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Please excuse the longish question, it just needs some explanation to get down to the actual problem. Those familiar with the mentioned algorithms probably could jump directly to the first simplex tablau.
To solve least absolute deviation problems (a.k.a $L_1$-optimization), the Barrodale-Roberts-algorithm is a special purpose simplex method that needs far less storage and computational efforts to find a suitable minimum.
My implementation of the algorithm terminates at a simple example before a proper minimum is reached. However, probably let me state the problem in a more elaborated manner first:
Given data $(x_i,y_i)$, $L_1$-optimization tries to find $c\in m$ that minimizes $$ \sum_{i=1}^n |y_i-f(x_i)| \quad\text{with}\quad f(x):=A_x\cdot \phi $$ where $A_x$ is a $n\times m$ matrix that depends in some way on $x$. This problem can be stated as a linear program and therefore among others be solved using simplex-like methods.
Barrodale and Roberts suggested an (apparently widely used) modification of the simplex method that radically simplifies the simplex method using the special structure of $L_1$-problems. Most notably, this is that an optimal solution interpolates at least $\mathop{rank}(A)$ of the given datapoints. Those with Jstor access may find the corresponding article here.
Lei and Anderson in 2002 proposed a small modification that is supposed to increase numerical stability and therefore to overcome known problems with the simplex algorithm.
Basically, this algorithm assumes that you start with a given set of points that have to be interpolated, use the given procedures to build a simplex tableau and then use the rules of Barrodale and Roberts to decide on which basis variables to change and therefore modify the set of datapoints that is approximated.
Barrodale and Roberts give a small example that I tried to reproduce. It tries to approximate the points $\{(1,1), (2,1), (3,2), (4,3), (5,2)\}$ by a function $a_1+a_2x$. The finish their algorithm with the following condensed simplex tableau:
$$ \begin{array}{l|l|ll} \hline \text{Basis}&R& u_1 & u_3\\ \hline b_1 & 1/2 & 3/2 & -1/2 \\ v_2 & 1/2 & 1/2 & 1/2 \\ b_2 & 1/2 & -1/2 & 1/2 \\ u_4 & 1/2 & 1/2 & -3/2 \\ v_5 & 1 & -1 & 2 \\ \hline \text{Marginal cost} &2 & -1 & 0\\ \hline \end{array} $$
Most importantly, the first and third point are interpolated and the overall error is equal to $2$. They conclude that
Since all of the nonbasic vectors have nonpositive marginal cost [...]
the iteration is finished and and optimum is reached.
If I use the algorithm of Lei and Anderson, I can reproduce that simplex tableau for the interpolation set {1,3}, as it is expected. However, if I start the algorithm with the set $\{2, 5\}$ (which clearly is not optimal), I get the following simplex tableau:
\begin{array}{l|l|ll} \hline \text{Basis}&R& u_2 & u_5\\ \hline u_1 & 1/3 & -4/3 & 1/3 \\ b_1 & 1/3 & 5/3 & -2/3 \\ u_3 & 2/3 & -2/3 & -1/3 \\ u_4 & 4/3 & -1/3 & -2/3 \\ b_2 & 1/3 & -1/3 & 1/3 \\ \hline \text{Marginal cost} &7/3 & -10/3 & -5/3\\ \hline \end{array}
This result is puzzling me, though. If I understand the quote above correctly, having no positive marginal cost indicates that the optimum is reached. The function value of about 2.33 certainly is not optimal, though. Exchanging $u_2$ with $u_1$ would yield a result that is on par with the solution of Barrodale and Roberts and therefore optimal.
Additional info: If I start with the initial tableau given by Barrodale and Roberts, I am also able to reproduce the tableau above by ordinary simplex steps, thus I am fairly confident that the actual numerical values are correct and my interpretation of the pivot selection rule is faulty.
Any thoughts on this?
I realize that the question itself is quite complicated and probably requires knowledge of at least the Barrodale and Roberts algorithm to be answered sufficiently. The algorithm as a whole is to long to repeat it here in full detail. However, if you have additional questions on steps I took or on missing bits of information, feel free to ask and I will gladly augment the question. |
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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 ...
