zero and at infinity. We first discuss how such conditions govern the behavior of the subordinator and the corresponding subordinate Brownian motion for both large and small time and space. Then we establish a global uniform boundary Harnack principle in (unbounded) open sets for the subordinate Brownian motion. When the open set satisfies the interior and exterior ball conditions with radius R>0, we get a global uniform boundary Harnack principle with explicit decay rate. Our boundary Harnack principle is global in the sense that it holds for all R>0 and the comparison constant does not depend on R, and it is uniform in the sense that it holds for all balls with radii r≤R and the comparison constant depends neither on D nor on r. As an application, we give sharp two-sided estimates for the transition densities and Green functions of such subordinate Brownian motions in the half-space.
Original language English (US) 235-267 33 Stochastic Processes and their Applications 124 1 https://doi.org/10.1016/j.spa.2013.07.007 Published - 2014
## Keywords
• Boundary Harnack principle
• Green function
• Harmonic functions
• Heat kernel
• Lévy processes
• Poisson kernel
• Subordinate Brownian motions
## ASJC Scopus subject areas
• Statistics and Probability
• Modeling and Simulation
• Applied Mathematics
## Fingerprint
Dive into the research topics of 'Global uniform boundary Harnack principle with explicit decay rate and its application'. 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": 0.9428144097328186, "perplexity": 1201.5182209949683}, "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-2022-49/segments/1669446710962.65/warc/CC-MAIN-20221204040114-20221204070114-00461.warc.gz"} |
http://astro.uchicago.edu/~vikram/SN_Remnants.html | SUPERNOVA REMNANT RESEARCH
Interaction of Supernova Remnants with a Circumstellar Shell - V. V. Dwarkadas,1995, AAS 187th Meeting
ABSTRACT: We are studying the interaction of supernova remnants (SNRs) with circumstellar shells, with an emphasis on Type II supernovae (SNe). These supernovae arise from massive progenitor stars (> 8 Msun), which lose mass during their lifetime, primarily in the form of a stellar wind. Often the stellar wind creates a circumstellar bubble surrounded by a dense shell. When the star explodes as a supernova, the resulting shock wave eventually collides with this dense shell. In a recent paper on SN 1987A (Chevalier & Dwarkadas, ApJL, 452, L45) we have shown that from the radio and X-ray emission, one can infer the presence of a high density region interior to the dense circumstellar shell. This can be explained as an HII region photoionized by the flux from the pre-supernova star. Using the Zeus code and assuming spherical symmetry, we have studied the dynamics of the shock wave interacting first with the HII region and then the circumstellar shell in SN 1987A. Collision with the HII region results in a significant deceleration of the shock wave, forming a high-density shocked region that grows with time, and is primarily responsible for the X-ray emission. X-ray emission from the reflected shock may begin to dominate when the forward shock hits the dense circumstellar shell and is considerably slowed down. Simulations are in progress with parameters suited to other remnants such as Cas A and W44. Radio and X-ray images of Cas A show a shell structure, which may result from interaction with a stellar bubble. W44 also shows a double-shell structure that may have been produced by a SN explosion inside a pre-existing wind bubble. The interaction is subject to instabilities that may give rise to filamentary structure.
SUPERNOVA REMNANTS, PULSARS AND THE INTERSTELLAR MEDIUM - SUMMARY OF A WORKSHOP HELD AT U SYDNEY, MARCH 1999
PASA, Vol 17, #1
Vikram Dwarkadas , Lewis Ball , James Caswell , Anne Green , Simon Johnston , Brian Schmidt , Mark Wardle
ABSTRACT: We summarise the proceedings of the SRCfTA workshop on ``Supernova Remnants, Pulsars and the Interstellar Medium'' that was held at the University of Sydney on Mar 18 and 19, 1999.
Vikram V. Dwarkadas, 2001, JKAS, 34, 243
ABSTRACT: We summarize various aspects of the interaction of supernova remnants (SNRs) with the ambient medium. We discuss the evolution of SNRs in environments sculpted by the progenitor star, and summarize the factors on which this evolution depends. As a specific example, we consider the evolution of the medium around a 35Msun star, and the interaction of the shock wave with this medium when the star explodes as a SN. We also discuss the interaction of Type Ia SNe with the ambient medium, especially the formation and growth of hydrodynamic instabilities.
Vikram V. Dwarkadas, 2002, in Interacting Winds from Massive Stars. ASP Conference Proceedings, Vol. 260. Edited by Anthony F. J. Moffat and Nicole St-Louis, p.141
Abstract: During their lifetime, massive stars lose considerable mass in the form of stellar winds. These winds may evacuate a cavity in the surrounding medium, bordered by a dense shell. If the star ends its life in a supernova explosion, the resulting shock wave will interact with this shell. The subsequent evolution of the supernova remnant depends in particular on the ratio of the mass of the shell to the ejecta mass. Using numerical techniques this evolution is studied for a range of parameter values.
Vikram V. Dwarkadas, 2003, in procedings of 3-D Signatures in Stellar Explosions, A Workshop Honoring J. Craig Wheeler's 60th Birthday, Edited by Pawan Kumar, Craig Wheeler and Peter Hoeflich, Cambridge University Press
Abstract: We study the evolution of supernova remnants in the circumstellar medium formed by mass loss from the progenitor star. The properties of this interaction are investigated, and the specific case of a 35 \$\msun\$ star is studied in detail. The evolution of the SN shock wave in this case may have a bearing on other SNRs evolving in wind-blown bubbles, especially SN 1987A.
Abstract: Mass loss from massive stars (>~8 Msolar) can result in the formation of circumstellar wind-blown cavities surrounding the star, bordered by a thin, dense, cold shell. When the star explodes as a core-collapse supernova (SN), the resulting shock wave will interact with this modified medium around the star, rather than the interstellar medium. In this work we first explore the nature of the circumstellar medium around massive stars in various evolutionary stages. This is followed by a study of the evolution of SNe within these wind-blown bubbles. The evolution depends primarily on a single parameter ?, the ratio of the mass of the dense shell to that of the ejected material. We investigate the evolution for different values of this parameter. We also plot approximate X-ray surface brightness plots from the simulations. For very small values ?<<1 the effect of the shell is negligible, as one would expect. Values of ?<~1 affect the SN evolution, but the SN ``forgets'' about the existence of the shell in about 10 doubling times or so. The remnant density profile changes, and consequently the X-ray emission from the remnant will also change. The initial X-ray luminosity of the remnant is quite low, but interaction of the shock wave with the dense circumstellar shell can increase the luminosity by 2-3 orders of magnitude. As the reflected shock begins to move inward, X-ray images will show the presence of a double-shelled structure. Larger values result in more SN energy being expended to the shell. The resulting reflected shock moves quickly back to the origin, and the ejecta are thermalized rapidly. The evolution of the remnant is speeded up, and the entire remnant may appear bright in X-rays. If ?>>1, then a substantial amount of energy may be expended in the shell. In the extreme case the SN may go directly from the free expansion to the adiabatic stage, bypassing the Sedov stage. Our results show that in many cases the SNR spends a significant amount of time within the bubble. The low density within the bubble can delay the onset of the Sedov stage and may end up reducing the amount of time spent in the Sedov stage. The complicated density profile within the bubble makes it difficult to infer the mass-loss properties of the pre-SN star by studying the evolution of the resulting SNR.
Vikram V Dwarkadas, 2007, Astrophysics and Space Science
Abstract: Core-Collapse supernovae arise from stars greater than 8 \$\msun\$. These stars lose a considerable amount of mass during their lifetime, which accumulates around the star forming wind-blown bubbles. Upon the death of the star in a spectacular explosion, the resulting SN shock wave will interact with this modified medium. We study the evolution of the shock wave, and investigate the properties of this interaction. We concentrate on the evolution of the SN shock wave in the medium around a 35 solar mass star. We discuss the hydrodynamics of the resulting interaction, the formation and growth of instabilities, and deviations from sphericity.
