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https://www.physicsforums.com/threads/understanding-of-higher-order-derivatives.332073/ | # Understanding of Higher-Order Derivatives
• #1
26
0
Hey guys, so this may be a really silly question, but I'm trying to grasp a subtle point about higher-order derivatives of multivariable functions. In particular, suppose we have an infinitely differentiable function
$$f: \mathbb{R}^{n} \rightarrow \mathbb{R}$$
I know that the first derivative of this function is a linear map $$\lambda: \mathbb{R}^{n}\rightarrow\mathbb{R}$$. However, when we take the second-derivative of $$\lambda$$, some questions arise for me:
1.) If we are taking this derivative when considering $$\lambda$$ as a linear function, then we'd just get back $$\lambda$$, which isn't the case. So how are we interpreting the first derivative when taking a second?
2.) In general, why do we say that $$D^{k}f:\mathbb{R}^{n^{k}}\rightarrow\mathbb{R}$$ and not $$D^{k}f:\mathbb{R}^{n}\rightarrow\mathbb{R}$$ ??
• #2
139
12
If $$f : \mathbb{R}^n \to \mathbb{R}$$, then the derivative of $$f$$ is a linear map $$\lambda : \mathbb{R}^n \to \mathbb{R}$$ at each point in $$\mathbb{R}^n$$. That is to say, the derivative of $$f$$, properly considered, is a map $$Df : \mathbb{R}^n \to L(\mathbb{R}^n, \mathbb{R})$$, where $$L(\mathbb{R}^n, \mathbb{R})$$ denotes the space of all linear maps $$\lambda : \mathbb{R}^n \to \mathbb{R}$$, which is just the dual of $$\mathbb{R}^n$$ (and is thus isomorphic to $$\mathbb{R}^n$$). The second derivative of $$f$$ is then a map $$D^2 f : \mathbb{R}^n \to L(\mathbb{R}^n, L(\mathbb{R}^n, \mathbb{R})) \cong L(\mathbb{R}^n, \mathbb{R}^n)$$, where $$L(\mathbb{R}^n, \mathbb{R}^n)$$ is the space of all $$n \times n$$ matrices, and is isomorphic to $$\mathbb{R}^{n^2}$$. (The output of the second derivative is usually called the Hessian matrix of $$f$$.) Continuing in this vein, you can show that $$D^k f$$ is a map from $$\mathbb{R}^n$$ to $$\mathbb{R}^{n^k}$$, not a map from $$\mathbb{R}^{n^k} \to \mathbb{R}$$ as you suggest in #2.
Basically, what's going on here is that a derivative, properly defined, is a best linear approximation to a function. Thus, at some point $$\mathbf{p} \in \mathbb{R}^n$$, the derivative $$Df$$ takes the value of the linear map $$\lambda : \mathbb{R}^n \to \mathbb{R}$$ which most closely resembles $$f$$ near $$\mathbf{p}$$. Thus, $$Df$$ is actually a map from $$\mathbb{R}^n$$ into the space of all possible such approximations, and $$D^k f$$ is a map from $$\mathbb{R}^n$$ into some higher tensor product of $$\mathbb{R}^n$$ and its dual space. Your answer to #1 is thus that, while elements of the range of $$Df$$ must be linear maps and have trivial derivatives, $$Df$$ itself is not necessarily linear. This is why it is necessary to specify two arguments when evaluating $$Df$$: a location $$\mathbf{a}$$, and a direction $$\mathbf{h}$$. The location specifies a linear map, i.e., there is some linear map $$\lambda$$ for which $$Df : \mathbf{a} \mapsto \lambda$$. The direction then serves as the argument for $$\lambda$$, and, in a slight abuse of notation, we usually write $$\lambda(\mathbf{h}) \equiv Df(\mathbf{a})(\mathbf{h})$$ or $$Df(\mathbf{a}, \mathbf{h})$$.
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4K | {"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": 0.9894880652427673, "perplexity": 112.17346536923355}, "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/1600400209999.57/warc/CC-MAIN-20200923050545-20200923080545-00243.warc.gz"} |
http://www.koreascience.or.kr/article/ArticleFullRecord.jsp?cn=HNSHCY_2013_v35n2_311 | HYPONORMALITY OF TOEPLITZ OPERATORS ON THE WEIGHTED BERGMAN SPACES
• Journal title : Honam Mathematical Journal
• Volume 35, Issue 2, 2013, pp.311-317
• Publisher : The Honam Mathematical Society
• DOI : 10.5831/HMJ.2013.35.2.311
Title & Authors
HYPONORMALITY OF TOEPLITZ OPERATORS ON THE WEIGHTED BERGMAN SPACES
Lee, Jongrak; Lee, Youho;
Abstract
In this note we consider the hyponormality of Toeplitz operators $\small{T_{\varphi}}$ on the Weighted Bergman space $\small{A^2_{\alpha}(\mathbb{D})}$ with symbol in the class of functions $\small{f+\bar{g}}$ with polynomials $\small{f}$ and $\small{g}$ of degree 2.
Keywords
Toeplitz operators;hyponormal;weighted Bergman space;
Language
English
Cited by
1.
Hyponormal Toeplitz operators with polynomial symbols on weighted Bergman spaces, Journal of Inequalities and Applications, 2014, 2014, 1, 335
References
1.
C. Cowen, Hyponormality of Toeplitz operators, Proc. Amer. Math. Soc. 103 (1988), 809-812.
2.
R.E. Curto, I.S. Hwang and W.Y. Lee, Hyponormality and subnormality of block Toeplitz operators, Adv. Math. 230 (2012), 2094-2151.
3.
R.E. Curto and W.Y. Lee, Joint hyponormality of Toeplitz pairs, Memoirs Amer. Math. Soc. 150(712) (2001).
4.
I.S. Hwang, I.H. Kim and W.Y. Lee, Hyponormality of Toeplitz operators with polynomial symbol, Math. Ann 313 (1999), 247-261.
5.
I.S. Hwang and W.Y. Lee, Hyponormality of trigonometric Toeplitz operators, Trans. Amer. Math. 354 (2002), 2461-2474.
6.
I.S. Hwang and J.R. Lee, Hyponormal Toeplitz operators on the weighted Bergman spaces, Math. Ineq. and App. 15 (2012), 323-330.
7.
Yufeng Lu and Chaomei Liu, Commutativity and hyponormality of Toeplitz operators on the weighted Bergman space, J. Korean Math. Soc. 46 (2009), 621-642.
8.
T. Nakazi and K. Takahashi, Hyponormal Toeplitz operators and extremal problems of Hardy spaces, Trans. Amer. Math. Soc. 338 (1993), 753-769.
9.
D. Sarason, Generalized interpolation in $H^{\infty}$, Trans. Amer. Math. Soc. 127 (1967), 179-203.
10.
K. Zhu, Hyponormal Toeplitz operators with polynomial symbols, Integral Equations and Operator Theory 21 (1995), 376-381. | {"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": 5, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9551909565925598, "perplexity": 2795.3052133767587}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-39/segments/1537267155413.17/warc/CC-MAIN-20180918130631-20180918150631-00655.warc.gz"} |
http://mathhelpforum.com/calculus/14173-simpson-s-rule-calculus.html | 1. ## Simpson's Rule (Calculus)
If anyone could explain how the following is done, it would be greatly appreciated!
Use Simpson's rule with delta x = .1 to obtain an approximation for the integral of cos(x^2) dx from .6 to 1.
2. Originally Posted by faure72
If anyone could explain how the following is done, it would be greatly appreciated!
Use Simpson's rule with delta x = .1 to obtain an approximation for the integral of cos(x^2) dx from .6 to 1.
Simpsons rule tells us that the integral from a to b of f(x) may be
approximated as follows.
Divide the interval [a,b] into n equal parts (n even) with end points:
a, a+h, a+2h, ... a+nh
where h=(b-a)/n. Then:
integral_{x=a to b} f(x) dx ~= [h/3] (f(a) + 4f(a+h) + 2f(a+2h) + 4f(a+3h) +
.................. ... 4f(a+(n-1)h) + f(a+nh))
In the case here we have a=0.6, b=1, h=0.1, so n=4
integral_{x=a to b} cos(x^2) dx = [0.1/3] [cos(0.6^2) + 4cos(0.7^2) + 2cos(0.8^2) + 4cos(0.9^2) + cos(1^2)]
............... = [0.033333..][0.936+4(0.882)+2(0.802)+4(0.689)+0.540]
...............~= 0.312
RonL | {"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": 0.9970518946647644, "perplexity": 3650.0712679268604}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-34/segments/1502886110774.86/warc/CC-MAIN-20170822123737-20170822143737-00383.warc.gz"} |
http://www.perimeterinstitute.ca/seminar/self-dual-n4-theories-four-dimensions | A A
Connect with us:
# Self-dual N=4 theories in four dimensions
Known N=4 theories in four dimensions are characterized by a choice of gauge group, and in some cases some "discrete theta angles", as classified by Aharony, Seiberg and Tachikawa. I will review how this data, for the theories with algebra su(N), is encoded in various familiar realizations of the theory, in particular in the holographic AdS_5 \times S^5 dual and in the compactification of the (2,0) A_N theory on T^2. I will then show how the resulting structure, given by a choice of polarization of an appropriate cohomology group, admits additional choices that, unlike known theories, generically preserve SL(2,Z) invariance in four dimensions.
Collection/Series:
Event Type:
Seminar
Scientific Area(s):
Speaker(s):
Event Date:
Tuesday, October 24, 2017 - 14:30 to 16:00
Location:
Space Room
Room #:
400 | {"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": 0.9206892848014832, "perplexity": 2676.378780155152}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-47/segments/1510934808133.70/warc/CC-MAIN-20171124123222-20171124143222-00704.warc.gz"} |
https://www.physicsforums.com/threads/transfer-function-to-differential-equation.178298/ | # Homework Help: Transfer function to differential equation
1. Jul 25, 2007
### barneygumble742
1. The problem statement, all variables and given/known data
Write the differential equation that is equivalent to the transfer function given below. Plot y(t). Assume that r(t) = 4t$$^{2}$$
Y(s) = 2s$$^{4}$$+3s$$^{3}$$+2s$$^{2}$$+s+1
R(s) = 2s$$^{5}$$+3s$$^{4}$$+2s$$^{3}$$+2s$$^{2}$$+4s+2
The transfer function is Y(s)/R(s).
2. Relevant equations
3. The attempt at a solution
Given r(t), I thought of converting it to LaPlace and then multiplying it with the numberator so I would be left with Y(s) = numerator / denominator. After that I'll have a mess that I don't think will factor without imaginary numbers. I'm thinking of using partial fraction expansion.
OR
I could have it in this form:
Y(s) [2s$$^{5}$$+3s$$^{4}$$+2s$$^{3}$$+2s$$^{2}$$+4s+2] = R(s) [2s$$^{4}$$+3s$$^{3}$$+2s$$^{2}$$+s+1]
and then convert each item to the time domain and then put it back in the transfer function form. However if I did this, then what about the final r(t) = 4t$$^{2}$$ that's left over?
Thanks,
BG742
2. Jul 26, 2007
### Mindscrape
I don't quite understand the point of making up a differential equation for which you have the solution (or really that part at all, is there more to the problem?), but as far as I can tell you have a transfer function in the Fourier domain, and you need to find the differential equation that it represents in the time domain. There is some response to a certain function given by the transfer function Y(s) = R(s)T(s), where T(s) is the transfer function. You know that y(t) = F(R(s)*T(s)), and you know that T(s) = Y(s)/R(s) with r(t) = 4t^2. So with all this you can figure out what y(t) is.
Whether you want to use Laplace Transforms or Fourier Transforms depends on the context of the problem. If this is a question about filters, which it seems like it is, then you should use Laplace. Actually dividing the two equations out will probably make the Laplace transform a lot easier.
3. Jul 26, 2007
### barneygumble742
The problem asks me to convert the transfer function to the time domain and plot it in terms of y(t). We have to use Laplace Transforms because of the nature of the course. Could you please explain what you meant by dividing the two equations out?
4. Jul 26, 2007
### Mindscrape
Actually go through the division of $$\frac{2s^4+3s^3++2s^2+s+1}{2s^5+3s^4+2s^3+2s^2+4s+2}$$. Matlab or Mathematica can do the job quick, but by hand won't take long either.
You might be over-analyzing the whole thing. The problems intent seems like it is to have you become more familiar with transfer functions and how the convolution theorem works.
Last edited: Jul 26, 2007
5. Jun 28, 2008
### suki
I think it's worth doing because if you ever have to implement a transfer function in anything less transparent than matlab you'll need to know.
-factor the numerator and denominator, make sure there isn't anything you can cancel between the two (there's not in this case).
-Convert any complex conjugate pairs back into 2nd order polynomials.
x=r1=(a+bi)
x=r2=(a-bi)
(x-r1)(x-r2) = x2-(r1+r2)x+r1*r2 = x2+2ax+(a2+b2)
-Now we don't have to deal with imaginary numbers.
-Think of the system not as one transfer function but a series of small ones.
-you can factor any polynomial into a series multiplication of first and second order polynomials.
-A first or second order denominator in a transfer function is a low pass filter or an under damped oscillator respectively.
-A multiplication of transfer functions, in the time domain means the output from one, is the input for the next.
- it doesn't matter what order you multiply things in-> it doesn't matter what order you chain these sub-systems together in if all you're interested in is the original input and output variables, but it will make a difference on the hidden variables.
-it's not difficult to turn a low pass filter back into it's differential equation. 1/(Ts+1) => X'=(X-Xi)/T
X is the output,
Xi is the input,
T is the time constant
-an oscilator isn't dificult either: 1/(ms2+bs+k) => V'= F-bV-KX, X'=V
F is the input or 'external force'
X is the output 'mass location'
V is the velocity of the mass
b is the 'damping coeficient'
K is the 'spring constant'.
(of course the physical analogy depends on your system. it's also possible to have negative spring constants and damping factors here).
- multiplying by 's' is taking the derivative. multiplying by (3s+2) takes x and outputs (3x'+2x). To include the numerator all you have to do is take all the derivatives that will be necessary (up to the maximum power of 's'), weight them (with the coeficients in the polynomial), and add them back together. I prefer to put the numerator at the end, so that you never have to take the derivative (or in this case the 4th derivative) of a step input.
6. Jun 29, 2008
### suki
actually now that I think a little more : you don't need to factor the denominator. You can get a differential equation directly from it using the same pattern as for the second order system. the max power of s in the denominator, put that many integrators in series, after each integrator put a negative feedback link, with a constant coefficient, to before the first integrator except for the coefficient on the highest power, the feedback coefficients go directly to the coefficients of the polynomial. and it'll be obvious:
Xin=2X(5)+3X(4)+2X(3)+2X''+4X'+2X
I just prefer to have it in terms of first and order systems, which I understand. | {"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": 0.8782700896263123, "perplexity": 554.3344444794117}, "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-2018-47/segments/1542039742963.17/warc/CC-MAIN-20181115223739-20181116005739-00145.warc.gz"} |
http://mathhelpforum.com/algebra/112915-find-make-c-subject.html | # Math Help - find/make C subject
1. ## find/make C subject
a= 3
b= 2
d= 5
a=4b + (6/ c-1)- 15d
okey so il show how far i am. could someone please tell me where im goin wrong. many thanks
3= (8) + (6/ C-1) - 75
75+3 = 8+ (6/ C-1)
78 = 8 + (6/ C-1)
78-8 = (6/C-1)
70 = (6/C-1)
C-1 (70) = 6
C (70) = 7
C= (7/70)
C= 0.1
2. You're right until the fourth line up from the bottom. If (c-1) definitely has brackets round then 6=70(c-1) you need to multiply out to get 6=70c-70 then rearrange. You should get c=76/70 = 1.1 approx
3. but if i then subsitute 1.1 value for C then the answer should equal a= (3)?
4. don't sub in 1.1 (the answer is really 1.0857....) as this is the rounded up answer. if you put c=76/70 into the original equation then you will get a=3.
5. Originally Posted by decoy808
a=4b + (6/ c-1)- 15d
You were told by Willow T that your bracketing is faulty; should be:
a = 4b + 6 / (c-1) - 15d
Now re-arrange:
6 / (c-1) = a - 4b + 15d
c - 1 = 6 / (a - 4b + 15d)
c = 6 / (a - 4b + 15d) + 1
6. ## Analysis
Hi ineed to solve this exercise. Who can help me?
Let A subset of R (real numbers) A no empty. A is annotated if and only if exist M (positive rel) such that absolute value of x is less than or equal that M, for all x in 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": 0.8754324913024902, "perplexity": 3109.0782990075}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-30/segments/1469257824853.47/warc/CC-MAIN-20160723071024-00238-ip-10-185-27-174.ec2.internal.warc.gz"} |
http://openstudy.com/updates/4f18bd3fe4b04992dd21bf1c | Here's the question you clicked on:
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## anonymous 4 years ago A construction truck collides with the back of a subcompact car. Which vehicle has the larger ch ange in momentum during the collision? The small car has a larger change in its momentum. The huge truck has a larger change in its momentum. They have the same change in momentum. Neither one changes its momentum. Delete Cancel Submit
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1. anonymous
• 4 years ago
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The huge truck. Two objects of different mass are moving at the same speed; the more massive object will have the greatest momentum. A less massive object can never have more momentum than a more massive object.
2. anonymous
• 4 years ago
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chiki22. What is we have a mass of x with a velocity 2v, and a mass of 2x and a velocity of x? These will have the same momentum, because $$p = mv$$ JazzyCaz. This answer has several answers depending on what type of collision occurs (elastic or inelastic) and the initial conditions of the car and truck. Let's take a look at the math here. We know that momentum is always conserved regardless of which type of collision occurs. Therefore, $p_{t,i} + p_{c,i} = p_{t,f} + p_{c,f}$
3. TuringTest
• 4 years ago
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@ eashmore What makes you think that the velocity and masses of the respective objects have those proportions? Both are unspecified. Did you just invent them hypothetically? You are certainly right about the elastic/inelastic collision making a difference in something, but it is in conservation of kinetic energy only. Momentum is always conserved as you said, so the only answer is$p_{ti} + p_{ci} = p_{tf} + p_{cf}\to p_{ti}-p_{tf}=p_{ci}-p_{cf}\to \Delta p_t=\Delta p_c$
4. TuringTest
• 4 years ago
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@eashmore I'm pretty sure you knew that, but your answer confused me a bit, so I wanted to clarify for the asker.
5. anonymous
• 4 years ago
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I gave those proportions as a counter-point to chiki22's statement that "a less massive object can never have more momentum than a more massive object." In regards to the type of collision, I was trying to be thorough. Then present the idea that using conservation of momentum eliminates this discrepancy.
6. TuringTest
• 4 years ago
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Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy. | {"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": 0.9698296189308167, "perplexity": 1476.4567421228398}, "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/1476988719041.14/warc/CC-MAIN-20161020183839-00050-ip-10-171-6-4.ec2.internal.warc.gz"} |
https://www.zbmath.org/?q=an%3A0581.62028 | ×
# zbMATH — the first resource for mathematics
Pitman’s measure of closeness. (English) Zbl 0581.62028
Let $${\hat \theta}{}_ 1,{\hat \theta}_ 2,...,{\hat \theta}_ k$$ be k real-valued estimators of a real parameter $$\theta$$. Denote the absolute error loss of the i-th estimator by $$L_ i$$ where $$L_ i=| {\hat \theta}_ i-\theta |$$, $$i=1,2,...,k$$. For $$k=2$$ E. J. G. Pitman [Proc. Camb. Philos. Soc. 33, 212-222 (1937; Zbl 0016.36404)] introduced $$\Pr (L_ 1<L_ 2)$$ as a pairwise measure of closeness for the estimators $${\hat \theta}{}_ 1$$ and $${\hat \theta}{}_ 2.$$
In the presence of $$k>2$$ estimators of $$\theta$$, the authors obtain a generalization of Pitman’s measure of closeness by using a closest approach partition. A procedure for the calculation of this simultaneous Pitman measure is given for a specific class of estimators and illustrated by examination of a one-parameter exponential failure model. Finally, Pitman’s measure is shown to provide information that cannot be ignored when several estimators are being compared.
Reviewer: J.Melamed
##### MSC:
62F10 Point estimation | {"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": 0.9434748291969299, "perplexity": 638.4976924648424}, "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/1627046152085.13/warc/CC-MAIN-20210805224801-20210806014801-00503.warc.gz"} |
https://santafe.edu/research/results/papers/1511-optimal-design-robustness-and-risk-aversion | #### Newman, M. E. J., M. Girvan, and J. D. Farmer
Highly optimized tolerance is a model of optimization in engineered systems, which gives rise to power-law distributions of failure events in such systems. The archetypal example is the highly optimized forest re model. Here we give an analytic solution for this model which explains the origin of the power laws. We also generalize the model to incorporate risk aversion, which results in truncation of the tails of the power law so that the probability of disastrously large events is dramatically lowered, giving the system more robustness. | {"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": 0.9263150095939636, "perplexity": 758.9126692522501}, "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-2021-43/segments/1634323585201.94/warc/CC-MAIN-20211018093606-20211018123606-00379.warc.gz"} |
https://www.physicsforums.com/threads/help-with-kinematics.327068/ | # Help with Kinematics
1. Jul 26, 2009
### harini_5
An object is moving in 3D space with position vector vectot r=(x)i+y(j)+z(k) and velocity vector v=a(i)+b(j)+c(k)
i,j,k are unit vectors along x,y,z axes
If the object is always moving towards origin,
then
(a) a/x =b/y = c/z
(b) (a/x)(b/y)(c/z)<0
(c)ax+by+cz<0
(d)ab/xy >0
x,y,z,a,b,c are functions of time
question may have one or more correct answer(s)
MY APPROACH
integrating velocity eqn,
r=at(i)+bt(y)+ct(z)
which implies,
x=at
y=bt
z=ct
I dont get the correct ans if I proceed this way
wats wrong?
2. Jul 26, 2009
### rl.bhat
Re: kinematics
Angle between position vector and velocity vector is π.
So the dot product of r and v i.e. r.v = - ( xa + yb + zc) < 0
3. Jul 26, 2009
### harini_5
Re: kinematics
plz explain me,how the angle is pi
what does the statement the object is always moving towards origin
imply?
4. Jul 26, 2009
### RoyalCat
Re: kinematics
For the sake of argument, let's simplify it to a one dimensional problem.
The x axis is positive to the right of the origin, and negative to the left of the origin.
The object starts with an initial displacement $$x_0 > 0$$
If the object is to move towards the origin, is its velocity negative or positive?
The dot product is the scalar product of two vectors.
$$\vec i\cdot \vec j \equiv |i|*|j|*\cos{\theta}$$ where $$\theta$$ is the angle between the two vectors.
So for your velocity and displacement vectors, make an analogy from the one-dimensional case.
The velocity runs opposite the displacement vector for each of the axes. Can you see why the dot product is necessarily negative, now?
Last edited: Jul 26, 2009
Similar Discussions: Help with Kinematics | {"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": 0.8715255260467529, "perplexity": 1647.385868773346}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-05/segments/1516084890991.69/warc/CC-MAIN-20180122034327-20180122054327-00384.warc.gz"} |
https://deepai.org/publication/computing-height-persistence-and-homology-generators-in-r-3-efficiently | # Computing Height Persistence and Homology Generators in R^3 Efficiently
Recently it has been shown that computing the dimension of the first homology group H_1(K) of a simplicial 2-complex K embedded linearly in R^4 is as hard as computing the rank of a sparse 0-1 matrix. This puts a major roadblock to computing persistence and a homology basis (generators) for complexes embedded in R^4 and beyond in less than quadratic or even near-quadratic time. But, what about dimension three? It is known that persistence for piecewise linear functions on a complex K with n simplices can be computed in O(n n) time and a set of generators of total size k can be computed in O(n+k) time when K is a graph or a surface linearly embedded in R^3. But, the question for general simplicial complexes K linearly embedded in R^3 is not completely settled. No algorithm with a complexity better than that of the matrix multiplication is known for this important case. We show that the persistence for height functions on such complexes, hence called height persistence, can be computed in O(n n) time. This allows us to compute a basis (generators) of H_i(K), i=1,2, in O(n n+k) time where k is the size of the output. This improves significantly the current best bound of O(n^ω), ω being the matrix multiplication exponent. We achieve these improved bounds by leveraging recent results on zigzag persistence in computational topology, new observations about Reeb graphs, and some efficient geometric data structures.
## Authors
• 15 publications
• ### Computing Zigzag Persistence on Graphs in Near-Linear Time
Graphs model real-world circumstances in many applications where they ma...
03/12/2021 ∙ by Tamal K. Dey, et al. ∙ 0
• ### Generalized Persistence Algorithm for Decomposing Multi-parameter Persistence Modules
The classical persistence algorithm virtually computes the unique decomp...
04/07/2019 ∙ by Tamal K. Dey, et al. ∙ 0
• ### Computing with quasiseparable matrices
The class of quasiseparable matrices is defined by a pair of bounds, cal...
02/03/2016 ∙ by Clement Pernet, et al. ∙ 0
• ### Computing syzygies in finite dimension using fast linear algebra
We consider the computation of syzygies of multivariate polynomials in a...
12/04/2019 ∙ by Vincent Neiger, et al. ∙ 0
• ### Stabilizer Circuits, Quadratic Forms, and Computing Matrix Rank
We show that a form of strong simulation for n-qubit quantum stabilizer ...
03/29/2019 ∙ by Chaowen Guan, et al. ∙ 0
• ### Efficient algorithms for computing a minimal homology basis
Efficient computation of shortest cycles which form a homology basis und...
01/21/2018 ∙ by Tamal K. Dey, et al. ∙ 0
• ### Multifractal Description of Streamflow and Suspended Sediment Concentration Data from Indian River Basins
This study investigates the multifractality of streamflow data of 192 st...
09/04/2019 ∙ by Adarsh Sankaran, et al. ∙ 0
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## 1 Introduction
Topological persistence for a filtration or a piecewise linear function on a simplicial complex is known to be computable in time [15] where is the number of simplices in and is the exponent of matrix multiplication. The question regarding the lower bound on its computation was largely open until Edelsbrunner and Parsa [12] showed that computing the rank of the first homology group of a simplicial complex linearly embedded in is as hard as the rank computation of a sparse - matrix. The current upper bound for matrix rank computation is super-quadratic [7] and lowering it is a well-recognized hard problem. Consequently, computing the dimension of the homology groups and hence the topological persistence for functions on general complexes in better than super-quadratic time is difficult, if not impossible. But, what about the special cases that are still interesting? The complexes embedded in three dimensions which arise in plenty of applications present such cases.
It is easy to see that the Betti numbers , the rank of the th homology group defined over a finite field for a simplicial complex linearly embedded in can be computed in time. For this, compute with a walk over the boundaries of the voids, compute as the number of components of , and then compute from the Euler characteristics of obtained as the alternating sum of the numbers of simplices of each dimension. Unfortunately, computation of other topological properties such as persistence and homology generators (basis) for such a complex is not known to be any easier than that of matrix multiplication ( time). In the special case when is a graph or a surface, the persistence for a PL function or a filtration on can be computed in time [1, 10]. In this paper, we show that when is more general, that is, a simplicial complex linearly embedded in , the persistence of a height function on it can be computed in time. This special type of persistence which we term as the height persistence is not as general as the standard persistence. Nonetheless, it provides an avenue to compute a set of basis cycles in time where is the total size of the output. Also, the height persistence provides a window to the topological features of the domain , the need for which arises in various applications.
To arrive at our result, we first observe a connection between the standard sublevel-set persistence [11, 17] and the level-set zigzag persistence [6] from the recent work in [3, 4, 6]. Then, with a sweep-plane algorithm that treats the level sets as planar graphs embedded in a plane, we compute a barcode graph in time. A barcode is extracted from this graph using a slight but important modification of an algorithm in [1]. The barcode extracted from this graph provides a part of the height persistence. We show that the missing piece can be recovered from the Reeb graph which can be computed again in time [16]. We make other observations that allow us to extract the actual basis cycles from both pieces in time as claimed.
## 2 Background
A zigzag diagram of topological spaces is a sequence
X:X0↔X1↔⋯↔Xm (2.1)
where each is a topological space and each bidirectional arrow ‘’ is either a forward or a backward continuous map. Applying the homology functor with coefficient in a field
, we obtain a sequence of vector spaces connected by forward or backward linear maps, also called a zigzag module:
Hp(X):Hp(X0)↔Hp(X1)↔⋯↔Hp(Xm)
When all vector spaces in are finite dimensional, the Gabriel’s theorem in quiver theory [13] says that is a direct sum of a finite number of interval modules which are of the form
I[b,d]:I1↔I2⋯↔Im
where for and otherwise with the maps and being identities. The decomposition provides a barcode (set of interval modules) for topological persistence when the topological spaces originate as sublevel or level sets of a real-valued function defined on a space . As shown in [6], classical persistence [11, 17], its extended version [8], and the more general zigzag persistence [6] arise as a consequence of choosing variants of the module in 2.1 that are derived from .
### 2.1 Standard persistence
Standard persistence [11, 17] is defined by considering the sublevel sets of , that is, is for some . These values are taken as the critical values of so that the barcode captures the evolution of the homology classes of the sub-level sets across the critical values of , which are defined below precisely.
For an interval , let denote the interval set. Following [3, 6], we assume that is tame. It means that it has finitely many homological critical values so that for each open interval , is homeomorphic to a product space , with . This homeomorphism should extend to a continuous function , with being the closure of and each interval set should have finitely generated homology groups.
It turns out that the description of the interval modules assumes one more subtle aspect when it comes to describing the standard persistence and zigzag persistence in general. Specifically, the interval modules can be open or closed at their end points. To elucidate this, consider a set of values of interleaving with its critical values:
s0
Assuming and , one can write the sub-level sets as . For standard persistence, we consider the sublevel set diagram and its corresponding homology module for dimension :
SL(f,X):X[0,a1]→X[0,s1]→X[0,a2]⋯→X[0,sm]→X[0,am+1] Hp(SL(f,X)):Hp(X[0,a1])→Hp(X[0,s1])→Hp(X[0,a2])⋯→Hp(X[0,sm])→Hp(X[0,am+1])
The summand interval modules, or the so called bars, for this case has the form . This means that a -dimensional homology class is born at the critical value and it dies at the value . The right end point of is an artifact of our choice of the intermediate value . Because of our assumption that is tame, homology classes cannot die in any open interval between the critical values. In fact, they remain alive in the interval and may die entering the critical value . To accommodate this fact, we convert each bar of the standard persistence to a bar that is open on the right end point.
One can see that there are two types of bars in the standard persistence, one of the type , , which is bounded (finite) on the right, and the other of the type which is unbounded (infinite) on the right. The unbounded bars represent the essential homology classes since . The work of [3, 4, 6] implies that both types of bars of the standard persistence can be recovered from those of the level set zigzag persistence as described next. This observation leads to an efficient algorithm for computing the standard persistence in .
### 2.2 Level set zigzag
In level set zigzag persistence, we track the changes in the homology classes in the level sets instead of the sub-level sets. We need maps connecting individual level sets, which is achieved by including the level sets into the adjacent interval sets. For this purpose we use the notation for the interval set between the two non-critical level sets. We have a zigzag sequence of interval and level sets connected by inclusions producing a level set zigzag diagram:
L(f,X):X00→X10←X11→X21⋯→Xmm−1←Xmm.
Applying the homology functor with coefficients in a field , we obtain the zigzag persistence module for any dimension
Hp(L(f,X)):Hp(X00)→Hp(X10)←Hp(X11)→⋯→Hp(Xmm−1)←Hp(Xmm). (2.2)
The zigzag persistence of is given by the summand interval modules of . Each interval module is of the type where and can be or for some . Just as in the sub-level set persistence, we identify the end points of the interval modules with the critical values [r] that were used to define the level set zigzag in the first place. In keeping with the understanding that even the level set homology classes do not change in the open interval sets, we convert an endpoint to an adjacent critical value and make the interval module open at that critical value. Precisely we modify the interval modules as (i) , (ii) (iii) (ii) . The intervals in (i)-(iv) are referred as closed-closed, closed-open, open-closed, and open-open bars respectively. The figure above shows the two bar codes, one for and another for for a height function on a torus. The rightmost picture shows the barcode graph of which we explain later.
Using the results in [4, 6], we can connect the standard persistence with the level set zigzag persistence as follows:
###### Theorem 1.
1. is a bar for iff it is so for ,
2. is a bar for iff either is a closed-closed bar for for some , or is an open-open bar for for some .
###### Proof.
We know . The first summand given by the finite intervals is isomorphic to a similar summand in the level set zigzag module ; see [6](Table 1, Type I). The second summand is isomorphic to , which by a result in [4] is isomorphic to where the open-open interval modules in generate and the closed-closed interval modules in generate . Then, the claimed result follows again from [6](Table 1, Type III and IV). ∎
Overview and main results. Let be a simplicial complex consisting of simplices that are linearly embedded in . Let denote the geometric realization arising out of this embedding. First, assume that is a pure -complex, that is, its highest dimensional simplices are triangles and all vertices and edges are faces of at least one triangle. The algorithm for the case when it has tetrahedra and possibly edges and vertices that are not faces of triangles follows straightforwardly from the case when is pure, and is remarked upon at the end. Another assumption we make for our algorithm is that the coefficient field of the homology groups is .
A function is called a height function if there is an affine transformation of the coordinate frame so that for all points with -coordinate being . Without loss of generality, assume that is indeed the -coordinate function and is proper, that is, its values on the vertices are distinct. The standard topological persistence of on is called the height persistence which we aim to compute. Theorem 1 says that we can compute the barcode of the height persistence by computing the same for the level set zigzag persistence using the same height function. Precisely, we first compute the barcode for from which we obtain a partial set of bars for and the complete set of bars for . This is achieved by maintaining a level set data structure and tracking a set of primary cycles in them as we sweep through along increasing . At the same time, we build a barcode graph that registers the birth, death, split, and merge of the primary cycles. We show that this can be done in time. The bars of are extracted from this graph again in time by adapting an algorithm of [1] to our case after a slight but important modification. According to Theorem 1, the closed-open and closed-closed bars of constitute a partial set of bars for . The open-open bars of , on the other hand, constitute a complete list of bars for the second homology module because the other summands for are trivial.
The rest of the bars of which are the open-open bars of (Theorem 1) are shown to be captured by the Reeb graph of on which can be computed in time [16]. We show that the basis cycles for the first and second homology groups can be computed as part of the level set persistence and Reeb graph computations.
###### Theorem 2.
Let be a simplicial complex embedded in with simplices. Let be a height function defined on it. One can compute the barcode for for , in time where is the number of simplices in . Furthermore, a set of basis cycles for , , can be computed in time where is the total size of the output cycles.
Similar statement holds for standard persistence.
###### Theorem 3.
Let be a simplicial complex embedded in with simplices. Let be a height function defined on it. One can compute the barcode for for , in time where is the number of simplices in .
## 3 Level set data structure
Let be the set of vertices of ordered by increasing -values, that is, for . Consider sweeping in the increasing order of -values. A level set , , viewed as a graph embedded in the plane , does not change its adjacency structure in any open interval . This structure, however, may change as the level set sweeps through a vertex of . Consequently, for every vertex , it suffices to track the changes when the level set jumps from the intermediate level to the level and then to the intermediate level where , and . All three level sets , , and are plane graphs embedded linearly in the planes , , and respectively. Let denote any such generic level set graph at a level , where the vertex set is the restrictions of the level set to the edges of and the edge set is the restriction of the level set to the triangles of . To avoid confusions, we will say complex edges and complex triangles to refer to the edges and triangles of respectively.
Level set graph and homology basis. We need to track a set of cycles representing a homology basis of to that of and then to that of as we sweep through the vertex . Consider any such generic level set graph representing . Primary and secondary cycles. [r] The embedding of in the plane produces a partition of into -dimensional faces, -dimensional edges, and -dimensional vertices. The faces are the connected components of . Let denote the collection of all -faces in this partition. A face has boundary cycle consisting of possibly multiple components, each being a cycle. We orient by orienting its boundary and denote it with . The orientation is such that has the face on its right. In Figure 3, the face has two boundaries, one around the outer curve (shown solid) and another around the inner circle (shown dotted). The unique face in that is unbounded plays a special role and is denoted .
###### Observation 3.1.
For a bounded face , there is a unique oriented cycle that bounds a bounded face of on its right. By definition, the unbounded face has no such . In the figure above, is the solid curve around outer boundary.
Because of the uniqueness of the cycles , we give them the special name of primary cycles. All other cycles are secondary. In Figure 3, the primary cycles are rendered solid and the secondary ones are rendered dotted. Recall that the elements of the first homology group are classes of cycles denoted for a cycle . It turns out that the classes of unoriented primary cycles form a basis for and thus tracking the primary cycles across the levels become the key to computing the level set zigzag persistence.
###### Proposition 4.
The classes of unoriented cycles form a basis of .
###### Proof.
We observe the following facts:
• The classes of unoriented primary cycles form a sub-basis of .
• where denotes the reduced zero-dimensional homology group.
• The faces in form a basis of .
For the first fact, observe that the set of such cycles are independent meaning that there is no unoriented primary cycle that can be written as the sum of other unoriented primary cycles. If it were true, let . Then, the boundary of is empty. But, that is impossible unless . Since , we have .
The second fact follows from Alexander duality because is embedded in the plane . The third fact follows from the definition of reduced homology groups.
Consider a map that sends each face to its unoriented primary cycle . This map is bijective due to Observation 3.1. Therefore, by the first and third facts, is isomorphic to the summand of generated by the classes of unoriented primary cycles. Indeed, this summand is itself since is isomorphic to by the second fact. ∎
The following Proposition complements Proposition 4. We do not use it, but remark about its connection to Reeb graphs at the end.
###### Proposition 5.
The -classes of unoriented secondary cycles form a basis of .
Representing level set graphs. Proposition 4 implies that we can maintain a basis of by maintaining the primary cycles alone. However, for realizing the zigzag maps that connect across the level sets (Eqn. 2.2), we need a different basis involving both primary and secondary cycles. For each bounded face , let be the boundary cycle for the face which is the -addition of the primary cycle with the secondary ones in . The next assertion follows from Proposition 4 immediately.
###### Proposition 6.
The classes form a basis of .
The importance of the boundary cycles in realizing the zigzag maps needed for the persistence module in Eqn. 2.2 is due to the following observation.
###### Observation 3.2.
For every and for every boundary cycle in the intermediate level , there are sum of boundary cycles and at the critical levels and respectively with so that the inclusions of , and into the interval space induce linear maps at the homology levels given by , .
By Proposition 6 and the above observation, the zigzag maps of the persistence module in Eqn. 2.2 can be tracked if we track the boundary cycles for each face. However, this requires additional bookkeeping for maintaining the primary and secondary cycles of a face together. Instead, we maintain each individual primary and secondary cycle independently being oblivious to their correspondence to a particular face though this information is maintained implicitly. Due to Proposition 4, it becomes sufficient to register the changes in the primary cycles for tracking the boundary cycles.
The primary and secondary cycles change as we sweep over vertices. Figure 1 illustrates some of these changes. A secondary cycle may split into two cycles one of which is primary and the other is not ( in Fig.), it may split into two secondary cycles ( in Fig.), or two primary cycles may merge (, in Fig.). Therefore, we need to maintain all oriented cycles in , and keep track of the primary ones among them.
We consider a directed version of where each edge is converted into two directed edges in that are oriented oppositely. The graph is represented with a set of oriented cycles that bound the faces in on right. These cycles are represented with a sequence of directed edges. Connection rules. [r] A vertex in either lies on a vertex , or in the interior of a complex edge in which case we denote it as the vertex . Any edge in is an intersection of the level set with a complex triangle , which we also denote as an edge . Let be any edge adjoining a vertex . We have two directed copies and of in . Assume that is directed away from and is directed toward .
We follow a connection rule for deciding the connections among the directed edges around to construct the cycles in as follows. Let and be a pair of directed edges, where the head of is the tail of . The directed path locally separates the plane around the meeting point of and . The region to the right of is called its right wedge, and the region to the left is called its left wedge. We have three cases for deciding the connections:
• has only one edge ( in Figure 1): connect to .
• has exactly two edges and ( in Figure 1): connect to , and connect to .
• has three or more edges ( in Figure 1): consider a circular order of all edges adjoining . Let be this circularly ordered edges around . For any consecutive pairs of edges , , determine if the right wedge of contains the edge . If so, connect to . If not, connect to .
The choice of our orientations and connections leads to the following observation:
###### Observation 3.3.
Let be any pair of directed edges in . They are consecutive directed edges on the oriented boundary of a face if and only if connects to by the connection rule around some vertex .
The observation above relates the directed cycles in with a local connection rule. We exploit this fact to update the cycles locally in our algorithm.
Cycle trees. The directed cycles in are represented with balanced trees that help implementing certain operations on them efficiently. We explain this data structure now.
A directed edge where or is represented with a node that has three fields; points to the complex triangle , and point to the complex edges and respectively where is directed from to . A cycle of directed edges is represented with a balanced tree , namely a 2-3 tree [2] where the directed edges of constitute the leaf nodes of with the constraint that the leaves of any subtree of represent a path (directed) in . The leaves of are joined with a linked list in the order they appear on the directed cycle . A pointer in a leaf node implements this link list. The node also maintains another pointer to access the previous node on the linked list in time. However, it is important to keep in mind that it is the pointers that provide the orientation of the cycle . Furthermore, the last node in both linked lists connected by and pointers respectively is assumed to connect to the first one. This creates the necessary circularity without actually making the list circular. We denote the linked list of leaves of a tree as . The 2-3 trees built on top of the paths support the following operations.
find(): returns the root of the tree belongs to.
split(): splits a tree into two trees and where is the sublist of that contains all elements in before , and is the sublist that contains all elements in after and including .
join(): takes two trees and and produces a single tree with as the concatenation of and in this order.
permute(): makes the first node in the cycle represented with . It is implemented by calling split() that produces and , and then returning join(,).
insert(): inserts the element after in where find().
delete(): deletes from where find().
All of the above operations maintain the trees well balanced allowing traversal of a path from a leaf to the root in time where is the total number of elements in the lists of the trees involved. This in turn allows each of these operations to be carried out in time. Using these basic operations, we implement two key operations, splitting and merging of cycles.
splitcycle(): This splits a directed cycle into two. A cycle may get first pinched and then splits into more cycles as we sweep through a vertex. This operation is designed to implement this event. Given a tree , it returns two trees and where represents the path from to in the directed cycle given by , and represents the path from to in the same cycle. See Figure 2, bottom row.
It is implemented as follows: Let permute(). Call split() which returns two trees and as required.
mergecycle(): This merges the two cycles that and belong to. The new cycle has after and after . This is implemented as follows: Let find() and find(). Let permute() and permute(). Then, return join().
## 4 Updating level sets
Now we describe how we update the graph to and then to . As we sweep through , only the cycles in these graphs containing a vertex on a complex edge with as an endpoint may change combinatorially. We only update the cycles for combinatorial changes to make sure that the combinatorics of the level set graphs are maintained correctly though their geometry is updated only when needed to infer the correct adjacencies. This allows us to inspect only simplices where is the number of simplices adjoining in . Summing over all vertices, this provides an bound which gets multiplied with the complexity for the tree operations that we perform for each such simplex. Also, local circular sorting of edges around each vertex and complex edges connected to it accounts for time in total.
Primary cycle detection. The cycles in that change combinatorially may experience splitting, merging, edge contraction, edge expansion, or a combination of such events. Specifically, during splitting and merging, new cycles are generated which need to be characterized as primary or not. Figure 2 illustrates two cases of a secondary cycle splitting. Two similar cases arise for the primary cycle splitting. For merging also we have four cases mirroring the splitting case. It turns out that we can determine if the new cycles are primary or not by the orientations of the edges around the ‘pinching’ vertex if we know the type (primary or not) of the original cycles. We explain this for the case of splitting.
Let be a cycle in which splits at . Let and be any two non-consecutive directed edges in that meet at in . Assume that we know that is secondary. The case when is primary is similar. We need to distinguish the case when one of the two new cycles nests inside the other. This can be checked in time by determining if the right wedge of contains or not. If not, both new cycles remain secondary. Otherwise, we have a nesting, and exactly one of the two new cycles becomes primary. We can determine again which of the two becomes primary in time. For this consider a ray with tail at and entering the left wedge of . If this ray enters the left wedge of , we declare the new cycle containing and to be secondary and the other cycle containing and to be primary. If the ray enters the right wedge, we flip the assignment for the type of the two new cycles.
With these local checks, we design the two routines below that decide the type of the new cycle(s) in both the splitting and merging cases assuming that we know if the input cycle(s) are primary or not.
splitPrim(,,): This routine assumes that indicates if the cycle to be split which contains and is primary (true or false), and returns a pair of booleans where is true if and only if the new cycle containing is primary.
mrgPrim(,,,): This routine assumes that the input boolean variables indicates if the cycle containing is primary, and returns a boolean variable which is true if and only if the new merged cycle is primary.
Now we describe the actual updates of the graphs when the sweep goes through a vertex . For convenience, we designate a complex triangle as top, middle, or bottom if it has as the lowest, middle, or highest vertex respectively w.r.t. the height . Similarly, a complex edge is called top, or bottom if it has as the lowest or highest vertex respectively. As we continue with the sweep, we keep on recording the birth, death, splitting and merging of primary cycles by creating a barcode graph. Current primary cycles are represented by current edges in the barcdoe graph whose one endpoint is already determined, but the other one is yet to be determined. The nodes in the barcode graph are created when a primary cycle is born, dies, splits, or merges with another cycle. It is important to note that the nodes of the barcode graph are created only at the intermediate levels . Each tree maintains a pointer that points to a current edge in the barcode graph if its cycle is primary. Otherwise, this pointer is assumed to be a null pointer. Additionally, we assume that there is a boolean field which is set true if and only if represents a primary cycle. The barcode graph at level is denoted . As we move from level to the next level , we keep updating this barcode graph by recording the birth, death, splitting and merging of primary cycles and still denote it as till we finish processing level at which point we denote it as .