Transverse sphericity of primary charged particles in minimum bias proton-proton collisions at $\sqrt{s}$=0.9, 2.76 and 7 TeV
(Springer, 2012-09)
Measurements of the sphericity of primary charged particles in minimum bias proton--proton collisions at $\sqrt{s}$=0.9, 2.76 and 7 TeV with the ALICE detector at the LHC are presented. The observable is linearized to be ...
Pion, Kaon, and Proton Production in Central Pb--Pb Collisions at $\sqrt{s_{NN}}$ = 2.76 TeV
(American Physical Society, 2012-12)
In this Letter we report the first results on $\pi^\pm$, K$^\pm$, p and pbar production at mid-rapidity (|y|<0.5) in central Pb-Pb collisions at $\sqrt{s_{NN}}$ = 2.76 TeV, measured by the ALICE experiment at the LHC. The ...
Measurement of prompt J/psi and beauty hadron production cross sections at mid-rapidity in pp collisions at root s=7 TeV
(Springer-verlag, 2012-11)
The ALICE experiment at the LHC has studied J/ψ production at mid-rapidity in pp collisions at s√=7 TeV through its electron pair decay on a data sample corresponding to an integrated luminosity Lint = 5.6 nb−1. The fraction ...
Suppression of high transverse momentum D mesons in central Pb--Pb collisions at $\sqrt{s_{NN}}=2.76$ TeV
(Springer, 2012-09)
The production of the prompt charm mesons $D^0$, $D^+$, $D^{*+}$, and their antiparticles, was measured with the ALICE detector in Pb-Pb collisions at the LHC, at a centre-of-mass energy $\sqrt{s_{NN}}=2.76$ TeV per ...
J/$\psi$ suppression at forward rapidity in Pb-Pb collisions at $\sqrt{s_{NN}}$ = 2.76 TeV
(American Physical Society, 2012)
The ALICE experiment has measured the inclusive J/ψ production in Pb-Pb collisions at √sNN = 2.76 TeV down to pt = 0 in the rapidity range 2.5 < y < 4. A suppression of the inclusive J/ψ yield in Pb-Pb is observed with ...
Production of muons from heavy flavour decays at forward rapidity in pp and Pb-Pb collisions at $\sqrt {s_{NN}}$ = 2.76 TeV
(American Physical Society, 2012)
The ALICE Collaboration has measured the inclusive production of muons from heavy flavour decays at forward rapidity, 2.5 < y < 4, in pp and Pb-Pb collisions at $\sqrt {s_{NN}}$ = 2.76 TeV. The pt-differential inclusive ...
Particle-yield modification in jet-like azimuthal dihadron correlations in Pb-Pb collisions at $\sqrt{s_{NN}}$ = 2.76 TeV
(American Physical Society, 2012-03)
The yield of charged particles associated with high-pT trigger particles (8 < pT < 15 GeV/c) is measured with the ALICE detector in Pb-Pb collisions at $\sqrt{s_{NN}}$ = 2.76 TeV relative to proton-proton collisions at the ...
Measurement of the Cross Section for Electromagnetic Dissociation with Neutron Emission in Pb-Pb Collisions at √sNN = 2.76 TeV
(American Physical Society, 2012-12)
The first measurement of neutron emission in electromagnetic dissociation of 208Pb nuclei at the LHC is presented. The measurement is performed using the neutron Zero Degree Calorimeters of the ALICE experiment, which ... |
You surely know about the Brahmagupta–Fibonacci identity,
$$(a_1^2 + b_1^2)(a_2^2 + b_2^2) = (a_1a_2 \pm b_1b_2)^2 + (a_1b_2 \mp a_2b_1)^2$$
which tells us that the product of two numbers, each of which is the sum of two squares, is itself a sum of two squares. That is to say, the set of all sums of two squares is closed under multiplication. If $a_k$, $b_k \in ℝ $, this is equivalent to the multiplication property for absolute values of complex numbers.
Analogous identities, are, for example, Euler's four-square identity and Degen's eight-square identity, which give the same result for the set of all sums of four and eight squares. Those can be related to quaternions and octonions, respectively (not really sure if this is relevant to the question, but I'm citing it just in case it can be useful).
