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: Saw you edited the question and added some comments at the end of my answer to address. Edit
There can't possibly be any factors between $\frac n2$ and $n$. Suppose $\frac n2 < a < n$ and $a \cdot b = n$. What could $b$ be? If $b=1$, then $a \cdot b = a < n$. If $b \geq 2$, then $a \cdot b > \frac n2 \cdot 2 = n$. So $b$ can't be any positive integer; thus $a$ isn't a factor of $n$.
In fact, as long as you list
both factors when you do the division, you can stop testing at $\sqrt n$. That's because it's impossible for both factors to be greater than $\sqrt n$: if $a,b > \sqrt n$, then $a \cdot b > \sqrt n \cdot \sqrt n = n$. So factors always come in pairs $a \cdot b = n$ with $a < \sqrt n$ and $b > \sqrt n$ (with the exception of $\sqrt n$ itself, if it's an integer). Here's how this method works to find the factors of $10$: $3 < \sqrt 10 < 4$, so we can stop testing at $3$ Test $1$: $\frac{10}1 = 10$, so $10 = 1 \cdot 10$, giving us the two factors $1$ and $10$ Test $2$: $\frac{10}2 = 5$, so $10 = 2 \cdot 5$, giving us the two factors $2$ and $5$ Test $3$: $\frac{10}3$ is not an integer so we don't get any factors
Thus the full list of factors is $1,10,2,5$ (or, reordered, $1,2,5,10$)
This method works to find all the factors, not just the prime factors, so it's a perfect method for testing whether a number is perfect (in our example, $1+2+5=8\neq10$, so $10$ is not perfect).
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Let $1<p_1,\dots,p_k<\infty$ such that $\sum_{i=1}^k\frac{1}{p_i}=1$.
Moreover, let $a_1,\dots,a_k\geq 0$. I have to show that $$\prod_{i=1}^k a_i\leq\max_i a_i^{p_i}$$ I want to use induction over $k$, but I am struggling with it. For $k=2$ I only got to here:
Since $p_1$ and $p_2$ are Hölder conjugates, we have $\frac{1}{p_1}+\frac{1}{p_2}=1\Leftrightarrow p_1=\frac{p_2}{p_2-1}$. I have been experimenting with the case when $a_1 a_2>a_2^{p_2}$, but I didn't get anywhere \begin{align*} 1<a_2^{p_2-1}a_1^{-1} \Leftrightarrow 1<a_2 a_1^{-1/(p_2-1)} \Leftrightarrow 1<a_2\frac{a_1^{p_1}}{a_1} \end{align*} ...
I thought maybe case distinction between the four cases $a_i\leq 1$ and $a_i>1$ could work, but I don't see how this works, since we also have to take care of the $p_i$...
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I'm not totally sure if this is answering your question, but let $\boldsymbol{\lambda}$ be a vector that you can think of as containing the eigenvalues of $\mathbf{X}^T\mathbf{X}$. I'm assuming that the Frobenius norm constraint you mentioned in your comment is of the form $||X||_{F}^2 \le c$. Let $\mathbf{1}$ be a vector of all 1s. Assume $\mathbf{X} \in \mathbb{R}^{M\times N}$. I think you should start with the program:
$$\max_{\lambda,t} t$$$$\text{st. } \mathbf{1}t \preccurlyeq \boldsymbol{\lambda}$$$$\boldsymbol{\lambda} \succcurlyeq 0$$$$\mathbf{1}^T\boldsymbol{\lambda} \le c$$
If, by chance, you wanted a constraint of the form $||X||_{F}^2 = c$ you should start with:
$$\max_{\lambda,t} t$$$$\text{st. } \mathbf{1}t \preccurlyeq \boldsymbol{\lambda}$$$$\boldsymbol{\lambda} \succcurlyeq 0$$$$\mathbf{1}^T\boldsymbol{\lambda} = c$$
Both of these are clearly linear programs. You can find an $\mathbf{X}$ that achieves the solution easily enough. First put $\boldsymbol{\lambda}$ into a diagonal matrix, say $\mathbf{S}$ and augment it with the appropriate number of zeros such that it has the same dimensions of $\mathbf{X}$, ie$$ \begin{bmatrix}\lambda_1 &0 &0 \\ 0& \ddots &0 \\ 0& 0 & \lambda_N \\ 0 & 0 & 0\\ \vdots & \vdots & \vdots \end{bmatrix}.$$ Pick your favorite orthogonal matrix $\mathbf{V}\in\mathbb{R}^{M\times M}$ and your second favorite orthogonal matrix $\mathbf{U}\in\mathbb{R}^{N\times N}$. Let $\mathbf{R} = \sqrt{\mathbf{S}}$ (elementwise I guess, idk if the matrix square root is technically allowed for non-square matrices). Let $\mathbf{X} = \mathbf{V}\mathbf{R}\mathbf{U}^T$ then clearly $$\mathbf{X}^T\mathbf{X} = \mathbf{U}\mathbf{S}\mathbf{U}^T.$$ This matrix has the eigenvalues that you want.
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Now showing items 1-5 of 5
Forward-backward multiplicity correlations in pp collisions at √s = 0.9, 2.76 and 7 TeV
(Springer, 2015-05-20)
The strength of forward-backward (FB) multiplicity correlations is measured by the ALICE detector in proton-proton (pp) collisions at s√ = 0.9, 2.76 and 7 TeV. The measurement is performed in the central pseudorapidity ...
Rapidity and transverse-momentum dependence of the inclusive J/$\mathbf{\psi}$ nuclear modification factor in p-Pb collisions at $\mathbf{\sqrt{\textit{s}_{NN}}}=5.02$ TeV
(Springer, 2015-06)
We have studied the transverse-momentum ($p_{\rm T}$) dependence of the inclusive J/$\psi$ production in p-Pb collisions at $\sqrt{s_{\rm NN}} = 5.02$ TeV, in three center-of-mass rapidity ($y_{\rm cms}$) regions, down to ...
Measurement of charm and beauty production at central rapidity versus charged-particle multiplicity in proton-proton collisions at $\sqrt{s}$ = 7 TeV
(Springer, 2015-09)
Prompt D meson and non-prompt J/$\psi$ yields are studied as a function of the multiplicity of charged particles produced in inelastic proton-proton collisions at a centre-of-mass energy of $\sqrt{s}=7$ TeV. The results ...
Coherent $\rho^0$ photoproduction in ultra-peripheral Pb-Pb collisions at $\mathbf{\sqrt{\textit{s}_{\rm NN}}} = 2.76$ TeV
(Springer, 2015-09)
We report the first measurement at the LHC of coherent photoproduction of $\rho^0$ mesons in ultra-peripheral Pb-Pb collisions. The invariant mass and transverse momentum distributions for $\rho^0$ production are studied ...
Inclusive, prompt and non-prompt J/ψ production at mid-rapidity in Pb-Pb collisions at √sNN = 2.76 TeV
(Springer, 2015-07-10)
The transverse momentum (p T) dependence of the nuclear modification factor R AA and the centrality dependence of the average transverse momentum 〈p T〉 for inclusive J/ψ have been measured with ALICE for Pb-Pb collisions ...
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I'm using the Limit Definition to find the derivative,
$$f'(x)=\lim_{\Delta x \to 0} {f(x+\Delta x) - f(x) \over \Delta x}$$ $$$$ Now, I want to find the derivative for the function,
$$f(x)={1 \over x+1}$$
So, here's what I did.
$$\lim_{\Delta x \to 0} {{1 \over (x+\Delta x) +1} - {1 \over x+1}\over \Delta x}$$
Now, I think I can multiply the numerator and the denominator by the least common multiple to get rid of the denominator in the numerator??
I'm not sure what to do from here. Thanks
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This question already has an answer here:
Background: This is part b of problem 12.4.3 from Arfken, Weber, Harris Math Methods for Physicists to show that $\int_0^\infty \frac{\ln^2(z)}{1+z^2}$dz$=4(1-\frac{1}{3^3}+\frac{1}{5^3}-\frac{1}{7^3}+\dots)$.
Part b of the question asks to show that this series evaluates to $\frac{\pi^3}{8}$ by contour integration. Where is my mistake: $\lim_{z \to 0}zf(z)=0$ and $\lim_{z \to \infty}zf(z)=0$ so the big and little circle equal 0.$
Assume $I=\int_0^\infty \frac{\ln^2(x)}{1+x^2}\text{dx}$
We can add the components of along the contour and set that equal to the value of $2\pi i \text{Res}[f(z),i]$ evaluated at the poles $\pm i$ $$\int_0^\infty \frac{\ln^2(z)}{1+z^2}\text{dz}+\int_{\infty}^0 \frac{\ln^2(z)}{1+z^2}\text{dz}=2\pi i \text{Res}[f(z),\pm i]\tag{1}$$
$$\int_0^\infty \frac{(\ln^2 \mid x\mid}{1+x^2}\text{dx}-\int_0^{\infty} \frac{(\ln\mid x\mid+2i\pi)^2}{1+x^2}\text{dx}=2\pi i \left (\lim_{z \to i}\frac{\ln^2(z)}{2z}+\lim_{z \to -i}\frac{\ln^2(z)}{2z}\right )\tag{2}$$
$$\int_0^\infty \frac{(\ln^2\mid x\mid}{1+x^2}\text{dx}-\int_0^{\infty} \frac{(\ln^2\mid x\mid+\color{red}{4\ln|x|i\pi}-4\pi^2)}{1+x^2}\text{dx}=2\pi i \left (\lim_{z \to i}\frac{\ln^2(z)}{2z}+\lim_{z \to -i}\frac{\ln^2(z)}{2z}\right )\tag{3}$$ $$0I+\color{red}{0}-\left[\tan^{-1}(x)\right]\mid^{\infty}_0(4\pi^2)\text{dx}=(2\pi i) \left (\frac{-\pi^2/4+9\pi^2/4}{2i}\right )\tag{4}$$ $$0I+2\pi^3=\frac{8\pi^3}{4}\tag{5}$$
I found my error. It was a negative sign, and the two sides cancel to zero so you can't evaluate it this way, but I found an answer which evaluates it from negative to positive infinity so I'm marking the question as a duplicate. See dustin's answer at the link for the contour integration.
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I am trying to understand a modal logic countermodel, illustrating that the
QK theorem,$$ \Box (\phi(x) \wedge \forall x \phi (x)) \rightarrow \Box \forall x \phi (x), $$ fails in the alternate semantics for modal logic given by David Lewis's counterpart theory.The countermodel is the following:
It comes from Oliver Kutz's
Kripke-Typ Semantiken für die modale Prädikatenlogik, p. 31. I don't understand German, but based on what I could glean from context and Google Translate it doesn't seem like the diagram is explained at all.
Here's what I do understand:
The large circles represent the two worlds, $w_1$ and $w_2$.
The smaller dotted circles, which I would prefer to be loops, indicate that $C (u_1, u_1)$ and $C (u_2, u_2)$ (where $C$ is the "counterpart relation").
At the bottom of each circle is an indication that $\phi/\lnot \phi$ hold of $u_1/u_2$ in their respective worlds.
What I don't get are the dotted arcs. Based on context I believe the quantifiers below the arcs are quantifiers over worlds. This makes sense because the original presentation of counterpart theory utilized a translation procedure to link statements of quantified modal logic to statements in the two-sorted (worlds and individuals) framework of counterpart theory. So, e.g., the formula $\Box \phi (x)$ gets translated as $$\forall v \forall y [W (v) \wedge I (y, v) \wedge C (y, x) \rightarrow \phi^v (y)],$$ which is read as "for every world $v$ if $y$ inhabits $v$ and $y$ is a counterpart of $x$ then $\phi$ is true of $y$ in $v$". Similarly, a statement like $\Diamond \phi (x)$ gets translated as $$\exists v \exists y (W (v) \wedge I (y, v) \wedge C (y, x) \wedge \phi^v (b)).$$
As a result there are various completion I could imagine to the quantification over worlds. It will be true that there is some world where $\exists x \lnot \phi (x)$ is true, namely $w_2$, and so $\Diamond \exists x \lnot \phi (x)$ is true in $w_1$ (and therefore $\lnot \Box \forall x \phi (x)$ is also true in $w_1$, falsifying the consequent of the conditional).
Additionally, both $\phi (x)$ and $\forall x \phi (x)$ are true in $w_1$ since the only counterpart of $u_1$ is itself it also follows that $\Box (\phi (x) \wedge \forall x \phi (x))$ is true in $w_1$. From this and the preceding we have that the conditional
QK theorem is false at $w_1$ in this counterpart theory model.
So obviously there is some quantification over worlds in the box and diamond statements, but I can't quite figure out what sort of a relationship the dotted arcs are attempting to indicate.
Any help would be appreciated.
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Give it a circumstellar cloud of oxygen.
Some planetary nebulae, such as NGC 6826, appear green because of ionized oxygen.
Image in the public domain. Yes, this is a true-color image.
I see no reason why you couldn't surround the star with an extremely dense cloud of hydrogen, containing a relatively high fraction of oxygen, which would absorb light and reradiate it at green wavelengths. This is the same effect that we see in auroras. The emission of light at the 557.7 nm wavelength is the cause of the green tinge.
The stability of such a cloud is, of course, a problem. Radiation pressure, responsible for the dissipation of molecular clouds surrounding newborn stars, can disperse quite a lot of gas. In fact, planetary nebulae can last for only about $\sim10,000$ years, an extremely short time relative to the lifespan of stars.
Obviously, the optimal way to combat this would be either a continuous resupply of gas, possible via accretion from a companion (unlikely, in the case of oxygen, although I don't have a source for that) or an extremely large gas reservoir. Extreme mass loss, possibly due to extremely strong stellar winds, is a possibility. Walmswell & Eldridge (2012) suggested that "superwinds" from red giants could be responsible for self-sustaining circumstellar envelopes of gas, which would actually decrease the star's luminosity (they were searching for a solution to the red supergiant problem).
Mass loss in the case of our star would require there to be a large amount of ionized oxygen in the stellar atmosphere and outer layers of the star - possible, given that red supergiants should be fusing heavier elements, and given also that higher-metallicity stars may have substantial quantities of these elements, known as "metals". I'm not confident that such a supply is totally realistic, but I see no reason to dismiss it entirely. While hydrogen dominates photospheric spectra in most stars, chemically peculiar atmospheres have been observed in many other stars nonetheless.
Technical details about forbidden lines
For anyone more curious, here's a bit more detail on how a cloud of oxygen can turn green.
An emission nebula gets its distinctive colors from photons emitted by different elements in its gas. Hydrogen is, of course, the most plentiful, and so $\text{H}\alpha$ emission often dominates the spectra of such nebulae. $\text{H}\alpha$ occurs when an electron in a hydrogen atom becomes excited and jumps from the third energy level to the second. The transition leads to the emission of a photon, which in the case of $\text{H}\alpha$ is red. (The process is actually more complicated than this, and evolves ionization and recombination, but the key issue here is that there is no perturbation by another electron).
Oxygen, however, emits light through a different process - the poorly named
forbidden transitions. Emission here is the result of the collision of a free electron with an electron in an atom of doubly ionized oxygen, denoted $[\text{O III}]$ (not a typo - there are indeed three $\text{I}$s). This collisional process happens more often at higher temperatures, as the mean velocity of electrons increases as temperatures increase. Therefore, hotter nebulae are more likely to be green than cooler nebulae with the same concentration of $[\text{O III}]$. Oxygen can then become strong in the nebula's spectrum, often almost as strong as $\text{H}\alpha$.
There are plenty of other emission lines (about 263, to be exact, for oxygen alone) that cause photons with green wavelengths ($500\text{ nm}<\lambda<565\text{ nm}$) to be emitted. However, this particular transition is preferred because of $[\text{O III}]$'s abundance in space and because of the high probability of this particular transition.
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@Mathphile I found no prime of the form $$n^{n+1}+(n+1)^{n+2}$$ for $n>392$ yet and neither a reason why the expression cannot be prime for odd n, although there are far more even cases without a known factor than odd cases.
@TheSimpliFire That´s what I´m thinking about, I had some "vague feeling" that there must be some elementary proof, so I decided to find it, and then I found it, it is really "too elementary", but I like surprises, if they´re good.
It is in fact difficult, I did not understand all the details either. But the ECM-method is analogue to the p-1-method which works well, then there is a factor p such that p-1 is smooth (has only small prime factors)
Brocard's problem is a problem in mathematics that asks to find integer values of n and m for whichn!+1=m2,{\displaystyle n!+1=m^{2},}where n! is the factorial. It was posed by Henri Brocard in a pair of articles in 1876 and 1885, and independently in 1913 by Srinivasa Ramanujan.== Brown numbers ==Pairs of the numbers (n, m) that solve Brocard's problem are called Brown numbers. There are only three known pairs of Brown numbers:(4,5), (5,11...
$\textbf{Corollary.}$ No solutions to Brocard's problem (with $n>10$) occur when $n$ that satisfies either \begin{equation}n!=[2\cdot 5^{2^k}-1\pmod{10^k}]^2-1\end{equation} or \begin{equation}n!=[2\cdot 16^{5^k}-1\pmod{10^k}]^2-1\end{equation} for a positive integer $k$. These are the OEIS sequences A224473 and A224474.
Proof: First, note that since $(10^k\pm1)^2-1\equiv((-1)^k\pm1)^2-1\equiv1\pm2(-1)^k\not\equiv0\pmod{11}$, $m\ne 10^k\pm1$ for $n>10$. If $k$ denotes the number of trailing zeros of $n!$, Legendre's formula implies that \begin{equation}k=\min\left\{\sum_{i=1}^\infty\left\lfloor\frac n{2^i}\right\rfloor,\sum_{i=1}^\infty\left\lfloor\frac n{5^i}\right\rfloor\right\}=\sum_{i=1}^\infty\left\lfloor\frac n{5^i}\right\rfloor\end{equation} where $\lfloor\cdot\rfloor$ denotes the floor function.
The upper limit can be replaced by $\lfloor\log_5n\rfloor$ since for $i>\lfloor\log_5n\rfloor$, $\left\lfloor\frac n{5^i}\right\rfloor=0$. An upper bound can be found using geometric series and the fact that $\lfloor x\rfloor\le x$: \begin{equation}k=\sum_{i=1}^{\lfloor\log_5n\rfloor}\left\lfloor\frac n{5^i}\right\rfloor\le\sum_{i=1}^{\lfloor\log_5n\rfloor}\frac n{5^i}=\frac n4\left(1-\frac1{5^{\lfloor\log_5n\rfloor}}\right)<\frac n4.\end{equation}
Thus $n!$ has $k$ zeroes for some $n\in(4k,\infty)$. Since $m=2\cdot5^{2^k}-1\pmod{10^k}$ and $2\cdot16^{5^k}-1\pmod{10^k}$ has at most $k$ digits, $m^2-1$ has only at most $2k$ digits by the conditions in the Corollary. The Corollary if $n!$ has more than $2k$ digits for $n>10$. From equation $(4)$, $n!$ has at least the same number of digits as $(4k)!$. Stirling's formula implies that \begin{equation}(4k)!>\frac{\sqrt{2\pi}\left(4k\right)^{4k+\frac{1}{2}}}{e^{4k}}\end{equation}
Since the number of digits of an integer $t$ is $1+\lfloor\log t\rfloor$ where $\log$ denotes the logarithm in base $10$, the number of digits of $n!$ is at least \begin{equation}1+\left\lfloor\log\left(\frac{\sqrt{2\pi}\left(4k\right)^{4k+\frac{1}{2}}}{e^{4k}}\right)\right\rfloor\ge\log\left(\frac{\sqrt{2\pi}\left(4k\right)^{4k+\frac{1}{2}}}{e^{4k}}\right).\end{equation}
Therefore it suffices to show that for $k\ge2$ (since $n>10$ and $k<n/4$), \begin{equation}\log\left(\frac{\sqrt{2\pi}\left(4k\right)^{4k+\frac{1}{2}}}{e^{4k}}\right)>2k\iff8\pi k\left(\frac{4k}e\right)^{8k}>10^{4k}\end{equation} which holds if and only if \begin{equation}\left(\frac{10}{\left(\frac{4k}e\right)}\right)^{4k}<8\pi k\iff k^2(8\pi k)^{\frac1{4k}}>\frac58e^2.\end{equation}
Now consider the function $f(x)=x^2(8\pi x)^{\frac1{4x}}$ over the domain $\Bbb R^+$, which is clearly positive there. Then after considerable algebra it is found that \begin{align*}f'(x)&=2x(8\pi x)^{\frac1{4x}}+\frac14(8\pi x)^{\frac1{4x}}(1-\ln(8\pi x))\\\implies f'(x)&=\frac{2f(x)}{x^2}\left(x-\frac18\ln(8\pi x)\right)>0\end{align*} for $x>0$ as $\min\{x-\frac18\ln(8\pi x)\}>0$ in the domain.
Thus $f$ is monotonically increasing in $(0,\infty)$, and since $2^2(8\pi\cdot2)^{\frac18}>\frac58e^2$, the inequality in equation $(8)$ holds. This means that the number of digits of $n!$ exceeds $2k$, proving the Corollary. $\square$
We get $n^n+3\equiv 0\pmod 4$ for odd $n$, so we can see from here that it is even (or, we could have used @TheSimpliFire's one-or-two-step method to derive this without any contradiction - which is better)
@TheSimpliFire Hey! with $4\pmod {10}$ and $0\pmod 4$ then this is the same as $10m_1+4$ and $4m_2$. If we set them equal to each other, we have that $5m_1=2(m_2-m_1)$ which means $m_1$ is even. We get $4\pmod {20}$ now :P
Yet again a conjecture!Motivated by Catalan's conjecture and a recent question of mine, I conjecture thatFor distinct, positive integers $a,b$, the only solution to this equation $$a^b-b^a=a+b\tag1$$ is $(a,b)=(2,5).$It is of anticipation that there will be much fewer solutions for incr...
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Wikipedia said
The description of most manifolds requires more than one chart (a single chart is adequate for only the simplest manifolds). A specific collection of charts which covers a manifold is called an atlas. An atlas is not unique as all manifolds can be covered multiple ways using different combinations of charts. Two atlases are said to be equivalent if their union is also an atlas.
I think it means: Let $X$ be a topological space and $\Phi_1=\{\phi_\alpha:U_{\alpha}\to\Bbb R^{n_\alpha}\mid\alpha\in A\}$, $\Phi_2=\{\phi_\beta:U_{\beta}\to\Bbb R^{n_\beta}\mid\beta\in B\}$ be two of its(i.e., $X$'s) topological atlas. Then $\Phi_1$ and $\Phi_2$ are
equivalent iff $\Phi_1\cup\Phi_2$ is also a topological atlas of $X$.
My question is: if $\Phi_1$ and $\Phi_2$ are atlases of $X$, which means they both can "cover" $X$, then it is definitely true that $\Phi_1\cup\Phi_2$ can "cover" $X$, isn't it? I can't see the meaning of such definition of equivalence. Where did I make the mistake?
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I've recently found an article (referred somewhere on this site) criticizing the use of common rules of algebra on infinite series. To be honest, the video referred is one of the videos of Numberphile I liked the most. I mean, informally, to say a rule doesn't hold, I think one should find an example (in modern logic, a $\forall$ statement is true by default, and a $\exists$ false); say, associativity of addition for infinite series:
$$ S_1=(1-1)+(1-1)+(1-1)+(1-1)\cdots=0\\ S_2=1+(-1+1)+(-1+1)+(-1+1)+\cdots=1\\ \therefore S_1\neq S_2 $$
But what inconsistency does:
$$ \begin{align} S=1&-1+1-1+\cdots\\ S+S=1&-1+1-1+\cdots+\\ &+1-1+1-1+\cdots=1\iff\\ \iff2S=1&\iff S=\frac12 \end{align} $$
create? I'm not even getting into Cesàro summation. Why does the limit of a sum have to equal the sum itself?
Why can't we have
$$ \frac12=\sum_{n=0}^\infty\ (-1)^n\neq \lim_{x\to\infty}\sum_{n=0}^x\ (-1)^n= \text{st}\sum_{n=0}^H\ (-1)^n= \text{undefined} $$
After all, in here $x$ is an arbitrarily big real
, $H$ is a positive infinite hyperinteger number and $\infty$ is number , not a number. Where is the inconsistency? NaN Edit:
There are already a lot of comments, and I feel I haven't made myself clear. Maybe the question is more philosophical than I thought. Here is an attempt to make my still developing points clearer:
$\cdots$ means the continuation to infinity of a series that continues the most simple pattern.
Example: $\displaystyle\sum_{n=0}^\infty\ (-1)^n$ means that for whatever number you have taken the partial sum, you are as far from the result as you were in the beginning. As by $6.$, such non-converging sum cannot be computed directly.
In the first example, it is proven that associativity does not hold for all infinite series, at least for divergent series, the same way $\sqrt a \sqrt b=\sqrt{ab}$ does not hold in $\mathbb C$. However, non-contradicting laws for associativity can be found:
Associativity may not work infinitely for numbers within the same series, as $S\neq S_1\neq S_2\neq S$ shows. However, it works pairwise between infinite series. $$ \begin{align} S+S=1&-1+1-1+\cdots+\\ &+1-1+1-1+\cdots \end{align} $$ is the same as $$ S+S=1+(-1+1)+(+1-1)+(-1+1)+\cdots $$ which is $1$, as we have seen. This rule is consistent. Other pairwise associations for this will either give the same, or $2-2+2-2+\cdots$, which is also $1$. In other words, for any numbers, associativity works. So it works pairwise (2 series, 2 numbers per application), an infinite number of times. But it doesn't work within the same series, as each set of numbers would be a finite series, infinity is not a number, so it can't have not-a-number of times per application. That is, $$ S_1=(1-1)+(1-1)+(1-1)+\cdots $$ is the same as $$ \begin{align} S_1=1+&1+1+\cdots\\ -1-&1-1+\cdots \end{align} $$ and not the same as $$ \begin{align} S=1-&1+\\ +1-&1+\\ +1-&1+\\ +\cdots \end{align} $$
The limit of a sum equals the sum when the sum converges. Example: $$ \sum_{n=0}^\infty 2^{-n}=\lim_{x\to\infty}\sum_{n=0}^x 2^{-n}=2 $$
The limit of a non-converging sum does not exist or is infinity, not a number. All sums have a value, even thought their limits might not have one, or the value of the sum is infinity.
If that is the case, infinity, as not a number, cannot be directly summed with another sum (eliminating problems as $\infty-\infty$ by reason of lack of information). If according to non-contradictory rules, a value can be assigned to a sum, that is the value of the sum. See the example above for $1-1+1-1+\cdots$
To distinguish between the values of two non-convergent sums, they first must be computed according to non-contradictory rules. Then their values can be compared, by transitivity of equality.
A divergent sum cannot be computed directly (the reason why $S\neq S_1\neq S_2\neq S$), as by definition of infinity, one cannot reach it. Again, use non-contradictory rules, making a finite number of changes that maintain the value of the divergent sum (see the examples' consistency).
Thank you for reading,
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Keywords
Quaternion matrix equations, Real representations, $\eta$-conjugates, $\eta$-conjugate transposes, Ordered units triple.
Abstract
For a given ordered units triple $\{q_1, q_2, q_3\}$, the solutions to the quaternion matrix equations $AX^{\star}-XB=C$ and $X-AX^{\star}B=C$, $X^{\star} \in \{ X , X^{\eta} , X^* , X^{\eta*}\}$, where $X^*$ is the conjugate transpose of $X$, $X^{\eta}=-\eta X \eta$ and $X^{\eta*}=-\eta X^* \eta$, $\eta \in \{q_1, q_2, q_3\}$, are discussed. Some new real representations of quaternion matrices are used, which enable one to convert $\eta$-conjugate (transpose) matrix equations into some real matrix equations. By using this idea, conditions for the existence and uniqueness of solutions to the above quaternion matrix equations are derived. Also, methods to construct the solutions from some related real matrix equations are presented.
Recommended Citation
Liu, Xin; Wang, QingWen; and Zhang, Yang.(2019),"Consistency of Quaternion Matrix Equations $AX^{\star}-XB=C$ and $X-AX^\star B=C$",
Electronic Journal of Linear Algebra,Volume 35, pp. 394-407. DOI: https://doi.org/10.13001/1081-3810.3950
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Now showing items 1-1 of 1
Production of charged pions, kaons and protons at large transverse momenta in pp and Pb-Pb collisions at $\sqrt{s_{NN}}$ = 2.76 TeV
(Elsevier, 2014-09)
Transverse momentum spectra of $\pi^{\pm}, K^{\pm}$ and $p(\bar{p})$ up to $p_T$ = 20 GeV/c at mid-rapidity, |y| $\le$ 0.8, in pp and Pb-Pb collisions at $\sqrt{s_{NN}}$ = 2.76 TeV have been measured using the ALICE detector ...
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The Canonical energy momentum tensor is given by $$T_{\mu\nu} = \frac{\partial {\cal L}}{\partial (\partial^\mu \phi_s)} \partial_\nu \phi_s - g_{\mu\nu} {\cal L}. $$ A priori, there is no reason to believe that the EM tensor above is symmetric. To symmetrize it we do the following trick.
To any EM tensor we can add the following term without changing its divergence and the conserved charges: $${\tilde T}_{\mu\nu} = T_{\mu\nu} + \partial^\beta \chi_{\beta\mu\nu}, $$ where $\chi_{\beta\mu\nu} = - \chi_{\mu\beta\nu}$. The antisymmetry of $\chi$ in its $\mu\beta$ indices implies that ${\tilde T}_{\mu\nu}$ is conserved. Also, all the conserved charges stay the same.
Now even though $T_{\mu\nu}$ is not a symmetric tensor, it is possible to choose $\chi_{\beta\mu\nu}$ in such a way so as to make ${\tilde T}_{\mu\nu}$ symmetric. It can be shown that choosing
$$\chi_{\lambda\mu\nu} = - \frac{i}{2}\left[ \frac{\delta {\cal L}}{\delta (\partial^\mu \phi_r) } (I_{\nu\lambda})_{rs} \phi_s + \frac{\delta {\cal L}}{\delta (\partial^\lambda \phi_r) } (I_{\mu\nu})_{rs} \phi_s + \frac{\delta {\cal L} }{\delta (\partial^\nu \phi_r) } (I_{\mu\lambda})_{rs} \phi_s \right]$$ makes the new EM tensor symmetric. Here $(I_{\mu\nu})_{rs}$ is the representation of the Lorentz Algebra under which the fields $\phi_s$ transform.
Here's my question - Is it possible to obtain the symmetric EM tensor directly from variational principles by adding a total derivative term to the Lagrangian. In other words, by shifting ${\cal L} \to {\cal L} + \partial_\mu X^\mu$, and choosing $X^\mu$ appropriately, can we exactly get the shift in the EM tensor required, in order to make the canonical EM tensor symmetric? What I've done so far - It is possible to show that under a shift in the Lagrangian by a total derivative, one shifts the EM tensor by $T_{\mu\nu} \to T_{\mu\nu} + \partial^\lambda \chi_{\lambda\mu\nu}$ where
$$\chi_{\lambda\mu\nu} = \frac{1}{2} \frac{\delta X_\lambda}{\delta (\partial^\mu \phi_r)} \partial_\nu \phi_r - \frac{1}{2} \frac{ \delta X_\mu }{\delta (\partial_\lambda \phi_r)} \partial_\nu \phi_r + X_\mu g_{\lambda\nu} - X_\lambda g_{\mu\nu} \,. $$
What I wish to do next - I now have a differential equation that I wish to solve:
\begin{align} &\frac{1}{2} \frac{\delta X_\lambda}{\delta (\partial^\mu \phi_r)} \partial_\nu \phi_r - \frac{1}{2} \frac{ \delta X_\mu }{\delta (\partial_\lambda \phi_r)} \partial_\nu \phi_r + X_\mu g_{\lambda\nu} - X_\lambda g_{\mu\nu} \\ &~~~~~~= - \frac{i}{2}\left[ \frac{\delta {\cal L}}{\delta (\partial^\mu \phi_r) } (I_{\nu\lambda})_{rs} \phi_s + \frac{\delta {\cal L}}{\delta (\partial^\lambda \phi_r) } (I_{\mu\nu})_{rs} \phi_s + \frac{\delta {\cal L} }{\delta (\partial^\nu \phi_r) } (I_{\mu\lambda})_{rs} \phi_s \right] \,. \end{align}
Any ideas on how to solve this?
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Analytic operator at a point $x_0$
An operator $A$, acting from one Banach space into another, that admits a representation of the form \begin{equation} A(x_0+h) - Ax_0 = \sum_{k=1}^{\infty}C_kh^k, \end{equation} where $C_k$ is a form of degree $k$ and the series converges uniformly in some ball $\|h\|<r$. An operator is called analytic in a domain $G$ if it is an analytic operator at all points of this domain. An analytic operator is infinitely differentiable. In the case of complex spaces, analyticity of an operator in a domain is a consequence of its differentiability (according to Gâteaux) at each point of this domain. Examples of analytic operators are Lyapunov's integro-power series, and the Hammerstein and Urysohn operators with smooth kernels on the space $\mathcal C$ of continuous functions.
References
[1] E. Hille, R.S. Phillips, "Functional analysis and semi-groups" , Amer. Math. Soc. (1957) [2] M.A. Krasnosel'skii, G.M. Vainikko, P.P. Zabreiko, et al., "Approximate solution of operator equations" , Wolters-Noordhoff (1972) (Translated from Russian) How to Cite This Entry:
Analytic operator.
Encyclopedia of Mathematics.URL: http://www.encyclopediaofmath.org/index.php?title=Analytic_operator&oldid=29225
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In mathematics, logarithmic functions is an inverse function to exponentiation. The logarithmic function is defined as
For x > 0 , a > 0, and a\(\neq\)1,
y= log
a x if and only if x = a y
Then the function is given by
f(x) = log
a x
The base of the logarithm is a. This can be read it as log base a of x. The most 2 common bases used in logarithmic functions are base 10 and base e.
Common Logarithmic Function: The logarithmic function with base 10 is called the common logarithmic function and it is denoted by log 10or simply log.
f(x) = log
10 x Natural Logarithmic Function :The logarithmic function to the base e is called the natural logarithmic function and it is denoted by log e .
f(x) = log
e x Logarithmic Functions Properties
Logarithmic Functions have some of the properties that allows you to simplify the logarithms when the input is in the form of product, quotient or the value taken to the power. Some of the properties are listed below.
Product Rule: log bMN = log bM + log bN
Multiply two numbers with the same base, then add the exponents.
Example : log 30 + log 2 = log 60
Quotient Rule :log bM/N = log bM – log bN
Divide two numbers with the same base, subtract the exponents.
Example : log
8 56 – log 8 7 = log 8(56/7)=log 88 = 1 Power Rule :Raise an exponential expression to a power and multiply the exponents.
Log
b M p = P log b M
Example : log 100
3 = 3. Log 100 = 3 x 2 = 6 Zero Exponent Rule :log a1 = 0. Change of Base Rule :log b(x) = ln x / ln b or log b(x) = log 10x / log 10b Log bb = 1 Example : log 1010 = 1 Log bb x= x Example : log 1010 x= x \(b^{\log _{b}x}=x\) . Substitute y= log bx , it becomes b y= x
There are also some of the logarithmic function with fractions. It has a useful property to find the log of a fraction by applying the identities
ln(ab)= ln(a)+ln(b) ln(a x) = x ln (a)
We also can have logarithmic function with fractional base.
Consider an example,\(3\log _{\frac{4}{9}}\sqrt[4]{\frac{27}{8}}=\frac{3}{4}\log _{\frac{4}{9}}\frac{27}{8}\)
By the definition, log
a b = y becomes a y = b
(4/9)
y = 27/8
(2
2/3 2) y = 3 3 / 2 3
(⅔)
2y = (3/2) 3 Sample Example
Here you are provided with some logarithmic functions example.
Question 1 :
Use the properties of logarithms to write as a single logarithm for the given equation: 5 log
9 x + 7 log 9 y – 3 log 9 z Solution :
By using the power rule , Log
b M p = P log b M, we can write the given equation as
5 log
9 x + 7 log 9 y – 3 log 9 z = log 9 x 5 + log 9 y 7 – log 9 z 3
From product rule, log
b MN = log b M + log b N
5 log
9 x + 7 log 9 y – 3 log 9 z = log 9 x 5y 7 – log 9 z 3
From Quotient rule, log
b M/N = log b M – log b N
5 log
9 x + 7 log 9 y – 3 log 9 z = log 9 (x 5y 7 / z 3 )
Therefore, the single logarithm is 5 log
9 x + 7 log 9 y – 3 log 9 z = log 9 (x 5y 7 / z 3 ) Question 2 :
Use the properties of logarithms to write as a single logarithm for the given equation: 1/2 log
2 x – 8 log 2 y – 5 log 2 z Solution :
By using the power rule , Log
b M p = P log b M, we can write the given equation as
1/2 log
2 x – 8 log 2 y – 5 log 2 z = log 2 x 1/2 – log 2 y 8 – log 2 z 5
From product rule, log
b MN = log b M + log b N
Take minus ‘- ‘ as common
1/2 log
2 x – 8 log 2 y – 5 log 2 z = log 2 x 1/2 – log 2 y 8z 5
From Quotient rule, log
b M/N = log b M – log b N
1/2 log
2 x – 8 log 2 y – 5 log 2 z = log 2 (x 1/2 / y 8z 5 )
The solution is
1/2 log
2 x – 8 log 2 y – 5 log 2 z = \(\log _{2}\left ( \frac{\sqrt{x}}{y^{8}z^{5}} \right )\)
For more related articles on logarithmic function and its properties, register with BYJU’S – The Learning app and watch interactive videos.
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I know that if something is thrown up with a velocity
v, it comes down with the same velocity as long as its motion faces no resistance. So is it possible for me to fire a bullet and be killed if it drops down on me directly? Assuming there is no air resistance.
I know that if something is thrown up with a velocity
Yes the bullets can fall down and injure or kill you. In fact in countries were celebratory gun firing is possible people are often injured by falling projectiles.
Shooting straight up is less dangerous than at an angle because the terminal velocity is much lower than the muzzle velocity of the projectile.
When shooting at an angle some of the horizontal motion tends to be conserved making the bullets velocity when it falls back substantially higher.
This article covers this in more detail
Now how fast would the bullet hit you when coming straigt down? Assuming it reached terminal velocity, its speed would depend on its mass and surface area.
$v_{terminal}=\sqrt{2mg/C \rho A}$
The density and drag coefficient of air are assumed to be constant at 1.29 $kg/m^3$ and 0.5 resepectively. If the bullet weighs 30 grams* and has a classic caliber of 9 mm, then the terminal velocity would be
$v_{terminal}=\sqrt{2*0.03*9.18/0.5*1.29* 0.009^2\pi}$
$=58.9 ms^{-1}$
Even a weak a very standard rifle has a muzzle velocity twice that value as seen here:
*averaged mass of bullets as given by the source: http://hypertextbook.com/facts/2000/ShantayArmstrong.shtml
There is always air resistance, unless you are in space. If there was no air resistance, a bullet would land at the same speed it was shot at.
On Earth, a spent bullet is rarely lethal. In big wars (like World War II) it was common for soldiers to get hit by spent bullets. Normally they will hit you and fall away, causing just a bruise,
An unlucky hit could kill a person. For example, if a falling bullet hit a person directly in their head and it was a hard jacketed bullet and the point happened to contact the skull first, it might penetrate and kill. Usually this will not happen, either because it will hit at an angle, or the bullet (which tumbles) makes contact on its side, not the point...
protected by Qmechanic♦ Apr 14 '16 at 13:42
Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).
Would you like to answer one of these unanswered questions instead?
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1. We can think about what $A_{5}$ actually is, that is, all even permutations of 5 symbols. If you think about it, what element in $A_{5}$ could possibly have order 60? For example if we look at something like the permutation $(1 2 3)$ which can be seen as $(13)(12)$ what is the order of this? Knowing that the order of disjoint cycles is the least common multiple of each cycle, how could we possibly get a permutation of order 60? And is this possible?
2. $|G| = 7$, then how is it possible to have elements of order either 14 or 10? There is a known fact that $\forall a \in G$ it is necessary that $ord(a)| |G|$ (Lagrange's theorem), so since 7 is prime, the only possibilities here are 1 and 7.
3. No, we can't make a homomorphism. First of all this $G$ is a cyclic group because its of prime order, and last time I checked $A_{5}$ is not even abelian. So if we have a homomorphism $\varphi: G \to A_{4}$ if we let $a,b \in G$ we'll see that $ab = ba$ (since $G$ is cyclic). Now then since $\varphi$ is a homomorphism we have $\varphi(ab) = \varphi(a)\varphi(b)$ but we also have $\varphi(ba) = \varphi(b) \varphi(a)$, this implies that $\varphi(a)\varphi(b) = \varphi(b)\varphi(a)$ but since $\varphi(a),\varphi(b)\in A_{4}$ and $A_{4}$ is not abelian then $\varphi$ cannot possibly be a homomorphism.
I hope this answers some of your questions. You may want to find or justify the facts I used for some of these answers because it's certainly the type of common knowledge you would need for a group theory midterm.