Updating to . The combinatorics of change only by the edges where is a bottom or middle triangle. If the edge has both vertices on bottom complex edges, then is contracted to in . Otherwise, the edge remains in , but its adjacency at the vertex which becomes in changes. Also, in both cases classes in may die or be born. We perform the combinatorial changes and detect the birth and deaths of homology classes as follows:
Contracting edges: When we contract edges, a cycle may simply contract and nothing else happens. But, we may also detect that a primary cycle of three edges is collapsed to two directed edges corresponding to a single undirected edge. This indicates a death of a class in which occurs entering the level but not exactly at . So, we operate as follows.
Let be any bottom complex triangle for , and let and be two directed edges associated with . Let find() and find(). For , we call delete(). If and has two leaves, we terminate the current edge pointed by with a closed node at level in and remove completely. The closed node indicates that the cycle dying entering the level is still alive at the level .
Cycle updates: The edges of a cycle in can come together at to create new cycles. After the edge contractions, the only edges that we need to update for possible combinatorial changes correspond to middle complex triangles. Let be such a triangle and let and be its edges that are top and bottom edges for respectively. For each directed edge with , we update or if originally we had or respectively.
Next, we update the cycles that may combinatorially change due to splitting or merging at , and also record new births as a result. We consider every directed edge so that the triangle is a middle triangle and determine a circular order of their undirected versions around . For every such directed edge , we determine its pair directed edge using the connection rule that we described before. Observe that plane embedding of the level set graph is used here. Actually, the lack of such canonical ordering of edges around a vertex for level set graphs becomes the roadblock for extending this algorithm to persistence of functions that are not heights. Let find() and find(). We have two cases: the splitting case when (see and in Figure 3) and the merging case when .
Splitting Case, : If , the cycle containing and and represented by does not change and we do nothing. Otherwise, the cycle splits into two new cycles whose type needs to be determined. So, we call splitPrim(,,) which returns a pair of boolean values indicating if the two new cycles are primary or not. We split to create the representations of the two new cycles. But, this operation destroys whose type (primary or not) and barcode pointer are needed for assigning the same for the two new trees. So, we save and first, and call splitcycle(,,) which returns two trees and representing the two cycles. Geometric constraints allow only the following two cases:
Case(i): : A new primary cycle is born at the level . This is an open-ended birth at the level because the cycle exists at the level but not at the level . If , we set and , . Then, we set where is a current edge created with an open end at level in .
Case(ii): : A new primary cycle is born at the level due to a split of the cycle represented by the saved pointer . We set , for , and call splitbar(,) which splits at level and returns two current edge pointers and . We set and .
Merging Case, : two cycles and represented by and respectively merge to become one. As before, we first store aside the type of and and associated current edge pointers by setting and for . Next, we call mergecycle(,) which merges the two cycles containing and and returns a tree representing this new cycle, say . A call to mrgPrim(,,,) returns a boolean variable which is true if and only if is primary. Again, we have only the following two cases.
Case(i): . In this case no primary cycle dies, but the new cycle remains primary. So, no current edge is terminated and the current edge associated to the primary cycle among and is continued by . If , we set , .
Case(ii):: No primary cycle dies and the new cycle is also not primary. We set and .
Updating to . To update to , we need to create directed edges corresponding to top triangles, that is, the complex triangles with as the bottom vertex. These new edges
change the combinatorics of in four ways: they may (i) expand the existing cycles without creating or destroying any primary class, (ii) create a new cycle giving birth to a new class, (iii) split a cycle pinched at (it turns out that no new class is born in this case), (iv) merge two cycles meeting at ; in this case, a primary cycle dies. Two cycles containing directed edges and in in Figure 3 get merged into one cycle in . The details of the merge is shown in Figure 4. It is preceded by an insertion of a sequence of edges that connect and . Similarly, another sequence connects and .
Expanding cycles: We iterate over all directed edges corresponding to the middle triangles. Let be any such directed edge in . The directed edge belongs to a unique cycle in the directed graph . Starting from , we aim to create the missing edges in . For this, we create a routine nextLink() that takes a directed edge and creates all missing directed edges in that lie between and the next directed edge with being a middle triangle.
nextLink(): Consider the complex edge and the circular order of directed edges of around . In time we determine the adjacent directed edge using the connection rule described before. If does not exist already, we create a directed edge for and insert it into the tree containing by calling insert(,). Replacing the role of with , we continue. If is a middle triangle, we stop and return .
To complete updating containing the directed edge , we call nextlink() which returns, say . Let find() and find(). We have two cases:
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http://www.maa.org/publications/maa-reviews/working-analysis | # Working Analysis
###### Jeffery Cooper
Publisher:
Elsevier
Publication Date:
2005
Number of Pages:
663
Format:
Hardcover
Price:
99.95
ISBN:
0-12-187604-7
Category:
Textbook
[Reviewed by
Warren Johnson
, on
01/27/2006
]
The preface begins "Over the years, I have become more and more dissatisfied with our advanced calculus course [at the University of Maryland]. In most books used for this type of course, theorems are proved to prove more theorems. An 'application' of a theorem is either a trivial calculation or a piece of the proof for another theorem. There are no examples or exercises that use the methods of analysis to solve a real problem. The traditional advanced calculus course has little or no contact with the world outside of mathematics." After a paragraph about computing (Cooper is in favor of it), it continues "My goals in writing this book are to teach the techniques and results of analysis and to show how they can be applied to solve problems. I have gone outside of the usual range of applications in physics to include examples from biology, sociology, chemistry, and economics."
When I first started browsing through the book it occurred to me that I might find the preface to be the most interesting part, but the more closely I looked at it the more I came to admire it. The author's taste in analysis is not quite orthogonal to my own, but the projections of each on the other are not large. Cooper does what he does carefully and well, with a few exceptions, and I always like to see a mathematics book with a point of view, even one I do not share.
The book is divided into two parts, one for each semester. Part I presents the one-dimensional theory, while Part II treats functions of several variables. Most of the applications are in Part II, but Part I also has some unusual features. Chapter 1 discusses supremums and infimums, inequalities, and induction. The section on induction seems a little skimpy to me, but the author does derive one of his favorite tools there, the Bernoulli inequality: (1 + x)n is at least as large as 1 + nx when n is at least 1 and x is at least -1.
Chapter 2 is about sequences. In Example 2.3 Cooper uses Bernoulli's inequality to get an upper bound on nrn for 0n tends to 0 as n tends to infinity. In Example 2.5 he shows that r+2r2+...+nrn is bounded for the same range of r's, by using the bound from Example 2.3 and the sum of a finite geometric series (which he does not derive). Since this sequence is clearly increasing with n, he concludes that it converges, "although we do not know the value of the limit." He does not mention that it could be derived by differentiating the geometric series formula, either because he has not discussed differentiation yet, or because he is not interested in the question, or possibly both. This might at least have made a good exercise, and there are many similar missed opportunities in the book.
A nice feature in chapter 2 is the arithmetic-geometric mean iteration, and Example 2.8 is also good. It contains a rare instance of the author using something that he doesn't prove until later, namely that |sin x – sin y| is at most |x – y|, which he wants to get from the mean value theorem. One might set x = u + v and y = u – v instead, to reduce to the statement that |(cos u)(sin v)| does not exceed |v|, and then recall the geometric proof that |sin x| does not exceed |x| that is usually done in first semester calculus. This would also come in handy in Example 3.2. Although the student is supposed to have had three semesters of calculus, Cooper actually assumes very little specific knowledge of it, perhaps because of his years of teaching experience. The finite geometric series, which one might ascribe to precalculus instead, is a rare exception. Another, in Example 1.4, is the sum of the first n positive integers.
There are two minor but significant errors in chapter 4, on differentiation. (Chapter 3 is on continuity.) In Example 4.3 the statement that f'(x) > 0 for all x is obviously false at x= 0, and if it were true then the last sentence in the example would make no sense. In exercise 11 four pages later v' should be sin u, not –sin u. (As exercise 12 shows, it's really –(–sin u).) A highlight of this chapter is an unusually careful treatment of L'Hôpital's rule, including not only the 0/0 case (via Cauchy's extension of the mean value theorem) but also infinity/infinity, but there are only a few exercises on it.
Chapter 5 is on higher derivatives and polynomial approximation. I was happy to see Lagrange interpolation in section 5.3. Cooper says that one can't get closed form solutions of the pendulum equation x'' + sin x = 0, and thereby passes up an opportunity to mention elliptic functions. He is less than usually careful in Example 5.4, where he should say at the outset that lambda is positive and, more importantly, he should have u'(0)= 0 rather than u(0)= 0.
Cooper discusses convex functions in section 5.4, but neglects to say that a calculus student would think of them as "concave up". What a calculus student would call "concave down" he calls "concave" in this section and "convex down" on page 207. In the absence of any explanation, I would find this pretty annoying if I was a student.
Cooper has much more on numerical analysis, a subject that has always left me cold, than most analysis textbooks. Chapter 6 is on solving equations numerically. Greater use is made of Lipschitz continuity here than in most books, but Definition 6.1 and the sentence that follows it only defines a Lipschitz constant, not the Lipschitz constant, and Definition 10.5 (the multivariable case) has the same problem.
Chapter 7 is on integration, and about 23% of it is on numerical methods. Perhaps the biggest divergence between Cooper's tastes and mine is that he does not like to do integrals, although he is happy to estimate them. (He is also not very interested in the history of his subject.) Problem 7 in section 7.4 is substantial: Cooper defines a function f(c) as the integral on (0,1) with respect to x of 1/(1 + exp(cx)), and the point is to find, by numerical methods, a positive value of c such that f(c) = 1/4. No hint is given that the problem can be solved exactly. The integral evaluates easily after multiplication by exp(-cx)/exp(-cx), and one then has to solve u5 – 2u4 + u = 0, where u = exp(c/4). Dividing out the extraneous roots u = 0 and u = 1 we are left with u3 – u2 – u – 1 = 0, which has one real root, evidently positive, that may be found by whatever method you like for solving a cubic equation. Similarly, the double integral in Example 14.5, about which Cooper says only that it "can be estimated easily using a two-dimensional version of Simpson's rule", can be found exactly by elementary methods (granted that it takes a while). I understand that integration in closed form is not one of the central concerns of modern analysis, but why not at least add this as an exercise? Even if the author would never assign it, someone else might.
Section 7.5 is on improper integrals. Part e of exercise 1 is very nice, and the hint in the back is fine (approximate the integrand near the doubtful endpoint by Taylor's theorem), but the integral can be evaluated exactly: it equals √2log(tan u), from 0 to π/8 (after substituting x = (π/2) – 4u), so it is infinite. Exercise 5 is on the gamma function. Part d uses the fact that the integral of exp(–x2) on the real line is the square root of π, but gives no hint of a proof. Cooper actually does prove this in section 14.6, but there is no reference from one section to the other.
Except for an appendix on transcendental functions, Part I concludes with chapter 8, on series. I found this chapter to be dull, and this is the main reason why I had a negative first impression of the book. There is a brief treatment of Euler's constant on page 242, but the first displayed formula there is wrong (it could be fixed by inserting two sets of parentheses), and I would rather get Euler's constant from the standard integral test argument, as in section 8.1. Applied to 1/x on (1,n), it shows that
1 + 1/2 + … + 1/(n-1) > log n > 1/2 + 1/3 + … + 1/n for n>1
and that these expressions get farther apart (though not much farther) as n increases. This not only gives the result of exercise 7 on page 230, it implies that
1 + 1/2 + … + 1/n – log n is positive and decreasing for n>0, and
1 + 1/2 + … + 1/n – log(n+1) is positive and increasing for n>0.
Obviously the former is larger, and evidently they have the same limit, which is Euler's constant gamma. It follows that
log n + γ < 1+1/2+...+1/n < log(n+1) + γ for n>0,
and this is a better result than that of exercise 7 except when n = 1. (This is also a good excuse to mention Julian Havil's beautiful book Gamma.)
The definition of interval of convergence on page 246 is sloppy. It could be improved by moving forward from page 247 a sentence about convergence at the endpoints.
The Weierstrass approximation theorem is mentioned already in chapter 5, but not proved until section 15.2. In the course of the proof Cooper derives the lower bound 2/(3√k) for a definite integral by using Bernoulli's inequality, which is good enough for his purposes, but he could have reduced the integral to essentially part c of exercise 9 in section 14.5 by substituting x= cos u. This gives the integral exactly, and by estimating this answer we could get the better bound 2/√(2k+1).
Something on the front cover bothers me. One of the key features claimed for the book on the back cover is "an informal and lively writing style." Cooper is, among mathematicians, a reasonably good writer, as the quotes from the preface show — not Halmos, Hardy, Rota or Truesdell, but comfortably above average, 80th percentile maybe. For a textbook author I would say he is about average. (Perhaps the worst writing in the book is in Example 1.4, where "formula" is used 4 times in the space of 19 words.) But "lively" is not the word that best describes the virtues of his style; "clear" is better. In a textbook, I would rather have clear: for example, Sylvester (who never wrote a book) is great fun to read in small doses, but he is often so excited about his ideas that he never gets around to explaining them very well.
"Informal" is nearly the last adjective I would use to describe the author's style. Proofs are always clearly set off, whereas in a less formal style they might often blend in smoothly with the rest of the exposition. (Many students prefer a little formality here.) They typically conclude with "The proof is complete" (11 times in the first 3 chapters) or "The theorem is proved" (14 times in the next 5 chapters). When they don't, it is often because Cooper (sensibly enough) proves only half of a theorem and says that the other half is similar. He is fond of certain constructions that are used almost exclusively in mathematical writing, such as "by an appeal to" Theorem X. I like this phrase, when used sparingly, but I wouldn't call it informal. He also typically writes "the Bernoulli inequality" or "the Euler constant" or "the Fubini theorem" or "the example of Runge" instead of using the possessive. He even refers to his own earlier book on partial differential equations as "the book of Cooper". The style is not as formal as it could be — when Cooper makes a remark, he doesn't say "Remark" — but I would say it is rather more formal than not.
This finally brings me back to the front cover, a (not unattractive) montage of numbers, text and figures. The numbers at the top come from page 414, the text below it from page 506, and the horrendous formula at the bottom is (11.16) from page 341. (At least two of the figures are also in the book, but I didn't notice the others.) The text from page 506 has been cut off at the left, and consequently Cooper's "Then by the Fubini theorem, we have" (followed by blank space and a display) appears as "by Fubini, we have" etc. I had not heard of the author before I read his book, but I am pretty sure that he would never write "by Fubini" in anything he intended to publish. I wouldn't either. Since the original sentence would fit, what am I to conclude, but that it was changed in an attempt to foster the illusion of an informal writing style?
On the other hand, I didn't find any applications to sociology in the book, so in that regard the back cover is more honest than the preface.
There is some small effort at informality in Cooper's treatment of epsilons and deltas. He sneaks in an epsilon already on pages 10-11, which I think is a good idea. Epsilons come in more formally with N's at the bottom of page 30, in the definition of a convergent sequence, and on page 35, he writes "Now we are ready for the ε-N drill" in the middle of a proof. By my count, this is the ninth ε-N argument since the top of page 31, so it was a little strange to see this phrase all of a sudden. The sentence "Let ε > 0 be given" occurs in most proofs involving ε from page 37 on, although Cooper seems to have tried to avoid it before that. Not infrequently, it is the first sentence of the proof.
Epsilon appears with delta for the first time at the top of page 61: "An alternate, equivalent, formulation of the limit of a function is stated in terms of the dread epsilons and deltas." This is not one of Cooper's better sentences, but even so, one is heartened; one expects (particularly if still looking for an informal style) that more than the usual amount of care will now be taken to motivate and explain these challenging ideas. But no, Cooper just dives right in. He is no worse than the average analyst at explaining epsilons and deltas, but his few attempts to be more engaging are so half-hearted as to be practically worthless.
The minimal amount of topology that one usually sees in analysis (open and closed sets, compactness, connectedness) occurs here in chapter 9, at the beginning of Part II, so if one were using the book for a one semester course then one might want to push this material forward.
Countability is not in the book at all. As I write this, I am just starting to teach out of Stephen Abbott's Understanding Analysis , which received an enthusiastic review from Steve Kennedy here . Except for countability, the content of Abbott's first chapter and Cooper's is almost identical. While Cooper is not without his advantages (for example, Bernoulli's inequality), these chapters are a marvelous case study of the difference between competence and excellence in textbook writing.
Chapter 10 is on the derivative in several variables. Chapter 11 is on solving systems of equations, and contains the Contraction Mapping Theorem as well as the Inverse and Implicit Function Theorems. Chapters 12 and 13 are on optimization, with an unusually careful treatment of Lagrange multipliers in chapter 13; I found this to be the most interesting material in Part II. Chapter 14 is on integration in several variables, and chapter 15 on applications of integration to differential equations. From chapter 11 on, much of the book discusses applications.
This is not a book that I would be likely to select for my analysis course, but I could live with it, and there is something to be said for using a book with a different emphasis than one's own. Real analysis is such a vast subject that one has to pick and choose even in a two semester course, and Cooper has made choices that are interesting and defensible. If you found yourself in strong agreement with the quotations in my first paragraph, then his book is definitely worth looking at.
Warren Johnson ([email protected] ) is visiting assistant professor of mathematics at Connecticut College.
Preface
Part I:
1. Foundations
1.1 Ordered Fields
1.2 Completeness
1.3 Using Inequalities
1.4 Induction
1.5 Sets and Functions
2. Sequences of Real Numbers
2.1 Limits of Sequences
2.2 Criteria for Convergence
2.3 Cauchy Sequences
3. Continuity
3.1 Limits of Functions
3.2 Continuous Functions
3.3 Further Properties of Continuous Functions
3.4 Golden-Section Search
3.5 The Intermediate Value Theorem
4. The Derivative
4.1 The Derivative and Approximation
4.2 The Mean Value Theorem
4.3 The Cauchy Mean Value Theorem and l’Hopital’s Rule
4.4 The Second Derivative Test
5. Higher Derivatives and Polynomial Approximation
5.1 Taylor Polynomials
5.2 Numerical Differentiation
5.3 Polynomial Inerpolation
5.4 Convex Funtions
6. Solving Equations in One Dimension
6.1 Fixed Point Problems
6.2 Computation with Functional Iteration
6.3 Newton’s Method
7. Integration
7.1 The Definition of the Integral
7.2 Properties of the Integral
7.3 The Fundamental Theorem of Calculus and Further Properties of the Integral
7.4 Numerical Methods of Integration
7.5 Improper Integrals
8. Series
8.1 Infinite Series
8.2 Sequences and Series of Functions
8.3 Power Series and Analytic Functions
Appendix I
I.1 The Logarithm Functions and Exponential Functions
I.2 The Trigonometric Funtions
Part II:
9. Convergence and Continuity in Rn
9.1 Norms
9.2 A Little Topology
9.3 Continuous Functions of Several Variables
10. The Derivative in Rn
10.1 The Derivative and Approximation in Rn
10.2 Linear Transformations and Matrix Norms
10.3 Vector-Values Mappings
11. Solving Systems of Equations
11.1 Linear Systems
11.2 The Contraction Mapping Theorem
11.3 Newton’s Method
11.4 The Inverse Function Theorem
11.5 The Implicit Function Theorem
11.6 An Application in Mechanics
12.1 Higher Derivatives and Quadratic Approximation
12.2 Convex Functions
12.3 Potentials and Dynamical Systems
12.4 The Method of Steepest Descent
12.6 Some Optimization Problems
13. Constrained Optimization
13.1 Lagrange Multipliers
13.2 Dependence on Parameters and Second-order Conditions
13.3 Constrained Optimization with Inequalities
13.4 Applications in Economics
14. Integration in Rn
14.1 Integration Over Generalized Rectangles
14.2 Integration Over Jordan Domains
14.3 Numerical Methods
14.4 Change of Variable in Multiple Integrals
14.5 Applications of the Change of Variable Theorem
14.6 Improper Integrals in Several Variables
14.7 Applications in Probability
15. Applications of Integration to Differential Equations
15.1 Interchanging Limits and Integrals
15.2 Approximation by Smooth Functions
15.3 Diffusion
15.4 Fluid Flow
Appendix II
A Matrix Factorization
Solutions to Selected Exercises
References
Index | {"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": 0.873079240322113, "perplexity": 695.2159657792331}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-15/segments/1397609536300.49/warc/CC-MAIN-20140416005216-00648-ip-10-147-4-33.ec2.internal.warc.gz"} |
https://www.physicsforums.com/threads/how-do-you-find-the-derivative-of.172736/ | # How do you find the derivative of
1. Jun 3, 2007
### AznBoi
How do you find the derivative of $$f(x)=5$$?? I got 0/0... but shouldn't there be no derivative then? I know that I can visualize the function f(x) because it is just a straight horizontal line and there is no slope at any point on the line... but how do you solve it algebraically so that it actually makes sense?
Last edited: Jun 3, 2007
2. Jun 3, 2007
### morphism
How did you get 0/0? The derivative of a constant function is 0. This is plausible because the slope isn't changing (and the derivative is the rate of change of slope). You can prove this pretty easily using the limit definition of a derivative.
3. Jun 3, 2007
### theperthvan
No, it's plausible because the slope is flat. What about a linear function, y=mx+c? The slope isn't changing, it's constant...but the deriv isn't 0.
So the bit in brackets should say "and the second derivative is the rate of change of slope"
4. Jun 4, 2007
### prasannapakkiam
D=capital delta
d=lowercase delta
y=5
therefore: y+Dy=5
therefore: y/Dx + Dy/Dx = 5/Dx
y=5, hence: y/Dx=5/Dx
therefore: Dy/Dx = 0
taking the limits: Dx-->0
therefore: dy/dx = 0 ==>
This is the algebraic proof you wanted...
5. Jun 4, 2007
### prasannapakkiam
I see that you got 0/0 probably using the l'hopital's rule. when you get (5-5)/h, it becomes 0/h which equals 0. You only take the limits once all simplifications are complete...
6. Jun 4, 2007
### VietDao29
I think you have messed up a little. A pure "0" differs from a fake "0" a lot. A pure "0" means that it's really 0, while a fake one means that it only tends to 0, it's not 0, it just tends towards 0.
You should notice that:
$$\lim_{x \rightarrow \infty} 0 x = 0$$, since 0 multiply by any number is 0, even if x is large.
Whereas:
$$\lim_{x \rightarrow \infty} \frac{1}{x} \times x = \lim_{x \rightarrow \infty} 1 = 1$$
Some may argue that when x tends to infinity, 1 / x tends to 0, so 0 * anything would be 0. But that's not true. 1 / x only tends to 0, it's not 0. So, you cannot claim that.
The same thing apply here:
$$f'(x) = \lim_{h \rightarrow 0} \frac{f(x + h) - f(x)}{h} = \lim_{h \rightarrow 0} \frac{5 - 5}{h} = \lim_{h \rightarrow 0} \frac{0}{h}$$
Now the numerator is a pure 0, a real one. It's 0, so 0 divided by anything will give out 0, so f'(x) = 0.
Can you get it? :)
7. Jun 4, 2007
### cepheid
Staff Emeritus
I don't know if "real zero" and "fake zero" are very good terms to use. But what VietDao is describing is essentially the difference between "exactly zero" and "really small". Anyway, the good thing about his post is that he proves that the derivative of a constant function is zero using the *definition* of a derivative. It just occured to me to try using the power rule and see what happens:
$$f(x) = C$$
C is a constant.
$$\frac{d}{dx}C = \frac{d}{dx}Cx^0 = 0(Cx^{-1}) = 0$$
The result is zero for all x. Is this valid?
8. Jun 4, 2007
### C0nfused
I will just add that if you look at the definition of the limit (-> http://mathworld.wolfram.com/Limit.html ) it says "for any ε>0 there exists δ .... with 0<|x-x1|<δ" and not just |x-x1|<δ. So we don't care about the case x=x1. This actually shows the "fake" and "real" zero concept. If you use this definition and the definition of the derivative, then you can prove, as VietDao29 did, that the derivative of the constant funtion is 0
C=Cx^0 for x<>0. So, if you prove the rule for the derivative of x^n with n natural (zero in our case), you will still have to check the derivative at x=0 using the definition
Last edited: Jun 4, 2007 | {"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": 0.9175750613212585, "perplexity": 911.9039928561234}, "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/1476988721606.94/warc/CC-MAIN-20161020183841-00430-ip-10-171-6-4.ec2.internal.warc.gz"} |
https://tex.stackexchange.com/questions/254749/texmaker-compiling-a-parent-file | Texmaker: compiling a parent file
I'm writing up some old math notes that closely follow a textbook, and I have a tex file for each section of the book I worked through. I keep all these section files in a directory `tex`, and then on the level of that same directory I have a `main.tex` file that `\input`s each section file into a book.
I'm using Texmaker to edit the individual section files. Is there some way to configure Texmaker to compile `main.tex` whenever I hit Quick Build while working on a section file, and then have the resulting `main.pdf` file show up in the PDF viewer and jump to the section I was working on?
2) Select from the top menu bar, `Options`, and then `Make Current Document the Master Document` | {"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": 0.9926027655601501, "perplexity": 1278.3994200221268}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-35/segments/1566027330962.67/warc/CC-MAIN-20190826022215-20190826044215-00208.warc.gz"} |
http://nrich.maths.org/7052/solution | ### Trigger
Can you sketch this tricky trig function?
How would you design the tiering of seats in a stadium so that all spectators have a good view?
# Geometric Trig
##### Stage: 5 Short Challenge Level:
There are many ways to create the solution as it is something like a jigsaw. We used this method
1. Mark the right angles (use the fact the a radius and tangent at a point are at right angles)
2. Make all occurrences of the angle a
3. Mark the unit length
The diagram then becomes
To work out all of the areas we need to decide what unit of measurement the angle $a$ is in. Of course, we choose radians: there are $2\pi$ radians in a circle. We will also need to know the formula for the area of a circle and the area of a triangle. In this case, the areas are given as
The two largest areas are equal for around 0.40523 radians (23.2 degrees). | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 0.9095545411109924, "perplexity": 560.6043119978015}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-26/segments/1498128323908.87/warc/CC-MAIN-20170629103036-20170629123036-00223.warc.gz"} |
https://www.physicsforums.com/threads/limits-when-finding-area-of-polar-curves.233259/ | # Limits when finding area of polar curves.
1. ### apples
167
The problem is related to polar curves. most of the topics i need to do are easy (finding the slope, finding the area etc.)
What i'm facing problems with is that when I find the area, I don't know how to find the limits.
1. The problem statement, all variables and given/known data
Sample problem:
Find the area of the region in the plane enclosed by the cardioid r=4+(4sinθ)
2. Relevant equations
b
A = (1/2)(r^2) dθ
a
3. The attempt at a solution
Graphing the curve is no biggie, I use my calculator. The problem is when I use the equation to find the Area, I don't know what the interval is I don't know 'b' and 'a' are.
I have a calculus book, an AP calculus book, and a pre-calculus book. None of the books tell you how to find the interval. In their solved questions, they tell you that in the solution. The problem I wrote above is an example also.
They tell you the interval is (for the q above) 0 to 2π(pi)
but how do i know that what about other questions.
Another example:
Find the area inside the smaller loop of the limacon r=1+(2cosθ)
Here they say the limits are (2π(pi))/(3) to (4π(pi))/3
The exact explanation to this in the book is:
(for first question) "Because r swoops out the region as θ goes from 0 to 2π(pi), these are our limits of integration.
(For 2nd q) Because in the inner loop, r sweeps out the region as θ goes from (2π(pi))/(3) to (4π(pi))/3, these are our limits of integration.
But why!? What does r swoops mean. How do I know the limits!?
2. ### Hootenanny
9,679
Staff Emeritus
Okay let's take your first example. So looking at your sketch you should be able to see that the cardioid is a closed path, which means that there are no breaks in the line. Now imagine that you are stood on the point where the cardioid intersects the x-axis and you have hold of a piece of string that is anchored at the origin. Now imagine walking around the cardioid whilst keeping hold of the string, every time you take a step you measure the angle between the x-axis and the string and write it down. Once you get back to where you started, what will be the range of the angles that you measured?
3. ### apples
167
Thank you for the help
I think it would be 0-2π(pi) in this case.
Can we make this a bit quick. If I explain the reason, you people might be disgusted, but...
...I am serious this is very urgent.
I need to have mastered this by today night at most.
4. ### Hootenanny
9,679
Staff Emeritus
I'm guessing that you've got your final tomorrow .
You're correct, $\theta$ can vary between zero and $2\pi$, this is the angle swept out. Hence, your angular limits are zero and $2\pi$. Do you follow?
Now for the radial limits. The question states that the area is bounded by the cardioid. So now imagine that the cardioid is a fence and you are inside it, you're allowed to walk anywhere inside the fence, but you can't jump over it. Now, the radius is simply the distance from the origin, so given the boundary, what is the minimum distance you can be from the origin?
5. ### apples
167
No, I have my AP Calculus BC exam day after tomorrow.
I follow the first part. The angle "sweeps out" creating an interval. (i know this might be incorrect terminology).
The second part, I don't fully understand. I understand that the area is bounded/enclosed by the cardioid.
The radius is the distance from the origin, so the minimum distance can be zero? right? because the cardioid intersects at the origin?
OK, so radius=distance, what's the interval?
Last edited: May 5, 2008
6. ### Hootenanny
9,679
Staff Emeritus
Sounds good to me.
Correct! So we now know that the lower limit is r=0. What about the upper limit? What's the further distance that you can go away from the origin?
7. ### apples
167
At the y-intercept? Other than the origin?
EDIT: in angular limits, where do we start from, the x-axis? always?
Last edited: May 5, 2008
8. ### Hootenanny
9,679
Staff Emeritus
Forget the mathematics for the moment. Your in a fenced area, what is the furthest you can get away from the centre of the area? What stops you getting out?
9. ### apples
167
I don't know. The farthest from the origin is the top pointy corner of the cardioid
10. ### Hootenanny
9,679
Staff Emeritus
If the area is bounded by the curve $r = 4\left(1+\sin\theta\right)$ then you know that $r < 4\left(1+\sin\theta\right)$ so the upper bound is...
11. ### apples
167
I don't understand.
The curve?
Last edited: May 5, 2008
12. ### Hootenanny
9,679
Staff Emeritus
What exactly don't you understand?
Correct
13. ### apples
167
now what are the limits?
14. ### Hootenanny
9,679
Staff Emeritus
I've just told you,
15. ### DavidWhitbeck
352
Putting aside the multivariable calc ways of doing it ($$r dr d\theta$$) which is what some people seem to be discussing here-- in AP calc you will generally see rose type curves and you set the bounds of integration by finding the values of $$\theta$$ where $$r=0$$ (because the curves you usually see loop back on itself at the origin) and integrating that gives you the area of one leaf, and you multiply by the number of leaves to get the total area.
The more general method is to find the points where the curve intersects itself, and use that to find the bounds of integration.
16. ### apples
167
OK, I'm confused.
I understood angular and radial limits.
b
A = ∫ (1/2)(r^2) dθ
a
But which one do I use in the formula above? What happens to the other limit.
If I make r=0, and find the value of θ, which limit will i get?
How do I find the other one.
Could you please explain in a bit more detail, and tell me the process.
I just simply need to know one thing.
How to find the limits when finding the area of polar curves.
17. ### DavidWhitbeck
352
If you find $$\theta$$ there will be multiple solutions. You take the smallest for a, and the second smallest for b so that $$a < b$$ and there are no other zeroes inbetween them. For example if you had $$r=a\sin4\theta$$ then the values are $$\theta=0,\pi/4,\pi/2,\hdots$$ and one loop would be from $$\theta=0$$ to $$\theta=\pi/4$$. And there are eight such leaves for that rose.
18. ### Hootenanny
9,679
Staff Emeritus
Your area integral isn't correct. To find the area enclosed by a curve one needs to evaluate,
$$A = \iint_A dA$$
Notice that the integral is a double integral. So in cylindrical coordinates,
$$A = \iint_A rdrd\theta$$
Now for the limits. Notice that the radial limits depend on $\theta$ and therefore, we must evaluate the radial integral first,
$$A = \int^{2\pi}_{0}\left[ \int^{4\left(1+\sin\theta\right)}_0 r dr\right]d\theta$$
Does that make sense?
19. ### DavidWhitbeck
352
I have to say that it's standard to cover area of polar curves *before* doing area integrals in general. Starting with $$\int \frac{1}{2}r^2 d\theta$$ is what the OP is listing because that's what he has in his textbook. Forcing him to evaluate the area as a double integral is probably confusing because he hasn't learned about double integrals yet. And throwing in the term "cylindrical coordinates" is even worse-- they don't see that jargon until multivariable calculus.
My point is that the standard textbook treatment of finding areas of polar curves is seen as an application of single variable calculus. | {"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": 0.9096014499664307, "perplexity": 614.8578121095497}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-22/segments/1432207928423.12/warc/CC-MAIN-20150521113208-00061-ip-10-180-206-219.ec2.internal.warc.gz"} |
https://mathoverflow.net/questions/225026/faithful-exact-functors-to-tensor-categories | # Faithful exact functors to tensor categories
Let $P$ be a "nice" $k$-linear abelian tensor category (e.g. A tannakian category or a fusion category over a field $k$) and $F: M\to P$ an additive $k$-linear exact and faithful functor. I want to know when $F$ is equivalent to the forgetful functor from the category of co-modules over a co-algebra in $P$ to $P$. What are the known results in this direction?
A necessary condition for the existence of this equivalence is the existence of a "box product": $$\boxtimes : P\times M \to M,$$ which admits natural isomorphisms: $$F(A\boxtimes X) \simeq A \otimes F(X), \\ 1\boxtimes X\simeq X,\\ (A\otimes B)\boxtimes X\simeq A\boxtimes (B\boxtimes X).$$ Satisfying some expected coherence diagrams. Is this condition sufficient? What about the nice tensor categories mentioned above?
(For a co-module $X$ over a co-algebra $A$, with the structure map $\rho: X \to X\otimes A$,the box product $Y\boxtimes X$ is defined to be $Y\otimes X$ with co-action $$Y\otimes \rho:Y\otimes X\to Y\otimes X\otimes A.$$)
• Have you tried applying the (co)monadicity theorem? – Qiaochu Yuan Dec 2 '15 at 4:05
• @QiaochuYuan I know the special case $P=Vect_k$ can be proved using comonadicity. But in general case I can't find an adjoint for $F$ and can't write $F$ as a tensor product with a certain object. – Mostafa Dec 2 '15 at 5:32
If for example $P$ is a fusion category, then such a box product is what is usually referred to as a module category structure on $M$. In case $M$ is nice enough- for example semisimple with finitely many simple objects, then a result of Ostrik says that $M$ is necessarily the representation category of some algebra $A$ in $P$ (the fact that it is an algebra and not a coalgebra is not very important here). About the functor $F$: if you assume that you have such natural isomorphisms with Coherence conditions, then $F$ is necessarily given by taking tensor product over $A$ with a left $A$-module $M$. The question is whether or not this module is isomorphic with $A$ itself. This answer can be answered by applying the inner hom construction and checking it by hand. See also http://arxiv.org/pdf/math/0111139.pdf
• Thank you very much for the answer and the link. But in the mentioned result of Ostrik, the semisimplicity of M is very restrictive and dose not satisfied in my interested examples. There is a result of Gabber which says that every finitely generated locally finite $k$-linear abelian category is equivalent to a category of R_modules (R finite k-algebra). This result can be interpreted as a result on module categories on the category of vector spaces. Does there exist a similar result in the general case? – Mostafa Dec 16 '15 at 19:43 | {"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": 0.963656485080719, "perplexity": 149.28611873561186}, "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/1593655891884.11/warc/CC-MAIN-20200707080206-20200707110206-00451.warc.gz"} |
https://electronics.stackexchange.com/questions/215752/the-relation-between-a-link-speed-and-fourier-coefficients | # The relation between a link speed and Fourier coefficients
I am reading Computer Networks by Tanenbaum and the the chapter about the physical link and there is an example that I don't understand.
The author states that given a bit rate of b bits/sec the time required to send 8 bits is 8/b bits/sec. thus the frequency of the first harmonic is 8/b. Then there is an example of a telephone line which have a cutoff at 3000 Hz and claims that the highest harmonic passed is 3000 /(b/8) = 24000/b.
An Example is given for a 300 Bps, the transmission time is 26.67 msec, the first harmonic is 37.5 Hz and the number of harmonics sent is 80.
I understand that a real signal is finite and we can regard it as having a period of some T. I noted that the first harmonic is at 1/T.
I have a few questions:
1. When sending a stream of data, do we do modulation for some of the bits up to a point and this is the signal we calculate its period etc' ? we can't read all the data to transmit before transmission.
2. I get that we can reconstruct the signal from the Fourier coefficients. I didn't understand the relation between harmonics and coefficients (does sending k harmonics means sending k coefficients ?)
3. I don't understand the calculation that was made to get that we can send 80 harmonics in the above example, why does it taking the bandwidth and dividing it to the frequency of the first harmonic ?
4. Is it possible or wished for that the harmonics sent would not be the first harmonic and all of its multiples up to a certain point ? e.g sending the first harmonic and sending the third one but not the second one.
I'd appreciate an answer to any of those questions, I have studied Fourier analysis but with no relation to signals and real world applications so I'm having difficulties putting theory to use.
I'll try to group all your questions together because they are related.
To find the spectrum of a signal (frequency content) one has to look at a finite period of time. Let's say one looks at the 1st 10 seconds of a signal. The fundamental frequency is the sinusoidal signal that can make 1 revolution in the measured time. The period is 10 seconds, the fundamental frequency is 0.1 Hz. If one is looking at the discrete Fourier transform of a signal they are looking to see how many multiples of this fundamental frequency there are. The 1st frequency is 0.1 Hz, the 2nd is 0.2 Hz, the 13 is 1.3 Hz, etc. The coefficients are the amount of each frequency present (the 0.5 Hz signal is the 5th harmonic of the fundamental frequency, which is 0.1 Hz). One could say they have 0.3 of the 1st harmonic, 0.2 of the 8th harmonic, etc. Different signals are made by having different amounts of each harmonic.
Certain signals might not have all harmonics. Ideal square waves are made of odd harmonics (0.1 Hz, 0.3 Hz, 0.5 Hz, etc, but no 0.2 Hz, 0.4 Hz, etc). Sawtooth waves are made of only even harmonics.
When using the DFT there is a midpoint where signals start aliasing. If you have 100 samples of your signal over 10 seconds then the highest frequency that should be present in your original signal is a 5 [Hz] signal. This is due to aliasing where 4.9 Hz looks like 5.1 Hz, 0.1 Hz looks like 9.9 Hz, etc. | {"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": 0.8609996438026428, "perplexity": 505.06499319304174}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-35/segments/1566027319082.81/warc/CC-MAIN-20190823214536-20190824000536-00155.warc.gz"} |
https://www.physicsforums.com/threads/homework-help-signal-processing.212394/ | # Homework Help (Signal Processing)
1. Feb 1, 2008
Hello All, I am currently taking Signal Processing & Linear Systems. Ive come across my last problem for homework but cant find a way to do it. If someone can show me in the right direction it will be very helpful.
Thanks
The Problem:
The voltage f(t) = 2u(t) is applied to the circuit shown in the figure below. The
initial inductor current is i(0) = 2mA.
250mH
|-----mmm------| +
f(t) ( ~ ) i(t)--> Z 500 Ohms y(t)
|______________| _
a) Find the zero-input response yzi(t) of the system.
b) Find the zero-state response yzs(t) of the system.
c) What is the total response y(t) of the system?
so far all i can get is the differential needed, but im not even sure if its right:
f(t) = 250mH*Ldi/dt + 500*i(t) --> 250mH D + 500y(t)
any help trying to finish this problem would be great thanks again!
Ajay
Last edited by a moderator: Feb 1, 2008
2. Feb 1, 2008
### CEL
The value of the inductor L is 250mH = 0.25H, so your first term is redundant. You should have:
$$0.25\frac{di}{dt}+500i=f(t)$$
with the initial condition $$i(0)=2\times10^{-3}$$
To find the zero input response you solve the homogeneous equation (f(t)=0) and replace the initial condition $$i(0)=2\times10^{-3}$$ in order to eliminate the integration constant.
To find the zero state response you solve the non-homogeneous equation (f(t)=2u(t)) and replace the initial condition $$i(0)=0$$ in order to eliminate the integration constant.
The total response is the sum of yzi and yzs. | {"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": 0.8970168828964233, "perplexity": 1317.6854445107533}, "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/1484560279923.28/warc/CC-MAIN-20170116095119-00503-ip-10-171-10-70.ec2.internal.warc.gz"} |
http://math.stackexchange.com/questions/297819/if-fx-gt-0-and-fx-gt-0-then-lim-x-to-inftyfx-infty | If $f''(x)\gt 0$ and $f'(x)\gt 0$, then $\lim_{x\to +\infty}f(x)=+\infty$
I'm trying to solve this question:
Suppose $f''(x)\gt 0$ in $(a,+\infty)$ and there is $x_0\gt a$ such that $f'(x_0)\gt 0$. Prove that $\lim_{x\to +\infty}f(x)=+\infty$.
My attempt:
I know that there is $\alpha \in (a,x_0)$ such that $f''(\alpha)=\frac{f'(x_0)-f'(a)}{x_0 - a}\gt0$, then $f'(x_0)\gt f'(a).$
I don't know what to do with this information, I can't follow from that, I feel that I'm close to the solution, I need help.
Thanks a lot
-
Is $f(\bullet)$ twice continuously differentiable? – Inquest Feb 8 '13 at 6:34
It doesn't need to be. Twice differentiable is all that is needed. – copper.hat Feb 8 '13 at 6:35
Hints:
• For every $x\geqslant x_0$, $f'(x)\geqslant f'(x_0)$.
• For every $x\geqslant x_0$, $f(x)\geqslant f(x_0)+f'(x_0)(x-x_0)$.
• $\lim\limits_{x\to+\infty}f(x_0)+f'(x_0)(x-x_0)=+\infty$.
-
Thank you very much for your answer, it was very helpful! – user42912 Feb 8 '13 at 6:46
If $f''(x) >0$ on $I=[x_0, \infty)$, then $f'(x)$ is strictly increasing on $I$. In particular, $f'(x) \geq f'(x_0)$.
Since $f(x) = f(x_0)+\int_{x_0}^x f'(t) dt$, we have $f(x) \geq f(x_0)+\int_{x_0}^x f'(x_0) dt = f(x_0)+f'(x_0)(x-x_0)$. Since $f'(x_0)>0$, we see that $\lim_{x \to \infty} f(x) = \infty$.
-
You know that $f'$ is increasing and that $f'(x_0) = m > 0$. Prove that $f(x) > m(x-x_0) + f(x_0)$ for all $x>x_0$.
-
We have $f''(x)>0$ for all $x\in(a,\infty)$, so that $f'(x)>f'(x_0)>0$ for all $x\in (x_0,\infty)$. Thus $f(x)=f(x_0)+\frac{f(x)-f(x_0)}{x-x_0}(x-x_0)=f(x_0)+f'(b)(x-x_0)$ (for some $b>x_0$ by mean value theorem) $>f(x_0)+f'(x_0)(x-x_0)\to\infty$ as $x\to\infty$.
-
Let's denote $f'(x_0)$ by $\beta$. First, prove that $f'(x) \ge \beta$ for all $x \ge x_0$. Then prove that $f(x) \ge f(x_0) + \beta (x - x_0)$ for all $x \ge x_0$, and you are done. Each of the two steps can be done using Lagrange's theorem.
- | {"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": 0.992190957069397, "perplexity": 102.74467973791317}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-06/segments/1422122192267.50/warc/CC-MAIN-20150124175632-00003-ip-10-180-212-252.ec2.internal.warc.gz"} |
http://clay6.com/qa/42950/the-time-period-of-oscillation-of-simple-pendulum-is-given-by-t-2-pi-sqrt-w | # The time period of oscillation of simple pendulum is given by $t = 2 \pi \sqrt{\large\frac{l}{g}}$ . What is the accuracy in the determination of ‘g’ if 10cm length is known to 1mm accuracy and 0.5 s time period is measured from time of 100 oscillations with a watch of 1sec. resolution?
$\pm5\%$
Hence (B) is the correct answer | {"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": 0.9697056412696838, "perplexity": 453.2432867305644}, "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/1505818690016.68/warc/CC-MAIN-20170924115333-20170924135333-00057.warc.gz"} |
https://www.physicsforums.com/threads/invariant-the-hell-no.18798/ | # Invariant? The hell NO!
1. Apr 8, 2004
### deda
I say the only viewpoint available that provides us with the truth is the center of mass. You go ahead be my guest and choose another but be ware: Your physics will not remain the same. Lets fix the coordinate system on the sun and watch the earth moving. The sun is truly immovable thus has no force on it because of Newton 2. Because of Newton’s gravity sun’s mass is also zero but also and earth’s force will be zero. So how can earth move relatively to the sun? It will probably dilate time and contract space in order to move!
I'm off to Bermuda!
2. Apr 8, 2004
### matt grime
The centre of mass is not a view point. How do you know the sun isn't moving? In another thread you say it is. The sun has zero mass? Bermuda? The triangle, try not to get lost, please.
3. Apr 9, 2004
### deda
If the system is fixed on the sun then the sun in this system is immovable.
If the sun is immovable or has constant velocity due to Newton 1 it is subjected to no force.
If the sun is subjected to no force then due to Newton's gravity the sun has zero mass.
If the sun is subjected to zero force and Newton 3 then the earth is also subjected to zero force.
If the earth is subjected to zero force then its velocity is constant or zero.
In few words we get totally different physics.
4. Apr 9, 2004
### matt grime
The system being fixed on the sun and the sun not moving are, if one assigns meaning to the words that are ambiguous, contradictory statements (the orbits are elliptic).
How do you know the sun is not moving, or moving with no acceleration? In what larger iniertial frame is this observation taking place? Why are you ignoring the other planets in the system?
5. Apr 9, 2004
### Severian596
Many of those few words are "if," and if you can't get past the first statement you have a hard time implying the rest of them
6. Apr 9, 2004
### jdavel
So many errors, so little time!
