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https://www.physicsforums.com/threads/question-about-electric-potential.45910/ | 1. Oct 3, 2004
### scissors
How do I calculate the potential at the point P? Assuming the rod is charged with 8 micro Coloumbs? I'm looking at equations, Kq/r, and I think there is some integrating involved, but I'm stumped. Here is a diagram....any help is greatly appreciated!
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| 10 cm.............35 cm
P - - - - ====================
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2. Oct 3, 2004
### Gokul43201
Staff Emeritus
Okay, I guess the double line refers to the rod (which is 35 cm long?), and P is 10cm from the end of the rod.
Divide your rod into little segments of length dx. Assume the rod has uniform charge density and write down the charge on this element. Treat each element as a point charge, at distance x from P and write down the potential due to this point charge at P. Now integrate this between the values of x that the rod occupies.
3. Oct 3, 2004
### scissors
Thanks very much! I was able to get the answer by integrating dr/r from (.10 to .45), and multiplying by lamda (Q/.45) times k.
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https://www.physicsforums.com/threads/number-of-sf-to-be-carried-forward.411245/ | # Number of sf to be carried forward
1. Jun 20, 2010
### hasan_researc
1. The problem statement, all variables and given/known data
My problem is with the number of significant figures I have to carry forward in intermediate calculations. I know that if the given data has a minimum of n sf, then the final ans should be to n+1 sf. But then should the intermediate answers be to n+2 sf??
2. Relevant equations
3. The attempt at a solution
I have no idea how to solve this problem.
2. Jun 20, 2010
### Chewy0087
If it's simply part of your working i'd of thought as many significant figures as possible, you won't lose any marks having all intermediate answers to n+3 SF as long as you have an accurate rounded answer. Much better that than having all intermediate answers to n SF which will almost inevitably affect the accuracy of your final answer to n+1 SF.
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https://www.physicsforums.com/threads/second-order-equations-can-anybody-help-me-greatly-appreciated.149619/ | # Second Order Equations Can Anybody help me? Greatly appreciated!
1. Dec 30, 2006
### xinerz
Second Order Equations!! Can Anybody help me?? Greatly appreciated!
1. The problem statement, all variables and given/known data
Interpret x(t) as the position of a mass on a spring at time t where x(t) satisfies
x'' + 4x' + 3x = 0.
Suppose the mass is pulled out, stretching the spring one unit from its equilibrium position, and given an initial velocity of +2 units per second.
(A) Find the position of the mass at time t.
(B) Determine whether or not the mass ever crosses the equilibrium position of x = 0.
(C) When (at what time) is the mass furthest from its equilibrium position? Approximately how far from the equilibrium position does it get?
2. Relevant equations
Previous problems on this homework set include transforming the initial value problem into a solution that looks partially like the following:
(example):
y = (1/3)e^(-4t) + (2/3)e^(-4t)
3. The attempt at a solution
I've attempted the following:
x" + 4x' + 3x = 0 --> r^2 + 4r + 3 = 0, solved for r
r = -3 or -1
therefore y=e^(-3t) , y=e^(-t)
y(t) = Ae^(-3t) + Be^(-t)
solved for A and B both = -1/2
however, I'm not sure that this is right.
THANK YOU!
2. Dec 30, 2006
### StatusX
Does it satisfy the differential equation and boundary conditions? If so, it's probably right.
3. Dec 30, 2006
### xinerz
so to find the position at t, i just solve for y in terms of t? like
t = something
also, how would i show if the mass crossed equilibrium at x = 0?
4. Dec 31, 2006
### StatusX
No, y(t) is the position at t. And you seem to be using x and y to refer to the same thing. Try graphing the function to see if it crosses 0.
5. Dec 31, 2006
### Mindscrape
Or just set it equal to zero and see if there is a solution.
6. Dec 31, 2006
### xinerz
thanks! i got it you guys :)
thanks for all the help!
HAPPY NEW YEAR!
Have something to add?
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https://chemistry.stackexchange.com/questions/118116/finding-emf-of-a-galvanic-cell-without-standard-potentials/118123 | # Finding EMF of a galvanic cell without standard potentials
For the galvanic cell
$$\ce{Ag | AgCl(s), KCl (\pu{0.2 M}) || KBr (\pu{0.001 M}), AgBr(s) | Ag}$$
find the EMF generated given $$K_\mathrm{sp}(\ce{AgCl}) = \pu{2.8e-10},$$ $$K_\mathrm{sp}(\ce{AgBr}) = \pu{3.3e-13}.$$
This is the question from JEE exam (1992).
How to start solving the problem since the $$E^°$$ of individual half reactions is not given? How to write the $$E^°$$ for the cell without it? Or is it not needed?
• I am upvoting this because several of us learned something today! Thanks for the nice question! – Ed V Jul 17 at 3:00
How to start solving the problem since the $$E^\circ$$ of individual half reactions is not given?
This is a concentration cell, i.e. the half reactions at the anode and at the cathode are the same (except for the direction).
AgCl(s) electrode $$\ce{AgCl(s) <=> Ag+(aq) + Cl-(aq)}$$ $$\ce{Ag+(aq) + e- -> Ag(s)}$$
AgBr(s) electrode $$\ce{Ag(s) -> Ag+(aq) + e-}$$
$$\ce{Ag+(aq) + Br-(aq) <=> AgBr(s)}$$
The standard reduction potentials will cancel out, i.e. $$E^\circ (\mathrm{cell}) = 0$$.
Further thoughts
[Comment by EdV] The Ag|AgCl electrodes I have are commercial, but making them in the lab is just a matter of oxidizing Ag wire in a chloride solution, so the electrode is Ag wire with an adherent coating of AgCl... I have never seen one of these made by just sticking an Ag wire in the Ag halide, but I guess it would work.
[...my own comment] I pictured the silver electrode submerged in the solution, with the solid halide on the bottom. I am puzzled now too. Does it make a difference if the electrode touches the solid, the liquid, or both?
What are the actual reduction potentials?
$$\ce{AgCl(s) + e− <=> Ag(s) + Cl−}\ \ \ \ E^\circ_\mathrm{red} = \pu{+0.22233 V}\tag{1}$$ $$\ce{AgBr(s) + e− <=> Ag(s) + Br−}\ \ \ \ E^\circ_\mathrm{red} = \pu{ +0.07133 V}\tag{2}$$ $$\ce{Ag+ + e− <=> Ag(s)}\ \ \ \ E^\circ_\mathrm{red} = \pu{ +0.7996 V}\tag{3}$$
Are they related?
If you subtract (3) from (1), you get the dissolution reaction of AgCl, if you subtract (3) from (2), you get the dissolution reaction of AgBr. Thus, standard reduction potentials for (1) and (3) should be different by
$$-\frac{RT}{zF} \ln K_\mathrm{sp}(\ce{AgCl})$$
and standard reduction potentials of (2) and (3) should be different by
$$-\frac{RT}{zF} \ln K_\mathrm{sp}(\ce{AgBr})$$
Finally, standard reduction potentials (1) and (2) should be different by
$$-\frac{RT}{zF} \ln \frac{K_\mathrm{sp}(\ce{AgBr})}{K_\mathrm{sp}(\ce{AgCl})}$$
Numerical answer using half reaction (1) and (2)
$$\ce{AgCl(s) + Br-(aq) <=> AgBr(s) + Cl-(aq)}$$
$$Q = \frac{[\ce{Cl-}]}{[\ce{Br-}]} = 200$$
$$E_\mathrm{cell} = E^\circ_\mathrm{cell} - \frac{R T}{z F} \ln(Q)$$
$$= \pu{(0.22233 V− 0.07133 V) - 0.13612 V = 0.0149 V}$$
Numerical answer using half reaction (3) twice
$$\ce{Ag+(c) + Ag(b) <=> Ag(c) + Ag+(b)}$$
"c" stands for chloride side, and "b" stands for bromide side. For consistency, I am using the following values for the solubility products (derived from difference of standard reduction potentials of half reactions (1), (2) and (3)).
$$K_\mathrm{sp}(\ce{AgCl}) = \pu{1.74e−10}$$ $$K_\mathrm{sp}(\ce{AgBr}) = \pu{4.89e−13}$$
$$[\ce{Ag+}]_c = K_\mathrm{sp}(\ce{AgCl}) / [\ce{Cl-}] = \pu{8,27e−10}$$ $$[\ce{Ag+}]_b = K_\mathrm{sp}(\ce{AgBr}) / [\ce{Br-}] = \pu{4.89e−10}$$
$$Q = \frac{[\ce{Ag+}]_c}{[\ce{Ag+}]_b} = 0.560$$
$$E_\mathrm{cell} = E^\circ_\mathrm{cell} - \frac{R T}{z F} \ln(Q)$$
$$\pu{= 0 - (-0.0149 V) = 0.0149 V}$$
• @EdV I wrote up the numerical answer calculated both ways. I gave myself some extra significant figures for the calculation of the solubility products. When you say "your expression", I think there was some misunderstanding. With a standard potential of 0 V and silver ion concentrations similar in both cells, I'm not sure how you got -0.173 V. In any case, you can see that the chloride and bromide concentrations are used both times because the reaction quotient contains the silver ion concentrations in one case, and the bromide and chloride ion concentrations in the other case. – Karsten Theis Jul 17 at 2:42
• Excellent, we got the same answer! The -0.173 V was using the OP's solubility product constants. Thanks for the clarification! – Ed V Jul 17 at 2:53
• @KarstenTheis Sir, you just said that E° = 0 but now by putting values you are saying that E°≠0, also this was an exam type situations so the E° was not given. Also, how did you write the equation 3 directly ie.$$\ce{Ag+(c) + Ag(b) <=> Ag(c) + Ag+(b)}$$ why did you ignore the Chloride/Bromide radicals from the equation ?? What I tried making was the equation from reaction 1 and 2 but since E° was not given I couldn't proceed. Please explain the equation 3. :) – RandomAspirant Jul 17 at 5:03
• @DivMit $$\ce{Ag+(aq) + e- <=> Ag(s)}$$ is a balanced half reaction, and if it describes both half-cells, you get the chemical equation you have in your comment. The concentration of silver cations in the two half cells indirectly depends on the concentration of chloride or bromide (as you can see in the calculation). Writing it this way, however, chloride and bromide are not part of the electrochemical equation and so they don't appear in the reaction quotient Q of the Nernst equation directly. – Karsten Theis Jul 17 at 6:10
Even though this question 1) has an answer with multiple upvotes (and I was the first upvote), 2) the OP has accepted the answer and 3) I have great respect for @Karsten Theis, having co-taught a quant class with him back in 2008 and knowing, first hand, that he is an excellent scientist and teacher, nonetheless, I have several problems with this trick exam question.
First, the $$K_{sp}(AgCl)$$ is about $$1.8 x 10^{-10}$$. This is the trivial problem, nothing more than a typo. More importantly, the $$K_{sp}$$ solution does not tell the whole story. So suppose T = 298.15K, i.e., standard temperature, n = 1 equivalent/mole, and all activity coefficients are assumed to be unity, so molar concentrations can be used in place of (tossing the molarity units) the unitless activities. Then, under standard state conditions, we have the following voltaic cell:
Thus AgCl(s) will be reduced to Ag(s) plus $$Cl^-$$ ions, on the right hand side (RHS) of fig. 1, and Ag(s) will be oxidized to AgBr(s), on the left hand side (LHS) of fig. 1.
But what happens if [$$Cl^-$$] = 0.2 M and [$$Br^-$$] = 0.001 M? Then both reduction potentials increase (become more positive), but the new Ag|AgCl reduction potential is still the most positive, so that is where reduction is spontaneous wrt (with respect to) the other half cell reaction reduction potential. This is shown is fig. 2 below:
In this figure, n = 1 equivalent/mole, T = 298.15K, R = 8.314472 J/(mole K), F = 96485.3383 C/equivalent, so RT/nF = 25.6926 mV and (ln10)•RT/nF = 59.1594 mV. Plugging in the numbers, as per the $$E_{cell}$$ equation in fig. 2, yields +0.0149V. If [$$Br^-$$] = 0.0001 M, then the cell potential would be -0.044V. This means the Ag|AgCl electrode would be the anode and the Ag|AgBr electrode would be the cathode. Note that the OP’s question actually showed the Ag|AgCl electrode as the anode, since this is the standard convention. But it wasn’t! So this was a rather nasty trick question, in my professional judgment.
• Thanks for finding all the problems in my answer. I think we have converged on a single result. Of course, because the OPs question has slightly different numbers, the numerical result to that question is a bit different. I learned a lot today. – Karsten Theis Jul 17 at 2:48
• I learned a lot as well! The way you did this problem is very sweet! Many thanks! – Ed V Jul 17 at 2:58 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 37, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8335055112838745, "perplexity": 1219.3623082231372}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-35/segments/1566027317113.27/warc/CC-MAIN-20190822110215-20190822132215-00205.warc.gz"} |
https://www.physicsforums.com/threads/4-vector-momentum-2.193894/ | # 4-vector - momentum^2
1. Oct 25, 2007
### Ene Dene
I'm having a problem understanding this:
$$P^2=P_{\mu}P^\mu=m^2$$
If we take c=1.
Here is what bothers me:
$$P(E, \vec{p})=E^2-(\vec{p})^2$$
Now, I assume that E=mc^2, and for c=1, E^2=m^2? Is that correct?
And I don't know what p^2 is, I look at it as:
$$(\vec{p})^2=m^2(\vec{v})^2$$
What am I doing wrong?
2. Oct 25, 2007
### nrqed
I am not exactly sure what bothers you but one point: $$E = \gamma m c^2$$, not mc^2 (which is valid only in the rest frame of the particle). Also the momentum is the relativistic three-momentum so it's $\gamma m \vec{v}$
3. Oct 25, 2007
### Meir Achuz
4. Oct 25, 2007
### dynamicsolo
The symbol 'E' in this equation is the total energy of the particle. The popularly-known equation E = mc^2 refers to the "rest mass-energy of the particle" and is really an incorrect use of the symbol.
'p' here is just the regular ol' 3-momentum, mass times the 3-dimensional velocity vector for the particle. The so-called "4-momentum" $$P^{\mu}$$ is a 4-dimensional vector whose components are
( iE, p_x, p_y, p_z ) [or flip signs depending on whose notational convention you use].
$$P^2=P_{\mu}P^{\mu}$$
is then just taking the "dot product" of P with itself to get the square of the magnitude,
$$P^2 = E^2 - p^2 = (mc^2)^2$$ , again with appropriate adjustments for local notational practice.
Last edited: Oct 25, 2007
5. Oct 26, 2007
### CompuChip
As already said, the popular formula E = mc^2 -- mostly misquoted -- is a) not generally applicable and b) not even a main result of special relativity, it's more like a small remark buried somewhere deep inside the text.
To stick with Ene Dene's approach: if you plug in the correct formula
$$E = \sqrt{ (\gamma m c^2)^2 + (m p^2)^2 }$$
you will get a consistent result.
In units where c = 1, it'd be
$$E^2 = \gamma m^2 + p^4$$
and
$$p = \gamma m v$$.
Also note that you can deduce where E = mc^2 is applicable; the general formula reduces to it in the rest frame (p = 0) at non-relativistic speeds ($\gamma \approx 1$).
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https://icml.cc/Conferences/2019/ScheduleMultitrack?event=4437 | Oral
When Samples Are Strategically Selected
Hanrui Zhang · Yu Cheng · Vincent Conitzer
Tue Jun 11th 11:30 -- 11:35 AM @ Room 102
In standard classification problems, the assumption is that the entity making the decision (the {\em principal}) has access to {\em all} the samples. However, in many contexts, she either does not have direct access to the samples, or can inspect only a limited set of samples and does not know which are the most relevant ones. In such cases, she must rely on another party (the {\em agent}) to either provide the samples or point out the most relevant ones. If the agent has a different objective, then the principal cannot trust the submitted samples to be representative. She must set a {\em policy} for how she makes decisions, keeping in mind the agent's incentives. In this paper, we introduce a theoretical framework for this problem and provide key structural and computational results. | {"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": 0.938117504119873, "perplexity": 1025.5899632783141}, "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/1570986668994.39/warc/CC-MAIN-20191016135759-20191016163259-00374.warc.gz"} |
http://mathhelpforum.com/geometry/45239-prism.html | 1. ## Prism
A rectangular prism, of which the base is a square, has a diagonal of length d, and the total area of all its faces is b. In terms of b and d, what is the total length of all its edges?
2. Originally Posted by atreyyu
A rectangular prism, of which the base is a square, has a diagonal of length d, and the total area of all its faces is b. In terms of b and d, what is the total length of all its edges?
Let x = side of square base,
and y = height of prism
And assume that the prism is a right rectangular prism, or the the base and the height are perpendicular.
In the square base:
diagonal = d .....given
So, by Pythagorean theorem, d^2 = x^ +x^2
x^2 = (d^2) /2
x = d / (sqrt(2))
Area of all the faces of the prism:
A = 2(x^2) +4(x*y) = b .....given
2x^2 +4xy = b
4xy = b -2x^2
y = (b -2x^2) / (4x)
y = [b -2(d^2 / 2)] / [4(d / sqrt(2)]
y = [b -d^2] / [2sqrt(2) *d]
Simplify further if you like...
Therefore, the sum of all the edges of this prism is
= 8x +4y
= [8(d / sqrt(2)] +[4(b -d^2) / (2sqrt(2) *d)]
= [4sqrt(2) *d] + [2sqrt(2) *(b -d^2)/d]
= 2sqrt(2)*[2d +(b -d^2)/d]
= 2sqrt(2)[(2d^2 +b -d^2) / d] | {"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": 0.8106510043144226, "perplexity": 3520.1388551668665}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-30/segments/1500549423927.54/warc/CC-MAIN-20170722082709-20170722102709-00518.warc.gz"} |
https://yart.me/blog/discrete-derivative/ | # Discrete Calculus: Definition of Discrete Derivative
May 30, 2020 · 16:45 pm · Discrete CalculusMath
We have learned about the operator $\Delta$ being used for a change in a variable, probably in a science class. However, your teachers probably handwaved them as something that means “a change in”. But what does it really mean?
According to Wikipedia, this is known as a finite difference. The formal notation uses a subscript to show the size of the difference between arguments like so:
$\Delta_h f(x) = f(x+h) - f(x)$
There are alternative definitions since we can start and end anywhere. The above is known as a forward difference since we substract the current value $f(x)$ from the next value $f(x+h)$.
There is also the backward difference:
$\nabla_h f(x) = f(x) - f(x - h)$
And the central difference:
$\delta_h f(x) = f\left(x + \frac{h}{2}\right) - f\left(x - \frac{h}{2}\right)$
The Calculus Wiki suggests that we can omit the subscript if $h = 1$:
$\Delta f(x) = f(x + 1) - f(x)$
However, omitting the subscript is also used for other cases, e.g. in physics to denote the change from the initial state to the final state, so it’s probably a good idea to clarify. Note that we can define conventional derivatives using finite differences, and along the way also get an informal intuition about derivatives. We can start from the definition and go from there:
\begin{aligned} \frac{df}{dx} &= \lim_{h \to 0} \frac{f(x + h) - f(x)}{h} \\ &= \lim_{h \to 0} \frac{\Delta_h f(x)}{h} \\ \end{aligned}
Let $I(x)=x$, or if you’re feeling frisky, $I \colon x \mapsto x$. Then,
\begin{aligned} \frac{df}{dx} &= \lim_{h \to 0} \frac{\Delta_h f(x)}{h} \\ &= \lim_{h \to 0} \frac{\Delta_h f(x)}{h} \\ &= \lim_{h \to 0} \frac{\Delta_h f(x)}{0+h - 0} \\ &= \lim_{h \to 0} \frac{\Delta_h f(x)}{I(0 + h) - I(0)} \\ &= \lim_{h \to 0} \frac{\Delta_h f(x)}{\Delta_h I(0)} \\ \end{aligned}
We can establish a few things with the final equation, $\frac{df}{dx} = \lim_{h \to 0} \frac{\Delta_h f(x)}{\Delta_h I(0)}$. Firstly, since $\lim$ doesn’t distribute, we can’t directly define $df$ or $dx$ with $\Delta_h f$ or $\Delta_h x$. That is, the differential operator, $d$, inherently compares the change in rate of multiple functions. Therefore, it makes sense to talk about $\frac{df}{dx}$, but not $df$.1
Secondly, we can have an informal intuition about what $df$ and $dx$ really means. $df$ is associated with $\Delta_h f(x)$ and $dx$ is associated with $\Delta_h I(0) = h$ as $h \to 0$.
Lastly, we can define the “discrete derivative” as when $h=1$, leading to
\begin{aligned} \frac{\Delta_1 f(x)}{\Delta_1 I(0)} &= \frac{\Delta f(x)}{I(1) - I(0)} \\ &= \frac{\Delta f}{1 - 0} \\ &= \Delta f \\ \end{aligned}
Interestingly, for discrete derivatives, we don’t need a “denominator” in our differential expression.
1. Of course, you can always be working in a system that explicitly defines differentials.
Streaming direct thought dumps from Yuto Nishida.
Connect with me on LinkedIn! I'm currently listening to: | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 30, "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": 1.0000094175338745, "perplexity": 607.1501246607068}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-04/segments/1610703581888.64/warc/CC-MAIN-20210125123120-20210125153120-00062.warc.gz"} |
http://math.stackexchange.com/questions/171319/is-vector-arithmetic-compatible-between-2d-and-3d-vectors | # Is Vector arithmetic compatible between 2D and 3D Vectors?
As the title suggests, is Vector arithmetic (including Cross and Dot Products and Length Calculations) compatible between 2D and 3D Vectors where a "2D Vector" is a 3D Vector with a third parameter that is always one (1)?
That is, is $\vec{A}$(x, y, 1) compatible with $\vec{B}$(x, y, z) where $\vec{A} + \vec{B} = \vec{C}$ and $\vec{C} = (Ax + Bx, Ay + By, 1)$; (etc for all basic arithmetic); $\vec{A} \cdot \vec{B} = \vec{C}$ and $\vec{C} = Ay \times 1 - 1 \times By, 1 \times Bx - Ax \times 1$; and $\vec{A} \times \vec{B} = x$ where $x = Ax * Bx + Ay * By + 1 * 1$ as well as $\lVert A \rVert = \sqrt{(Ax^2) + (Ay^2) + (1^2)}$ and $\lVert B \rVert = \sqrt{(Bx^2) + (By^2) + (1^2)}$
Will this produce inaccurate results?
-
Yes and no. Yes because you can technically do this all you want, but no because when we use 2D vectors we don't typically mean $(x,y,1)$. We actually mean $(x,y,0)$. As in, "it's 2D because there's no z-component". These are just the vectors that sit in the $xy$-plane, and they behave as you'd expect. For example, dot products can show you that they're perpendicular to the $z$-axis. The vectors of the form $(x,y,1)$, on the other hand, are sitting on the plane $z=1$, which is somewhat unsatistfactory, largely because you've lost your $\vec{0}$-vector. They exist as $3D$ vectors, but they don't sit in the $xy$-plane as we generally say $2D$ vectors do, so they're probably not what you want.
EXAMPLE:
$$(x,y,0)\cdot(0,0,z)=0$$ $$(x,y,1)\cdot(0,0,z)=z$$
the vectors you suggest all stick up slightly (pointing from the origin to $z=1$), so they don't actually lie in a plane. They point to one, but don't lie in it, which means they lose a lot of nice properties. As another example:
$$(x,0,1)\cdot(0,y,1)=1$$
when you would clearly want pure $x$ and $y$ vectors to be perpendicular.
EDIT: with cross product:
$$(x,0,1)\times(0,y,1)=(-y,-x,xy)$$
Again, you would want this vector to have length $xy$ and stick out along the $z$-axis, but it does neither. On the other hand, $$(x,0,0)\times(0,y,0)=(0,0,xy)$$ which is what you want.
-
Would the Cross Product fail spectacularly by returning A(x, y, 0) x B(x, y, 0) = C(Ay * 0 - 0 * By, 0 * Bx - Ax * 0, 0) or is the math different for the cross product of a 2D vector? (or even possible?) – Casey Jul 15 '12 at 22:53
You can't define a cross product with literal $2D$ vectors, but you can with your construction (since they are $3D$). I've updated with more information. – Robert Mastragostino Jul 15 '12 at 22:59
If I understand you correctly, you want to "re-use" 3D vector "arithmetic" and do 2D vector arithmetic? Then you need to set the 3rd coordinate to zero. Here is why.
You won't have a subspace unless you include the zero vector $(0, 0, 0).$ Now it's easy to verify that the set $V$ of all vectors $(x, y, 0) \in \Bbb R^3$ forms a subspace of $\Bbb R^3$ of dimension $2.$ Also, the set $V$ is isomorphic to $\Bbb R^2$ in the natural way $(x, y, 0) \mapsto (x, y).$
So whenever you have two 2D vectors: $(x, y)$ and $(z, w),$ extend them to $(x, y, 0)$ and $(z, w, 0).$ Perform the arithmetic in $\Bbb R^3.$ Say the resulting vector is $(u, v, \ell).$ Extract the results back as $(u, v).$ Note: $\ell \color{blue}{=} 0,$ by the fact that $V$ is a subspace closed under linear operations.
-
Oh if you use $1$ in the 3rd coordinate is invalid. For example, adding two vectors on the plane $z = 1$ will result in vector on the plane $z = 2.$ Also the euclidean norm (i.e. length) is no longer the same norm in $\Bbb R^2.$ – user2468 Jul 15 '12 at 22:37
If you replace "$A=(x, y,1)$ with $A=(x,y,0)$, then a 2D vector will work with 3D arithmetic just fine.
- | {"extraction_info": {"found_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": 0.869572639465332, "perplexity": 357.7140928817519}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-27/segments/1435375097861.54/warc/CC-MAIN-20150627031817-00110-ip-10-179-60-89.ec2.internal.warc.gz"} |
http://math.stackexchange.com/questions/97954/when-is-f-1-tildem-a-quasi-coherent-sheaf | # When is $f^{-1}(\tilde{M})$ a quasi-coherent sheaf?
To be precise, I want to know whether the following statement is true or false:
Let $A$ be a ring (it can be reduced), $f:\rm{Spec}(A/I) \to \rm{Spec}(A)$ is a closed immersion, $\tilde{M}$ is a quasi-coherent sheaf of $\mathcal{O}_{\rm{Spec}(A)}$-module, is it true $f^{-1}(\tilde{M})$ is also a quasi-coherent sheaf of $\mathcal{O}_{\rm{Spec}(A/I)}$-module?
-
No, it is even not a module over $O_{\mathrm{Spec}(A/I)}$ in general. Actually let $F=\tilde{M}$. The stalk $f^{-1}(F)_x=F_{f(x)}$ and the RHS is not a $A/I$-module (not killed by $I$) in general. – user18119 Jan 10 '12 at 20:43
Yes, I see. Thank you! – Li Zhan Jan 11 '12 at 17:58 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8380568027496338, "perplexity": 170.58383200935367}, "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-2015-18/segments/1429246639414.6/warc/CC-MAIN-20150417045719-00114-ip-10-235-10-82.ec2.internal.warc.gz"} |
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Issue A&A Volume 521, October 2010 A9 10 Stellar structure and evolution https://doi.org/10.1051/0004-6361/201014305 14 October 2010
A&A 521, A9 (2010)
## Thermohaline mixing in evolved low-mass stars
M. Cantiello - N. Langer
Argelander-Institut für Astronomie der Universität Bonn, Auf dem Hügel 71, 53121 Bonn, Germany
Received 23 February 2010 / Accepted 27 May 2010
Abstract
Context. Thermohaline mixing has recently been proposed to occur in low-mass red giants, with large consequence for the chemical yields of low-mass stars.
Aims. We investigate the role of thermohaline mixing during the evolution of stars between 1 and 3 , in comparison with other mixing processes acting in these stars.
Methods. We use a stellar evolution code which includes rotational mixing, internal magnetic fields and thermohaline mixing.
Results. We confirm that during the red giant stage, thermohaline mixing has the potential to decrease the abundance of 3He, which is produced earlier on the main sequence. In our models we find that this process is working on the RGB only in stars with initial mass . Moreover we report that thermohaline mixing is also present during core He-burning and beyond, and has the potential to change the surface abundances of AGB stars. While we find rotational and magnetic mixing to be negligible compared to the thermohaline mixing in the relevant layers, the interaction of thermohaline motions with the differential rotation may be essential to establish the timescale of thermohaline mixing in red giants.
Conclusions. To explain the surface abundances observed at the bump in the luminosity function, the speed of the mixing process needs to be more than two orders of magnitude higher than in our models. However it is not clear if thermohaline mixing is the only physical process responsible for these surface-abundance anomalies. Therefore it is not possible at this stage to calibrate the efficiency of thermohaline mixing against the observations.
Key words: stars: abundances - stars: evolution - stars: magnetic field - nuclear reactions, nucleosynthesis, abundances - stars: AGB and post-AGB - stars: rotation
## 1 Introduction
Stars are rotating, self-gravitating balls of hot plasma. Due to thermonuclear reactions in the deep stellar interior, stars, which presumably start out chemically homogeneous, develop chemical inhomogeneities. At the densities typically achieved in stars, thermal diffusion, or Brownian motion, is not able to lead to chemical mixing. However, various turbulent mixing processes are thought to act inside stars, leading to transport of chemical species, heat, angular momentum, and magnetic fields (Heger et al. 2000; Pinsonneault 1997; Heger et al. 2005).
Thermohaline mixing is usually not considered to be an important mixing process in single stars, because the ashes of thermonuclear fusion consist of heavier nuclei than its fuel, and stars usually burn from the inside out. The condition for thermohaline mixing, however, is that the mean molecular weight () decreases inward. This can occur in accreting binaries, and the importance of thermohaline mixing has long been recognized by the binary community (e.g., de Greve & Cugier 1989; Sarna 1992; Wellstein et al. 2001). Recently Charbonnel & Zahn (2007b, CZ07) identified thermohaline mixing as an important mixing process, which significantly modifies the surface composition of red giants after the first dredge-up. The work by CZ07 was initiated by the paper of Eggleton et al. (2006, EDL06), who found an mean molecular weight () inversion - i.e., - below the red giant convective envelope in a 1D-stellar evolution calculation. While EDL06 then investigated the stability of the zone containing the -inversion with a 3D hydro-code and found these layers to be Rayleigh-Taylor-unstable, CZ07 could not confirm this, but found the layers to be unstable due to thermohaline mixing.
Eggleton et al. (2006) found a -inversion in their stellar evolution model, occurring after the so-called luminosity bump on the red giant branch, which is produced after the first dredge-up, when the H-burning shell source enters the chemically homogeneous part of the envelope. The -inversion is produced by the reaction 3He(3He, 2p)4He, as predicted by Ulrich (1972). It does not occur earlier, because the magnitude of the -inversion is small and negligible compared to a stabilizing -stratification.
Mixing processes below the convective envelope in models of low-mass stars turn out to be essential for the prediction of their chemical yield of 3He (EDL06), and are essential to understand the surface abundances of red giants - in particular the 12C/13C ratio, 7Li and the carbon and nitrogen abundances (CZ07). This may also be important for other occurrences of thermohaline mixing in stars, i.e., in single stars when a -inversion is produced by off-center ignition in semi-degenerate cores (Siess 2009) or in stars which accrete chemically enriched matter from a companion in a close binary (e.g., Stancliffe et al. 2007). Accreted metal-rich matter during the phases of planetary formation also leads to thermohaline mixing, which can reconcile the observed metallicity distribution of the central stars of planetary systems (Vauclair 2004).
In the present paper we investigate the evolution of solar metallicity stars between and from the ZAMS up to the thermally-pulsing AGB stage, based on models computed during the last years. We show for which initial mass range and during which evolutionary phase thermohaline mixing occurs and what consequences it has. Besides thermohaline mixing, our models include convection, rotation-induced mixing, and internal magnetic fields, and we compare the significance of these processes in relation to the thermohaline mixing.
## 2 The speed of thermohaline mixing
### 2.1 Thermohaline mixing VS Rayleigh-Taylor
As pointed out by CZ07, a -inversion inside a star should give rise to thermohaline mixing, which is a slow mixing process acting on the local thermal timescale. Could a Rayleigh-Taylor instability be present in these layers? Indeed, EDL06 interpret the origin of the instability which they find in their 3D models as due to the buoyancy produced by the -inversion, i.e. a dynamical effect. But a dynamical instability should only occur if the -inversion were to lead to a density inversion. However, this would only be possible if the considered layers were convectively unstable in the hydrostatic 1D stellar evolution models. As pointed out by CZ07 and as confirmed by our models, the -inversion produced by the 3He(3He, 2p)4He reaction does not induce convection. We conclude that the Rayleigh-Taylor instability may not be a likely explanation of the hydrodynamic motions found by EDL06. A similar conclusion was also reached by Denissenkov & Pinsonneault (2008), who studied in detail the instability driven by the -inversion in both the adiabatic and the radiative limit.
Because smaller blobs have a smaller thermal timescale (Kippenhahn et al. 1980), could EDL06 have found the high wavenumber tale of the thermohaline instability? They found the instability to occur within 2000 s. The size of a blob with such a short thermal timescale above the H-burning shell of a red giant is on the order of 50 km. This is too small to be resolved in the 3D model shown by ELD06. Furthermore, an inspection of their Fig. 5 reveals that the length scale of the instability they found is on the order of 103...104 km, which corresponds to thermal timescales of about 1 yr. Therefore, it seems unlikely that EDL06 actually picked up the thermohaline instability in their 3D hydrodynamic model, unless its non-linear manifestation involves a timescale much shorter than the thermal timescale.
### 2.2 The efficiency of thermohaline mixing
In Sect. 3 we explain the details of our implementation of thermohaline mixing in 1D stellar evolution calculations. The diffusion coefficient for the mixing process contains a parameter which depends on the geometrical configuration of the fluid elements. This parameter ( ) is very important to understand the role of thermohaline mixing. It determines the timescale of the mixing (the velocity of the fingers/blobs) that we show in Sects. 4 and 5 plays a role not only in determining how fast the surface abundances of redgiants can change, but also if thermohaline mixing is present in stars of different mass and at different evolutionary phases.
Charbonnel & Zahn (2007b) showed that a high value of is needed to match the surface abundances of field stars after the luminosity bump (Gratton et al. 2000). Similar to the value adopted by Ulrich (1972) they use an efficiency factor corresponding to in our diffusion coefficient. This value corresponds to the diffusion process involving fingers with an aspect ratio (length/width) of 5. Denissenkov & Pinsonneault (2008) claim that to explain the observed mixing pattern in low-mass RGB stars, fluid elements have to travel over length scales exceeding their diameters by a factor of 10 or more.
On the other hand, the order of magnitude of our efficiency parameter corresponds to the prescription of Kippenhahn et al. (1980), where the diffusion process involves blobs of size L traveling a mean free path L before dissolving. The same prescription has been used by Stancliffe et al. (2007), who dealt with the problem of thermohaline mixing in accreting binaries. The sensitivity of thermohaline mixing to a change of the efficiency parameter is shown in Fig. 1, where the change of the surface abundance of at the luminosity bump is shown for different values of .
That a prescription can reproduce the observed surface abundances may not be sufficient to prefer it over others. It is possible that other mixing processes are at work. The resulting observed abundances could still be mainly due to thermohaline mixing, as proposed by CZ07, but at this stage it is not possible to exclude that magnetic buoyancy (Nordhaus et al. 2008; Busso et al. 2007), or the interaction of different mixing processes (e.g., thermohaline mixing and magnetic buoyancy, Denissenkov et al. 2009), could play an even more important role.
Figure 1: Evolution of the surface abundance of 3He for (blue solid line), (red dotted line) and (green dashed line) from before the onset of thermohaline mixing to the core He-flash in a star. Open with DEXTER
To clarify the picture here we discuss the main differences between the two physical prescriptions for thermohaline mixing. Experiments of thermohaline mixing show slender fingers in the linear regime (e.g., Krishnamurti 2003), supporting the picture of Ulrich (1972). On the other hand the physical conditions inside a star are quite different from those in the laboratory. In particular the Prandtl number , defined as the ratio of the kinematic viscosity to the thermal diffusivity , is very small in stars ( ). This number is about 7 in water, where most of thermohaline mixing experiments have been performed. The question arises if for such small values of a finger-like structure can be stable, especially in layers where shear and horizontal turbulence is present. 2D hydrodynamic simulations of double-diffusive phenomena, without external perturbations have been performed in the past (e.g., Merryfield 1995; Bascoul 2007b,a). While the resolution required by the physical conditions in stellar interiors is computationally not accessible, lowering the Prandtl number to values of about 10-2 always results in increasingly unstable structures (Merryfield 1995; Bascoul 2007a). Therefore it might be dangerous to assume that the same configuration of thermohaline mixing, as observed in water, is occurring in stars.
As we show in Sect. 6, the radiative buffer between the H-burning shell and the convective envelope is a region of the star where the angular velocity is rapidly changing. Even if the order of magnitude of the rotationally-induced instabilities is much lower than the one from thermohaline mixing (cf. Fig. 9), it is possible that the interaction of the shear motions with the thermohaline diffusion prevents a relatively ordered flow to be stable, contrary to Ulrich's assumption. That shear can decrease the efficiency of thermohaline mixing was already pointed out by Canuto (1999). The effect of strong horizontal turbulence in stellar layers has also been discussed by Denissenkov & Pinsonneault (2008), and we conclude in Sect. 6.3 that this effect could work against the fingers and in favor of blobs.
The thermohaline mixing prescription proposed by Kippenhahn et al. (1980) has been used for the calculations presented here. It is clear from the discussion above that a better calibration of the mixing speed requires realistic hydrodynamic calculations of the instability.
## 3 Method
We use a 1D hydrodynamic stellar evolution code (Yoon et al. 2006, and references therein). Mixing is treated as a diffusive process and is implemented by solving the diffusion equation
(1)
where D is the diffusion coefficient constructed from the sum of individual mixing processes and Xn the mass fraction of species n. The second term on the right hand side accounts for nuclear reactions. The contributions to the diffusion coefficient are convection, semiconvection, thermohaline mixing, rotationally induced mixing, and magnetic diffusion. The code includes the effect of centrifugal force on the stellar structure, and the transport of angular momentum is also treated as a diffusive process (Pinsonneault et al. 1989; Endal & Sofia 1978).
The condition for the occurrence of thermohaline mixing is
(2)
i.e. the instability operates in regions that are stable against convection (according to the Ledoux criterion) and where an inversion in the mean molecular weight is present. Here , , , , and . Numerically, we treat thermohaline mixing through a diffusion scheme (Wellstein et al. 2001; Braun 1997). The corresponding diffusion coefficient is based on the work of Stern (1960), Ulrich (1972), and Kippenhahn et al. (1980); it reads
(3)
where is the density, the thermal conductivity, and the specific heat capacity. The quantity is a efficiency parameter for the thermohaline mixing. The value of this parameter depends on the geometry of the fingers arising from the instability and is still a matter of debate (Ulrich 1972; Denissenkov & Pinsonneault 2008; Kippenhahn et al. 1980; Charbonnel & Zahn 2007b). As explained in Sect. 2.2 unless otherwise specified we assume a value for the efficiency of thermohaline mixing. This value roughly corresponds to the prescription of Kippenhahn et al. (1980), in which fluid elements (blobs) travel over length scales comparable to their diameter.
For rotational mixing, four different diffusion coefficients are calculated for dynamical shear, secular shear, Eddington-Sweet circulation and Goldreich-Schubert-Fricke instability. Details on the physics of these instabilities and their implementation in the code can be found in Heger et al. (2000).
Chemical mixing and transport of angular momentum due to magnetic fields (Spruit 2002) is included as in Heger et al. (2005). The contribution of magnetic fields to the mixing is also calculated as a diffusion coefficient ( ), which is added to the total diffusion coefficient D that enters Eq. (1).
We compute evolutionary models of , , and at solar metallicity (Z=0.02). The initial equatorial velocities of these models were chosen to be 10, 45, 140 and 250 (Tassoul 2000); we assume the stars to be rigidly rotating at the zero-age main sequence. Throughout the evolution of all models, the mass-loss rate of Reimers (1975) was used.
## 4 Thermohaline mixing on the giant branch
We compute the stellar models of 1.0, 1.5, 2.0 and 3.0 with solar metallicity. The evolutionary calculations presented are the same as in Suijs et al. (2008), to which we refer for the details of their main sequence evolution.
The surface composition of low-mass stars is substantially changed during the first dredge-up: lithium and carbon abundances as well as the carbon isotopic ratio decline, 3He and nitrogen abundances increase. After the first dredge-up the H-burning shell is advancing while the convective envelope retreats; the shell source then enters the chemically homogeneous part of the envelope. Eggleton et al. (2006) and CZ07 have shown that in this situation an inversion of the molecular weight is created by the reaction 3He(3He, 2p)4He in the outer wing of the H-burning shell in models of 1.0 and 0.9 . This inversion was already predicted by Ulrich (1972).
We confirm an inversion in the mean molecular weight in the outer wing of the H-burning shell. This inversion occurs after the luminosity bump on the red giant branch in the 1.0, 1.5 and 2.0 models. The size of the -inversion depends on the local amount of 3He and in the studied mass range decreases with increasing initial mass. According to Inequality 2 this inversion causes thermohaline mixing in the radiative buffer layer, the radiative region between the H-burning shell and the convective envelope. We emphasize that the extension of the region in which the mixing process is active is not chosen arbitrarily, but is calculated self-consistently by the code. This is done at each time step of the evolutionary calculation by checking which grid points fulfill condition 2. This is a major difference between models including thermohaline mixing and models where the extra mixing is provided by magnetic buoyancy (Denissenkov et al. 2009; Nordhaus et al. 2008; Busso et al. 2007). Indeed for the latter a self-consistent implementation is still not available in 1D stellar evolution codes, and the extension of the extra mixing has to be set arbitrarily.
Figure 2: Evolution of the region between the H-burning shell source and the convective envelope in the RGB phase after the onset of thermohaline mixing for a star. Green hatched regions indicate convection and red cross hatched regions indicate thermohaline mixing, as displayed in the legend. Blue shading shows regions of nuclear energy generation, tracing the H-burning shell. Open with DEXTER
Figure 3: Evolution of the surface abundance profiles of the 12C/13C ratio (dotted red line) and 3He (dashed green line), and of the luminosity (solid blue line) from the onset of thermohaline mixing up to the AGB for a star. Open with DEXTER
In our 1 model thermohaline mixing develops at the luminosity bump and transports chemical species between the H-burning shell and the convective envelope (see Fig. 2). This results in a change of the stellar surface abundances. Figure 3 shows the evolution of the 3He surface-abundance and of the ratio , qualitatively confirming the result of EDL06 and CZ07, namely that thermohaline mixing is depleting 3He and lowering the ratio on the giant branch. As already observed by CZ07, the surface abundance of 16O is not affected because thermohaline mixing does not transport chemical species deep enough the H-burning shell.
Unlike the 1.0 and the 1.5 model, in the 2.0 model thermohaline mixing starts but never connects the H-burning shell to the convective envelope. This is a direct consequence of the lower 3He abundance, which results in smaller -inversion and therefore in a slower thermohaline mixing, according to Eq. (3). It is surprising that thermohaline mixing, once started in the outer wing of the H-burning shell, does not spread through the whole radiative buffer layer. In fact the H-shell burns in a chemically homogeneous region, meaning that no compositional barrier is expected to stop the instability. The reason is that the region unstable to thermohaline mixing moves too slowly in the mass coordinates and never catches-up with the quicker receding envelope. This situation is shown in Fig. 4. As a result no change in the stellar surface composition due to thermohaline mixing is observed during the RGB phase of the 2.0 model.
In our 3.0 model the H-burning shell never penetrates the homogeneous region left by the 1DUP. Accordingly thermohaline mixing does not occur during this phase.
In conclusion our models predict that before the He-core burning thermohaline mixing is able to change surface abundances only in stars with .
During the RGB evolution the choice of , roughly corresponding to the prescription of Kippenhahn et al. (1980) for the thermohaline mixing, allows our stellar models to reach helium ignition without having depleted too much 3He in the envelope. The presence of leftover 3He allows thermohaline mixing to play a role also during a later evolutionary phase, as we show below.
Figure 4: Evolution of the region between the H-burning shell source and the convective envelope in the RGB phase after the onset of thermohaline mixing for a star. Green hatched regions indicate convection and red cross hatched regions indicate thermohaline mixing, as displayed in the legend. Blue shading shows regions of nuclear energy generation. Open with DEXTER
Figure 5: Evolution of the internal structure of a star from the onset of thermohaline mixing to the asymptotic giant branch. Green hatched regions indicate convection, yellow filled regions represent semiconvection and red cross hatched regions indicate thermohaline mixing, as displayed in the legend. Blue shading shows regions of nuclear energy generation. Open with DEXTER
## 5 Thermohaline mixing on the horizontal branch and during the AGB stage
While CZ07 and EDL07 investigate thermohaline mixing only during the RGB, we followed the evolution of our models until the thermally-pulsing AGB stage (TP-AGB). Indeed a -inversion is always created if a H-burning shell is active in a chemically homogeneous layer, the size of the inversion depending on the local abundance of 3He. This happens not only during the RGB, but also during the horizontal branch (HB) and the AGB phase. As a result thermohaline mixing can also operate during these evolutionary phases (Stancliffe 2010; Cantiello & Langer 2008).
Depending on the efficiency of thermohaline mixing during the RGB, the can be exhausted at the end of this phase (e.g. in the models of CZ07). However, stars that avoid extra mixing during the RGB are observed (Charbonnel & Do Nascimento 1998). For these stars the 3He reservoir is intact at He ignition, and thermohaline mixing has the potential to play an important role during the HB and AGB phases. This is confirmed by the evolutionary calculations presented in Sects. 5.1 and 5.2.
### 5.1 Horizontal branch
After the core He-flash, helium is burned in the core, while a H-burning shell is still active below the convective envelope. In our 1 model we found that during this phase thermohaline mixing is present and can spread through the whole radiative buffer layer. This is clear in Fig. 5 where thermohaline mixing (red, cross hatched region) extends from the H-shell to the convective envelope also after ignition of the core He-burning (HB label in the plot). Accordingly surface abundances change during this phase, as shown in Fig. 3. Here a change of surface abundances is also visible after the luminosity peak corresponding to the core He-flash.
Contrary to the 1 model, in our 1.5 and 2.0 models thermohaline mixing does not change the surface abundances during the HB phase. In the 1.5 model the instability succeeds in connecting the H-shell and the convective envelope only at the end of the core He-burning (Fig. 6), while in the 2.0 model this is never achieved (Fig. 7). In the latter case thermohaline diffusion is confined to a tiny layer on top of the H-burning shell, never spreading through the radiative layer (the red cross-hatched region in Fig. 7). This is due to a -barrier, which stops the development of the instability. In Fig. 8 we show the profile of for the 2.0 model at three successive times during core He-burning: the initial peak created by the reaction 3He(3He, 2p)4He gets smaller, while a dip begins to be visible at slightly higher mass coordinate, i.e. at a lower temperature. This -barrier is responsible for stopping the instability; this process is discussed in greater detail in Appendix A.
In the 3.0 model the H-burning shell enters for the first time the chemically homogeneous region after igniting He in the core. However, in this case also thermohaline mixing does not change the surface abundances because is not able to connect the H-burning shell with the convective envelope.
We conclude that in our models thermohaline mixing during the HB changes the surface abundances only in stars with M < .
Figure 6: Evolution of the internal structure of a star from the onset of thermohaline mixing to the AGB phase. Green hatched regions indicate convection, yellow regions represent semiconvection and regions of thermohaline mixing are red cross hatched, as is displayed in the legend. Blue shading shows regions of nuclear energy generation. Open with DEXTER
Figure 7: Evolution of the internal structure of a star from the onset of thermohaline mixing to the AGB phase. Green hatched regions indicate convection, yellow regions represent semiconvection and regions of thermohaline mixing are red cross hatched, as is displayed in the legend. Blue shading shows regions of nuclear energy generation. Open with DEXTER
Figure 8: Profiles of the reciprocal mean molecular weight () in the region above the H-burning shell. The plot shows three successive times in a 2 model during the horizontal branch. The black, continuous line represents the model at t =1.13 109; the green, dashed line shows the same model at t =1.16 109, while the blue, dotted line is the profile at t =1.21 109. Open with DEXTER
### 5.2 Asymptotic giant branch
The subsequent evolutionary phase is characterized by two burning shells and a degenerate core. The star burns H in a shell and the ashes of this process feed an underlying He-burning shell. This is referred to as the asymptotic giant branch (AGB) phase.
During the low-luminosity part of the AGB thermohaline mixing works under the same conditions present in the last part of the HB phase (see Fig. 5, label AGB). In 1.0 and 1.5 models, thermohaline mixing connects the shell source to the envelope. As a consequence surface abundances change, as shown for our 1.0 model in Fig. 3 (label AGB). Similarly to the RGB and HB phases, no thermohaline mixing is present in models with an initial mass higher than .
During the most luminous part of the AGB the He shell periodically experiences thermal pulses (TPs); in stars more massive than these thermal pulses are associated with a deep penetration of the convective envelope, the so-called third dredge-up (3DUP). In our model we find thermohaline mixing to be present also in the TP-AGB. The instability propagates through the thin radiative buffer region (thin'' in mass coordinates), and reaches the convective envelope. This situation is illustrated in Fig. 11. But there thermohaline mixing only leads to negligible changes in the surface abundances. This because of to the very short timescale of this evolutionary stage and because most of the has already been burned in previous evolutionary phases. Overall in our models we found no impact of thermohaline mixing on the surface abundances of and on the ratio during the TP-AGB phase. On the other hand thermohaline mixing can affect the surface abundance of lithium, as we discuss in Sect. 7.
We want to stress here that the presence and impact on surface abundances of thermohaline mixing during the TP-AGB, critically depends on the local abundance and on the value of the efficiency factor . This is because the local abundance is related to the previous history of mixing, which in turn also depends on the efficiency of the diffusion process.
We do not know the correct value of in stellar interiors. Indeed could also depend on stellar parameters such as rotation, metallicity or magnetic fields (see Sect. 6), and it could well be that it changes in the same star through different evolutionary phases. Therefore our predictions for the changes of surface abundances due to thermohaline mixing, especially during the TP-AGB phase, are strongly affected by these uncertainties. Further study is needed to clarify the picture.
## 6 Other mixing processes
### 6.1 Other mixing processes in our models
In our and models we found that in the relevant layers thermohaline mixing has generally higher diffusion coefficients than rotational instabilities and magnetic diffusion. Figure 9 clearly shows that rotational and magnetically induced chemical diffusion is negligible compared to the thermohaline mixing in our model. The only rotational instability acting on a shorter timescale is the dynamical shear instability, visible in Fig. 9 as a spike at the lower boundary of the convective envelope. This instability works on the dynamical timescale in regions of a star where a high degree of differential rotation is present; it sets in if the energy that can be gained from the shear flow becomes comparable to the work which has to be done against the potential for an adiabatic turn-over of a mass element (eddy'') (Heger 1998). However, if present, this instability acts only in a very small region (in mass coordinates) at the bottom of the convective envelope. As a result thermohaline mixing is still setting the timescale for the diffusion of chemical species from the convective envelope to the H-burning shell.
Figure 9: Diffusion coefficients in the region between the H-burning shell and the convective envelope for the model during the HB (t=1.267 1010). The initial equatorial velocity of the model is . The black, continuous line shows convective and thermohaline mixing diffusion coefficients, the green, dashed line is the sum of the diffusion coefficients due to rotational instabilities, while the blue, dot-dashed line shows the magnitude of the magnetic diffusion coefficient. Open with DEXTER
In models of and thermohaline mixing is less efficient due to the lower abundance of 3He. At the same time rotational instabilities and magnetic diffusion have bigger diffusion coefficients, mainly because these models have initial equatorial velocities of 140 and 250 respectively. Figure 10 shows how during core He-burning rotational mixing and magnetic diffusion become more important than thermohaline mixing in the model. The radiative buffer layer is dominated by the Eddington-Sweet circulation, dynamical shear, and magnetic diffusion. Yet the rotational mixing diffusion coefficient is still too small to allow the surface abundances to change appreciably in this phase, in agreement with results from Palacios et al. (2006). The same conclusion is valid for the magnetic diffusion, which has the same order of magnitude as the rotational diffusion in the radiative buffer layer. Our models are calculated with the Kippenhahn et al. (1980) prescription for thermohaline mixing, which implies a smaller diffusion coefficient with respect to that proposed by Ulrich (1972). As a consequence the result that thermohaline mixing has in general a higher impact than rotational mixing and magnetic diffusion in the relevant layers is valid regardless of which of the two prescriptions was chosen.
### 6.2 Critical model ingredients
The results described above are obtained with a particular model for rotational mixing and angular momentum transport, for which several assumptions need to be made. Here we discuss the two most important assumptions in the present context. The first assumption is that angular momentum transport in convection zones can be described with a diffusion approximation and a diffusion coefficient derived from the mixing length theory (Sect. 3). The result is near rigid rotation in convection zones. Recent 3D hydrodynamic models of rotating red giant convective zones (Steffen & Freytag 2007; Brun & Palacios 2009) indicate that the picture may be more complex. Brun & Palacios (2009) find radial profiles intermediate between constant angular momentum and constant angular velocity, with a large dependence on the rotation rate of the star. While no general conclusion can easily be drawn from these studies, we may wonder how a reduced angular momentum transport efficiency in the convective envelope might affect our results. While detailed models would be required to exhaustively answer this question, we can expect that a more rapidly rotating base of the convective envelope would lead to less shear, and would thus render shear mixing in the layers below the envelope less relevant.
A second crucial assumption is the adoption of magnetic angular momentum transport according to Spruit (2002). Even though the Spruit-Taylor dynamo has been criticized (Zahn et al. 2007; Denissenkov & Pinsonneault 2007), an effecient angular momentum transport mechanism like the Spruit-Taylor dynamo is clearly needed to understand the observed slow rotation of stellar remnants (Suijs et al. 2008; Heger et al. 2005). While angular momentum transport through gravity waves has been advocated as an interesting alternative (Talon & Charbonnel 2008,2003), it remains to be demonstrated that this mechanism can work to break the rotation of the helium core in the post-main sequence stages of stellar evolution. Therefore, while the reader should be aware of the related uncertainties, using the Spruit-Taylor dynamo at this time appears reasonable.
Figure 10: Diffusion coefficients in the region between the H burning shell and the convective envelope for the model during core He-burning (t=1.124 109). The initial equatorial velocity of the model is . The black, continuous line shows convective and thermohaline mixing diffusion coefficients, the green, dashed line is the sum of the diffusion coefficients due to rotational instabilities while the blue, dot-dashed line shows the magnitude of magnetic diffusion coefficient. Open with DEXTER
### 6.3 Interaction of instabilities
The discussion of the interactions of thermohaline motions with the rotational instabilities and magnetic fields is complex. In this respect Canuto (1999) argues that shear due to differential rotation decreases the efficiency of thermohaline mixing. Not only Denissenkov & Pinsonneault (2008) claim that rotation-induced horizontal turbulent diffusion may suppress thermohaline mixing. This is because horizontal diffusion (molecular plus turbulent) may change the mean molecular weight of the fluid element during its motion. They argue that this horizontal diffusion is able to halt thermohaline mixing. We think this argument is correct in an ideal situation, in which a single blob of material is crossing an infinite, parallel slab. Yet in a star the horizontal turbulence is acting on a shell, which can be locally approximated to a parallel slab with periodic boundary conditions in the horizontal direction. This horizontal layer (shell) is rapidly homogenized by the horizontal turbulence. Fingers trying to cross this horizontal layer are quickly disrupted and mixed. This results in a rapid increase of the mean molecular weight in the shell, so that the region will become unstable to thermohaline mixing. A new generation of fingers is therefore expected. But the presence of horizontal turbulence is probably making fingers an unlikely geometrical configuration: blobs that travel a small distance before the turbulence is mixing them on a horizontal layer are more likely. This way thermohaline mixing is not stopped, but only slowed down. This scenario would favor the Kippenhahn et al. (1980) prescription, which actually predicts blobs traveling a distance comparable to their size.
Figure 11: Evolution of the region between the H-burning shell source and the convective envelope during a thermal pulse in a star. Green hatched regions indicate convection, and regions of thermohaline mixing are red-cross hatched, as displayed in the legend. Blue shading shows regions of nuclear energy generation. This model is evolved from the zero-age main sequence to the TP-AGB with . Open with DEXTER
Another interesting idea has been proposed by Charbonnel & Zahn (2007a). They claim that internal magnetic fields can play a stabilizing role, trying to counteract the destabilizing effect of the inverse gradient. Their conclusion is that thermohaline mixing can be inhibited by a magnetic field stronger than 104-105 Gauss. But they warn that their analysis ignores both stellar rotation and the spatial variation of B, which results in neglecting any possible instability of the magnetic field itself (e.g., Spruit 1999).
The instability of magnetic fields below the convective envelope of RGB and AGB stars has been discussed by Busso et al. (2007) and Nordhaus et al. (2008). They argue that dynamo-produced buoyant magnetic fields could provide the source of extra mixing in these stars.
## 7 Lithium-rich giants
Lithium is a fragile element, which is destroyed at temperatures higher than about 3 106 K. For this reason it is expected that lithium should decrease from its initial value during the evolution of stars. On the other hand, observations have shown that about of giants show strong Li lines (e.g., Brown et al. 1989; Wallerstein & Sneden 1982). Some of these stars even show surface Li-abundances higher than the interstellar values.
For intermediate mass stars a possible solution was proposed by Cameron & Fowler (1971), who showed how a net production of 7Li can be achieved during hot-bottom burning (HBB). During HBB the convective envelope penetrates into the H-shell burning, where 7Be is produced by the pp-chain. In this situation the unstable isotope 7Be can be transported to cooler temperatures by the convective motions, decaying into 7Li in regions of the envelope where the temperature is low enough for lithium to survive. This results in Li-enrichment at the surface.
At solar metallicity stars below do not experience hot-bottom burning (Forestini & Charbonnel 1997), whereas at Z=0 hot-bottom burning is found down to (Siess et al. 2002). For stars avoiding hot-bottom burning, some other mechanism is needed in order to increase the Li surface-abundance. A possibility is that some kind of extra mixing connects the H-burning shell and the convective envelope, which in the literature is often referred to as the cool bottom process (CBP). The work of Charbonnel & Balachandran (2000) supports this hypothesis. Indeed they found Li-rich stars to be either red giants at the luminosity bump or early-AGB stars before the second dredge-up, in agreement with the idea that some internal mixing occurs when the H-burning shell enters a homogeneous region. A lithium production during the RGB evolution is also supported by the recent observations of Gonzalez et al. (2009), who find lithium enriched red giants at the luminosity bump or at higher luminosities.
Uttenthaler et al. (2007) reported the detection of low-mass, Li-rich AGB stars in the galactic bulge. Interestingly two of the four stars which show surface-Li enhancement present no evidence for third dredge-up, and thermohaline mixing is advocated as a possible source for the extra mixing.
In our calculations we found that the Li surface-abundance is affected by thermohaline mixing during the evolution of low-mass stars. While Li is burned during the RGB and HB, thermohaline mixing has the potential to enhance the Li surface-abundance during the TP-AGB phase. To show this, we computed stellar evolution calculations of the TP-AGB phase in 1 and 3 with different values of . An example of the evolution of the Li surface-abundance in the 3 model during one thermal pulse is shown in Fig. 12. Our models qualitatively confirm that this instability can enhance the surface Li abundances in low-mass AGB stars, even if we can not quantitatively reproduce the high level of enrichment observed by Uttenthaler et al. (2007). To reach the values of Uttenthaler et al. (2007) a value of much higher than those proposed by Kippenhahn et al. (1980) and Ulrich (1972) is needed. As discussed in Sect. 5.2, a quantitative study requires a better knowledge of the efficiency parameter for thermohaline mixing .
The observations of Uttenthaler et al. (2007) show that only 4 out of 27 galactic bulge stars are Li-enriched. If thermohaline mixing is the physical process providing the high Li-enrichment observed, we still have to understand why only 15% of the sample show this strong enhancement. One possibility is that these stars did not experience thermohaline mixing in previous evolutionary phases. This would leave the reservoir intact, leading to a much more efficient mixing during the TP-AGB phase.
Figure 12: Evolution of Li surface-abundance during one thermal pulse in a 3 model. The black, continuous line shows a model evolved with = 1000; the blue, dotted line refers to the same model evolved with a thermohaline mixing efficiency = 200. In both cases the model experiences third dredge-up. The evolution of the star prior to the TP-AGB has been calculated with = 2. Open with DEXTER
This scenario requires a way to prevent the extra mixing during the RGB and HB phases. Charbonnel & Zahn (2007a) have proposed that strong magnetic fields stop thermohaline mixing in those red giants stars that are the descendants of Ap stars. They call these stars thermohaline deviant stars''.
Because the fraction of Ap stars relative to A stars (5-10%), the number of red giants that seem to avoid the extra mixing () and the observed fraction of Li-enriched AGB stars (15%) are similar, it may be possible that we are looking at the same group of stars at different evolutionary stages. If this is the case, it remains to be understood why the process that inhibits the mixing during the RGB and HB phases is not at work during the AGB.
A further complication arises from the observations of Drake et al. (2002), showing that the incidence of Li-rich giants is much higher among fast-rotating objects. They consider single-K giants and find that among rapid rotators ( ) a very large proportion ( ) is Li-rich, in contrast with a very low proportion () of Li-rich stars among the much more common slowly rotating giants. Thermohaline mixing is not driven by rotational energy, and if any effect would be expected, it would be a lower efficiency of the mixing with increasing shear and horizontal turbulence (Denissenkov & Pinsonneault 2008; Canuto 1999). On the other hand, an increase in the mixing efficiency with the rotation rate is expected if the physical mechanism behind the extra mixing is magnetic buoyancy (Denissenkov et al. 2009; Nordhaus et al. 2008; Busso et al. 2007). In this case rotation is necessary to amplify the magnetic field below the convective envelope.
Another possibility is that lithium has an external origin, resulting from accretion and ingestion of planets or a brown dwarf by an expanding red giant (e.g., Siess & Livio 1999b,a). Mass transfer or wind accretion in a binary system is also a possible scenario.
The far-IR excess, which is observed in all fast rotating, Li enriched giants, is another interesting piece of the puzzle (Drake et al. 2002; Reddy & Lambert 2005). While models in which some kind of accretion process occurs could explain the IR excess, the internal production of lithium cannot reproduce these observations (but see Palacios et al. 2001). We refer to Drake et al. (2002) for an accurate review of the proposed mechanism for the formation of Li-rich giants.
## 8 Conclusion
We qualitatively confirm the results of CL07: thermohaline mixing in low-mass giants is capable of destroying large quantities of 3He, as well as decreasing the ratio . Thermohaline mixing indeed starts when the H-burning shell source moves into the chemically homogeneous layers established by the first dredge-up. At solar metallicity we find that this process is working only in stars with a mass below . This result is sensitive to the choice of the parameter, which regulates the speed of thermohaline mixing.
Our models show further that thermohaline mixing remains important during core He-burning and can also operate on the AGB - including the thermally-pulsing AGB stage. Depending on the efficiency of the mixing process, this can result in considerable lithium enrichment.
Our calculations show that in the relevant layers thermohaline mixing generally has a higher diffusion coefficient than rotational instabilities and magnetic diffusion. However, we cannot address the interaction of thermohaline motions with differential rotation and magnetic fields, for which hydrodynamic calculations are required.
In stellar evolution codes thermohaline mixing is implemented as a diffusive process. This process acts on a thermal timescale, but the exact velocity of the motion depends on a parameter . This parameter is related to the geometry of the fingers (or blobs) displacing the stellar material and is still a matter of debate. The two widely used prescriptions have a parameter that differs by two orders of magnitude. We used the Kippenhahn et al. (1980) prescription, even though we also investigated the effect of using different values of in a few calculations. Charbonnel & Zahn (2007b) used a much more efficient thermohaline mixing (Ulrich 1972), justifying their choice on the basis of laboratory experiments of thermohaline mixing performed in water, and on the observations of surface abundances of red giants.
But the physical conditions inside a star are very different from these laboratory experiments, which clearly cannot be used for a quantitative study of this hydrodynamic instability. Moreover it is not clear if thermohaline mixing is the only physical process responsible for the extra mixing, and therefore it is not possible to calibrate its efficiency against the observations.
We argue that is not possible at this stage to firmly identify thermohaline mixing as the cause of the observed surface abundances in low-mass giants (Gratton et al. 2000). In particular the long standing problem cannot be considered as solved.
In agreement with CZ07 we claim that to clarify the picture it would be desirable to have realistic hydrodynamic simulations of thermohaline mixing.
Acknowledgements
M.C. thanks Steve N. Shore, Maria Lugaro, Onno Pols, Evert Glebbeek, Selma de Mink, Jonathan Braithwaite, Miro Mocák and John Lattanzio for helpful discussions. M.C. acknowledges support from the International Astronomical Union and from the Leids Kerkhoven-Bosscha Fonds.
## Appendix A: how to stop thermohaline mixing
For models of we found two situations in which thermohaline mixing fails to connect the H-burning shell with the convective envelope:
1.
The envelope is receding in mass coordinates and the thermohaline mixing is not fast enough to catch-up. This situation is shown in Fig. 4.
2.
A chain of reactions rising the mean molecular weight can create a barrier that stops the mixing process.
The second scenario occurs depending on the efficiency of the reactions rising the molecular weight in the outer wing of the H-burning shell. The reactions responsible for creating this compositional barrier are 3He (4He, 7Be (e-7Li (1H, 4He) 4He.
In the first one 3He and 4He produce 7Be that rapidly decays into 7Li. The lithium easily reacts with a proton producing two -particles; this way the initial molecular weight of 8/9 rises to the value 4/3. This occurs in the outer wing of the H-burning shell, at a lower temperature with respect to the region where the mean molecular weight inversion discussed by EDL06 and CZ07 is present. As a consequence thermohaline mixing is halted (see Figs. 7 and 8).
Therefore two sets of reactions play a role in the evolution of thermohaline mixing in the region between the H-burning shell and the convective envelope:
(A.1)
(A.2)
As a consequence, the efficiency in starting and stopping thermohaline mixing is regulated by the local abundance of 3He and 4He. The first reaction in the chain A.2 has a lower rate than reaction A.1 at the temperatures in the region of interest. However, 3He (4He, 7Be depends linearly on the local abundance of 3He and 4He, while reaction A.1 depends quadratically on the abundance of 3He. Increasing the initial mass of the stellar model, the ratio 3He/4He decreases in the radiative buffer layer (e.g., Boothroyd & Sackmann 1999). Consequently if one increases the initial mass, the set of reactions A.2 at some point becomes more important than A.1, meaning that a compositional barrier is created. This explains why we observe a threshold in mass at about , above which thermohaline mixing is not efficient anymore.
Note that both scenarios 1. and 2. depend on , i.e. on the speed of the mixing process. Increasing the value of makes thermohaline mixing more difficult to be stopped; therefore the value of the mass threshold depends on the choice of .