Vikram V Dwarkadas, 2007, ApJ, 667, 226
Abstract: Mass-loss from massive stars leads to the formation of circumstellar wind-blown bubbles surrounding the star, bordered by a dense shell. When the star ends its life in a supernova (SN) explosion, the resulting shock wave will interact with this modified medium. In a previous paper \citep{d05} we discussed the basic parameters of this interaction with idealized models. In this paper we go a step further and study the evolution of SNe in the wind blown bubble formed by a 35 \$\msun\$ star that starts off as an O star, goes through a red supergiant phase, and ends its life as a Wolf-Rayet star. We model the evolution of the circumstellar medium throughout its lifetime, and then the expansion of the SN shock wave within this medium. Our simulations clearly reveal fluctuations in density and pressure within the surrounding medium, due to the changing mass-loss parameters over the star's evolution. The SN shock interacting with these fluctuations, and then with the dense shell surrounding the wind-blown cavity, gives rise to a variety of transmitted and reflected shocks in the wind bubble. The interactions between these various shocks and discontinuities is examined, and its effects on the emission from the remnant, especially in the X-ray regime, is noted. In this particular case the shock wave is trapped in the dense shell for a large number of doubling times, and the remnant size is restricted by the size of the surrounding circumstellar bubble. Our multi-dimensional simulations reveal the presence of several hydrodynamic instabilities. They show that the turbulent interior, coupled with the large fluctuations in density and pressure, gives rise to an extremely corrugated SN shock wave. The shock shows considerable wrinkles as it impacts the dense shell, and the impact occurs in a piecemeal fashion, with some parts of the shock wave interacting with the shell before the others. As each interaction is accompanied by an increase in the X-ray and optical emission, different parts of the shell will `light-up' at different times. The reflected shock that is formed upon shell impact will comprise of several smaller shocks with different velocities, and which are not necessarily moving radially inwards. The non-spherical nature of the interaction means that it will occur over a prolonged period of time, and the spherical symmetry of the initial shock wave is completely destroyed by the end of the simulation.
Supernova propagation in the circumstellar and interstellar medium
Memorie della Societa Astronomica Italiana, v.82, p.781 (2011)
We describe the propagation of supernova shocks within the surrounding medium, which may be due to mass-loss from the progenitor star. The structure and density profile of the ejected material and surrounding medium are considered. Shock wave interaction with clouds and wind-bubbles, and issues relevant to cosmic rays are briefly discussed.
Supernova remnant evolution in wind bubbles: A closer look at Kes 27 | {"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": 0.8700370788574219, "perplexity": 1085.8474344054582}, "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/1418802776996.17/warc/CC-MAIN-20141217075256-00066-ip-10-231-17-201.ec2.internal.warc.gz"} |
https://www.physicsforums.com/threads/higgs-decay.615062/ | # Higgs decay
1. Jun 19, 2012
### dingo_d
1. The problem statement, all variables and given/known data
I have decay of Higgs to fermion and antifermion and I need to find out the invariant, averaged amplitude.
And I wrote down the Feynman diagram, and calculated everything and I came to this part:
$\langle|M|^2\rangle=\frac{g_w^2}{4}\frac{m_f^2}{m_w^2}(4p_1\cdot p_2-4m_f^2)$
Now to calculate $p_1\cdot p_2$ I sit in Higgs rest frame, so that the impulses of the fermions are the same (magnitude), but have different sign: $\vec{p}_1=-\vec{p}_2$.
So what I need to end up with is:
$\langle|M|^2\rangle=\frac{g_w^2}{4}\frac{m_f^2}{m_w^2}(2m_h^2-8m_f^2)$
But I just cannot get the right result!
$p_1\cdot p_2=(E_1+E_2)^2-(\vec{p}_1+\vec{p}_2)^2=(E_1+E_2)^2$
If the magnitude of the impulses of the two fermions are the same, and if the masses are the same the energies should also be the same, right? That is, $E_1=E_2=E$, then the scalar product of two 4-vectors is:
$p_1\cdot p_2=4E^2$, where E is the energy of the fermion. Since the energy is conserved: $E_H=E_1+E_2\Rightarrow E=\frac{E_H}{2}$, but since the Higgs is at rest it's mass is equal to it's energy so $E=\frac{m_H}{2}$, but if I put that back into the scalar product I have:
$p_1\cdot p_2=m_H$, but that doesn't give me the right answer :\
Where is my mistake?!
2. Jun 19, 2012
### dingo_d
Ok I think I solved it, I was messing up the kinematics. I feel like an idiot -.-"
Similar Discussions: Higgs decay | {"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": 0.9752264618873596, "perplexity": 338.3531133774726}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "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-2017-43/segments/1508187826049.46/warc/CC-MAIN-20171023130351-20171023150351-00402.warc.gz"} |
https://www.physicsforums.com/threads/force-acting-on-struts.342280/ | # Force acting on struts
1. Oct 2, 2009
### spotty
the problem is actually for a MATLAB programming exercise. the math is really simple but I want to know if my alternate solution works compared to the "hints" the prof gave us. (sorry if we aren't supposed to put images directly into threads but I'm in a hurry and I didn't see anything relevant in the rules)
1. The problem statement, all variables and given/known data
"As shown in the following figure, we know the angle between the two struts, denoted by ∂, in the range 0-90 degrees. Determine the angle ø so that a 500-lb horizontal force has a component of 600-lb directed from A towards C. Also, you need to determine how much the component of force acting along member BA is."
2. The attempt at a solution
I figure that the horizontal force must form a right triangle with the force along AC, so the angle between them (phi) must be cos (x) = 500 / 600 = 33.56 degrees.
the solution given to us (below) was to use the law of sines, which makes sense to me, but what is wrong about my solution?
the difference seems to be that when using the right-angle method, a steeper angle = greater force. The closer the angle is to being horizontal, the closer the force is to 500- makes sense to me. My real question is, what effect would a 500lb-horizontal force have on the strut if it hit it at 90 degrees? 0lb (0/500), or 500(500/1)?
File size:
8 KB
Views:
8
2. Oct 2, 2009
### Delphi51
The key to the problem is to realize that the three forces acting on point A must add up to zero, because that point is not accelerating. In the second solution, these forces are arranged in head-to-tail fashion beginning and ending at A so the total is zero. It is unfortunate that the arrowhead on the A to C vector is backwards so it is hard to see this.
In your first solution, I do not see anything to indicate that the 3 forces are adding to zero. The AB vector is not even shown. It does not seem to have anything to do with the problem at all.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook
Similar Discussions: Force acting on struts | {"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": 0.8834048509597778, "perplexity": 460.7963604649449}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-43/segments/1508187825264.94/warc/CC-MAIN-20171022132026-20171022152026-00890.warc.gz"} |
http://arxiv.org/abs/astro-ph/0410165 | astro-ph
(what is this?)
# Title: The Spatial Clustering of Star-forming Galaxies at Redshifts 1.4<z<3.5
Abstract: We analyzed the spatial distribution of 28500 photometrically selected galaxies with AB magnitude 23.5<R<25.5 and redshift 1.4<z<3.5 in 21 fields with a total area of 0.81 square degrees. The galaxies were divided into three subsamples, with mean redshifts z=1.7, 2.2, 2.9, according to the UGR selection criteria of Adelberger et al. (2004) and Steidel et al. (2003). Combining the galaxies' measured angular clustering with redshift distributions inferred from 1600 spectroscopic redshifts, we find comoving correlation lengths at the three redshifts of r_0 = 4.5+-0.6, 4.2+-0.5, and 4.0+-0.6 h^-1 Mpc, respectively, and infer a roughly constant correlation function slope of gamma=1.6+-0.1. We derive similar numbers from the 1600 object spectroscopic sample itself with a new statistic, K, that is insensitive to many possible systematics. Galaxies that are bright in R (lambda_rest ~ 1500-2500A) cluster more strongly than fainter galaxies at z=2.9 and z=2.2 but not, apparently, at z=1.7. Comparison to a numerical simulation that is consistent with recent WMAP observations suggests that galaxies in our samples are associated with dark matter halos of mass ~ 10^12, and that a small fraction of the halos contain more than one galaxy that satisfies our selection criteria. Adding recent observations of galaxy clustering at z~0 and z~1 to the simulation results, we conclude that the typical object in our samples will evolve into an elliptical galaxy by redshift z=0 and will already have an early-type spectrum by redshift z=1. We comment briefly on the implied relationship between galaxies in our survey and those selected with other techniques. (slightly abridged)
Comments: 17 pages, accepted for publication in the ApJ Subjects: Astrophysics (astro-ph) Journal reference: Astrophys.J. 619 (2005) 697-713 DOI: 10.1086/426580 Cite as: arXiv:astro-ph/0410165 (or arXiv:astro-ph/0410165v1 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": 0.8976762890815735, "perplexity": 3413.5123259829984}, "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-2016-40/segments/1474738659680.65/warc/CC-MAIN-20160924173739-00075-ip-10-143-35-109.ec2.internal.warc.gz"} |
http://www.physicsforums.com/showthread.php?t=754795 | # What would gravity be in these 2 conditions
Tags: conditions, gravity
P: 6 I was wondering about earth and it's gravity and I came up with these 2 questions... 1.) What would earth's gravity be if the earth stopped spining? 2.) What would earth's gravity be if the earth stopped spining and there was no other stars or planets that interfered with there gravity? As I understand the average standard for gravity is 9.8m/s^square
Math Emeritus Sci Advisor Thanks PF Gold P: 39,310 In Newton's theory, the force of gravity is $-GmM/r^2$. That has nothing to do with the rotation of the earth. And other stars and bodies do NOT "interfere" with gravity.