I'll just straighten you out on this one: "If the sun is subjected to no force then due to Newton's gravity the sun has zero mass."
You need to review the equation for "Newton's gravity" (or maybe 8th grade math). There are two masses in the equation. Only one of them (not necessarily the sun's) needs to be zero for there to be "no force".
7. Apr 9, 2004
### FrankM
How lost am I? That is one of the stangest things I've ever heard. Just because something is not subjected to a "NET" force (notice what I did there with the quotes and capitalizing it?) doesn't mean it has no mass. I am not being subjected to a net force right now (meaning that I am sitting in my chair and not moving, at least my center of gravity is not moving), if I was being subjected to a force I would accelerate. OK so I am not being subjected to a net force (just really bad reasoning) but this in no way means I have no mass. I have mass (a little too much mass actually but such is life in a cube), I can tell by the fact that my fingers actually come in contact with the keys of my keyboard (even though it is mostly e/m fields that are keeping my fingers from going through the keyboard, they would not be there if I had no electrons).
Here is most of the misconception: misinterpreting the math without understanding what is going on. Equating Newton's formula for gravitational force with his laws of motion without a clear understanding of what they mean is ..... meaningless. Sorry but you should probably take a look at a beginning text in Physics.
8. Apr 9, 2004
### deda
Then the earth is the one with the zero mass and you can eat my shorts!
9. Apr 9, 2004
### FrankM
Sorry forgot something
By the way, the sun is not fixed in space, it wobbles because of the pull from the planets. | {"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": 0.8542478680610657, "perplexity": 896.9497155500791}, "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-2018-47/segments/1542039746227.72/warc/CC-MAIN-20181120035814-20181120061814-00406.warc.gz"} |
http://tex.stackexchange.com/questions/192334/miktex-on-windows-8-fatal-format-file-error | # Miktex on windows 8: fatal format file error
I've just freshly installed miktex 2.9 on windows 8. Plain tex seems to work but running latex on a simple file produces
This is pdfTeX, Version 3.1415926-2.4-1.40.13 (MiKTeX 2.9)
(Fatal format file error; I'm stymied)
Searching for this error suggests that I should regenerate format files via miktex's options program, which I have done. Indeed I observe that miktex options->update formats touches the files latex.fmt and pdflatex.fmt (and others) in the following directory:
C:\Users\Jenny\AppData\Local\MiKTeX\2.9\miktex\data\le\pdftex
However the error still occurs. (I also tried miktex options->format->build, which seems to do the same thing but one file at a time).
I'm not sure how to verify that latex is looking in the right place for these files. In miktex options->roots I see the path C:\Users\Jenny\AppData\Local\MiKTeX\2.9 with description "UserInstall, UserConfig". However I can't see any environment variables pointing to this directory. Following this answer I tried defining
TEXMFCACHE=C:\Users\Jenny\AppData\Local\MiKTeX\2.9
to no avail. I also tried copying the .fmt files to both of these directories:
C:\Users\Jenny\AppData\Roaming\MiKTeX\2.9\miktex\data\le\pdftex
C:\ProgramData\MiKTeX\2.9\miktex\data\le\pdftex
-
Welcome to TeX.SX! – Adam Liter Jul 20 '14 at 6:53
I see two possible reasons: either you had another version of MiKTeX already installed (I doubt that on a Windows 8 system) or some files got corrupted at installation, and you should uninstall and re-install. I never had to set environment variables myself on a MiKTeX system. – Bernard Jul 20 '14 at 8:49
Don't copy files around. Remove the fmt-files you have placed everywhere. They only can confuse the search order. Open a command line in the folder of your test file and check the location of the format file with kpsewhich --engine=pdftex pdflatex.fmt. Then go to miktex settings (user mode), tab format, select the pdflatex entry and click on build. Do you get an error? Does this recreate the fmt you found with kpsewhich? – Ulrike Fischer Jul 20 '14 at 10:12
I did also try reinstalling. kpsewhich points to C:/Users/Jenny/AppData/Local/MiKTeX/2.9/miktex/data/le/pdftex/pdflatex.fmt which is recreated without error from the format tab. – stewbasic Jul 21 '14 at 0:01
The "Fatal format file error" normally happens when the format file has not been created by same pdftex that is trying to use it. I can e.g. force it by creating a format in the current folder with texlive and then trying to compile with miktex. So check very carefully if there are other pdflatex.fmt around. Check also if your editor points to some other pdflatex executable. Check with "set" your environment variables. – Ulrike Fischer Jul 21 '14 at 9:36 | {"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": 0.9718374013900757, "perplexity": 4665.6273456047265}, "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-2016-30/segments/1469257824109.37/warc/CC-MAIN-20160723071024-00289-ip-10-185-27-174.ec2.internal.warc.gz"} |
http://math.stackexchange.com/questions/337053/second-part-of-the-factorial-sum-divisibility-question?answertab=votes | # Second part of the factorial sum divisibility question
Which primes $p$ divide the sum of factorials $1! + 2! + 3! + 4! + 5! + \cdots + (p-1)!$? This is related to my previous question.
-
Have you ever considered beginning one of your contributions "I'm just a simple caveman. Your prime numbers frighten and confuse me. But one thing I DO know is,..." etc... ? – Stephen Mar 25 '13 at 1:38
Up to 100,000, the only values of $p$ (not necessarily prime) that satisfy this are $3, 9, 11, 33, 99$. – Yoni Rozenshein Mar 25 '13 at 15:09
What's motivating the question? Fun or work (or both)? Surely this isn't some demented homework problem! – Douglas B. Staple Mar 26 '13 at 12:12
@Thus, please don't tell that the margin of math.stackexchange.com is too small for your beautiful proof... – vonbrand Mar 26 '13 at 16:23
I must somehow correct it! – user67878 Mar 26 '13 at 17:10
Note that heuristically, if these numbers were equidistributed mod $p$ we would expect each one to be zero (i.e., expect $p$ to divide the result) with probability $\approx 1/p$; presuming that all the values are independent (which seems a reasonable assumption), we shoud expect to have $\sum_{p\lt n}\frac{1}{p}\approx M+\ln \ln n$ of them less than $n$, where $M$ is the Meissel-Mertens Constant — this sum is approximately 2.887 for $n=10^6$, so while we would heuristically 'expect' another one in that range it's not surprising that there aren't any, and despite the apparent lack of any more solutions, because of the divergence of the series the 'expected' number of solutions is still infinite!
Added: I put together my own software implementation in C++, confirming the results (that only $p=3$ and $p=11$ work) for $p\lt 5\times 10^5$ but also looking at the distribution of results $\mod p$ to see if any patterns arose. I computed the value of $\dfrac{\sum_{1}^{p-1}i!\bmod p}{p}$ (the division by $p$ serving to 'normalize' values into the range $[0, 1)$ ) for all $p$ in that range and then grouped them in bins; these are the plots with 100 bins and 256 bins. (These bin counts should be small enough relative to the prime values in consideration that there shouldn't be any substantial aliasing effects in binning the data.) The results are a bit scattershot but there doesn't appear to be any skew towards particular values (e.g. $\frac{p-1}2$) which would show up as a spike in the bins. If I have more time I may take a look at the square of the sum and particularly even the inverse of the sum $\pmod p$, but since the latter would actually require me writing a quick GCD algorithm to find the inverse it'll have to wait for later.
(Note: the sharp drop at the last element is an artifact of my adding a 0 value to the end of the bin data so my spreadsheet didn't automatically normalize the range; it's not an actual value.)
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Interesting. Reminds me of an estimation of the number of Wieferich primes... – Gottfried Helms Mar 26 '13 at 6:27
+1 for your heuristic. – Douglas B. Staple Mar 26 '13 at 12:59
+1 But "we would expect each one to be prime with prob..." Shouldn't that be "to be a multiple of $p$"? – leonbloy Mar 26 '13 at 18:27
@leonbloy Whoops, good catch - fixed now. – Steven Stadnicki Mar 26 '13 at 18:40
Also, $M + \ln \ln n \ge 4$ for $n$ around $10^{18}$, so perhaps finding the next prime satisfying the property will require us to look at all primes up to that much. – ShreevatsaR Mar 27 '13 at 10:42
I have a few small analytic comments and a relatively aggressive numerical search.
Firstly, I checked all odd $p$ (not necessarily prime) up to $10^6$ and found only the same five as Yoni above, namely 3, 9, 11, 33, and 99. This took ~1 hour using 1 core of my MacBook Pro, with an early unoptimized version of my code.
Secondly, of course we have: $$p\nmid\sum_{n=1}^{p-1}n!\quad \forall p\in\mathbb{N}, p \textrm{ even}$$ because all of the terms in the sum are even except the first one, so the sum is always odd.
Thirdly, a numerical search is sped-up slightly using the identity: $$\sum_{n=1}^{p-1} n! = 1+2[1+3[1+4[...[1+(p-1)]]]]$$ I've written software to check primes $p$ for $p|\sum_{n=1}^{p-1} n!$ using the above identity, and set it up to use multiple threads in an embarrassingly-parallel fashion. Using this software I've shown: $$\textrm{3 and 11 are the only primes }p<10^7\textrm{ that divide }\sum_{n=1}^{p-1} n!$$ If you want to compile it yourself then you'll have to link with the primesieve library.
#include <stdio.h>
#include <stdlib.h>
#include <primesieve/soe/PrimeSieve.h>
// To compile: g++ -O3 factorialSum.cpp -lprimesieve -o factorialSum
// or: g++ -O3 -static factorialSum.cpp -lprimesieve -o factorialSum
// We perform arithmetic at u1 precision; p is restricted to u2 precision
typedef uint_fast64_t u1;
typedef uint_fast32_t u2;
// This evaluates $\sum_{n=1}^{p-1} n!$ based on the definition.
// We don't actually use this in our program below.
u2 factorialSum_modp_method1(u2 p){
u1 factorial_modp=1;
u1 factorialSum_modp=0;
for(u1 n=1; n<=p-1; n++){
factorial_modp*=n;
factorial_modp%=p;
factorialSum_modp+=factorial_modp;
factorialSum_modp%=p;
}
return factorialSum_modp;
}
// This evaluates $\sum_{n=1}^{p-1} n! = 1+2[1+3[1+4[...[1+(p-1)]]]]$ via the RHS of the expression.
// This is somewhat faster than 'method1' above.
u2 factorialSum_modp_method2(u2 p){
u1 factorialSum_modp=1;
for(u1 n=p-1; n>1; n--)
factorialSum_modp=(factorialSum_modp*n+1)%p;
return factorialSum_modp;
}
// This prints out p iff p divides $\sum_{n=1}^{p-1} n!$.
// We pass this as an arument to PrimeSieve.generatePrimes().
void check_factorialSum_modp(u2 p){
if(factorialSum_modp_method2(p)==0)
printf("%u\n", (unsigned)p);
}
void error_incorrectUsage(char *args[]){
printf(" numThreads is a natural number, 1, 2, ...\n");
printf(" blockSize is the interleave size between threads\n");
printf(" pMin and pMax are integers, the upper and lower limits on candidate primes p\n\n");
printf("e.g.: %s 0 1 10000 0 100000\n", args[0]);
printf(" will check [0, 100000] for primes p that divide $\\sum_{n=1}^{p-1} n!$ \n");
printf(" (expression written LaTeX).\n\n");
printf("In order to use multiple threads, run multiple instances with different values of 'thread'.\n");
exit(1);
}
int main(int nargs, char *args[]){
// Parse command line arguments and check for proper usage
if( !(nargs==6) ) error_incorrectUsage(args);
char *error1, *error2, *error3, *error4, *error5;
u2 thread = strtoul(args[1], &error1, 10);
u2 numThreads = strtoul(args[2], &error2, 10);
u2 blockSize = strtoul(args[3], &error3, 10);
u2 pMin = strtoul(args[4], &error4, 10);
u2 pMax = strtoul(args[5], &error5, 10);
if(*error1 || *error2 || *error3 || *error4 || *error5) error_incorrectUsage(args);
// Sieve intervals [pLower, pUpper] that belong to this thread
PrimeSieve ps;
u2 pUpper=pLower+blockSize;
ps.generatePrimes(pLower, pUpper, check_factorialSum_modp);
}
}
-
Thank you kindly it will be very interesting to see whether there are more or not. – user58512 Mar 26 '13 at 3:51
You're welcome. I had fun looking at this problem. Computing up to $10^7$ took 45 minutes using 16 threads on an 8-core Opteron. Pushing the limit up to $10^8$ would take about three days on that machine, which we can do if you're using this for research. Otherwise I think we've learned all we're going to learn from computation. – Douglas B. Staple Mar 26 '13 at 17:44
The following is more a comment than a usable answer, but too long for the comment-box
Another oberservation which has a surprising symmetry, which seems to occur only if the modulus is a prime (and this is thus an observation in its own right).
We define the lower triangular matrix of the Eulerian numbers:
$\qquad \qquad \displaystyle E= \Tiny \left[ \begin{array} {rrrrr} 1 & . & . & . & . & . \\ 1 & 0 & . & . & . & . \\ 1 & 1 & 0 & . & . & . \\ 1 & 4 & 1 & 0 & . & . \\ 1 & 11 & 11 & 1 & 0 & . \\ 1 & 26 & 66 & 26 & 1 & 0 \end{array} \right]$
We observe, that the row sums accumulate to the factorials so $E \cdot V(1) = F$ where $V(1)$ is the column-vector of ones, gives the column-vector of factorials $F$. If we now add the rows up to the index of the prime $p$ in question and take this modulo $p$ then we have another expression of the sum. If we now change the order of computation: we compute first the column-sums (modulo p) then we get a rowvector with symmetric entries, seemingly iff p is prime. For instance, for $p=13$ (and note, that in Pari/GP the index begins at 1 and not at 0)
$\qquad \qquad V(1)^T_{p+1} ~ \cdot E_{p+1,p+1} \pmod p = \\\qquad \qquad \left[ \begin{array} {rrrrrrrrrrrrr} 1 & 2 & 3 & 11 & 6 & 12 & 5 & 12 & 6 & 11 & 3 & 2 & 1 & 0 \end{array} \right]$
and then we sum the resulting vector modulo(p) and subtract 1 because we have also the 0! in the sum.
I don't see actually what this might give us - and I can't go deeper in this - but it may also be significant (and possibly on the way to a -more- closed form) that the columns of the Eulerian triangle are expressible as sequences of leading terms of geometric series and their derivatives and so might have a closed form for their leading segments.
Anyway -perhaps we can make the resulting row and its modular sum simpler by some pattern.
Here are the resulting vectors for the first few primes: $\Tiny \begin{matrix} 2: & 1 & 1 & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . \\ 3: & 1 & 2 & 1 & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . \\ 5: & 1 & 2 & 3 & 2 & 1 & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . \\ 7: & 1 & 2 & 3 & 1 & 3 & 2 & 1 & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . \\ 11: & 1 & 2 & 3 & 10 & 8 & 8 & 8 & 10 & 3 & 2 & 1 & . & . & . & . & . & . & . & . & . & . & . & . & . \\ 13: & 1 & 2 & 3 & 11 & 6 & 12 & 5 & 12 & 6 & 11 & 3 & 2 & 1 & . & . & . & . & . & . & . & . & . & . & . \\ 17: & 1 & 2 & 3 & 13 & . & 13 & 15 & 9 & 3 & 9 & 15 & 13 & . & 13 & 3 & 2 & 1 & . & . & . & . & . & . & . \\ 19: & 1 & 2 & 3 & 14 & 15 & 1 & 7 & 8 & 15 & 10 & 15 & 8 & 7 & 1 & 15 & 14 & 3 & 2 & 1 & . & . & . & . & . \\ 23: & 1 & 2 & 3 & 16 & 9 & . & 15 & 12 & 18 & 12 & 6 & 17 & 6 & 12 & 18 & 12 & 15 & . & 9 & 16 & 3 & 2 & 1 & . \\ 29: & 1 & 2 & 3 & 19 & 24 & 9 & 18 & . & 28 & 23 & 7 & 24 & 22 & 14 & 6 & 14 & 22 & 24 & 7 & 23 & 28 & . & 18 & 9 \end{matrix}$
[update] If we decompose the columns of the Eulerian matrix in their component-series we're getting involved with the leading segments of geometric series and their derivatives - and that all modulo p. Because in the end we need the sum of the column-sums I've expressed the final partial sum in terms of its (truncated) series-components, here for the example prime p=5 . We get as the final result the sum of all elements of the following matrix $$\small \begin{bmatrix} 1\cdot 5^0 & 0\cdot 4^0 & 0\cdot 3^0 & 0\cdot 2^0 & 0\cdot 1^0 \\ 1\cdot 5^1 & -1\cdot 4^1 & 0\cdot 3^1 & 0\cdot 2^1 & 0\cdot 1^1 \\ 1\cdot 5^2 & -2\cdot 4^2 & 1\cdot 3^2 & 0\cdot 2^2 & 0\cdot 1^2 \\ 1\cdot 5^3 & -3\cdot 4^3 & 3\cdot 3^3 & -1\cdot 2^3 & 0\cdot 1^3 \\ 1\cdot 5^4 & -4\cdot 4^4 & 6\cdot 3^4 & -4\cdot 2^4 & 1\cdot 1^4\\ 1\cdot 5^5 & -5\cdot 4^5 & 10\cdot 3^5 & -10\cdot 2^5 & 5\cdot 1^5 \end{bmatrix} \pmod 5$$ where -because p is a prime- we have systematic zeros and congruences expressible in negative numbers $\pmod p$ and get this:
$$\small \left[ \begin{array} {rrrrrrrrrrr} 1 & . & .& . & . \\ . & 1\cdot 1^1 & . & . & . \\ .& -2\cdot 1^2 & 1\cdot 2^2 & . & . \\ . & 3\cdot 1^3 & -3\cdot 2^3 & 1\cdot 3^3 & . \\ . & -4\cdot 1^4 & 6\cdot 2^4 & -4\cdot 3^4 & 1\cdot 4^4\\ . & . & . & . & . \end{array} \right] \pmod 5$$ which can be generalized to other prime moduli p in the obvious way.
The idea with all this is to be of some advantage for a better/more intuitive expression of the sum perhaps by closed forms of the truncated geometric series (modulo p). Still we can recognize the row-sums as factorials (while modulo p), but the change of order of computation and the hopefully advantageous closed forms of the column sums might be helpful.
We can even more make from the modularity. The following represntation can analoguously be made even for higher primes p, but I use just the example before. The above matrix can be rewritten in complete congruence as $$\small \left[ \begin{array} {rrrrrrrrrrr} 1 & . & .& . & . \\ . & 1\cdot 1^1 & . & . & . \\ .& 3\cdot 1^2 & 1\cdot 2^2 & . & . \\ . & 3\cdot 1^3 & 2\cdot 2^3 & 1\cdot 3^3 & . \\ . & 1\cdot 1^4 & 1\cdot 2^4 & 1\cdot 3^4 & 1\cdot 4^4\\ . & . & . & . & . \end{array} \right] \pmod 5$$ and those simple binomials give then as columnsums $$\small \left[ \begin{array} {rrrrrrrrrrr} 1 & 1^1\cdot 2^3 & 2^2\cdot 3^2 & 3^3\cdot 4^1 & 1\\ \end{array} \right] \pmod 5$$ and if I put together each two terms reading from left from right then this is
$$\small \begin{array} {rrrrrrrrrrr} 2 \cdot 1 & + 2 \cdot 2^3 & + 2^2\cdot 2^2 \\ \end{array} \equiv 4\pmod 5$$ where by the definition in the OP we had to decrement by 1 to delete the $0!$ term.
If we write the odd prime $p=2q+1$ then the general formula (including the $0!$) becomes now: $$S_p \equiv \sum_{k=0}^{p-1} k! \equiv 2 \sum _{k=0}^{q-1} k^k (k+1)^{p-1-k} + (q \cdot -q)^q \pmod p$$
which might be finally accessible by a bit better knowledge about primitive roots modulo p.
-
Gottfried's answer and Steven's replies bring up the point that there's a choice to be made in a direct numerical search: either you can compute $(p-1)!$ $(\textrm{mod }p)$ separately for each $p$, or you can store $(p-1)!$ and then compute $p! = p(p-1)!$ to test the next $p$. This same tradeoff comes up in other similar problems, such as the search for Brown Numbers.
In order to check every $p<N$, the first approach requires $O(N^2)$ multiplies of numbers of order $p$; the second approach requires $N$ multiplies, but each multiply is very expensive, being between a number of order $p$ and one of order $p!$. Another consideration is that the second solution, storing $(p-1)!$ un-reduced, requires memory linear in $p$. I don't really want to see what happens when $(p-1)!$ no longer fits in cache. Furthermore, it's not obvious how to parallelize the second algorithm.
My impression is that if you want to check all $p<N$ and not just the primes $p$ it's fastest to store $(p-1)!$ for small $N \lesssim 10^6$, but for larger $N$ it's faster to compute $(p-1)!$ $(\textrm{mod }p)$ separately for each $p$. If you want to only check primes $p$, then the "first approach" gets a $O(\log(n))$ speedup and the "second approach" hardly benefits at all, meaning that the first approach is faster: see the "Update:" section below the code for an explanation.
Here is a C code that uses the GMP to implement this "second approach"; I implemented the "first approach" in my first answer above.
#include <gmp.h>
#include <time.h>
#define NMAX 100000
int main(){
// Print begin time & date
time_t beginTime;
struct tm * timeinfo;
time ( &beginTime );
timeinfo = localtime ( &beginTime );
gmp_printf("# Begin time/date is: %s", asctime (timeinfo) );
// Stores n! and sum(n!)
mpz_t cur_factorial;
mpz_t sum_factorial;
mpz_t sum_factorial_modn;
mpz_init_set_ui(cur_factorial, 1);
mpz_init_set_ui(sum_factorial, 1);
mpz_init(sum_factorial_modn);
for(unsigned long int n=2; n<=NMAX; n++){
// Check to see if n divides sum(m=1..n-1) m!
if( mpz_divisible_ui_p(sum_factorial, n) )
gmp_printf("%u\n", n);
// Update n!
mpz_mul_ui(cur_factorial, cur_factorial, n);
// Update sum(n!)
}
// Print end time & date; total run time
time_t endTime;
time ( &endTime );
timeinfo = localtime ( &endTime );
gmp_printf("# End time/date is: %s", asctime (timeinfo) );
gmp_printf("# Total runtime of %.0f seconds.\n", difftime(endTime, beginTime) );
// Release and return
mpz_clear(cur_factorial);
mpz_clear(sum_factorial);
}
Update:
This code implementing the "second approach" is strictly slower than the one in my previous answer. The above code takes ~9s to for an upper limit to $10^5$, which only takes ~5s using the code in my previous answer. A search up to $10^6$ takes ~19 min with this code, compared to ~8 min with the other one. Part of the reason for this is that, with the other code, you get a huge speedup if you only test primes $p$. This code is set up to check all $n$, even the pointless even ones. This might not seem like a fair comparison, but modifying the above code to only check odd $n$ doesn't give any speed improvement; the point is that you anyway have to calculate $(n-1)!$ for every $n$ with this method, so you've already paid the cost; the divisibility check doesn't take any time compared with the multiplies.
-
In addition to @Douglas post, here is a very short (and I think: efficient) implementation in Pari/GP - however, I didn't run it to high N myself: [Update] I added an improved computation method at the end[/update]
MaxN=101 \\ set some upper limit for the computation
S=[1,1] \\ contains current factorial and current sum
for(k=3,MaxN, S *= [k-1,k-1;0,1]; if(isprime(k),print([k,S[1] % k,S[2]%k])))
Result :
p | (p-1)! %p | S %p
--+-----------+-----
[3, 2, 0]
[5, 4, 3]
[7, 6, 5]
[11, 10, 0]
[13, 12, 9]
[17, 16, 12]
[19, 18, 8]
[23, 22, 20]
[29, 28, 16]
[31, 30, 1]
[37, 36, 4]
[41, 40, 3]
...
Up to the 1000'th prime MaxN=7919
MaxN = prime(1000)
...
...
[7919, 7918, 2882]
this needed 281 msec.
A Version using the modulo-computation is the following
gettime()
{for(j=1,1000,p=prime(j);
S=[1,1];
for(k=3,p,
S*=[k-1,k-1;0,1];
S=S % p
);
print([p,S[1] % p,S[2]%p])
)}
gettime()
%1124 = 9219 \\ this is msecs
[update] The following is only relevant, if we want a routine, which produces the list over all primes up to some limit.
It has then some advantage to do a compromise between the first method where we need all factorials and sums only once but which needs many bits for the representation of the numbers and the complete modulo-computation, which operates on small numbers but must do the whole run of computing factorials (residuals) for each prime such that we have a double loop.
If we compute and store the required factorials and partial sums in chunks of, say, $t=10$ terms, such that we have a list of precomputed values $$\small \begin{array} {r|l|l|} & f & S \\ \hline \\ c_1=& 10! & \sum_{k=1}^{10} k! \\ c_2=& {20!\over 10!} & {\sum_{k=11}^{20} k! \over 10!} \\ c_3=& {30!\over 20!} & {\sum_{k=21}^{30} k! \over 20!} \\ \ldots \end{array}$$ then we can compose our results for each prime by modular multiplication of that constants and a short remainder. Finetuning the parameter $t$ for higher allows then to find a compromise between storage/size of the stored numbers and number of steps in the inner loop, which contains the modular computation per prime.
Using some empirical data I have done an estimation with a quadratic curve in Excel, which says, for the list of primes up to 1000000 I need 90 min with blocksize $t=20$ , 30 min with $t=100$ and 15 min with $t=500$ .
(If it is wished I can append the Pari/GP code here)
-
This isn't likely to be very efficient for long because (AFAICT) it computes the exact values of the current factorial and current sum, and these will become very large numbers very quickly. It should be much better to work mod $p$ for each prime $p$ individually. – Steven Stadnicki Mar 26 '13 at 5:44
@Steven: yes, you may be correct. However, to compute the list up to some MaxPrime this needs then time quadratic in the list length. Up to the 400'th prime (MaxN=2741) that modulo-method needed 3478 msec. I didn't check, whether the ratio of the timings change with higher MaxN/MaxPrime – Gottfried Helms Mar 26 '13 at 6:12
That's a very good point - there's a lot to be said for the linear time. If I'm figuring right, though, note that your multiplications and additions will start to take time proportional to the length of the values which are roughly proportional to $n\log n$, so your solution is also quadratic in the list length in terms of bit-operations. – Steven Stadnicki Mar 26 '13 at 6:25
@Steven: agreed! – Gottfried Helms Mar 26 '13 at 6:29
@Steven You know what... I've encountered exactly this tradeoff between $n^2$ multiplies and reductions mod $n$ versus $n$ multiplies without reduction in the factorial part of the search from Brown Numbers. It came up somewhere on mersenneforum.org. The answer isn't at all obvious; I'm going to post an answer addressing this. – Douglas B. Staple Mar 26 '13 at 12:08
This is what I thought about, it should be some comment, but this exceeds the character limit for comment.
If you take Wilson Theorem that
$(p-1)!\equiv -1 \mod{p}$, then
$(p-2)!\equiv 1 \mod{p}$.
If we take inverse of $a$ in $F_p$ as $\dfrac{1}{a}$, we know that $\dfrac{1}{a} + \dfrac{1}{b} = \dfrac{a+b}{ab}$ in $F_p$, it can taken as usual addition.
Then
$(p-3)! \equiv -\dfrac{1}{2}\mod{p}$
$(p-4)! \equiv -\dfrac{1}{2}\cdot\left(-\dfrac{1}{3}\right)\mod{p}$
$\cdots$
$1! \equiv (-1)^{p-3}\dfrac{1}{(p-2)!} \mod{p}$
Thus
$1!+2!+\cdots(p-3)! \equiv -\sum_{k=0}^{p-2}(-1)^k\dfrac{1}{k!}\mod{p}$
And we state without proof.
Lemma:
Consider the fraction $\dfrac{p}{q}$ with $q = m!$, then the closest number to $\exp(-1)$ with form $\dfrac{p}{q}$ should satisfy
$$\dfrac{p}{q} = \sum_{j=0}^m (-1)^j\dfrac{1}{j!}$$
Then
$-\sum_{k=0}^{p-2}(-1)^k\dfrac{1}{k!} \equiv - \mathrm{floor}\left[\dfrac{(p-2)!}{\exp(1)}\right]\mod{p}$
However, this cannot be checked for large $p$, since the constant $\exp(1)$ is not accurate enough. I simply checked for smaller primes.
However, I felt that there should be other solutions than 3 and 11, otherwise, this can be a very interesting property for the constant.
-
Unfortunately, this doesn't really provide much help because it's very hard to translate back from the 'analytical' description to anything useful about the residues; even as simple a question as 'is $\frac{1}{n} \bmod p$ less than or greater than $\frac{p}{2}$?' doesn't really (AFAIK) have a clear large-scale 'analytical' structure to it. – Steven Stadnicki Mar 26 '13 at 21:07
Exactly, thus I said this is rather a comment than an answer. However , whether there are more solutions are still unclear. – Yimin Mar 26 '13 at 21:20
As far as for me - I like the nice try and would like if we could do more progress with an ansatz like this... (I've come across such an ansatz sometimes elsewhere and got stuck for the same reason of (principally) lack of accuracy. For instance in a treatize on the lucas-lehmer-test: go.helms-net.de/math/expdioph/lucasLehmer.pdf ). The additional argument of @Steven makes it even more complicated... – Gottfried Helms Mar 28 '13 at 6:46 | {"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": 3, "x-ck12": 0, "texerror": 0, "math_score": 0.8468052744865417, "perplexity": 1092.5788312770655}, "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/1454701153323.32/warc/CC-MAIN-20160205193913-00299-ip-10-236-182-209.ec2.internal.warc.gz"} |
https://hal.inria.fr/hal-01626604v2 | # Derivation of the stochastic Burgers equation with Dirichlet boundary conditions from the WASEP
Abstract : We consider the weakly asymmetric simple exclusion process on the discrete space $\{1,...,n-1\}$, in contact with stochastic reservoirs, both with density $\rho\in{(0,1)}$ at the extremity points, and starting from the invariant state, namely the Bernoulli product measure of parameter $\rho$. Under time diffusive scaling $tn^2$ and for $\rho=\frac12$, when the asymmetry parameter is taken of order $1/ \sqrt n$, we prove that the density fluctuations at stationarity are macroscopically governed by the energy solution of the stochastic Burgers equation with Dirichlet boundary conditions, which is shown to be unique and different from the Cole-Hopf solution.
Document type :
Preprints, Working Papers, ...
Domain :
https://hal.inria.fr/hal-01626604
Contributor : Marielle Simon <>
Submitted on : Thursday, March 28, 2019 - 3:15:34 PM
Last modification on : Friday, March 29, 2019 - 2:37:33 AM
### File
SBE_DBC_AHL_corrections-black....
Files produced by the author(s)
### Identifiers
• HAL Id : hal-01626604, version 2
• ARXIV : 1710.11011
### Citation
Patricia Gonçalves, Nicolas Perkowski, Marielle Simon. Derivation of the stochastic Burgers equation with Dirichlet boundary conditions from the WASEP. 2019. ⟨hal-01626604v2⟩
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": 0.8840509653091431, "perplexity": 1616.366871997437}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-26/segments/1560627998513.14/warc/CC-MAIN-20190617163111-20190617185111-00438.warc.gz"} |
https://www.physicsforums.com/threads/determine-distance-with-constant-acceleration.512822/ | # Determine distance with constant acceleration
1. Jul 8, 2011
### brunettegurl
1. The problem statement, all variables and given/known data
A particle moving at 5m/s reverses its direction in 1 s to move at 5m/s in the opposite direction. If its acceleration is constant, what distance does it travel?
2. Relevant equations
x=vot+0.5at^2; a=(vf-vi)/t
3. The attempt at a solution
total time is 1s therefore acceleration is 10m/s^2 and then i'm stuck because when I input the value into the equation I'm not getting the answer of 2.5m which is the correct answer.
2. Jul 8, 2011
### tiny-tim
hi brunettegurl!
i think it means that it goes 1.25m in 0.5s, and then 1.25m back again, = 2.5m
try again
3. Jul 8, 2011
### Timo
That's slightly tricky because in the equation you gave x is not the distance traveled but merely a coordinate describing the location. To highlight the difference: if I go for a stroll and come back home then my location will be the same as before, but I obviously traveled some distance. In your case, if you properly plugged in all values (taking into account the relative signs, because the acceleration is in opposite direction than the starting velocity) you should have ended up with x=0, which is completely correct but not the sought-for variable.
The solution to your problem is to realize that the distance traveled consists of a part that the particle moved to positive x-direction and a part where it traveled into negative x-directions. Treat these two parts separately and you get the correct result.
btw.: In the future, please also give the incorrect result you got, because that can greatly help understanding what went wrong.
4. Jul 8, 2011
### brunettegurl
Hi
I'm still a little confused my book states that to find distance we use half the total time in the equation w/accleration and then solve for x by doubling the answer for total distance by doing this ::
a=(5-(-5))/0.5= 20m/s^2
and when plugged into the equation x=(-5)(1)+ (0.50)(20)(1)^2
x(displacemnt)=5m and doubling this answer gives 10m
tiny-tim how did you get 1.25 as the distance travelled in 0.5s??
Thanks
5. Jul 8, 2011
### tiny-tim
hi brunettegurl!
no, it's not 20 m/s2, it's still 10, isn't it?
6. Jul 8, 2011
### brunettegurl
Im sorry how is it still 10 when you divide by time which is 0.5 now according to the explanation (1/2 the time) making the 10 now a 20 as acceleration
7. Jul 8, 2011
### tiny-tim
it goes from +5 to -5 in 1 s (or from +5 to 0 in 0.5 s) …
that's 10 !
8. Jul 8, 2011
### brunettegurl
ohhh so for acceleration the vf would be 0 and vi would be +5 giving the acceleration of 10 .. correct?
so then when we sub in the values in x=(5)(1)+ (0.50)(10)(1)^2
i still get an x(displacemnt) value of =10m and doubling this answer gives 20m
do i continue to use 0.5 as my time value and then since it is reversing in the same scalar value of 5 just dbl the new x-value??
9. Jul 8, 2011
### tiny-tim
(try using the X2 icon just above the Reply box )
that 1 should be 0.5, and that + should be -
10. Jul 8, 2011
### brunettegurl
Ok Thanks I understand now ... I think what confused me when I was reading the explanation in my book was the implication that 2 different time values were to be used... Thank you :)
Similar Discussions: Determine distance with constant acceleration | {"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": 0.903171181678772, "perplexity": 1234.2882055188563}, "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-2017-47/segments/1510934806832.87/warc/CC-MAIN-20171123123458-20171123143458-00591.warc.gz"} |
https://physics.stackexchange.com/questions/102578/electron-speed-in-atoms | # Electron speed in Atoms
Is the speed of electrons in Atoms consistent in all Atoms or clusters/groups/individual elements, and if it is consistent in any atom, is that speed constant?
Quantum mechanically bound particles (like electrons in an atom or molecule) do not have well defined momentum.1 What they do have is a well defined distribution of momenta. This is one of the reasons that we say electrons are bound in "orbitals" and not in "orbits".
That distribution is the same for the orbital around the same type of atom (or in the same type of molecule). Now, for atoms heavier than helium, there are more than one type of occupied orbital in the atom even in the ground state. So, it is not true that every electron in the atom has the same momentum distribution.2 All the 1s in carbon electrons have the same distribution. All the 1p electrons in carbon have the same distribution as each other but a different one than the 1s electrons. The 1s electrons in oxygen have a different distribution than the 1s electrons in carbon. And so on.
Molecular orbitals are distinct from atomic orbitals.
1 Physicists usually talk about momentum instead of speed in this context, but as almost all of these speeds remain much lower than the speed of light we can connect the two ideas by $p = mv$.
2 I've been talking as if you can tell electrons apart, but in reality you can't so there is room for some hemming and hawing about the way I am writing this. The distributions are associated with the state and not with the particle.
No, the speed is not consistent from atom to atom. Electrons nearest the nucleus of highly charged nuclei move the fastest.
In the simplistic Bohr model of a single electron atom (H, He+, Li2+...), speed is proportional to charge of the nucleus.
As a consequence, relativistic effects are much more pronounced in heavy atoms. | {"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": 0.8772225379943848, "perplexity": 466.4634913483849}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-43/segments/1570986718918.77/warc/CC-MAIN-20191020183709-20191020211209-00313.warc.gz"} |
https://community.arlo.com/t5/Arlo-Idea-Exchange/jump-arrows/idi-p/1708080 | Arlo|Smart Home Security|Wireless HD Security Cameras|NETGEAR
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## jump arrows
hello all
my idea is to add jump arrows that would allow you to jump forward or backwards on the history time line on the arlo ultra. i think they sould be set up so that one tap of the jump arrow would allow you to jump forward or backwards on the time line 10 seconds, for a double tap on the jump arrow would allow for a 30 second time jump, 3 taps on the jump arrow would allow for a 60 second time jump, 4 taps on the jump arrow would allow for a 2 minute time jump forward or backeard depending on the arrow you slect.
the other thing that would be nice is if you could download the entier day of history time line to a external hard drive. that way you could edit or trim the video on windows. this would be the cleanest way to cut out the parts of the video that you want and get rid of the parts that you don't need.
1 Comment
Arlo Moderator
The Arlo development team routinely reviews posts in the Arlo Idea Exchange to assess which features the community would like to see implemented. We greatly appreciate the community’s contribution and will keep the status of this idea updated as we get new information on its potential implementation. Thank you for posting your idea! | {"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": 0.8275701403617859, "perplexity": 1194.9280587557676}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-39/segments/1568514572516.46/warc/CC-MAIN-20190916080044-20190916102044-00162.warc.gz"} |
https://windowscommand-line.com/set-path-from-command-line/ | # Set Path from Command Line [Add to Path Windows]
By | April 7, 2022
Do you know, how to set the patch from the command line? The Windows command prompt allows users to run executable files by supplying the absolute path to the file. Regardless of whether it’s just an executable file name. Depending on which environment variable is associated with the executable, Windows will search across a list of folders. The following variables are part of the environment.
## How to Add to Path Windows & Set Path from Command Line?
A. System path &
B. User path
In the system properties, you can check the values of these variables (run the `sysdm.cpl` from computer properties or Run). User-specific paths are initially empty in the environment variable. This variable can be used to specify paths to directories containing executables. Administrators can also modify the environment variable indicating the path to the system.
### Using the command line, how do you set a path?
Set paths from the command line in Windows 7 and Windows 10 by using the “setx” command.
`setx path "%path%;c:\directoryPath"`
The following command can be run to add `c:/dir1/dir2` to the path variable.
`setx path "%path%;c:\dir1\dir2"`
It is also possible to use “pathman.exe” from the Windows resource kit. In addition to removing a directory from the path variable, we are also able to use the command described above. The same applies to Windows 7 as well.
### Add directory to system path to set environment variable Windows cmd:
Step-1: Open the command prompt from the administrator.
Step-2: Then you need to run the following command below.
`pathman /as directoryPath`
### Environment variable for system path should be removed:
From an elevated command prompt, execute the following command.
`pathman /rs directoryPath`
### An environment variable to set the user path:
Users do not need admin privileges to access user environment variables. Add the following command to the environment variable user path to add a directory.
`pathman /au directoryPath`
Here is the command you can run to remove the directories from the user path.
`pathman /ru directoryPath`
#### As a default, ‘2’ times is the maximum amount of time you can allow each user:
Having no double quotes around ‘path’ results in this error. For an example of setting the Firefox path, see the following.
```C:\Users\>setx path %path%;"c:\Program Files (x86)\Mozilla Firefox\"
ERROR: Invalid syntax. Default option is not allowed more than '2' time(s).
Type "SETX /?" for usage.```
When you move `%path%` inside double quotes it will look like this:
```C:\Users\>setx path "%path%;c:\Program Files (x86)\Mozilla Firefox\"
SUCCESS: Specified value was saved.
C:\Users\>```
This site uses Akismet to reduce spam. Learn how your comment data is processed. | {"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": 0.9484498500823975, "perplexity": 1886.782815358092}, "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/1652662517245.1/warc/CC-MAIN-20220517095022-20220517125022-00768.warc.gz"} |
http://mathhelpforum.com/differential-geometry/130036-finding-roots-polynomial.html | # Math Help - Finding roots of a polynomial
1. ## Finding roots of a polynomial
Prove the polynomial x^6 + x^4 +5x^2 + 1 has at least four real roots.
This is what I have so far:
1. f(x)=x^6 + x^4 +5x^2 + 1 is continuous because it is a polynomial
2. f(0)= 1>0
3. f(-1)= -2<0
Since 1-3 are satisfied, there exists x in (0,1) such that for f(x)=0
Using f(-0.5), f(-0.75), f(-0.875) we can find a root.
My question is: Is there another way of find the roots besides the way I'm doing it since I not only have to find one root, i have to find four?
2. Originally Posted by summerset353
Prove the polynomial x^6 + x^4 +5x^2 + 1 has at least four real roots.
This is what I have so far:
1. f(x)=x^6 + x^4 +5x^2 + 1 is continuous because it is a polynomial
2. f(0)= 1>0
3. f(-1)= -2<0
Since 1-3 are satisfied, there exists x in (0,1) such that for f(x)=0
Using f(-0.5), f(-0.75), f(-0.875) we can find a root.
My question is: Is there another way of find the roots besides the way I'm doing it since I not only have to find one root, i have to find four?
First , $f(-1)=8\neq -2$ , second the polynomial you wrote has no real roots at all since all the powers of x are even and thus its minimal value is $f(0)=1$ ...
Tonio | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 2, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8889188766479492, "perplexity": 577.8364923817763}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-10/segments/1394678694248/warc/CC-MAIN-20140313024454-00037-ip-10-183-142-35.ec2.internal.warc.gz"} |
http://mathhelpforum.com/calculus/40804-two-problems-about-openess.html | 1. ## Two problems about openess
Show that $S = \{ (x,y) \in \mathbb {R}^2 : xy > 1 \}$ and $C = \{ x \in \mathbb {R}^n : d(x,y) < 1 \ , \ y \in B$, B being any set, are open.
I understand that for both problems I need to pove that $D(x, t )$ are subsets of S and C for some positive number t. But, in the first one, I don't know how to get the y involve.
For the second one, I have w in D(x,a), then |w-y| = |w-x+x-y| < |w-x| + |x-y| = |w-x| + 1, but then... how do I get it < 1?
We don't know what $B$ is. Please tell us. | {"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": 0.8089102506637573, "perplexity": 621.2288806246631}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-50/segments/1480698542932.99/warc/CC-MAIN-20161202170902-00409-ip-10-31-129-80.ec2.internal.warc.gz"} |
https://math.stackexchange.com/questions/2793085/prove-that-there-is-no-subspace-which-is-invariant-under-t-and-complementary-t | # Prove that there is no subspace which is invariant under $T$ and complementary to $W_1$
This is problem 6.7.2 from Hoffman and Kunze.
Let $T$ be a linear operator on $\mathbb{R}^2$, which in the standard basis is given by, $$\begin{bmatrix}2 & 1 \\ 0 & 2\end{bmatrix}.$$ Let $W_1$ be the subspace spanned by the vector $\epsilon_1=(1,0)$. First, prove that $W_1$ is invariant under $T$. Second, prove that there is no subspace $W_2$ which is invariant under $T$ and complementary to $W_1$, such that $W_1\oplus W_2=\mathbb{R}^2$.
The first part is trivial. Any vector in $W_1$ must be of the form $(a,0)$. We note that $T(a,0)=(2a,0)$, which is still in the desired form. So, $W_1$ is invariant under $T$.
Here is my attempt at the second part:
First, we note that $T$ is not diagonalizable.
Next, we have from Theorem 9 that if $V=W_1\oplus W_2$ then there exist projection operators, $E_1,E_2$ satisfying the following properties:
• $E_1E_2=0$.
• $I=E_1+E_2$.
• The range of $E_i$ is $W_i$.
We note then that because the two subspaces are invariant under $T$, that their direct sum must also be invariant under $T$. That is, for $v\in\mathbb{R}^2$, we have that $Tv\in\mathbb{R}^2$. Any $v$ can be written in the form $c_1\epsilon_1+c_2\epsilon_2$. So we also have that $Tv=c_1'\epsilon_1+c_2'\epsilon_2$. From this we have, $T=c_1'E_1+c_2'E_2$. By Theorem 11 then, $T$ is diagonalizable, from where the contradiction arises.
• "$T$ is not diagonalizable" is the key element here. The rest is just expanding on what "diagonalizable" means (and is correct, though it could be tightened-up). But I think you need to do more that just say that $T$ is not diagonalizable. – Paul Sinclair May 23 '18 at 23:49
Suppose $$\mathbb R^2 = W_1 \oplus W_2$$. Then to span the second coordinate of $$\mathbb R^2$$, vectors in $$W_2$$ must be multiples of $$(b, c)$$ where $$c \ne 0$$. The coordinate matrix of $$T(b, c)$$ is $$\begin{bmatrix}2 & 1\\ 0 & 2\end{bmatrix} \begin{bmatrix}b\\ c\end{bmatrix} = \begin{bmatrix}2b + c\\ 2c\end{bmatrix} = 2\begin{bmatrix} b\\ c\end{bmatrix} + c\begin{bmatrix} 1\\ 0\end{bmatrix},$$ so $$T(b, c) = 2(b, c) + c(1, 0)$$ which cannot be in $$W_2$$ because $$c \ne 0$$ prevents $$2(b, c) + c(1, 0)$$ from being a multiple of $$(b, c)$$. Thus, there is no such subspace $$W_2$$ that is invariant under $$T$$. | {"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": 14, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9915109872817993, "perplexity": 76.61511719310106}, "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/1579251796127.92/warc/CC-MAIN-20200129102701-20200129132701-00495.warc.gz"} |
https://richardzach.org/2004/04/ | # Decidability of quantified propositional intuitionistic logic and S4 on trees of height and arity ≤ ω
Journal of Philosophical Logic 33 (2004) 155–164.