## References
1. Bascoul, G. P. 2007a, Ph.D. Thesis [Google Scholar]
2. Bascoul, G. P. 2007b, in IAU Symp., ed. T. Kuroda, H. Sugama, R. Kanno, & M. Okamoto, 239, 317 [Google Scholar]
3. Boothroyd, A. I., & Sackmann, I.-J. 1999, ApJ, 510, 232 [Google Scholar]
4. Braun, H. 1997, Ph.D. Thesis, Ludwig-Maximilians-Univ. München [Google Scholar]
5. Brown, J. A., Sneden, C., Lambert, D. L., & Dutchover, E. J. 1989, ApJS, 71, 293 [CrossRef] [Google Scholar]
6. Brun, A. S., & Palacios, A. 2009, ApJ, 702, 1078 [NASA ADS] [CrossRef] [Google Scholar]
7. Busso, M., Wasserburg, G. J., Nollett, K. M., & Calandra, A. 2007, ApJ, 671, 802 [NASA ADS] [CrossRef] [Google Scholar]
8. Cameron, A. G. W., & Fowler, W. A. 1971, ApJ, 164, 111 [Google Scholar]
9. Cantiello, M., & Langer, N. 2008, in IAU Symp., ed. L. Deng, & K. L. Chan, 252, 103 [Google Scholar]
10. Canuto, V. M. 1999, ApJ, 524, 311 [NASA ADS] [CrossRef] [Google Scholar]
11. Charbonnel, C., & Balachandran, S. C. 2000, A&A, 359, 563 [NASA ADS] [Google Scholar]
12. Charbonnel, C., & Do Nascimento, Jr., J. D. 1998, A&A, 336, 915 [NASA ADS] [Google Scholar]
13. Charbonnel, C., & Zahn, J.-P. 2007a, A&A, 476, L29 [NASA ADS] [CrossRef] [EDP Sciences] [Google Scholar]
14. Charbonnel, C., & Zahn, J.-P. 2007b, A&A, 467, L15 [NASA ADS] [CrossRef] [EDP Sciences] [Google Scholar]
15. de Greve, J. P., & Cugier, H. 1989, A&A, 211, 356 [NASA ADS] [Google Scholar]
16. Denissenkov, P. A., & Pinsonneault, M. 2007, ApJ, 655, 1157 [NASA ADS] [CrossRef] [Google Scholar]
17. Denissenkov, P. A., & Pinsonneault, M. 2008, ApJ, 684, 626 [NASA ADS] [CrossRef] [Google Scholar]
18. Denissenkov, P. A., Pinsonneault, M., & MacGregor, K. B. 2009, ApJ, 696, 1823 [NASA ADS] [CrossRef] [Google Scholar]
19. Drake, N. A., de la Reza, R., da Silva, L., & Lambert, D. L. 2002, AJ, 123, 2703 [NASA ADS] [CrossRef] [Google Scholar]
20. Eggleton, P. P., Dearborn, D. S. P., & Lattanzio, J. C. 2006, Science, 314, 1580 [NASA ADS] [CrossRef] [PubMed] [Google Scholar]
21. Endal, A. S., & Sofia, S. 1978, ApJ, 220, 279 [NASA ADS] [CrossRef] [Google Scholar]
22. Forestini, M., & Charbonnel, C. 1997, A&AS, 123, 241 [NASA ADS] [CrossRef] [EDP Sciences] [Google Scholar]
23. Gonzalez, O. A., Zoccali, M., Monaco, L., et al. 2009, A&A, 508, 289 [NASA ADS] [CrossRef] [EDP Sciences] [Google Scholar]
24. Gratton, R. G., Sneden, C., Carretta, E., & Bragaglia, A. 2000, A&A, 354, 169 [NASA ADS] [Google Scholar]
25. Heger, A. 1998, Ph.D. Thesis [Google Scholar]
26. Heger, A., Langer, N., & Woosley, S. E. 2000, ApJ, 528, 368 [NASA ADS] [CrossRef] [Google Scholar]
27. Heger, A., Woosley, S. E., & Spruit, H. C. 2005, ApJ, 626, 350 [NASA ADS] [CrossRef] [Google Scholar]
28. Kippenhahn, R., Ruschenplatt, G., & Thomas, H.-C. 1980, A&A, 91, 175 [NASA ADS] [Google Scholar]
29. Krishnamurti, R. 2003, J. Fluid Mech., 483, 287 [NASA ADS] [CrossRef] [Google Scholar]
30. Merryfield, W. J. 1995, ApJ, 444, 318 [NASA ADS] [CrossRef] [Google Scholar]
31. Nordhaus, J., Busso, M., Wasserburg, G. J., Blackman, E. G., & Palmerini, S. 2008, ApJ, 684, L29 [NASA ADS] [CrossRef] [Google Scholar]
32. Palacios, A., Charbonnel, C., & Forestini, M. 2001, A&A, 375, L9 [NASA ADS] [CrossRef] [EDP Sciences] [Google Scholar]
33. Palacios, A., Charbonnel, C., Talon, S., & Siess, L. 2006, A&A, 453, 261 [NASA ADS] [CrossRef] [EDP Sciences] [Google Scholar]
34. Pinsonneault, M. 1997, ARA&A, 35, 557 [Google Scholar]
35. Pinsonneault, M. H., Kawaler, S. D., Sofia, S., & Demarque, P. 1989, ApJ, 338, 424 [NASA ADS] [CrossRef] [Google Scholar]
36. Reddy, B. E., & Lambert, D. L. 2005, AJ, 129, 2831 [Google Scholar]
37. Reimers, D. 1975, Mem. Soc. R. Sci. de Liege, 8, 369 [Google Scholar]
38. Sarna, M. J. 1992, MNRAS, 259, 17 [NASA ADS] [CrossRef] [Google Scholar]
39. Siess, L. 2009, A&A, 497, 463 [NASA ADS] [CrossRef] [EDP Sciences] [Google Scholar]
40. Siess, L., & Livio, M. 1999a, MNRAS, 304, 925 [NASA ADS] [CrossRef] [Google Scholar]
41. Siess, L., & Livio, M. 1999b, MNRAS, 308, 1133 [NASA ADS] [CrossRef] [Google Scholar]
42. Siess, L., Livio, M., & Lattanzio, J. 2002, ApJ, 570, 329 [NASA ADS] [CrossRef] [Google Scholar]
43. Spruit, H. C. 1999, A&A, 349, 189 [NASA ADS] [Google Scholar]
44. Spruit, H. C. 2002, A&A, 381, 923 [NASA ADS] [CrossRef] [EDP Sciences] [Google Scholar]
45. Stancliffe, R. J. 2010, MNRAS, 403, 505 [NASA ADS] [CrossRef] [Google Scholar]
46. Stancliffe, R. J., Glebbeek, E., Izzard, R. G., & Pols, O. R. 2007, A&A, 464, L57 [NASA ADS] [CrossRef] [EDP Sciences] [Google Scholar]
47. Steffen, M., & Freytag, B. 2007, Astron. Nachr., 328, 1054 [NASA ADS] [CrossRef] [Google Scholar]
48. Stern, M. E. 1960, Tellus, 12, 172 [NASA ADS] [CrossRef] [Google Scholar]
49. Suijs, M. P. L., Langer, N., Poelarends, A.-J., et al. 2008, A&A, 481, L87 [NASA ADS] [CrossRef] [EDP Sciences] [Google Scholar]
50. Talon, S., & Charbonnel, C. 2003, A&A, 405, 1025 [NASA ADS] [CrossRef] [EDP Sciences] [Google Scholar]
51. Talon, S., & Charbonnel, C. 2008, A&A, 482, 597 [NASA ADS] [CrossRef] [EDP Sciences] [Google Scholar]
52. Tassoul, J.-L. 2000, Stellar Rotation (Cambridge University Press) [Google Scholar]
53. Ulrich, R. K. 1972, ApJ, 172, 165 [NASA ADS] [CrossRef] [Google Scholar]
54. Uttenthaler, S., Lebzelter, T., Palmerini, S., et al. 2007, A&A, 471, L41 [NASA ADS] [CrossRef] [EDP Sciences] [Google Scholar]
55. Vauclair, S. 2004, ApJ, 605, 874 [NASA ADS] [CrossRef] [Google Scholar]
56. Wallerstein, G., & Sneden, C. 1982, ApJ, 255, 577 [NASA ADS] [CrossRef] [Google Scholar]
57. Wellstein, S., Langer, N., & Braun, H. 2001, A&A, 369, 939 [NASA ADS] [CrossRef] [EDP Sciences] [Google Scholar]
58. Yoon, S.-C., Langer, N., & Norman, C. 2006, A&A, 460, 199 [NASA ADS] [CrossRef] [EDP Sciences] [Google Scholar]
59. Zahn, J., Brun, A. S., & Mathis, S. 2007, A&A, 474, 145 [NASA ADS] [CrossRef] [EDP Sciences] [Google Scholar]
## Footnotes
... mass
During the main sequence of these stars, the pp chain operates partially burning hydrogen to 3He, but not beyond, into a wide zone outside the main energy-producing region. At the end of the core H-burning the first dredge-up mixes this 3He with the stellar envelope. Because the main sequence lifetime is longer for lower mass stars, these are able to produce bigger amounts of 3He.
## All Figures
Figure 1: Evolution of the surface abundance of 3He for (blue solid line), (red dotted line) and (green dashed line) from before the onset of thermohaline mixing to the core He-flash in a star. Open with DEXTER In the text
Figure 2: Evolution of the region between the H-burning shell source and the convective envelope in the RGB phase after the onset of thermohaline mixing for a star. Green hatched regions indicate convection and red cross hatched regions indicate thermohaline mixing, as displayed in the legend. Blue shading shows regions of nuclear energy generation, tracing the H-burning shell. Open with DEXTER In the text
Figure 3: Evolution of the surface abundance profiles of the 12C/13C ratio (dotted red line) and 3He (dashed green line), and of the luminosity (solid blue line) from the onset of thermohaline mixing up to the AGB for a star. Open with DEXTER In the text
Figure 4: Evolution of the region between the H-burning shell source and the convective envelope in the RGB phase after the onset of thermohaline mixing for a star. Green hatched regions indicate convection and red cross hatched regions indicate thermohaline mixing, as displayed in the legend. Blue shading shows regions of nuclear energy generation. Open with DEXTER In the text
Figure 5: Evolution of the internal structure of a star from the onset of thermohaline mixing to the asymptotic giant branch. Green hatched regions indicate convection, yellow filled regions represent semiconvection and red cross hatched regions indicate thermohaline mixing, as displayed in the legend. Blue shading shows regions of nuclear energy generation. Open with DEXTER In the text
Figure 6: Evolution of the internal structure of a star from the onset of thermohaline mixing to the AGB phase. Green hatched regions indicate convection, yellow regions represent semiconvection and regions of thermohaline mixing are red cross hatched, as is displayed in the legend. Blue shading shows regions of nuclear energy generation. Open with DEXTER In the text
Figure 7: Evolution of the internal structure of a star from the onset of thermohaline mixing to the AGB phase. Green hatched regions indicate convection, yellow regions represent semiconvection and regions of thermohaline mixing are red cross hatched, as is displayed in the legend. Blue shading shows regions of nuclear energy generation. Open with DEXTER In the text
Figure 8: Profiles of the reciprocal mean molecular weight () in the region above the H-burning shell. The plot shows three successive times in a 2 model during the horizontal branch. The black, continuous line represents the model at t =1.13 109; the green, dashed line shows the same model at t =1.16 109, while the blue, dotted line is the profile at t =1.21 109. Open with DEXTER In the text
Figure 9: Diffusion coefficients in the region between the H-burning shell and the convective envelope for the model during the HB (t=1.267 1010). The initial equatorial velocity of the model is . The black, continuous line shows convective and thermohaline mixing diffusion coefficients, the green, dashed line is the sum of the diffusion coefficients due to rotational instabilities, while the blue, dot-dashed line shows the magnitude of the magnetic diffusion coefficient. Open with DEXTER In the text
Figure 10: Diffusion coefficients in the region between the H burning shell and the convective envelope for the model during core He-burning (t=1.124 109). The initial equatorial velocity of the model is . The black, continuous line shows convective and thermohaline mixing diffusion coefficients, the green, dashed line is the sum of the diffusion coefficients due to rotational instabilities while the blue, dot-dashed line shows the magnitude of magnetic diffusion coefficient. Open with DEXTER In the text
Figure 11: Evolution of the region between the H-burning shell source and the convective envelope during a thermal pulse in a star. Green hatched regions indicate convection, and regions of thermohaline mixing are red-cross hatched, as displayed in the legend. Blue shading shows regions of nuclear energy generation. This model is evolved from the zero-age main sequence to the TP-AGB with . Open with DEXTER In the text
Figure 12: Evolution of Li surface-abundance during one thermal pulse in a 3 model. The black, continuous line shows a model evolved with = 1000; the blue, dotted line refers to the same model evolved with a thermohaline mixing efficiency = 200. In both cases the model experiences third dredge-up. The evolution of the star prior to the TP-AGB has been calculated with = 2. Open with DEXTER In the text
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https://www.physicsforums.com/threads/semantics-question-about-this-angular-momentum-problem.277769/ | 1. Dec 7, 2008
### FisherDude
1. The problem statement, all variables and given/known data
Heading straight toward the summit of Pikes Peak, an airplane of mass 12,000 kg flies over the plains of Kansas at nearly constant altitude 4.30 km with constant velocity 175 m/s west. a) What is the airplane's vector angular momentum relative to a wheat farmer on the ground directly below the airplane? b) Does this value change as the airplane continues its motion along a straight line?
2. Relevant equations
mag. of angular momentum = position*mass*velocity*angle between
3. The attempt at a solution
The answer to part a) is (-9.03 x 10^9 kg x m^2/s) j, which I have no problem with. But the answer to part b) is "No, L = |r||p|sin(Θ) = mv(rsinΘ), and r*sinΘ is the altitude of the plane. Therefore, L = constant as the plane moves in level flight with constant velocity."
But the problem asks for the plane's angular momentum relative to the wheat farmer. So if the plane keeps on moving west, wouldn't r (the distance between the farmer and the plane) keep on increasing?
The only way the answer makes sense to me is if they're really asking for the angular momentum of the plane relative to the ground, not the wheat farmer, because then, the distance between the plane and the ground would be constant.
Any help would be great...
Last edited: Dec 7, 2008
2. Dec 7, 2008
### Staff: Mentor
Sure, the distance r from farmer to plane increases. But r*sinΘ does not. Angular momentum is not just mv*r, but mvr*sinΘ. (Only when the plane is directly overhead does sinΘ = 1.)
Note that r*sinΘ is the distance between plane and ground.
3. Dec 7, 2008
### flatmaster
You're correct, L remains constant even with the farmer. You realized that r*sin(theta) is simply the altitute, which is constant. You're correct that r will increase because the plane-farmer distance increases. However, sin(theta) will decrease at such a rate to keep L constant. Remember what the angle theta is defined as. It's the angle between the vector r and the vector v. You were allowed to ignore this in your origional calculation because theta = 90 Sin(theta) = 1
4. Dec 7, 2008
### FisherDude
Wow, I wish i had caught that.
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https://nalinkpithwa.com/2015/02/19/analysis-real-variables-chapter-1-examples-ii/ | # Analysis — Real Variables — Chapter 1 — Examples II
Examples II.
1) Show that no rational number can have its cube equal to 2.
Proof #1.
Let, if possible, $p/q, q \neq 0$, and p and q do not have any common factor and are integers. Then, if $(p/q)^{3}=2$, we have
$p^{3}=2q^{3}$. So, p contains a factor of 2. So, let $p=2k$. So, q contains a factor of 2. Hence, both p and q have a common factor have a common factor 2, contradictory to out assumption. Hence, the proof.
2) Prove generally that a rational function $p/q$ in its lowest terms cannot be the cube of a rational number unless p and q are both perfect cubes.
Proof #2.
Let, if possible, $p/q = (m/n)^{3}$ where m,n,p,q are integers, with n and q non-zero and p and q are in lowest terms. This implies that m and n have no common factor. Hence, $p=m^{3}, q=n^{3}$.
3) A more general proposition, which is due to Gauss and includes those which precede as particular cases, is the following: an algebraical equation
$z^{n}+p_{1}z^{n-1}+p_{2}z^{n-2}+ \ldots + p_{n}=0$ with integral coefficients, cannot have rational, but non-integral root.
Proof #3.
For suppose that, the equation has a root $a/b$, where a and b are integers without a common factor, and b is positive. Writing
$a/b$ for z, and multiplying by $b^{n-1}$, we obtain
$-(a^{n}/b)=p_{1}a^{n-1}+p_{2}a^{n-2}b+ \ldots + p_{n}b^{n-1}$,
a function in its lowest terms equal to an integer, which is absurd. Thus, $b=1$, and the root is a. It is evident that a must be a divisor of $p_{n}$.
4) Show that if $p_{n}=1$ and neither of
$1+p_{1}+p_{2}+p_{3}+\ldots$ and $1-p_{1}+p_{2}-p_{3}+\ldots$
is zero, then the equation cannot have a rational root.
Proof #4. Please try this and send me a solution.. I do not have a solution yet 🙂
5) Find the rational roots, if any, of $x^{4}-4x^{3}-8x^{2}+13x+10=0$.
Solution #5.
Use problem #3.
The roots can only be integral, and so find the roots by trial and error. It is clear that we can in this way determine the rational roots of any such equation.
More later,
Nalin Pithwa
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https://mathoverflow.net/users/3365/alexod | # alexod
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5 (Un)Decidability of the root existence problem for functions with bounded domain 3 About the well ordering of finite sequences of numbers 2 Does the Feferman-Schutte analysis give a precise characterization of Predicative Second-Order Arithmetic?
# 360 Reputation
+35 (Un)Decidability of the root existence problem for functions with bounded domain +5 Applications of infinite Ramsey's Theorem (on N)? +20 Does the Feferman-Schutte analysis give a precise characterization of Predicative Second-Order Arithmetic? +5 Partitions of central sets via dynamical systems
# 3 Questions
15 Applications of infinite Ramsey's Theorem (on N)? 9 Sequence that converge if they have an accumulation point 1 Partitions of central sets via dynamical systems
# 14 Tags
10 lo.logic × 3 2 foundations 5 reverse-math × 2 0 examples 5 decidability 0 applications 2 peano-arithmetic 0 ca.analysis-and-odes 2 constructive-mathematics 0 co.combinatorics
# 4 Accounts
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https://www.efgs.info/a-scalable-analytical-framework-for-spatio-temporal-analysis-of-neighbourhood-change-a-sequence-analysis-approach/ | Select Page
# A Scalable Analytical Framework for Spatio-Temporal Analysis of Neighbourhood Change: A Sequence Analysis Approach
## Nikos Patias (University of Liverpool)
Spatio-temporal changes reflect the complexity of real-life events. Changes in the spatial distribution of population and consumer demand at urban and rural areas are expected to trigger changes in future housing and infrastructure needs. This study presents a scalable analytical framework for understanding spatio-temporal population change, using a sequence analysis approach. We use gridded cell Census data for Great Britain from 1971 to 2011 with 10-year intervals, creating neighbourhood typologies for each Census year. These typologies are then used to analyse transitions of grid cells between different types of neighbourhoods and define representative trajectories of neighbourhood change. The results reveal seven prevalent trajectories of neighbourhood change across Great Britain, identifying neighbourhoods which have experienced stable, upward and downward pathways through the national socioeconomic hierarchy over the last four decades. | {"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": 0.8064286112785339, "perplexity": 3261.9007734266065}, "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/1593657176116.96/warc/CC-MAIN-20200715230447-20200716020447-00329.warc.gz"} |
http://math.stackexchange.com/questions/241006/is-there-a-closed-form-equation-for-fibonaccin-modulo-m | # Is there a closed form equation for fibonacci(n) modulo m?
Basically I am curious if there's a direct way to calculate fibonacci(n) modulo m with a closed form formula so I don't have to bother with matrix exponentials.
-
This is not yet a complete solution; just a heuristical approach. I'm considering the cyclicity of residues modulo some m, here only for $m \in \mathbb P$ . After this we need only the sequences along one cycle.
I give a couple of empricial results, organized such that they shall give a hint how to proceed:
$$\small \begin{array} {lll} &m & \operatorname{fibcyclelength}(m) & \text{ general rule,estimated }\\ \hline \\ \hline \\ p=2& 2,4,8,16,32,... &3,6,12,24,48,...& a=(2+1),a_k=a \cdot 2^{k-1} \\ p=5& 5,25,125,625,...&20,100,500,... & a=5(5-1),a_k=a \cdot 5^{k-1}\\ \hline \\ p=\pm 2 \pmod 5 & 3,9,27,81,... &8,24,72,216,... &a=2(3+1),a_k=a \cdot 3^{k-1} \\ & 13,169,... &28,364,...&a=2(13+1),a_k=a \cdot 13^{k-1} \\ & 23,23^2 &48,&a=2(23+1), a_k=a \cdot 23^{k-1}\\ \hline \\ & 7,49,343 &16,112,...&a=2(7+1),a_k=a \cdot 7^{k-1} \\ & 17,289 &36,612,...&a=2(17+1),a_k=a \cdot 17^{k-1} \\ \hline \\ p= \pm 1 \pmod 5& 11,121, &10,110, & a=(11-1),a_k=a \cdot 11^{k-1} \\ & 31, &30,...&a=(31-1), a_k=a \cdot 31^{k-1} \\ \hline \\ & 19,... &18,...&a=(19-1), a_k=a \cdot 19^{k-1} \\ \end{array}$$ I'm pretty confident that this is generalizable in the obvious way. Thus -if that above scheme is correct- we can compute the required results for that m without matrix-exponentiation after we have the entries over one cycle only..
Next step is then to look at the squarefree n with two primefactors, and their powers. Their sequences seem to begin with some irregularity, but finally seem to become regular in the above way. I've looked at n=6, n=10, n=15 so far, and if I get some idea of the characteristic of the irregularity I'll post it here.
-
I think this is old news, but it is straightforward to say what I know about this, in terms which I think there is some chance of addressing the intent of the question.
That is, as hinted-at by the question, the recursion $\pmatrix{F_{n+1}\cr F_n}=\pmatrix{1 & 1 \cr 1 & 0}\cdot \pmatrix{F_n\cr F_{n-1}}$ can be usefully dissected by thinking about eigenvectors and eigenvalues. Namely, the minimal (also characteristic) equation is $(x-1)x-1=0$, which has roots more-or-less the golden ratio. Thus, doing easy computations which I'm too lazy/tired to do on this day at this time, $F_n=A\cdot a^n + B\cdot b^n$ for some constants $a,b,A,B$. These constants are algebraic numbers, lying in the field extension of $\mathbb Q$ obtained by adjoining the "golden ratio"... This expression might seem not to make sense mod $m$, but, perhaps excepting $m$ divisible by $2$ or $5$, the finite field $\mathbb Z/p$ allows sense to be made of algebraic extensions, even with denominators dividing $2$ or $5$, the salient trouble-makers here.
So, except possibly for $m$ divisible by $2$ or $5$, the characteristic-zero formula for the $n$-th Fibonacci number makes sense.
The further question, raised in the other answer, of the precise period mod $m$ is probably as intractable (currently) as questions about primitive roots... Indeed. But, perhaps, the given question is not that hard...?
-
Keep on truckin', bro... – TonyK Nov 20 '12 at 20:10
Mod 19 there are two solutions to $x^2=x+1$, the generating function for Fibonacci numbers. These are $x=5, x=15$. On a whim (based on some of the above comments), I thought I would try these to get a closed form for fibonacci numbers mod 19. I tried $$f(n) = \frac{15^n-5^n}{15-5},$$ where the computation is to be done mod 19. And the formula worked, at least as far as I tried it.
(I'm just including this answer because it surprised me; perhaps the other responders above already know such a thing has to work.)
ADDED: The above seems to work fine for any prime for which $x^2=x+1$ has two solutions. I got that this is primes for which 5 is a quadratic residue. (it didn't work for $p=5$ since the only solution there is $x=3$ and then cannot imitate the above formula.)
Also I found that it seems to work even for $209 = 11 \cdot 19$, where there are four solutions, namely $15,81,129,195$; however one needs to choose a pair so that their difference is coprime to 209. Thus for mod 209 I tried the formula $$f(n)=\frac{129^n-81^n}{129-81}$$ and that seems to work to give fibonacci numbers mod 209.
- | {"extraction_info": {"found_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": 0.9651786684989929, "perplexity": 274.7966519446223}, "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/1469257824995.51/warc/CC-MAIN-20160723071024-00182-ip-10-185-27-174.ec2.internal.warc.gz"} |
https://www.physicsforums.com/threads/thermodynamics-homework.58834/ | # Thermodynamics homework
1. Jan 6, 2005
Hello all
Given
(a) N2(g) + O2(g) --> 2NO(g) ∆H = + 180. 7 kJ
(b) 2NO(g) +O2(g) --> 2NO2(g) ∆H = -113.1 kJ
(c) 2N20(g) --> 2N2(g) --> O2(g) ∆H = -163.2 kJ
Desired Reaction: N2O (g) + NO2(g) --> 3NO (g)
Can someone please tell me where to start? How do I apply Hess's Law? I know I have to work backwards. However, how do I do this in a systematic way?
Any help is appreciated
Thanks
Last edited: Jan 6, 2005
2. Jan 6, 2005
### apchemstudent
please rewrite the equations. Just look at B), for instance, it's not even balanced... and C makes no sense since how can you get O2 from 2 N2...
3. Jan 6, 2005
### dextercioby
A good place would be to state the equations correctly:
$$N_{2}+O_{2}\rightarrow 2NO$$ $$(\Delta H)_{1}=+180.7 kJ$$
$$2NO+O_{2}\rightarrow 2NO_{2}$$ $$(\Delta H)_{2}=-113.1 kJ$$
$$2NO_{2}\rightarrow N_{2}+2O_{2}$$ $$(\Delta H)_{3}=-67.6 kJ$$
Now,which is the quantity u wanna compute and what's the reaction u wish to get??
Daniel.
4. Jan 6, 2005
I want to compute ∆H for
N2O (g) + NO2(g) --> 3NO (g)
5. Jan 6, 2005
### apchemstudent
I think C should be 2 N2O(g) -> 2N2(g) + O2(g)...
Balance the reactions so that you will get rid of the unwanted reactants or products.... For instance, N2 + O2 -> 2NO
you you might want to multiply the reaction by 2 since N2 is not wanted and there's an N2 in reaction C of the product. Im assuming you know they will cancel out when added together...
Last edited: Jan 6, 2005
6. Jan 7, 2005
yes but is there an actual method to solve the problem? Or do you have to just guess and check? | {"extraction_info": {"found_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": 0.8409082889556885, "perplexity": 4116.998791753363}, "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/1476988720154.20/warc/CC-MAIN-20161020183840-00500-ip-10-171-6-4.ec2.internal.warc.gz"} |
https://www.bas.ac.uk/data/our-data/publication/on-the-conditional-frazil-ice-instability-in-seawater/ | # On the conditional frazil ice instability in seawater
It has been suggested that the presence of frazil ice can lead to a conditional instability in seawater. Any frazil forming in the water column reduces the bulk density of a parcel of frazil-seawater mixture, causing it to rise. Due to the pressure-decrease in the freezing point, this causes more frazil to form, causing the parcel to accelerate, and so on. We use linear stability analysis and a non-hydrostatic ocean model to study this instability. We find that frazil ice growth caused by the rising of supercooled water is indeed able to generate a buoyancy-driven instability. Even in a gravitationally stable water column, the frazil ice mechanism can still generate convection. The instability does not operate in the presence of strong density stratification, high thermal driving (warm water), a small initial perturbation, high background mixing or the prevalence of large frazil ice crystals. In an unstable water column the instability is not necessarily expressed in frazil ice at all times; an initial frazil perturbation may melt and refreeze. Given a large enough initial perturbation this instability can allow significant ice growth. A model shows frazil ice growth in an Ice Shelf Water plume several kilometres from an ice shelf, under similar conditions to observations of frazil ice growth under sea ice. The presence of this instability could be a factor affecting the growth of sea ice near ice shelves, with implications for Antarctic bottom water formation.
### Details
Publication status:
Published
Author(s):
Authors: Jordan, James R., Kimura, Satoshi, Holland, Paul R., Jenkins, Adrian, Piggott, Matthew D.
On this site: Adrian Jenkins, James Jordan, Paul Holland, Satoshi Kimura
Date:
1 April, 2015
Journal/Source:
Journal of Physical Oceanography / 45
Page(s):
1121-1138
Digital Object Identifier (DOI):
https://doi.org/10.1175/JPO-D-14-0159.1 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8628138899803162, "perplexity": 4772.822178503831}, "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-2019-22/segments/1558232262311.80/warc/CC-MAIN-20190527085702-20190527111702-00334.warc.gz"} |
https://byjus.com/statistical-significance-formula/ | # Statistical Significance Formula
Statistical hypothesis testing is the a result that is attained when a p – value is lesser than the significance level, denoted by , alpha. p – value is the probability of getting at least as extreme results that is provided that the null hypothesis is true. Statistical significance is the mean to get sure that the statistic is reliable.
If there is a large sample size, then small difference in the research findings can be negligible if you are very sure that the differences did not arise out of fluke. This formula helps us determine that there is a relationship in the differences or variations. Depending upon the sample size, to know how moderate, weak or strong is the relationship, statistical significance is used.
Statistical significance is also referred to as type 1 error. The formula and terminologies related to this formula is given as:
$\large Z=\frac{\overline{x}-\mu}{\frac{\sigma}{\sqrt{n}}}$
### Solved Examples
Question: Find out the statistical significance using the z test if the sample mean is 16, is $\mu$ 12, σ is 4 and the sample size is 30?
Solution
Given parameters are
$\overline{x}=15$
$\mu =12$
$\sigma =4$
$n = 30$
With the formula we can say that: $Z=\frac{\overline{x}-\mu}{\frac{\sigma}{\sqrt{n}}}$
$Z=\frac{15-12}{\frac{5}{\sqrt{30}}}$
$=\frac{3}{0.73}$
$=4.10$ | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9734819531440735, "perplexity": 406.7799080118017}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-47/segments/1573496669755.17/warc/CC-MAIN-20191118104047-20191118132047-00550.warc.gz"} |
https://www.baryudin.com/blog/entry/django-and-wagtail-math-formulas/ | Since moving my blog to Django and Wagtail, I have been adding new features to it on a regular basis. The most recent addition is the new ability to typeset mathematical equations and formulas. To be fair, all the heavy lifting is courtesy of the MathJax project. These guys did an amazing job - they implemented, in JavaScript, support for LaTeX syntax. Quite astonishing, if you ask me. Very easy to use, assuming you know LaTeX. Works in any modern browser, including on mobile, i.e. Android or iPhone. See for yourself:
$$\LaTeX \, \text{equations for the masses!}$$ \begin{align*} ax^2 + bx + c= 0 & \quad\text{Remember the good old school days?} \\ E=mc^2 & \quad\text{Looks familiar, doesn't it?} \\ \int_e^\pi\! \cos{\sqrt[3.2]{\log_{2\pi}x_1}} \,\mathrm{d}x_1 & \quad \text{Something a bit fancier} \end{align*}
More complicated layouts can be typeset as well, of course, all free of charge (other than the CPU cycles of the viewer's computer):
$$\left\langle \begin{matrix} 1 & 2 & 3\\ a & b & c \\ t^x & t^y & t^z \\ \sqrt{-1} & \sqrt{-7\times e} & \sqrt{-20/\pi} \end{matrix} \right\rangle$$
I remember, that some time ago displaying neatly set mathematical equations in a browser seemed like an impossible dream to achieve. There were next to zero built in support for math, and alternative solutions meant using pictures most of the time. How quickly, and, occasionally, positively, things change nowadays!
By the way, this is how math looks like in a mobile browser:
Categories: None | | {"extraction_info": {"found_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": 0.9649307727813721, "perplexity": 2765.0511512506573}, "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/1669446711126.30/warc/CC-MAIN-20221207021130-20221207051130-00774.warc.gz"} |
https://chemistry.stackexchange.com/questions/86841/is-the-haber-process-here-proceeding-at-positive-gibbs-free-energy-change | # Is the Haber Process here proceeding at positive Gibbs free energy change?
$\ce{3H2 + N2 -> 2NH3}$ is the forward reaction used in the Haber process, the industrial production of ammonia. After studying Gibbs energy and how it is just another way of saying that the 2nd law of thermodynamics is obeyed if $\Delta G < 0$, I decided to apply it to some common reactions that I know, just to confirm that my understanding of the topic is correct.
When applying it to the Haber process forward reaction, I worked out change in entropy to be $\pu{-199 J K^{-1} mol^{-1}}$. The enthalpy change for the forward reaction is exothermic and has a value of $\pu{-92 kJ mol^{-1}}$. Looking up common reaction conditions for the Haber process, I found that a compromise temperature of $\pu{650K}$ is used.
Applying this to the Gibbs energy equation $-92 - 650(-199/1000)$, I calculated a positive change in Gibbs energy for the reaction of which had a value of $\pu{37.35 kJ mol^{-1}}$.
As a result of this, I am now quite confused as to how the forward reaction can be allowed to proceed by the 2nd law of thermodynamics at a temperature of $\pu{650K}$, and I was wondering where my understanding of the Gibbs energy and thermodynamics falls short. I need help to explain the pitfalls in my calculations, understanding or even whether there is another factor that I am not considering.
• This source has it as -32.7 kJ/mole N2: surfguppy.com/thermodynamics/…
– Gert
Dec 4, 2017 at 20:49
• Did you adjust the values for $\Delta H$ for the higher temperature?
– Gert
Dec 4, 2017 at 20:58
• @Gert That source uses 298K as the temperature which results in change in Gibbs energy being negative. However in industry, a higher compromise temperature of 650K is used for the forward reaction to achieve higher rates of reaction. At this higher compromise temperature, change in Gibbs energy is positive. I assumed that the value for ΔH does not change with temperature. I know that it does but I thought it would only be very negligible. chemguide.co.uk/physical/equilibria/haber.html This source takes ΔH = -92kJmol<sup>-1</sup> and states conditions used are 673K-723K. Dec 4, 2017 at 21:00
• You need to do the same thing for the $\Delta S$. Dec 4, 2017 at 21:29
• Just because the change of standard free energy for a reaction is positive, that does not mean that the reaction will not go at all. It just means that the equilibrium conversion will be low. Dec 5, 2017 at 13:05
It's all about the pressure! The Haber process requires a pressure of about 200 atm. This is how the process becomes favorable, since LeChatelie's principle says that increasing pressure shifts toward the side with less moles of gas.
As long as Q is smaller than K, the reaction will go forward. For example, if nitrogen and hydrogen are supplied at stoichiometric ratio and 1% of a given reaction mixture is product, we would have the following partial pressures:
$$P_\ce{N2} = 0.2475 \times \pu{200 atm}$$
$$P_\ce{H2} = 0.7425 \times \pu{200 atm}$$
$$P_\ce{NH3} = 0.01 \times \pu{200 atm}$$
Accordingly, Q would be
$$Q = \frac{2^2}{50 \times 75^3} = \pu{1.9e-7}$$
In the Haber process, the concentration of product is kept small by continuously removing product (liquifying ammonia at low temperature) and feeding the remaining reactants back into the reaction mixture.
• This is nice, could you expand on your explanation? Apr 13, 2018 at 17:49
The tendency of a reaction to proceed in a certain direction depends on the state of the concentration of each reactant. If you calculate the standard Gibbs free energy of the reaction and it's a positive number, that only means that - if the composition of the reaction medium corresponds to all components at standard concentration ($$1 \ \textrm{mol/L}$$) - then the reaction will shift in the reverse direction (to which you supposed when you calculated $$\Delta G$$).
I am not sure though if this is the case for the Haber Process as you presented.
It must be noted that often one will increase the temperature of a reaction in a way that is non-intuitive from a thermodynamic perspective: one might conclude that the reaction is being made less spontaneous in the desired reaction. However, in a practical setting, the kinetics are also important, that is, how fast the reaction proceeds towards equilibrium. A temperature increase will always increase the reaction rate, which may get us to an acceptable yield earlier, even if the maximum achievable yield (dictated by thermodynamics) will be compromised.
the ΔH and ΔS don't change too much on temperature, but they change on pressure! when the pressure is 200 atm, the ΔH and ΔS are changed and the calculated ΔG is negative! I couldn't find the equation for ΔH and ΔS to P. But I check the results from Aspen Plus directly. | {"extraction_info": {"found_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": 6, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.837773859500885, "perplexity": 541.3036121329145}, "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/1664030335058.80/warc/CC-MAIN-20220927194248-20220927224248-00469.warc.gz"} |
https://science.sciencemag.org/content/329/5991/542?ijkey=d0d70aea9945169a777a8be001ae72aa2f866ee7&keytype2=tf_ipsecsha | Report
# Single-Shot Readout of a Single Nuclear Spin
See allHide authors and affiliations
Science 30 Jul 2010:
Vol. 329, Issue 5991, pp. 542-544
DOI: 10.1126/science.1189075
## Abstract
Projective measurement of single electron and nuclear spins has evolved from a gedanken experiment to a problem relevant for applications in atomic-scale technologies like quantum computing. Although several approaches allow for detection of a spin of single atoms and molecules, multiple repetitions of the experiment that are usually required for achieving a detectable signal obscure the intrinsic quantum nature of the spin’s behavior. We demonstrated single-shot, projective measurement of a single nuclear spin in diamond using a quantum nondemolition measurement scheme, which allows real-time observation of an individual nuclear spin’s state in a room-temperature solid. Such an ideal measurement is crucial for realization of, for example, quantum error correction protocols in a quantum register.
Since the birth of quantum computing, researchers have sought scalable room-temperature systems that could be incorporated as quantum coprocessors. Much enthusiasm arose when room-temperature nuclear magnetic resonance (NMR) quantum computers were developed (1). However, these are essentially classical as they lack the ability to initialize and read out individual spins at room temperature (2). Recent efforts have focused on the development of ultracold quantum processors like trapped ions and superconducting qubits which operate at millikelvin temperatures (3). Electronic and nuclear spins associated with nitrogen-vacancy (NV) centers in diamond have been shown to be a room-temperature solid-state system with exceptionally long coherence times that fulfills most of the requirements needed to build a quantum computer (47). However, it lacked single-shot readout (8), and hence only the cryogenic version was considered to be applicable for most quantum information applications. For example, projective readout enables testing Bell-type inequalities and active feedback in quantum error correction protocols. Here, we experimentally showed single-shot readout of a single nuclear spin in diamond. Our technique is based on the repetitive readout of nuclear spins (9) and the essential decoupling of the nuclear from the electronic spin dynamics by means of a strong magnetic field (10).
The fluorescence time trace of a single NV center shown in Fig. 1B represents the real-time dynamics of a single nuclear spin and exhibits well-defined jumps attributed to abrupt, discontinuous evolution of the nuclear spin state (quantum jumps). The spin used in our experiments belongs to the nucleus of the nitrogen atom [14N; nuclear spin I = 1 (11)] of a single NV defect in diamond (Fig. 1B). In essence, the measurement sequence consists of a correlation of the electron spin state of the NV color center with the nuclear spin state and a subsequent optical readout of the electron spin, which exhibits the nuclear spin state. Therefore, initially the electron spin is optically pumped into the electron spin sublevel |0e (mS = 0) of its triplet ground state (S = 1) (8), leaving the nuclear spin in an incoherent mixture of its eigenstates (|mI=)|1n, |0n, and |1n (here and below states are defined according to electron and nuclear magnetic quantum numbers, mS and mI). The application of a narrowband, nuclear spin state–selective microwave (MW) π pulse flips the electron spin into the |1e state conditional on the state of the nuclear spin. This operation is equivalent to a controlled not (CNOT) operation (Fig. 1A), in that it maps a specific nuclear spin state onto the electron spin (e.g., |1n|0e|1n|1e, |0n|0e|0n|0e). This is possible because of the long coherence time of the NV center, providing a spectral linewidth of the electron spin transitions narrow enough to resolve the hyperfine structure. Because the fluorescence intensity differs by roughly a factor of 2 for electron spin states |0e and |1e (8, 12), these target states can be distinguished by shining a short laser pulse. This destroys the electron spin state but leaves the nuclear spin state population almost undisturbed under the experimental conditions. Thus, repeated application of this scheme allows nondestructive accumulation of fluorescence signal in order to determine the nuclear spin state optically.
The fidelity F to detect a given state in a single shot [reaching F = 92 ± 2% in our experiments (13)] can be extracted from the photon-counting histograms (Fig. 2A), which show distinguishable peaks corresponding to different nuclear spin states. The fidelity is limited by the measurement time (bounded by relaxation time of the nuclear spin), fluorescence count rate, and magnetic resonance signal contrast. Further improvement in readout speed can be achieved by engineering of photon emission into photonic nanostructures (14). A consecutive measurement of the same spin state gives an identical result with a probability of (F2) of ≈82.5% (Fig. 2C). Such a correlation between consecutive measurements is the signature of so-called quantum nondemolition (QND) protocols (15). For the nitrogen nuclear spin qubit initially in a superposition of two states, the measurement affects its state by projection into one of the eigenstates, but does not demolish it (as happens with photons arriving at a photomultiplier tube or fluorescent atoms that are shelved in a dark state, which is not a qubit state). Hence, the same nuclear spin eigenstate can be redetected in consecutive measurements.
The difference between projective measurement and a practical QND has been analyzed in detail (16, 17) and can be summarized as three conditions that must be simultaneously fulfilled in order to have a true QND measurement. Our system observable is the nuclear spin I^z, our probe observable is the electron spin S^z, and their Hamiltonians are Hn and He, respectively (13). The interaction Hamiltonian Hi for our case is separable Hi = HA + Hp, where HA describes the hyperfine interaction and Hp represents the MW field applied in the experiment.
The first condition for QND is simply that the probe observable S^z must be measurably influenced by the system observable I^z that we desire to measure. Therefore, the interaction Hamiltonian Hi has to depend on Iz and must not commute with the probe observable S^z ([S^z,Hi]0) (16, 17). These demands are met by the CNOT gate. The corresponding HamiltonianHp=Ωexp(iωt)S^x|1n1n| acts for a time τ and flips the electron spin by an angle β = Ωτ only for the nuclear spin |1n subspace (Ω, Rabi frequency; ω, MW frequency). The strength of the QND measurement can by tuned by preparing the electron spin in a superposition state rather than in an eigenstate before the action of Hp (18).
The second QND condition requires that the system observable state Iz be stable with respect to back action of the measurement. This translates to the requirement that the system Hamiltonian must not be a function of the observable’s conjugate (I^x or I^y) in order to avoid back action of the measurement, which imposes a large uncertainty on the conjugates. In our case, this condition is fulfilled as long as the applied magnetic field is exactly parallel to the NV center symmetry axis (13).
The third condition is that the probe and system observables, S^z and I^zin our case, should not be mixed by any interactions that are neither intrinsic to the material nor created by the action of the MW or laser probes (i.e., that the nuclear spin is well isolated from the environment). In other words (16, 17), the interaction Hamiltonian must commute with the observable ([I^z,Hi]=0). Fulfilling this condition perfectly is an impossible task for any experimental system, particularly in the solid state. However, defect center spins in diamond are very close to an ideal system for QND measurements. In the case of the NV center, the nuclear spin–selective MW pulse on the electron spin does not act on the nuclear spin subspace (hence[I^z,Hp]=0). However, the hyperfine coupling tensor A¯¯ contains contributions parallel and perpendicular to the symmetry axis of the NV center (A and A), and the perpendicular component is responsible for an undesirable mixing. The first term of the hyperfine Hamiltonian HA=(S^+I^+S^I^+)A/2+S^zI^zA is noncommuting with I^z and therefore induces nuclear-electron spin flip-flop processes. This mixing is responsible for the quantum jumps in Fig. 1B. The key to succeeding at QND measurements is therefore to make this jump time longer than the measurement time.
To quantify the hyperfine induced flip-flop rate, assume an isotropic case (AAA) and use the measured A= 40 MHz in the excited state (19, 20). Electron-nuclear spin dynamics occur on a time scale of 2/A ~ 50 ns in the vicinity of excited-state level anticrossing at magnetic field B = 50 mT (19, 21) (Fig. 3A). Relaxation in the ground state is expected to be slower owing to a much weaker hyperfine coupling (13) and can be neglected here. The relaxation process slows down when the magnetic field along the NV symmetry axis is increased owing to the growing energy mismatch between electron and nuclear spin transitions due to increasing Zeeman shifts (Fig. 3A). A detailed analysis (13) and experimental data (Fig. 3B) show that the relaxation rate γ depends on the detuning δ from the level anticrossing (1.42 GHz) as γ(A2/2)/[(A2/2)+δ2] (i.e., like a Lorentzian lineshape). Hence, we expect a quadratic dependence of T1 on the detuning from the excited-state level anticrossing (T1=1/γδ2 for δA2). Experimental data confirm this behavior (Fig. 3B). This dependence also explains why quantum jumps were not observed in previous experiments with NV centers performed at low magnetic fields [similar magnetic field–enabled decoupling of nuclear spin was proposed recently for alkaline earth metal ions (10, 22)]. The dominance of flip-flop processes is also visible in the quantum state trajectory of the nuclear spin shown in Fig. 3C (top). Here, jumps obey the selection rule ΔmI=±1 imposed by the flip-flop term HA. From analyzing the whole quantum state trajectory, a matrix showing the transition probabilities can be obtained (Fig. 3C, bottom).
Single-shot measurement of a single nuclear spin places diamond among leading quantum computer technologies. The high readout fidelity (92%) demonstrated in this work is already close to the threshold for enabling error correction (23), although the experiments were carried out in a moderate-strength magnetic field. Even though the optical excitation induces complex dynamics in the NV center (including passage into singlet electronic state), the nuclear spin relaxation rates are defined solely by electron-nuclear flip-flop processes induced by hyperfine interaction. Therefore, we expect improvement of T1 by two orders of magnitude (reaching seconds under illumination) when a magnetic field of 5 T is used. This will potentially allow readout fidelities comparable with that achieved for single ions in traps (24). The present technique can be applied to multiqubit quantum registers (5, 6, 25), enabling tests of nonclassical correlations. Finally, single-shot measurements open new perspectives for solid-state sensing technologies. Spins in diamond are considered to be among the promising candidates for nanoscale magnetic field sensing (26, 27). Currently their performance is limited by photon shot noise (26): “Digital” QND will provide improvement over conventional photon counting in the case of short acquisition time. This requires that the electron spin state used for magnetic field sensing can be mapped onto the nuclear spin with high accuracy, but this was already shown to be practical in NV diamond (5).
## Supporting Online Material
www.sciencemag.org/cgi/content/full/science.1189075/DC1
Methods
SOM Text
Figs. S1 to S6
References
## References and Notes
1. The presented single-shot readout works in the same way and shows a similar fidelity for the nuclear spin of the 15N isotope.
2. Supporting material is available on Science Online.
3. We thank F. Dolde for fabrication of microwave structures; N. Zarrabi for assistance with data analysis; J. Mayer and P. Bertet for helpful information on QND measurements in superconducting qubits; and M. D. Lukin, J. Twamley, F. Y. Khalili, and J. O’Brien for comments and discussions. We thank G. Denninger for the loan of a X-band microwave synthesizer. This work was supported by the European Union, Deutsche Forschungsgemeinschaft (SFB/TR21 and FOR1482), Bundesministerium für Bildung und Forschung, and Landesstiftung BW.
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http://wfqr.4-motion.fr/poiseuille-flow-formula.html | # Poiseuille Flow Formula
For this example, we resolve the plane poiseuille flow problem we previously solved in Post 878 with the builtin solver bvp5c, and in Post 1036 by the shooting method. Poiseuille's equation only applies to liquid flow at low pressure, in relatively short tubes with relatively narrow radii. GUYON(**) and P. Cite this Article: S. , RBCs drift to the center so velocity profile flattens from ideal parabolic). Learning Objectives: By the end of this lecture, students will be able to be able to 1) Employ Navier-Stokes equations to solve general fluid mechanics problems, such as, general velocity profile, calculate volumetric flow rate of poiseuille flow between parallel plates or in an Annular Die (important for blow molding). Poiseuille’s Law Combo With Khan Academy May 9, 2014 aarongtrips 3 Comments While watching a Khan Academy video on blood pressure and flow, I came across this equation,. Couette-Poisseuille Flow Couette-Poiseuille flow is a steady, one-dimensional flow between two plates with constant gap; the flow is along the plates or along the xˆ direction. Poiseuille Flow Poiseuille law describes laminar flow of a Newtonian fluid in a round tube (case 1). In addition, D and [u. A laboratory experiment on inferring Poiseuille's law for undergraduate students 1085 Figure 1. cz ABSTRACT The flow of non-Newtonian fluids through an annulus is. His equation is the basis for measurement of viscosity hence his nam e has been used for the unit of viscosity. The rate of flow (v) of liquid through a horizontal pipe for steady flow is given by. The greater the pressure differential between two points, the greater the. The stability of Poiseuille flow in a pipe of circular cross-section to azimuthally varying as well as axisymmetric disturbances has been studied. We report on a study of heat flow in bulk black phos-phorus between 0. Because Couette and Poiseuille flow types are independent solutions of the dynamic balance equation for viscous flow, they can occur together within Earth's asthenosphere. The flow can be pressure or viscosity driven, or a combination of both. The fluid flow will be turbulent for velocities and pipe diameters above a threshold, leading to larger pressure drops than would be expected according to the Hagen-Poiseuille equation. Poiseuille flow. Experimental set-up. Poiseuille's Law - Pressure Difference, Volume Flow Rate, Fluid Power Physics Problems - Duration: 17:21. 1 3D Poiseuille Flow Over the next two lectures we will be going over the stabilization of the 3-D Poiseuille flow. In 1838 he experimentally derived, and in 1840 and 1846 formulated and published, Poiseuille's law (now commonly known as the Hagen-Poiseuille equation, crediting Gotthilf Hagen as well), which applies to laminar flow, that is, non-turbulent flow of liquids through pipes of uniform section, such as blood flow in capillaries and veins. Does anybody know where this formula comes from? Are there exact solutions of the Navier-Stokes equation for compressible Poiseuille flow?. Turbulent Poiseuille flow with near‐critical wall transpiration Turbulent Poiseuille flow with near‐critical wall transpiration Vigdorovich, Igor; Oberlack, Martin 2010-12-01 00:00:00 An incompressible, pressure‐driven, fully developed turbulent flow between two parallel walls, with an extra constant transverse velocity component, is considered. In some situations, such as that of water flowing in a riverbed, calculating A is difficult, and the best you can do is an approximation. In nonideal fluid dynamics, the Hagen-Poiseuille equation, also known as the Hagen-Poiseuille law, Poiseuille law or Poiseuille equation, is a physical law that gives the pressure drop in an incompressible and Newtonian fluid in laminar flow flowing through a long cylindrical pipe of constant cross section. Does anybody know where this formula comes from? Are there exact solutions of the Navier-Stokes equation for compressible Poiseuille flow?. Navier-Stokes Equations {2d case NSE (A) Equation analysis Equation analysis Equation analysis Equation analysis Equation analysis Laminar ow between plates (A) Flow dwno inclined plane (A) Tips (A) The NSE are Non-linear { terms involving u x @ u x @ x Partial di erential equations { u x, p functions of x , y , t 2nd order { highest order. Jean Leonard Marie Poiseuille (I 797-1869). al(2012) studied steady MHD Poiseuille flow between two infinite parallel porous plates in an inclined magnetic field and discover that high magnetic field strength decreases the velocity. The poiseuille's equation is: V = π * R 4 * ΔP / (8η * L). See Hagen-Poiseuille Equation. References. In this paper we considered one dimensional poiseuille flow of an electrically conducting fluid between. Ansarib, A. Poiseuille's law applies to laminar flow of an incompressible fluid of viscosity η through a tube of length l and radius r. * and David J. In addition to its merit of a closed system in the spanwise (azimuthal) direction, the aPf may be an ideal flow system to understand canonical wall-bounded shear flows in a comprehensive manner based on the radius ratio, denoted by η = r i / r o (where r i and r o are the inner and outer radii. First, an internal restriction is created. Burton, Physiology and Biophysics of the Circulation, Yearbook, 1965. This allows us to investigate the influence of the external force on the non-Newtonian properties of the Couette flow. Poiseuille's equation. Poiseuille formula equation for the laminar flow regime in the hydraulic pipess. In others, such as that of a fluid flowing in a closed pipe, it's. Use k as the constant of proportionality. Relation of Hagen-Poiseuille the volumic flowrate can be calculated thanks to the Hagen-Poiseuille equation. We will derive Poiseuille law for a Newtonian fluid and leave the flow of a power-law fluid as an assignment. 1 3D Poiseuille Flow Over the next two lectures we will be going over the stabilization of the 3-D Poiseuille flow. Polarization. Schwalbe,*a Frederick R. Poiseuille Flow Up: Incompressible Viscous Flow Previous: Flow Between Parallel Plates Flow Down an Inclined Plane Consider steady, two-dimensional, viscous flow down a plane that is inclined at an angle to the horizontal. Poiseuille’s law applies to laminar flow of an incompressible fluid of viscosity through a tube of length and radius. The greater the pressure differential between two points, the greater the flow. Flow rate Q is in the direction from high to low pressure. The steady planar Poiseuille flow generated by a constant external force is analyzed in the context of the nonlinear Bhatnagar-Gross-Krook kinetic equation for a gas of Maxwell molecules. For laminar flow, resistance is quite low. Poiseuille flow is pressure-induced flow ( Channel Flow) in a long duct, usually a pipe. Abstract At low Reynolds numbers for which the flow through a jet viscometer orifice strictly obeys the Poiseuille equation, the effective hydrodynamic length L 0 which may be calculated from the volume flow rate, the applied pressure difference, the radius of the orifice, and the density and low rate of shear viscosity of the liquid is much larger than the length L of 'constant diameter' of. 029, and pressure difference 4000 dynes/cm{eq}^2 {/eq}. Hagen Poiseuille Equation Derivation Pdf 12 -- DOWNLOAD (Mirror #1) 3b9d4819c4 20120903 P620 13C Lec 09 (Work) Mod2 PtrPhy 03 Perm Dev. The flow rate formula, in general, is Q = A × v, where Q is the flow rate, A is the cross-sectional area at a point in the path of the flow and v is the velocity of the liquid at that point. It can be successfully applied to air flow in lung alveoli, for the flow through a drinking straw or through a hypodermic needle. However, some limitations of these models have motivated. As the diagram shows, and as the formula has stated, Poiseuille's law relates the flow rate with the pressure, viscosity, vessel radius and length. The method is applied to the stability of plane Poiseuille flow; it is found that the critical Reynolds number is 5772. It is an analytical equation that applies regardless of. Note that for a given pressure gradient and distance from the center of the flow, the pipe flow, equation 8, predicts lower velocities than the flow in the gap between parallel plates, equation 4. So it is not correct to use the Poiseuille’s equation for flow in the porous media. Poisson's partial differential equation Saint-Venant solution was used, to calculate Poiseuille number values whatever is rectangles aspect ratio. The flow rate of an incompressible fluid undergoing laminar flow* in a cylindrical tube can be expressed in Poiseuille’s equation. presenting viscosities between about 0. Poiseuille developed an equation relating to the volume V of an out flowed water during the time t in a horizontal tube to the inside tube radius r (m), the fluid viscosity η, the length L of the tube and the fluid pressure p difference between the ends of capillary. dV/dt = constant, where dV/dt is volume flow rate. Some of the fundamental solutions for fully developed viscous flow are shown next. In nonideal fluid dynamics, the Hagen–Poiseuille equation, also known as the Hagen–Poiseuille law, Poiseuille law or Poiseuille equation, is a physical law that gives the pressure drop in an incompressible and Newtonian fluid in laminar flow flowing through a long cylindrical pipe of constant cross section. My problem is I cannot get the right figure Poiseuille flow using MATLAB -- CFD Online Discussion Forums. Poiseuille flow is pressure-induced flow ( Channel Flow) in a long duct, usually a pipe. simplify the continuity equation (mass balance) 4. The pressure drop through the length in the simulation was 1421070 Pa. The flow is driven by a pressure gradient in the direction. Dynamic NMR Microscopy of Gas Phase Poiseuille Flow Lana G. If the width of a sample is too large, the excess. The direction of flow is from greater to lower pressure. (or other fluid) is moving at each point within the vessel. solving the momentum Equation (1). Numerical description of start-up viscoelastic plane Poiseuille flow Korea-Australia Rheology Journal March 2009 Vol. 1 Poiseuille °ow in a rectangular pipe { the structure of the velocity fleld. Anaesthesia, 1976, Volume 3 1, pages 273-275 HISTORICAL NOTE Poiseuille and his law J. The flow is caused by a pressure gradient, dp/dx, in the axial direction,x. As per the theory, the following conditions must be retained while deriving the equation. More specifically, we measure the distance of the point from the center of the tube to be at a specific radius (r), at which point the speed is given by the formula. presenting viscosities between about 0. Poiseuille's equation for laminar flow measurement. Poiseuille's equation only holds for fully developed flow. Flow rate is directly proportional to the pressure difference , and inversely proportional to the length of the tube and viscosity of the fluid. Poiseuille's Law (pronounced a bit like Pwah-soy's) describes the volume rate of flow (the volume of fluid passing a point along the tube per second) in terms of the fluid's viscosity, the tube's radius and length, and the pressure difference along the tube: Notice that this equation is an example of the Current = Potential / Resistance form. Poiseuille's law is found to be in reasonable agreement with experiment for uniform liquids (called Newtonian fluids) in. Processing. Schwalbe,*a Frederick R. A decrease in radius has an equally dramatic effect, as shown in blood flow examples. POISEUILLE FLOW OF POWER-LAW FLUIDS IN CONCENTRIC ANNULI - LIMITING CASES Filip P. Haroona, A. COUETTE AND PLANAR POISEUILLE FLOW Couette and planar Poiseuilleflow are both steady flows between two infinitely long, parallel plates a fixed distance, h, apart as sketched in Figures 1 and 2. First, notice that the blood is not moving when r=a. If the unknown quantity is stored in a large column vector, then the above approximation can be represented as a large sparse block matrix being applied from the left. Flow rate Q is in the direction from high to low pressure. This relationship (Poiseuille's equation) was first described by the 19th century French physician Poiseuille. It says the volume that will flow per time is dependent on delta P times pi, times R to the fourth, divided by eight eta, times L. Changes in cross sectional area directly affect velocity in order to keep bulk flow constant (Q=Av, If Q1 = Q2, then A1v1 = A2v2) Poiseuille’s equation revolves around the factors that change bulk flow. The reason we can't use an initial value solver for a BVP is that there is not enough information at the initial value to start. In the framework of physically justified scaling of velocity and length, an analysis of energy and linear critical Reynolds numbers was carried out in a practically important range of groove heights, sharpness and spacing. The flow of fluid through a pipe of uniform (circular) cross-section is known as Hagen-Poiseuille flow. al(2012) studied steady MHD Poiseuille flow between two infinite parallel porous plates in an inclined magnetic field and discover that high magnetic field strength decreases the velocity. So it is not correct to use the Poiseuille’s equation for flow in the porous media. Stroke’s Law When a small spherical body is dropped in a viscous medium, the layer in contact with it starts moving with the same velocity as that of the body whereas the layer at a considerable far distance will be at rest. Adjustments to blood flow are primarily made by varying the size of the vessels, since the resistance is so sensitive to the radius. Media in category "Hagen-Poiseuille equation" The following 10 files are in this category, out of 10 total. 8Ln/pi r^4. Double flow by doubling pressure as long as the flow pattern remains laminar. The travel of heat in insulators is commonly pictured as a flow of phonons scattered along their individual trajectory. This means that if we multiply Bernoulli’s equation by flow rate , we get power. In fact, there is a very simple relationship between horizontal flow and pressure. In equation form, this is. Poiseuille's Law - Pressure Difference, Volume Flow Rate, Fluid Power Physics Problems - Duration: 17:21. When the blood flows around and around and around, The flow rate through a given vessel can be found. The flow rate #F# is proportional to the pressure drop #Δp = p_1 – p_2# divided by #R#, the resistance to flow. Poisson's partial differential equation Saint-Venant solution was used, to calculate Poiseuille number values whatever is rectangles aspect ratio. Blood, with which Poiseuille was concerned, is not a simple Newtonian fluid, and its flow is a complicated problem, but water, oil and such obey the equation very well. The primary unidirectional flow is between two infinite parallel plates, one of which moves relative to the other. Flow rate $$Q$$ is in the direction from high to low pressure. We investigated these relationships in an ex vivo model and aimed to offer some rationale for equipment selection. As it can be seen in the contours image, the velocity profile throughout the pipe and towards the end of the pipe follows the same profile as what is predict by the Hagen-Poiseuille Theory. The direction of flow is from greater to lower pressure. Interfacial effects on droplet dynamics in Poiseuille flow† Jonathan T. •Concentration, distribution, shape, and rigidity of the suspended particles (e. As the diagram shows, and as the formula has stated, Poiseuille's law relates the flow rate with the pressure, viscosity, vessel radius and length. We have extended our study of shear flow instabilities in nematics, aligned perpen dicular to the velocity and velocity gradient, to the plane Poiseuille flow case. gravity-driven Poiseuille flow can be observable on granular gases under laboratory conditions. Synonyms for Poiseuille flow in Free Thesaurus. Poiseuilles Law, also known as the Hagen-Poiseuille equation, gives us the relationship between airway resistance and the diameter of the. In respiratory physiology, airway resistance is the resistance of the respiratory tract to airflow during inspiration and expiration. Previously, the phenomenon has been studied with models of the Boltzmann equation, but results for the Boltzmann equation itself have not been reported. The radius of drainage to be used in the radial flow equation and the productivity index expression is where A is the well's drainage area in square feet and the radius is in feet. Laminar Flow Confined to Tubes—Poiseuille's Law. Hagen-Poiseuille equation Pressure drop for laminar flow in pipe. Seton Hall University eRepository @ Seton Hall Seton Hall University Dissertations and Theses (ETDs) Seton Hall University Dissertations and Theses Spring 5-15-2017 Apparent reten. When you wanna think-a like Poiseuille, There's a formula you employ. Let us make a few initial observations. Poise-Stokes conversion Kinematic and dynamic viscosity converter. If this equation is substituted into the Pressure loss equation above it is also known as Poiseuille’s law or the Hagen–Poiseuille law. List and explain the assumptions behind the classical equations of fluid dynamics 3. Some of the fundamental solutions for fully developed viscous flow are shown next. Consider a liquid of co-efficient of viscosity η flowing, steadily through a horizontal capillary tube of length l and radius r. The fluid flow will be turbulent for velocities and pipe diameters above a threshold, leading to larger pressure drops than would be expected according to the Hagen-Poiseuille equation. is viscosity (uniform), and the continuity equation ∇·U = 0 (2) is an additional constraint representing the conservation of mass. Haroona, A. These make sense. drbeen has specialized online and on demand programs for nurse practitioners, physician assistants, and other advanced practitioners seeking continuing medical education (CME) as well as medical students seeking USMLE Steps 1 and 2 training. TOP RESULTS. Continuity equation revolves around the premise that the bulk flow is constant. This is the charge that flows through the cross section per unit time, i. Poiseuille °ow, in which an applied pressure difierence causes °uid motion between. * Re:poiseuille's equation!!!!! #726773 : enticerguyin - 04/02/07 13:08 `just keep in mind that R is proportionate to viscosity and length. In the case of smooth flow (laminar flow), the volume flowrate is given by the pressure difference divided by the viscous resistance. Poiseuille Flow Up: Incompressible Viscous Flow Previous: Flow Between Parallel Plates Flow Down an Inclined Plane Consider steady, two-dimensional, viscous flow down a plane that is inclined at an angle to the horizontal. Finlayson Professor Emeritus of Chemical Engineering University of Washington Seattle, WA 98195‐1750 Abstract An analytic solution is derived for fully developed. For steady flow of an incompressible fluid in a constant diameter horizontal pipe using the Darcy-Weisbach friction loss equation, the energy equation from location 1 to 2 is expressed in terms of pressure drop as:. 1 Couette-flow Consider the steady-state 2D-flow of an incompressible Newtonian fluid in a long horizontal rectangular channel. Some of the fundamental solutions for fully developed viscous flow are shown next. In this paper, we develop a thermodynamic formalism for studying this problem. Poiseuille’s equation states that fluid flow rate through a tube is inversely proportional to tube length and fluid viscosity and is proportional to the pressure drop across the tube and the tube radius to the fourth power : However, Poiseuille’s equation only applies to fluids with a constant viscosity regardless of the fluid velocity. The Hagen-Poiseuille equation has been widely applied to the study of fluid feeding by insects that have sucking (haustellate) mouthparts. The History of Poiseuille's Law Simulation of the flow around a pickup truck using a ghost-cell method (Kalitzin et al. The assumptions of the equation are that the flow is laminar, viscous and incompressible and the flow is through a constant circular cross-section that is substantially longer than its diameter. 1 word related to laminar flow: streamline flow. Head loss The poiseuille formula is used to evaluate the coefficient of pressure losses of laminar flow. Fluid Flow Hydrodynamics Aerodynamics Bernoulli’s Principle Poiseuille’s Law Wind tunnel visualization of air flow AIR FLOW streamlines The black lines are the paths that the fluid takes as it flows. From the velocity gradient equation above, and using the empirical velocity gradient limits, an integration can be made to get an expression for the velocity. However, the equation is valid only when the length of the cylinder is much longer than the entrance length (the length of the entrance region within which the flow is not fully developed). We're asked to find how the flow rate will differ between the two pipes. Flow rate Q is directly proportional to the pressure difference P 2 −P 1, and inversely proportional to the length l of the tube and viscosity η of. These make sense. Poiseuille's Law In the case of smooth flow (laminar flow), the volume flowrate is given by the pressure difference divided by the viscous resistance. Units of Measurement Pressure. Figure 2: Planar Poiseuille flow. determine the dissipation function for a system undergoing thermostatted Poiseuille ow and nd that it is an important physical property of the system. GUYON(**) and P. I try to write script for channel flow using Lattice Bolzmann method in MATLAB. In fluid dynamics, the Hagen–Poiseuille equation is a physical law that gives the pressure drop in a fluid flowing through a long cylindrical pipe. Poiseuille's Law Derivation. 1 Couette–flow Consider the steady-state 2D-flow of an incompressible Newtonian fluid in a long horizontal rectangular channel. He suggests that one should instead use the Poiseuille number: Clearly, a unit Poiseuille number is more convenient than a varying friction coefficient. The instability of shear flows, of which the Poiseuille flow is a canonical example, is among the most classical and most challenging problems in fluid mechanics, and a huge amount of effort has been devoted to it (1 -13). annualreviews. In fluid dynamics, the Hagen–Poiseuille equation is a physical law that gives the pressure drop in a fluid flowing through a long cylindrical pipe. Suppose that the fluid forms a uniform layer of. 6 of the channel half-width from the centerline which is in good agreement with Segrè and Silberberg5,6. This restriction is known as a Laminar Flow Element (LFE). 5 mm/s and vessels are generally smaller than 0. It is distinguished from drag-induced flow such as Couette Flow. I propose that Hagen-Poiseuille flow from the Navier-Stokes equations be merged into Hagen-Poiseuille equation. Turbulent Poiseuille flow with near‐critical wall transpiration Turbulent Poiseuille flow with near‐critical wall transpiration Vigdorovich, Igor; Oberlack, Martin 2010-12-01 00:00:00 An incompressible, pressure‐driven, fully developed turbulent flow between two parallel walls, with an extra constant transverse velocity component, is considered. The poiseuille (symbol Pl) has been proposed as a derived SI unit of dynamic viscosity, named after the French physicist Jean Léonard Marie Poiseuille (1797–1869). This article deals with the origins of this relationship and the assumptions and limitations inherent for Poiseuille flow. When individuals are doing so, one come across many concepts, problems etc. Some texts also discuss the Poiseuille equation, which deals only with viscous flow. Discharge is directly. how much of an increase in flow rate would be expected if the radius of the pipe increased by a factor of 2. The entrance length for fully developed flow can be found for turbulent flow and for laminar flow. Categorize solutions to fluids problems by their fundamental assumptions 2. Poiseuille's Formula. " "College physics texts present the Bernoulli equation as the most useful equation in fluid dynamics. Poiseuille's law applies to laminar flow of an incompressible fluid of viscosity through a tube of length and radius. First, we assume that the fluid is in a steady state. Viscosity: A definition as well as some values for assorted fluids. It is a description of how flow is related to perfusion pressure, radius, length, and viscosity. Poiseuille flow synonyms, Poiseuille flow pronunciation, Poiseuille flow translation, English dictionary definition of Poiseuille flow. Above 105, however, the Blasius equation diverges substantially from experiment. The instability of shear flows, of which the Poiseuille flow is a canonical example, is among the most classical and most challenging problems in fluid mechanics, and a huge amount of effort has been devoted to it (1 –13). 5 Compare Hagen-Poiseuille Relationship for Laminar Flow and the Average Velocity and Pressure Drop in Turbulent Flow. Plane Poiseuille flow. The steady planar Poiseuille flow generated by a constant external force is analyzed in the context of the nonlinear Bhatnagar-Gross-Krook kinetic equation for a gas of Maxwell molecules. I propose that Hagen–Poiseuille flow from the Navier–Stokes equations be merged into Hagen–Poiseuille equation. Poiseuille (1799-1869) was a French scientist interested in the physics behind blood circulation. The emergence of a liquid-like electronic flow from ballistic flow in graphene is imaged, and an almost-ideal viscous hydrodynamic fluid of electrons exhibiting a parabolic Poiseuille flow profile. The Hagen–Poiseuille Equation (or Poiseuille equation) is a fluidic law to calculate flow pressure drop in a long cylindrical pipe and it was derived separately by Poiseuille and Hagen in 1838 and 1839, respectively. Vessel radius, vessel density, stem transverse area occupied by vessel lumina, and volume flow rate of stems predicted by the Poiseuille flow equation differed among families. The circulatory system provides many examples of Poiseuille's law in action—with blood flow regulated by changes in vessel size and blood pressure. 029, and pressure difference 4000 dynes/cm{eq}^2 {/eq}. For laminar flow (Reynolds number, R ≤ 2100), the friction factor is linearly dependent on R, and calculated from the well-known Hagen-Poiseuille equation: R 64 λ= (2) Where, R, the Reynolds number, is defined as ūD/ν. al(2012) studied steady MHD Poiseuille flow between two infinite parallel porous plates in an inclined magnetic field and discover that high magnetic field strength decreases the velocity. 26, 1869, Paris), French physician and physiologist who formulated a mathematical expression for the flow rate for the laminar (nonturbulent) flow of fluids in circular tubes. From real icicles, we measure the net ux to be Qˇ0:01 cm3/s, a typical radius to be R 0 ˇ1 10 cm, and a typical wall angle as ˇ5. The most definitive advance has been the recent experimental work by Avila et al. Hudsona Received 27th January 2011, Accepted 10th March 2011. Poiseuille's equation pertains to moving incompressible fluids exhibiting laminar flow. [1][2] Derivation The laminar flow through a pipe of uniform (circular) cross-section is known as Hagen-Poiseuille flow. Experimental set-up. The assumptions of the equation are that the flow is laminar viscous and incompressible and the flow is through a constant circular cross-section that is substantially longer than its diameter. microfluidic-nanofluidic-Hagen-Poiseuille flow-equation 1. The Poiseuille’s law states that the flow of liquid depends on following factors like the pressure gradient (∆P), the length of the narrow tube (L) of radius (r) and the viscosity of the fluid (η) along with relationship among them. The result is a formula, which gives a mean velocity of flow in the direction at right angles to the layer plane in terms of its thickness and other parameters. Ansarib, A. This resistance depends linearly upon the viscosity and the length, but the fourth power dependence upon the radius is dramatically different. In fluid mechanics: Stresses in laminar motion …famous result is known as Poiseuille's equation, and the type of flow to which it refers is called Poiseuille flow. Sign up now to enroll in courses, follow best educators, interact with the community and track your progress. In 1838 he experimentally derived, and in 1840 and 1846 formulated and published, Poiseuille's law (now commonly known as the Hagen-Poiseuille equation, crediting Gotthilf Hagen as well), which applies to laminar flow, that is, non-turbulent flow of liquids through pipes of uniform section, such as blood flow in capillaries and veins. We report on a study of heat flow in bulk black phos-phorus between 0. r = radius of the tube, n = coefficient of viscosity and 1 = length of the tube. In Poiseuille flow, the soft particle migrates away from the wall to an off-center position dependent on the particle deformation and inertia, in contrast to hard sphere migration where the steady state position is independent of the shear rate. Darrigol, World of Flow, A history of Hydrodynamics from the Bernoullis to Prandtl (Oxford U. Flow rate is directly proportional to the pressure difference , and inversely proportional to the length of the tube and viscosity of the fluid. where, p = pressure difference across the two ends of the tube. 6 ¤ A viscous fluid flow upward through a small circular tube and then downward in laminar flow on the outside. The Darcy equation describes the Darcy friction factor for laminar flow. If the pipe is too short, the Hagen—Poiseuille equation may result in unphysically high flow rates; the flow is bounded by Bernoulli's principleunder less restrictive conditions, by. Now this is a crazy equation. Increasing the cannula size from 14 to 20 Fr increased flow rate by a mean (SD) of 13. Laminar Flow Confined to Tubes—Poiseuille’s Law. A laboratory experiment on inferring Poiseuille’s law for undergraduate students 1085 Figure 1. Poiseuille's Law calculation: Index Poiseuille's law concepts. It relates the difference in pressure at different spatial points to volumetric flow rate for fluids in motion in certain cases, such as in the flow of fluid through a rigid pipe. If a water pipe is 15 mm diameter and the water pressure is 3 bar, assuming the pipe is open ended, is it possible to calculate the flow rate or water velocity in the pipe? Most of the calculation. Poiseuille's Law (also Hagen-Poiseuille equation) calculates the fluid flow through a cylindrical pipe of length L and radius R. On the other hand, for R and M being not necessarily small, an instability criterion for plane Poiseuille flow is known; and the criterion says that, when R increases, a pair of complex conjugate eigenvalues of the linearized operator cross the imaginary axis. However, Churchill (1988) points out that Reynolds number is unsuitable for this nonaccelerating flow, since density does not play a part. This type of result was conjectured in [17] for non-rotating combined Couette-Poiseuille flow. (flow rate) 2. Poiseuille (1799–1869), who derived it in an attempt to understand the flow of blood, an often turbulent fluid. The Poiseuille Equation in Viscosity of Liquids. This derivation can be removed and a derivation. The contribution of these airways toward airways resistance is explained by the Poiseuille 's equation for laminar flow of gas or liquid in cylindrical tubes of different diameter. Poiseuille's Law gives the rate of flow, R, of a gas through a cylindrical pipe in terms of the radius of the pipe, r, for a fixed drop in pressure between the two ends of the pipe. Liquid flow through a pipe. Module 6: Navier-Stokes Equation Lecture 16: Couette and Poiseuille flows Ex. Ansarib, A. However, if you can measure the fluid pressure – which is usually easy to do, using a pressure gauge – you can use Poiseuille's Law to calculate flow rate. [1][2] Derivation The laminar flow through a pipe of uniform (circular) cross-section is known as Hagen-Poiseuille flow. The equations governing the Hagen-Poiseuille flow can be derived directly from the Navier-Stokes momentum equations in 3D cylindrical coordinates by making the following set of assumptions: The flow is steady ( ∂ (. Continuity equation (A·v = constant) The volume flow rate of a fluid is constant. Although geosynthetic clay liners (GCLs) have gained advantage over compacted clay liners regarding the ability to withstand large differential settlement in cover systems, the ability of strained. Does anybody know where this formula comes from? Are there exact solutions of the Navier-Stokes equation for compressible Poiseuille flow?. If this equation is substituted into the Pressure loss equation above it is also known as Poiseuille’s law or the Hagen–Poiseuille law. Module 6: Navier-Stokes Equation Lecture 16: Couette and Poiseuille flows Ex. if n and l will increase R will increase in same proportion. Poise-Stokes conversion Kinematic and dynamic viscosity converter. Naseem UddinMechanical Engineering Department NED University of Engineering & Technology Hagen Poiseuille Flow Problem. Srinivas, Effect of Elasticity on Hagen-Poiseuille Flow of a Jeffrey Fluid in a Tube, International Journal of Mechanical Engineering and Technology 8(8), 2017, pp. If this equation is substituted into the Pressure loss equation above it is also known as Poiseuille's law or the Hagen-Poiseuille law. The Poiseuille's formula express the disharged streamlined volume flow through a smooth-walled circular pipe: V = π p r 4 / 8 η l (1). Normally, Hagen-Poiseuille flow implies not just the relation for the pressure drop, above, but also the full solution for the laminar flow profile, which is parabolic. Referencias. The equation is: airflow = pressure gradient / resistance. If the pipe is too short, the Hagen—Poiseuille equation may result in unphysically high flow rates; the flow is bounded by Bernoulli's principleunder less restrictive conditions, by. • Another equation was developed to compute hL under Laminar flow conditions only called the Hagen-Poiseuille equation 16. The stability of Poiseuille flow in a pipe of circular cross-section to azimuthally varying as well as axisymmetric disturbances has been studied. Poiseuille (1799–1869), who derived it in an attempt to understand the flow of blood, an often turbulent fluid. For the solution of the problem they used OHAM. A laminar flow element (LFE) inside the meter forces the gas into laminar (streamlined) flow. The aim of this test case is to validate the following parameters of incompressible steady-state laminar fluid flow through a pipe: Velocity. The Darcy equation describes the Darcy friction factor for laminar flow. The results for the case of Poiseuille flow. Stroke’s Law When a small spherical body is dropped in a viscous medium, the layer in contact with it starts moving with the same velocity as that of the body whereas the layer at a considerable far distance will be at rest. In nonideal fluid dynamics, the Hagen-Poiseuille equation, also known as the Hagen-Poiseuille law, Poiseuille law or Poiseuille equation, is a physical law that gives the pressure drop in an incompressible and Newtonian fluid in laminar flow flowing through a long cylindrical pipe of constant cross section. An equation expressing the relation between the volume V of fluid flowing per second through a long narrow cylinder under conditions of Poiseuille flow, the viscosity of the fluid, and the dimensions of the cylinder, namely V = π r 4 p /8η l, where p is the difference in pressure between the ends of the cylinder, η is the viscosity of the fluid, l is the length of the cylinder, and r is its. Laminar Flow Confined to Tubes—Poiseuille’s Law. where η is the dynamic viscosity of the fluid. [1][2] Derivation The laminar flow through a pipe of uniform (circular) cross-section is known as Hagen-Poiseuille flow. The Hagen-Poiseuille equation states that $$\Delta p = \frac{8\mu LQ}{\pi R^4}$$ where $\mu$ is the dynamic viscosity of the fluid. This derivation can be removed and a derivation. A laboratory experiment on inferring Poiseuille’s law for undergraduate students 1085 Figure 1. V = discharge volume flow (m 3 /s). PHYSICAL REVIEW E VOLUME 60, NUMBER 4 OCTOBER 1999 Burnett description for plane Poiseuille flow F. A decrease in radius has an equally dramatic effect, as shown in blood flow examples. Poiseuille's Law calculation: Index Poiseuille's law concepts. The Poiseuille's formula express the disharged streamlined volume flow through a smooth-walled circular pipe: V = π p r 4 / 8 η l (1). He derived an expression for the volume of the liquid flowing per second through the capillary tube. It is defined as the ratio of driving pressure to the rate of air flow. Laminar flow is a flow regime characterized by high momentum diffusion and low momentum convection. We have seen that when the flow is turbulent it is necessary to resort to experiment to find f as a function of Re. Darrigol, World of Flow, A history of Hydrodynamics from the Bernoullis to Prandtl (Oxford U. Airway resistance is the opposition to flow caused by the forces of friction. It is interesting that warm-blooded animals regulate the heat loss from their bodies by changing the diameter of their blood vessels (varying r) and hence controlling the rate of blood flow. n (Poiseuille equation) The equation of steady, laminar, Newtonian flow through circular tubes:. Burton, Physiology and Biophysics of the Circulation, Yearbook, 1965. Figure 2-21: Metallic cast of pore space in a consolidated sand. Kolmogorov microscales Finds length, time and velocity. Double flow by doubling pressure as long as the flow pattern remains laminar. 8QLV/pi r^4. There is a point far from the entrance of the tube at which the radial velocity distribution is identical for all points farther downstream; this is Poiseuille's flow and the mean velocity is given by:. In this paper we considered one dimensional poiseuille flow of an electrically conducting fluid between. presenting viscosities between about 0. Discovered independently by Gotthilf Hagen, a German hydraulic engineer, this relation is also known as the Hagen-Poiseuille equation. Specifically, it is assumed that there is Laminar Flow of an incompressible Newtonian Fluid of viscosity η) induced by a constant positive pressure difference or pressure drop Δp in a pipe of length L and radius R << L. It outputs the flow type you can expect (laminar, transitional, or turbulent) based on the Reynolds Number result. Poiseuille's equation. To determine the driving height of the liquid level 2. Srinivas, Effect of Elasticity on Hagen-Poiseuille Flow of a Jeffrey Fluid in a Tube, International Journal of Mechanical Engineering and Technology 8(8), 2017, pp. Granular Poiseuille flowGranular Poiseuille flow Andrés Santos* University of ExtremaduraUniversity of Extremadura Badajoz (Spain) *In collaboration with Mohamed Tij, Université Moulay Ismaïl, Meknès (Morocco). We investigated these relationships in an ex vivo model and aimed to offer some rationale for equipment selection. References Used For some references to the applicability of Poiseuille's law to blood flow, I looked at the following books: A. Viscosity: A definition as well as some values for assorted fluids. In fluid mechanics: Stresses in laminar motion …famous result is known as Poiseuille’s equation, and the type of flow to which it refers is called Poiseuille flow. b) A cross section of the tube shows the lamina moving at different speeds. | {"extraction_info": {"found_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": 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": 0.907917857170105, "perplexity": 891.1643821917726}, "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/1585370506673.7/warc/CC-MAIN-20200402045741-20200402075741-00539.warc.gz"} |
https://physics.stackexchange.com/questions/467844/what-is-the-difference-between-real-and-complex-instantons-mathemtically-and-t | # What is the difference between real and complex instantons (mathemtically, and their physical significance), and connection to Wick rotation
I am struggling to understand the difference and physical significance between real and complex instantons- I think these are also sometimes called ghost instantons? There are also anti-instantons. Partly, I think this is due to the fact that some authors make a distinction between these, and others just say 'instantons/anti-instantons' for the 'ordinary' type, and ghosts/complex (i'm not sure if these are the same).