P: 684 I'm pretty sure HadanIdea means acceleration experienced by objects on Earth's surface. If you took away those two factors, you wouldn't notice any difference. Certainly not if you're only measuring it with accuracy of two significant figures. It'd still be 9.8 m/s^2.
PF Gold
P: 1,459
What would gravity be in these 2 conditions
Quote by Bandersnatch I'm pretty sure HadanIdea means acceleration experienced by objects on Earth's surface.
If we were to account for all the forces in the universe,it would be impossible to even think of building structures like Skyscrapers.
P: 6 Would it still be 9.8 m/s^2? I thought that it would be more, don't we need to take into account the rotating force outwards as the earth spins?
Mentor P: 15,061 Centrifugal force due to the Earth's rotation is a tiny effect, about 0.034 m/s2 at the equator (and yes, centrifugal force is a part of g). There is another indirect effect of the Earth's rotation, the equatorial bulge. The Earth would change shape if the Earth stopped spinning, eventually stabilizing to a spherical shape. This reshaping would also result in a slight change in the local gravitational acceleration. Right now, g at sea level at the North Pole is about 9.832 m/s2 while its only 9.780 m/s2 at sea level at the equator. If the Earth stopped spinning, gravitation acceleration at sea level would be about 9.807 m/s2 everywhere after the Earth has reshaped itself to that spherical shape.
P: 558 Does the value of g typically include the centrifugal force?
Mentor
P: 15,061
Quote by paisiello2 Does the value of g typically include the centrifugal force?
Yes. The quantity g is defined from the perspective of an Earth-fixed frame, a rotating with the Earth. This means g is a combination of the acceleration due to gravitation and the acceleration due to the fictitious centrifugal force. Also see this post.
P: 140 We should also add that the amount of mass underneath you varies from place to place, even at sea level. Some places have dense rock underneath them, while on the ocean it might be miles of water underneath. Plumes of higher density "lava", mountain ranges, all affect local gravity. Wikipedia has an interesting map of local variations in gravity (at pretty coarse resolution).https://en.wikipedia.org/wiki/Earth_gravity It is easy for even a grade school child to calculate the force of attraction of any planet (or moon) on us using the simple formula F = G × m × M ÷ d² which can be compared to g, the attraction at sea level by dividing by m giving you g' = G × M ÷ d² where M is the mass of the planet or Moon, d is the distance and G is the Universal Gravitational Constant. The only tricky part is making sure the units of measure are all consistent. We can compare g' with g by taking their ratio: (g' ÷ g)×100% The nearest astronomical object to us is the Moon. The closest it gets is 350,000,000 meters, its mass is 7.35E22 kg and G is 6.67E-11 (m²/kg)(m/s²) [where E stands for "× 10^ " like 1.23E4 = 1.23× 10^4 =12,300 or like 5.67E-1 = 0.567] that means g' = 4.00E-5 m/s² or 0.00004. Compare that with g and you get 0.00004÷9.8 is about 0.000004x100% or 0.0004%. The Sun's g on us is about a third of that, and the next significant acceleration is from Venus, but I forget how much that is, its really, really tiny.
P: 5 The gravity would be more because centrifugal force that pulls us of the planet, but the gravity keeps us on the plant
PF Gold
P: 6,082
Quote by nitro21345 The gravity would be more because centrifugal force that pulls us of the planet, but the gravity keeps us on the plant
Which part of post #6 did you not understand? As stated in that post, yes, it would be more but only by a negligible amount.
P: 5
Quote by phinds Which part of post #6 did you not understand? As stated in that post, yes, it would be more but only by a negligible amount.
what part did I not understand
P: 5 What about centrifugal force though because the roller coaster ride has a lot of centrifugal force the PULLS you against the sides or when you stop newtons second law of motion is at play. Gravity pushes on you not pull so if the planet stopped spinning the gravity would be more because to stay on the planet you have to be pushed on harder than pushed with no spinning, therefore if the world stopped gravity would be more.
PF Gold
P: 6,082
Quote by nitro21345 What about centrifugal force though because the roller coaster ride has a lot of centrifugal force the PULLS you against the sides or when you stop newtons second law of motion is at play. Gravity pushes on you not pull so if the planet stopped spinning the gravity would be more because to stay on the planet you have to be pushed on harder than pushed with no spinning, therefore if the world stopped gravity would be more.
As I said before, yes, it would be more, but by a negligible amount.
P: 5 No you need more force to keep you on the planet than if it was not spinning | {"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": 0.8003947138786316, "perplexity": 811.6110164123662}, "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-2014-23/segments/1405997891176.68/warc/CC-MAIN-20140722025811-00069-ip-10-33-131-23.ec2.internal.warc.gz"} |
http://sciforums.com/threads/the-mm-experiment-is-wrong.165432/#post-3697438 | # The MM experiment is wrong!
Discussion in 'Physics & Math' started by tomtushey, May 3, 2022.
1. ### tomtusheyRegistered Member
Messages:
79
The MM experiments - i.e. the Michaelson-Morley experiment - are flawed not only in their interpretation but also in their design. It is very important to set the record straight because this is the origin of the profoundly mistaken physical law that light travels at the speed of light relative to any moving object. I apologize if I do not write here with the usual brevity, but in some cases, I will expand on the lack of knowledge, the many related delusions, and philosophical fantasies that actually eclipse real physics. (Tom)
Preliminary
Relativity is a strongly abstract but universally accepted theory. He has dominated physics for 100 years, since 1921, when Einsten received in his hands the Nobel Prize. The theory conflicts perfectly with quantum theory, although some still hope that the two can be reconciled. Certainly not, and quantum theory is certainly true. In what follows, I will show that relativity is flawed and contains logical internal contradictions, regardless of the contradiction I have just mentioned. I will also base this proof on my reactivated aether theory, see later. By the way, Einstein agreed with the aether theory, saying that nature cannot exist without a connecting substance, the aether. I may be able to obtain the relevant audio material (BBC 1923) and publish it (in English, with a thick German accent) for those who believe in authority, for whom the question is settled. There are legitimate questions about the physical properties and natural behaviour of the aether. On the first subject I have written in detail elsewhere, now a dozen papers. I will publish some more initial thoughts on its behaviour below.
If the aether exists, then there is much question about its connection with the big bang, the accelerating expansion of the universe, and the relationship of moving, orbiting planets. It is aether related to material bodies, as it expands more rapidly when saturated with energy and somewhat entrains celestial bodies in the form of gravitational pressure (the aether energy gradients). Forming a relatively attractive relationship between solar systems and planets.
The central celestial body attracts the orbiting planets, so they used to say. In fact, it does not attract, but only repels less. In light of this, Newtonian mechanics tells us that the planets must orbit the Sun. The inner planets orbit faster than the outer planets, and in proportion to their own speed, they drag the aether around them with them. Curiously, the matter in the aether sea around the Sun is subject to the same rule as the planets, i.e. they orbit at a speed inversely proportional to the square root of the distance, as if in synchrony with the planets. The aether sea follows the movement of a planet everywhere. This idea is Dr. K.G.'s + theory, but I had to add a lot of other things before a sound, relativity-free cosmology coalesced. Near celestial bodies, the aether sticks tightly to the surface, moving away from it a transient velocity state occurs. So, standing on rotating celestial bodies and moving away from the surface, the aether wind blows.
I apologize for putting all this upfront, but this is a preliminary clarification of the basic current misconceptions that dominate the so-called "today's, modern" physics. There always have been, are, and always will be misconceptions the physics. But it still feels dad me sick to think that in 500 years' time future physicists will be laughing or even crying at 21st-century physics.