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"half sliced unit disk"
Can somebody tell me how to map this conformally to the upper half plane? I think the symmetry principle should be applied here but stuck on that for hours. Pardon my hasty sketch, this is a unit disk cut at every 45 degree.
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"half sliced unit disk"
Can somebody tell me how to map this conformally to the upper half plane? I think the symmetry principle should be applied here but stuck on that for hours. Pardon my hasty sketch, this is a unit disk cut at every 45 degree.
It is easier to map onto the unit disk, because it has the same sort of symmetry as this domain. (Then of course you can map it onto halfplane.) So, the goal is to somehow "retract" these spikes into the boundary of the disk.
Take one of these eight pieces, say the sector $0<\theta<\pi/4$. This means we make additional cuts along $re^{i\theta }$ with $0\le r<1/2$, $\theta=0,\pi/4$. Mark them with dashed lines. The sector can be mapped on upper halfdisk by $z\mapsto z^4$, and then onto lower halfplane by $z\mapsto \left(z+\frac{1}{z}\right)$ (Joukowski map). In the process, the dashed lines turned into $(-\infty, a)$ and $(a,\infty)$ where $a$ is some positive number that you can find yourself.
Apply $z\mapsto 2z/a$, so that the dashed line turn to $(-\infty,-2)$ and $(2,\infty)$. Then apply the
inverse of the Joukowki map, thus mapping the lower halfplane back to upper half-disk. The dashed lines become radii $(-1,0)$ and $(0,1)$. Finally, apply $z\mapsto z^{1/4}$ to return to the sector of the opening angle $\pi/4$. The only thing we achieved so far was to map all solid lines to the circular boundary, leaving the radii dashed. Finally, use reflection across dashed lines, in effect erasing them. This means that, without any changes in the formula, you now have a map of the entire domain onto the unit disk.
Did you look at the Cayley Transform,
$$f(z) = \frac{z- i}{z + i}$$
which maps (conformally) the upper half-plane onto the unit disk?
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Agglomerative clustering methods that rely on a multiple sequence alignment and a matrix of pairwise distances can be computationally infeasible for large DNA and amino acid datasets. Alternative k-mer based clustering methods involve enumerating all
k-letter words in a sequence through a sliding window of length k. The \(n \times 4^k\) matrix of k-mer counts (where \(n\) is the number of sequences) can then be used in place of a multiple sequence alignment to calculate distances and/or build a phylogenetic tree. kmer is an R package for clustering large sequence datasets using fast alignment-free k-mer counting. This can be achieved with or without a multiple sequence alignment, and with or without a matrix of pairwise distances. These functions are detailed below with examples of their utility.
The function
kcount is used to enumerate all
k-mers within a sequence or set of sequences, by sliding a window of length k along each sequence and counting the number of times each k-mer appears (for example, the \(4^3 = 64\) possible DNA 3-mers: AAA, AAC, AAG, …, TTT). The
kdistance function can then compute an alignment-free distance matrix, using a matrix of
k-mer counts to derive the pairwise distances. The default distance metric used by
kdistance is the
k-mer ( k-tuple) distance measure outlined in Edgar (2004). For two DNA sequences \(a\) and \(b\), the fractional common k-mer count over the \(4^k\) possible words of length \(k\) is calculated as: \[F = \sum\limits_{\tau}\frac{min (n_a(\tau), n_b (\tau))}{min (L_a , L_b ) - k + 1} \tag{1}\]
where \(\tau\) represents each possible
k-mer, \(n_a(\tau)\) and \(n_b(\tau)\) are the number of times \(\tau\) appears in each sequence, \(k\) is the k-mer length and \(L\) is the sequence length. The pairwise distance between \(a\) and \(b\) is then calculated as:
\[d = \frac{log(0.1 + F) - log(1.1)}{log(0.1)} \tag{2}\]
For \(n\) sequences, the
kdistance operation has time and memory complexity \(O(n^2)\) and thus can become computationally infeasible when the sequence set is large (e.g. > 10,000 sequences). As such, the
kmer package also offers the function
mbed, that only computes the distances from each sequence to a smaller (or equal) sized subset of ‘seed’ sequences (Blackshields
et al., 2010). The default behavior of the
mbed function is to select \(t = (log_2n)^2\) seeds by clustering the sequences (
k-means algorithm with \(k = t\)), and selecting one representative sequence from each cluster.
DNA and amino acid sequences can be passed to
kcount,
kdistance and
mbed either as a list of non-aligned sequences or a matrix of aligned sequences, preferably in either the “DNAbin” or “AAbin” raw-byte format (see the
ape package documentation for more information on these S3 classes). Character sequences are supported; however ambiguity codes may not be recognized or treated appropriately, since raw ambiguities are counted according to their underlying residue frequencies (e.g. the 5-mer “ACRGT” would contribute 0.5 to the tally for “ACAGT” and 0.5 to that of “ACGGT”). This excludes the ambiguity code “N”, which is ignored.
The
ape R package (Paradis et al., 2004) contains a dataset of 15 aligned mitochondrial cytochrome b gene DNA sequences from the woodmouse Apodemus sylvaticus, originally published in Michaux et al. (2003). While the kmer distance functions do not require sequences to be aligned, this example will enable us to compare the performance of the k-mer distances with the alignment-dependent distances produced by
ape::dist.dna. First, load the dataset and view the first few rows and columns as follows:
data(woodmouse, package = "ape")ape::as.character.DNAbin(woodmouse[1:5, 1:5])#> [,1] [,2] [,3] [,4] [,5]#> No305 "n" "t" "t" "c" "g" #> No304 "a" "t" "t" "c" "g" #> No306 "a" "t" "t" "c" "g" #> No0906S "a" "t" "t" "c" "g" #> No0908S "a" "t" "t" "c" "g"
This is a semi-global (‘glocal’) alignment featuring some incomplete sequences, with unknown characters represented by the ambiguity code “n” (e.g. No305). To avoid artificially inflating the distances between these partial sequences and the others, we first trim the gappy ends by subsetting the global alignment (note that the
ape function
dist.dna also removes columns with ambiguity codes prior to distance computation by default).
woodmouse <- woodmouse[, apply(woodmouse, 2, function(v) !any(v == 0xf0))]
The following code first computes the full \(n \times n\) distance matrix, and then the embedded distances of each sequence to three randomly selected seed sequences. In both cases the
k-mer size is set to 6.
### Compute the full distance matrix and print the first few rows and columnslibrary(kmer)woodmouse.kdist <- kdistance(woodmouse, k = 6)print(as.matrix(woodmouse.kdist)[1:7, 1:7], digits = 2)#> No305 No304 No306 No0906S No0908S No0909S No0910S#> No305 0.000 0.0322 0.0295 0.033 0.036 0.037 0.037#> No304 0.032 0.0000 0.0051 0.020 0.022 0.032 0.023#> No306 0.030 0.0051 0.0000 0.016 0.017 0.026 0.018#> No0906S 0.033 0.0202 0.0162 0.000 0.024 0.033 0.014#> No0908S 0.036 0.0224 0.0171 0.024 0.000 0.033 0.025#> No0909S 0.037 0.0322 0.0264 0.033 0.033 0.000 0.034#> No0910S 0.037 0.0233 0.0176 0.014 0.025 0.034 0.000### Compute and print the embedded distance matrixsuppressWarnings(RNGversion("3.5.0"))set.seed(999)seeds <- sample(1:15, size = 3)woodmouse.mbed <- mbed(woodmouse, seeds = seeds, k = 6)#> Registered S3 method overwritten by 'openssl':#> method from#> print.bytes Rcppprint(woodmouse.mbed[,], digits = 2)#> No0909S No0913S No304#> No305 0.0368 0.0391 0.0322#> No304 0.0322 0.0102 0.0000#> No306 0.0264 0.0098 0.0051#> No0906S 0.0332 0.0215 0.0202#> No0908S 0.0332 0.0273 0.0224#> No0909S 0.0000 0.0368 0.0322#> No0910S 0.0341 0.0176 0.0233#> No0912S 0.0242 0.0322 0.0273#> No0913S 0.0368 0.0000 0.0102#> No1103S 0.0171 0.0251 0.0202#> No1007S 0.0046 0.0368 0.0322#> No1114S 0.0451 0.0428 0.0373#> No1202S 0.0345 0.0176 0.0233#> No1206S 0.0304 0.0251 0.0202#> No1208S 0.0046 0.0409 0.0359
In this example the alignment-free
k-mer distances calculated in Example 1 are compared with the Kimura (1980) distance metric as featured in the ape package examples. The resulting neighbor-joining trees are visualized using the
tanglegram function from the
dendextend package.
## compute pairwise distance matricesdist1 <- ape::dist.dna(woodmouse, model = "K80") dist2 <- kdistance(woodmouse, k = 7) ## build neighbor-joining treesphy1 <- ape::nj(dist1)phy2 <- ape::nj(dist2)## rearrange trees in ladderized fashionphy1 <- ape::ladderize(phy1)phy2 <- ape::ladderize(phy2)## convert phylo objects to dendrogramsdnd1 <- as.dendrogram(phy1)dnd2 <- as.dendrogram(phy2)## plot the tanglegramdndlist <- dendextend::dendlist(dnd1, dnd2)dendextend::tanglegram(dndlist, fast = TRUE, margin_inner = 5)
Figure 1: Tanglegram comparing distance measures for the woodmouse sequences. Neighbor-joining trees derived from the alignment-dependent (left) and alignment-free (right) distances show congruent topologies.
To avoid excessive time and memory use when building large trees (e.g.
n > 10,000), the kmer package features the function
cluster for fast divisive clustering, free of both alignment and distance matrix computation. This function first generates a matrix of
k-mer counts, and then recursively partitions the matrix row-wise using successive k-means clustering ( k = 2). While this method may not necessarily reconstruct sufficiently accurate phylogenetic trees for taxonomic purposes, it offers a fast and efficient means of producing large trees for a variety of other applications such as tree-based sequence weighting (e.g. Gerstein et al. (1994)), guide trees for progressive multiple sequence alignment (e.g. Sievers et al. (2011)), and other recursive operations such as classification and regression tree (CART) learning.
The package also features the function
otu for rapid clustering of sequences into operational taxonomic units based on a genetic distance (k-mer distance) threshold. This function performs a similar operation to
cluster in that it recursively partitions a k-mer count matrix to assign sequences to groups. However, the top-down splitting only continues while the highest k-mer distance within each cluster is above a defined threshold value. Rather than returning a dendrogram,
otu returns a named integer vector of cluster membership, with asterisks indicating the representative sequences within each cluster.
In this final example, the woodmouse dataset is clustered into operational taxonomic units (OTUs) with a maximum within-cluster
k-mer distance of 0.03 and with 20 random starts per k-means split (recommended for improved accuracy).
suppressWarnings(RNGversion("3.5.0"))set.seed(999)woodmouse.OTUs <- otu(woodmouse, k = 5, threshold = 0.97, method = "farthest", nstart = 20)woodmouse.OTUs#> No305* No304 No306* No0906S No0908S No0909S* No0910S No0912S #> 3 1 1 1 1 2 1 2 #> No0913S No1103S No1007S No1114S No1202S No1206S No1208S #> 1 2 2 3 1 1 2
The function outputs a named integer vector of OTU membership, with asterisks indicating the representative sequence from each cluster (i.e. the most “central” sequence). In this case, three distinct OTUs were found, with No305 and N01114S forming one cluster (3), No0909S, No0912S, No1103S, No1007S and No1208S forming another (2) and the remainder belonging to cluster 1 in concordance with the consensus topology of Figure 1.
The
kmer package is released under the GPL-3 license. Please direct bug reports to the GitHub issues page at http://github.com/shaunpwilkinson/kmer/issues. Any feedback is greatly appreciated.
This software was developed with funding from a Rutherford Foundation Postdoctoral Research Fellowship from the Royal Society of New Zealand.
Blackshields,G.
et al. (2010) Sequence embedding for fast construction of guide trees for multiple sequence alignment. Algorithms for Molecular Biology, 5, 21.
Edgar,R.C. (2004) Local homology recognition and distance measures in linear time using compressed amino acid alphabets.
Nucleic Acids Research, 32, 380–385.
Gerstein,M.
et al. (1994) Volume changes in protein evolution. Journal of Molecular Biology, 236, 1067–1078.
Kimura,M. (1980) A simple method for estimating evolutionary rates of base substitutions through comparative studies of nucleotide sequences.
Journal of Molecular Evolution, 16, 111–120.
Michaux,J.R.
et al. (2003) Mitochondrial phylogeography of the woodmouse (Apodemus sylvaticus) in the Western Palearctic region. Molecular Ecology, 12, 685–697.
Paradis,E.
et al. (2004) APE: analyses of phylogenetics and evolution in R language. Bioinformatics, 20, 289–290.
Sievers,F.
et al. (2011) Fast, scalable generation of high-quality protein multiple sequence alignments using Clustal Omega. Molecular Systems Biology, 7, 539.
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Given an ellipse with foci $F_1, F_2$ and a point $P$. Let $T_1, T_2$ the points of tangency on the ellipse determined by the tangent lines through $P$. Show that $\widehat {T_1 P F_1} = \widehat {T_2 P F_2}$.
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Let the ellipse be: $$\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$$ Set $P(a,t)$, the tangent line $PT_1$ is $y=k(x-a)+t$, $k=\tan A_1$, since it is tangent line, so put in ellipse equation, we have: $$ b^2x^2+a^2(kx-ka+t)^2=a^2b^2$$ i.e., $$(b^2+a^2k^2)x^2+2a^2k(t-ka)x+a^2(t-ka)^2-a^2b^2=0$$ Solving for the point of tangency, the discriminant $\Delta$ must be zero: $$\Delta$=$4a^4k^2(t-ka)^2-4(b^2+a^2k^2)[a^2(t-ka)^2-a^2b^2] $$ At $\Delta=0$, we get $ 2kta-t^2+b^2=0$, that is, $k=\dfrac{t^2-b^2}{2ta}$.
Let $k_1$ be $PF_1$'s slope, $k_1=\dfrac{t}{a+c}=\tan A_2$, the $ \angle T_1PF_1=|A_1-A_2|$, now we calculate $\tan(A_1-A_2)$:
$$\tan(A_1-A_2)=\dfrac{\tan A_1-\tan A_2}{1+\tan A_1 \tan A_2}=\dfrac{k-k_1}{1+k k_1}=\dfrac{\dfrac{t^2-b^2}{2ta}-\dfrac{t}{a+c}}{1+\dfrac{t^2-b^2}{2ta}\dfrac{t}{a+c}}=\dfrac{(t^2-b^2)(a+c)-2at^2}{2at^2+2act+t^3-tb^2}$$
Since $b^2=a^2-c^2$, we get:
$$ RHS=\dfrac{(t^2-a^2+c^2)(a+c)-2at^2}{t(2a^2+2ac+t^2-a^2+c^2)}=\dfrac{c-a}{t}$$
Let $k_2$ be line $PF_2$'s slope, then $k_2=\dfrac{t}{a-c}=\tan\angle PF_2T_2$. Since $\angle T_2PF_2=\dfrac{\pi}{2}-\angle PF_2T_2$, so $\tan \angle T_2PF_2=\dfrac{1}{\tan\angle PF_2T_2}=\dfrac{a-c}{t}=\tan(A_2-A_1)$. Since the angles are acute, we have:
$\angle T_2PF_2=A_2-A_1=\angle T_1PF_1$
From Geometry Expressions
Equal by inspection.
Let $F_1'$ be the image of $F_1$ under reflection in the tangent at $T_1$. Since the tangent at $T_1$ is the external bisector of $\angle F_1T_1F_2$, the points $F_2T_1F_1'$ are collinear. Likewise, let $F_1''$ be the image of $F_1$ under reflection in the tangent at $T_2$; then $F_2T_2F_1''$ are collinear.
Now,
By SSS, $\triangle F_2PF_1' \cong \triangle F_2PF_1''$. In particular, $$ \angle F_2PF_1' = \angle F_2PF_1'' = \tfrac12\angle F_1'PF_1'' \tag{1} $$
Next, by construction, reflecting in the tangent at $T_1$ and then in the tangent at $T_2$ sends $F_1'$ to $F_1$ and then to $F_1''$. But the composition of reflections in two lines is a rotation around their point of intersection, by twice the (directed) angle between the lines. Thus $F_1''$ is the image of $F_1'$ under a rotation around $P$ by $2\angle T_1PT_2$; in particular, $$ \angle F_1'PF_1'' = 2\angle T_1PT_2 \tag{2} $$
Combining (1) and (2) yields $$ \angle T_1PT_2 = \angle F_1'PF_2 $$ Subtracting $\angle T_1PF_2$ from both sides yields $$ \angle F_2PT_2 = \angle F_1'PT_1 $$ Finally, by reflection, $\angle F_1'PT_1 = \angle F_1PT_1$, so $$ \angle F_2PT_2 = \angle F_1PT_1 $$ as desired.
Incidentally, from $\triangle F_2PF_1' \cong \triangle F_2PF_1''$ it also follows that, letting $Q$ be the intersection of $F_2T_1$ and $F_1T_2$, the triangle $F_1T_1Q$ and $F_2T_2Q$ have a common excircle, whose centre is $P$.
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According to the first and second law for a closed system containing different chemicals we have
\begin{align} &\delta Q - \delta W = dU = T dS - p dV +\sum_i \mu_i d N_i\\ &\Rightarrow\;\delta W - p dV + \sum_i \mu_i dN_i = \delta Q - T d S \le 0\qquad\because\text{2nd law}\\ &\Rightarrow\;\delta W - p dV + dG\bigr|_{p,T} \le 0\\ &\Rightarrow\;\delta W - p dV \le -dG\bigr|_{p,T}\\ \end{align}
If $\delta W = p dV$ then $dG\bigr|_{p,T}\le 0$, that is, the condition $dG\bigr|_{p,T}\le 0$ coincides with the second law only if the only work done by the system is the pressure work and no other kind.
In addition, if $\delta W = p dV + \text{``other works"}$, then $\text{``other works"}\le -dG\bigr|_{p,T}$. This means the change in the "minus Gibbs function" is the maximum work attainable from the system beside the pressure work. This extra work can be positive or negative, in the form of electric work, friction work etc.
Therefore, it is clear that:
A system cannot at the same be derived by $dG\bigr|_{p,T}\le 0$ and does e.g. an electric work $\delta W_{Electric}$;
If the system does have a $\delta W_{Electric}$ then its maximum value would be equal to $dG\bigr|_{p,T}$ but if so, then the process is already assumed reversible and all the inequalities should be substituted by equalities.
However, according to this Mc Graw-Hill link during a spontaneous reduction-oxidation chemical reaction the Gibbs Free Enthalpy must decrease and at the same time the change in the Gibbs Free Enthalpy is the maximum
electric work that the reaction can do:$$\Delta G = W_{max}\le 0$$ I cannot understand this and have a number of problems with this derivation:
First of all, the maximum work that a system can do on its surrounding equals minus the change of the Gibbs Free Enthalpy, so the equality $\Delta G = W_{max}$ doesn't hold?
If the work has attained its maximum value, then the process must be assumed as reversible, but the inequality in the formula above holds for irreversible processes!
If there is an Electric work done by the system then decreasing Gibbs free energy is no longer necessary due to the second law?
When both the donor and acceptor of electron in the chemical reaction lay inside the system the electric work will nowhere enter the formulation as the electric work is an internal work that does not crosses the boundary of the system!? But the whole formulation is here to study $W_{electric}$ in the system as spontaneity of the reaction should be related to the electromotive force of the reaction which has a chance to appear in $W_{electric}$ only! So what should be taken as system here?
Hint. In the books on thermodynamics that I have seen that discuss the electric works they are usually dealing with the problem of Electrochemical cells. But electrochemical cells work in an outer electric circuit and so if one assume the cell as the closed system yet the electric work will enter the discussion. Only one book was talking about open-circuit Emf but again I have the problems listed above with that as well.
[
I have asked the same question at the Chemistry.SE here but have not been convinced with the single answer given there.]
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Hi I'm having trouble with this homework question on wave mechanics. The question is as follows:
"State the boundary condition which must be met at a point where the string of question $2$ is fixed.
Hence find the real standing wave solutions to the wave equation, and determine the allowed oscillation frequencies, when such a string of length $L$ is fixed at its ends.
If the bottom string of a guitar has a mass of $5.4 g m^{-1}$ and its length is determined by the distance $0.648 m$ from the bridge to the nut, find the tension required to tune the string to the note known as $E_2$ (a frequency of $82.4 Hz$)."
Question 2 describes a wave on a guitar string with the wave equation:
$\frac {\partial^2 \psi}{\partial t^2}=\frac WM \frac {\partial^2 \psi}{\partial x^2}$
Where $W$ is the tension, $M$ is the mass per unit length and $\psi$ is the displacement. It has been shown that the velocity of the wave is equal to $\sqrt{W/M}$
My attempt:
So for the first part I stated that $\psi$ must be $0$ at the points where the sting is fixed as it has no displacement.
Next I stated that the real standing wave solution is given as follows: $\psi=\psi_0cos(kx+\phi_x)cos(\omega t+\phi_t)$.
For the allowed oscillation frequencies I did the following:
$\psi=0$ at $x=0$ and $x=L$
Hence, $cos(kx+\phi_t)=0$ at these two points.
Setting $x=0$ yields that $\phi_x=\pi/2$
Setting $x=L$ gives $k=n\pi/L$
Since $k$ is the wavenumber $\lambda=2\pi/k$ so $\lambda=2L/n$
$v=f\lambda$ gives the allowed oscillation frequencies as $f= \frac{n}{2L}\sqrt{W/M}$.
When I come to try to calculate the tension however I rearrange f above to get: $W=\frac{4L^2f^2}{n^2}M$ and I've been given all the numbers to calculate $W$ except I don't know what to pick for the value of $n$. If I had to guess I would say $n=2$ just because in the question it asks for $E_2$ but I have no clue.
Any explanation or help would be very much appreciated.
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I have a problem with this exercise because I really don't know how to proceed. It's related with the "S-matrix".
In class we saw this example:
Consider the spherically symmetric potential:
$$V(r)=\begin{cases} -V_{0}, & 0\leq r\leq a\\ 0, & r>a \end{cases} $$
We know that for $r>a$ and $l=0$ we have
$$U_>''(r)+k^{2}U_>(r)=0 \qquad (1)$$
where $U(r):=rR(r)$, and $R(r)$ satisfies the radial part of the Schrödinger equation in spherical coordinates (for $l=0$):
$$-\frac{\hbar^{2}}{2m}\frac{1}{r^{2}}\frac{d}{dr}\left(r^{2}\frac{dR}{dr}\right)+\left[V(r)+\frac{\hbar^{2}l(l+1)}{2mr^{2}}\right]R=ER$$
In this case ($l=0$), from (1) we have the solution (for $r>a$)
$$U_>(r)=A e^{ikr}+Be^{-ikr} \qquad (2)$$
Or
$$U_{>}(r)=\frac{i}{2}\left[e^{-ikr}-S(k)e^{ikr}\right] \qquad (3)$$
Since we saw this really quick in class, I have 3
questions: 1 How do I manipulate (2) in order to obtain (3)?
After writing this, my teacher said the classical "it can be shown that:"
$$S(k)=\frac{-ik\sin qa-q\cos qa}{ik\sin qa-q\cos qa}e^{-2ika} \qquad (4)$$
then, my second question is:
2 How do I obtain (4)? I've been searching for this topic in my books (Griffiths, Schaum's Outlines) but I don't find anything related. Our reacher said that the function $S(k)$ is related with the "S-Matrix". 3 Could you suggest me a book where I can study this, please?
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If $Y_n$ is limited in probability and $X_n \to 0$ in probability then $X_nY_n \to 0$ in probability.
Attempt: I would go like:
for every $\epsilon$, $\mathbb P(|X_nY_n|\leq \epsilon) = \mathbb P(|Y_n|\leq \frac{ \epsilon}{ |X_n|})\geq\mathbb P(|Y_n|\leq C_{\delta}) \geq 1-\delta \to 1$ but the truth is that $\frac{ \epsilon}{ |X_n|}\geq C_{\delta}$ is not always true but almost certainly, so this strategy should be wrong!
Thanks!
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A Statistical Background A.1 Basic statistical terms
Note that all the following statistical terms apply only to
numerical variables, except the distribution which can exist for both numerical and categorical variables. A.1.1 Mean
The
mean is the most commonly reported measure of center. It is commonly called the average though this term can be a little ambiguous. The mean is the sum of all of the data elements divided by how many elements there are. If we have \(n\) data points, the mean is given by:
\[Mean = \frac{x_1 + x_2 + \cdots + x_n}{n}\]
A.1.2 Median
The median is calculated by first sorting a variable’s data from smallest to largest. After sorting the data, the middle element in the list is the
median. If the middle falls between two values, then the median is the mean of those two middle values. A.1.3 Standard deviation
We will next discuss the
standard deviation of a variable. The formula can be a little intimidating at first but it is important to remember that it is essentially a measure of how far we expect a given data value will be from its mean:
\[Standard \, deviation = \sqrt{\frac{(x_1 - Mean)^2 + (x_2 - Mean)^2 + \cdots + (x_n - Mean)^2}{n - 1}}\]
A.1.4 Five-number summary
The
five-number summary consists of five summary statistics: the minimum, the first quantile AKA 25 th percentile, the second quantile AKA median AKA 50 th percentile, the third quantile AKA 75 th, and the maximum. The five-number summary of a variable is used when constructing boxplots, as seen in Section 2.7.
The quantiles are calculated as
first quantile (\(Q_1\)): the median of the first half of the sorted data third quantile (\(Q_3\)): the median of the second half of the sorted data
The
interquartile range (IQR) is defined as \(Q_3 - Q_1\) and is a measure of how spread out the middle 50% of values are. The IQR corresponds to the length of a box in a boxplot.
The median and the interquartile range are not influenced by the presence of outliers in the ways that the mean and standard deviation are. It is, thus, recommended for skewed datasets. We say in this case that the median and interquartile range are more
robust to outliers. A.1.5 Distribution
The
distribution of a variable shows how frequently different values of a variable occur. Looking at visualization of a distribution can show where the values are centered, show how the values vary, and give some information about where a typical value might fall. It can also alert you to the presence of outliers.
Recall from Chapter 2 that we can visualize the distribution of a numerical variable using a histogram and that we can visualize the distribution of a categorical variable using a barplot.
A.1.6 Outliers Outliers correspond to values in the dataset that fall far outside the range of “ordinary” values. In context of a boxplot, by default they correspond to values below \(Q_1 - (1.5 * IQR)\) or above \(Q_3 + (1.5 * IQR)\). A.2 Normal distribution
Let’s discuss one particular kind of distribution:
normal distributions . Such bell-shaped distributions are defined by two values: 1) the mean \(\mu\) (“mu”) which locates the center of the distribution and 2) the standard deviation \(\sigma\) (“sigma”) which determines the variation of the distribution. In Figure A.1, we plot three normal distributions where: The solid normal curve has mean \(\mu\) = 5 and standard deviation \(\sigma\) = 2. The dashed normal curve has mean \(\mu\) = 5 and standard deviation \(\sigma\) = 5. The dotted normal curve has mean \(\mu\) = 15 and standard deviation \(\sigma\) = 2.
Notice how the solid and dashed line normal curves have the same center due to their common mean \(\mu\) = 5. However the dashed line normal curve is wider due to its larger standard deviation of \(\sigma\) = 5. On the other hand, the solid and dotted line normal curves have the same variation due to their common standard deviation \(\sigma\) = 2. However, they are centered at different locations.
When the mean \(\mu\) = 0 and the standard deviation \(\sigma\) = 1, the normal distribution has a special name: the
standard normal distribution or the \(z\)-curve.
Furthermore, if a variable follows a normal curve, there are
three rules of thumb we can use: 68% of values will lie within \(\pm\) 1 standard deviation of the mean. 95% of values will lie within \(\pm\) 1.96 \(\approx\) 2 standard deviations of the mean. 99.7% of values will lie within \(\pm\) 3 standard deviations of the mean.
Let’s illustrate this on a standard normal curve in Figure A.2. The dashed lines are at -3, -1.96, -1, 0, 1, 1.96, and 3. These 7 lines cut up the x-axis into 8 segments. The areas under the normal curve for each of the 8 segments are marked and add up to 100%. For example:
The middle two segments represent the interval -1 to 1. The shaded area above this interval represents 34% + 34% = 68% of the area under the curve. In other words, 68% of values. The middle four segments represent the interval -1.96 to 1.96. The shaded area above this interval represents 13.5% + 34% + 34% + 13.5%= 95% of the area under the curve. In other words, 95% of values. The middle six segments represent the interval -3 to 3. The shaded area above this interval represents 2.35% + 13.5% + 34% + 34% + 13.5% + 2.35% = 99.7% of the area under the curve. In other words, 99.7% of values. Learning check
Say you have a normal distribution with mean \(\mu\) = 6 and standard deviation \(\sigma\) = 3.
(LC11.3) What proportion of the area under the normal curve is less than 3? Greater than 12? Between 0 and 12? (LC11.4) What is the 2.5th percentile of the area under the normal curve? The 95th percentile? The 100th percentile? A.3 log10 transformations
At its simplest, log10 transformations return base 10
logarithms. For example, since \(1000 = 10^3\), running
log10(1000) returns
3 in R. To undo a log10-transformation, we raise 10 to this value. For example, to undo the previous log10-transformation and return the original value of 1000, we raise 10 to this value to the power of 3 by running
10^(3) = 1000 in R.
Log-transformations allow us to focus on changes in
orders of magnitude. In other words, they allow us to focus on multiplicative changes instead of additive ones. Let’s illustrate this idea in Table A.1 with examples of prices of consumer goods in US dollars.
Price log10(Price) Order of magnitude Examples $1 0 Singles Cups of coffee $10 1 Tens Books $100 2 Hundreds Mobile phones $1,000 3 Thousands High definition TV’s $10,000 4 Tens of thousands Cars $100,000 5 Hundreds of thousands Luxury cars & houses $1,000,000 6 Millions Luxury houses
Let’s make some remarks about log10-transformations based on Table A.1:
When purchasing a cup of coffee, we tend to think of prices ranging in single dollars. Ex: $2 or $3. However when purchasing a mobile phone, we don’t tend to think of their prices in units of single dollars such as $313 or $727. Instead, we tend to think of their prices in units of hundreds of dollars. Ex: $300 or $700. Thus cups of coffee and mobile phones are of different orders of magnitudeof price. Let’s say we want to know the log10-transformed value of $76. This would be hard to compute exactly without a calculator. However, since $76 is between $10 and $100 and since log10(10) = 1 and log10(100) = 2, we know log10(76) will be between 1 and 2. In fact, log10(76) is 1.880814. log10-transformations are monotonic, meaning they preserve orders. So if Price A is lower than Price B, then log10(Price A) will also be lower than log10(Price B). Most importantly, increments of one in log10-scale correspond to relative multiplicative changesin the original scale and not absolute additive changes. For example, increasing a log10(Price) from 3 to 4 corresponds to a multiplicative increase by a factor of x10: $100 to $1000.
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Gravitational force $\vec{F_g}$ is trying to rotate the cylinder around point A.
Lever arm has length $r_c= r\,\sin\theta$ Torque $\tau$ is $$\tau = F_g \, r_c = m \, g\, r \, \sin\theta$$Torque causes angular acceleration $\alpha$ of the cylinder. We are interested in acceleration $\vec{a}$ of point C.Absolute value of point C acceleration is $|\vec{a}| = \alpha \, r$. We need to compute angular acceleration $\alpha$ which is quotient of torque $\tau$ and moment of inertia $I$. $$\alpha = \frac{\tau}{I}$$
Moment of inertia of a cylinder is $I_{cylinder}=\frac{1}{2}\,m\,r^2$. Axis of rotation doesn't intersect center of mass, we will use parallel axis theorem.$$I = I_{cylinder} + m \, r^2 = \frac{1}{2}\,m\,r^2 + m\,r^2 = \frac{3}{2}\,m\,r^2$$
And from moment of inertia we compute acceleration of point C$$a=\alpha\,r=\frac{\tau}{I}\,r=\frac{m\,g\,r\,\sin\theta}{\frac{3}{2}\,m\,r^2}\,r=\frac{2}{3}\,\frac{m\,g\,r^2\,\sin\theta}{m\,r^2}$$
And finally$$a=\frac{2}{3}\,g\,\sin\theta$$
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It is plain, from looking at the question geometrically, that
the expected distance between two independent, uniform, random points within a convex set is going to be a little less than half its diameter. (It should be less because it's relatively rare for the two points to be located within extreme areas like corners and more often the case they will be near the center, where they are close.) Since the diameter of this rectangle is $50$, by this reasoning alone we would anticipate the answer to be a little less than $25$.
An exact answer is obtained from the definition of expectation as the probability-weighted value of the distance. In general, consider a rectangle of sides $1$ and $\lambda$; we will scale it up to the correct size afterwards (by setting $\lambda = 40/30$ and multiplying the expectation by $30$). For this rectangle, using coordinates $(x,y)$, the uniform probability density is $\frac{1}{\lambda}dx dy$. The mean distance within this rectangle then is given by
$$\int_0^\lambda\int_0^1\int_0^\lambda\int_0^1 \sqrt{(x_1-x_2)^2+(y_1-y_2)^2} \frac{1}{\lambda}dx_1 dy_1 \frac{1}{\lambda} dx_2 dy_2.$$
Using elementary integration methods this is straightforward but painful to do; I employed a computer algebra system (
Mathematica) to obtain the answer
$$[2+2 \lambda ^5-2 \sqrt{1+\lambda ^2}+6 \lambda ^2 \sqrt{1+\lambda ^2}-2 \lambda ^4 \sqrt{1+\lambda ^2} +5 \lambda \text{ArcSinh}(\lambda)+5 \lambda ^4 \log\left(\frac{1+\sqrt{1+\lambda ^2}}{\lambda }\right)]/(30\lambda^2).$$
The presence of $\sqrt{1+\lambda^2}$ in many of these terms is no surprise: it is the diameter of the rectangle (the maximum possible distance between any two points within it). The appearance of logarithms (which includes the arcsinh) is also unsurprising if you have ever investigated average distances within simple plane figures: somehow it always shows up (a hint of this appears in integral of the secant function). Incidentally, the presence of $30$ in the denominator has nothing to do with the specifics of the problem involving a rectangle of sides $30$ and $40$: it's a universal constant.)
With $\lambda=4/3$ and scaling up by a factor of $30$, this evaluates to $\frac{1}{108} (871+960 \log(2)+405 \log(3)) \approx 18.345919\ldots$.
One way to understand the situation more deeply is to plot the mean distance
relative to the diameter of $\sqrt{1+\lambda^2}$ for varying values of $\lambda$. For extreme values (near $0$ or much greater than $1$), the rectangle becomes essentially one-dimensional and a more elementary integration indicates the mean distance should reduce to one-third the diameter. Also, because the shapes of rectangles with $\lambda$ and $1/\lambda$ are the same, it is natural to plot the result on a logarithmic scale of $\lambda$, where it must be symmetric about $\lambda=1$ (the square). Here it is:
With this we learn a
rule of thumb: the mean distance within a rectangle is between $1/3 \approx 0.33$ and (approximately) $0.37$ of its diameter, with the larger values associated with squarish rectangles and the smaller values associated with long skinny (linear) rectangles. The midpoint between these extremes is achieved roughly for rectangles with aspect ratios of $3:1$. With this rule in mind, you can just glance at a rectangle and estimate its mean distance to two significant figures.
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Difference between revisions of "Probability Seminar"
(→February 14, TBA)
(→April 11, Eviatar Proccia, Texas A&M)
Line 59: Line 59:
== April 4, TBA ==
== April 4, TBA ==
−
== April 11, [https://sites.google.com/site/ebprocaccia/ Eviatar
+
== April 11, [https://sites.google.com/site/ebprocaccia/ Eviatar ], [http://www.math.tamu.edu/index.html Texas A&M] ==
== April 18, [https://services.math.duke.edu/~agazzi/index.html Andrea Agazzi], [https://math.duke.edu/ Duke] ==
== April 18, [https://services.math.duke.edu/~agazzi/index.html Andrea Agazzi], [https://math.duke.edu/ Duke] ==
Revision as of 16:02, 9 February 2019 Spring 2019 Thursdays in 901 Van Vleck Hall at 2:25 PM, unless otherwise noted. We usually end for questions at 3:15 PM.
If you would like to sign up for the email list to receive seminar announcements then please send an email to join-probsem@lists.wisc.edu
January 31, Oanh Nguyen, Princeton
Title:
Survival and extinction of epidemics on random graphs with general degrees
Abstract: We establish the necessary and sufficient criterion for the contact process on Galton-Watson trees (resp. random graphs) to exhibit the phase of extinction (resp. short survival). We prove that the survival threshold $\lambda_1$ for a Galton-Watson tree is strictly positive if and only if its offspring distribution has an exponential tail, settling a conjecture by Huang and Durrett. On the random graph with degree distribution $D$, we show that if $D$ has an exponential tail, then for small enough $\lambda$ the contact process with the all-infected initial condition survives for polynomial time with high probability, while for large enough $\lambda$ it runs over exponential time with high probability. When $D$ is subexponential, the contact process typically displays long survival for any fixed $\lambda>0$. Joint work with Shankar Bhamidi, Danny Nam, and Allan Sly.
Wednesday, February 6 at 4:00pm in Van Vleck 911 , Li-Cheng Tsai, Columbia University
Title:
When particle systems meet PDEs
Abstract: Interacting particle systems are models that involve many randomly evolving agents (i.e., particles). These systems are widely used in describing real-world phenomena. In this talk we will walk through three facets of interacting particle systems, namely the law of large numbers, random fluctuations, and large deviations. Within each facet, I will explain how Partial Differential Equations (PDEs) play a role in understanding the systems..
Title:
Fluctuations of the KPZ equation in d\geq 2 in a weak disorder regime
Abstract: We will discuss some recent work on the Edwards-Wilkinson limit of the KPZ equation with a small coupling constant in d\geq 2.
February 21, TBA Wednesday, February 27 at 1:10pm Jon Peterson, Purdue March 7, TBA March 14, TBA March 21, Spring Break, No seminar March 28, Shamgar Gurevitch UW-Madison
Title:
Harmonic Analysis on GLn over finite fields, and Random Walks
Abstract: There are many formulas that express interesting properties of a group G in terms of sums over its characters. For evaluating or estimating these sums, one of the most salient quantities to understand is the {\it character ratio}:
$$ trace(\rho(g))/dim(\rho), $$
for an irreducible representation $\rho$ of G and an element g of G. For example, Diaconis and Shahshahani stated a formula of this type for analyzing G-biinvariant random walks on G. It turns out that, for classical groups G over finite fields (which provide most examples of finite simple groups), there is a natural invariant of representations that provides strong information on the character ratio. We call this invariant {\it rank}. This talk will discuss the notion of rank for GLn over finite fields, and apply the results to random walks. This is joint work with Roger Howe (Yale and Texas AM).
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Katok/ Hasselblatt, "Modern Theory of Dynamical Systems", p. 156:
Every probability distribution $p=(p_0,\ldots,p_{N-1})$ where $0\leq p_i\leq 1$ for $i=0,\ldots,N-1$ and $\sum_{i=0}^{N-1}p_i=1$, determines the product measure $\mu_p$ on the space $$ \Omega_n=\{\omega=(\ldots,\omega_{-1},\omega_0,\omega_1,\ldots)|\omega_i\in\{0,1,\ldots,n-1\}\text{ for }i\in\mathbb{Z}\}. $$ Namely, for any cylinder $$ C_{\alpha_1,\ldots,\alpha_k}^{n_1,\ldots,n_k}=\{\omega\in\Omega_n|\omega_{n_i}=\alpha_i, i=1,\ldots,k\} $$ we set $$ \mu_p(C_{\alpha_1,\ldots,\alpha_k}^{n_1,\ldots,n_k})=\prod_{i=1}^kp_i $$
and then extend $\mu_p$ in the standard fashion to the $\sigma$-algebra of all Borel sets.
I have some problems to understand the bold marked text, because my understanding was the following:
The set consisting of all the cylinders, I denote it by $C$, is invariant under intersections, which means that $\mu_p$ can be extended uniquely to $\sigma(C)$, the smallest $\sigma$-algebra containing the cylinders which is here the product $\sigma$-algebra (and sometimes called cylinder $\sigma$-algebra).
But in general, $\sigma(C)\subseteq\mathcal{B}(\tau)$, where $\tau$ is the topology on $\Omega_n$ with base consisting of the cylinders and $\mathcal{B}(\tau)$ being the Borel $\sigma$-algebra.
So my question is: How to extend $\mu_p$ to all Borel sets as the quote says, i.e. how to extend $\mu_p$ to $\mathcal{B}(\tau)$?
(If this is what the quote means...)
The only explanation I have is that maybe in this context $\sigma(C)=\mathcal{B}(\tau)$, however I cannot prove that $\mathcal{B}(\tau)\subseteq\sigma(C)$.
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Let $f:\mathbb{R}\to\mathbb{R}$ be a function satisfying $|f(x+y)-f(x-y)-y|$$\le y^2$
for all $x,y\in\mathbb{R}$. Show that $f(x)={x\over2}+c$, where $c$ is a constant.