I was wondering, do similar identities exist for polynomials of degree $n>2$? That is, can we find polynomials of degree $n>2$ closed under multiplication?* (In case there are none, why?) I'm not at all knowledgeable in this subject and asking this just out of curiosity, so I'd appreciate any recommendation on where to look for more information.
*(I was specially curious about integer polynomials) |
There's a little bit of calculation you need to do here to make sure Cauchy's Residue Theorem is applicable here (you need to make sure that certain integrals are bounded etc) but this is a sketch:
Let
$$f(z) = \frac {\pi} {(1+z^2)\tan(\pi z)}$$
Then $f$ has simple poles $\forall n \in \mathbb{Z}$ and also at $\pm i$.
You can calculate the residues as
$$ \text{Res}(f(z), n\pi) = \frac {1} {1 + n^2} $$
and
$$ \text{Res}(f(z), \pm i) = \frac {-\pi} {2tanh(\pi)}$$
If these are hard to calculate for you, I can give you more detail.
Now, let $\Gamma _N$ be the square contour with vertices $(N + \frac 1 2) (\pm 1 \pm i)$
Then Cauchy's Residue Theorem tells us
$$ \int_{\Gamma N} f(z) dz = 2\pi i \sum \text{Res}(f(z), z)$$
Where the sum is across all the poles inside the contour.
Now, we see that the simple poles inside the contour are all the ones at integers $n$ s.t. $|n| < N$ and the ones at $\pm i$.
So
$$\int_{\Gamma N} f(z) dz = 2\pi i \left [ \frac {-2\pi} {2 \tanh(\pi)} + \sum_{n = -N}^{N} \frac{1} {1 + n^2} \right ]$$
Now we need to show that the integral on the left goes to zero as $N$ tends to infinity. Now, I'm going to leave this as an exercise for you (unlucky) but the basic idea is to find a constant $C_1$ which bounds $\frac {\pi} {tan(\pi z)}$ on the top and bottom of the square and another constant $C_2$ on the sides of the square and take $C$ to be the maximum of these.
Then by the Estimation Theorem we would have that
$$ \left | \int_{\Gamma_N} f(z) dz \right | \leq \text{length}(\Gamma_N) \text{sup}_{z \in \Gamma _N} |f(z)| \leq 4(2N + 1) C \text{sup}_{z \in \Gamma_N} \left\| \frac {1} {1 + z^2}\right \| \leq \frac {4C(2N + 1)} { 1 + N^2} = O(\frac 1 N)$$
So we let $N \to \infty$ then we get
$$0 = 2\pi i \left [ \frac {-2\pi} {2 \tanh(\pi)} + \sum_{n = -\infty}^{\infty} \frac{1} {1 + n^2} \right ]$$So then we have$$0 = \frac {-\pi} { \tanh(\pi)} + 2\sum_{n = 1}^{\infty} \frac{1} {1 + n^2} + 1$$ Where the 1 that has randomly appeared is the $n = 0 $ term
So $$\sum_{n = 1}^{\infty} \frac{1} {1 + n^2} = \frac 1 2 \left [\frac {\pi} { \tanh(\pi)} - 1 \right] = \frac 1 2 (\pi \coth (\pi) - 1).$$
I hope you know some complex analysis otherwise this might have meant nothing to you...
This method works for most sums though (as long as you get the right things tending to zero, which you do in this case) and I don't think I've ever had it not work.
If you want to evaluate
$$ \sum_{n = 1} ^ {\infty} \phi(n)$$
Where \phi can easily be extended to all of $\mathbb {C}$ you just take $$f(z) = \frac {\pi} {\phi(z) \tan(\pi z)}$$And do the same thing, and if you want to evaluate $$\sum_{n = 1} ^ {\infty} (-1)^n \phi(n)$$You just take $$f(z) = \frac {\pi} {\phi(z) \sin(\pi z)}$$In each case integrating over the same contour. Hope this helps, and if you don't know much complex analysis, you should learn more, it is a very interesting and powerful area. Sorry this was the longest answer ever. |
As is well-known (see here for a M.O. question) all Kahler manifolds are $spin^c$. I would like to ask which are in fact $spin$.