Quantified propositional intuitionistic logic is obtained from propositional intuitionistic logic by adding quantifiers $$\forall p$$, $$\exists p$$, where the propositional variables range over upward-closed subsets of the set of worlds in a Kripke structure. If the permitted accessibility relations are arbitrary partial orders, the resulting logic is known to be recursively isomorphic to full second-order logic (Kremer, 1997). It is shown that if the Kripke structures are restricted to trees of at height and width at most $$\omega$$, the resulting logics are decidable. This provides a partial answer to a question by Kremer. The result also transfers to modal S4 and some Gödel-Dummett logics with quantifiers over propositions.
Review: M. Yasuhara (Zentralblatt 1054.03011)
Preprint
# Hilbert’s “Verunglückter Beweis,” the first epsilon theorem, and consistency proofs
History and Philosophy of Logic 25(2) (2004) 79–94.
In the 1920s, Ackermann and von Neumann, in pursuit of Hilbert’s Programme, were working on consistency proofs for arithmetical systems. One proposed method of giving such proofs is Hilbert’s epsilon-substitution method. There was, however, a second approach which was not reflected in the publications of the Hilbert school in the 1920s, and which is a direct precursor of Hilbert’s first epsilon theorem and a certain ‘general consistency result’ due to Bernays. An analysis of the form of this so-called ‘failed proof’ sheds further light on an interpretation of Hilbert’s Program as an instrumentalist enterprise with the aim of showing that whenever a ‘real’ proposition can be proved by ‘ideal’ means, it can also be proved by ‘real’, finitary means.
Review: Dirk Schlimm (Bulletin of Symbolic Logic 11/2 (2005) 247–248)
Preprint | {"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": 0.9486478567123413, "perplexity": 1167.0188148470831}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-39/segments/1568514573331.86/warc/CC-MAIN-20190918193432-20190918215432-00532.warc.gz"} |
https://par.nsf.gov/search/author:%22Woods,%20N.%20L.%22 | # Search for:All records
Creators/Authors contains: "Woods, N. L."
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1. Abstract The accurate simulation of additional interactions at the ATLAS experiment for the analysis of proton–proton collisions delivered by the Large Hadron Collider presents a significant challenge to the computing resources. During the LHC Run 2 (2015–2018), there were up to 70 inelastic interactions per bunch crossing, which need to be accounted for in Monte Carlo (MC) production. In this document, a new method to account for these additional interactions in the simulation chain is described. Instead of sampling the inelastic interactions and adding their energy deposits to a hard-scatter interaction one-by-one, the inelastic interactions are presampled, independent of the hard scatter, and stored as combined events. Consequently, for each hard-scatter interaction, only one such presampled event needs to be added as part of the simulation chain. For the Run 2 simulation chain, with an average of 35 interactions per bunch crossing, this new method provides a substantial reduction in MC production CPU needs of around 20%, while reproducing the properties of the reconstructed quantities relevant for physics analyses with good accuracy.
Free, publicly-accessible full text available December 1, 2023
2. Abstract During LHC Run 2 (2015–2018) the ATLAS Level-1 topological trigger allowed efficient data-taking by the ATLAS experiment at luminosities up to 2.1 $$\times$$ × 10 $$^{34}$$ 34 cm $$^{-2}$$ - 2 s $$^{-1}$$ - 1 , which exceeds the design value by a factor of two. The system was installed in 2016 and operated in 2017 and 2018. It uses Field Programmable Gate Array processors to select interesting events by placing kinematic and angular requirements on electromagnetic clusters, jets, $$\tau$$ τ -leptons, muons and the missing transverse energy. It allowed to significantly improve the background event rejection and signal event acceptance, in particular for Higgs and B -physics processes.
Free, publicly-accessible full text available January 1, 2023
3. Abstract Several improvements to the ATLAS triggers used to identify jets containing b -hadrons ( b -jets) were implemented for data-taking during Run 2 of the Large Hadron Collider from 2016 to 2018. These changes include reconfiguring the b -jet trigger software to improve primary-vertex finding and allow more stable running in conditions with high pile-up, and the implementation of the functionality needed to run sophisticated taggers used by the offline reconstruction in an online environment. These improvements yielded an order of magnitude better light-flavour jet rejection for the same b -jet identification efficiency compared to the performance in Run 1 (2011–2012). The efficiency to identify b -jets in the trigger, and the conditional efficiency for b -jets that satisfy offline b -tagging requirements to pass the trigger are also measured. Correction factors are derived to calibrate the b -tagging efficiency in simulation to match that observed in data. The associated systematic uncertainties are substantially smaller than in previous measurements. In addition, b -jet triggers were operated for the first time during heavy-ion data-taking, using dedicated triggers that were developed to identify semileptonic b -hadron decays by selecting events with geometrically overlapping muons and jets.
Free, publicly-accessible full text available December 1, 2022
4. A correction to this paper has been published: 10.1140/epjc/s10052-021-09344-w
5. A bstract A search for dark-matter particles in events with large missing transverse momentum and a Higgs boson candidate decaying into two photons is reported. The search uses 139 fb − 1 of proton-proton collision data collected at $$\sqrt{s}$$ s = 13 TeV with the ATLAS detector at the CERN LHC between 2015 and 2018. No significant excess of events over the Standard Model predictions is observed. The results are interpreted by extracting limits on three simplified models that include either vector or pseudoscalar mediators and predict a final state with a pair of dark-matter candidates and a Higgs boson decaying into two photons.
6. Abstract This paper presents a search for dark matter in the context of a two-Higgs-doublet model together with an additional pseudoscalar mediator, a , which decays into the dark-matter particles. Processes where the pseudoscalar mediator is produced in association with a single top quark in the 2HDM+ a model are explored for the first time at the LHC. Several final states which include either one or two charged leptons (electrons or muons) and a significant amount of missing transverse momentum are considered. The analysis is based on proton–proton collision data collected with the ATLAS experiment at $$\sqrt{s} = 13$$ s = 13 TeV during LHC Run 2 (2015–2018), corresponding to an integrated luminosity of 139 $$\hbox {fb}^{-1}$$ fb - 1 . No significant excess above the Standard Model predictions is found. The results are expressed as 95% confidence-level limits on the parameters of the signal models considered.
7. Abstract Jet energy scale and resolution measurements with their associated uncertainties are reported for jets using 36–81 fb $$^{-1}$$ - 1 of proton–proton collision data with a centre-of-mass energy of $$\sqrt{s}=13$$ s = 13 $${\text {Te}}{\text {V}}$$ TeV collected by the ATLAS detector at the LHC. Jets are reconstructed using two different input types: topo-clusters formed from energy deposits in calorimeter cells, as well as an algorithmic combination of charged-particle tracks with those topo-clusters, referred to as the ATLAS particle-flow reconstruction method. The anti- $$k_t$$ k t jet algorithm with radius parameter $$R=0.4$$ R = 0.4 is the primary jet definition used for both jet types. This result presents new jet energy scale and resolution measurements in the high pile-up conditions of late LHC Run 2 as well as a full calibration of particle-flow jets in ATLAS. Jets are initially calibrated using a sequence of simulation-based corrections. Next, several in situ techniques are employed to correct for differences between data and simulation and to measure the resolution of jets. The systematic uncertainties in the jet energy scale for central jets ( $$|\eta |<1.2$$ | η | < 1.2 ) vary from 1% for a wide range of high- $$p_{{\text {T}}}$$ p Tmore » | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 2, "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": 0.918694257736206, "perplexity": 1626.9461369017893}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-49/segments/1669446710719.4/warc/CC-MAIN-20221130024541-20221130054541-00379.warc.gz"} |
http://math.stackexchange.com/questions/37108/what-kind-of-series-is-this-and-how-do-i-sum-it | # What kind of series is this, and how do I sum it?
$\displaystyle\sum{\frac1{a_n}}$, e.g. $$\frac 1 1 + \frac 1 2 + \frac 1 3 + \frac 1 4 + \dots + \frac 1 n$$ or
$$\frac 1 2 + \frac 1 4 + \frac 1 6 + \frac 1 8 + \dots + \frac 1 {2n}$$
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You are looking for harmonic numbers. An excellent article can be found here. The corresponding (diverging) series is the harmonic series. To sum it, use the integral representation given in the first article:
$$\int_0^1 \frac{1 - x^n}{1 - x}\,{\rm d}x = \int_0^1 (1 + x + \cdots + x^{n-1})\ {\rm d}x = 1 + \frac12 + \cdots + \frac1{n} = H_n$$
Maybe if it's not clear the left one you mentioned is $H_n$, the right one is obviously $\frac12\cdot H_n$.
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That is a harmonic series. You can approximate it by integration, as explained in the link. Note that if $n \to \infty$, the sum diverges.
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A nice approximation for $n \to \infty$ can be derived from
\eqalign{ & \log \left( {1 + x} \right) = \sum\limits_{k = 0}^\infty {{{\left( { - 1} \right)}^k}\frac{{{x^{k + 1}}}}{{k + 1}}} = x - \frac{{{x^2}}}{2} + \frac{{{x^3}}}{3} - + \cdots \cr & \log \left( {1 + \frac{1}{n}} \right) = \sum\limits_{n = 0}^\infty {{{\left( { - 1} \right)}^k}\frac{1}{{{n^{k + 1}}}}\frac{1}{{k + 1}}} = \frac{1}{n} - \frac{1}{{2{n^2}}} + \frac{1}{{3{n^3}}} - + \cdots \cr}
$$\log \left( {n + 1} \right) - \log n = \sum\limits_{n = 0}^\infty {{{\left( { - 1} \right)}^k}\frac{1}{{{n^{k + 1}}}}\frac{1}{{k + 1}}} = \frac{1}{n} - \frac{1}{{2{n^2}}} + \frac{1}{{3{n^3}}} - + \cdots$$
If we sum for $n=1,2,3,\dots,m$ we have
$$\log \left( {m + 1} \right) = \sum\limits_{n = 1}^m {\frac{1}{n}} - \frac{1}{2}\sum\limits_{n = 1}^m {\frac{1}{{{n^2}}}} + \frac{1}{3}\sum\limits_{n = 1}^m {\frac{1}{{{n^3}}}} - + \cdots$$ or
$$\sum\limits_{n = 1}^m {\frac{1}{n}} - \log \left( {m + 1} \right) = \frac{1}{2}\sum\limits_{n = 1}^m {\frac{1}{{{n^2}}}} - \frac{1}{3}\sum\limits_{n = 1}^m {\frac{1}{{{n^3}}}} + - \cdots$$
If we let $m\to \infty$ , we have that
$$\mathop {\lim }\limits_{m \to \infty } \left[ {\sum\limits_{n = 1}^m {\frac{1}{n}} - \log \left( {m + 1} \right)} \right] = \frac{1}{2}\zeta \left( 2 \right) - \frac{1}{3}\zeta \left( 3 \right) + \frac{1}{4}\zeta \left( 4 \right) - + = \sum\limits_{n = 2}^\infty {{{\left( { - 1} \right)}^n}\frac{{\zeta \left( n \right)}}{n}}$$
where $$\zeta \left( n \right) = \sum\limits_{k = 1}^\infty {\frac{1}{{{k^n}}}}$$
The value of the right hand side is know as Euler's constant, $\gamma \approx 0.577$, and this result tells us that for large values of $m$, we can approximate the $m$th harmonic number by
$$\sum\limits_{n = 1}^m {\frac{1}{n}} \sim \log \left( {m + 1} \right) + \gamma$$
- | {"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": 0.9946698546409607, "perplexity": 298.9292945271397}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-48/segments/1448398460263.61/warc/CC-MAIN-20151124205420-00290-ip-10-71-132-137.ec2.internal.warc.gz"} |
https://www.gradesaver.com/textbooks/math/algebra/intermediate-algebra-for-college-students-7th-edition/chapter-5-section-5-5-factoring-special-forms-concept-and-vocabulary-check-page-371/3 | ## Intermediate Algebra for College Students (7th Edition)
$(A-B)^2$
RECALL: A perfect square trinomial can be factored using either of the following formulas: (i) $a^2-2ab+b^2=(a-b)^2$ (ii) $a^2+2ab+b^2=(a+b)^2$ The given perfect square trinomial is similar in form to the one in formula (i) above. Thus, the missing expression in the given statement is: $(A-B)^2$ | {"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": 0.8855484127998352, "perplexity": 341.83424709616014}, "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-2018-30/segments/1531676590901.10/warc/CC-MAIN-20180719125339-20180719145339-00110.warc.gz"} |
http://mathoverflow.net/questions/19210/holomorphic-vector-fields-acting-on-dolbeault-cohomology | # Holomorphic vector fields acting on Dolbeault cohomology
## The question.
Let $(X, J)$ be a complex manifold and $u$ a holomorphic vector field, i.e. $L_uJ = 0$. The holomorphicity of $u$ implies that the Lie derivative $L_u$ on forms preserves the (p,q) decomposition and also that it commutes with $\bar{\partial}$. From this it follows that $u$ acts infinitesimally on the Dolbeault cohomology groups $H^{p,q}(X)$ of $X$. My question is, does anyone know of an example in which this action is non-trivial?
## Some context.
To give some context, first note that the analgous action for de Rham cohomology is always trivial: If $M$ is any smooth manifold and $v$ any vector field, then the formula $L_v = d \circ i_v + i_v \circ d$ shows that the infinitesimal action of $v$ on de Rham cohmology is trivial. (This is an instance of the more general fact that homotopic maps induce the same homomorphisms on singular cohomology. The field $v$ generates diffeomorphisms which are by construction isotopic to the identity map.)
Returning to Dolbeault cohomology, suppose we know that each Dolbeault class is represented by a $d$-closed form. (For example, this is true if $X$ is a compact Kähler manifold, by Hodge theory.) Then the action is necessarily trivial. The proof is as follows. Let $\alpha$ be a $\bar{\partial}$-closed (p,q)-form which is also $d$-closed. Then we know that $L_u \alpha = d(i_u \alpha)$ is also of type (p,q). So, $$L_u\alpha = \bar{\partial}\left((i_u\alpha)^{p, q-1}\right) + \partial\left((i_u \alpha)^{p-1, q}\right)$$ and the other contributions $\bar{\partial}((i_u\alpha)^{p-1,q}$) and $\partial((i_u\alpha)^{p,q-1})$ vanish. Now the fact that $\bar\partial((i_u\alpha)^{p-1,q}) = 0$ and our hypothesis imply that there is a (p-1, q-1)-form $\beta$ such that $$(i_u\alpha)^{p-1,q}+ \bar\partial \beta$$ is closed. Hence $$\partial \left((i_u\alpha)^{p-1,q}\right) = \bar\partial \partial \beta$$ and so $$L_u\alpha = \bar \partial \left( (i_u \alpha)^{p,q-1} + \partial \beta\right)$$ which proves the action of $u$ on $H^{p,q}(X)$ is trivial.
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Mayebe I m totally wrong, but does the Lefschetz fix point formula help: Integrate up your vector field u until some fix points occur. Hopefully the action on $H^{p,q}$ of $u$ integrate up gives one the same. – Sebastian Mar 25 '10 at 7:16
@Sebastian. Are you hoping to prove that the action is necessarily trivial? As far as I understand, the Lefschetz fixed point formula gives the number of fixed points of a (suitable) map (counted appropriately) in terms of the trace of its action on cohomology. There is certainly a Dolbeault version - due to Atiyah-Bott (ams.org/bull/1966-72-02/S0002-9904-1966-11483-0/…) - but I don't quite see how it helps, since it will only ever tell you about the trace of the action on Hp;q – Joel Fine Mar 25 '10 at 15:27
Take a complex nilpotent or solvable group $G$ with the right action by a co-compact lattice $\Gamma$ and conisder the quotient $G/\Gamma$. On this quotient right-invariant $1$-forms give a subspace of $H^{1,0}$. The group $G$ is acting on $G/\Gamma$ on the left and if it would presrve all the $1$-forms, $G$ would be abelian.
Torsten Ekedahl expained that what is following IS NOT CORRECT (the article of Hasegawa tells something different)
In fact, the simplest example of this kind is given by primary Kodaira surfaces (http://en.wikipedia.org/wiki/Kodaira_surface), they have two holomorphic $1$-forms. These surfaces are described as quotinets of sovlable groups, for example, in an article of Keizo Hasegawa http://arxiv.org/PS_cache/math/pdf/0401/0401413v1.pdf
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A comment to Dmitri's answer is given below but is too long to give as a comment. I write here as I believe that this will mean that he will be informed of my comments. – Torsten Ekedahl Apr 19 '10 at 18:19
Dear Torsten, thanks a lot for your comment! Indeed what you have said about surfaces shows that this can not work, I must have misinterpreted Hasegawa's article... – Dmitri Apr 19 '10 at 19:02
I don't think the surface example can work as all holomorphic forms on a compact surface are closed (a result due to Kodaira I believe). The Cartan formula $L_v = d\iota_v+\iota_vd$ shows that if all holomorphic forms are closed then vector fields act trivially.
If I understand the Hasegawa paper correctly he does not claim that the primary Kodaire surfaces are of the form $G/\Gamma$ for a complex Lie group $G$, only that the manifold underlying $G$ can be given a complex structure such that $\Gamma$ acts holomorphically.
Dmiri's method does work however. One example is given by the complex Heisenberg group (i.e., strict upper triangular $3\times3$-matrices) divided by the subgroup whose matrices have entries in the Gaussian integers. This then also gives an example of a compact complex manifold with non-closed holomorphic $1$-forms. In fact, for translation invariant $1$-forms the exterior differential, which is the map $d\colon \mathfrak g^\ast \to \Lambda^2\mathfrak g^\ast$ dual to the Lie bracket.
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Thank you very much for clarifying Dmitri's answer. I hope you don't mind, but I felt I should accept his, since it was first and I guess made the main point, even if the specific example he gave wasn't right. If I could accept both answers, I would have done! – Joel Fine Apr 21 '10 at 9:36
Both the above examples are for compact $X$, which the original problem didn't specify. If you allow $X$ not to be compact, the question is much simpler. Take $X = \mathbb{C}$. Then $H^{0,0}(X)$ is the vector space of entire holomorphic functions, and $\partial/\partial z$ acts nontrivially.
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Yes, sorry, I forgot to specify that I was really looking for compact examples. – Joel Fine Feb 8 '12 at 16:13 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 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": 0.9649287462234497, "perplexity": 288.028745054906}, "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/1427131297172.60/warc/CC-MAIN-20150323172137-00172-ip-10-168-14-71.ec2.internal.warc.gz"} |
https://byjus.com/jee/energy-stored-capacitor/ | # Energy Stored in a Capacitor
The energy stored on a capacitor is expressed in terms of the work performed by the battery. The Voltage shows energy per unit charge, hence the work performed to move a charge element DQ from negative plate to the positive plate equals to V DQ where V is the voltage of the capacitor. The voltage is proportional to a number of charges on the capacitor.
When the capacitor is connected across a battery, the charges enter from the battery and are stored in the plates of capacitors. At the beginning, the capacitor does not hold any potential or charge, i.e, q = 0 C and V = 0 volts.
At the time to switch, the complete battery voltage falls across the capacitor. A +ve charge (q) enters at the capacitor’s positive plate but no work is done in bringing the 1st charge (q) towards the positive plate from the battery.
This occurs because capacitor does not contain any voltage across its plates and the previous voltage exists because of the battery.
## Capacitor Working
First, the charge initiates little voltage across the capacitor’s plate and second +ve charge enters the capacitor’s positive plate but replaced by the 1st charge. Since the battery’s voltage is higher than the capacitor’s voltage, the second charge is stored in positive plate. In this condition, little work has to be performed to store 2nd charge in the capacitor. A similar phenomenon occurs for the 3rd charge. Gradually the charges are stored against pre-stored charges and their small amount of work done hikes up.
The capacitor’s voltage is not constant from the start. It is maximum when the potential of the capacitor is equal to that of a battery. The voltage of capacitor rises and so does the capacitor’s energy as the storage of charges increases.
As the voltage rises the electric field in the capacitor dielectric hikes gradually but from positive to negative plate, i.e. in opposite direction.
$E=-frac{dV}{dX}$
Here, dx indicates the distance between the two plates.
The charge flows from battery to capacitor plate till the capacitor acquires potency same as the battery. Hence we need to determine the capacitor’s energy to the final moment of getting the complete charge.
Suppose a charge is stored in positive plate according to the battery’s voltage and the work done is DW.
If the whole charging time is considered, it can be written as:
Now let’s consider the energy lost by a capacitor during charging time.
Since the battery has a constant voltage, the energy lost always obeys the equation W=V.q. The equation doesn’t apply to the capacitor since it doesn’t carry constant voltage from the point it has started charging.
The collected charge by from the battery is given by:
The charge lost by the battery is given by:
The half energy from the total amount moves to the capacitor and the rest gets off from the battery.
#### Practise This Question
H.C.F of two co prime numbers is | {"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": 0.8904133439064026, "perplexity": 518.4986376580803}, "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-2018-47/segments/1542039745281.79/warc/CC-MAIN-20181119043725-20181119065725-00322.warc.gz"} |
http://mathhelpforum.com/pre-calculus/95974-number-sign.html | 1. ## Number and sign
m is the real parameter.
Find the real number and sign for the equation :
By two methods : graphic , algebric.
2. solve the equation in $\mathbb{C}$ perhaps ?
x^2+6x+9m=0
$\delta=36(1-m)$
for m>1 , m=1, m<1
3. Originally Posted by J.R
solve the equation in $\mathbb{C}$ perhaps ?
x^2+6x+9m=0
$\delta=36(1-m)$
for m>1 , m=1, m<1
Hello : Thank you this equation in R
4. so what happened when m<1 ?
m=1 ?
and m>1 ?
5. Graphic.
We have $m=-\frac{1}{9}x^2-\frac{2}{3}x$
Intersect the graph of $f(x)=-\frac{1}{9}x^2-\frac{2}{3}x$ with parallel lines to x-axis with equation $y=m$
If $m>1$ then the line doesn't intersect the graph.
If $m=1$ the line is tangent to the graph and the equation has a double solution $x=-3$
If $m\in(0,1)$ the line intersects the graph in 2 points and the equation has two negative solutions.
If $m=0$ then the equation has the solutions $x_1=-6, \ x_2=0$
If $m<0$ then the equation has one negative solution and one positive solution.
Attached Thumbnails
6. Originally Posted by red_dog
Graphic.
We have $m=-\frac{1}{9}x^2-\frac{2}{3}x$
Intersect the graph of $f(x)=-\frac{1}{9}x^2-\frac{2}{3}x$ with parallel lines to x-axis with equation $y=m$
If $m>1$ then the line doesn't intersect the graph.
If $m=1$ the line is tangent to the graph and the equation has a double solution $x=-3$
If $m\in(0,1)$ the line intersects the graph in 2 points and the equation has two negative solutions.
If $m=0$ then the equation has the solutions $x_1=-6, \ x_2=0$
If $m<0$ then the equation has one negative solution and one positive solution.
HELLO : ALGEBRIC SOLUTION
Student sign of : delta , p=x1 ×x2 and s=x1 + x 2 | {"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": 24, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8968866467475891, "perplexity": 691.5343448517684}, "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/1368711005985/warc/CC-MAIN-20130516133005-00001-ip-10-60-113-184.ec2.internal.warc.gz"} |
https://mathoverflow.net/questions/37737/is-the-grothendieck-ring-of-varieties-reduced | # Is the Grothendieck ring of varieties reduced?
A neat construction of Bjorn Poonen shows that the Grothendieck ring of varieties (over a field of char. 0) is not a domain: http://arxiv.org/abs/math/0204306
Is the Grothendieck ring of varieties reduced? (My guess: the answer is yes, the proof is easy enough that several people have observed this without writing it up anywhere. But I don't know how to show it.)
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In positive characteristic $p$, if you take two supersingular elliptic curves $E_1, E_2$, then $E_i\times E_j$ is isomorphic to $E_1^2$ for any pair $i,j$, so $([E_1]-[E_2])^2=0$. But it is unclear for me why should $[E_1]=[E_2]$. – Qing Liu Sep 9 '10 at 7:41
Qing Liu's example probably works, only we don't know if an abelian variety in positive characteric is determined by its class in $K_0(\mathrm{Var}_k)$. However, we do know that in characteristic zero (this is what Bjorn Poonen uses in his examples) and the non-cancellation is a purely arithmetic phenomenon and hence can be realised in characteristic zero.
Hence, we let $\mathcal A$ be a maximal order in a definite (i.e., $\mathcal A\otimes\mathbb R$ is non-split) quaternion algebra over $\mathbb Q$. There is an abelian variety $A$ over some field $k$ of characteristic zero with $\mathcal A=\mathrm{End}(A)$ (Bjorn works hard to get his example defined over $\mathbb Q$, here I make no such claim). For any (right) f.g. projective (i.e., torsion free) $\mathcal A$-module $M$ we may define an abelian variety $M\bigotimes_{\mathcal A}A$ characterised by $\mathrm{Hom}(M\bigotimes_{\mathcal A}A,B)=\mathrm{Hom}_{\mathcal A}(M,\mathrm{Hom}(A,B))$ for all abelian varieties (concretely it is constructed by realising $M$ is the kernel of an idempotent of some $\mathcal A^n$ and then taking the kernel of the same idempotent acting on $A^n$). In any case we see that $M$ and $N$ are isomorphic precisely when $M\bigotimes_{\mathcal A}A$ is isomorphic to $N\bigotimes_{\mathcal A}A$.
Now (all the arithmetic results used below can be found in for instance Irving Reiner: Maximal orders, Academic Press, London-New York), the class group of $\mathcal A$ is equal to the ray class group of $\mathbb Q$ with respect to the infinite prime, i.e., the group of fractional ideals of $\mathbb Q$ modulo ideals with a strictly positive generators. As that is all ideals we find that the class group is trivial. Furthermore, we have the Eichler stability theorem which says that projective modules of rank $\geq2$ are determined by their rank and image in the class group and hence are determined by their rank (the rank condition comes in in that $\mathrm{M}_k(\mathcal A)$ is a central simple algebra which is indefinite at the infinite prime). In particular if $M_1$ and $M_2$ are two rank $1$ modules over $\mathcal A$ and $A_1$ and $A_2$ are the corresponding abelian varieties we get that $A_1\bigoplus A_2\cong A\bigoplus A$ as the left (resp. right) hand side is associated to $M_1\bigoplus M_2$ (resp. $\mathcal A^2$). Therefore, to get an example it is enough to give an example of an $\mathcal A$ for which there exist $M_1\not\cong M_2$. The number (or more easily the mass) of isomorphism classes of ideals can be computed using mass formulas and tends to infinity with the discriminant of $\mathcal A$. It is interesting to note that when the discriminant is a prime $p$ we can go backwards using supersingular elliptic curves: The mass is equal to the mass of supersingular elliptic curves in characteristic $p$ and the latter mass can be computed geometrically to be equal to $(p-1)/24$.
-
Very nice example ! So in characteristic 0, the Grothendieck ring of varieties is not reduced (at least over some field $k$). Probably neither in postive characteristic. – Qing Liu Sep 27 '10 at 20:37
Just a clarification: $A$ exists over some number field and hence over all larger fields. – Torsten Ekedahl Sep 28 '10 at 4:23 | {"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": 0.9645591974258423, "perplexity": 159.98174381360127}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-18/segments/1461863352151.52/warc/CC-MAIN-20160428170912-00041-ip-10-239-7-51.ec2.internal.warc.gz"} |
http://math.stackexchange.com/questions/420658/why-does-this-limit-not-exist | # Why does this limit not exist?
Working through some limit exercises. The answer sheet says the limit below does not exist. Is this correct. Shouldn't it be $-\infty$? $$\lim_{x \to 0^+} \left( \frac{1}{\sqrt{x^2+1}} - \frac{1}{x} \right)\ \ \$$
-
Are you sure the question wasn't about just $\lim_{x\to 0}$? As it is, you're right. – tomasz Jun 15 '13 at 0:17
Perhaps the book means it doesn't exist finitely... – DonAntonio Jun 15 '13 at 0:17
## 4 Answers
It is true that as $x$ approaches through positive values, $\frac{1}{\sqrt{1+x^2}}-\frac{1}{x}$ becomes large negative. However, some people do not allow $\infty$ or $-\infty$ as answers to a limit problem. As a simpler example, some would say that $\lim_{x\to\infty}x^2$ does not exist, and some would say $\lim_{x\to\infty} x^2=\infty$.
Remark: Mathematical English has dialects. When you answer a question, it may be necessary to conform to the local dialect. (On a test, I would accept either answer if proper justification were given, but cannot guarantee that someone else would.)
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+1 Love the remark about dialects. It is a pain to deal with it ... – Calvin Lin Jun 15 '13 at 0:50
The same book states that $lim_{x \to \left(\frac{\pi^-}{2}\right)} \left(\tan x\right)^x = \infty$ – hondaman Jun 15 '13 at 1:16
$hondaman: Then they are being very inconsistent, and one one of the answers is incorrect. – André Nicolas Jun 15 '13 at 1:21 @hondaman: See the reply to your comment (if you haven't already). – Cameron Buie Jun 15 '13 at 3:50 $$\lim_{x \to 0^+} \left( \frac{1}{\sqrt{x^2+1}} - \frac{1}{x} \right)=\lim_{x\to 0^+}\frac{x-\sqrt{x^2+1}}{x\sqrt{x^2+1}}=$$ $$=-\lim_{x\to 0^+}\frac1{x\sqrt{x^2+1}(x+\sqrt{x^2+1})}=\lim_{x\to 0^+}-\frac1{x^2\sqrt{x^2+1}+x^3+x}=-\infty$$ so you're right...unless the book meant "doesn't exists finitely " , say. - That is correct. Since$-\infty$is not a real number, the limit does not exist. - since $$\lim_{x \to 0^+} \frac{1}{\sqrt{x^2+1}} =1$$ and $$\lim_{x \to 0^+} = - \frac{1}{x} =-\infty$$ hence you can say the given limit is$-\infty\$ or the limit doesn't exist since it's not finite.
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Needs an upvote! – amWhy May 30 '14 at 16:52 | {"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": 0.892343282699585, "perplexity": 897.315580474403}, "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-2015-40/segments/1443738006497.89/warc/CC-MAIN-20151001222006-00105-ip-10-137-6-227.ec2.internal.warc.gz"} |
https://homework.cpm.org/category/CCI_CT/textbook/calc/chapter/8/lesson/8.3.2/problem/8-116 | ### Home > CALC > Chapter 8 > Lesson 8.3.2 > Problem8-116
8-116.
Suppose f(g(x)) = x, such that f and g are differentiable. If f(1) = 3 and f ′(1) = 2, determine the value of g′(3). Homework Help ✎
Recall that f(g(x)) = x is the definition of inverse functions. In other words, you are being told that f(x) and g(x) are inverses.
Inverse functions have reciprocal slopes at corresponding (x, y)→(y, x) values.
$\text{For example, if }f(x)\text { has coordinate point (2, 3) and }f'(2)=\frac{4}{5},\text{ then }g'(3)=\frac{5}{4}.$ | {"extraction_info": {"found_math": true, "script_math_tex": 1, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.930107831954956, "perplexity": 1263.4320907108724}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-16/segments/1585371883359.91/warc/CC-MAIN-20200410012405-20200410042905-00324.warc.gz"} |
http://physics.stackexchange.com/questions/41321/intuition-of-impulse-formula-j-sum-f-delta-t/41322 | # Intuition of Impulse Formula $J = \sum F \Delta t$
I understand that
\begin{align} J = \sum F \Delta t &= \Delta p \\ \sum F &= \frac{\Delta p}{\Delta t} \\ &= \frac{mv_2 - mv_1}{\Delta t} \\ &= m \cdot \frac{\Delta v}{\Delta t} \\ &= ma \end{align}
$J$ is impulse, $p$ is momentum
However, by just looking at the equation
$$J = \sum F \Delta t$$
I seem to think that impluse is high if for same force, time is higher. But I think that's wrong? I think for a certain force, if I exert it in a short time theres a larger impluse? How can I then understand the above equation more intuitively without expanding it out?
- | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 1, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 1.0000100135803223, "perplexity": 1383.3043152096814}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-18/segments/1461860112228.39/warc/CC-MAIN-20160428161512-00002-ip-10-239-7-51.ec2.internal.warc.gz"} |
https://dsp.stackexchange.com/questions/11262/can-two-nonzero-signals-xn-and-yn-give-a-zero-convolution | # Can two nonzero signals $x[n]$ and $y[n]$ give a zero convolution
Suppose $x[n]$ and $y[n]$ are two nonzero signals(i.e., $x[n] \neq 0$ for at least one value of n and similarly for $y[n]$).Can the convolution between $x[n]$ and $y[n]$ result in an identically zero signal? In other words, is it possible that $\displaystyle\sum_{k = -\infty}^{k = +\infty}x[k]y[n-k] = 0$ for all n.
$$x[k]=1$$
for all $k$ and
$$y[k] = \begin{cases}1 & k=0\\-1 & k=1\\0 & otherwise \end{cases}$$
It is easy to see that in case of a convolution, the result will be zero for all values of $n$.
• It is worth noting that this result only holds if the signal $x[k]$ has infinite length. In practice you will see non-zero convolution tails. Oct 24, 2013 at 22:40 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 0.9423137903213501, "perplexity": 153.31137178915264}, "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/1664030337398.52/warc/CC-MAIN-20221003035124-20221003065124-00743.warc.gz"} |
https://thiseven.blogspot.com/2018/02/blog-post_17.html | 2/17/2018 03:54:00 下午
via here
Often, beginning students in logic have this problem. They read about fallacies like the ones we cover here and then think they see them everywhere. (p. 185) | {"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": 0.8636338710784912, "perplexity": 4887.903238100886}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-43/segments/1634323585504.90/warc/CC-MAIN-20211022084005-20211022114005-00602.warc.gz"} |
https://www.physicsforums.com/threads/net-force-exerted-on-a-object.205187/ | # Net Force Exerted On A Object?
1. Dec 18, 2007
### cpuboye11
Hi,
I would like to first open this post by saying that I am not a slacker, and I try my hardest to figure things out to the best of my abilities.
I am in Physics 1 right now. We are studying what our book calls Forces in One Dimension.
I have a problem, the problem being I have a chapter test tomorrow and no hope of passing.
One of the problems that seem to hunt me and can't figure out for the life of me is finding out the net force that is exerted on a object; such as this:
A race car has a mass of 710 kg. It starts from rest and travels 40.0 m in 3 secs. The car is uniformly accelerated during the entire time. What net force is exerted on it?
It seem it is a easy problem for most people in Physics but I just can't figure it out. Little help please?
Thanks Kyle
2. Dec 18, 2007
### arildno
Determine the acceleration by means of 3 pieces of information given:
1. The acceleration is uniform, i.e, equal to some constant "a"
2. Its initial velocity is zero
3. It travelled 40 meters in 3 seconds.
Set up the relevant kinematic equation, and solve for "a"
Then find the net force exerted upon the car at any given moment by means of Newton's 2.law of motion.
3. Dec 18, 2007
### cpuboye11
*Sorry for putting the topic in the wrong spot*
So this is what i understand or it seems,
Vi = 0
Vf = 40**********
a = ?
t = 3
So i did this: 40-0 / 3 = 13.33
Now, maybe I am just being dumb; but don't I just * 13.33 by the Kg's which is 710kg?
And get 9464?
4. Dec 19, 2007
### arildno
It is not the final velocity that equals 40, don't mix up distance and velocity!!
We have that, since the initial velocity is 0, the distance s travelled in t seconds under uniform acceleration "a" is given by:
$$s=\frac{a}{2}t^{2}$$
Now, solve for "a", knowing that s=40, t=3.
5. Nov 13, 2009
### mause82
hi i have a problem and i need you help...
2.0 kg mud ball drops from rest at height ok 15m. If the impact between the ball and the ground lasts 0.50s. what is the average net force exerted by the ball on the ground.
this is the problem. can you help me
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https://www.physics.uoguelph.ca/problem-6-7-friction-coefficient-part-2 | # Problem 6-7 Friction coefficient - Part 2
A hockey puck travels $33\; m$ along an ice rink before coming to rest. Its initial speed was $11\;m/s$.
(a) What is the magnitude of the acceleration of the puck?
(b) What is the coefficient of kinetic friction between the puck and the ice?
Accumulated Solution
$v_x{^2} = v_{0x}{^2} + 2a_xx \\ a_x = -1.83\; m/s^2 \; \text{ (answer to (a))}$
$v_x{^2} = v_{0x}{^2} + 2a_xx \\ 0^2 = 11^2 + 2a_x(33) \\ a_x = -1.83\; m/s^2$
The minus sign means:
(A) $\text{The object is slowing down.}$
(B) $\text{The object is speeding up.}$
(C) $\text{You can't have a negative acceleration.}$ | {"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": 0.912699818611145, "perplexity": 1188.4020133695903}, "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/1627046152129.33/warc/CC-MAIN-20210726120442-20210726150442-00496.warc.gz"} |
https://socratic.org/questions/what-is-the-distance-between-3-13-10-and-3-17-1 | Algebra
Topics
# What is the distance between (3,13,10) and (3,-17,-1)?
Apr 7, 2016
Distance between $\left(3 , 13 , 10\right)$ and $\left(3 , - 17 , - 1\right)$ is $31.95$ units.
#### Explanation:
Distance between two points $\left({x}_{1} , {y}_{1} , {z}_{1}\right)$ and $\left({x}_{2} , {y}_{2} , {z}_{2}\right)$ is given by $\sqrt{{\left({x}_{2} - {x}_{1}\right)}^{2} + {\left({y}_{2} - {y}_{1}\right)}^{2} + {\left({z}_{2} - {z}_{1}\right)}^{2}}$.
Hence distance between $\left(3 , 13 , 10\right)$ and $\left(3 , - 17 , - 1\right)$ is
$\sqrt{{\left(3 - 3\right)}^{2} + {\left(\left(- 17\right) - 13\right)}^{2} + {\left(\left(- 1\right) - 10\right)}^{2}} = \sqrt{{\left(0\right)}^{2} + {\left(- 17 - 13\right)}^{2} + {\left(- 1 - 10\right)}^{2}}$
= $\sqrt{{\left(0\right)}^{2} + {\left(- 30\right)}^{2} + {\left(- 11\right)}^{2}} = \sqrt{0 + 900 + 121}$
= $\sqrt{1021} = 31.95$
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https://export.arxiv.org/abs/2107.09719?context=physics.space-ph | physics.space-ph
(what is this?)
# Title: Asymmetric One-Dimensional Slow Electron Holes
Abstract: Slow solitary positive-potential peaks sustained by trapped electron deficit in a plasma with asymmetric ion velocity distributions are in principle asymmetric, involving a potential change across the hole. It is shown theoretically how to construct such asymmetric electron holes, thus providing fully consistent solutions of the one-dimensional Vlasov-Poisson equation for a wide variety of prescribed background ion velocity distributions. Because of ion reflection forces experienced by the hole, there is generally only one discrete slow hole velocity that is in equilibrium. Moreover the equilibrium is unstable unless there is a local minimum in the ion velocity distribution, in which the hole velocity then resides. For stable equilibria with Maxwellian electrons, the potential drop across the hole is shown to be $\Delta\phi\simeq{2\over 9}f'''{T_e\over e}({e\psi\over m_i})^2$, where $\psi$ is the hole peak potential, $f'''$ is the third derivative of the background ion velocity distribution function at the hole velocity, and $T_e$ the electron temperature. Potential asymmetry is small for holes of the amplitudes usually observed, $\psi\lesssim 0.5T_e/e$.
Subjects: Plasma Physics (physics.plasm-ph); Space Physics (physics.space-ph) DOI: 10.1103/PhysRevE.104.055207 Cite as: arXiv:2107.09719 [physics.plasm-ph] (or arXiv:2107.09719v1 [physics.plasm-ph] for this version)
## Submission history
From: Ian Hutchinson [view email]
[v1] Tue, 20 Jul 2021 18:40:57 GMT (638kb,D)
Link back to: arXiv, form interface, contact. | {"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": 0.805681049823761, "perplexity": 2792.5572151093684}, "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-2022-05/segments/1642320300722.91/warc/CC-MAIN-20220118032342-20220118062342-00121.warc.gz"} |
http://math.stackexchange.com/users/1778/charles?tab=activity | Charles
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1d revised Is it allowed to define a number system where a number has more than 1 representation? second example 1d comment define the “optimal” automatic theorem prover I don't think that any extant theorem provers work like that. As I said earlier, HR is the only system I know that works along those lines -- but it's not a theorem prover, I think it uses Otter (and MACE?) to do that. 1d comment define the “optimal” automatic theorem prover See pages.cs.miami.edu/~tptp/CASC/J7/SystemDescriptions.html for descriptions of the strategies employed by some of the leading ATP systems. 1d comment define the “optimal” automatic theorem prover The state of the art is much smarter than merely randomly exploring a tree! 1d comment define the “optimal” automatic theorem prover But we don't have any automated provers which understand concepts in that sense. They're much smarter than the random prover but not at all insightful. 1d comment define the “optimal” automatic theorem prover Most theorem provers are of type (1), and usually they are quite limited -- in a practical sense you ask them to prove small steps toward a theorem, but you still have to map out the path. The only example I know of for (2) is Colton's HR. 1d comment define the “optimal” automatic theorem prover I don't think that (2) is equivalent to (1). A theorem prover might have a thousand or even a million sub-results along the way to prove a certain theorem, but not know which are interesting and which are boring. 2d answered Is it allowed to define a number system where a number has more than 1 representation? 2d comment Union of a null set and a non-measurable set @Bey: Yes, that is precisely my reason for asking! 2d accepted Union of a null set and a non-measurable set 2d asked Union of a null set and a non-measurable set 2d comment Where $ax + b$ prime infinitely often, is $ax + b - 2$ semiprime at least once? @RobertGross: Right, to say that you'd need Dickson's conjecture, as I mentioned. It doesn't follow from Dirichlet alone. Jun18 answered Is this a valid equivalent expression of the twin prime conjecture? Jun18 answered Estimating total number of twin primes Jun17 answered How big is the chance that a arbitrary man is taller than a arbitrary woman? Jun17 comment Where $ax + b$ prime infinitely often, is $ax + b - 2$ semiprime at least once? @RobertGross: Yes, why don;t you post that specific case as a new question? If $a$ is a product of primes (not necessarily twins) then there's at least a chance it will work. Jun16 answered When $\pi(x) \leqslant 0.4x +1$? Jun16 comment Where $ax + b$ prime infinitely often, is $ax + b - 2$ semiprime at least once? @RobertGross: OK, so choose $a=315,\ b = 107$. Jun16 awarded Nice Answer Jun16 comment Where $ax + b$ prime infinitely often, is $ax + b - 2$ semiprime at least once? @RobertGross: Clearly that's not possible, given my counterexample of $a=105,\ b=2.$ | {"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": 0.9037420749664307, "perplexity": 876.8286390215372}, "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-2015-27/segments/1435375094957.74/warc/CC-MAIN-20150627031814-00287-ip-10-179-60-89.ec2.internal.warc.gz"} |
https://www.gregschool.org/gregschoollessons/2017/6/21/the-dependency-of-theta-in-the-dot-product-a6rdd-m5x4z | # The Dependency of theta in the dot product
How the work done depends on the angle $$θ$$
The work $$W$$ done on an object acted upon by a constant force $$\vec{F}$$ is given by
$$W=\vec{F}·Δ\vec{R},\tag{1}$$
where $$Δ\vec{R}$$ is the displacement of the object. As a reminder, the dot product between any two vectors $$\vec{A}$$ and $$\vec{B}$$ is defined as
$$\vec{A}·\vec{B}≡ABcosθ,$$
where $$θ$$ is the angle between those two vectors. From this definition, you can see that if the two vectors $$\vec{A}$$ and $$\vec{B}$$ are perpendicular to each other, then $$cos(±90°)=0$$ and, thus, $$\vec{A}·\vec{B}=0$$.
This basically means that if an object moves from $$\vec{R}(t_0)$$ to $$\vec{R}(t)$$ with a displacement of $$Δ\vec{R}$$, if a constant force $$\vec{F}$$ was acting on that object during its displacement and if $$\vec{F}$$ was perpendicular to $$Δ\vec{R}$$, then that force would have done no work on the object. For example, the Moon revolves around the Earth in a roughly circular path and the force of gravity $$\vec{F}_g$$ exerted on the Moon by the Earth is always towards the center of Earth. Since this force is always acting on the Moon in a direction perpendicular to its displacement, it follows that the force $$\vec{F}_g$$ doesn't do any work on the Moon.
We can rewrite Equation (1) in the following way:
$$W=\vec{F}·Δ\vec{R}=FΔRcosθ=(Fcosθ)ΔR=F_RΔR,\tag{2}$$
where $$F_R$$ is the component of the force $$\vec{F}$$ which is acting in the same direction as $$Δ\vec{R}$$. Equation (2) tells us that it is only the component of force that is parallel to $$Δ\vec{R}$$ which does work on the object. Therefore, if I apply the force $$\vec{F}$$ to a box moving to the right as in Figure 1, the y-component of force $$F_y$$ will contribute zero work to the object. It is only the component of force $$F_x=Fcosθ$$ which is doing any work on the box. By changing the angle $$θ$$ at which $$\vec{F}$$ (while keeping the magnitude $$F$$ the same) is applied in a way that increases $$cosθ$$ (in this example, this can be accomplished by rotating $$\vec{F}$$ clockwise), the same force $$\vec{F}$$ will do more work on the object. | {"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": 0.9446856379508972, "perplexity": 58.58329090971545}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-26/segments/1560627998325.55/warc/CC-MAIN-20190616222856-20190617004856-00417.warc.gz"} |
https://www.physicsforums.com/threads/stationary-or-not.576604/ | # Stationary or not?
1. Feb 11, 2012
### Grimstone
stationary or not?
I understand that satellites orbit the earth at a speed and angle that allows them to "free fall" the entire time. That is they are going so fast that they are always cresting the edge of the planet and always in a state of free fall.
isn't it possible to place a satellite or station that is pretty much stationary over the poles?
yes the orbit of the earth is not round. yes the earth tilts.