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It only takes a minute to sign up.Sign up to join this community
Let $f:\mathbb{R}\to\mathbb{R}$ be a function satisfying $|f(x+y)-f(x-y)-y|$$\le y^2$
for all $x,y\in\mathbb{R}$. Show that $f(x)={x\over2}+c$, where $c$ is a constant.
Hint:
Set $x-y = z$ and $2y = h$.
Let $\epsilon > 0$ be given.
We have $\bigg|\dfrac{f(z+ h) - f(z)}{h} - \dfrac{1}{2} \bigg| \le \dfrac{h^2}{4}$.
So if we take $|h| < 2 \sqrt{\epsilon}$, then $\bigg|\dfrac{f(z+ h) - f(z)}{h} - \dfrac{1}{2}\bigg | < \epsilon$, this means $f'(z)$ exists and equals $\dfrac{1}{2}$. Rest is trivial.
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How do I solve this system of equations?
$$\begin{cases} 7(a+b)=b-a \\4(3a+2b)=b-8\end{cases}$$
Progress
I tried both substitution and elimination, but when I set $a$ or $b$ free on one side, I keep getting $a$ or $b$ also on the other side.
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It only takes a minute to sign up.Sign up to join this community
How do I solve this system of equations?
$$\begin{cases} 7(a+b)=b-a \\4(3a+2b)=b-8\end{cases}$$
I tried both substitution and elimination, but when I set $a$ or $b$ free on one side, I keep getting $a$ or $b$ also on the other side.
$$7 \cdot (a+b) = b - a \Rightarrow a = - \dfrac{6}{8} b$$
Substitute for a in the second equation
$$4 \cdot (3a+ 2b) = b-8 \Rightarrow 12a + 7b = -8$$
$$7b + 12 \cdot \left( - \dfrac{6}{8}b \right) = -8$$
This equation has only $b$ unknown so solve for $b$ then use $b$ to find $a$
Hint:
Try to perform algebra on both the equations till you have the $a$ and $b$ on one side and a number on the other. Then see if you can "combine" them together.
$$7(a+b) = b-a \Rightarrow 7a+7b=b-a \Rightarrow 7a+a=b-7b \Rightarrow 8a=-6b \Rightarrow a=-\frac{6}{8}b \\ \Rightarrow a=-\frac{3}{4}b\\ 4(3a+2b)=b-8 \Rightarrow 12a+8b=b-8 \Rightarrow 12a=b-8b-8 \Rightarrow 12a=-7b-8 \overset{a=-\frac{3}{4}b}{\Rightarrow} 12 \left (-\frac{3}{4}b \right ) =-7b-8 \Rightarrow -9b=-7b-8 \Rightarrow -9b+7b=-8 \Rightarrow -2b=-8 \\ \Rightarrow b=4$$
Replacing at $a=-\frac{3}{4}b$ we get $a=-\frac{3}{4}4 \Rightarrow a=-3$
CAUTION: this is a non-standard approach.
In both equations, isolate the variable $a$ in the LHS: $$7(a+b)=b-a\implies 8a=-6b\implies 4a=-3b,$$ $$4(3a+2b)=b-8\implies12a=-7b-8.$$ Now equate the two: $$(12a=)-9b=-7b-8.$$ This is an equation in a single unknown ($b$). $$-2b=-8\implies b=4,$$ and $$4a=-3b\implies a=-3.$$
Hint: from the first equation, we have $$a=-\frac{3}{4}b.$$
Now substitute this into the second equation and you've now got an equation in $b$. Solve this (for $b$), then find $a$ (again) using the fact that $$a=-\frac{3}{4}b.$$
Here is a different approach:
$\begin{cases} 7(a+b)=b-a \\ 4(3a+2b) = b-8 \end{cases}$ $\Leftrightarrow$ $\begin{cases}8a +6b=0\\12a+7b=-8 \end{cases}$
Then
$\begin{bmatrix} 8&6&0\\12&7&-8\end{bmatrix}$ $\Leftrightarrow$ $\begin{bmatrix} 1&3/4&0\\12&7&-8 \end{bmatrix}$ $\Leftrightarrow$ $\begin{bmatrix}1&3/4&0\\0&-2&-8\end{bmatrix}$ $\Leftrightarrow$ $\begin{bmatrix}1&3/4&0\\0&1&4\end{bmatrix}$ $\Leftrightarrow$ $\begin{bmatrix}1&0&-3\\0&1&4\end{bmatrix}$
Resulting in $a=-3$ aand $b=4$.
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Lasing characteristics of heavily doped single-crystal Fe:ZnSe Abstract
Characteristics of a Fe:ZnSe laser are studied at room temperature. The laser active elements are heavily doped single crystals with the \(\hbox {Fe}^{2+}\) ion concentration \(n=0.64\times 10^{19}-5.7\times 10^{19}\hbox {cm}^{-3}\), grown from melt by the Bridgman method. The generated energy of 870 mJ is obtained at the total efficiencies with respect to the absorbed and incident energies \(\eta _{\mathrm{{abs}}}=43\%\) and \(\eta _{\mathrm{{inc}}}\approx 35\%\), respectively. The laser slope efficiency with respect to the absorbed energy is \(\eta _\mathrm{{slope}}\approx 50\%\). In a heavily doped active element with the \(\hbox {Fe}^{2+}\) concentration \(n=5.7\times 10^{19}\hbox {cm}^{-3}\), in which the medium excitation depth is just a part of the total element dimension along the optical axis (the element is completely non-transparent for the pumping radiation), the radiation spectrum of the Fe:ZnSe laser shifts to the long-wavelength range by more than 300 nm as compared to spectra of the laser on crystals excited along the whole element length. It is shown that Fe:ZnSe lasers on heavily doped single-crystal elements can be efficiently excited by a radiation of a Cr:ZnSe laser without tuning the spectrum of the latter to the longer wavelength range.
KeywordsFe:ZnSe laser HF laser Single crystal Room temperature Zinc chalcogenides Bridgman method Cr:ZnSe laser Notes Acknowledgements
This work was supported by the Russian Science Foundation, Grant no. 19-13-00205 (development of the technique for creating Fe:ZnSe samples and preparation of active elements for experiments) and by the Russian Foundation for Basic Research, project no. 18-08-00793 (study of laser characteristics).
References 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16.H. Jelinkova, M.E. Doroshenko, M. Jelinek, J. Sulc, M. Nemec, V. Kubecek, Y.A. Zagoruiko, N.O. Kovalenko, A.S. Gerasimenko, V.M. Puzikov, V.K. Komar, Fe:ZnSe and Fe:ZnMgSe lasers pumped by Er:YSGG radiation. International Society for Optics and Photonics. In: Solid State Lasers XXIV: Technology and Devices 934293421VGoogle Scholar 17. 18. 19.S.B. Mirov, V. Fedorov, D. Martyshkin, I. Moskalev, M. Mirov, S. Vasilyev, Advanced Solid State Lasers Conference, Berlin Germany, (2015) AW4A.1Google Scholar 20. 21. 22. 23. 24.R.I. Avetisov, S.S. Balabanov, K.N. Firsov, E.M. Gavrishchuk, A.A. Gladilin, V.B. Ikonnikov, V.P. Kalinushkin, SYu. Kazantsev, I.G. Kononov, M.P. Zykova, E.N. Mozhevitina, A.V. Khomyakov, D.V. Savin, N.A. Timofeeva, O.V. Uvarov, ICh. Avetissov, J. Cryst. Growth 491, 36–41 (2018)ADSCrossRefGoogle Scholar 25. 26. 27. 28. 29. 30. 31. 32. 33.A.A. Voronov, Generatsionnye i spektral’no-kineticheskie kharakteristiki lazera na kristalle \(\text{Fe}^{2+}\):ZnSe (Emission and Spectral-Kinetic Characteristics of Laser on \(\text{ Fe }^{2+}\):ZnSe crystal). PhD thesis (2009) (in Russian)Google Scholar 34. 35. 36.A.A. Davydov, V.N. Ermolov, S.V. Neustroev, L.P. Pavlova, Neorganicheskie Materialy 28(1–6), 42–48 (1992). [in Russian]Google Scholar 37.
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The
Frobenius norm of a matrix is the square root of the sum of the squares of its elements; it provides a rough measure of the “size” of the matrix.
Wikipedia gives three ways of expressing the Frobenius norm of a matrix whose elements are real numbers:
\|A\|_{\mathrm{F}}\equiv \sqrt{\sum_{i=1}^{m} \sum_{j=1}^{n}\left|a_{i j}\right|^{2}}=\sqrt{\operatorname{trace}\left(A^{T} A\right)}=\sqrt{\sum_{i=1}^{\min \{m, n\}} \sigma_{i}^{2}(A)}
Here, the superscript ^{T} refers to the transpose matrix, and the singular values \sigma_{i}(A) are the square roots of the eigenvalues of the matrix A A^T
The third expression is the square root of the sum of the eigenvalues of A A^T; although mathematically interesting, it’s not the best way to compute the Frobenius norm.
Let’s demonstrate the equality of the 1st and 2nd expressions. Suppose A is an n\times m matrix, i.e. A has n rows and m columns.
Then
A^{T} A is an m\times m matrix whose diagonal elements are
A_{j} \cdot A_{j}, where j runs from 1 to m, and
A_{j} \equiv (a_{1j}, ..., a_{nj}) is the jth column of A.
The diagonal elements of A^{T} A are evidently the set of dot products of each column of A with itself.
\operatorname{trace}\left(A^{T} A\right) is defined as the sum of the diagonal elements of A^{T} A,
which is just the sum of the squares of all the elements of A.
QED
It’s interesting to note that the
order of the terms in the product of the matrix with its transpose doesn’t matter, giving rise to an alternate form for the Frobenius norm:
\operatorname{trace}\left(A^{T} A\right) = \operatorname{trace}\left(A A^{T} \right);
\operatorname{trace}\left( A^{T}A \right) is the sum of the m dot products of the n\times1 column vectors of A with themselves, while \operatorname{trace}\left(A A^{T} \right) is the sum of the n dot products of the m\times1 row vectors of A with themselves. The expressions are equivalent because each reduces to the sum of the squares of the elements of the matrix A. Their computational complexity is O(mn), which is lower than that of the singular value decomposition required to compute \sigma.
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I am currently working on a research project for a pairs trading strategy and would like to know the correct positions to take when a signal has been triggered.
Say we are using this equation to generate signals: \begin{equation} z = y - \beta x \end{equation} $\mu_z$ = 0, $\sigma_z$ = 0.5, and $\beta$ = 1, for simplicity. And our signals to open are: (z >= $\mu_z$+$\sigma_z$ & z <= $\mu_z$-$\sigma_z$) Then a signal will be triggered when : 1) y = 10, x = 9.5 2) y = 10.5, x = 10 3) y = 10, x = 10.5 4) y = 9.5, x = 10 Should the positions taken in these scenarios be different? Should we short 1 share y and go long $\beta$ shares of x for 1 and 2? Should we go long 1 share y and short $\beta$ shares of x for 3 and 4? Or should we actually be calculating pct change in y and $\beta$x and then short which ever had the greatest pct change and go long the other?
I am currently working on a research project for a pairs trading strategy and would like to know the correct positions to take when a signal has been triggered.
It should be consistent with the way you calculate $\beta:$ if you use stock returns to compute it, then you should be using returns to compute the spread and your signal, and aim to be cash neutral: N $\it{dollars}$ long of x and $\beta\times N$ $\it{dollars}$ short of y. If you regress the prices of x and y to obtain $\beta,$ then pretty much, what you wrote - N $\it{shares}$ of x and $\beta\times N$ $\it{shares}$ of y.
This doesn't answer your question directly, but it may be interesting for you to see a visualization of a pair trading strategy.
I'm the developer of an interactive chart tool for technical analysis that also has a pair strategy function. In my pair trading implementation I'm calculating the mean ratio between the stocks and generating entry signals when the ratio is 2 std. deviations from the mean ratio and exit when 1 stddev. I'm also running cointegration (ADF) and correlation tests that has to pass for the entry signal to be generated.
An "random" example of the strategy is shown in the live interactive chart (link below). The white area shows when a trade is active - and "Algo" shows the return for all trades in total. As shown in the chart for the past 9 months the automated strategy generated 6 trades and all of them gave positive returns including the one that is currently open.
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Supercells¶
Supercell is a widely used crystallographic concept
1. A general theoretical background is offered in the expandable section below for the readers unfamiliar with the topic. The rest of this documentation page explains how supercell generation is implemented in the Materials Designer interface. Theoretical Background¶ What are supercells?¶
A material has a crystal structure associated with it and the latter is described by a unit cell. There are an infinite number of unit cells U with different shapes and sizes which can describe the same crystal. The supercell S of unit cell U is defined as a cell which describes the same crystal, but has larger volume than cell U
1. 2D examples¶
Examples of different supercells for the same underlying 2D cubic crystal are shown in the figure below. The relatively large size of such supercells has to be compared to the smaller surface area of the most basic (primitive) unit cell of this crystal containing just one lattice point, example of which is shown towards the center of the figure. Both diagonal and non-diagonal supercells are included in this picture to reflect the general scope of the supercell definition.
Why are supercells useful?¶
There are many instances where the construction of supercells affords for an easier ascertainment of useful crystal properties and visual symmetric qualities, which could otherwise be difficult to determine by looking at just the initial cell.
Phonon calculations¶
There are many occurrences of methods in computational materials science for determining crystal structure properties which rely on small perturbations of the supercells away from their original equilibrium configurations. For example, during phonon calculations by the small displacement method ("frozen phonon"), the frequencies are calculated using the finite force values computed on slightly displaced atoms in a supercell of the original structure.
Conventional vs primitive cells¶
The conventional cells of body-centered (bcc) or face-centered (fcc) cubic crystals, containing two atoms and four atoms respectively, reflect more intuitively the overall symmetry of such systems. The figure below illustrates this for the example of an fcc lattice. In this image, the volume marked in red represents the primitive unit cell, whereas the over-arching cubic conventional supercell exhibits the full face-centered cubic symmetry from which the corresponding crystal structure gets its name:
Defects¶
Finally, supercells are also commonly used in computational models of crystal defects, in order to allow for the use of periodic boundary conditions.
How are supercells defined?¶
The basis vectors of unit cell U ({\vec {a}},{\vec {b}},{\vec {c}}) can be transformed to basis vectors of supercell S ({\vec {a}}',{\vec {b}}',{\vec {c}}') by way of the following linear transformation:
where {\hat {P}} is the corresponding linear transformation matrix. All items P_{ij} should be integer numbers, and it is furthermore required that \det({\hat {P}})>1 (with \det({\hat {P}})=1 the transformation preserves the volume of the original unit cell). For example, the matrix
transforms the primitive cell of a body-centered cubic lattice to its fully-symmetric conventional unit cell.
Another particular case of the transformation is a diagonal form P_{i\neq j}=0 of the matrix. This type of transformation is referred to as diagonal supercell expansion, and can be interpreted as a simple repetition of the initial cell over its crystallographic axes.
The Generate Dialog¶
Choose the
Supercell option in the
Advanced menu to opend the "Supercell Generation Dialog".
Supercell Transformation Matrix¶
The overlay, highlighted in the image below, will allow the reader to set the parameters of the supercell transformation matrix (with its elements positioned exactly as explained in the previous paragraph).
Generate Supercell¶
When finished setting up the transformation matrix, click
Submit button and both the crystal data and the 3D representation will update accordingly to reflect the newly-generated supercell.
Animation¶
Click on the animation below to see the above in action. In this example, we generate a 2 x 2 x 2 supercell of a cubic structure containing two atoms by inserting the diagonal elements of the corresponding transformation matrix, whilst leaving the off-diagonal elements to zero.
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Now showing items 11-20 of 76
Anisotropic flow of inclusive and identified particles in Pb–Pb collisions at $\sqrt{{s}_{NN}}=$ 5.02 TeV with ALICE
(Elsevier, 2017-11)
Anisotropic flow measurements constrain the shear $(\eta/s)$ and bulk ($\zeta/s$) viscosity of the quark-gluon plasma created in heavy-ion collisions, as well as give insight into the initial state of such collisions and ...
Multiplicity dependence of identified particle production in proton-proton collisions with ALICE
(Elsevier, 2017-11)
The study of identified particle production as a function of transverse momentum ($p_{\text{T}}$) and event multiplicity in proton-proton (pp) collisions at different center-of-mass energies ($\sqrt{s}$) is a key tool for ...
Probing non-linearity of higher order anisotropic flow in Pb-Pb collisions
(Elsevier, 2017-11)
The second and the third order anisotropic flow, $V_{2}$ and $V_3$, are determined by the corresponding initial spatial anisotropy coefficients, $\varepsilon_{2}$ and $\varepsilon_{3}$, in the initial density distribution. ...
The new Inner Tracking System of the ALICE experiment
(Elsevier, 2017-11)
The ALICE experiment will undergo a major upgrade during the next LHC Long Shutdown scheduled in 2019–20 that will enable a detailed study of the properties of the QGP, exploiting the increased Pb-Pb luminosity ...
Neutral meson production and correlation with charged hadrons in pp and Pb-Pb collisions with the ALICE experiment at the LHC
(Elsevier, 2017-11)
Among the probes used to investigate the properties of the Quark-Gluon Plasma, the measurement of the energy loss of high-energy partons can be used to put constraints on energy-loss models and to ultimately access medium ...
Direct photon measurements in pp and Pb-Pb collisions with the ALICE experiment
(Elsevier, 2017-11)
Direct photon production in heavy-ion collisions provides a valuable set of observables to study the hot QCD medium. The direct photons are produced at different stages of the collision and escape the medium unaffected. ...
Azimuthally differential pion femtoscopy relative to the second and thrid harmonic in Pb-Pb 2.76 TeV collisions from ALICE
(Elsevier, 2017-11)
Azimuthally differential femtoscopic measurements, being sensitive to spatio-temporal characteristics of the source as well as to the collective velocity fields at freeze-out, provide very important information on the ...
D-meson nuclear modification factor and elliptic flow measurements in Pb–Pb collisions at $\sqrt {s_{NN}}$ = 5.02TeV with ALICE at the LHC
(Elsevier, 2017-11)
ALICE measured the nuclear modification factor ($R_{AA}$) and elliptic flow ($\nu_{2}$) of D mesons ($D^{0}$, $D^{+}$, $D^{⁎+}$ and $D^{s+}$) in semi-central Pb–Pb collisions at $\sqrt{s_{NN}} =5.02$ TeV. The increased ...
ALICE measurement of the $J/\psi$ nuclear modification factor at mid-rapidity in Pb–Pb collisions at $\sqrt{{s}_{NN}}=$ 5.02 TeV
(Elsevier, 2017-11)
ALICE at the LHC provides unique capabilities to study charmonium production at low transverse momenta ( p T ). At central rapidity, ( |y|<0.8 ), ALICE can reconstruct J/ ψ via their decay into two electrons down to zero ...
Charmonium production in Pb–Pb and p–Pb collisions at forward rapidity measured with ALICE
(Elsevier, 2017-11)
The ALICE collaboration has measured the inclusive charmonium production at forward rapidity in Pb–Pb and p–Pb collisions at sNN=5.02TeV and sNN=8.16TeV , respectively. In Pb–Pb collisions, the J/ ψ and ψ (2S) nuclear ...
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Now showing items 1-10 of 55
J/Ψ production and nuclear effects in p-Pb collisions at √sNN=5.02 TeV
(Springer, 2014-02)
Inclusive J/ψ production has been studied with the ALICE detector in p-Pb collisions at the nucleon–nucleon center of mass energy √sNN = 5.02TeV at the CERN LHC. The measurement is performed in the center of mass rapidity ...
Measurement of electrons from beauty hadron decays in pp collisions at root √s=7 TeV
(Elsevier, 2013-04-10)
The production cross section of electrons from semileptonic decays of beauty hadrons was measured at mid-rapidity (|y| < 0.8) in the transverse momentum range 1 < pT <8 GeV/c with the ALICE experiment at the CERN LHC in ...
Suppression of ψ(2S) production in p-Pb collisions at √sNN=5.02 TeV
(Springer, 2014-12)
The ALICE Collaboration has studied the inclusive production of the charmonium state ψ(2S) in proton-lead (p-Pb) collisions at the nucleon-nucleon centre of mass energy √sNN = 5.02TeV at the CERN LHC. The measurement was ...
Event-by-event mean pT fluctuations in pp and Pb–Pb collisions at the LHC
(Springer, 2014-10)
Event-by-event fluctuations of the mean transverse momentum of charged particles produced in pp collisions at s√ = 0.9, 2.76 and 7 TeV, and Pb–Pb collisions at √sNN = 2.76 TeV are studied as a function of the ...
Multiplicity dependence of the average transverse momentum in pp, p-Pb, and Pb-Pb collisions at the LHC
(Elsevier, 2013-12)
The average transverse momentum <$p_T$> versus the charged-particle multiplicity $N_{ch}$ was measured in p-Pb collisions at a collision energy per nucleon-nucleon pair $\sqrt{s_{NN}}$ = 5.02 TeV and in pp collisions at ...
Production of $K*(892)^0$ and $\phi$(1020) in pp collisions at $\sqrt{s}$ =7 TeV
(Springer, 2012-10)
The production of K*(892)$^0$ and $\phi$(1020) in pp collisions at $\sqrt{s}$=7 TeV was measured by the ALICE experiment at the LHC. The yields and the transverse momentum spectra $d^2 N/dydp_T$ at midrapidity |y|<0.5 in ...
Directed flow of charged particles at mid-rapidity relative to the spectator plane in Pb-Pb collisions at $\sqrt{s_{NN}}$=2.76 TeV
(American Physical Society, 2013-12)
The directed flow of charged particles at midrapidity is measured in Pb-Pb collisions at $\sqrt{s_{NN}}$=2.76 TeV relative to the collision plane defined by the spectator nucleons. Both, the rapidity odd ($v_1^{odd}$) and ...
Centrality, rapidity and transverse momentum dependence of the J/$\psi$ suppression in Pb-Pb collisions at $\sqrt{s_{NN}}$=2.76 TeV
(Elsevier, 2014-06)
The inclusive J/$\psi$ nuclear modification factor ($R_{AA}$) in Pb-Pb collisions at $\sqrt{s_{NN}}$=2.76TeV has been measured by ALICE as a function of centrality in the $e^+e^-$ decay channel at mid-rapidity |y| < 0.8 ...
Production of charged pions, kaons and protons at large transverse momenta in pp and Pb-Pb collisions at $\sqrt{s_{NN}}$ = 2.76 TeV
(Elsevier, 2014-09)
Transverse momentum spectra of $\pi^{\pm}, K^{\pm}$ and $p(\bar{p})$ up to $p_T$ = 20 GeV/c at mid-rapidity, |y| $\le$ 0.8, in pp and Pb-Pb collisions at $\sqrt{s_{NN}}$ = 2.76 TeV have been measured using the ALICE detector ...
Multiplicity Dependence of Pion, Kaon, Proton and Lambda Production in p-Pb Collisions at $\sqrt{s_{NN}}$ = 5.02 TeV
(Elsevier, 2014-01)
In this Letter, comprehensive results on $\pi^{\pm}, K^{\pm}, K^0_S$, $p(\bar{p})$ and $\Lambda (\bar{\Lambda})$ production at mid-rapidity (0 < $y_{CMS}$ < 0.5) in p-Pb collisions at $\sqrt{s_{NN}}$ = 5.02 TeV, measured ...
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Prove that $ {(x+y)}^n-x^n-y^n$ is divisible by $ xy(x+y) \times (x^2+xy+y^2)$ if $ n$ is an odd number not divisible by $ 3$ . Prove that $ {(x+y)}^n-x^n-y^n$ is divisible by $ xy(x+y) \times {(x^2+xy+y^2)}^2$ if $ n \equiv \pmod{6}1$ Solution
1.Considering the given expression as a polynomial in $ y$ , let us put $ y=0 $ . We see that at $ y=0 $ the polynomial vanishes (for any $ x$ ). Therefore our polynomial is divisible by $ y$ . Similarly, it is divisible by $ x$ as well. Thus the polynomial is divisible by $ xy$ .
To prove that it is divisible by $ x+y $ , put $ x+y=0 \ {or} \ y=-x $ . It is evident that for odd n we have : $ {(x+(-x)}^n-x^n-{(-x)}^n = 0 $ for $ y=-x $ . Consequently, our polynomial is divisible by $ x+y $ . It only remains to prove the divisibility of the polynomial by $ x^2 +xy+y^2$ , which also be written as $ (y-x\epsilon)(y-x{\epsilon}^2 ) $ where $ \epsilon^2+\epsilon+1=0 $ . For this purpose it only remains to replace $ y $ first by $ x \epsilon $ and then by $ x\epsilon^2 $ to make sure that with these substitutions the polynomial vanishes. Since, by hypothesis, $ n$ is not divisible by 3, it follows that $ n=3l+1 \ or \ 3l+2 $ , for every $ l \in \mathbb{Z} $ , in which $ 3l+1$ is not acceptable since $ n$ is odd from the problem. At $ y=x\epsilon $ the polynomial attains the following value $ {(x+x\epsilon)}^n-x^n-{(x\epsilon)}^n=x^n [{(1+\epsilon)}^n-1-\epsilon^n] \\ =x^n {(-\epsilon^2)}^n -1 -\epsilon^n …. $ since ($ 1+\epsilon + \epsilon^2=0 $ ) substituting $ n=3l+2 $ we get $ 1+\epsilon+\epsilon^2 =0 $ Likewise we prove that at $ y=x\epsilon^2$ the polynomial vanishes as well, and consequently, its by divisibility by $ xy(x+y) \times (x^2+xy+y^2) $ is proved.
2.To prove the second statement, let us proceed as follows. Let the quantities $ {-x, -y, \, and \, x+y} $ be the roots of a cubic equation $ X^3-rX^2-pX-q=0 $ . Then by virtue of the known relations between the roots of an equation and its coefficients we have $ r=-x-y-(x+y)=0 \\ -p=xy-x(x+y)-y(x+y)$ or $ p=x^2+xy+y^2$ and $ q=xy(x+y)$ .
Thus, $ -x, \, -y \, x+y$ are the roots of the equation $ X^3-pX-q=0 $ where $ p=x^2+xy+y^2$ and $ q=xy(x+y) $ Put $ {(-x)}^n-{(-y)}^n+{(x+y)}^n=S_n$ . Among successive values of $ S_n$ , there exist the relationship $ S_{n+3}=pS_{n+1}+qS_n$ ,: $ S_1$ being equal to zero. Let us prove that $ S_n$ is divisible by $ p^2$ if $ n \equiv 1 \pmod{6}$ using the method of mathematical induction. Suppose $ S_n $ is divisible by $ p^2 $ and prove that then $ S_{n+6} $ is also divisible by $ p^2$ . So, using this relation we get that $ S_{n+6}=p(pS_{n+2} + qS_{n+1}) + q(pS_{n+1}+qS_n) \\ =p^2S_{n+2}+2pqS_{n+1}+q^2S_n$ . Since, by supposition, $ S_n$ is divisible by $ p^2$ , it suffices to prove that $ S_{n+1}$ is divisible by $ p$ . Thus we only have to prove than $ S_n={(x+y)}^n+(-x)^n+(-y)^n$ is divisible by $ p=x^2+xy+y^2$ if $ n \equiv 2 \pmod{6}$ , we easily prove our assertion. And so, assuming that $ S_n$ is divisible by $ p^2$ , we have proved that (from induction) $ S_{n+6}$ is also divisible by $ p^2$ . Consequently $ S_n ={(x+y)}^n+(-x)^n+(-y)^n={(x+y)}^n-x^n-y^n$ for any $ n \equiv 1 \pmod{6}$ is divisible by $ p^2={(x^2+xy+y^2)}^2 $ . Now it only remains to prove its divisibility by $ x+y $ and by $ xy$ , which is quite elementary.
Feel free to ask questions, send feedback and even point out mistakes. Great conversations start with just a single word.
How to write better comments?
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I am looking to describe all functions of type $f:\mathbb{R} \rightarrow \mathbb{R}$ as a vector space. Is this possible? What is a basis? Can I write any function of this type as $$g(x) = \int^{\infty}_{-\infty}a(k)sin(kx) + b(k)cos(kx)dk$$ for $a(k), b(k) \in \mathbb{R}$, and $k, \in \mathbb{R}$. I am assuming that $sin(kx)$ is a basis for this function space.
Let's suppose we have a basis of cardinality $Card(\mathbb{R})$, call it $B = \{b_1(kx), b_2(kx), \ldots \}, k \in \mathbb{R}$.
I don't understand why people are saying that the vector is a finite linear combination, but I can write a general function as
$$g(x) = \int^{\infty}_{-\infty}a_1(k)b_1(kx)+ a_2(k)b_2(kx) + \ldots dk$$
This seems like a continuous combination. Can someone explain this?
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Ah, the smell of a new seismic interpretation project. All those traces, all that geology — perhaps unseen by humans or indeed any multicellular organism at all since the Triassic. The temptation is to Just Start Interpreting, why, you could have a map by lunchtime tomorrow! But wait. There are some things to do first.
Once I've made sure all is present and correct (see
How to QC a seismic volume), I spend a bit of time making some helpful horizons... The surface.One of the fundamental horizons, the seafloor or ground surface is a must-have. You may have received it from the processor (did you ask for it?) or it may be hidden in the SEG-Y headers — ask whoever received or loaded the data. If not, ground elevation is usually easy enough to get from your friendly GIS guru. If you have to interpret the seafloor, at least it should autotrack quite well. Seafloor multiple model.In marine data, I always make a seafloor multiple model — just multiply the seafloor pick by 2. This will help you make sense of any anomalous reflectors or amplitudes at that two-way time. Maybe make a 3× version too if things look really bad. Remember, the 2× multiple will be reverse polarity. Other multiples.You can model the surface multiple of any strong reflectors with the same arithmetic — but the chances are that any residual multiple energy is quite subtle. You may want to seek help modeling them properly, once you have a 3D velocity model. Water depth markers.I like to make flat horizons* at important water depths, eg shelf edge (usually about 100–200 m), plus 1000 m, 2000 m, etc. This mainly helps to keep track of where you are, and also to think about prospectivity, accessibility, well cost, etc. You only want these to exist in the water, so delete them anywhere they are deeper than the seafloor horizon. Your software should have an easy way to implement a simple model for time tin ms, given depth din m and velocity** Vin m/s, e.g.
$$ t = \frac{2000 d}{V} \approx \frac{2000 d}{1490} \qquad \qquad \mathrm{e.g.}\ \frac{2000 \times 1000}{1490} = 1342\ \mathrm{ms} $$
Hydrate stability zone.In marine data and in the Arctic you may want to model the bottom of the gas hydrate stability zone (GHSZ) to help interpret bottom-simulating reflectors, or BSRs. I usually do this by scanning the literature for reports of BSRs in the area, or data on hydrate encounters in wells. In the figure above, I just used the seafloor plus 400 ms. If you want to try for more precision, Bale et al. (2014) provided several models for computing the position of the GHSZ — thank you to Murray Hoggett at Birmingham for that tip. Fold.It's very useful to be able to see seismic fold on a map along with your data, so I like to load fold maps at some strategic depths or, better yet, load the entire fold volume. That way you can check that anomalies (especially semblance) don't have a simple, non-geological explanation. Gravity and magnetics.These datasets are often readily available. You will have to shift and scale them to some sensible numbers, either at the top or the bottom of your sections. Gravity can be especially useful for interpreting rifted margins. Important boundaries.Your software may display these for you, but if not, you can fake it. Simply make a horizon that only exists within the polygon — a lease boundary perhaps — by interpolating within a polygon. Make this horizon flat and deep (deeper than the seismic), then merge it with a horizon that is flat and shallow (–1 ms, or anything shallower than the seismic). You should end up with almost-vertical lines at the edges of the feature. Section headings.I like to organize horizons into groups — stratigraphy, attributes, models, markers, etc. I make empty horizons to act only as headings so I can see at a glance what's going on. Whether you need do this, and how you achieve it, depends on your software.
Most of these horizons don't take long to make, and I promise you'll find uses for them throughout the interpretation project.
If you have other helpful horizon hacks, I'd love to hear about them — put your favourites in the comments. Footnotes
* It's not always obvious how to make a flat horizon. A quick way is to take some ubiquitous horizon — the seafloor maybe — and multiply it by zero.
The 2D seismic dataset shown is from the Laurentian Basin, offshore Newfoundland. The dataset is copyright of Natural Resources Canada, and subject to the Open Government License – Canada. You can download it from the OpendTect Open Seismic Repository. The cultural boundary and gravity data is fictitious — I made them up for the purposes of illustration.
References
Bale, Sean, Tiago M. Alves, Gregory F. Moore (2014). Distribution of gas hydrates on continental margins by means of a mathematical envelope: A method applied to the interpretation of 3D seismic data.
Geochem. Geophys. Geosyst. 15, 52–68, doi:10.1002/2013GC004938. Note: the equations are in the Supporting Information.
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Probability Seminar Spring 2019 Thursdays in 901 Van Vleck Hall at 2:25 PM, unless otherwise noted. We usually end for questions at 3:15 PM.
If you would like to sign up for the email list to receive seminar announcements then please send an email to join-probsem@lists.wisc.edu
January 31, Oanh Nguyen, Princeton
Title:
Survival and extinction of epidemics on random graphs with general degrees
Abstract: We establish the necessary and sufficient criterion for the contact process on Galton-Watson trees (resp. random graphs) to exhibit the phase of extinction (resp. short survival). We prove that the survival threshold $\lambda_1$ for a Galton-Watson tree is strictly positive if and only if its offspring distribution has an exponential tail, settling a conjecture by Huang and Durrett. On the random graph with degree distribution $D$, we show that if $D$ has an exponential tail, then for small enough $\lambda$ the contact process with the all-infected initial condition survives for polynomial time with high probability, while for large enough $\lambda$ it runs over exponential time with high probability. When $D$ is subexponential, the contact process typically displays long survival for any fixed $\lambda>0$. Joint work with Shankar Bhamidi, Danny Nam, and Allan Sly.
Wednesday, February 6 at 4:00pm in Van Vleck 911 , Li-Cheng Tsai, Columbia University
Title:
When particle systems meet PDEs
Abstract: Interacting particle systems are models that involve many randomly evolving agents (i.e., particles). These systems are widely used in describing real-world phenomena. In this talk we will walk through three facets of interacting particle systems, namely the law of large numbers, random fluctuations, and large deviations. Within each facet, I will explain how Partial Differential Equations (PDEs) play a role in understanding the systems..
Title:
Fluctuations of the KPZ equation in d\geq 2 in a weak disorder regime
Abstract: We will discuss some recent work on the Edwards-Wilkinson limit of the KPZ equation with a small coupling constant in d\geq 2.
February 14, Timo Seppäläinen, UW-Madison
Title:
Geometry of the corner growth model
Abstract: The corner growth model is a last-passage percolation model of random growth on the square lattice. It lies at the nexus of several branches of mathematics: probability, statistical physics, queueing theory, combinatorics, and integrable systems. It has been studied intensely for almost 40 years. This talk reviews properties of the geodesics, Busemann functions and competition interfaces of the corner growth model, and presents some new qualitative and quantitative results. Based on joint projects with Louis Fan (Indiana), Firas Rassoul-Agha and Chris Janjigian (Utah).
February 21, Diane Holcomb, KTH
Title:
On the centered maximum of the Sine beta process Abstract: There has been a great deal or recent work on the asymptotics of the maximum of characteristic polynomials or random matrices. Other recent work studies the analogous result for log-correlated Gaussian fields. Here we will discuss a maximum result for the centered counting function of the Sine beta process. The Sine beta process arises as the local limit in the bulk of a beta-ensemble, and was originally described as the limit of a generalization of the Gaussian Unitary Ensemble by Valko and Virag with an equivalent process identified as a limit of the circular beta ensembles by Killip and Stoiciu. A brief introduction to the Sine process as well as some ideas from the proof of the maximum will be covered. This talk is on joint work with Elliot Paquette.
Title: Quantitative homogenization in a balanced random environment
Abstract: Stochastic homogenization of discrete difference operators is closely related to the convergence of random walk in a random environment (RWRE) to its limiting process. In this talk we discuss non-divergence form difference operators in an i.i.d random environment and the corresponding process—a random walk in a balanced random environment in the integer lattice Z^d. We first quantify the ergodicity of the environment viewed from the point of view of the particle. As consequences, we obtain algebraic rates of convergence for the quenched central limit theorem of the RWRE and for the homogenization of both elliptic and parabolic non-divergence form difference operators. Joint work with J. Peterson (Purdue) and H. V. Tran (UW-Madison).
Wednesday, February 27 at 1:10pm Jon Peterson, Purdue
Title:
Functional Limit Laws for Recurrent Excited Random Walks
Abstract:
Excited random walks (also called cookie random walks) are model for self-interacting random motion where the transition probabilities are dependent on the local time at the current location. While self-interacting random walks are typically very difficult to study, many results for (one-dimensional) excited random walks are remarkably explicit. In particular, one can easily (by hand) calculate a parameter of the model that will determine many features of the random walk: recurrence/transience, non-zero limiting speed, limiting distributions and more. In this talk I will prove functional limit laws for one-dimensional excited random walks that are recurrent. For certain values of the parameters in the model the random walks under diffusive scaling converge to a Brownian motion perturbed at its extremum. This was known previously for the case of excited random walks with boundedly many cookies per site, but we are able to generalize this to excited random walks with periodic cookie stacks. In this more general case, it is much less clear why perturbed Brownian motion should be the correct scaling limit. This is joint work with Elena Kosygina.
March 21, Spring Break, No seminar March 28, Shamgar Gurevitch UW-Madison
Title:
Harmonic Analysis on GLn over finite fields, and Random Walks
Abstract: There are many formulas that express interesting properties of a group G in terms of sums over its characters. For evaluating or estimating these sums, one of the most salient quantities to understand is the
character ratio:
$$ \text{trace}(\rho(g))/\text{dim}(\rho), $$
for an irreducible representation $\rho$ of G and an element g of G. For example, Diaconis and Shahshahani stated a formula of this type for analyzing G-biinvariant random walks on G. It turns out that, for classical groups G over finite fields (which provide most examples of finite simple groups), there is a natural invariant of representations that provides strong information on the character ratio. We call this invariant
rank. This talk will discuss the notion of rank for $GL_n$ over finite fields, and apply the results to random walks. This is joint work with Roger Howe (Yale and Texas AM). April 4, Philip Matchett Wood, UW-Madison
Title:
Outliers in the spectrum for products of independent random matrices
Abstract: For fixed positive integers m, we consider the product of m independent n by n random matrices with iid entries as in the limit as n tends to infinity. Under suitable assumptions on the entries of each matrix, it is known that the limiting empirical distribution of the eigenvalues is described by the m-th power of the circular law. Moreover, this same limiting distribution continues to hold if each iid random matrix is additively perturbed by a bounded rank deterministic error. However, the bounded rank perturbations may create one or more outlier eigenvalues. We describe the asymptotic location of the outlier eigenvalues, which extends a result of Terence Tao for the case of a single iid matrix. Our methods also allow us to consider several other types of perturbations, including multiplicative perturbations. Joint work with Natalie Coston and Sean O'Rourke.
April 11, Eviatar Procaccia, Texas A&M Title: Stabilization of Diffusion Limited Aggregation in a Wedge.
Abstract: We prove a discrete Beurling estimate for the harmonic measure in a wedge in $\mathbf{Z}^2$, and use it to show that Diffusion Limited Aggregation (DLA) in a wedge of angle smaller than $\pi/4$ stabilizes. This allows to consider the infinite DLA and questions about the number of arms, growth and dimension. I will present some conjectures and open problems.
April 18, Andrea Agazzi, Duke
Title:
Large Deviations Theory for Chemical Reaction Networks
Abstract: The microscopic dynamics of well-stirred networks of chemical reactions are modeled as jump Markov processes. At large volume, one may expect in this framework to have a straightforward application of large deviation theory. This is not at all true, for the jump rates of this class of models are typically neither globally Lipschitz, nor bounded away from zero, with both blowup and absorption as quite possible scenarios. In joint work with Amir Dembo and Jean-Pierre Eckmann, we utilize Lyapunov stability theory to bypass this challenges and to characterize a large class of network topologies that satisfy the full Wentzell-Freidlin theory of asymptotic rates of exit from domains of attraction. Under the assumption of positive recurrence these results also allow for the estimation of transitions times between metastable states of this class of processes.
April 25, Kavita Ramanan, Brown
Title:
Beyond Mean-Field Limits: Local Dynamics on Sparse Graphs
Abstract: Many applications can be modeled as a large system of homogeneous interacting particle systems on a graph in which the infinitesimal evolution of each particle depends on its own state and the empirical distribution of the states of neighboring particles. When the graph is a clique, it is well known that the dynamics of a typical particle converges in the limit, as the number of vertices goes to infinity, to a nonlinear Markov process, often referred to as the McKean-Vlasov or mean-field limit. In this talk, we focus on the complementary case of scaling limits of dynamics on certain sequences of sparse graphs, including regular trees and sparse Erdos-Renyi graphs, and obtain a novel characterization of the dynamics of the neighborhood of a typical particle. This is based on various joint works with Ankan Ganguly, Dan Lacker and Ruoyu Wu.