Taking my motivation from the case of complex projective space, I make the following (naive) conjecture:
Conjecture: A compact $2n$-dimensional Kahler manifold $M$ is spin, if there exists a line bundle $L$ over $M$ such that $$ L \otimes L \simeq \Omega^{(0,n)}, $$ where $\Omega^{(a,b)}$ denotes the space of $(a,b)$-forms.
Can someone tell me if this is true or nor, and if not, what is an easy counter-example. |
I think that no. For example, Let $x=\{1,2,3\}$ and $z=\{1,2,3,4\}$. So, $\{\{1,2,3\}\}=\{\{1,2,3,4\}\}$. Yet, $x\neq z$.
Can you explain?
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I think that no. For example, Let $x=\{1,2,3\}$ and $z=\{1,2,3,4\}$. So, $\{\{1,2,3\}\}=\{\{1,2,3,4\}\}$. Yet, $x\neq z$.
Can you explain?
By definition two sets are equal if and only if they have the same members: $x=y$ if and only if $\forall z(z\in x\leftrightarrow z\in y)$. In your example the sets $\{x\}$ and $\{z\}$ are assumed to be equal, so they must have exactly the same elements. The only element of $\{x\}$ is $x$, and the only element of $\{z\}$ is $z$, so it must be the case that $x=z$.
As was noted in the comments, your example is incorrect: if $x=\{1,2,3\}$ and $z=\{1,2,3,4\}$, then $x\ne z$, so $\{x\}\ne\{z\}$.
Notice that in order to prove that two set $A$ and $B$ are equal, we need to show: $$A\subseteq B~~~~ \text{and}~~~~ B\subseteq A$$
In this case we have the set $x=\{1,2,3\}$ and $z=\{1,2,3,4\}$.
We can clearly see that $x \subset z$, since every element of $x$ is contained in $z$. In order to show equality, we now also need $z\subset x$.
However, since $4 \in z$, but $4 \not\in x$ we have that $z \not\subset x$.
So $x \neq z$.
I am going to hazard a guess that when the OP says
Let x={1,2,3} and z={1,2,3,4}. So, {{1,2,3}}={{1,2,3,4}}
that he or she is confusing "equal sets" with "equal size sets". It is definitely
true that $\{ \{1,2,3 \}\} $ and $\{ \{1,2,3,4\}\}$ are both sets containing a single element, and they are therefore equal in size. But they are not equal sets. "Equal sets" means "the same sets", which means "sets containing the same elements". If you want to indicate that the two sets are the same size, use the notation $ \left| \{ \{1,2,3 \} \} \right| = \left| \{ \{ 1,2,3,4 \} \} \right|$.
That is correct. x $\neq z$. $$x=\{1,2,3\}\ and\ z=\{1,2,3,4\}\ then\ x \neq z. $$
But because all of $x$ is in $z$ and $z$ has items $x$ does not have, $x$ $\subseteq$ $z$. Meaning x is a proper subset of $z$.
$x$ consists of $4$ elements, $z$ consists of $3$ elements. Obviously, they
cannot be equal.