2. Feb 11, 2012
### mathman
Satellites can be placed stationary over a point on the equator, not at the poles. The satellite must orbit the earth - over the equator, the orbital period can be made coincident with the earth's rotation.
3. Feb 12, 2012
### Chronos
A stable orbit is impossible unless the satellite is traveling at least at escape velocity wrt to the body it orbits - this is orbital mechancis 101.
4. Feb 12, 2012
### SW VandeCarr
No. As mathman said, an orbiting body can only remain stationary with respect to a point on the Earth's surface when it is in the geocentric orbit (more specifically geostationary). That is, an orbit in the equatorial plane at a distance of about 40,000 km where the orbital speed matches the Earth's rotation in terms of a fixed point on the surface.
Any other orbit must lie in a plane which contains the Earth's center of gravity (mass) so polar orbits are possible, but they cannot be geostationary.
Last edited: Feb 12, 2012
5. Feb 12, 2012
### Grimstone
thank you.
6. Feb 12, 2012
### Staff: Mentor
Er, what? No it isn't! Orbital speed is of course lower than "escape" velocity!
7. Feb 12, 2012
### D H
Staff Emeritus
8. Feb 12, 2012
### cepheid
Staff Emeritus
Like russ, I don't understand either. If you just apply the virial theorem to the simple system of a satellite in a circular orbit around some parent body, you get that the speed at the orbital radius is exactly half of what the escape speed would be at that radius. You can also derive this result just by using centripetal force, of course.
So, regardless of what a hyperbolic orbit is (and I haven't googled it), the statement that the escape speed is a minimum speed required for a stable orbit seems manifestly wrong.
9. Feb 13, 2012
### Chronos
Agreed, I was thinking of escape velocity at the surface of earth vs orbital velocity around earth. Obviously orbital velocity is always less than escape velocity by a factor of sqrt 2.
10. Feb 13, 2012
### D H
Staff Emeritus
I misread. I thought Chronos was talking about a radial orbit directly away from the pole. The satellite isn't stationary in terms of distance, but it is in terms of angular position. Such an orbit is of course of no use for communications or Earth observations.
Still not right. (Closed) orbital velocity is always less than escape velocity. There's no factor of sqrt(2). Escape velocity by definition is that speed at which a trajectory changes from a closed orbit to an open orbit (or escape trajectory, if you don't like using the word "orbit" for things that aren't "orbiting").
11. Feb 13, 2012
### Janus
Staff Emeritus
The sqrt 2 factor holds when comparing orbital velocity for a circular orbit to escape velocity. For an elliptical orbit, orbital velocity changes throughout the orbit, and approaches escape velocity at periapsis as the eccentricity approaches 1. This leads to an interesting fact: If you are already in an elliptical orbit, it takes less delta v to achieve escape velocity at periapsis than it does apoapsis
12. Feb 15, 2012
### BobG
It's not possible to put a single satellite into an orbit that's stationary over the poles, but it is possible to accomplish the same result using several satellites.
With a constellation of satellites in Molniya orbits, you can stagger their orbits so there's always some satellite in the same given location all of the time. From the point of view of the tracking station, they point their antenna roughly in one direction and each satellite moves into that location in a kind of relay race. The antenna barely has to move to pick up the next satellite coming along.
Typically, this is done with six satellites in a semi-synchronous orbit (two orbits per day, which means each satellite is used for two periods per day, giving you the equivalent of a 12 satellite relay race). You could do this with a minimum of four satellites if you were willing to move your antenna just a bit further to pick up the next satellite coming along.
And, the satellites wouldn't be directly over the poles. They have to have an inclination angle of 63.4 degrees, so you'd actually always have a satellite at about 63.4 degrees latitude with some nearly constant longitude.
The 63.4 degree requirement is because you create this relay race with satellites in highly elliptical orbits. The satellite is used when its at apogee and is moving very slowly. Because of the equatorial bulge, perigee and apogee will move forwards or backwards depending on the inclination angle, with 63.4 degrees being the angle where perigee and apogee remain stationary. | {"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": 0.9308968782424927, "perplexity": 768.2378925524491}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-05/segments/1516084887077.23/warc/CC-MAIN-20180118071706-20180118091706-00205.warc.gz"} |
https://physics.stackexchange.com/questions/721945/does-the-unruh-effect-assume-its-conclusion?noredirect=1 | # Does the Unruh effect assume its conclusion?
Unruh effect says that accelerating observers see the single particle states of inertial frames as thermal baths.
But it proves it by defining the particle states in the accelerating observer's frame according to $$b|0\rangle =0$$, where $$b$$ is a different operator from the $$a$$ used to define the vacuum in inertial frames.
But this definition is the same as assuming the conclusion of Unruh effect. We've given different definition of the particle in both frames, hence both frames disagree on the number of particles.
What justifies the different definition of a real-world particle in an accelerating frame? Why can't we just postulate that all real-world particles correspond to the states defined in inertial frames?
Or maybe we're trying to define a particle in terms of something physical : the Eigenstates of the field Hamiltonian. Different frames have different Hamiltonians of the field, hence different states will be defined as particles. If this is true, what about the frames where the Hamiltonian is time-dependent?
Your last paragraph gets it right: the Hamiltonians are different, and they're not always available.
The Unruh effect is a prediction of Quantum Field Theory in Curved Spacetime, so I'll first discuss how one formulates QFT in general spacetimes before focusing on the Minkowski case.
In flat spacetimes, we typically understand states in terms of particles due to Poincaré invariance. The fact that Minkowski spacetime is Poincaré invariant lets us single out a quantum state that is Poincaré invariant, which we call the vacuum. We can then interpret other states as excitations of this vacuum, and hence understand them as particle states. This is the short version. Notice this is always done in an inertial frame of reference, and (orthochronous) Poincaré transformations will keep this vacuum invariant (i.e., all inertial observers agree on what is the vacuum).
Consider now a generic spacetime. Poincaré invariance now means nothing: general spacetimes lack Poincaré symmetry. Hence, there is no way of defining a preferred state. One can no longer choose a vacuum. And that's it. Without a vacuum, you have no way of defining a notion of particle. This is not a problem, because QFT is a theory about fields. Particles are just a convenience we can use on some situations.
However, suppose the particular case in which you have a timelike Killing vector field. In other words, you have a generalized notion of time-translation symmetry. In this situation, you can (roughly speaking) understand what would be the Hamiltonian for the theory as seen by the observers whose trajectories follow that Killing field. Using this Hamiltonian, one can then once again speak of a vacuum that minimizes energy. Notice that this will lead you to a time-independent Hamiltonian (with respect to the "time" defined by the Killing field).
What happens in the Unruh effect is that you have two possibilities in Minkowski spacetime. Usually, we pick the Poincaré time-translation as a Killing field. However, another option is to pick the Killing field associated with boosts along some Cartesian direction (say $$x$$), and restrict it to the region $$x > |t|$$ of the spacetime (known as the right Rindler wedge). In this region of spacetime (which can be understood as a spacetime of its own right), we can then define a vacuum by minimizing the Hamiltonian defined in "boost time", and get a different notion of particle.
It then remains to understand what this means. It turns out that the observers that follow the boost Killing field are accelerated observers, so we can interpret this result as telling us how accelerated observers define the notion of particle, which disagrees with the definition of inertial observers. If we leave it like this, it would definitely sound sketchy. However, another way of studying these themes is by using particle detectors (couple a 2-level system to the field, so it can work as a quantum system that tells you when a particle is seen). The accelerating and inertial particle detectors will then react differently to the field, in perfect agreement with the prediction that the accelerated detector should be interacting with a thermal state.
It might also be worth pointing out that, while there is no direct experimental evidence for the Unruh effect, I should mention two papers:
• arXiv: gr-qc/0205078 "illustrate[s] how the Fulling-Davies-Unruh effect is indeed mandatory to maintain the consistency of standard Quantum Field Theory" by studying the decay of accelerated protons (while inertial protons don't have enough energy to decay, accelerated protons have an external source of energy)
• arXiv: 1701.03446 proposes an experimental procedure to measure the Unruh effect and predicts its result with standard classical electrodynamics by relating it to radiation emitted by accelerated charges. It mentions that "Unless one is willing to question the validity of classical electrodynamics, this must be seen as a virtual observation of the Unruh effect".
For a more technical post on how to define QFT on curved spacetimes, you might be interested in this answer I wrote a while ago. For a more "SciComm" view (still a bit technical though), there's also this answer, also by me. For more information on these topics, I recommend Wald's Quantum Field Theory in Curved Spacetime and Black Hole Thermodynamics. For more on the particle detector view, you might be interested in the paper by Unruh, W. G. and Wald, R. M. (1984), "What happens when an accelerating observer detects a Rindler particle", Phys. Rev. D 29, 1047.
• Does all of this mean that we've solved the "problem of time" discrepancy between QM and GR? I read about a "problem of time" here. If we're able to formulate QFT in a non-stationary spacetime, then I think there's no problem of time. The only problem left is to make the metric respond to other fields in such a way that Einstein's field equation is recovered. Aug 8 at 13:30
• @RyderRude I believe that deserves a post in its own right. I don't think this solves the problem of time. As far as I know, the problem of time concerns quantum gravity, and how a quantum theory of gravity should interpret what is meant by "time" (QM suggests a parameter, GR suggests a coordinate). The Unruh effect is a prediction of QFT in curved spacetime, which assumes a GR background and does not attempt to quantize the gravitational field. In this sense, it is not a fundamental theory. In a quantum gravity theory, you might not have a background spacetime that allows you these + Aug 8 at 13:41
• constructions, which makes the problem persist (as far as I can tell). Aug 8 at 13:41
• I don't know much about the problem of time, but this paper by Isham was recommended to me a while ago and perhaps could be a nice start. I skimmed over it and it seems to discuss semiclassical gravity (QFTCS + effects on the background metric due to the stress-energy of the quantum fields), so it might touch on how (and if) the problem of time is related to QFTCs. Aug 8 at 13:45 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 5, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8852106332778931, "perplexity": 230.70225629835926}, "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/1664030334987.39/warc/CC-MAIN-20220927033539-20220927063539-00503.warc.gz"} |
http://www.wikihow.com/Calculate-the-Volume-of-a-Cone | Edit Article
# wikiHow to Calculate the Volume of a Cone
You can calculate the volume of a cone easily once you know its height and radius and can plug those measurements into the formula for finding the volume of a cone. The formula for finding the volume of a cone is v = hπr2/3.
### Calculating the Volume of a Cone
1. 1
Find the radius. If you already know the radius, then you can move on to the next step. If you know the diameter, divide it by 2 to get the radius. If you know the circumference, divide it by 2π to get the diameter. And if you don't know any of the measurements of the shape, just use a ruler to measure the widest pie circular base (the diameter) and divide that number by 2 to get the radius. Let's say the radius of this cone's circular base is .5 inches (1.3 cm).
2. 2
Use the radius to find the area of the base circle. To find the area of the base circle, you can just use the same formula you'd use to find the area of a circle: A = πr2. Plug ".5" in for r to get A = π(.5)2 and square the radius and multiply it by the value of π to find the area of the circular base. π(.5)2 = .79 in.2.
3. 3
Find the height of the cone. If you know it already, write it down. If you don't know it, use a ruler to measure it. Let's say the height of the cone is 1.5 inches (1.3 cm). Make sure that the height of the cone is written in the same measurements as the radius.
4. 4
Multiply the area of the base by the height of the cone. Multiply the area of the base, .79 in.2, by the height, 1.5 in. So, .79 in.2 x 1.5 in = 1.19 in.3
5. 5
Divide the product by three. Simply divide 1.19 in.3 by 3 to find the volume of a cone. 1.19 in.3/3 = .40 in.3. Always state the volume in cubic units because it's a measure of three-dimensional space.
## Community Q&A
Search
• How do I find the height of a cone if I know the radius and slope angle?
wikiHow Community Q&A
Look up the tangent of the angle. Set the tangent equal to the height divided by half the radius. Then solve for the height.
• How do I find the volume of a cone if the diameter is given, not the radius?
wikiHow Contributor
Since the volume of a cone is pi times the radius squared times the height divided by 3, if you know the diameter, you simply divide by two to find the radius, which you then plug into the formula.
• How do I calculate the radius of a cone when I already have the volume?
wikiHow Contributor
You would do: V = 1/3 *π*r^2. Divide volume by 3. Divide volume by π, then square root that number. And there is your radius.
• What if the given is slant height and base? How to find the volume of cone?
wikiHow Contributor
Use the Pythagorean Theorem to find the height, using the slant length as the hypotenuse and half the base as one side.
• If I double the height, will I double the volume?
Yes.
• How do I calculate the radius of a base of a cone?
You must be given either the diameter of the base, the area of the base, the circumference of the base, or the volume of the cone. Then work backwards to the radius from the formula for the volume, base area or base circumference of a cone.
• Where did the 1/3 in the formula come from?
wikiHow Contributor
The volume of a cone is one-third the volume of a cylinder of the same height and radius.
• How do I work out the perpendicular height of a cone without any radii or volume measures?
You can't do it.
• If the quadrant of a circle of radius 14 cm is folded to a cone, how can I find the radius of this cone?
Double the radius, and multiply that by pi: that's the circumference of the circle. Divide by 4: that's the arc of the quadrant and the circumference of the cone. Divide by twice pi: that's the radius of the cone.
• Given the height and radius of a cone are 3 cm and 4 cm respectively, calculate the total surface area of the cone.
With the Pythagorean theorem, use the radius and the height to calculate the slant height of the cone, then multiply the slant height by the radius by pi. That gives you the lateral area of the cone. To that you add the base area of the cone, which is found by multiplying pi by the square of the radius. The total surface area is found by adding the lateral surface area to the base area.
• How do I segment a cone?
200 characters left
If this question (or a similar one) is answered twice in this section, please click here to let us know.
## Tips
• Make sure you have accurate measurements.
• Don't do this while there's still ice cream in the cone.
• Make sure the measurements are all in the same type/unit of measurement.
• How it Works:
• In this method, you are basically calculating the volume of the cone as if it was a cylinder. When you calculate the area of the base circle, and multiply it by the height, you are "stacking" the area up until it reaches the height, thus creating a cylinder. And because a cylinder can fit three cones of its matching measurements, you multiply it by one third so that it's the volume of a cone. This provides you with the volume of the cone.
• The radius, the height, and the slant height ---slant height is measured down the sloping side of the cone, while the true height is measured through the middle from the tip to the center of the circular base --- form a right triangle. Therefore, they are related by the Pythagorean Theorem: (radius)2+ (height)2 = (slant height)2
## Warnings
• Make sure to divide by 3
## Article Info
Categories: Volume
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Español: calcular el volumen de un cono, Português: Calcular o Volume de um Cone, Italiano: Calcolare il Volume di un Cono, Français: calculer le volume d'un cône, Deutsch: Das Volumen eines Kegels ausrechnen, Nederlands: De inhoud van een kegel berekenen, Русский: посчитать объем конуса, 中文: 计算一个圆锥体的体积, Bahasa Indonesia: Menghitung Volume Kerucut, Čeština: Jak vypočítat objem kuželu, العربية: حساب حجم المخروط, Tiếng Việt: Tính Thể tích Hình nón, हिन्दी: शंकु का आयतन ज्ञात करें, ไทย: คำนวณหาปริมาตรของทรงกรวย, 한국어: 원뿔의 부피 구하는 법
Thanks to all authors for creating a page that has been read 557,660 times. | {"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": 0.9438152313232422, "perplexity": 435.77865878389656}, "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-2017-22/segments/1495463608004.38/warc/CC-MAIN-20170525063740-20170525083740-00321.warc.gz"} |
http://tex.stackexchange.com/questions/63558/how-to-get-microtype-to-work-with-mathematical-minus-sign-and-package-lmodern?answertab=active | # How to get microtype to work with mathematical minus sign and package lmodern?
Using the package microtype by including the line
\usepackage{microtype}
in the preamble subtly alters the appearance of a typeset document by, amongst other things, letting optically less dense characters, like dashes or punctuations, protrude slightly in the margins. When I use a mathematical minus sign in a document written in the class article, I write it like this:
$-$
If I include \usepackage{lmodern} in the preamble, the minus sign does not get moved out a little bit if it happens to become the first character of a line, but stays at the same horizontal position as for example an "m" letter. Microtype does work for the plus sign and dashes, including when the plus sign is written in math mode:
$+$
Does anyone know how to enable microtype also for the mathematical minus sign when using lmodern?
When not using lmodern, the problem can be fixed as suggested by Mico by using \usepackage[protrusion=allmath]{microtype}.
edit: MWE, for which the fix to use \usepackage[protrusion=allmath]{microtype} does not work.
Here is some example code that generates the incorrectly placed mathematical minus sign:
\documentclass{article}
\usepackage{lmodern}
\usepackage[protrusion=allmath]{microtype}
\begin{document}
\section{Test Section}
\label{Test Label}
Test text to reproduce the microtype mathematical minus sign problem that I experienced. I am putting in a few more words to get some text to demonstrate the effect. This is just some text with mathematical minusaaaaaaaaaaaa signs $-$inserted (that was a mathematical minus sign in front of inserted'')aaaaa. A +plus sign looks like that. In a longer text that I had, the plus sign looked fine, but the minus sign clearly looked like it was to far in. I am now putting in some more text, just to get one more line to show the look of the left margin.
\end{document}
Compiled result (first an enlargement to show the left margin, and then the entire page):
-
Have you tried providing the protrusion=allmath option to microtype? – Mico Jul 16 '12 at 23:27
I was not aware of that. I tried it (just now), but unfortunately it did not help. It only rendered several warning about microtype not finding protrusion list for font ... – hjb981 Jul 17 '12 at 0:06
Aaah, now we're starting to get somewhere: Which (math) fonts do you use? According to the microtype manual, protrusion should be enabled for all of the "main" math font packages (Computer Modern (obviously), mathptmx, mathpazo, eulervm, and probably a few more). Please consider editing your question to include a standalone MWE (minimum working example) that generates the problem behavior you're trying to correct. – Mico Jul 17 '12 at 0:09
Ah, protrusion=allmath does solve the problem for your MWE. It still doesn't make sense that + and - are treated differently without the allmath option. (Ping, @Mico.) – Hendrik Vogt Jul 17 '12 at 13:51
@hjb981: Please try and remove anything from the preamble that does not contribute to the problem. You could do that by removing things piece by piece (and recompiling of course), or by removing 1. the first half of the preamble (+recompile), 2. the 2nd half of the preamble (+recompile), hoping that with either 1. or 2. the problem persists. Then repeat until you get some MWE. – Hendrik Vogt Jul 17 '12 at 14:36
\DeclareMicrotypeAlias{lmsy}{cmsy}
This seems to be an oversight in the microtype package since there is already a line
\DeclareMicrotypeAlias{lmr} {cmr}
so maybe report this to microtype author...
-
That worked. I still get several errors of the kind: I cannot not find a protrusion list for font 'OML/lmm/m/it/10.95' Switching of protrusion for this font on line ... The numbers can also be exchanged for 6, 7, 8, 10, 12 and a few more---I guess that it refers to different sizes. Is there a way to enable microtype for those fonts also? – hjb981 Jul 17 '12 at 15:18
I guess \DeclareMicrotypeAlias{lmm}{cmm} should do it. – Lev Bishop Jul 17 '12 at 15:36
Yes, it worked. What does that line mean? Does one always write {l abc}{c abc}, where abc is the name of the font? – hjb981 Jul 17 '12 at 15:53
@hjb981 it means "use the settings from this font for that one". Since lmodern mostly the same as computer modern, and the fonts are (mostly?) named the same just with interchange of c and l, it should work ok. However computer modern and lmodern are not identical. Microtype already knows to treat the \textquotedblleft et al slightly differently for example, so this quick fix may not be good enough (perhaps this is why it is currently missing from microtype). – Lev Bishop Jul 17 '12 at 23:17
indeed, these two alias declarations were missing in microtype.cfg. I'll add them for the next version (coming "soon"). Thanks for bringing this to my attention. – Robert Jul 18 '12 at 1:59 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8116475939750671, "perplexity": 1502.377968408746}, "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-2016-18/segments/1461860111396.55/warc/CC-MAIN-20160428161511-00200-ip-10-239-7-51.ec2.internal.warc.gz"} |
https://mathoverflow.net/questions/229725/a-gaussian-integral-over-complex-variables-by-a-defined-greens-function-for-a-g | # A Gaussian integral over complex variables by a defined Green's function for a Gaussian ensemble of random matrix
We construct an $N\times N$ matrix $J$ whose elements are drawn from Gaussian distribution with zero mean and variance $\frac{1}{N}$. Since we want to have different variances for different columns, so that we express the elements as $J_{ij}\sigma_j$ such that the matrix satisfies $\mathbb{Var}[J_{ij}\sigma_j] = \frac{\sigma_j^2}{N}$.
Given a Green function (analogous electrostatic potential) $$\Phi(\omega) = -\frac{1}{N}\mathbb{E}_J\big[ \ln\det((I\omega^* - J^T)(I\omega - J)) \big]$$, where $\omega\in \mathbb{C}$ and $J$ is an $N\times N$ random matrix. Since $J$ is positive semi-definite, by adding diagonal matrix $\epsilon\delta_{ij}$, determinant can be represented by a Gaussian integral over complex variables, we have $$\Phi(\omega) = \frac{1}{N}\ln\mathbb{E}_J\big[ \int\big( \prod_i\frac{d^2 z_i}{\pi}\big)\exp\big\{ -\epsilon\sum_i|z_i|^2 - \sum_{i, j, k} z_i^*(\omega^*\delta_{ik}-J_{ik}^T)(\omega\delta_{kj}-J_{kj})z_j \big\} \big]$$
Then the potential $\phi$ is determined by
$$\exp(-N\Phi) = \int\prod_{i = 1}^N \frac{dz_i^* dz_i}{\pi}\exp(-N\hat{Q})$$ with the average over the distribution of $J$ values given by $$\exp(-N\hat{Q}) = \int\prod_{i,j = 1}^N \sqrt{\frac{N}{2\pi}}dJ_{ij}\exp(-NQ)$$ where $$Q = \sum_i\big( \frac{|\omega|^2}{\sigma_i^2} + \epsilon_i \big)\frac{z_i^*z_i}{N} + \frac{1}{2}\sum_{i, j, k} J_{ki}A_{ij}J_{kj} - \sum_{k, j}B_{kj}J_{kj}$$ and $$A_{ij} = \frac{z_i^*z_j}{N} + \frac{z_j^*z_i}{N} + \delta_{ij}\text{ and } B_{kj} = \frac{\omega^*z_k^*z_j}{\sigma_kN}+ \frac{\omega z_j^*z_k}{\sigma_kN}$$
Question:
It confuses me how this later expression expends the previous formula for $\Phi(\omega)$ ? Could someone help me to show some detailed calculations to explain ?
How the formula later with $Q, \hat{Q}, A, B$ etc. to be derived from previous formula of $\Phi(\omega)$ ? They are supposed to be equivalent.
Remark: The question is extracted from two papers for understanding of gap calculation.
Sommers, H. J., et al. "Spectrum of large random asymmetric matrices." Physical Review Letters 60.19 (1988): 1895-1898.
Rajan, Kanaka, and L. F. Abbott. "Eigenvalue spectra of random matrices for neural networks." Physical review letters 97.18 (2006): 188104.
## 1 Answer
Your first expression for the potential $$\Phi(\omega)=\frac{1}{N}\log E_J \int d^2z...$$ is equivalent to $$e^{N\Phi(\omega)}=E_J \int d^2z...$$ Ok? Writing the expectation in $J$ according to its definition gives $$e^{N\Phi(\omega)}=\int dJ e^{-N\sum_{ij}J_{ij}^2} \int \prod_i\frac{d^2 z_i}{\pi}e^{-\epsilon z^\dagger z}e^{-z^\dagger(\omega^*-(J\sigma)^T)(\omega-J\sigma)z}$$ Notice that the matrix they are interested in is $J\sigma$ (in the second paper you mentioned).
Now redefine the complex integration variables as $z_i\to z_i/\sigma_i$. This will lead to your final expressions. | {"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": 0.9915985465049744, "perplexity": 227.87824326222253}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-50/segments/1606141681209.60/warc/CC-MAIN-20201201170219-20201201200219-00628.warc.gz"} |
https://db0nus869y26v.cloudfront.net/en/Dihedral_angle | Angle between two half-planes (α, β, pale blue) in a third plane (red) which cuts the line of intersection at right angles
A dihedral angle is the angle between two intersecting planes or half-planes. In chemistry, it is the clockwise angle between half-planes through two sets of three atoms, having two atoms in common. In solid geometry, it is defined as the union of a line and two half-planes that have this line as a common edge. In higher dimensions, a dihedral angle represents the angle between two hyperplanes. The planes of a flying machine are said to be at positive dihedral angle when both starboard and port main planes (commonly called wings) are upwardly inclined to the lateral axis. When downwardly inclined they are said to be at a negative dihedral angle.
Mathematical background
When the two intersecting planes are described in terms of Cartesian coordinates by the two equations
${\displaystyle a_{1}x+b_{1}y+c_{1}z+d_{1}=0}$
${\displaystyle a_{2}x+b_{2}y+c_{2}z+d_{2}=0}$
the dihedral angle, ${\displaystyle \varphi }$ between them is given by:
${\displaystyle \cos \varphi ={\frac {\left\vert a_{1}a_{2}+b_{1}b_{2}+c_{1}c_{2}\right\vert }((\sqrt {a_{1}^{2}+b_{1}^{2}+c_{1}^{2))}{\sqrt {a_{2}^{2}+b_{2}^{2}+c_{2}^{2))))))$
and satisfies ${\displaystyle 0\leq \varphi \leq \pi /2.}$
Alternatively, if nA and nB are normal vector to the planes, one has
${\displaystyle \cos \varphi ={\frac {\left\vert \mathbf {n} _{\mathrm {A} }\cdot \mathbf {n} _{\mathrm {B} }\right\vert }{|\mathbf {n} _{\mathrm {A} }||\mathbf {n} _{\mathrm {B} }|))}$
where nA · nB is the dot product of the vectors and |nA| |nB| is the product of their lengths.[1]
The absolute value is required in above formulas, as the planes are not changed when changing all coefficient signs in one equation, or replacing one normal vector by its opposite.
However the absolute values can be and should be avoided when considering the dihedral angle of two half planes whose boundaries are the same line. In this case, the half planes can be described by a point P of their intersection, and three vectors b0, b1 and b2 such that P + b0, P + b1 and P + b2 belong respectively to the intersection line, the first half plane, and the second half plane. The dihedral angle of these two half planes is defined by
${\displaystyle \cos \varphi ={\frac {(\mathbf {b} _{0}\times \mathbf {b} _{1})\cdot (\mathbf {b} _{0}\times \mathbf {b} _{2})}{|\mathbf {b} _{0}\times \mathbf {b} _{1}||\mathbf {b} _{0}\times \mathbf {b} _{2}|))}$,
and satisfies ${\displaystyle 0\leq \varphi <\pi .}$ In this case, switching the two half-planes gives the same result, and so does replacing ${\displaystyle \mathbf {b} _{0))$ with ${\displaystyle -\mathbf {b} _{0}.}$ In chemistry (see below), we define a dihedral angle such that replacing ${\displaystyle \mathbf {b} _{0))$ with ${\displaystyle -\mathbf {b} _{0))$ changes the sign of the angle, which can be between π and π.
In polymer physics
In some scientific areas such as polymer physics, one may consider a chain of points and links between consecutive points. If the points are sequentially numbered and located at positions r1, r2, r3, etc. then bond vectors are defined by u1=r2r1, u2=r3r2, and ui=ri+1ri, more generally.[2] This is the case for kinematic chains or amino acids in a protein structure. In these cases, one is often interested in the half-planes defined by three consecutive points, and the dihedral angle between two consecutive such half-planes. If u1, u2 and u3 are three consecutive bond vectors, the intersection of the half-planes is oriented, which allows defining a dihedral angle that belongs to the interval (−π, π]. This dihedral angle is defined by[3]
{\displaystyle {\begin{aligned}\cos \varphi &={\frac {(\mathbf {u} _{1}\times \mathbf {u} _{2})\cdot (\mathbf {u} _{2}\times \mathbf {u} _{3})}{|\mathbf {u} _{1}\times \mathbf {u} _{2}|\,|\mathbf {u} _{2}\times \mathbf {u} _{3}|))\\\sin \varphi &={\frac {\mathbf {u} _{2}\cdot ((\mathbf {u} _{1}\times \mathbf {u} _{2})\times (\mathbf {u} _{2}\times \mathbf {u} _{3}))}{|\mathbf {u} _{2}|\,|\mathbf {u} _{1}\times \mathbf {u} _{2}|\,|\mathbf {u} _{2}\times \mathbf {u} _{3}|)),\end{aligned))}
or, using the function atan2,
${\displaystyle \varphi =\operatorname {atan2} (\mathbf {u} _{2}\cdot ((\mathbf {u} _{1}\times \mathbf {u} _{2})\times (\mathbf {u} _{2}\times \mathbf {u} _{3})),|\mathbf {u} _{2}|\,(\mathbf {u} _{1}\times \mathbf {u} _{2})\cdot (\mathbf {u} _{2}\times \mathbf {u} _{3})).}$
This dihedral angle does not depend on the orientation of the chain (order in which the point are considered) — reversing this ordering consists of replacing each vector by its opposite vector, and exchanging the indices 1 and 3. Both operations do not change the cosine, but change the sign of the sine. Thus, together, they do not change the angle.
A simpler formula for the same dihedral angle is the following (the proof is given below)
{\displaystyle {\begin{aligned}\cos \varphi &={\frac {(\mathbf {u} _{1}\times \mathbf {u} _{2})\cdot (\mathbf {u} _{2}\times \mathbf {u} _{3})}{|\mathbf {u} _{1}\times \mathbf {u} _{2}|\,|\mathbf {u} _{2}\times \mathbf {u} _{3}|))\\\sin \varphi &={\frac {|\mathbf {u} _{2}|\,\mathbf {u} _{1}\cdot (\mathbf {u} _{2}\times \mathbf {u} _{3})}{|\mathbf {u} _{1}\times \mathbf {u} _{2}|\,|\mathbf {u} _{2}\times \mathbf {u} _{3}|)),\end{aligned))}
or equivalently,
${\displaystyle \varphi =\operatorname {atan2} (|\mathbf {u} _{2}|\,\mathbf {u} _{1}\cdot (\mathbf {u} _{2}\times \mathbf {u} _{3}),(\mathbf {u} _{1}\times \mathbf {u} _{2})\cdot (\mathbf {u} _{2}\times \mathbf {u} _{3})).}$
This can be deduced from previous formulas by using the vector quadruple product formula, and the fact that a scalar triple product is zero if it contains twice the same vector:
${\displaystyle (\mathbf {u} _{1}\times \mathbf {u} _{2})\times (\mathbf {u} _{2}\times \mathbf {u} _{3})=[(\mathbf {u} _{2}\times \mathbf {u} _{3})\cdot \mathbf {u} _{1}]\mathbf {u} _{2}-[(\mathbf {u} _{2}\times \mathbf {u} _{3})\cdot \mathbf {u} _{2}]\mathbf {u} _{1}=[(\mathbf {u} _{2}\times \mathbf {u} _{3})\cdot \mathbf {u} _{1}]\mathbf {u} _{2))$
Given the definition of the cross product, this means that ${\displaystyle \varphi }$ is the angle in the clockwise direction of the fourth atom compared to the first atom, while looking down the axis from the second atom to the third. Special cases (one may say the usual cases) are ${\displaystyle \varphi =\pi }$, ${\displaystyle \varphi =+\pi /3}$ and ${\displaystyle \varphi =-\pi /3}$, which are called the trans, gauche+, and gauche conformations.
In stereochemistry
Configuration namesaccording to dihedral angle syn n-Butane in thegauche− conformation (−60°) Newman projection syn n-Butane sawhorse projection
Free energy diagram of n-butane as a function of dihedral angle.
In stereochemistry, a torsion angle is defined as a particular example of a dihedral angle, describing the geometric relation of two parts of a molecule joined by a chemical bond.[4][5] Every set of three non-colinear atoms of a molecule defines a half-plane. As explained above, when two such half-planes intersect (i.e., a set of four consecutively-bonded atoms), the angle between them is a dihedral angle. Dihedral angles are used to specify the molecular conformation.[6] Stereochemical arrangements corresponding to angles between 0° and ±90° are called syn (s), those corresponding to angles between ±90° and 180° anti (a). Similarly, arrangements corresponding to angles between 30° and 150° or between −30° and −150° are called clinal (c) and those between 0° and ±30° or ±150° and 180° are called periplanar (p).
The two types of terms can be combined so as to define four ranges of angle; 0° to ±30° synperiplanar (sp); 30° to 90° and −30° to −90° synclinal (sc); 90° to 150° and −90° to −150° anticlinal (ac); ±150° to 180° antiperiplanar (ap). The synperiplanar conformation is also known as the syn- or cis-conformation; antiperiplanar as anti or trans; and synclinal as gauche or skew.
For example, with n-butane two planes can be specified in terms of the two central carbon atoms and either of the methyl carbon atoms. The syn-conformation shown above, with a dihedral angle of 60° is less stable than the anti-conformation with a dihedral angle of 180°.
For macromolecular usage the symbols T, C, G+, G, A+ and A are recommended (ap, sp, +sc, −sc, +ac and −ac respectively).
Proteins
Depiction of a protein, showing where ω, φ, & ψ refer to.
A Ramachandran plot (also known as a Ramachandran diagram or a [φ,ψ] plot), originally developed in 1963 by G. N. Ramachandran, C. Ramakrishnan, and V. Sasisekharan,[7] is a way to visualize energetically allowed regions for backbone dihedral angles ψ against φ of amino acid residues in protein structure. In a protein chain three dihedral angles are defined:
• ω (omega) is the angle in the chain Cα − C' − N − Cα,
• φ (phi) is the angle in the chain C' − N − Cα − C'
• ψ (psi) is the angle in the chain N − Cα − C' − N (called φ′ by Ramachandran)
The figure at right illustrates the location of each of these angles (but it does not show correctly the way they are defined).[8]
The planarity of the peptide bond usually restricts ω to be 180° (the typical trans case) or 0° (the rare cis case). The distance between the Cα atoms in the trans and cis isomers is approximately 3.8 and 2.9 Å, respectively. The vast majority of the peptide bonds in proteins are trans, though the peptide bond to the nitrogen of proline has an increased prevalence of cis compared to other amino-acid pairs.[9]
The side chain dihedral angles are designated with χn (chi-n).[10] They tend to cluster near 180°, 60°, and −60°, which are called the trans, gauche, and gauche+ conformations. The stability of certain sidechain dihedral angles is affected by the values φ and ψ.[11] For instance, there are direct steric interactions between the Cγ of the side chain in the gauche+ rotamer and the backbone nitrogen of the next residue when ψ is near -60°.[12] This is evident from statistical distributions in backbone-dependent rotamer libraries.
Converting from dihedral angles to Cartesian coordinates in chains
It is common to represent polymers backbones, notably proteins, in internal coordinates; that is, a list of consecutive dihedral angles and bond lengths. However, some types of computational chemistry instead use cartesian coordinates. In computational structure optimization, some programs need to flip back and forth between these representations during their iterations. This task can dominate the calculation time. For processes with many iterations or with long chains, it can also introduce cumulative numerical inaccuracy. While all conversion algorithms produce mathematically identical results, they differ in speed and numerical accuracy.[13][non-primary source needed]
Geometry
Every polyhedron has a dihedral angle at every edge describing the relationship of the two faces that share that edge. This dihedral angle, also called the face angle, is measured as the internal angle with respect to the polyhedron. An angle of 0° means the face normal vectors are antiparallel and the faces overlap each other, which implies that it is part of a degenerate polyhedron. An angle of 180° means the faces are parallel, as in a tiling. An angle greater than 180° exists on concave portions of a polyhedron.
Every dihedral angle in an edge-transitive polyhedron has the same value. This includes the 5 Platonic solids, the 13 Catalan solids, the 4 Kepler–Poinsot polyhedra, the two quasiregular solids, and two quasiregular dual solids.
Law of cosines for dihedral angle
Given 3 faces of a polyhedron which meet at a common vertex P and have edges AP, BP and CP, the cosine of the dihedral angle between the faces containing APC and BPC is:[14]
${\displaystyle \cos \varphi ={\frac {\cos(\angle \mathrm {APB} )-\cos(\angle \mathrm {APC} )\cos(\angle \mathrm {BPC} )}{\sin(\angle \mathrm {APC} )\sin(\angle \mathrm {BPC} )))}$
This can be deduced from Spherical law of cosines
References
1. ^ "Angle Between Two Planes". TutorVista.com. Retrieved 2018-07-06.
2. ^ Kröger, Martin (2005). Models for polymeric and anisotropic liquids. Springer. ISBN 3540262105.
3. ^ Blondel, Arnaud; Karplus, Martin (7 Dec 1998). "New formulation for derivatives of torsion angles and improper torsion angles in molecular mechanics: Elimination of singularities". Journal of Computational Chemistry. 17 (9): 1132–1141. doi:10.1002/(SICI)1096-987X(19960715)17:9<1132::AID-JCC5>3.0.CO;2-T.
4. ^ IUPAC, Compendium of Chemical Terminology, 2nd ed. (the "Gold Book") (1997). Online corrected version: (2006–) "Torsion angle". doi:10.1351/goldbook.T06406
5. ^ IUPAC, Compendium of Chemical Terminology, 2nd ed. (the "Gold Book") (1997). Online corrected version: (2006–) "Dihedral angle". doi:10.1351/goldbook.D01730
6. ^ Anslyn, Eric; Dennis Dougherty (2006). Modern Physical Organic Chemistry. University Science. p. 95. ISBN 978-1891389313.
7. ^ Ramachandran, G. N.; Ramakrishnan, C.; Sasisekharan, V. (1963). "Stereochemistry of polypeptide chain configurations". Journal of Molecular Biology. 7: 95–9. doi:10.1016/S0022-2836(63)80023-6. PMID 13990617.
8. ^ Richardson, J. S. (1981). "The Anatomy and Taxonomy of Protein Structure". Anatomy and Taxonomy of Protein Structures. Advances in Protein Chemistry. Vol. 34. pp. 167–339. doi:10.1016/S0065-3233(08)60520-3. ISBN 9780120342341. PMID 7020376.
9. ^ Singh J, Hanson J, Heffernan R, Paliwal K, Yang Y, Zhou Y (August 2018). "Detecting Proline and Non-Proline Cis Isomers in Protein Structures from Sequences Using Deep Residual Ensemble Learning". Journal of Chemical Information and Modeling. 58 (9): 2033–2042. doi:10.1021/acs.jcim.8b00442. PMID 30118602. S2CID 52031431.
10. ^
11. ^ Dunbrack, RL Jr.; Karplus, M (20 March 1993). "Backbone-dependent rotamer library for proteins. Application to side-chain prediction". Journal of Molecular Biology. 230 (2): 543–74. doi:10.1006/jmbi.1993.1170. PMID 8464064.
12. ^ Dunbrack, RL Jr; Karplus, M (May 1994). "Conformational analysis of the backbone-dependent rotamer preferences of protein sidechains". Nature Structural Biology. 1 (5): 334–40. doi:10.1038/nsb0594-334. PMID 7664040. S2CID 9157373.
13. ^ Parsons, J.; Holmes, J. B.; Rojas, J. M.; Tsai, J.; Strauss, C. E. (2005), "Practical conversion from torsion space to cartesian space for in silico protein synthesis", Journal of Computational Chemistry, 26 (10): 1063–1068, doi:10.1002/jcc.20237, PMID 15898109, S2CID 2279574
14. ^ "dihedral angle calculator polyhedron". www.had2know.com. Archived from the original on 25 November 2015. Retrieved 25 October 2015. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 22, "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": 0.9275999665260315, "perplexity": 3357.5667114516414}, "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/1674764499966.43/warc/CC-MAIN-20230209112510-20230209142510-00829.warc.gz"} |
https://dsp.stackexchange.com/questions/24056/what-is-the-reason-for-filtering-out-the-negative-frequencies-of-a-signal | # What is the reason for filtering out the negative frequencies of a signal?
I am reading this tutorial. Quoting the lines from the topic "Analytic signals and Hilbert transform filters":
the corresponding analytic signal $z(t)=x(t) + j {\cal H}_t\{x\}$ has the property that all negative frequencies'' of $x(t)$ have been "filtered out''
1. what is the purpose of filtering out negative frequencies?
2. What is the interpretation of filtering out negative frequency?
3. What are negative frequencies in physical world? I read that negative frequencies doesn't exist and are introduced to simplify the mathematics behind theory.
4. If negative frequencies don't exist, then why do we all plot two sided spectrum?
5. If negative frequencies exist, why do we only consider the first half of an N-point FFT?
3. Negative frequencies do exist. There is nothing controversial about them.
• For a cosine signal: since $\cos(2\pi(-f)t)=\cos(2\pi ft)$, a cosine of negative frequency $-f$ is equal to a cosine of frequency $f$.
• Since $\sin(2\pi(-f)t)=-\sin(2\pi ft)$, a sine of frequency $-f$ is $\pi$ radians out of phase with respect to a sine of frequency $f$.
• More importantly, a complex exponential $e^{2\pi ft}$ can be represented by a point in the complex plane, which rotates counter-clockwise if $f$ is positive and clockwise if $f$ is negative.
5. It turns out that, for all physical signals (whose imaginary part is zero), the magnitude spectrum is even (the negative frequencies are a mirror image of the positive frequencies). There is no need to display or calculate them. That's the reason a spectrum analyzer will only display positive frequencies. Complex signals do not have an even magnitude spectrum, and you need to calculate it for negative as well as positive frequencies.
2. Filtering out the negative frequencies is just that: remove the negative frequencies of a signal. The result is a complex signal, because the resulting spectrum is not even. The Hilbert transform you mention at the start of your question is an easy way to implement such a filter. Without it, you'd need a complex filter, which are not trivial to implement and use.
1. There can be many reasons to filter out the negative frequencies of a signal. In digital communications, it is used as one step in obtaining the complex envelope of a modulated signal. In general, a modulated signal looks like $s(t)=\Re[a(t)e^{2\pi f_ct}]$, where $a(t)$ is a quadrature signal, and is complex, and where $f_c$ is very large (up to several gigahertz). However, the information is contained in $a(t)$, so we would like to recover it from $s(t)$. This is accomplished by:
• Filtering out the negative frequencies of $s(t)$
• Downconverting the result to baseband
• Thanks . Thats great answered. I'd like to add another question, Do we have complex signals (or pseudo-complex signals) in physical world ? Does it arise in any domains like seismic, biological etc.,etc., signal processing ? – nmxprime Jun 12 '15 at 4:53
• And, I dont understand, specifically This is accomplished by: Filtering out the negative frequencies of s(t) . Can you point any reference to study relating to this ? – nmxprime Jun 12 '15 at 5:11
• @nmxprime Complex signals do not exist physically. They are very useful, though; for example, they are used to design quadrature signals. First you design your quadrature complex signal in the DSP domain (that is, in a computer). Then, before transmission, it is converted to a real, physical signal. – MBaz Jun 12 '15 at 14:30
• @nmxprime Regarding filtering negative frequencies, I'd suggest reading this: en.wikipedia.org/wiki/Analytic_signal – MBaz Jun 12 '15 at 14:31 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 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": 0.8697984218597412, "perplexity": 521.9941557667444}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-26/segments/1560628001014.85/warc/CC-MAIN-20190627075525-20190627101525-00052.warc.gz"} |
https://www.gradesaver.com/textbooks/math/calculus/university-calculus-early-transcendentals-3rd-edition/chapter-3-practice-exercises-page-202/70 | ## University Calculus: Early Transcendentals (3rd Edition)
$$y'=-\frac{y}{x}$$
$$x^2y^2=1$$ Using implicit differentiation, we differentiate both sides of the equation with respect to $x$: $$2xy^2+2x^2y\times y'=0$$ $$xy^2+x^2y\times y'=0$$ Now we set all elements with $y'$ to one side and the rest to the other side: $$x^2y\times y'=-xy^2$$ Finally, we calculate $y'$: $$y'=-\frac{xy^2}{x^2y}=-\frac{y}{x}$$ | {"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": 0.8650634288787842, "perplexity": 143.35900514799008}, "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/1618039594341.91/warc/CC-MAIN-20210422160833-20210422190833-00198.warc.gz"} |
http://math.stackexchange.com/questions/167674/partially-ordered-group-positive-cone-quotient-exercise | # partially ordered group, positive cone, quotient (exercise)
Definitions: A partially ordered group or po-group is a po-set $(G,\leq)$, such that $G$ is a group and $\forall x,y,a,b\!\in\!G\!:x\!\leq\!y\Rightarrow axb\!\leq\!ayb$, i.e. a po-set that is a group in which left and right translations are isotone (= order preserving). A directed group or do-group is a po-group that is an (up and down) directed set. A lattice ordered group or lo-group is a po-group that is a lattice. A totally ordered group or to-group is a po-group that is a totally ordered set. The (positive) cone of $G$ is $G_{\geq1}\!=\!G_+\!:=\!\{g\!\in\!G;\, g\!\geq\!1\}$.
For a group $G$ and subsets $A,B\!\subseteq\!G$, let $\langle A\rangle$ denote the subgroup generated by $A$, and let $AB\!=\!\{ab; a\!\in\!A,b\!\in\!B\}$ and $A^{-1}\!=\!\{a^{-1}; a\!\in\!A\}$.