Friday, April 26, Colloquium, Van Vleck 911 from 4pm to 5pm, Kavita Ramanan, Brown
Title:
Tales of Random Projections
Abstract: The interplay between geometry and probability in high-dimensional spaces is a subject of active research. Classical theorems in probability theory such as the central limit theorem and Cramer’s theorem can be viewed as providing information about certain scalar projections of high-dimensional product measures. In this talk we will describe the behavior of random projections of more general (possibly non-product) high-dimensional measures, which are of interest in diverse fields, ranging from asymptotic convex geometry to high-dimensional statistics. Although the study of (typical) projections of high-dimensional measures dates back to Borel, only recently has a theory begun to emerge, which in particular identifies the role of certain geometric assumptions that lead to better behaved projections. A particular question of interest is to identify what properties of the high-dimensional measure are captured by its lower-dimensional projections. While fluctuations of these projections have been studied over the past decade, we describe more recent work on the tail behavior of multidimensional projections, and associated conditional limit theorems.
May 7, Tuesday Van Vleck 901, 2:25pm,, Duncan Dauvergne (Toronto)
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Line 25: Line 25: == q-Integral == == q-Integral == - For a function $f$, the q-integral from $0$ to $1$ is defined as + For a function $f$, the q-integral from $0$ to $1$ ("$z$-integral" if we stick to our notation above) is defined as - $\sum_{k=0}^\infty f(z^k)\,z^k=\dfrac{1}{1-z}\int_0^1 f(s){\mathrm d}_zs$ + $\sum_{k=0}^\infty f(z^k)\,z^k=\dfrac{1}{1-z} \cdot\int_0^1 f(s) \,{\mathrm d}_zs$ == Related notes == == Related notes ==
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User:Nikita2/sandbox A mapping $\varphi:D\to D'$ possesses Luzin's $\mathcal N$-property if the image of every set of measure zero is a set of measure zero. A mapping $\varphi$ possesses Luzin's $\mathcal N{}^{-1}$-property if the preimage of every set of measure zero is a set of measure zero. Briefly\begin{equation*}\mathcal N\text{-property:}\quad \Sigma\subset D, |\Sigma| = 0 \Rightarrow |\varphi(\Sigma)|=0,\end{equation*}\begin{equation*}\mathcal N{}^{-1}\text{-property:} \quad M \subset D, |M| = 0 \Rightarrow |\varphi^{-1}(M)|=0.\end{equation*} Contents $\mathcal N$-property of a function $f$ on an interval $[a,b]$
Let $f:[a,b]\to \mathbb R$ be a measurable function. In this case the definition is following:
For any set $E\subset[a,b]$ of measure zero ($|E|=0$), the image of this set, $f(E)$, also has measure zero.It was introduced by N.N. Luzin in 1915 (see [1]). The following assertions hold. A function $f\not\equiv \operatorname{const}$ on $[a,b]$ such that $f'(x)=0 $ almost-everywhere on $[a,b]$ (see for example Cantor ternary function) does not have the Luzin $\mathcal N$-property. If $f$ does not have the Luzin $\mathcal N$-property, then on $[a,b]$ there is a perfect set $P$ of measure zero such that $|f(P)|>0$. An absolutely continuous function has the Luzin $\mathcal N$-property. If $f$ has the Luzin $\mathcal N$-property and has bounded variation on $[a,b]$ (as well as being continuous on $[a,b]$), then $f$ is absolutely continuous on $[a,b]$ (the Banach–Zaretskii theorem). If $f$ does not decrease on $[a,b]$ and $f'$ is finite on $[a,b]$, then $f$ has the Luzin $\mathcal N$-property. In order that $f(E)$ be measurable for every measurable set $E\subset[a,b]$ it is necessary and sufficient that $f$ have the Luzin $\mathcal N$-property on $[a,b]$. A function $f$ that has the Luzin $\mathcal N$-property has a derivative $f'$ on the set for which any non-empty portion of it has positive measure. For any perfect nowhere-dense set $P\subset[a,b]$ there is a function $f$ having the Luzin $\mathcal N$-property on $[a,b]$ and such that $f'$ does not exist at any point of $P$.
The concept of Luzin's -property can be generalized to functions of several variables and functions of a more general nature, defined on measure spaces.
References
[1] N.N. Luzin, "The integral and trigonometric series" , Moscow-Leningrad (1915) (In Russian) (Thesis; also: Collected Works, Vol. 1, Moscow, 1953, pp. 48–212) Comments
There is another property intimately related to the Luzin -property. A function continuous on an interval has the Banach -property if for all Lebesgue-measurable sets and all is a such that
This is clearly stronger than the -property. S. Banach proved that a function has the -property (respectively, the -property) if and only if (respectively, only if — see below for the missing "if" ) the inverse image is finite (respectively, is at most countable) for almost-all in . For classical results on the - and -properties, see [a3].
Recently a powerful extension of these results has been given by G. Mokobodzki (cf. [a1], [a2]), allowing one to prove deep results in potential theory. Let and be two compact metrizable spaces, being equipped with a probability measure . Let be a Borel subset of and, for any Borel subset of , define the subset of by (if is the graph of a mapping , then ). The set is said to have the property (N) (respectively, the property (S)) if there exists a measure on (here depending on ) such that for all ,
(respectively, for all there is a such that for all one has
Now has the property (N) (respectively, the property (S)) if and only if the section of is at most countable (respectively, is finite) for almost-all .
References
[a1] C. Dellacherie, D. Feyel, G. Mokobodzki, "Intégrales de capacités fortement sous-additives" , Sem. Probab. Strasbourg XVI , Lect. notes in math. , 920 , Springer (1982) pp. 8–28 MR0658670 Zbl 0496.60076 [a2] A. Louveau, "Minceur et continuité séquentielle des sous-mesures analytiques fortement sous-additives" , Sem. Initiation à l'Analyse , 66 , Univ. P. et M. Curie (1983–1984) Zbl 0587.28003 [a3] S. Saks, "Theory of the integral" , Hafner (1952) (Translated from French) MR0167578 Zbl 1196.28001 Zbl 0017.30004 Zbl 63.0183.05 [a4] E. Hewitt, K.R. Stromberg, "Real and abstract analysis" , Springer (1965) MR0188387 Zbl 0137.03202 AsymptotePlay
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How to Cite This Entry:
Nikita2/sandbox.
Encyclopedia of Mathematics.URL: http://www.encyclopediaofmath.org/index.php?title=Nikita2/sandbox&oldid=29412
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Define the function $f:\mathbb{N}\to\mathbb{R}$ as
$$ f(n)=e^\gamma n \log \log n, $$
where $e$ is Euler's constant and $\gamma$ is the Euler-Mascheroni constant. Then Robin's criterion states that for all positive integers $n>5040$, the statement
$$ \sigma(n) < f(n), $$
where $\sigma$ denotes the divisor sum function, is equivalent to the Riemann Hypothesis.
It is then natural to study the minima of the function $g$ defined by
$$ g(n)=f(n)-\sigma(n). $$
To my surprise, after (
briefly) googling the subject, I found no reference to "record integers" $n$ that locally minimize $g$. Perhaps a more intuitive way to ask the same question is by studying another function $h$ defined as the quotient $\frac{\sigma(n)}{f(n)}$. However, I found no references for such studies either.
It is obvious that colossally abundant numbers attain such minima, however I am interested in whether there exists a "list" of record integers $n$ for which $\sigma(n)$ approaches $f(n)$ the most. Is there such a list (perhaps in the OEIS)?
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This is the first in a series of articles in which I want to transmit what I learned (or what I think I learned) from the books, papers and lectures of Alexander Stepanov.
These are the lessons that Alex gives us, and I want to show them in this series:
Specify our algorithms correctly Programming must be based on a solid mathematical foundation Designing our API’s consistently Not always the library implementations provided by the programming languages we use are correct, even though they are designed by “experts”. The concept of Stability Generic programming[1], of course!
And… the following lesson is mine:
Please don’t blindly accept what it is expressed on this blog. In case of doubt you should go to the source, the Elements of Programming book [2]
In this article I want to avoid using any programming language, I want to focus on the algorithms and the specifications. In subsequent articles, I will try to implement what we learned using several mainstream programming languages.
Writing min
I will try to write the function
min, that is, a function that returns the minimum of two things.
At this time you may be wondering, this guy is writing an entire blog post about a two-line function, is this serious?The answer is yes. As Alex says:
“Simple things are beautiful”, and believe it or not, we can learn a lot in the process of writing min.
The objective is to learn to correctly determine what are the
requirements that a function must impose on types used in it. “It is better to design our Components (algorithms and data structures) not in terms of concrete types, but in terms of requirements on types expressed as syntactic and semantic properties”
Alex calls a collection of requirements a
Concept[3].
Despite having no support for
Concepts in programming languages, he has been using them for decades, not in code, but in his mind and in the documentation of the components developed by him [4]. First Attempt
Well, let’s start writing the specification and then, the code:
Spec: Given two objects[5], a and b, return the smaller of both.
The above function is written in pseudo-code (which looks like a mix between
C and Python), it has some flaws, but we will see them later.
The most important question is… What are the requirements of
min function must impose to the arguments a and b? That is… Which are the Concepts?
For some programmers, especially those advocates of duck-typing, imposing requirements to arguments may be something uninteresting. They simply use the arguments in the function body, and they hope to at least get a runtime error.
I strongly disagree!
“Even if we do not have Concepts in the language, they should exists in our mind. You have to learn to think in terms of Concepts whichever language you use.”
Forget for a while about programming languages, let’s see what the requirements are. The problem arises from this code snippet
a < b
What does this mean?
Someone could say that the requirement is that arguments a and b must be compared using the less-than-operator.
But this is just a syntactic requirement, we have to go further in order to correctly specify our function. We need to include
in our type requirements! But… how to do it? semantics Mathematics to the rescue!
What is the less-than-operator?
It is a way for comparing two objects, returning a boolean value.Is this enough for defining
min function?
No, and to ilustrate that, see what happened if the less-than-operator is defined this way:
This function returns
true if the number of seconds of the system time is even, otherwise returns false. With this code I want to emphasize that the less_than_operator could be defined using a random behaviour, but we need to define an specific behaviour.
Mathematically the less-than-operator is a
Relation[6]. A relation is a binary Predicate[6].
That is, a predicate that takes two parameters of the same type.
“If you look of two things, is either true or false. The relation holds, or not.”The difference between the code above and a relation is that the relation is considered a FunctionalProcedure[6], that is, a function in which by replacing its inputs with equal objects results in equal output objects.
But the
relation concept is too weak, we need a stronger concept: Ordering. “What is an ordering? What do mathematicians call ordering? The only absolute rule for ordering is the requirement of transitivity [6]. A relation is transitive if, whenever it holds between a and b, and between b and c, it holds between a and c. A transitive relation is the most basic notion of ordering, but it is still too weak for our needs.”
Let’s review what kinds of Ordering Relations exist:
: A Partial Ordering Partial Orderingis an ordering relationin which not every pair of elements need to be related. Examples: “The canonical example of Partial Ordering is the Subset Relation” Subset are ordered, one subset could be a Subset of another subset, for example, the subset {2, 4} Is a Subset Ofthe subset {1, 2, 3, 4}. But it also happens that there are subsets which you could said nothing about, for example, given {2, 4} and {3, 5}. Which one is greater? Which one includes the other? It is not defined! We have two kinds of Partial Ordering: (or Non-Strict Partial Ordering): A relation is a Reflexive Partial Ordering if it is Reflexive Partial Ordering transitive, reflexive[6] and antisymmetric[6]. (or Non-Reflexive Partial Ordering): A relation is a Strict Partial Ordering if it is Strict Partial Ordering transitiveand ireflexive[6] (it is also asymmetric[6], but this axiom is implied by irreflexivity and transitivity) : a Total Ordering Total Orderingis an Ordering Relation in which any pair of elements in the set of the relation are comparable under the relation. Total Ordering is a specialization of Partial Ordering. Examples: The Real numbers ordered by the less-than relation (<) (also Rational, Integers and Natural numbers) The letters of the alphabet ordered by the natural dictionary order. We have two kinds of Total Ordering: (or Non-Strict Total Ordering): A relation is a Reflexive Total Ordering if it is transitive, antisymmetric and Reflexive Total Ordering total[6]. (it is also reflexive, but is implied by totally) [6] (or Non-Reflexive Total Ordering): A relation is a Strict Total Ordering if it is Strict Total Ordering transitiveand obeys the trichotomy law, whereby for every pair of elements, exactly one of the following holds: the relation, its converse, or equality. (It is also irreflexive, but this axiom is implied by the trichotomy law)
(Note: There are more ordering relations, but we will see them later)
Writing min using Concepts
Well, now we know about ordering relations, let’s look at how we can use them to specify the
min function.
But first, what should we use? Partial or Total Ordering?
“If our relation is the Subset relation on a set, then, _min_ and _max_ of two sets doesn’t make sense.”
Then, Partial Ordering is too weak, because the relation doesn’t hold for every pair of elements of the set.
We need to use Total Ordering for define the requirements of
min, let’s do it:
Note that the requirements were expressed as code comments. Later we will see what the programming languages provide us to express them as code.
Well, this is enough for a single post.
In the next articles of the series, we will:
complete and fix the errors of the implementation of min write the maxfunction refine the requirements of minand max implement them in real programming languages analyze the API provided by some programming languages.
Stay tuned!
Acknowledgments
Thanks in particular to the following for their feedback to improve this article: Mario Dal Lago, Andrzej Krzemienski, Dean Michael Berris, Javier Centurión, Alejandro Santos, Ezequiel Reyno.
The Series
Part 1: The rise of Concepts
Part 2: Understanding Concepts Part 3: Weakening the ordering Part 4: Const-Correctness Part 5: Stabilizing the algorithm References
[1] Original definition of “Generic Programming”:
David R. Musser and Alexander A. Stepanov: Generic Programming. ISSAC 1988, pages 13-25. http://www.stepanovpapers.com/genprog.pdf [2] Elements of Programming of Alexander Stepanov and Paul McJones: http://www.elementsofprogramming.com Buy it at Amazon [3] Concept definition: Stepanov and McJones [2009, page 10] [4] SGI’s STL using Concepts in Documentation: https://www.sgi.com/tech/stl/min.html [5] Object definition: The definition used in this article has nothing to do with an OOP-like definition of object [7]. The definition used here is a practical definition of what an object is: “Object is a sequence of bits in memory” or “Object is a value residing in memory” See Stepanov and McJones [2009, page 4] for a complete definition. [6] See Appendix A [7] Object-Oriented Software Construction (2nd Ed) by Bertrand Meyer [1997, page 1198] Appendix A: Definitions
Some of the definitions presented here are based on: http://www.elementsofprogramming.com/eop-concepts.pdf
$$Relation(\texttt{Op}) \triangleq \\\qquad HomogeneousPredicate(\texttt{Op}) \\\quad \land \texttt{ Arity(Op) = 2}$$
$$HomogeneousPredicate(\texttt{P}) \triangleq \\\qquad Predicate(\texttt{P}) \\\quad \land HomogeneousFunction(\texttt{P})$$
$$Predicate(\texttt{P}) \triangleq \\\qquad FunctionalProcedure(\texttt{P}) \\\quad \land \texttt{Codomain(P) = bool}$$
$$\textbf{property}\text{(R : Relation)} \\\text{transitive : R} \\\quad r \mapsto (\forall a, b, c \in \texttt{Domain(R)}) (r(a, b) \land r(b, c) \Rightarrow r(a, c))$$
$$\textbf{property}\text{(R : Relation)} \\\text{refexive : R} \\\quad r \mapsto (\forall a \in \texttt{Domain(R)}) (r(a, a))$$
$$\textbf{property}\text{(R : Relation)} \\\text{antisymmetric : R} \\\quad r \mapsto (\forall a, b \in \texttt{Domain(R)}) (r(a, b) \land r(b, a) \Rightarrow a = b)$$
$$\textbf{property}\text{(R : Relation)} \\\text{irreflexive : R} \\\quad r \mapsto (\forall a \in \texttt{Domain(R)}) (\lnot r(a, a))$$
$$\textbf{property}\text{(R : Relation)} \\\text{asymmetric : R} \\\quad r \mapsto (\forall a, b \in \texttt{Domain(R)}) (r(a, b) \Rightarrow \lnot r(b, a))$$
$$\textbf{property}\text{(R : Relation)} \\\text{total : R} \\\quad r \mapsto (\forall a, b \in \texttt{Domain(R)}) (r(a, b) \lor r(b, a))$$
$$\textbf{property}\text{(R : Relation)} \\\text{total_ordering : R} \\\quad \text{r} \mapsto \text{transitive(r) } \land \\\quad \text{ (} \forall \text{ a, b} \in \texttt{Domain(R)} \text{) exactly one of the following holds: r(a, b), r(b, a), or a = b}$$
$$TotallyOrdered(\texttt{T}) \triangleq \\\qquad \texttt{Regular(T)} \\\quad \land <\texttt{: T x T} \rightarrow \text{bool} \\\quad \land total\_ordering(<)$$
For definitions of: HomogeneousFunction, FunctionalProcedure, and Regular, see http://www.elementsofprogramming.com/eop-concepts.pdf [page 1]Share
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How many solutions do for the following equation exists?
\begin{align} \varphi(n)=\pi(n), n\in\mathbb{N}, \end{align}
where $\varphi(n)$ is Euler's totient function and $\pi(n)$ is the prime-counting function.
Thank you very much
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Surprisingly, the list of solutions is finite: 2, 3, 4, 8, 10, 14, 20, 90. Moreover $\varphi(n)>\pi(n)$ for $n>90$.
A proof is given at page 179 in the following paper by Leo Moser: "On the equation $\varphi(n)=\pi(n)$", Pi Mu Epsilon J. 1951, 177–180.
Sketch of Moser's proof. We have that $\varphi(n)-\pi(n)=B(n) - A(n)$ where $A(n)$ is the number of prime divisors of $n$ and $B(n)$ is the number of non-primes, which do not exceed $n$ and are relatively prime to $n$.
Now i) $\pi(\sqrt{n})\geq 2A(n)$ for $n>360$ (lemma 3 where Bertrand's postulate is used) and ii) $B(n)>\pi(\sqrt{n})-A(n)$ (lemma 4).
Hence by for $n>360$, $$\varphi(n)-\pi(n)=B(n) - A(n)>\pi(\sqrt{n})-A(n)-A(n)\geq 0.$$
P.S. According http://oeis.org/A037171, the result has been proved by David W. Wilson and Jeffrey Shallit, but unfortunately no reference is provided.
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This question is somewhat naive. Please see the following proof before reading the question itself.
Let $L$ be a vector space over a field $\Bbb{F}.$ Assume there is an associative multiplication on $L$ so that $L$ is an associative ring, and denote this multiplication simply as $xy$ for $x,y\in L$. Furthermore, suppose this multiplication to be compatible with scaling, that is $\alpha\in\Bbb{F},$ then $\alpha x = x\alpha\in L$ for all $x\in L$ ($L$ should be a left and right $\Bbb{F}$ vector space). Define $[x,y]=xy-yx$ for all $x,y\in L.$ Then $[-,-]:L\times L \longrightarrow L$ is a Lie bracket.
Proof: It is clear that $[-,-]$ is anti-commutative, and bilinear. We only need to show that it satisfies the Jacobi identity. That is
$$[[x,y],z]+[[y,z],x]+[[z,x],y]=0.$$
One readily sees that this is equivalent to
$$(xy)z-(yx)z-z(xy)+z(yx)+(yz)x-(zy)x-x(yz)+x(zy)+(zx)y-(xz)y-y(zx)+y(xz)=0.$$
The previous equation is certainly true when the multiplication defined earlier is associative. Therefore $L$ is a Lie algebra. $\Box$
So here's the question:
1) Is it true that an algebra defined in this way is a Lie algebra?
An example of such an associative algebra is the $M_{n\times n}(\Bbb{F}).$
2) If so, is there any name which separates these from the classical Lie algebras?
There is rich theory surrounding classical Lie algebras, especially their connection to Lie groups. This motivates the next question.
3) Is there any significance in viewing an associative algebra in this way?
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This is a long question, so I thank anyone willing to help me sort it out. That said, the material is elementary (to anyone who's taken real analysis) and I'm just long winded. The catalyst of the question is the following theorem and proof, which I repeat here verbatim:
Compact subsets of metric spaces are closed.
\begin{proof}
Let $K$ be a compact subset of a metric space $X$. We shall prove that the complement of $K$ is an open subset of $X$.
Suppose $p \in X$, $p \not\in K$. If $q \in K$, let $V_q$ and $W_q$ be the neighborhoods of $p$ and $q$, respectively, of radius less than $\frac{1}{2} d(p,q)$. Since $K$ is compact, there are finitely many points $q_1, \ldots, q_n$ in $K$ such that $$ K \subset W_{q_1} \cup\cdots\cup W_{q_n} = W. $$ If $V= V_{q_1} \cap\cdots\cap V_{q_n}$, then $V$ is a neighborhood of $p$ which does not intersect $W$. Hence $V \subset K^c$, so that $p$ is an interior point of $K^c$. The theorem follows. \end{proof}
Yes, $W_q$ of a neighborhood with center $q$ while $V_q$ is a neighborhood of center $p$, not great notation IMHO. Here, I have included the entire proof for the sake of completeness, but what I do not understand is simply the line: ''Since $K$ is compact, there are finitely many points $q_1, \ldots, q_n$ in $K$ such that $ K \subset W_{q_1} \cup\cdots\cup W_{q_n} = W. $''
This confuses me for two unrelated (I believe) reasons:
As far as I can understand, the definition of a compact subset $K$ of a metric space $X$ is not that
there existsa finite cover $W_1 \ldots , W_n$ of $K$ in $X$ (which is what the author seems to say) but, rather, that ifthere exists an open cover of $K$ in $X$ thenthere exists a finite sub-cover of $K$ in $X$. So, the question is: Before considering the finite open cover$W_{q_1}, \ldots, W_{q_n}$ of$K$ in$X$, is it not necessary to, first, justify the existence of any open cover?
It seems strange to me that the author asserts that the set $\{q_1, \ldots, q_n\}$ which indexes the open cover $W_{q_1}, \ldots, W_{q_n}$ of the compact set $K$ is, itself, a subset of $K$. This is important because the author defines the neighborhoods in terms of these $\{q_1, \ldots, q_n\}$: $\forall q \in K, \hspace{1mm} W_q := \left\{x \in X : d(x,q)< r \right\}, \hspace{1mm} r <\frac{1}{2} d(p,q)$. Here, I have written out the definition of a neighborhood with center $p$ and radius $r <\frac{1}{2} d(p,q)$ for clarity; $d$ is the (arbitrary) metric in the definition of the (arbitrary) metric space $X$. So, the question is:
Taking for given that there exists a finite open cover$W_{q_1}, \ldots, W_{q_n}$ of$K$ in$X$, how does one justify that$q_1, \ldots, q_n \in K$?
For reference, the book I'm using is
Principles of Mathematical Analysis 3e by Walter Rudin: one of the most popular texts at the undergraduate level, if I understand correctly.The above is Theorem 2.34.
For reference, I will also give this author's definition of a compact subset of a metric space:
''A subset $K$ of a metric space $X$ is said to be compactif every open cover of $K$ contains a finitesubcover. More explicitly, the requirement is that if $\{G_\alpha\}$ is an open cover of $K$, then there are finitely many indices $\alpha_1, \ldots, \alpha_n$ such that $K \subset G_{\alpha_1} \cup\cdots\cup G_{\alpha_n}$.''
Notice that he does not say 'indeces in $K$'. Also, this requires an open cover, as well as the notation $\{G_\alpha\}$ to be defined.
''By an
open coverof a set $E$ in a metric space $X$ we mean a collection $\{G_\alpha\}$ of open subsets of $X$ such that $E \subset\bigcup_\alpha G_\alpha.$''
''Let $A$ and $\Omega$ be sets, and suppose that with each element $\alpha$ of $A$ there is associated a subset of $\Omega$ which we denote by $E_\alpha$. The set whose elements are the sets $E_\alpha$ will be denoted by $\{E_\alpha\}$.''
Notice that the set $\Omega$ is, in general, not the same as the set $A$ which contains the indices $\alpha$. Hence, my second question.
Thanks, already, if you made it this far! I tried to make that to be not excessively dense reading material.
Edit: I mistook a $p$ for a $q$ and fixed it (I think)
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Suppose I am asked to show that some topology is not metrizable. What I have to prove exactly for that ?
Since you've used the tag
Functional analysis, you might be interested in non-metrisability of certain topologies ubiquitous in analysis: Two questions on the Zariski topology on $\mathbb{R}$
Some other examples can be found here.
Recall that if $(X,d)$ is a metric space, then the metric induces a topology. But different metrics can induce the same topology, for example $(X,d')$ where $d'(x,y)=2d(x,y)$, is a different metric, but induces the same topology.
We say that a topological space is
metrizable, if there is a metric which induces the topology.
So to show that a space is not metrizable, you have to show that there is no metric which can induce this topology. This is often done by refuting certain consequences of metrizability. For example we can prove that metric induced topology is always Hausdorff, and first-countable.
If a space is not first-countable, it's not a metric space. If it is not Hausdorff it's not a metric space.
If you have more information on the space, then you can use other conditions as well, e.g. connected metric space with two points is uncountable. If the space is countable and connected then it is not metrizable.
And so on and so forth.
If a topology is metrizable, then the "diagonal sequence trick" is available. This means that if you have a sequence $$ x_{(n)} \to x,\qquad n \to \infty $$ and each term of the sequence is the limit of another sequence, belonging to a "good" set $G$: $$ G\ni x^{k}_{(n)}\to x_{(n)}, \qquad k\to \infty $$ then you can construct a "diagonal" sequence $x^{k(n)}_{(n)}\in G$ which converges to the original limit point $x$. In particular, $x$ is in the sequential closure of the good set $G$.
This fact is usually phrased in topological terms as "topological closure and sequential closure are the same". So the first thing to try when proving that a topology is not metrizable is showing that the diagonal sequence trick does not work.
P.S.: I have in mind von Neumann's proof that the weak topology on $L^2(\mathbb{S}^1)$-space is not metrizable. It goes as follows. Let $$G=\left\{ e^{int}+n e^{imt}\ | n,m\in \mathbb{N}_{\ge 1}\right\}$$ be the "good" set in $L^2(\mathbb{S}^1)$. The question is whether we can or we cannot reach $0$ by taking one weak limit in the good set. The answer is negative because, roughly speaking, to reach $0$ we need to let $n$ go to infinity, otherwise the term $e^{int}$ would never vanish in the weak limit. But letting $n$ go to infinity the term $ne^{imt}$ diverges in norm, hence it is not bounded in weak sense as well. So $0$ is not in the sequential closure of $G$.
However, if we let $n$ be fixed and we take a limit in $m$, then the term $ne^{imt}$ vanishes in the weak limit. With the above terminology we have $$ e^{int}+ne^{imt}=x^m_{(n)}\rightharpoonup x_{(n)}=e^{int}, \qquad m\to \infty.$$ If we now let $n\to \infty$ we get $$ x_{(n)}=e^{int}\rightharpoonup 0,$$ but we have already seen that $0$ is not in the sequential closure of $G$. So the diagonal sequence trick is not available in $L^2(\mathbb{S}^1)$-space with weak topology. Hence that topology is not metrizable.
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Math question on Newton's method and detecting actual zeros - Printable Version
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/thread-7742.html) Math question on Newton's method and detecting actual zeros - Han - 02-07-2017 05:04 PM
(Admins: If this is in the wrong forum, please feel free to move it)
This came up during a debugging process in which Newton's method (using backtracking linesearch) gave me a solution to the system
\[ \frac{x\cdot y}{x+y} = 127\times 10^{-12}, \quad \left( \frac{x+y}{x} \right)^2 = 8.377 \]
(This problem was posed on the HP Prime subforum: http://hpmuseum.org/forum/thread-7677.html)
One solution I found was: \( x=1.94043067156\times 10^{-10}, \
y=3.67576704293\times 10^{-10} \) (hopefully no typos).
On the Prime, the error for the equations are in the order of \(10^{-19} \) and \(10^{-11}\) for the first and second equations, respectively (again, assuming I made no typos copying). So my question is: should a numerical solver should treat \(1.27\times 10^{-10}\) as "significant" or 0 (especially when it comes time to check for convergence, when the tolerance for \( |f_i| \) might be set to, say, \( 10^{-10} \) -- here \( f_i \) is the i-th equation in the system, set equal to 0)?
RE: Math question on Newton's method and detecting actual zeros - Valentin Albillo - 02-07-2017 06:45 PM
.
Hi, Han:
(02-07-2017 05:04 PM)Han Wrote: (Admins: If this is in the wrong forum, please feel free to move it)
Your system is trivial to solve by hand, like this:
1) Parameterize:
y = t*x
2) Substitute y=t*x into the first equation (a = 127E-12):
x*t*x = a*(x+t*x) -> t*x^2 = a*(1+t)*x -> (assuming x is not 0, which would make the second equation meaningless) t*x = a*(1+t) -> x = a*(1+t)/t
3) Substitute y=t*x in the second equation (b=8.377)
(1+t)^2 = b -> 1+t= sqr(b) -> t = sqr(b)-1 or t = -sqr(b)-1
4) let's consider the first case (the second is likewise):
t = sqr(b)-1 = 1.8943047524405580466334231771918
5) substitute the value of t in the first equation above in (2):
x = a*(1+t)/t = 1.9404306676968291608003859882111e-10
6) now, y=t*x, so:
y = t*x = 3.6757670355995087192244474350336e-10
which gives your solution. Taking the negative sqrt would give another.
As for your question, the best way to check for convergence is not to rely on some tolerance for the purported zero value when evaluating both equations for the computed x,y approximations in every iteration but rather to stop when consecutive approximations differ in less than a user-set tolerance expressed in ulps, i.e. units in the last place.
For instance, if you're making your computation with 10 digits and you set your tolerance to 2 ulps you would stop iterating as soon as consecutive approximations for both x and y have 8 digits in common (mantissa digits, regardeless of the exponents which of course should be the same).
Once you stop the iterations you should then check the values of f(x,y) and g(x,y) to determine whether you've found a root, a pole, or an extremum (maximum, minimum) but as far as stopping the iterations is concerned, the tolerance in ulps is the one to use for best results as it is completely independent of the magnitude of the roots, they might be of the order of 1E38 or of 1E-69 and it wouldn't matter.
Regards.
V.
.
RE: Math question on Newton's method and detecting actual zeros - Han - 02-07-2017 08:03 PM (02-07-2017 06:45 PM)Valentin Albillo Wrote: .
Thank you for the detailed solution; though in truth it was merely to present a case where a function might itself produce outputs that are extremely tiny. The math I understand quite well; it's the computer science part of implementing Newton's method that was giving me trouble. Your explanation above regarding ulps was precisely the answer I was looking for.
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In general, an algebraic closure of a field $K$ is denoted by $\overline{K}$. Typical examples arising in number theory are $K=\mathbb{Q}$, $K=\mathbb{F}_p(t)$, $K=\mathbb{Q}_p$. Usually one needs the axiom of choice in order to prove the existence of algebraic closures. There are (at least) two exceptions: For $K=\mathbb{R}$ we have $\overline{K}=\mathbb{C}$. Then, for every subfield $K \subseteq \mathbb{R}$, we may realize $\overline{K}$ as the subfield of $\mathbb{C}$ which consists of complex numbers which are algebraic over $K$. It also exists without AC. We may also replace $\mathbb{R}$ by a real closed field, one only has to adjoin $\sqrt{-1}$. For $K=\mathbb{F}_q$, a finite field, we have for every $n$ an extension $\mathbb{F}_{q^n}$ of degree $n$ and every divisibility relation $n|m$ induces a canonical $\mathbb{F}_q$-homomorphism $\mathbb{F}_{q^n} \to \mathbb{F}_{q^m}$. It follows that we may consider the
colimit $\mathbb{F}_{q^{\infty}} := \varinjlim_{n} \mathbb{F}_{q^n}$ (often this is written as a union, which is not quite correct). This turns out to be an algebraic closure of $\mathbb{F}_q$.
Let me also share a quite nice construction of an algebraic closure: Consider the (infinite) tensor product $A$ of all the $K$-algebras $K[x]/(f)$, where $f \in K[x] \setminus \{0\}$. By linear algebra it is non-zero, hence has a maximal ideal $\mathfrak{m}$ (Zorn's Lemma!). Then $K' := A/\mathfrak{m}$ is a field extension of $K$, and by construction every $f \in K[x] \setminus \{0\}$ has a root in $K'$. It is a nontrivial result that this already
is the algebraic closure; but even if we don't use this, we can just repeat this process $K \hookrightarrow K' \hookrightarrow K'' \hookrightarrow K''' \hookrightarrow \dotsc$ and observe that the colimit $\overline{K}$ is an algebraic closure of $K$. A similar reasoning can be obtained to show that every two algebraic closures are isomorphic (but not in a canonical way).
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On a radial positive solution to a nonlocal elliptic equation
DOI: http://dx.doi.org/10.12775/TMNA.2003.017
Abstract
The existence of a solution to the Dirichlet boundary value problem for
nonlinear Poisson equations with the nonlocal nonlinear term $$ -\Delta u=f\bigg(u,\int (g\circ u)\bigg),\quad u\vert\partial U=0, $$ is proved by means of fixed point theorems for increasing compact operators.
nonlinear Poisson equations with the nonlocal nonlinear term
$$
-\Delta u=f\bigg(u,\int (g\circ u)\bigg),\quad u\vert\partial U=0,
$$
is proved by means of fixed point theorems for increasing compact operators.
Keywords
35J65
Full Text:FULL TEXT Refbacks There are currently no refbacks.
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Denote by $X := [\mathbb{N}]^\infty$ the set of infinite subsets of $\mathbb{N}$. Recall that the Ellentuck topology is a topology on $X$ generated by sets of the form $\{A\text{ infinite} \mid s\text{ is an initial segment of }A\text{ and }A\setminus s\subset B\}$ for some finite $s\subset \mathbb{N}$ and (infinite) $B\subset \mathbb{N}$.
In the paper
On completely Ramsey sets by Szymon Plewik, it was proved that the Ellentuck topology is not normal. The argument essentially reduced to a construction of a closed separable subspace, say $Y\subset X$, containing a discrete closed subset $Z$ of cardinality $2^{\aleph_0}$. Suppose that $X$ is normal. Because $Y$ is a closed subset of $X$, $Y$ is normal as well. On the one hand, since $Y$ is separable, the set of continuous function on $Y$, denoted by $C(Y)$, is of cardinality $\le 2^{\aleph_0}$. On the other hand, by the Tietze extension theorem, every (continuous) function on $Z$ can be extended to a continuous function on $Y$, and so $C(Y)$ is of cardinality at least $2^{2^{\aleph_0}}$. A contradiction.
I am wondering if one can demonstrate an explicit construction of two disjoint closed subsets of $X$ that cannot be separated by two disjoint open neighborhoods?
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Introduction Energy and Power Basic Operations Practice Problems Transformation of signals defined piecewise Even and Odd Signals Commonly encountered signals
A variety of operations can be carried out on signals to obtain new signals. These operations can be classified into two categories - operations that are performed on the dependent variable and operations that can be performed on the independent variable. Recall that for a signal $x(t)$, $t$ is the independent variable and $x(t)$ is the dependent variable.
Operations performed on the dependent variable
Amplitude Scaling: Signals are often amplified or attenuated, i.e., the amplitude of the signal is scaled. Mathematically, this corresponds to multiplying the signal $x(t)$ by a (possibly complex constant) $c$ to obtain a new signal, say $y(t)$ given by(1)
Addition, Subtraction, Multiplication, and Division
Take two signals $x_1(t)$ and $x_2(t)$. We can add, subtract, multiply and divide these signals. These operations are performed for every value of $t$. This is similar to performing operations on functions of $t$. Thus we can obtain a new signal $y(t)$ from two signals $x_1(t)$ and $x_2(t)$ according to,
Derivatives and Integrals of CT signals
For continuous-time signals (CT) signals, we define the derivative and integral of the signal as follows. It is important to note the notation used in defining the integral. Notice that the variable of integration is $\tau$ and $t$ appears in the limit of the integral making the result of the integral a function of $t$.(3)
Difference and Partial sums of DT signals
For discrete-time signals, the operation of derivative and integrals are not well defined. The equivalent notion of derivative and integral for DT signals are given by backward difference and partial sum as given below Operations performed on the Independent Variable
The various operations are:
1. Time Scaling:
Given a signal $x(t)$,Time scaling is defined as-(5)
If $a > 1$,
Let $\hspace{0.5cm}a=2$(6)
If $a < 1$,
Suppose we have a DT signal $x[n]$,(10)
2.
Reflection:
Given a signal $x(t)$,Time scaling is defined as-(11)
3.
Time Shifting:
Given a signal $x(t)$,Time shifting is defined as-(12)
Example:Cellular Communication System
For a DT signal $x[n]$,(13)
If $n_0 > 0 \Rightarrow x[n-n_0]$ is a right shift and if $n_0 < 0 \Rightarrow x[n-n_0]$ is a left shift.
4.
Multiple Operations on a Signal: Given a signal $x(t)$, $y(t)=x(at-b)$ Given a signal $x(t)$, $y(t)=x\bigg(\frac{t-b}{a}\bigg)$ Given a signal $x(t)$, $y(t)=x(-at-b)$
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Stochastic 3D Modeling of Three-Phase Microstructures for Predicting Transport Properties: A Case Study 143 Downloads Abstract
We compare two conceptually different stochastic microstructure models, i.e., a graph-based model and a pluri-Gaussian model, that have been introduced to model the transport properties of three-phase microstructures occurring, e.g., in solid oxide fuel cell electrodes. Besides comparing both models, we present new results regarding the relationship between model parameters and certain microstructure characteristics. In particular, an analytical expression is obtained for the expected length of triple phase boundary per unit volume in the pluri-Gaussian model. As a case study, we consider 3D image data which show a representative cutout of a solid oxide fuel cell anode obtained by FIB-SEM tomography. The two models are fitted to image data and compared in terms of morphological characteristics (like mean geodesic tortuosity and constrictivity) as well as in terms of effective transport properties. The Stokes flow in the pore phase and effective conductivities in the solid phases are computed numerically for realizations of the two models as well as for the 3D image data using Fourier methods. The local and effective physical responses of the model realizations are compared to those obtained from 3D image data. Finally, we assess the accuracy of the two methods to predict permeability as well as electronic and ionic conductivities of the anode.