My present understanding is as follows:
When we have a path integral in real/Minkowski time:
$$\int \mathcal{D}[\phi] e^{-\frac{iS[\phi]}{\hbar}}$$ with $$S$$ the classical action, there can be more than one classical solution to the equations of motion. i.e. real functions that extremise the action. These correspond to real instantons.
Then when it comes to complex instantons, I thought these were just stationary points of the action if one analytically continues the domain of integration to complex $$\phi$$ as well? I am struggling with seeing the significance of this in the quantum case. I agree that in the real case, we only seek real $$\phi$$ solutions (it may be that $$\phi(t) = x(t)$$), but in the complex case would be not expect to be integrating over the domain of complex paths $$\phi$$, so there is no 'analytic continuation' involved, so to speak?
If complex instantons are indeed complex $$\phi$$ solutions to make the action stationary, how does this correspond to Wick rotation to Euclidean time? i.e. I have seen authors use the Wick rotated path integral $$\int \mathcal{D}[\phi] e^{\frac{S[\phi]}{\hbar}}$$ with $$S$$ now the Eulcidean action, and refer to stationary points of this action as 'instantons'. How does making time imaginary relate with finding complex $$\phi$$ solutions, and thus with a classification of real or complex instantons? How do these notions relate to the physical interpretations of these contributions to the path integral?
Note: an additional thing that may be related. I have some familiarity with Piccard Lefshetz theory and thimble decompostion of an integration path. I know that the imaginary part of the exponential argument is constant on a downwards flow, so one of the advatages of Wick rotation is that real action solutions can be the
I couldn't find any stack exchange posts answering these questions, but relevant papers that I have been trying (with limited success) to digest are:
Tying up instantons with antiinstantons, N. Nekrasov
Resurgence Theory, Ghost instantons, and the analytic continuation of path integrals, G. Basar, G. Dunne, M. Unsal
• These complex saddles have to do with complexifying field space not time. Most of the discussion of these takes place in Euclidean path integrals. The notion of an ordinary instanton is also in the Euclidean path integral by the way (not Minkowski as you claim) – octonion Mar 21 at 14:17
• The way I saw it was, the solutions that extremise the action in the Minkowski time will conserve energy. But in quantum mechanics, you also want to have paths with tunneling events which doesn't conserve the energy (when the particle cross the potential barrier). To take those paths into account, you do a Wick rotation and now the extremal paths will be those tunneling events. Not sure it answers your question. – E. Bellec Apr 8 at 9:39 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 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": 10, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9572068452835083, "perplexity": 488.70013842537355}, "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-2019-35/segments/1566027318952.90/warc/CC-MAIN-20190823172507-20190823194507-00082.warc.gz"} |
https://www.physicsforums.com/threads/mechanical-energy.160046/ | # Mechanical Energy
• Start date
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## Homework Statement
A thin rod, of length L and negligible mass, that can pivot about one end to rotate in a vertical circle. A heavy ball of mass m is attached to the other end. The rod is pulled aside through an angle and released.
What is the speed of the ball at the lowest point if L = 2.20 m, = 19.0°, and m = 500 kg?
W=mgd(cos theta)
KE=(1/2)mv^2
## The Attempt at a Solution
How do I get the velocity? Can I get it by using the above equations if I solve for W? Where do I plug the answer for W in to find the velocity?
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Do you know what W is as defined in your post ? Can you give it a name ?
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W=work done
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2K | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8297339677810669, "perplexity": 1860.760624097547}, "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/1618038460648.48/warc/CC-MAIN-20210417132441-20210417162441-00528.warc.gz"} |
https://brilliant.org/discussions/thread/rmo-1990/?sort=new | # RMO #1990
After seeing that my friends on Brilliant are eagerly preparing for RMO, I wish to help them and therefore I am posting the RMO$(1990)$ question paper, I also want to know about the correct way of solving these problems (my second purpose for posting these questions). Please post solution also.
$1)$Two boxes contain between them $65$ balls of several different sizes. Each ball is white, black, red or yellow. If you take any five balls of the same colour, at least $2$ of them will always be of the same size (radius). Prove that there are at least three balls which lie in the same box have the same colour and have the same size (radius).
$2)$For all positive real numbers $a, b, c$ prove that $\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b} \geq \frac{3}{2}$.
$3)$A square sheet of paper $ABCD$ is so folded that $B$ falls on the mid-point of $M$ of $CD$. Prove that the crease will divide $BC$ in the ratio $5:3$.
$4)$Find the remainder when $2^{1990}$ is divided by $1990$.
$5)$$P$ is any point inside a $\triangle ABC$. The perimeter of the triangle $AB +BC +CA=2s$. Prove that $s < AP + BP + CP < 2s.$
$6)$$N$ is a $50$-digit number (in a decimal scale). All digits except the $26^{th}$ digit (from the left) are 1.If $N$ is divisible by $13$, find the $26^{th}$ digit.
$7)$A census-man on duty visited a house in which the lady inmates declined to reveal their individual ages, but said “We do not mind giving you the sum of the ages of any two ladies you may choose”. There upon the census man said, “In that case, please give me the sum of the ages of every possible pair of you”. They gave the sums as follows: $30, 33, 41, 58, 66, 69$. The census-man took these figures and happily went away. How did he calculate the individual ages of the ladies from these figures?
$8)$If the circumcenter and centroid of a triangle coincide, prove that the triangle must be equilateral.
Note by Akshat Sharda
4 years, 11 months ago
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Solution of Q.5
We know that in a triangle (sum of two sides) $>$ (third side)
In $\triangle APB, AP + PB>AB$
similarly, In $\triangle APC, AP + PC>AC$ and In $\triangle BPC, BP + PC>BC$
Adding the above inequalitiees we get $2(AP+BP+CP)>(AB+BC+AC) \Rightarrow (AP+BP+CP)>\frac{(AB+BC+AC)}{2} \Rightarrow (AP+BP+CP)>s$
Now, it is easy to see that
$\angle APB>\angle BAP , \angle APB>\angle ABP$
$\Rightarrow AB>AP$ and $AB>BP$
Similar reasoning gives us
$AC>AP , AC>PC$ and $BC>BP, BC>PC$
Adding we get $2(AP+BP+CP)<2(AB+BC+AC) \Rightarrow (AP+BP+CP)<2s$
$\therefore s<(AP+BP+CP)<2s$, as desired
- 3 years ago
Q7 Ages are 47 ,22 ,19 and 11
- 3 years, 12 months ago
Q6 it is 3
- 3 years, 12 months ago
It is easy to use the collory of C-S for Q2. however here is the solution with AM-HM: $\dfrac{a}{b+c}+\dfrac{b}{a+c}+\dfrac{c}{a+b} \geq \dfrac{3}{2}$, $\dfrac{a}{b+c}+1+\dfrac{b}{a+c}+1+\dfrac{c}{a+b}+1-3$ $(a+b+c)(\dfrac{1}{b+c}+\dfrac{1}{a+c}+\dfrac{1}{a+b}) -3$ $\dfrac{1}{2}[(a+b)+(b+c)+(a+c)][\dfrac{1}{b+c}+\dfrac{1}{a+c}+\dfrac{1}{a+b}]-3$ Apply AM-HM to find that $\frac{1}{2}[(a+b)+(b+c)+(a+c)][\dfrac{1}{b+c}+\dfrac{1}{a+c}+\dfrac{1}{a+b}]-3 \geq \dfrac{1}{2}[3^2]-3=\dfrac{9}{2}-3=\dfrac{3}{2}$
- 4 years, 11 months ago
This is probably one of the easiest RMO.
- 4 years, 11 months ago
2) Immediate using Titu's Lemma
4) Remainder is $1024$
- 4 years, 11 months ago
@Akshat Sharda Thanks for posting! And of course keep posting! :)
- 4 years, 11 months ago
RMO then contained 8 questions!
- 4 years, 11 months ago | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 58, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9431655406951904, "perplexity": 1900.0266600416003}, "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/1600400227524.63/warc/CC-MAIN-20200925150904-20200925180904-00568.warc.gz"} |
https://brilliant.org/practice/parametric-equations-arc-length-basic/ | Calculus
# Parametric Equations - Arc Length
The location of a dot $P$ at a given time $t$ in the $xy$ plane is given by $(x,y) = (t - \sin t, 1 - \cos t)$. What is the distance traveled by $P$ in the interval $0 \leq t \leq 2\pi$?
What is the length of the curve parametrized by the equations $\begin{array}{c}\displaystyle x=e^{2t}\cos t, & y=e^{2t}\sin t,\end{array}$ in the domain $0 \leq t \leq 4 ?$
If $x=4\sin^2 t$ and $y=4\cos^2 t,$ what is the distance traveled by the point $P=(x,y)$ during the time interval $0 \leq t \leq 5\pi?$
Given the curve $H(t) = \frac{2}{3} (t+4)^{3/2}$, the arc length of the graph between $t=4$ and $t=12$ can be expressed in the form $\frac {a\sqrt{b}}{c} - d$ where $a$, $b$, $c$, and $d$ are positive integers, $a$ and $c$ are coprime, and $b$ is not divisible by the square of any prime. What is $a+b+c+d$?
Given the curve defined by $x = t^3$ and $y = t^2,$ what is the length of the curve from $t=0$ to $t= 10?$
× | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 40, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9696633815765381, "perplexity": 38.50782875937404}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-40/segments/1600400283990.75/warc/CC-MAIN-20200927152349-20200927182349-00620.warc.gz"} |
https://math.stackexchange.com/questions/1841777/how-to-show-convergence-and-evaluate-sum-limits-n-0-infty-fracnn1 | # How to show convergence and evaluate $\sum\limits_{n=0}^{\infty} \frac{n}{(n+1)!}$ [duplicate]
I try to evaluate series ${n\over(n+1)!}$ obviously the term $s_n = {n\over (n+1)!} = {1\over (n-1)! (n+1)}$ converges, which has a limit = $0$ when $n\to \infty$ .Then I was stuck and I don't know how to sum this series and whether it converges.
• $$\frac{n}{(n+1)!}=\frac{n+1-1}{(n+1)!}=\frac{1}{n!}-\frac{1}{(n+1)!}$$ – Did Jun 27 '16 at 20:58
Hint. One may write $$\frac{n}{(n+1)!}=\frac{(n+1)-1}{(n+1)!}=\frac1{n!}-\frac1{(n+1)!}$$ then terms telescope, we get $$\sum_{n=0}^N\frac{n}{(n+1)!}=1-\frac1{(N+1)!}$$ then we make $N \to \infty$.
To prove it the sum converges, compare it to something like $\frac 1{n!}$, which you may know converges. To evaluate the sum, define $t_n=\frac 1{(n+1)!}$ Can you sum $(s_n+t_n)$ and $t_n$ and subtract?
Your logic for convergence is a little faulty. Merely showing that the sequence decreases to $0$ as $n\to \infty$ is not sufficient. To see this, consider the series $\sum \frac{1}{n}$.
Another way to determine convergence: using the ratio test, one may show $\lim_\limits{ n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|<1$ with $a_n={n\over(n+1)!}$
Thus, we write:
$$\lim_\limits{ n\to\infty}\left|\frac{{n+1\over(n+2)!}}{{n\over(n+1)!}}\right|=\lim_\limits{ n\to\infty}\frac{n+1}{(n+2)!}\cdot\frac{(n+1)!}{n}=\lim_\limits{ n\to\infty}\frac{n+1}{n(n+2)}=0<1$$
Thus, the sum converges. For the exact sum, you may use the telescoping method, as described in the other excellent answers.
Here's an alternate way to find the sum. Recall that $$e^x = \sum_{n=0}^{\infty}\frac{x^n}{n!}.$$ We can write the above as: $$e^x = \sum_{n=0}^{\infty}\frac{x^n}{n!} = 1+\sum_{n=0}^{\infty}\frac{x^{n+1}}{(n+1)!} = 1+x\sum_{n=0}^{\infty}\frac{x^n}{(n+1)!},$$ and so: $$\frac{e^x-1}{x} = \sum_{n=0}^{\infty}\frac{x^n}{(n+1)!}.$$ Differentiating both sides gives: $$\frac{xe^x-e^x+1}{x^2} = \sum_{n=1}^{\infty}\frac{nx^{n-1}}{(n+1)!} = \sum_{n=0}^{\infty}\frac{nx^{n-1}}{(n+1)!}.$$ Evaluating both sides at $x=1$ implies: $$\sum_{n=0}^{\infty}\frac{n}{(n+1)!} = 1.$$ | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9641372561454773, "perplexity": 211.85760222182302}, "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/1593655886865.30/warc/CC-MAIN-20200705023910-20200705053910-00271.warc.gz"} |
https://infoscience.epfl.ch/record/213529?ln=en | ## Sur quelques foncteurs de bi-ensembles
This thesis is in the context of representation theory of finite groups. More specifically, it studies biset functors. In this thesis, I focus on two biset functors: the Burnside functor and the functor of p-permutation modules. For the Burnside functor we first give a result that characterize some B-groups; B-groups being the essential ingredient in the classification of composition factors of the Burnside functor. The second result compares the Burnside functor and the functor of free modules. Note that the functor of free modules is not a biset functor since the inflation of a free module is not necessarily free. To compare those functors we will work on an adjunction between the category of biset functors and the category of functors that do not have inflation. An aspect of the work done on the functor of p-permutation module is to compare the functor of p-permutation modules and the functor of ordinary representations. On the other hand, because of the classification of p-permutation modules, we try to express the functor o p-permutation modules in terms of the functor of projective modules (which is not a biset functor). We will use an adjunction between the category of biset functors and a category that contains the functor of projective modules.
Thévenaz, Jacques
Year:
2015
Publisher:
Lausanne, EPFL
Keywords:
Note:
La date de soutenance figurant sur la p. de titre (11 nov. 2015) est erronée
Other identifiers:
urn: urn:nbn:ch:bel-epfl-thesis6753-7
Laboratories:
Note: The status of this file is: Anyone | {"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": 0.9132217764854431, "perplexity": 600.9576606529531}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-34/segments/1596439738425.43/warc/CC-MAIN-20200809043422-20200809073422-00054.warc.gz"} |
http://mathhelpforum.com/differential-geometry/212397-mean-curvature-surface.html | # Thread: Mean curvature of a surface
1. ## Mean curvature of a surface
Hi everyone, new here and trying to calculate the mean curvature for the following surface:
r(u,v)=ui+f(u)cos(v)j+f(u)sin(v)k
I have been through the process of finding tangent vectors and a unit normal vector to the surface, so up to the point where I have had to substitute 6 scalars into:
h(p)=
aG-2bF-cE
2(EG-F2)
The only problem I seem to be having is not being able to simplify it... This is what I have:
h(p)=
± f''(u)f(u)2-f'(u)2-1
2f(u)(1+f'(u)2)√(cos2(2v)f'(u)+1)
Maybe missing something really simple here or I have made a mistake in calculating the scalars.
Any help, much appreciated!
2. ## Re: Mean curvature of a surface
I have worked through this again and shamefully the original post had some errors in. I have attached a correct answer, however still stuck with having to derive the differential equation!
3. ## Re: Mean curvature of a surface
From this link, because it would be a lot of typesetting:
Replace S(x,y) with your r(u,v). Respond with calculated partial derivatives as in the formula so we can see where you are going wrong. | {"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": 0.9300121665000916, "perplexity": 492.4072500933967}, "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/1474738662133.5/warc/CC-MAIN-20160924173742-00023-ip-10-143-35-109.ec2.internal.warc.gz"} |
http://beta.briefideas.org/search?tags=test | #### Report a relative proportion of patients ruled-out or ruled-in with a diagnostic test
One measure of the performance of a diagnostic test is the proportion of patients the test stratifies as likely not to have the condition (ruled-out:RO) or to have the condition (ruled-in:RI). This proportion depends on the prevalence of the disease in the cohort. This means that it is not valid...
#### Pre-registration of a test for statistical significance of a GALFA-HI -- WMAP ILC correlation
In 2007 Gerrit Verschuur [proposed](http://dx.doi.org/10.1086/522685) that HI structures as seen in the Leiden-Argentina-Bonn 36' HI survey of the sky are correlated with structures in the WMAP sky, which are typically ascribed to the CMB. He has recently proposed to see WMAP correlations in... | {"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": 0.923050045967102, "perplexity": 2436.1318988157577}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-39/segments/1537267158001.44/warc/CC-MAIN-20180922005340-20180922025740-00537.warc.gz"} |
https://www.physicsforums.com/threads/plz-help-motion-in-a-plane-vectors-etc.349115/ | # Plz help! Motion in a plane. (vectors, etc)
1. Oct 26, 2009
### unknownplaya
Hello. I have a question on this problem.
What are two quantities of an object launched parallel to the ground which are equal to zero?
I think one of the two is vertical velocity, but I'm not sure.
2. Oct 26, 2009
### Staff: Mentor
Welcome to the PF. You are correct that the initial vertical velocity is zero. What could the second quantity be? I'm guessing that you are to neglect air resistance in this problem?
Similar Discussions: Plz help! Motion in a plane. (vectors, 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": 0.8941138386726379, "perplexity": 1370.769811407772}, "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/1516084886739.5/warc/CC-MAIN-20180116204303-20180116224303-00352.warc.gz"} |
http://www-old.newton.ac.uk/programmes/GMR/seminars/2005121615001.html | GMR
Seminar
Why we should not take the Liu-Yau quasi-local mass seriously
Friday 16 December 2005, 15:00-15:30
Seminar Room 1, Newton Institute
Abstract
The Liu-Yau mass is a true mass, it is frame independent. However, the Liu-Yau mass is bigger than the Brown-York energy on any surface for which both can be defined. Further, if I take a sequence of coordinate spheres' on any spacelike slice, both the Liu-Yau mass and the Brown-York energy asymptote to the ADM mass (which is really an energy, the 0'-th component of a Lorentz covariant 4-vector). This means that in any asymptotically flat spacetime, I can find a 2-surface with unboundedly large Liu-Yau mass. This is even true for Minkowski space. | {"extraction_info": {"found_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": 0.9869118928909302, "perplexity": 1209.9908673262157}, "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-2015-27/segments/1435375099361.57/warc/CC-MAIN-20150627031819-00069-ip-10-179-60-89.ec2.internal.warc.gz"} |
http://math.stackexchange.com/questions/147068/a-sum-involving-permutation | # A sum involving permutation
Does there exist a nice closed form formula for the sum $$\sum_{k=0}^m P(m,k)x^k$$ where $P(m,k)=C(m,k)*k!$, $C(m,k)$ being the "m choose k" number. Formula given by Maple 11 is complicated. I thought there may be an easier one. Of course $$\sum_{k=0}^m C(m,k)x^k=(1+x)^m.$$
-
We can write \begin{align} \sum_{k=0}^m\binom mkk!x^k&=\sum_{k=0}^m\frac{m!}{(m-k)!}x^k\\ &=m!\sum_{j=0}^m\frac{x^{m-j}}{j!}\\ &=m!x^mS_m\left(\frac 1x\right), \end{align} where $S_n(t):=\sum_{j=0}^n\frac{t^j}{j!}$. But the $S_n$ are hard to simplify.
Your variables in your definition of $S_n$ seem to be mixed up... – Ben Millwood May 19 '12 at 17:46
We could use the integral representation $$S_m(t) = \frac{e^t}{m!} \int_t^\infty u^m e^{-u}\,du = \frac{e^t}{m!}\Gamma(m+1,t)$$ to get $$\sum_{k=0}^m\binom mkk!x^k = e^{1/x} x^m \Gamma(m+1,1/x),$$ where $\Gamma(s,t)$ is the incomplete gamma function. – Antonio Vargas May 19 '12 at 20:14 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 1, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9980493187904358, "perplexity": 569.7163237688702}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-18/segments/1429246661916.33/warc/CC-MAIN-20150417045741-00066-ip-10-235-10-82.ec2.internal.warc.gz"} |
https://www.physicsforums.com/threads/density-speed-in-relation-to-power.619162/ | # Density & Speed in Relation to Power
1. Jul 7, 2012
Hi,
Just out of curiosity, why is the density and volume used to find the power of a flow of water for example. I know the equation for this is P=0.5ρAv^3 where ρ=density and v=speed of the water cubed.
Any suggestions?
Thank you
2. Jul 8, 2012
### sankalpmittal
You've put Volume as area times velocity. Instead of volume , you've put volumetric flow of fluid per second. Your equation is dimensionally correct.
Work done by fluid in motion = Kinetic energy = mv2/2
Putting m = ρV
we have
Work done by fluid in motion = ρVv2/2
Now putting V= Ah where h is height
Work done = ρAhv2/2/2
Dividing by time , we have
Power of flow of fluid = ρAv3/2
Through this derivation I think its clear why density and velocity of fluid are related with power.
Also Power is force times velocity. So power depends on velocity of fluid. Work done here is the kinetic energy possessed by the fluid which of course depends on mass and hence depends on density of fluid. So Since power depends on work , so here in this case , power has to depend on density as well.
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https://rd.springer.com/article/10.1007%2Fs10208-018-9404-1 | Foundations of Computational Mathematics
, Volume 19, Issue 6, pp 1315–1361
Rational Invariants of Even Ternary Forms Under the Orthogonal Group
• Paul Görlach
• Evelyne Hubert
Article
Abstract
In this article we determine a generating set of rational invariants of minimal cardinality for the action of the orthogonal group $$\mathrm {O}_{3}$$ on the space $$\mathbb {R}[x,y,z]_{2d}$$ of ternary forms of even degree 2d. The construction relies on two key ingredients: on the one hand, the Slice Lemma allows us to reduce the problem to determining the invariants for the action on a subspace of the finite subgroup $$\mathrm {B}_{3}$$ of signed permutations. On the other hand, our construction relies in a fundamental way on specific bases of harmonic polynomials. These bases provide maps with prescribed $$\mathrm {B}_{3}$$-equivariance properties. Our explicit construction of these bases should be relevant well beyond the scope of this paper. The expression of the $$\mathrm {B}_{3}$$-invariants can then be given in a compact form as the composition of two equivariant maps. Instead of providing (cumbersome) explicit expressions for the $$\mathrm {O}_{3}$$-invariants, we provide efficient algorithms for their evaluation and rewriting. We also use the constructed $$\mathrm {B}_{3}$$-invariants to determine the $$\mathrm {O}_{3}$$-orbit locus and provide an algorithm for the inverse problem of finding an element in $$\mathbb {R}[x,y,z]_{2d}$$ with prescribed values for its invariants. These computational issues are relevant in brain imaging.
Keywords
Computational invariant theory Harmonic polynomials Orthogonal group Slice Rational invariants Diffusion MRI Neuroimaging
Mathematics Subject Classification
12Y05 13A50 13P25 14L24 14Q99 20B30 20C30 33C55 42C05 68U10 68W30
Notes
Acknowledgements
Paul Görlach was partly funded by INRIA Mediterranée Action Transverse. Evelyne Hubert wishes to thank Rachid Deriche, Frank Grosshans, Boris Kolev for discussions and valuable pointers. Théo Papadopoulo receives funding from the ERC Advanced Grant No. 694665 : CoBCoM—Computational Brain Connectivity Mapping.
References
1. 1.
S. Axler, P. Bourdon, and W. Ramey. Harmonic function theory, volume 137 of Graduate Texts in Mathematics. Springer-Verlag, New York, second edition, 2001.Google Scholar
2. 2.
K. Atkinson and W. Han. Spherical harmonics and approximations on the unit sphere: an introduction, volume 2044 of Lecture Notes in Mathematics. Springer, Heidelberg, 2012.Google Scholar
3. 3.
P. Basser and S. Pajevic. Spectral decomposition of a 4th-order covariance tensor: Applications to diffusion tensor mri. Signal Processing, 87(2):220–236, 2007.
4. 4.
J.-L. Colliot-Thélène and J.-J. Sansuc. The rationality problem for fields of invariants under linear algebraic groups (with special regards to the Brauer group). In Algebraic groups and homogeneous spaces, volume 19 of Tata Inst. Fund. Res. Stud. Math., pages 113–186. Tata Inst. Fund. Res., Mumbai, 2007.Google Scholar
5. 5.
E. Caruyer and R. Verma. On facilitating the use of Hardi in population studies by creating rotation-invariant markers. Medical Image Analysis, 20(1):87 – 96, 2015.
6. 6.
H. Derksen and G. Kemper. Computational invariant theory. Springer-Verlag, 2 edition, 2015.Google Scholar
7. 7.
C. Delmaire, M. Vidailhet, D. Wassermann, M. Descoteaux, R. Valabregue, F. Bourdain, C. Lenglet, S. Sangla, A. Terrier, R. Deriche, and S. Lehéricy. Diffusion abnormalities in the primary sensorimotor pathways in writer’s cramp. Archives of Neurology, 66(4), 2009.Google Scholar
8. 8.
K. Fox and B. Krohn. Computation of cubic harmonics. J. Computational Phys., 25(4):386–408, 1977.
9. 9.
A. Ghosh and R. Deriche. A survey of current trends in diffusion mri for structural brain connectivity. J. Neural Eng., 13, 2016.
10. 10.
A. Ghosh, T. Papadopoulo, and R. Deriche. Biomarkers for Hardi: 2nd & 4th order tensor invariants. In IEEE International Symposium on Biomedical Imaging: From Nano to Macro - 2012, Barcelona, Spain, May 2012.Google Scholar
11. 11.
G. Golub and C. Van Loan. Matrix computations. Johns Hopkins Studies in the Mathematical Sciences. Johns Hopkins University Press, Baltimore, MD, fourth edition, 2013.
12. 12.
R. Goodman and N. R. Wallach. Symmetry, representations, and invariants, volume 255 of Graduate Texts in Mathematics. Springer, Dordrecht, 2009.Google Scholar
13. 13.
J. Grace and A. Young. The algebra of invariants. Cambridge Library Collection. Cambridge University Press, Cambridge, 2010. Reprint of the 1903 original.Google Scholar
14. 14.
E. Hubert and I. Kogan. Rational invariants of a group action. Construction and rewriting. J. Symbolic Comput., 42(1-2):203–217, 2007.
15. 15.
E. Hubert and I. Kogan. Smooth and algebraic invariants of a group action. Local and global constructions. Foundations of Computational Mathematics, 7(4):355–393, 2007.
16. 16.
E. Hubert and G. Labahn. Rational invariants of scalings from Hermite normal forms. In ISSAC 2012, pages 219–226. ACM Press, 2012.Google Scholar
17. 17.
E. Hubert and G. Labahn. Scaling invariants and symmetry reduction of dynamical systems. Foundations of Computational Mathematics, 13(4):479–516, 2013.
18. 18.
E. Hubert and G. Labahn. Computation of the invariants of finite abelian groups. Mathematics of Computations, 85(302):3029–3050, 2016.
19. 19.
E. Hubert. Algebraic and differential invariants. In F. Cucker, T. Krick, A. Pinkus, and A. Szanto, editors, Foundations of computational mathematics, Budapest 2011, number 403 in London Mathematical Society Lecture Note Series, pages 168–190. Cambridge University Press, 2012.Google Scholar
20. 20.
I. Isaacs. Algebra: a graduate course, volume 100 of Graduate Studies in Mathematics. American Mathematical Society, Providence, RI, 2009.Google Scholar
21. 21.
H. Johansen-Berg and T. Behrens, editors. Diffusion MRI. From Quantitative Measurement to In vivo Neuroanatomy. Academic Press, second edition. edition, 2014.Google Scholar
22. 22.
D. Jones, editor. Diffusion MRI. Theory, Methods, and Applications. Oxford University Press, 2011.Google Scholar
23. 23.
D. Kressner. Numerical methods for general and structured eigenvalue problems, volume 46 of Lecture Notes in Computational Science and Engineering. Springer-Verlag, Berlin, 2005.Google Scholar
24. 24.
D. Le Bihan, E. Breton, D. Lallemand, P. Grenier, E. Cabanis, and M. Laval-Jeantet. Mr imaging of intravoxel incoherent motions: Application to diffusion and perfusion in neurologic disorders. Radiology, 161(2):401–407, 1986.
25. 25.
X. Li, A. Messé, G. Marrelec, M. Pélégrini-Issac, and H. Benali. An enhanced voxel-based morphometry method to investigate structural changes: application to Alzheimer’s disease. Neuroradiology, 52(3):203–213, 2010.
26. 26.
J. Muggli. Cubic harmonics as linear combinations of spherical harmonics. Z. Angew. Math. Phys., 23:311–317, 1972.
27. 27.
M. Olive, B. Kolev, and N. Auffray. A minimal integrity basis for the elasticity tensor. Archive for Rational Mechanics and Analysis, 2017.Google Scholar
28. 28.
M. Olive. About Gordan’s algorithm for binary forms. Found. Comput. Math., 17(6):1407–1466, 2017.
29. 29.
T. Papadopoulo, A. Ghosh, and R. Deriche. Complete Set of Invariants of a 4th Order Tensor: The 12 Tasks of HARDI from Ternary Quartics. In Polina Golland, Nobuhiko Hata, Christian Barillot, Joachim Hornegger, and Robert Howe, editors, Medical Image Computing and Computer-Assisted Intervention – MICCAI 2014, volume 8675 of Lecture Notes in Computer Science, pages 233 – 240, Boston, United States, September 2014.
30. 30.
V. Popov. Sections in invariant theory. In The Sophus Lie Memorial Conference (Oslo, 1992), pages 315–361. Scand. Univ. Press, Oslo, 1994.Google Scholar
31. 31.
V. Popov and È. Vinberg. Invariant theory, volume 55 of Encyclopaedia of Mathematical Sciences. Springer-Verlag, Berlin, 1994. A translation of ıt Algebraic geometry. 4 (Russian), Akad. Nauk SSSR Vsesoyuz. Inst. Nauchn. i Tekhn. Inform., Moscow, 1989, Translation edited by A. N. Parshin and I. R. Shafarevich.Google Scholar
32. 32.
M. Rosenlicht. Some basic theorems on algebraic groups. American Journal of Mathematics, 78:401–443, 1956.
33. 33.
G. Schwarz. Algebraic quotients of compact group actions. J. Algebra, 244(2):365–378, 2001.
34. 34.
C. Seshadri. On a theorem of Weitzenböck in invariant theory. J. Math. Kyoto Univ., 1:403–409, 1961/1962.
35. 35.
T. Schultz, A. Fuster, A. Ghosh, R. Deriche, L. Florack, and L. Lek-Heng. Higher-Order Tensors in Diffusion Imaging. In B. Burgeth, A. Vilanova, and C. F. Westin, editors, Visualization and Processing of Tensors and Higher Order Descriptors for Multi-Valued Data. Dagstuhl Seminar 2011. Springer, 2013.Google Scholar
36. 36.
B. Sturmfels. Algorithms in invariant theory. Texts and Monographs in Symbolic Computation. Springer, second edition, 2008.Google Scholar
Authors and Affiliations
• Paul Görlach
• 1
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• Evelyne Hubert
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Email author | {"extraction_info": {"found_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": 0.9622475504875183, "perplexity": 2512.3497943351695}, "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-2019-51/segments/1575541315293.87/warc/CC-MAIN-20191216013805-20191216041805-00299.warc.gz"} |
https://proofwiki.org/wiki/Combination_Theorem_for_Continuous_Mappings/Topological_Ring/Combined_Rule | # Combination Theorem for Continuous Mappings/Topological Ring/Combined Rule
## Theorem
Let $\struct {S, \tau_S}$ be a topological space.
Let $\struct {R, +, *, \tau_R}$ be a topological ring.
Let $\lambda, \mu \in R$ be arbitrary element in $R$.
Let $f, g : \struct {S, \tau_S} \to \struct {R, \tau_R}$ be continuous mappings.
Let $\lambda * f + \mu * g: S \to R$ be the mapping defined by:
$\forall x \in S: \map {\paren{\lambda * f + \mu * g}} x = \lambda * \map f x + \mu * \map g x$
Let $f * \lambda + g * \mu : S \to R$ be the mapping defined by:
$\forall x \in S: \map {\paren{f *\lambda + g * \mu}} x = \map f x * \lambda + \map g x * \mu$
Then:
$\lambda * f + \mu * g: \struct{S, \tau_{_S}} \to \struct{R, \tau_{_R}}$ is a continuous mapping
$f * \lambda + g * \mu: \struct{S, \tau_{_S}} \to \struct{R, \tau_{_R}}$ is a continuous mapping. | {"extraction_info": {"found_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": 0.9742661118507385, "perplexity": 221.92657563666125}, "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/1631780057564.48/warc/CC-MAIN-20210924171348-20210924201348-00660.warc.gz"} |
https://stats.stackexchange.com/questions/337319/estimator-for-pareto-distribution-delta-method | # Estimator for Pareto Distribution & Delta Method
Assume that $Y$ has a Pareto distribution with parameters ($\theta, t$ = 1). An estimator of $\theta$ is $\tilde{\theta}$ where $\bar{Y} = \frac{\tilde{\theta}t}{\tilde{\theta} - 1}$. Solve for $\tilde{\theta}$ and then use the delta method to derive the asymptotic distribution of $\sqrt{n}(\tilde{\theta} - \theta)$, assuming $\theta$ > 2.
A good place to begin, if I am correct, is to assume $\bar{Y}$ = $\frac{\theta t}{\theta - 1}$. From there, we can solve for $\tilde{\theta}$. Except, this doesn't necessarily get rid of our random variable. Additionally, I'm confused about how this relates to the delta method. As someone utterly lost, can I get any help for this? Thanks!
Since $t=1$,
$$\bar{Y} = \frac{\tilde{\theta}}{\tilde{\theta} - 1} \implies \tilde \theta = \frac {\bar Y}{\bar Y-1}$$
We know the limiting distribution of
$$\sqrt{n}\left(\bar Y - \frac{\theta}{\theta-1}\right)$$
since it is the sample mean. Then we need the Delta method to find the limiting distribution of $\sqrt{n}\left(\tilde \theta - \theta\right)$, since $\theta$ is a non-linear function of $\bar Y$. The $g$ function here is $g(z) = z/(z-1)$.
I guess the rest are up to the OP.
• Limiting distribution of a sample mean being a N(0,variance) distribution? – Brendan G Mar 28 '18 at 22:35
• @BrendanG Yes. The "i.i.d. sample" assumption is usually implied when not mentioned. – Alecos Papadopoulos Mar 28 '18 at 23:13 | {"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": 0.9512685537338257, "perplexity": 264.989295699066}, "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/1570986655735.13/warc/CC-MAIN-20191015005905-20191015033405-00107.warc.gz"} |
https://www.varsitytutors.com/ap_physics_b-help/understanding-resistors-and-resistance | AP Physics B : Understanding Resistors and Resistance
Example Questions
Example Question #1 : Electric Circuits
Which of the following changes to a copper wire will lead to the greatest decrease in voltage?
Replacing the copper wire with a more conductive material
Increasing the cross-sectional area of the copper wire by a factor of two
Increasing the length of the copper wire by a factor of two
Increasing the current through the copper wire by a factor of two
Increasing the cross-sectional area of the copper wire by a factor of two
Explanation:
According to Ohm’s law a decrease in current and/or resistance will lead to a decrease in voltage, since voltage is directly proportional to both current and resistance.
Increasing the current will not decrease voltage. Remember that resistance is defined as:
In this formula, is the resistivity, is the length of the wire, and is the cross-sectional area of the wire. Increasing length will lead to an increase in resistance and voltage; however, increasing the area will lead to a decrease in resistance and, subsequently, a decrease in voltage.
The only answer that will lead to a decrease in voltage is the choice to increase the cross-sectional area of the wire.
Replacing the copper wire with a more conductive material will increase the resistivity, which will subsequently increase resistance and voltage.
Example Question #1 : Electric Circuits
A student assembles a circuit made up of a voltage source and two resistors. All three circuit elements are connected in parallel. The voltage across the voltage source is and the resistance of the resistors are and respectively. Which of the following is true of this circuit?
The voltage across the resistor will be less than the voltage across the resistor
The current through the resistor will be less than the current through the resistor
The current through the resistor will equal the current through the resistor
The voltage across the resistor will equal the voltage across the resistor | {"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": 0.8086860775947571, "perplexity": 507.79691944181883}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-43/segments/1539583512332.36/warc/CC-MAIN-20181019062113-20181019083613-00238.warc.gz"} |
http://www.physicsforums.com/showthread.php?p=3787480 | ## Why does comoving Hubble radius increase with time?
I am looking at inflation at the moment, and it says in my textbook that (aH)^(-1) is constantly increasing in matter or radiation dominated epochs.
a is always positive and always increasing. This tells me that da/dt is positive. I think that setting the universe to MD/RD means that da/dt is decreasing with time (eg a decelerating universe as there is no cosmological constant driving expansion). So dt/da (which is another expression for comoving Hubble radius) is increasing with time.
Have I got this right?
PhysOrg.com science news on PhysOrg.com >> Heat-related deaths in Manhattan projected to rise>> Dire outlook despite global warming 'pause': study>> Sea level influenced tropical climate during the last ice age
Recognitions: Science Advisor da/dt decreasing in time is not a decelerating universe, it's a contracting one. EDIT: This statement is obviously incorrect. See below for my efforts to redeem myself...
If todays value of da/dt is lower than yesterdays, but they are both greater than 0, doesn't that mean the universe is expanding, but that the rate of expansion is slowing down?
Recognitions:
## Why does comoving Hubble radius increase with time?
Quote by ck99 If todays value of da/dt is lower than yesterdays, but they are both greater than 0, doesn't that mean the universe is expanding, but that the rate of expansion is slowing down?
Errr...of course. My apologies. Yes, the comoving Hubble radius is indeed increasing in time during RD/MD, because as you say the universe is decelerating. You've undoubtedly noticed this is not the case during inflation.
Ummm...no one has answered the question correctly so far. Comoving Coordinates (and Comoving distances) do NOT increase with the expansion of the Universe, and do not increase in time. That is the whole point of the Comoving coordinate system. Proper distances increase in time. The Hubble Paramater is measuring the rate of change of Scale Factor (da/dt) divided by the Scale factor (a). The Scale factor is time-dependent, and is directly related to the Proper distance.
Recognitions:
Quote by Deuterium2H Ummm...no one has answered the question correctly so far. Comoving Coordinates (and Comoving distances) do NOT increase with the expansion of the Universe, and do not increase in time. That is the whole point of the Comoving coordinate system. Proper distances increase in time. The Hubble Paramater is measuring the rate of change of Proper Distance divided by the Proper Distance.
I think you mean to say that comoving coordinate systems do expand with the universe, so that comoving distances have constant coordinates.
It does not follow that all quantities measured in comoving coordinates are constant -- what would be the point of using them then?? Any proper distance that is not increasing with the expansion will have non-constant comoving coordinates.
The Hubble radius, $H^{-1}$, measured with respect to comoving coordinates is the comoving Hubble radius, $(Ha)^{-1}$. It very much depends on time.
Quote by bapowell I think you mean to say that comoving coordinate systems do expand with the universe, so that comoving distances have constant coordinates. It does not follow that all quantities measured in comoving coordinates are constant -- what would be the point of using them then?? Any proper distance that is not increasing with the expansion will have non-constant comoving coordinates. The Hubble radius, $H^{-1}$, measured with respect to comoving coordinates is the comoving Hubble radius, $(Ha)^{-1}$. It very much depends on time.
I didn't say that "all quantities measured in comoving coordinates are constant". I specifically said that comoving distances are constant. And any equation involving the Hubble paramater (which involves the scale factor) is time-dependent, because it is based upon proper distance at a given (fixed) instant in time.
Recognitions:
Quote by Deuterium2H I specifically said that comoving distances are constant.
OK, well then what does this have to do with the OP? He's asking about the comoving Hubble parameter, which is the Hubble parameter in comoving coordinates. It is not a comoving distance!
And any equation involving the Hubble paramater (which involves the scale factor) is time-dependent, because it is based upon proper distance.
I don't know what this has to do with the OP. Looks like you're making things more confused than they need to be. He's asking about the comoving Hubble radius. It is increasing in an RD/MD universe. So please tell me where we've gone wrong here?
Quote by bapowell OK, well then what does this have to do with the OP? He's asking about the comoving Hubble parameter, which is the Hubble parameter in comoving coordinates. It is not a comoving distance! I don't know what this has to do with the OP. Looks like you're making things more confused than they need to be. He's asking about the comoving Hubble radius. It is increasing in an RD/MD universe. So please tell me where we've gone wrong here?
Perhaps I misunderstood the question of the OP. The term Comoving Hubble radius only makes sense when measured at a particular instant of cosmological time, and it is dependent upon the coordinate (proper) distance at the time of measuement.
And I agree, I have probably needlessly confused the question in the original post.
Recognitions:
Yes, it is the quantity $(aH)^{-1}$, which is a function of coordinate time, $t$.
Yeah, your reasoning sounds alright to me. As you've already pointed out, just from the definition of the Hubble parameter $(aH)^{-1} = (\dot{a})^{-1}$. For the rad-dominated and matter-dominated cases, a~t1/2 and a~t2/3 respectively. (I believe that these only apply for models with no cosmological constant). Differentiating those, you get da/dt ~ t-1/2 or t-1/3 respectively. So a-dot decreases with time, which means that its reciprocal increases with time. | {"extraction_info": {"found_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": 0.8992881178855896, "perplexity": 534.4926817845258}, "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/1368699186520/warc/CC-MAIN-20130516101306-00065-ip-10-60-113-184.ec2.internal.warc.gz"} |
http://physics.stackexchange.com/questions/18572/near-field-around-parabolic-antenna | # Near-field around parabolic antenna?
Having a discussion at work about the $H$-field around a big parabolic antenna.
All of the safety tests done around the antenna only mention $E$-Field. They state in the radiating near-field the $E$ and $H$ are related so no need to measure them both. On the other hand all of the wire antenna have $H$-field included in their measurments.
I would have thought this was the wrong way around. Simple wire antenna have near-fields extending only a couple of wavelength but paraboic near-field can extend to a few $km$.
Only thing I can think of is that the near-field definition only applies to the front of the antenna - the back and surrounds are shielded by the parabola.
Is there a $H$-field, independent of the $E$-field in the immediate vicinity of a parabolic antenna?
-
As soon as you do not touch the wire (the source), you are in a free space where the relationship between $E$ and $H$ is unique. It does not mean there is no $H$. The magnetic field can be independent if it is static. The static part (if any) should be measured separately. – Vladimir Kalitvianski Dec 21 '11 at 10:47
I think the reason you are confused is because of the different notion of "near field". The near field that extends for kilometers in the parabolic antenna case is not a "near field" like an electrostatic field, it is just a radiative field which is collimated and hasn't travelled far enough to diffract over an appreciable area.
If you have a wire whose voltage alternates, you have an electrostatic field near the wire which changes with time, which follows the electrostatic law of $1/r^2$ decay, so $1/r^4$ energy density. The same holds for currents which change with time. Near enough, the law of the magnetic field is just by magnetostatics. This field falls away fast, and further away, you have a radiation field which falls off as 1/r, so that the energy density goes as $1/r^2$. The two cross over at a certain scale.
The notion of "near field" you are using in your case of a parabolic antenna is different--- it is not a local electrostatic or magnetostatic field, it is just the idea that the field is collimated, and so all the energy is going in the same direction, and it hasn't had time to spread away. This field is still purely radiative, just directional, and E and B are proportional. There is a crossover to a different power-law for falling off, but this crossover is purely in the radiation field, and has nothing to do with the non-radiative components which are important only very close to the antenna itself.
This radiative "near field" is only going forward from the antenna, because it is reflected by the metal. Only radiative fields are reflected by a metal, non-radiative electric fields are screened, while magnetic fields just go through, (changed perhaps by the magnetic characteristics of the metal).
So you will see some residual magnetic field near the antenna, but it will be due to either ferromagnetic materials, or the currents driving the antenna itself. It will fall off as $1/r^3$, and it will be the same as for a short wire antenna ignoring the parabolic reflector entirely (unless you have an iron reflector!).
-
So reactive fields are present around a parabolic antenna, not just in front of the antenna. In this reactive region E and H are independent. Is there a boundary where this reactive component can be ignored?
It looks like my concerns about not measuring H in the immediate vicinity of the antenna are valid.
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http://geotech.chinaxiv.org/user/search.htm?pageId=1610623344256&type=filter&filterField=authors&value=Dong,%20Xiaolong | Current Location:home > Browse
1. chinaXiv:201703.00309 [pdf]
Subjects: Geosciences >> Space Physics
Dual-frequency polarized scatterometer (DFPSCAT) is a pencil-beam rotating scatterometer which is designed for snow water equivalent (SWE) measurement, and Doppler beam sharpening (DBS) technique is proposed for DFPSCAT to achieve the azimuth resolution. However, the DBS technique is inapplicable for the forward-looking and afterward-looking regions. Based on an approximate aperiodic model of scatterometer echo signal, an improved adaptive regularization deconvolution algorithm with gradient histogram preservation (GHP) constraint is implemented to settle the problem. To investigate its performance of resolution enhancement and resulted accuracy, both a synthetic backscattering coefficient (σ0field reconstruction and SWE σ0reconstruction are carried out. The results show that the proposed method can recover the truth signal and achieve azimuth resolution of 2 km with the designed scatterometer system, which is required by the SWE retrieval. Moreover, the relative errors of reconstructed σ0are less than 0.5 dB that satisfy the accuracy requirement for SWE retrieval, and comparisons with observed results show that the error reduction is more than 0.03 dB. Meanwhile, a comparison between the proposed algorithm and some existing resolution enhancement methods is analyzed, which concludes that the proposed method can obtain a comparable resolution enhancement as L1method and has less noise. The technique is also verified with advanced scatterometer (ASCAT) scatterometer data.
2. chinaXiv:201703.00306 [pdf]
Subjects: Geosciences >> Space Physics
In order to satisfy a relatively high resolution for the retrieval of snow water equivalent, an X/Ku-band dual-frequency full-polarized SCATterometer (DFPSCAT) onboard Water Cycle Observation Mission (WCOM) satellite is designed for high-resolution observations. However, given the following situations, the method called “rotating azimuth Doppler discrimination” is proposed, which can satisfy the resolution requirement and real-time processing: 1) the conically rotation rate of antenna is relatively fast; 2) the swath width is larger than 1000?km; and 3) day or night observation capabilities are required. Considering the complexity of the system's design and the improvement of azimuth resolution capability, a burst pulsing scheme is addressed to satisfy the numbers of azimuth sampling. The simulation model is used to analyze the feasibility of azimuth discrimination method based on geometry and system parameters. It is shown that the achievable azimuth resolution is about 2–5?km at far end of the swath and only 5km at near end of the swath. The results show that when the size of a slice is set as 2–5?km, the Kpcis about less than 0.4 as snow depth varies, and the Kpcof combined slices is smaller than a single slice.
3. chinaXiv:201703.00294 [pdf]
Subjects: Geosciences >> Space Physics
Dual-Frequency Polarized Scatterometer (DFPSCAT) is a pencil-beam rotating scatterometer which is used to measure snow water equivalence (SWE). Respecting the low azimuth resolution of its forward-looking region, an adaptive regularization deconvolution super-resolution method, based on the scatterometer echo signal model, is proposed. Compared with the classical SIR and MAP algorithms, the proposed method can better reconstruct the original signal, and has less noise amplification. The algorithm processing accuracy with different Kpcis also studied, and the results show that when the value of Kpcis less than 0.1, nearly the entire restored data can satisfy the requirement of 0.5dB accuracy.
4. chinaXiv:201703.00290 [pdf]
Subjects: Geosciences >> Space Physics
Dual-Frequency Polarized Scatterometer (DFPSCAT) is a new system utilizing Doppler beam sharpening (DBS) technology for azimuthal resolution enhancement. Considering the DBS technology is inapplicable for the middle areas of the swath, a theoretical framework of deconvolution signal processing is proposed to improve resolution. A deconvolution method of the nonlocally centralized sparse representation (NCSR) is adopted to verify its feasibility, and simulation results show that the deconvolution method have obviously better resolution enhancement and higher recovery accuracy than these of the classical scatterometer image reconstruction (SIR) method. ?2016 IEEE.
5. chinaXiv:201605.00291 [pdf]
Subjects: Geosciences >> Space Physics
The RFSCAT(Rotating Fan-beam SCATterometer) is one of the two payloads of CFOSAT, the China-France Oceanography Satellite. It is a radar scatterometer designed to measure the electromagnetic back-scatter from wind roughened ocean surface. The operating frequency of the scatterometer is 13.256GHz (Ku-band) and has a swath about 1,000 kilometers. In this paper, based on the characteristics of echo signal of RFSCAT, the on-board processing of RFSCAT signal is introduced. A Doppler pre-compensation LUT and A slice division LUT are developed, and the signal processing algorithms are validated and the wind retrieval performances of RFSCAT are analyzed based on the well-developed data simulation system of CFOSAT RFSCAT.
6. chinaXiv:201605.00289 [pdf]
Subjects: Geosciences >> Space Physics
In order to satisfy a high resolution for the measurement of snow water equivalent, we use the system of using the dual frequency(X-band 9.6GHz and Ku-band 17GHz) and full polarization. This paper discusses the various system options(scanning pencil-beam, high PRF, high SNR) and tradeoffs considered for improving the azimuth resolution of scatterometer are required.
7. chinaXiv:201605.00288 [pdf]
Subjects: Geosciences >> Space Physics
Ocean surface current is a very important parameter of ocean dynamic environment, which has been connected to global climate change, marine environment forecasting, marine navigation, engineering security and so on. Doppler scatterometer is a new type of scatterometer that can be used to measure ocean surface current as well as the ocean wind vector in space. The Doppler Scatterometer is based on a real aperture radar, which can achieve a very wide swath. It can provide the ocean surface current and wind vector information in a certain resolution and achieve global coverage quickly, which is very important for the marine environment forecasting and climate changes research. | {"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": 0.8872119784355164, "perplexity": 3361.908382279476}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-04/segments/1610704799741.85/warc/CC-MAIN-20210126104721-20210126134721-00473.warc.gz"} |
http://slideplayer.com/slide/5070945/ | # Weather.
## Presentation on theme: "Weather."— Presentation transcript:
Weather
Weather is the condition of the atmosphere at a particular time and place.
Earth's Atmosphere
Earth’s atmosphere is a thin layer of gases that surround the planet
Earth’s atmosphere is a thin layer of gases that surround the planet. The atmosphere is made up of 78% nitrogen, 21% oxygen, 1% other gases, as well as water vapor, dust particles, smoke, salt, and other chemicals. We call the combination of gases in our atmosphere air.
The atmosphere is divided into layers that are determined by temperature change.
Troposphere The layer we live in is the Troposphere. The troposphere starts at Earth’s surface and extends upward for 9 to 20 km. This layer is where water vapor and clouds are and where weather happens. Even though this is the thinnest layer, it contains most of the air, 80% of the total mass of the atmosphere. The air temperature drops as you go higher up in the troposphere. At its upper limits, the temperature of the troposphere is about -60°C. The average temperature is 25°C. Mt. Everest, the highest landform on earth, rises 8.85 km. in the troposphere.
Jet Stream The jet stream, a fast-flowing river of wind, travels generally west to east between the lower stratosphere and the upper troposphere. Many military and commercial jet aircraft fly in this zone when going west to east.
Stratosphere The stratosphere is located 10 to 50 km above earth’s surface. This layer contains a layer of ozone (O3), a form of oxygen, that absorbs high-energy ultraviolet (UV) radiation from the sun. The temperature stays cold in the stratosphere except in the ozone layer where it rises to 0°C.
Mesosphere The mesosphere is located 50 to 80 km above Earth’s surface. The temperature plunges here reaching its coldest temperature of around -90°C. This is the layer in which meteors burn up while entering Earth’s atmosphere, producing what we call shooting stars.
Thermosphere The thermosphere is located 80 to 300 km above Earth’s surface. This is an extremely thin layer of air. It is the first layer of the atmosphere to get heated by the sun. When the sun is extra active with sunspots or flares, the temperature in the thermosphere can surge up to 1500°C. The thermosphere is divided into two layers, the ionosphere and the exosphere.
Ionosphere The ionosphere is located in the lower part of the thermosphere from about 60 to 300 km above Earth’s surface. The ionosphere has a large number of electrically charged ions which create the Aurora Borealis in the north and the Aurora Australis in the south.
Exosphere The exosphere is the uppermost layer of Earth’s atmosphere, 500 to 10,000 km above Earth’s surface. Atmospheric gases, atoms, and molecules can escape into space from this layer. Satellites that are used to monitor the weather, send telephone and T.V. signals, and that carry telescopes orbit the earth in the exosphere.
What Affects the Atmosphere?
Air has the properties of mass, density, and pressure.
Altitude is the distance above sea level
Altitude is the distance above sea level. Air pressure decreases as altitude increases. As air pressure decreases, so does density. The composition of the atmosphere remains the same until the exosphere.
Barometer A barometer is an instrument used to measure changes in atmospheric air pressure.
The troposphere is heated by the sun’s energy in 3 ways: radiation, conduction, and convection.
Radiation Radiation is the direct transfer of energy through empty space by electromagnetic waves.
Convection Convection is the transfer of heat by movement of currents within a fluid.
Conduction Conduction is the direct transfer of heat from one substance to another substance that it is touching.
The Earth’s surface absorbs solar energy becoming warmer than the air
The Earth’s surface absorbs solar energy becoming warmer than the air. Air near the Earth’s surface is warmed by radiation and conduction of the heat from the surface. Then convection causes the air near the ground that is heated to have more energy. The air molecules move faster, bumping into other air molecules, and moving farther apart. This air becomes less dense. The warmer air moves upward and the cooler air moves downward.
A wind is the horizontal movement of air from an area of high pressure to an area of lower pressure.
Coriolis Effect The way Earth’s rotation makes winds in the Northern Hemisphere curve to the right and winds in the Southern Hemisphere curve to the left.
The water cycle also effects weather
The water cycle also effects weather. The water cycle is the movement of water through the Earth’s atmosphere. Water on Earth’s surface absorbs energy from the sun. When water has enough energy, it changes from a liquid to a gas called water vapor. This process is called evaporation. The water vapor rises up through the troposphere where it cools and condenses forming clouds.
Precipitation is any form of water that falls from clouds and reaches Earth’s surface. Common types of precipitation include rain, sleet, freezing rain, hail, and snow. | {"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": 0.8347384333610535, "perplexity": 1015.4016282048979}, "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-2020-24/segments/1590347399830.24/warc/CC-MAIN-20200528170840-20200528200840-00458.warc.gz"} |
https://physics.stackexchange.com/questions/127391/how-can-i-calculate-the-divergence-of-the-lienard-wiechert-eletric-field | # How can I calculate the divergence of the lienard wiechert eletric field?
I was reading Introduction to Eletrodynamics by Griffiths and I see that´s nothing there about to prove the gauss law for charges in arbitrary motion and non constant velocity . So I try to calculate the divergence of the lienard wiechert fields to show that the divergence of the lienard wiechert eletric field is equal to the charge density but I am getting stuck in the calculations ! Somebody know how to do these calculations ??? | {"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": 0.9725763201713562, "perplexity": 256.1048413881954}, "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/1627046150266.65/warc/CC-MAIN-20210724125655-20210724155655-00543.warc.gz"} |
http://mathhelpforum.com/advanced-algebra/38585-matrices.html | Math Help - Matrices
1. Matrices
Which of the matrices below are diagonalizable and which are not?Explain your answer.
i)
ii)
HINT: In both cases the only eigenvalues are λ=-1 and λ=2.
2. Hello,
Originally Posted by matty888
Which of the matrices below are diagonalizable and which are not?Explain your answer.
i)
ii)
HINT: In both cases the only eigenvalues are λ=-1 and λ=2.
Find the eigenspaces associated to each eigenvalue.
If the sum of their dimension is 3, then it's diagonalisable.
3. Originally Posted by matty888
Which of the matrices below are diagonalizable and which are not?Explain your answer.
i)
ii)
HINT: In both cases the only eigenvalues are λ=-1 and λ=2.
An n x n matrix A is diagonalizable if and only if it has n linearly independent eigenvectors ..... | {"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": 0.89341801404953, "perplexity": 1172.6918036927727}, "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-2014-52/segments/1419447548066.118/warc/CC-MAIN-20141224185908-00002-ip-10-231-17-201.ec2.internal.warc.gz"} |
http://tex.stackexchange.com/users/119/willie-wong?tab=activity&sort=comments | Willie Wong
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Feb 12 comment Broken words and alignments in LaTeX For first question: en.wikibooks.org/wiki/LaTeX/Text_Formatting#Hyphenation // For second question: fully justified is the default. You have to issue \raggedleft or \raggedright (usually) to even make the ends of lines not justified. But whatever your cause ctan.org/pkg/ragged2e?lang=en may help. Please, however, consider writing a Minimal Working Example to illustrate the undesireable behavior so we know better how to answer your question. Dec 22 comment Having regular text within a lemma It is much more common to use \mathrm{Ric} instead, FWIW. (Most geometers would not consider it an operator.) Dec 15 comment How can I write a corrected format of reference for a target journal Proof stage is after the journal accepts the paper and is preparing the publication. I've never heard of any serious journal in mathematics or physics that rejects a manuscript due to citation formatting. Even for non-adherence of journal style for the main body it is rare. Dec 15 comment What is this math symbol and its latex representation? Looks a lot like the capital C is the calligraphic Euler fonts. See texdoc.net/texmf-dist/doc/fonts/amsfonts/euscript.pdf But seems also slightly different. Dec 15 comment How can I write a corrected format of reference for a target journal I can't help but wonder "Why?" Your excerpt clearly states that the journal will apply its on formatting at the proof stage, so why waste your time doing this when the journal has volunteered theirs? // That said, basically you are looking at building, by yourself, a collection of biblatex styles. Then perhaps you should write your own class file that allows you to set in the preamble once and for all the target journal. I know that Loop Space has done something like that once, maybe he can comment. Dec 14 comment Delete commas within numbers @Joe: David's answer is what I would try (by recursion), if limited to the world of TeX. Dec 14 comment Delete commas within numbers I am not well-versed enough in TeX internals to explain to you exactly what went wrong with your original try. This 'solution' is arrived at from the "guess-and-try" department. Dec 11 comment How to hyperref prefixed URLs? @KonradHöffner: Ah, I sort-of see the problem. Try the version I am about to put up in 5 minutes. Some veterans however may come by and chide me for my poor code form. Dec 10 comment How to hyperref prefixed URLs? @KonradHöffner: it works for me. Can you past your code? The command \urlprefixabbrv@ should never be called by itself: the defined command should be \urlprefixabbrv@string where string is the class of prefix (in the example above, the command defined is \urlprefixabbrv@dbo). If you can construct a minimum non-working example, please edit it into the question and ping me again, and I'll see if I can debug. Right now I cannot reproduce your error so I don't know what it is that went wrong. Oct 26 comment minimal looking chapter dependency wheel If you are not looking for fancy, you can even just use an array environment and make use of the various arrows available in math environments. Oct 26 comment minimal looking chapter dependency wheel What you are looking for is a "flow chart". For really complicated ones you can use dia. For less-so ones options abound: any of the common graphics producing methods for LaTeX can work: asymptote or TikZ being common choices. You can also use/abuse packages designed to draw commutative diagrams such as amscd or xypic. Oct 22 comment How to hyperref prefixed URLs? Is your syntax for \prefixurl flexible? It would be somewhat easier to code if you allow the syntax \prefixurl{dbo}{City}. Sep 16 comment biblatex set author font based on bib field @moewe: go ahead. I don't have much time to write up a complete answer (hence the comment). I agree that \ifkeyword would be cleaner. After reading the comments I personally prefer Alan's option of creating a new entry type. But I can see situations where having the keyword based answer can be helpful (for example, multiple different keywords that can stack). Sep 15 comment biblatex set author font based on bib field You are looking for the \iffieldequalstr command. In the current version of the biblatex documentation it is in section 4.6.2 on page 182. Jun 26 comment Write diagrams without a separate style file Congrats! (And I will file this info away too in case I need it in the future.) Jun 23 comment Write diagrams without a separate style file Try first using the AMScd package; that has the highest chance of being accepted. Jun 16 comment Write diagrams without a separate style file Hmm... I should note that in many other cases the easiest thing would be to just re-submit the article with the relevant .sty file included in your submission. But TikZ has so many of them and it may be more hassle than it is worth. Jun 12 comment Temporarily changing catcode of % Thanks, that's exactly the sort of comment that I am looking for, and thinking about it I can see the advantage to what you wrote. +1 Jun 12 comment Temporarily changing catcode of % @Andrew: that's exactly my intended approach. Jun 12 comment Temporarily changing catcode of % +1: that's a solution I didn't think of. I'll have to check whether showing white space instead of %d is acceptable, but it does solve the problem of latex not running. | {"extraction_info": {"found_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": 0.8253101706504822, "perplexity": 1553.7292175423497}, "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/1461860118790.25/warc/CC-MAIN-20160428161518-00025-ip-10-239-7-51.ec2.internal.warc.gz"} |
http://www.maplesoft.com/support/help/MapleSim/view.aspx?path=Svd(deprecated) | compute the singular values/vectors of a numeric matrix - Maple Help
Svd - compute the singular values/vectors of a numeric matrix
Calling Sequence Svd(X) Svd(X, U,left) Svd(X, V,right) Svd(X, U, V)
Parameters
X - n x p matrix U - (optional) the left singular vectors are to be returned in U V - (optional) the right singular vectors are to be returned in V
Description
• Important: The command Svd has been deprecated. Use the superseding command LinearAlgebra[SingularValues] instead.