Sincerely Tom Tushey
Mech. Engineer
Hobby Physicist
Hobby Astronomer
Science Writer
Relativity Expert
www.aether-tom.com (Eng.)
www.reactivated-aether.hupont.hu (Eng.)
www.aparadox.hupont.hu(Hu, vut it is the best.)
3. ### SarkusHippomonstrosesquippedalo phobeValued Senior Member
Messages:
9,817
Alas, I'm not sure many people here speak Hungarian. Do you have an English version, please?
5. ### originHeading towards oblivionValued Senior Member
Messages:
11,637
How about the thousands of other experiments that confirm there results? Are all those other experiments also wrong?
Einstein did not get a Nobel Prize for Relativity.
Experiments specifically designed to detect aether dragging have all failed to show the hypothesized dragging.
7. ### tomtusheyRegistered Member
Messages:
79
I fixed it because for some reason it didn't put in English. Is it still in Hungarian?
Messages:
79
9. ### Motor DaddyValued Senior Member
Messages:
5,425
SR is wrong because the 2nd postulate is wrong.
10. ### SarkusHippomonstrosesquippedalo phobeValued Senior Member
Messages:
9,817
No - it's in English now, thanks.
Bizarrely, what you have me quoted as saying I am seeing in Hungarian!
11. ### originHeading towards oblivionValued Senior Member
Messages:
11,637
You cannot prove a theory. That's is basic science.
DaveC426913 likes this.
12. ### tomtusheyRegistered Member
Messages:
79
Little Einstein heard about the speed of light and its extraordinary value as a child. He imagined running fast after a wave of light and seeing frozen waves when he caught up with it. And that's impossible, he told himself, so reaching the speed of light is impossible. Too bad he lost his courage at this point and didn't try to run even faster in his mind. Then he would have seen with his mind's eye that the waves were lagging behind him. This would have shown him that light is not a border speed.It's even more unfortunate that there was a physicist who took the 12-year-old's dream seriously. It was Einstein himself, aged 26. [Tt]
"Postulate 2: The speed of light in a vacuum, usually denoted by c, is constant, whatever the inertial frame and whatever the direction, regardless of the frequency of the light, the speed of the detector, or the speed of the light source. [Wikipedia EN]"
The definition is correct in that the measured speed of light is independent of the frequency of the light and the speed of motion of the light source. (It is independent the light source is stationary, moving away, or approaching.) However, it is not independent of the relative speed of motion of the light source and the aether. For large celestial bodies (Earth), the aether adheres to the surface, while the speed of light c is determined by the aether. The aether must always be considered stationary! So with a detector placed on the ground, we cannot expect a non-zero signal, which would be a logical absurdity. Unfortunately, all the MM experiments so far have been done in this wrong setup, with the detector stationary and the aether stationary. In fact, the error could have been avoided if the detector had been moved. But they tried to tell themselves that, although the detector is stationary, it is moving. They could not have known that the aether also was stationary (its speed relative to the detector is zero). So "the aether winds were not blowing" at the site of the experiment. [Tt]
Ether down adhesion has become a fact ever since atomic clocks were invented and set in motion. (See more later!) But the impact of this on the MM experiment would be too delicate to admit. [Tt]
13. ### tomtusheyRegistered Member
Messages:
79
But the fallacy of the theory can be proven. Take Einstein's first postulate, for example. When Galileo, onboard the ship, made one forward jump and then a second backward jump, he assumed that the two jumps were equal. He then stated that motion with uniform velocity cannot be detected. However, consider that when jumping, the speed of the boat and the jumper are added together. His speed increases, and so his mass increases, according to the Lorentz factor: L=(1+(vs+vG)2/c2)-0.5. This allows the forward jump to be distinguished. [Tt]
14. ### originHeading towards oblivionValued Senior Member
Messages:
11,637
Correct.
No, that is not correct. No matter how fast my velocity is, I will never detect an increase in my mass. This is basic relativity.
15. ### Janus58Valued Senior Member
Messages:
2,312
You are tying to assume absolute motion in order to refute relative motion, which is a logical fallacy.
What you have really "proven" here is that you don't actually grasp the theory of Relativity.
For the moment, we will put aside the fact that most present day physicists would take issue with the "mass increases" statement, due how "mass" is defined in modern physics.
That being said, all such measurements are "frame dependent". They depend on which inertial frame the measurement is made from, and are not absolute. So, from the inertial frame of the boat*, any "mass" gain by the jumper only depends on his speed relative to the boat, and not the direction he jumps.
This will not be the case as measured from a different inertial frame, like that of the ocean itself. In this frame, the boat is moving in one direction, and if the person jumps in that direction, his "mass" would increase, but if he jumps in the other it would decrease. In addition, because of the way velocities add in Relativity, his speed with respect to the boat when jumping one way would be less than when jumping the other (even if they were equal in the boat frame)
This "frame dependency" extents to measurements of time rate, length along the axis of relative motion, and whether or not given events are simultaneous or not.
The upshot is that you cannot disprove Relativity by assuming concepts ( such as absolute motion) that are not assumed in the theory.
* when I say "the inertial frame of the boat", this does not mean that there is an inertial frame physically assigned to the boat, but is shorthand for "the inertial frame that the boat is considered to be at rest with respect to"
16. ### SsssssssRegistered Senior Member
Messages:
302
Oh hey an ether dragging hypothesis. They were popular about 130 years ago but then people figured out that light coming from space would have to be affected by passing through the boundary between the dragged ether and the undragged interstellar ether and started looking at stellar parallax and found no such effect so discarded the idea because it had been falsified by experiment.
17. ### DaveC426913Valued Senior Member
Messages:
17,302
Seriously?
You haven't even bothered to read a primer on relativity - as witnessed by the basic misunderstandings you have about it - and you call yourself and expert?
To think outside the box, you must first understand what's in the box.
18. ### billvonValued Senior Member
Messages:
20,787
Not to an observer on the boat.
19. ### James RJust this guy, you know?Staff Member
Messages:
37,550
Controversial claims should be supported by appropriate argument or evidence.
Please avoid making blanket claims without trying to support them.
If you think that you can show that SR is wrong, start a new thread posting your proof. If you're not going to do that, don't make empty claims.
ajanta likes this.
20. ### tomtusheyRegistered Member
Messages:
79
I think it matters how fast the speed is. Consider a fog chamber into which an electron is shot at speed c'. A perpendicular and homogeneous magnetic field will set this electron in a circular orbit of a given radius. It is true that this is an inward spiraling orbit because it is swirling with the fog and also radiating electromagnetically. Take a larger electron with a velocity of c'', which has a larger mass and starts its orbit in a circle of a larger radius and moves inwards in a similar spiral shape. During these experiments, it does not matter how fast the solar system is traveling through space and in what direction it is moving relative to the experimental device.
Around our experimental instruments, the nearby aether is at zero absolute velocity. Relative velocity was introduced into physics by Einstein, but it is not useful for calculations, at most it is a hindsight aid in everyday life. I can show you an example where a railway worker (v0) is standing on the embankment of a railway track (v1), a train is passing by, a passenger is walking (v2) on the train and a beetle is crawling (v3) on his hat. If I remember correctly, this implies a relative speed of 6 different types, maybe you should do the resulting. I'll do it later and an English translation, but you can check the original here. Relative speed always causes chaos, and I'm interested to see counterexamples of it, I like to untangle such things.
21. ### tomtusheyRegistered Member
Messages:
79
Thank you for the warning. I will try to justify everything, although I have tried to do so in the past. Which is the one, do you think my justification was incomplete? Only the introduction is on the opening page, where I will add further explanatory details shortly.
22. ### tomtusheyRegistered Member
Messages:
79
The aether clings to all nearby celestial bodies. As bodies move, the ether sea moves flows and swirls without sharp transitions. Similar to a river that sticks to two banks while avoiding obstacles, boats, and not exactly following the curves. Einstein's theory of relativity was not taken seriously by any scientist in 1905. The turning point was 1921.
Messages:
19,227
False. | {"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": 0.8520858883857727, "perplexity": 1007.4544595785088}, "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/1669446710916.40/warc/CC-MAIN-20221202183117-20221202213117-00783.warc.gz"} |
http://en.wikipedia.org/wiki/Steiner_surface | # Roman surface
(Redirected from Steiner surface)
An animation of the Roman Surface
The Roman surface or Steiner surface (so called because Jakob Steiner was in Rome when he thought of it) is a self-intersecting mapping of the real projective plane into three-dimensional space, with an unusually high degree of symmetry. This mapping is not an immersion of the projective plane; however the figure resulting from removing six singular points is one.