KeywordsStochastic microstructure modeling Effective conductivity Permeability Solid oxide fuel cells 3D image data Nomenclature \(\beta _1, \beta _2, \beta _3\)
Constrictivities of the three phases
\(\varepsilon _1, \varepsilon _2, \varepsilon _3\)
Volume fractions of the three phases
\(\widehat{\varepsilon }\)
Estimator for the volume fraction of a stationary random closed sets
\(\widehat{\varepsilon }^{\star }\)
Estimator for the volume fraction in the graph-based microstructure model
\(\gamma _1, \gamma _2, \gamma _3\)
Parameters of the distance measure used for the graph-based microstructure model
\(\varGamma \)
Pore–solid interface
\(\kappa \,{(}\mathrm {m}^2{)}\)
Permeability
\(\widehat{\kappa }~{(}\mathrm {m}^2{)}\)
Geometrical predictor of permeability
\(\lambda _1, \lambda _2, \lambda _3~{(}\mathrm {m}^{-3}{)}\)
Intensities of the Poisson point processes
\(\mu _f~{(}{{\mathrm {kg}}}\cdot \mathrm {m}^{-1} \cdot \mathrm {s}^{-1}{)}\)
Viscosity of an incompressible Newtonian fluid
\(\nu _3\)
3-dimensional Lebesgue measure
\(\varPhi \)
probability distribution function of the standard normal distribution
\(\phi ~{(}{\mathrm {kg}}\cdot \mathrm {m}^2 \cdot \mathrm {s}^{-3} \cdot \mathrm {A}^{-1}{)}\)
Electrical potential (or ionic concentration)
\(\rho _Y, \rho _Z\)
Covariance functions of the Gaussian random fields
Yand Z \(\sigma ~{(}{\mathrm {kg}}^{-1}\cdot \mathrm {m}^{-2} \cdot \mathrm {s}^{3} \cdot \mathrm {A}^{2}{)}\)
Effective conductivity
\(\sigma _{\mathrm {sol}}~{(}{\mathrm {kg}}^{-1}\cdot \mathrm {m}^{-2} \cdot \mathrm {s}^{3} \cdot \mathrm {A}^{2}{)}\)
Intrinsic conductivity
\(\tau _1, \tau _2, \tau _3\)
Mean geodesic tortuosities of the three phases
\(\varTheta \)
Parameter space
\(\theta _{ij}~{(}\mathrm {m}^{-1}{)}\)
Parameters for modeling two-point coverage probability functions, \(i,j \in \lbrace 1,2 \rbrace \)
\(\vartheta _0~{(}\mathrm {m}^{-2}{)}, \vartheta _1~{(}\mathrm {m}^{-1}{)}\)
Intensities of point processes related to the triple phase boundary
\(\varXi _1, \varXi _2, \varXi _3\)
Random closed sets denoting the three different phases
\(b_1, b_2, b_3\)
Parameters of the beta-skeletons
\(C_1, C_2, C_3\)
Two-point coverage probability functions of the three phases
d( x, A)
Euclidean distance between a point \(x \in \mathbb {R}^3\) and a set \(A \subset \mathbb {R}^3\)
\(d_\gamma (x,A)\)
Distance measure with parameter \(\gamma \) between a point \(x \in \mathbb {R}^3\) and a set \(A \subset \mathbb {R}^3\)
\(\mathbf {E}~{(}{\mathrm {kg}}\cdot \mathrm {m} \cdot \mathrm {s}^{-3} \cdot \mathrm {A}^{-1}{)}\)
Electrical vector field (or opposite gradient of ionic concentration)
\(\mathbf {G}~{(}{\mathrm {kg}}\cdot \mathrm {m}^{-2} \cdot \mathrm {s}^{-2}{)}\)
Macroscopic pressure gradient
\(\mathcal {G}_1, \mathcal {G}_2, \mathcal {G}_3\)
Beta-skeletons of the three phases
h
Function used to estimate the volume fraction in the graph-based model
\(\mathcal {H}_{k}\) k-dimensional Hausdorff measure for \(k \in \lbrace 1,2,3 \rbrace \) \(\mathbf {J}~{(}A{)}\)
Electrical current (or particle current)
\(L_{\mathrm {TPB}}~{(}{\mathrm {m}}^{-2}{)}\)
Expected length of the triple phase boundary per unit volume
M M-factor, i.e., the ratio of effective and intrinsic conductivity \(\widehat{M}\)
Geometrical predictor of the
M-factor o
Origin in the 3-dimensional Euclidean space
\(p~{(}{\mathrm {kg}}\cdot \mathrm {m}^{-1} \cdot \mathrm {s}^{-2}{)}\)
Pressure field
\(\mathbb {R}^3\)
3-dimensional Euclidean space
\(R^2\)
Coefficient of determination
\(r_{\mathrm {max}}~{(}\mathrm {m}{)}\)
Median of the volume equivalent particle radius distribution
\(r_{\mathrm {min}}~{(}\mathrm {m}{)}\)
Median radius of the characteristic bottleneck in a microstructure
\({\mathcal {S}}\)
Conductive phase
\(S_1, S_2, S_3\)
Specific surface area of the three phases
\(s_{\mathrm {GBM}}~{(}\mathrm {m}{)}\)
Smoothing parameter of the graph-based microstructure model
\(s_{\mathrm {PGM}}~{(}\mathrm {m}{)}\)
Smoothing parameter of the pluri-Gaussian microstructure model
\(u_Y, u_Z\)
Thresholds defining the excursion sets of the Gaussian random fields
Yand Z \(\mathbf {v}~{(}\mathrm {m} \cdot \mathrm {s}^{-1}{)}\)
Velocity of an incompressible Newtonian fluid
\(X_1, X_2, X_3\)
Homogeneous Poisson point processes
Y, Z
Gaussian random fields
\(\varDelta \)
Laplacian operator
\(\nabla \)
Gradient operator
\(\partial A\)
Boundary of a set \(A \subset \mathbb {R}^3\)
Notes Supplementary material References Adler, R.J.: The Geometry of Random Fields. Wiley, Chichester (1981)Google Scholar Holzer, L., Pecho, O., Schumacher, J., Marmet, P., Stenzel, O., Büchi, F.N., Lamibrac, A., Münch, B.: Microstructure-property relationships in a gas diffusion layer (GDL) for polymer electrolyte fuel cells, part I: effect of compression and anisotropy of dry GDL. Electrochim. Acta 227, 419–434 (2017)CrossRefGoogle Scholar Lantuéjoul, C.: Geostatistical Simulation: Models and Algorithms. Springer, Berlin (2013)Google Scholar Matheron, G.: Random Sets and Integral Geometry. Wiley, New York (1975)Google Scholar MATLAB 2015b, The MathWorks. www.matlab.com (2015) Molchanov, I.: Statistics of the Boolean Model for Practitioners and Mathematicians. Wiley, Chichester (1997)Google Scholar Møller, J., Waagepetersen, R.P.: Statistical Inference and Simulation for Spatial Point Processes. Chapman & Hall/CRC, Boca Raton (2004)Google Scholar Moulinec, H., Suquet, P.: A fast numerical method for computing the linear and non linear mechanical properties of the composites. Comptes Rendus de l’Académie des Sciences Série II(318), 1417–1423 (1994)Google Scholar Neumann, M., Hirsch, C., Staněk, J., Beneš, V., Schmidt, V.: Estimation of geodesic tortuosity and constrictivity in stationary random closed sets. Scand. J. Stat. (2019). https://doi.org/10.1111/sjos.12375 Torquato, S.: Random Heterogeneous Materials: Microstructure and Macroscopic Properties. Springer, New York (2013)Google Scholar Westhoff, D., Van Franeker, J.J., Brereton, T., Kroese, D.P., Janssen, R.A.J., Schmidt, V.: Stochastic modeling and predictive simulations for the microstructure of organic semiconductor films processed with different spin coating velocities. Modell. Simul. Mater. Sci. Eng. 23(4), 045003 (2015)CrossRefGoogle Scholar Wiegmann, A.: Computation of the permeability of porous materials from their microstructure by FFF-Stokes (2007). http://kluedo.ub.uni-kl.de/files/1984/bericht129.pdf. Accessed 22 July 2015
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Define $f: \ell^2(\mathbb{N}) \to \Bbb C$ by: $$f(x) = \sum_{n \geq 1}\frac{x_n}{n^2},$$where $x = (x_n)_{n \geq 1} \in \ell^2(\mathbb{N})$. It is pretty clear that $f$ is linear. Also, since $\ell^2(\mathbb{N}) \subset \ell^\infty(\mathbb{N})$ and $\|x\|_\infty \leq \|x\|_2$, we have: $$\begin{align}|f(x)| &= \left|\sum_{n \geq 1}\frac{x_n}{n^2}\right| \leq \sum_{n \geq 1}\frac{|x_n|}{n^2} \\ &\leq \sum_{n \geq 1}\frac{\|x\|_\infty}{n^2} = \left(\sum_{n \geq 1}\frac{1}{n^2}\right)\|x\|_\infty \\ &\leq \frac{\pi^2}{6}\|x\|_2. \end{align} $$So $f$ is bounded and $\|f\| \leq \pi^2/6$. I would like to compute what exactly is this norm. The professor said that the book from where he took the exercise says that the norm is $\pi^2/(3\sqrt{10})$, which is indeed possible, but there's no comments on it. He thinks that the norm is actually that $\pi^2/6$ we have up here, but I can't come up with convenient sequences to prove or refute this. If the domain were $\ell^\infty(\mathbb{N})$ instead of $\ell^2(\mathbb{N})$ we could take $(1,1,1,\cdots)$ and the norm would be $\pi^2/6$, but this sequence is not in $\ell^2(\mathbb{N})$. Help?
By Cauchy Schwarz,
$$|f(x)| =\left| \sum \frac{x_n}{n^2}\right|\le \sqrt{\sum \frac{1}{n^4}} \|x\|_2 = \sqrt{\frac{\pi^4}{90}}\|x\|_2$$
(See here for the calculation). Note that equality holds when $x_n = \frac{1}{n^2}$.
(Indeed, if you think of $f\in \ell^2(\mathbb N)^*$, then $f(x) = \langle x, y\rangle$, where $y_n= \frac{1}{n^2}$. Thus $\|f\| = \|y\|$)
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Remark: Throughout this post, I will be assuming the Axiom of Choice. This won't always be necessary (in fact, in most of the instances where we'll use it, we could've instead used weaker choice principles), but it makes things far less messy to prove. Let me know if you're interested in a more choice-free version (or if there's any claim I make herein that you don't believe/understand), and I'll see what I can do. Some choice will be necessary, though, as discussed in this answer.
Given a Vitali set $V$, we get a whole new Vitali set $V'$ if we pick one equivalence class, then swap the unique element of $V$ lying in that equivalence class for another member of that equivalence class, yes? Furthermore, if we'd picked a
different equivalence class to perform that swap in--to get a set $V''$, say--then we'd have $V''\neq V$ by the same reasoning, but we'd also have $V''\neq V'$. (Why?) Thus, for each equivalence class--of which there are uncountably many--we can obtain a different Vitali set by performing a swap as described above, and swaps performed in different equivalence classes yield different Vitali sets, so in this fashion, we see that there are uncountably many Vitali sets.
It's worth noting that any collection of
pairwise disjoint Vitali sets will be at most countably infinite, but that's another story.
If you're interested in the precise cardinality of the set of Vitali sets--let's call it $\Bbb V$--we can use the following approach. In the following, by $\mathfrak{c}$ I denote the cardinality of $\Bbb R$, and by $\aleph_0$ I denote the countably infinite cardinality. Given any two sets $A,B$, I let $B^A$ denote the set of functions from $A$ to $B$. By definition, we have $|B|^{|A|}:=\bigl|B^A\bigr|.$
First, observe that every Vitali set is, of course, a subset of $\Bbb R$--an element of the power set of $\Bbb R$. There is a ready bijection from the power set of $\Bbb R$ to $\{0,1\}^{\Bbb R}$, given by $A\mapsto\chi_A$--where $\chi_A$ is the characteristic function of $A$. Hence, the power set of $\Bbb R$ has cardinality $\left|\{0,1\}^{\Bbb R}\right|=\left|\{0,1\}\right|^{|\Bbb R|}=2^\mathfrak{c},$ and since $\Bbb V$ is a subset of the power set of $\Bbb R$, then $$|\Bbb V|\leq 2^\mathfrak{c}.\tag{1}$$
Now, $[0,1]$ also cardinality $\mathfrak{c}$, and so
cannot be the union of less than $\mathfrak{c}$-many sets of cardinality $\aleph_0$. Since $[0,1]$ is the union of the equivalence classes described above, then there must be at least $\mathfrak{c}$-many of them. On the other hand, there can't be more than $\mathfrak{c}$-many such classes, since they're disjoint, and so their union would have cardinality greater than $\mathfrak{c}$ if there were more than $\mathfrak{c}$-many. Hence, there are precisely $\mathfrak{c}$-many equivalence classes. Let $f$ be any bijection from $\Bbb R$ to the set of equivalence classes.
The rationals are countable, so let the sequence $\{q_n\}_{n\in\Bbb N}$ be an enumeration of the rationals. Fix some Vitali set $V$. For any $x\in\Bbb R$, let $g(x)$ be the unique member of $V$ in the equivalence class $f(x)$--that is, $g(x)$ is the unique element of $V\cap f(x)$. Observe that, given any $x\in\Bbb R$, the sequence $\{g(x)+q_n\}_{n\in\Bbb N}$ is an enumeration of the equivalence class $f(x)$.
Now, given any $h\in \Bbb N^{\Bbb R}$ (any function $h:\Bbb R\to\Bbb N$), we'll let $$V_h=\left\{g(x)+q_{h(x)}:x\in\Bbb R\right\}.$$ Then $h\mapsto V_h$ is an injection from $\Bbb N^{\Bbb R}$ to $\Bbb V$, so $$\aleph_0^\mathfrak{c}=\left|\Bbb N\right|^{|\Bbb R|}=\left|\Bbb N^{\Bbb R}\right|\le|\Bbb V|.\tag{2}$$
Finally, observe that $\{1,2\}^{\Bbb R}$ is a subset of $\Bbb N^{\Bbb R}$--that is, every function $\Bbb R\to\{1,2\}$ is a function $\Bbb R\to\Bbb N$--regardless of whether your natural numbers are the positive integers or the nonnegative integers. (If you define your natural numbers any other way...you're just a jerk, lol.) Therefore, $$2^\mathfrak{c}=\left|\{1,2\}\right|^{|\Bbb R|}=\left|\{1,2\}^{\Bbb R}\right|\le\left|\Bbb N^{\Bbb R}\right|=\left|\Bbb N\right|^{|\Bbb R|}=\aleph_0^\mathfrak{c}.\tag{3}$$ Therefore, we have $$|\Bbb V|\overset{(1)}\le 2^\mathfrak{c}\overset{(3)}\le\aleph_0^\mathfrak{c}\overset{(2)}\le|\Bbb V|,$$ so by Schroeder-Bernstein theorem, we have $$|\Bbb V|=2^\mathfrak{c}=\aleph_0^\mathfrak{c}.$$
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The other answers so far focus on how to transform the integral so that you don't have a singularity. But that's not always viable so it's important to know how to deal with a singularity when one exists.
The trapezoid rule is a poor choice for this case, because it uses the values of the function at the endpoints, which can be infinite. Simposon's rule will have a similar problem.
More appropriate is the more primitive rectangle rule. That is, divide the domain to segments, and multiply each segment's width by the value of the function at the midpoint of the segment. This will work since you never try to sample the function at the infinite endpoints.
Alternatively, if you want faster convergence, you can use the more sophisticated and extremely powerful Gaussian Quadrature. More complicated but well worth knowing. Here, too, you divide the domain to intervals. For each interval, you sample the function at a few carefully chosen points to obtain surprisingly high accuracy.
There are different methods depending on the number of points in each interval, but the simplest is the 2-point method. And again, it will work with singularities, since the endpoints of intervals aren't sampled.
There is even a variant of Gaussian quadrature tailored specifically for a well-behaved function multiplied by $\frac{1}{\sqrt{1-x^2}}$, called Chebyshev–Gauss quadrature. To use it, you don't even have to divide to intervals - you just choose $n$ (the higher $n$, the better the accuracy) and you have
$$\int_{-1}^1\frac{f(x)}{\sqrt{1-x^2}}\approx\frac{\pi}{n}\sum_{i=1}^nf\left(\cos\left(\frac{2i-1}{2n}\pi\right)\right)$$
Which is very similar to transforming the integral as some other answers suggested. It will maximize the accuracy you can have for a given number of samples from the function $f$, and convergence is exponential.
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Monday-Sept. 30 Tuesday-Oct. 1 Wednesday-Oct. 2 9:00-9:30 Registration/Coffee Coffee Coffee 9:30 - 10:30 Lehalleur Wendt Sechin 10:30-11:00 Coffee Coffee Coffee 11:00 - 12:00 Yakerson Dotto Xu 12:00-14:00 Lunch Lunch Lunch 14:00 - 15:00 Drew Ozornova Jin 15:00-16:00 Coffee Coffee Coffee 16:00 - 17:00 D’Addezio Wang Bachman 19:00 Conference Dinner
For a suitable algebraic group G acting on a scheme S, various authors have defined the motivic stable homotopy theory of smooth S-schemes with a G-action. While this category has excellent formal properties, the construction is very involved. In this talk I will explain that if G is finite, the construction can be simplified: one can just work with the category obtained from motivic G-spaces with finite étale transfers (over S) by inverting the *trivial* motivic representation sphere. This is essentially a corollary of the motivic tom Dieck splitting theorem, established recently by David Gepner and Jeremiah Heller.
I will report on a joint work with Emiliano Ambrosi. Let k be a field which is finitely generated over the algebraic closure of a finite field. As a consequence of the theorem of Lang-Néron, for every abelian variety over k which does not admit any isotrivial abelian subvariety the group of k-rational torsion points is finite. We showed that the same is true for the group of torsion points defined on a perfect closure of k. This gives a positive answer to a question posed by Hélène Esnault in 2011. To prove the theorem we translated the problem into another one on morphisms of F-isocrystals. During the talk, I will explain some ideas of our proof.
We will introduce the Witt vectors of a ring with coefficients in a bimodule and use them to calculate the components of the Hill-Hopkins-Ravenel norm for cyclic p-groups. This algebraic construction generalizes Hesselholt's Witt vectors for non-commutative rings and Kaledin's polynomial Witt vectors over perfect fields. We will discuss applications to the characteristic polynomial over non-commutative rings and to the Dieudonné determinant. This is all joint work with Krause, Nikolaus and Patchkoria.
The stable motivic homotopy category $\mathrm{SH}(S)$ over a scheme $S$ is characterized by an $\infty$-categorical universal property: each homology theory defined on the category of smooth $S$-schemes satisfying analogues of the Eilenberg-Steenrod axioms factors essentially uniquely through $\mathrm{SH}(S)$. Ayoub has equipped the categories $\mathrm{SH}(S)$ with a six-functor formalism, i.e., pullback, pushforward, and tensor operations satisfying the same formal properties as derived categories of $\ell$-adic sheaves. In this talk, we will promote the universal property of $\mathrm{SH}(S)$ to a universal property of the associated six-functor formalism: each six-functor formalism $S \mapsto \mathrm{D}(S)$ satisfying a reasonable list of axioms admits an essentially unique family of realization functors $\mathrm{SH}(S) \to \mathrm{D}(S)$ compatible with the six functors. As an application, we will discuss the construction of motivic realization functors with values in arithmetic D-modules. This is joint work with Martin Gallauer.
We prove several Künneth formulas in motivic homotopy theory, and use them to define the characteristic class of a motive, an invariant closely related to the Euler characteristic. For cdh motives, this class lives in the Chow group of 0-cycles. We show its additivity along distinguished triangles, and deduce that the characteristic class for cdh motives is characterized by some elementary operations. If time permits, we also discuss the relative characteristic class. This is a joint work with E. Yang.
Grothendieck and Quillen introduced a notion of homotopy equivalences for categories using a by-now-standard tool called "nerve" of a category. This idea leads to various models of categories-up-to-homotopy. In a joint ongoing projects with Julie Bergner and Martina Rovelli, we study variants of the Roberts-Street-nerve for 2-categories and notions of homotopy equivalences arising from this nerve, with an eye towards 2-categories-up-to-homotopy.
Moduli spaces of stable Higgs bundles on curves are central objects of study in non-abelian Hodge theory, geometric representation theory and mathematical physics, and their cohomology has been intensely studied. We consider the motives with rational coefficients of these moduli spaces, and prove that they are Chow motives which lie in the tensor subcategory generated by the motive of the curve, and hence in particular of abelian type. Although this is a statement about motives of smooth varieties, the proof relies crucially on the study of motives of various moduli stacks. This is joint work with Victoria Hoskins (Freie Universität Berlin).
For a fixed prime $p$ algebraic Morava K-theories $K(n)$ (where $n\ge 1$) are oriented cohomology theories which play an intermediate role between the Grothendieck K-theory and Chow groups with $p$-local coefficients. I will show that if the (pure) $K(n)$-motive of a smooth projective variety is a sum of Tate motives, then the same is true for the $K(m)$-motive, for $m < n$. As for particular examples, the obstructions for the splitting of the $K(n)$-motive of a quadric are known: they are precisely the cohomological invariants lying in degrees no greater than $n+1$. This allows to obtain bounds on the torsion of Chow groups of these quadrics in low codimensions. I will also explain how the gamma filtration on Morava K-theories and Chern classes from them can be useful for such computations.
For complete intersection rings over p-adic intergers, we use the relative THH to construct resolutions of topological cyclic homology and its periodic analogue. The resulting spectral sequences are believed to be isomorphic with the BMS spectral sequence, whose $E_2$ terms are the motivic cohomolpgy and prismatic cohomology respectively. We show that the $E_2$ terms are the Ext groups of a Hopt algebroid, with explicit formulae for the structure maps. As an example, we will give concrete descriptions of the computations in the case of local fields and compare them with known results on their TC and Galois cohomlogies.
In recent work, Asok and Fasel have applied motivic homotopy methods to make significant progress on the very classical question when a vector bundle over a smooth affine scheme splits off a trivial line bundle. The goal of the talk is to show that the same methods can also be applied to similar questions for quadratic forms. On the one hand, it is actually possible to get a full isometry classification of generically split quadratic forms over smooth affine schemes of dimension at most 3. On the other hand, the computation of the first non-vanishing homotopy sheaf of the orthogonal Stiefel varieties gives rise to Euler classes controlling when generically split quadratic forms split off hyperbolic planes. For the talk, I'll formulate some of the general results that can be obtained and try to illustrate some of the phenomena with low-dimensional examples.
I will discuss joint work with Bogdan Gheorghe and Guozhen Wang on the equivalence of stable infinity categories, between the motivic cellular $S\tau$-modules over the complex numbers and the derived category of $BP_*BP$-comodules. As a consequence, the motivic Adams spectral sequence for $S\tau$ is isomorphic to the algebraic Novikov spectral sequence for the sphere. In joint work with Dan Isaksen and Guozhen Wang, we use this isomorphism of spectral sequences to compute classical stable stems at least to the 90-stem, with ongoing computations into even higher dimensions. If time permits, I will also discuss some connections to the New Doomsday Conjecture and the Kervaire invariant problem.
When k is a field with resolution of singularities, it is known that Voevodsky’s category of motives DM(k) is equivalent to the category of modules over the motivic cohomology spectrum HZ. This means that a structure of an HZ-module on a motivic spectrum is equivalent to a structure of transfers in the sense of Voevodsky. In this talk, we will discuss an analogous result for modules over the algebraic cobordism spectrum MGL. Concretely, a structure of an MGL-module is equivalent to a structure of coherent transfers along finite syntomic maps, over arbitrary base scheme. Time permitting, we will see a generalization of this result to modules over other motivic Thom spectra, such as the algebraic special linear cobordism spectrum MSL. This is joint work with Elden Elmanto, Marc Hoyois, Adeel Khan and Vladimir Sosnilo.
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Large Eddy Simulation ( LES) is one of the most promising methods for computing industry-relevant turbulent flows. It is used to predict unsteady flow behaviors with lower computational cost as compared to Direct Numerical Simulation (DNS).
The essential idea of LES dates back to
Joseph Smagorinsky (1963) in meteorology and to James W. Deardorff (1970) in engineering. Keywords Inertial subrange, Wall-modeled LES, Wall-resolved LES, Smagorinsky, Taylor microscale Nomenclature – Abbreviations LES: Large Eddy Simulation ILES: Implicit Large Eddy Simulation MILES: Monotone Integrated Large Eddy Simulation VLES: Very Large Eddy Simulation WMLES: Wall-Modeled Large Eddy Simulation SGS: subgrid-scale WALE: Wall-Adapting Local Eddy-Viscosity What is Large Eddy Simulation? Professor Parviz Moin of Stanford University briefly explains what LES is in his interview. I’m surprised that he received his Masters and Ph.D. degrees in Mathematics and Mechanical Engineering from Stanford in the same year (1978)!
The verification of LES simulations is difficult because both the error induced by the SGS model and the numerical discretization error are dependent on the grid resolution. In the following presentation (37:40-42:06), he introduces one method using explicit filtering to distinguish the errors [6]:
If the grid-independent solution of the explicit filtered LES equations fails to accurately predict the filtered DNS flow field, its failure can be solely attributed to the capability of the subgrid stress model employed.
Categorization of LES
LES can be classified in terms of the following points as shown in Fig. 1:
How the effect of the SGS stress is modeled How the spatial filtering is applied in the solution procedure.
In OpenFOAM, LES with the
implicit filtering is implemented, in which only the filter width is specified and the filter shape is not. There exists this ambiguity in the definition of the filter. Lund [7] provides a clear description of the implicit filtering in LES:
The nearly universal approach is to simply write down the filtered Navier-Stokes equations together with an assumed model for the subgrid-scale stresses and then apply the desired spatial discretization to this “filtered” system. Although it is rarely mentioned, what one is doing by adopting this procedure is to imagine that the finite support of the computational mesh together with the low-pass characteristics of the discrete differentiating operators act as an effective filter. One then directly associates the computed velocity field with the filtered velocity. This procedure will be referred to as implicit filtering since an explicit filtering operation never appears in the solution procedure.
The explicit filtering methodology aims to damp the error-prone length scales smaller than the filter width that can be generated by the nonlinear convective term. In the case of the implicit filtering, we hope that this error has little effect on the resolved scales.
Subgrid-scale (SGS) model SGS models in OpenFOAM Smagorinsky
Smagorinsky SGS model
kEqn
One equation eddy-viscosity model
dynamicLagrangian
Dynamic SGS model with Lagrangian averaging
dynamicKEqn
Dynamic one equation eddy-viscosity model
WALE
Wall-adapting local eddy-viscosity (WALE) SGS model
DeardorffDiffStress
Differential SGS Stress Equation Model
For the dynamic SGS models, the spatial averaging operations of the coefficients are often performed to stabilize the calculation. The
model that had been implemented in older versions takes the average of the coefficient in the whole computational domain. homogeneousDynSmagorinsky Other SGS models Vreman SGS model[3]
\begin{equation}
\nu_{sgs} = c \sqrt{\frac{B_{\beta}}{\alpha_{ij}\alpha_{ij}}} \end{equation} Calculation of filter width in OpenFOAM
The method for calculating the filter width \(\Delta\) is specified in the
file. The available options in OpenFOAM are as follows: turbulenceProperties cubeRootVol maxDeltaxyz maxDeltaxyzCubeRoot smooth vanDriest Prandtl IDDESDelta Grid resolution measures
The choice of the computational grid size has sensible impact on the quality of the LES simulations and there are several techniques to assess the grid resolution in the LES computations [4].
Estimations based on prior RANS results Single-grid estimators Multi-grid estimators Principal use Aeroacoustics Aerodynamics Combustion Refereces
[1] W. Rodi, J. H. Ferziger, M. Breuer and M. Pourquié, Status of Large Eddy Simulation: Results of a Workshop. Journal of Fluids Engineering, 119(2), 248-262, 1997.
[2] P. Comte, Large eddy simulations and subgrid scale modelling of turbulent shear flows [3] A. W. Vreman, An eddy-viscosity subgrid-scale model for turbulent shear flow: Algebraic theory and applications. Physics of Fluids, 16(10), 2004. [4] S. E. Gant, Reliability Issues of LES-Related Approaches in an Industrial Context. Flow, turbulence and combustion, 84(2), 325-335, 2010. [5] U. Piomelli, Large eddy simulations in 2030 and beyond. Philosophical Transactions of the Royal Society A: Mathematical, Physical and Engineering Sciences, 372(2022), 2014. [6] S. T. Bose, P. Moin and D. You, Grid-independent large-eddy simulation using explicit filtering. Center for Turbulence Research Annual Research Briefs 2008. [7] T. S. Lund, The Use of Explicit Filters in Large Eddy Simulation. Computers and Mathematics with Applications 46, 603-616, 2003. Refereces (Japanese) List of Laboratories Center for Turbulence Research (Stanford University)
Professor Parviz Moin is the founding director of the Center for Turbulence Research at Stanford and the co-founder of Cascade Technologies that develops CHARLES, a high-fidelity unstructured compressible flow solver for Large Eddy Simulation.
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In this work, Wald & Zoupas developed a framework to define the "conserved quantities" in a diffeomorphism-invariant theory using the covariant phase space formalism.
Let $\xi^a$ be a vector field on the $n$-dimensional spacetime manifold $M$. Physical fields are collectively represented by $\phi$, which could be the metric $g_{ab}$ and the 4-potential $A_a$, etc.. $\xi^a$ will generate an infinitesimal diffeomorphism, which is a local symmetry of the theory, so the variation of the Lagrangian $n$-form $\boldsymbol L$ is
$$\mathscr L_\xi\boldsymbol L=\boldsymbol E\hat\delta\phi+d\boldsymbol\Theta(\phi,\mathscr L_\xi\phi)=d(i_\xi\boldsymbol L),$$
where $\hat\delta\phi=\mathscr L_\xi\phi$, $\boldsymbol E$ represent the equations of motion, $d$ means the exterior derivative, $\boldsymbol\Theta$ is a $(n-1)$-form called the presymplectic potential current, $i_\xi$ refers to the interior product.
The Noether current $(n-1)$-form $\boldsymbol j$ is given by
$$\boldsymbol j[\xi]=\boldsymbol\Theta(\phi,\mathscr L_\xi\phi)-i_\xi\boldsymbol L.\tag{9}$$
On shell, one can show that $d\boldsymbol j=0$ and locally, one can find a $(n-2)$-form $\boldsymbol Q[\xi]$ such that $\boldsymbol j[\xi]=d\boldsymbol Q[\xi]$. $\boldsymbol Q$ is the Noether charge $(n-2)$-form. In this reference, the general form of $\boldsymbol Q$ was given in Proposition 4.1, which is really complicated.
Now, in Wald & Zoupas, a different vector field $\eta^a$ is considered. Its effect on $\boldsymbol Q$ is given by Eq. (33),
$$\delta\boldsymbol Q[\xi]=\mathscr L_\eta\boldsymbol Q[\xi]-\boldsymbol Q[\mathscr L_\eta\xi].\tag{33}$$
However, I do not understand how this relation was obtained. It looks to me that the left-hand side is nothing but
$$\delta\boldsymbol Q[\xi]=\mathscr L_\eta(\boldsymbol Q[\xi]).$$
Clearly, this is not the option.
Please help me with this problem.
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Fix a sequence $a_n={n+2\choose 2}$ of triangular numbers with the initial condition $a_0=1$,such that
$1,3,6,10,15,21,\dots$
given by
$F(x)=\frac{1}{(1-x)^3}=\sum_{n=0}^{\infty} a_n x^n\tag1$
Now if we consider the following generating function
$G(x)=-\frac{\frac{1}{x^5}\Big(1+\frac{1}{x}\Big)\Big(1-\frac{1}{x^2}\Big)}{\Big(\Big(1-\frac{1}{x}\Big)\Big(1-\frac{1}{x^3}\Big)\Big)^2}\tag2$
How do we prove
$$G(x)\overset{\color{red}?}=\sum_{n=0}^{\infty} a_n(4x^{3n+2}+x^{(6n+(-1)^{n}-1)/4})$$
Using the simplified form of $(2)$,we have the relation
$$\frac{G(x)}{F(x)}=\Big(\frac{1+x}{1+x+x^2}\Big)^2$$
$$\sum_{n=0}^{\infty} a_n(4x^{3n+2}+x^{(6n+(-1)^{n}-1)/4})=\Big(\sum_{n=0}^{\infty} a_n x^n\Big)\Big(\sum_{n=0}^{\infty} (-1)^{n}x^{(6n-(-1)^n+1)/4}\tag3\Big)^2$$
which relates the number of partitions of $6n$ into two odd parts $\frac{(6n-(-1)^n+1)}{4}$ A007494 oeis to the number of partitions of $6n$ into two even parts $\frac{(6n+(-1)^n-1)}{4}$ A032766 oeis and the number of partitions of 6n into at most 2 parts $3n+1$.
Evidently there's an interesting combinatorial information about partitions of $6n$ encoded in the exponents of the identity
If we introduce the notation $P(6n)=\frac{(6n-(-1)^n+1)}{4}$,$Q(6n)=\frac{(6n+(-1)^n-1)}{4}$ and $R(6n)=3n+1$
the identity can be written succinctly as follows
$$\sum_{n=0}^{\infty} a_n(4x^{R(6n)+1}+x^{Q(6n)})=\Big(\sum_{n=0}^{\infty} a_n x^n\Big)\Big(\sum_{n=0}^{\infty} (-1)^{n}x^{P(6n)}\Big)^2$$
whereby the ff. algebraic equation is satisfied
$R(6n)-Q(6n)-P(6n)=1\tag4$ for natural number $n$
which is an equation of a triangular plane with equal sides $x-y-z=1$ on a 3D space
The connection between triangular numbers $a_n={n+2\choose 2}$(which count objects arranged in an equilateral triangle) as coefficients in the identity $(3)$ and the equation $R(6n)-Q(6n)-P(6n)=1$ for natural number $n$, of a triangular plane with equal sides $(4)$ is quite interesting, since both deal with equilateral triangles
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Your mistake was forgetting that when you square a non-$0$ equality, the result satisfies two possible equations. In particular, both $x = y$ and $x = -y$ gives that $x^2 = y^2$. From what you describe, you did the following:
$$\sqrt{3}\sin(x) = 2 - \cos(x) \tag{1}\label{eq1}$$
then square both sides to get
$$3\sin^2(x) = 4 - 4\cos(x) + \cos^2(x) \tag{2}\label{eq2}$$
Next, you made certain manipulations to get
$3 - 3\cos^2(x) = 4 - 4\cos(x) + \cos^2(x) \; \Rightarrow \; 4\cos^2(x) - 4\cos(x) + 1 = 0$
Using $\cos(x) = t$ then gives
$$4t^2 - 4t + 1 = 0 \; \Rightarrow \; (2t - 1)^2 = 0 \; \Rightarrow \; t = \frac{1}{2} \tag{3}\label{eq3}$$
The full set of solutions for $\cos(x) = \frac{1}{2}$ is $x = \pm\frac{\pi}{3} + 2k\pi, k \in \mathbb{Z}$. However, note the RHS of \eqref{eq1} is always $\frac{3}{2}$, but for $x = \frac{\pi}{3} + 2k\pi$ the LHS is also $\frac{3}{2}$, but for $x = -\frac{\pi}{3} + 2k\pi$ the LHS is $-\frac{3}{2}$ instead. Squaring with either case still gives \eqref{eq2}.
Whenever you use a non-reversible operation, like squaring, it's important you check to remove any extraneous results you may have got from your manipulations. However, checking by substituting your results into the original equation is generally always a good idea in case you made some mistake.
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For an entangled pure state, the Schmidt decomposition is such that there are at least two non-zero Schmidt coefficients. Tracing out one subsystem implies that the other subsystem is mixed.
Explicitly, we have
$$\psi = \sum_k\sqrt{\lambda_k}\vert k\rangle_A \vert k\rangle_B$$
$$\rho = \vert\psi\rangle\langle\psi\vert = \sum_{k,k'}\sqrt{\lambda_k}\sqrt\lambda_{k'}\vert k\rangle_A \vert k\rangle_B\langle k\vert_A\langle k'\vert_B$$
Taking the partial trace gives us a sum of the form below with at least two nonzero terms. $$\rho_A = \sum_k \lambda_k \vert k\rangle\langle k\vert$$
This is a diagonal matrix with rank 2 or more and is hence a mixed state.
Is there a similar argument one can make for the case where $\rho$ is a mixed entangled state? Alternatively, if this is not true, can one provide a counter example for which the reduced state is pure but the state $\rho$ is still entangled?
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I think that the source is Richard Dedekind's Was sind und was sollen die Zahlen ? (Vieweg: Braunschweig, 1888).
I quote from the English translation : THE NATURE AND MEANING OF NUMBERS, (The Open Court Publishing Co., 1901) :
98. Definition [page 37]. If $n$ is any number, then will we denote by $Z_n$ the system [set] of all numbers that are not greater than $n$ [...].
106. Theorem [page 38]. If $m < n$, then is $Z_m$ proper part of $Z_n$ and conversely.
I've found it through :
Gregory Moore, Zermelo's Axiom of Choice : Its Origin Development and Influence (1982), page 26.
We can easily generalise it to a poset $M$ whatever, defining the mapping :
$\pi : a \to M_a$, for any $a \in M.$
where $M_a = \{ x \in M : x \le a \}$
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We will examine situations where quantities are increasing exponentially. This situation is known as $\textit{exponential growth modelling}$, and occur frequently in our real life around us.
Population of species, people, bacteria and investment usually $\textit{growth}$ in an exponential way.
Growth is exponential when the quantity present is multiplied by a constant for each unit time interval. this constant is called the $\textit{growth}$ or $\textit{compounding factor}$ is greater than one.
Considering a situation where the changes to a certain population of bacteria over a period of time is a good example of $\textit{exponential growth}$.
It can be seen that the population is increasing but also that the rate of growth is increasing; that is, the graph is getting steeper.
Consider a population of $10$ bacteria which under favourable conditions it increasingly by $20\%$ each day.
To increase a quantity by $20\%$, it is know that to multiply it by $1.2$.
If $B_n$ is the population of bacteria after $n$ days, then:
\( \begin{align} \displaystyle
B_0 &= 10 &\textit{the original population} \\ B_1 &= B_0 \times 1.2 = 10 \times 1.2 \\ B_2 &= B_1 \times 1.2 = 10 \times 1.2 \times 1.2 = 10 \times 1.2^2 \\ B_3 &= B_2 \times 1.2 = 10 \times 1.2^2 \times 1.2 = 10 \times 1.2^3 \\ \end{align} \) and so on.
The population is $1.2$ times every day, so the $\textit{growth}$ or $\textit{compounding factor}$ is $2$.
From the pattern above, we see that $B_n = 10 \times 1.2^n$.
In general, the exponential growth function has an equation of the form: $y=Aa^{kx}$
$A$, $a$ and $k$ are constants. $a>1$ and $k>0$ are the growth or compounding factor. $A$ is the initial value of $y$ (when $x=0$) Example 1
The weight $B_n$ of bacteria in a colony $n$ hours after establishment is given by $B_n = 100 \times 5^{0.2n}$ grams.
(a) Find the initial weight.
\( \begin{align} \displaystyle B_0 &= 100 \times 5^{0.2 \times 0} \\ &= 100 \times 5^0 \\ &= 100 \times 1 \\ &= 100 \\ \end{align} \) (b) Find the weight after $5$ hours. \( \begin{align} \displaystyle B_5 &= 100 \times 5^{0.2 \times 5} \\ &= 100 \times 5^1 \\ &= 100 \times 5 \\ &= 500 \\ \end{align} \) (c) Find the percentage increase from from $n=10$ to $n=20$. \( \begin{align} \displaystyle \dfrac{B_{20} – B_{10}}{B_{10}} \times 100\% &= \dfrac{100 \times 5^{0.2 \times 20} – 100 \times 5^{0.2 \times 10}}{100 \times 5^{0.2 \times 10}} \times 100 \% \\ &= \dfrac{5^{0.2 \times 20} – 5^{0.2 \times 10}}{5^{0.2 \times 10}} \times 100 \% \\ &= \dfrac{5^4 – 5^2}{5^2} \times 100 \% \\ &= \dfrac{5^2(5^2 – 1)}{5^2} \times 100 \% \\ &= (5^2 – 1) \times 100 \% \\ &= 2400 \% \\ \end{align} \) Example 2
The speed $V_t$ is given by $V_t = 8 \times 2^{0.5t}$, where $t$ is the temperature in $^{\circ}C$. Find the temperature when the speed is $24$, correcting to $3$ significant figures.
\( \begin{align} \displaystyle 8 \times 2^{0.5t} &= 24 \\ 2^{0.5t} &= 3 \\ 0.5t &= \log_2{3} \\ t &= \dfrac{1}{0.5} \log_2{3} \\ &= 3.1699 \cdots \\ &= 3.17 ^{\circ}C\\ \end{align} \)
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Here is a report of the Ray Tracer written by myself Christopher Chedeau. I've taken the file format and most of the examples from the Ray Tracer of our friends Maxime Mouial and Clément Bœsch. The source is available on Github.
It is powered by Open Source technologies: glMatrix, CodeMirror, CoffeeScript, Twitter Bootstrap, jQuery and Web Workers.
Check out the
demo, or click on any of the images. Objects
Our Ray Tracer supports 4 object types: Plane, Sphere, Cylinder and Cone.
The core idea of the Ray Tracer is to send rays that will be reflected on items. Given a ray (origin and direction), we need to know if it intersect an object on the scene, and if it does, how to get a ray' that will be reflected on the object.
Knowing that, we open up our high school math book and come up with all the following formulas.
Legend: Ray Origin \(O\), Ray Direction \(D\), Intersection Position \(O'\), Intersection Normal \(N\) and Item Radius \(r\).
Intersection Normal Plane \[t = \frac{O_z}{D_z}\] \[ N = \left\{ \begin{array}{l} x = 0 \\ y = 0 \\ z = -sign(D_z) \end{array} \right. \] Sphere \[ \begin{array}{l l l} & t^2 & (O \cdot O) \\ + & 2t & (O \cdot D) \\ + & & (O \cdot O) - r^2 \end{array} = 0\] \[ N = \left\{ \begin{array}{l} x = O'_x \\ y = O'_y \\ z = O'_z \end{array} \right. \] Cylinder \[ \begin{array}{l l l} & t^2 & (D_x D_x + D_y D_y) \\ + & 2t & (O_x D_x + O_y D_y) \\ + & & (O_x O_x + O_y O_y - r^2) \end{array} = 0\] \[ N = \left\{ \begin{array}{l} x = O'_x \\ y = O'_y \\ z = 0 \end{array} \right. \] Cone \[ \begin{array}{l l l} & t^2 & (D_x D_x + D_y D_y - r^2 D_z D_z) \\ + & 2t & (O_x D_x + O_y D_y - r^2 O_z D_z) \\ + & & (O_x O_x + O_y O_y - r^2 O_z O_z) \end{array} = 0\] \[ N = \left\{ \begin{array}{l} x = O'_x \\ y = O'_y \\ z = - O'_z * tan(r^2) \end{array} \right. \]
In order to solve the equation \(at^2 + bt + c = 0\), we use
\[\Delta = b^2 - 4ac \]\[ \begin{array}{c c c} \Delta \geq 0 & t_1 = \frac{-b - \sqrt{\Delta}}{2a} & t_2 = \frac{-b + \sqrt{\Delta}}{2a} \end{array} \]
And here is the formula for the reflected ray:
\[
\left\{ \begin{array}{l} O' = O + tD + \varepsilon D' \\ D' = D - 2 (D \cdot N) * N \end{array} \right. \]
In order to fight numerical precision errors, we are going to move the origin of the reflected point a little bit in the direction of the reflected ray (\(\varepsilon D'\)). It will avoid to falsely detect a collision with the current object.