By extensionality of sets, there is an equivalence
$$\{ x \} = \{ z \} \quad \Longleftrightarrow \quad \forall t: (t \in \{ x \} \Leftrightarrow t \in \{ z \}) $$
By definition of set enumeration notation, there is an equivalence
$$ a \in \{ b \} \quad \Longleftrightarrow \quad a = b $$
Substituting this into the above equivalence,
$$\{ x \} = \{ z \} \quad \Longleftrightarrow \quad \forall t: (t = x \Leftrightarrow t =z) $$
The right hand side is equivalent to $x=z$. But if you need to see that proved, I will instead prove something simpler: by substituting $t = x$, we get
$$ \forall t: (t = x \Leftrightarrow t =z) \implies (x=x \Leftrightarrow x=z) \implies x=z $$
and consequently,
$$\{ x \} = \{ z \} \quad \implies \quad x=z $$
The other direction
$$ \{ x \} = \{ z \} \quad \Longleftarrow \quad x=z $$
is proven by substitution, thus
$$ \{ x \} = \{ z \} \quad \Longleftrightarrow \quad x=z $$
Your error is that $\{\{1,2,3\}\} \neq \{\{1,2,3,4\}\}$. |
Let $X_1,X_2,X_3$ be three discrete (integer and non-negative valued) random variables with local probabilities $a_k:=\mathbb{P}(X_1=k)$, $b_k:=\mathbb{P}(X_2=k)$, $c_k:=\mathbb{P}(X_3=k)$ and $s_k:=\mathbb{P}(X_1+X_2+X_3=k)$, $k=0,1,2,\dots$
Let us define two recurrent sequences. The first $(\alpha_n)$ by $$\alpha_0:=1,\\ \alpha_1:=0,\\ \alpha_2:=-\frac{1}{b_0c_0},\\ \alpha_n:=\frac{1}{s_0}\left(\alpha_{n-3}-a_{n-2}-\sum_{k=1}^{n-1}s_k\,\alpha_{n-k}\right), n=3,4,\ldots $$ The second $(\beta_n)$ by $$ \beta_0:=0,\\ \beta_1:=1,\\ \beta_2:=-\frac{c_1}{c_0}-\frac{1}{b_0},\\ \beta_n:=\frac{1}{s_0}\left(\beta_{n-3}-\sum_{k=1}^{n-1}s_k\,\beta_{n-k}-a_{n-2}\,c_0+c_0\sum_{k=0}^{n-1}a_k\,b_{n-1-k}\right), n=3,4,\ldots $$ Let us also define a determinant $$ D_n:=\begin{vmatrix} \alpha_{n}& \beta_{n}\\ \alpha_{n+1}& \beta_{n+1} \end{vmatrix}. $$
Can you show that $D_{n+1}>D_{n}>0$ for every $n=0,1,2,\ldots$?
Such a problem arises in insurance mathematics (calculating a ruin probability if a certain claim appears) and has quite a long context behind it.
Since we have a term $1/s_0$, we have to add that $s_0>0$. This implies that $a_0>0$, $b_0>0$ and $c_0>0$. On the other hand, if $\mathbb{P}(X_i=0)=1$ for $ i=1,2,3$, then (in the context of insurance mathematics) this means that there are no claims at all and that's not a problem to solve.
Numerical calculations show that $D_{n+1}>D_{n}>0$ is verified with certain distributions. Moreover, if we define $$ \widetilde{D}_n:=\begin{vmatrix} \alpha_{n}& \beta_{n}\\ \alpha_{n+2}& \beta_{n+2} \end{vmatrix}, $$ then it seems (by numerical calculations) that $\widetilde{D}_{n+1}<\widetilde{D}_n<0$ for every $n=0,1,2,\ldots$
Thanks in advance. |
Is it true that the cardinality of every maximal linearly independent subset of a finitely generated free module $A^{n}$ is equal to $n$ (not just at most $n$, but in fact $n$)? Here $A$ is a nonzero commutative ring. I know that it's true if $A$ is Noetherian or integral domain. I thought it was not true in general but I came up with something that looks like a proof and I can't figure out where it went wrong.
I think I have a counter-example. Let $A$ be the ring of functions $f$ from $\mathbb{C}^2 \setminus (0,0) \to \mathbb{C}$ such there is a polynomial $\widetilde{f} \in \mathbb{C}[x,y]$ such that $\widetilde{f}(x,y)=f(x,y)$ for all but finitely many $(x,y)$ in $\mathbb{C}^2$.
Map $A$ into $A^2$ by $f \mapsto (fx, fy)$. We check that this is injective: If $fx=0$ then $f$ is zero off of the $x$-axis. Similarly, if $fy=0$, then $f$ is zero off of the $y$-axis. So $(fx, fy) = (0,0)$ implies that $f$ is zero everywhere on $\mathbb{C}^2 \setminus (0,0)$.