Theorem (order/cone correspondence for groups) [Steinberg: Lattice-ordered Rings and Modules p.34-35; Blyth: Lattices and Ordered Algebraic Structures p.144-145]: Let $G$ be a group and $P\!\subseteq\!G$ a subset. Consider the following conditions on $P$: $$\text{(i) } PP\!\subseteq\!P; \hspace{3mm} \text{(ii) } \forall g\!\in\!G\!: gPg^{-1}\!\subseteq\!P; \hspace{3mm} \text{(iii) } P\!\cap\!P^{-1}\!=\!\{1\}; \hspace{3mm} \text{(iv) } PP^{-1}\!=\!G; \hspace{3mm} \text{(v) } P\!\cup\!P^{-1}\!=\!G.$$
a) Define the relation $x\leq_P\,y\Leftrightarrow x^{-1}y\!\in\!P$ on $G$. Call $P$ a po-cone / do-cone / lo-cone / to-cone on $G$ when (i)-(iii) hold / (i)-(iii),(iv) hold / (i)-(iii),(iv) hold and $(P,\leq_P\!)$ is a lattice / (i)-(iii),(v) hold. Then the maps $\begin{smallmatrix} \scriptstyle\leq &\!\scriptstyle \mapsto \! &\scriptstyle G_{\geq1}\\ \scriptstyle\leq_P &\!\scriptstyle \leftarrow\! &\scriptstyle P \end{smallmatrix}$ are mutually inverse bijections on the following pairs of sets: $$\begin{array}{r @{\hspace{1mm}} c @{\hspace{1mm}} l} \{\leq\subseteq\!G^2; \text{ the pair }(G,\leq)\text{ is a po-group}\} & \rightleftarrows & \{P\!\subseteq\!G; \text{ the subset }P\text{ is a po-cone on }G\}\\ \bigcup\!\mathbf{|}\hspace{2.6cm} & & \hspace{0.0cm}\bigcup\!\mathbf{|}\\ \{\leq\subseteq\!G^2; \text{ the pair }(G,\leq)\text{ is a do-group}\} & \rightleftarrows & \{P\!\subseteq\!G; \text{ the subset }P\text{ is a do-cone on }G\}\\ \bigcup\!\mathbf{|}\hspace{2.6cm} & & \hspace{0.0cm}\bigcup\!\mathbf{|}\\ \{\leq\subseteq\!G^2; \text{ the pair }(G,\leq)\text{ is a lo-group}\} & \rightleftarrows & \{P\!\subseteq\!G; \text{ the subset }P\text{ is a lo-cone on }G\}\\ \bigcup\!\mathbf{|}\hspace{2.6cm} & & \hspace{0.0cm}\bigcup\!\mathbf{|}\\ \{\leq\subseteq\!G^2; \text{ the pair }(G,\leq)\text{ is a to-group}\} & \rightleftarrows & \{P\!\subseteq\!G; \text{ the subset }P\text{ is a to-cone on }G\}. \end{array}$$
b) Let $(G,\leq)$ be a po-group with $P\!=\!G_{\geq1}$. T.f.a.e.: $(G,\leq)$ is a do-group; $(G,\leq)$ is directed up; $(G,\leq)$ is directed down; $\langle P\rangle \!=\!G$. T.f.a.e.: $(G,\leq)$ is an lo-group; $\forall g\!\in\!G\,\exists g\!\wedge\!1 \in G$; $\forall g\!\in\!G\,\exists g\!\vee\!1 \in G$.
Exercise: Let $G$ be a group with $H\!\unlhd\!G$, and suppose that $H$ and $G/H$ are po-groups. Prove that $$P:=H_+\cup\{g\!\in\!G;\, H\!\neq\!gH\!\in\!(G/H)_+\}$$ is a po-cone on $G$ iff $\forall g\!\in\!G\!: gH_+g^{-1}\!\subseteq\!H_+$. From now on, assume that $P$ is a po-cone on $G$, and $H,G/H$ have po-cones $H_+,(G/H)_+$. Prove: a) $H$ is convex in $G$ [and $P$ induces $H_+,(G/H)_+$]. b) $H\!<\!P\!\setminus\!H$. c) $G$ is a to-group iff $H$ and $G/H$ are to-groups. d) If $H\!\neq\!\{1\}$, then $G$ is an lo-group iff ($H$ is an lo-[sub]group and $G/H$ is a to-group). e) If $G$ is an lo-group, then the inclusion $H\!\rightarrow\!G$ is a complete lattice morphism.
Comment: I added the [...] parts (they were not originally part of the exercise).
Partial Solution: $(\Rightarrow)$: If $P$ is a po-cone, then $\forall g\!\in\!G\!: gPg^{-1}\!\subseteq\!P$, so for $h\!\in\!H_+$, we have $ghg^{-1}\!\in\!P$. From $H\!\unlhd\!G$ we get $ghg^{-1}\!\in\!H$, so we cannot have $H\!\neq\!ghg^{-1}H\!\in\!(G/H)_+$, hence $ghg^{-1}\!\in\!H_+$. $(\Leftarrow)$: (i): If $h,h'\!\in\!H_+$, then $hh'\!\in\!H_+$. If $h\!\in\!H_+$ and $H\!\neq\!gH\!\in\!(G/H)_+$, then $H\!\neq\!hgH\!\in\!(G/H)_+$ (if $hg\!\in\!H$, then $h^{-1}hg\!=\!g\!\in\!H$, $\rightarrow\leftarrow$; since $gH\!\in\!(G/H)_+$ and $hH\!=\!1H\!\in\!(G/H)_+$, we have $hHgH\!=\!hgH\!\in\!(G/H)_+$), and similarly, $H\!\neq\!ghH\!\in\!(G/H)_+$. If $H\!\neq\!gH\!\in\!(G/H)_+$ and $H\!\neq\!g'H\!\in\!(G/H)_+$, then $gg'H\!=\!gHg'H\!\in\!(G/H)_+$, and if $gg'\!\in\!H$, then ???. (ii): By assumption, $gH_+g^{-1}\!\subseteq\!H_+$. Furthermore, if $H\!\neq\!g'H\!\in\!(G/H)_+$, then $H\!\neq\!gg'g^{-1}H$ (since $H\!\unlhd\!G$) and $gg'g^{-1}H\!=\!gH(g'H)g^{-1}H\!\in\!(G/H)_+$. Thus $gPg^{-1}\!\subseteq\!P$. (iii): Assume that $g,g^{-1}\!\in\!P$. If $g,g^{-1}\!\in\!H_+$, then $g\!=\!1$. If $g\!\in\!H_+$ and $H\!\neq\!g^{-1}H\!\in\!(G/H)_+$, then $g^{-1}\!\in\!H$, $\rightarrow\leftarrow$. If $H\!\neq\!gH\!\in\!(G/H)_+$ and $H\!\neq\!g^{-1}H\!\in\!(G/H)_+$, then $gH\!\in\!(G/H)_+\!\cap\!(G/H)_+^{-1}$, so $gH\!=\!H$, $\rightarrow\leftarrow$.
a) If $h\!\leq\!g\!\leq\!h'$ and $h,h'\!\in\!H$, then $h^{-1}g, g^{-1}h'\!\in\!P$. If $h^{-1}g\!\in\!H_+$ or $g^{-1}h'\!\in\!H_+$, then $g\!\in\!H$. But if $h^{-1}gH\!\in\!(G/H)_+$ and $g^{-1}h'H\!\in\!(G/H)_+$, then $gH\!=\!hHh^{-1}gH\!\in\!(G/H)_+$ and $g^{-1}H\!=\!g^{-1}h'Hh'^{-1}H\!\in\!(G/H)_+$, i.e. $gH\!\in\!(G/H)_+\!\cap\!(G/H)_+^{-1}$, which implies $gH\!=\!H$, i.e. $g\!\in\!H$.
If $h,h'\!\in\!H$, then $h\!\leq_{H_+}\!h' \Leftrightarrow h^{-1}h'\!\in\!H_+ \Leftrightarrow h^{-1}h'\!\in\!P \Leftrightarrow h\!\leq_{P}\!h'$. If $gH,g'H\!\in\!G/H$, then $gH\!\leq_{\!(G/H)_+}\!\!g'H \Leftrightarrow g^{-1}g'H\!\in\!(G/H)_+ \Leftrightarrow \big((\exists h\!\in\!H\!:g^{-1}g'h\!\in\!H_+)\text{ or }(H\!\neq\!g^{-1}g'H\!\in\!(G/H)_+)\big) \Leftrightarrow \exists h\!\in\!H\!:\big(g^{-1}g'h\!\in\!H_+\text{ or }H\!\neq\!g^{-1}g'hH\!\in\!(G/H)_+)\big) \Leftrightarrow \exists h\!\in\!H\!: g\!\leq_P\!g'h$. In the second equivalence, $\Leftarrow$ is clear, but for $\Rightarrow$, either $g^{-1}g'H\!\neq\!H$ or $g^{-1}g'\!=\!h^{-1}$ for some $h\!\in\!H$, and then $g^{-1}g'h\!=\!1\!\in\!H_+$.
b) Let $h\!\in\!H$ and $H\!\neq\!gH\!\in\!(G/H)_+$. To prove $h\!<\!g$, we must show that $h^{-1}g\!\in\!P$. We have $h^{-1}gH\!=\!gH\!\neq\!H$ and $h^{-1}gH\!=\!gH\!\in\!(G/H)_+$.
c) $(\Rightarrow)$: Assume $P\!\cup\!P^{-1}\!=\!G$. For any $h\!\in\!H$, either $h\!\in\!P$ or $h^{-1}\!\in\!P$, but since $H\!=\!hH\!=\!h^{-1}H$, either $h\!\in\!H_+$ or $h^{-1}\!\in\!H_+$, i.e. $h\!\in\!H_+\!\cup\!H_+^{-1}$. For any $gH\!\in\!G/H$, either $g\!\in\!P$ or $g^{-1}\!\in\!P$. If $g\!\in\!H_+$ or $g^{-1}\!\in\!H_+$, then $gH\!=\!H\!\in\!(G/H)_+$ or $g^{-1}H\!=\!H\!\in\!(G/H)_+$. Otherwise $gH\!\in\!(G/H)_+$ or $g^{-1}H\!\in\!(G/H)_+$. $(\Leftarrow)$: Assume $H_+\!\cup\!H_+^{-1}\!=\!H$ and $(G/H)_+\!\cup\!(G/H)_+^{-1}\!=\!G/H$. For any $g\!\in\!G$, either $g\!\in\!H$ (then $g\!\in\!H_+\!\cup\!H_+^{-1}\!\subseteq\!P\!\cup\!P^{-1}$) or $gH\!\neq\!H$ (then $g^{-1}H\!\neq\!H$, and either $gH\!\in\!(G/H)_+$ or $g^{-1}H\!\in\!(G/H)_+$), so either $g\!\in\!P$ or $g^{-1}\!\in\!P$.
d) $(\Rightarrow)$: Let $G$ be an lo-group. If $h\!\in\!H$, then $\exists h\!\vee_P\!1\!=:\!g\!\in\!P$, and ???, so $g\!\in\!H$, hence $g\!=\!h\!\vee_{H_+}\!1$. $(\Leftarrow)$: Let $H$ be an lo-group and $G/H$ a to-group. It suffices to prove that $\forall g\!\in\!G\!\setminus\!H\!: \exists g\!\vee_P\!1$. We have either $gH\!\in\!(G/H)_+$ or $g^{-1}H\!\in\!(G/H)_+$, i.e. $hg\!\in\!P$ or $g^{-1}h\!\in\!P$ for some $h\!\in\!H$. ???
e) We must prove that $\exists\inf_H\{h_i; i\!\in\!I\}\!=:\!h_0\Rightarrow \inf_G\{h_i; i\!\in\!I\}\!=\!h_0$ and $\exists\sup_H\{h_i; i\!\in\!I\}\!=:\!h_1\Rightarrow \sup_G\{h_i; i\!\in\!I\}\!=\!h_1$. If $\exists g\!\in\!G\!\setminus\!H\!: g\!\leq\!\{h_i; i\!\in\!I\}$, then ???, so $g\!\leq\!h_0$.
Question: How can I finish the '???' parts? I'm stuck at $PP\!\subseteq\!P$, d), e).
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For the first gap: if $gg'\in H$, then $g'\in g^{-1}H$, so $g'H=(gH)^{-1}$, $g'H\in(G/H)_+\cap\big((G/H)_+\big)^{-1}$, and therefore $g'H=H$, $\to\leftarrow$.
(d) Let $\pi:G\to G/H:g\mapsto gH$. For $g,g'\in G$, $g\le_G g'$ iff $g^{-1}g'\in P$; that that this is the case iff either
• $\pi(g)<_{G/H}\pi(g')$, or
• $\pi(g)=\pi(g')$ and $1\le_H g^{-1}g'$.
Note in particular that if $g,g'\in H$, then $g\le_G g'$ iff $g\le_H g'$.
Suppose that $G/H$ is a to-group, and let $g\in G$. Then exactly one of the following holds: $\pi(g)=\pi(1)$, $\pi(g)<_{G/H}\pi(1)$, or $\pi(g)>_{G/H}\pi(1)$. If $\pi(g)>_{G/H}\pi(1)$, then $g>_G 1$, and $g\lor_G 1=g$. If $\pi(g)<_{G/H}\pi(1)$, then $g<_G 1$, and $g\lor_G 1=1$. And if $\pi(g)=\pi(1)$, then $g\in H$.
We’ll show now that $G$ is a lo-group iff $H$ is a lo-group. Assume first that $H$ is a lo-group, and let $g\in G$. We just saw that if $g\notin H$, then $g\lor_G 1$ exists and is either $g$ or $1$, so assume now that $g\in H$, and let $u=g\lor_H 1$. Then if $g'\in G$ is any upper bound for $g$ and $1$ in $G$, either $\pi(g')>_{G/H}\pi(1)$, in which case $u<_G g'$, or $g'\in H$, in which case $u\le_G g'$, so $u=g\lor_G 1$.
Now suppose that $H$ is not a lo-group, and let $h\in H$ be such that $h$ and $1$ have no supremum in $H$; I claim that $h$ and $1$ can have no supremum in $G$, either. This is clear if there is a $u\in H$ such that $h,1\le_H u$: if $g$ is any upper bound of $h$ and $1$ in $G\setminus H$, $u<_G g$, so $g\ne h\lor_G 1$. Thus, we may assume that $\{h,1\}$ has no upper bound in $H$. Suppose that $g$ is an upper bound for $\{h,1\}$ in $G$; clearly $\pi(g)>_{G/H}\pi(1)$. If $H_+\ne\{1\}$, choose $p\in H_+\setminus\{1\}$; then $h,1<_G gp^{-1}<_G g$, so $g\ne h\lor_G 1$. The only remaining possibility is that $H_+=\{1\}$. In that case $\pi(gh)=\pi(g)$, so $gh$ is an upper bound for $\{h,1\}$. However, it’s clear that $h\ne 1$, so $g^{-1}(gh)=h\notin P$, $g\not\le_G gh$, and $g\ne h\lor_G 1$. Thus, $h$ and $1$ have no supremum in $G$, and $G$ is not a lo-group.
To complete the proof of (d) we must show that if $G/H$ is not a to-group, then $G$ is not a lo-group. Suppose, then, that $G$ is not a to-group; then there is a $g\in G$ such that $\pi(g)\not\le_{G/H}\pi(1)\not\le_{G/H}\pi(g)$. Suppose that $u$ an upper bound for $\{g,1\}$ in $G$; then $\pi(u)>_{G/H}\pi(g),\pi(1)$. As before, if $H$ has a strictly positive element $p$, then $up^{-1}$ is an upper bound for $\{g,1\}$ strictly smaller than $u$. If $H_+=\{1\}$, let $h\in H\setminus\{1\}$, and observe that $uh$ is an upper bound for $\{g,1\}$ such that $u\not\le_G uh$. In either case $u\ne g\lor_G 1$, and $G$ is not a lo-group.
(e) Recall from (d) that if $g,g'\in H$, then $g\le_G g'$ iff $g\le_H g'$.
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Thank you for your reply. I do not understand your 5th (longest) paragraph. Why do we even need that "$h,1$ has no upper bound in $H$"? However, using your ideas, I found an analogous solution to (d) and (e) which I post below, since it's too long for a comment. Thank you again. – Leon Jul 27 '12 at 0:23
This is just a solution to (d) and (e) which I understand easier, but due to Brian M. Scott. It's too long for a comment.
d) Let $\pi:G\!\rightarrow\!G/H$, $g\!\mapsto\!gH$. For $g,g'\!\in\!G$, we have $g\!\leq_G\!g'$ iff $g^{-1}g'\!\in\!P$, iff either $\pi(g)\!<_{G/H}\!\pi(g')$ or ($\pi(g)\!=\!\pi(g')$ and $1\!\leq_H\!g^{-1}g'$).
$(\Leftarrow)$: Suppose that $G/H$ is a to-group. For $g\!\in\!G$, exactly one of the following holds: $\pi(g)\!=\!\pi(1)$, $\pi(g)\!<_{G/H}\!\pi(1)$, $\pi(g)\!>_{G/H}\!\pi(1)$. In the first case, $g\!\in\!H$. In the second, $g\!<_G\!1$, hence $g\!\vee_G\!1\!=\!1$. In the third, $g\!>_G\!1$, hence $g\!\vee_G\!1\!=\!g$. Thus if $g\!\notin\!H$, then $\exists\,g\!\vee_G\!1$. But if $g\!\in\!H$, let $s\!:=\!g\!\vee_H\!1$, so $\pi(s)\!=\!\pi(1)$. For any $g'\!\in\!G$ with $g,1\!\leq\!g'$, either $\pi(1)\!<_{G/H}\!\pi(g')$ (then $s\!<_G\!g'$), or $g'\!\in\!H$ (then $s\!\leq\!g'$). Thus $s\!=\!g\!\vee_G\!1$. We have proved that $g\!\vee_G\!1$ always exists, so by the correspondence theorem, $G$ is an lo-group.
$(\Rightarrow)$: Suppose that $G$ is an lo-group, but $H$ is not, i.e. $\exists h\!\in\!H\!:\nexists\,h\!\vee_H\!1$. Take $s\!:=\!h\!\vee_G\!1$. We see that $s\!\in\!G\!\setminus\!H$. By a), we have $H\!<\!s$, so $H\!<\!Hs$. If $H_+\!\neq\!\{1\}$, choose $p\!\in\!H_+\!\setminus\!\{1\}$, and then $h,1\!<\!p^{-1}\!s\!<\!s$, a contradiction with the minimality of $s$. But if $H_+\!=\!\{1\}$, then $h,1\!\leq\!hs$ gives (by minimality of $s$) $s\!\leq\!hs$, i.e. $h\!\in\!H_+$, so $h\!=\!1$, a contradiction.
Now suppose that $G$ is an lo-group, but $G/H$ is not a to-group, i.e. $\exists g\!: \pi(g)\!\nleq_{G/H}\!\pi(1)\!\nleq_{G/H}\!\pi(g)$. Take $s\!:=\!g\!\vee_G\!1$. Then $\forall h\!\in\!H\!:\pi(g),\pi(1)<_{G/H}\pi(s)=\pi(hs)$. If $H_+\!\neq\!\{1\}$, choose $p\!\in\!H_+\!\setminus\!\{1\}$, and then $g,1 < p^{-1}u < u$, a contradiction. If $H_+\!=\!\{1\}$, let $h\!\in\!H\!\setminus\!\{1\}$, and then $g,1\!\leq\!hs$, so $s\!\leq\!hs$, i.e. $h\!\in\!H_+$, a contradiction.
e) We must prove that $\exists\sup_H\{h_i;i\!\in\!I\}$ implies $\sup_H\{h_i;i\!\in\!I\}\!=\!\sup_G\{h_i;i\!\in\!I\}$, and that $\exists\inf_H\{h_i;i\!\in\!I\}$ implies $\inf_H\{h_i;i\!\in\!I\}\!=\!\inf_G\{h_i;i\!\in\!I\}$. By d), $H$ is an lo-group and $G/H$ is a to-group. We use the criterium from d). Suppose that $s\!:=\!\sup_H\{h_i\}\!\in\!H$ exists. For any $g\!\in\!G$ with $\{h_i\}\!\leq\!g$, either $\pi(s)\!=\!\pi(h_i)\!=\!\pi(g)$ (then $g\!\in\!H$, so $s\!\leq\!g$), or $\pi(s)\!=\!\pi(h_i)\!<\!\pi(g)$ (then $s\!<\!g$). Thus $s\!=\!\sup_G\{h_i\}$. The equality for infimums is proved analogously. $\blacksquare$
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http://media.nips.cc/nipsbooks/nipspapers/paper_files/nips31/reviews/3717.html | NIPS 2018
Sun Dec 2nd through Sat the 8th, 2018 at Palais des Congrès de Montréal
Paper ID: 3717 Multi-Agent Generative Adversarial Imitation Learning
### Reviewer 1
This paper proposed a novel framework for multiagent imitation learning for general Markov games. The main contribution the proposal of a general multiagent Inverse RL framework, MAGAIL that bridges the gap between existing multiagent RL methods and implicit generative models, such as GANs. MAGAIL is rigorous extension of a previous work on generalized single agent inverse reinforcement learning, GAIL. MAGAIL allows incorporation of prior knowledge into the discriminators, including the presence of cooperative or competitive agents. The paper also proposed a novel multiagent natural policy gradient algorithm that addresses the high variance issue. MAGAIL is evaluated on both cooperative and competitive tasks from the Particle environments, and a cooperative control task where only sub-optimal expert demonstrations are available. Three variants of MAGIAL (centralized, decentralized, zero-sum) are evaluated and outperform several baselines, including behavior cloning and GAIL IRL baseline that operates on each agent separately. This paper is well-written and reasonably accessible (given the mathematical rigor), and the topic is of importance to the machine learning community (both multiagent systems and reinforcement learning). The contributions are nicely laid out, and the structure of the paper leads to a good flow for the reader. The contributions are clear. The experiments are convincing. Minor points: The advantage function in equation (11) is wrongly specified. There is a minus sign missing. In the definition of occupancy measures, it might be better to change T to P to avoid the confusion of transition function.
### Reviewer 2
This paper proposes several alternative extensions of GAIL to multi-agent imitation learning settings. The paper includes strong, positive results on a wide range of environments against a suitable selection of baselines. However, insufficient details of the environments is provided to reproduce or fully appreciate the complexity of these environments. If accepted, I would request the authors add these to the appendix and would appreciate details (space permitting) to be discussed in the rebuttal - particularly the state representation. The more pressing point I would like to raise for discussion in the rebuttal is with regard to the MACK algorithm proposed for the generator. The authors make a justified argument for the novelty of the algorithm, but do not thoroughly justify why they used this algorithm instead of an established MARL algorithm (e.g. Multi-Agent DDPG). It seems that MACK is in itself also a contribution to the literature, but it is not directly compared to a suitable existing baseline. Its use also convolutes whether the gains seen by multi-agent GAIL were due to the different priors used or the different RL algorithm used for the generator. Ideally it would be interesting to see an ablation study showing how multi-agent GAIL performs with the different priors proposed in Figure 1 but with TRPO as in the original paper. Additionally, given the choice to use MACK instead of an established multi-agent RL algorithm, a comparison of multi-agent GAIL with MACK to multi-agent GAIL with multi-agent DDPG would provide empirical evidence of the benefit of MACK. For future work, I would suggest the authors revisit an assumption of the paper that the number of experts matches directly the number of agents. However, I don't think this assumption is strictly necessary for some of the multi agent extensions proposed and would be an interesting further empirical study as it enables unique use cases where expert data is limited but could generalise across agents or provide partial insight into how some of the agents should act.
### Reviewer 3
This paper proposes a general multi-agent IRL framework or multi-agent imitation learning for general Markov games, which generalizes Generative Adversarial Imitation Learning to multi-agent cases. The learning procedure corresponds to a two-player game between a generator and a discriminator. The generator controls the policies of agents in a distributive fashion, and the discriminator contains a classifier for each agent that is trained to distinguish that agent’s behavior from that of the corresponding expert. Experimental results on a particle environment and a cooperative control task demonstrate that it is able to imitate complex behaviors in high-dimensional environments with both cooperative and adversarial interactions. Unlike most existing work in multi-agent imitation learning where the agents have very specific reward structures, the proposed framework can handle both cooperative and competitive scenarios. However, the most difficult part is the case when the agents are self-interested and the payoffs are general sum. In this case, there may be multiple Nash equilibriums. When the agents learn in a decentralized manner, they may not be converging and jump between equilibriums. It is not clear how such cases are handled in the proposed method. Update: Thanks for the rebuttal. I have read it carefully.
### Reviewer 4
This paper proposes a Multi Agent Inverse Reinforcement Learning paradigm by finding connections of multi-agent reinforcement learning algorithms and implicit generative models when working with the occupancy measure. The problem is very important and the solution provided looks interesting. Experiments on cooperative and completive environments establish the effectiveness of the proposed framework. The paper is very well written and easy to follow and read. It could improve by a proof check especially towards the end of the paper. Intuitively, it looks that the generalization over reference [16] is achieved by considering other agents as part of the environment in the training. Is that the case? It’s good to be elaborated and the connections become more clear. The reader may wonder how the method performs without the V_\phi and how the non-stationery of the environment will affect the variance. How much of the success is due to V_\phi and how much is the real contribution of the proposed method. A minor comment is that T is used both for transition function and time horizon. Also, notation-wise D is overloaded. It’s better to change one of them. It’s gonna help the reader grasp the connection to generative adversarial models if the max or min operators in important equations (like 9) are interpreted and their role in the optimization become more clear. The experimental results look convincing. Though it could be more comprehensive to have an empirical ablation study to see how much each part is contributing to the performance. ----------------------- Update: I've read other reviews and the author response. The author response is satisfactory. I'm still in favor of accepting this paper. | {"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": 0.8239740133285522, "perplexity": 723.6452542407283}, "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-2019-22/segments/1558232256163.40/warc/CC-MAIN-20190520222102-20190521004102-00528.warc.gz"} |
https://chemistry.stackexchange.com/tags/atomic-radius/hot | # Tag Info
15
The thing is that you need to know the coordination environment in the first place as ionic radii are C.N.-dependent. From Shannon's canonical paper "Revised effective ionic radii and systematic studies of interatomic distances in halides and chalcogenides"[1]: \begin{array}{cc} \hline \text{Ion} & \text{C.N.} & \text{C.R., Å} & \text{I.R., Å} \...
13
You are mixing apples and oranges. Or, to be more precise, an ionic radius for $\ce{Be^2+}$ with coordination number (C.N.) 6 and van der Waals radius of $\ce{He}$. To make it clear I compiled data for van der Waals and covalent radii [1, p. 9-58] as well as ionic radii [1, p. 12-13]: $$\begin{array}{lccccc} \hline \text{Element} & R_\mathrm{vdW}/\pu{Å}... 10 As you may know, atomic orbitals are wave functions, solutions of the Schrödinger equation for an atomic system. In a perfectly spherical system you may express an orbital as a function depending on the distance from the nucleus (r) and two angles (\phi and \theta). [If you pick a particular r, the angles \phi and \theta work in a way similar to ... 9 Let me start by providing a reference to a nice collection of van der Waals' data (actually the site provides a lot of other interesting data about the elements, you might want to poke around). As to your questions, Yes, that is a general trend, but as you can see there are many exceptions. The trend exists for the same reason that covalent radii generally ... 8 The whole concept of ionic radius is not sharply defined one. Note, that the same is true for the notion of atomic radius, as well as for any other kind of radius or any other notion of size. In general, the notion of size of an object looses its usual (i.e. classical) meaning in the microscopic realm since objects there do not have well-defined physical ... 8 As you told it is not possible to measure the inter nuclear distance of single atom. But here internuclear distance does not mean diameter of a single atom but it means distance between nucleus of two atom of same element. This internuclear distance can be determined by two methods: X-rays method Spectroscopy method Note: Atomic radius is not a set ... 7 There is no such thing as unified atomic radius. An atomic radius is a class consisting of van der Waals radii R_\mathrm{vdW} (steric interactions), covalent radii R_\mathrm{cov}, and ionic radii R_\mathrm{i} (and some other as well). From the recent edition of CRC Handbook [1, p. 9-57]:$$ \begin{array}{llrr} \hline \text{Element} & \text{...
6
The short answer: no, there is no such a general distance. The long answer: atoms and molecules are not just hanging is the vacuum and waiting someone to grab them. They actually have rather high energy in translation, rotation and vibration modes and particles actually collide during a reaction some of these energies can be involved. The science that study ...
6
6
I completely agree with Wildcat's answer. Excellent stuff. This part, though, got me thinking: Anyway, values of the ionic radii are based on crystallographic data, but what is actually determined in X-ray crystallography is the distance between two ions, not their radii. Then it is basically up to you how to divide this distance into the radii of ions ...
6
Iron and sulfur ions bond by electrostatic attraction. This is the right answer. I see a hard time maneuvering around it. Breaking an ionic bond between $\ce{Fe^2+}$ and $\ce{S^2-}$ releases energy. Around the basic definition of forces and bonds, there are usually two big beliefs. One is a big misconception, and the other one is right. The ...
5
Atomic radii cannot really be uniquely nor accurately defined. However, there is generally a very big difference between covalent, van der Waals, and ionic radii. So, by most any definition of the respective terms the covalent radius of fluorine will be smaller than the van der Waals radius of neon. Similarly, the ionic radius of fluoride ion will be ...
5
We do not know. Physicists THINK that there ought to be a fundamental limit in scale for space-time that occurs near $10^{-33}$ centimeters and $10^{-43}$ seconds; often called the Planck Scale. There is also a unit of mass associated with this scale which is about $10^{-5}$ GRAMS or $10^{19}$ Billion Electron Volts (BeV or GeV). It is a simple matter to ...
5
To answer the second part: We know $M=208m_e$, $Z=3$, $\hbar=\frac{h}{2\pi}$. Part one has a mistake, as it is \begin{align} &&\frac{Mv^2}{r}&=\mathcal{k_e}\cdot\frac{(\mathcal{e})(Z\mathcal{e})}{r^2}\\ &&Mvr&=n\hbar\\ \implies&& r &=\frac{n^2\hbar^2}{M\cdot\mathcal{k_e}\cdot Z \mathcal{e}^2} \end{align} We also ...
5
You are right: the outer boundary is quite arbitrary. There is no any intrinsic threshold; the probability just gradually decreases lower and lower, but never reaches 0. You may draw a sphere so as to have the electron inside with 90% probability, or 95%, or 99%, or any other value as you see fit. This, in particular, is the reason why atomic radii are ...
4
The series you cited belongs to so known 'metallic' radius, and it depends on crystal structure of the element, which changes through the row. In short, you cited series, that is not suited for consideration of isolated tendencies. There are, indeed, several types of atomic radii (covalent with different valur for bonds of different order, van-der-waals ...
4
There are different notions of atomic radius; the one you're using seems to be the metallic radius, which is half the distance between nearest neighbors in the metal. This notion is very sensitive to the number of electrons per atom involved in bonding. Scandium has only 3 valence electrons, while $\ce{Ti}$ has 4. These all participate, in some extent, in ...
4
definitely $\ce{O-}$ In $\ce{O-}$ 7 outer electrons are tied to nuclei with 6 positive charges (number 8 minus 2 electons of inner shell), while in $\ce{F-}$ 8 outer electrons are tied to nuclei with 7 charges. The difference, however, should be quite small.
4
I think the point of this question is for you to realise that options 1 and 2 can't be the correct answer. As you go down a group, new electron shells are occupied which extend further from the nucleus, increasing the atomic radius. Therefore option 1 must be wrong. Effective nuclear charge increases across a period because the nuclear charge increases but ...
4
There are several definitions of atomic radius. Guessing from context of your answer, I assume you are talking about van der Waals atomic radius. By definition, it is a radius of atom in respect of equilibrium position with other atoms when only van der Waals forces are acting between the atoms. Since atoms of inert gases do not tend to make covalent bonds, ...
4
The size of the 1s orbital in $\ce{Li+}$ and $\ce{Al^3+}$ is not necessarily the same. In fact, they are quite different because of the much larger effective nuclear charge in $\ce{Al^3+}$. One can easily look up the wavefunction for the 1s orbital and see the radial dependence on $Z_\mathrm{eff}$. Therefore, merely looking at the electronic configuration ...
4
This is something students often forget when comparing radii: shells are not simply additive. If you move along eight elements to go from lithium to sodium, you are not only filling the second shell and putting one electron into the third, you are also adding a proton to the nucleus with every additional electron. The larger the nuclear charge is, the ...
4
First, you have to look at the definition of metallic radius, which is the half distance between two atoms in a lattice. It has a significant dependence on crystal structure. Tanget relevant to the question and other answers: Gallium has an orthorhombic crystal structure (CN = 6) whereas as aluminum has a face-centered cubic crystal structure (CN = 12)....
4
The first thing you should realize is that there are various definitions of the size of the atom. One distinguishes for example the covalent radius, the Van der Waals radius and the ionic radius. The covalent radius is based on the binding of atoms into molecules and on the resulting bond length. The van der Waals radius is half the minimum distance between ...
4
As Paul has pointed out, there are many different ways of defining atomic radii. ron has also pointed out that the trend goes in the opposite direction from what you said. For instance, according to Ptable, |Calculated radii | | |r(pm)| |r(pm)| |H | 53 |He| 31 | |F | 42 |Ne| 38 | |Cl| 79 |Ar| 71 | |Br| 94 |Kr| 88 | I have added hydrogen here as ...
4
You're right about the expected order. From largest to smallest the expected order would be: $\ce{_{53}I^-}\quad$ = [Kr] $4d^{10}\text{ }5s^2\text{ }5p^6$ = [Xe] $\ce{_{34}Se^{2-}}$ = [Ar] $3d^{10}\text{ }4s^2\text{ }4p^6$ = [Kr] $\ce{_{35}Br^-}\text{ }$ = [Ar] $3d^{10}\text{ }4s^2\text{ }4p^6$ = [Kr] $\ce{_8O^{2-}}\text{ }\text{ }$ = [He] $2s^2\text{ ... 4 The Bohr model is a comparatively simple model, whereas the quantum description is part of the huge theory of quantum mechanics, so to list all the differences including implications would need to explain quantum mechanics itself. But I think quite a good starting point would be to say, the Bohr model has electrons travel on certain, specific, classical ... 4 Unlike a ball, an atom doesn't have a fixed radius. The radius of atom can only be found by measuring the distance between the nuclei of two touching atoms and then halving that distance. According to my findings, I think the reasoning lies in the way in which the measurements are done: i. Metallic/Covalent radius (left) ii. Van der Waal radius (right)... 4 Covalent radius is measured as the distance ($r_\text{cov} = d/2$) between the nuclei of two bonded atoms (covalent). But, if you try to do the same for noble gases/inert gases (good luck!), as they have fully filled$np\$ orbitals, they will repel each other, hence the closest distance between the two atoms is taken (high pressure, low temperature) as the ...
4
This is how I would do the calculation... Given data: $$\text {Weight of crystal} = \pu{92.90638 g/mole} \ce{->[Round] \pu{92.906} g/mole}$$ $$\text {Density} = \pu{8.57 g/cm^3}$$ Ok, the density is good to only 3 significant figures so the answer shouldn't have any more than that. But doing the whole problem, I'll carry 5 significant figures ...
Only top voted, non community-wiki answers of a minimum length are eligible | {"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": 0.8598310351371765, "perplexity": 706.8493746549598}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-10/segments/1581875145708.59/warc/CC-MAIN-20200222150029-20200222180029-00327.warc.gz"} |
https://www.physicsforums.com/threads/binding-energy-per-nucleon-and-fission-fusion.819296/ | # Binding energy per nucleon and fission/fusion
Tags:
1. Jun 16, 2015
### Hashiramasenju
1. The problem statement, all variables and given/known data
Ok so i know that the binding energy per nucleon inceases after fission and fusion and the difference in the binding energy is given out as energy. But if the binding enrgy increases shouldnt there be energy taken in(rather than given out)? Also why is the net loss in my = the gain in Binding energy of the fused(OR FISSION EITHRER WAY) nuclei ?
2. Relevant equations
3. The attempt at a solution
2. Jun 16, 2015
### Staff: Mentor
Think of binding energy as negative: it's energy that's "missing" from the nucleus, compared to the separated nucleons; it's not energy that the nucleus "has".
3. Jun 16, 2015
### Hashiramasenju
Thanks for the reply kinda got it. But in the exam i get question like calculate the difference in mass and hence the energy released. Does this equal to the difference in binding energy.
And one question was"explain fission and fusion in terms of the binding energy" How do you Answer that? Thanx alot.
4. Jun 16, 2015
### Staff: Mentor
No, you have to convert from mass to energy. Try adding up the mass of 2 protons and 2 neutrons and subtracting the mass of an alpha particle from that sum. Then convert that difference in mass to energy.
Well, if binding energy is negative, meaning that an increase in binding energy gives off energy, what happens during fission and fusion?
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https://math.stackexchange.com/questions/1965914/power-of-a-graph-proof | # Power of a graph proof
Prove that if $G$ is a connected graph, with at least three vertices, then it’s square graph $G^2$ is 2-connected.
My attempt: Let $G=(V,E)$. For a graph to be 2-connected, I understand that for every vertex $x \in V(G)$, $G-x$ is connected.
Now, I can see why the above statement is true through different examples for vertices greater than 3. However, I'm not sure how to prove it for the general case. My guess is that since the power is 2, we connect every $x_{i}$ vertice to $x_{i+2}$ and by doing this we're essentially putting all the vertices on cycles in the graph.
I'm assuming that $G^2$ is the Cartesian product of $G$ with itself.
• For any path $P = (v_1,v_2,\ldots,v_n)$ in $G$ and vertex $u \in V$ let $\langle P,u\rangle$ be the path in $G^2$ defined as $$\langle P,u\rangle = \Big(\langle v_1,u\rangle,\langle v_2,u\rangle,\ldots,\langle v_n,u\rangle\Big).$$
• If $u$ and $v$ are connected in $G$, then there exists a simple path $P$ that connects them.
• Thus, in the $G^2$ path $\langle P,u\rangle$ connects $\langle u,u\rangle$ with $\langle v,u\rangle$, and similarly \begin{align} \langle u,v\rangle \xrightarrow{\langle P,v\rangle} \langle v,v\rangle,\\ \langle u,u\rangle \xrightarrow{\langle u,P\rangle} \langle u,v\rangle,\\ \langle v,u\rangle \xrightarrow{\langle v,P\rangle} \langle v,v\rangle. \end{align}
I hope this helps $\ddot\smile$ | {"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": 0.9776400327682495, "perplexity": 93.37007364163296}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-43/segments/1570986675316.51/warc/CC-MAIN-20191017122657-20191017150157-00238.warc.gz"} |
http://mathhelpforum.com/algebra/171828-inequality-problem-print.html | # Inequality Problem
• February 19th 2011, 04:21 AM
dd86
Inequality Problem
I can't solve this question:
Solve for k where y^2-4k>=2(k-1)y where y=[(x^2-2x+6)/(x-3)]. (Ans. k=5)
Your tips/ hints are sufficient. Please don't give the specific steps to solve this question.
• February 19th 2011, 04:28 AM
Prove It
Try completing the square on $\displaystyle y^2 - 2(k - 1)y - 4k \geq 0$.
Then you will be able to solve for $\displaystyle k$.
• February 19th 2011, 05:15 AM
dd86
I did attempt that. But I guess I messed up somewhere.
I'll have another go at it.
• February 19th 2011, 05:55 AM
dd86
I've completed the square correctly. But I think I'm still missing out on something very obvious but I can't seem to get it.
I substituted y for the equation it stood for but I still came to a dead end.... :@
• February 19th 2011, 06:06 AM
Prove It
Assuming you have completed the square properly you should have
$\displaystyle y^2 - 2(k - 1)y - 4k \geq 0$
$\displaystyle y^2 - 2(k - 1)y + [-(k-1)]^2 - [-(k-1)]^2 - 4k \geq 0$
$\displaystyle [y - (k-1)]^2 - (k - 1)^2 - 4k \geq 0$
$\displaystyle (y - k +1)^2 \geq (k - 1)^2 + 4k$
$\displaystyle (y - k + 1)^2 \geq k^2 - 2k + 1 + 4k$
$\displaystyle (y - k + 1)^2 \geq k^2 + 2k + 1$
$\displaystyle (y - k + 1)^2 \geq (k + 1)^2$
$\displaystyle |y - k + 1| \geq |k + 1|$.
This should give you four inequalities to solve for $\displaystyle k$.
• February 19th 2011, 06:46 AM
dd86
Four inequalities? Hmm...
OK, I've only come across one way on how to solve inequalities with absolute values on both sides. And that's by squaring both sides.
However, it's now the other way.
And I'm boggled by the fact that I have two unknowns on the LHS. So, I need more guidance, please.
• February 22nd 2011, 03:09 AM
dd86
Hi
The problem still can't be solved yet. Any one else has got an idea how to? | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 11, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8849191069602966, "perplexity": 1448.2563408758444}, "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-2016-07/segments/1454701149548.13/warc/CC-MAIN-20160205193909-00201-ip-10-236-182-209.ec2.internal.warc.gz"} |
http://eprint.iacr.org/2003/133/20040124:101614 | ## Cryptology ePrint Archive: Report 2003/133
Minimum Distance between Bent and 1-resilient Boolean Functions
Soumen Maity and Subhamoy Maitra
Abstract: In this paper we study the minimum distance between the set of bent functions and the set of 1-resilient Boolean functions and present a lower bound on that. The bound is proved to be tight for functions up to $10$ input variables. As a consequence, we present a strategy to modify the bent functions, by toggling some of its outputs, in getting a large class of $1$-resilient functions with very good nonlinearity and autocorrelation. In particular, the technique is applied upto $12$-variable functions and we show that the construction provides a large class of $1$-resilient functions reaching currently best known nonlinearity and achieving very low autocorrelation values which were not known earlier. The technique is sound enough to theoretically solve some of the mysteries of $8$-variable, $1$-resilient functions with maximum possible nonlinearity. However, the situation becomes complicated from $10$ variables and above, where we need to go for complicated combinatorial analysis with trial and error using computational facility.
Category / Keywords: secret-key cryptography / Boolean Functions
Publication Info: Accepted in FSE 2004
Date: received 12 Jul 2003, last revised 24 Jan 2004
Contact author: subho at isical ac in
Available format(s): Postscript (PS) | Compressed Postscript (PS.GZ) | PDF | BibTeX Citation
Note: Some proofs are now written more carefully after getting the review comments from FSE 2004.
Short URL: ia.cr/2003/133
[ Cryptology ePrint archive ] | {"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": 0.9251703023910522, "perplexity": 2682.479445848754}, "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/1469257824994.73/warc/CC-MAIN-20160723071024-00237-ip-10-185-27-174.ec2.internal.warc.gz"} |
http://psychology.wikia.com/wiki/F_test?direction=prev&oldid=89130 | # F test
34,147pages on
this wiki
An F-test is any statistical test in which the test statistic has an F-distribution if the null hypothesis is true. A great variety of hypotheses in applied statistics are tested by F-tests. Among these are given below:
• The hypothesis that the means of multiple normally distributed populations, all having the same standard deviation, are equal. This is perhaps the most well-known of hypotheses tested by means of an F-test, and the simplest problem in the analysis of variance.
• The hypothesis that the standard deviations of two normally distributed populations are equal, and thus that they are of comparable origin.
In many cases, the F-test statistic can be calculated through a straightforward process. Two regression models are required, one of which constrains one or more of the regression coefficients according to the null hypothesis. The test statistic is then based on a modified ratio of the sum of squares of residuals of the two models as follows:
Given n observations, where model 1 has k unrestricted coefficients, and model 0 restricts m of the coefficients (typically to zero), the F-test statistic can be calculated as
$\frac{\left(\frac{RSS_0 - RSS_1 }{m}\right)}{\left(\frac{RSS_1}{n - k}\right)}.$
The resulting test statistic value would then be compared to the corresponding entry on a table of F-test critical values, which is included in most statistical texts. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 1, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9199714660644531, "perplexity": 396.11598576939997}, "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-2015-06/segments/1422122080417.25/warc/CC-MAIN-20150124175440-00032-ip-10-180-212-252.ec2.internal.warc.gz"} |
https://tex.stackexchange.com/questions/309100/arial-font-baseline-for-letters-i-and-l | # Arial font baseline for letters 'i' and 'L'
I am attempting to use the uarial font in latex. But the vertical alignment of characters appears to be off. If
\documentclass{article}
\usepackage[english]{babel}
\usepackage[T1]{fontenc}
\usepackage{uarial}
\renewcommand{\familydefault}{\sfdefault}
\begin{document}
\Huge{tial}
\end{document}
Produces the following:
Notice that the letters 'i' and 'l' have a different baseline than do 't' and 'a'. Is there a way to adjust this? That is not a feature of Arial
• Take or leave, I'm afraid. May 11, 2016 at 16:54
I don't think that uarial is a good choice. It is a rather curious mix between Arial and Helvetica. As you can see in the following picture the "C", "t" and "a" are from Helvetica, while the "G" and "R" is from arial. Also as you discovered the metrics are not really good. It is naturally possible to correct this by manipulating the tfm, but I don't think that it is worth the trouble.
%needs lualatex or xelatex
\documentclass{article}
\usepackage{fontspec}
\setmainfont{Arial}
\setsansfont{TeX Gyre Heros}
\begin{document}
\Huge
CGRtial (Arial)\par
{\sffamily CGRtial} (Helvetica/TeX Gyre Heros)\par
\fontencoding{T1}\fontfamily{ua1}\selectfont CGRtial (uarial)
\end{document} | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 0.960293710231781, "perplexity": 2442.232813818032}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-14/segments/1679296943695.23/warc/CC-MAIN-20230321095704-20230321125704-00522.warc.gz"} |
https://www.physicsforums.com/threads/a-generic-quantum-state-and-the-charge-of-an-electron.950416/ | # A generic quantum state and the charge of an electron
• Start date
• Tags
• #1
138
2
I'm watching a lecture and the professor is talking about generic quantum states as
$$|\psi>$$
He's making the point that this state is very generic. It can represent anything. He references some examples like the polarization of a photon and the path of a photon and the spin of an electron. Those (I think) make sense to me as quantum states.
But then he says that the "charge on an electron" could be represented by this generic state, ## |\psi> ##.
This example I didn't understand. How could talking about the charge of an electron by itself by represented by ## |\psi> ##. For one, what would a superposition of states look like in this case? The charge can't be fractional or be 2e or 3e. It's just e. What am I missing?