The simplest construction is as the image of a sphere centered at the origin under the map f(x,y,z) = (yz,xz,xy). This gives an implicit formula of
$x^2 y^2 + y^2 z^2 + z^2 x^2 - r^2 x y z = 0. \,$
Also, taking a parametrization of the sphere in terms of longitude (θ) and latitude (φ), gives parametric equations for the Roman surface as follows:
x = r2 cos θ cos φ sin φ
y = r2 sin θ cos φ sin φ
z = r2 cos θ sin θ cos2 φ.
The origin is a triple point, and each of the xy-, yz-, and xz-planes are tangential to the surface there. The other places of self-intersection are double points, defining segments along each coordinate axis which terminate in six pinch points. The entire surface has tetrahedral symmetry. It is a particular type (called type 1) of Steiner surface, that is, a 3-dimensional linear projection of the Veronese surface.
## Derivation of implicit formula
For simplicity we consider only the case r = 1. Given the sphere defined by the points (x, y, z) such that
$x^2 + y^2 + z^2 = 1,\,$
we apply to these points the transformation T defined by
$T(x, y, z) = (y z, z x, x y) = (U,V,W),\,$
say.
But then we have
\begin{align} U^2 V^2 + V^2 W^2 + W^2 U^2 & = z^2 x^2 y^4 + x^2 y^2 z^4 + y^2 z^2 x^4 = (x^2 + y^2 + z^2)(x^2 y^2 z^2) \\[8pt] & = (1)(x^2 y^2 z^2) = (xy) (yz) (zx) = U V W, \end{align}
and so
$U^2 V^2 + V^2 W^2 + W^2 U^2 - U V W = 0\,$
as desired.
Conversely, suppose we are given (U, V, W) satisfying
(*) $U^2 V^2 + V^2 W^2 + W^2 U^2 - U V W = 0.\,$
We prove that there exists (x,y,z) such that
(**) $x^2 + y^2 + z^2 = 1,\,$
for which
$U = x y, V = y z, W = z x,\,$
with one exception: In case 3.b. below, we show this cannot be proved.
1. In the case where none of U, V, W is 0, we can set
$x = \sqrt{\frac{WU}{V}},\ y = \sqrt{\frac{UV}{W}},\ z = \sqrt{\frac{VW}{U}}.\,$
(Note that (*) guarantees that either all three of U, V, W are positive, or else exactly two are negative. So these square roots are of positive numbers.)
It is easy to use (*) to confirm that (**) holds for x, y, z defined this way.
2. Suppose that W is 0. From (*) this implies
$U^2 V^2 = 0\,$
and hence at least one of U, V must be 0 also. This shows that is it impossible for exactly one of U, V, W to be 0.
3. Suppose that exactly two of U, V, W are 0. Without loss of generality we assume
(***)$U \neq 0, V = W = 0.\,$
It follows that
$z = 0,\,$
(since
$z \neq 0,\,$
implies that
$x = y = 0,\,$
and hence
$U = 0,\,$
a. In the subcase where
$|U| \leq \frac{1}{2},$
if we determine x and y by
$x^2 = \frac{1 + \sqrt{1 - 4 U^2}}{2}$
and
$y^2 = \frac{1 - \sqrt{1 - 4 U^2}}{2},$
this ensures that (*) holds. It is easy to verify that
$x^2 y^2 = U^2,\,$
and hence choosing the signs of x and y appropriately will guarantee
$x y = U.\,$
Since also
$y z = 0 = V\text{ and }z x = 0 = W,\,$
this shows that this subcase leads to the desired converse.
b. In this remaining subcase of the case 3., we have
$|U| > \frac{1}{2}.$
Since
$x^2 + y^2 = 1,\,$
it is easy to check that
$xy \leq \frac{1}{2},$
and thus in this case, where
$|U| >1/2,\ V = W = 0,$
there is no (x, y, z) satisfying
$U = xy,\ V = yz,\ W =zx.$
Hence the solutions (U, 0, 0) of the equation (*) with
$|U| > \frac12$
and likewise, (0, V, 0) with
$|V| > \frac12$
and (0, 0, W) with
$|W| > \frac12$
(each of which is a noncompact portion of a coordinate axis, in two pieces) do not correspond to any point on the Roman surface.
4. If (U, V, W) is the point (0, 0, 0), then if any two of x, y, z are zero and the third one has absolute value 1, clearly
$(xy, yz, zx) = (0, 0, 0) = (U, V, W)\,$
as desired.
This covers all possible cases.
## Derivation of parametric equations
Let a sphere have radius r, longitude φ, and latitude θ. Then its parametric equations are
$x = r \, \cos \theta \, \cos \phi,$
$y = r \, \cos \theta \, \sin \phi,$
$z = r \, \sin \theta.$
Then, applying transformation T to all the points on this sphere yields
$x' = y z = r^2 \, \cos \theta \, \sin \theta \, \sin \phi,$
$y' = z x = r^2 \, \cos \theta \, \sin \theta \, \cos \phi,$
$z' = x y = r^2 \, \cos^2 \theta \, \cos \phi \, \sin \phi,$
which are the points on the Roman surface. Let φ range from 0 to 2π, and let θ range from 0 to π/2.
## Relation to the real projective plane
The sphere, before being transformed, is not homeomorphic to the real projective plane, RP2. But the sphere centered at the origin has this property, that if point (x,y,z) belongs to the sphere, then so does the antipodal point (-x,-y,-z) and these two points are different: they lie on opposite sides of the center of the sphere.
The transformation T converts both of these antipodal points into the same point,
$T : (x, y, z) \rightarrow (y z, z x, x y),$
$T : (-x, -y, -z) \rightarrow ((-y) (-z), (-z) (-x), (-x) (-y)) = (y z, z x, x y).$
Since this is true of all points of S2, then it is clear that the Roman surface is a continuous image of a "sphere modulo antipodes". Because some distinct pairs of antipodes are all taken to identical points in the Roman surface, it is not homeomorphic to RP2, but is instead a quotient of the real projective plane RP2 = S2 / (x~-x). Furthermore, the map T (above) from S2 to this quotient has the special property that it is locally injective away from six pairs of antipodal points. Or from RP2 the resulting map making this an immersion of RP2 — minus six points — into 3-space.
(It was previously stated that the Roman surface is a homeomorphic to RP2, but this was in error. It was subsequently stated that the Roman surface is an immersion of RP2 into R3, but that too was in error.)
## Structure of the Roman surface
The Roman surface has four bulbous "lobes", each one on a different corner of a tetrahedron.
A Roman surface can be constructed by splicing together three hyperbolic paraboloids and then smoothing out the edges as necessary so that it will fit a desired shape (e.g. parametrization).
Let there be these three hyperbolic paraboloids:
• x = yz,
• y = zx,
• z = xy.
These three hyperbolic paraboloids intersect externally along the six edges of a tetrahedron and internally along the three axes. The internal intersections are loci of double points. The three loci of double points: x = 0, y = 0, and z = 0, intersect at a triple point at the origin.
For example, given x = yz and y = zx, the second paraboloid is equivalent to x = y/z. Then
$y z = {y \over z}$
and either y = 0 or z2 = 1 so that z = ±1. Their two external intersections are
• x = y, z = 1;
• x = −y, z = −1.
Likewise, the other external intersections are
• x = z, y = 1;
• x = −z, y = −1;
• y = z, x = 1;
• y = −z, x = −1.
Let us see the pieces being put together. Join the paraboloids y = xz and x = yz. The result is shown in Figure 1.
Figure 1.
The paraboloid y = x z is shown in blue and orange. The paraboloid x = y z is shown in cyan and purple. In the image the paraboloids are seen to intersect along the z = 0 axis. If the paraboloids are extended, they should also be seen to intersect along the lines
• z = 1, y = x;
• z = −1, y = −x.
The two paraboloids together look like a pair of orchids joined back-to-back.
Now run the third hyperbolic paraboloid, z = xy, through them. The result is shown in Figure 2.
Figure 2.
On the west-southwest and east-northeast directions in Figure 2 there are a pair of openings. These openings are lobes and need to be closed up. When the openings are closed up, the result is the Roman surface shown in Figure 3.