Coordinates, Groups and Rotations
We want to move and rotate objects. In order to do that, we compute a transformation matrix (and it's inverse) for each object in the scene using the following code:
\[
T = \begin{array}{l} (Identity * Translate_g * RotateX_g * RotateY_g * RotateZ_g) * \\ (Identity * Translate_i * RotateX_i * RotateY_i * RotateZ_i) \end{array} \]\[ I = T^{-1} \]
\[Translate(x, y, z) = \left(\begin{array}{c c c c} 1 & 0 & 0 & x \\ 0 & 1 & 0 & y \\ 0 & 0 & 1 & z \\ 0 & 0 & 0 & 1 \end{array}\right)\] \[RotateX(\alpha) = \left(\begin{array}{c c c c} 1 & 0 & 0 & 0 \\ 0 & cos(\alpha) & -sin(\alpha) & 0 \\ 0 & sin(\alpha) & cos(\alpha) & 0 \\ 0 & 0 & 0 & 1 \end{array}\right)\] \[RotateY(\alpha) = \left(\begin{array}{c c c c} cos(\alpha) & 0 & sin(\alpha) & 0 \\ 0 & 1 & 0 & 0 \\ -sin(\alpha) & 0 & cos(\alpha) & 0 \\ 0 & 0 & 0 & 1 \end{array}\right)\] \[RotateZ(\alpha) = \left(\begin{array}{c c c c} cos(\alpha) & -sin(\alpha) & 0 & 0 \\ sin(\alpha) & cos(\alpha) & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right)\]
We have written the intersection and normal calculations in the object's coordinate system instead of the world's coordinate system. It makes them easier to write. We use the transformation matrix to do object -> world and the inverse matrix to do world -> object.
\[ \left\{\begin{array}{l} O_{world} = T * O_{object} \\ D_{world} = (T * D_{object}) - (T * 0_4) \end{array}\right. \] \[ \left\{\begin{array}{l} O_{object} = I * O_{world} \\ D_{object} = (I * D_{world}) - (I * 0_4) \end{array}\right. \] \[0_4 = \left(\begin{array}{c} 0 \\ 0 \\ 0 \\ 1 \end{array}\right) \]
Bounding Box
The previous equations give us objects with infinite dimensions (except for the sphere) whereas objects in real life have finite dimensions. To simulate this, it is possible to provide two points that will form a bounding box around the object. On the intersection test, we are going to use the nearest point that is inside the bounding box.
This gives us the ability to do various objects such as mirrors, table surface and legs, light bubbles and even a Pokeball!
Light
An object is composed of an Intensity \(I_o\), a Color \(C_o\) and a Brightness \(B_o\). Each light has a Color \(C_l\) and there is an ambient color \(C_a\). Using all those properties, we can calculate the color of a point using the following formula:
\[
I_o * (C_o + B_o) * \left(C_a + \sum_{l}{(N \cdot D) * C_l}\right) \]
Only the lights visible from the intersection point are used in the sum. In order to check this, we send a shadow ray from the intersection point to the light and see if it intersects any object.
The following images are examples to demonstrate the lights.
Textures
In order to put a texture on an object, we need to map a point \((x, y, z)\) in the object's coordinate system into a point \((x, y)\) in the texture's coordinate system. For planes, it is straightforward, we just the \(z\) coordinate (which is equal to zero anyway). For spheres, cylinders and cones it is a bit more involved. Here is the formula where \(w\) and \(h\) are the width and height of the texture.
\[
\begin{array}{c c} \phi = acos(\frac{O'_y}{r}) & \theta = \frac{acos\left(\frac{O'_x}{r * sin(\phi)}\right)}{2\pi} \end{array} \]\[ \begin{array}{c c} x = w * \left\{\begin{array}{l l} \theta & \text{if } O'_x < 0 \\1 - \theta & \text{else}\end{array}\right. & y = h * \frac{\phi}{\pi}\end{array}\]Once we have the texture coordinates, we can easily create a checkerboard or put a texture. We added options such as scaling and repeat in order to control how the texture is placed.
We also support the alpha mask in order to make a color from a texture transparent.
Progressive Rendering
Ray tracing is a slow technique. At first, I generated pixels line by line, but I found out that the first few lines do not hold much information.
Instead, what we want to do is to have a fast overview of the scene and then improve on the details. In order to do that, during the first iteration we are only generating 1 pixel for a 32x32 square. Then we generate 1 pixel for a 16x16 square and so on ... We generate the top-left pixel and fill all the unknown pixels with it.
In order not to regenerate pixels we already seen, I came up with a condition to know if a pixel has already been generated. \(size\) is the current square size (32, 16, ...).
\[\left\{\begin{array}{l}
x \equiv 0 \pmod{size * 2}\\ y \equiv 0 \pmod{size * 2} \end{array}\right. \] Supersampling
Aliasing is a problem with Ray Tracing and we solve this issue using supersampling. Basically, we send more than one ray for each pixel. We have to chose representative points from a square. There are multiple strategies: in the middle, in a grid or random. Check the result of various combinations in the following image:
1x: Basic 4x: 4 random 1 fixed, 3 random 2x upscale 16x: 4x upscale 2x upscale, 1 fixed, 3 random Perlin Noise
We can generate random textures using Perlin Noise. We can control several parameters such as \(octaves\), the number of basic noise, the initial scale \(f\) and the factor of contribution \(p\) of the high frequency noises.
\[ noise(x, y, z) = \sum_{i = 0}^{octaves}{p^i * PerlinNoise(\frac{2^i}{f}x, \frac{2^i}{f}y, \frac{2^i}{f}z)} \]
\[noise\] \[noise * 20 - \lfloor noise * 20 \rfloor\] \[\frac{cos(noise) + 1}{2}\]
As seen in the example, we can apply additional functions after the noise has been generated to make interesting effects.
Portal
Last but not least, Portals from the self-titled game. They are easy to reproduce in a Ray Tracer and yet, I haven't seen any done.
If a ray enters portal A, it will go out from portal B. It is trivial to implement it, it is just a coordinates system transformation. Like we did for world and object transformation, we do it between A and B using their transformation matrix.
\[ \left\{\begin{array}{l} O_{a}' = T * O_{b} \\ D_{a}' = (T * D_{b}) - (T * 0_4) \end{array}\right. \] \[ \left\{\begin{array}{l} O_{b}' = T * O_{a} \\ D_{b}' = (T * D_{a}) - (T * 0_4) \end{array}\right. \]
Scene Editor
In order to create scenes more easily, we have defined a scene description language. We developed a basic CodeMirror syntax highlighting script. Just enter write your scene down and press Ray Trace 🙂
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You mean that functions are formally defined as
sets of ordered pairs. Thus, if $I=\{0,1,2\}$ and $X_0=X_1=X_2=\{a,b,c\}$, one member of $\prod\limits_{i\in I}X_i$ is the function $\{\langle 0,a\rangle,\langle 1,b\rangle,\langle 2,a\rangle\}$. You're asking, I take it, how to square this with the idea that a member of $\prod\limits_{i\in I}X_i$ 'ought' to be an ordered triple, for instance $\langle a,b,a\rangle$.
The answer depends on how we choose to define
ordered n-tuple. One way, for $n>2$, is to define it as a function whose domain is $\{0,1,\dots,n-1\}$. If we do that, $\{\langle 0,a\rangle,\langle 1,b\rangle,\langle 2,a\rangle\}$ is an ordered triple.
Another approach is to define ordered triples to be ordered pairs of the form $\langle \langle a_0,a_1\rangle,a_2\rangle$, ordered $4$-tuples to be ordered pairs of the form $\langle\langle\langle a_0,a_1\rangle,a_2\rangle,a_3\rangle$, and in general to define ordered $(n+1)$-tuples to be ordered pairs of the form $\langle\pi,a\rangle$, where $\pi$ is an ordered $n$-tuple. If you adopt this definition, then the members of my little product $\prod\limits_{i\in I}X_i$
aren't actually ordered triples. However, there is a natural bijection between them and the 'real' ordered triples, given by $$\{\langle 0,x\rangle,\langle 1,y\rangle,\langle 2,z\rangle\}\leftrightarrow\langle\langle x,y\rangle, z\rangle\;,$$ by means of which any statement about the one corresponds trivially to a corresponding statement about the other.
Should ordered $n$-tuples be defined in any other reasonable way, a similar situation will obtain.
The function definition may look odd at first, especially when you're dealing only with finite index sets, but it's the only really convenient way to deal with products over infinite index sets.
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ABC is equilateral triangle inscribed in circle with Radius R. D point on the circle. I want to proof that: $DA^2+DB^2+DC^2=6R^2$ (in three ways- the proof can be in anyway not just geometry)
closed as off-topic by Namaste, Qwerty, Daniel W. Farlow, MathOverview, Watson Dec 29 '16 at 21:38
This question appears to be off-topic. The users who voted to close gave this specific reason:
" This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Namaste, Qwerty, Daniel W. Farlow, MathOverview, Watson
You can solve this by using the coordinates of your points.
Without any loss of generality, you have:
$$ A(0, R), \ B(\frac{\sqrt{3}}{2}R, -\frac{1}{2}R), \ C(\frac{-\sqrt{3}}{2}R, -\frac{1}{2}R) \textrm{ and } D(x,y)$$ with $x^2+y^2 = R^2$.
Then $$DA^2 = x^2 + (y - R)^2 = 2R^2-2Ry$$ $$DB^2 = (x- \frac{\sqrt{3}}{2}R)^2+(y+\frac{1}{2}R)^2 = 2R^2+Ry-\sqrt{3}xR$$ $$DC^2 = (x+ \frac{\sqrt{3}}{2}R)^2+(y+\frac{1}{2}R)^2 = 2R^2+Ry+\sqrt{3}xR$$ Finally you get $$DA^2+DB^2+DC^2=6R^2$$
You can use the law of sines:
$${DB\over\sin(60-\angle DAC)}={DC\over\sin(\angle DBC)}={DA\over\sin(60+\angle DBC)}=2R$$
Now we know that $\angle DBC=\angle DAC=\beta$ (both of them are over the same arc $\overset{\frown}{DC}$), so we have:
$$DB=2R\sin(60-\beta)=2R\left({\sqrt{3}\over2}\cos\beta-{\sin\beta\over2}\right)$$ $$DC=2R\sin\beta$$ $$DA=2R\sin(60+\beta)=2R\left({\sqrt{3}\over2}\cos\beta+{\sin\beta\over2}\right)$$
so we get:
$$DB^2+DC^2+DA^2=\\4R^2\left({3\over4}\cos^2\beta+{\sin^2\beta\over4}-{\sqrt3\over2}\sin\beta\cos\beta+\sin^2\beta+{3\over4}\cos^2\beta+{\sin^2\beta\over4}+{\sqrt3\over2}\sin\beta\cos\beta\right)=4R^2{3\over2}=6R^2$$
Without loss of generality, let the circle have unit radius and rotate the image so $A$ $B$ and $C$ are the cube roots of unity, and let $D=e^{i\theta}$
Then the expression on the left hand side is $$|e^{i\theta}-1|^2+|e^{i\theta}-\omega|^2+|e^{i\theta}-\omega^2|^2$$
Noting that $1+\omega+\omega^2=0$ and that $|z|^2=zz^*$ you can expand this expression and it quickly reduces to $6$
I leave the details to you.
WLOG, assume D is located on the shorter circular arc between $B$ and $C$ (as in the picture). Let $\theta=\angle CBD$. If $O$ is the center of the circle, one can show that $\angle BOD=\frac{2\pi}{3}-2\theta$, $\angle AOD=\frac{2\pi}{3}+2\theta$, and $\angle COD=2\theta$.
From the law of cosines on triangles $BOD$, $AOD$, and $COD$, we get:
$$DA^2=2R^2-2R^2\cos(\frac{2\pi}{3}+2\theta)$$ $$DB^2=2R^2-2R^2\cos(\frac{2\pi}{3}-2\theta)$$ $$DC^2=2R^2-2R^2\cos(2\theta)$$ However, note that
$$\cos(\frac{2\pi}{3}+2\theta)+\cos(\frac{2\pi}{3}-2\theta)+\cos(2\theta)=0$$
So, adding the three equations above, we get:
$$DA^2+DB^2+DC^2=6R^2$$
Using triangles $AOD, BOD, COD$ and angle names $A=DBA, B=BAD, C=CAD$ we have $A=60^{\circ}+C$ and $B=60^{\circ}-C$ (noting that angles $DBC$ and $DAC$ are equal, because they are on the same arc) and using the cosine formula in triangles $OAD, OBD, OCD$
$$DA^2+DB^2+DC^2=2R^2-2R^2\cos2A+2R^2-2R^2\cos 2B+2R^2-2R^2\cos 2C$$
Now,using $\cos (P+Q)=\cos P \cos Q-\sin P\sin Q$, we have $$\cos 2A+\cos 2B+\cos 2C= \cos (120^{\circ}+2C)+\cos (120^{\circ}-2C)+\cos 2C =$$$$=(2\cos120^{\circ}+1)\cos 2C -\sin 120^{\circ} (\sin 2C-\sin 2C)$$
and the terms in brackets reduce to zero.
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Simplest Calculation of Half-band Filter Coefficients
Half-band filters are lowpass FIR filters with cut-off frequency of one-quarter of sampling frequency f
s and odd symmetry about f s/4 [1]*. And it so happens that almost half of the coefficients are zero. The passband and stopband bandwiths are equal, making these filters useful for decimation-by-2 and interpolation-by-2. Since the zero coefficients make them computationally efficient, these filters are ubiquitous in DSP systems.
Here we will compute half-band coefficients using the window method. While the window method typically does not yield the fewest taps for a given performance, it is useful for learning about half-band filters. Efficient equiripple half-band filters can be designed using the Matlab function firhalfband [2].
Coefficients by the Window Method
The impulse response of an ideal lowpass filter with cut-off frequency ω
c = 2πf c/f s is [3]:
$$h(n)=\frac{sin(\omega_c n)}{\pi n}, \quad -\infty < n < \infty $$
This is the familiar sinx/x or sinc function (scaled by ω
c/π). To create a filter using the window method, we truncate h(n) to N+1 samples and then apply a window. For a halfband filter, ω c = 2π*1/4 = π/2. So the truncated version of h(n) is:
$$h(n)=\frac{sin(n\pi /2)}{n\pi}, \quad -N/2 < n < N/2 $$
Now apply a window function w(n) of length N+1 to obtain the
filter coefficients b:
$$b(n)=h(n)w(n), \quad -N/2 < n < N/2 $$
$$b(n)=\frac{sin(n\pi /2)}{n\pi}w(n), \quad -N/2 < n < N/2 \qquad (1)$$
Choice of a particular window function w(n) is based on a
compromise between transition bandwidth and stopband attenuation. As a simple example, the Hamming window
function of length N+1 is**:
$$w(n)= 0.54+0.46cos(\frac{2\pi n}{N}),\quad -N/2 < n < N/2 $$
And that’s it!
Equation 1 is all you need to compute the coefficients. Note that N is even and the filter length is
N+1 (odd length).
Examples using Matlab
Here is Matlab code that implements Equation 1 for ntaps = 19:
ntaps= 19; N= ntaps-1; n= -N/2:N/2; sinc= sin(n*pi/2)./(n*pi+eps); % truncated impulse response; eps= 2E-16 sinc(N/2 +1)= 1/2; % value for n --> 0 win= kaiser(ntaps,6); % window function b= sinc.*win';
I used the Matlab Kaiser window function b= kaiser(ntaps,beta), where beta determines the sidelobe attenuation in the frequency domain [5]. Higher beta gives better stopband attenuation in the halfband filter, at the expense of wider transition band. Figure 1 shows the truncated sinc function, the window function, and the coefficients. Every other coefficient is zero, except for the main tap. This follows from the fact that every other sample of the truncated sinc function is zero. The scale of the plot obscures the values of the end coefficients; b(-9) and b(9) are equal to .000526.
Figure 2 shows the magnitude response for f
s= 100 Hz, with a linear amplitude scale. The response of the ideal lowpass filter is also shown. The amplitude is 0.5 at f s/4 = 25 Hz. Note the odd symmetry with respect to (x,y) = (25, 0.5).
As already mentioned, ntaps must be odd. It is possible to create a halfband frequency response with ntaps even, but in that case, there will be no zero-valued coefficients. Also, for ntaps = 9, 13, … etc, the end coefficients are zero. For this reason, we should choose ntaps = 4m + 3, where m is an integer. Thus the useful values of ntaps are 7, 11, 15, etc. The number of zero coefficients is ½(ntaps-3).
Now let’s look at the response for different values of ntaps (Figure 3). The upper graph uses a linear amplitude scale, while the lower one uses dB. As you can see, the stopband attenuation does not change much vs. ntaps, but the transition band becomes sharper as ntaps is increased. The trade-off between attenuation and transition band can be changed by adjusting the Kaiser window’s beta value.
Footnotes
* Mathematically, $H(e^{jω}) + H(e^{j(π-ω)}) = 1 $ , where ω = 2πf/f
s.
** If we let n = 0:N instead of –N/2:N/2, the expression for the Hamming window is [4]:
$$w(n)= 0.54-0.46cos(\frac{2\pi n}{N}),\quad n= 0:N $$
Figure 1. Coefficients of Halfband Filter with ntaps = 19
top: Truncated sinc function center: Kaiser window with beta = 6 bottom: Filter coefficients
Figure 2. Magnitude response of halfband filter with ntaps = 19, f
s= 100 Hz (linear amplitude scale)
Figure 3. Halfband filter magnitude response for different length filters, f
s = 100 Hz
top: linear amplitude scale bottom: dB amplitude scale
Other DSPRelated.com Posts on Halfband filters
Here are a couple of other posts about halfband filters on DSPRelated.com:
Rick Lyons, “Optimizing the Half-band Filters in Multistage Decimation and Interpolation”, Jan 2016. https://www.dsprelated.com/showarticle/903.php
Christopher Felton, “Halfband Filter Design with Python/Scipy, Feb 2012. https://www.dsprelated.com/showcode/270.php
References
1. Sanjit K. Mitra,
Digital Signal Processing, 2 nd Ed., McGraw-Hill, 2001, p 701-702.
2. Mathworks website https://www.mathworks.com/help/dsp/ref/firhalfband.html
3. Mitra, p 448.
4. Mathworks website https://www.mathworks.com/help/signal/ref/hamming.html
5. Mathworks website https://www.mathworks.com/help/signal/ref/kaiser.html
Neil Robertson November, 2017
Appendix A Matlab Function for Halfband Filter Coefficients
This program is provided as-is without any guarantees or warranty. The author is not responsible for any damage or losses of any kind caused by the use or misuse of the program.
%hbsynth.m 11/13/17 Neil Robertson % Halfband filter synthesis by the window method, using a Kaiser window. % input argument ntaps must be odd. function b= hbsynth(ntaps) if mod(ntaps,2)==0 error(' ntaps must be odd') end N= ntaps-1; n= -N/2:N/2; sinc= sin(n*pi/2)./(n*pi+eps); % truncated impulse response; eps= 2E-16 sinc(N/2 +1)= 1/2; % value for n --> 0 win= kaiser(ntaps,6); % window function b= sinc.*win'; % apply window function Appendix B Matlab Code to Create Figure 3 %hb_ex1.m 11/13/17 Neil Robertson fs= 100; b1= hbsynth(15); b2= hbsynth(23); b3= hbsynth(39); [h1,f]= freqz(b1,1,512,fs); H1= 20*log10(abs(h1)); [h2,f]= freqz(b2,1,512,fs); H2= 20*log10(abs(h2)); [h3,f]= freqz(b3,1,512,fs); H3= 20*log10(abs(h3)); subplot(211),plot(f,abs(h1),f,abs(h2),f,abs(h3)),grid axis([0 fs/2 -.1 1.1]),xlabel('Hz') subplot(212),plot(f,H1,f,H2,f,H3),grid axis([0 fs/2 -80 5]),xlabel('Hz'),ylabel('dB') text(37,-38,'ntaps= 15') text(34,-48,'ntaps= 23') text(23,-52,'ntaps= 39') Previous post by Neil Robertson:
There's No End to It -- Matlab Code Plots Frequency Response above the Unit Circle
Next post by Neil Robertson:
Design IIR Butterworth Filters Using 12 Lines of Code
Hi Neil,
great article ! Do you think that a half band filter designed with this method is suitable for oversampling decimation purposes (low pass filtering oversampled signal).
Thank you !
Andreas
Andreas,
Yes, it is appropriate. For example, you could use a single HB filter for decimation by 2, two HB's for decimation by 4, etc. Note in the case of decimation by 4, 8, etc., the stopband frequency range of each stage is different, so the HB filters are not normally identical to each other. See article by Rick Lyons: https://www.dsprelated.com/showarticle/903.php
For decimation by a large factor, a CIC filter may be more appropriate.
regards,
Neil
Andreas,
I should also mention again that the window method does not typically result in the most efficient design; a Parks- McClellan optimizer like Matlab's firhalfband or firpm would be desirable for getting fewest taps.
regards,
Neil
Fantasic, maybe the algorithm is not the most efficient, but it's easy to implement. The other article is also great, obviously the first (downsampling) filter doesn't need many taps in the fir, the last one needs. There are other approaches e.g. overlap/add in the frequency domain. I will give this easy sinc filter a try and will have a look what the performance drop of my synth plug-in is.
This is also very interesting:
Neil, I have implemented the algorithm into my synth plug-in using C++, optimized it and tested a lot of FIR tap sizes using the great online designer here: https://fiiir.com/ and looking at the frequency response of my software at different sampling rates. I'm now using 7, 11 and 15 taps (in case of 8 x downsampling) and it seems that this is enough for my purposes. It filters the most portions of a) generated gaussion noise aliasing and b) distortion/overdrive effect aliasing. I come to the conclusion that I have a performance drop of 6-10% in comparison to the other synth functionality. For what it does, that's a moderate value but I'm thinking about additional optimizations. Another remark is, that every second tap value should really be 0 (instead of very very small float/double values). Modern CPUs seem to automatically "skip" multiplications by 0., my performance profiling showed that.
I should add that the 6-10% performance drop was measured with an easy synth patch, with more complex patches and more polyphonic voices, the drop is about 2-3%.
Andreas,
Good work! Thanks for the link.
regards,
Neil
Very intuitive and good tutorial. Thanks very much.
You're welcome!
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Let $G,G_1$ and $G_2$ are three abelian groups with group homomorphisms $\phi_i:G\to G_i$. This gives $k$-algebra homomorphisms $k[\phi_i]:k[G]\to k[G_i]$. So we can consider $k[G_i]'s$ as $k[G]$-module via the homomorphisms $k[\phi_i]$. We can consider the tensor product $k[G_1]\otimes_{k[G]}k[G_2]$ and this will be again $k$-algebras.
My question is there a simpler way to describe the $k$-algebra: $k[G_1]\otimes_{k[G]}k[G_2]$?
For example take $G=\{e\}$, the identity group; then $k[G]=k$ and hence $$k[G_1]\otimes_{k[G]}k[G_2]=k[G_1]\otimes_kk[G_2]\cong k[G_1\times G_2].$$
So I was wondering if there exists any simpler way to express $k[G_1]\otimes_{k[G]}k[G_2]$ like above.
Note that here the groups are abelian and hence the group algebras are commutative rings. Therefore the tensor product makes sense.
Thank you in advance.
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There are many properties of circle geometry with semi circles, such as equal arcs on circles of equal radii subtend equal angles at the centres equal angles at the centre stand on equal chords the angles at the centre is twice an angle at the circumference subtended ay the same arc the perpendicular from the centre of a circle to a chord bisects the chord equal chords in equal circles are equidistant from the centres angles in the same segment are equal the angle in a semi-circle is a right angle opposite angles of a cyclic quadrilateral are supplementary Worked Example of Circle Geometry with Semi Circles (a) Explain why \( CTDS \) is a rectangle.
\( \begin{aligned} \require{color}
\angle SDT &= 90^{\circ} &\color{green} \text{angle in semi circle, diameter } AB \\
\angle ASC &= 90^{\circ} &\color{green} \text{angle in semi circle, diameter } AC \\
\angle DSC &= 90^{\circ} &\color{green} AD \text{ is a straight line} \\
\angle CTB &= 90^{\circ} &\color{green} \text{angle in semi circle, diameter } CB \\
\angle CTD &= 90^{\circ} &\color{green} DB \text{ is a straight line} \\
\angle SCT &= 90^{\circ} &\color{green} \text{angle sum of quadrilateral} \\
\therefore CTDS &\text{ is a rectangle} &\color{green} \text{quadrilateral with all angles right angles} \\
\end{aligned} \\ \)
(b) Show that \( \triangle MXS \) and \( \triangle MXC \) are congruent.
\( \begin{aligned} \require{color}
MX &= MX &\color{green} \text{common side} \\
MS &= MC &\color{green} \text{same radii} \\
SX &= XC &\color{green} \text{diagonals bisect each other} \\
\therefore \triangle MXS &\equiv \triangle MXC &\color{green} \text{SSS} \\
\end{aligned} \\ \)
(c) Show that the line \( ST \) is a tangent to the semicircle with diameter \( AC \).
\( \begin{aligned} \require{color}
\angle MCX &= 90^{\circ} &\color{green} \text{given} \\
\angle MSX &= \angle MCX &\color{green} \text{corresponding angles of congruent triangles} \\
\angle MCX &= 90^{\circ} \\
\therefore ST \text{ is a } &\text{tangent to the circle} &\color{green} \text{meets radius on the circumference at right angles} \\
\end{aligned} \\ \)
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*All sequences considered are non-negative.
Suppose we are given sequences $a_n,b_n$ such that $\sum a_n=1$ and $b_n$ is a sequence going to infinity. I am searching for any sequence $c_n$ with the properties that
(1) $c_n\to \infty$
(2) $c_n\le b_n$
(3) $c_n\cdot\max\lbrace a_k:b_n\le k\le 2b_n\rbrace\to 0$
Does such a sequence $c_n$ exists?
The main difficulty I am having is due to the fact that $na_n$ need not go to zero, so $b_n \max\lbrace a_k:b_n\le k\le 2b_n\rbrace$ need not go to zero (otherwise I would just take $c_n=b_n)$. I am not sure if it is true that there is a $c_n$ (e.g., $c_n= \log b_n$) that satisfies (3).
I have tried to consider $c_n=b_n^a$ with $0<a<1$, however I am not sure if (even for small $a$) we have $b_n^a\max\lbrace a_k:b_n\le k\le 2b_n\rbrace\to 0$. Perhaps the choice of $c_n$ should depend on $a_n$ as well.
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Let $x, y \in \mathbb{R}$ be arbitrary, and without loss of generality, assume $y < x$.
The function $f: \mathbb{R} \to \mathbb{R}$ defined by $f(t) = \cos(t)$ is continuous and differentiable in $\mathbb{R}$, so it is obviously continuous in $[y, x]$ and differentiable in $(y, x)$.
By the Mean Value Theorem, there exists a point $c \in (y, x)$ such that$$f^{\prime}(c) = -\sin(c) = \dfrac{\cos(x)-\cos(y)}{x-y}\text{.}$$However, recalling that $|-\sin(w)| = |\sin(w)| \leq 1$ for all $w \in \mathbb{R}$, it follows that $$\left|\dfrac{\cos(x)-\cos(y)}{x-y}\right| \leq 1$$hence$$|\cos(x)-\cos(y)| \leq |x-y|$$and the claim follows since $x$ and $y$ were arbitrarily chosen.
If $y > x$, then a nearly-identical argument to the above leads to the same conclusion.
If $y = x$, equality holds (since $|\cos(x)-\cos(y)| = |x-y| = 0$).
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Let $X$ and $Y$ be normed spaces, either both real or both complex. Let $f \colon X \to Y$ be a linear operator. Then $f$ is said to be
bounded if there exists a real number $r > 0$ such that $$ \lVert f(x) \rVert_Y \leq r \lVert x \rVert_X \mbox{ for every } x \in X. $$If there is no such $r$, then $f$ is said to be unbounded.
Now my question is, can we find an example of an unbounded linear operator (i) $f \colon \ell^\infty \to \ell^\infty$? (ii) $f \colon \ell^p \to \ell^p$, where $p$ is such that $1 \leq p < +\infty$? (iii) $f \colon \ell^\infty \to \ell^p$? (iv) $f \colon \ell^p \to \ell^\infty$? (v) $f \colon \mathrm{C}[a, b] \to \mathrm{C}[a, b]$?
Or, examples of unbounded linear operators between any other pairs of these normed spaces?
By definition, $\ell^\infty$ is the normed space of all the bounded sequences of (real or complex) numbers, with the normed defined by $$ \left\lVert \left( \xi_n \right)_{n \in \mathbb{N} } \right\rVert \colon= \sup \left\{ \ \left\lvert \xi_n \right\rvert \ \colon \ n \in \mathbb{N} \ \right\}. $$
For any real number $p$ such that $1 \leq p < +\infty$, the normed space $\ell^p$, by definition, is the vector space of all the sequences $\left( \xi_n \right)_{n \in \mathbb{N} }$ of (real or complex) numbers, for which the series $\sum \left\lvert \xi_n \right\rvert^p$ converges, that is, $$ \sum_{n=1}^\infty \left\lvert \xi_n \right\rvert^p < +\infty, $$ with the norm given by the formula $$ \left\lVert \left( \xi_n \right)_{n \in \mathbb{N} } \right\rVert \colon= \sqrt[p]{ \sum_{n=1}^\infty \left\lvert \xi_n \right\rvert^p }. $$
For any real numbers $a$ and $b$ such that $a < b$, the space $\mathrm{C}[a, b]$ is the normed space of all the real or complex-valued functions defined and continuous on the closed interval $[a, b]$ of the real line, with the norm defined by $$ \lVert x \rVert \colon= \max \{ \ \lvert x(t) \rvert \ \colon \ a \leq t \leq b \ \}. $$
Right now, the only example of an unbounded linear operator that I can recall is that of the differentiation operator of the normed space of all the continuously differentiable functions on a closed interval $[a, b]$ with the maximum norm into this normed space itself.
So any other examples, please?
I would appreciate references to some elementary-level text on analysis where examples of such unbounded linear operators can be found.
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Consider the following integral,
$$I(n)=\int_0^{\pi/2}\cos^nx\cos(nx)dx $$
I tried taking one $\cos x$ out and then integrating by parts. I also tried integrating by parts using $\cos(nx)$ as the other function but didn't get any relation between $I(n)$ and $I(n-1)$.
In both the ways, $\cos(nx)$ was creating a problem as it wasn't getting converted to $\cos[(n-1)x]$ which was required for getting a term of $I(n-1)$.
Any hint/partial solution would be appreciated.
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Let's be perfectly clear about the assumptions. I assume that "randomly selecting 4 points" A,B,C,D in the unit square means selecting uniformly and independently at random their coordinates $x_A,y_A,x_B...$ in the $[0,1]$ range; and that one of the 6 possible pairings between points to form the segments is also chosen uniformly and independently at random -- so the problem is equivalent to asking what is the probability that, e.g. AC and BD intersect.
It's easy to see that the solution is $\frac{25}{108}$, or about $23\%$, if one is willing to take as a given the solution for Sylvester's four-point problem for the square -- the probability that the convex hull of four points chosen uniformly and independently at random in a square is a quadrilateral, which is $\frac{25}{36}$. It is immediate to see that two segments chosen by a random pairing of four points intersect if, and only if, the convex hull of the four points is indeed a quadrilateral, and the two segments are its diagonals. If the former condition is satisfied, the latter happens with probability $\frac{1}{3}$ (the probability that a given point is paired with "middle one" of the remaining three), yielding a solution of $\frac{25}{36}\cdot\frac{1}{3}=\frac{25}{108}$ for the original problem.
Sylvester's four-point problems allows one to obtain the answer in exactly the same way even if the points are chosen uniformly at random in another convex region, simply by multiplication by $\frac{1}{3}$. The general formula for Sylvester's four-point problem in a convex region $R$, as given in the above link, equals $1-4\frac{\bar{A_R}}{A(R)}$ where $A(R)$ is the area of the region (in the case of the unit square, $1$) and $\bar{A}_R$ equals the expected area of a triangle obtained from $3$ points chosen uniformly at random within it. The formula is easily obtained noting that probability that one
given point lies within the triangle formed by the other $3$ is $\frac{\bar{A_R}}{A(R)}$, and since the events of this happening for different points are mutually exclusive, the probability that no point lies within the triangle formed by the other three is indeed $1-4\frac{\bar{A}_R}{A(R)}$.
Computing $\frac{\bar{A_R}}{A(R)}$ is non-trivial, in general. If R is a square (our case), a simple proof that $\frac{\bar{A_R}}{A(R)}=\frac{11}{144}$ can be found here. More in general if R is regular polygon of $n$ sides the solution is given by Alikosky's formula, $\frac{\bar{A_R}}{A(R)}=\frac{9\cos^2 (2\pi/n)+52\cos(2\pi/n)+44}{36n^2\sin^2(2\pi/n)}$. Note that this is a strictly decreasing function of $n$, so the intersection probability is strictly increasing with $n$, and the limit for $n\to\infty$, i.e. $\frac{35}{48\pi^2}$, yields the solution when picking points in the circle.
So, in general, the probability of intersection of $2$ segments, each made by $2$ points chosen uniformly and independently at random from a regular $n$-agon or any of its affine transformations (such as a rectangle or parallelogram) is:
$\frac{1}{3}\left(1-4\cdot\frac{9\cos^2(2\pi/n) + 52\cos(2\pi/n) + 44}{36 n^2\sin^2 (2\pi/n)}\right)$
with the limit for $n\to\infty$, i.e. $\frac{1}{3}-\frac{35}{36\pi^2}$, yielding the probability when choosing the points from a circle (or any affine transformation, such as an ellipse).
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Nevertheless, I wonder whether there are conditions for existence of a complement $M$ to the kernel $N$ of a bounded linear operator $T:V\to Q$. That is, under which conditions there is a closed subspace $M\subset V$ such that $N \oplus M = V$?
In my particular case, the operator $T\colon V \to Q$ fulfills these equivalent properties:
$T'\colon Q' \to V'$ is an homeomorphism on its range $T'$ is injective and has a closed range $T$ is surjective $T$ has a bounded right inverse(I was wrong here, see comments below)
Any ideas?
Disclaimer: This relates to the problem I have posted the day before.
EDIT: I additionally assume that $V$ and $Q$ are reflexive and separable.
UPDATE: I have answered the questions, based on the comments.
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Suppose you have a vector function of space: $\vec{E}(x,y,z)=Ex(x,y,z)\hat{x}+Ey(x,y,z)\hat{y}+Ez(x,y,z)\hat{z}$.
Suppose now you want to rotate the whole vector function by using a unitary rotation matrix $\mathbf{R}(\theta)$ but you still want to describe it in terms of the original coordinates. That is, a bunch of little arrows are firmly connected to some type of frame and you keep your head steady and rotate the frame with your hands. What will it look like now? I claim that it can be done in the following way:
The rotated vector field, $\vec{E}_r(x,y,z)$, is given by $$\vec{E}_r(x,y,z)=\mathbf{R(\theta)}\vec{E}\left ( \left [ \begin{array}{ccc} 1 \\ 0 \\ 0 \end{array} \right ]\cdot \mathbf{R}^T(\theta)\vec{r}, \left [ \begin{array}{ccc} 0 \\ 1 \\ 0 \end{array} \right ]\cdot \mathbf{R}^T(\theta) \vec{r},\left [ \begin{array}{ccc} 0 \\ 0 \\ 1 \end{array} \right ]\cdot \mathbf{R}^T(\theta) \vec{r} \right ),$$ where $\vec{r}$ = $\left ( \begin{array}{ccc} x \\ y \\ z \end{array} \right )$, $T$ indicates a matrix transpose, and $\cdot$ indicates a dot product.
My question: Is this correct?
It seems good enough in two dimensions. Suppose $\vec{E}(x,y)=x\hat{x}+\hat{y}$:
Using the rotation matrix $\mathbf{R}(\theta)=\left( \begin{array}{ccc} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \end{array} \right)$, with $\theta=\pi/2$ gives
This looks correct. It is as though I just took the image and rotated it 90 degrees clockwise. I checked it in three dimensions for some specific cases but nothing rigorous. I have not been able to find a reference regarding the rotation of vector fields.
Consider a specific case of a rotation of a vector field around the x axis by 180 degrees. Given that $\vec{E}$ is as described at the top, this would mean $\vec{E}_r(x,y,z)=Ex(x,-y,-z)\hat{x}-Ey(x,-y,-z)\hat{y}-Ez(x,-y,-z)\hat{z}$. Again, is this correct?
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Following appendix A of
"Ergoregions in Magnetised Black Hole Spacetimes" by G. W. Gibbons, A. H. Mujtaba and C. N. Pope, starting from the Lagrangian
$$\mathcal{L} = \hat{R} - \hat{F}_{\mu\nu}\hat{F}^{\mu\nu},$$ metric and field
$$\mathrm{d}\hat{s}^2 = e^{2\phi} \mathrm{d}s^2 + e^{-2\phi}(\mathrm{d}z + 2\mathcal{A})^2,\\ \hat{A} = A + \chi(\mathrm{d}z +2 \mathcal{A}),$$
after Kaluza-Klein reduction, with Killing vector $K\equiv\partial_z$ corresponding to a spatial dimension, we obtain the reduced lagrangian. After using Lagrange multipliers and dualizing the fields, we obtain $$\hat{F} = -e^{2\phi} \star \mathrm{d}\psi + d \chi \wedge (\mathrm{d}z + 2\mathcal{A}),\quad e^{-2\phi} \star F = \mathrm{d} \psi, \quad F \equiv \mathrm{d} A + 2\chi \mathrm{d}\mathcal{A},$$
where the hatted quantities are 4-dimensional, none of the fields depends on $z$ and $\star$ is the Hodge dual with respect to $\mathrm{d}s^2$. At the end, they define the complex Ernst potential by $d \Phi = i_K(\hat{\star}\hat{F} + \mathrm{i} \hat{F})$, $\Phi = \psi + \mathrm{i}\chi$, where $i_K$ is the interior product by $K$. $i_K \hat{F} = -\mathrm{d} \chi$ is easy to derive, but $\mathrm{d}\psi = i_K \hat{\star} \hat{F}$ has proved to be not so easy.
My question is about this last equation. Considering that $\det \hat{g}_{MN} = e^{4\phi} \det g_{\mu\nu}$, I obtain $$i_K \hat{\star} \hat{F} = - e^{2\phi} \star(F + 2 \mathrm{d} \chi \wedge \mathcal{A}),$$ which is obviously wrong. Could somebody provide some hint about this last equation? It must be easy to derive, but I cannot see the way at the moment.
This post imported from StackExchange Physics at 2015-03-31 11:34 (UTC), posted by SE-user auxsvr
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The details about what an RSA key is made up of are explained succinctly here.
Is it possible to reduce the amount of data that's usually packaged with the (private) key and then derive it later?
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If we want to compact an
existing RSA private key expressed as $(N,e,d,p,q,d_p,d_q,q_\text{inv})$, we can reduce it to $(e,p,q)$ and easily recompute the rest as:
$\begin{align} N&=p\cdot q\\ d&=e^{-1}\bmod\operatorname{lcm}(p-1,q-1)\;\text{ or }\;d=e^{-1}\bmod((p-1)\cdot(q-1))\\ d_p&=d\bmod(p-1)\;\text{ or equivalently }\;d_p=e^{-1}\bmod(p-1)\\ d_q&=d\bmod(q-1)\;\text{ or equivalently }\;d_q=e^{-1}\bmod(q-1)\\ q_\text{inv}&=q^{-1}\bmod p \end{align}$
It is possible to gain a few more bits; for example the low order bits of $p$, $q$ and $e$ are known to be set and need not be stored; further we know that $p\bmod6$ is either $1$ or $5$, thus it is enough to store $\lfloor p/6\rfloor$ and an extra bit, etc.. All in all, any RSA private key with $k$-bit public modulus $N$ and common (small) $e$ can be stored in about $k$ bits.
If we want a compact representation of private keys that we are free to choose, we can fix $e$ (removing need to store it) and decide to generate keys using some well defined deterministic procedure employing some Cryptographically Secure Pseudo Random Number Generator, and store the seeds of that CSPRNG, rather than the private keys. Whenever we need a private key, we (re)generate it from its seed. That has a performance issue, with workaround, see kasperd's answer.
If we'll generate $k$ keys of a certain size, without salt, we want to use use a (truly random) seed of at least $n+\log_2(k)$ bits, where $n$ is the security level corresponding to the public key size (perhaps $n\approx 112$ for 2048-bit RSA): we need to guard against the adversary enumerating seeds, generating the corresponding public modulus, and testing if it matches one of the public keys, which is expected to succeed after enumerating about $1/k$ of the seeds.
We can also use a passphrase, salt (user identifier), and a password-based key generation function, see this answer.
You can use a seed to start a PRNG. Then you can use that PRNG to generate the two (or more) primes required to generate the key pair. Now if you save that seed you can regenerate the key pair, which means you don't have the store the modulus, CRT components or private exponent.
So yes, it is possible to reduce the size, but this approach does have drawbacks:
Using this approach you can just store the seed, some 128 bits should be sufficient.