• #2
Staff Emeritus
2021 Award
28,215
12,857
Can you write exactly what he says, word for word, in context?
Likes Dale
• #3
Dale
Mentor
2021 Award
32,844
10,034
I'm watching a lecture
• #4
Staff Emeritus
2021 Award
28,215
12,857
I'd rather he post the text than "here's an hour long video and somewhere in it he says this".
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1K | {"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": 0.8556987643241882, "perplexity": 2709.318929598106}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-27/segments/1656103645173.39/warc/CC-MAIN-20220629211420-20220630001420-00335.warc.gz"} |
https://deepai.org/publication/a-maximum-likelihood-approach-to-speed-estimation-of-foreground-objects-in-video-signals | # A Maximum Likelihood Approach to Speed Estimation of Foreground Objects in Video Signals
Motion and speed estimation play a key role in computer vision and video processing for various application scenarios. Existing algorithms are mainly based on projected and apparent motion models and are currently used in many contexts, such as automotive security and driver assistance, industrial automation and inspection systems, video surveillance, human activity tracking techniques and biomedical solutions, including monitoring of vital signs. In this paper, a general Maximum Likelihood (ML) approach to speed estimation of foreground objects in video streams is proposed. Application examples are presented and the performance of the proposed algorithms is discussed and compared with more conventional solutions.
## Authors
• 1 publication
• 2 publications
• 4 publications
• ### Motion trails from time-lapse video
From an image sequence captured by a stationary camera, background subtr...
12/04/2015 ∙ by Camille Goudeseune, et al. ∙ 0
• ### On Flow Profile Image for Video Representation
Video representation is a key challenge in many computer vision applicat...
05/12/2019 ∙ by Mohammadreza Babaee, et al. ∙ 0
• ### Reciprocal Maximum Likelihood Degrees of Brownian Motion Tree Models
We give an explicit formula for the reciprocal maximum likelihood degree...
09/24/2020 ∙ by Tobias Boege, et al. ∙ 0
• ### Appearance-free Tripartite Matching for Multiple Object Tracking
Multiple Object Tracking (MOT) detects the trajectories of multiple obje...
08/09/2020 ∙ by Lijun Wang, et al. ∙ 0
• ### Context-aware learning for finite mixture models
This work introduces algorithms able to exploit contextual information i...
07/29/2015 ∙ by Serafeim Perdikis, et al. ∙ 0
• ### MAS for video objects segmentation and tracking based on active contours and SURF descriptor
In computer vision, video segmentation and tracking is an important chal...
08/01/2013 ∙ by Mohamed Chakroun, et al. ∙ 0
• ### A Maximum A Posteriori Estimation Framework for Robust High Dynamic Range Video Synthesis
High dynamic range (HDR) image synthesis from multiple low dynamic range...
12/08/2016 ∙ by Yuelong Li, et al. ∙ 0
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## I Introduction
Motion and speed estimation can be necessary processes in order to recover specific real-world related information arising from the movement of an object in a three-dimensional (3D) scene captured by an image acquisition system, e.g. a camera. When a video is captured, a 3D real-world scene is projected onto the two-dimensional (2D) camera plane, resulting in a sequence of digital images. Hence, motion in a video stream can be considered as the consequence of the projection of objects moving in the 3D scene [pesquet]. This phenomenon may arise when objects are moving and the camera is still, when the camera is moving and the objects are still or when both are moving.
Various motion estimation algorithms are widely documented in the literature, as reported, e.g., in [pesquet], [chen] and [VideoProcessing, Ch. 4]. In particular, the most popular techniques include differential methods and matching methods.
Differential methods aim to estimate the optical flow, defined as the apparent motion perceived from the variations in the pixel intensity patterns in the 2D image, that may be due either to a true motion in the 3D space or to illumination changes [optflow]. These approaches are based on gradient techniques that exploit the properties of spatial and temporal partial derivatives. The most popular solutions in this category are the Lucas-Kanade [kanade] and the Horn-Schunck methods [schunck]
, that rely on the brightness constancy model and the smoothness constraint of the velocity vector
[optflow], respectively. Due to these specific assumptions, the two methods may fail in properly describing some realistic scenarios, which may represent their main limitation.
Block matching approaches, the most widely employed matching methods, are based on the partition of the considered video frame into several blocks of pixels, that may or may not overlap, where each block is associated with a motion vector [VideoProcessing, Ch. 4]. Block matching criteria, despite being very straightforward and basic, are highly sensitive to noise and parameter setting, especially the block size. As a consequence, the estimated motion vector may not coincide with the true motion. In this category, algorithms to estimate the mean speed of a group of vehicles from Motion Pictures Experts Group (MPEG) video streams are presented in [yu] and [hu], where motion vectors are directly extracted from the considered stream.
To avoid problems related to differential and matching methods, we wish to investigate the application of a more fundamental estimation principle, namely the Maximum Likelihood (ML) approach [Kay], to estimate the speed of a foreground object in considered video sequences.
The reminder of the paper is organized as follows. In Section II the motion estimation model of a framed object in a video sequence is described and a brief discussion on background removal techniques is proposed. The ML estimation approach is presented in Section III. The performance of the proposed algorithms is discussed in Section IV on the basis of several experimental results. Finally, conclusions are drawn in Section IV.
## Ii Observation Model
### Ii-a Preliminary Definitions
A video can be defined as a time sequence of digital images, which are referred to as frames, whose spatial-intensity pattern may vary over time. It is hence a time varying multidimensional signal where two spatial components identify a pixel position within the frame, which can also be thought of as a two-dimensional image (per colour component) formed by projecting a real world 3D scene onto a 2D image plane. The motion and variation of an object in the real world correspond to changes in the pixel intensity values in the 2D image plane. Motion estimation is therefore a process that allows to recover real-world related information by analysing the time evolution of the framed area.
Before presenting the formulation of the motion model and discussing the ML approach to speed estimation of a framed foreground object, the following simplifying assumptions are introduced:
• only framed areas including a single moving object are considered;
• the capturing camera is still;
• object transformations due to perspectival issues arising from the projection of the 3D real world scene onto the 2D image plane are not accounted for.
The first condition allows to simplify the model and test the effectiveness of the proposed algorithm in the simplest possible scenario. The camera is considered still to avoid superposed motions other than the considered object motion. Finally, the last condition assumes that no perspectival transformation affects the object. These simplifying assumptions allow us to concentrate on the main goal of this paper, which is motion estimation, and will be relaxed by further investigation currently undergoing.
### Ii-B Object Motion Model
A video signal is composed by a set of frames sampled at instants , where is the discrete time index corresponding to the frame number, and is the sampling time, with being the sampling frequency, or frame rate, of the camera. A digital grayscale video signal can be described as a two-dimensional discrete function , where indicates the pixel position within the frame,111 denotes the transpose operator. with and , whose value corresponds to the pixel intensity. In particular, is the frame size, being and its height and width, respectively. As customary, the dynamic range of the pixel intensity is limited to the interval . In the case of coloured videos, a fourth dimension can be added to this function in order to specify the colour channel index.
Considering a framed object shifting with a simple translation in the two-dimensional projected camera plane, its motion can be described by a displacement vector , with horizontal () and vertical () components. Defining as the image of the still object of interest, the pixel intensities of a video stream can be modelled as the summation of a few main elements [Bri2]:
x[m,n]=b[m]+s[m−δ[n]]+vb[m,n]+w[m,n] (1)
where is the still background, the object is shifted with a displacement , is a term which takes into account the occluded/un-occluded parts of the background and are samples of independent and identically distributed (i.i.d.) zero-mean Gaussian noise. The background is assumed static, i.e., time-invariant. Equation (1) can be considered as the most general model to describe a single framed moving object in a video stream. This model applies directly to a group of objects subject to the same displacement.
Consider now the special case of constant speed for further simplification. The displacement term can be explicitly written as , where is the uniform speed motion vector in pixel/frame, that we wish to estimate. Hence, the observation model in (1) can be rewritten as:
x[m,n]=b[m]+s[m−vn]+vb[m,n]+w[m,n] (2)
where the direction and the speed of the object displacement have been considered time-invariant for the sake of simplicity. This model can be easily applied to a slowly time varying scenario, provided can be considered approximately constant over a sufficiently long time window.
Models (1) and (2
) are formulated in the time domain. Using a frequency domain representation may result more convenient, as discussed in
[Bri2]
. The main advantage of operating in the frequency domain relies on the properties of the Fourier Transform (FT). In particular, thanks to the shift theorem it is possible to describe a displacement in the time domain as a linear phase term in the frequency domain. Hence, the dependence of
on the parameter can be factored out when working in the frequency domain. Defining the Discrete Fourier Transform (DFT) of a generic two-dimensional discrete function of size as in [Sol], the observation model in (2) can be expressed as:
X[k,n]=B[k]+S[k]e−j2πukTvn+Vb[k,n]+W[k,n] (3)
where is the vector collecting the two discrete indeces of the two-dimensional DFT, with , and is the vector of the normalized spatial frequencies.
Equations (2) and (3) describe Gaussian observations that are independent both in discrete time and frequency domains. Focusing on the model in (3), the speed vector represents the only unknown parameter to be estimated.
The ML criterion can be applied to the model in (3) to obtain an expression for the speed vector estimator . Before detailing the estimation procedures, we briefly discuss the role of background removal techniques that allow to detect the foreground object and that may be useful to further simplify the models in (2) and (3).
### Ii-C Background Removal
Background removal techniques aim to separate the background from the foreground moving objects. The basic requirement is that the background is static. Nevertheless, this condition is often violated especially in outdoor scenes that are subject to illumination changes and other variations, e.g., due to wind effects or shadows. For this reason, adaptive algorithms for background removal, such as the Gaussian Mixture Model (GMM)
[stauffer], are usually preferred. The concept of background removal relies on the analysis and the comparison of two frames in the considered video stream: the background frame, also referred to as the reference frame where no moving objects are present, and the frame at a chosen instant where moving objects are present. The difference between these two frames represents the foreground. A comprehensive review on background removal approaches is proposed in [piccardi].
In this paper, we use the reliable GMM that adaptively models each pixel belonging to the background as a mixture of Gaussian distributions. The interested reader is referred to
[stauffer] for an accurate description of the proposed method.
Using the GMM technique, it is possible to remove the background-related terms and to concentrate on the foreground and simplify the observation model in (2) as:
x[m,n]=s[m−vn]+w[m,n]. (4)
Likewise the frequency domain observation model in (3) simplifies as:
X[k,n]=S[k]e−j2πukTvn+W[k,n]. (5)
## Iii Maximum Likelihood Speed Estimation
### Iii-a Speed Estimation with Included Background
The observation model (2) and its frequency-domain counterpart (3) include the terms and
, respectively, to account for the background occlusion by the moving foreground object. This masking operation requires the extraction of the foreground and is equivalent to background removal discussed in the next subsection. As a heuristic approach, the term
in (3) can be simply neglected [Bri1]. The resulting observation model is obtained by setting in (3).
Following standard methods described in [Kay], the ML criterion can be applied to obtain an expression for the log-likelihood function to be maximized. A likelihood function of the observed data in (3), with , can be defined on the basis of a window of observed frames as:
p(X[k,0]⋯X[k,N−1];v) (6)
where
is the standard deviation of the additive Gaussian noise elements. A log-likelihood function can be derived from (
6) as:
ln(p(X[k,0]⋯X[k,N−1];v)) (7) =−M1M2N2ln(2πσ2)(b% )−12σ2M1−1∑k1=0M2−1∑k2=0N−1∑n=0∣∣X[k,n]−S[k]e−j2πukTvn−B[k]∣∣2(a).
The operating principle of the ML approach is to find the value of that maximizes the log-likelihood function in (7). Equivalently, the term (a) can be minimized by observing that the term (b) and the multiplicative coefficient are irrelevant and constant, being independent from . The term (a) in (7) can be written explicitly as:
M1−1∑k1=0M2−1∑k2=0N−1∑n=0 (8) −2Re{X[k,n]B∗[k]}+2Re{S∗[k]ej2πukTvnB[k]}}
where indicates the real part. The terms , and can be neglected because irrelevant, since independent of . Hence, minimizing (8) corresponds to maximizing the following quantity, where three terms are highlighted:
(9)
Using the linearity of and sum operators, can be factored out from (T1) and (T3) to obtain:
M1−1∑k1=0M2−1∑k2=0{Re{N−1∑n=0(X[k,n]−B[k])S∗[k]ej2πukTvn+X[k,n]B∗[k]}}. (10)
The continuous-frequency FT of the temporal sequence can be now defined as:
Y[k,f]=N−1∑n=0(X[k,n]−B[k])S∗[k]e−j2πfTsn. (11)
Setting , equation (10) can be expressed as:
M1−1∑k1=0M2−1∑k2=0Re{Y[k,−ukTvTs]+N¯¯¯¯¯X[k]B∗[k]}, (12)
where is the temporal mean of the sequence and is defined as:
¯¯¯¯¯X[k]=1NN−1∑n=0X[k,n]. (13)
Finally, an expression for the estimator of the speed vector with included background is:
^v= argmaxvM1−1∑k1=0M2−1∑k2=0Re{Y[k,−ukTvTs]+N¯¯¯¯¯X[k]B∗[k]}. (14)
### Iii-B Speed Estimation with Omitted Background
Following the afore described procedure, it is possible to derive an estimator of the speed vector in the case of omitted background according to model (5), by setting .
In this case, (11) becomes
Y[k,f]=N−1∑n=0X[k,n]S∗[k]e−j2πfTsn (15)
and (14) simplifies as:
^v=argmaxvM1−1∑k1=0M2−1∑k2=0Re{Y[k,−ukTvTs]}. (16)
## Iv Numerical Results
In this section, the performance of the presented ML speed estimation algorithms is discussed. In particular, the results are presented in terms of Root Mean Square Error (RMSE) between and the correct speed of the framed object for two different categories of videos. A set of synthetic and software-generated videos is considered to preliminarily test the effectiveness of the derived methods in a controlled environment, then a number of real-world videos, specifically recorded for this purpose, is analysed. Comparisons with the well-known block-matching method [VideoProcessing, Ch.4] are also presented.
### Iv-a Software-Generated Videos
As a first experiment, we apply the proposed speed estimation methods to synthetic videos obtained by inserting real pictures of moving objects in an artificial environment (i.e. a static background) and by adding white Gaussian noise to simulate the behaviour of a real image acquisition system. A sample frame of a grayscale video considered for this purpose is shown in Figure 1(a), where the moving object is the (highlighted) bird, placed in the upper part.222The bird has been highlighted to ease visualization in Figure 1(a) only, not in the processed video. The main advantage of synthetic videos relies on the possibility to manually or automatically set a number of parameters, including the speed components, thus allowing a direct assessment of the estimation performance.
In particular, we consider a set of videos in order to test as many values of speed components, measured in pixel/frame. The duration of each video is about s with a frame rate Hz and a frame size of pixels. The total number of frames is , of which the first are background frames, exploited by the GMM algorithm, and in (11) and (15
). The initial variance of the Gaussian distributions in the mixture (
in this case) is set to [stauffer], as a good compromise to account for the variations due to the additive white Gaussian noise (AWGN) and the motion of the foreground object.
Using (14) and (16) the speed components are estimated in the cases of included and omitted background. Results are shown in Figure 1(b), along with a comparison with the well-known block-matching approach, for which a block size of pixels is chosen by trial and error to take into account the foreground size. For each video, the RMSE normalized to the correct values of the speed components for different noise realizations is measured. The obtained normalized RMSE values are then averaged with respect to the various videos and the result is plotted against increasing values of the noise variance for all the assessed algorithms.
Figure 1(b) shows the robustness of the proposed methods against noise variations for both the considered cases. The average RMSE is zero for low values of noise variance because the correct and estimated speed components are quantized to integer values. In particular, the case of omitted background gives lower errors up to a noise variance of about , while the case of included background is more efficient for higher values of noise variance. On the other hand, the high noise sensitivity of the block-matching criterion is confirmed by observing the rapidly increasing trend of its RMSE curve.
### Iv-B Real-World Videos
As realistic examples, real videos of moving cars are considered. In particular we apply the proposed algorithms to a set of videos with various duration that ranges between about and s and with various numbers of background frames. The videos are recorded with a frame rate Hz and have a frame size of pixels. The number of frames in (11) and (15) ranges between and . Furthermore, the initial variance of the Gaussian distributions in the mixture model is set to to track scene modifications due to the lightning changes and wind effects in addition to the the AWGN and motion of the foreground object, as in the previous scenario. The car reference speeds are manually measured and can be considered approximately constant over the entire video duration. A sample frame of a video in this category is shown in Figure 1(c).
Following the previous approach, our speed estimation methods are applied with included and omitted background, and compared with the performance of the block-matching algorithm, for which the block size is again set to pixels.
As in the previous example, for each video the RMSE normalized to the correct speed components is measured for different noise realizations and the obtained values are again averaged with respect to the various videos. In Figure 1(d) the results are plotted against increasing values of the noise variance for the three algorithms. For very low noise, e.g. for , the performance of the block-matching approach provides a low error, nevertheless its high noise sensitivity is again confirmed by the rapidly increasing trend of its RMSE curve. This demonstrates the robustness of the proposed ML estimation techniques, which exhibit lower values of RMSE even at much higher values of noise variance.
In this example, the average RMSE with included background settles about , but for small values of noise variance, because speed estimation consistently fails in some of the considered videos. However, the performance of the algorithm with omitted background is significantly better. We also note that the RMSE values for low noise variance are not exactly zero, unlike the previous case. This is due to the used quantization of the speed component estimates to integer values, unlike the non integer values of the correct speed components. Furthermore, the nature of real videos, including artifacts such as illumination changes and wind effects, may affect the operation of background removal and foreground detection performed by means of the GMM algorithm.
## V Conclusion
In this paper, we presented a novel approach for speed estimation in video signals. We derived an observation model for included and omitted background and we applied the ML criterion to estimate the speed components of a foreground object shifting with a nearly constant speed. By considering synthetic and real videos as examples of application, we demonstrated the effectiveness of the proposed algorithms by comparison with the well-known block-matching algorithm. | {"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": 0.8278645873069763, "perplexity": 684.4246244919352}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-49/segments/1637964363301.3/warc/CC-MAIN-20211206133552-20211206163552-00417.warc.gz"} |
http://mathhelpforum.com/advanced-statistics/11945-binomial-probability.html | Math Help - binomial probability
1. binomial probability
HI ,
I would like to have some help with this problems:
A problem in a metal casting process of a part, called severe flashing results in scrapping the part that is being cast. A particular casting operation has in the past scrapped 10% of its parts due to severe flashing. Assume that scrapping a part or not scrapping a part in future operations is a binomial experiment with p=0.10.
1-Find the probability that no part is scrapped in the next 10 experiments.
My solution:
b(0;10,0.1)
b(x;n;p) is the binomial distribution.
where x =number os success
n=number of trial
p=probability of success
2- find the probability that at most two parts will be scrapped in the next operations.
My solution: B(2;10,0.10)
B(x;n,p) is the cumulative probabilities
3- Find the expected number of scrapped parts in the next 25 operations.
NO solution
4- If the scrapping cost in n operation is the square of the number of parts scrapped, find the expected scrapping cost in the next 25 operations.
No solutions
5- Find the probability that the third part scrapped is the 20th part cast.
No solution
To compute the expected value was supposed to have some some kinds of frequencies. That confused me a little since I don't see any.
Thank you
B
HI ,
I would like to have some help with this problems:
A problem in a metal casting process of a part, called severe flashing results in scrapping the part that is being cast. A particular casting operation has in the past scrapped 10% of its parts due to severe flashing. Assume that scrapping a part or not scrapping a part in future operations is a binomial experiment with p=0.10.
1-Find the probability that no part is scrapped in the next 10 experiments.
My solution:
b(0;10,0.1)
b(x;n;p) is the binomial distribution.
where x =number os success
n=number of trial
p=probability of success
b(0;10,0.1) = 10!/(0! (10-0)!) (0.1)^0 (0.9)^10 = 0.9^10 ~= 0.349
2- find the probability that at most two parts will be scrapped in the next operations.
My solution: B(2;10,0.10)
B(x;n,p) is the cumulative probabilities
B(2;10,0.1) = b(0;10,0.1) + b(1;10,0.1) + b(2;10,0.1)
................= 0.9^10 + 10 0.1 0.9^9 + 45 0.1^2 0.9^8 ~= 0.930
3- Find the expected number of scrapped parts in the next 25 operations.
NO solution
4- If the scrapping cost in n operation is the square of the number of parts scrapped, find the expected scrapping cost in the next 25 operations.
No solutions
5- Find the probability that the third part scrapped is the 20th part cast.
No solution
To compute the expected value was supposed to have some some kinds of frequencies. That confused me a little since I don't see any.
Thank you
B
HI ,
I would like to have some help with this problems:
A problem in a metal casting process of a part, called severe flashing results in scrapping the part that is being cast. A particular casting operation has in the past scrapped 10% of its parts due to severe flashing. Assume that scrapping a part or not scrapping a part in future operations is a binomial experiment with p=0.10.
1-Find the probability that no part is scrapped in the next 10 experiments.
My solution:
b(0;10,0.1)
b(x;n;p) is the binomial distribution.
where x =number os success
n=number of trial
p=probability of success
2- find the probability that at most two parts will be scrapped in the next operations.
My solution: B(2;10,0.10)
B(x;n,p) is the cumulative probabilities
3- Find the expected number of scrapped parts in the next 25 operations.
NO solution
The expected number of scrapped parts in 25 trials is:
E(n) = Sum n*pr(n) n=0, .., 25 = Sum n*b(n;25,0.1) n=0,..,25
Now we can do this sum, or we can use the fact that we know the expected
number of failures in N Bernoulli trials with a probability of failur on a single
trial of p is N*p, so here it is 25*0.1 = 2.5.
4- If the scrapping cost in n operation is the square of the number of parts scrapped, find the expected scrapping cost in the next 25 operations.
No solutions
The expected cost of scrapped parts in 25 trials is:
E(n) = Sum (n^2)*pr(n) n=0, .., 25 = Sum (n^2)*b(n;25,0.1) n=0,..,25
but I don't know a short cut to calculate this.
5- Find the probability that the third part scrapped is the 20th part cast.
No solution
To compute the expected value was supposed to have some some kinds of frequencies. That confused me a little since I don't see any.
Thank you
B
HI ,
I would like to have some help with this problems:
A problem in a metal casting process of a part, called severe flashing results in scrapping the part that is being cast. A particular casting operation has in the past scrapped 10% of its parts due to severe flashing. Assume that scrapping a part or not scrapping a part in future operations is a binomial experiment with p=0.10.
5- Find the probability that the third part scrapped is the 20th part cast.
No solution
This probabability is equal to the probability of exactly two scrapped in the
first 19 times the probability that the 20th is scrapped:
P = b(2;19,0.1)*0.1 = [19*18/2 * 0.1^2*0.9^17 ]* 0.1 ~= 0.0285
RonL
5. Originally Posted by CaptainBlack
The expected cost of scrapped parts in 25 trials is:
E(n) = Sum (n^2)*pr(n) n=0, .., 25 = Sum (n^2)*b(n;25,0.1) n=0,..,25
but I don't know a short cut to calculate this.
On second thoughts I do know a short cut for this:
Var(n) = E((n - m)^2) = E(n^2) - m^2 =
but the variance of the number of success of a Binomial random variable in N trials is
Var(n)=N*p*(1-p)
and the mean m=Np, so:
N*p*(1-p) = E(n^2) - N^2p^2,
or:
E(n^2) = N*p*(1-p) + N^2*p^2 = 8.5
RonL | {"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": 0.8454566597938538, "perplexity": 2188.7162147412805}, "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-2015-06/segments/1422122102237.39/warc/CC-MAIN-20150124175502-00252-ip-10-180-212-252.ec2.internal.warc.gz"} |
http://sepwww.stanford.edu/sep/prof/waves/fgdp8/paper_html/node2.html | Next: THE LAYER MATRIX Up: Waves in layered media Previous: Waves in layered media
# REFLECTION AND TRANSMISSION COEFFICIENTS
Consider two halfspaces (deep ocean on top of earth, for example). If a wave of unit amplitude is incident onto the boundary, there is a transmitted wave of amplitude t and a reflected wave of amplitude c as depicted in Figure 1.
8-1 Figure 1 Waves incident, reflected c, and transmitted t at an interface.
A very simple relationship exists between t and c. The wave amplitudes have a physical meaning of something like pressure, material displacement, traction, or tangential electric or magnetic fields. These physical variables must be the same value on either side of the boundary. This means the transmitted wave must equal the sum of the incident plus reflected waves.
(1) (2)
The reflection coefficient c may be positive or negative so the transmission coefficient t may be greater than unity. It may seem surprising that t can be greater than unity. This does not violate any physical laws. At the seashore we see waves approaching the shore and they get larger as they arrive. Energy is not determined by wave height alone. Energy is equal to the squared wave amplitude multiplied by a proportionality factor Y depending upon the medium in which the wave is measured. If we denote the factor of the top medium by Y1 and the bottom by Y2, then the statement that the energy before incidence equals the energy after incidence is
(3)
solving for c leads us to
(4)
In acoustics the up- and downgoing wave variables may be normalized to either pressure or velocity. When they measure velocity, the scale factor multiplying velocity squared is called the impedance I. When they measure pressure, the scale factor is called the admittance Y.
The wave c' which reflects when energy is incident from the other side is obtained from (4) if Y1 and Y2 are interchanged. Thus
(5)
A perfectly reflecting interface is one which does not allow energy through. This comes about not only when t = 0 or c = -1, but also when t = 2 or c = +1. To see this, note that on the left in Figure 1
(6)
Equation (6) says that 100 percent of the incident energy is transmitted when Y1 = Y2, but the percentage of transmission is very small when Y1 and Y2 are very different.
Ordinarily there are two kinds of variables used to describe waves, and both of these can be continuous at a material discontinuity. One is a scalar like pressure, tension, voltage, potential, stress, or temperature. The other is a vector which we use the vertical component. Examples of the latter are velocity, stretch, electric current, dislacement, and heat flow. Occasionally a wave variable is a tensor. When a boundary condition is the vanishing of one of the motion components, then the boundary is often said to be rigid. When it is the pressure or potential which vanishes, then the boundary is often said to be free. Rigid and free boundaries reflect waves with unit magnitude reflection coefficients.
A goal here is to establish fundamental mathematical properties of waves in layers while minimizing specialization to any particular physical type of waves. Each physical problem has its own differential equations. These equations are Fourier transformed over x and y leading to coupled ordinary differential equations in depth z. This analytical process is explained in more detail in FGDP. The next step is an eigenvector analysis that relates the physical variables to our abstract up- and down-going waves U and D. In order to better understand boundary conditions we will examine one concrete example, acoustic waves.
In acoustics we have pressure P and vertical component of parcel velocity W (not to be confused with wave velocity v). The acoustic pressure P is the sum of U and D. Vertical velocity W obviously changes sign when the z axis changes sign (interchange of up and down) and that accounts for the minus sign in the definition of W. (The eigenvector analysis provides us with the scaling factor Y.)
(7) (8)
These definitions are easily inverted.
(9) (10)
For sound waves in the ocean, the sea surface is nearly a perfect reflector because of the great contrast between air and water. If this interface is idealized to a perfect reflector, then it is a free surface. Since the pressure vanishes on a free surface, we have D = -U at the surface so the reflection coefficient is -1 as shown in Figure 2.
8-2 Figure 2 A waveform R(Z) reflecting at the surface of the sea. Pressure equal to U + D vanishes at the surface. The vertical velocity of the surface is proportional to D - U. Theoretically, waves are observed by measuring W at the surface.
In principle we should measure velocity W at the water surface. In practice, we generally measure pressure P a few meters below the free surface.
The pressure normally vanishes at the sea surface, but if we wish to initiate an impulsive disturbance, the pressure may momentarily take on some other value, say 1. This 1 denotes a constant function of frequency which is an impulsive function of time at t=0. Ensuing waves are depicted in Figure 3 where the upcoming wave -R(Z) is a consequence of both the downgoing 1 and the downgoing +R(Z). The vertical component of velocity W of the sea surface due to the source and to the resulting acoustic wave is D - U = 1 + 2R(Z).
8-3 Figure 3 An initial downgoing disturbance 1 results in a later upgoing reflected wave -R(Z) which reflects back down as R(Z). The pressure at the surface is D + U = 1 + R - R = 1.
## EXERCISES:
1. In a certain application continuity is expressed by saying that D-U is the same on either side of the interface. This implies that t = 1 - c. Derive an equation like (4) for the reflection coefficient in terms of the admittance Y.
2. What are reflection and transmission coefficients in terms of the impedance I? (Clear fractions from your result.)
3. From the principle of energy conservation we showed that c' = -c. It may also be deduced from time reversal. To do this, copy Figure 1 with arrows reversed. Scale and linearly superpose various figures in an attempt to create a situation where a figure like the right-hand side of Figure 1 has -c' for the reflected wave. (HINT: Draw arrows at normal incidence.)
Next: THE LAYER MATRIX Up: Waves in layered media Previous: Waves in layered media
Stanford Exploration Project
3/1/2001 | {"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": 0.9332273006439209, "perplexity": 580.641338857581}, "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-2017-30/segments/1500549423764.11/warc/CC-MAIN-20170721082219-20170721102219-00150.warc.gz"} |
http://mathhelpforum.com/calculus/60285-nondifferentiable-function.html | Math Help - Nondifferentiable function
1. Nondifferentiable function
For the function f(x) = (x-6)^7/9 find the x-value at which the derivative does not exist.
2. Originally Posted by BOOTYBOY
For the function f(x) = (x-6)^7/9 find the x-value at which the derivative does not exist.
$f'(x)=\frac{7}{9(x-6)^{\frac{2}{9}}}$. Now where is that undefined? | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 1, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9647862911224365, "perplexity": 1803.1183629855968}, "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-2016-22/segments/1464049276964.14/warc/CC-MAIN-20160524002116-00105-ip-10-185-217-139.ec2.internal.warc.gz"} |
https://socratic.org/questions/58b1f427b72cff77f33ebb7d#399434 | Physics
Topics
# Question #ebb7d
Mar 31, 2017
$4.9 \text{m}$
#### Explanation:
Assuming that the ball is initially at rest and that air resistance can be neglected, the distance covered by the ball in $1 \text{s}$ is
$s = \frac{1}{2} a {t}^{2}$
$= \frac{1}{2} {\left(9.8 \text{ms"^{-2}) (1"s}\right)}^{2}$
$= 4.9 \text{m}$
##### Impact of this question
573 views around the world
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https://www.physicsforums.com/threads/hamilton-and-lagrange-functions.362354/ | # Homework Help: Hamilton and Lagrange functions
1. Dec 10, 2009
### Shafikae
The hamilton function of a particle in two dimensions is given by
H = (p$$\stackrel{2}{x}$$)/2m + (p$$\stackrel{2}{y}$$)/2m + apxpy + U(x,y)
Obtain the Hamiltonian equations of motion. Find the corresponding Lagrange function and Lagrange equations.
Would it be px = dH/dpy (of course it would be partial)
and py = - dH/dpx ?
and how do we take into account the potential?
2. Dec 11, 2009
### gabbagabbahey
No, Hamilton's equations of motion are $$\dot{p_i}=-\frac{\partial H}{\partial q_i}$$ qnd $$\dot{q_i}=\frac{\partial H}{\partial p_i}$$, where $q_i$ are the generalized coordinates and $p_i$ are there corresponding momenta.
In this case, your generalized coordinates are $x$ and $y$ (i.e. $q_1=x$ and $q_2=y$) )and there corresponding momenta are $p_x$ and $p_y$ (i.e. $p_1=p_x$ and $p_2=p_y$).... | {"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": 0.9691802263259888, "perplexity": 1175.8109297179321}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-26/segments/1529267860684.14/warc/CC-MAIN-20180618164208-20180618184208-00622.warc.gz"} |
https://www.physicsforums.com/threads/excuse-me-while-i-drill-a-hole-my-head.1351/ | # Excuse Me While I Drill A Hole My Head
1. Apr 19, 2003
### eNtRopY
Maybe I'm a little slow, but the true meaning of pi just occured to me.
tan(pi/4) = 1
arctan(x) = [sum]n=0[oo] (-1)n * [ x2n+1 / (2n + 1) ]
pi = 4 * arctan(1) = 4 * [sum]n=0[oo] (-1)n * [ 1 / (2n + 1) ]
= 4 * ( 1 - 1/3 + 1/5 - 1/7 + 1/9 - 1/11 + ... )
= 4 * ( 2/3 + 2/35 + 2/99 + ... )
= 4 * [(1 * 2/3) + (4 * 1/70) + (8 * 1/396) + ...]
Remember the equation for the area of a circle.
A = pi * R2
= (2*R)2 * [(1 * 2/3) + (4 * 1/70) + (8 * 1/396) + ...]
= D2 * [(1 * 2/3) + (4 * 1/70) + (8 * 1/396) + ...]
Now, draw a circle with diameter D. In that circle, place a square with area (2/3)*D2. Notice that there is still some empty space left on the sides square (where the square and the circle do not both exist). Place four squares with area 1/70*D2 in each of these four empty spaces. Continue to fill up the unoccupied space of the circle with progressively smaller squares ad infinitum.
Pi is simply an infinite series sum of scaling factors needed for filling a circle with squares in order to perform a numerical integration method.
eNtRopY
2. Apr 20, 2003
### ranyart
Pi
Its a bit more complicated, or simple really., glad to see someone else has found the Antiqity Keys to the Universe!
There is an obvious scaling factor, and when you invert the 'outside'square into the 'inside' circle, then the dimensional scale is maintained!
http://groups.msn.com/Youcanseehomefromhere/consciouswaves.msnw
The movement, or transition from a Square to a Circle in '2-dimensions
', involves the inversion of area external to internal.
Below is representative of such an invertion:
http://groups.msn.com/Youcanseehomefromhere/consciouswaves.msnw?action=ShowPhoto&PhotoID=54
And the hidden relationships of interconnective Pi and geometric shapes can be clearly seen, as the shape expands, it expands to precise geometric formulas.
Click on the image for a larger pic.
http://groups.msn.com/Youcanseehomefromhere/consciouswaves.msnw?Page=2
3. Apr 21, 2003
### climbhi
Well, yeah that's a neat way to get pi, but excuse me if I am wrong isnt the true meaning of pi the ratio of a circles circumference to it's diameter? I think the most simple and elegant way to find pi is to just inscribe a polygon inside a unit circle, we know how to find the length of the sum of the sides of the polygon so pi is the limit as the number of sides goes to infinity. This is what makes the most sense to me. But there are many various other ways to get it. The two others that I like best are Wallis forumula and a monte carlo method where you have a unit circle inscribed inside a unit box and then you start randomly dropping points inside the box and can get pi as a ratio of the number of points inside the circle versus the total number of points (this ones fun to see a computer program do, the applet I saw do it also had a timeline above it which showed what the current value of pi was and how it changed in time as you added dots, it was neat to see it converge very quickly and accurately to pi!) | {"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": 0.829606294631958, "perplexity": 562.9765298036439}, "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-2016-36/segments/1471982290634.12/warc/CC-MAIN-20160823195810-00250-ip-10-153-172-175.ec2.internal.warc.gz"} |
https://www.yaclass.in/p/mathematics-cbse/class-9/constructions-8412/re-f808212c-95b2-4930-a69c-b51532675be4 | ### Theory:
Construct a triangle $$ABC$$ in which $$\angle B = 60^\circ$$, $$\angle C = 45^\circ$$ and $$AB + BC + CA = 13 \ cm$$.
Step 1: Draw a line segment $$PQ = 13 \ cm$$.
Step 2: Construct $$\angle QPX = \angle B = 60^\circ$$, and $$\angle PQY = \angle C = 45^\circ$$.
Step 3: Now, draw bisectors of $$\angle QPX$$ and $$\angle PQY$$. Let these bisectors intersect at a point $$A$$.
Step 4: Construct perpendicular bisectors $$RS$$ of line segment $$AP$$ and $$TU$$ of line segment $$AQ$$. The points where $$RS$$ intersects $$PQ$$ as $$B$$ and $$TS$$ intersects $$AQ$$ as $$C$$.
Step 5: Join $$AB$$ and $$AC$$. Thus, $$ABC$$ is the required triangle.
JUSTIFICATION:
The point at which $$RS$$ cuts $$AP$$ as $$M$$ and $$TU$$ cuts $$AQ$$ as $$N$$.
We need to prove $$AB + BC + CA = PQ$$, $$\angle QPX = \angle B$$ and $$\angle PQY = \angle C$$.
In triangles $$BMP$$ and $$BMA$$:
$$PM = AM$$ [Since $$RS$$ is the bisector of $$AP$$]
$$\angle BMP = \angle BMA$$ [Since $$RS$$ is perpendicular to $$AP$$]
$$MB = MB$$ [Common side]
Therefore, $$\Delta BMP \cong \Delta BMA$$ [by $$SAS$$ congruence rule]
$$\Rightarrow PB = AB$$ [by CPCT] - - - - - (I)
Similarly, for triangles $$CNA$$ and $$CNQ$$, $$CQ = CA$$ - - - - - (II)
$$AB + CA = PB + CQ$$
Add $$BC$$ on both sides.
$$AB + BC + CA = PB + BC + CQ$$
$$AB + BC + CA = PQ$$
In triangle $$ABP$$, $$PB = AB$$
Angles opposite to equal sides are equal.
$$\Rightarrow \angle APB = \angle BAP$$
The exterior angle is equal to the sum of opposite interior angles.
$$\Rightarrow \angle ABC = \angle APB + \angle BAP$$
$$\Rightarrow \angle ABC = \angle APB + \angle APB$$
$$\Rightarrow \angle ABC = 2 \angle APB$$
$$\Rightarrow \angle ABC = \angle XPB = \angle XPQ$$ [Since $$AP$$ is an angle bisector of $$\angle QPX$$]
Similarly, $$\angle ACB = \angle PQY$$.
Hence, we justified. | {"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": 0.8725372552871704, "perplexity": 509.9958935375916}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-21/segments/1620243989756.81/warc/CC-MAIN-20210518063944-20210518093944-00181.warc.gz"} |
https://nyuscholars.nyu.edu/en/publications/cutting-a-graph-into-two-dissimilar-halves-2 | # Cutting a graph into two dissimilar halves
Paul Erdós, Mark Goldberg, János Pach, Joel Spencer
Research output: Contribution to journalArticlepeer-review
## Abstract
Given a graph G and a subset S of the vertex set of G, the discrepancy of S is defined as the difference between the actual and expected numbers of the edges in the subgraph induced on S. We show that for every graph with n vertices and e edges, n < e < n(n − 1)/4, there is an n/2‐element subset with the discrepancy of the order of magnitude of \documentclass{article}\pagestyle{empty}\begin{document}$\sqrt {ne}$\end{document} For graphs with fewer than n edges, we calculate the asymptotics for the maximum guaranteed discrepancy of an n/2‐element subset. We also introduce a new notion called “bipartite discrepancy” and discuss related results and open problems.
Original language English (US) 121-131 11 Journal of Graph Theory 12 1 https://doi.org/10.1002/jgt.3190120113 Published - 1988
## ASJC Scopus subject areas
• Geometry and Topology
## Fingerprint
Dive into the research topics of 'Cutting a graph into two dissimilar halves'. Together they form a unique fingerprint. | {"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": 0.811972439289093, "perplexity": 1255.2855526325984}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-39/segments/1631780056892.13/warc/CC-MAIN-20210919160038-20210919190038-00710.warc.gz"} |
http://physics.stackexchange.com/questions/13325/what-is-the-minimum-amount-of-fissile-mass-required-to-acheive-criticality/64975 | # What is the minimum amount of fissile mass required to acheive criticality?
Wikipedia gives the following definition for critical mass.
A critical mass is the smallest amount of fissile material needed for a sustained nuclear chain reaction.
No mention is made of a neutron moderator in this definition. If critical mass is defined like this, then that should allow infinite moderator material to be used if doing so would lead to the minimum amount of fissile material.
My question: What is the universally minimum amount of fissile mass needed to achieve a critical configuration (allowing a neutron moderator and no restriction on the moderator) for something like Uranium-235 and what would its configuration be?
Uranium-235 has a bare sphere critical mass (BSCM) of around $52 kg$, which is the mass of a critical sphere containing only the fissile material where no neutrons are reflected back into the sphere after leaving the surface. Wikipedia has a good illustration of a bare sphere versus a sphere surrounded by a moderator.
Image: First item is a BSCM illustration, 2nd item is a critical sphere that uses less fissile mass than the BSCM due to the introduction of a neutron moderator blanket
This example illustrates how a critical configuration can be made with U-235 that uses less than the BSCM, or $52 kg$. It is possible, although unlikely, that the above sphere surrounded by a moderator could be the configuration that answers my question. Alternatives would include a homogenous mixture of the moderator and the fissile material, a mixture of the two that varies radially, or something I have not thought of.
The hard part is showing that the particular fissile material/moderator mix can not be improved by any small change. It would also be necessary to show that it can not be improved by adding more than 1 type of moderator (I suspect this could be done with a reasonably short argument).
Technical mumbo junbo
Everything I write here is just a suggestion, answer however you want to or can
Both the moderator and fissile material have a certain density. I would make the simplifying assumption that the density of a homogenous mixture of the fuel and moderator would be a linear combination of their specific volumes, with the understanding that this is not true in real life. Changing the mix changes the macroscopic cross section of both. It might help to note that the BSCM almost exactly determines the macroscopic cross section of the pure fissile material, one over the macroscopic cross section is the path length, which is on the same order of magnitude of the radius. Generally fast and thermal scattering from U-235 is small compared to the fission cross section.
I would use 2 neutron energy groups, and assume either diffusion or immediate absorption after scattering to thermal energies. In fact, I would just assume immediate absorption. At that point, all you would need is the scatter to absorption ratio and microscopic cross section at both thermal and fast energies for the moderator and fissile material, in addition to the densities of course. Even then some type of calculus of variations may be necessary (if you're looking for a radially varying mixture) in addition to the fact that it would be hard to describe the fast group without a fairly complicated form of neutron transport. The BSCM is comparatively simple since it has a constant number density of the fissile material and even that is actually pretty complicated.
-
I don't know enough about critical mass calculations to answer your question definitively, but the minimum critical mass is less than 780 g of U-235 (this example is a solution-type core). – mmc Aug 9 '11 at 17:18
@mmc Your comment was exactly the type of answer I was looking for, thanks! I understand definitively answering it is a very BIG problem, but your answer knocks it down a whole lot. – Alan Rominger Aug 9 '11 at 17:23
Somewhat off topic, but I've seen speculation that if isotopes at the island of stability en.wikipedia.org/wiki/Island_of_stability had half-lives long enough to allow macroscopic samples to exist, then you could make a nuclear bomb the size of a pencil eraser. Luckily for world peace, these isotopes are both incredibly hard to produce and unlikely to have such long half-lives. – Ben Crowell Aug 9 '11 at 17:52
@mmc could you make your comment an answer so I can select it? Thanks. – Alan Rominger Aug 9 '11 at 18:53
@Zassounotsukushi Done. – mmc Aug 9 '11 at 22:27
I'm not an expert in critical mass calculation, but the minimum critical mass for U-235 must be less than (or equal to) 780 g as shown in "Mass Estimates of Very Small Reactor Cores Fueled by Uranium-235, U-233 and Cm-245" for a solution type core.
-
to add to this answer, the naturally occurring uranium has very small percentage of U235, hence you need at least 100kg of nat. Uranium to achieve critical mass. – Vineet Menon Jan 3 '12 at 4:53
To further add to your confusion, you might want to distinguish between "prompt critical" and "delayed" critical. Reactors rely on delayed neutrons, which is why they must be cooled even after they drop below critical. Bombs use prompt critical because they operate much faster. One rule of thumb is that an external relfector drops the spherical critical mass by about a factor of three. But obviously the lowest mass is achieved by homogeneous moderated systems (eg. U-235 hydride).
-
While the factor of 3 is an attractive rule of thumb, I should add that it depends on the absorption and scattering cross section ratios which depends on the moderator material. – Alan Rominger Jan 3 '12 at 6:18
You are correct in assuming that the presence of a moderator (or reflector) can change the mass required for a critical configuration. But there are too many other variables to consider to provide a concise answer to your question. All I can offer is a qualitative explanation:
The smallest critical mass would be for a high density spherical core of fissile fuel with a low neutron absorption cross section and high fission cross section surrounded by an infinite (or very thick) high density moderator with a high neutron scatter cross section and low neutron absorption cross section.
The critical mass will depend on the geometry and material properties of the fissile fuel and moderator. A spherical mass of pure fissile fuel surrounded by an infinite moderator will generally require the lowest critical mass. This is for 2 main reasons: 1) The spherical configuration is the most efficient when it comes to the use of neutrons in a simple homogenious reactor (spherical configurations give the highest 'non-leakage factor', which means that fewer neutrons escape the system without causing a fission). 2) The moderator surrounding the spherical fuel configuration will scatter some of the neutrons escaping from the fissioning core back into the core where they can cause more fissions, effectively acting a a reflector to improve the usage of neutrons in the system.
1. Fissile fuel composition (pure U-235, 90%/10% U-235/U-238, etc.)
2. Fuel configuration (geometry, density)
3. Moderator/reflector material (water, U-238, etc.)
4. Moderator/reflector configuration(geometry, density)
If you then had the mass of the fuel, you would be able to compute the 'neutron multiplication factor' (most often represented as 'k') for the system. Methods for computing k depend on the system, application, and assumptions made and can be quite complex, so I won't list a particular equation here. For a 'critical' system, k = 1 (sub-critical is k < 1, super-critical is k > 1). To compute the 'critical mass' you hold k = 1 and can then solve for mass.
-
I'm just a layman, but if a nuclear reactor depends on minimum sustained criticallity, then isn't it possible to "burn-out" nuclear waste by combining just enough of it to achieve "near critical mass" ? If the "unlined ditch" accident at Hanniford was a real event, (and not just a Stephen King invention), then it seems that this process would be possible in a very controlled environment. It would result in the half-life of that nuclear material being made much longer, and the nuclear material's radioactivity being reduced very quickly. Soon there would be no more nuclear waste, because in order to continue to achieve even the most minimal sustained criticality, all the nuclear waste of that isotope would have to be combined to achieve it.