• Svd(X) returns a 1 by $\mathrm{min}\left(n,p\right)$ array of the singular values of X.
• The entries of X must be all numerical.
• Svd(X,U,left) returns the singular values and the left singular vectors in U.
• Svd(X,V,right) returns the singular values and the right singular vectors in V.
• Svd(X,U,V) returns the singular values and the left and right singular vectors in U and V respectively. The singular vectors together with the singular values satisfy $U'\mathrm{XV}=\mathrm{D}$ where U' is the transpose of U and U is n by n, V is p by p, X is n by p, and D is n by p where ${\mathrm{D}}_{i,i}$ is/are the singular value/values of X.
• This procedure Svd is compatible with the Fortran library linpack.
• Note that nothing happens when the user invokes Svd(X) (same for other calling sequences); the user must use evalf(Svd(X)) to actually compute the singular values and singular vectors.
Examples
Important: The command Svd has been deprecated. Use the superseding command LinearAlgebra[SingularValues] instead.
> $A:=\mathrm{linalg}[\mathrm{matrix}]\left(2,2,\left[1,2,3,4\right]\right)$
${A}{:=}\left[\begin{array}{rr}{1}& {2}\\ {3}& {4}\end{array}\right]$ (1)
> $\mathrm{evalf}\left(\mathrm{Svd}\left(A\right)\right)$
$\left[\begin{array}{cc}{5.46498570421904}& {0.365966190626257}\end{array}\right]$ (2) | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 7, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9447541236877441, "perplexity": 2278.9529106062614}, "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/1454701163729.14/warc/CC-MAIN-20160205193923-00271-ip-10-236-182-209.ec2.internal.warc.gz"} |
https://cdsweb.cern.ch/collection/ATLAS%20Preprints?ln=hr | # ATLAS Preprints
Najnovije dodano:
2018-04-25
16:55
A search for lepton-flavor-violating decays of the $Z$ boson into a $\tau$-lepton and a light lepton with the ATLAS detector Direct searches for lepton flavor violation in decays of the $Z$ boson with the ATLAS detector at the LHC are presented. [...] CERN-EP-2018-052. - 2018. Fulltext - Previous draft version
2018-04-17
14:27
Search for pair production of Higgs bosons in the $b\bar{b}b\bar{b}$ final state using proton-proton collisions at $\sqrt{s} = 13$ TeV with the ATLAS detector / ATLAS Collaboration A search for Higgs boson pair production in the $b\bar{b}b\bar{b}$ final state is carried out with up to 36.1 $\mathrm{fb}^{-1}$ of LHC proton--proton collision data collected at $\sqrt{s} = 13$ TeV with the ATLAS detector in 2015 and 2016. [...] CERN-EP-2018-029 ; arXiv:1804.06174 ; CERN-EP-2018-029. - 2018. - 50 p. Fulltext - Previous draft version - Fulltext
2018-04-16
14:21
Hard QCD Measurements at LHC / Pasztor, Gabriella (Eotvos U.) /ATLAS, CMS and LHCB Collaborations The rich proton-proton collision data of the LHC allow to study QCD processes in a previously unexplored region with ever improving precision. This paper summarises recent results of the ATLAS, CMS and LHCb Collaborations using primarily multi-jet and vector boson plus jet data collected at $\sqrt s$ = 8 and 13 TeV. [...] CMS-CR-2018-029.- Geneva : CERN, 2018 - 21 p. Fulltext: PDF; In : The 28th International Symposium on Lepton-Photon Interactions at High Energies, Guangzhou, Guangdong Sheng, China, 7 - 12 Aug 2017
2018-04-13
10:01
Top physics in ATLAS and CMS / Diez, Carmen (DESY) /ATLAS and CMS Collaborations A selection of recent results of top quark production performed by the ATLAS and CMS Collaborations at the LHC is presented. The results include measurements of top quark pair production, including inclusive and differential cross sections, as well as cross sections for ${\rm t\bar{t}}$ production in association with additional heavy-quark jets or additional bosons. [...] CMS-CR-2018-027.- Geneva : CERN, 2018 - 13 p. Fulltext: PDF; In : Corfu2017: 17th Hellenic School and Workshops on Elementary Particle Physics and Gravity, Corfu, Attiki, Greece, 2 - 28 Sep 2017
2018-04-10
18:31
Search for low-mass dijet resonances using trigger-level jets with the ATLAS detector in $pp$ collisions at $\sqrt{s}=13$ TeV / ATLAS Collaboration Searches for dijet resonances with sub-TeV masses using the ATLAS detector at the Large Hadron Collider can be statistically limited by the bandwidth available to inclusive single-jet triggers, whose data-collection rates at low transverse momentum are much lower than the rate from Standard Model multijet production. [...] CERN-EP-2018-033 ; arXiv:1804.03496 ; CERN-EP-2018-033. - 2018. - 30 p. Fulltext - Previous draft version - Fulltext
2018-04-10
18:22
Search for R-parity-violating supersymmetric particles in multi-jet final states produced in $p$--$p$ collisions at $\sqrt{s} =$ 13 TeV using the ATLAS detector at the LHC / ATLAS Collaboration Results of a search for gluino pair production with subsequent R-parity-violating decays to quarks are presented. [...] CERN-EP-2017-298 ; arXiv:1804.03568 ; CERN-EP-2017-298. - 2018. - 39 p. Fulltext - Previous draft version - Fulltext
2018-04-10
18:07
Search for supersymmetry in events with four or more leptons in $\sqrt{s}=13$ TeV $pp$ collisions with ATLAS / ATLAS Collaboration Results from a search for supersymmetry in events with four or more charged leptons (electrons, muons and taus) are presented. [...] CERN-EP-2017-300 ; arXiv:1804.03602 ; CERN-EP-2017-300. - 2018. - 45 p. Fulltext - Previous draft version - Fulltext
2018-04-07
08:05
Model-independent constraints on the Abelian $Z'$ couplings within the ATLAS data on the dilepton production processes at $\sqrt{s} =$ 13 TeV / Pevzner, A.O. (Dnepropetrovsk Natl. U.) ; Skalozub, V.V. (Dnepropetrovsk Natl. U.) ; Gulov, A.V. (Dnepropetrovsk Natl. U.) ; Pankov, A.A. (Dubna, JINR ; Minsk, Inst. Nucl. Problems ; Gomel State Tech. U., ICTP) The study of lepton pair production is a powerful test of the Standard Model (SM) and can be used to search for phenomena beyond the SM. [...] arXiv:1803.07532. - 6 p. Fulltext
2018-04-03
22:54
Search for a heavy Higgs boson decaying into a $Z$ boson and another heavy Higgs boson in the $\ell\ell bb$ final state in $pp$ collisions at $\sqrt{s}=13$ TeV with the ATLAS detector / ATLAS Collaboration A search for a heavy neutral Higgs boson, $A$, decaying into a $Z$ boson and another heavy Higgs boson, $H$, is performed using a data sample corresponding to an integrated luminosity of 36.1 fb$^{-1}$ from proton--proton collisions at $\sqrt{s} = 13$ TeV recorded in 2015 and 2016 by the ATLAS detector at the Large Hadron Collider. [...] CERN-EP-2018-030 ; arXiv:1804.01126 ; CERN-EP-2018-030. - 2018. - 39 p. Fulltext - Previous draft version - Fulltext
2018-03-28
18:16
Search for Higgs boson decays into pairs of light (pseudo)scalar particles in the $\gamma\gamma jj$ final state in $pp$ collisions at $\sqrt{s}=13$ TeV with the ATLAS detector / ATLAS Collaboration This Letter presents a search for exotic decays of the Higgs boson to a pair of new (pseudo)scalar particles, $H\to aa$, with a mass in the range 20--60 GeV, and where one of the $a$ bosons decays into a pair of photons and the other to a pair of gluons. [...] CERN-EP-2017-295 ; arXiv:1803.11145. - 2018. - 31 p. Fulltext - Previous draft version - Fulltext | {"extraction_info": {"found_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": 0.9942055344581604, "perplexity": 3257.9863048407246}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-17/segments/1524125948549.21/warc/CC-MAIN-20180426203132-20180426223132-00348.warc.gz"} |
http://mathcs.chapman.edu/~jipsen/structures/doku.php/boolean_semilattices | Boolean semilattices
Abbreviation: BSlat
Definition
A Boolean semilattice is a structure $\mathbf{A}=\langle A,\vee,0, \wedge,1,\neg,\cdot\rangle$ such that
$\mathbf{A}$ is in the variety generated by complex algebras of semilattices
Let $\mathbf{S}=\langle S,\cdot\rangle$ be a semilattice. The complex algebra of $\mathbf{S}$ is $Cm(\mathbf{S})=\langle P(S),\cup,\emptyset,\cap,S,-,\cdot\rangle$, where $\langle P(S),\cup,\emptyset, \cap,S,-\rangle$ is the Boolean algebra of subsets of $S$, and
$X\cdot Y=\{x\cdot y\mid x\in X,\ y\in Y\}$.
Morphisms
Let $\mathbf{A}$ and $\mathbf{B}$ be Boolean semilattices. A morphism from $\mathbf{A}$ to $\mathbf{B}$ is a function $h:A\rightarrow B$ that is a Boolean homomorphism and preserves $\cdot$:
$h(x\cdot y)=h(x)\cdot h(y)$
Example 1:
Properties
Classtype variety open no unbounded yes yes yes, $n=2$ yes yes
Finite members
$\begin{array}{lr} f(1)= &1\\ f(2)= &1\\ f(3)= &0\\ f(4)= &5\\ f(5)= &0\\ f(6)= &0\\ f(7)= &0\\ f(8)= &\ge 97\text{ out of }104\\ \end{array}$
References
\end{document} %</pre> | {"extraction_info": {"found_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": 0.9850621223449707, "perplexity": 648.6328496764169}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-04/segments/1610703519984.9/warc/CC-MAIN-20210120085204-20210120115204-00399.warc.gz"} |
https://eng.libretexts.org/Sandboxes/CareySmith_at_vcccd.edu/MATLAB_Commands_and_Functions/06%3A_Mathematical_Functions/6.05%3A_Statistical_Functions | Skip to main content
# 6.5: Statistical Functions
$$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$ $$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$$$\newcommand{\AA}{\unicode[.8,0]{x212B}}$$
Function Result
erf(x)
Computes the error function erf (x).
mean
Calculates the average.
median
Calculates the median.
std
Calculates the standard deviation.
6.5: Statistical Functions is shared under a CC BY 1.3 license and was authored, remixed, and/or curated by Brian Vick, Virginia Tech.
• Was this article helpful? | {"extraction_info": {"found_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": 0.997208833694458, "perplexity": 2892.0532785306664}, "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-2022-49/segments/1669446710719.4/warc/CC-MAIN-20221130024541-20221130054541-00227.warc.gz"} |
http://homtools.lma.cnrs-mrs.fr/spip.php?article2 | ]
### Heterogeneous Materials
Homtools can be used to estimate the average mechanical response of a heterogeneous material by the way of Finite Element Analysis on a Representative Volume Element (RVE). It allows to compute the relationship between the average stress and the average strain, according to three different methods of localization: kinematic uniform boundary conditions (KUBC), periodic boundary conditions (PBC) and static uniform boundary conditions (SUBC). For all these methods, the average loading can be either the average stress or the average strain (or an independent combination of average strains and stresses components). The constitutive materials of the RVE can be either linear or non-linear and the analysis can be in small or finite-strain. The presence of interaction properties (such as contact and friction) is naturally taken into account.
# Localisation boundary conditions
The following figures illustrate a simple 2D test case of a stiff elastic cylindrical filler in a soft elastic matrix, treated with the three available methods in Homtools: KUBC (Kinematic Uniform Boundary Conditions), PBC (Periodic Boundary Conditions), SUBC (Static Uniform Boundary Conditions).
These methods require specific boundary conditions.
KUBC (click for a description)
This method consists in applying on the boundary the displacement field that would occur if the strain were uniform inside the RVE.
In small strains, the boundary conditions are : in which : is the displacement field, is the strain field, is the position vector of the point on the boundary. is the averaging operator over the RVE: is the inner product. In finite strains, the boundary conditions are : in which : is the deformation gradient, is the identity tensor. is the position vector of the point on the boundary in the reference configuration. is the averaging operator over the RVE in the reference configuration: is the inner product.
There is no restriction concerning the use of this method, except that no rigid part must intersect the boundary (holes are permitted).
SUBC (click for a description)
This method consists in applying on the boundary the stress vector field that would occur if the stress were uniform inside the RVE.
In small strains, the boundary conditions are : in which : is the cauchy stress tensor, is the outwarding normal. is the averaging operator over the RVE: is the inner product. In finite strains, the boundary conditions are : in which : is the Piolla Kirchoff II stress tensor, is the outwarding normal in the reference configuration. is the averaging operator over the RVE in the reference configuration: is the inner product.
There is no restriction concerning the use of this method, except that no holes must intersect the boundary (rigid parts are permitted but a specific treatment is required).
PBC (click for a description)
The method is theoretically relevant for periodic media, which can be defined by a periodicity cell and the associated periodicity vector of translation. The periodic homogenisation process consists in assuming that the strains and stresses are periodic at the level of the periodicity cell (which is defined as the RVE). The periodicity of stresses and strains leads to specific periodic boundary conditions for the localisation problem on the RVE.
In small strains, the boundary conditions are : periodic and antiperiodic. A quantity is said to be "periodic" (resp. "antiperiodic") when it takes the same value (resp. opposite value) at two opposite points on the boundary (one of them being the image of the other by translation of a periodicity vector). In finite strains, the boundary conditions are : periodic and antiperiodic. A quantity is in this case said to be "periodic" (resp. "antiperiodic") when it takes the same value (resp. opposite value) at two opposite points on the boundary (one of them being the image of the other by translation of a periodicity vector in the initial configuration).
There is no restriction concerning the use of this method (periodic holes and rigid parts intersecting the boundary are permitted but a specific treatment is required for the rigid parts).
The difficulty of applying these methods resides in the creation of the corresponding non classical boundary conditions, that can be written as linear constraints between the displacements at each point on the boundary of the RVE and the average strains.
In Homtools, we use Reference Points to prescribe the average loading and these constraints.
# Reference Points to apply the average loading and the specific boundary conditions
The average strains are introduced as degrees of freedom of Reference Points and Homtools automatically generates the linear constraints (*equation) between these additionnal degrees of freedom and the displacements on the boundary of the RVE. Thanks to duality properties, the reaction forces at the Reference Points are the average stresses multiplied by the volume of the RVE. With this method, prescribing average strains or stresses becomes as easy as prescribing displacements or concentrated forces at a node. Furthermore, obtaining the relationship between the average strain and the average stress does not require any specific post-treatment.
For fiinte strain problems, the degrees of freedom of the Reference Points are the components of the average deformation gradient tensor and the reaction forces are the components of the avrage Piola-Kirchoff II stress tensor.
# How to use Homtools
The main steps are :
mesh the RVE
in the "Interaction" module, create the Reference Points (2 for 2D problems in plane stress or strain or 3D problems in small strains, 3 for 3D problems in finite strains)
in the "Interaction" module, choose "Homtools" in the drop off menu "Plug-in" and answer the questions.
in the "Load" module, prescribe the average strains components or/and the stresses components by prescribing the degrees of freedom (dsplacements) or the nodal forces at the reference points.
after solving the problem, in the "Vizualisation" module, you can obtain the effective response of the RVE by the postreatment of the displacements and yhe nodal forces at the Reference Points.
Check the following video to see the steps required to define a complete homogenization problem (in french).
# Remarks
for the method SUBC, no hole should intersect the boundary of the RVE (inconsistency of the method in this case). Furthermore, the displacement field is not unique for this method (rigid translations and rotations are not fixed)
the method PBC is theoretically only relevant for periodic materials. The linear constraints couple the displacements at nodes on opposite faces of the boundary. The mesh of opposite faces must be identical: at each node of a face must correspond a node on the opposite face defined by translation of a periodicity vector. In Homtools, you must define the couple of faces together with a periodicity vectors : the mesh of the second set must result from the translation of the mesh of the first set according to the periodicity vector. You must repeat the operation as many times as necessary for all the couple of faces. Fot this method, the displacement field is not unique: the rigid translation is not fixed by the periodicity conditions and you can make the solution unique by fixing the dispalcements at a node in the RVE. | {"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": 0.8500941395759583, "perplexity": 881.8152190375139}, "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/1585371883359.91/warc/CC-MAIN-20200410012405-20200410042905-00133.warc.gz"} |
https://people.maths.bris.ac.uk/~matyd/GroupNames/449/D7xC34.html | Copied to
clipboard
## G = D7×C34order 476 = 22·7·17
### Direct product of C34 and D7
Aliases: D7×C34, C14⋊C34, C2383C2, C1194C22, C7⋊(C2×C34), SmallGroup(476,9)
Series: Derived Chief Lower central Upper central
Derived series C1 — C7 — D7×C34
Chief series C1 — C7 — C119 — D7×C17 — D7×C34
Lower central C7 — D7×C34
Upper central C1 — C34
Generators and relations for D7×C34
G = < a,b,c | a34=b7=c2=1, ab=ba, ac=ca, cbc=b-1 >
Smallest permutation representation of D7×C34
On 238 points
Generators in S238
(1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34)(35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68)(69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102)(103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136)(137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155 156 157 158 159 160 161 162 163 164 165 166 167 168 169 170)(171 172 173 174 175 176 177 178 179 180 181 182 183 184 185 186 187 188 189 190 191 192 193 194 195 196 197 198 199 200 201 202 203 204)(205 206 207 208 209 210 211 212 213 214 215 216 217 218 219 220 221 222 223 224 225 226 227 228 229 230 231 232 233 234 235 236 237 238)
(1 193 124 86 170 233 42)(2 194 125 87 137 234 43)(3 195 126 88 138 235 44)(4 196 127 89 139 236 45)(5 197 128 90 140 237 46)(6 198 129 91 141 238 47)(7 199 130 92 142 205 48)(8 200 131 93 143 206 49)(9 201 132 94 144 207 50)(10 202 133 95 145 208 51)(11 203 134 96 146 209 52)(12 204 135 97 147 210 53)(13 171 136 98 148 211 54)(14 172 103 99 149 212 55)(15 173 104 100 150 213 56)(16 174 105 101 151 214 57)(17 175 106 102 152 215 58)(18 176 107 69 153 216 59)(19 177 108 70 154 217 60)(20 178 109 71 155 218 61)(21 179 110 72 156 219 62)(22 180 111 73 157 220 63)(23 181 112 74 158 221 64)(24 182 113 75 159 222 65)(25 183 114 76 160 223 66)(26 184 115 77 161 224 67)(27 185 116 78 162 225 68)(28 186 117 79 163 226 35)(29 187 118 80 164 227 36)(30 188 119 81 165 228 37)(31 189 120 82 166 229 38)(32 190 121 83 167 230 39)(33 191 122 84 168 231 40)(34 192 123 85 169 232 41)
(1 59)(2 60)(3 61)(4 62)(5 63)(6 64)(7 65)(8 66)(9 67)(10 68)(11 35)(12 36)(13 37)(14 38)(15 39)(16 40)(17 41)(18 42)(19 43)(20 44)(21 45)(22 46)(23 47)(24 48)(25 49)(26 50)(27 51)(28 52)(29 53)(30 54)(31 55)(32 56)(33 57)(34 58)(69 86)(70 87)(71 88)(72 89)(73 90)(74 91)(75 92)(76 93)(77 94)(78 95)(79 96)(80 97)(81 98)(82 99)(83 100)(84 101)(85 102)(103 166)(104 167)(105 168)(106 169)(107 170)(108 137)(109 138)(110 139)(111 140)(112 141)(113 142)(114 143)(115 144)(116 145)(117 146)(118 147)(119 148)(120 149)(121 150)(122 151)(123 152)(124 153)(125 154)(126 155)(127 156)(128 157)(129 158)(130 159)(131 160)(132 161)(133 162)(134 163)(135 164)(136 165)(171 228)(172 229)(173 230)(174 231)(175 232)(176 233)(177 234)(178 235)(179 236)(180 237)(181 238)(182 205)(183 206)(184 207)(185 208)(186 209)(187 210)(188 211)(189 212)(190 213)(191 214)(192 215)(193 216)(194 217)(195 218)(196 219)(197 220)(198 221)(199 222)(200 223)(201 224)(202 225)(203 226)(204 227)
G:=sub<Sym(238)| (1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34)(35,36,37,38,39,40,41,42,43,44,45,46,47,48,49,50,51,52,53,54,55,56,57,58,59,60,61,62,63,64,65,66,67,68)(69,70,71,72,73,74,75,76,77,78,79,80,81,82,83,84,85,86,87,88,89,90,91,92,93,94,95,96,97,98,99,100,101,102)(103,104,105,106,107,108,109,110,111,112,113,114,115,116,117,118,119,120,121,122,123,124,125,126,127,128,129,130,131,132,133,134,135,136)(137,138,139,140,141,142,143,144,145,146,147,148,149,150,151,152,153,154,155,156,157,158,159,160,161,162,163,164,165,166,167,168,169,170)(171,172,173,174,175,176,177,178,179,180,181,182,183,184,185,186,187,188,189,190,191,192,193,194,195,196,197,198,199,200,201,202,203,204)(205,206,207,208,209,210,211,212,213,214,215,216,217,218,219,220,221,222,223,224,225,226,227,228,229,230,231,232,233,234,235,236,237,238), (1,193,124,86,170,233,42)(2,194,125,87,137,234,43)(3,195,126,88,138,235,44)(4,196,127,89,139,236,45)(5,197,128,90,140,237,46)(6,198,129,91,141,238,47)(7,199,130,92,142,205,48)(8,200,131,93,143,206,49)(9,201,132,94,144,207,50)(10,202,133,95,145,208,51)(11,203,134,96,146,209,52)(12,204,135,97,147,210,53)(13,171,136,98,148,211,54)(14,172,103,99,149,212,55)(15,173,104,100,150,213,56)(16,174,105,101,151,214,57)(17,175,106,102,152,215,58)(18,176,107,69,153,216,59)(19,177,108,70,154,217,60)(20,178,109,71,155,218,61)(21,179,110,72,156,219,62)(22,180,111,73,157,220,63)(23,181,112,74,158,221,64)(24,182,113,75,159,222,65)(25,183,114,76,160,223,66)(26,184,115,77,161,224,67)(27,185,116,78,162,225,68)(28,186,117,79,163,226,35)(29,187,118,80,164,227,36)(30,188,119,81,165,228,37)(31,189,120,82,166,229,38)(32,190,121,83,167,230,39)(33,191,122,84,168,231,40)(34,192,123,85,169,232,41), (1,59)(2,60)(3,61)(4,62)(5,63)(6,64)(7,65)(8,66)(9,67)(10,68)(11,35)(12,36)(13,37)(14,38)(15,39)(16,40)(17,41)(18,42)(19,43)(20,44)(21,45)(22,46)(23,47)(24,48)(25,49)(26,50)(27,51)(28,52)(29,53)(30,54)(31,55)(32,56)(33,57)(34,58)(69,86)(70,87)(71,88)(72,89)(73,90)(74,91)(75,92)(76,93)(77,94)(78,95)(79,96)(80,97)(81,98)(82,99)(83,100)(84,101)(85,102)(103,166)(104,167)(105,168)(106,169)(107,170)(108,137)(109,138)(110,139)(111,140)(112,141)(113,142)(114,143)(115,144)(116,145)(117,146)(118,147)(119,148)(120,149)(121,150)(122,151)(123,152)(124,153)(125,154)(126,155)(127,156)(128,157)(129,158)(130,159)(131,160)(132,161)(133,162)(134,163)(135,164)(136,165)(171,228)(172,229)(173,230)(174,231)(175,232)(176,233)(177,234)(178,235)(179,236)(180,237)(181,238)(182,205)(183,206)(184,207)(185,208)(186,209)(187,210)(188,211)(189,212)(190,213)(191,214)(192,215)(193,216)(194,217)(195,218)(196,219)(197,220)(198,221)(199,222)(200,223)(201,224)(202,225)(203,226)(204,227)>;
G:=Group( (1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34)(35,36,37,38,39,40,41,42,43,44,45,46,47,48,49,50,51,52,53,54,55,56,57,58,59,60,61,62,63,64,65,66,67,68)(69,70,71,72,73,74,75,76,77,78,79,80,81,82,83,84,85,86,87,88,89,90,91,92,93,94,95,96,97,98,99,100,101,102)(103,104,105,106,107,108,109,110,111,112,113,114,115,116,117,118,119,120,121,122,123,124,125,126,127,128,129,130,131,132,133,134,135,136)(137,138,139,140,141,142,143,144,145,146,147,148,149,150,151,152,153,154,155,156,157,158,159,160,161,162,163,164,165,166,167,168,169,170)(171,172,173,174,175,176,177,178,179,180,181,182,183,184,185,186,187,188,189,190,191,192,193,194,195,196,197,198,199,200,201,202,203,204)(205,206,207,208,209,210,211,212,213,214,215,216,217,218,219,220,221,222,223,224,225,226,227,228,229,230,231,232,233,234,235,236,237,238), (1,193,124,86,170,233,42)(2,194,125,87,137,234,43)(3,195,126,88,138,235,44)(4,196,127,89,139,236,45)(5,197,128,90,140,237,46)(6,198,129,91,141,238,47)(7,199,130,92,142,205,48)(8,200,131,93,143,206,49)(9,201,132,94,144,207,50)(10,202,133,95,145,208,51)(11,203,134,96,146,209,52)(12,204,135,97,147,210,53)(13,171,136,98,148,211,54)(14,172,103,99,149,212,55)(15,173,104,100,150,213,56)(16,174,105,101,151,214,57)(17,175,106,102,152,215,58)(18,176,107,69,153,216,59)(19,177,108,70,154,217,60)(20,178,109,71,155,218,61)(21,179,110,72,156,219,62)(22,180,111,73,157,220,63)(23,181,112,74,158,221,64)(24,182,113,75,159,222,65)(25,183,114,76,160,223,66)(26,184,115,77,161,224,67)(27,185,116,78,162,225,68)(28,186,117,79,163,226,35)(29,187,118,80,164,227,36)(30,188,119,81,165,228,37)(31,189,120,82,166,229,38)(32,190,121,83,167,230,39)(33,191,122,84,168,231,40)(34,192,123,85,169,232,41), (1,59)(2,60)(3,61)(4,62)(5,63)(6,64)(7,65)(8,66)(9,67)(10,68)(11,35)(12,36)(13,37)(14,38)(15,39)(16,40)(17,41)(18,42)(19,43)(20,44)(21,45)(22,46)(23,47)(24,48)(25,49)(26,50)(27,51)(28,52)(29,53)(30,54)(31,55)(32,56)(33,57)(34,58)(69,86)(70,87)(71,88)(72,89)(73,90)(74,91)(75,92)(76,93)(77,94)(78,95)(79,96)(80,97)(81,98)(82,99)(83,100)(84,101)(85,102)(103,166)(104,167)(105,168)(106,169)(107,170)(108,137)(109,138)(110,139)(111,140)(112,141)(113,142)(114,143)(115,144)(116,145)(117,146)(118,147)(119,148)(120,149)(121,150)(122,151)(123,152)(124,153)(125,154)(126,155)(127,156)(128,157)(129,158)(130,159)(131,160)(132,161)(133,162)(134,163)(135,164)(136,165)(171,228)(172,229)(173,230)(174,231)(175,232)(176,233)(177,234)(178,235)(179,236)(180,237)(181,238)(182,205)(183,206)(184,207)(185,208)(186,209)(187,210)(188,211)(189,212)(190,213)(191,214)(192,215)(193,216)(194,217)(195,218)(196,219)(197,220)(198,221)(199,222)(200,223)(201,224)(202,225)(203,226)(204,227) );
G=PermutationGroup([(1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34),(35,36,37,38,39,40,41,42,43,44,45,46,47,48,49,50,51,52,53,54,55,56,57,58,59,60,61,62,63,64,65,66,67,68),(69,70,71,72,73,74,75,76,77,78,79,80,81,82,83,84,85,86,87,88,89,90,91,92,93,94,95,96,97,98,99,100,101,102),(103,104,105,106,107,108,109,110,111,112,113,114,115,116,117,118,119,120,121,122,123,124,125,126,127,128,129,130,131,132,133,134,135,136),(137,138,139,140,141,142,143,144,145,146,147,148,149,150,151,152,153,154,155,156,157,158,159,160,161,162,163,164,165,166,167,168,169,170),(171,172,173,174,175,176,177,178,179,180,181,182,183,184,185,186,187,188,189,190,191,192,193,194,195,196,197,198,199,200,201,202,203,204),(205,206,207,208,209,210,211,212,213,214,215,216,217,218,219,220,221,222,223,224,225,226,227,228,229,230,231,232,233,234,235,236,237,238)], [(1,193,124,86,170,233,42),(2,194,125,87,137,234,43),(3,195,126,88,138,235,44),(4,196,127,89,139,236,45),(5,197,128,90,140,237,46),(6,198,129,91,141,238,47),(7,199,130,92,142,205,48),(8,200,131,93,143,206,49),(9,201,132,94,144,207,50),(10,202,133,95,145,208,51),(11,203,134,96,146,209,52),(12,204,135,97,147,210,53),(13,171,136,98,148,211,54),(14,172,103,99,149,212,55),(15,173,104,100,150,213,56),(16,174,105,101,151,214,57),(17,175,106,102,152,215,58),(18,176,107,69,153,216,59),(19,177,108,70,154,217,60),(20,178,109,71,155,218,61),(21,179,110,72,156,219,62),(22,180,111,73,157,220,63),(23,181,112,74,158,221,64),(24,182,113,75,159,222,65),(25,183,114,76,160,223,66),(26,184,115,77,161,224,67),(27,185,116,78,162,225,68),(28,186,117,79,163,226,35),(29,187,118,80,164,227,36),(30,188,119,81,165,228,37),(31,189,120,82,166,229,38),(32,190,121,83,167,230,39),(33,191,122,84,168,231,40),(34,192,123,85,169,232,41)], [(1,59),(2,60),(3,61),(4,62),(5,63),(6,64),(7,65),(8,66),(9,67),(10,68),(11,35),(12,36),(13,37),(14,38),(15,39),(16,40),(17,41),(18,42),(19,43),(20,44),(21,45),(22,46),(23,47),(24,48),(25,49),(26,50),(27,51),(28,52),(29,53),(30,54),(31,55),(32,56),(33,57),(34,58),(69,86),(70,87),(71,88),(72,89),(73,90),(74,91),(75,92),(76,93),(77,94),(78,95),(79,96),(80,97),(81,98),(82,99),(83,100),(84,101),(85,102),(103,166),(104,167),(105,168),(106,169),(107,170),(108,137),(109,138),(110,139),(111,140),(112,141),(113,142),(114,143),(115,144),(116,145),(117,146),(118,147),(119,148),(120,149),(121,150),(122,151),(123,152),(124,153),(125,154),(126,155),(127,156),(128,157),(129,158),(130,159),(131,160),(132,161),(133,162),(134,163),(135,164),(136,165),(171,228),(172,229),(173,230),(174,231),(175,232),(176,233),(177,234),(178,235),(179,236),(180,237),(181,238),(182,205),(183,206),(184,207),(185,208),(186,209),(187,210),(188,211),(189,212),(190,213),(191,214),(192,215),(193,216),(194,217),(195,218),(196,219),(197,220),(198,221),(199,222),(200,223),(201,224),(202,225),(203,226),(204,227)])
170 conjugacy classes
class 1 2A 2B 2C 7A 7B 7C 14A 14B 14C 17A ··· 17P 34A ··· 34P 34Q ··· 34AV 119A ··· 119AV 238A ··· 238AV order 1 2 2 2 7 7 7 14 14 14 17 ··· 17 34 ··· 34 34 ··· 34 119 ··· 119 238 ··· 238 size 1 1 7 7 2 2 2 2 2 2 1 ··· 1 1 ··· 1 7 ··· 7 2 ··· 2 2 ··· 2
170 irreducible representations
dim 1 1 1 1 1 1 2 2 2 2 type + + + + + image C1 C2 C2 C17 C34 C34 D7 D14 D7×C17 D7×C34 kernel D7×C34 D7×C17 C238 D14 D7 C14 C34 C17 C2 C1 # reps 1 2 1 16 32 16 3 3 48 48
Matrix representation of D7×C34 in GL3(𝔽239) generated by
238 0 0 0 163 0 0 0 163
,
1 0 0 0 0 1 0 238 198
,
1 0 0 0 0 1 0 1 0
G:=sub<GL(3,GF(239))| [238,0,0,0,163,0,0,0,163],[1,0,0,0,0,238,0,1,198],[1,0,0,0,0,1,0,1,0] >;
D7×C34 in GAP, Magma, Sage, TeX
D_7\times C_{34}
% in TeX
G:=Group("D7xC34");
// GroupNames label
G:=SmallGroup(476,9);
// by ID
G=gap.SmallGroup(476,9);
# by ID
G:=PCGroup([4,-2,-2,-17,-7,6531]);
// Polycyclic
G:=Group<a,b,c|a^34=b^7=c^2=1,a*b=b*a,a*c=c*a,c*b*c=b^-1>;
// generators/relations
Export
×
𝔽 | {"extraction_info": {"found_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": 0.9972802996635437, "perplexity": 1333.3420030494815}, "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/1579251690379.95/warc/CC-MAIN-20200126195918-20200126225918-00193.warc.gz"} |
http://www.chegg.com/homework-help/questions-and-answers/find-total-admittance-networks-fig--15142-identify-values-total-conductance-admittance-dra-q2550033 | Find the total admittance for the networks of Fig below. 15.142. Identify the values of total conductance and admittance, and draw the admittance diagram. | {"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": 0.9487490057945251, "perplexity": 2285.5881765388867}, "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/1416400378431.55/warc/CC-MAIN-20141119123258-00032-ip-10-235-23-156.ec2.internal.warc.gz"} |
http://mathoverflow.net/questions/83438/cesaro-bounded-operator-which-is-not-power-bounded?sort=votes | # Cesaro bounded Operator which is not power bounded
Good evening!
Let X be a banachspace and T a bounded linear operator on X. The cesaro avearges of T are defined as:
$A_n:=\frac{1}{n} \sum\limits_{j=0}^{n-1}T^j$
We call T cesaro bounded if: $\sup_{n \geq 0}\Vert A_n \Vert<\infty$.
We call T power bounded if: $\sup_{n \geq 0}\Vert T^n \Vert<\infty$.
E. Hille showed in "Remarks on ergodic theorems, Trans. Amer. Math. Soc. 57, 1945, 246-269" that one can find a cesaro bounded Operator in $\mathcal{L}(L_1[0,1])$ which is not power bounded.
Here is my question: can this be achieved in a finite dimesional setting?
With best regards,
Matthias
-
Consider $T = \pmatrix{-1 & 1\cr 0 & -1\cr}$. Then $T^n = \pmatrix{(-1)^n & (-1)^{n+1} n\cr 0 & (-1)^n\cr}$ so $T$ is not power-bounded. But $A_n = \pmatrix{\frac{1-(-1)^n}{2n} & \frac{(-1)^n}{2} + \frac{1-(-1)^n}{4n}\cr 0 & \frac{1-(-1)^n}{2n}\cr}$ so it is cesaro-bounded.
You could replace $-1$ by any $\lambda \ne 1$ with $|\lambda|=1$.
-
Thank you for this nice example! – Matthias Dec 14 '11 at 20:37
I think you can read off from the Jordan canonical form that both conditions are equivalent to the spectral radius being at most one and every eigenvalue of modulus one having algebraic and geometric multiplict the same.
EDIT: As Robert Israel points out in his answer, my answer is wrong.
-
That eigenvalues of modulus 1 have equal algebraic and geometric multiplicities is necessary for it to be power-bounded, but not to be cesaro-bounded: see my example. – Robert Israel Dec 14 '11 at 18:40 | {"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": 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": 0.942172646522522, "perplexity": 609.1917711309043}, "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-2015-18/segments/1429246639482.79/warc/CC-MAIN-20150417045719-00074-ip-10-235-10-82.ec2.internal.warc.gz"} |
https://www.splashlearn.com/math-vocabulary/algebra/brackets | # Brackets in Math – Definition with Examples
## What are Brackets?
You must have seen different symbols like these: (, ), [, ], {, and } in your math books. These symbols are called brackets. Brackets in mathematics serve a very important purpose; these symbols help us group different expressions or numbers together. Brackets imply that the thing or expression enclosed by them is to be given higher precedence over other things.
## Different kinds of Brackets
In math, you will often have to use brackets while creating or solving equations. They help in grouping numbers and defining the order of operations. Generally, three kinds of brackets are used in mathematics,
• Parentheses or Round Brackets, ( )
• Curly or Brace Brackets { }
• Square or Box Brackets [ ]
Brackets always come in pairs, and if there is an opening bracket, there has to be a closing bracket. The opening brackets are (, [ and {. Their corresponding closing brackets are ), ] and }.
## Parentheses Brackets
These are also known as the round brackets and written as ( ). These are the most common types of brackets. They are used for grouping different values and equations together. Parentheses or “round brackets” are used to group terms together or specify the order of operations in an equation.
## How to Use Parentheses in Math?
1. In math, you can use parentheses in math to separate numbers. For instance, you can use them to mention negative numbers when writing an addition equation.
Here is an example to understand this better:
$3 + ( −5) = −2$
1. The second use of parentheses in math is to multiply numbers. If there is no arithmetic operation present in an equation, the presence of parentheses means you have to apply multiplication.
Let us understand this with an example:
$6 (4 + 2)$
can be written as $6 \times (4 + 2)$
Therefore, the answer is $6 \times 6 = 36$.
1. The third and final use of parentheses in math is to group numbers and define the order of operations.
2. When used simply around numbers, the round brackets denote multiplication.
For example : $(3)(4) = 12$
1. They can also be used to write negative integers in mathematical expressions.
For example $5 + ( −4) = 1$
1. Parentheses can also be used to separate out numbers from their exponents. For example: $(2)^{-3}$
Examples: $(2 + 4), 5(111), 25 − (12 + 8)$, etc.
## Curly Brackets
Braces in math are symbols that are used twice, once to open “{“ and once to close “}” an argument, expression, or equation. These are commonly referred to as curly brackets and written as { }.
In general, we use braces in math for two purposes:
• For grouping a large equation, in which the second-last bracket is braces or curly brackets. For example, $7[2 + \left\{3(1 + 1) + 1\right\}]$
• For denoting a set, such as {x, y, z,…}
Like Parentheses, curly brackets are also used to group various mathematical components; however, curly brackets are also used to depict sets or to write nested expressions. Examples:
$[\left\{4+[3 \times ( −2)\right\}] − [\left\{(4 \times 6)+(14 \div 7)\right\} − ( −3)]$,
$[\left\{12 − (12 − 2)\right\} + (5 − 7)] + 9$, etc.
### How Do We Use Braces in Math?
Braces in math are frequently used in mathematical expressions when we have two or more than two nested groups for calculations.
So, in the first nested group, we use parentheses. In the second nested group, we use braces, and in the third nested group, we use box brackets, which contain both parentheses and braces.
For example: $3[2 − \left\{4(2 + 2) + 2\right\}]$
Here, we have three nested groups with appropriate brackets.
So, the order of solving would be:
Fun Fact: Some conventions differentiate the order of solving brackets, which is:
We will use the first convention with curly brackets in the second position throughout this article.
You need to know the BODMAS or order of operations to simplify and solve a problem.
## Square Brackets
Square brackets are generally used to distinguish between sub-expressions of a complex mathematical expression.
Examples: $[100 − (3 − 1) + (7 \times 8)], 10 \times [(4 − 2) \times ( 4 \times 2)]$, etc.
## Order of Operations of Brackets
When we evaluate a mathematical expression that is made up of different brackets, we have to follow certain rules. This is called the rules of operation or order of operation of brackets.
When we have a long equation for multiplication, division, addition, and subtraction, we solve each function in order to find the right answer. If the problem is solved without this order, then the chances of getting a wrong answer are high!
• The general order of operation of the bracket can be illustrated as $[ \left\{ ( ) \right\} ]$; this means that in a given problem, you would have to simplify the values in the innermost bracket first. This means that first $( )$ brackets will be solved, following which, $\left\{ \right\}$ brackets are solved and finally $[ ]$ brackets.
• The second step in solving these problems is to look for an exponent; if there is any, solve it first.
• In the third step, we look for expressions with multiplication or division operators. If both the operators are present, we check the expression from left to right. Whichever operator comes first, we solve that operator first.
For example, in the expression, $10 \times 6 \div 5$, we check from left to right, since multiplication comes first so we solve multiplication first and then division.
$10 \times 6 \div 5$
$=60 \div 5$
$= 12$
• In the fourth and last step, we look for numbers that need to be added or subtracted. We follow the same instruction if both the operators are present, we look from left to right in the expression, and whichever operator comes first, we solve that expression first. But if the operations are in brackets, we always solve the brackets first since brackets have the utmost precedence.
To remember the above mention steps, we can use the acronym PEMDAS,
P – Parentheses,(or brackets)
E – Exponents, (or order)
M – Multiplication
D – Division
S – Subtraction.
Example 1: Let’s use pemdas to evaluate the expression
$100 − [(3 − 1) + (7 \times 8)]$
Step 1: Solve the brackets. Follow the order of solving round brackets $( )$ first, then curly brackets $\left\{ \right\}$, and then square brackets $[ ]$.
$= 100 − [(2) + (56)]$
$= 100 − 58$
Step 2: No exponent in the given expression.
Step 3: No multiplication or division in the given expression.
Step 4: Solve the subtraction.
$= 100 − 58$
$= 42$
Example 2: While we write the order in the above form, division or multiplication and addition or subtraction hold equal importance. This means that you can either take up multiplication first or division first.
Similarly, you can take either addition first or subtraction first. The answer will be the same. So, we usually try to solve these two from left to right.
Let’s solve the above example:
$4[2 + \left\{3(1 + 1) + 2\right\}]$
$= 4[2 + \left\{3(2) + 2\right\}]$
Now, we solve the braces or curly brackets.
$= 4[2 + \left\{6 + 2\right\}]$
$= 4[2 + 8]$
Then, we solve the square brackets.
$= 4[10]$
$= 40$
In summary:
Here is the order you can follow when multiple symbols are present in an equation:
If you come across parentheses in an equation, you will first look at the terms present within them.
Let us understand this better with an example.
Take the problem: $9 − 10 \div 5 – 3 \times 2 + 7$
Let us solve this using the order of operations you have learned.
$= 9 − 10 \div 5 – 3 \times 2 + 7$
$= 9 − 2 − 3 \times 2 + 7$ (First, you divide)
$= 9 − 2 − 6 + 7$ (Then, you multiply)
$= 7 − 6 + 7$ (Then, you subtract)
$= 1 + 7$ (Then, you subtract)
$= 8$ (And finally, you add)
Now, let us look at the same problem with parentheses:
$9 − 10 \div (5 − 3) \times 2 + 7$
You need to calculate the numbers within the parentheses first.
$= 9 − 10 \div 2 \times 2 + 7$ (Solve the expression inside the parentheses)
$= 9 − 5 \times 2 + 7$ (Divide)
$= 9 − 10 + 7$ (Multiply)
$= −1 + 7$ (Add)
$= 6$
Did you notice? The answer to the same equation changed because parentheses were present in the equation!
Point to Remember: If there are parentheses inside other parentheses, you solve the inner expression first.
Let us understand this with an example:
Simplify the expression $(2 + (3 \times 4))$
Here, we will solve the inner bracket first.
So, the expression will become $(2 + 12) = 14$
Note that it is highly recommended to write any mathematical equation or expression with proper use of parentheses, leaving no place for ambiguity. It is important to convey the intention behind writing the math operations and indicate which operations should be carried out first.
## Solved Examples
Question 1: Find the value of the expression: $(5 + 4) − (3 − 2)$.
$(5 + 4) − (3 − 2)$,
Step 1: Solving the values in the brackets,
$(9) − (1)$,
Thus, the answer is $(9) − (1) = 8$.
Question 2: Find the value of the expression: $\left\{(7 − 2) \times 3\right\} \div 5$
$\left\{(7 − 2) \times 3\right\} \div 5$
Step 1: Solving the parentheses
$\left\{(7 − 2) \times 3\right\} \div 5$
$= \left\{5 \times 3\right\} \div 5$
Solving the curly bracket
$= \left\{15\right\} \div 5$
$= 15 \div 5$
$= 3$
Question 3: Find the value of the expression: $(12 \div 6) \times (4 − 2)$
Solution:
The given equation is,
$(12 \div 6) \times (4 − 2)$
Solving the values in the brackets,
$(2) \times (2)$
Thus, the answer is $(2) \times (2) = 4$
Question 4: Find the value of the expression: $[120 + \left\{ (3 \times 4) + (4 − 2) − 1 \right\} + 20 ]$
Answer: Following the PEMDAS rule, first,
Step 1: We solve the values in ( ) brackets,
$[120 + \left\{ (3 \times 4) + (4 − 2) − 1 \right\} + 20 ]$
$= [ 120 + \left\{ (12 ) + ( 2 ) − 1 \right\} + 20 ]$,
Now we solve the values inside the { } brackets,
$= [ 120 + \left\{ 13 \right\} + 20 ]$,
Finally, add all the values in the [ ] bracket,
Example 5: Simplify the expression: $(2 + 4 \times 6) − 4 + (2 \times 3)$
Solution: Start by solving the expressions inside the parentheses.
$= (2 + 24) − 4 + 6$ (Multiply inside the parentheses)
$= 26 − 4 + 6$ (Solve the terms inside the parentheses)
$= 22 + 6$ (Add)
$= 28$
Example 6: Simplify the expression: $( 2 \times (7 − 5)) − ((6 \div 3) + 4)$
Start by solving the innermost parentheses
$= (2 \times 2) − (2 + 4)$
$= 4 − 6$
$= − 2$
Example 7: Simplify the expression: $2 (3 + 5) + 8 (4 − 1)$
First, solve the expressions within the parentheses.
Here, the parentheses also denote a multiplication sign.
$= 2 \times 8 + 8 \times 3$
$= 16 + 24$
$= 40$
Example 8: If you have to solve the following equation, how will you proceed?
$2[1 − \left\{2(2 + 2) + 2\right\}]$
Solution: We solve the parentheses first:
$= 2[1 − \left\{2(4) + 2\right\}]$
$= 2[1 − \left\{8 + 2\right\}]$
Now, we solve the braces:
$= 2[1 − \left\{10\right\}]$
Finally, we solve the square brackets:
$= 2[ −9]$
$= −18$
Example 9: How would you solve the following equation?
$4\left\{5(4 + 2) + 1\right\}$
Solution: First, we solve the parentheses:
$= 4\left\{5(6) + 1\right\}$
Now, we need to solve the curly brackets. But within these brackets, we have to solve multiplication and addition.
So, we multiply first and then add:
$= 4 \left\{30 + 1\right\}$
$= 4 \left\{31\right\}$
Finally, we multiply 4 with the value inside the braces:
$= 124$
Example 10: What is the process you will follow to solve an equation with more than one parentheses?
$20 \div \left\{1(2 + 2) + (3 + 3)\right\}$
Solution: We will start by solving the equations within the parentheses:
$= 20 \div \left\{1(4) + (3 + 3)\right\}$
$= 20 \div \left\{1(4) + (6)\right\}$
Now, we have to solve the equation within the braces, but we have multiplication within the curly brackets, so we will solve that first:
$= 20 \div \left\{4 + (6)\right\}$
$= 20 \div \left\{10\right\}$
$= 2 \div 1$
$= 2$
## Practice Problems
1
### Solve: $[\left\{(2^{2} + 3^{3}) \times 4^{2}\right\} − (20 \div 5)]$
490
492
494
500
CorrectIncorrect
Step 1: Solve all brackets keeping the precedence in mind.
$[\left\{(2^{2} + 3^{3}) \times 42\right\} − (20 \div 5)]$
$= [\left\{(4 + 27) \times 16\right\} − (4)]$
$= [\left\{(31) \times 16\right\} − (4)]$
$= [{31 \times 16} − 4]$
$= [496 − 4]$
$= 492$
2
### What is the right representation of the order of operation in brackets?
$( \left\{ [ ] \right\} )$
$[ ( \left\{ \right\} ) ]$
$\left\{ [ ( ) ] \right\}$
$[ \left\{ ( ) \right\} ]$
CorrectIncorrect
Correct answer is: $[ \left\{ ( ) \right\} ]$
$[ \left\{ ( ) \right\} ]$ is the correct representation of order of operation in brackets.
3
### $\Biggr [\bigg\{ \bigg( \frac{1}{2} \bigg)^2 \bigg\}^{-3} \Biggr]^2$
4,096
64
256
1,024
CorrectIncorrect
$\Biggr[\bigg\{ \bigg(\frac{1}{2}\bigg)^{2}\bigg\}^{-3}\Biggr]^{2} = \Biggr[\bigg\{\frac{1}{4}\bigg\}^{-3}\Biggr]^{2} = \Biggr[\bigg\{\frac{4}{1}\bigg\}^{3}\Biggr]^{2} = [64]^{2} = 4,096$
4
### Solve this expression, $12 + (5 + 3)$,
18
20
16
8
CorrectIncorrect
$12 + (5 + 3) = 12 + 8 = 20$
5
### Simplify the expression: $(3 + 2 \times 8) – 4 + (5 \times 7)$
45
50
24
40
CorrectIncorrect
We know that the equation within the parentheses is solved first.
So, $19 – 4 + 35 = 50$
6
### Simplify the expression: $( 4 \times (6 – 2)) – ((8 \div 2) + 5 )$
7
2
17
10
CorrectIncorrect
We know that the equation within the parentheses is solved first.
So, $( 4 \times 4) – ( 4 + 5)$
$16 – 9 = 7$
7
### Simplify the expression: $4 (3 + 2) + 4 (7 – 2)$
10
50
20
40
CorrectIncorrect
We know that parentheses also denote multiplication.
So, $4 \times 5 + 4 \times 5$
$20 + 20 = 40$
8
### Solve the equation containing braces in math. $57 \div \left\{5 + (4 \times 2) + (3 + 3)\right\}$
3
4
13
4
CorrectIncorrect
After solving the $( )$, we perform addition within the $\left\{ \right\}$, and then divide. $57 \div {5 + (4 2) + (3 + 3)} = 57 {5 + 8 + 6} = 57 19 = 3$
9
### Which of the following examples use braces, brackets, and parentheses correctly?
60 $\div$ [(2 $\times$ 2) + (3 + 3)}
60 $\div$ {(2 $\times$ 2) + (3 + 3)}
60 $\div$ {[2 $\times$ 2] + (3 + 3)}
(60 $\div$ {[2 $\times$ 2] + (3 + 3})
CorrectIncorrect
Correct answer is: 60 $\div$ {(2 $\times$ 2) + (3 + 3)}
It uses the braces, brackets, and parentheses correctly because the innermost brackets have parentheses and then braces.
10
### If we have the following expressions inside the curly brackets, which of the expressions would you solve first? $10\left\{(\frac{4}{2}) + (6 \times 2) – (3 + 3) + (7 – 2)\right\}$
$(\frac{4}{2})$
$(\frac{4}{2}) \text{or} (6 \times 2)$
Any parentheses inside the $\left\{ \right\}, (\frac{4}{2}), (6 \times 2), (3 + 3), (7 – 2)$}
None of the above
CorrectIncorrect
Correct answer is: Any parentheses inside the $\left\{ \right\}, (\frac{4}{2}), (6 \times 2), (3 + 3), (7 – 2)$}
We can solve any of the parentheses inside the curly brackets first. Once these parentheses are solved, we have to simply add and subtract, which can be done in any order.
Brackets are very important parts of a mathematical equation; they separate different mathematical expressions from each other and help set the priority for expressions that need to be solved first.
BODMAS is a different acronym for PEMDAS, where B stands for Bracket, O for Of or Exponents, D for Division, M for Multiplication, A for Addition, and S for Subtraction. Any expression is considered correctly solved if they have followed the PEMDAS or BODMAS rule.
Angle Brackets are also used in various mathematical expressions; they are represented with〈 〉. The angle brackets are used to represent a list of numbers or a sequence of numbers.
Brackets are also used to define the coordinates of a point on a map or to describe the variable of a function.
No. Parentheses denoted by ( ) are different from braces { }. They have distinct uses in math. They are used in nesting expressions. You will learn more about them later.
Yes. Sometimes, parentheses are also called round brackets.
These are curly brackets, also known as braces in math. Braces are used in math equations when we are making at least two nested groups for calculation.
Braces are also used to define a set.
For example, $\left\{3, 5, 7, 9, 10\right\}$ means a set containing the numbers 3, 5, 7, 9, 10.
Yes, braces can also mean multiplication. You need to multiply the value outside the braces by the value inside the braces.
Take this equation as an example: $2\left\{2(4 + 2) + 1\right\}$
Here, 2 will be multiplied by the answer inside the curly brackets or braces. | {"extraction_info": {"found_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": 0.9484250545501709, "perplexity": 915.0933767688094}, "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/1679296945333.53/warc/CC-MAIN-20230325130029-20230325160029-00252.warc.gz"} |
http://mathoverflow.net/questions/22648/area-of-cross-section-at-midpoint-perpendicular-to-longest-diagonal-in-the-uni?sort=newest | # Area of cross-section (at midpoint perpendicular to longest diagonal) in the unit cube of dimension N
Take a unit cube (of side 1) in N dimensions. Construct the cross-section at the midpoint of the longest diagonal. What is the area of this N-1 dimensional region? I can compute this, but it would be nice to have a reference and a formula to cross-check.
-
do you mean the cross-section which is perpendicular to the longest diagonal? – Yemon Choi Apr 26 '10 at 21:45
@Ila: You might expand the problem by posing your solutions (at least the related integral). There could be references for (more general!) formulas like yours without the geometric background. – Wadim Zudilin Apr 26 '10 at 22:44
For $n$ even, I get $$\frac{\sqrt{n}}{(n-1)!} \left\langle \begin{matrix} n-1 \\ n/2-1 \end{matrix} \right\rangle.$$
Here $\left\langle \begin{matrix} a \\ b \end{matrix} \right\rangle$ is the Eulerian number: the number of permutations of $a$ elements with $b$ descents.
Proof sketch: We are dealing with the convex hull of the $\binom{n}{n/2}$ zero-one vectors which have $n/2$ zeroes and $n/2$ ones. This is better known as the $(n/2,n)$-hypersimplex. The hypersimplex is well known to have a triangulation with $\left\langle \begin{matrix} n-1 \\ n/2-1 \end{matrix} \right\rangle$ tetrahedra. There are many references for this; my favorite, which discusses a number of prior references, is the early parts of Lam and Postnikov.
All the tetrahedra in this triangulation have the same volume, which is the volume of the convex hull of $e_1$, $e_2$, \ldots, $e_{n-1}$. If I didn't screw up, that volume is $\frac{\sqrt{n}}{(n-1)!}$.
-
This is a very old problem and there is a classical analytic approach to it. You can express the volume of sections of a convex body in terms of the Fourier transform of powers of the Minkowski functional.
Let $Q_n=[-1/2,1/2]^n$ be the unit cube in $\mathbb R^n$ and let $[\xi^\perp]$ denote the hyperplane orthogonal to the vector $\xi\in S^{n-1}$. Then $$Vol_{n-1}(Q_n\cap [\xi^\perp])=\frac{1}{\pi}\int_{-\infty}^{\infty} \prod\limits_{k=1}^n\frac{\sin(r \xi_k)}{r\xi_k}dr.$$
In our case $\xi=\xi_*=n^{-1/2}(1,1,\dots,1)$ so the integral becomes $$Vol_{n-1}(Q_n\cap [\xi_*^\perp])=\frac{1}{\pi}\int_{-\infty}^{\infty} \left(\frac{\sin(r/\sqrt n)}{r/\sqrt n}\right)^ndr.$$ In fact, the latter formula was already known to Laplace! The integral tends to $\sqrt {6/\pi}$, as $n\to\infty$ (e.g. by Laplace's method). This can be also justified via the probabilistic interpretation suggested by Michael Lugo.
The Fourier analytic approach to sections of convex bodies is nicely presented in The Interface between Convex Geometry and Harmonic Analysis by A. Koldobsky and V. Yaskin. The derivation of the formula for volumes is available here.
EDIT (15.01.2011). In fact both integrals can be calculated explicitly. The sinc integrals were studied by Borwein & Borwein (see also Multi-Variable sinc Integrals and the Volumes of Polyhedra).
For any $n\in \mathbb N$, $n>1$, we have $$\frac{1}{\pi}\int_{-\infty}^{\infty} \left(\frac{\sin(r/\sqrt n)}{r/\sqrt n}\right)^ndr =\frac{\sqrt n}{2^{n-1}(n-1)!}\sum_{k=0}^{n/2}(-1)^k{n \choose k}(n-2k)^{n-1}$$
-
Note that $\int_{-\infty}^{\infty} \ (sin(t)/t)^n dt$ can be computed exactly by residues for any specific positive integer $n$. – David Speyer May 1 '10 at 14:05
Thanks, David. I should have mentioned this in the answer. – Andrey Rekalo May 1 '10 at 22:28
Let the cube be $[0,1]^n$. I'm assuming you mean the cross-section perpendicular to the longest diagonal, from the all-zeros corner to the all-ones corner. This is the cross-section corresponding to the plane $x_1 + \cdots + x_n = n/2$.
Alternatively, then, you want the probability density function of the sum $S_n$, of $n$ independent uniform(0,1) random variables, evaluated at $n/2$.
By the central limit theorem, the $S_n$ have asymptotic normal distribution with mean $n/2$ and variance $n/12$. So the answer should behave asymptotically like $1/\sqrt{2\pi n/12}$, i. e. like $\sqrt{\pi/6n}$.
It appears that this distribution is called the Irwin-Hall distribution.
-
Michael, I think that the answer is rather $6/\sqrt \pi$ for large n. – Andrey Rekalo May 1 '10 at 3:50
You're right -- I didn't do the algebra correctly. – Michael Lugo May 1 '10 at 4:00
I'm sorry, the final answer is $\sqrt{6/\pi}$. – Andrey Rekalo May 1 '10 at 22:33 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9816926717758179, "perplexity": 327.3370656046323}, "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/1393999651896/warc/CC-MAIN-20140305060731-00046-ip-10-183-142-35.ec2.internal.warc.gz"} |
https://de.maplesoft.com/support/help/maple/view.aspx?path=content | content - Maple Programming Help
Home : Support : Online Help : Mathematics : Algebra : Polynomials : content
content
content of a multivariate polynomial
primpart
primpart of a multivariate polynomial
Calling Sequence content(a, x, 'pp') primpart(a, x, 'co')
Parameters
a - multivariate polynomial in x x - (optional) name or set or list of names pp - (optional) unevaluated name co - (optional) unevaluated name
Description
• If a is a multivariate polynomial with integer coefficients, content returns the content of a with respect to x, thus returning the greatest common divisor of the coefficients of a with respect to the indeterminate(s) x. The indeterminate(s) x can be a name, list, or set of names.
• The third argument pp, if present, will be assigned the primitive part of a, namely a divided by the content of a.
• If the coefficients of a in x are rational functions then the content computed will be such that the primitive part is a multivariate polynomial over the integers whose content is 1.
• Similarly, primpart returns a/content(a, x). The third argument co, if present, will be assigned the content. Note: Whereas the sign is removed from the content, it is not removed from the primitive part.
Examples
> $\mathrm{content}\left(3-3x,x\right)$
${3}$ (1)
> $\mathrm{content}\left(3xy+6{y}^{2},x\right)$
${3}{}{y}$ (2)
> $\mathrm{content}\left(3xy+6{y}^{2},\left[x,y\right]\right)$
${3}$ (3)
> $\mathrm{content}\left(-4xy+6{y}^{2},x,'\mathrm{pp}'\right)$
${2}{}{y}$ (4)
> $\mathrm{pp}$
${3}{}{y}{-}{2}{}{x}$ (5)
> $\mathrm{content}\left(\frac{x}{a}-\frac{1}{2},x,'\mathrm{pp}'\right)$
$\frac{{1}}{{2}{}{a}}$ (6)
> $\mathrm{pp}$
${2}{}{x}{-}{a}$ (7)
> $\mathrm{primpart}\left(-4xy+6{y}^{2},x\right)$
${3}{}{y}{-}{2}{}{x}$ (8)
> $\mathrm{primpart}\left(\frac{x}{a}-\frac{1}{2},x\right)$
${2}{}{x}{-}{a}$ (9) | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 18, "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": 0.9747987985610962, "perplexity": 2279.1614980058803}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-30/segments/1563195524679.39/warc/CC-MAIN-20190716160315-20190716182315-00459.warc.gz"} |
https://en.formulasearchengine.com/wiki/Localization_of_a_category | Localization of a category
In mathematics, localization of a category consists of adding to a category inverse morphisms for some collection of morphisms, constraining them to become isomorphisms. This is formally similar to the process of localization of a ring; it in general makes objects isomorphic that were not so before. In homotopy theory, for example, there are many examples of mappings that are invertible up to homotopy; and so large classes of homotopy equivalent spaces. Calculus of fractions is another name for working in a localized category.
Introduction and motivation
A category C consists of objects and morphisms between these objects. The morphisms reflect relations between the objects. In many situations, it is meaningful to replace C by another category C' in which certain morphisms are forced to be isomorphisms. This process is called localization.
For example, in the category of R-modules (for some fixed commutative ring R) the multiplication by a fixed element r of R is typically (i.e., unless r is a unit) not an isomorphism:
${\displaystyle M\to M\quad m\mapsto r\cdot m.}$
The category that is most closely related to R-modules, but where this map is an isomorphism turns out to be the category of ${\displaystyle R[S^{-1}]}$-modules. Here ${\displaystyle R[S^{-1}]}$ is the localization of R with respect to the (multiplicatively closed) subset S consisting of all powers of r, ${\displaystyle S=\{1,r,r^{2},r^{3},\dots \}}$ The expression "most closely related" is formalized by two conditions: first, there is a functor
${\displaystyle \varphi :{\text{Mod}}_{R}\to {\text{Mod}}_{R[S^{-1}]}\quad M\mapsto M[S^{-1}]}$
sending any R-module to its localization with respect to S. Moreover, given any category C and any functor
${\displaystyle F:{\text{Mod}}_{R}\to C}$
sending the multiplication map by r on any R-module (see above) to an isomorphism of C, there is a unique functor
${\displaystyle G:{\text{Mod}}_{R[S^{-1}]}\to C}$
Localization of categories
The above examples of localization of R-modules is abstracted in the following definition. In this shape, it applies in many more examples, some of which are sketched below.
Given a category C and some class W of morphisms in C, the localization C[W−1] is another category which is obtained by inverting all the morphisms in W. More formally, it is characterized by a universal property: there is a natural localization functor CC[W−1] and given another category D, a functor F: CD factors uniquely over C[W−1] if and only if F sends all arrows in W to isomorphisms.
Thus, the localization of the category is unique provided that it exists. One construction of the localization is done by declaring that its objects are the same as those in C, but the morphisms are enhanced by adding a formal inverse for each morphism in C. Under suitable hypotheses on W, the morphisms between two objects X, Y are given by roofs
${\displaystyle X{\stackrel {f}{\leftarrow }}X'\rightarrow Y}$
(where X' is an arbitrary object of C and f is in the given class w of morphisms), modulo certain equivalence relations. These relations turn the map going in the "wrong" direction into an inverse of f. This procedure, however, in general yields a proper class of morphisms between them. Typically, the morphisms in a category are only allowed to form a set. Some authors simply ignore such set-theoretic issues.
Model categories
A rigorous construction of localization of categories, avoiding these set-theoretic issues, was one of the initial reasons for the development of the theory of model categories: a model category M is a category in which there are three classes of maps; one of classes is a class of weak equivalences. The homotopy category Ho(M) is then the localization with respect to the weak equivalences. The axioms of a model category ensure that this localization can be defined without set-theoretical difficulties.
Alternative definition
Some authors also define a localization of a category C to be an idempotent and coaugmented functor. A coaugmented functor is a pair (L,l) where L:C → C is an endofunctor and l:Id → L is a natural transformation from the identity functor to L (called the coaugmentation). A coaugmented functor is idempotent if, for every X, both maps L(lX),lL(X):L(X) → LL(X) are isomorphisms. It can be proven that in this case, both maps are equal.{{ safesubst:#invoke:Unsubst||date=__DATE__ |\$B= {{#invoke:Category handler|main}}{{#invoke:Category handler|main}}[citation needed] }}
This definition is related to the one given above as follows: applying the first definition, there is, in many situations, not only a canonical functor ${\displaystyle C\to C[W^{-1}]}$, but also a functor in the opposite direction,
${\displaystyle C[W^{-1}]\to C}$
For example, modules over the localization ${\displaystyle R[S^{-1}]}$ of a ring are also modules over R itself, giving a functor
${\displaystyle R[S^{-1}]-Mod\to R-Mod.}$
In this case, the composition
${\displaystyle L:C\to C[W^{-1}]\to C}$
is a localization of C in the sense of an idempotent and coaugmented functor.
Examples
Serre's C-theory
Serre introduced the idea of working in homotopy theory modulo some class C of abelian groups. This meant that groups A and B were treated as isomorphic, if for example A/B lay in C. Later Dennis Sullivan had the bold idea instead of using the localization of a topological space, which took effect on the underlying topological spaces.
Module theory
In the theory of modules over a commutative ring R, when R has Krull dimension ≥ 2, it can be useful to treat modules M and N as pseudo-isomorphic if M/N has support of codimension at least two. This idea is much used in Iwasawa theory.
Derived categories
The derived category of an abelian category is much used in homological algebra. It is the localization of the category of chain complexes (up to homotopy) with respect to the quasi-isomorphisms.
Abelian varieties up to isogeny
An isogeny from an abelian variety A to another one B is a surjective morphism with finite kernel. Some theorems on abelian varieties require the idea of abelian variety up to isogeny for their convenient statement. For example, given an abelian subvariety A1 of A, there is another subvariety A2 of A such that
A1 × A2
is isogenous to A (Poincaré's theorem: see for example Abelian Varieties by David Mumford). To call this a direct sum decomposition, we should work in the category of abelian varieties up to isogeny.
Related concepts
The localization of a topological space produces another topological space whose homology is a localization of the homology of the original space.
A much more general concept from homotopical algebra, including as special cases both the localization of spaces and of categories, is the Bousfield localization of a model category. Bousfield localization forces certain maps to become weak equivalences, which is in general weaker than forcing them to become isomorphisms.[1] | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 14, "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": 0.9510639905929565, "perplexity": 418.1877671483796}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-25/segments/1623488504969.64/warc/CC-MAIN-20210622002655-20210622032655-00412.warc.gz"} |
https://www.intechopen.com/books/statistical-methodologies/a-study-on-the-comparison-of-the-effectiveness-of-the-jackknife-method-in-the-biased-estimators | Open access peer-reviewed chapter
# A Study on the Comparison of the Effectiveness of the Jackknife Method in the Biased Estimators
By Nilgün Yıldız
Submitted: September 10th 2018Reviewed: November 2nd 2018Published: December 18th 2018
DOI: 10.5772/intechopen.82366
## Abstract
In this study, we proposed an alternative biased estimator. The linear regression model might lead to ill-conditioned design matrices because of the multicollinearity and thus result in inadequacy of the ordinary least squares estimator (OLS). Scientists have developed alternative estimation techniques that would eradicate the instability in the estimates. Several biased estimators such as Stein estimator, the ordinary ridge regression (ORR) estimator, the principal components regression (PCR) estimator. Liu developed a Liu estimator (LE) by combining the Stein estimator with the ORR estimator. Since both ORR and LE depend on OLS estimator, multicollinearity affects them both. Therefore, the ORR and LE may give misleading information in the presence of multicollinearity. To overcome this problem, Liu introduced a new estimator, which is based on k and d biasing parameters, the authors worked on developing an estimator that would still have the valuable characteristics of the Liu-type estimator (LTE) but have a smaller bias. We are proposing a modified jackknife Liu-type estimator (MJLTE) that was created by combining the ideas underlying both the LTE and JLTE. Under mean square error matrix criteria, the MJLTE is superior to Liu-type estimator (LTE) and jackknifed Liu-type estimator (JLTE). Finally, a real data example and a Monte Carlo simulation are also given to illustrate theoretical results.