We now claim that there do not exist $(u,v)$ in $A^2$ such that $(f,g) \mapsto (fx+gu, \ fy+gv)$ is injective. Suppose such a $(u,v)$ exists. Let $\widetilde{u}$ and $\widetilde{v}$ be the polynomials in $\mathbb{C}[x,y]$ which coincide with $u$ and $v$ at all but finitely many points. Let $\Delta=\widetilde{u} y - \widetilde{v} x$. Since $\Delta$ is a polynomial which vanishes at $(0,0)$, it is not a non-zero constant. Thus, $\Delta$ vanishes on an entire infinite subset of $\mathbb{C}^2$. Let $(p,q)$ be a point in $\mathbb{C}^2 \setminus (0,0)$ such that $\Delta(p,q)=0$, $\widetilde{u}(p,q)= u(p,q)$ and $\widetilde{v}(p,q)=v(p,q)$.
So $q u(p,q) - p v(p,q) =0$. Since $(p,q) \neq (0,0)$, there is some $k \in \mathbb{C}$ such that $(u(p,q), v(p,q)) = (kp, kq)$. Take $f$ to be $-k$ at $(p,q)$ and $0$ elsewhere; let $g$ be $1$ at $(p,q)$ and $0$ elsewhere. So $(fx+gu, fy+gv)=0$, and the map $(f,g) \mapsto (fx+gu, \ fy+gv)$ is not injective.
We have to prove that $m \leq n$ if there is a monomorphism $A^m \to A^n$. Since this is given by a $n \times m$ matrix with entries in $A$ and every finitely generated ring is noetherian, it is enough to consider the case that $A$ is noetherian.
Now you already know the proof for this case, but I just add it. Pick a minimal prime ideal $\mathfrak{p} \subseteq A$. This exists since $A \neq 0$. Now localize at $\mathfrak{p}$. Then we may replace $A$ by $A_{\mathfrak{p}}$, and thereby assume that $A$ is a $0$-dimensional noetherian ring, thus artinian. For such a ring it is known that the length of finitely generated modules is finite, and additive on short exact sequences. In particular $m * l(A) \leq n * l(A)$. Since $l(A) \neq 0$ is finite, we get $m \leq n$.
By the way, the assertion can be generalized to the infinite case:
Let $M$ be a free module with basis $B$ and $L \subseteq M$ a linearly independent subset. Then $|L| \leq |B|$.
Proof: Let $B$ be infinite. Representing elements of $L$ as linear combinations of elements in $B$ yields a map $f : L \to E(B)$, where $E(B)$ denotes the set of finite subsets of $B$. Now let $F$ be such a finite subset with $n$ elements. The finite case yields that there are at most $n$ linearly independent elements in $\langle F \rangle$, thus also in $f^{-1}(F)$. Now we use cardinal arithmetics:
$|L| = \sum_{n > 0} \sum_{F \in E(B), |F|=n} |f^{-1}(F)| \leq \sum_{n > 0} |B^n| = \sum_{n > 0} |B| = |B|.$
EDIT: See the comments; this does not answer kwan's question yet.
I have spotted the mistake in my proof. So here is the "wrong" proof:
Let $v_{1},\ldots,v_{m}$ be linearly independent elements in $A^{n}$, where $m\lt n$. Write them as $n$-tuples of elements in $A$, thereby forming an $n$-by-$m$ matrix. Linear independence of $v_{i}\ $s means that the rank of this matrix is $m$. So there is an $m$-by-$m$ minor with non-zero determinant. By exchanging rows if necessary, bring these $m$ rows to the top part of the matrix. Now add a colomn to the right side of the matrix whose entries are $0$ except at the $n+1$ th position, where the entry is $1$. Then the new $n$-by-$(m+1)$ matrix has rank $(m+1)$ and hence the $(m+1)$ columns, the first $m$ of which are the $v_{i}\ $s, are linearly independent.
The mistake was the notion of rank of a matrix. When the entries are not from integral domain, the proper definition should be the largest integer $m$ such that there is no nonzero element in the ring annihilating the determinant of every $m$-by-$m$ minor. In the above example, I can't conclude that the rank of the $n$-by-$(m+1)$ matrix is $(m+1)$.