-
You might theoretically be able to get rid of all the fissile isotopes, but there are an enormous number of non-fissile highly radioactive fission reaction products. How do you plan to get rid of all the Cesium 137 and Strontium 90, for example? – Peter Shor Jan 2 '12 at 23:17
Hi Mark - I've merged your new account into your old one. You should now be able to edit this answer. Please do that if you have more to add, instead of posting separate answers. – David Z Jan 2 '12 at 23:27
## protected by David Z♦Aug 20 '13 at 5:03
Thank you for your interest in this question. Because it has attracted low-quality answers, posting an answer now requires 10 reputation on this 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": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8838361501693726, "perplexity": 645.3037570856623}, "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/1416931006855.76/warc/CC-MAIN-20141125155646-00102-ip-10-235-23-156.ec2.internal.warc.gz"} |
http://www.perimeterinstitute.ca/videos/dark-matter-and-dark-energy-fact-or-fiction | # Dark matter and dark energy - Fact or fiction?
Playing this video requires the latest flash player from Adobe. | {"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": 0.891218364238739, "perplexity": 4177.258681434865}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-13/segments/1521257649627.3/warc/CC-MAIN-20180324015136-20180324035136-00222.warc.gz"} |
https://linearalgebras.com/solution-abstract-algebra-exercise-3-2-15.html | If you find any mistakes, please make a comment! Thank you.
## Compute the stabilizer of an element under a given action of Sym(n)
Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 3.2 Exercise 3.2.15
Fix $i \in \{ 1, \ldots, n \} = A$, and let $S_n$ act on $A$ in the natural way. Show that $\mathsf{stab}_{S_n}(i) \cong S_{n-1}$.
Solution: It is clear that $$\mathsf{stab}_{S_n}(i) = \{ \sigma \in S_n \ |\ \sigma(i) = i \}.$$ Removing from each function in this set the pair $(i,i)$, we have the full symmetric group on $\{ 1, 2, \ldots, i-1,i+1,\ldots,n\}$. We saw in a previous exercise that the isomorphism type of $S_A$ depends only on the cardinality of $A$, so that the conclusion holds. | {"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": 0.9745757579803467, "perplexity": 89.91209044982406}, "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/1642320304261.85/warc/CC-MAIN-20220123111431-20220123141431-00019.warc.gz"} |
https://www.zora.uzh.ch/id/eprint/75836/ | # Search for new physics in events with same-sign dileptons and b jets in pp collisions at s√=8 TeV
CMS Collaboration; Amsler, C; Chiochia, V; Kilminster, B; et al (2013). Search for new physics in events with same-sign dileptons and b jets in pp collisions at s√=8 TeV. Journal of High Energy Physics, 2013:37.
## Abstract
A search for new physics is performed using events with isolated same-sign leptons and at least two bottom-quark jets in the final state. Results are based on a sample of proton-proton collisions collected at a center-of-mass energy of 8 TeV with the CMS detector and corresponding to an integrated luminosity of 10.5 fb−1. No excess above the standard model background is observed. Upper limits are set on the number of events from non-standard-model sources and are used to constrain a number of new physics models. Information on acceptance and efficiencies is also provided so that the results can be used to confront an even broader class of new physics models.
## Abstract
A search for new physics is performed using events with isolated same-sign leptons and at least two bottom-quark jets in the final state. Results are based on a sample of proton-proton collisions collected at a center-of-mass energy of 8 TeV with the CMS detector and corresponding to an integrated luminosity of 10.5 fb−1. No excess above the standard model background is observed. Upper limits are set on the number of events from non-standard-model sources and are used to constrain a number of new physics models. Information on acceptance and efficiencies is also provided so that the results can be used to confront an even broader class of new physics models.
## Statistics
### Citations
Dimensions.ai Metrics
8 citations in Web of Science®
36 citations in Scopus® | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9406636357307434, "perplexity": 1157.0855038932136}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-43/segments/1570986684226.55/warc/CC-MAIN-20191018154409-20191018181909-00401.warc.gz"} |
http://mathhelpforum.com/statistics/146244-probability-breakdowns-print.html | # probability of breakdowns
• May 24th 2010, 11:47 AM
Suchy
probability of breakdowns
Assuming that number of breakdowns per hour follows a poisson distribution with mean ( λ) equal 0,4 answer the following:
a) calculate the probability that the peeler does not have any breakdowns within an hour.
b) calculate the probability that a maximum of 1 breakdown happens within 8 hours.
• May 24th 2010, 12:42 PM
galactus
Quote:
Originally Posted by Suchy
Assuming that number of breakdowns per hour follows a poisson distribution with mean ( λ) equal 0,4 answer the following:
Is it safe to assume that 0,4 means 0.4?. Are you in one of those countries where they use a comma for a decimal point?. (Lipssealed)
$f(x;{\lambda})=\frac{{\lambda}^{x}e^{-\lambda}}{x!}$
Quote:
a) calculate the probability that the peeler does not have any breakdowns within an hour.
In the above formula, x=0. So, we have
$f(0;0.4)=\frac{{\lambda}^{0} e^{-0.4}}{0!}$
Quote:
b) calculate the probability that a maximum of 1 breakdown happens within 8 hours.
Quote:
${\lambda}=(0.4)(8)=3.2$
$f(\text{at most 1}; 3.2)=\frac{(3.2)^{0}e^{-3.2}}{0!}+\frac{(3.2)^{1}e^{-3.2}}{1!}$ | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 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": 0.9726579785346985, "perplexity": 1624.6480848985739}, "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/1474738661555.40/warc/CC-MAIN-20160924173741-00196-ip-10-143-35-109.ec2.internal.warc.gz"} |
https://tensorflow.google.cn/versions/r2.0/api_docs/python/tf/math/atan2?hl=bn | # tf.math.atan2
Computes arctangent of y/x element-wise, respecting signs of the arguments.
This is the angle ( \theta \in [-\pi, \pi] ) such that [ x = r \cos(\theta) ] and [ y = r \sin(\theta) ] where (r = \sqrt(x^2 + y^2) ).
y A Tensor. Must be one of the following types: bfloat16, half, float32, float64.
x A Tensor. Must have the same type as y.
name A name for the operation (optional).
A Tensor. Has the same type as y.
[]
[] | {"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": 0.9926355481147766, "perplexity": 3368.265914934636}, "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-2022-33/segments/1659882572215.27/warc/CC-MAIN-20220815235954-20220816025954-00363.warc.gz"} |
https://quantumcomputing.stackexchange.com/questions/2059/are-there-connections-between-long-range-entanglement-and-topological-quantum-co | # Are there connections between long-range entanglement and topological quantum computation?
Long-range entanglement is characterized by topological order (some kinds of global entanglement properties), and the "modern" definition of topological order is the ground state of the system cannot be prepared by a constant-depth circuit from a product state, instead of ground states dependency and boundary excitations in traditional. Essentially, a quantum state which can be prepared by a constant-depth circuit is called trivial state.
On the other hand, quantum states with long-range entanglement are "robust". One of the most famous corollaries of quantum PCP conjecture which proposed by Matt Hastings is the No Low-energy Trivial States conjecture, and the weaker case proved by Eldar and Harrow two years ago (i.e. NLETS theorem: https://arxiv.org/abs/1510.02082). Intuitively, the probability of a series of the random errors are exactly some log-depth quantum circuit are very small, so it makes sense that the entanglement here is "robust".
It seems that this phenomenon is some kinds of similar to topological quantum computation. Topological quantum computation is robust for any local error since the quantum gate here is implemented by braiding operators which is connected to some global topological properties. However, it needs to point that "robust entanglement" in the NLTS conjecture setting only involved the amount of entanglement, so the quantum state itself maybe changed -- it does not deduce a quantum error-correction code from non-trivial states automatically.
Definitely, long-range entanglement is related to homological quantum error-correction codes, such as the Toric code (it seems that it is related to abelian anyons). However, my question is that are there some connections between long-range entanglement (or "robust entanglement" in the NLTS conjecture setting) and topological quantum computation? Perhaps there exists some conditions regarding when the correspondent Hamiltonian can deduce a quantum error-correction code.
• It is worth mentioning that long-range and robust entanglement are not the same thing. Think of the GHZ state: it requires an $O(N)$ depth circuit to prepare, but certainly is not robust! May 15 '18 at 14:49
• @DaftWullie I clarified the statement. "rubost entanglement" in the NLTS conjecture setting is a confusing term since the state itself may be changed, even the amount of entanglement remains same. May 22 '18 at 21:37
These use the area law of entanglement seen by states that can be expressed as ground states of a Hamiltonian with only local interactions.
Specifically, suppose you have a 2D system of interacting particles in a pure state. You then single out some region, and calculate the von Neumann entropy of the reduced density matrix for that region. This will essentially be a measure of how entangled the region is with its complement. The area law tells us that this entropy, $S$, should obey
$S = \alpha L - \gamma + \ldots$
Here $L$ is the length of the perimeter of the region. The first term accounts for the fact that correlations in these systems are typically short range, and so the entanglement is mostly composed of correlations between particles on each side of the boundary.
The $\gamma$ term is unaffected by the size or shape of the region, and so represents a contribution of global and topological effects. Whether this is non-zero, and what the value is, tells you about the topologically ordered nature of your entangled system.
The $\ldots$ term just represents contributions that decay as the region increases, and so can be ignored as $L\rightarrow \infty$.
The two papers, and ones based upon them, then find ways to isolate and calculate $\gamma$ for different entangled states. The value is shown to depend on the anyon model for which these entangled states represent the vacuum. | {"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": 0.8835853934288025, "perplexity": 499.81739160333706}, "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/1642320306335.77/warc/CC-MAIN-20220128182552-20220128212552-00060.warc.gz"} |
https://brilliant.org/problems/1-d-2-d-simultaneously/ | # 1-D, 2-D simultaneously!
Two bodies were thrown simultaneously from same point. One straight up and other at an angle $$60^\circ$$ with the horizontal. Initial velocity is $$25\text{ m/s}$$ for each body. Find the distance between the two bodies after 1.7 seconds. | {"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": 0.9195488691329956, "perplexity": 410.0975660851827}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-51/segments/1512948568283.66/warc/CC-MAIN-20171215095015-20171215115015-00245.warc.gz"} |
https://2022.ieeerasse.org/tutorials/ | ## Tutorials
IRT SystemX
### Complex simulation specification: Do your simulations meet their needs?
A simulation can be a complex architecture of simulation models, simulation tools, and computing hardware. However, its development often relies on informal procedures and can begin without a clear, complete, and formal definition of simulation needs. Simulation traceability is then compromised, which prevents from easily validating whether a simulation meets the needs, or understanding the purpose of a simulation model that can be reused. In this tutorial, we will discuss in light of current standards all the aspects of simulation needs: (1) the part of the system to be simulated; (2) the objective of the simulation; (3) the simulation quality, cost, and delivery; (4) the test scenarios; (5) the data for simulation calibration and validation; and (6) the verification and validation of the simulation. Using the design of an autonomous car as an example (thanks to the contribution of several automotive companies), we will see how MBSE and artificial intelligence can improve agility by allowing the partial automatization of the definition and analysis of simulations needs. This tutorial is a 60 minute lecture. Come and let’s make sure that you start your simulations with clear, complete, and formal simulation needs which can then be efficiently used in simulation validation and reuse. | {"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": 0.9247434139251709, "perplexity": 1629.073013669081}, "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/1652662531779.10/warc/CC-MAIN-20220520093441-20220520123441-00185.warc.gz"} |
http://math.stackexchange.com/questions/218057/stratified-random-sampling | # Stratified random sampling
1. Which of the following is NOT a characteristic of stratified random sampling?
(A) Random sampling is part of the sampling procedure.
(B) The population is divided into groups of units that are similar on some characteristic.
(C) The strata are based on facts known before the sample is selected.
(D) Each individual unit in the population belongs to one and only one of the strata.
(E) Every possible subset of the population, of the desired sample size, has an equal chance of being selected.
The correct answer is (e), but why (e)? isn't every subset of the population has an equal chance of being selected. the idea of stratified sampling?
-
Stratified sampling allocates a part of the sample to each stratum. So, for example, subsets where you allocate the whole sample to one stratum has probability zero of being selected. – Stefan Hansen Oct 21 '12 at 14:56
The correct answer is $E$.
It seems like you're thinking of 'simple random sampling'. That has the property of independence, which means that every possible subset of the sample size has the same chance of occurring. While simple random sampling is used in stratified random sampling, the extra leeway in picking the sample sizes for each strata changes the independence property, and means choice $E$ is not true. | {"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": 0.9666450619697571, "perplexity": 395.1806609138205}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-23/segments/1406510261958.8/warc/CC-MAIN-20140728011741-00394-ip-10-146-231-18.ec2.internal.warc.gz"} |
http://ec.citizendium.org/wiki/LaTeX | # LaTeX
Main Article
Discussion
Related Articles [?]
Bibliography [?]
Citable Version [?]
This editable Main Article is under development and subject to a disclaimer.
LaTeX is a markup language for generating print-quality typesetting. It was originally written by Leslie Lamport as a package of macros extending Donald Knuth's TeX system. TeX and LaTeX are Free Software.
A file with formatting tags is compiled to create a high-quality print-ready graphical file, observing the rules of style defined by typesetting professionals. It is mostly used in academic circles, primarily in the natural sciences, for creating material for scientific publications. LaTeX is particularly well-equipped for displaying formulas and diagrams. Citizendium and Wikipedia use LaTeX for the presentation of mathematical formulas.
While critics claim that using a markup language structure is dated compared to using WYSIWYG word processors, LaTeX still plays an important part in the industry and science. This is especially true for areas where formulae play an important role. The LaTeX system is continually under development, and it is extensible by packages. Such packages can found at (and submitted to) the Comprehensive TeX Archive Network[1].
## How it works
A LaTeX document structure is split in two: There is a preamble, which indicates some basic, overall features of the document, and there is the main document with the relevant markup for formatting.
The document is initially created as a flat file, meaning that one can use any non-formatting editor of choice. Editors like Notepad++,[2] TeXmaker[3] or XEmacs with the AUCTeX package[4] will syntax color the LaTeX code, making it more intuitive to write.
Also, if one would like a more WYSIWYG approach, one would consider using LyX[5] or TeXmacs[6].
As the code is a flat file, the output file will be small compared to files from many standard word processing applications.
### Code example
\documentclass[12pt,a4paper,notitlepage]{article}
\usepackage{amsmath}
\usepackage{amsfonts}
\author{Morten Juhl Johansen}
\title{LaTeX article}
\begin{document}
This is the beginning of an article on \LaTeX!\\
There are \textbf{bold text}, \textit{italics}, \underline{underscoring},
footnotes\footnote{this is a footnote}\\
- and lots of other goodies. A math example:
\begin{equation*}
\sum_{n=1}^\infty
\frac{1}{n^2}= \frac{\pi^2}{6}.
\end{equation*}
\end{document}
Output: | {"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": 0.9840163588523865, "perplexity": 3761.5164551858047}, "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/1656104669950.91/warc/CC-MAIN-20220706090857-20220706120857-00262.warc.gz"} |
https://hal-ens-lyon.archives-ouvertes.fr/ensl-01141697 | Statistical properties of the coarse-grained velocity gradient tensor in turbulence: Monte-Carlo simulations of the tetrad model - Archive ouverte HAL Access content directly
Journal Articles New Journal of Physics Year : 2010
## Statistical properties of the coarse-grained velocity gradient tensor in turbulence: Monte-Carlo simulations of the tetrad model
(1) , (1)
1
Alain Pumir
• Function : Author
• PersonId : 830872
Aurore Naso
#### Abstract
A proper description of the velocity gradient tensor is crucial for understanding the dynamics of turbulent flows, in particular the energy transfer from large to small scales. Insight into the statistical properties of the velocity gradient tensor and into its coarse-grained generalization can be obtained with the help of a stochastic 'tetrad model' that describes the coarse-grained velocity gradient tensor based on the evolution of four points. Although the solution of the stochastic model can be formally expressed in terms of path integrals, its numerical determination in terms of the Monte-Carlo method is very challenging, as very few configurations contribute effectively to the statistical weight. Here, we discuss a strategy that allows us to solve the tetrad model numerically. The algorithm is based on the importance sampling method, which consists here of identifying and sampling preferentially the configurations that are likely to correspond to a large statistical weight, and selectively rejecting configurations with a small statistical weight. The algorithm leads to an efficient numerical determination of the solutions of the model and allows us to determine their qualitative behavior as a function of scale. We find that the moments of order n≤4 of the solutions of the model scale with the coarse-graining scale and that the scaling exponents are very close to the predictions of the Kolmogorov theory. The model qualitatively reproduces quite well the statistics concerning the local structure of the flow. However, we find that the model generally tends to predict an excess of strain compared to vorticity. Thus, our results show that while some physical aspects are not fully captured by the model, our approach leads to a very good description of several important qualitative properties of real turbulent flows.
### Dates and versions
ensl-01141697 , version 1 (13-04-2015)
### Identifiers
• HAL Id : ensl-01141697 , version 1
### Cite
Alain Pumir, Aurore Naso. Statistical properties of the coarse-grained velocity gradient tensor in turbulence: Monte-Carlo simulations of the tetrad model. New Journal of Physics, 2010, 12, pp.25. ⟨ensl-01141697⟩
### Export
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42 View | {"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": 0.8552433848381042, "perplexity": 775.9839138610236}, "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/1674764499953.47/warc/CC-MAIN-20230201211725-20230202001725-00593.warc.gz"} |
https://sciencing.com/estimate-derivative-graph-5580.html | # How to Estimate a Derivative from a Graph
••• Worawee Meepian/iStock/GettyImages
Print
Rates of change show up all over in science, and especially in physics through quantities like speed and acceleration. Derivatives describe the rate of change of one quantity with respect to another mathematically, but calculating them can be complicated sometimes, and you might be presented with a graph rather than a function in equation form. If you’re presented with a graph of a curve and have to find the derivative from it, you might not be able to be as accurate as with an equation, but you can easily make a solid estimate.
#### TL;DR (Too Long; Didn't Read)
Choose a point on the graph to find the value of the derivative at.
Draw a straight line tangent to the curve of the graph at this point.
Take the slope of this line to find the value of the derivative at your chosen point on the graph.
### What Is a Derivative?
Outside of the abstract setting of differentiating an equation, you might be a little confused about what a derivative really is. In algebra, a derivative of a function is an equation that tells you the value of the “slope” of the function at any point. In other words, it tells you how much one quantity changes given a small change in the other. On a graph, the gradient or slope of the line tells you how much the dependent variable (placed on the y-axis) changes with the independent variable (on the x-axis).
For straight-line graphs, you determine the (constant) rate of change by calculating the slope of the graph. Relationships described by curves aren’t as easy to deal with, but the principle that the derivative just means the slope (at that specific point) still holds true.
For relationships described by curves, the derivative takes a different value at every point along the curve. To estimate the derivative of the graph, you need to choose a point to take the derivative at. For example, if you have a graph showing distance traveled against time, on a straight-line graph, the slope would tell you the constant speed. For speeds that change with time, the graph would be a curve, but a straight line that just touches the curve at one point (a line tangential to the curve) represents the rate of change at that specific point.
Choose a spot that you need to know the derivative at. Using the distance traveled vs. time example, select the time at which you want to know the speed of travel. If you need to know the speed at several different points, you can run through this process for each individual point. If you want to know the speed 15 seconds after the start of the motion, choose the spot on the curve at 15 seconds on the x-axis.
Draw a line tangential to the curve at the point you’re interested in. Take your time when doing this, because it’s the most important and most challenging part of the process. Your estimate will be better if you draw a more accurate tangent line. Hold a ruler up to the point on the curve and adjust its orientation so the line you draw will only touch the curve at the single point you’re interested in.
Draw your line as long as the graph will allow. Make sure you can easily read two values for both the x and y coordinates, one near the start of your line and one near the end. You don’t absolutely need to draw a long line (technically any straight line is suitable), but longer lines tend to be easier to measure the slope of.
Locate two places on your line and make a note of the x and y coordinates for them. For example, imagine your tangent line as two notable spots at x = 1, y = 3 and x = 10, y = 30, which you can call Point 1 and Point 2. Using the symbols x1 and y1 to represent the coordinates of the first point and x2 and y2 to represent the coordinates of the second point, the slope m is given by:
m = \frac{y_2 - y_1}{x_2 - x_1}
This tells you the derivative of the curve at the point where the line touches the curve. In the example, x1 = 1, x2 = 10, y1 = 3 and y2 = 30, so:
\begin{aligned} m &= \frac{30 - 3}{10 - 1} \\ \,\\ &= \frac{27}{9} \\ \,\\ &=9 \end{aligned}
In the example, this result would be the speed at the chosen point. So if the x-axis was measured in seconds and the y-axis was measured in meters, the result would mean that the vehicle in question was travelling at 3 meters per second. Regardless of the specific quantity you’re calculating, the process of estimating the derivative is the same. | {"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": 0.9416845440864563, "perplexity": 273.9963342490263}, "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/1618038078021.18/warc/CC-MAIN-20210414185709-20210414215709-00598.warc.gz"} |
https://chemistry.stackexchange.com/questions/119807/neutralization-between-caoh2-h2so4-in-a-30-hydrogen-peroxide-solution | # Neutralization between Ca(OH)2 + H2SO4 in a 30% hydrogen peroxide solution
I have performed an experiment where I added an excess of $$\ce{Ca(OH)2}$$ base to a solution consisting of 5 mL of 30 % hydrogen peroxide (buffered at pH 5) and a very small amount of sulfuric acid (such that the pH of the original solution of 30 % $$\ce{H2O2}$$ and acid was 1.7).
I am having trouble understanding what reactions maybe occurring between these three substances to give a final pH of 10.5.
I know that $$\ce{Ca(OH)2}$$ and $$\ce{H2O2}$$ form $$\ce{CaO2}$$ when reacted, but shouldn't the final pH be equal to that of $$\ce{Ca(OH)2}$$ considering it is in excess? What reactions could be occurring here and why are they not allowing the pH to reach that of $$\ce{Ca(OH)2}$$?
Any help at all would be very much appreciated. Thank you in advance.
• There is no CaO2. – Ivan Neretin Sep 1 '19 at 3:48
• @Ivan Neretin en.wikipedia.org/wiki/Calcium_peroxide Calcium peroxide is produced by combining calcium salts and hydrogen peroxide: Ca(OH)2 + H2O2 → CaO2 + 2 H2O The octahydrate precipitates upon the reaction of calcium hydroxide with dilute hydrogen peroxide. Upon heating it dehydrates. – Poutnik Sep 1 '19 at 4:18
• All the same, I don't believe it will form in the said conditions. – Ivan Neretin Sep 1 '19 at 4:21
• It is not the same as "there is no CaO2". – Poutnik Sep 1 '19 at 4:24
• Let me add that, if pH is 10.5, it's clearly not "neutralized pH". As for the final question, you should read about buffers. – The_Vinz Sep 1 '19 at 4:49
in your question formulation, you have forgotten to take into account $$\ce{H2O2}$$ is a weak acid.
The title should rather be: Neutralisation between calcium hydroxide and 30% hydrogen peroxide"
Unless $$\ce{Ca(OH)2}$$ was in excess over $$\ce{H2O2}$$ - and it was said it was not - $$\mathrm{pH}$$ would be always significantly lower than pH of the hydroxide.
$$\mathrm{pH}=\mathrm{p}K_ \mathrm{a,\ce{H2O2}} + \log \frac{[\ce{HO2-}]}{[\ce{H2O2}]}$$
where $\mathrm{p}K_ \mathrm{a,\ce{H2O2}}=11.75$\$ by Wikipedia, but see the links below.
If we consider reaction
$$\ce{Ca(OH)2 + H2O2 -> H2O + Ca(OH)(HO2)}$$
we need to neutralize 50% of $$\ce{H2O2}$$ to reach $$\mathrm{pH}=\mathrm{p}K_ \mathrm{a,\ce{H2O2}}$$
The hydroxide forms from $$\ce{H2O2}$$ the $$\mathrm{pH}$$ buffer solution of a weak acid and it's salt.
\begin{align} \ce{Ca(OH)2 &<=>> CaOH+ + OH-}\\ \ce{CaOH+ &<=>> Ca^2+ + OH- }\\ \ce{H2O2 &<<=> H+ + HO2-}\\ \ce{H+ + OH- &<=>> H2O}\\ \end{align}
$$\ce{Ca(OH)2}$$: $$\mathrm{p}K_\mathrm{b1} =1.37$$, $$\mathrm{p}K_\mathrm{b2} =2.43$$ ( Wikipedia )
Additionally, $$\ce{HO2-}$$ is partially eliminated by precipitation, therefore ratio $$\frac{[\ce{HO2-}]}{[\ce{H2O2}]}$$ is kept low and so does $$\mathrm{pH}$$.
$$\ce{CaOH+ + HO2- + 7 H2O <=>> CaO2 \cdot 8 H2O v}$$
Note also the hydrogen peroxide is weakly acidic even without addition of sulphuric acid and that it's $$\mathrm{p}K_ \mathrm{a}$$ depends on $$\ce{H2O2}$$ concentration.
H2O2 pH-and-Ionization-Constant
The solubility constant of calcium peroxide octahydrate in relation to temperature; its influence on radiolysis in cement-based materials
• Does the pKa in the pH equation refer to H2O2 or Ca(OH)2? Thank you. – BigDog12 Sep 9 '19 at 5:12
• H2O2. for Ca(OH)2, there would calcium compounds instead. – Poutnik Sep 9 '19 at 5:14
• Okay. Thank you very much for all of your help. – BigDog12 Sep 9 '19 at 8:06 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 27, "wp-katex-eq": 0, "align": 1, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.864730715751648, "perplexity": 3261.391763771678}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-16/segments/1585371810807.81/warc/CC-MAIN-20200408072713-20200408103213-00555.warc.gz"} |
https://www.chapters.indigo.ca/en-ca/books/a-primer-of-lebesgue-integration/9780120839711-item.html | # A Primer Of Lebesgue Integration
## byH. S. BearEditorH. S. Bear
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The Lebesgue integral is now standard for both applications and advanced mathematics. This books starts with a review of the familiar calculus integral and then constructs the Lebesgue integral from the ground up using the same ideas. A Primer of Lebesgue Integration has been used successfully both in the classroom and for individual study.Bear presents a clear and simple introduction for those intent on further study in higher mathematics. Additionally, this book serves as a refresher providing new insight for those in the field. The author writes with an engaging, commonsense style that appeals to readers at all levels.
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The Lebesgue integral is now standard for both applications and advanced mathematics. This books starts with a review of the familiar calculus integral and then constructs the Lebesgue integral from the ground up using the same ideas. A Primer of Lebesgue Integration has been used successfully both in the classroom and for individual s...
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This successful text offers a reader-friendly approach to Lebesgue integration. It is designed for advanced undergraduates, beginning graduate students, or advanced readers who may have forgotten one or two details from their real analysis courses."The Lebesgue integral has been around for almost a century. Most authors prefer to blast...
H.S. Bear is a professor at the University of Hawaii, Manoa and a member of both the American Mathematical Society and the Mathematical Association of America.
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Format:HardcoverDimensions:164 pages, 9 × 6 × 0.98 inPublished:October 1, 2001Publisher:Academic PressLanguage:English
The following ISBNs are associated with this title:
ISBN - 10:0120839717
ISBN - 13:9780120839711
Extra Content | {"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": 0.8537635803222656, "perplexity": 2433.518571433296}, "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/1484560281659.81/warc/CC-MAIN-20170116095121-00531-ip-10-171-10-70.ec2.internal.warc.gz"} |
http://mathhelpforum.com/calculus/130461-proof-involving-integrals-equal-0-a.html | # Thread: Proof involving integrals equal to 0.
1. ## Proof involving integrals equal to 0.
Let
f be a bounded integrable non-negative function continous on [a, b].
Prove:
integral from a to b f(x) dx = 0 implies f(x) = 0 for every x
I believe I understand why it is true, I'm just not sure how to write the formal proof.
2. Originally Posted by amm345
Let
f be a bounded integrable non-negative function continous on [a, b].
Prove:
integral from a to b f(x) dx = 0 implies f(x) = 0 for every x
I believe I understand why it is true, I'm just not sure how to write the formal proof.
What have you tried? Here is a weird approach.
Suppose that $f\ne 0$ then there exists some $x\in[a,b]$ such $f(x)\in(0,\infty)$ and since this set is open there exists some open ball $B_{\varepsilon}(f(x))\subseteq(0,\infty)$ and by continuity there exists some $B_{\delta}(x)$ such that $f\left(B_{\delta}(x)\right)\subseteq B_{\varepsilon}(f(x))\subseteq(0,\infty)$. In other words, there exists some open ball around the point where $f(x)\ne0$ which is entirely positive. Take the concentric closed ball $B_{\frac{\delta}{2}}$. This is a closed and bounded interval and thus, by the Heine-Borel theorem, compact. Thus, since $f:B_{\frac{\delta}{2}}[x]\mapsto\mathbb{R}$ is continuous we know that $f$ assumes a minimum on $B_{\frac{\delta}{2}}[x]$. In particular, $\inf_{x\in B_{\frac{\delta}{2}}[x]}f(x)=\varepsilon>0$. Thus, $\int_a^b\text{ }f=\int_a^{x-\frac{\delta}{2}}\text{ }f+\int_{x-\frac{\delta}{2}}^{x+\frac{\delta}{2}}\text{ }f+\int_{x+\frac{\delta}{2}}^b \text{ }f\geqslant 0+\frac{\varepsilon \delta}{2}+0>0$. This of course is a contradiction.
3. That was sort of my original approach, but when I checked with my instructer he said we could not use that theorem and I can't think of anything else.
4. Originally Posted by amm345
That was sort of my original approach, but when I checked with my instructer he said we could not use that theorem and I can't think of anything else.
Which theorem is that?
5. Oh sorry, the Heine-Borel theorem.
6. Originally Posted by amm345
Oh sorry, the Heine-Borel theorem.
Are you serious? What's the point of having tools if you can't use them! What if I proved it first?
7. I have no idea.. I couldn't think of anything else it is kinda ridiculous!
8. Originally Posted by amm345
I have no idea.. I couldn't think of anything else it is kinda ridiculous!
Ok. I can think of one or two other ways....what are you currently doing in class?
9. We started integration last week and we've mostly just covered the riemann integral, basic principles, the fundamental theorem and continuity.
10. Originally Posted by amm345
We started integration last week and we've mostly just covered the riemann integral, basic principles, the fundamental theorem and continuity.
Hmm...Ok. I see an angle. Make a modified argument of mine and use the FTC.
11. Hello,
Why can't we just say that (without HB theorem) :
if there exists $m\in[a,b]$ such that $f(m)>0$, then by definition of the continuity, $\forall \epsilon>0, \exists \delta>0,\forall x\in[a,b], (|x-m|<\delta)\Rightarrow (|f(x)-f(m)|<\epsilon)$
In particular, if we let $\epsilon=\frac{f(m)}{2}$, $|f(x)-f(m)|<\epsilon \Leftrightarrow -\tfrac{f(m)}{2}
So there exists $\delta>0$ such that $\forall x\in]m-\delta,m+\delta[$, $f(x)>\tfrac{f(m)}{2}>0$
Hence $\int_a^b f(x) ~dx=\int_a^{m-\delta} \underbrace{f(x)}_{\geqslant 0} ~dx+\int_{m+\delta}^b \underbrace{f(x)}_{\geqslant 0} ~dx+\int_{m-\delta}^{m+\delta} \underbrace{f(x)}_{> \frac{f(m)}{2}} ~dx$
So $\int_a^b f(x) ~dx>0+0+2\delta \cdot\tfrac{f(m)}{2}=\delta f(m)>0$
So m can't exist, and hence $f(x)=0 ~,~ \forall x\in[a,b]$ | {"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": 24, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9689199328422546, "perplexity": 477.0922924568279}, "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-2016-36/segments/1471982293615.23/warc/CC-MAIN-20160823195813-00202-ip-10-153-172-175.ec2.internal.warc.gz"} |
http://mathhelpforum.com/calculus/174673-applications-integration-force-due-fluid-pressure.html | ## Applications of Integration: Force Due to Fluid Pressure
I'm working on a particular problem that involves the force on a face of a triangular prism. What I am wondering is if someone could explain how to find dA for the prism? If the prism is an equilateral triangle, then $dA=\sqrt3dy$ doesn't it? My book is showing that the area is $2\sqrt3ydy$ and I really cannot see how... | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 2, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8133677244186401, "perplexity": 173.02617577846007}, "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/1408500830903.34/warc/CC-MAIN-20140820021350-00446-ip-10-180-136-8.ec2.internal.warc.gz"} |
https://stats.stackexchange.com/questions/230345/waiting-time-for-the-first-event-in-an-exponential-distribution-with-different-r | # Waiting time for the first event in an Exponential Distribution with different rate parameters
I have a random variable $t$ that follows an exponential distribution with rate $\lambda_1$ in the interval $[0,t_1]$ and an exponential distribution with rate $\lambda_2$ in the interval $[t_1,\infty]$. I want to estimate the probability that there is no event in the time interval $[0,t_2]$, where $t_2>t_1$.
Here is how I have approached this problem.
$$$P(t>t_1+t_2)\\ =\{1-\int_{-\infty}^{t_1}f_{\lambda_1}(t)\} \{1-\int_{t_1}^{t_2}f_{\lambda_2}(t)\}$$$
Is the correct way to approach this problem? Should the second integral be $\{1-\int_{0}^{t_2}f_{\lambda_1}(t)\}$?
In general, what is the procedure to find the waiting time to the 1st event in a case where the inter-arrival process is non-homogenous?
• Suppose $\lambda_1$ happened to equal $\lambda_2$. You would likely use a simpler expression for the waiting time involving just one integral. Is your formula equivalent in that case to the simple expression? – whuber Aug 17 '16 at 18:53
• That is a good point. It is not. – statBeginner Aug 17 '16 at 18:58
• OK. The simple expression can be written as an integral. There is a natural way to break it into an integral over two regions separated at the value $t_1$, which re-expresses it as a sum of two integrals. Consider why that is so and apply that insight to the case $\lambda_1 \ne \lambda_2$. – whuber Aug 17 '16 at 19:02
• Unless $\lambda_1 = \lambda_2$, the density of the "random variable" does not integrate to 1. If $\lambda_1 \ge \lambda_2$, it would integrate out to 1 if the $\lambda_1$ rate section ended at $t_1$, then there was a gap of 0 density, and the $\lambda_2$ rate section started at $t_3 = (\lambda_1/\lambda_2) t_1$. Barring that, a time-inhomogeneos Possion Process, yes, a proper random variable, no. – Mark L. Stone Aug 17 '16 at 19:49
• I think Math1000 has given a nice demonstration. It differs from your approach by recognizing you have to compute the probability by multiplying a marginal by a conditional probability rather than adding those two probabilities. – whuber Aug 18 '16 at 19:44
Let $\{N(t):t\in\mathbb R_+\}$ be a (non-homogeneous) Poisson process with intensity $$\lambda(t) = \lambda_1\mathsf 1_{[0,t_1)}(t) + \lambda_2\mathsf 1_{[t_1,\infty)}(t).$$ Then for $t_2>t_1$, we have \begin{align} \mathbb P(N(t_2)=0) &= \mathbb P(N(t_1)=0,N(t_2)-N(t_1)=0)\\ &= \mathbb P(N(t_1)=0)\mathbb P(N(t_2)-N(t_1)=0)\\ &= e^{-\lambda_1 t_1}e^{-\lambda_2(t_2-t_1)}. \end{align} | {"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": 1, "x-ck12": 0, "texerror": 0, "math_score": 0.9999653100967407, "perplexity": 622.470325263605}, "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/1627046153791.41/warc/CC-MAIN-20210728185528-20210728215528-00200.warc.gz"} |
https://chem.libretexts.org/Bookshelves/Ancillary_Materials/Reference/Organic_Chemistry_Glossary/E1_Mechanism | # E1 Mechanism
E1 mechanism (E-elimination, 1-first order) is one of the three limiting mechanisms of 1,2-elimination. It is a two-step mechanism.
Step 1:
* Only the leaving group and one beta hydrogen are shown for clarity.
Step 2:
A 1,2-elimination occurring via E1 mechanism is called an E1 reaction. The rate law of an E1 reaction is
rate = k [substrate]
According to the rate law, an E1 reaction is first order overall, and the concentration of base does not affect the rate of reaction. The implication is that the base does not participate in the rate-limiting step or any prior steps, which suggests that the first step is the rate-limiting step. Since the base is not involved in the rate-limiting first step, the nature of base does not affect the rate of E1 reactions. | {"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": 0.854465126991272, "perplexity": 1644.1383561116186}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-16/segments/1585370505359.23/warc/CC-MAIN-20200401003422-20200401033422-00461.warc.gz"} |
https://documen.tv/question/a-stone-with-a-weight-of-5-30-n-is-launched-vertically-from-ground-level-with-an-initial-speed-o-15072139-74/ | ## A stone with a weight of 5.30 N is launched vertically from ground level with an initial speed of 23.0 m/s, and the air drag on it is 0.266
Question
A stone with a weight of 5.30 N is launched vertically from ground level with an initial speed of 23.0 m/s, and the air drag on it is 0.266 N throughout the flight. What are (a) the maximum height reached by the stone and (b) its speed just before it hits the ground
in progress 0
1 year 2021-08-29T20:36:45+00:00 1 Answers 6 views 0
a) $$h=25.7\ m$$
b) $$v’=21.8733\ m.s^{-1}$$
Explanation:
Given:
• weight of the stone, $$w=5.3\ N$$
• initial velocity of vertical projection, $$u=23\ m.s^{-1}$$
• air drag acting opposite to the motion of the stone, $$D=0.266\ N$$
The mass of the stone:
$$m=\frac{w}{g}$$
$$m=\frac{5.3}{9.8}$$
$$m=0.5408\ kg$$
Now the acceleration of the stone opposite of the motion:
$$D=m.d$$
where:
d = deceleration
$$0.266=0.5408\times d$$
$$d=0.4918\ m.s^{-2}$$
In course of going up the net acceleration on the stone will be:
$$g’=g+d$$
$$g’=9.8+0.4918$$
$$g’=10.2918\ m.s^{-2}$$
a)
Now using the equation of motion:
$$v^2=u^2-2 g’.h$$
where:
$$v=$$ final velocity when the stone reaches at the top of the projectile = 0
h = height attained by the stone before starting to fall down
$$0^2=23^2-2\times 10.2918\times h$$
$$h=25.7\ m$$
b)
during the course of descend from the top height of the projectile:
initial velocity, $$v=0\ m.s^{-1}$$
The acceleration will be:
$$g”=g-d$$
$$g”=9.8-0.4918$$
$$g”=9.3082\ m.s^{-2}$$
here the gravity still acts downwards but the drag acceleration acts in the direction opposite to the motion of the stone, now the stone is falling down hence the drag acts upwards.
Using equation of motion:
$$v’^2=v^2+2g”.h$$ (+ve acceleration because it acts in the direction of motion)
$$v’^2=0^2+2\times 9.3082\times 25.7$$
$$v’=21.8733\ m.s^{-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": 0.8806219696998596, "perplexity": 727.1624750782355}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-49/segments/1669446711417.46/warc/CC-MAIN-20221209144722-20221209174722-00017.warc.gz"} |
http://cms.math.ca/cjm/msc/46B28 | location: Publications → journals
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Search: MSC category 46B28 ( Spaces of operators; tensor products; approximation properties [See also 46A32, 46M05, 47L05, 47L20] )
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1. CJM 2011 (vol 63 pp. 1161)
Neuwirth, Stefan; Ricard, Éric
Transfer of Fourier Multipliers into Schur Multipliers and Sumsets in a Discrete Group We inspect the relationship between relative Fourier multipliers on noncommutative Lebesgue-Orlicz spaces of a discrete group $\varGamma$ and relative Toeplitz-Schur multipliers on Schatten-von-Neumann-Orlicz classes. Four applications are given: lacunary sets, unconditional Schauder bases for the subspace of a Lebesgue space determined by a given spectrum $\varLambda\subseteq\varGamma$, the norm of the Hilbert transform and the Riesz projection on Schatten-von-Neumann classes with exponent a power of 2, and the norm of Toeplitz Schur multipliers on Schatten-von-Neumann classes with exponent less than 1. Keywords:Fourier multiplier, Toeplitz Schur multiplier, lacunary set, unconditional approximation property, Hilbert transform, Riesz projectionCategories:47B49, 43A22, 43A46, 46B28
2. CJM 2007 (vol 59 pp. 614)
Labuschagne, C. C. A.
Preduals and Nuclear Operators Associated with Bounded, $p$-Convex, $p$-Concave and Positive $p$-Summing Operators We use Krivine's form of the Grothendieck inequality to renorm the space of bounded linear maps acting between Banach lattices. We construct preduals and describe the nuclear operators associated with these preduals for this renormed space of bounded operators as well as for the spaces of $p$-convex, $p$-concave and positive $p$-summing operators acting between Banach lattices and Banach spaces. The nuclear operators obtained are described in terms of factorizations through classical Banach spaces via positive operators. Keywords:$p$-convex operator, $p$-concave operator, $p$-summing operator, Banach space, Banach lattice, nuclear operator, sequence spaceCategories:46B28, 47B10, 46B42, 46B45
3. CJM 2005 (vol 57 pp. 1249)
Lindström, Mikael; Saksman, Eero; Tylli, Hans-Olav
Strictly Singular and Cosingular Multiplications Let $L(X)$ be the space of bounded linear operators on the Banach space $X$. We study the strict singularity andcosingularity of the two-sided multiplication operators $S \mapsto ASB$ on $L(X)$, where $A,B \in L(X)$ are fixed bounded operators and $X$ is a classical Banach space. Let \$1 Categories:47B47, 46B28
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https://tex.stackexchange.com/questions/178706/is-there-a-pre-defined-width-that-covers-the-text-area-and-the-margin | # Is there a pre-defined width that covers the text area and the margin?
Is there a pre-defined width that covers both the text and margin areas? Correct me if I am wrong, but it would basically be equal to \textwidth+\marginparsep+\parginparwidth.
Is that provided by LaTeX somewhere?
If not, I am currently defining it this way in the preamble:
\newlength{\widewidth}
\setlength{\widewidth}{\textwidth+\marginparsep+\marginparwidth}
Is that the right approach? Are there any caveats? For example, if I change a page's orientation, would \widelength change its size to reflect that?
• Have you tried \linewidth? – John Kormylo May 18 '14 at 13:40
• @JohnKormylo Yes, \linewidth is the same as \textwidth. Should that not have been the case? – sudosensei May 18 '14 at 17:55
• @sudosensei I haven't confirmed it, but I seem to recall that \linewidth and \textwidth are the same in the default configuration, but if you change the page's orientation, \textwidth does not change, but \linewidth does account for the revised orientation. Also, keep in mind that the \marginpar can be changed from the one side of the page to the other. While the overall width may stay the same, the beginning/ending locations on the paper will change. – Steven B. Segletes May 19 '14 at 2:23
• @StevenB.Segletes Thanks for the explanation, Steven. The beginning and ending positions are decided by the figure* environment (from the sidenotes package) that places figures across the margin and text block. The length I am defining is only a convenience variable, really, so that I can say \includegraphics[width=\widewidth] - \textwidth and linewidth are the same as outside the figure* environment. I guess the question is whether this is the right way to define the length, assuming that starting/ending points have ben taken care of. – sudosensei May 19 '14 at 9:30
Once a user length is defined, it is fixed in physical units until reset by the user (see Defining a Length that Scales with Fontsize Changes). So the way you define a "length" that can adjust with page dimension changes is to define it as a macro \def in terms of the system lengths that are updated with format changes, rather than as a \setlength which will fix the length. In this manner, the \def is re-evaluated to its "current" value at every invocation.
Once that lesson is learned, the next aspect to your question is that \linewidth rather than \textwidth is the appropriate length to use when, for example, changing into landscape mode, etc. On page 1 of my MWE, it makes no difference which is used, but on page 2, one can see that the \linewidth definition gives the proper result, not the \textwidth result.
And finally, if \reversemarginpar is used, as is done on page 3 of the MWE, then extra provisions must be made
\documentclass{article}
\usepackage[demo]{graphicx}
\usepackage{pdflscape}
\usepackage[nopar]{lipsum}
\def\wrongwidewidth{\dimexpr\textwidth+\marginparsep+\marginparwidth}%
\def\rightwidewidth{\dimexpr\linewidth+\marginparsep+\marginparwidth}%
\begin{document}
\lipsum[1]\marginpar{This is a test of a margin par}\vspace{2cm}\par
Doesn't matter\par
\noindent\includegraphics[width=\wrongwidewidth]{test}
Doesn't matter\par
\noindent\includegraphics[width=\rightwidewidth]{test}
\begin{landscape}
\lipsum[2]\marginpar{This is a test of a margin par}\vspace{2cm}\par
Wrong\par
\noindent\includegraphics[width=\wrongwidewidth]{test}
Right\par
\noindent\includegraphics[width=\rightwidewidth]{test}
\end{landscape}
With reverse marginpar, tricks must be done\vspace{2cm}\par\reversemarginpar
\lipsum[1]\marginpar{This is a test of a margin par}\vspace{2cm}\par
Doesn't matter\par
\noindent\makebox[\linewidth][r]{\includegraphics[width=\wrongwidewidth]{test}}
Doesn't matter\par
\noindent\makebox[\linewidth][r]{\includegraphics[width=\rightwidewidth]{test}}
\end{document}
• Very well explained. I wasn't aware of the difference between \textwidth and \linewidth. Thank you, Steven! – sudosensei May 19 '14 at 11:08 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 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": 0.9095351099967957, "perplexity": 1404.1246687775708}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-10/segments/1581875145859.65/warc/CC-MAIN-20200223215635-20200224005635-00474.warc.gz"} |
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