### Keywords
• jackknifed estimators
• jackknified Liu-type estimator
• multicollinearity
• MSE
• Liu-type estimator
## 1. Introduction
With regression analysis; Is there a relationship between dependent and independent variables? If there is a relationship, what is the power of this relationship? What is the relationship between variables? Is it possible to predict prospective variables and how should they be estimated? What is the effect of a particular variable or group of variables on other variables or variables in the event that certain conditions are checked? Try to search for answers to questions such as. Linear regression is very important, popular method in statistics. According to Web of Science, the number of publications about linear regression between 2014 and 2018 is given in Figure 1.
According to Figure 1, the number of studies conducted in 2014 is 12,381, while the number of studies conducted in 2018 is 13,137.
The number of publications about linear regression by document types is given in Figure 2.
The most common type of document about linear regression is the article. This is followed by proceeding paper, review, and editorial material.
The number of publications about linear regression by research area is given in Figure 3.
The most widely published area related to linear regression is engineering, followed by mathematics, computer science, environmental sciences, ecology and other scientific fields.
The number of publications about linear regression by countries is given in Figure 4.
The countries with the most publications on linear regression are USA, China, England, Germany, Canada, Australia, respectively.
In regression analysis, the most commonly used method for estimating coefficients is ordinary least squares (OLS). We considered the multiple linear regression model given as
y=+εE1
where yis n×1observable random vector, Xis a n×pmatrix of non-stochastic (independent) variables of rank p; βis p×1vector of unknown parameters associated with X, and εis a n×1vector of error terms with
Ee=0,Cove=σ2IE2
In regression analysis, there are several methods to estimate unknown parameters. The most frequently used method is the least squares method (OLS). Apart from this method, there are three general estimation methods: maximum likelihood, generalized least squares, and best linear unbiased estimator BLUE [1].
Since the use of once very popular estimators such as the ordinary least squares (OLS) estimation has become limited due to multicollinearity, which makes them unstable and results in bias and reduced variance of the regression coefficients.
We can give, it is a linear (or close to linear) relationship between the independent variables as the definition of multicollinearity. In the regression analysis, multicollinearity leads to the following problems:
• In the case of multicollinearity, linear regression coefficients are uncertain and the standard errors of these coefficients are infinite.
• The regression coefficients of the multicollinearity increase the variance and covariance of OLS.
• The value of the model R2is high but none of the independent variables is significant compared to the partial ttest.
• The direction of the related independent variables’ relations with the dependent variable may contradict the theoretical and empirical expectations.
• If independent variables are interrelated, some of them may need to be removed from the model. But what variables will be extracted? Removing an incorrect variable from the model will result in a model error. On the other hand, there are no simple rules that we can use to include and subtract the arguments in the model.
Methods for dealing with multicollinearity are collecting additional data, model respecification. Instead of two related variables, the sum of these two variables (as a single variable) can be taken and use of biased estimators. In this book provides information on biased estimators used as OLS alternatives. In literature many researchers have developed biased regression estimators [2, 3].
Examples of such biased estimators are the ordinary ridge regression (ORR) estimator introduced by Hoerl and Kennard [4].
β̂k=XX+kI1Xyk0E3
where kis a biasing parameter, in later years researchers combined various estimators to obtain better results. For example, Baye and Parker [5] introduced rkclass estimator, which combines the ORR and principal component regression (PCR). In addition, Baye and Parker also showed that rkclass estimator is superior to PCR estimator based on the scalar mean square error (SMSE) criterion.
Since both ORR and LE depend on OLS estimator, multicollinearity affects both of them. Therefore, the ORR and LE may give misleading information in the presence of multicollinearity. Liu estimator (LE) was developed by Liu [6] by combining the Stein [7] estimator with the ORR estimator.
β̂d=XX+I1Xy+dβ̂0<d<1E4
To overcome this problem, Liu [8] introduced a new estimator, which is based on kand dbiasing parameters as follows
β̂LTE=XX+kI1Xy+dβ̂k>0,<d<E5
Next, the authors worked on developing an estimator that would still have valuable characteristics of the Liu-type estimator (LTE), but have a smaller bias. In 1956, Quenouille [9] suggested that it is possible to reduce bias by applying a jackknife procedure to a biased estimator.
This procedure enables processing of experimental data to get statistical estimator for unknown parameters. A truncated sample is used calculate specific function of estimators. The advantage of jackknife procedure is that it presents an estimator that has a small bias while still providing beneficial properties of large samples. In this article, we applied the jackknife technique to the LTE. Further, we established the mean squared error superiority of the proposed estimator over both the LTE and the jackknifed Liu-type estimator (JLTE).
The article is organized as follows: The model as well as LTE and the JLTE are described in Section 2. The proposed new estimator is introduced in Section 3. Superiority of the new estimator vis-a-vis the LTE and the JLTE are studied and the performance of the modified Jackknife Liu-type estimator (MJLTE) is compared to that of the JLTE in Section 4. Sections 5 and 6 consider a real data example and a simulation study to justify the superiority of the suggested estimator.
## 2. The model
We assume that two or more regressors in Xare closely linearly related, therefore model suffers from multicollinearity problem. A symmetric matrix S=XXhas an eigenvalue–eigenvector decomposition of the form S=TΛT, where Tis an orthogonal matrix and Λis (real) a diagonal matrix. The diagonal elements of Λare the eigenvalues of Sand the column vectors of Tare the eigenvectors of S. The orthogonal version of the standard multiple linear regression models is
y=XTTβ+ε=+εE6
where Z=XT, γ=Tβand ZZ=Λ. The ordinary least squares estimator (OLSE) of γis given by
γ̂=ZZ1Zy=Λ1ZyE7
Liu [8] proposed a new biased estimator for γ, called the Liu-type estimators (LTE), and defined as
γ̂LTEkd=Λ+kI1Zydγ̂fork0andd+=Λ+kI1ZydΛ1Zy=IΛ+kI1k+dγ̂=Fkdγ̂E8
where
Fkd=Λ+kI1ΛdIE10
γ̂LTEhas bias vector
Biasγ̂LTE=FkdIγE11
and covariance matrix
Covγ̂LTE=σ2FkdΛ1FkdE12
By using Hinkley [10], Singh et al. [11], Nyquist [12], and Batah et al. [13] we can propose the jackknifed form of γ̂LTE. Quenouille [9] and Tukey [14] introduced the jackknife technique to reduce the bias. Hinkley [10] stated that with few exceptions, the jackknife had been applied to balanced models. After some algebraic manipulations, the corresponding jackknife estimator is obtained by deleting the ith observation ziyias
Azizi1=A1+A1ziziA11ziA1zi
γ̂LTEikd=Azizi1Zyziyi=A1+A1ziziA11ziA1ziZyziyi=A1ZyA1ziyi+A1ziziA11ziA1ziZyA1ziziA11ziA1ziziyi
=γ̂LTEkd+A1ziyi1+ziA1zi1ziA1zi+A1zizi1ziA1ziγ̂LTEkd=γ̂LTEkdA1ziA1ziyiziγ̂LTEkd1ziA1zi=γ̂LTEkdA1ziei1wiE13
where ziis the ithrow of Z, ei=yiziγ̂LTEkdis the Liu-type residual, wi=ziA1ziis the distance factor and A1=Λ+kI1IdΛ1=FkdΛ1. In the view of the non-zero value of wireflecting the lack of balance in the model, we use the weighted jackknife procedure. Thus, weighted pseudo values are defined as
Qi=γ̂LTEkd+n1wiγ̂LTEkdγ̂LTEi(kd)
the weighted jackknifed estimator of γ is obtained as
γ̂JLTEkd=1ni=1nQi=γ̂LTEkd+A1i=1nzieiE14
i=1nziei=i=1nziyiziγ̂LTEkd=IA1Zy
γ̂JLTEkd=γ̂LTEkd+A1ZyA1ΛA1Zy=2IA1Λγ̂LTEkdE15
However, since IA1Λ=IΛ+kI1ΛdI=IFkd, we obtain
γ̂JLTEkd=2IFkdγ̂LTEkdE16
From (9) we have
γ̂JLTEkd=2IFkdFkdγ̂E17
Biasγ̂JLTEkd=IFkd2γE18
Variance of the JLTE as,
Covγ̂JLTEkd=σ22IFkdFkdΛ1Fkd2IFkdE19
MSEMs of the JLTE and LTE as
MSEMγ̂JLTEkd=Covγ̂JLTEkd+Biasγ̂JLTEkdγ̂JLTEkd=σ22IFkdFkdΛ1Fkd2IFkd+ Fkd2γγIFkd2E20
MSEMγ̂LTEkd=σ2FkdΛ1Fkd+FkdIββFkdIE21
## 3. Our novel MJLTE estimator
In this section, Yıldız [15] propose a new estimator for γ. The proposed estimator is designated as the modified jackknifed Liu-type estimator (MJLTE) denoted by γ̂MJLTEkd
γ̂MJLTEkd=Ik+d2Λ+kI2Ik+dΛ+kI1γ̂E22
It may be noted that the proposed estimator MJLTE in (22) is obtained as in the case of JLTE but by plugging in the LTE instead of the OLSE. The expressions for bias, covariance and mean squared error matrix (MSEM) of γ̂MJLTEkdare obtained as
Biasγ̂MJLTEkd=k+dΛ+kI1WΛ+kI1γE23
Covγ̂MJLTEkd=σ2ΦΛ1ΦE24
MSEMγ̂MJLTEkd=σ2ΦΛ1Φ+k+d2Λ+kI1WΛ+kI1γγΛ+kI1WΛ+kI1E25
where W=I+k+dΛ+kI1k+d2Λ+kI2=I+FkdFkd2and Φ=2IFkdFkd2
## 4. Properties of the MJLTE
One of the most prominent features of our novel MJLTE estimator is that its bias, under some conditions, is less than LTE estimator from which it originates from.
Theorem 4.1.Under the model (1) with the assumptions (2), the inequality
Biasγ̂MJLTEkd2<Biasγ̂LTEkd2holds true for d>0and k>d
Proof. From 11 and 23, we can obtain that
Biasγ̂MJLTEkd2Biasγ̂LTEkd2=k+d2Λ+kI2W2Λ+kI2Λ+kI2>0
It is obvious that the difference is greater than 0, because it consists of the product of the squares in the expression above. Thus, the proof is completed.
Corollary 4.1.The bias of the absolute value of the ithcomponent of MJLTE is smaller than that of LTE, namelyBias(γ̂MJLTEkdi<Bias(γ̂LTEkdi.
Theorem 4.2.The MJLTE has smaller variance than the LTE
Proof. From 12 and 24, it can be shown that
Covγ̂LTEkdCovγ̂MJLTEkd=σ2H
where
H=I+k+dΛ+kI1Λ1I+k+dΛ+kI1ΦΛ1Φ=IFkdΛ1IFkd2Λ1(IF(kd)2IFkd
His a diagonal matrix and ith element
hii=λi+ki4λi+2ki+di2λidi2λidi2λiλi+ki6
is a positive number. Thus we conclude that H is a positive definite matrix. This completes the proof.
Next, we prove necessary and sufficient condition for the MJLTE to outperform the LTE using the MSEM condition. The proof requires the following lemma.
Lemma 4.1.Let Mbe a positive definite matrix, namely M>0, αbe some vector, then
Mαα0if and only if αM1α1
Proof. see Farebrother[16]
Theorem 4.3.MJLTE is superior to the LTE in the MSEM sense, namelyMSEMγ̂LTEkdMSEMγ̂MJLTEkd>0, if the inequality
Δ1=MSEMγ̂LTEkdMSEMγ̂MJLTEkdis nonnegative definite matrix if and if the inequality
γL1σ2H+FkdγγFkdL11γ1E26
is satisfied withL=FkdW,Fkd=FkdIandW=I+FkdFkd2
Proof.
We consider the difference from (21, 25) we have
Δ1=MSEMγ̂LTEkdMSEMγ̂MJLTEkd=σ2H+FkdγγFkdLγγLE27
where
H=I+k+dΛ+kI1Λ1I+k+dΛ+kI1ΦΛ1Φ=IFkdΛ1IFkd2Λ1(IF(kd)2IFkd
W=I+FkdFkd2is a positive definite matrix. We have seen His a positive definite matrix from Theorem 2. Therefore, the difference Δ1is a nonnegative definite, if and only if L1Δ1L1is a nonnegative definite. The matrix L1Δ1L1can be written as
L1Δ1L1=L1σ2H+FkdγγFkdL1γγE28
Since the matrix σ2H+FkdγγFkdis symmetric and positive definite, using Lemma 4.1, we may conclude that L1Δ1L1is a nonnegative definite, if and only if the inequality
γL1σ2H+FkdγγFkdL11γ1
is satisfied.
### 4.1 Comparison between the JLTE and the MJLTE
Here, we show that the MJLTE outperforms the JLTE in terms of the sampling variance.
Theorem 4.4.The variance ofMJLTE has a smaller variance than that of the JLTE ford>0andk>d
Proof.From (19, 24) it can be written as
Covγ̂JLTEkd=σ22IFkdFkdΛ1Fkd2IFkd=σ2VUΛ1UVE29
and
Covγ̂MJLTEkd=σ2ΦΛ1Φ=σ2VUVΛ1VUVE30
where V=IFkdand U=I+Fkd, respectively. It can be shown that
Covγ̂JLTEkdCovγ̂MJLTEkd=σ2ΣE31
where Σ=VUΛ1VΛ1VUV,Σis a diagonal matrix. Then ith the diagonal element of Covγ̂JLTEkdCovγ̂MJLTEkdis
σ2λi+k+2di2λidi2k+di2λi+k+diλi+λi+ki6
Hence of Covγ̂JLTEkdCovγ̂MJLTEkd>0which completes the proof.
In the following theorem, we have obtained a necessary and sufficient condition for the MJLTE to outperform the JLTE in terms of matrix mean square error. The proof of the theorem is similar to that of Theorem 4.3.
Theorem 4.5.
Δ2=MSEMγ̂JLTEkdMSEMγ̂MJLTEkdis a nonnegative definite matrix, if and if the inequality
γL1σ2Σ+Fkd2γγFkd2L11γ1E32
is satisfied.
Proof. From (20, 25) we have
Δ2=MSEMγ̂JLTEkdMSEMγ̂MJLTEkd=σ2Σ+Fkd2γγFkd2FkdWγγWFkd
We have seen from Theorem 4.4 that Σis a positive definite matrix. Therefore, the difference Δ2is a nonnegative definite, if and only if L1Δ2L1is a nonnegative definite. The matrix L1Δ2L1can be written as
L1Δ2L1=L1σ2Σ+Fkd2γγFkd2L1γγ
The difference Δ2is a nonnegative definite matrix, if and only if L1Δ2L1is a nonnegative definite matrix. Since the matrix σ2Σ+Fkd2γγFkd2is symmetric and positive definite, using Lemma 4.1, we may conclude that L1Δ2L1is nonnegative definite, if and only if the inequality
γL1σ2Σ+Fkd2γγFkd2L11γ1
is satisfied. This confirms our validation. Theorems 4–6 showed that the estimator we proposed was superior to the LTE estimator and JLTE estimator. Accordingly, we can easily say that the MJLTE estimator is better than other estimators LTE, JLTE.
## 5. Numerical example
To motivate the problem of estimation in the linear regression model, we consider the hedonic prices of housing attributes. The data consists of 92 detached homes in the Ottawa area sold during 1987 (see Yatchew [17]).
Let y be the sale price (sp)of the house, Xbe a 92 × 9 observation matrix consisting of the variables: frplc: dummy forfireplace(s), grge: dummy forgarage, lux: dummy forluxury appointment, avginc:average neighborhood income, dhwy:distance to highway, lot area: area of lot, nrbed:number of bedrooms, usespc: usable space. The data are given in Table 1.
sellprixfireplacgarageluxbathavginccrowdistncrosdstdisthwylotareanrbedusespacesouthwestnsouthnwest
18001032.31630.9342800.638073.6329731.233090.840.40900
13501031.50162.186240.139420.664526.530.845921.441.6450.0871710.16962
165.901032.06542.311480.153370.434225.7230.875081.6731.5950.121020.16276
10100037.83482.543810.179240.268723.13620.714452.2521.1830.205140.10622
12700037.83482.54580.179470.166212.730.737892.1931.2930.196570.12131
23510167.00562.771470.20460.229356.69541.35182.3521.4660.219670.14505
19511065.82783.087470.239790.302954.2331.168292.5641.720.250470.17991
184.510062.30533.48440.283990.330094.22430.982282.7852.0940.282580.23123
10611038.49463.830860.322580.870563.23420.795072.1483.1720.190030.37917
15611052.35523.853060.325050.39006530.902392.5632.8770.250330.33869
19510052.35523.882830.328360.672114.841.294652.3633.0810.221270.36668
20611052.35523.924930.333050.694954.7541.71112.383.1210.223740.37217
15701052.35523.958880.336830.43005530.9012.6212.9670.258750.35104
18010079.45833.962360.337220.12078430.955273.0152.5710.3160.29669
19311079.45833.96260.337250.195036.3554531.504363.0632.5140.322970.28887
23011052.35523.972840.338390.39307531.084212.6612.950.264560.3487
21201153.86474.043290.346230.22395541.087242.8452.8730.29130.33814
10210059.57744.114940.354211.18386421.248132.1133.5310.184950.42843
13711059.57744.131410.356051.011872.430.7532.2853.4420.209940.41622
18701059.57744.137230.356691.004136.1068630.919362.2973.4410.211680.41608
10300059.57744.163380.359611.14419420.688932.1933.5390.196570.42953
10000059.57744.225210.366491.22908431.074832.1693.6260.193080.44147
15211039.76524.296880.374471.020719.941.016683.7932.0190.429030.22094
12711039.52294.528790.40030.69035.1641.647432.8973.4810.298850.42157
119.501035.64.596490.407841.393924.920.97684.2061.8540.489030.1983
10300039.52294.638410.412510.647744.615821.143583.0233.5180.317160.42665
9901030.65144.695990.418921.00164540.86554.1122.2680.475370.25511
7500039.54.699410.41930.493821.89131.14363.1923.4490.341710.41718
12811039.52294.733530.42310.7881531.039112.9923.6680.312650.44723
13210035.64.759380.425981.235045.089231.325794.2732.0960.498770.23151
13200038.12164.807010.431280.353736.541541.253123.8512.8770.437450.33869
13400039.55.047630.458080.274294.72530.805363.643.4970.40680.42377
12011039.87325.080060.461690.58519531.24524.2082.8460.489320.33443
12501039.87325.218550.477120.71216.9620.624724.3912.820.515910.33086
13511025.95455.235990.479061.017165.430.882734.5632.5680.54090.29628
13911038.245.265720.482370.922314.0841.444143.3354.0750.362490.50309
15101034.40495.343160.490990.962856.631.032053.3684.1480.367280.51311
116.501038.12165.537040.512580.519615.330.993254.5443.1640.538140.37807
13711050.05485.567670.5160.706376.630.87733.7574.1090.42380.50775
14911055.86675.754110.536761.382965.631.192553.374.6640.367570.58392
16711058.47635.80550.542481.203416.415241.486243.5654.5820.39590.57266
16311058.47635.82130.544241.033754.9241.600653.7164.4810.417840.5588
147.501058.47636.07490.572480.733265.182530.9724.164.4270.482350.55139
237.711058.47636.07520.572520.623635.449541.999674.2414.350.494120.54083
16811044.24756.25990.593091.49998531.302753.7065.0450.416390.6362
18011058.47636.31730.599481.032635.301441.597824.1354.7760.478720.59929
15610057.74466.63980.635391.09965531.410064.3545.0130.510530.63181
14511044.24756.64690.636181.673996.57241.221693.8915.3890.443270.68341
14411057.68616.68540.640470.760676.82541.6895.6163.6270.693880.44161
14011057.68616.697520.641820.633766.331.147785.5563.740.685170.45712
23611144.24756.73980.646531.596214.72541.631844.0375.3970.464480.68451
17211053.59076.82310.655810.98439541.229564.5965.0430.545690.63593
14810071.02697.11470.688281.42051631.142764.5015.510.531890.70001
153.511071.02697.12240.689141.545597.383241.4334.4065.5960.518090.71182
15411071.02697.1570.692991.218925.29231.440254.6965.4010.560220.68506
145.511050.71017.30230.709170.62336.73541.492976.0444.0980.756070.50624
14910050.71017.352720.714790.47386.088841.2668764.250.749670.5271
13810050.71017.35380.714910.580635.335241.503396.0624.1630.758680.51516
141.510052.63787.446720.725251.072276.5441.034626.4033.8020.808220.46562
12510052.63787.468920.727730.96967.0241.167726.3683.9030.803140.47948
13001052.63787.522660.733710.939246.541.097576.3963.960.807210.48731
13201052.63787.557570.73760.913615.141.203776.4114.0020.809390.49307
132.911052.63787.587830.740970.88591631.060596.4214.0430.810840.4987
12211052.63787.62210.744790.85984530.804946.4354.0850.812870.50446
16211051.50877.636280.746370.359164.99241.43185.7055.0760.706810.64046
127.510073.44647.65030.747930.687576.78530.846925.4885.330.675290.67531
8700042.31387.67850.751071.679194.640.740174.7696.0180.570830.76973
139.911040.92467.903530.776130.401385.531.192356.3664.6840.802850.58666
24011141.957.91940.777891.160655.00441.39725.3165.870.65030.74942
13411052.63787.925040.778521.080984.258131.22986.7764.110.862410.50789
136.510042.31387.93810.779981.581844.62531.066945.0086.1590.605550.78908
143.511051.50877.985520.785260.19433541.449726.0385.2260.755190.66104
14011057.57067.993080.78610.940967.85631.031676.7434.2920.857620.53287
12310051.50878.020720.789180.76856.75741.287825.6655.6780.7010.72307
14711057.57068.068750.794530.55185.0430.960546.5654.6910.831760.58762
134.911057.57068.166780.805440.776845.18541.357816.7634.5780.860530.57211
15411057.57068.238680.813450.853445.2741.609386.8554.570.873890.57102
143.910040.92468.567690.850090.53129530.911126.8775.110.877090.64512
12611053.80298.625040.856480.48411540.888386.8855.1950.878250.65679
118.510040.06188.94860.892510.103175.131.058856.855.7580.873170.73405
15811139.52628.95520.893251.864325.03531.11925.4257.1250.666130.92164
11800037.14579.29640.931240.186115.03530.935957.1275.9690.913410.763
109.2500030.97049.45720.949150.703295.430.674497.565.6820.976320.72362
12400049.02979.54450.958870.43645631.466836.8636.6330.875050.85412
13710035.51889.57950.962771.7827530.973885.8827.5610.732530.98147
14211030.58449.62730.968091.59895541.47616.0567.4840.757810.97091
120.500026.99479.64830.970431.51784530.699716.1327.4490.768850.9661
12301051.15699.74010.980661.79905520.685745.977.6960.745311
157.510051.15699.7440.981091.71459541.072266.0397.6470.755340.99328
11510047.86889.75860.982720.92055530.901636.6517.1410.844250.92384
126.511055.29019.86280.994320.412530.9027.6376.2410.987510.80033
15511055.29019.913810.49461531.413197.7236.21610.7969
### Table 1.
Data set.
The eigenvalues of the matrix XX: 9 × 9 are given by λ1=1.47, λ2=3.77, λ3=4.52, λ4=15.33, λ5=18.57, λ6=20.97, λ7=41.79, λ8=271.15and λ9=239153.68.
If we use the spectral norm, then the corresponding measure of conditioning of Xis the number κX=λmaxXX/λminXXwhereκ.1. We obtained κX=403.27, which is large and so Xmay be considered as being ill-conditioned.
In this case, the regression coefficients become insignificant and therefore, it is hard to make a valid inference or prediction using OLS method. To overcome many of the difficulties associated with the OLS estimates, the LTE. When β̂=XX1Xyand kand dare biasing parameters the use of β̂LTE=XX+kI1Xy+dβ̂, k>0, <d<+has become conventional. The LTE estimator will be used for the following example. The original model was used to reconstruct a canonical form as shown in (6) y=+ε. Estimators γ̂LTE,γ̂JLTEand γ̂MJLTEused data d=0.10,0.30,0.70,1and k=0.30,0.50,0.70,1. Then, the original variable scale was obtained by using the coefficients estimated by these estimators. The individual values of dand kfor the scalar MSE (SMSE = trace (MSEM)) of the estimators are shown in Tables 25. The effects of different values of don MSE can be seen in Figures 58 that clearly show that the proposed estimator (MJLTE) has smaller estimated MSE values compared to those of the LTE, JLTE.
d = 0.10d = 0.30d = 0.70d = 1
MSE(LTE)810.45111037.64541900.689712905.5467
MSE(JLTE)733.5563729.5050977.13821688.9649
MSE(MJLTE)631.2267669.0754967.79051289.1137
### Table 2.
The estimated MSE values of LTE, JLTE and MJLTE k = 0.30.
d = 0.10d = 0.30d = 0.70d = 1
MSE(LTE)957.66231245.72432157.46933134.9466
MSE(JLTE)725.6311752.21251102.89701872.6471
MSE(MJLTE)608.2459656.6023892.22141115.3394
### Table 3.
The estimated MSE values of LTE, JLTE and MJLTE k = 0.50.
d = 0.10d = 0.30d = 0.70d = 1
MSE(LTE)1133.25671459.73602393.90792042.7127
MSE(JLTE)734.5155795.83111234.97203340.5986
MSE(MJLTE)587.0096633.9972815.8143973.5845
### Table 4.
The estimated MSE values of LTE, JLTE and MJLTE k = 0.70.
d = 0.10d = 0.30d = 0.70d = 1
MSE(LTE)1415.1481774.12221774.12223613.8006
MSE(JLTE)779.0405891.0250891.02502274.5162
MSE(MJLTE)551.0494588.8484588.8484807.4456
### Table 5.
The estimated MSE values of LTE, JLTE and MJLTE k = 1.
We observed that for all values of d SMSE(MJLTE) assumed smaller values compared to both SMSE(JLTE) and SMSE(LTE). The estimators’ SMSE values are affected by increasing values of k, however the estimator that is affected the least by these changes is our proposed MJLTE estimator. When compared to the other two estimators, the SMSE values of MJLTE gave the best results for both the small and large values of k and d.
## 6. A simulation study
We want to illustrate the behavior of the proposed parameter estimator by a Monte Carlo simulation. The main purpose of this article is to demonstrate the construction and the details of the simulation which is designed to evaluate the performances of the estimators LTE, JLTE and MJLTE when the regressors are highly intercorrelated. According to Liu [8] and Kibria [18] the explanatory variables and response variable are generated by using the following equations
xij=1γ21/2zij+γzip,yi=1γ21/2zi+γzipi=1,2,,n,j=1,2,,p
where zijis an independent standard normal pseudo-random number and pis specified so that correlation between any two explanatory variables is given by γ2. In this study, we used γ=0.90,0.95,0.99to investigate the effects of different degrees of collinearity with sample sizes n=20,50and 100, while four different combinations for kdare taken as (0.8, 0.5), (1, 0.7), (1.5, 0.9), (2, 1). The standard deviations considered in the simulation study are σ=0.1;1.0;10. For each choice of γ, σ2and n, the experiment was replicated 1000 times by generating new error terms. The average SMSE was computed using the following formula
SMSEβ̂=11000j=11000βjββjβ
Let us consider the LTE, JLTE and MJLTE and compute their respective estimated MSE values with the different levels of multicollinearity. According to the simulation results shown in Tables 4 and 5 for LTE, JLTE and MJLTE with increasing levels of multicollinearity there was a general increase in the estimated MSE values Moreover, increasing level of multicollinearity also lead to the increase in the MSE estimators for fixed dand k.
In Table 4, the MSE values of the estimators corresponding to different values of dare given for k = 0.70. For all values of d, the smallest MSE value appears to belong to the MJLTE estimator. The least affected by multicollinearity is MJLTE according to MSE criteria.
In Table 5, the MSE values of the estimators corresponding to different values of dare given for k = 1.
For all values of d, the smallest MSE value appears to belong to the MJLTE estimator. The least affected by multicollinearity is MJLTE according to MSE criteria.
We can see that MJLTE is much better than the competing estimator when the explanatory variables are severely collinear. Moreover, we can see that for all cases of LTE, JLTE and MJLTE in MSE criterion the MJLTE has smaller estimated MSE values than those of the LTE and JLTE.
## 7. Conclusion
In this paper, we combined the LTE and JLTE estimators to introduce a new estimator, which we called MJLTE. Combining the underlying criteria of LTE and JLTE estimators enabled us to create a new estimator for regression coefficients of a linear regression model that is affected by multicollinearity. Moreover, the use of jackknife procedure enabled as to produce an estimator with a smaller bias. We compared our MJLTE to its originators LTE and JLTE in terms of MSEM and found that MJLTE has a smaller variance compared to both LTE and JLTE. Thus, MJLTE is superior to both LTE and JLTE under certain conditions.
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Nilgün Yıldız (December 18th 2018). A Study on the Comparison of the Effectiveness of the Jackknife Method in the Biased Estimators, Statistical Methodologies, Jan Peter Hessling, IntechOpen, DOI: 10.5772/intechopen.82366. Available from:
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We are IntechOpen, the world's leading publisher of Open Access books. Built by scientists, for scientists. Our readership spans scientists, professors, researchers, librarians, and students, as well as business professionals. We share our knowledge and peer-reveiwed research papers with libraries, scientific and engineering societies, and also work with corporate R&D departments and government entities. | {"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": 0.8724576234817505, "perplexity": 1650.8627291082896}, "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/1620243989614.9/warc/CC-MAIN-20210511122905-20210511152905-00473.warc.gz"} |
https://brilliant.org/problems/a-classical-mechanics-problem-by-swapnil-maiti/ | A classical mechanics problem by Swapnil Maiti
An archer pulls back 0.75 m on a bow which has a stiffness of 200 N/m. The arrow weighs 50 g. What is the velocity of the arrow immediately after it crosses the bowstring (traveled 0.75 m)?
× | {"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": 0.8599380254745483, "perplexity": 2181.251425978152}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-26/segments/1529267867666.97/warc/CC-MAIN-20180625111632-20180625131632-00313.warc.gz"} |
http://euclidlab.org/unsolved/336-pascals-triangle | # Does any number appear exactly 5 times in Pascal's triangle?
Pascal's triangle is a triangle of whole number formed by starting with a 1 at the top, then putting two 1's in the next row below, and then continuing as in the figure, such that each entry is the sum of the two numbers above to the left and the right, except that each row starts and ends with 1's. Continuing this pattern forever, is there any number that you will encounter exactly 5 times, no more, no less? | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8301398158073425, "perplexity": 154.98455301779575}, "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-2018-17/segments/1524125945669.54/warc/CC-MAIN-20180423011954-20180423031954-00421.warc.gz"} |
https://en.wikipedia.org/wiki/Vapor_pressure | Vapor pressure
The picture shows the particle transition, as a result of their vapor pressure, from the liquid phase to the gas phase and converse.
Vapor pressure or equilibrium vapor pressure is defined as the pressure exerted by a vapor in thermodynamic equilibrium with its condensed phases (solid or liquid) at a given temperature in a closed system. The equilibrium vapor pressure is an indication of a liquid's evaporation rate. It relates to the tendency of particles to escape from the liquid (or a solid). A substance with a high vapor pressure at normal temperatures is often referred to as volatile. The pressure exhibited by vapor present above a liquid surface is known as vapor pressure. As the temperature of a liquid increases, the kinetic energy of its molecules also increases. As the kinetic energy of the molecules increases, the number of molecules transitioning into a vapor also increases, thereby increasing the vapor pressure.
The vapor pressure of any substance increases non-linearly with temperature according to the Clausius–Clapeyron relation. The atmospheric pressure boiling point of a liquid (also known as the normal boiling point) is the temperature at which the vapor pressure equals the ambient atmospheric pressure. With any incremental increase in that temperature, the vapor pressure becomes sufficient to overcome atmospheric pressure and lift the liquid to form vapor bubbles inside the bulk of the substance. Bubble formation deeper in the liquid requires a higher pressure, and therefore higher temperature, because the fluid pressure increases above the atmospheric pressure as the depth increases.
The vapor pressure that a single component in a mixture contributes to the total pressure in the system is called partial pressure. For example, air at sea level, and saturated with water vapor at 20 °C, has partial pressures of about 2.3 kPa of water, 78 kPa of nitrogen, 21 kPa of oxygen and 0.9 kPa of argon, totaling 102.2 kPa, making the basis for standard atmospheric pressure.
Measurement and units
Vapor pressure is measured in the standard units of pressure. The International System of Units (SI) recognizes pressure as a derived unit with the dimension of force per area and designates the pascal (Pa) as its standard unit. One pascal is one newton per square meter (N·m−2 or kg·m−1·s−2).
Experimental measurement of vapor pressure is a simple procedure for common pressures between 1 and 200 kPa.[1] Most accurate results are obtained near the boiling point of substances and large errors result for measurements smaller than 1kPa. Procedures often consist of purifying the test substance, isolating it in a container, evacuating any foreign gas, then measuring the equilibrium pressure of the gaseous phase of the substance in the container at different temperatures. Better accuracy is achieved when care is taken to ensure that the entire substance and its vapor are at the prescribed temperature. This is often done, as with the use of an isoteniscope, by submerging the containment area in a liquid bath.
Estimating vapor pressures with Antoine equation
The Antoine equation[2][3] is a mathematical expression of the relation between the vapor pressure and the temperature of pure liquid or solid substances. The basic form of the equation is:
${\displaystyle \log P=A-{\frac {B}{C+T}}}$
and it can be transformed into this temperature-explicit form:
${\displaystyle T={\frac {B}{A-\log P}}-C}$
where: ${\displaystyle P}$ is the absolute vapor pressure of a substance
${\displaystyle T}$ is the temperature of the substance
${\displaystyle A}$, ${\displaystyle B}$ and ${\displaystyle C}$ are substance-specific coefficients (i.e., constants or parameters)
${\displaystyle \log }$ is typically either ${\displaystyle \log _{10}}$ or ${\displaystyle \log _{e}}$[3]
A simpler form of the equation with only two coefficients is sometimes used:
${\displaystyle \log P=A-{\frac {B}{T}}}$
which can be transformed to:
${\displaystyle T={\frac {B}{A-\log P}}}$
Sublimations and vaporizations of the same substance have separate sets of Antoine coefficients, as do components in mixtures.[2] Each parameter set for a specific compound is only applicable over a specified temperature range. Generally, temperature ranges are chosen to maintain the equation's accuracy of a few up to 8–10 percent. For many volatile substances, several different sets of parameters are available and used for different temperature ranges. The Antoine equation has poor accuracy with any single parameter set when used from a compound's melting point to its critical temperature. Accuracy is also usually poor when vapor pressure is under 10 Torr because of the limitations of the apparatus used to establish the Antoine parameter values.
The Wagner Equation[4] gives "one of the best"[5] fits to experimental data but is quite complex. It expresses reduced vapor pressure as a function of reduced temperature.
Relation to boiling point of liquids
Further information: Boiling point
A log-lin vapor pressure chart for various liquids
As a general trend, vapor pressures of liquids at ambient temperatures increase with decreasing boiling points. This is illustrated in the vapor pressure chart (see right) that shows graphs of the vapor pressures versus temperatures for a variety of liquids.[6]
For example, at any given temperature, methyl chloride has the highest vapor pressure of any of the liquids in the chart. It also has the lowest normal boiling point (−24.2 °C), which is where the vapor pressure curve of methyl chloride (the blue line) intersects the horizontal pressure line of one atmosphere (atm) of absolute vapor pressure.
Although the relation between vapor pressure and temperature is non-linear, the chart uses a logarithmic vertical axis to produce slightly curved lines, so one chart can graph many liquids. A nearly straight line is obtained when the logarithm of the vapor pressure is plotted against 1/(T+230)[7] where T is the temperature in degrees Celsius. The vapor pressure of a liquid at its boiling point equals the pressure of its surrounding environment.
Liquid mixtures
Raoult's law gives an approximation to the vapor pressure of mixtures of liquids. It states that the activity (pressure or fugacity) of a single-phase mixture is equal to the mole-fraction-weighted sum of the components' vapor pressures:
${\displaystyle p_{\text{tot}}=\sum _{i}p_{i}\chi _{i}\,}$
where p tot is the mixture's vapor pressure, i is one of the components of the mixture and Χi is the mole fraction of that component in the liquid mixture. The term piΧi is the partial pressure of component i in the mixture. Raoult's Law is applicable only to non-electrolytes (uncharged species); it is most appropriate for non-polar molecules with only weak intermolecular attractions (such as London forces).
Systems that have vapor pressures higher than indicated by the above formula are said to have positive deviations. Such a deviation suggests weaker intermolecular attraction than in the pure components, so that the molecules can be thought of as being "held in" the liquid phase less strongly than in the pure liquid. An example is the azeotrope of approximately 95% ethanol and water. Because the azeotrope's vapor pressure is higher than predicted by Raoult's law, it boils at a temperature below that of either pure component.
There are also systems with negative deviations that have vapor pressures that are lower than expected. Such a deviation is evidence for stronger intermolecular attraction between the constituents of the mixture than exists in the pure components. Thus, the molecules are "held in" the liquid more strongly when a second molecule is present. An example is a mixture of trichloromethane (chloroform) and 2-propanone (acetone), which boils above the boiling point of either pure component.
The negative and positive deviations can be used to determine thermodynamic activity coefficients of the components of mixtures.
Solids
Vapor pressure of liquid and solid benzene
Equilibrium vapor pressure can be defined as the pressure reached when a condensed phase is in equilibrium with its own vapor. In the case of an equilibrium solid, such as a crystal, this can be defined as the pressure when the rate of sublimation of a solid matches the rate of deposition of its vapor phase. For most solids this pressure is very low, but some notable exceptions are naphthalene, dry ice (the vapor pressure of dry ice is 5.73 MPa (831 psi, 56.5 atm) at 20 degrees Celsius, which causes most sealed containers to rupture), and ice. All solid materials have a vapor pressure. However, due to their often extremely low values, measurement can be rather difficult. Typical techniques include the use of thermogravimetry and gas transpiration.
There are a number of methods for calculating the sublimation pressure (i.e., the vapor pressure) of a solid. One method is to estimate the sublimation pressure from extrapolated liquid vapor pressures (of the supercooled liquid), if the heat of fusion is known, by using this particular form of the Clausius–Clapeyron relation:[8]
${\displaystyle \ln \,P_{solid}^{S}=\ln \,P_{liquid}^{S}-{\frac {\Delta H_{m}}{R}}\left({\frac {1}{T}}-{\frac {1}{T_{m}}}\right)}$
with:
${\displaystyle P_{solid}^{S}}$ = Sublimation pressure of the solid component at the temperature ${\displaystyle T = Extrapolated vapor pressure of the liquid component at the temperature ${\displaystyle T = Heat of fusion = Gas constant = Sublimation temperature = Melting point temperature
This method assumes that the heat of fusion is temperature-independent, ignores additional transition temperatures between different solid phases, and it gives a fair estimation for temperatures not too far from the melting point. It also shows that the sublimation pressure is lower than the extrapolated liquid vapor pressure (ΔHm is positive) and the difference grows with increased distance from the melting point.
Boiling point of water
Graph of water vapor pressure versus temperature. At the normal boiling point of 100 °C, it equals the standard atmospheric pressure of 760 Torr or 101.325 kPa.
Like all liquids, water boils when its vapor pressure reaches its surrounding pressure. In nature, the atmospheric pressure is lower at higher elevations and water boils at a lower temperature. The boiling temperature of water for atmospheric pressures can be approximated by the Antoine equation:
${\displaystyle \log _{10}P=8.07131-{\frac {1730.63}{233.426+T_{b}}}}$
or transformed into this temperature-explicit form:
${\displaystyle T_{b}={\frac {1730.63}{8.07131-\log _{10}P}}-233.426}$
where the temperature ${\displaystyle T_{b}}$ is the boiling point in degrees Celsius and the pressure ${\displaystyle P_{}}$ is in Torr.
Dühring's rule
Main article: Dühring's rule
Dühring's rule states that a linear relationship exists between the temperatures at which two solutions exert the same vapor pressure.
Examples
The following table is a list of a variety of substances ordered by increasing vapor pressure (in absolute units).
Substance Vapor Pressure
(SI units)
Vapor Pressure
(Bar);
Vapor Pressure
(mmHg);
Temperature
Tungsten 100 Pa 0.001 0.75 3203 °C
Ethylene glycol 500 Pa 0.005 3.75 20 °C
Xenon difluoride 600 Pa 0.006 4.50 25 °C
Water (H2O) 2.3 kPa 0.023 17.5 20 °C
Propanol 2.4 kPa 0.024 18.0 20 °C
Ethanol 5.83 kPa 0.0583 43.7 20 °C
Methyl isobutyl ketone 2.66 kPa 0.0266 19.95 25 °C
Freon 113 37.9 kPa 0.379 284 20 °C
Acetaldehyde 98.7 kPa 0.987 740 20 °C
Butane 220 kPa 2.2 1650 20 °C
Formaldehyde 435.7 kPa 4.357 3268 20 °C
Propane[9] 997.8 kPa 9.978 7584 26.85 °C
Carbonyl sulfide 1.255 MPa 12.55 9412 25 °C
Nitrous oxide[10] 5.660 MPa 56.60 42453 25 °C
Carbon dioxide 5.7 MPa 57 42753 20 °C
Estimating vapor pressure from molecular structure
Several empirical methods exist to estimate liquid vapor pressure from molecular structure for organic molecules. Some examples are SIMPOL,[11] the method of Moller et al.,[8] and EVAPORATION.[12][13]
Meaning in meteorology
In meteorology, the term vapor pressure is used to mean the partial pressure of water vapor in the atmosphere, even if it is not in equilibrium,[14] and the equilibrium vapor pressure is specified otherwise. Meteorologists also use the term saturation vapor pressure to refer to the equilibrium vapor pressure of water or brine above a flat surface, to distinguish it from equilibrium vapor pressure, which takes into account the shape and size of water droplets and particulates in the atmosphere.[15]
References
1. ^ Růžička, K.; Fulem, M. & Růžička, V. "Vapor Pressure of Organic Compounds. Measurement and Correlation" (PDF).
2. ^ a b What is the Antoine Equation? (Chemistry Department, Frostburg State University, Maryland)
3. ^ a b Sinnot, R.K. (2005). Chemical Engineering Design] (4th ed.). Butterworth-Heinemann. p. 331. ISBN 0-7506-6538-6.
4. ^ Wagner, W. (1973), "New vapour pressure measurements for argon and nitrogen and a new method for establishing rational vapour pressure equations", Cryogenics 13 (8): 470–482, Bibcode:1973Cryo...13..470W, doi:10.1016/0011-2275(73)90003-9
5. ^ Perry's Chemical Engineers' Handbook, 7th Ed. pp. 4–15
6. ^ Perry, R.H.; Green, D.W., eds. (1997). Perry's Chemical Engineers' Handbook (7th ed.). McGraw-Hill. ISBN 0-07-049841-5.
7. ^ Dreisbach, R. R. & Spencer, R. S. (1949). "Infinite Points of Cox Chart Families and dt/dP Values at any Pressure". Industrial and Engineering Chemistry, 41 (1). p. 176. doi:10.1021/ie50469a040.
8. ^ a b Moller B.; Rarey J.; Ramjugernath D. (2008). "Estimation of the vapour pressure of non-electrolyte organic compounds via group contributions and group interactions". Journal of Molecular Liquids 143: 52. doi:10.1016/j.molliq.2008.04.020.
9. ^ "Thermophysical Properties Of Fluids II – Methane, Ethane, Propane, Isobutane, And Normal Butane" (page 110 of PDF, page 686 of original document), BA Younglove and JF Ely.
10. ^ "Thermophysical Properties Of Nitrous Oxide" (page 14 of PDF, page 10 of original document), ESDU.
11. ^ Pankow, J. F.; et al. (2008). "SIMPOL.1: a simple group contribution method for predicting vapor pressures and enthalpies of vaporization of multifunctional organic compounds". Atmos. Chem. Phys. 8 (10): 2773–2796. doi:10.5194/acp-8-2773-2008.
12. ^ "Vapour pressure of pure liquid compounds. Estimation by EVAPORATION". tropo.aeronomie.be
13. ^ Compernolle, S.; et al. (2011). "EVAPORATION: a new vapour pressure estimation method for organic molecules including non-additivity and intramolecular interactions". Atmos. Chem. Phys. 11 (18): 9431–9450. Bibcode:2011ACP....11.9431C. doi:10.5194/acp-11-9431-2011.
14. ^ Glossary (Developed by the American Meteorological Society)
15. ^ A Brief Tutorial. jhuapl.edu (An article about the definition of equilibrium vapor pressure) | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 26, "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": 0.9241490960121155, "perplexity": 1463.086893911324}, "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-2016-30/segments/1469257832939.78/warc/CC-MAIN-20160723071032-00135-ip-10-185-27-174.ec2.internal.warc.gz"} |
https://www.polar.ox.ac.uk/publication/understanding-the-anomalously-cold-european-winter-of-200506-using-relaxation-experiments/ | # Understanding the Anomalously Cold European Winter of 2005/06 Using Relaxation Experiments
Peer Reviewed
Jung T, Palmer TN, Rodwell MJ, & Serrar S
Monthly Weather Review 138, Issue 8, pages 3157-3174, 2010, 10.1175/2010MWR3258.1.
Experiments with the atmospheric component of the ECMWF Integrated Forecasting System (IFS) have been carried out to study the origin of the atmospheric circulation anomalies that led to the unusually cold European winter of 2005/06. Experiments with prescribed sea surface temperature (SST) and sea ice fields fail to reproduce the observed atmospheric circulation anomalies suggesting that the role of SST and sea ice was either not very important or the atmospheric response to SST and sea ice was not very well captured by the ECMWF model. Additional experiments are carried out in which certain regions of the atmosphere are relaxed toward analysis data thereby artificially suppressing the development of forecast error. The relaxation experiments suggest that both tropospheric circulation anomalies in the Euro–Atlantic region and the anomalously weak stratospheric polar vortex can be explained by tropical circulation anomalies. Separate relaxation experiments for the tropical stratosphere and tropical troposphere highlight the role of the easterly phase of quasi-biennial oscillation (QBO) and, most importantly, tropospheric circulation anomalies, especially over South America and the tropical Atlantic. From the results presented in this study, it is argued that the relaxation technique is a powerful diagnostic tool to understand possible remote origins of seasonal-mean anomalies.
Keywords: Europe, Winter/cool season, Quasibiennial oscillation, Diagnostics, Seasonal effects, Teleconnections
Categories: Arctic, Natural Science | {"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": 0.8449739217758179, "perplexity": 4666.662010663436}, "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-2022-05/segments/1642320304686.15/warc/CC-MAIN-20220124220008-20220125010008-00643.warc.gz"} |
http://nicf.net/2018/02/17/hamiltonian-mechanics.html | ## Introduction
This article is the first in a series I plan to write about physics for a mathematically trained audience. We’re going to start by talking about classical mechanics, the stuff that your first physics class was probably about if you’ve ever taken one. The formulation of classical physics usually presented in introductory physics classes is called Newtonian mechanics; it talks about things like masses and forces and Newton’s laws of motion. Newtonian mechanics is easy to teach and to work with without much machinery, but it has some features that can make it difficult to analyze mathematically. Physical systems and their interactions are described in terms of coordinates with velocity and force vectors all over the place, and it can be difficult to know how to deal with things like symmetries and constraints.
There are other, equivalent ways of describing classical mechanics, sometimes collectively called “analytical mechanics,” which are much easier to describe in a coordinate-free way. At the cost of a bit more abstraction, the analytical formulations have two big advantages: they make it easier to set up and solve some very complicated mechanics problems and, probably more importantly for our purposes, they make the relationship between classical mechanics and its generalizations most clear. The two most prominent such formulations are called Hamiltonian and Lagrangian mechanics, and they’re what we’re going to discuss in this article. They are, as we’ll see, two different ways of saying the same thing, but they highlight different enough aspects of the situation that they’re worth talking about separately.
This article assumes some mathematical background beyond what’s usually used to present these ideas in a physics class that covers them. In particular, the reader is expected to be familiar with the basics of the theory of smooth manifolds to the level of someone who’s finished a one-semester class on the subject. I will assume you remember a little bit about physics, but that you have never seen the Hamiltonian and Lagrangian frameworks discussed here.
Some of the examples and a couple ideas about the presentation are adapted from Gerald Folland’s Quantum Field Theory: A Tourist Guide for Mathematicians and Michael Spivak’s Physics for Mathematicians: Mechanics I, both of which I recommend.
## Hamiltonian Mechanics
### The Newtonian Setup
We’ll start by briefly describing, in coordinates, the sort of Newtonian mechanics problem we’re eventually going to be describing in a coordinate-free way. The prototypical example to keep in mind is that of a collection of $$N$$ particles moving in $$\mathbb{R}^3$$ where particle $$i$$ has mass $$m_i$$. We’ll write the position of particle $$i$$ as $$\mathbf q_i$$, with the boldface there to remind you that it’s an element of $$\mathbb{R}^3$$ and not an $$\mathbb{R}$$-valued coordinate on $$\mathbb{R}^{3N}$$. We’ll write $$\mathbf p_i=m_i( d \mathbf q_i/ d t)$$ for the momentum of particle $$i$$.
In the Newtonian setup, we describe physics in terms of forces; we imagine that there is some vector $$\mathbf F_i$$ we can compute for each particle for all time which tells us how that particle is accelerating, or equivalently, how its momentum is changing. Specifically, the relationship is given by “Newton’s second law”: $\mathbf F_i=\frac{ d \mathbf p_i}{ d t}=m_i\frac{d^2\mathbf q_i}{ d t^2}.$
In general one could imagine these forces depending on any data whatsoever about the physical system, but we’re going to be most interested in the case of conservative forces. This is the case where there is a function $$V$$ on $$\mathbb{R}^{3N}$$ called a potential for which the force on particle $$i$$ is given by $\mathbf F_i=-\frac{ \partial V}{ \partial \mathbf q_i}.$ So the force is given by the gradient of a function which depends only on positions, not on momenta. This condition is equivalent to saying that the integral of the force vector field around a closed loop — a quantity called the work done by the force while traveling around the loop — is always zero.
The name “conservative” comes from the fact that, suitably interpreted, this last condition is what we mean by saying that energy is conserved. There are many physical phenomena that are often modeled as nonconservative forces; friction is probably the most familiar example. But an overwhelming amount of physical evidence points toward the belief that the fundamental laws of physics do conserve energy, and that physical models of things like friction are merely “neglecting” the energy that leaks into forms like heat and sound that are more difficult to model. It is possible, but somewhat painful, to set up Hamiltonian mechanics in a way that allows for things like friction, but we’re going to focus on the conservative case in this article.
Throughout this short description we’ve already done things that make it difficult to keep track of what needs to be done with all these quantities when we change coordinates. The force on a particle is given by a gradient, and each momentum coordinate is “attached” to both a mass and a particular spatial coordinate. It will often be convenient to switch to a coordinate system that does not isolate each particle so neatly in its own triple of coordinates, or even one that mixes what we are now calling position and momentum coordinates. This all cries out for a description that describes the physical system in terms of points on a manifold, to which we can assign coordinates only once we know how all the mathematical objects involved are defined intrinsically.
### Configuration Space and Phase Space
We’ll start our quest for a coordinate-free description of mechanics by fixing a smooth manifold $$Q$$ which we’ll call configuration space. You should think of a point in $$Q$$ as corresponding the “position” of each component of a physical system at some fixed time. Some examples worth keeping in mind are:
1. A particle moving in $$\mathbb{R}^3$$. In this case, $$Q$$ is just $$\mathbb{R}^3$$.
2. $$N$$ particles moving in $$\mathbb{R}^3$$. We specify the configuration of this system by specifying the position of each particle, which we can do using a point in $$\mathbb{R}^{3N}$$.
3. Two particles connected by a rigid rod of length $$\ell$$. We could describe the configuration of this system using a point in $$\{(a,b)\in\mathbb{R}^3\times\mathbb{R}^3:||a-b||=\ell\}$$.
4. A rigid body moving through space. We could describe its configuration using a point in $$\mathbb{R}^3\times SO(3)$$, specifying the location of the object’s center of mass and its orientation.
In particular, specifying a point $$q\in Q$$ gives you an instantaneous snapshot of the system, but it doesn’t tell you anything about it’s changing. Even if you have a complete description of the physics, this doesn’t provide enough information to predict how the system will evolve in the future. (Imagine a ball rolling on a table; if you just know its position and not its velocity you don’t know where it’s about to move.)
So if we want to describe the state of a physical system in a way that allows us to do physics, the state needs to carry some additional information. Different formulations of analytical mechanics do this in different ways, and unfortunately the version used by the Hamiltonian formulation is one of the more opaque choices: the state of a physical system is given by specifying a point in the cotangent bundle of $$Q$$, which to match with physicists’ conventions we will call phase space. We’ll usually use the coordinates $$(q,p)$$ to refer to a point in phase space. (So $$q$$ is a point in $$Q$$ and $$p$$ is a cotangent vector at $$q$$.) When the system is in the state $$(q,p)$$, we’ll call $$p$$ the momentum.
The first time I encountered this setup I was confused by the fact that momentum is represented by a cotangent vector rather than a tangent vector — after all, the velocity of a particle is definitely a tangent vector, and momentum is supposed to be a multiple of it.
It will be easier to talk about this once we have the finished picture in front of us, but we can say a bit right now. While velocities should inarguably be tangent vectors — a velocity is literally the time derivative along the path that a particle is following — it’s actually not clear that this extends to momenta. When we use the word “momentum” we will mean something more general than “mass times velocity”; the two will coincide for Newtonian mechanics in rectangular coordinates but they can be different in general. For example, if we have a particle of mass $$m$$ moving in $$\mathbb{R}^2$$ and use polar coordinates, the momentum corresponding to the $$\theta$$ coordinate turns out to be the angular momentum $$xp_y-yp_x$$, which is not $$m( d \theta/ d t)=(xp_y-yp_x)/(x^2+y^2)$$.
Of course I have not yet said what it means for one expression or another to be the “right” generalization of momentum to a given coordinate system, but the point is that the relationship between momentum coordinates and derivatives of the corresponding position coordinates depends on the physical meaning of those coordinates; it’s not something you can extract just by looking at configuration space. (Indeed, this is true even in rectangular coordinates: the relationship depends on the mass of the particle, which is a physical quantity.) We’ll return to this question later.
### Symplectic Geometry
To properly describe Hamiltonian mechanics, we’ll need some basic facts about symplectic geometry, which we’ll briefly go over now in case they aren’t familiar.
A symplectic manifold is a smooth manifold $$M$$ together with a choice of a nondegenerate closed 2-form $$\omega$$ on $$M$$. (That is, the antisymmetric bilinear form $$\omega$$ defines on each tangent space is nondegenerate and $$d\omega=0$$. We’ll see soon how each of these two conditions is relevant.) A diffeomorphism between two symplectic manifolds that preserves the symplectic form is called a symplectomorphism.
The main thing that will turn out to make Hamiltonian mechanics go is the fact that the cotangent bundle of a manifold naturally has the structure of a symplectic manifold. The cotangent bundle of any manifold comes with a canonical symplectic form which can be described pretty simply. We start by defining the tautological 1-form on $$T^*Q$$. Given a tangent vector $$v$$ at a point $$(q,p)\in T^*Q$$, we’ll write $$\theta(v)=p(\pi_*(v))$$, where $$\pi:T^*Q\to Q$$ is the projection map. Given a local coordinate system $$q_1,\ldots,q_n$$ on a chart on $$Q$$, we also get coordinates $$p_1,\ldots,p_n$$ on each cotangent space. I encourage you to check that in these coordinates $\theta=\sum_{i=1}^n p_i d q_i.$ We then define $$\omega=d\theta$$, so that in these same coordinates $\omega=\sum_{i=1}^n d p_i\wedge d q_i,$ which is clearly nondegenerate.
One very striking difference between Riemannian and symplectic geometry is that in a neighborhood of any point on any symplectic manifold (even if it’s not a cotangent bundle) there is a coordinate system $$q_1,\ldots,q_n,p_1,\ldots,p_n$$ for which $$\omega=\sum_i d p_i\wedge d q_i$$. This result is called “Darboux’s theorem” and the $$q$$’s and $$p$$’s are said to provide canonical coordinates. This means that, very unlike on a Riemannian manifold, a symplectic manifold has no local geometry, so there’s no symplectic analogue of anything like curvature.
Even though phase space will end up being the only symplectic manifold we’ll use to do physics, it’s actually cleaner to describe the required machinery in more generality, so for now $$M$$ will be an arbitrary symplectic manifold. We’ll return to the case of phase space soon.
Since $$\omega$$ puts a nondegenerate bilinear form on each tangent space, it gives an isomorphism between the tangent and cotangent spaces at each point of $$M$$, and therefore an isomorphism between vector fields and 1-forms. We will especially be interested in this isomorphism in the case where the 1-form is $$d f$$ for some function $$f$$. In this case, we’ll write $$X_f$$ for the unique vector field for which $$\omega(Y,X_f)= d f(Y)$$ for all $$Y$$. (There is an arbitrary sign choice to make here — I could have said that $$\omega(X_f,Y)= d f(Y)$$. As always happens with such things, this decision seems to about evenly split the authors of books on this subject. Hamilton’s equations, discussed below, do have an arrangement of signs that everyone agrees on, and I’ve made choices in this section that are consistent with that.)
This vector field is sometimes called the symplectic gradient of $$f$$; if we had a Riemannian metric instead of $$\omega$$ here then this construction would of course give the usual gradient. It’s worth emphasizing, though, that while a Riemannian gradient of $$f$$ (usually) gives a direction in which $$f$$ is increasing, the symplectic gradient gives a direction in which $$f$$ is constant, since $$X_f(f)= d f(X_f)=\omega(X_f,X_f)=0$$.
Given any vector field $$X$$ at all on a smooth manifold, the existence and uniqueness of solutions of ODE’s lets us define a flow, that is, a one-parameter family of diffeomorphisms $$\phi^t:M\to M$$ for which $\left.\frac{d}{ d t}\right|_{t=0}\phi^t(a)=X|_a$ for any point $$a\in M$$, where the notation $$X|_a$$ means the tangent vector we get by restricting $$X$$ to $$a$$. (In general the flow might only be defined for $$t$$ in some neighborhood of 0, but this will always be enough for our purposes.) The flow is used to construct the Lie derivative of a tensor field with respect to a vector field. In order to take any sort of derivative of a tensor field on a manifold it’s necessary to be able to compare values of the tensor field at different points, and the flow gives us a way to do this. We define $\mathcal{L}_X(T)=\left.\frac{d}{ d t}\right|_{t=0}(\phi^t)^*(T).$
A vector field that arises as a symplectic gradient — that is, as $$X_f$$ for some $$f$$ — is called a Hamiltonian vector field and the corresponding flow is called a Hamiltonian flow. Note that since the definition of $$X_f$$ depends on $$\omega$$, in order for the Hamiltonian flow corresponding to a function to make sense, it’s necessary for $$\omega$$ to be preserved by the flow. Otherwise after running time forward using the flow our vector field won’t be $$X_f$$ for the same $$f$$ anymore!
So we’d like to characterize the $$X$$ for which $$\mathcal{L}_X\omega=0$$. To do this we invoke Cartan’s magic formula, which says that $$\mathcal{L}_X=\iota_X\circ d+d\circ\iota_X$$. (Here $$\iota_X$$ is the interior product with $$X$$, which is the map from $$d$$-forms to $$(d-1)$$-forms defined by $$\iota_X\alpha(Y_1,\ldots,Y_{d-1})=\alpha(X,Y_1,\ldots,Y_{d-1})$$.) This is where we use the fact that $$\omega$$ is closed: we see that $\mathcal{L}_X\omega=\iota_X( d \omega)+d(\iota_X\omega)=d(\iota_X\omega).$ If $$X$$ corresponds to $$\alpha$$ under the isomorphism between vector fields and 1-forms given by $$\omega$$, then $$\iota_X\omega=-\alpha$$ by definition, so we see that flowing along $$X_\alpha$$ preserves $$\omega$$ if and only if $$\alpha$$ is closed. In particular, since $$X_f$$ corresponds to $$d f$$, all Hamiltonian flows preserve $$\omega$$.
It will be important for us to analyze how functions change along Hamiltonian flows; we will, in fact, basically be translating all the physical questions this framework can address into what values functions take along a Hamiltonian flow. That is, if $$X_f$$ is a Hamiltonian vector field, $$a$$ is a point in $$M$$, and $$g$$ is a function on $$M$$, we’d like to compute $$d g/ d t$$ along the flow of $$X_f$$ through $$a$$. By definition, this is just $$X_f(g)$$, so by the definition of $$X_g$$, $\frac{ d g}{ d t}=X_f(g)= d g(X_f)=\omega(X_g,X_f).$ This fact will turn out to be important enough to warrant a definition: we’ll write $$\{g,f\}=\omega(X_g,X_f)$$ and call it the Poisson bracket of $$g$$ and $$f$$. As we just saw, the Poisson bracket measures how $$g$$ changes along the Hamiltonian flow corresponding to $$f$$. In particular, $$\{g,f\}=0$$ if and only if $$f$$’s Hamiltonian flow preserves $$g$$. Note also that the Poisson bracket is antisymmetric (because $$\omega$$ is), which means that $$f$$’s Hamiltonian flow preserves $$g$$ if and only if $$g$$’s Hamiltonian flow preserves $$f$$. (The Poisson bracket in fact turns out to put a Lie algebra structure on $$C^\infty(M)$$ — that is, it also satisfies the Jacobi identity — but we won’t need this fact here.)
So, to summarize:
• Phase space, being the cotangent bundle of configuration space, has a natural symplectic structure. In coordinates, the symplectic form is given by $$\omega=\sum d p_i\wedge d q_i$$.
• On any symplectic manifold, we can associate to each function $$f$$ a vector field $$X_f$$, and vector fields arising in this way are called Hamiltonian vector fields. Flowing along a Hamiltonian vector field always preserves the symplectic form.
• This construction lets us define the Poisson bracket $$\{g,f\}=\omega(X_g,X_f)$$, which measures both how $$f$$ changes when flowing along $$X_g$$ and how $$g$$ changes when flowing along $$X_f$$.
• Since flowing along a vector field $$X$$ preserves $$\omega$$ if and only if the corresponding 1-form $$\alpha$$ is closed, we can reverse this entire process if $$M$$ is simply connected. In that case, $$\alpha= d f$$ for some $$f$$, so $$X=X_f$$, and $$f$$ is uniquely determined up to adding a constant. So if $$M$$ is simply connected (or if not, then in an open neighborhood of any point), there is a one-to-one correspondence between vector fields whose flow preserves $$\omega$$ and smooth functions on $$M$$ modulo constants.
### Phase Space and Hamiltonians
We’re now ready to see how this machinery can allow us to do physics. We fix a manifold $$Q$$ called configuration space, and we write $$P=T^*Q$$ for its cotangent bundle, which we’ll call phase space. The basic assumption of Hamiltonian mechanics is that the way we “run time forward” in our physical system is by following the Hamiltonian flow corresponding to a distinguished function $$H$$, which we’ll call the Hamiltonian. That is, if our system is in state $$(q,p)$$ at time $$t_0$$ and $$\phi^t$$ is the flow along $$X_H$$, then our system is in state $$\phi^t(q,p)$$ at time $$t+t_0$$.
Suppose we are using local coordinates $$q_1,\ldots,q_n,p_1,\ldots,p_n$$ in which the symplectic form can be written as $$\omega=\sum_i d p_i\wedge d q_i$$. Given two vector fields $X=\sum(a_i\partial_{q_i}+b_i\partial_{p_i}),\quad X'=\sum(a'_i\partial_{q_i}+b'_i\partial_{p_i}),$ we get that $$\omega(X,X')=\sum(b_ia'_i-a_ib'_i)$$. I encourage the reader to verify that this means that for a function $$f$$, $X_f=\sum_i\left(\frac{ \partial f}{ \partial p_i}\partial_{q_i}-\frac{ \partial f}{ \partial q_i}\partial_{p_i}\right),$ and that the Poisson bracket is given by $\{f,g\}=\sum_i\left(\frac{ \partial f}{ \partial q_i}\frac{ \partial g}{ \partial p_i}-\frac{ \partial f}{ \partial p_i}\frac{ \partial g}{ \partial q_i}\right).$
If we’ve chosen a Hamiltonian $$H$$, then the value of a function $$f$$ evolves through time according to solutions of the differential equation $$df/ d t=\{f,H\}$$. Plugging in $$q_i$$ and $$p_i$$ for $$f$$, we get Hamilton’s equations: $\frac{ d q_i}{ d t}=\frac{ \partial H}{ \partial p_i},\qquad\frac{ d p_i}{ d t}=-\frac{ \partial H}{ \partial q_i}.$
As we saw in the last section, Hamiltonian flows always preserve their corresponding function, so the Hamiltonian itself ought to measure some scalar quantity that doesn’t change as time moves forward. In classical physics there’s really only one such quantity to choose: the value Hamiltonian at a point in $$P$$ ought to be physically interpreted as the total energy of the system when it is in that state.
In particular, consider the case where $$\{f,H\}=0$$. This happens exactly when $$H$$’s Hamiltonian flow preserves $$f$$, that is, $$f$$ is conserved by the laws of physics. But it is also equivalent to the claim that $$f$$’s Hamiltonian flow preserves $$H$$, that is, flowing along $$X_f$$ preserves $$H$$. This phenomenon gives us the Hamiltonian mechanics version of a result called Noether’s theorem: going between $$X_f$$ and $$f$$ gives us a one-to-one correspondence between Hamiltonian vector fields whose flow preserves $$H$$ (that is, vector fields whose flow preserves both $$H$$ and $$\omega$$) and scalar functions which are conserved by the laws of physics.
Let’s see how to recover Newtonian mechanics. In mechanics problems, energy is usually given as a sum of two terms, one representing kinetic energy, written $$T$$, and one representing potential energy, written $$V$$. In our Newtonian example from above, the kinetic energy is the usual $T=\sum_i\frac12 m_i\left|\frac{ d \mathbf{q}_i}{ d t}\right|^2=\sum_i\frac{|\mathbf p_i|^2}{2m_i},$ and the potential energy is simply our potential function $$V$$. So our Hamiltonian all together is: $H(q,p)=T(p)+V(q)=\sum_i\frac{|\mathbf p_i|^2}{2m_i}+V(q),$ and then, combining the three coordinates for each particle into a single vector, Hamilton’s equations give us $\frac{ d \mathbf q_i}{ d t}=\frac{ \partial H}{ \partial \mathbf p_i}=\frac{\mathbf p_i}{m_i}$ $\frac{ d \mathbf p_i}{ d t}=-\frac{ \partial H}{ \partial \mathbf q_i}=-\frac{ \partial V}{ \partial \mathbf q_i}.$
Note that Hamilton’s first equation exactly tells you how to compute the velocity of a particle once you know its momentum, which does something to address the concern we had earlier. Importantly, we see that this relationship depends on Hamiltonian; asking which velocity corresponds to a given momentum is meaningless until you’ve specified the laws of physics.
For our mechanical Hamiltonian, since the kinetic energy term is a homogeneous quadratic function of the momentum, we think of it as corresponding to a Riemannian metric on configuration space. In order to get agreement between the two ways of translating between velocity and momentum — using the inner product or going through Hamilton’s first equation — we need to include the masses of the particles in the metric, so that in our case for two tangent vectors $$v,v'$$ we have $\langle v,v'\rangle_T=\sum_im_i\langle\mathbf{v}_i,\mathbf{v}'_i\rangle$ where $$\langle\cdot,\cdot\rangle$$ is the usual inner product on $$\mathbb R^3$$. This induces a metric on the cotangent space given by $\langle p,p'\rangle_T=\sum_i\frac{\langle\mathbf{p}_i,\mathbf{p}'_i\rangle}{m_i},$ so following this convention the Hamiltonian would be written $H(q,p)=\frac12\langle p,p\rangle_T+V(q).$
### Examples
#### The Harmonic Oscillator
First, let’s consider a harmonic oscillator. This is a physical system with one degree of freedom $$q$$ in which the potential energy has the form $$\frac12kq^2$$ for some $$k$$. (The factor of $$\frac12$$ is of course purely for convenience.) This is a decent model for, for example, a mass attached to a light, frictionless spring.
If the mass of this particle is $$m$$, then our Hamiltonian is $$H=p^2/2m+kq^2/2$$, and Hamilton’s equations are $\frac{ d q}{ d t}=\frac pm\qquad\frac{ d p}{ d t}=-kq.$ This is of course a very easy pair of differential equations to solve: you get, writing $$\alpha=\sqrt{k/m}$$, that $$q=A\sin(\alpha(t-t_0))$$ and $$p=A\alpha\cos(\alpha(t-t_0))$$ for some $$A$$ and $$t_0$$.
So far this analysis is basically identical to what we would have gotten using regular Newtonian mechanics. Still, even though we just found a solution, we can get some practice with this machinery by performing a change of coordinates that makes the solution even easier. These solutions lie on the ellipse $$(\alpha q)^2+p^2=A^2$$, which suggests that we ought to rescale $$q$$ and $$p$$ and switch to polar coordinates.