With this, I can now exhale a sigh of relief and continue believing that this is not true in general. (By the way I also know that it is true for free modules of infinite rank)
Assuming that $A$ has a maximal ideal $\mathfrak{m}$ (for example, by using Zorn's Lemma), one can proceed as follows: if $M$ is a free $A$-module with basis $(v_i)_{i\in I}$, then $M \cong A^I$, whence $M / \mathfrak{m} M \cong A^I / \mathfrak{m} A^I \cong (A / \mathfrak{m} A)^I$. This is a vector space over $k := A / \mathfrak{m} A$ of dimension $|I|$. Since over fields, all vector space bases of the same vector space have the same length, and since the $k$-vector space structure of $M / \mathfrak{m} M$ is independent of the choice of the basis, this shows that all $A$-bases of $M$ have the same cardinality.
I don't remember where I first saw this though... maybe someone else has a reference? I saw this first in the case that $A = \mathbb{Z}$ and $\mathfrak{m} = (2)$ for free abelian groups $M$, to show that the rank is well-defined. |
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Hi I'm stuck on an integration problem where I need to use the method of cylindrical shells to calculate a volume.
Q. Using the method of cylindrical shells, find the volume generated when the area bounded by the curve y = x^2 - 3 and the line y = 2x is revolved about the line x = 7.
The main thing that I'm having trouble with is setting up the integral. I think it's due to a limited conceptual understanding of what's going on with questions of this type. I started by drawing a quick sketch and finding the x values of intersection, x = +/- sqrt(3).
I drew a cylinder, with the vertical axis x = 7 going through its centre, around the region.
Volume(cylinder) = (circumference)(height)(thickness) = CHT
[tex] C = 2\pi r = 2\pi \left( {7 - x} \right) [/tex]
[tex] H = 2x - \left( {x^2 - 3} \right) = - x^2 + 2x + 3 [/tex]
T = dx
[tex] V = 2\pi \int\limits_{ - \sqrt 3 }^{\sqrt 3 } {\left( {7 - x} \right)\left( { - x^2 + 3x + 3} \right)dx} [/tex]
That's what I came up with. I can't check to see if my answer is correct because I don't have the solution to the question. Any help with this question would be good thanks.
Q. Using the method of cylindrical shells, find the volume generated when the area bounded by the curve y = x^2 - 3 and the line y = 2x is revolved about the line x = 7.
The main thing that I'm having trouble with is setting up the integral. I think it's due to a limited conceptual understanding of what's going on with questions of this type. I started by drawing a quick sketch and finding the x values of intersection, x = +/- sqrt(3).
I drew a cylinder, with the vertical axis x = 7 going through its centre, around the region.
Volume(cylinder) = (circumference)(height)(thickness) = CHT
[tex]
C = 2\pi r = 2\pi \left( {7 - x} \right)
[/tex]
[tex]
H = 2x - \left( {x^2 - 3} \right) = - x^2 + 2x + 3
[/tex]
T = dx
[tex]
V = 2\pi \int\limits_{ - \sqrt 3 }^{\sqrt 3 } {\left( {7 - x} \right)\left( { - x^2 + 3x + 3} \right)dx}
[/tex]
That's what I came up with. I can't check to see if my answer is correct because I don't have the solution to the question. Any help with this question would be good thanks. |
Elementary number theory¶ Taking modular powers¶
How do I compute modular powers in Sage?
To compute \(51^{2006} \pmod{97}\) in Sage, type
sage: R = Integers(97)sage: a = R(51)sage: a^200612
Instead of
R = Integers(97) you can also type
R = IntegerModRing(97). Another option is to use the interfacewith GMP:
sage: 51.powermod(99203843984,97)96
Discrete logs¶
To find a number \(x\) such that\(b^x\equiv a \pmod m\) (the discrete log of\(a \pmod m\)), you can call ‘s
log command:
sage: r = Integers(125)sage: b = r.multiplicative_generator()^3sage: a = b^17sage: a.log(b)17
This also works over finite fields:
sage: FF = FiniteField(16,"a")sage: a = FF.gen()sage: c = a^7sage: c.log(a)7
Prime numbers¶
How do you construct prime numbers in Sage?