All in all, you may be better off choosing Elliptic Curve Cryptography, which uses a pretty small private key value to begin with (you can just use a specific curve, so you would not need to store all the parameters).
The drawback is the performance. Each time you need to use the key you need to spend as much CPU time for regenerating the key as you used for generating it the first time. However by analyzing how the CPU time is spent during RSA key generation, it is possible to improve performance.
In order to find a suitable prime the RSA key generation tries many different numbers - most of which have to be discarded because they are not primes. And each candidate requires CPU time to decide whether it is prime.
Once the first candidate has been generated using a specific seed and found not to be prime, you can continue generating another candidate using the next random numbers from the PRNG. A better approach however is to simply discard the seed which did not produce a suitable prime immediately. Instead continue trying new seeds until a suitable prime has been found.
This procedure can be repeated until two seeds have been found, which each produce a suitable prime as the very first candidate. Now storing those two seeds will be sufficient to work as secret key.
The advantage is that when using the private key there is no need to run the primality test because the seeds are already known to produce primes as the first candidate. This will not be much slower to use than an ordinary RSA secret key.
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June 17th, 2019, 01:49 PM
# 1
Senior Member
Joined: Oct 2015
From: Greece
Posts: 137
Thanks: 8
How to solve this simple Laplace integral?
Check this Image. I tried to post it as an image but it takes the entire space of the post...
It seems I have forgotten a lot about solving integrals... But I really need to remember quickly how to solve this kind of fractional integrals since I haven't memorized the Laplace "trick" formulas (like 1 over Laplace is $\displaystyle \frac{1}{s}$) and tomorrow I have a test.
Last edited by babaliaris; June 17th, 2019 at 01:53 PM.
June 18th, 2019, 05:37 AM
# 4
Senior Member
Joined: Oct 2015
From: Greece
Posts: 137
Thanks: 8
By the way I finished the test today, and it was 90% of Fourier transformation exercises while in the past you had like one exercise only... I probably failed the test and the weird thing is that i knew how to solve the integrals...
The problem is that the textbook says that $\displaystyle \mathcal{L}[1] = \frac{1}{s}$ but if I try to calculate it,
$\displaystyle
\mathcal{L}[1] = \lim_{T->\infty} \int_{-T}^{T} 1 \cdot e^{-st} dt =
\lim_{T->\infty} [\frac{e^{-st}}{-s}]_{-T}^{T} =
\lim_{T->\infty} [\frac{e^{-sT}}{-s} - \frac{e^{sT}}{-s}] =
\frac{1}{s} \lim_{T->\infty} [-e^{-sT} + e^{sT}]
$
I believe that $\displaystyle \lim_{T->\infty} [-e^{-sT}] = 0 $ and $\displaystyle \lim_{T->\infty} [e^{sT}] = \infty$ by trying to "see it" using the graph below.
But of course this is not true, because the limit of them both must be 1 in order to get $\displaystyle \frac{1}{s}$
This is a really great problem that is keeping me from calculating Laplace integrals in general.
I certainly need to review calculus all over again, but I have no time... I will try this summer though.... This limit is extremely easy and yet I don't remember how to solve it.
Offtopic: How did you put that image into a thumbnail? When I try to add a big image in a post usually it takes over the whole screen.
Last edited by babaliaris; June 18th, 2019 at 05:43 AM.
June 18th, 2019, 11:14 AM
# 5
Global Moderator
Joined: Oct 2008
From: London, Ontario, Canada - The Forest City
Posts: 7,963
Thanks: 1148
Math Focus: Elementary mathematics and beyond
Tags integral, laplace, simple, solve
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When can we do a linear regression without fixed or random effects and when do we need to use those in the regression analysis? I have tried studying a lot but have got only a vague idea. I would be extremely thankful if anyone can explain me with examples.
Here is a standard linear panel data model: $$ y_{it}=X_{it}\delta+\alpha_i+\eta_{it}, $$ the so-called error component model. Here, $\alpha_i$ is what is sometimes called individual-specific heterogeneity, the error component that is constant over time. The other error component $\eta_{it}$ is "idiosyncratic", varying both over units and over time.
A reason to use a random or fixed effects approach instead of pooled OLS is that the presence of $\alpha_i$ will lead to an error covariance matrix that is not "spherical" (so not a multiple of the identity matrix(, so that a GLS-type approach like random effects will be more efficient than OLS.
If, however, the $\alpha_i$ correlate with the regressors $X_{it}$ - as will be the case in many typical applications - omitting these individual-specific intercepts will lead to omitted variable bias. Then, a fixed effect approach which effectively fits such intercepts will be more convincing.
The following figure aims to illustrate this point. The raw correlation between $y$ and $X$ is positive. But, the observations belonging to one unit (color) exhibit a negative relationship - this is what we would like to identify, because this is the reaction of $y_{it}$ to a change in $X_{it}$.
Also, there is correlation between the $\alpha_i$ and $X_{it}$: If the former are individual-specific intercepts (i.e., expected values for unit $i$ when $X_{it}=0$), we see that the intercept for, e.g., the lightblue panel unit is much smaller than that for the brown unit. At the same time, the lightblue panel unit has much smaller regressor values $X_{it}$.
So, random effects or pooled OLS would be the wrong strategy here, because it would result in a positive esimate of $\delta$, as these two estimators basically ignore the colors (RE only incroporates the colors for the estimate of the variance covariance matrix).
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One recovers Einstein's equations by considering the bosonic closed string theory with a zero B-field, a constant dilaton $\phi$ and by taking the limit $\alpha' \rightarrow 0$, $g_s \rightarrow +\infty$ such that $g_s \sqrt{\alpha'}$ is some constant $l_p$. Here $\alpha'$ is the Regge's slope i.e. the inverse of the string tension up to a normalization, $g_s$ is the string coupling constant related to the dilaton background by $g_s = e^\phi$. In this limit, the only excitation of the theory is a massless symmetric rank 2 tensor field $h_{\mu \nu}$ and the statement is that its $n$-point correlation functions coincide with the $n$-points correlations functions of the graviton computed at tree level from the Einstein-Hilbert action with the Planck length equals to $l_p$.
The original papers for this result are by Yoneya
http://ptp.oxfordjournals.org/content/51/6/1907.refs
and Scherk, Schwarz
http://ccdb5fs.kek.jp/cgi-bin/img/allpdf?196800234
The relation between this derivation of the Einstein's equation and the one through the $\beta$-function presented in the answer of joshphysics is nicely explained in the section 10.1 of the Polyakov's book "Gauge fields and strings". The low energy effective action for the massless excitations of the string can be computed from the correlation functions of the corresponding vertex operators. These vertex operators are constructed form the weight 2 local operators of the world-sheet CFT and the spacetime correlation functions of the massless excitations can be expressed in terms of the correlation functions of these operators in the 2d CFT. But in the other hand, weight 2 local operators define the marginal deformations of the 2d CFT and the beta function with respect to one of the massless field is obtained from the variation of the theory under the corresponding deformation. To compute this variation, one can perturbatively expand the deformation of the action and one has to compute n-point correlations functions of the weight 2 local operators in the 2d CFT, i.e. the same object as for the spacetime amplitudes. One is simply saying that the deformations of the space-time background by turning on massless fields correspond to the marginal deformations of the 2d worldsheet CFT.
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June 18th, 2019, 01:53 AM
# 1
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Joined: Apr 2017
From: India
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Function of two variables and differentiability
I read the following three statements for a real valued two variable function 'f' which are true:
1. I know that when the function(of two variables) is differentiable, then it implies that the function is continuous and partial derivatives of the function exists.
2. I also know that if one of the partial derivatives of the function is continuous at a point (a,b) and other partial derivative merely exists at point (a,b) then also the function is differentiable at (a,b)
3. Also, a continuously differentiable function is always differentiable.
After reading all the three, a question aroused:
If it is given that the function is continuous and whose partial derivatives exist at (a,b). Is it enough for me deduce that the function will be differentiable or I also need the partial derivatives to be continuous at the point (a,b) to deduce that.
Please help me.
Last edited by shashank dwivedi; June 18th, 2019 at 01:55 AM.
June 18th, 2019, 04:59 AM
# 2
Senior Member
Joined: Sep 2016
From: USA
Posts: 647
Thanks: 412
Math Focus: Dynamical systems, analytic function theory, numerics
This topic is a bit subtle the first time through so I'll include some extra details. Assume $f: D \subset \mathbb{R}^n \to \mathbb{R}$ is a scalar function of $n$ variables where $n > 1$. Let $\partial_j f$ denote the partial derivivative with respect to the $j^{\rm th}$ input. Now, we define a few sets:
1. Let $A$ denote the set of functions whose partial derivatives exist for every $x \in D$.
2. Let $B$ denote the set of differentiable functions defined on $D$.
3. Let $C$ denote the set of functions whose partial derivatives are continuous for every $x \in D$.
Then, we have the following inclusions
\[
A \subset B \subset C
\]
and both inclusions are strict.
The reason this can be confusing is typically due to the fact that the derivative of $f$ (when it exists) is a linear functional which is perfectly fine to talk about. However, when one needs to compute it in practice, you need to choose a basis and the partial derivatives are a highly convenient choice.
To be more specific, the derivative of $f$ should be a linear approximation of $f$ which means, for a fixed $x \in D$, we want a linear function $A: \mathbb{R}^n \to \mathbb{R}$ such that $f(x+h) \approx f(x) + Ah$ when $\left| \left| h \right| \right| \approx 0$. First notice that $A$ is a linear functional. Now, what do we mean by approximation? Well, we require the following:
Let $h \in \mathbb{R}^n$ be an arbitrary unit vector, then $A$ is the derivative of $f$ at $x$, denoted by $A = Df(x)$ if
\[
\lim\limits_{t \to 0} \frac{f(x + th) - f(x) - tAh}{t} = 0
\]
for every unit vector $h$. There are of course many ways to write the same thing here so don't be thrown off if my definition differs slightly from another one. What is important is that a derivative is a linear functional.
Now, if we have a specific $f$, how can we get our hands on $Df(x)$? Even simpler, how can we even determine whether such a linear functional exists or not (i.e. how can we decide if $f \in B$?) This is much harder in the abstract setting.
However, we have the following theorem which is used very often in practice. Roughly, it says if we ask for something a little bit stronger than differentiability, then we get an easy condition to check which is sufficient for differentiability. Moroever, we get an easy way to compute these derivatives in the basis, $\{dx_1, \dotsc, dx_n \}$.
Theorem: Fix $x \in D$ and suppose $\partial_1f, \dotsc, \partial_nf$ exist. Even more, assume that in some open neighborhood of $x$, $\{\partial_1f, \dotsc, \partial_n f\}$ are all continuous. Then, $Df(x)$ exists and moreover, the Jacobian matrix (matrix of partial derivatives) is a representation for it.
I have to run but I might add more to this later.
June 18th, 2019, 08:17 PM
# 3
Member
Joined: Apr 2017
From: India
Posts: 73
Thanks: 0
Thanks for the elaborate explanation on this topic.
So, The conclusion I reached after your explanation is this:
If the function is given to be continuous and it's said that partial derivatives exist at (a,b)(nothing is said about the continuity of partial derivatives). Above you mentioned that A is the subset B, then accordingly the function can be differentiable or cannot be differentiable, that entirely depends on the definition of the domain in this case. (Did I get that right?)
However, the existence and continuity of the partial derivatives at (a,b) ensures the differentiability of the function at that point. (This is true by the theorem above)
Last edited by shashank dwivedi; June 18th, 2019 at 08:21 PM.
Tags differentiability, function, variables
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Colloquia/Fall18 Contents 1 Mathematics Colloquium 1.1 Spring 2018 1.2 Spring Abstracts 1.2.1 January 29 Li Chao (Columbia) 1.2.2 February 2 Thomas Fai (Harvard) 1.2.3 February 5 Alex Lubotzky (Hebrew University) 1.2.4 February 6 Alex Lubotzky (Hebrew University) 1.2.5 February 9 Wes Pegden (CMU) 1.2.6 March 2 Aaron Bertram (Utah) 1.2.7 March 16 Anne Gelb (Dartmouth) 1.2.8 April 6 Edray Goins (Purdue) 1.3 Past Colloquia Mathematics Colloquium
All colloquia are on Fridays at 4:00 pm in Van Vleck B239,
unless otherwise indicated. Spring 2018
date speaker title host(s) January 29 (Monday) Li Chao (Columbia) Elliptic curves and Goldfeld's conjecture Jordan Ellenberg February 2 (Room: 911) Thomas Fai (Harvard) The Lubricated Immersed Boundary Method Spagnolie, Smith February 5 (Monday, Room: 911) Alex Lubotzky (Hebrew University) High dimensional expanders: From Ramanujan graphs to Ramanujan complexes Ellenberg, Gurevitch February 6 (Tuesday 2 pm, Room 911) Alex Lubotzky (Hebrew University) Groups' approximation, stability and high dimensional expanders Ellenberg, Gurevitch February 9 Wes Pegden (CMU) The fractal nature of the Abelian Sandpile Roch March 2 Aaron Bertram (University of Utah) Stability in Algebraic Geometry Caldararu March 16 (Room: 911) Anne Gelb (Dartmouth) Reducing the effects of bad data measurements using variance based weighted joint sparsity WIMAW April 4 (Wednesday) John Baez (UC Riverside) TBA Craciun April 6 Edray Goins (Purdue) Toroidal Belyĭ Pairs, Toroidal Graphs, and their Monodromy Groups Melanie April 13 Jill Pipher (Brown) TBA WIMAW April 16 (Monday) Christine Berkesch Zamaere (University of Minnesota) TBA Erman, Sam April 25 (Wednesday) Hitoshi Ishii (Waseda University) Wasow lecture TBA Tran date person (institution) TBA hosting faculty date person (institution) TBA hosting faculty date person (institution) TBA hosting faculty date person (institution) TBA hosting faculty date person (institution) TBA hosting faculty date person (institution) TBA hosting faculty date person (institution) TBA hosting faculty date person (institution) TBA hosting faculty date person (institution) TBA hosting faculty Spring Abstracts January 29 Li Chao (Columbia)
Title: Elliptic curves and Goldfeld's conjecture
Abstract: An elliptic curve is a plane curve defined by a cubic equation. Determining whether such an equation has infinitely many rational solutions has been a central problem in number theory for centuries, which lead to the celebrated conjecture of Birch and Swinnerton-Dyer. Within a family of elliptic curves (such as the Mordell curve family y^2=x^3-d), a conjecture of Goldfeld further predicts that there should be infinitely many rational solutions exactly half of the time. We will start with a history of this problem, discuss our recent work (with D. Kriz) towards Goldfeld's conjecture and illustrate the key ideas and ingredients behind these new progresses.
February 2 Thomas Fai (Harvard)
Title: The Lubricated Immersed Boundary Method
Abstract: Many real-world examples of fluid-structure interaction, including the transit of red blood cells through the narrow slits in the spleen, involve the near-contact of elastic structures separated by thin layers of fluid. The separation of length scales between these fine lubrication layers and the larger elastic objects poses significant computational challenges. Motivated by the challenge of resolving such multiscale problems, we introduce an immersed boundary method that uses elements of lubrication theory to resolve thin fluid layers between immersed boundaries. We apply this method to two-dimensional flows of increasing complexity, including eccentric rotating cylinders and elastic vesicles near walls in shear flow, to show its increased accuracy compared to the classical immersed boundary method. We present preliminary simulation results of cell suspensions, a problem in which near-contact occurs at multiple levels, such as cell-wall, cell-cell, and intracellular interactions, to highlight the importance of resolving thin fluid layers in order to obtain the correct overall dynamics.
February 5 Alex Lubotzky (Hebrew University)
Title: High dimensional expanders: From Ramanujan graphs to Ramanujan complexes
Abstract:
Expander graphs in general, and Ramanujan graphs , in particular, have played a major role in computer science in the last 5 decades and more recently also in pure math. The first explicit construction of bounded degree expanding graphs was given by Margulis in the early 70's. In mid 80' Margulis and Lubotzky-Phillips-Sarnak provided Ramanujan graphs which are optimal such expanders.
In recent years a high dimensional theory of expanders is emerging. A notion of topological expanders was defined by Gromov in 2010 who proved that the complete d-dimensional simplical complexes are such. He raised the basic question of existence of such bounded degree complexes of dimension d>1.
This question was answered recently affirmatively (by T. Kaufman, D. Kazdhan and A. Lubotzky for d=2 and by S. Evra and T. Kaufman for general d) by showing that the d-skeleton of (d+1)-dimensional Ramanujan complexes provide such topological expanders. We will describe these developments and the general area of high dimensional expanders.
February 6 Alex Lubotzky (Hebrew University)
Title: Groups' approximation, stability and high dimensional expanders
Abstract:
Several well-known open questions, such as: are all groups sofic or hyperlinear?, have a common form: can all groups be approximated by asymptotic homomorphisms into the symmetric groups Sym(n) (in the sofic case) or the unitary groups U(n) (in the hyperlinear case)? In the case of U(n), the question can be asked with respect to different metrics and norms. We answer, for the first time, one of these versions, showing that there exist fintely presented groups which are not approximated by U(n) with respect to the Frobenius (=L_2) norm.
The strategy is via the notion of "stability": some higher dimensional cohomology vanishing phenomena is proven to imply stability and using high dimensional expanders, it is shown that some non-residually finite groups (central extensions of some lattices in p-adic Lie groups) are Frobenious stable and hence cannot be Frobenius approximated.
All notions will be explained. Joint work with M, De Chiffre, L. Glebsky and A. Thom.
February 9 Wes Pegden (CMU)
Title: The fractal nature of the Abelian Sandpile
Abstract: The Abelian Sandpile is a simple diffusion process on the integer lattice, in which configurations of chips disperse according to a simple rule: when a vertex has at least 4 chips, it can distribute one chip to each neighbor.
Introduced in the statistical physics community in the 1980s, the Abelian sandpile exhibits striking fractal behavior which long resisted rigorous mathematical analysis (or even a plausible explanation). We now have a relatively robust mathematical understanding of this fractal nature of the sandpile, which involves surprising connections between integer superharmonic functions on the lattice, discrete tilings of the plane, and Apollonian circle packings. In this talk, we will survey our work in this area, and discuss avenues of current and future research.
March 2 Aaron Bertram (Utah)
Title: Stability in Algebraic Geometry
Abstract: Stability was originally introduced in algebraic geometry in the context of finding a projective quotient space for the action of an algebraic group on a projective manifold. This, in turn, led in the 1960s to a notion of slope-stability for vector bundles on a Riemann surface, which was an important tool in the classification of vector bundles. In the 1990s, mirror symmetry considerations led Michael Douglas to notions of stability for "D-branes" (on a higher-dimensional manifold) that corresponded to no previously known mathematical definition. We now understand each of these notions of stability as a distinct point of a complex "stability manifold" that is an important invariant of the (derived) category of complexes of vector bundles of a projective manifold. In this talk I want to give some examples to illustrate the various stabilities, and also to describe some current work in the area.
March 16 Anne Gelb (Dartmouth)
Title: Reducing the effects of bad data measurements using variance based weighted joint sparsity
Abstract: We introduce the variance based joint sparsity (VBJS) method for sparse signal recovery and image reconstruction from multiple measurement vectors. Joint sparsity techniques employing $\ell_{2,1}$ minimization are typically used, but the algorithm is computationally intensive and requires fine tuning of parameters. The VBJS method uses a weighted $\ell_1$ joint sparsity algorithm, where the weights depend on the pixel-wise variance. The VBJS method is accurate, robust, cost efficient and also reduces the effects of false data.
April 6 Edray Goins (Purdue)
Title: Toroidal Belyĭ Pairs, Toroidal Graphs, and their Monodromy Groups
Abstract: A Belyĭ map [math] \beta: \mathbb P^1(\mathbb C) \to \mathbb P^1(\mathbb C) [/math] is a rational function with at most three critical values; we may assume these values are [math] \{ 0, \, 1, \, \infty \}. [/math] A Dessin d'Enfant is a planar bipartite graph obtained by considering the preimage of a path between two of these critical values, usually taken to be the line segment from 0 to 1. Such graphs can be drawn on the sphere by composing with stereographic projection: [math] \beta^{-1} \bigl( [0,1] \bigr) \subseteq \mathbb P^1(\mathbb C) \simeq S^2(\mathbb R). [/math] Replacing [math] \mathbb P^1 [/math] with an elliptic curve [math]E [/math], there is a similar definition of a Belyĭ map [math] \beta: E(\mathbb C) \to \mathbb P^1(\mathbb C). [/math] Since [math] E(\mathbb C) \simeq \mathbb T^2(\mathbb R) [/math] is a torus, we call [math] (E, \beta) [/math] a toroidal Belyĭ pair. The corresponding Dessin d'Enfant can be drawn on the torus by composing with an elliptic logarithm: [math] \beta^{-1} \bigl( [0,1] \bigr) \subseteq E(\mathbb C) \simeq \mathbb T^2(\mathbb R). [/math]
This project seeks to create a database of such Belyĭ pairs, their corresponding Dessins d'Enfant, and their monodromy groups. For each positive integer [math] N [/math], there are only finitely many toroidal Belyĭ pairs [math] (E, \beta) [/math] with [math] \deg \, \beta = N. [/math] Using the Hurwitz Genus formula, we can begin this database by considering all possible degree sequences [math] \mathcal D [/math] on the ramification indices as multisets on three partitions of N. For each degree sequence, we compute all possible monodromy groups [math] G = \text{im} \, \bigl[ \pi_1 \bigl( \mathbb P^1(\mathbb C) - \{ 0, \, 1, \, \infty \} \bigr) \to S_N \bigr]; [/math] they are the ``Galois closure
of the group of automorphisms of the graph. Finally, for each possible monodromy group, we compute explicit formulas for Belyĭ maps [math] \beta: E(\mathbb C) \to \mathbb P^1(\mathbb C) [/math] associated to some elliptic curve [math] E: \ y^2 = x^3 + A \, x + B. [/math] We will discuss some of the challenges of determining the structure of these groups, and present visualizations of group actions on the torus.
This work is part of PRiME (Purdue Research in Mathematics Experience) with Chineze Christopher, Robert Dicks, Gina Ferolito, Joseph Sauder, and Danika Van Niel with assistance by Edray Goins and Abhishek Parab.
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While reading up on quantile regression I found a really nice hack described in Bayesian Quantile Regression Methods (Lancaster & Jae Jun, 2010). It is called
Jeffreys’ substitution posterior for the median, first described by Harold Jeffreys in his Theory of Probability, and is a non-parametric method for approximating the posterior of the median. What makes it cool is that it is really easy to understand and pretty simple to compute, while making no assumptions about the underlying distribution of the data. The method does not strictly produce a posterior distribution, but has be shown to produce a conservative approximation to a valid posterior (Lavine, 1995). In this post I will try to explain Jeffreys’ substitution posterior, give R-code that implements it and finally compare it with a classical non-parametric test, the Wilcoxon signed-rank test. But first a picture of Sir Harold Jeffreys:
Say we have a vector $x$ and we are interested in the median of the underlying distribution.
The property of the median is that it sits smack in the middle of the distribution. We would expect that, on average, half of the datapoints of a vector generated from a distribution would fall to the left of the median (and so the other half would fall to the right). Now, given the vector $x$ of length $n$ and a candidate for the median $m$, let $n_l$ be the number of values that fall to the left of $m$ and $n_r$ be the number of values that fall to the right. If $m$ was the actual median the probability of getting the $n_l$ and $n_r$ that we actually got could be calculated as
$$ \text{The number of ways of selecting } n_l \text{ out of } n \text{ values in } x\over \text{The number of ways of placing the } n \text{ values in x into two categories } (n_l\text{ and } n_r) $$
Or using notation:
$${{n}\choose{n_l}} \big / 2^n$$
Here is an example:
Jeffreys proposed to use the probability ${{n}\choose{n_l}} \big / 2^n$ as a substitution for the likelihood, $p(x \mid m)$, and then calculate $p(m \mid x) \propto p(x \mid m) \cdot p(m)$.
By doing this we get something posterior-like we can do the usual stuff with, calculate credible intervals, calculate the probability that the median is smaller than this and that, etc. Before doing this we will just rewrite the substitution likelihood $p(x \mid m)$ to make it easier to work with. We can rewrite the $n_l$-combinations statement as
$${{n}\choose{n_l}} = {n! \over n_l!n_r!}$$
resulting in
$${{{n}\choose{n_l}} \big / 2^n} = {{{n! \over n_l!n_r!2^n}} }$$
If we change the candidate for the median $m$ the only numbers that are going to change are $n_l!$ and $n_r!$, $n!$ and $2^n$ stays the same. So, if we are interested in an expression that is proportional to the likelihood we could just use
$$ p(x \mid m) \propto {1 \over n_l!n_r!}$$
If we assume a uniform prior over $m$ we get that
$$p(m \mid x) \propto p(x \mid m) \propto {1 \over n_l!n_r!}$$
Great! Now we have a method to go from a vector of data, $x$, to a posterior probability for the median, $p(m \mid x)$.
Doing it in R
Using Jeffreys’ substitution posterior in R is pretty simple, but we have to be cautious not to run into numerical trouble. Directly calculating $1 / (n_l!n_r!)$ will only work for small datasets, already
factorial(180) results in values so large that you’ll get an error in R. To overcome this problem we’ll do all the calculations on the log-scale as far as possible, using
lfactorial instead. We’ll also use the fact that all candidate $m$ between two datapoints have the same likelihood, and so it suffices to calculate the likelihood once for each interval between two datapoints. First, here is the code for plotting the posterior:
# Generating some test data x <- rnorm(20, mean=0, sd=1) x <- sort(x) n <- length(x) # Now calculating the "likelihood" for the median being # in each of the intervals between the xs. Easily done with the following: # 1 / (factorial(1:(n - 1)) * factorial((n - 1):1)) # but on the log-scale to avoid numerical problems loglike <- 1 - ( lfactorial(1:(n - 1)) + lfactorial((n - 1):1) ) # Going back to the likelihood scale, subtracting max(loglike) to avoid # floating-point underflow. like <- exp(loglike - max(loglike)) # Plotting with type="s" which plots a step function. plot(x, c(like, 0), type="s", xlim=c(-2, 2), ylab="Likelihood (sort of)", xlab="median")
Nice, we got a plot of the posterior which looks sort of like a skyscraper, but this is alright, the posterior is supposed to be a step function. Looking at the plot we can see that, most likely, the median is between -0.5 and 1.0 (which seems alright as we actually know that the true median is 0.0). However, the representation of the posterior that we currently have in R is not very useful if we would want to calculate credible intervals or the probability of the median being higher than, for example 0.0. A more useful format would be to have a vector of random samples drawn from the posterior. This is also pretty easy to do in R, but first we need a helper function:
# A more numerically stable way of calculating log( sum( exp( x ))) Source: # http://r.789695.n4.nabble.com/logsumexp-function-in-R-td3310119.html logsumexp <- function(x) { xmax <- which.max(x) log1p(sum(exp(x[-xmax] - x[xmax]))) + x[xmax] }
Then we calculate the proportional likelihood of the median being in each interval, transforms this into a probability, and finally use these probabilities to sample from the posterior. I’m going to be slightly sloppy and disregard the posterior that lies outside the range of the data, if we have more than a dozen datapoints this sloppiness is negligible.
# like * diff(x), but on the log scale. Calculates the proportional # likelihood of the median being in each interval like: # c( like[1] * (x[2] - x[1]), like[2] * (x[3] - x[2]), ... ) interval_loglike <- loglike + log(diff(x)) # Normalizing the likelihood to get to the probability scale interval_prob <- exp(interval_loglike - logsumexp(interval_loglike)) # Sampling intervals proportional to their probability n_samples <- 10000 samp_inter <- sample.int(n - 1, n_samples, replace=TRUE, prob = interval_prob) # Then, within each sampled interval, draw a uniform sample. s <- runif(n_samples, x[samp_inter], x[samp_inter + 1])
Now that we have a couple of samples in
swe can do all the usual stuff such as calculating a 95% credible interval:
quantile(s, c(0.025, 0.975))
## 2.5% 97.5% ## -0.4378 0.6978
or plotting the posterior as a histogram with a superimposed 95% CI:
hist(s, 30, freq = FALSE, main = "Posterior", xlab = "median") lines(quantile(s, c(0.025, 0.975)), c(0, 0), col = "red", lwd = 6) legend("topright", legend = "95% CI", col = "red", lwd = 6)
A Handy Function to Copy-n-Paste
Here is a handy function you should be able to copy-n-paste directly into R. It packages all the calculations done above into one function that plots the posterior, calculates a 95% CI and returns a vector of samples that can be further inspected. I haven’t tested this function extensively so use it at your own risk and please tell me if you find any bugs.
# A more numerically stable way of calculating log( sum( exp( x ))) Source: # http://r.789695.n4.nabble.com/logsumexp-function-in-R-td3310119.html logsumexp <- function(x) { xmax <- which.max(x) log1p(sum(exp(x[-xmax] - x[xmax]))) + x[xmax] } # Silently returns samples from Jeffreys’ substitution posterior given a # vector of data (x). Also produces a histogram of the posterior and prints # out a 95% quantile credible interval. Assumes a non-informative uniform # [-Inf, Inf] prior over the median. jeffreys_median <- function(x, n_samples = 10000, draw_plot = TRUE) { x <- sort(x) n <- length(x) loglike <- 1 - lfactorial(1:(n - 1)) - lfactorial((n - 1):1) interval_loglike <- loglike + log(diff(x)) interval_prob <- exp(interval_loglike - logsumexp(interval_loglike)) samp_inter <- sample.int(n - 1, n_samples, replace = TRUE, prob = interval_prob) s <- runif(n_samples, x[samp_inter], x[samp_inter + 1]) cat("\n Jeffreys’ Substitution Posterior for the median\n\n") cat("median\n ", median(x), "\n") cat("95% CI\n ", quantile(s, c(0.025, 0.975)), "\n") if (draw_plot) { hist(s, 30, freq = FALSE, main = "Posterior of the median", xlab = "median") lines(quantile(s, c(0.025, 0.975)), c(0, 0), col = "red", lwd = 5) legend("topright", legend = "95% CI", col = "red", lwd = 5) } invisible(s) }
jeffreys_medianvs
wilcox.test
Finally, I thought I would compare Jeffreys’ substitution posterior with one of the most common classical non-parametric tests used, Wilcoxon signed-rank test. From the documentation for
wilcox.test:
## Hollander & Wolfe (1973), 29f. Hamilton depression scale factor ## measurements in 9 patients with mixed anxiety and depression, taken at the ## first (x) and second (y) visit after initiation of a therapy ## (administration of a tranquilizer). x <- c(1.83, 0.5, 1.62, 2.48, 1.68, 1.88, 1.55, 3.06, 1.3) y <- c(0.878, 0.647, 0.598, 2.05, 1.06, 1.29, 1.06, 3.14, 1.29) wilcox.test(y - x, alternative = "less")
## ## Wilcoxon signed rank test ## ## data: y - x ## V = 5, p-value = 0.01953 ## alternative hypothesis: true location is less than 0
So
wilcox.testgives us a pretty small p-value, which could be taken to support that the distribution of
y - xis not symmetric about 0.0, but not much more. Let’s rerun the analysis but using the R function we defined earlier.
s <- jeffreys_median(y - x)
## ## Jeffreys’ Substitution Posterior for the median ## ## median ## -0.49 ## 95% CI ## -0.9133 0.04514
Looking at the posterior we see that the most likely median difference is around -0.5, but the difference is also likely to be as negative as -1 and as positive as 0.1. Using the samples
swe can calculate the probability that the median difference is less than 0:
mean(s < 0)
## [1] 0.9523
So, given the data and Jeffreys’ substitution posterior there is more than 95% probability that the median difference is less than 0. Compared with Wilcoxon signed-rank test I think Jeffreys’ substitution posterior is a good alternative when you are interested in the median but otherwise want to make as few assumptions as possible. However, I don’t want to
generallyrecommend Jeffreys’ substitution posterior, as it doesn’t make strong assumptions almost any other model based on reasonable assumptions will be more powerful and will make better use of the data. Still, I think Jeffreys’ substitution posterior is a cool method with a clean, intuitive definition, and I just wonder: Why isn’t it better known? References
Jeffreys, H. (1961).
Theory of Probability. Clarendon Press: Oxford. Amazon link
Lancaster, T., & Jae Jun, S. (2010). Bayesian quantile regression methods.
Journal of Applied Econometrics, 25(2), 287-307.
Lavine, M. (1995). On an approximate likelihood for quantiles.
Biometrika, 82(1), 220-222. link (unfortunately behind paywall)
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For the wave equation $u_{tt} =c^2u_{xx}$ subject to the Robin boundary conditions $$a_0u(0, t)-b_0u_x (0, t) = 0$$ $$a_1u(1, t)+b_1u_x (1, t) = 0$$ for constants $a_0, b_0, a_1, b_1$ with $b_0\neq0$ and $b_1\neq 0$. Calculate the derivative $E'(t)$ of the energy given in the example.
How should $E(t)$ be modified in order to remain constant in time with these new boundary conditions?
What constraints should be imposed on $a_0, b_0, a_1, b_1$ to ensure the modified energy is a nonnegative function of $u$?
Attempt:
The energy $E$ associated with the wave equation $u_{tt} =c^2u_{xx}$ for $0<x<1,t>0$ is given by $$E(t)=\int_0^1 (u_t)^2+c^2(u_x)^2dx.$$
By differentiating with respect to $t$, we find, $$\frac{dE}{dt}=\int_0^1((u_t)^2+c^2(u_x)^2)dx.$$ Next integrating the second term on the right side by parts gives \begin{align*} \frac{dE}{dt} &=2\int_0^1 u_t(u_{tt}-c^2u_{xx})dx+c^2u_xu_t\Big|_0^1 \\ \end{align*} Given our PDE and boundary conditions, \begin{align*} a_0u(0,t)-b_0u_x(0,t)=0 \implies u_x(0,t)=\frac{a_0}{b_0}u(0,t) \\ a_1u(1,t)+b_1u_x(1,t)=0 \implies u_x(1,t)=-\frac{a_1}{b_1}u(1,t) \end{align*} we get \begin{align*} \frac{dE}{dt} &= 2\int_0^1 u_t(0)dx-c^2\Big[\frac{a_1}{b_1}+\frac{a_0}{b_0}\Big]uu_t\Big|_0^1 \\ &= -c^2\Big[\frac{a_1}{b_1}+\frac{a_0}{b_0}\Big]uu_t\Big|_0^1. \end{align*}
Assuming my $E'(t)$ is correct, the answer to the second question would be, let $c^2=\frac{a_1}{b_1}-\frac{a_0}{b_0}$. This will ensure $E'(t)$ to vanish - making energy constant in time.
My intuition is to show that at the boundaries the max and min are both $\ge0$. Invoking max principle, we can argue that E(t) will be non-negative. But I'm not sure what constraints to impose to the modified Energy.
EDIT: For Question 3, since I let $c^2=\frac{a_1}{b_1}-\frac{a_0}{b_0}$, we just have to let this term be positive (i.e. $\frac{a_1}{b_1}>\frac{a_0}{b_0})$
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For proving the quadratic reciprocity, Gauss sums are very useful. However this seems an ad-hoc construction. Is this useful in a wider context? What are some other uses for Gauss sums?
Gauss sums are not an ad-hoc construction! I know two ways to motivate the definition, one of which requires that you know a little Galois theory and the other which is totally mysterious to me.
Here is the Galois-theoretic explanation. Let $\zeta_p$ be a primitive $p^{th}$ root of unity, for $p$ prime. The cyclotomic field $\mathbb{Q}(\zeta_p)$ is Galois, so one can define its Galois group, the group of all field automorphisms which preserve $\mathbb{Q}$. Such an automorphism is determined by what it does to $\zeta_p$, and it must send $\zeta_p$ to another primitive $p^{th}$ root of unity. It follows that the Galois group $G = \text{Gal}(\mathbb{Q}(\zeta_p)/\mathbb{Q})$ is isomorphic to $(\mathbb{Z}/p\mathbb{Z})^{\times}$, which is cyclic of order $p-1$.
Now suppose $p$ is odd. As a cyclic group of even order, $G$ has a unique subgroup $H$ of index two given precisely by the multiplicative group of quadratic residues $\bmod p$, so by the fundamental theorem of Galois theory the fixed field $\mathbb{Q}(\zeta_p)^H$ is the unique quadratic subextension of $\mathbb{Q}(\zeta_p)$. And it's not hard to see that this unique quadratic subextension must be generated by
$$\sum_{\sigma \in H} \sigma(\zeta_p) = \sum_{a \text{ is a QR}} \zeta_p^a = \frac{1}{2} \left( \sum_{a=1}^{p-1} \zeta_p^{a^2} \right)$$
which you will of course recognize as a Gauss sum! So the Gauss sum generates a quadratic subextension, and any of various methods will tell you that this subextension is precisely $\mathbb{Q}(\sqrt{p^{\ast}})$ where $p^{\ast} = (-1)^{ \frac{p-1}{2} } p$. (This does not actually require any computation: if you know enough algebraic number theory, it follows from a consideration of which primes ramify in cyclotomic extensions.)
The totally mysterious explanation is that Gauss sums naturally appear when you start thinking about the discrete Fourier transform. For example, the trace of the DFT matrix is a Gauss sum. But more mysteriously, Gauss sums are eigenfunctions of the DFT in a certain sense. (I sketch how this works here.) There is a sort of mysterious connection here to the Gaussian distribution, which is an eigenfunction of the continuous Fourier transform; see this MO question. Again, I don't know what to make of this. There is a book by Berg called The Fourier-analytic proof of quadratic reciprocity and it may or may not be about this construction.
Not just
quadratic reciprocity, one can use them to prove higherreciprocity laws: see Ireland and Rosen's A Classical Introductionto Modern Number Theory. They also turn up in the functionalequation for Dirichlet L-functions (and are massively generalized in thetopic of root numbers).
They are also used to describe something called the Talbot Effect:
look at #8 in the list. I attended a seminar by Mike Berry about 12 years ago where he claimed that the Talbot Effect was a physical manifestation of Gauss Sums.
Srinivasa Ramanujan actually had discovered some definite integral formulas related to the Gauss sums. Please see the below article:
Some definite integrals connected with Gauss sums. Messenger of MathematicsXLIV, $1915$, $75-85$
From Wikipedia: ( Sorry, I can't explain this.)
The absolute value of Gauss sumsis usually found as an application of Plancherel's theorem on finite groups.
Another application of the Gauss sum: How to prove that: $\tan(3\pi/11) + 4\sin(2\pi/11) = \sqrt{11}$
Gauss sums and exponential sums in general are particularly useful for determining the size of certain algebraic varieties in finite fields or even in general abelian groups. If one defines
$$ A_t = \{x \in \mathbb{F}_q^d : f(x) = t\} $$
where $t \in \mathbb{F}_q\setminus\{0\}$, then by orthogonality we have
$$ |A_t| = q^{-1} \sum_{s \in \mathbb{F}_q} \sum_{x \in \mathbb{F}_q^d} \chi(s (f(x) - t)), $$
where $\chi$ is any nontrivial additive character on $\mathbb{F}_q$.
For example, if one considers $x = (x_1, \dots , x_d) \in \mathbb{F}_q^d$ and defines $f(x) = x_1^2 + \dots + x_d^2$, then $A_t$ would be some finite field analogue of a sphere. Bounding such a set would then be equivalent to bounding
$$ q^{-1}\sum_{s \in \mathbb{F}_q} \left(\sum_{x \in \mathbb{F}_q} \chi(sx^2) \right)^d \chi(-st). $$
Gauss Sums and in particular well-known bounds for Gauss sums imply that such a sum is of size $q^{d-1}(1 + o_d(1))$ as $q \to \infty$.
As Qiaochu points out above, such bounds are nice to have when one works with the discrete Fourier transform.
A small additional note, in line with an earlier answer: Gauss sums are, literally, the Lagrange resolvents obtained in the course of expressing roots of unity in terms of radicals. (Yes, then the Kummer-Stickelberger business can be used to effectively obtain the actual radical expressions...: here .)
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Article Title Keywords
Unitarily invariant norms, the Frobenius norm, Singular values
Abstract
Let $A$ and $B$ be complex square matrices. Some inequalities between $\mid A \mid + \mid B \mid$ and $\mid A^{*} \mid + \mid B^{*} \mid$ are established. Applications of these inequalities are also given. For example, in the Frobenius norm, $$ \parallel\, A+B \,\parallel_{F} \leq \sqrt[4]{2} \parallel \mid A\mid + \mid B\mid \, \parallel_{F}. $$
Recommended Citation
Zhang, Yun.(2018),"Inequalities between $\mid A\mid + \mid B\mid $ and $\mid A^{*} \mid + \mid B^{*} \mid$",
Electronic Journal of Linear Algebra,Volume 34, pp. 561-565. DOI: https://doi.org/10.13001/1081-3810.3878
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Recently I read an interesting paper: A Contextual-Bandit Approach to Personalized News Article Recommendation. I noticed its dataset is available, so I thought to play with it. Here, I share a background theory and basic intermediate experimetnal results.