Figure 3. Roman surface.
A pair of lobes can be seen in the West and East directions of Figure 3. Another pair of lobes are hidden underneath the third (z = xy) paraboloid and lie in the North and South directions.
If the three intersecting hyperbolic paraboloids are drawn far enough that they intersect along the edges of a tetrahedron, then the result is as shown in Figure 4.
Figure 4.
One of the lobes is seen frontally—head on—in Figure 4. The lobe can be seen to be one of the four corners of the tetrahedron.
If the continuous surface in Figure 4 has its sharp edges rounded out—smoothed out—then the result is the Roman surface in Figure 5.
Figure 5. Roman surface.
One of the lobes of the Roman surface is seen frontally in Figure 5, and its bulbous – balloon-like—shape is evident.
If the surface in Figure 5 is turned around 180 degrees and then turned upside down, the result is as shown in Figure 6.
Figure 6. Roman surface.
Figure 6 shows three lobes seen sideways. Between each pair of lobes there is a locus of double points corresponding to a coordinate axis. The three loci intersect at a triple point at the origin. The fourth lobe is hidden and points in the direction directly opposite from the viewer. The Roman surface shown at the top of this article also has three lobes in sideways view.
## One-sidedness
The Roman surface is non-orientable, i.e. one-sided. This is not quite obvious. To see this, look again at Figure 3.
Imagine an ant on top of the "third" hyperbolic paraboloid, z = x y. Let this ant move North. As it moves, it will pass through the other two paraboloids, like a ghost passing through a wall. These other paraboloids only seem like obstacles due to the self-intersecting nature of the immersion. Let the ant ignore all double and triple points and pass right through them. So the ant moves to the North and falls off the edge of the world, so to speak. It now finds itself on the northern lobe, hidden underneath the third paraboloid of Figure 3. The ant is standing upside-down, on the "outside" of the Roman surface.
Let the ant move towards the Southwest. It will climb a slope (upside-down) until it finds itself "inside" the Western lobe. Now let the ant move in a Southeastern direction along the inside of the Western lobe towards the z = 0 axis, always above the x-y plane. As soon as it passes through the z = 0 axis the ant will be on the "outside" of the Eastern lobe, standing rightside-up.
Then let it move Northwards, over "the hill", then towards the Northwest so that it starts sliding down towards the x = 0 axis. As soon as the ant crosses this axis it will find itself "inside" the Northern lobe, standing right side up. Now let the ant walk towards the North. It will climb up the wall, then along the "roof" of the Northern lobe. The ant is back on the third hyperbolic paraboloid, but this time under it and standing upside-down. (Compare with Klein bottle.)
## Double, triple, and pinching points
The Roman surface has four "lobes". The boundaries of each lobe are a set of three lines of double points. Between each pair of lobes there is a line of double points. The surface has a total of three lines of double points, which lie (in the parametrization given earlier) on the coordinate axes. The three lines of double points intersect at a triple point which lies on the origin. The triple point cuts the lines of double points into a pair of half-lines, and each half-line lies between a pair of lobes. One might expect from the preceding statements that there could be up to eight lobes, one in each octant of space which has been divided by the coordinate planes. But the lobes occupy alternating octants: four octants are empty and four are occupied by lobes.
If the Roman surface were to be inscribed inside the tetrahedron with least possible volume, one would find that each edge of the tetrahedron is tangent to the Roman surface at a point, and that each of these six points happens to be a Whitney singularity. These singularities, or pinching points, all lie at the edges of the three lines of double points, and they are defined by this property: that there is no plane tangent to surface at the singularity. | {"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": 39, "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": 0.8710902333259583, "perplexity": 699.3729039412316}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2013-48/segments/1386164737564/warc/CC-MAIN-20131204134537-00015-ip-10-33-133-15.ec2.internal.warc.gz"} |
https://epsilon-delta-limit.com/en/chap2e/ | # 2 Epsilon-Delta Definition
## Limits of Functions
We can easily understand the limits of functions if we clearly understand the limits of sequences. The intuitive picture of the limit of a sequence is that
the value of $a_n$ infinitely approaches a constant value $\alpha$ as the index $n$ increases.
As was explained in the previous chapter, this picture is precisely characterized by the following definition:
For any $\varepsilon > 0$, there exists a natural number $N(\varepsilon)$ such that for all $n \geq N(\varepsilon)$, we have $| \, a_n \, - \, \alpha \, | < \varepsilon$.
On the other hand, the intuitive picture of the limit of a function is that
the value of $f(x)$ infinitely approaches a constant value $\alpha$ as the value of $x$ with $x \neq a$ approaches $a$.
If we regard the variable $x$ as corresponding to the index $n$, we find that this picture of the limit of a function is essentially the same as that of a sequence.
The difference between these pictures is in the phrase “as the value of $x$ with $x \neq a$ approaches $a$.”
In the case of sequences, “as the index $n$ increases” is expressed by the phrase “for all $n \geq N(\varepsilon)$” since the value of $N(\varepsilon)$ is generally large.
In contrast, in the case of functions, “as the value of $x$ with $x \neq a$ approaches $a$” is expressed by:
$$0 < | \, x \, - \, a \, | < \delta(\varepsilon),$$
where $\delta$ denotes the Greek letter called “delta.”
Notice that $x$ is close to $a$ if $\delta(\varepsilon)$ is small. In fact, recalling that $| \, x \, - \, a \, |$ means the distance between $x$ and $a$, the inequality $| \, x \, - \, a \, | < \delta(\varepsilon)$ implies that $x$ is in a ball (interval) with radius $\delta(\varepsilon)$ and center $a$. In addition, we have $x \neq a$ because of $0 < | \, x \, - \, a \, |$.
Thus, we arrive at the following definition of the limit of a function:
For any $\varepsilon > 0$, there exists a positive real number $\delta(\varepsilon)$ such that for all $x$ satisfying $0 < | \, x \, - \, a \, | < \delta(\varepsilon)$, we have $| \, f(x) \, - \, \alpha \, | < \varepsilon$.
When $f(x)$ satisfies this condition, we say that $f(x)$ converges to $\alpha$ as $x \to a$, which is denoted by $\displaystyle\lim_{ x \to a} f(x) = \alpha$. Moreover, $\alpha$ is called the limit value of $f(x)$ as $x \to a$. This is the definition of the limit of a function by the epsilon-delta argument. The key point of this definition is that $\delta$ is a function of $\varepsilon$.
The existence of the function $\delta(\varepsilon)$ guarantees the convergence of a function in the same way as that of a sequence.
I understand that $\delta$ is a function of $\varepsilon$, and that the existence of $\delta(\varepsilon)$ guarantees the convergence of a function. However, I wonder whether the value of $\delta(\varepsilon)$ decreases as $\varepsilon$ decreases. In the case of sequences, the value of $N(\varepsilon)$ increases as $\varepsilon$ decreases.
In general, the value of $\delta(\varepsilon)$ decreases as $\varepsilon$ decreases. I think that the following figure is helpful for understanding this property intuitively. Suppose that the value of $\varepsilon$ in Figure 1 is smaller than that in Figure 2. Then, the value of $\delta(\varepsilon)$ in Figure 1 is smaller than that in Figure 2.
In order to understand the definition of the limit of a function, we will show $\displaystyle\lim_{x \to 2} f(x) = 4$, where $f(x) = x^2$. That is,
for any $\varepsilon > 0$, there exists a positive real number $\delta(\varepsilon)$ such that for all $x$ satisfying $0 < | \, x \, - \, 2 \, | < \delta(\varepsilon)$, we have $| \, f(x) \, - \, 4 \, | < \varepsilon$.