So let’s first try setting $$r=\sqrt{kq^2+p^2/m}$$ and $$\theta=\arctan(\sqrt{km}q/p)$$; these are the polar coordinates corresponding to $$\tilde p=p/\sqrt m$$ and $$\tilde q=\sqrt k q$$. Sadly, this doesn’t quite do what we want: these aren’t canonical coordinates, that is, the symplectic form isn’t $$d r\wedge d \theta$$. Indeed, $\omega= d p\wedge d q=\sqrt{\frac mk} d \tilde p\wedge d \tilde q=\sqrt{\frac mk}r d r\wedge d \theta.$
It would be possible to work out the form of the Poisson bracket in these coordinates and see what equations we get, but it’s even easier to just find coordinates that are canonical and use those. We can do this by replacing $$r$$ with $$s=\frac12\sqrt{m/k}r^2=r^2/2\alpha$$. We then have $$\omega= d s\wedge d \theta$$ and $$H=\alpha s$$, and so Hamilton’s equations are $\frac{ d \theta}{ d t}=\frac{ \partial H}{ \partial s}=\alpha\qquad\frac{ d s}{ d t}=-\frac{ \partial H}{ \partial \theta}=0.$
This analysis makes it obvious that $$s$$ is a conserved quantity — that’s literally what the second equation says. This is equivalent to saying that $$\{s,H\}=0$$, which we could have checked in the original coordinates if we wanted. In this case this is all kind of silly, since $$s$$ is just a constant multiple of $$H$$; the next example will feature a less silly version of this phenomenon.
#### The Two-Body Problem
Consider two particles, with masses $$m_1$$ and $$m_2$$, moving under the influence of a conservative force that depends only on their relative positions, that is, on the difference $$\mathbf q_1-\mathbf q_2$$. (You might imagine for example two celestial bodies moving under the influence of gravity.) So our configuration space is $$\mathbb R^3\times\mathbb R^3$$, and our Hamiltonian is $H=\frac{|\mathbf p_1|^2}{2m_1}+\frac{|\mathbf p_2|^2}{2m_2}+V(\mathbf q_1-\mathbf q_2)$ for some function $$V$$.
We can already see another case of Noether’s theorem here. The fact that $$V$$ depends only on $$\mathbf q_1-\mathbf q_2$$ means that if we translate both particles by the same vector and leave their momenta fixed, $$H$$ is unchanged. For concreteness let’s consider translating in the positive $$x$$ direction; this corresponds to flowing along the vector field $$\partial_{x_1}+\partial_{x_2}$$ (writing $$x_i$$ for the $$x$$ component of $$\mathbf q_i$$). These translations also self-evidently preserve the symplectic form, and so our vector field must be Hamiltonian. And indeed, it’s $$X_f$$ for $$f=(p_1)_x+(p_2)_x$$, the $$x$$ component of the total momentum of the system. You could also check directly that the Poisson bracket $$\{f,H\}$$ is zero. So we see that a Hamiltonian that is preserved by translations in some direction corresponds to physics that preserve the component of total momentum in that direction.
There is a common change of coordinates that makes this system a bit easier to analyze: write $\mathbf Q=\frac{m_1\mathbf q_1+m_2\mathbf q_2}{m_1+m_2}\qquad\mathbf q=\mathbf q_1-\mathbf q_2.$ Now, given any diffeomorphism $$f$$ from a manifold $$Q$$ to itself, we can lift it to a diffeomorphism on the cotangent bundle by setting $$f^\sharp(q,p)=(f(q),(f^{-1})^*(p))$$. We call $$f^\sharp$$ the cotangent lift of $$f$$. It turns out that a diffeomorphism on a cotangent bundle has the form of a cotangent lift if and only if it preserves the canonical 1-form $$\theta$$. To compute $$f^\sharp$$ in coordinates, first note that $$(f^{-1})^*(p)(v)=p((f^{-1})_*(v))=p((f_*)^{-1}(v))$$ by definition, so the matrix for $$(f^{-1})^*$$ is the transpose of the inverse of the Jacobian of $$f$$.
So in particular, the cotangent lift gives us a natural way to turn any diffeomorphism on configuration space into a symplectomorphism on phase space. Once can check that doing this for our change of coordinates here gives us the momentum coordinates $\mathbf P=\mathbf p_1+\mathbf p_2\qquad\mathbf p=\frac{m_2\mathbf p_1-m_1\mathbf p_2}{m_1+m_2},$ and our Hamiltonian becomes $H=\frac{|\mathbf P|^2}{2M}+\frac{|\mathbf p|^2}{2m}+V(\mathbf q),$ where $$M=m_1+m_2$$ and $$m=m_1m_2/(m_1+m_2)$$. (The reader is encouraged to verify these computations; it’s good practice!)
The point of this change of coordinates was to “decouple” the two parts of the Hamiltonian. The coordinate $$\mathbf Q$$ is called the center of mass of the system; what we’ve shown is that our original system is equivalent to one with a free particle of mass $$M$$ moving with the center of mass and a particle of mass $$m$$ moving under the influence of the potential $$V$$.
#### Central Potentials
As one more example of the relationship between symmetries and conservation laws, let’s consider a particle moving in a potential that depends only on the distance of that particle from the origin. That is, $H=\frac{|\mathbf p|^2}{2m}+V(|\mathbf q|).$ This is a good model for a planet moving around the sun under the influence of Newtonian gravity; in this case we’ll have $$V(r)=-GMm/r$$, where $$M$$ is the mass of the sun and $$G$$ is the gravitational constant.
But no matter what $$V$$ is, the fact that it depends only on the length of $$\mathbf q$$ means that the physics is preserved by any rotation about the origin. It’s worth being precise about what we mean by this: rotation about the origin is a diffeomorphism on configuration space, and to extend it to a symplectomorphism on phase space we need to take its cotangent lift.
If $$R_\theta$$ is the rotation by $$\theta$$ around the $$z$$ axis, then \begin{aligned} (R_\theta)^\sharp(x,y,z,p_x,p_y,p_z)=(&\cos\theta x-\sin\theta y,\ \sin\theta x+\cos\theta y,\ z,\\ &\cos\theta p_x-\sin\theta p_y,\ \sin\theta p_x+\cos\theta p_y,\ p_z).\end{aligned} (The transpose of the inverse of $$R_\theta$$ is just $$R_\theta$$ itself, since $$R_\theta$$ is orthogonal.) This is the map that has to preserve the Hamiltonian if our analysis is to go through, which means it’s important that $$H$$ depends only on $$|\mathbf p|$$ and $$|\mathbf q|$$.
To get the vector field whose flow produces this symmetry, we take the derivative of this with respect to $$\theta$$ at $$\theta=0$$. We get $X=-y\partial_x+x\partial_y-p_y\partial_{p_x}+p_x\partial_{p_y},$ which is $$X_{L_z}$$ where $$L_z=xp_y-yp_x$$.
We call $$L_z$$ the angular momentum of our particle about the $$z$$ axis, and this analysis shows that any physics arising from a Hamiltonian which is symmetric under rotations about the $$z$$ axis.
#### A Note
It is easy to construct Hamiltonians which aren’t invariant under rotations or translations. Indeed, the one from the last example isn’t preserved by translations, and correspondingly we shouldn’t expect momentum to be conserved by, say, Newtonian gravity. Nonetheless, it’s believed by most physicists that the fundamental laws that the universe runs on, whatever they are, do have these two symmetries — the results of a physical experiment don’t depend on where you do it or which way you were facing — and that therefore conservation of linear and angular momentum hold in general.
If you are presented with a Hamiltonian that doesn’t have this symmetry, like the one in the last example, the assumption is that there’s some part of the physics that you’re neglecting, and that if you included it the symmetry would appear again. For example, if we imagine the last example to be about a planet moving around the sun, we are neglecting the influence of the planet’s gravity on the sun, and if we included it we would be in the situation from the previous example about the two-body problem.
There is another symmetry that classical physics obeys: it also shouldn’t matter when an experiment is performed. Under our formalism, time translation comes from flowing along the vector field given by $$H$$ itself, so this symmetry corresponds to the conservation of energy. This example is a bit different from the others, though, because the relationship is true by definition! This is an artifact of the way we set up the Hamiltonian formalism: it picks out time translation as “special,” as the flow that corresponds to the Hamiltonian, and specifying the Hamiltonian is the way we specify the laws of physics.
Like with momentum, it is possible to “break” the time translation symmetry (and therefore energy conservation) by using a Hamiltonian that depends explicitly on time. This is useful when the forces acting on the particles or the constraints of the physical system change over time. (An example of the latter that’s often trotted out in physics classes is a bead attached to a spinning circle of wire.) I’ve chosen not to consider the case of time-dependent Hamiltonians or Lagrangians in this article for simplicity, but the theory does continue to work just fine in that setting.
## Lagrangian Mechanics
Recall that Hamilton’s equations are given by $\frac{ d q_i}{ d t}=\frac{ \partial H}{ \partial p_i},\qquad\frac{ d p_i}{ d t}=-\frac{ \partial H}{ \partial q_i}.$ As I mentioned briefly in that section, the first equation can give us a sort of answer to the question we had earlier about the relationship between momentum and velocity: it supplies, for every point $$(q,p)\in T^*Q$$ a tangent vector $$v\in T_qQ$$, and it tells us to interpret that tangent vector as a velocity. Provided that this procedure is invertible, which it will be in all of the cases we care about, we can think of it as giving us a “change of coordinates” from the cotangent bundle to the tangent bundle. This will turn out to give us another formulation of mechanics, called Lagrangian mechanics, which, while formally equivalent to everything we’ve done so far, sheds light on different aspects of mechanics than the Hamiltonian picture.
### The Legendre Transform
We’ll take this assignment of tangent vectors to points in phase space as our starting point. If the tangent vector $$v$$ comes from $$(q,p)$$ in this way, our goal will be to rewrite Hamilton’s equations in a way that depends on $$q$$ and $$v$$ rather than on $$q$$ and $$p$$. Put another way, the Hamiltonian gave us a way to turn paths in the cotangent bundle into paths in the tangent bundle, and we’d like to see what restrictions Hamilton’s equations impose directly on these new paths.
Hamilton’s first equation was, in some sense, “used up already” in the definition of $$v$$. By switching coordinates to $$v$$ and interpreting $$v$$ as a velocity, this equation tells us simply that $$v= d q/ d t$$, that is, at every point $$(q,v)=\gamma(t)$$ along our path, we should have $$v=\gamma'(t)$$. This means that we might as well just talk about paths in configuration space rather than its tangent bundle; we can identify such a path with its lift to the tangent bundle and do away with one of our two equations.
So it remains to translate the second equation into something to do with $$v$$. Note that the $$q$$ coordinate has very little to do with our goal here: we’re trying to turn a statement about the cotangent bundle into a statement about the tangent bundle, and all of the action is happening in the fibers of these two bundles. So it will be cleaner to ignore $$q$$ for now and examine how our coordinate change procedure behaves on a general vector space.
Suppose we have a smooth function $$H$$ on a vector space $$U$$. ($$U$$ will end up being the cotangent space at a point of configuration space.) Then for each point $$p\in U$$, $$d H$$ gives a linear map from the tangent space $$T_pU$$ to $$\mathbb R$$. But since $$U$$ is a vector space, there is a canonical identification of each of its tangent spaces with $$U$$ itself, so we can in fact think of $$d H$$ as giving us a way of assigning, to each $$p\in U$$, a linear map from $$U$$ to $$\mathbb R$$, that is, an element of $$U^*$$.
So $$H$$ gives us a map $$W_H:U\to U^*$$. (It’s important to emphasize that $$W_H$$ has no reason to be linear, so this is not any sort of inner product.) Geometrically, $$W_H$$ takes $$p$$ to the element of $$U^*$$ corresponding to the linear part of the linear approximation to $$H$$ at $$p$$.
Suppose now that $$W_H$$ is invertible. (This will happen, for example, if $$H$$ is a convex function of $$p$$, as is the case for our mechanical Hamiltonians from the last section.) Then it turns out that $$W_H^{-1}$$ arises in the same way as $$W_H$$: there is a function $$L$$ so that $$W_L=W_H^{-1}$$. We call $$L$$ the Legendre transform of $$H$$.
In fact, $$L$$ can be computed explicitly: one can show that when $$W_H$$ is invertible, $$W_L=W_H^{-1}$$ if and only if $L(W_H(p))+H(p)=\langle W_H(p),p\rangle$ up to an additive constant, where $$\langle\cdot,\cdot\rangle$$ is the pairing between $$U^*$$ and $$U$$. (Usually we let the constant be zero.) In particular, this makes it clear that the Legendre transform is an involution: $$H$$ is also the Legendre transform of $$L$$.
To see all this, first note that by definition, for any $$p\in U$$, we have $$\langle W_H(p),p'\rangle=dH_p(p')$$, where on the right hand side we think of $$p'$$ as living in the tangent space at $$p$$. So, taking the derivative of both sides and pairing with an arbitrary $$p'$$, we get that they are equal if and only if $\langle(DW_H)_p(p'),W_L(W_H(p))\rangle=\langle(DW_H)_p(p'),p\rangle,$ where $$(DW_H)_p$$ is the total derivative of $$W_H$$ at $$p$$. Since $$W_H$$ is invertible, the left side of this pairing can be anything, so this is true if and only if $$W_L(W_H(p))=p$$. The reader is encouraged to fill in the missing steps of this argument; it’s a good exercise in following all the relevant definitions.
So what did this giant mess of symbols get us? To any function $$H$$ on $$U$$ we’ve associated a new function $$L$$ on $$U^*$$ so that their “coordinate-change functions” are inverses of each other. In coordinates $$p_1,\ldots,p_n$$ on $$U$$ and $$v_1,\ldots,v_n$$ on $$U^*$$, this means that $v_i=\frac{ \partial H}{ \partial p_i},\qquad p_i=\frac{ \partial L}{ \partial v_i},$ and $L=\langle v,p\rangle-H.$ In the case we’re interested in, when we’re doing this in every fiber of the cotangent bundle and $$H$$ is the Hamiltonian of some physical system, we call $$L$$ the Lagrangian of that same system.
What does this look like for the mechanical Hamiltonians we were working with before? There we had $$H=T+V=\sum_i\frac{|\mathbf p_i|^2}{2m_i}+V(q).$$ So we get $$\mathbf v_i= \partial H/ \partial \mathbf p_i=\mathbf p_i/m_i$$ and $$\langle v,p\rangle=\sum_i\frac{|\mathbf p_i|^2}{m_i}=2T$$. This means the Lagrangian turns out to be $$L=2T-(T+V)=T-V$$.
It’s worth stressing again that despite the fact that we’ve arrived at an expression for $$L$$ that looks very similar to $$H$$, there is an additional important difference between the two aside from the fact that the sign on $$V$$ has flipped: $$H$$ is a function on the cotangent bundle and $$L$$ is a function on the tangent bundle! The relationship between momenta and velocities — that is, between $$p$$ and $$v$$ — depends entirely on the physics being modeled, so unless you’ve picked a Hamiltonian or a Lagrangian this relationship remains unspecified. Only when this relationship has been established does it even make sense to write something like $$L+H=2T$$; if we were being more careful we would actually write something like $$L(q,W_H(p))+H(q,p)=2T(p)$$.
Recall that Hamilton’s first equation now just tells that the tangent vector we pick at every point of our path should be the time derivative of the path at that point, so we are just left with translating the second into a statement about Lagrangians. That equation was $\frac{ d p_i}{ d t}=-\frac{ \partial H}{ \partial q_i}.$ Now, since we performed our Legendre transform just in the fibers of the cotangent and tangent bundles, nothing interesting happened to derivatives with respect to $$q$$ coordinates, so the fact that $$L=\langle v,p\rangle-H$$ means that $$\partial L/ \partial q_i=- \partial H/ \partial q_i$$. This, combined with the fact that $$p_i= \partial L/ \partial v_i$$ gives us Lagrange’s equation: $\frac{d}{ d t}\left(\frac{ \partial L}{ \partial v_i}\right)=\frac{ \partial L}{ \partial q_i}.$
It is more common for authors to write $$\dot q_i$$ where I’ve written $$v_i$$ here, using the usual physicists’ convention of dots to indicate time derivatives. The reason I didn’t do this was to avoid a common confusion: when you write expressions like $$\partial L/ \partial \dot q_i$$ it’s tempting to assume that one is supposed to compute $$\dot q_i$$ from $$q_i$$ or something. But the Lagrangian is a function on the tangent bundle, not just on configuration space, and $$\partial L/ \partial v_i$$ is just a derivative with respect to one of the coordinates on the tangent bundle. Given a smooth path in configuration space there is a natural way to lift it and obtain a path in the tangent bundle, and the physical assumption we are making is that the paths that happen physically are exactly the ones whose lifts satisfy Lagrange’s equation.
Note that we end up with half as many equations as before — we have one for every position coordinate, rather than one for every position or momentum coordinate. But because the Lagrangian depends on both $$q$$ and $$v$$ and we fix $$v_i$$ to be the time derivative of $$q_i$$, these end up being second-order differential equations rather than the first-order Hamilton’s equations, so we still need the same amount of information in our initial conditions to solve them as before.
### An Example
Let’s look at a concrete example of an at least somewhat nontrivial mechanics problem and see how to solve it using the Lagrangian formalism. This problem still would be feasible to tackle using the techniques from a Newtonian mechanics class, but the Lagrangian approach makes it quite straightforward.
Consider a two-dimensional world with a mass hanging from a very light spring. The end of the spring without the mass is fixed in place and the other end is free to swing around. We’ll pick coordinates $$r,\theta$$ for our configuration space, where $$r$$ is the current length of the spring and $$\theta$$ is the angle the spring makes with the vertical. We’ll write the corresponding time derivatives as $$v_r$$ and $$v_\theta$$. These coordinates have the nice property that their time derivatives are always perpendicular, so the speed of the particle is $$\sqrt{v_r^2+(rv_\theta)^2}$$. Therefore, the kinetic energy is simply $$T=\frac12mv_r^2+\frac12mr^2v_\theta^2$$.
There are two contributions to the potential energy: gravity and the restoring force from the spring. Springs are well modeled by potentials of the form $$V_{\mathrm{spring}}=\frac12 k(r-\ell)^2$$ for some constant $$k$$, where $$\ell$$ is the “natural length” of the spring. We encountered a potential of this form when we discussed the harmonic oscillator. Gravity (at least in cases like this where we can neglect the varying distances from the center of the earth) produces a constant acceleration in all falling bodies, and I encourage you to check that this is the same as asserting that $$V_{\mathrm{gravity}}=mgh$$ where $$g$$ is that constant acceleration and $$h$$ is the height of a particle above an arbitrary reference height.
Putting this all together, we get $L=\frac12m(v_r^2+r^2v_\theta^2)-\frac12k(r-\ell)^2+mgr\cos\theta.$ We can plug this into Lagrange’s equation and get, simplifying a bit and using physicists’ dot notation for derivatives, $m\ddot{r}=mr\dot\theta^2+mg\cos\theta-k(r-\ell)$ and $mr\ddot{\theta}+2m\dot r\dot\theta=-mgr\sin\theta.$
### The Calculus of Variations
There is another, very different way to obtain the Lagrange’s equation involving a technique called the calculus of variations. The calculus of variations is a sort of infinite-dimensional calculus performed on spaces of functions rather than finite-dimensional vector spaces. It provides tools for doing things like finding local minima or maxima of some functional on such a space of functions. I’ll sketch the part of this story that produces Lagrangian mechanics.
Suppose we are given an arbitrary function $$L:\mathbb{R}\times TQ\to\mathbb{R}$$, where we think of the first $$\mathbb{R}$$ as representing a time coordinate. Given any path $$\gamma:[0,1]\to Q$$, we consider the following quantity, called the action: $S(\gamma)=\int_0^1L(t,\gamma(t),\gamma'(t)) d t.$ (Here again $$t\mapsto (\gamma(t),\gamma'(t))$$ is the natural lift of $$\gamma$$ to a path in the tangent bundle of $$Q$$; we are abusing notation slightly writing the components the way we are here.) We want to find, out of all paths $$\gamma$$ with a fixed starting and ending point, the ones that locally minimize or maximize $$S$$.
We’ll answer this by considering smooth homotopies $$h:(-\epsilon,\epsilon)\times[0,1]\to Q$$ for which $$h(0,t)=\gamma(t)$$ and which leave the endpoints fixed. As usual, we’ll write $$h_u(t)=h(u,t)$$ If $$\gamma$$ minimizes the action, then it ought to in particular be a local minimum within any such homotopy. That is, 0 should be a critical point of the function $$u\mapsto S(h_u)$$.
For ease of notation from here on out, I’ll write $$\bar\gamma(t)=(t,\gamma(t),\gamma'(t))$$ for the argument to $$S$$, so $$S(\gamma)=\int_0^1 L(\bar\gamma(t)) d t$$.
So we need $\left.\frac{d}{ d u}\right|_{u=0}\int_0^1 L\left(t,h(u,t),\frac{ \partial h}{ \partial t}(u,t)\right) d t=0,$ and we can turn this into a condition which doesn’t mention $$h$$ with a bit of computation. We can pull the derivative inside the integral sign and use the chain rule to get that the left side is $\int_0^1\sum_{i=1}^n\left[\frac{ \partial h_i}{ \partial u}(0,t)\frac{ \partial L}{ \partial q_i}(\bar\gamma(t))+\frac{\partial^2 h_i}{ \partial u \partial t}(0,t)\frac{ \partial L}{ \partial v_i}(\bar\gamma(t))\right] d t$ for any coordinate system $$(q_1,\ldots,q_n,v_1,\ldots,v_n)$$ on the tangent bundle; here $$h_i$$ is the $$q_i$$ component of $$h$$. Then, using integration by parts, we can transform the second term a bit more, turning it into: $\sum_{i=1}^n\left[\left.\left(\frac{ \partial h_i}{ \partial u}(0,t)\frac{ \partial L}{ \partial v_i}(\bar\gamma(t))\right)\right|_{t=0}^1-\int_0^1\frac{ \partial h_i}{ \partial u}(0,t)\frac{d}{ d t}\left(\frac{ \partial L}{ \partial v_i}(\bar\gamma(t))\right) d t\right].$ The fact that the endpoints are fixed means that the boundary term on the left is zero: at $$t=0$$ and $$t=1$$, $$h_u(t)$$ doesn’t vary as I change $$u$$, so $$\partial h/ \partial u=0$$. So we’re left with just the second term which we combine with the missing term above to get that we want $\int_0^1\sum_{i=1}^n\frac{ \partial h_i}{ \partial u}(0,t)\left[\frac{ \partial L}{ \partial q_i}(\bar\gamma(t))-\frac{d}{ d t}\left(\frac{ \partial L}{ \partial v_i}(\bar\gamma(t))\right)\right] d t=0.$
Finally, we note that, since $$\partial h/ \partial u$$ could be any vector at all at each $$t$$, the only way that integral is zero for every $$h$$ is if, for each $$i$$, the part in square brackets is identically zero as a function of $$t$$. But that is exactly the same as saying that $$\gamma$$ must satisfy Lagrange’s equation!
So we get a description of the laws of physics that looks much more “global” than the one from symplectic geometry above: a physical system gets from one point in configuration space to another by following a path that is a critical point of the action functional $$S$$. The original description told us how the position and momentum (or velocity) evolve through time if we know them at one moment, whereas this new description puts a restriction on the entire path at once. We’ll have more to say about this in the next section.
## A Little Philosophy
The first time I encountered Noether’s theorem gave me a particular feeling about physics that I’ll do my best to explain here. There is a sense in which, for example, the fact that the laws of physics are invariant under rotation is “obvious” and fact that they conserve angular momentum isn’t. After all, an expression like $$L_z=xp_y-yp_x$$ looks a little arbitrary. Why should it be a fundamental of nature that that’s conserved rather than, say, $$xp_y+yp_x$$?
Now that we’ve built up the machinery that connects rotation and angular momentum, it’s very tempting to say that we’ve “reduced” the fact that angular momentum is conserved to “merely” the fact that physics is rotationally symmetric. This is, of course, not true: we have reduced it to the fact that physics is rotationally symmetric and the fact that physics can be described using Hamiltonian mechanics. (There is of course a Lagrangian version of Noether’s theorem too, and in fact this is the form in which is was originally stated.) Indeed, trying to prove Noether’s theorem from Hamiltonian mechanics gets things a bit backwards: the fact that something like Noether’s theorem is true is baked into the fact that all the flows we consider are along Hamiltonian vector fields.
This is emphatically a physical and not a mathematical assertion — it is certainly possible to imagine a universe where physics doesn’t work this way, but this does not seem to be the universe we live in. We can’t hope to prove that conservation of angular momentum is “the only way it could logically have been” without making some assumptions about the nature of the laws of physics. What we can do, and what in some sense is the main goal of all of physics, is to try to find a way to state those laws with as few moving parts as possible, and I think that at least is something that this whole discussion manages to do. In either math or physics, any time you can take an arbitrary-looking algebraic expression and show that it falls out naturally from less arbitrary-looking mathematical objects you’ve made some progress toward making the world more intelligible.
This is partly why I took the particular path through the material that I did. It’s actually much more common in mechanics texts to do the Lagrangian version first and derive the Hamiltonian picture from it using a Legendre transform. I always had some trouble making this approach fit in my head without seeming like magic. Until you’ve seen how it fits into the story it’s unclear why the Lagrangian has anything to do with anything, and especially unclear why one would want to minimize the action. Even if, as I keep saying, it’s necessary to make up some physical assumption at some point to make this all go, I feel like the assumptions made here feel at least a little less made up; the reader is of course free to disagree.
As I sort of said when discussing it above, it isn’t quite as “magical” as it might first appear — the claim is not that the universe searches over all possible paths for the global minimum of the action, but rather that the one that occurs is a critical point of the action under any perturbation. Sometimes a path with this property is called “stationary.” So the rule is not quite as global as it might be — it is global in the sense of pertaining to the whole path at once rather than a particular point along it, but not in the sense of pertaining to all possible paths, even those far away from the one under consideration.
Still, it was very striking to people in the early 19th century that Newtonian mechanics could be recast in such a strange way, and I think this reaction makes a lot of sense. The fact that the action-principle version is mathematically equivalent to the more local description shouldn’t detract from this; even in pure mathematics one can often learn a great deal from expressing the same fact in two different ways. A lot of this work happened around the end of the 18th century, and, in keeping with the Enlightenment-era times, many authors took the action principle as evidence of the benevolent guiding hand of Nature making sure the particles don’t waste too much of their precious action on frivolous non-stationary paths or something.
This is, of course, not the way most modern physicists talk, but there is a different sense in which we now understand the story told in this article to be offering us a hint about something beyond Newtonian mechanics. Basically all the theories of modern physics, including quantum mechanics, quantum field theory, general relativity, and more speculative extensions to them like string theory, are most naturally expressed not in terms of forces and equal and opposite reactions but in terms of Lagrangians and Hamiltonians. | {"extraction_info": {"found_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": 0.9518768191337585, "perplexity": 117.00224138379734}, "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/1521257647706.69/warc/CC-MAIN-20180321215410-20180321235410-00498.warc.gz"} |
http://ssk.im/blog/schrodinger-density/ | Random Schrodinger Operators
Consider the one-dimensional, discrete Schrodinger operator on $\ell^2(\mathbb{Z})$:
This is a toy model for the behavior of an electron in a large domain. In order to model electron dynamics in the presence of random disordered environment, we can place assumptions on the randomness of the potential $V$. From the point of view of matrices, $H$ is an infinite Jacobi matrix with random elements along the diagonal. While this model is very simplified, random Schrodinger operators and their spectra turn out to have many interesting mathematical properties, including the prediction/replication of physically observed phenomena – see Chapter 9 of Cycon, Froese, Kirsch, Simon for an introduction which is fairly gentle, aside from a few typos. I will focus on two nice results: the integrated density of states and the Thouless formula.
Integrated Density of States
Let $(\Omega,\mathcal{F},\mathbb{P})$ be the canonical probability space associated with $V$, and suppose the potential $V$ is stationary and ergodic. For a given outcome $\omega\in\Omega$, let $V_\omega$ denote the potential, and $H_\omega$ the associated operator.
By elementary manipulations and applications of the ergodic property, it is possible to show that $\sigma(H_\omega)$, the spectrum of $H_\omega$, equals a deterministic set, call it $\Sigma$, $\mathbb{P}$-a.s. However, it is interesting to ask how the spectrum is distributed along this set $\Sigma$. Let $\delta_0\in \ell^2(\mathbb{Z})$ be the unit vector with value 1 in the 0th coordinate, and 0 elsewhere. Define the measure $dk$ by
Theorem. The support of $dk$ is $\Sigma$.
From just this definition and claim above, it is not clear why $dk$ is in any way related to the “distribution” of the spectrum. Note that the spectrum of $H_\omega$ is an infinite set, so it is not possibly to naively bucket and histogram to describe the distribution… but it essentially is! That is, we will restrict $H_\omega$ to a finite interval $[-L,L]$, compute the empirical density of the eigenvalues in this interval, and then see what happens as we take $L\rightarrow\infty$.
Let $\{\mathcal{E}_\Delta(\omega)\}_{\Delta\subset\mathbb{R}}$ represent the family of spectral projections associated with $H_\omega$, let $\chi_L$ be the indicator function of $[-L,L]$, and define the measure $dk_L$ by
Theorem. As $L\rightarrow\infty$, $dk_L$ converges vaguely to $dk$, $\mathbb{P}$-a.s.
Idea of Proof. First, prove that for a given bounded measurable function $f$, then $\int f dk_L \rightarrow \int f dk$, $\mathbb{P}$-a.s. To do so requires an application of Birkhoff’s ergodic theorem. Then, for each bounded measurable function $f$, we have a set $\Omega_f$ of measure 1 on which the desired behavior occurs. The conclusion of the proof is a classical approximation argument which exploits the separability of $C_0$ to stitch together countably many $\Omega_f$ to get a set of measure 1 on which the statement is true for all $f\in C_0$. $\square$.
Note that the prelimit measures $dk_L$ are random, so it is not a priori obvious (at least, to me) that $dk$ should be a deterministic measure! This offers a nice parallel to other results in random matrix theory, where taking the limit of empirical spectral distributions can give an unexpectedly explicit (and universal!) limiting measure – e.g., the semicircle law for Wigner matrices, or the circular law for matrices with iid elements.
Thouless Formula
Part of the joy of the one-dimensional assumption is that solutions to the eigenvalue problem $(H-E)u=0$ can be written in terms of $2\times 2$ transfer matrices since any solution is determined by its value at two adjacent points in $\mathbb{Z}$. That is, let $\mathbf{u}(n) = (u(n+1), u(n))$, and define
Then,
Note that $\mathbf{u}(n)$ can be written in terms of the product of the random matrices $A_n(E)$ applied to some initial condition. Then, Furstenberg’s theorem tells us that for $E\in\mathbb{R}$ and $\mathbb{P}$-a.s. $\omega\in\Omega$, there exists $\gamma(E)$ such that
Theorem. $\gamma(E) = \int \log |E - E'| dk(E')$.
To me, this relationship is incredible! Unfortunately, I don’t have much intuition as to why it should be true, but the proof involves showing that a similar result holds in the finite $N$ case, and then exploiting the subharmonicity of $\gamma$. Moreover, this is not merely a nice connection between $\gamma$ and $k$, but also a fundamental ingredient of the proof that under certain conditions on $V$, the spectrum $\sigma(H_\omega)$ has no absolutely continuous part. | {"extraction_info": {"found_math": true, "script_math_tex": 9, "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": 0.983253538608551, "perplexity": 130.28593301340672}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-39/segments/1537267160641.81/warc/CC-MAIN-20180924185233-20180924205633-00042.warc.gz"} |
http://www.econometricsbysimulation.com/2013/02/fractional-response-estimation.html | ## Tuesday, February 12, 2013
### Fractional Response Estimation
* Often times, many questions that we would like to answer take the form of fractional response data.
* What percentage of students by school meet international standards, what percentage of losing military commanders are brought up on war crimes charges, or what percentage of those in different cities have of getting the flu.
* Using standard OLS methods without accounting for the natural upper and lower bound may lead to biased and inconsistent estimates (if there are build ups around the upper or lower bounds).
* This is due to corr(u,y)!=0 by the same reasoning as with binary response, censored response, and truncated response data.
* For example:
* Simulation 1
clear
set obs 1000
gen x=rnormal()
gen u=rnormal()
* Unobserved heterogeniety
gen y=normal(0+6*x+u)
label var y "Observed outcome"
* Such a large coefficient on x will drive most of the observations into the extremes.
hist y, frac title(We can see large buildups at 0 and 1)
reg y x
local B_OLS = _b[x]
* Let's see our predicted outcomes
predict y_hat
label var y_hat "Predicted from OLS"
* One of the problems we realize immediately is that it is hard to discuss bias or consistency when we don't really know that the true expected change in y is as a result of a change in x.
* Fortunately, this is easy to calculate.
* y=NormalCDF(B0 + xB + u)
* dy/dx = B * NormalPDF(B0 + xB + u) ... by chain rule
* dy/dx = 2 * NormalPDF(-2 + x*2 + u)
gen partial_x_y = 6 * normalden(0+x*6 + u)
* Of course each effect is unique to each observation currently.
* To find the average we just average.
sum partial_x_y
* The mean of the partial_x_y shows us that the OLS estimator is too large.
local B = r(mean)
* To find the "true" linear projection of x on y we need first to identify the B0 coefficient.
sum y
local y_bar = r(mean)
sum x
local x_bar = r(mean)
local B0 = y_bar'-(x_bar'*B')
gen true_XB = B0' + B'*x
label var true_XB "Predicted from 'true' LM"
gen linear_u = y-true_XB
corr linear_u x
* We can see that the linear error terms are correlated with x, which is potentially what we expect if x has a positive coefficient.
* This is because as x gets large (approaches infitity for example) then the max(linear_u) must approach 0 since x is driving y to be 1 and u cannot make y larger than 1.
* The argument is identical for as x gets small. Thus when x is large E(u)<0 e="" is="" small="" u="" when="" while="" x="">0.
* We can see our OLS estimator is about half that of the true (though when the data was more evenly dispursed it seemed the OLS estimator was working well.)
* A greater concern for many might be not only the bias of the estimates but the feasibility of the predictions.
two (scatter y x, sort) (scatter y_hat x, sort) (line true_XB x, sort)
* We can see that the true Average Partial Effect is steeper, being less vulnerable to being adjusted by the outlierers created by Y being bound.
* This is not to say that the true APE is ignoring that for a large portion of the observations changes in x has very little effect on y.
* Only that OLS assumes a particular relationship in the errors which is violated.
* Alternatively we can use a estimator that better fits our data generating process.
glm y x, family(binomial) link(probit) robust
predict y_glm
label var y_glm "Predicted from GLM"
two (scatter y_glm y) (lfit y_glm y), title("A perfect fit would be a line that goes from 0,0 to 1,1")
* This looks pretty good.
two (scatter y x, sort) (scatter y_hat x, sort) (line y_glm x, sort col(red))
* The fitted data is much more satisfying as well.
* Notice though, that even if you ignored the fractional responses and just looked at the binary responses, a probit model would probably fit pretty well.
* We should be able to estimate the marginal effects with the margins command
margins, dydx(x)
di "True APE = B'"
di "OLS APE = B_OLS'"
* Unfortunately, it does not appear that the glm command is doing worse than the OLS estimator at estimating the APEs.
* This is probably because of the model misspecification. That is, when generating the data, the error term should not be a separate term within the normalCDF.
* However, I am not sure how best to do this otherwise. If anybody has any ideas I am open to them.
* An alternative might be the following:
* Simulation 2
clear
set obs 1000
gen x=rnormal()
gen yp=normal(3*x+2)
gen ob = rpoisson(100)
label var ob "Number of individuals"
gen y=rbinomial(ob,yp)/ob
label var y "Observed outcome"
* This formulation fits better into our concept of how fractional response data works.
* That is there is some number of individual units (in this case 100) which each are being evaluated for either 0 or 1 outcomes.
* By dividing by 100 we reduce the data to a fractional response problem which could be used to evaluate different sized groups since the fractional reponse maintains the same scale.
hist y, frac title(Fractional Responses Look Very Similar)
* Perhaps a little less build up on the tails.
reg y x
local B_OLS=_b[x]
* Let's see our predicted outcomes
predict y_hat
label var y_hat "Predicted from OLS"
* Now to esimate the partial effect we use the same formulation as before because
* for the binomial E(y|n,p)=np
* But we transformed y=E(y/n|n,p)=np/n=p
* Thus:
* y=NormalCDF(xB) ... but now we do not have to worry about the error u.
* dy/dx = B * NormalPDF(xB) ... by chain rule
* dy/dx = 2 * NormalPDF(x*2)
gen partial_x_y = 3 * normalden(x*3+2)
* To find the average we just average.
sum partial_x_y
local B = r(mean)
* We can see that the OLS regressor now is too large.
sum y
local y_bar = r(mean)
sum x
local x_bar = r(mean)
local B0 = y_bar'-(x_bar'*B')
gen true_XB = B0' + B'*x
label var true_XB "Predicted from 'true' LM"
gen linear_u = y-true_XB
corr linear_u x
two (scatter y x, sort) (line y_hat x, sort) (line true_XB x, sort)
* The only problem with generating data this way is that our data looks too perfect.
* How do we add greater variability into such a formulation?
* A method many might think of would be to add some kind of fixed effect per observation.
* This would suggest that the individuals within each observation are more or less likely to participate based on some unobserved factor which is unique to that observation.
* This is probably not a stretch to think about: People in a city are more likely to get the flu if the flue is already common in the city, Student are more likely to pass an exam if the school is particularly strong or attracts strong students, etc.
* However, if this is the case then the previous data generating process we discussed is more appropriate since u is in effect an unobserved observation level effect.
* In the extreme, as the number of individuals gets large the binomial formulation converges on having the response be the pdf (by the law of large numbers).
glm y x, family(binomial) link(probit) robust
di "Notice the standard errors are about " e(dispers_p)^.5 " fraction of the those if there were a single draw of the binomail"
* Which is appropriate since there are one average 100 draws, thus the estimated standard error should be about 100 times too large
predict y_glm
margins, dydx(x)
di "True APE = B'"
di "OLS APE = B_OLS'"
* We can see that now the glm estimator is working better.
* A researcher might be instead tempted to estimate total raw responses as a function of x via OLS.
* In order to do this the researcher would not calculate the fraction of responses.
gen yraw = y*ob
hist yraw
* The data does not look like it would be crazy to estimate some kind of truncated regression (tobit for example).
* However, given how we know the data has been generated, this would be a mistake.
* The zeros in the data are not truncated values but rather meant to be zeros.
tobit yraw x, ll(0)
* We can adjust the results to our previous scale by dividing the coefficient by 100.
di "Our censored regression estimate is: " _b[x]/100
* This is larger than our previous estimate and biased because it is making the assumption that if our data were not bottom coded at 0 then we would have actually observed negative values for y.
* This is clearly impossible.
* We can further aggravate our mistake by specifying an upper limit as well.
* However, we will make this mistake looking at the original ratios:
tobit y x, ll(0) ul(1)
predict y_tobit
* This yeilds a large and even more biased estimate.
two (scatter y x) (line y_tobit x)
* Graphically it is easy to see how strong the assumptions would need be to justify use of a censored tobit.
* It is similar to what would have been estimated if we just dropped all values that were equal to 0 or 1.
reg y x if y>0 & y<1 p=""> | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 0.8711980581283569, "perplexity": 2603.701688127222}, "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/1454701166261.11/warc/CC-MAIN-20160205193926-00195-ip-10-236-182-209.ec2.internal.warc.gz"} |
http://tex.stackexchange.com/questions/16127/add-text-to-toc-without-dotted-lines-and-line-number | # Add text to TOC without dotted-lines and line number
Despite following many suggestions on the net, I have the following issue. I am using TeXnicCenter and MiKTeX. [I have posted the same on Latex Community Forum]. I have not used/tried the tocloft package yet. I am currently using the thesis2 class.
Objective: In my table of contents, I want a text "Chapter" without the dotted lines and the page number. Eg.,
============================================
Acknowledgments ................ ii
Abstract ........................ v
Chapter
1 Introduction .................. 1
1.1 Network Design .............. 1
1.1.1 Steiner Tree Problems ..... 2
============================================
Trial-1: When I do \addcontentsline{toc}{chapter}{\protect Chapter}, it adds the required text (Chapter) to toc. But, the dotted lines and page number is present.
Trial-2: The same happens when I do \addcontentsline{toc}{chapter}{\protect \numberline{}Chapter}.
Q1) Also, I want only one-line spacing between "Chapter" and "1 Introduction". Currently, I get 2 line-spacing (like between Acknowledgments and Abstract). How do I get rid of that 2 line spacing and make it 1-line spacing?
-
Did you have a look at this question tex.stackexchange.com/questions/9289/toc-without-page-numbers? – Thorsten Apr 19 '11 at 5:24
Use \addtocontents instead of \addcontentsline, and add positive and negative vertical space as appropriate. (Note: I used the thesis2 class file available here.)
\documentclass{thesis2}
\begin{document}
\tableofcontents
\frontmatter
\chapter{Acknowledgments}
\chapter{Abstract}
\mainmatter
\protect\vspace{1em}%
\protect\noindent Chapter\protect\par
\protect\vspace{-1em}%
}
\chapter{Introduction}
\section{Network design}
\end{document}
-
Lockstep - Many thanks for this neat solution. Really appreciate. – srinivas Apr 19 '11 at 13:40
@srinivas: Please consider marking my answer as ‘Accepted’ by clicking on the tickmark below its vote count. – lockstep Apr 19 '11 at 15:36
@lockstep Do you know how to do the reversed case? I want the dotted line after acknowledgements, abstract etc, but by default of tocloft they don't have the dotted leaders. Thanks. – xslittlegrass Jul 14 '15 at 15:38
Here's an alternative solution based on the suggestions here. It's 'cleaner' in the sense that you don't set spacing and indentation explicitly:
\documentclass{article}
\newcommand{\notocnumsection}[1]{
\bgroup%
\section{#1}%
\egroup
}
\begin{document}
\tableofcontents
\section{foo}
This is the foo section.
\notocnumsection{bar}
This is the bar section.
\section{baz}
This is the baz section.
\end{document}
It does not, however, address your wish regarding spacing. But I believe you should think about changing the spacing once for all chapter-level toc entries. For that, I think the tocloft package should be useful (documentation here).
- | {"extraction_info": {"found_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": 0.8532712459564209, "perplexity": 3955.4834910314353}, "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/1461860116173.76/warc/CC-MAIN-20160428161516-00174-ip-10-239-7-51.ec2.internal.warc.gz"} |
https://math.stackexchange.com/questions/998246/hilbert-space-product-and-tensor-product-space | # Hilbert space: product and tensor product space
Let $H_1$ and $H_2$ be Hilbert spaces, then I would intuitively define the inner product on $H_1 \times H_2$ by $\langle (x_1,x_2),(y_1,y_2)\rangle = \langle x_1,y_1 \rangle + \langle x_2,y_2 \rangle$.
Now the inner product of the tensor space is given by $\langle (x_1 \otimes x_2),(y_1 \otimes y_2)\rangle = \langle x_1 y_1 \rangle \cdot \langle x_2,y_2 \rangle$ and I was wondering how this fits together? Could anybody comment on the issue why we have to define the inner products exactly in that way for the two different spaces?
The way the cartesian and tensor product combine can be seen in the isomorphism of complex valued functions $$L^{2}(R\times R)\cong L^{2}(R)\otimes L^{2}(R)$$ The inner product on the left hand side is a double integral while that on the right is your product of inner products (integrals in one variable) . This identifies functions on $R\times R$ with limits of sums of functions of the form f(x)g(y) with respect to the norm generated by the appropriate inner product.
First you treat Hilbert spaces as algebraic objects, vector spaces. Thus you may take an algebraic tensor product $H_1 \underline{\otimes} H_2$, however an algebraic tensor product of Hilbert spaces is not a Hilbert space. Therefore, we have to fix this obstacle, we define the inner product (as you done it in your question) on their algebraic tensor product and take the completion of $H_1 \underline{\otimes} H_2$ with respect to this inner product. This object equipped with the inner product which you have pointed out is a Hilbert space and we denote it as $H_1 \otimes H_2$ and call a (Hilbert space) tensor product of $H_1$ and $H_2$. | {"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": 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": 0.9893577098846436, "perplexity": 108.7865147021474}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-30/segments/1563195526517.67/warc/CC-MAIN-20190720132039-20190720154039-00038.warc.gz"} |
http://cms.math.ca/cmb/msc/16R50 | location: Publications → journals
Search results
Search: MSC category 16R50 ( Other kinds of identities (generalized polynomial, rational, involution) )
Expand all Collapse all Results 1 - 5 of 5
1. CMB 2012 (vol 56 pp. 584)
Liau, Pao-Kuei; Liu, Cheng-Kai
On Automorphisms and Commutativity in Semiprime Rings Let $R$ be a semiprime ring with center $Z(R)$. For $x,y\in R$, we denote by $[x,y]=xy-yx$ the commutator of $x$ and $y$. If $\sigma$ is a non-identity automorphism of $R$ such that $$\Big[\big[\dots\big[[\sigma(x^{n_0}),x^{n_1}],x^{n_2}\big],\dots\big],x^{n_k}\Big]=0$$ for all $x \in R$, where $n_{0},n_{1},n_{2},\dots,n_{k}$ are fixed positive integers, then there exists a map $\mu\colon R\rightarrow Z(R)$ such that $\sigma(x)=x+\mu(x)$ for all $x\in R$. In particular, when $R$ is a prime ring, $R$ is commutative. Keywords:automorphism, generalized polynomial identity (GPI)Categories:16N60, 16W20, 16R50
2. CMB 2011 (vol 55 pp. 271)
Di Vincenzo, M. Onofrio; Nardozza, Vincenzo
On the Existence of the Graded Exponent for Finite Dimensional $\mathbb{Z}_p$-graded Algebras Let $F$ be an algebraically closed field of characteristic zero, and let $A$ be an associative unitary $F$-algebra graded by a group of prime order. We prove that if $A$ is finite dimensional then the graded exponent of $A$ exists and is an integer. Keywords:exponent, polynomial identities, graded algebrasCategories:16R50, 16R10, 16W50
3. CMB 2009 (vol 52 pp. 267)
Ko\c{s}an, Muhammet Tamer
Extensions of Rings Having McCoy Condition Let $R$ be an associative ring with unity. Then $R$ is said to be a {\it right McCoy ring} when the equation $f(x)g(x)=0$ (over $R[x]$), where $0\neq f(x),g(x) \in R[x]$, implies that there exists a nonzero element $c\in R$ such that $f(x)c=0$. In this paper, we characterize some basic ring extensions of right McCoy rings and we prove that if $R$ is a right McCoy ring, then $R[x]/(x^n)$ is a right McCoy ring for any positive integer $n\geq 2$ . Keywords:right McCoy ring, Armendariz ring, reduced ring, reversible ring, semicommutative ringCategories:16D10, 16D80, 16R50
4. CMB 2003 (vol 46 pp. 14)
Bahturin, Yu. A.; Parmenter, M. M.
Generalized Commutativity in Group Algebras We study group algebras $FG$ which can be graded by a finite abelian group $\Gamma$ such that $FG$ is $\beta$-commutative for a skew-symmetric bicharacter $\beta$ on $\Gamma$ with values in $F^*$. Categories:16S34, 16R50, 16U80, 16W10, 16W55
5. CMB 1998 (vol 41 pp. 118)
Valenti, Angela
On permanental identities of symmetric and skew-symmetric matrices in characteristic \lowercase{$p$} Let $M_n(F)$ be the algebra of $n \times n$ matrices over a field $F$ of characteristic $p>2$ and let $\ast$ be an involution on $M_n(F)$. If $s_1, \ldots, s_r$ are symmetric variables we determine the smallest $r$ such that the polynomial $$P_{r}(s_1, \ldots, s_{r}) = \sum_{\sigma \in {\cal S}_r}s_{\sigma(1)}\cdots s_{\sigma(r)}$$ is a $\ast$-polynomial identity of $M_n(F)$ under either the symplectic or the transpose involution. We also prove an analogous result for the polynomial $$C_r(k_1, \ldots, k_r, k'_1, \ldots, k'_r) = \sum_ {\sigma, \tau \in {\cal S}_r}k_{\sigma(1)}k'_{\tau(1)}\cdots k_{\sigma(r)}k'_{\tau(r)}$$ where $k_1, \ldots, k_r, k'_1, \ldots, k'_r$ are skew variables under the transpose involution. Category:16R50
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https://scicomp.stackexchange.com/users/3980/nico-schl%C3%B6mer?tab=summary | Nico Schlömer
11 Using fixed point iteration to decouple a system of pde's 11 What is the Complexity of MATLAB operations 10 Tanh-sinh quadrature numerical integration method converging to wrong value 10 Are there any numerical advantages in solving symmetric matrix compared to matrices without symmetry? 9 Is variable scaling essential when solving some PDE problems numerically?
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25 pde × 6 14 iterative-method × 2 18 sparse-matrix × 3 13 mesh × 5 18 finite-difference × 2 13 linear-algebra × 3 17 quadrature × 7 13 matlab × 2 17 python × 6 12 optimization × 4 | {"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": 0.8319922089576721, "perplexity": 4755.897483670419}, "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/1579251788528.85/warc/CC-MAIN-20200129041149-20200129071149-00094.warc.gz"} |
http://math.stackexchange.com/questions/226496/is-every-planar-graph-without-triangles-3-colorable/226504 | # Is every planar graph without triangles 3-colorable?
In other words, can a planar graph without k3 have a chromatic number larger than 3?
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Yes, every triangle-free planar graph is 3-colorable. This is known as Grötzsch's theorem.
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It is perhaps worth noting that the theorem does not generalize in the natural way, as evidenced in answers to this question: math.stackexchange.com/questions/225872/… – Austin Mohr Nov 1 '12 at 2:58 | {"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": 0.8960883021354675, "perplexity": 709.0241986459213}, "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-2014-52/segments/1418802775656.66/warc/CC-MAIN-20141217075255-00159-ip-10-231-17-201.ec2.internal.warc.gz"} |
https://hal-insu.archives-ouvertes.fr/insu-01288769 | Studying Galactic interstellar turbulence through fluctuations in synchrotron emission
Abstract : Aims: The characteristic outer scale of turbulence (i.e. the scale at which the dominant source of turbulence injects energy to the interstellar medium) and the ratio of the random to ordered components of the magnetic field are key parameters to characterise magnetic turbulence in the interstellar gas, which affects the propagation of cosmic rays within the Galaxy. We provide new constraints to those two parameters. Methods: We use the LOw Frequency ARray (LOFAR) to image the diffuse continuum emission in the Fan region at (l,b) ~ (137.0°, +7.0°) at 80' × 70' resolution in the range [146, 174] MHz. We detect multi-scale fluctuations in the Galactic synchrotron emission and compute their power spectrum. Applying theoretical estimates and derivations from the literature for the first time, we derive the outer scale of turbulence and the ratio of random to ordered magnetic field from the characteristics of these fluctuations. Results: We obtain the deepest image of the Fan region to date and find diffuse continuum emission within the primary beam. The power spectrum displays a power law behaviour for scales between 100 and 8 arcmin with a slope α = -1.84 ± 0.19. We find an upper limit of ~20 pc for the outer scale of the magnetic interstellar turbulence toward the Fan region, which is in agreement with previous estimates in literature. We also find a variation of the ratio of random to ordered field as a function of Galactic coordinates, supporting different turbulent regimes. Conclusions. We present the first LOFAR detection and imaging of the Galactic diffuse synchrotron emission around 160 MHz from the highly polarized Fan region. The power spectrum of the foreground synchrotron fluctuations is approximately a power law with a slope α ≈ -1.84 up to angular multipoles of ≤1300, corresponding to an angular scale of ~8 arcmin. We use power spectra fluctuations from LOFAR as well as earlier GMRT and WSRT observations to constrain the outer scale of turbulence (Lout) of the Galactic synchrotron foreground, finding a range of plausible values of 10-20 pc. Then, we use this information to deduce lower limits of the ratio of ordered to random magnetic field strength. These are found to be 0.3, 0.3, and 0.5 for the LOFAR, WSRT and GMRT fields considered respectively. Both these constraints are in agreement with previous estimates.
Keywords :
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Article dans une revue
Astronomy and Astrophysics - A&A, EDP Sciences, 2013, 558, pp.A72. 〈10.1051/0004-6361/201322013〉
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Littérature citée [29 références]
https://hal-insu.archives-ouvertes.fr/insu-01288769
Contributeur : Nathalie Rouchon <>
Soumis le : jeudi 6 juillet 2017 - 11:14:13
Dernière modification le : vendredi 10 août 2018 - 15:36:02
Document(s) archivé(s) le : mardi 23 janvier 2018 - 23:17:44
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Citation
M. Iacobelli, M. Haverkorn, E. Orrú, R. F. Pizzo, J. Anderson, et al.. Studying Galactic interstellar turbulence through fluctuations in synchrotron emission. Astronomy and Astrophysics - A&A, EDP Sciences, 2013, 558, pp.A72. 〈10.1051/0004-6361/201322013〉. 〈insu-01288769〉
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http://kleine.mat.uniroma3.it/mp_arc-bin/mpa?yn=96-42 | 96-42 Dell'Antonio G.F., Figari R., Teta A.
Statistics in Space Dimension Two (58K, TeX) Feb 17, 96
Abstract , Paper (src), View paper (auto. generated ps), Index of related papers
Abstract. We construct as a selfadjoint operator the Schroedinger hamiltonian for a system of $N$ identical particles on a plane, obeying the statistics defined by a representation $\pi_1$ of the braid group. We use quadratic forms and potential theory, and give details only for the free case; standard arguments provide the extension of our approach to the case of potentials which are small in the sense of forms with respect to the laplacian. We also comment on the relation between the analysis given here and other approaches to the problem, and also on the connection with the description of a quantum particle on a plane under the influence of a shielded magnetic field (Aharanov-Bohm effect).
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https://arxiv.org/abs/1902.11152 | cs.IT
Title:Diffusive Molecular Communications with Reactive Molecules: Channel Modeling and Signal Design
Abstract: This paper focuses on molecular communication (MC) systems using two types of signaling molecules which may participate in a reversible bimolecular reaction in the channel. The motivation for studying these MC systems is that they can realize the concept of constructive and destructive signal superposition, which leads to favorable properties such as inter-symbol interference (ISI) reduction and avoiding environmental contamination due to continuous release of signaling molecules into the channel. This work first presents a general formulation for binary modulation schemes that employ two types of signaling molecules and proposes several modulation schemes as special cases. Moreover, two types of receivers are considered: a receiver that is able to observe both types of signaling molecules (2TM), and a simpler receiver that can observe only one type of signaling molecules (1TM). For both of these receivers, the maximum likelihood (ML) detector for general binary modulation is derived under the assumption that the detector has perfect knowledge of the ISI-causing sequence. In addition, two suboptimal detectors of different complexity are proposed, namely an ML-based detector that employs an estimate of the ISI-causing sequence and a detector that neglects the effect of ISI. The proposed detectors, except for the detector that neglects ISI for the 2TM receiver, require knowledge of the channel response (CR) of the considered MC system. Moreover, the CR is needed for performance evaluation of all proposed detectors. However, deriving the CR of MC systems with reactive signaling molecules is challenging since the underlying partial differential equations that describe the reaction-diffusion mechanism are coupled and non-linear. Therefore, we develop an algorithm for efficient computation of the CR and validate its accuracy via particle-based simulation.
Comments: This paper has been accepted for publication in the IEEE Transactions on Molecular, Biological, and Multi-Scale Communications(Tmbmc). arXiv admin note: text overlap with arXiv:1711.00131 Subjects: Information Theory (cs.IT); Emerging Technologies (cs.ET) Cite as: arXiv:1902.11152 [cs.IT] (or arXiv:1902.11152v2 [cs.IT] for this version)
Submission history
From: Vahid Jamali [view email]
[v1] Wed, 27 Feb 2019 10:14:57 UTC (2,627 KB)
[v2] Mon, 22 Jul 2019 09:42:21 UTC (3,348 KB) | {"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": 0.9171410799026489, "perplexity": 1346.141640459919}, "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-2019-43/segments/1570986675316.51/warc/CC-MAIN-20191017122657-20191017150157-00188.warc.gz"} |
https://kluedo.ub.uni-kl.de/frontdoor/index/index/docId/351 | ## Analogies between Proofs - A Case Study
• This case study examines in detail the theorems and proofs that are shownby analogy in a mathematical textbook on semigroups and automata, thatis widely used as an undergraduate textbook in theoretical computer scienceat German universities (P. Deussen, Halbgruppen und Automaten, Springer1971). The study shows the important role of restructuring a proof for findinganalogous subproofs, and of reformulating a proof for the analogical trans-formation. It also emphasizes the importance of the relevant assumptions ofa known proof, i.e., of those assumptions actually used in the proof. In thisdocument we show the theorems, the proof structure, the subproblems andthe proofs of subproblems and their analogues with the purpose to providean empirical test set of cases for automated analogy-driven theorem proving.Theorems and their proofs are given in natural language augmented by theusual set of mathematical symbols in the studied textbook. As a first step weencode the theorems in logic and show the actual restructuring. Secondly, wecode the proofs in a Natural Deduction calculus such that a formal analysisbecomes possible and mention reformulations that are necessary in order toreveal the analogy.
$Rev: 13581$ | {"extraction_info": {"found_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": 0.950169563293457, "perplexity": 1700.0880069675688}, "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-2015-48/segments/1448398462686.42/warc/CC-MAIN-20151124205422-00066-ip-10-71-132-137.ec2.internal.warc.gz"} |
https://stats.stackexchange.com/questions/203608/discrepency-between-confidence-intervals-of-individual-estimates-and-differences | # Discrepency between confidence intervals of individual estimates and differences of estimates
I carried out a GEE Poisson regression on my dependent variable, number of days, and my independent variables are binary categories, including a high/low treatment indicator of interest. I obtained the following 95% confidence intervals for a high treatment group and low treatment group (the other binary covariates are held fixed):
High Treatment Group Point Estimate, (95% CI): 10.7099, (10.2836, 11.1540)
Low Treatment Group Point Estimate, (95% CI): 8.9673, (6.3788, 12.6062)
I noted that the two confidence intervals overlap, which leads me to believe that the two treatment groups means are not statistically significantly different.
However, I then calculated an estimate of the predicted group differences and the corresponding confidence interval (e.g. $\hat{\mu_{high}}-\hat{\mu_{low}}$), using the Delta method, and I got the following:
1.7426, (0.1714438, 3.313756).
Now, I noted that the the point estimate makes perfect sense as it is $10.7099-8.9673=1.7426$, but the confidence interval does not contain zero, which implies that the difference in means is statistically significantly different fro zero.
Why are the two conclusions different? Is the confidence made smaller in and therefore significant in the mean difference calculations because maybe, I'm taking advantage of the covariances between the estimates or is something else maybe going on here? Which one of the conclusions should I believe?
UPDATE: I've found some articles that talk about how overlapping confidence intervals don't necessarily imply that the differences between estimated means are not significant in a t-test. The made reference to the sample size being larger (and hence smaller confidence intervals) in the two-sided t-test. That doesn't seem to be going on here since I'm using regression. Can anyone help?
You should believe you second conclusion. As you stated you need to take into account the covariance between the two values. The hypothesis you are interested in testing is $$H_0: \mu_{high} - \mu_{low} = \mu_{diff} = 0$$ So to construct a confidence interval for $\mu_{diff}$ we need the standard error of $\hat \mu_{diff}$, we have the variance is given by \begin{align*} Var(\hat \mu_{diff}) &= Var(\hat \mu_{high} - \hat \mu_{low}) \\ & = Var(\hat \mu_{high}) + Var(\hat \mu_{low}) - 2 \times Cov(\hat \mu_{high}, \hat \mu_{low}) \end{align*} You can take the square root to get the standard error and calculate the confidence interval from there to preform inference. | {"extraction_info": {"found_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": 1, "x-ck12": 0, "texerror": 0, "math_score": 0.9715318083763123, "perplexity": 654.3160985011738}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-30/segments/1563195529480.89/warc/CC-MAIN-20190723151547-20190723173547-00253.warc.gz"} |
http://mathoverflow.net/questions/57589/heuristic-behind-a-infty-algebras?sort=oldest | # Heuristic behind $A_{\infty}$ - algebras
How to think about the $A_{\infty}$-algebras ?
I am looking at the Bernhard Keller's introduction, he says a few words about the topological origin (not in details) and motivates by stating two problems in homological algebra but what I am looking for is the intuitive idea and some prototypical examples to keep in mind and justify the axioms of $A_{\infty}$- algebra and their morphisms.
Thanks
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I'd like to correct one mistake in the answers to this question. $A_{\infty}$ spaces and algebras are ones that are associative up to all higher homotopies; $E_{\infty}$ spaces and algebras are ones that are associative and commutative up to all higher homotopies. Neither notion has anything to do with groups, a priori. When connected, an $A_{\infty}$ space is equivalent to a loop space (Stasheff) and an $E_{\infty}$ space is equivalent to an infinite loop space (Boardman and Vogt, May). In the non-connected case, both assertions remain true provided that the monoid of components is a group. The $A$ can be thought of as standing for "Associative'', the $\infty$ standing for "up to the infinitude of relevant higher homotopies''. The $E$, due to Boardman, stands for homotopy Everything'', the $\infty$ again standing for "up to the infinititude of relevant higher homotopies''. In both cases there are notions of $A_n$ spaces and $E_n$ spaces and many examples of $A_n$-spaces that are not $A_{n+1}$ spaces. This is more striking for $E_n$-spaces, since those are equivalent to n-fold loop spaces (with the same proviso on components in the non-connected case). It is correct that $A_{\infty}$ spaces preceded operads. A key motivation for operads is to encode the infinitude of homotopies in the $E_{\infty}$ case since, unlike the $A_{\infty}$ case, it seems impossible to specify all of the relevant homotopies explicitly. The DG world works in exactly the same way, and passage to chains transports the topological examples to algebraic ones.
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I did not know the $E$ stood for Everything. Thanks! – Mariano Suárez-Alvarez Mar 7 '11 at 2:40
If you have a differential algebra $A$, a complex $B$ and a isomorphism of complexes $f:A\to B$, then you can transport the structure (i.e., the multiplication) on $A$ to turn $B$ into a differential algebra. If you only have a homotopy equivalence $f:A\to B$, then that does not work: but $f$ is enough to construct a $A_\infty$-algebra structure on $B$.
«$A_\infty$-algebra structures are what you get from transporting algebra structures along homotopy equivalences» is not a bad slogan.
(Notice that this really gives you a slogan for $X_\infty$ for most $X$s... We can talk about $\mathrm{Lie\\ triple\\ systems}\_\infty$ things, for example)
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The standard example is the based loop space $\Omega M$ thought of as the space of continuous maps from $[0,1]$ to $M$ where 0 and 1 are mapped to the base point. This has a product on it given by taking the loop you get from going around one loop and then the other. In other words, if you have loops $f$ and $g$, if $t\in [0,1/2]$, $(fg)(t) = g(2t)$, and if $t\in [1/2,1]$, $(fg)(t) = f(2t-1)$.
It's easy to see that this product is not associative, but it is associative up to a reparametrization of the circle. Thus, there's a homotopy between $f(gh)$ and $(fg)h$ which is a map
$$[0,1] \times \Omega M^{{}\times 3} \to \Omega M$$
For four loops, you can draw a pentagon of homotopies: $$f(g(hi)) \sim f((gh)i) \sim (f(gh))i \sim ((fg)h)i \sim (fg)(hi) \sim f(g(hi))$$
Let $K_4$ be the pentagon. This is a map:
$$\partial K_4 \times \Omega M^{{}\times 4} \to \Omega M$$
These homotopies are coherent which means that this extends to a map
$$K_4 \times \Omega M^{{}\times 4} \to \Omega M$$
This pattern continues and gives Stasheff's Associahedra. A space, $H$, possessing a set of maps (and my memory's not so great here, so I'll assume $i>1$ which might not be correct)
$$K_i \times H^{{}\times i} \to H$$
where $K_2 = pt$, $K_3 = [0,1]$, $K_4$ is as above, etc. is called an $A_\infty$-space. It's a theorem of Stasheff that any connected $A_\infty$-space is homotopic to a based loop space.
Now, pass to chains on the space and you get an $A_\infty$ algebra. The theorem of Stasheff is then the statement that the bar construction on an $A_\infty$ algebra is a dg-coalgebra.
This can all be thought of in terms of operads, of course, but I think this all came first. I don't know which introduction of Keller's you're using (he has several), but I believe this is all in this one.
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There is one aspect of $A_\infty$-algebras I haven't seen mentioned yet: Massey products and related structures.
Suppose you have a differential graded algebra $(A,d)$: a chain complex with a product such that the differential satisfies the Leibnitz rule: $d(xy) = d(x)y + (-1)^{|x|} x d(y)$, where $x$ is in degree $|x|$ in the chain complex. Then its cohomology $H^*(A,d)$ has the structure of an algebra. Suppose further that the product in the chain complex is associative (this is not strictly necessary, but it makes the rest of this easier to write down). Then $H^*(A,d)$ has the structure of an $A_\infty$-algebra. There are two important features:
• There is an $A_\infty$-algebra isomorphism between $A$ and $H^*(A,d)$. This is despite the fact that there doesn't have to be any nontrivial honest map of associative algebras between these two.
• $H^*(A,d)$ is associative on the nose, not just up to homotopy, so the $A_\infty$-structure conveys other information: it has to do with Massey products. Massey products are complicated, but they are an important piece of the structure in the cohomology of any differential graded algebra. To define them: suppose that $a_1$, $a_2$, and $a_3$ are cocycles, representing cohomology classes $\alpha_1$, $\alpha_2$, and $\alpha_3$. Suppose also that $a_1 a_2$ is a boundary: $d(b) = a_1 a_2$ for some $b$, and similarly $d(c) = a_2 a_3$. This means that in cohomology, the products $\alpha_1 \alpha_2$ and $\alpha_2 \alpha_3$ are zero. Then with an appropriate sign, $a_1 c \pm b a_3$ is a cocycle, because $d(a_1 c) = \pm a_1 a_2 a_3$ and the same for $d(b a_3)$. The element $a_1 c \pm ba_3$ represents a cohomology class, an element of the Massey product $\langle \alpha_1, \alpha_2, \alpha_3 \rangle$. (You can modify $b$ by adding any cocycle to it, and similarly for $c$, and the Massey product is the resulting set of cohomology classes.) You can also define Massey products of length 4, 5, etc., as long as suitable shorter products vanish. It turns out that the $A_\infty$ structure on $H^*(A,d)$ captures at least some of this information: the higher product $m_3(\alpha_1 \otimes \alpha_2 \otimes \alpha_3)$ is an element of $\langle \alpha_1, \alpha_2, \alpha_3 \rangle$, and similarly for higher products. The axioms for an $A_\infty$-algebra says that the $A_\infty$-structure on $H^*(A,d)$ is a way of choosing elements from all of the different Massey products according to some compatibility requirement.
-
Abstractly, you can think of an $A_\infty$-space (an $A_\infty$-algebra in the category of topological spaces) as being a homotopy theorist's substitute for a space with a product which is strictly associative (a monoid). An $E_\infty$-space would then be a homotopy theorist's (edit: abelian) group (this analogy is quoted from Classifying spaces made easy by John Baez). and an $L_\infty$-space $L_\infty$-algebra would be a homotopy theorist's Lie algebra.
More concretely, the archetype for an $A_\infty$-space is a loop space, I think. Indeed, a space $X$ is a loop-space if and only if it is an $A_\infty$-space whose $\pi_0$ is a group. For more on this point of view, I recommend Infinite Loop Spaces by J.F. Adams.
-
Isn't $L_\infty$ homotopy theorists Lie algebras, rather (whatever that may mean...) – Mariano Suárez-Alvarez Mar 7 '11 at 2:34
@Mariano: corrected – Daniel Moskovich Mar 7 '11 at 2:39
I'm confused about a couple of things here. Aren't $E_\infty$ spaces a homotopy theorists (very) commutative group? the closest to a "group" is a grouplike $A_\infty$ space, i.e. as you point out, a loop space. Also is there such a thing as an $L_\infty$-space? I was under the impression that $L_\infty$ is a stable notion (operad), and doesn't make sense in unstable settings like spaces. (In the stable setting it's the Koszul dual to $E_\infty$, but I don't think that notion makes sense unstably - then again would be very happy to learn otherwise.) – David Ben-Zvi Mar 7 '11 at 3:12
@David You're right- I wrote rubbish. Hopefully now it's correct. – Daniel Moskovich Mar 7 '11 at 3:49
I like to think of definition of $A_\infty$ algebras as what you get when you take a space with an action of Stasheff's $A_\infty$ operad of associahedra and you take cellular chains. The algebraic definition follows directly from the cellular structure of the associahedra - they are convex polytopes with poset of faces isomorphic to the opposite of the poset of planar rooted trees and edge contractions.
Similar to this, the view of $A_\infty$ that I enjoy the most (as opposed to the one that I find most productive as a mental model) is that it is what you get if you want to do homological algebra on algebraic theories: The $A_\infty$ operad is the free resolution of the associative operad. – Mikael Vejdemo-Johansson Mar 6 '11 at 22:48 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8788440227508545, "perplexity": 259.79971180071607}, "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/1394011025965/warc/CC-MAIN-20140305091705-00015-ip-10-183-142-35.ec2.internal.warc.gz"} |
http://perimeterinstitute.ca/fr/seminar/neorealism-and-internal-language-topoi | # Neorealism and the Internal Language of Topoi
Crudely formulated, the idea of neorealism, in the way that
Chris Isham and Andreas Doering use it, means that each theory of
physics, in its mathematical formulation should share certain structural
properties of classical physics. These properties are chosen to allow some degree of
realism in the interpretation (for example, physical variables always have values).