Suppose you are running a local news company that earns most of its advertisement impressions from your website. You want a system that makes a personalized recommendation for an article (item) to each site visitor (user). How do we design such a system? Well, Contextual Bandit (CB) algorithms could be a good option.
Exploration-Exploitation Dillemma
To motivate CB, let us add details to the story. Your workforce is quite big, so you have a constant influx of new items. Your site is also popular, so new users come and go often. Now, learning a recommender system in this setting is tricky. First, we would know very little about the new users and quite nothing about how they relate to existing users (similarity). The same for new items.
One way to resolve it is to passively wait and collect information about new users until we feel confident about their preferences. Then again, the call for a decision is upon us and we cannot serve them a blank page. Another way would be to make bold moves by carefully recommending articles to a new user such that each article effectivelydecreases the uncertainty about the user preference, all the while thinkingabout the best recommmendation for the user. Overall, this is called anexploration-exploitation dillemma. The user/item-changing nature makes it hardto apply popular recommendation algorithms like collaborative filtering orcontent-based filtering. Especially so, when there are not many data points (thecold-start problem). In contrast, Contextual Bandit
by design balances betweenexploration and exploitation. Multi-armed Bandit
Suppose you are given $k$ actions . At each timestep , you take one action and receive a non-negative scalar reward $r_t \sim P_\theta(\cdot \vert a_t)$ where $\theta$ is the unknown parameters of a stationary distribution over rewards. Define the value of taking an action to be . Define the value of the optimal action . Define the regret of taking . From this, the expected regret after $T$ steps can be defined: where $n_j$ is the number of times $a_j$ was chosen over $T$ steps. Define the action value estimator . We hope to achieve $\mu_j \approx Q(a_j)$. In order to compute an optimal action, we would need to $\arg\max_j Q(a_j)$ (exploitation).
Then, again how do we explore? The general principle of CB for exploration is optimism under uncertainty. That is, if we are not sure about our $Q$ estimate for an action, we intentionally overestimate its action value such that we end up trying it out.
$\epsilon_n$-greedy
We just average all the rewards collected from taking $a_j$ over $T$ steps. . This is a sample mean estimator ($n_j$ estimates from one sample) and as such, it is unbiased. Since the main goal is to discover an optimal action (with the maximum expected reward), we need to try all actions at least once. To explore, we take an action chosen uniformly at random with a small probability $\epsilon$ and otherwise take a greedy decision . As we explore, we will have reduced enough uncertainty to conclude the optimal action. At that point, exploration would not be needed. Hence, we decay $\epsilon_1, \epsilon_2, \dots$ by a small factor (hyperparameter).
Upper Confidence Bound (UCB)
Suppose for each timestep $t$, we want to construct a confidence interval that keeps with a high probability that we can control. Suppose we decide how many steps wewill play the game and let it be $n$. Furthermore, we assume in our formulation rewards are i.i.d. We assume the rewards are bounded. Then, usingHoeffding’s inequality, we canupper-bound the probability that our estimate deviates from theestimand by more than any constant $a > 0$ and show the bound can bemade arbitrarily small with a large $n$. This leads to $c_j(t) =\sqrt{\frac{\log n}{ n_j(t) }}$ and we solve: . Of course, it is awkward to assume we know $n$ a priori. Auer et al
1 proved amore natural bound $c_j(t) = \sqrt{\frac{2 \log t}{n_j(t)}}$. Thompson Sampling
Thompson Sampling is a Bayesian kid for Multi-Armed Bandit. It follows the typical routine of posterior inference: a) set up a hypothesis (likelihood model) that is assumed to generate observations, b)define a prior over the model parameters, c) using Bayes rule, compute the posterior or the posterior predictive. In our case, we would model $P_\theta (r_t \vert a_t) \approx P_w(r_t \vert a_t)$ and define a prior over $\theta$. For certain combinations of likelihood model and prior that we can write the posterior down in a closed form (conjugate models), the exact inference is tractable. That said, in general, we know an exact Bayesian inference is often intractable because of the evidence in the denominator. I refer the reader to a nice tutorial on Thompson Sampling
2. Contextual Bandit
Up until now, our formulation was context-free; it only conditioned on the action (index). Now, suppose for each timestep, we observe an additional random variable about either user or item (=action). Let us define the context at the timestep $t$, that depends both on the user context $u_t$ and action $a_t$. The aim is almost the same as before. We are to learn an estimator .
LinUCB Policy
The algorithm implements a ridge linear regression with UCB. Define the linear estimator . Let $D_a \in \mathbb{R}^{n_j \times d}$ denote the design matrix (the training data) and $d$ the dimension of $x_{t,a}$. Let $\mathbf{r}_j \in \mathbb{R}^{n_j}$ the observed rewards corresponding to $D_a$. Assuming we solve a least-squares problem with a ridge regularization ($\lambda=1$), we have a closed-form solution . When elements $r_j$ are conditionally independent given correponding rows in $D_a$, it holds that
.
where $A_a = (D_a^\top D_a + I_d)^{-1}$ and $\alpha = 1 + \sqrt{\log(2/\delta)/2}$. Thanks to $I_d$, is likely invertible. Inverting a matrix is quite $O(d^3)$ so in practice, we want to solve the linear system periodically as opposed to every step. Finally, we choose .
The original paper suggests two versions: a disjoint model that learns separate $w_a$ for each action and a hybrid model that has shared parameters across actions. I did not quite like the disjoint model; especially because one must maintain the set of valid actions that can change over time (e.g. new articles coming in and old articles perish). For the experiment (yahoo news), as a simple baseline, I wrote a model that shares all the parameters for all actions. Such a simple linear model may underperform in the high data regime but I expected an okay-level performance in the low data regime.
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 Linear Gaussian Thompson Sampling Policy (lgtsp)
This is a Thompson Sampling policy that implements Bayesian linear regression with a conjugate prior. We assume the underlying model (likelihood) satisfies where . We define a prior jointly on $w, \sigma^2$ such that where . Theposterior update is well written in wiki. By assumging the initial hyperparam $\mu_0=0$, we can zero out some terms which I did for the experiments. Although the exact posterior update is tractable in this formulation, evaluating a covariance matrix can be too expensive. I noticed it is indeed the bottleneck in my implementation and
multivariate_normal ran into a degenerate covariance matrix and failed SVD on it. A better approach would be find a diagonal covariance matrix approximation, which I have not tried out yet. For computational reasons, I collect every data point but update the posterior periodically. I built, again, a shared parameter model across all actions.
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 Neural $\epsilon_n$-Greedy Policy (nueralp)
We fit a simple fully-connected neural network wth stochastic-gradient-based optimizers like RMSProp to the data coming online. It is the same as $\epsilon_n$-Greedy Policy except we use a neural network to represent $Q(x_{t,a})$. Neural networks may be one good way to overcome the limited representation power of the linear models introduced so far. Instead of leaving the exploration to the hyperparameter $\epsilon$, one may keep the Bayesian linear regression formulation and fit a nueral network $g$ such that .
Other baselines random policy (rp): uniformly at random -greedy (egp): sample mean policy + annealing exploration. sample mean policy (smp): no exploration. optimal policy (opt_p): assumes an oracle that knows the optimal action. Experiments
Well, that was a review of basic Contextual Bandit algorithms. Now we move on the experiemnts.
Partially-observable Reward Dataset
Here’s a qustion. How do we train the algorithms on a dataset someone else collected for us? So far, our formulation assumed we are learning a policy in an on-policy setting. It means the policy we collect data from (behavior poilcy) is the same as the policy we optimizer (the target policy). In most practical situations, one would have a dataset sampled by a policy and train another policy using the dataset. This realistic setting makes it hard for us to evaluate the true performance of our policy. If a behavior policy always chooses one action among others, we would not have any samples for counterfactuals (=what happens if we had taken other action). This problem is called Off-policy Policy Evaluation and an important research topic.
In an on-policy setting, we can observe the rewards for all actions at will. All it takes is to try the actions we want. In an off-policy setting, however, we cannot directly observe rewards unless the behavior policy that collects data deliver them to us. Then, how do we use such
partially observable reward data for training? One naive way to reject any samples where our policy’s action does not match the ation empirically taken by the behavior policy. If we assume a uniform random behavior policy, indeed the performance evaluation would be unbiased. Or we could use an Importance Sampling to obtain an unbiased estimate for the performance. These methods are okay if there are a lot of samples. For the experiment we use the naive rejection method. Synthetic (fully observable reward, item context)
We consider a simple hypothetical contextual bandit problem. The true distribution that samples rewards is Gaussian with some predefined variance. Assuming no interactions between actions, we keep the model as an isotropic multivariate Gaussian. As the true model is Gaussian, we expect the thompson sampling poilcy (lgtsp) to perform well.
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As expected, lgtsp outperformed baselines.
Other baselines especially linucbp did not perform so well because they locked on to one action that it thinks the best and does not get out of it. You can observe that in the action distribution plot below.
Mushroom (fully observable rewards, item context)
Mushroom consists of 8,124 hypothetical mushroom datapoints that show features (item context) and whether each is poisonous/edible. I modeled the reward distribution similar to the paper where +10.0 for eating a good mushroom, -35.0 for eating a bad mushroom with a 30% chance, still +10.0 with 70% (bad mushroom but lucky), and 0.0 for not eating. The optimal policy would eat only good mushrooms and not take the risk with bad mushrooms. Good and bad mushrooms are in an almost equal proportion. Of course, whether a mushroom is good or bad is hidden (except for opt_p).
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We observe again lgtsp outperfoming baselines. It is quite surprising linucbp performed so poorly. Perhaps there is a bug in the code (recall, this is an intermediate report).
Yahoo Click Log Data (partially observable reward, user+item context)
Yahoo! Front Page Today Module User Click Log Dataset that features a log data for ~45 Million user visit events. For each user visit event, we have features available for the article shortlist (the candidate article pool) and user features. Crucially, the shortlist elements change over time so the algorithm must learn to adapt to a new action set. This is a partially observable reward problem because we do not have the data for the counterfactuals and instead know what article in the shortlist was displayed to the user and whether the user clicked it or not. If we use an on-policy algorithm, the data points we can use is only when our policy’s recommendation matches the chosen article in the data.
Since the behavior policy was a uniform random policy for roughly 20 actions at each time step, the effective sample rate (the rate at which samples could be used for training) was roughly at 5%. This means, in order to compute training over n=10000, we have to evaluate approximately 200,000 points and reject the rest.
Notice now the y-axis is cumulative
reward. Since this is a partially observable reward problem we cannot compute the regret.
Due to computational costs (both theoretically and practically—I run my experiments on EC2 Spot Instances), I ran neuralp for $n=100000$ for now. I observed a high variance in the performance of neuralp (the shaded region). It seemed there is a bad local optimum it tends to get stuck at.
There wasn’t a significant difference in the learning performance between Adam and RMSProp.
Source Code
Check out the source code for more details.
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For giving sequence $ A_1,A_2,A_3 ……A_n$ , find number of triples $ i,j,k$ such that $ 1<=i<j<=k<=N$ and $ A_i + A_{i+1} + … A_{j-1} = A_{j} + A_{j+1} ….. A_{k}$ .Where $ +$ is bitwise xor operation .
I tried to solve it using dynamic programming somewhat similar to https://www.geeksforgeeks.org/count-number-of-subsets-having-a-particular-xor-value/ , but it’s time complexity if $ O(n*m)$ , where m is maximum element in the array .Can we do better than $ O(n*n)$ or $ O(n*m)$ ?
Let $ A_n=\sum_{i=1}^n \frac{n}{n^2+i^2}$ , then $ \lim_{n\to\infty}A_n=\frac{\pi}{4}$ . My friend asks me about the limit $ \lim_{n\to\infty}n(\frac{\pi}{4} -A_n)$ . How to get it?
Give an example of a convergent series $ \sum {a_n}$ such that the series $ \sum {a_{3n}}$ is divergent.
Give an example of a divergent series $ \sum {b_n}$ such that the series $ \sum {b_{3n}}$ is convergent.
Attempt:
I am not sure if this is a valid forumla for a sequence : $ a_{3n-2} = \frac{1}{1+4(n-1)} ,a_{3n-1} = \frac{1}{3+4(n-1)}, a_{3n} = -\frac{1}{2n}$ . This series converges to $ \frac{3}{2} \log(2)$ . But, $ \sum {a_{3n}}$ diverges.
We define $ b_{3n-2}=1 , b_{3n-1}=1, b_{3n} = \frac{1}{n^2}$ . The series diverges, but $ \sum{b_{3n}}$ converges to $ \frac{\pi^2}{6} $
The problem is, I am not sure if the this type of “formula” works [unlike the sequence defined by $ 1/n$ or something. Is this valid to define the sequence “term-by-term” (here, three different types of indices)?].
There was an excercise About condensed sums. We should prove that:
If $ (a_n)$ is
monotonous then $ \sum_{n}a_n$ converges/diverges iff $ \sum_{k}2^ka_{2^k}$ converges/diverges
It is sufficient to look at a decreasing Zero convergent sequence $ (a_n)$
$ s_n=\sum_{k=1}^{n}a_k,v_n=\sum_{k=1}^n2^ka_{2^k}$
If $ K,N\in\mathbb{N}$ with $ 2^{K-1}<N\leq 2^K$ then
I am sorry for the duplicate please answer one specific Question About the Problem I have in the comment and I will delete the answer afterwards
$ S_N\le S_{2^K}\leq a_1+v_{K-1}$ and
$ S_N\ge S_{2^{K-1}}\geq a_1/2 +v_{k-1}/2$
I have understood that the partialsums are Always positive since every $ a_n$ must be positive. And if the partialsums can be bounded then because the partialsums are monotonously rising the sequence of partialsums is converging to its Supremum. And $ s_k$ is bounded if and only if $ v_k$ is bounded.
Again because $ a_n$ is decreasing and zeroconvergent the partialsums are always positive. If one cannot be boundet then also the other one cannot be bounded.
My Question is that in the excercise the Task was to Show that the Statement holds if $ (a_n)$ is monotonous. But we have only looked at decreasing monotonous $ (a_n)$ and what is more important we have assumed that $ (a_n)$ is Zero convergent.
What if $ (a_n)$ is not Zero convergent then $ \sum_n a_n$ necessarily diverges.
But to prove the Statement we also have to Show that $ \sum_n 2^na_{2^n}$ diverges. It would be sufficient in this case to Show that $ (2^na_{2^n})$ is not Zero convergent.
How can I do that?
If $ \{a_n^2\}$ is bounded then I know for a non-negative number $ M$ , $ \mid a_n^2 \mid \leq M, \forall n$ .
How do I show that this leads to $ \{a_n\}$ being bounded?
$ \mid a_n^2 \mid \leq M$
$ -M \leq a_n^2 \leq M$
but now I don’t think I can replace $ M$ with $ \sqrt{M}$ , can I?
$ -\sqrt{M} \leq a_n \leq \sqrt{M}$ is probably allowed, but how do I justify this?
If $ a_n=\displaystyle\prod_{k=1}^n \Bigg(1+\dfrac{k^2}{n^2}\Bigg)^k$ .
Then find the value of $ \displaystyle\lim_{n\rightarrow \infty} a_n^{\dfrac{-1}{n^2}}$ .
Let $ a_n $ complex sequence prove that if $ a_n\to \infty$ then $ |a_n|\to\infty$ . Note that $ a_n = x_n + y_ni$ i dont know how to write that mathmatically.
trial :
Can i say that for every $ M>0$ there exist $ N$ such that for every $ n>N$ ,
$ ~~|x_n|>M~~ OR ~~~|y_n|>M$ ( At least one of them goes to $ \infty$ )
because of that $ |an| = \sqrt{(x_n)^2+(y_n)^2} > M$ and so $ |a_n|\to\infty$ .
For $ n\in\mathbb{Z}_{\ge 1}$ , let $ A_n$ be the $ n\times n$ matrix given by $ (A_n)_{i,j}={n\choose |i-j|}$ . From this post it is clear that $ $ \det(A_n)=\prod_{k=0}^{n-1}\left[\left(\exp\left(\frac{2\pi k i}{n}\right)+1\right)^n-1\right]=\prod_{k=0}^{n-1}\left(2^n(-1)^k\cos^n\left(\frac{\pi k}n\right)-1\right).$ $ Also, $ \det(A_n)$ is obviously integer for all $ n$ .
Question: What is known about the (prime) divisors of $ \det(A_n)$ ?
If this is too broad, I am particularly interested in pairs $ (p,d)$ where $ p$ is prime with $ d\mid p-1$ and $ p\nmid\det(A_{(p-1)/d})$ .
Let $ A_n$ represent the number of integers that can be written as the product of two element of $ [[1,n]]$ .
I am looking for an asymptotic estimation of $ A_n$ .
First, I think it’s a good start to look at the exponent $ \alpha$ such that :
$ $ A_n = o(n^\alpha)$ $
I think we have : $ 2 < alpha $ . To prove this lower bound we use the fact that the number of primer numbers $ \leq n$ is about $ \frac{n}{\log n}$ . Hence we have the trivial lower bound (assuming $ n$ is big enough) :
$ $ \frac{n}{\log n} \cdot \binom{ E(\frac{n}{\log n})}{2} = o(n^3)$ $
Now is it possible to get a good asymptotic for $ A_n$ and not just this lower bound ? Is what I’ve done so far correct ?
Thank you !
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Production of charged pions, kaons and protons at large transverse momenta in pp and Pb-Pb collisions at $\sqrt{s_{NN}}$ = 2.76 TeV
(Elsevier, 2014-09)
Transverse momentum spectra of $\pi^{\pm}, K^{\pm}$ and $p(\bar{p})$ up to $p_T$ = 20 GeV/c at mid-rapidity, |y| $\le$ 0.8, in pp and Pb-Pb collisions at $\sqrt{s_{NN}}$ = 2.76 TeV have been measured using the ALICE detector ...
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Forward-backward multiplicity correlations in pp collisions at √s = 0.9, 2.76 and 7 TeV
(Springer, 2015-05-20)
The strength of forward-backward (FB) multiplicity correlations is measured by the ALICE detector in proton-proton (pp) collisions at s√ = 0.9, 2.76 and 7 TeV. The measurement is performed in the central pseudorapidity ...
Rapidity and transverse-momentum dependence of the inclusive J/$\mathbf{\psi}$ nuclear modification factor in p-Pb collisions at $\mathbf{\sqrt{\textit{s}_{NN}}}=5.02$ TeV
(Springer, 2015-06)
We have studied the transverse-momentum ($p_{\rm T}$) dependence of the inclusive J/$\psi$ production in p-Pb collisions at $\sqrt{s_{\rm NN}} = 5.02$ TeV, in three center-of-mass rapidity ($y_{\rm cms}$) regions, down to ...
Measurement of charm and beauty production at central rapidity versus charged-particle multiplicity in proton-proton collisions at $\sqrt{s}$ = 7 TeV
(Springer, 2015-09)
Prompt D meson and non-prompt J/$\psi$ yields are studied as a function of the multiplicity of charged particles produced in inelastic proton-proton collisions at a centre-of-mass energy of $\sqrt{s}=7$ TeV. The results ...
Coherent $\rho^0$ photoproduction in ultra-peripheral Pb-Pb collisions at $\mathbf{\sqrt{\textit{s}_{\rm NN}}} = 2.76$ TeV
(Springer, 2015-09)
We report the first measurement at the LHC of coherent photoproduction of $\rho^0$ mesons in ultra-peripheral Pb-Pb collisions. The invariant mass and transverse momentum distributions for $\rho^0$ production are studied ...
Inclusive, prompt and non-prompt J/ψ production at mid-rapidity in Pb-Pb collisions at √sNN = 2.76 TeV
(Springer, 2015-07-10)
The transverse momentum (p T) dependence of the nuclear modification factor R AA and the centrality dependence of the average transverse momentum 〈p T〉 for inclusive J/ψ have been measured with ALICE for Pb-Pb collisions ...
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I come from the world of precise satellite orbit determination, so my references also come from there. I realize this means most of the equations in my references relate to orbits and not general measurements, however, I do think you'll find them interesting.
I think a good reference is Fundamentals of Astrodynamics and Applications, by D.A. Vallado. The relevant section is Section 3.7 and onward.
Another good reference is Satellite Orbits, by Montebruck and Gill. Relevant sections are in Chapter 3 (could not find a link).
To summarize: quasi-inertial reference frames are frames such that Newton's laws and special relativity can be applied without any significant impact on accuracy. Usually, for applications where this accuracy is not sufficient, accuracy is most easily improved by adding corrective terms to Newton's equations of motion. Corrective terms account for planetary precession & nutation, GR effects, Coriolis effect, tidal effects, etc.
To give an impression of the magnitudes of such corrections; the second paragraph on the wiki is helpful:
ECI coordinate frames are not truly inertial since the Earth itself is
accelerating as it travels in its orbit about the Sun. In many cases,
it may be assumed that the ECI frame is inertial without adverse
effect. However, when computing the gravitational influence of a third
body such as the Moon on the dynamics of a spacecraft, the
acceleration of the ECI frame must be considered. For example, when
computing the acceleration of an Earth-orbiting spacecraft due to the
gravitational influence of the Moon, the acceleration of the Earth
itself due to the Moon's gravity must be subtracted
The main non-inertial effect to take into account is the acceleration of the ECI frame towards the Moon. If you assume the Earth and Moon are in circular orbit about their barycenter, this acceleration can be estimated;
by Newtonian gravity:
$$a_{\mathrm{Earth}} = \frac{\mu_{\mathrm{Moon}}}{r_{\mathrm{E-M}}^2} \approx \frac{4902.8\ \mathrm{km}^3\,\mathrm{s}^{-2}}{(384399\ \mathrm{km})^2} = 33.180\, \mu\mathrm{m\,s}^{-2}$$
from circular motion:
$$a = \frac{v_{\mathrm{Earth}}^2}{r_{\mathrm{E-M}}} \approx \frac{(1.022\ \mathrm{km\ s}^{-1})^2}{384399\ \mathrm{km}} = 33.180\, \mu\mathrm{m\,s}^{-2}$$
The same can be said for the Sun. Again, assuming circular orbit:
$$a_{\mathrm{Earth}} = \frac{\mu_{\mathrm{Sun}}}{r_{\mathrm{S-E}}^2} \approx \frac{132712440018\ \mathrm{km}^3\,\mathrm{s}^{-2}}{(149597871\ \mathrm{km})^2} = 5.9301\, \mathrm{mm\,s}^{-2}$$
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Let $K$ be a quadratic number field. Let $\mathcal{O}_K$ be the ring of algebraic integers in $K$. Let $R$ be an order of $K$, $D$ its discriminant. I am interested in the ideal theory on $R$ because it has a deep connection with the theory of binary quadratic forms as shown in this. By this question, $1, \omega = \frac{(D + \sqrt D)}{2}$ is a basis of $R$ as a $\mathbb{Z}$-module.
Let $x_1,\cdots, x_n$ be a sequence of elements of $R$. We denote by $[x_1,\cdots,x_n]$ the $\mathbb{Z}$-submodule of $R$ generated by $x_1,\cdots, x_n$.
Let $I$ be a non-zero ideal of $R$. I am interested in a decomposition of $I$ into a product of ideals. By this question, there exist unique rational integers $a, b, c$ such that $I = [a, b + c\omega], a \gt 0, c \gt 0, 0 \le b \lt a, a \equiv 0$ (mod $c$), $b \equiv 0$ (mod $c$). If $c = 1$, we say $I$ is a primitive ideal. Let $a = ca'$, $b = cb'$. Then $I = cJ$, where $J = [a', b' + \omega]$. Clearly $J$ is a primitive ideal. So the decomposition problem can be reduced to the case when $I$ is primitive.
Let $\frak{f}$ $= \{x \in R | x\mathcal{O}_K \subset R\}$. Let $I$ be an ideal of $R$. If $I + \mathfrak{f} = R$, we call $I$ regular.
My QuestionIs the following proposition correct?If yes, how do you prove it? PropositionLet $I = [a, r + \omega]$ be a primitive regular ideal of $R$, where $a \gt 0$ and $r$ are rational integers.Suppose $a = gh, g \gt 0, h \gt 0$Then $J_1 = [g, r + \omega], J_2 = [h, r + \omega]$ are primitive regular ideals and$I = J_1J_2$. RemarkI am not 100% sure of the correctness of the proposition, though I think it is likely to be true(see my method below).I would like to know if the proposition is correct. I also would like to know other proofs based on different ideas if the proposition is correct.I welcome you to provide as many different proofs of the result as possible.Please provide full proofs which can be understood by people who have basic knowledge of introductory algebraic number theory.
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We often use Ricker wavelets to model seismic, for example when making a synthetic seismogram with which to help tie a well. One simple way to guesstimate the peak or central frequency of the wavelet that will model a particlar seismic section is to count the peaks per unit time in the seismic. But this tends to overestimate the actual frequency because the maximum frequency of a Ricker wavelet is more than the peak frequency. The question is, how much more?
To investigate, let's make a Ricker wavelet and see what it looks like in the time and frequency domains.
>>> T, dt, f = 0.256, 0.001, 25>>> import bruges>>> w, t = bruges.filters.ricker(T, dt, f, return_t=True)>>> import scipy.signal>>> f_W, W = scipy.signal.welch(w, fs=1/dt, nperseg=256)
When we count the peaks in a section, the assumption is that this apparent frequency — that is, the reciprocal of apparent period or distance between the extrema — tells us the dominant or peak frequency.
To help see why this assumption is wrong, let's compare the Ricker with a signal whose apparent frequency does match its peak frequency: a pure cosine:
>>> c = np.cos(2 * 25 * np.pi * t)>>> f_C, C = scipy.signal.welch(c, fs=1/dt, nperseg=256)
Notice that the signal is much narrower in bandwidth. If we allowed more oscillations, it would be even narrower. If it lasted forever, it would be a spike in the frequency domain.
Let's overlay the signals to get a picture of the difference in the relative periods:
The practical consequence of this is that if we estimate the peak frequency to be \(f\ \mathrm{Hz}\), then we need to reduce \(f\) by some factor if we want to design a wavelet to match the data. To get this factor, we need to know the apparent period of the Ricker function, as given by the time difference between the two minima.
Let's look at a couple of different ways to find those minima: numerically and analytically.
Find minima numerically
We'll use
scipy.optimize.minimize to find a numerical solution. In order to use it, we'll need a slightly different expression for the Ricker function — casting it in terms of a time basis
t. We'll also keep
f as a variable, rather than hard-coding it in the expression, to give us the flexibility of computing the minima for different values of
f.
Here's the equation we're implementing:
$$ w(t, f) = (1 - 2\pi^2 f^2 t^2)\ e^{-\pi^2 f^2 t^2} $$
In Python:
>>> def ricker(t, f):>>> return (1 - 2*(np.pi*f*t)**2) * np.exp(-(np.pi*f*t)**2)
Check that the wavelet looks like it did before, by comparing the output of this function when
f is 25 with the wavelet
w we were using before:
>>> f = 25>>> np.allclose(w, ricker(t, f=25))True
Now we call SciPy's
minimize function on our
ricker function. It itertively searches for a minimum solution, then gives us the
x (which is really
t in our case) at that minimum:
>>> import scipy.optimize>>> f = 25>>> scipy.optimize.minimize(ricker, x0=0, args=(f))fun: -0.4462603202963996 hess_inv: array([[1]]) jac: array([-2.19792128e-07]) message: 'Optimization terminated successfully.' nfev: 30 nit: 1 njev: 10 status: 0 success: True x: array([0.01559393])
So the minimum amplitude, given by
fun, is -0.44626 and it occurs at an
x (time) of \(\pm 0.01559\ \mathrm{s}\).
In comparison, the minima of the cosine function occur at a time of \(\pm 0.02\ \mathrm{s}\). In other words, the period appears to be \(0.02 - 0.01559 = 0.00441\ \mathrm{s}\) shorter than the pure waveform, which is...
>>> (0.02 - 0.01559) / 0.020.22050000000000003
...about 22% shorter. This means that if we naively estimate frequency by counting peaks or zero crossings, we'll tend to overestimate the peak frequency of the wavelet by about 22% — assuming it is approximately Ricker-like; if it isn't we can use the same method to estimate the error for other functions.
This is good to know, but it would be interesting to know if this parameter depends on frequency, and also to have a more precise way to describe it than a decimal. To get at these questions, we need an analytic solution.
Find minima analytically
Python's SymPy package is a bit like Maple — it understands math symbolically. We'll use
sympy.solve to find an analytic solution. It turns out that it needs the Ricker function writing in yet another way, using SymPy symbols and expressions for \(\mathrm{e}\) and \(\pi\).
import sympy as spt, f = sp.Symbol('t'), sp.Symbol('f')r = (1 - 2*(sp.pi*f*t)**2) * sp.exp(-(sp.pi*f*t)**2)
Now we can easily find the solutions to the Ricker equation, that is, the times at which the function is equal to zero:
>>> sp.solvers.solve(r, t)[-sqrt(2)/(2*pi*f), sqrt(2)/(2*pi*f)]
But this is not quite what we want. We need the minima, not the zero-crossings.
Maybe there's a better way to do this, but here's one way. Note that the gradient (slope or derivative) of the Ricker function is zero at the minima, so let's just solve the first time derivative of the Ricker function. That will give us the three times at which the function has a gradient of zero.
>>> dwdt = sp.diff(r, t)>>> sp.solvers.solve(dwdt, t)[0, -sqrt(6)/(2*pi*f), sqrt(6)/(2*pi*f)]
In other words, the non-zero minima of the Ricker function are at:
$$ \pm \frac{\sqrt{6}}{2\pi f} $$
Let's just check that this evaluates to the same answer we got from
scipy.optimize, which was 0.01559.
>>> np.sqrt(6) / (2 * np.pi * 25)0.015593936024673521
The solutions agree.
While we're looking at this, we can also compute the analytic solution to the amplitude of the minima, which SciPy calculated as -0.446. We just plug one of the expressions for the minimum time into the expression for
r:
>>> r.subs({t: sp.sqrt(6)/(2*sp.pi*f)})-2*exp(-3/2)
Apparent frequency
So what's the result of all this? What's the correction we need to make?
The minima of the Ricker wavelet are \(\sqrt{6}\ /\ \pi f_\mathrm{actual}\ \mathrm{s}\) apart — this is the apparent period. If we're assuming a pure tone, this period corresponds to an apparent frequency of \(\pi f_\mathrm{actual}\ /\ \sqrt{6}\ \mathrm{Hz}\). For \(f = 25\ \mathrm{Hz}\), this apparent frequency is:
>>> (np.pi * 25) / np.sqrt(6)32.06374575404661
If we were to try to model the data with a Ricker of 32 Hz, the frequency will be too high. We need to multiply the frequency by a factor of \(\sqrt{6} / \pi\), like so:
>>> 32.064 * np.sqrt(6) / (np.pi)25.00019823475659
This gives the correct frequency of 25 Hz.
To sum up, rearranging the expression above:
$$ f_\mathrm{actual} = f_\mathrm{apparent} \frac{\sqrt{6}}{\pi} $$
Expressed as a decimal, the factor we were seeking is therefore \(\sqrt{6}\ /\ \pi\):
>>> np.sqrt(6) / np.pi0.779696801233676
That is, the reduction factor is 22%.
Curious coincidence: in the recent Pi Day post, I mentioned the Riemann zeta function of 2 as a way to compute \(\pi\). It evaluates to \((\pi / \sqrt{6})^2\). Is there a million-dollar connection between the humble Ricker wavelet and the Riemann hypothesis?
I doubt it.
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The number of of the cells touched by the line is (following Byron's answer) $N (|X_1-X_2| + |Y_1-Y_2| +1) $ . Calling $X =|X_1 - X_2|$ and $Y = |Y_1 - X_2|$ and leaving apart the $N$ factor, this is approximately equivalent, for large $N$, to $d_M=X+Y$, the Manhattan distance.
Hence, I'll attack the following problem: We throw two points at random on the unit square; we want to compute $$E(d_M | d)$$
where $d=\sqrt{X^2+Y^2}$ is the euclidean distance.
The expected number of touched cells, in this approximation, would be $N E(d_M | d) + 1$.
It's not difficult to see that $X$ (absolute value of the difference of two uniform v.a) has a triangular density : $f_X(x)=2(1-x)$ The same for $Y$, and both are independent, hence: $$f_{XY}(x y)=4(1-x)(1-y)$$ on the unit square.
UPDATE: Below I found a much simpler solution
--------------- begin ignore -----
Conditioning on a fixed value of $d$ implies that we must restrict to an arc, the intersection of $x^2+y^2=d^2$ and the unit square.
Let's assume first that $d<1$
Note that $d\le d_M \le \sqrt{2} d$
To compute $P(d_M \le \xi | d)$, we must integrate over two pieces of arcs begining on the axis (because of symmetry, we compute just one and multiply by two). (blue arcs in the figure) The first limit point is given by $$x_1=\frac{\xi - \sqrt{2 d^2-\xi^2}}{2}$$Further to change the integration over the arc for an integral over $x$, we note that $ds=\sqrt{1+\frac{x^2}{y^2}}{dx} = \frac{d}{y}dx$. So, leaving out the constants (that disappear on the division), we get that the cumulative distribution function (conditioned on $d$) is given by
$$F_{d_M|d}(\xi;d)=P(d_M \le \xi | d) = \frac{2 \int_{0}^{x_1}\left( 1-x\right) \,\left( \frac{1}{\sqrt{{d}^{2}-{x}^{2}}}-1\right) dx}{\int_{0}^{d}\left( 1-x\right) \,\left( \frac{1}{\sqrt{{d}^{2}-{x}^{2}}}-1\right) dx}$$
Finally, we must integrate to get the expected value
$$E(d_M |d) = d+ \int_{d}^{\sqrt{2} d} \left[1- F_{d_M|d}(\xi;d) \right] \; d\xi $$
This is all.. but it seems pretty complicated to get in close form. Let's see Maxima:
assume(x1>0);assume(d>0);assume(d>x1);
integrate((1-x)*(1/sqrt(d^2-x^2)-1), x,0,x1);
$$\frac{2\,\mathrm{asin}\left( \frac{x1}{d}\right) +2\,\sqrt{{d}^{2}-{x1}^{2}}+{x1}^{2}-2\,x1-2\,d}{2}$$
ix2(x1,d):=(2*asin(x1/d)+2*sqrt(d^2-x1^2)+x1^2-2*x1-2*d)/2;
x1(dm,d):=(dm-sqrt(2*d^2-dm^2))/2;
assume(dm>d); assume(dm<sqrt(2)*d);
fdist(dm,d):=2*(ix2(xx(dm,d),d))/ix2(d,d);
This gets a little messy. Let's try at least some numerical values:
dd:0.8;
dd+quad_qags(1-fdist(dm,dd), dm, dd, sqrt(2)*dd)
1.01798
Let's simulate to check: Matlab/Octave:
N=300000;
t=rand(N,4);
xy = [abs(t(:,1)-t(:,3)),abs(t(:,2)-t(:,4))];
t=[];
d2=xy(:,1).^2+xy(:,2).^2;
d=sqrt(d2);
dm=xy(:,1)+xy(:,2);
step=0.02;
dround=round(d/step)*step;
%size(dround(dround==0.8))
mean(dm(dround==0.8))
>>>ans = 1.0174
Not bad, I'd say. Some other values:
d Maxima Octave (simul)
0.9 1.15847 1.1569
0.8 1.01798 1.0174
0.7 0.88740 0.8863
0.6 0.75983 0.7579
0.5 0.63328 0.6331
0.4 0.50698 0.5063
0.3 0.38062 0.3808
The case $d>1$ should be solved in a similar way.
--------------- end ignore -----
A much simpler solution:
Let's change to polar coordinates: $x = r \cos(t)$,$y = r \sin(t)$. The joint density is
$$f_{r,t}(r,t) = 4 r (1- r \cos(t))(1- r \sin(t))$$
in the domain $0 \le r \le \sqrt{2}$, $0 \le t \le \pi/2$ for $r\le 1$, $ arccos(1/r) \le t \le arcsin(1/r)$ for $r > 1$
The conditional density is then
$$f_{t|r}(t|r) = \frac{1}{g(r)} (1- r \cos(t))(1- r \sin(t))$$
where $g(r)$ is the normalization constant (indepent of $t$).
Assuming first that $r<1$, then
$$g(r) = \int_0^{\pi/2} (1- r \cos(t))(1- r \sin(t)) dt = \frac{{r}^{2}}{2}-2\,r+\frac{\pi }{2}$$
Further, $d_M= x + y= r (\cos(t)+\sin(t))$, so the conditional expectation is given by
$$E[d_M | r] = r \frac{1}{g(r)} \int_0^{\pi/2} (1- r \cos(t))(1- r \sin(t)) (\cos(t)+\sin(t)) dt$$
which gives
$$ E[d_M | r] = r \frac{4\,{r}^{2}-3\,\left( \pi +2\right) \,r+12}{3\,{r}^{2}-12\,r+3\,\pi }$$
For $r>1$ it's more complicated. The result is
$$ E[d_M | r] = r \frac{\sqrt{{r}^{2}-1}\,\left( 4\,{r}^{2}+8\right) +\left( 6\,\mathrm{asin}\left( \frac{1}{r}\right) -6\,\mathrm{acos}\left( \frac{1}{r}\right) -6\right) \,{r}^{2}-4}{3\,{r}^{3}-12\,r\,\sqrt{{r}^{2}-1}-\left( 6\,\mathrm{asin}\left( \frac{1}{r}\right) -6\,\mathrm{acos}\left( \frac{1}{r}\right) -6\right) \,r} $$
The figure shows $E[d_M | r] / r$, ie. the factor by which the expected Manhattan distance exceeds the euclidean distance (we already knew that this must be in the $[1,\sqrt{2}]$ range).
(BTW: sorry if the formatting is not optimal, but I got sick of my Chorme crashing on the edition, lots of times, sometimes losing changes - am I the only one?)
Added: Notice that for small distances ($r \to 0$) the expected number of squares results $r N \frac{4}{\pi} +1 $, which agrees with Ronald's answer.
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Logical AND: Use the linear constraints $y_1 \ge x_1 + x_2 - 1$, $y_1 \le x_1$, $y_1 \le x_2$, $0 \le y_1 \le 1$, where $y_1$ is constrained to be an integer. This enforces the desired relationship. (Pretty neat that you can do it with just linear inequalities, huh?)
Logical OR: Use the linear constraints $y_2 \le x_1 + x_2$, $y_2 \ge x_1$, $y_2 \ge x_2$, $0 \le y_2 \le 1$, where $y_2$ is constrained to be an integer.
Logical NOT: Use $y_3 = 1-x_1$.
Logical implication: To express $y_4 = (x_1 \Rightarrow x_2)$ (i.e., $y_4 = \neg x_1 \lor x_2$), we can adapt the construction for logical OR. In particular, use the linear constraints $y_4 \le 1-x_1 + x_2$, $y_4 \ge 1-x_1$, $y_4 \ge x_2$, $0 \le y_4 \le 1$, where $y_4$ is constrained to be an integer.
Forced logical implication: To express that $x_1 \Rightarrow x_2$ must hold, simply use the linear constraint $x_1 \le x_2$ (assuming that $x_1$ and $x_2$ are already constrained to boolean values).
XOR: To express $y_5 = x_1 \oplus x_2$ (the exclusive-or of $x_1$ and $x_2$), use linear inequalities $y_5 \le x_1 + x_2$, $y_5 \ge x_1-x_2$, $y_5 \ge x_2-x_1$, $y_5 \le 2-x_1-x_2$, $0 \le y_5 \le 1$, where $y_5$ is constrained to be an integer.
And, as a bonus, one more technique that often helps when formulating problems that contain a mixture of zero-one (boolean) variables and integer variables:
Cast to boolean (version 1): Suppose you have an integer variable $x$, and you want to define $y$ so that $y=1$ if $x \ne 0$ and $y=0$ if $x=0$. If you additionally know that $0 \le x \le U$, then you can use the linear inequalities $0 \le y \le 1$, $y \le x$, $x \le Uy$; however, this only works if you know an upper and lower bound on $x$. Or, if you know that $|x| \le U$ (that is, $-U \le x \le U$) for some constant $U$, then you can use the method described here. This is only applicable if you know an upper bound on $|x|$.
Cast to boolean (version 2): Let's consider the same goal, but now we don't know an upper bound on $x$. However, assume we do know that $x \ge 0$. Here's how you might be able to express that constraint in a linear system. First, introduce a new integer variable $t$. Add inequalities $0 \le y \le 1$, $y \le x$, $t=x-y$. Then, choose the objective function so that you minimize $t$. This only works if you didn't already have an objective function. If you have $n$ non-negative integer variables $x_1,\dots,x_n$ and you want to cast all of them to booleans, so that $y_i=1$ if $x_i\ge 1$ and $y_i=0$ if $x_i=0$, then you can introduce $n$ variables $t_1,\dots,t_n$ with inequalities $0 \le y_i \le 1$, $y_i \le x_i$, $t_i=x_i-y_i$ and define the objective function to minimize $t_1+\dots + t_n$. Again, this only works nothing else needs to define an objective function (if, apart from the casts to boolean, you were planning to just check the feasibility of the resulting ILP, not try to minimize/maximize some function of the variables).
For some excellent practice problems and worked examples, I recommend Formulating Integer Linear Programs:A Rogues' Gallery.
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