Note the last inequality $| \, f(x) \, - \, 4 \, | < \varepsilon$ in the above condition. Since
$$| \, f(x) \, - \, 4 \, | = | \, x^2 \, - \, 4 \, | = | \, x \, - \, 2 \, | | \, x \, + \, 2 \, |,$$
we seek the range of $x$ where $| \, x \, - \, 2 \, | | \, x \, + \, 2 \, |< \varepsilon$ holds. We consider that the value of $| x + 2 |$ can be approximated by $4$ because we take $x$ in a small neighborhood of $2$ for examining the limit of $f(x)$ as $x \to 2$. Therefore, we assume that $$| x + 2 | < 5$$ holds, where we add $1$ to $4$ with a margin. Then, for all $x$ satisfying $| x \, - 2 | < \displaystyle\frac{\varepsilon}{5}$, we have $$|f(x) \, - \, 4| = |x+2| |x-2| < 5 | x-2| < 5 \cdot \frac{\varepsilon}{5} = \varepsilon$$
The above argument holds under the assumption $| x + 2 | < 5$. This inequality is equivalent to $-5 < x + 2 < 5$; i.e., $-7 < x < 3$. Hence, we can obtain $| x + 2 | < 5$ as long as we restrict the range of $x$ to satisfy $1 < x < 3$; i.e., $| x \, - 2 | < 1$. Thus, if $$| x \, - 2 | < 1 \ \ \ {\rm and} \ \ \ | x \, - 2 | < \displaystyle\frac{\varepsilon}{5},$$ then, noting $|x +2 | < 5$ by $|x -2 | <1$, we have $$|f(x) \, - \, 4| = |x+2| |x-2| < 5 | x-2| < 5 \cdot \frac{\varepsilon}{5} = \varepsilon.$$ Now, we define $\delta(\varepsilon)$ as follows:
$$\delta(\varepsilon) = \min( \frac{\varepsilon}{5}, \ 1 ),$$
where $\min(x, y)$ denotes the minimum value of $x$ and $y$. We note that $\delta(\varepsilon) \leq \displaystyle\frac{\varepsilon}{5}$ and $\delta(\varepsilon) \leq 1$ hold.
Let $x$ satisfy $0 < | \, x \, - \, 2 \, | < \delta(\varepsilon)$. Then, noting that $\delta(\varepsilon) \leq 1$, we have $| x \, - 2 | < 1$, which leads to $| x + 2 | < 5$. Therefore, we have
$$|f(x) \, - \, 4| = |x + 2| |x - 2| < 5 | x -2| < 5 \, \delta(\varepsilon) \leq 5 \cdot \frac{\varepsilon}{5} = \varepsilon.$$
Thus, for any $\varepsilon > 0$, there exists a real positive number $\delta(\varepsilon) = \min( \displaystyle\frac{\varepsilon}{5}, 1 )$ such that for all $x$ satisfying $0 < | \, x \, - \, 2 \, | < \delta(\varepsilon)$, we have $| \, f(x) \, - \, 4 \, | < \varepsilon$.
Exercise 2 Let $f(x) = \sqrt{x}$. Show that $\displaystyle\lim_{ x \to 1 }f(x) = 1$.
## Continuity of Functions
When the graph of a function is neither cut off nor has a hole, we say that the function is continuous. To be precise, we say that a function $f(x)$ is continuous at $x = a$ if
(A) the value of $f(x)$ is defined at $x = a$,
and
(B) the limit value $\displaystyle\lim_{x \to a} f(x)$ exists, and $\displaystyle\lim_{x \to a} f(x) = f(a)$ holds.
For example, the function presented in Figure 1 is continuous at $x = a$ because the limit value $\displaystyle\lim_{x \to a} f(x)$ exists and $\displaystyle\lim_{x \to a} f(x) = f(a)$ holds; the graph of the function is neither cut off nor has a hole at $x = a$.
In contrast, the functions presented in Figures 2 and 3 are not continuous at $x = a$. In the case of Figure 2, the limit value $\displaystyle\lim_{x \to a} f(x)$ does not exist; the graph of the function is cut off at $x = a$. In the case of Figure 3, $\displaystyle\lim_{x \to a} f(x)$ exists, but $\displaystyle\lim_{x \to a} f(x) \neq f(a)$ does not hold; the graph of the function has a hole at $x = a$.
By using the definition of the limit of a function, the above condition B that the limit value $\displaystyle\lim_{x \to a} f(x)$ exists and $\displaystyle\lim_{x \to a} f(x) = f(a)$ holds is given by:
For any $\varepsilon > 0$, there exists a real number $\delta(\varepsilon) > 0$ such that for all $x$ satisfying $0 < | \, x \, - \, a \, | < \delta(\varepsilon)$, we have $| \, f(x) \, - \, f(a) \, | < \varepsilon$.
However, noting that the value of $f(x)$ is defined at $x = a$, the last inequality in this statement $| \, f(x) \, - \, f(a) \, | < \varepsilon$ holds at $x = a$. In fact, we have $| \, f(a) \, - \, f(a) \, | = 0 < \varepsilon$. Therefore, the above statement can be slightly modified as follows:
For any $\varepsilon > 0$, there exists a real number $\delta(\varepsilon) > 0$ such that for all $x$ satisfying $| \, x \, - \, a \, | < \delta(\varepsilon)$, we have $| \, f(x) \, - \, f(a) \, | < \varepsilon$.
This is exactly the condition that $f(x)$ is continuous at $x = a$ when the value of $f(x)$ is defined at $x = a$.
In order to understand the definition of the continuity of a function, we will show that $f(x) = x^2$ is continuous at $x = a$. Since it is clear that the value of $f(x)$ is defined for all $x$, our goal is to show the following:
For any $\varepsilon > 0$, there exists a real number $\delta(\varepsilon) > 0$ such that for all $x$ satisfying $| \, x \, - \, a \, | < \delta(\varepsilon)$, we have $| \, f(x) \, - \, f(a) \, | < \varepsilon$.
As was explained before, noting the last inequality in the above assertion $| \, f(x) \, - \, f(a) \, | < \varepsilon$, we seek the range of $x$ where this inequality holds. Since
$$| \, f(x) \, - \, f(a) \, | = | \, x^2 \, - \, a^2 \, | = | \, x \, - \, a \,| | \, x \, + \, a \,|,$$
we must seek the range of $x$ where $|x \, - \, a| | x + a | < \varepsilon$ holds.
When $x$ is close to $a$, we can consider that $| x + a |$ is approximated by $2|a|$. Therefore, noting that $x$ is close to $a$, we assume that $$| x + a | < 2|a| + 1$$
holds, where we add $1$ to $2|a|$ with a margin. Then, for all $x$ satisfying $| x \, - \, a | < \displaystyle\frac{\varepsilon}{2|a| + 1}$, we have
$$| \, f(x) \, - \, f(a) \, | = | \, x \, - \, a \,| | \, x \, + \, a \, | < \displaystyle\frac{\varepsilon}{2|a| + 1} \cdot (2|a| + 1 ) = \varepsilon.$$
This result is obtained under the assumption $| x + a | < 2|a| + 1$. However, we have not yet specified the range of $x$ where this inequality holds.
Suppose $| x \, - \, a | < \rho$, where $\rho$ will be specified later. Then, we have $|x| < |a| + \rho$ because $x$ is in a ball (interval) with radius $\rho$ and center $a$. Therefore, we have $$\begin{array}{l} | x + a | \leq |x| + |a| \\[2ex] \ \ \ \ < |a| + \rho + |a| \\[2ex] \ \ \ \ = 2|a| + \rho \end{array}$$ Hence, when $\rho \leq 1$, we have
$$| x + a | < 2|a| + 1.$$ Consequently, noting the case that $\rho = 1$, we see that $| x \, - \, a | < 1$ implies $| x + a | < 2|a| + 1$.
Thus, if $$| x \, - a | < 1 \ \ \ {\rm and} \ \ \ | x \, - a | < \displaystyle\frac{\varepsilon}{2|a| + 1},$$ then, noting $| x + a | < 2|a| + 1$ by $| x - a | <1$, we have $$| \, f(x) \, - \, f(a) \, | = | \, x \, - \, a \,| | \, x \, + \, a \, | < \displaystyle\frac{\varepsilon}{2|a| + 1} \cdot (2|a| + 1 ) = \varepsilon.$$
Now, we define
$$\delta(\varepsilon) = \min\left( \frac{\varepsilon}{2|a| + 1}, \ 1 \right).$$
Notice that $| x \, - \, a | < \delta(\varepsilon)$ implies $| x \, - \, a | < 1$ and $| x \, - \, a | < \displaystyle\frac{\varepsilon}{2|a|+1}$ because $\delta(\varepsilon) \leq 1$ and $\delta(\varepsilon) \leq \displaystyle\frac{\varepsilon}{2|a|+1}$.