Apart from restricting the form of physical theories, neorealism does
increase freedom in the shape of physical theories in another
way. Theories of physics may be interpreted in other topoi than the category
of sets and functions.
In my talk I will concentrate on two topos models for quantum
theory. The contravariant model of Butterfield, Isham and Doering on the
one hand, and the covariant model of Heunen, Landsman and Spitters on the
other. I will argue that when we think of the topoi as generalized
categories of sets (i.e. when we use the internal perspective of the topoi at hand),
these two models are closely related, and both resemble classical physics.
I will assume no background knowledge in topos theory.
Collection/Series:
Event Type:
Seminar
Scientific Area(s):
Speaker(s):
Event Date:
Mardi, Mars 12, 2013 - 15:30 to 17:00
Location:
Time Room
Room #:
294 | {"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": 0.8352005481719971, "perplexity": 4454.228733862384}, "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-2018-09/segments/1518891815435.68/warc/CC-MAIN-20180224053236-20180224073236-00425.warc.gz"} |
https://amathew.wordpress.com/tag/non-wandering-points/ | So, let’s fix a compact metric space ${X}$ and a transformation ${T: X \rightarrow X}$ which is continuous. We defined the space ${M(X,T)}$ of probability Borel measures which are ${T}$-invariant, showed it was nonempty, and proved that the extreme points correspond to ergodic measures (i.e. measures with respect to which ${T}$ is ergodic). We are interested in knowing what ${M(X,T)}$ looks like, based solely on the topological properties of ${T}$. Here are some techniques we can use:
1) If ${T}$ has no fixed points, then ${\mu \in M(X,T)}$ cannot have any atoms (i.e. ${\mu(\{x\})=0, x \in X}$). Otherwise ${\{x, Tx , T^2x, \dots \}}$ would have infinite measure.
2) The set of recurrent points in ${X}$ (i.e. ${x \in X}$ such that there exists a sequence ${n_i \rightarrow \infty}$ with ${T^{n_i}x \rightarrow x}$) has ${\mu}$-measure one. We proved this earlier.
3) The set of non-wandering points has measure one. We define this notion now. Say that ${x \in X}$ is wandering if there is a neighborhood ${U}$ of ${X}$ such that ${T^{-n}(U) \cap U = \emptyset, \forall n \in \mathbb{N}}$. In other words, the family of sets ${T^{i}(U), i \in \mathbb{Z}_{\geq 0}}$ is disjoint. If not, say that ${x}$ is non-wandering. Any recurrent point, for instance, is non-wandering, which implies that the set of non-wandering points has measure one.
Here is an example. (more…) | {"extraction_info": {"found_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": 22, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9947907328605652, "perplexity": 112.74524599874852}, "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/1593655899931.31/warc/CC-MAIN-20200709100539-20200709130539-00289.warc.gz"} |
https://proofwiki.org/wiki/Combination_Theorem_for_Continuous_Mappings/Topological_Ring/Multiple_Rule | # Combination Theorem for Continuous Mappings/Topological Ring/Multiple Rule
## Theorem
Let $\struct{S, \tau_{_S}}$ be a topological space.
Let $\struct{R, +, *, \tau_{_R}}$ be a topological ring.
Let $\lambda \in R$.
Let $f: \struct{S, \tau_{_S}} \to \struct{R, \tau_{_R}}$ be a continuous mapping.
Let $\lambda * f : S \to R$ be the mapping defined by:
$\forall x \in S: \map {\paren{\lambda * f}} x = \lambda * \map f x$
Let $f * \lambda : S \to R$ be the mapping defined by:
$\forall x \in S: \map {\paren{f * \lambda}} x = \map f x * \lambda$
Then
$\lambda * f : \struct{S, \tau_{_S}} \to \struct{R, \tau_{_R}}$ is continuous.
$f * \lambda: \struct{S, \tau_{_S}} \to \struct{R, \tau_{_R}}$ is continuous.
## Proof
By definition of a topological ring, $\struct{R, *, \tau_{_R}}$ is a topological semigroup.
From Multiple Rule for Continuous Mappings to Topological Semigroup, $\lambda * f, f * \lambda : \struct{S, \tau_{_S}} \to \struct{R, \tau_{_R}}$ are continuous mappings.
$\blacksquare$ | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9138942956924438, "perplexity": 504.66553309858455}, "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/1581875148163.71/warc/CC-MAIN-20200228231614-20200229021614-00377.warc.gz"} |
https://math.stackexchange.com/questions/1706061/if-a-is-a-square-matrix-such-that-a27-a64-i-then-a-i | # If $A$ is a square matrix such that $A^{27}=A^{64}=I$ then $A=I$
If $A$ is a square matrix such that $A^{27}=A^{64}=I$ then $A=I$.
What I did is to subtract I from both sides of the equation:
$$A^{27}-I=A^{64}-I=0$$
then:
\begin{align*} A^{27}-I &= (A-I)(A+A^2+A^3+\dots+A^{26})=0\\ A^{64}-I &= (A-I)(A+A^2+A^3+\dots+A^{63})=0. \end{align*}
So from what I understand, either $A=I$ (as needed) or $A+A^2+A^3+\dots+A^{26}=0$ or $A+A^2+A^3+\dots+A^{63}=0$.
At this point I got stuck. By the way, I found out that $A$ is an invertible matrix because if $A^{27}=I$ then also $A^{26}A=AA^{26}=I$ then $A^{26}=A^{-1}$.
Also I thought to use the contradiction proving by assuming that $A+A^2+A^3+\dots+A^{63}=0$, but because $A^{27}=I$, then: $$A+A^2+A^3+\dots+A^{26}+I+A^{28}+\dots+A^{53}+I+A^{55}+\dots+A^{63}=0$$ but yet nothing.
Would appreciate your guidance, thanks!
• Note, if the product of two matrices is zero, it is not necessarily the case that one of the matrices is zero. That is, $AB = 0 \nRightarrow A = 0$ or $B = 0$. – Michael Albanese Mar 20 '16 at 18:18
• @MichaelAlbanese good point. wondering why I didn't notice. thanks – Ami Gold Mar 20 '16 at 18:20
• Hint: Compute $$\left(A^{64}\right)^8\cdot A=A^{64\times8+1}=A^{513}=A^{27\times19}=\left(A^{27}\right)^{19}.$$ – Did Mar 20 '16 at 18:41
\begin{align*} & I = A^{64} = A^{2(27) + 10} = (A^{27})^2A^{10} = A^{10}\\ \implies & I = A^{27} = A^{2(10) + 7} = (A^{10})^2A^7 = A^7\\ \implies & I = A^{10} = A^{1(7)+3} = (A^7)^1A^3 = A^3\\ \implies & I = A^7 = A^{2(3) + 1} = (A^3)^2A = A. \end{align*}
This is nothing more than the Euclidean algorithm applied to the exponents. The same procedure can be used to show that if $A^p = I$ and $A^q = I$ with $p$ and $q$ coprime, then $A = I$.
• wow this is tricky! liked your solution, however I wonder I could think about it all by myself. what was your motivation for such a solution? – Ami Gold Mar 20 '16 at 18:30
• Well, I knew I could write $A^{64}$ as $A^{64-27}A^{27} = A^{37}$, so $A^{37} = I$. But then I could just keep repeating this trick, each time getting a smaller power of $A$ which was the identity, until that power was $1$. – Michael Albanese Mar 20 '16 at 18:34
Firstly, since $A A^{26} = A^{26}A=I$, then $A$ is an invertible matrix.
Use the fact that $\gcd(27,64)=1$: hence there exist some $a,b \in \Bbb{Z}$ such that $1=27a+64b$. Now, compute $$A=A^1=A^{27a+64b}=(A^{27})^a(A^{64})^b=I^aI^b=I$$
• Nice. This is much quicker than the method I suggested. – Michael Albanese Mar 20 '16 at 18:19
Hint: if matrix $A$ has eigenpair $(\lambda, v )$, then $A^{27}$ and $A^{64}$ have eigenpairs $(\lambda^{27}, v)$ and $(\lambda^{64}, v)$. At the same time $A^{27} v = A^{64} v = 1 \cdot v$. Could you proceed from here? | {"extraction_info": {"found_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": 0.9681929349899292, "perplexity": 398.94288695225737}, "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-2019-35/segments/1566027316549.78/warc/CC-MAIN-20190821220456-20190822002456-00238.warc.gz"} |
https://www.physicsforums.com/threads/energy-the-ability-to-do-work.77442/ | # Homework Help: Energy - the ability to do work
1. May 31, 2005
### rolan
I'm fuzzy on what F actually is in the equation:
W= F d
is it only mg or the act of some force on an object
I was absent when we started this unit at school and now I'm having trouble with the rest of the unit because of my uncertainty of this equation.
Also, I'm having trouble with a specific question in my homework, I'll put down what I have, but it doesn't seem quite right.
A 1550kg car is travelling down the highway at a constant speed of 33.3m/s
a) If the car's engine is providing a power of 29, 500W how much force is it causing the ground to exert on the car?
P=W/t = F v
F= P/v = 29500/33.3 = ~885.8N
now I'm unsure what to do from there, any help hints or suggestions would be helpful
2. May 31, 2005
### OlderDan
The force in W=Fd is any force that is acting on anything that moves, provided the motion is in the direction of the force (or opposite). If the force and the motion are perpendicular, no work is done. If they are not perpendicular you need to find the component of the force in the direction of motion. Sometimes they are in the same direction and a simple product will do.
Your calculation is fine. When force is constant, power is force times velocity, with the same considerations about direction described above.
3. May 31, 2005
### rolan
Thanks very much!
If that first equation is fine then I guess my problem lies in the b part of the question.
It asks how much energy must be removed from the car to slow it to 13.9m/s
so,
F = P/(Vf-vi) = 29500/(33.3-13.9) = 1003N
but I'm fairly sure this isn't right because it would remove more energy than is present in a) part. (885.8)
4. May 31, 2005
### whozum
The answer is actually simpler, if you can find the kinetic energy of the car at 33m/s, and the kinetic energy at 13.9m/s then the difference between the two is the energy that needs to be removed.
5. May 31, 2005
### rolan
KE= 1/2 m (vf^2-vi^2) , right? At least for this equation
so,
Ke= .5(1550)(-33.3^2+13.9^2) = .5(1550)(-1108.89 + 193.21)
KE= .5(1550)(-915.68) = -709652N
That's quite the large number, that can't be the answer, can it?!!
I'm quite unsure what I'm supposed to do with this number, or with the b) part of the equation for that matter.
Last edited: May 31, 2005
6. May 31, 2005
### whozum
The first equation you have will give you the difference in kinetic energy. That is the answer to B. You've already solved A correctly with dan. It is alot of Joules, because a Joule is not a very large unit of energy. Its actually pretty small when dealing with very macro-scaled objects, such as a 1500kg car.
7. May 31, 2005
### OlderDan
You are really talking about a change in kinetic energy, so it would be best write your left side as (KE_f - KE_i) rather than just KE, or if you use LaTex $\Delta KE$. The units of energy are not Newtons. The units are Joules; a Joule is equal to one Newton*Meter or kg*m^2/s^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": 0.8888702988624573, "perplexity": 703.6728415545278}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-43/segments/1539583511365.50/warc/CC-MAIN-20181018001031-20181018022531-00374.warc.gz"} |
http://umj.imath.kiev.ua/article/?lang=en&article=7471 | 2018
Том 70
№ 2
# On locally perturbed equilibrium distribution functions
Abstract
We construct a new class of locally perturbed equilibrium distribution functions for which local (in time) solutions of the BBGKY equations can be extended onto the entire time axis.
English version (Springer): Ukrainian Mathematical Journal 56 (2004), no. 6, pp 919-928.
Citation Example: Malyshev D. V., Malyshev P. V. On locally perturbed equilibrium distribution functions // Ukr. Mat. Zh. - 2004. - 56, № 6. - pp. 774–781.
Full text | {"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": 0.8965402841567993, "perplexity": 2585.6393734778044}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-17/segments/1524125945484.58/warc/CC-MAIN-20180422022057-20180422042057-00612.warc.gz"} |
https://math.stackexchange.com/questions/1876571/how-is-the-cofinite-topology-a-topology-if-x-is-not-finite-or-doesnt-contain | # How is the cofinite topology a topology if $X$ is not finite or doesn't contain the empty set?
I've just started an introductory topology course, and I've come across the cofinite topology
$$\tau = \{\emptyset\} \cup \{U\subseteq X: X \setminus U \text{ is finite}\}$$
All the definitions I've found online say that this is for any non-empty set $X$, but I don't understand how this is a topology if $X$ is not finite, or if $X$ does not contain the empty set?
My current thinking is, say we have $X$ not finite, containing the empty set. If we have $U=\emptyset$, then $X\setminus U=X$, which is not finite and so is not in the topology?
Also, if $X$ does not contain the empty set, then how does one construct a $U$ such that $X\setminus U=X$, which would need to be the case for $\tau$ to be a topology?
I know I've got something twisted around in my head, I just can't figure out what!
Every set contains the empty set. And by that I mean that the empty set is a subset of every set, and not to be confused with the statement $\varnothing\in X$.
Since the empty set is a subset of every set, and a topology on $X$ is a collection of subsets of $X$, there is no problem with the fact that $\tau$ is non-empty to begin with.
If $X$ is any non-empty set, then we can fix some element $x\in X$ and then $U=X\setminus\{x\}$ will satisfy that $X\setminus U=\{x\}$ which is certainly a finite set. Also $X=U$ will satisfy that $X\setminus U$ is finite, because that would be the empty set itself, which is finite.
So certainly $\varnothing,X\in\tau$ and if $x\in X$, then $X\setminus\{x\}\in\tau$ as well. So again, certainly this topology is not empty.
If $X$ is infinite, then indeed $\varnothing$ is not a cofinite set. Which is exactly the reason we add it explicitly to the topology. Consequently, the cofinite topology is the topology where every non-empty open set is cofinite.
The open sets in the cofinite topology are, by definition, those for which the complement is finite plus the empty set.
As you noticed, if $X$ is infinite then the empty set would not be an open set in the sense of having a finite complement, which would imply that $\tau$ is not a topology, as a topology needs the empty set. Therefore one has to add it (so to say arbitrarily) to make $\tau$ a topology, as done in the process $$\color{red}{\{\emptyset \}\cup}\dots$$ | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9190381765365601, "perplexity": 52.70556832340343}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-31/segments/1627046154214.36/warc/CC-MAIN-20210801123745-20210801153745-00045.warc.gz"} |
https://in.mathworks.com/help/audio/ref/gtcc.html | # gtcc
Extract gammatone cepstral coefficients, log-energy, delta, and delta-delta
## Syntax
``coeffs = gtcc(audioIn,fs)``
``coeffs = gtcc(___,Name,Value)``
``[coeffs,delta,deltaDelta,loc] = gtcc(___)``
## Description
example
````coeffs = gtcc(audioIn,fs)` returns the gammatone cepstral coefficients (GTCCs) for the audio input, sampled at a frequency of `fs` Hz.```
example
````coeffs = gtcc(___,Name,Value)` specifies options using one or more `Name,Value` pair arguments.```
example
````[coeffs,delta,deltaDelta,loc] = gtcc(___)` returns the delta, delta-delta, and location in samples corresponding to each window of data. This output syntax can be used with any of the previous input syntaxes.```
## Examples
collapse all
Get the gammatone cepstral coefficients for an audio file using default settings. Plot the results.
```[audioIn,fs] = audioread('Counting-16-44p1-mono-15secs.wav'); [coeffs,~,~,loc] = gtcc(audioIn,fs); t = loc./fs; plot(t,coeffs) xlabel('Time (s)') title('Gammatone Cepstral Coefficients') legend('logE','0','1','2','3','4','5','6','7','8','9','10','11','12', ... 'Location','northeastoutside')```
`[audioIn,fs] = audioread('Turbine-16-44p1-mono-22secs.wav');`
Calculate 20 GTCC using filters equally spaced on the ERB scale between `hz2erb(62.5)` and `hz2erb(12000)`. Calculate the coefficients using 50 ms windows with 25 ms overlap. Replace the 0th coefficient with the log-energy. Use time-domain filtering.
```[coeffs,~,~,loc] = gtcc(audioIn,fs, ... 'NumCoeffs',20, ... 'FrequencyRange',[62.5,12000], ... 'WindowLength',round(0.05*fs), ... 'OverlapLength',round(0.025*fs), ... 'LogEnergy','Replace', ... 'FilterDomain','Time');```
Plot the results.
```t = loc./fs; plot(t,coeffs) xlabel('Time (s)') title('Gammatone Cepstral Coefficients') legend('logE','1','2','3','4','5','6','7','8','9','10','11','12','13', ... '14','15','16','17','18','19','Location','northeastoutside');```
Read in an audio file and convert it to a frequency representation.
```[audioIn,fs] = audioread("Rainbow-16-8-mono-114secs.wav"); win = hann(1024,"periodic"); S = stft(audioIn,"Window",win,"OverlapLength",512,"Centered",false);```
To extract the gammatone cepstral coefficients, call `gtcc` with the frequency-domain audio. Ignore the log-energy.
`coeffs = gtcc(S,fs,"LogEnergy","Ignore");`
In many applications, GTCC observations are converted to summary statistics for use in classification tasks. Plot probability density functions of each of the gammatone cepstral coefficients to observe their distributions.
```nbins = 60; for i = 1:size(coeffs,2) figure histogram(coeffs(:,i),nbins,'Normalization','pdf') title(sprintf("Coefficient %d",i-1)) end```
## Input Arguments
collapse all
Input signal, specified as a vector, matrix, or 3-D array.
If '`FilterDomain`' is set to `'Frequency'` (default), then `audioIn` can be real or complex.
• If `audioIn` is real, it is interpreted as a time-domain signal and must be a column vector or a matrix. Columns of the matrix are treated as independent audio channels.
• If `audioIn` is complex, it is interpreted as a frequency-domain signal. In this case, `audioIn` must be an L-by-M-by-N array, where L is the number of DFT points, M is the number of individual spectrums, and N is the number of individual channels.
If '`FilterDomain`' is set to `'Time'`, then `audioIn` must be a real column vector or matrix. Columns of the matrix are treated as independent audio channels.
Data Types: `single` | `double`
Complex Number Support: Yes
Sample rate of the input signal in Hz, specified as a positive scalar.
Data Types: `single` | `double`
### Name-Value Pair Arguments
Specify optional comma-separated pairs of `Name,Value` arguments. `Name` is the argument name and `Value` is the corresponding value. `Name` must appear inside quotes. You can specify several name and value pair arguments in any order as `Name1,Value1,...,NameN,ValueN`.
Example: `coeffs = gtcc(audioIn,fs,'LogEnergy','Replace')` returns gammatone cepstral coefficients for the audio input signal sampled at `fs` Hz. For each analysis window, the first coefficient in the `coeffs` vector is replaced with the log energy of the input signal.
Number of samples in analysis window used to calculate the coefficients, specified as the comma-separated pair consisting of `'WindowLength'` and an integer in the range [2, `size(audioIn,1)`]. If unspecified, `WindowLength` defaults to `round(0.03*fs)`.
Data Types: `single` | `double`
Number of samples overlapped between adjacent windows, specified as the comma-separated pair consisting of `'OverlapLength'` and an integer in the range [0, `WindowLength`). If unspecified, `OverlapLength` defaults to `round(0.02*fs)`.
Data Types: `single` | `double`
Number of coefficients returned for each window of data, specified as the comma-separated pair consisting of `'NumCoeffs'` and an integer in the range [2, v]. v is the number of valid passbands. If unspecified, `NumCoeffs` defaults to `13`.
The number of valid passbands is defined as the number of ERB steps (ERBN) in the frequency range of the filter bank. The frequency range of the filter bank is specified by `FrequencyRange`.
Data Types: `single` | `double`
Domain in which to apply filtering, specified as the comma-separated pair consisting of `'FilterDomain'` and `'Frequency'` or `'Time'`. If unspecified, `FilterDomain` defaults to `Frequency`.
Data Types: `string` | `char`
Frequency range of gammatone filter bank in Hz, specified as the comma-separated pair consisting of `'FrequencyRange'` and a two-element row vector of increasing values in the range [0, `fs`/2]. If unspecified, `FrequencyRange` defaults to ```[50, fs/2]```
Data Types: `single` | `double`
Number of bins used to calculate the DFT of windowed input samples, specified as the comma-separated pair consisting of `'FFTLength'` and a positive scalar integer. If unspecified, `FFTLength` defaults to `WindowLength`.
Data Types: `single` | `double`
Type of nonlinear rectification applied prior to the discrete cosine transform, specified as `'log'` or `'cubic-root'`.
Data Types: `char` | `string`
Number of coefficients used to calculate the delta and the delta-delta values, specified as the comma-separated pair consisting of `'DeltaWindowLength'` and two or an odd integer greater than two. If unspecified, `DeltaWindowLength` defaults to `2`.
If `DeltaWindowLength` is set to `2`, the `delta` is given by the difference between the current coefficients and the previous coefficients.
If `DeltaWindowLength` is set to an odd integer greater than `2`, the following equation defines their values:
$delta=\frac{\sum _{k=-M}^{M}k\cdot coeffs\left(k,:\right)}{\sum _{k=-M}^{M}{k}^{2}}$
The function uses a least-squares approximation of the local slope over a region around the coefficients of the current analysis window. The delta cepstral values are computed by fitting the cepstral coefficients of neighboring analysis windows (M analysis windows before the current analysis window and M analysis windows after the current analysis window) to a straight line. For details, see [3].
Data Types: `single` | `double`
Log energy usage, specified as the comma-separated pair consisting of `'LogEnergy'` and `'Append'`, `'Replace'`, or `'Ignore'`. If unspecified, `LogEnergy` defaults to `'Append'`.
• `'Append'` –– The function prepends the log energy to the coefficients vector. The length of the coefficients vector is 1 + `NumCoeffs`.
• `'Replace'` –– The function replaces the first coefficient with the log energy of the signal. The length of the coefficients vector is `NumCoeffs`.
• `'Ignore'` –– The function does not calculate or return the log energy.
Data Types: `char` | `string`
## Output Arguments
collapse all
Gammatone cepstral coefficients, returned as an L-by-M matrix or an L-by-M-by-N array, where:
• L –– Number of analysis windows the audio signal is partitioned into. The input size, `WindowLength`, and `OverlapLength` control this dimension: ```L = floor((size(audioIn,1) − WindowLength))/(WindowLength − OverlapLength) + 1```.
• M –– Number of coefficients returned per frame. This value is determined by `NumCoeffs` and `LogEnergy`.
When `LogEnergy` is set to:
• `'Append'` –– The object prepends the log energy value to the coefficients vector. The length of the coefficients vector is 1 + `NumCoeffs`.
• `'Replace'` –– The object replaces the first coefficient with the log energy of the signal. The length of the coefficients vector is `NumCoeffs`.
• `'Ignore'` –– The object does not calculate or return the log energy. The length of the coefficients vector is `NumCoeffs`.
• N –– Number of input channels (columns). This value is `size(audioIn,2)`.
Data Types: `single` | `double`
Change in coefficients from one analysis window to another, returned as an L-by-M matrix or an L-by-M-by-N array. The `delta` array is the same size and data type as the `coeffs` array. See `coeffs` for the definitions of L, M, and N.
The function uses a least-squares approximation of the local slope over a region around the current time sample. For details, see [3].
Data Types: `single` | `double`
Change in `delta` values, returned as an L-by-M matrix or an L-by-M-by-N array. The `deltaDelta` array is the same size and data type as the `coeffs` and `delta` arrays. See `coeffs` for the definitions of L, M, and N.
The function uses a least-squares approximation of the local slope over a region around the current time sample. For details, see [3].
Data Types: `single` | `double`
Location of last sample in each analysis window, returned as a column vector with the same number of rows as `coeffs`.
Data Types: `single` | `double`
## Algorithms
collapse all
The `gtcc` function splits the entire data into overlapping segments. The length of each analysis window is determined by `WindowLength`. The length of overlap between analysis windows is determined by `OverlapLength`. The algorithm to determine the gammatone cepstral coefficients depends on the filter domain, specified by `FilterDomain`. The default filter domain is frequency.
### Frequency-Domain Filtering
`gtcc` computes the gammatone cepstral coefficients, log energy values, delta, and delta-delta values for each analysis window as per the algorithm described in `cepstralFeatureExtractor`.
### Time-Domain Filtering
If `FilterDomain` is specified as `'Time'`, the `gtcc` function uses the `gammatoneFilterBank` to apply time-domain filtering. The basic steps of the `gtcc` algorithm are outlined by the diagram.
The `FrequencyRange` and sample rate (`fs`) parameters are set on the filter bank using the name-value pairs input to the `gtcc` function. The number of filters in the gammatone filter bank is defined as ```hz2erb(FrequencyRange(2)) − hz2erb(FrequencyRange(1))```.This roughly corresponds to placing a gammatone filter every 0.9 mm in the cochlea.
The output from the gammatone filter bank is a multichannel signal. Each channel output from the gammatone filter bank is buffered into overlapped analysis windows, as specified by `WindowLength` and `OverlapLength`. Then a periodic Hamming window is applied to each analysis window. The energy for each analysis window of data is calculated. The STE of the channels are concatenated. The concatenated signal is then passed through a logarithm function and transformed to the cepstral domain using a discrete cosine transform (DCT).
The log-energy is calculated on the original audio signal using the same buffering scheme applied to the gammatone filter bank output.
## References
[1] Shao, Yang, Zhaozhang Jin, Deliang Wang, and Soundararajan Srinivasan. "An Auditory-Based Feature for Robust Speech Recognition." IEEE International Conference on Acoustics, Speech and Signal Processing. 2009.
[2] Valero, X., and F. Alias. "Gammatone Cepstral Coefficients: Biologically Inspired Features for Non-Speech Audio Classification." IEEE Transactions on Multimedia. Vol. 14, Issue 6, 2012, pp. 1684–1689.
[3] Rabiner, Lawrence R., and Ronald W. Schafer. Theory and Applications of Digital Speech Processing. Upper Saddle River, NJ: Pearson, 2010. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 1, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8638471364974976, "perplexity": 2349.8479448117055}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-24/segments/1590348502204.93/warc/CC-MAIN-20200605174158-20200605204158-00436.warc.gz"} |
http://soft-matter.seas.harvard.edu/index.php?title=The_Role_of_Polymer_Polydispersity_in_Phase_Separation_and_Gelation_in_Colloid%E2%88%92Polymer_Mixtures&oldid=17518 | # The Role of Polymer Polydispersity in Phase Separation and Gelation in Colloid−Polymer Mixtures
Entry by Emily Redston, AP 225, Fall 2011
Work in progress
## Reference
The Role of Polymer Polydispersity in Phase Separation and Gelation in Colloid−Polymer Mixtures by J. J. Lietor-Santos, C. Kim, M. L. Lynch, A. Fernandez-Nieves, and D. A. Weitz. Langmuir 26 3174–3178 (2010)
## Introduction
Mixtures of non-adsorbing polymer and colloidal particles exhibit a very rich range of morphologies. These microstructures depend on the particle and polymer concentrations as well as the relative size of the particles and polymer. The addition of the polymer to a colloidal suspension leads to a depletion attraction that is capable of inducing a fluid-solid transition (i.e. forming a gel). A gel is defined as a connected network that spans space and can support weak stresses. Gels are extensively used in commercial applications, such as personal care or food products, where they are able to help stabilize the system against sedimentation. In this manner, gels can reduce phase separation, which will increase product shelf life. The nature of the fluid-solid transition depends on the range of the depletion attraction, which, in turn, depends on the ratio of the size of the polymer to the colloidal particle. For short-range interactions, gelation is induced by spinodal decomposition. A gel is formed because, when the system undergoes a gas-liquid phase separation, it is interrupted by the dynamical arrest of the particles in the colloid-rich region. By contrast, for larger ranges of the attraction, phase separation can proceed to completion, without being interrupted by dynamic arrest.
However, in the presence of gravitational effects, the structure may no longer be capable of sustaining its own weight; instead it collapses, disrupting the phase separation process or rupturing the gel that was previously formed. This is obviously an undesirable effect as it can dramatically shorten the shelf life of a commercial product. This suggests that the use of shorter polymers at sufficiently high concentration is of greatest practical interest. However, technological polymers are very rarely monodisperse, and thus the microstructures and their behavior may be drastically modified. However, despite the practical importance, the gravitational behavior of colloid-polymer mixtures using polydisperse polymers has never been investigated. It is not known how the polydispersity of polymers effects the phase behavior of the mixture.
In this paper, the authors investigate the behavior of model colloidal particles mixed with nonadsorbing polymer with a polydisperse size distribution, similar to that often found in commercial samples. Ultimately they find that the presence of even a small amount of large polymer in a distribution of nominally much smaller polymer can drastically modify the behavior.
## Sample Preparation
The authors used an aqueous dispersion of polystyrene particles with a density $\rho = 1.057 g/cm^3$ and an average radius $a = 1.5 \mu m$. They also used polyethyleneglycol (PEG) with an average molecular weight $M_w = 475500 g/mol$, polydispersity index $M_w/ M_n = 2.63$, and mean radius of gyration $r_g = 40 nm$. Here, $M_w$ and $M_n$ are the mass- and number-averaged molecular weights, respectively. Salt was also added to reduce electrostatic interactions.
## Experiments and Results
Figure 1. Experimental height profiles at different polymer concentrations: (\blacksquare) cp = 1 mg/mL, ($\circ$) cp = 7.5 mg/mL, and (▲) cp = 20 mg/mL. We measure the height from the bottom of the cell, as shown in the schematic, where h0 = (1.90 ± 0.05) cm is the initial height. (Upper inset) Expanded details of the initial evolution for a time window of 5 h.
Figure 2. Diagram for the behavior of mixtures of polystyrene colloidal particles and polyethylene glycol polymer as a function of the particle volume fraction and the polymer concentration cp: () sedimentation; () collapse; (▲) compression.
The rate of sedimentation of the colloidal particles is dependent on the microstructure of the colloid-polymer mixture. Sedimentation is driven by the density mismatch between the solvent and the colloidal particles, Δρ, that results in a body force on the individual particles, F=4/3πΔρg$a^3$. For a particle suspension, the settling rate is hindered by solvent backflow. However, settling in particle gels is more complicated as the gravitational forces act on and are transmitted through the entire network, resulting in a distinctive settling behavior. By monitoring the time-dependent height profile of the sample as it sediments, it is possible to distinguish between the two cases.
At low polymer concentration, the depletion-induced attraction between particles is small and the interface immediately begins to fall, with a constant rate, $v = (1.5 \pm 0.2) 10^{-7} m/s$, as shown in Figure 1 for a polymer concentration $c_p =1 mg/mL$ and a particle volume fraction $\phi$ = 0.1 (closed squares); this value is consistent with hindered settling models for dispersed particles, where
$v = \frac{2}{9} \frac{\Delta \rho g a^2}{\eta} (1-\phi)^{5.5} = 1.8 \times 10^{-7} m/s$
where $\eta$ is the viscosity of the solution. At high polymer concentration, the attraction between particles is large. As a result, the height of the interface evolves in a different fashion, as shown in Figure 1 for $c_p =20 mg/mL$ at the same $\phi$ (closed triangles). The height evolution in this case quantitatively agrees with poroelastic models for the gravitational compression of particle gels. In this case, the height evolution of the interface is determined by the balance of the gravitational and elastic stresses imposed on the network, and by the viscous stress due to solvent backflow through the network as it is compressed.
Surprisingly, at intermediate polymer concentrations, there is a fundamentally different behavior. Initially, the interface slowly moves but, after some time, abruptly and rapidly falls, as shown in Figure 1 for = 0.1 and cp = 7.5 mg/mL (open circles),(19) a behavior which is similar to transient gelation,(9-11) where a gel forms and subsequently collapses under the influence of gravitational stresses. Using the temporal evolution of the interface as a criterion, we summarize the behavior of our colloid/polymer system in a -cp diagram, shown in Figure 2. For low or low cp, the system undergoes hindered sedimentation (closed squares), consistent with the absence of gelation. By contrast, at high or high cp, the system exhibits a compression behavior (closed triangles) indicating the presence of a particle gel. Between these two behaviors, for intermediate and intermediate cp, we find a region in the -cp diagram corresponding to a behavior which is reminiscent of transient gelation (open circles).
Interestingly, this transient behavior is linked to a coarsening of the structure, as shown in Supporting Information, Movie S2, for cp= 5 mg/mL. This coarsening, which is reminiscent of “curding”, precedes the collapse of the structure. Thus, although collapse must result from gravitational stress, it is unclear whether this external stress is the ultimate cause of the collapse, or whether the coarsening behavior that precedes the collapse is, in some way, implicated in the collapse itself. To investigate this, we prepare samples at intermediate and cp and vary both the initial height and the density difference between the particles and the solvent, by using water/D2O mixtures, while keeping both and cp constant. In this way, we span 3 orders of magnitude in the magnitude of the total gravitationally induced stress at the bottom of the gel σg = Δρgh0, from 10−3 Pa to 1 Pa; this is the “weight” of the gel, which is supported at the bottom of the sample. We find that the structure of the colloid/polymer mixtures coarsens to some extent before it actually collapses due to the gravitational stress. Interestingly, the time for this coarsening to become appreciable in our images does not change with σg and always remains equal to 15 min.
The Role of Polymer Polydispersity in Phase Separation and Gelation in Colloid−Polymer Mixtures
Abstract
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Supporting Info
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References
Citing Articles
J. J. Lietor-Santos†§, C. Kim†, M. L. Lynch‡, A. Fernandez-Nieves†§* and D. A. Weitz† † School of Engineering and Applied Sciences and Department of Physics, Harvard University, Cambridge, Massachusetts 02138 ‡ Procter & Gamble Co., Corporation Research Division, Miami Valley Lab, Cincinnati, Ohio 45252 Langmuir, 2010, 26 (5), pp 3174–3178 DOI: 10.1021/la903127a Publication Date (Web): November 3, 2009 Copyright © 2009 American Chemical Society
• Corresponding author. E-mail: [email protected]., §
Present address: School of Physics, Georgia Institute of Technology, 837 State Street, Atlanta, GA 30332. ,
Present address: Chemical Physics Interdisciplinary Program, Liquid Crystal Institute, Kent State University, Kent, OH 44242. CASSection: Surface Chemistry and Colloids Abstract Abstract Image
Mixtures of nonadsorbing polymer and colloidal particles exhibit a range of different morphologies depending on the particle and polymer concentrations and their relative size ratios. These can be very important for technological applications, where gelation can produce a weak solidlike structure that can help reduce phase separation, extending product shelf life. However, industrial products are typically formulated with polydisperse polymers, and the consequences of this on the phase behavior of the mixture are not known. We investigate the role of polymer polydispersity and show that a small amount of larger polymer in a distribution of nominally much smaller polymer can drastically modify the behavior. It can induce formation of a solidlike gel structure, abetted by the small polymer, but still allow further evolution of the phase separation process, as is seen with a monodisperse distribution of larger polymer. This coarsening ultimately leads to gravitational collapse. We describe the full phase behavior for polydisperse polymer mixtures and account for the origin of the behavior through measurements of the structure and dynamics and by comparing to the behavior with monodisperse polymers.
Introduction Addition of nonadsorbing polymer to a colloidal suspension induces a very rich range of microstructures, whose nature depends on the particle and polymer concentrations and the relative size of the particles and polymer.(1, 2) The addition of the polymer leads to a depletion attraction that is capable of inducing a fluid−solid transition, resulting from the formation of a gel, a connected network that spans the space and can support weak stresses. Such gels are extensively used in commercial applications such as personal care or food products, to help stabilize the system against sedimentation, thereby precluding phase separation, and increasing product shelf life.(3-5) The nature of the fluid−solid transition depends on the range of the depletion attraction, which, in turn, depends on the ratio of the size of the polymer to the colloidal particle. For short-range interactions, gelation is induced by spinodal decomposition;(6)the system undergoes a gas−liquid phase separation which is interrupted by the dynamical arrest of the particles in the colloid-rich region, leading to formation of the gel. By contrast, for larger ranges of the attraction, phase separation can proceed to completion, without being interrupted by dynamic arrest leading to gelation.(7, 8) However, in the presence of gravitational effects the structure may no longer be capable of sustaining its own weight; instead, it collapses, disrupting the phase separation process(8) or rupturing the gel that was previously formed.(9-11) This is an undesirable effect technologically, as it can dramatically shorten the shelf life of a commercial product. This suggests that the use of shorter polymers at sufficiently high concentration is most likely of greatest practical use. However, technological polymers are very rarely monodisperse, and thus the microstructures and their behavior may be drastically modified. However, despite the practical importance, the gravitational behavior of colloid−polymer mixtures using polydisperse polymers has never been investigated. In this paper, we investigate in detail the behavior of model colloidal particles mixed with nonadsorbing polymer with a polydisperse size distribution, similar to that often found in commercial samples. We use a predominantly short-range interaction to induce gelation. However, we show that the presence of a small concentration of much larger polymers can have dramatic consequences on the behavior. We find that only those mixtures in the proximity of the liquid−gel boundary collapse under the influence of gravitational effects. Interestingly, however, we find that the gravitational stress imposed on the network is not the ultimate cause for this behavior. Instead, it results from coarsening of the particle network due to spinodal decomposition, which results from the long-range attraction induced by the fraction of the polymer distribution with long chain lengths. We verify this by comparison to the behavior of a monodisperse polymer mixture for the same range of attraction, where the coarsening of the structure disappears.
Experimental System and Methods We use an aqueous dispersion of polystyrene particles with a density ρ = 1.057 g/cm3 and an average radius a = 1.5 μm, as determined by dynamic light scattering.(12, 13) We use polyethyleneglycol (PEG) (American Polymer Standard Co.) with an average molecular weight Mw = 475 500 g/mol, polydispersity index Mw/Mn = 2.63, and mean radius of gyration r̅g = 40 nm, which we estimate from the average molecular weight.(14) Here, Mw and Mn are the mass- and number-averaged molecular weights, respectively. We also add a 1:1 salt, NaClO4, at 5 mM to reduce electrostatic interactions. We homogenize the samples by tumbling for 24 h.
Results and Discussion The rate of sedimentation of the colloidal particles is dependent on the microstructure of the colloid−polymer mixture. Sedimentation is driven by the density mismatch between the solvent and the colloidal particles, Δρ, that results in a body force on the individual particles, F = 4/3πΔρga3. For a particle suspension, the settling rate is hindered by solvent backflow.(4) However, settling in particle gels is more complicated as the gravitational forces act on and are transmitted through the entire network, resulting in a distinctive settling behavior. By monitoring the time-dependent height profile of the sample as it sediments, it is possible to distinguish between the two cases. figure
Figure 1. Experimental height profiles at different polymer concentrations: () cp = 1 mg/mL, () cp = 7.5 mg/mL, and (▲) cp = 20 mg/mL. We measure the height from the bottom of the cell, as shown in the schematic, where h0 = (1.90 ± 0.05) cm is the initial height. (Upper inset) Expanded details of the initial evolution for a time window of 5 h. At low polymer concentration, the depletion-induced attraction between particles is small and the interface immediately begins to fall, with a constant rate, v = (1.5 ± 0.2) 10−7 m/s, as shown in Figure 1 for a polymer concentration cp = 1 mg/mL and a particle volume fraction = 0.1 (closed squares); this value is consistent with hindered settling models for dispersed particles,(15) where with η the viscosity of the solution. At high polymer concentration, the attraction between particles is large. As a result, the height of the interface evolves in a different fashion, as shown in Figure 1 for cp = 20 mg/mL at the same (closed triangles). The height evolution in this case quantitatively agrees with poroelastic models for the gravitational compression of particle gels.(16-18) In this case, the height evolution of the interface is determined by the balance of the gravitational and elastic stresses imposed on the network, and by the viscous stress due to solvent backflow through the network as it is compressed. Surprisingly, at intermediate polymer concentrations, there is a fundamentally different behavior. Initially, the interface slowly moves but, after some time, abruptly and rapidly falls, as shown in Figure 1 for = 0.1 and cp = 7.5 mg/mL (open circles),(19) a behavior which is similar to transient gelation,(9-11) where a gel forms and subsequently collapses under the influence of gravitational stresses. Using the temporal evolution of the interface as a criterion, we summarize the behavior of our colloid/polymer system in a -cp diagram, shown in Figure 2. For low or low cp, the system undergoes hindered sedimentation (closed squares), consistent with the absence of gelation. By contrast, at high or high cp, the system exhibits a compression behavior (closed triangles) indicating the presence of a particle gel. Between these two behaviors, for intermediate and intermediate cp, we find a region in the -cp diagram corresponding to a behavior which is reminiscent of transient gelation (open circles). Interestingly, this transient behavior is linked to a coarsening of the structure, as shown in Supporting Information, Movie S2, for cp= 5 mg/mL. This coarsening, which is reminiscent of “curding”, precedes the collapse of the structure. Thus, although collapse must result from gravitational stress, it is unclear whether this external stress is the ultimate cause of the collapse, or whether the coarsening behavior that precedes the collapse is, in some way, implicated in the collapse itself. To investigate this, we prepare samples at intermediate and cp and vary both the initial height and the density difference between the particles and the solvent, by using water/D2O mixtures, while keeping both and cp constant. In this way, we span 3 orders of magnitude in the magnitude of the total gravitationally induced stress at the bottom of the gel σg = Δρgh0, from 10−3 Pa to 1 Pa; this is the “weight” of the gel, which is supported at the bottom of the sample. We find that the structure of the colloid/polymer mixtures coarsens to some extent before it actually collapses due to the gravitational stress. Interestingly, the time for this coarsening to become appreciable in our images does not change with σg and always remains equal to 15 min. figure
Figure 2. Diagram for the behavior of mixtures of polystyrene colloidal particles and polyethylene glycol polymer as a function of the particle volume fraction and the polymer concentration cp: () sedimentation; () collapse; (▲) compression. As a result, we conclude that the coarsening behavior is not dependent on the gravitational stress, thereby implying that the presence of this stress is ultimately not the mechanism responsible for the temporal evolution of our samples. To further investigate this, we prepare a density-matched sample inside a glass cuvette of height 4 cm and an optical path length of 1 mm, and follow the time evolution of the sample using a CCD camera and a high magnification objective, focused on a small region in the middle of the cuvette. We do not observe any collapse of the structure within the experimental time-window since σg is so small. However, as time proceeds, the system separates into colloid-poor and colloid-rich regions, as shown in the optical images of Figure 3 and in Supporting Information, Movie S3, where darker regions correspond to colloid-poor regions and brighter regions correspond to colloid-rich regions.
This suggests that the coarsening behavior, even when the densities of the colloids and solvent are not matched, is due to spinodal decomposition. However, if the densities are not matched, spinodal decomposition is interrupted by the presence of gravitational stresses. We therefore propose that the nonadsorbing polymer induces the requisite attraction to initiate spinodal decomposition of the mixtures. The initially homogeneous mixture then evolves into a bicontinuous network of colloid-poor and colloid-rich domains. The structure continues to coarsen and eventually looses mechanical strength; it fractures and collapses under the presence of the gravitational stress. Surprisingly, however, the typical range of the attractive interaction in our experiments is which is much lower than those that are typically required to induce phase separation. However, this value is based on the average molecular weight of the polymer, Mw; because of the polydispersity of the polymer, there is a sizable fraction of chains which are several orders of magnitude larger than the mean, as shown in Figure 5.(25) Despite the relatively smaller concentration of these larger chains, we hypothesize that they nevertheless determine both the magnitude and width of the attraction, resulting in a weaker and wider attraction than would be expected from the peak of the Mw distribution, and causing the mixtures to exhibit a transient gel behavior.
To test this hypothesis, we perform the same experiments with monodisperse polymer. Mixtures were prepared with the same colloidal particles but using a PEG (American Polymer Standard Company, Mentor, OH) of Mw = 438 000 g/mol, but with a much smaller polydispersity index, Mw/Mn = 1.18. In these mixtures, the coarsening behavior is not observed. Instead, we observe only hindered sedimentation at low polymer concentrations and gel compression at high polymer concentrations, as shown in Figure 6 for = 0.05 and cp ranging from 5 mg/mL (right vial) to 60 mg/mL (left vial). The transition between these two behaviors occurs between cp = 20 mg/mL and cp = 30 mg/mL; for cp = 20 mg/mL, the system exhibits hindered sedimentation, as shown by the squares in Figure 6b, while for cp = 30 mg/mL, the system exhibits gravitational compression, as shown by the circles in Figure 6b. This behavior is in striking contrast to that observed with the polydisperse polymer, emphasizing the important role of the largest polymer chains on the gravitational stability of colloid/polymer mixtures when the polymer is polydisperse, and confirming our hypothesis that it is these larger polymer chains that determine the behavior of the mixture. However, by comparing the behavior presented in Figure 6 with that of the polydisperse case (see Figure 2, for = 0.05), we observe that also in this case a polymer concentration of cp ≈ 20 mg/mL marks the formation of a gel that compresses in time. Thus, at high polymer concentrations the overall behavior of the system is mainly controlled by the smaller rather than by the larger polymer chains. It is thus apparent that to properly account for the behavior of colloid−polymer mixtures prepared with polydisperse polymer, the full nature of the polymer distribution must be considered. This is of particular importance in practical applications of these mixtures. | {"extraction_info": {"found_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": 0.8035209774971008, "perplexity": 1338.0178321941132}, "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/1659882571911.5/warc/CC-MAIN-20220813081639-20220813111639-00062.warc.gz"} |
https://www.semanticscholar.org/paper/A-hyper-reduced-MAC-scheme-for-the-parametric-and-Chen-Ji/100e2819f739af1c5c28eea24fc8090b8af39dd9 | # A hyper-reduced MAC scheme for the parametric Stokes and Navier-Stokes equations
@article{Chen2021AHM,
title={A hyper-reduced MAC scheme for the parametric Stokes and Navier-Stokes equations},
author={Yanlai Chen and Lijie Ji and Zhu Wang},
journal={ArXiv},
year={2021},
volume={abs/2110.11179}
}
• Published 21 October 2021
• Computer Science
• ArXiv
The need for accelerating the repeated solving of certain parametrized systems motivates the development of more efficient reduced order methods. The classical reduced basis method is popular due to an offline-online decomposition and a mathematically rigorous a posterior error estimator which guides a greedy algorithm offline. For nonlinear and nonaffine problems, hyper reduction techniques have been introduced to make this decomposition efficient. However, they may be tricky to implement and…
## References
SHOWING 1-10 OF 50 REFERENCES
L1-based reduced over collocation and hyper reduction for steady state and time-dependent nonlinear equations
• Computer Science
J. Sci. Comput.
• 2021
This paper augment and extend the EIM approach as a direct solver, as opposed to an assistant, for solving nonlinear pPDEs on the reduced level, and the resulting method, called Reduced Over-Collocation method (ROC), is stable and capable of avoiding the efficiency degradation inherent to a traditional application of EIM.
An EIM-degradation free reduced basis method via over collocation and residual hyper reduction-based error estimation
• Mathematics
J. Comput. Phys.
• 2021
A Reduced Basis Technique for Long-Time Unsteady Turbulent Flows
• Computer Science
• 2017
An \emph{a posteriori} error indicator, which corresponds to the dual norm of the residual associated with the time-averaged momentum equation, is proposed and demonstrated that the error indicator is highly-correlated with the error in mean flow prediction, and can be efficiently computed through an offline/online strategy.
A posteriori error bounds for reduced-basis approximations of parametrized parabolic partial differential equations
• Mathematics, Computer Science
• 2005
The reduced-basis methods and associated a posteriori error estimators developed earlier for elliptic partial differential equations to parabolic problems with affine parameter dependence are extended and time is treated as an additional, albeit special, parameter in the formulation and solution of the problem.
Efficient geometrical parametrization for finite‐volume‐based reduced order methods
• Mathematics
International Journal for Numerical Methods in Engineering
• 2020
This work presents an approach for the efficient treatment of parametrized geometries in the context of POD-Galerkin reduced order methods based on Finite Volume full order approximations, which relies on basis functions defined on an average deformed configuration and makes use of the Discrete Empirical Interpolation Method (D-EIM).
Generalized parametric solutions in Stokes flow
• Computer Science
• 2017
An LP empirical quadrature procedure for reduced basis treatment of parametrized nonlinear PDEs
• Mathematics
Computer Methods in Applied Mechanics and Engineering
• 2019
A Progressive Reduced Basis/Empirical Interpolation Method for Nonlinear Parabolic Problems
• Computer Science
SIAM J. Sci. Comput.
• 2018
A new methodology, the Progressive RB-EIM (PREIM) method, which is to reduce the offline cost while maintaining the accuracy of the RB approximation in the online stage, in contrast to the standard approach where the EIM approximation and the RB space are built separately. | {"extraction_info": {"found_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": 0.8725846409797668, "perplexity": 2057.164476603556}, "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/1652662604495.84/warc/CC-MAIN-20220526065603-20220526095603-00294.warc.gz"} |
http://blog.thematicmapping.org/2008/12/blog-post.html | ## Sunday, 7 December 2008
### Thematic mapping techniques - a summary
Thematic maps have become a primary mechanism for summarising and communicating the increased volumes of geographically related information. This blog post is a short summary of the most common thematic mapping techniques.
Jaques Bertin (1967) established a graphic system of visual variables, which represents an universally recognized theory of the cartographic transcription of geographical information (Koch, 2001). Visual variables describe the perceived differences in map symbols that are used to represent geographical phenomena (Slocum et al., 2005). Bertin’s system has been subsequently modified by various cartographers, and the visual variables presented below are based on Slocum et al. (2007), which add 3-D symbolisation.
(Image from InfoVis:Wiki)
Cartographers commonly distinguish between point, line, area and volume symbolisation (Robinson et al., 1995; Slocum et al., 2005). These distinctions may be summarised as follows:
A point symbol refer to a particular location in space, and is used when the geographical phenomena being mapped is located at a place or is aggregated to a given location (MacEachren, 1979). Differentiation among point symbols is achieved by using visual variables, like size, colour and shape. Common thematic mapping techniques using point symbols are dot maps and proportional symbol maps. On a dot map one dot represents a unit of some phenomena, and dots are placed at locations where the phenomenon is likely to occur (Slocum et al., 2005). A proportional symbol map is constructed by scaling symbols in proportion to the magnitude of data occurring at point locations. These locations can be true points or conceptual points, such as the centre of a country for which the data have been collected.
Line symbols are used to indicate connectivity or flow, equal values along a line and boundaries between unlike areas (MacEachren, 1979). Line symbols are differentiated on the basis of their form (e.g. solid line versus dotted line), colour and width. Common thematic mapping techniques using line symbols are flow maps and isarithmic maps. Flow maps utilise lines of differing width to depict the movement of phenomena between geographical locations (Slocum et al., 2005). Isarithmic maps depict smooth continuous phenomena, like rainfall or barometric pressure (Slocum etal., 2005).
Area symbols are used to assign a characteristic or value to a whole area on a map. Visual variables used for area symbols are colour, texture and perspective height (Slocum et al., 2005). The choropleth map is probably the most commonly employed method of thematic mapping, and is used to portray data collected for enumeration units, such as countries or statistical reporting units. While choropleth maps reflect the structure of data collection units, dasymetric maps assume areas of relative homogeneity, separated by zones of abrupt change. The country statistics used in the Thematic Mapping Engine can be considered as areal phenomena, because the statistical values are associated with political units specified as enclosed regions.
Volume symbols can be considered as 2½-D or true 3-D (Slocum et al. 2005). The first can be thought of as a surface, in which a geographical location is defined by x and y coordinate pairs and the value of the phenomenon is the height above a zero point. An example is prism maps which uses perspective height as the visual variable. 3-D symbols can be used to represent true 3-D phenomena, like the concentration of carbon dioxide (CO2) in the atmosphere or geological material underneath the earth’s surface (Slocum et al., 2005).
I'm interested in how these techniques can be represented in KML. You'll find several examples on this site. Please provide other examples by adding a comment. Especially, I would like to see examples of dot maps, flow maps, isarithmic maps and dasymetric maps in KML.
References:
• Bertin, J., 1967, "Semiologie Graphique", Paris
• Koch, W. G., 2001, "Jaques Bertin’s theory of graphics and its development and influence onmultimedia cartography", Information Design Journal 10(1), pp 37-43, John BenjaminPublishing Company
• MacEachren, A. M., 1979, "The Evolution of Thematic Cartography / A Research Methodology and Historical Review", The Canadion Cartographer Vol 16, No 1 June 1979, pp 17-33
• Robinson, A. H., Morrison, J.L., Muehrcke, P.C., Kimerling, A. J., Guptill, S. C., 1995, "Elements of Cartography", Sixth Edition, John Wiley & Sons
• Slocum, T. A., McMaster, R. B., Kessler, F. C., Howard, H. H., 2005, "Thematic Cartography and Geographic Visualization", Second Edition, Person Education Inc. | {"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": 0.8402366042137146, "perplexity": 2948.8239836386665}, "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/1397609539493.17/warc/CC-MAIN-20140416005219-00474-ip-10-147-4-33.ec2.internal.warc.gz"} |
https://math.stackexchange.com/questions/1450881/invariant-subspace-under-representation-phi | # Invariant subspace under representation $\phi$
Let $\phi: \Bbb{Z}/n\Bbb{Z}\to GL_{2}(\Bbb{C})$ be the representation which takes $\bar{m} \to \begin{bmatrix} \cos (\theta) & -\sin (\theta) \\ \sin (\theta) & \cos (\theta) \end{bmatrix}$ where $\theta = \frac{2\pi\ m}{n}$ which is matrix of rotation.
My book says that $W=\Bbb{C}e_1$, where $e_1=(1,0)$, is invariant under $\phi$ but I don't get it. Vectors on X-axis will be rotated. How does rotation does not affect them?
• It seems that your book is wrong – Omnomnomnom Sep 25 '15 at 19:48
• it is on page 15 of steinbergs' Rep theory of Finite groups – Departed Sep 26 '15 at 5:16 | {"extraction_info": {"found_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": 0.9313291907310486, "perplexity": 293.27292428559014}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-22/segments/1558232257920.67/warc/CC-MAIN-20190525064654-20190525090654-00299.warc.gz"} |
http://mathhelpforum.com/number-theory/164683-sum-digits-numbers-between-1-n.html | # Math Help - Sum of digits of numbers between 1 to n
1. ## Sum of digits of numbers between 1 to n
Hi, is there any general formula for finding the sum of digits from 1 to n (where n can be upto 10^9) .
I know it's got to do with some multiple of 45 (sum of digits from 1 to 9) but can't relate that to the required forumula.
Thanks .
2. The sum of the numbers from $\displaystyle 1$ to $\displaystyle n$ is $\displaystyle \frac{n}{2}(1+n)$.
3. Thanks but i was interested in sum of digits rather than the sum of numbers
4. Originally Posted by pranay
Hi, is there any general formula for finding the sum of digits from 1 to n (where n can be upto 10^9) .
I know it's got to do with some multiple of 45 (sum of digits from 1 to 9) but can't relate that to the required formula.
I don't think you are going to find a straightforward formula for this, except for some special cases of n.
For example, you can find the sum of the digits of all the numbers containing k digits, as follows. There are $9*10^{k-1}$ such numbers (9 possibilities for the first digit, 10 possibilities for each of the remaining k–1 digits). Each of these numbers has k digits, so there are $9*k*10^{k-1}$ digits altogether. Of these, $(k-1)*9*10^{k-2}$ will be zeros (each number stands a 1-in-10 chance of having a 0 in each position except the first). That leaves $9k*10^{k-1} - (k-1)*9*10^{k-2}$ other digits. Each of the nine other digits (1 to 9) is equally likely to occur, so each digit occurs $k*10^{k-1} - (k-1)*10^{k-2} = (9k+1)*10^{k-2}$ times. The sum of the numbers from 1 to 9 is 45, so the sum of all the digits in all the k-digit numbers is $45*(9k+1)*10^{k-2}$.
If you sum that result for k going from 1 to m, you find that the sum of all the digits in all the numbers from 1 to $10^m-1$ is $\boxed{45*m*10^{m-1}}$.
But if you want the result for a general value of n (not of the form $10^m-1$ ) then you will have a lot more work to do.
5. Thanks a lot for the great explanation , exaclty what i wanted . Thank you | {"extraction_info": {"found_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": 12, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8660460114479065, "perplexity": 118.30596592462024}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-40/segments/1443737899086.53/warc/CC-MAIN-20151001221819-00126-ip-10-137-6-227.ec2.internal.warc.gz"} |
https://math.stackexchange.com/questions/2844840/divergence-and-gradient-integration | Let $f$ a scalar function and $\vec{v}$ a vector function. We can write a product as:
$f \vec{v} = \vec{F}$
If we differentiate the LHS of above equation by the chain rule we can obtain
\begin{align} \nabla \cdot (f \vec{v}) = f \, \nabla \cdot \vec{v} + \vec{v} \, \nabla f \end{align}
Integrating we have:
\begin{align} \int \nabla \cdot (f \vec{v}) = \int f \, \nabla \cdot \vec{v} + \int \vec{v} \, \nabla f \end{align}
By inner product definition
\begin{align} \int \mathbf{D} (f \vec{v}) = \langle f, \mathbf{D} \vec{v} \rangle + \langle \vec{v} , \mathbf{G} f \rangle \end{align}
where $\mathbf{D}$ and $\mathbf{G}$ are the divergent and gradient operators. That lead us to
\begin{align} \int \mathbf{D} \vec{F} = \langle f, \mathbf{D} \vec{v} \rangle + \langle \vec{v} , \mathbf{G} f \rangle \end{align}
And by Fundamental Theorem of Calculus
\begin{align} \left. \vec{F} \right|_{\partial\Omega} = \langle f, \mathbf{D} \vec{v} \rangle + \langle \vec{v} , \mathbf{G} f \rangle \end{align}
where ${\partial\Omega}$ is the region boundary.
Is that correct?
• What inner product definition is that? – Jackozee Hakkiuz Jul 12 '18 at 18:56 | {"extraction_info": {"found_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": 5, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.999889612197876, "perplexity": 965.7387991346214}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-30/segments/1563195524290.60/warc/CC-MAIN-20190715235156-20190716021156-00492.warc.gz"} |
https://estudyassistant.com/mathematics/question515383471 | , 19.09.2022 14:20 kitttimothy55
# If g(x) is the inverse of f(x) and f(x) = 4x + 12, what is g(x)? O g(x) = 12x+4 O g(x) = 1/4x-12 O g(x) = x-3 O g(x)= 1/4x-3
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What are the solutions of the system? solve by graphing. y = x^2 + 3x + 2 y = 2x + 2
Mathematics, 21.06.2019 19:30
Acourt reporter is transcribing an accident report from germany. the report states that the driver was traveling 80 kilometers per hour. how many miles per hour was the driver traveling? | {"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": 0.9075506925582886, "perplexity": 2285.221983681033}, "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/1664030337631.84/warc/CC-MAIN-20221005140739-20221005170739-00392.warc.gz"} |
http://math.stackexchange.com/questions/205354/finding-supremum-of-real-numbers | # Finding Supremum of Real Numbers
Let $A$ be any non-empty subset of $\mathbb{R}$. Then $s = \sup A$ iff $s$ has the following properties:
1. $s \geq a$ for every $a \in A$,
2. if $t < s$, then there exists an $a \in A$ such that $a > t$.
Prove it? having problem in proving 2.
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What does "If tt" mean? – joriki Oct 1 '12 at 10:11
Huh? What is your condition (2) meant to say, and what is your definition of 'sup A'? – Billy Oct 1 '12 at 10:12
@robjohn: I'm pinging you for Brian's comment to Michael (since you were the one approving it). – Asaf Karagila Oct 1 '12 at 10:43
@BrianM.Scott: I didn't make any conscious changes to the mathematical content, I only Latexed the symbols and put (1) and (2) into a list. There was certainly a lot more than 'If tt' written for (2) when I edited the question. – Michael Albanese Oct 1 '12 at 10:47
@Michael: You’re right. Very weird. Anyhow, I’ve restored your edit. Sorry about all the confusion. – Brian M. Scott Oct 1 '12 at 11:02
Whatever it is, compare it to the definition given for the $\sup$, try to deduce one from the other.
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Step 1: write down the definition of $\sup$ of a set.
Step 2: Prove the $\implies$ direction. To do this, assume $s = \sup A$. Then show that $s$ satisfies 1. and 2.
Step 3: Prove the $\Longleftarrow$ direction. To do this, assume that $s$ satisfies 1. and 2. and show that $s = \sup A$.
Once you have done that you have solved the question, that's all you need to do.
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Hint: If $t<s$, then prove that there exists $h>0$(however small), such that $(t+h)<s$, and this $(t+h)$ will be your required $a$.You can refer Analysis by Terence Tao for detailed explanation.
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http://mathoverflow.net/questions/50703/defining-multiplication-in-polynomials-over-rings-of-matrices/50721 | # Defining Multiplication in Polynomials over Rings of Matrices
More explicitly, if $M_{2 \times 2}(\mathbb{R})[x]$ denotes the ring of polynomials over the ring of 2x2 matrices with real coefficients (with indeterminate x a 2 by 2 matrix with real coefficients), how do i properly define multiplication? e.g. suppose $A_0,A_1,B_0,B_1 \in M_{2 \times 2}(\mathbb{R})$, and let $f[x] = A_0 + A_1x$ and $g[x] = B_0 + B_1x$ be elements of $M_{2 \times 2}(\mathbb{R})[x]$. Then I would assume the product $fg[x]$ would be
$fg[x] = (A_0 + A_1x)(B_0 + B_1x) = A_0B_0 + A_0B_1x + A_1xB_0 + A_1xB_1x$
But then complications arise due to $M_{2 \times 2}(\mathbb{R})$ being non-commutative. As far as I know (I've only taken one course so far on abstract algebra), this ring is well-defined (in that a polynomial ring can have coefficients in any ring, not just commutative ones). I've checked google, wikipedia, etc and haven't found anything relevant to this topic. Is there any standard literature on this topic?
My plan was to eventually be able to investigate cases in which unique factorizations may hold (if any), or maybe polynomials having unique left or right inverses, etc.
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I won't vote to close, since no one else has yet, but I think this question could use a lot of sharpening. If you're just asking how the multiplication is defined, then I don't think this is an MO level question (one point you may be missing is that x commutes with matrices of scalars). You seem to just be confusing yourself by thinking about polynomials in matrices rather than matrices with polynomial entries, which are the same thing. – Ben Webster Dec 30 '10 at 7:59
If you want to ask for references for this stuff, that's a separate question; I'm sure a lot has been written about matrix rings over commutative rings, though I don't personally know of any good references. – Ben Webster Dec 30 '10 at 8:01
How does x commute with matrices of scalars if i define x as a 2x2 matrix as above? – Brian Hepler Dec 30 '10 at 8:08
Ah, now I see that I missed that part. Now I'm just confused by your definitions. The standard terminology is that for a ring A (not necessarily commutative), A[x] (called "polynomials in A") is the ring gotten by adding an indeterminate x which commutes with elements of A. It sounds like you want something else. – Ben Webster Dec 30 '10 at 8:20
The use of "polynomial" here is confusing (see my non-answer below). Perhaps you are really asking a more general question: Given a ring $R$, does there exist a ring $S$ containing $R$ as a subring and an indeterminate $x\in S$ (which does not necessarily commute with elements of $R$)? – Mark Grant Dec 30 '10 at 11:11
I have two answers for you, depending on what you have in mind.
You want to add an $x$ to the ring of 2x2 matrices, such that while $x$ commutes with multiples of the identity, it doesn't commute with anything else. You can adjoin a noncommuting indeterminate by using what's called the free product. You take the two rings $M_2(\mathbb{R})$ and $\mathbb{R}[x]$, and then you form the free product over $\mathbb{R}$. Wikipedia has an entry on the free product of groups. The construction for rings is fairly similar.
That construction has one weakness: $x$ will not satisfy any relations. There relations that all 2x2 matrices will satisfy, but $x$ in the free product won't. Rings all of whose elements satisfy identities are known as polynomial identity rings. For example, any four 2x2 matrices satisfy an identity of degree four (described in this blog post). So any three elements of $M_2(\mathbb{R})$ and $x$ should satisfy that relation. (I don't know if all possible relations are generated from specializing this one relation. There could be other relations that rely on specific properties of specific matrix elements.)
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Also, a good general reference for noncommutative ring theory is Louis Rowen's 2 volume Ring Theory. – arsmath Dec 30 '10 at 12:45
+1: This seems the right answer to the right question. – Mark Grant Dec 30 '10 at 14:16
Is it not possible to quotient out by a suitable ideal to force x to satisfy certain relations though? For instance, if we have the ideal J generated by $\{ (\lambda I ) x − x (\lambda I) : \lambda \in R \}$, where R is the underlying ring of the matrix ring, then take the quotient by J, would we necessarily get something with unwanted relations? – Zhen Lin Dec 30 '10 at 14:28
You definitely could. You just have to figure out what the relations are. The standard four-variable identity from the theory of polynomial identity rings gives a you a family of relations, but there may be others. – arsmath Dec 30 '10 at 15:17
Ben has already answered the question in the comments by noting that the indeterminate $x$ commutes with elements of $M_2(\mathbb{R})\subseteq M_2(\mathbb{R})[x]$. But I wanted to add some comments because the construction of polynomial rings is a subtle business.
The polynomial ring $R[x]$ should be thought of a ring containing $x$ and a subring isomorphic to $R$, such that $R$ commutes with $x$, and an element $a_0 + a_1x + \cdots +a_nx^n\in R[x]$ is zero if and only if each $a_i\in R$ is zero. Note that this implies that the indeterminate $x$ is transcendental over $R$, that is, $x$ is not a root of any polynomial with nonzero coefficients in $R$. In your example, this cannot be true if $x$ is an element of $M_2(\mathbb{R})$.
A good reference treating the construction of $R[x]$ for an arbitrary rings $R$ (and lots more besides) is Hungerford's Algebra book.
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The OP didn't actually want to talk about the usual polynomial ring, contrary to what the first phrase says, because he wants $x$ to be an indeterminate matrix. – Alex B. Dec 30 '10 at 9:39
There are already issues with $f[X] =X$ and $g[X] =B_0$. (I switched the variable to upper case, that seems less confusing in this context) Then $fg$ can not be written as $A_0 + A_1X$. So you have to allow "coefficents" on both sides and even between. Addition $AXB+CXD$ would also be a problem.
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http://math.stackexchange.com/questions/35851/hypothesis-testing | # Hypothesis testing
Suppose we have a random sample $X_1$, $X_2$ from the Beta($\theta$, $1$) distribution and we want to test $H_{\theta} :\theta \leq 1$ against $H_1:\theta > 1$. The following test issued: “Reject $H_0$ if and only if $3X_1 \leq 4X_2$.” How to show that the power function of the test is given by $$\beta(\theta) = 1-\frac12\left(\frac34\right)^\theta$$
My try :
This is not related to question, but I know how to solve if it is only one observation: let's say $X_1$ ~ Beta($\theta$, $1$) and the condition is $X_1 > \frac{1}{2}$ for same $H_0$ and $H_1$. To get the power function we have to solve for :
$$\beta(\theta) = P_\theta(X > 1/2) = \int_{1/2}^1 \frac{\Gamma(\theta + 1)}{\Gamma(\theta)\Gamma(1)}x^{\theta-1}(1-x)^{1-1}\mathrm dx$$
I do not know how to go about the original problem I mentioned in the question.
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What have you tried? – Ross Millikan Apr 29 '11 at 15:34
The test will reject if $X_{1} < 4X_{2}/3$. Therefore
$$\beta(\theta) = P(X_{1} < 4X_{2}/3 \ | \ \theta)$$
I'm assuming that $X_{1}, X_{2}$ are independent. By smoothing,
$$\beta(\theta) = E_{X_{2}} \left( P(X_{1} < 4X_{2}/3 \ | \ \theta, X_{2} = x) \right ) = E_{X_{2}} ( F_{\theta}(4x/3) ) = \int_{0}^{1} F_{\theta}(4x/3) p_{\theta}(x) dx$$
where $F_{\theta}, p_{\theta}$ are the ${\rm Beta}(\theta,1)$ CDF and PDF, respectively. This is not a very fun integral to compute but when you do, you will find it equals $1 - (1/2)(3/4)^{\theta}$. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 0.9272751808166504, "perplexity": 189.16649428792996}, "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/1432207929422.8/warc/CC-MAIN-20150521113209-00038-ip-10-180-206-219.ec2.internal.warc.gz"} |
http://www.physicsforums.com/showthread.php?t=316606 | # Surface charge density on the outer surface of a plate in a parallel plate capacitor
by dharm0us
Tags: capacitor, charge, density, outer, parallel, plate, surface
P: 7 Let me know if the following reasoning is correct : If the surface charge density on the internal surface of one of the plates in a parallel plate capacitor is Sigma, which becomes Sigma/K, when a dielectric material is inserted between the plates with dielectric constant K, so the charge density on the outer surface of the same plate will be Sigma - Sigma/K
P: 4,664 If you have a parallel plate capacitor with charge +/-Q on the inside of each plate, the voltage between the two plates is (d=separation, A = area) V=Q/C = Qd/e0A If you now stick a dielectric with relative dielectric constant k into the gap, the voltage becomes V=Q/C = Qd/k e0A The unchanged charge remains on the inside of both plates. The stored energy is now W = (1/2) Q V = Q2d/k e0A So the stored energy is reduced when you inserted the dielectric. Where did it go? If there is any unbalanced charge on the outer surface of either plate, it will remain there. Gauss law will tell you that the E field lines from charges on the outside of the plates are still there, and terminate at infinity.