Therefore, for all $x$ satisfying $| x \, - \, a | < \delta(\varepsilon)$, we have $| x + a | < 2|a| + 1$ by $| x \, - \, a | < 1$, which leads to $$\begin{array}{l} | f(x) \, - \, f(a) | = | x \, - \, a| | x \, + \, a | \\[2ex] \ \ \ \ < | x \, - \, a| (2|a| + 1) \\[2ex] \ \ \ \ < \delta(\varepsilon) (2|a|+ 1) \\[2ex] \ \ \ \ = \displaystyle\frac{\varepsilon}{2|a| + 1} \cdot (2|a| + 1) = \varepsilon \end{array}$$ Hence, we see that $f(x) = x^2$ is continuous at $x = a$.
From the definition of the continuity of a function, I understand that $\delta$ is a function of $\varepsilon$ because $\delta$ depends on $\varepsilon$. However, in the above example, we define $\delta = \min\left( \displaystyle\frac{\varepsilon}{2|a| + 1}, \ 1 \right)$. Therefore, I think that $\delta$ can be regarded as a function of $a$ as well as of $\varepsilon$.
That’s a very good point. You are completely right.
It is usual that a function is defined on a set on the real line ${\bf R}$. This set is called the “domain.” In many cases, the domain of a function is given by an interval on ${\bf R}$. For example, when the domain of a function $f(x)$ is the interval $I = [a, \, b]$, the interval $I$ is given by the set
$$[a, \, b] = \{ \, x \, | \, a \leq x \leq b \, \}.$$
In this case, the value of the function $f(x)$ is defined for $x$ satisfying $a \leq x \leq b$. We can define a function on an interval that does not include its endpoints, such as
$$(a, \, b) = \{ \, x \, | \, a < x < b \, \}.$$
Moreover, the real line denoted by ${\bf R}$ can be regarded as an interval, and is often expressed by ${\bf R } = (-\infty, \, \infty)$.
Suppose that the domain of a function $f(x)$ is an interval $I$. Then, we say that $f(x)$ is continuous on $I$ if for any $a \in I$, $f(x)$ is continuous at $x = a$. That is, the definition that $f(x)$ is continuous on $I$ is as follows:
For any $a \in I$ and for any $\varepsilon > 0$, there exists a real number $\delta(\varepsilon ; a) > 0$ such that for all $x$ satisfying $| \, x \, - \, a \, | < \delta(\varepsilon ; a)$, we have $| \, f(x) \, - \, f(a) \, | < \varepsilon$.
It should be noted that $\delta$ is a function of $\varepsilon$ and $a$ in this definition. For example, the function $f(x) = x^2$ is continuous on ${\bf R}$. In fact, as was explained above, for any $a \in {\bf R}$ and for any $\varepsilon > 0$, there exists a real positive number $\delta(\varepsilon ; a)$ defined by
$$\delta(\varepsilon ; a) = \min\left( \displaystyle\frac{\varepsilon}{2|a| + 1}, \ 1 \right)$$ such that for all $x$ satisfying $| \, x \, - \, a \, | < \delta(\varepsilon ; a)$, we have $| \, f(x) \, - \, f(a) \, | < \varepsilon$.
I understand that when a function $f (x)$ is continuous on an interval $I$, $\delta$ is a function of $a \in I$ and $\varepsilon > 0$. However, I wonder whether there exists a function such that $\delta$ is a function of only $\varepsilon$?
Indeed. When $\delta$ is a function of only $\varepsilon$, we say that $f(x)$ is uniformly continuous on $I$. As seen in the following exercise, there are many uniformly continuous functions. The uniform continuity of functions often plays an important role in advanced mathematical analysis.
Exercise 3 Show that $f(x) = x^2 -x$ is continuous at any $a \in {\bf R}$.
Exercise 4 Show that $f(x) = x^2$ is uniformly continuous on an interval $[a, \, b]$, where $a$ and $b$ are finite real numbers.
## Concluding Remarks
We come to the end of our explanation for the basis of the epsilon-delta argument. The idea of the epsilon-delta argument is to formulate our naive intuition for the limits of functions as the existence problem of a function $\delta = \delta(\varepsilon)$. Although this idea is quite simple, the epsilon-delta argument requires some technical training in using logic symbols, estimating inequalities, negating statements, and so on. We will be happy if this website can help calculus beginners to understand the basic underlying ideas and to overcome the technical barriers to using the epsilon-delta argument.
Can you explain the underlying basic idea of the epsilon-delta definition of the limits of functions (the continuity of functions) through the parable of an airplane approaching an airport in the previous chapter? I think that it is a good exercise to confirm your understanding for the epsilon-delta definition of limits of functions.
epsilon-delta
タイトルとURLをコピーしました | {"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": 0, "wp-katex-eq": 277, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9931631088256836, "perplexity": 51.43587591350058}, "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/1664030335530.56/warc/CC-MAIN-20221001035148-20221001065148-00491.warc.gz"} |
https://math.stackexchange.com/questions/1916557/different-lagrangians-defined-for-the-same-optimization-problem | # Different Lagrangians defined for the same optimization problem
In the book "Sparse and Redundant Representations", the following optimization problem is stated:
$$\min_x J(x) \quad \text{subject to} \quad b=Ax$$
then using Lagrange multipliers, the following Lagrangian has been defined:
$$\mathcal{L}(x)= \parallel x\parallel_2^2+\lambda^T (Ax-b)$$
Where $\lambda$ is the $Lagrange\;multipliers$ for the constraint set.
And the following requirement has been obtained by taking a derivative of $\mathcal{L}(x)$ with respect to $x$:
$$\frac{\partial\mathcal{L}(x)}{\partial(x)}=2x+A^T\lambda$$
1. My first question is, why in the derivative, we have $A^T\lambda$? Shouldn't it be $\lambda^TA$ instead?
The above optimization problem has been solved differently in the book "Convex Optimization by Stephen Boyd":
For the following optimization problem: $$\min_x f_0(x)\\\text{subject to}\quad h_i(x)=0, \quad i = 1, . . . , p$$
The Lagrangian associated with the above problem is:
$$L(x,\lambda,v)=f_0(x)+\sum_{i=1}^{p}v_ih_i(x)$$
Where $\lambda_i$ is referred to as $Lagrange\;multiplier$ associated with the $i$th inequality constraint $f_i(x) ≤ 0$
1. Why $Lagrangian$ has been defined differently in the two books?
• For question number 1: $\lambda^TA$ doesn't even have the proper dimensions; $2x+\lambda^TA$ would be ill-posed. – Michael Grant Sep 6 '16 at 22:14
• Why do you think they've been defined differently? In the first case all you have are equality constraints. In the second you have both inequalities and equations. – Michael Grant Sep 6 '16 at 22:15
• Thank you for your comments @MichaelGrant, great help. I still don't understand your answer to question 1, isn't the derivative of $\lambda^TAx$, $\lambda^TA$? – Gigili Sep 7 '16 at 1:32
• No, it is not. And again, how could it be added to $2x$ if it were? – Michael Grant Sep 7 '16 at 1:50
• Look at it this way: why is the derivative of $x^Tx$ equal to $2x$, and not $2x^T$? – Michael Grant Sep 7 '16 at 3:51
The derivative of $\langle \lambda ,A x\rangle$ is $A^T \lambda$. Maybe it becomes clearer if you look at the following:
$$\langle \lambda , Ax \rangle = \sum_{i=1}^n \lambda_i \sum_{j=1}^m A_{ij} x_j$$
so the $j$th element of the gradient, is the derivative of the above with respect to $x_j$ which is
$$\sum_{i=1} \lambda_i A_{ij} = (A^T\lambda)_j$$
Q2: For your second question, the two Lagrangians are essentially the same (except that in the first case they only consider an equality constraint while in the second I suspect they consider both equality and inequality (*)). Just write:
$$h_i(x) = (Ax -b)_i$$
and $\nu_i=\lambda_i$ then, in the first case, you have $\lambda^T(Ax-b) = \sum_{i=1}^p \lambda_i(Ax-b)_i$ and in the second you have $\nu^T(Ax-b) = \sum_{i=1}^p \nu_i(Ax-b)_i = \sum_{i=1}^p \nu_i h_i(x)$, indeed, exactly the same thing.
(*) by the way, your edits are a bit confusing here, (a) either add the sum corresponding to the inequality constraint and leave the sentence below it or (b) remove the sum (as you have done) and the sentence below, and the $\lambda$ in the Lagrangian. | {"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": 0.925923228263855, "perplexity": 297.81177226210025}, "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-2019-30/segments/1563195532251.99/warc/CC-MAIN-20190724082321-20190724104321-00458.warc.gz"} |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.