P: 7 Hi Bob S, Let me rephrase my question : When a dielectric is inserted, some charge is induced on the inside of the plate, so the opposite charge of equal amount will be induced on the outside of the plate. Is it correct? Actually, I am making a question, to be handed out to students, based on the above assumption, so I wanted to cross check if the above reasoning is correct indeed.
P: 4,513
## Surface charge density on the outer surface of a plate in a parallel plate capacitor
Charge-up a plate capacitor. Let things settle until there is no current. The electric potential on both sides of a plate will be equal to each other.
All the charge is on the surface of the conductor, so we talk about the surface charge density. Although the electric potential is equal throughout the volume of the plate, the charge density is not.
The charge will be redistributed upon the surface to equalize the electric potential.
P: 7 Hi Phrak, So, is the reasoning in my original post is correct?
P: 4,513
Quote by dharm0us Hi Phrak, So, is the reasoning in my original post is correct?
You have proposed that the surface charge density is the same on both surfaces-the one facing the oppositely charged plate and the outer plate.
What should draw electrons to the outward facing surface? You need to specify the environment or no answer can be given.
I presume you mean that there are no external charges present, in which case the charge density on the external side is zero.
P: 7 Hi Phrak, What I have proposed is this : Initially there is no charge on the outer surface of the plate, and the surface charge density on the inner surface is Sigma. When the dielectric is inserted, the charge density on the inner surface is = Sigma / K, where K is the dielectric constant of the material. So, rest of the charge moves on the outer surface. Hence the surface charge density on the outer surface becomes : Sigma - Sigma/K
Sci Advisor PF Gold P: 1,721 No, the charge density is just sigma/k. The rest of the charges will migrate back between the connected leads to the opposite plate. In essence, the stored energy does not change as the loss in power stored in the electric field between the plates due to the lessening of the charges is made up by energy stored in the polarization of the dielectric. This assumes that you still have a DC potential hooked up between the plates. If you have the parallel plate capacitor charged, at steady state, then disconnect the leads, and then move a dielectric in between the two plates, the charge on the plates is still sigma, the charge has nowhere to go (barring dielectric breakdown, etc). The difference here is that moving the dielectric between the plates will require extra work, this extra work is stored in the polarization of the dielectric. So since you have to add energy into the system, there is no violation of energy conservation.
Mentor
P: 16,469
Quote by dharm0us Actually, I am making a question, to be handed out to students, based on the above assumption, so I wanted to cross check if the above reasoning is correct indeed.
Is it really fair to your students to ask them a question that you are not confident enough yourself to answer it without assistance? I think you are getting into a potentially bad situation here.
That said, I don't think there is enough information. Are the two capacitor plates charged with some voltage source that remains connected? Are they charged and then disconnected? Etc.
P: 4,664 If there is a charge +/-sigma on the inside of both plates, and the plates are NOT connected to a power supply, when a dielectric with constant k is inserted into the gap, the charge sigma will remain the same, and the voltage between the plates will drop by a factor k. If there is a charge +/-sigma on the inside of both plates, and the plates ARE connected to a power supply, when a dielectric with constant k is inserted into the gap, the charge sigma will drop by a factor k, and the voltage between the plates will be unchanged. If there is no charge on the OUTSIDE of the capacitor at the beginning, meaning Gauss' Law surface integral outside the capacitor =0, there are no field lines to (and charges at) infinity, there will be no outside charges after the dielectric is inserted. In either case, the total energy stored in the capacitor is reduced because the stored energy W is W = (1/2) Q V = (1/2) C V2 = (1/2)Q2/C = (1/2)Q2d/k e0A.
Related Discussions General Physics 0 General Physics 2 Introductory Physics Homework 4 Introductory Physics Homework 16 Introductory Physics Homework 4 | {"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": 0.9612807631492615, "perplexity": 287.9596410679813}, "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/1397609530136.5/warc/CC-MAIN-20140416005210-00216-ip-10-147-4-33.ec2.internal.warc.gz"} |
http://www.theinfolist.com/html/ALL/s/integer.html | TheInfoList
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# Symbol
The symbol $\mathbb$ can be annotated to denote various sets, with varying usage amongst different authors: $\mathbb^+$,$\mathbb_+$ or $\mathbb^$ for the positive integers, $\mathbb^$ or $\mathbb^$ for non-negative integers, and $\mathbb^$ for non-zero integers. Some authors use $\mathbb^$ for non-zero integers, while others use it for non-negative integers, or for . Additionally, $\mathbb_$ is used to denote either the set of integers modulo (i.e., the set of congruence classes of integers), or the set of -adic integers.Keith Pledger and Dave Wilkins, "Edexcel AS and A Level Modular Mathematics: Core Mathematics 1" Pearson 2008
# Algebraic properties
Like the
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multiplication Multiplication (often denoted by the cross symbol , by the mid-line dot operator , by juxtaposition, or, on computers, by an asterisk ) is one of the four Elementary arithmetic, elementary Operation (mathematics), mathematical operations ...
, that is, the sum and product of any two integers is an integer. However, with the inclusion of the negative natural numbers (and importantly, ), $\mathbb$, unlike the natural numbers, is also closed under
subtraction Subtraction is an arithmetic operation that represents the operation of removing objects from a collection. Subtraction is signified by the minus sign, . For example, in the adjacent picture, there are peaches—meaning 5 peaches with 2 taken ...
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unital ring In mathematics Mathematics (from Ancient Greek, Greek: ) includes the study of such topics as quantity (number theory), mathematical structure, structure (algebra), space (geometry), and calculus, change (mathematical analysis, analysis). ...
which is the most basic one, in the following sense: for any unital ring, there is a unique
ring homomorphism In ring theory, a branch of abstract algebra, a ring homomorphism is a structure-preserving function (mathematics), function between two ring (algebra), rings. More explicitly, if ''R'' and ''S'' are rings, then a ring homomorphism is a function s ...
from the integers into this ring. This
universal property In category theory Category theory formalizes mathematical structure and its concepts in terms of a Graph labeling, labeled directed graph called a ''Category (mathematics), category'', whose nodes are called ''objects'', and whose labelled d ...
, namely to be an
initial object In category theory Category theory formalizes mathematical structure and its concepts in terms of a Graph labeling, labeled directed graph called a ''Category (mathematics), category'', whose nodes are called ''objects'', and whose labelled d ...
in the
category of rings In mathematics, the category of rings, denoted by Ring, is the category (mathematics), category whose objects are ring (mathematics), rings (with identity) and whose morphisms are ring homomorphisms (that preserve the identity). Like many categor ...
, characterizes the ring $\mathbb$. $\mathbb$ is not closed under
division Division or divider may refer to: Mathematics *Division (mathematics), the inverse of multiplication *Division algorithm, a method for computing the result of mathematical division Military *Division (military), a formation typically consisting o ...
, since the quotient of two integers (e.g., 1 divided by 2) need not be an integer. Although the natural numbers are closed under
exponentiation Exponentiation is a mathematical Mathematics (from Greek: ) includes the study of such topics as numbers (arithmetic and number theory), formulas and related structures (algebra), shapes and spaces in which they are contained (geometry) ...
, the integers are not (since the result can be a fraction when the exponent is negative). The following table lists some of the basic properties of addition and multiplication for any integers , and : The first five properties listed above for addition say that $\mathbb$, under addition, is an
abelian group In mathematics Mathematics (from Greek: ) includes the study of such topics as numbers ( and ), formulas and related structures (), shapes and spaces in which they are contained (), and quantities and their changes ( and ). There is no g ...
. It is also a
cyclic group In group theory The popular puzzle Rubik's cube invented in 1974 by Ernő Rubik has been used as an illustration of permutation group">Ernő_Rubik.html" ;"title="Rubik's cube invented in 1974 by Ernő Rubik">Rubik's cube invented in 1974 by Er ...
, since every non-zero integer can be written as a finite sum or . In fact, $\mathbb$ under addition is the ''only'' infinite cyclic group—in the sense that any infinite cyclic group is
isomorphic In mathematics Mathematics (from Ancient Greek, Greek: ) includes the study of such topics as quantity (number theory), mathematical structure, structure (algebra), space (geometry), and calculus, change (mathematical analysis, analysis). I ...
to $\mathbb$. The first four properties listed above for multiplication say that $\mathbb$ under multiplication is a
commutative monoid In abstract algebra, a branch of mathematics Mathematics (from Ancient Greek, Greek: ) includes the study of such topics as quantity (number theory), mathematical structure, structure (algebra), space (geometry), and calculus, change (ma ...
. However, not every integer has a multiplicative inverse (as is the case of the number 2), which means that $\mathbb$ under multiplication is not a group. All the rules from the above property table (except for the last), when taken together, say that $\mathbb$ together with addition and multiplication is a
commutative ring In , a branch of , a commutative ring is a in which the multiplication operation is . The study of commutative rings is called . Complementarily, is the study of s where multiplication is not required to be commutative. Definition and first e ...
with
unity Unity may refer to: Buildings * Unity Building The Unity Building, in Oregon, Illinois, is a historic building in that city's Oregon Commercial Historic District. As part of the district the Oregon Unity Building has been listed on the National R ...
. It is the prototype of all objects of such
algebraic structure In mathematics Mathematics (from Ancient Greek, Greek: ) includes the study of such topics as quantity (number theory), mathematical structure, structure (algebra), space (geometry), and calculus, change (mathematical analysis, analysis). It ...
. Only those equalities of
expressions Expression may refer to: Linguistics * Expression (linguistics), a word, phrase, or sentence * Fixed expression, a form of words with a specific meaning * Idiom, a type of fixed expression * Metaphor#Common types, Metaphorical expression, a parti ...
are true in $\mathbb$
for all In mathematical logic Mathematical logic, also called formal logic, is a subfield of mathematics Mathematics (from Ancient Greek, Greek: ) includes the study of such topics as quantity (number theory), mathematical structure, structure (alg ...
values of variables, which are true in any unital commutative ring. Certain non-zero integers map to
zero 0 (zero) is a number A number is a mathematical object A mathematical object is an abstract concept arising in mathematics. In the usual language of mathematics, an ''object'' is anything that has been (or could be) formally defined, and ...
in certain rings. The lack of
zero divisor In abstract algebra In algebra, which is a broad division of mathematics, abstract algebra (occasionally called modern algebra) is the study of algebraic structures. Algebraic structures include group (mathematics), groups, ring (mathematics), ...
s in the integers (last property in the table) means that the commutative ring $\mathbb$ is an
integral domain In mathematics Mathematics (from Greek: ) includes the study of such topics as numbers ( and ), formulas and related structures (), shapes and spaces in which they are contained (), and quantities and their changes ( and ). There is no g ...
. The lack of multiplicative inverses, which is equivalent to the fact that $\mathbb$ is not closed under division, means that $\mathbb$ is ''not'' a
field Field may refer to: Expanses of open ground * Field (agriculture), an area of land used for agricultural purposes * Airfield, an aerodrome that lacks the infrastructure of an airport * Battlefield * Lawn, an area of mowed grass * Meadow, a grassl ...
. The smallest field containing the integers as a
subring In mathematics, a subring of ''R'' is a subset of a ring (mathematics), ring that is itself a ring when binary operations of addition and multiplication on ''R'' are restricted to the subset, and which shares the same multiplicative identity as ' ...
is the field of
rational number In mathematics, a rational number is a number that can be expressed as the quotient or fraction (mathematics), fraction of two integers, a numerator and a non-zero denominator . For example, is a rational number, as is every integer (e.g. ) ...
s. The process of constructing the rationals from the integers can be mimicked to form the
field of fractions In abstract algebra In algebra, which is a broad division of mathematics, abstract algebra (occasionally called modern algebra) is the study of algebraic structures. Algebraic structures include group (mathematics), groups, ring (mathematics) ...
of any integral domain. And back, starting from an
algebraic number field In mathematics Mathematics (from Ancient Greek, Greek: ) includes the study of such topics as quantity (number theory), mathematical structure, structure (algebra), space (geometry), and calculus, change (mathematical analysis, analysis). It h ...
(an extension of rational numbers), its
ring of integersIn mathematics Mathematics (from Ancient Greek, Greek: ) includes the study of such topics as quantity (number theory), mathematical structure, structure (algebra), space (geometry), and calculus, change (mathematical analysis, analysis). It ha ...
can be extracted, which includes $\mathbb$ as its
subring In mathematics, a subring of ''R'' is a subset of a ring (mathematics), ring that is itself a ring when binary operations of addition and multiplication on ''R'' are restricted to the subset, and which shares the same multiplicative identity as ' ...
. Although ordinary division is not defined on $\mathbb$, the division "with remainder" is defined on them. It is called
Euclidean division In arithmetic, Euclidean division – or division with remainder – is the process of division (mathematics), dividing one integer (the dividend) by another (the divisor), in a way that produces a quotient and a remainder smaller than the divisor ...
, and possesses the following important property: given two integers and with , there exist unique integers and such that and , where denotes the
absolute value In mathematics Mathematics (from Greek: ) includes the study of such topics as numbers (arithmetic and number theory), formulas and related structures (algebra), shapes and spaces in which they are contained (geometry), and quantities an ...
of . The integer is called the ''quotient'' and is called the ''
remainder In mathematics Mathematics (from Ancient Greek, Greek: ) includes the study of such topics as quantity (number theory), mathematical structure, structure (algebra), space (geometry), and calculus, change (mathematical analysis, analysis). It ...
'' of the division of by . The
Euclidean algorithm In mathematics, the Euclidean algorithm,Some widely used textbooks, such as I. N. Herstein's ''Topics in Algebra'' and Serge Lang's ''Algebra'', use the term "Euclidean algorithm" to refer to Euclidean division or Euclid's algorithm, is an effi ...
for computing
greatest common divisor In mathematics Mathematics (from Greek: ) includes the study of such topics as numbers ( and ), formulas and related structures (), shapes and spaces in which they are contained (), and quantities and their changes ( and ). There is no gene ...
s works by a sequence of Euclidean divisions. The above says that $\mathbb$ is a
Euclidean domain In mathematics, more specifically in ring theory, a Euclidean domain (also called a Euclidean ring) is an integral domain that can be endowed with a #Definition, Euclidean function which allows a suitable generalization of the Euclidean division of ...
. This implies that $\mathbb$ is a
principal ideal domain In mathematics Mathematics (from Ancient Greek, Greek: ) includes the study of such topics as quantity (number theory), mathematical structure, structure (algebra), space (geometry), and calculus, change (mathematical analysis, analysis). It h ...
, and any positive integer can be written as the products of
primes A prime number (or a prime) is a natural number File:Three Baskets.svg, Natural numbers can be used for counting (one apple, two apples, three apples, ...) In mathematics, the natural numbers are those numbers used for counting (as in "the ...
in an
essentially uniqueIn mathematics Mathematics (from Ancient Greek, Greek: ) includes the study of such topics as quantity (number theory), mathematical structure, structure (algebra), space (geometry), and calculus, change (mathematical analysis, analysis). It ha ...
way. This is the
fundamental theorem of arithmetic In number theory, the fundamental theorem of arithmetic, also called the unique factorization theorem or the unique-prime-factorization theorem, states that every integer An integer (from the Latin wikt:integer#Latin, ''integer'' meaning "wh ...
.
# Order-theoretic properties
$\mathbb$ is a
totally ordered set In mathematics Mathematics (from Ancient Greek, Greek: ) includes the study of such topics as quantity (number theory), mathematical structure, structure (algebra), space (geometry), and calculus, change (mathematical analysis, analysis). I ...
without upper or lower bound. The ordering of $\mathbb$ is given by: An integer is ''positive'' if it is greater than
zero 0 (zero) is a number A number is a mathematical object A mathematical object is an abstract concept arising in mathematics. In the usual language of mathematics, an ''object'' is anything that has been (or could be) formally defined, and ...
, and ''negative'' if it is less than zero. Zero is defined as neither negative nor positive. The ordering of integers is compatible with the algebraic operations in the following way: # if and , then # if and , then . Thus it follows that $\mathbb$ together with the above ordering is an
ordered ring 350px, The real numbers are an ordered ring which is also an ordered field. The integers">ordered_field.html" ;"title="real numbers are an ordered ring which is also an ordered field">real numbers are an ordered ring which is also an ordered field. ...
. The integers are the only nontrivial
totally ordered In mathematics Mathematics (from Ancient Greek, Greek: ) includes the study of such topics as quantity (number theory), mathematical structure, structure (algebra), space (geometry), and calculus, change (mathematical analysis, analysis). I ...
abelian group In mathematics Mathematics (from Greek: ) includes the study of such topics as numbers ( and ), formulas and related structures (), shapes and spaces in which they are contained (), and quantities and their changes ( and ). There is no g ...
whose positive elements are
well-ordered In mathematics Mathematics (from Ancient Greek, Greek: ) includes the study of such topics as quantity (number theory), mathematical structure, structure (algebra), space (geometry), and calculus, change (mathematical analysis, analysis). It ...
. This is equivalent to the statement that any
NoetherianIn mathematics, the adjective In linguistics, an adjective (list of glossing abbreviations, abbreviated ) is a word that grammatical modifier, modifies a noun or noun phrase or describes its referent. Its Semantics, semantic role is to change inf ...
valuation ringIn abstract algebra, a valuation ring is an integral domain In mathematics Mathematics (from Ancient Greek, Greek: ) includes the study of such topics as quantity (number theory), mathematical structure, structure (algebra), space (geometr ...
is either a
field Field may refer to: Expanses of open ground * Field (agriculture), an area of land used for agricultural purposes * Airfield, an aerodrome that lacks the infrastructure of an airport * Battlefield * Lawn, an area of mowed grass * Meadow, a grassl ...
—or a
discrete valuation ringIn abstract algebra, a discrete valuation ring (DVR) is a principal ideal domain In mathematics Mathematics (from Ancient Greek, Greek: ) includes the study of such topics as quantity (number theory), mathematical structure, structure (algebra ...
.
# Construction
In elementary school teaching, integers are often intuitively defined as the (positive) natural numbers,
zero 0 (zero) is a number A number is a mathematical object A mathematical object is an abstract concept arising in mathematics. In the usual language of mathematics, an ''object'' is anything that has been (or could be) formally defined, and ...
, and the negations of the natural numbers. However, this style of definition leads to many different cases (each arithmetic operation needs to be defined on each combination of types of integer) and makes it tedious to prove that integers obey the various laws of arithmetic. Therefore, in modern set-theoretic mathematics, a more abstract construction allowing one to define arithmetical operations without any case distinction is often used instead. The integers can thus be formally constructed as the
equivalence class In mathematics Mathematics (from Ancient Greek, Greek: ) includes the study of such topics as quantity (number theory), mathematical structure, structure (algebra), space (geometry), and calculus, change (mathematical analysis, analysis). ...
es of
ordered pair In mathematics Mathematics (from Ancient Greek, Greek: ) includes the study of such topics as quantity (number theory), mathematical structure, structure (algebra), space (geometry), and calculus, change (mathematical analysis, analysis). It h ...
s of
natural number In mathematics, the natural numbers are those numbers used for counting (as in "there are ''six'' coins on the table") and total order, ordering (as in "this is the ''third'' largest city in the country"). In common mathematical terminology, w ...
s . The intuition is that stands for the result of subtracting from . To confirm our expectation that and denote the same number, we define an
equivalence relation In mathematics Mathematics (from Ancient Greek, Greek: ) includes the study of such topics as quantity (number theory), mathematical structure, structure (algebra), space (geometry), and calculus, change (mathematical analysis, analysis). ...
on these pairs with the following rule: :$\left(a,b\right) \sim \left(c,d\right)$ precisely when :$a + d = b + c.$ Addition and multiplication of integers can be defined in terms of the equivalent operations on the natural numbers; by using to denote the equivalence class having as a member, one has: : : The negation (or additive inverse) of an integer is obtained by reversing the order of the pair: : Hence subtraction can be defined as the addition of the additive inverse: : The standard ordering on the integers is given by: :
if and only if In logic Logic is an interdisciplinary field which studies truth and reasoning. Informal logic seeks to characterize Validity (logic), valid arguments informally, for instance by listing varieties of fallacies. Formal logic represents st ...
$a+d < b+c.$ It is easily verified that these definitions are independent of the choice of representatives of the equivalence classes. Every equivalence class has a unique member that is of the form or (or both at once). The natural number is identified with the class (i.e., the natural numbers are embedded into the integers by map sending to ), and the class is denoted (this covers all remaining classes, and gives the class a second time since Thus, is denoted by :$\begin a - b, & \mbox a \ge b \\ -\left(b - a\right), & \mbox a < b. \end$ If the natural numbers are identified with the corresponding integers (using the embedding mentioned above), this convention creates no ambiguity. This notation recovers the familiar
representation Representation may refer to: Law and politics *Representation (politics) Political representation is the activity of making citizens "present" in public policy making processes when political actors act in the best interest of citizens. This def ...
of the integers as . Some examples are: : In theoretical computer science, other approaches for the construction of integers are used by Automated theorem proving, automated theorem provers and Rewriting, term rewrite engines. Integers are represented as Term algebra, algebraic terms built using a few basic operations (e.g., zero, succ, pred) and, possibly, using
natural number In mathematics, the natural numbers are those numbers used for counting (as in "there are ''six'' coins on the table") and total order, ordering (as in "this is the ''third'' largest city in the country"). In common mathematical terminology, w ...
s, which are assumed to be already constructed (using, say, the Peano axioms, Peano approach). There exist at least ten such constructions of signed integers. These constructions differ in several ways: the number of basic operations used for the construction, the number (usually, between 0 and 2) and the types of arguments accepted by these operations; the presence or absence of natural numbers as arguments of some of these operations, and the fact that these operations are free constructors or not, i.e., that the same integer can be represented using only one or many algebraic terms. The technique for the construction of integers presented above in this section corresponds to the particular case where there is a single basic operation pair$\left(x,y\right)$ that takes as arguments two natural numbers $x$ and $y$, and returns an integer (equal to $x-y$). This operation is not free since the integer 0 can be written pair(0,0), or pair(1,1), or pair(2,2), etc. This technique of construction is used by the proof assistant Isabelle (proof assistant), Isabelle; however, many other tools use alternative construction techniques, notable those based upon free constructors, which are simpler and can be implemented more efficiently in computers.
# Computer science
An integer is often a primitive data type in computer languages. However, integer data types can only represent a
subset In mathematics Mathematics (from Ancient Greek, Greek: ) includes the study of such topics as quantity (number theory), mathematical structure, structure (algebra), space (geometry), and calculus, change (mathematical analysis, analysis). ...
of all integers, since practical computers are of finite capacity. Also, in the common two's complement representation, the inherent definition of sign (mathematics), sign distinguishes between "negative" and "non-negative" rather than "negative, positive, and 0". (It is, however, certainly possible for a computer to determine whether an integer value is truly positive.) Fixed length integer approximation data types (or subsets) are denoted ''int'' or Integer in several programming languages (such as Algol68, C (computer language), C, Java (programming language), Java, Object Pascal, Delphi, etc.). Variable-length representations of integers, such as bignums, can store any integer that fits in the computer's memory. Other integer data types are implemented with a fixed size, usually a number of bits which is a power of 2 (4, 8, 16, etc.) or a memorable number of decimal digits (e.g., 9 or 10).
# Cardinality
The cardinality of the set of integers is equal to (Aleph number, aleph-null). This is readily demonstrated by the construction of a bijection, that is, a function that is injective and surjective from $\mathbb$ to $\mathbb= \.$ Such a function may be defined as :$f\left(x\right) = \begin -2x, & \mbox x \leq 0\\ 2x-1, & \mbox x > 0, \end$ with graph of a function, graph (set of the pairs $\left(x, f\left(x\right)\right)$ is :. Its inverse function is defined by :$\beging\left(2x\right) = -x\\g\left(2x-1\right)=x, \end$ with graph :.
* Canonical representation of a positive integer, Canonical factorization of a positive integer * Hyperinteger * Integer complexity * Integer lattice * Integer part * Integer sequence * Integer-valued function * Mathematical symbols * Parity (mathematics) * Profinite integer
# Sources
* Eric Temple Bell, Bell, E.T., ''Men of Mathematics.'' New York: Simon & Schuster, 1986. (Hardcover; )/(Paperback; ) * Herstein, I.N., ''Topics in Algebra'', Wiley; 2 edition (June 20, 1975), . * Saunders Mac Lane, Mac Lane, Saunders, and Garrett Birkhoff; ''Algebra'', American Mathematical Society; 3rd edition (1999). . | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 48, "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": 0.918846845626831, "perplexity": 1521.6113117907994}, "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/1664030335514.65/warc/CC-MAIN-20221001003954-20221001033954-00194.warc.gz"} |
http://www.bellbajao.org/uec6wi/0991e2-dividing-radical-expressions-worksheet | Some of the worksheets displayed are Multiplying radical, Adding subtracting multiplying radicals, Multiplying dividing radicals, Multiplying and dividing radicals, Multiply and divide radical expressions, Supplemental work problems to accompany the algebra, Section multiplication and division of radicals, Radicals. simplify radical expressions online ; adding multiplying divide and subtracting integers free worksheet ; solving equations involving rational expressions ; wordbook of Mcdougal littell ; what are two ways you can do scale factor ; algabra formulas ; mathematics draw equation graphs free gratis ; worksheets on multiplication of algebraic identities We have, $$\Large \frac{5}{{\sqrt 5 + 3\sqrt 3 }} = \Large \frac{5}{{\sqrt 5 + 3\sqrt 3 }} \cdot \Large \frac{{\sqrt 5 - 3\sqrt 3 }}{{\sqrt 5 - 3\sqrt 3 }}$$, $$= \Large \frac{{5\sqrt 5 - 15\sqrt 3 }}{{5 - 3\sqrt {15} + 3\sqrt {15} - 9 \cdot 3}}$$, remembering to FOIL here, $$= \Large \frac{{5\sqrt 5 - 15\sqrt 3 }}{{ - 22}}$$, $$= \Large \frac{{ - 5\sqrt 5 + 15\sqrt 3 }}{{22}}$$. Below you can download some free math worksheets and practice. In the mean time we talk concerning Dividing Radical Expressions Worksheet, scroll down to see some related pictures to complete your ideas. You have most likely already rationalized denominators in simple radical expressions such as, The way you rationalize the denominator in the above expression is by multiplying the expression by a fancy form of the number 1 that eliminates the radical in the denominator. 6. Major Operations 5. I changed the division into a multiplication and I flipped this guy right here. 4. It contains plenty of examples and practice problems. ©o 6KCuAtCav QSMoMfAtIw0akrLeD nLrLDCj.r m 0A0lsls 1r6i4gwh9tWsx 2rieAsKeLrFvpe9dc.c G 3Mfa0dZe7 UwBixtxhr AIunyfVi2nLimtqel bAmlCgQeNbarwaj w1Q.V-6-Worksheet by Kuta Software LLC Answers to Multiplying and Dividing Radicals 1) 3 2) −30 3) 8 4) 48 5 5) 33 + 15 6) 10 5 − 50 7) 33 + 32 8) 20 3 + 530 9) 30 1) 9 25 2) 4 36 3) 15 12 4) 4 Also included in: Algebra 1 Bundle ~ All My Algebra Products for 1 Low Price. A common way of dividing the radical expression is to have the denominator that contain no radicals. The radicand contains no fractions. There . Then simplify and combine all like radicals. When you divide by a fraction or a rational expression, it's the same thing as multiplying by the inverse. When dividing radical expressions, we use the quotient rule. Step 2 : Identify the common factor at both numerator and denominator. When dividing radical expressions, use the quotient rule. Some of the worksheets for this concept are Dividing radical, Dividing radicals period, Grade 9 simplifying radical expressions, Section multiplication and division of radicals, Multiplying dividing rational expressions, Dividing radicals period, Divide and reduce the radicals, Simplifying radicals 020316. It is common practice to write radical expressions without radicals in the denominator. The converse of this definition is also true. That's a mathematical symbols way of saying that when the index is even there can be no negative number in the radicand, but when the index is odd, there can be. Quiz & Worksheet - Dividing Radical Expressions | Study.com #117518 Help students … Sometimes in mathematics there exist conventions that have lasted throughout the years. Definition: If $$a\sqrt b + c\sqrt d$$ is a radical expression, then the conjugate is $$a\sqrt b - c\sqrt d$$. Subjects: Math, Algebra, Algebra 2. There’s nothing we can do about that. Showing top 8 worksheets in the category - Multiplying Dividing With Radicals. Simplify the expression, Here we will multiply both the numerator and denominator by the conjugate of the denominator, which is $$5\sqrt 3 - 2$$. If n is even, and a ≥ 0, b > 0, then . One of those conventions is that we must always rationalize the denominator in a radical expression. You can & download or print using the browser document reader options. Dividing Radical Expressions. Showing top 8 worksheets in the category - Dividing Radical Expressions. I can add and subtract radical expressions. Identify the radicand and index 2. Access these printable radical worksheets, carefully designed and proposed for students of grade 8 and high school. Multiplying and dividing radical expressions worksheet with answers Collection. The pdf worksheets cover topics such as identifying the radicand and index in an expression, converting the radical form to exponential form and the other way around, reducing radicals to its simplest form, rationalizing the denominators, and simplifying the radical expressions. Found worksheet you are looking for? As you become more familiar with dividing and simplifying radical expressions, make sure you continue to pay attention to the roots of the radicals that you are dividing. Simplifying Radicals Worksheet Algebra 2 Elegant Simplify Radicals Worksheet via aiasonline.org. The problems will ask you to simplify radical expressions. We have, $$\Large \frac{4}{{5\sqrt 3 + 2}} = \Large \frac{4}{{5\sqrt 3 + 2}} \cdot \Large \frac{{5\sqrt 3 - 2}}{{5\sqrt 3 - 2}}$$, $$= \Large \frac{{20\sqrt 3 - 8}}{{25 \cdot 3 - 10\sqrt 3 + 10\sqrt 3 - 4}}$$, remembering to FOIL here, $$= \Large \frac{{20\sqrt 3 - 8}}{{71}}$$, Look what happened when we multiplied the denominator by the conjugate. Given real numbers A n and B n, A n B n = A B n. Example 8. Multiplying a two-term radical expression involving square roots by its conjugate results in a rational expression. Simplify the expression, $$\Large \frac{5}{{\sqrt 5 + 3\sqrt 3 }}$$, Again we will multiply both numerator and denominator by the conjugate of the denominator. If n is odd, and b ≠ 0, then . I can multiply radical expressions. To multiply radical expressions, we follow the typical rules of multiplication, including such rules as the distributive property, etc. We must be careful and remember to FOIL in the second step. Worksheet will open in a new window. It is a self-worksheet that allows students to strengthen their skills at using multiplication to simplify radical expressions.All radical expressions in this maze are numerical radical expressions. When dividing radical expressions, the rules governing quotients are similar: $\sqrt[x]{\frac{a}{b}}=\frac{\sqrt[x]{a}}{\sqrt[x]{b}}$. Divide: 96 3 6 3. DIVIDING RATIONAL EXPRESSIONS. Note that a radical still remains in the expression. Definition: If $$a\sqrt b + c\sqrt d$$ is a radical expression, then the conjugate is $$a\sqrt b - c\sqrt d$$. Multiplying and Dividing 3. Adding Subtracting Multiplying And Dividing Radicals - Displaying top 8 worksheets found for this concept.. adding subtracting rational expressions worksheet answer, kuta software infinite algebra 1 answers and adding and subtracting radical expressions are some main things we will show you based on the post title. There is one property of radicals in multiplication that is important to remember. Simplifying Radical Expressions 2. Let’s try one more example. Apply the distributive property when multiplying a radical expression with multiple terms. And so we have rationalized the denominator. Solution: In this case, we can see that 6 and 96 have common factors. Thus division of one rational expression by other is equivalent to the product of first and reciprocal of the second expression. We have, $$\Large \frac{2}{{\sqrt 3 }} = \Large \frac{2}{{\sqrt 3 }} \cdot \Large \frac{{\sqrt 3 }}{{\sqrt 3 }}$$. To download/print, click on pop-out icon or print icon to worksheet to print or download. Now we bring you various awesome photos that we've gathered just for you, in this gallery we are pay more attention about Algebra 1 Radicals Worksheet. Dividing radical expressions worksheet. There’s nothing we can do about that. Now when dealing with more complicated expressions involving radicals, we employ what is known as the conjugate. So let’s look at it. Grades: 8 th, 9 th, 10 th. Fractions With Radicals - Displaying top 8 worksheets found for this concept.. In this case, we would multiply by $$\Large \frac{{\sqrt 3 }}{{\sqrt 3 }}$$. Multiplying and Dividing Radical Expressions #117517. Rationalizing is the process of starting with a fraction containing a radical in its denominator and determining fraction with no radical in its denominator. If the resulting expression is not in its lowest form then reduce to its lowest form. These simplifying radicals worksheest are appropriate for 3rd grade 4th grade 5th grade 6th grade and 7th grade. This lesson will describe the quotient rule and how to use it to solve these radical expressions. Let me just rewrite this thing over here. Dividing radical is based on rationalizing the denominator. The radical disappeared, and the denominator was rationalized. We can do this because we are in effect multiplying the expression by 1, which changes nothing. Dividing Radical Expressions. Simplify.This free worksheet contains 10 assignments each with 24 questions with answers.Example of one question: Completing the square by finding the constant, Solving equations by completing the square, Solving equations with The Quadratic Formula, Copyright © 2008-2020 math-worksheet.org All Rights Reserved. If we apply the quotient rule for radicals and write it as a single cube root, we will be able to reduce the fractional radicand. Some of the worksheets below are simplifying radical expressions worksheet steps to simplify radical combining radicals simplify radical algebraic expressions multiply radical expressions divide radical expressions solving radical equations graphing radicals. Simplifying radical expressions worksheets; Square roots; Ordering square roots from least to greatest; Operations with radicals; How to simplify radical expressions ; After having gone through the stuff given above, we hope that the students would have understood "Divide radical expressions". In the day and age of the super-powered calculators, these conventions do not always make sense anymore. Note that a radical still remains in the expression. Step 1 : Factor both numerator and denominator, if possible. This particular quiz is a series of math problems. Combining radicals is possible when the index and the radicand of two or more radicals are the same. Displaying top 8 worksheets found for - Dividing Radical Expressions. radical simplifying radicals worksheet algebra 2 worksheets 1 practice equations with answers factoring answer key via tusfacturas.co. Well, what if you are dealing with a quotient instead of a product? About This Quiz & Worksheet. Let’s look at an example of using the conjugate to rationalize the denominator. Let’s see what happens when we do that. You can use the same ideas to help you figure out how to simplify and divide radical expressions. Types: Worksheets, Activities, Printables. ©G 32v071 d2N 2KOuutiaG MSHoyfNt4wGagr 5ec JL 7L pC W.f H pAQlRlB BrGiAgvh4t Rsd 4rgeUseSr tvye Rdy. In this case, we are multiplying both numerator and denominator by $$\sqrt 5 - 3\sqrt 3$$. I can divide radical expressions (and rationalize a denominator). For example, while you can think of as equivalent to since both the numerator and the denominator are square roots, notice that you cannot express as . Basic radical worksheets for kids to identify radicand and index. G O XAfl wlv ur di 2g Uh2tWsF jrZe csse 2r8v kezdT.R 8 bM fa CdNeh 7wZiQtchS tI Pnsf gi4nDi6tye T DARljgReOb0rHad a2 Y.5 Worksheet by Kuta Software LLC Kuta Software - Infinite Algebra 2 Name_____ Dividing Radicals Period____ Simplify. 2p plus 6 over p plus 5 divided by 10 over 4p plus 20 is the same thing as multiplying by the reciprocal here, multiplying by 4p plus 20 over 10. o 9 lM da gdCes Fwoi5toh l 5IGnJf dian9i Ztwe2 HAHl Rgveob3r na4 61 J.U Worksheet by Kuta Software LLC Kuta Software - Infinite Algebra 1 Name_____ Dividing Radical Expressions Date_____ Period____ Simplify. Radicals with the same index and radicand are known as like radicals. For all real values, a and b, b ≠ 0 . ©u 32f0 t1u2 j 9Kxu Vt8a5 sS8onfet8w 4a Ir 8e3 CLlLfCj. This algebra video tutorial explains how to divide radical expressions with variables and exponents. I can simplify radical algebraic expressions. Some of the worksheets for this concept are Dividing radical, Dividing radicals period, Multiply and divide radical expressions, Multiplying dividing radicals, Dividing radical expressions of index 2, Multiplying dividing rational expressions, Multiplying radical expressions of index 2 with variable, Section multiplication and division of radicals. There is a rule for that, too. Talking related with Algebra 1 Radicals Worksheet, below we will see various related pictures to give you more ideas. But we continue to use them anyway. Radical expressions are written in simplest terms when . Recall that the Product Raised to a Power Rule states that $\sqrt[x]{ab}=\sqrt[x]{a}\cdot \sqrt[x]{b}$. To divide radical expressions with the same index, we use the quotient rule for radicals. Now when dealing with more complicated expressions involving radicals, we employ what is known as the conjugate. Figure out how to use it to solve these radical expressions thing multiplying! 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Let ’ s see what happens when we do that the product of and. | {"extraction_info": {"found_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": 0.9247016310691833, "perplexity": 1598.2440712630403}, "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-2021-17/segments/1618038879305.68/warc/CC-MAIN-20210419080654-20210419110654-00482.warc.gz"} |
http://mathoverflow.net/questions/126464/is-there-an-analog-of-determinant-for-linear-operators-in-infinite-dimensions-as/126483 | # Is there an analog of determinant for linear operators in infinite dimensions as that of finite dimensions?
I am trying to find out the essence of what a determinant is. Besides, in finite dimensions, determinant is the kind of numerical invariant that determines the invertibility of a linear operator, but what about infinite case? Is there a similar invariant? Why if not?
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If $V$ is a vector space and $f$ is an endomorphism of $V$, then sometimes a determinant $\det f$ can be defined even without requiring $V$ to be finite-dimensional. For example, if there is a finite-dimensional vector subspace $W$ of $V$ such that $f\left(V\right)\subseteq W$, then we can define $\det f$ to be the determinant of the map $W\to W$ obtained by restricting $f$; this will actually be independent on $W$ (as long as $W$ satisfies $f\left(V\right)\subseteq W$). But this is a rather trivial case (although surprisingly useful in K-theory). – darij grinberg Apr 4 '13 at 1:33
If $V$ is Banach(?), there might be a more general type of endomorphisms $f$ for which $\det f$ could be defined in some reasonable way; "determinant-class" endomorphisms, so to speak (in analogy to trace-class endomorphisms). In the general case, however, determinants make no sense in infinite dimensions. This shouldn't be very surprising since, e. g., an injective endomorphism of an infinite-dimensional space can fail to be surjective, and vice versa. – darij grinberg Apr 4 '13 at 1:36
For me, "the essence of what a determinant is" is the measure of the distortion of volume. But volume is a complicated notion in infinite dimensions. – Claudio Gorodski Apr 4 '13 at 1:51
@darij: The condition you described only lets you define $\det (1-Tf)$. This will be a polynomial whose roots are the reciprocals of the nonzero eigenvalues of $f$. – Kevin Ventullo Apr 4 '13 at 2:42
Oh, right! Yeah, that's what I meant. – darij grinberg Apr 4 '13 at 3:53
We can define the Fredholm determinant on operators on Hilbert space which differ from the identity by a trace-class operator. This satisfies $$\det(\exp(T)) = \exp(\text{Tr}(T))$$ for trace-class operators $T$. See http://en.wikipedia.org/wiki/Fredholm_determinant
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Thank you Professor Israel. Now everything is perfectly understandable! – xuxuzhu Apr 6 '13 at 2:21
Attempting to define a well-behaved "infinite-dimensional determinant" for all operators will get us into trouble fairly quickly. Consider, for example, a vector space $V$ of countably infinite dimension and let $A$ be multiplication by some nonzero scalar $\lambda\neq1$. Then $A$ is certainly invertible so its putative determinant should be nonzero. But now upon fixing a basis for $V$ we get the following matrix identities $$A = \begin{pmatrix} \lambda \\ & \lambda \\ & & \lambda \\ & & & \ddots \end{pmatrix} = \begin{pmatrix} \lambda \\ & 1 \\ & & 1\\ & & & \ddots \end{pmatrix} \begin{pmatrix} 1 \\ & \lambda \\ & & \lambda \\ & & & \ddots \end{pmatrix} = \begin{pmatrix} \lambda \\ & I \end{pmatrix} \begin{pmatrix} 1 \\ & A \end{pmatrix}.$$ If we expect our determinant to behave like its finite-dimensional counterpart, then the above would yield $\det A = \lambda \det A$ and consequently that $\det A = 0$, counter to what we expect from the invertibility of $A$.
So in attempting to define $\det$ for operators on infinite-dimensional spaces, you either have to restrict the class of operators under consideration or lower your expectations of how your $\det$ will be analogous to the finite-dimensional one.
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I am trying to find out the essence of what a determinant is.
The abstract way to define the determinant of a linear operator $T : V \to V$ of a finite-dimensional vector space is that it is the induced action of $T$ on the top exterior power $\Lambda^n(T) : \Lambda^n(V) \to \Lambda^n(V)$, where $n = \dim V$. The top exterior power is $1$-dimensional, so $\Lambda^n(T)$ is canonically a scalar. By functoriality of the exterior power, $T \mapsto \Lambda^n(T)$ is a monoid homomorphism, which is why it detects invertibility.
So you can see what the problem is in infinite dimensions: if $V$ is infinite-dimensional, there is no top exterior power! All the exterior powers are also infinite-dimensional.
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For $T$ sufficiently close to the identity, one might hope for a reasonable way to define an action of $T$ on an "infinite exterior power" $\Lambda^{\dim V}(V)$ (there is an infinite product that has a hope of converging), and I guess this is what the Fredholm determinant makes precise. – Qiaochu Yuan Apr 4 '13 at 6:17
Thank you so much Qiaochu. It just occurred to me that I've run into the stuff of viewing determinant as actions on top exterior forms when studying smooth manifolds, but your comments for infinite case is really inspiring and I really appreciate your help! – xuxuzhu Apr 6 '13 at 2:19
There are modifications of the notion of Fredholm determinant for operators on Hilbert space which differ from the identity by an operator from a von Neumann-Schatten ideal. A related notion is the one of a von Koch determinant defined for some classes of infinite matrices. For all this see
Gohberg, I.C.; Krein, M.G. Introduction to the theory of linear nonselfadjoint operators. Translations of Mathematical Monographs. 18. Providence, RI: American Mathematical Society, 1969.
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The answers above point out that one cannot define a determinant in a meaningful way on the algebra of bounded operators on a Banach space, unless finite-dimensional. However, this does not preclude the possibility of doing this for suitable subclasses and this is precisely what Alexander Grothendieck did in his celebrated (amongst functional analysts) article "Théorie de Fredholm" (Bull. Soc. Math. vol. 84---freely available online). This is one of those articles which changed the face of mathematics forever. The closely related question of which operators have a trace has been investigated in great detail, by, for example, Albrecht Pietsch and Hermann König.
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I believe this is related to the reference Robert Israel gives for the Fredholm determinant? (And I guess this was what led Grothendick to the eponymous AP?) – Yemon Choi Apr 4 '13 at 6:41
To the references already given for the definition and the properties of the determinant $\det(I+T)$, where $T$ is of trace class, I would add:
Reed; Simon: Methods of Modern Mathematical Physics IV, 1978. Chapter 13
Taylor, M. E.: Partial Differential Equations I (Basic Theory), 1996. Appendix A.
Schmüdgen, K.: Unbounded Self-adjoint Operators on Hilbert Space. Springer GTM, 2011. Ch. 9, sect. 5.
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https://fr.maplesoft.com/support/help/errors/view.aspx?path=Physics/ToCovariant&L=F | Physics/ToCovariant - Maple Help
Home : Support : Online Help : Physics/ToCovariant
Physics[ToCovariant] - Rewrite in covariant form the indices of the tensors of a given expression
Physics[ToContravariant] - Rewrite in covariant form the indices of the tensors of a given expression
Calling Sequence ToCovariant(tensorial_expression, optional ...) ToContravariant(tensorial_expression, optional ...)
Parameters
tensorial_expression - any tensorial expression, or a set, list, equation or matrix of them, onto which the operation is performed
Options
• changecharacteroffreeindices = ... : synonym changecharacteroffreeindices, the right-hand side can be true or false (default), to change or not (flip covariant <-> contravariant) the character of the free indices
• evaluateexpression = ... : can be true or false (default); to evaluate or not the expression after having manipulated its tensor indices
• evaluatetensor = ... : can be true or false (default); to evaluate or not the tensors after manipulating their indices
• onlytheseindices = ... : can be any symbol representing a tensor index, or a set or list of them possibly found in tensorial_expressions, to restrict the operation to only those indices
• changerepeatedindices = ... : can be true (default) or false, in which case the repeated indices are returned unchanged
• quiet = ... : the right-hand side can be true or false (default), to display or not information related to matching keywords
Description
• When working with tensors in spaces where the covariant and contravariant tensors' components have a different value (the underlying metric is not Euclidean) one frequently wants to express formulations with some or all of the tensors's indices expressed either in covariant or contravariant form. In previous Maple releases, also in Maple 2021, you can raise or lower free indices multiplying by the metric and performing the contraction. That, however, involves a whole simplification process not always desired, and does not result in flipping the character of repeated indices. The SubstituteTensorIndices is also useful for that purpose but requires changing the indices one by one. Instead, to handle the whole manipulation operation, you can use ToCovariant and ToContravariant.
• Several options are available to adjust the operation in different ways, as explained in the Options section above. Perhaps two more relevant ones are changecharacteroffreeindices (default value is false), that can be used to receive an expression where you get all free indices flipping their character, and onlytheseindices = ... to restrict the operation to only some of the indices.
• Note that closely related to ToCovariant and ToContravariant, the Physics package includes a SubstituteTensorIndices command.
Examples
> $\mathrm{with}\left(\mathrm{Physics}\right):$
Set coordinates and a tensor for experimentation; avoid redundant display of functionality using CompactDisplay
> $\mathrm{Setup}\left(\mathrm{coordinates}=\mathrm{Cartesian},\mathrm{tensors}=A\left[\mathrm{\mu }\right]\right)$
${}{}\mathrm{Systems of spacetime coordinates are:}{}{}{}\left\{X=\left(x{,}y{,}z{,}t\right)\right\}$
$\mathrm{_______________________________________________________}$
$\left[{\mathrm{coordinatesystems}}{=}\left\{{X}\right\}{,}{\mathrm{tensors}}{=}\left\{{{A}}_{{\mathrm{\mu }}}{,}{{\mathrm{\gamma }}}_{{\mathrm{\mu }}}{,}{{\mathrm{\sigma }}}_{{\mathrm{\mu }}}{,}{{\partial }}_{{\mathrm{\mu }}}{,}{{g}}_{{\mathrm{\mu }}{,}{\mathrm{\nu }}}{,}{{\mathrm{\epsilon }}}_{{\mathrm{\alpha }}{,}{\mathrm{\beta }}{,}{\mathrm{\mu }}{,}{\mathrm{\nu }}}{,}{{X}}_{{\mathrm{\mu }}}\right\}\right]$ (1)
> $\mathrm{CompactDisplay}\left(A\left(X\right)\right)$
${A}{}\left({X}\right){}{\mathrm{will now be displayed as}}{}{A}$ (2)
Consider the following tensorial expression, define it as the components of a new tensor ${F}_{\mathrm{\mu },\mathrm{\nu }}$
> $F\left[\mathrm{\mu },\mathrm{\nu }\right]=\mathrm{d_}\left[\mathrm{\mu }\right]\left(A\left[\mathrm{\nu }\right]\left(X\right)\right)-\mathrm{d_}\left[\mathrm{\nu }\right]\left(A\left[\mathrm{\mu }\right]\left(X\right)\right)$
${{F}}_{{\mathrm{μ}}{,}{\mathrm{ν}}}{=}{{\mathrm{d_}}}_{{\mathrm{μ}}}{}\left({{A}}_{{\mathrm{ν}}}{}\left({X}\right){,}\left[{X}\right]\right){-}{{\mathrm{d_}}}_{{\mathrm{ν}}}{}\left({{A}}_{{\mathrm{μ}}}{}\left({X}\right){,}\left[{X}\right]\right)$ (3)
> $\mathrm{Define}\left(\right)$
$\mathrm{Defined objects with tensor properties}$
$\left\{{{A}}_{{\mathrm{\mu }}}{,}{{\mathrm{\gamma }}}_{{\mathrm{\mu }}}{,}{{F}}_{{\mathrm{\mu }}{,}{\mathrm{\nu }}}{,}{{\mathrm{\sigma }}}_{{\mathrm{\mu }}}{,}{{\partial }}_{{\mathrm{\mu }}}{,}{{g}}_{{\mathrm{\mu }}{,}{\mathrm{\nu }}}{,}{{\mathrm{\epsilon }}}_{{\mathrm{\alpha }}{,}{\mathrm{\beta }}{,}{\mathrm{\mu }}{,}{\mathrm{\nu }}}{,}{{X}}_{{\mathrm{\mu }}}\right\}$ (4)
The covariant components of ${F}_{\mathrm{\mu },\mathrm{\nu }}$ are
> $F\left[\right]$
${{F}}_{{\mathrm{μ}}{,}{\mathrm{ν}}}{=}\left(\left[\begin{array}{cccc}0& \frac{\partial }{\partial x}{}{A}_{2}{}\left(x,y,z,t\right)-\left(\frac{\partial }{\partial y}{}{A}_{1}{}\left(x,y,z,t\right)\right)& \frac{\partial }{\partial x}{}{A}_{3}{}\left(x,y,z,t\right)-\left(\frac{\partial }{\partial z}{}{A}_{1}{}\left(x,y,z,t\right)\right)& \frac{\partial }{\partial x}{}{A}_{4}{}\left(x,y,z,t\right)-\left(\frac{\partial }{\partial t}{}{A}_{1}{}\left(x,y,z,t\right)\right)\\ \frac{\partial }{\partial y}{}{A}_{1}{}\left(x,y,z,t\right)-\left(\frac{\partial }{\partial x}{}{A}_{2}{}\left(x,y,z,t\right)\right)& 0& \frac{\partial }{\partial y}{}{A}_{3}{}\left(x,y,z,t\right)-\left(\frac{\partial }{\partial z}{}{A}_{2}{}\left(x,y,z,t\right)\right)& \frac{\partial }{\partial y}{}{A}_{4}{}\left(x,y,z,t\right)-\left(\frac{\partial }{\partial t}{}{A}_{2}{}\left(x,y,z,t\right)\right)\\ \frac{\partial }{\partial z}{}{A}_{1}{}\left(x,y,z,t\right)-\left(\frac{\partial }{\partial x}{}{A}_{3}{}\left(x,y,z,t\right)\right)& \frac{\partial }{\partial z}{}{A}_{2}{}\left(x,y,z,t\right)-\left(\frac{\partial }{\partial y}{}{A}_{3}{}\left(x,y,z,t\right)\right)& 0& \frac{\partial }{\partial z}{}{A}_{4}{}\left(x,y,z,t\right)-\left(\frac{\partial }{\partial t}{}{A}_{3}{}\left(x,y,z,t\right)\right)\\ \frac{\partial }{\partial t}{}{A}_{1}{}\left(x,y,z,t\right)-\left(\frac{\partial }{\partial x}{}{A}_{4}{}\left(x,y,z,t\right)\right)& \frac{\partial }{\partial t}{}{A}_{2}{}\left(x,y,z,t\right)-\left(\frac{\partial }{\partial y}{}{A}_{4}{}\left(x,y,z,t\right)\right)& \frac{\partial }{\partial t}{}{A}_{3}{}\left(x,y,z,t\right)-\left(\frac{\partial }{\partial z}{}{A}_{4}{}\left(x,y,z,t\right)\right)& 0\end{array}\right]\right)$ (5)
Related to the character of indices, note first that, since Maple 2021, when you request the contravariant components of a tensor definition as ${F}_{}^{\mathrm{\mu },\mathrm{\nu }}$, the output is already expressed using the contravariant components of the tensors involved, in this case ${A}^{\mathrm{\mu }}$
> $F\left[\mathrm{~}\right]$
${{F}}_{{\mathrm{~mu}}{,}{\mathrm{~nu}}}{=}\left(\left[\begin{array}{cccc}0& -\left(\frac{\partial }{\partial x}{}{A}_{\mathrm{~2}}{}\left(x,y,z,t\right)\right)+\frac{\partial }{\partial y}{}{A}_{\mathrm{~1}}{}\left(x,y,z,t\right)& -\left(\frac{\partial }{\partial x}{}{A}_{\mathrm{~3}}{}\left(x,y,z,t\right)\right)+\frac{\partial }{\partial z}{}{A}_{\mathrm{~1}}{}\left(x,y,z,t\right)& -\left(\frac{\partial }{\partial x}{}{A}_{\mathrm{~4}}{}\left(x,y,z,t\right)\right)-\left(\frac{\partial }{\partial t}{}{A}_{\mathrm{~1}}{}\left(x,y,z,t\right)\right)\\ -\left(\frac{\partial }{\partial y}{}{A}_{\mathrm{~1}}{}\left(x,y,z,t\right)\right)+\frac{\partial }{\partial x}{}{A}_{\mathrm{~2}}{}\left(x,y,z,t\right)& 0& -\left(\frac{\partial }{\partial y}{}{A}_{\mathrm{~3}}{}\left(x,y,z,t\right)\right)+\frac{\partial }{\partial z}{}{A}_{\mathrm{~2}}{}\left(x,y,z,t\right)& -\left(\frac{\partial }{\partial y}{}{A}_{\mathrm{~4}}{}\left(x,y,z,t\right)\right)-\left(\frac{\partial }{\partial t}{}{A}_{\mathrm{~2}}{}\left(x,y,z,t\right)\right)\\ -\left(\frac{\partial }{\partial z}{}{A}_{\mathrm{~1}}{}\left(x,y,z,t\right)\right)+\frac{\partial }{\partial x}{}{A}_{\mathrm{~3}}{}\left(x,y,z,t\right)& -\left(\frac{\partial }{\partial z}{}{A}_{\mathrm{~2}}{}\left(x,y,z,t\right)\right)+\frac{\partial }{\partial y}{}{A}_{\mathrm{~3}}{}\left(x,y,z,t\right)& 0& -\left(\frac{\partial }{\partial z}{}{A}_{\mathrm{~4}}{}\left(x,y,z,t\right)\right)-\left(\frac{\partial }{\partial t}{}{A}_{\mathrm{~3}}{}\left(x,y,z,t\right)\right)\\ \frac{\partial }{\partial t}{}{A}_{\mathrm{~1}}{}\left(x,y,z,t\right)+\frac{\partial }{\partial x}{}{A}_{\mathrm{~4}}{}\left(x,y,z,t\right)& \frac{\partial }{\partial t}{}{A}_{\mathrm{~2}}{}\left(x,y,z,t\right)+\frac{\partial }{\partial y}{}{A}_{\mathrm{~4}}{}\left(x,y,z,t\right)& \frac{\partial }{\partial t}{}{A}_{\mathrm{~3}}{}\left(x,y,z,t\right)+\frac{\partial }{\partial z}{}{A}_{\mathrm{~4}}{}\left(x,y,z,t\right)& 0\end{array}\right]\right)$ (6)
The definition of ${F}_{\mathrm{\mu },\mathrm{\nu }}$ involves only free and covariant indices; make all the tensors be expressed using covariant indices without changing the mathematical value of the expression
> $\mathrm{ToContravariant}\left(\right)$
${{\mathrm{g_}}}_{{\mathrm{α}}{,}{\mathrm{μ}}}{}{{\mathrm{g_}}}_{{\mathrm{β}}{,}{\mathrm{ν}}}{}{{F}}_{{\mathrm{~alpha}}{,}{\mathrm{~beta}}}{=}{{\mathrm{g_}}}_{{\mathrm{α}}{,}{\mathrm{μ}}}{}{{\mathrm{g_}}}_{{\mathrm{β}}{,}{\mathrm{ν}}}{}{{\mathrm{d_}}}_{{\mathrm{~alpha}}}{}\left({{A}}_{{\mathrm{~beta}}}{}\left({X}\right){,}\left[{X}\right]\right){-}{{\mathrm{g_}}}_{{\mathrm{α}}{,}{\mathrm{ν}}}{}{{\mathrm{g_}}}_{{\mathrm{β}}{,}{\mathrm{μ}}}{}{{\mathrm{d_}}}_{{\mathrm{~alpha}}}{}\left({{A}}_{{\mathrm{~beta}}}{}\left({X}\right){,}\left[{X}\right]\right)$ (7)
> $\mathrm{indets}\left(,\mathrm{Or}\left(\mathrm{specindex}\left(F\right),\mathrm{specfunc}\left(A\right)\right)\right)$
$\left\{{{F}}_{{\mathrm{~alpha}}{,}{\mathrm{~beta}}}{,}{{A}}_{{\mathrm{~beta}}}{}\left({X}\right)\right\}$ (8)
Restrict that operation to only $\mathrm{\mu }$
> $\mathrm{ToContravariant}\left(,\mathrm{only}=\mathrm{\mu }\right)$
$\mathrm{* Partial match of \text{'}}{}\mathrm{only}{}\mathrm{\text{'} against keyword \text{'}}{}\mathrm{onlytheseindices}{}\text{'}$
${{\mathrm{g_}}}_{{\mathrm{α}}{,}{\mathrm{μ}}}{}{{F}}_{{\mathrm{~alpha}}{,}{\mathrm{ν}}}{=}{{\mathrm{g_}}}_{{\mathrm{α}}{,}{\mathrm{μ}}}{}{{\mathrm{d_}}}_{{\mathrm{~alpha}}}{}\left({{A}}_{{\mathrm{ν}}}{}\left({X}\right){,}\left[{X}\right]\right){-}{{\mathrm{g_}}}_{{\mathrm{α}}{,}{\mathrm{μ}}}{}{{\mathrm{d_}}}_{{\mathrm{ν}}}{}\left({{A}}_{{\mathrm{~alpha}}}{}\left({X}\right){,}\left[{X}\right]\right)$ (9)
> $\mathrm{indets}\left(,\mathrm{Or}\left(\mathrm{specindex}\left(F\right),\mathrm{specfunc}\left(A\right)\right)\right)$
$\left\{{{F}}_{{\mathrm{ν}}{,}{\mathrm{~alpha}}}{,}{{A}}_{{\mathrm{ν}}}{}\left({X}\right){,}{{A}}_{{\mathrm{~alpha}}}{}\left({X}\right)\right\}$ (10)
In the above we see the tensors originally having $\mathrm{\mu }$ covariant, now with the index in that position contravariant. To achieve the other possible meaning of to contravariant use the option changecharacteroffreeindices
> $\mathrm{ToContravariant}\left(,\mathrm{only}=\mathrm{\mu },\mathrm{changecharacter}\right)$
$\mathrm{* Partial match of \text{'}}{}\mathrm{only}{}\mathrm{\text{'} against keyword \text{'}}{}\mathrm{onlytheseindices}{}\text{'}$
$\mathrm{* Partial match of \text{'}}{}\mathrm{changecharacter}{}\mathrm{\text{'} against keyword \text{'}}{}\mathrm{changecharacteroffreeindices}{}\text{'}$
${{F}}_{{\mathrm{~mu}}{,}{\mathrm{ν}}}{=}{{\mathrm{d_}}}_{{\mathrm{~mu}}}{}\left({{A}}_{{\mathrm{ν}}}{}\left({X}\right){,}\left[{X}\right]\right){-}{{\mathrm{d_}}}_{{\mathrm{ν}}}{}\left({{A}}_{{\mathrm{~mu}}}{}\left({X}\right){,}\left[{X}\right]\right)$ (11)
> $\mathrm{indets}\left(,\mathrm{Or}\left(\mathrm{specindex}\left(F\right),\mathrm{specfunc}\left(A\right)\right)\right)$
$\left\{{{F}}_{{\mathrm{ν}}{,}{\mathrm{~mu}}}{,}{{A}}_{{\mathrm{ν}}}{}\left({X}\right){,}{{A}}_{{\mathrm{~mu}}}{}\left({X}\right)\right\}$ (12)
An expression that has free and repeated indices:
> $\mathrm{Define}\left(A,B,G\right)$
$\mathrm{Defined objects with tensor properties}$
$\left\{{B}{,}{G}{,}{{A}}_{{\mathrm{\mu }}}{,}{{\mathrm{\gamma }}}_{{\mathrm{\mu }}}{,}{{F}}_{{\mathrm{\mu }}{,}{\mathrm{\nu }}}{,}{{\mathrm{\sigma }}}_{{\mathrm{\mu }}}{,}{{\partial }}_{{\mathrm{\mu }}}{,}{{g}}_{{\mathrm{\mu }}{,}{\mathrm{\nu }}}{,}{{\mathrm{\epsilon }}}_{{\mathrm{\alpha }}{,}{\mathrm{\beta }}{,}{\mathrm{\mu }}{,}{\mathrm{\nu }}}{,}{{X}}_{{\mathrm{\mu }}}\right\}$ (13)
> $A\left[\mathrm{\alpha }\right]B\left[\mathrm{\beta }\right]F\left[\mathrm{\mu },\mathrm{\nu }\right]G\left[\mathrm{\nu },\mathrm{\alpha }\right]+A\left[\mathrm{\beta }\right]B\left[\mathrm{\alpha }\right]F\left[\mathrm{\mu },\mathrm{\rho }\right]G\left[\mathrm{\rho },\mathrm{\alpha }\right]$
${{A}}_{{\mathrm{\beta }}}{}{{G}}_{{\mathrm{\rho }}{,}{\mathrm{\alpha }}}{}{{F}}_{{\mathrm{\mu }}\phantom{{\mathrm{\rho }}}}^{\phantom{{\mathrm{\mu }}}{\mathrm{\rho }}}{}{{B}}_{\phantom{{}}\phantom{{\mathrm{\alpha }}}}^{\phantom{{}}{\mathrm{\alpha }}}{+}{{B}}_{{\mathrm{\beta }}}{}{{G}}_{{\mathrm{\nu }}{,}{\mathrm{\alpha }}}{}{{A}}_{\phantom{{}}\phantom{{\mathrm{\alpha }}}}^{\phantom{{}}{\mathrm{\alpha }}}{}{{F}}_{{\mathrm{\mu }}\phantom{{\mathrm{\nu }}}}^{\phantom{{\mathrm{\mu }}}{\mathrm{\nu }}}$ (14)
There are several tensors with covariant indices, and the free and repeated indices can be determined using Check
> $\mathrm{Check}\left(,\mathrm{all}\right)$
$\mathrm{The repeated indices per term are:}{}\left[\left\{\mathrm{...}\right\}{,}\left\{\mathrm{...}\right\}{,}\mathrm{...}\right]{}\mathrm{, the free indices are:}{}\left\{\mathrm{...}\right\}$
$\left[\left\{{\mathrm{\alpha }}{,}{\mathrm{\rho }}\right\}{,}\left\{{\mathrm{\alpha }}{,}{\mathrm{\nu }}\right\}\right]{,}\left\{{\mathrm{\beta }}{,}{\mathrm{\mu }}\right\}$ (15)
We see the free indices are $\mathrm{\mu }$ and $\mathrm{\beta }$, both covariant. To have all the tensors of this expression (but for the metric) with all their indices contravariant, use
> $\mathrm{ToContravariant}\left(\right)$
${{g}}_{{\mathrm{\beta }}{,}{\mathrm{\kappa }}}{}{{B}}_{\phantom{{}}\phantom{{\mathrm{\kappa }}}}^{\phantom{{}}{\mathrm{\kappa }}}{}{{g}}_{{\mathrm{\nu }}{,}{\mathrm{\sigma }}}{}{{g}}_{{\mathrm{\alpha }}{,}{\mathrm{\tau }}}{}{{G}}_{\phantom{{}}\phantom{{\mathrm{\sigma }}}\phantom{{,}}\phantom{{\mathrm{\tau }}}}^{\phantom{{}}{\mathrm{\sigma }}{,}{\mathrm{\tau }}}{}{{A}}_{\phantom{{}}\phantom{{\mathrm{\alpha }}}}^{\phantom{{}}{\mathrm{\alpha }}}{}{{g}}_{{\mathrm{\lambda }}{,}{\mathrm{\mu }}}{}{{F}}_{\phantom{{}}\phantom{{\mathrm{\lambda }}}\phantom{{,}}\phantom{{\mathrm{\nu }}}}^{\phantom{{}}{\mathrm{\lambda }}{,}{\mathrm{\nu }}}{+}{{g}}_{{\mathrm{\beta }}{,}{\mathrm{\nu }}}{}{{A}}_{\phantom{{}}\phantom{{\mathrm{\nu }}}}^{\phantom{{}}{\mathrm{\nu }}}{}{{g}}_{{\mathrm{\lambda }}{,}{\mathrm{\rho }}}{}{{g}}_{{\mathrm{\alpha }}{,}{\mathrm{\sigma }}}{}{{G}}_{\phantom{{}}\phantom{{\mathrm{\lambda }}}\phantom{{,}}\phantom{{\mathrm{\sigma }}}}^{\phantom{{}}{\mathrm{\lambda }}{,}{\mathrm{\sigma }}}{}{{g}}_{{\mathrm{\kappa }}{,}{\mathrm{\mu }}}{}{{F}}_{\phantom{{}}\phantom{{\mathrm{\kappa }}}\phantom{{,}}\phantom{{\mathrm{\rho }}}}^{\phantom{{}}{\mathrm{\kappa }}{,}{\mathrm{\rho }}}{}{{B}}_{\phantom{{}}\phantom{{\mathrm{\alpha }}}}^{\phantom{{}}{\mathrm{\alpha }}}$ (16)
> $\mathrm{indets}\left(,\mathrm{specindex}\left(\left[A,B,F,G\right]\right)\right)$
$\left\{{{A}}_{\phantom{{}}\phantom{{\mathrm{\alpha }}}}^{\phantom{{}}{\mathrm{\alpha }}}{,}{{A}}_{\phantom{{}}\phantom{{\mathrm{\nu }}}}^{\phantom{{}}{\mathrm{\nu }}}{,}{{B}}_{\phantom{{}}\phantom{{\mathrm{\alpha }}}}^{\phantom{{}}{\mathrm{\alpha }}}{,}{{B}}_{\phantom{{}}\phantom{{\mathrm{\kappa }}}}^{\phantom{{}}{\mathrm{\kappa }}}{,}{{F}}_{\phantom{{}}\phantom{{\mathrm{\kappa }}}\phantom{{,}}\phantom{{\mathrm{\rho }}}}^{\phantom{{}}{\mathrm{\kappa }}{,}{\mathrm{\rho }}}{,}{{F}}_{\phantom{{}}\phantom{{\mathrm{\lambda }}}\phantom{{,}}\phantom{{\mathrm{\nu }}}}^{\phantom{{}}{\mathrm{\lambda }}{,}{\mathrm{\nu }}}{,}{{G}}_{\phantom{{}}\phantom{{\mathrm{\lambda }}}\phantom{{,}}\phantom{{\mathrm{\sigma }}}}^{\phantom{{}}{\mathrm{\lambda }}{,}{\mathrm{\sigma }}}{,}{{G}}_{\phantom{{}}\phantom{{\mathrm{\sigma }}}\phantom{{,}}\phantom{{\mathrm{\tau }}}}^{\phantom{{}}{\mathrm{\sigma }}{,}{\mathrm{\tau }}}\right\}$ (17)
This result is mathematically equal to the starting expression - all that happened is that the covariant versions of the indices were replaced by the contravariant ones at the cost of adding metric factors, so
> $\mathrm{Simplify}\left(-\right)$
${0}$ (18)
In addition to the onlytheseindices option, to perform these operations only on the free indices, you can also use changerepeatedindices = false
> $\mathrm{ToContravariant}\left(,\mathrm{changerepeatedindices}=\mathrm{false}\right)$
${{g}}_{{\mathrm{\beta }}{,}{\mathrm{\kappa }}}{}{{B}}_{\phantom{{}}\phantom{{\mathrm{\kappa }}}}^{\phantom{{}}{\mathrm{\kappa }}}{}{{G}}_{{\mathrm{\nu }}{,}{\mathrm{\alpha }}}{}{{A}}_{\phantom{{}}\phantom{{\mathrm{\alpha }}}}^{\phantom{{}}{\mathrm{\alpha }}}{}{{g}}_{{\mathrm{\lambda }}{,}{\mathrm{\mu }}}{}{{F}}_{\phantom{{}}\phantom{{\mathrm{\lambda }}}\phantom{{,}}\phantom{{\mathrm{\nu }}}}^{\phantom{{}}{\mathrm{\lambda }}{,}{\mathrm{\nu }}}{+}{{g}}_{{\mathrm{\beta }}{,}{\mathrm{\nu }}}{}{{A}}_{\phantom{{}}\phantom{{\mathrm{\nu }}}}^{\phantom{{}}{\mathrm{\nu }}}{}{{G}}_{{\mathrm{\rho }}{,}{\mathrm{\alpha }}}{}{{g}}_{{\mathrm{\kappa }}{,}{\mathrm{\mu }}}{}{{F}}_{\phantom{{}}\phantom{{\mathrm{\kappa }}}\phantom{{,}}\phantom{{\mathrm{\rho }}}}^{\phantom{{}}{\mathrm{\kappa }}{,}{\mathrm{\rho }}}{}{{B}}_{\phantom{{}}\phantom{{\mathrm{\alpha }}}}^{\phantom{{}}{\mathrm{\alpha }}}$ (19)
>
References
Landau, L.D., and Lifshitz, E.M. The Classical Theory of Fields, Course of Theoretical Physics Volume 2, fourth revised English edition. Elsevier, 1975.
Compatibility
• The Physics[ToCovariant] command was introduced in Maple 2021.
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