The class
Primes allows for primality testing:
sage: 2^(2^12)+1 in Primes()Falsesage: 11 in Primes()True
The usage of
next_prime is self-explanatory:
sage: next_prime(2005) 2011
The Pari command
primepi is used via the command
pari(x).primepi(). This returns the number of primes\(\leq x\), for example:
sage: pari(10).primepi() 4
Using
primes_first_n or
primes one can check that, indeed,there are \(4\) primes up to \(10\):
sage: primes_first_n(5)[2, 3, 5, 7, 11]sage: list(primes(1, 10))[2, 3, 5, 7]
Divisors¶
How do you compute the sum of the divisors of an integer in Sage?
Sage uses
divisors(n) for the list of divisors of \(n\),
number_of_divisors(n) for the number of divisors of \(n\)and
sigma(n,k) for the sum of the \(k\)-th powers of the divisorsof \(n\) (so
number_of_divisors(n) and
sigma(n,0) are the same).
For example:
sage: divisors(28); sum(divisors(28)); 2*28[1, 2, 4, 7, 14, 28]5656sage: sigma(28,0); sigma(28,1); sigma(28,2)6561050
Quadratic residues¶
Try this:
sage: Q = quadratic_residues(23); Q[0, 1, 2, 3, 4, 6, 8, 9, 12, 13, 16, 18]sage: N = [x for x in range(22) if kronecker(x,23)==-1]; N[5, 7, 10, 11, 14, 15, 17, 19, 20, 21]
Q is the set of quadratic residues mod 23 and N is the set of non-residues.
Here is another way to construct these using the
kroneckercommand (which is also called the “Legendre symbol”):
sage: [x for x in range(22) if kronecker(x,23)==1][1, 2, 3, 4, 6, 8, 9, 12, 13, 16, 18]sage: [x for x in range(22) if kronecker(x,23)==-1][5, 7, 10, 11, 14, 15, 17, 19, 20, 21] |
Of course you can restore the original equation from dispersion relation, at least sometimes. But I think your equations of $k = -i\nabla$ is not correct, because $k$ is a number the other is an operator, you cannot put = between them. Be careful when you abuse the notation and know what you are doing.
But you are on the right track.
The first condition your equation should obey is the superposition principle. That is the linear combinations of the solutions should be also a solution. In order to do that you need to bring your equation in the form like this:
$$A\Psi = 0$$
in the entire domain of $\Psi$ (see why). There $A$ is an expression that evaluates to 0, and $\Psi(x, t)$ is your wave function. $x$ is a spatial coordinate, $t$ is time.
The wave function is a linear combination of elementary plane waves which can be expressed in the form (and is a solution to the wave equation in itself):
$$\Psi(x, t) = e^{i(kx - \omega t)}$$
There $k$ is the wave number $\omega$ is the frequency.
The second ingredient is seeing how this function react to the differential operators.
Wave equations usually contain a Laplacian operator. Since that connects the points in the space and allow the disturbance to spread. If you differentiate $\Psi$ with respect to $x$ twice you get:
$$\nabla^2 \Psi = -k^2\Psi$$
And time derivatives will be in the form:
$$\partial_t^n \Psi = \left( -i \omega \right)^n \Psi$$
(There $\partial_t^n$ is a shorthand for $\frac{\partial^n}{\partial t^n}$).
Now all you need to do is, combine these two equations, substitute the dispersion relation for $\omega$, and fiddle with it till the right side equals to zero. One important thing that $k$ shouldn't appear in the left side, otherwise, your equation will be true for a single particular $k$ which is obviously not what you want.
There is no standard method how to fiddle with these equations. It's a mathematical puzzle which probably doesn't always have a solution.
My examples are in 1 dimension but it's straightforward to generalize it to multiple dimensions.
Example
Consider light which have a very simple dispersion relation:
$$\omega = ck$$
In that case the time derivatives are:
$$\partial_t^n \Psi = \left( -i c k \right)^n \Psi$$
The second derivative looks promising:
$$\partial_t^2 \Psi = -c^2 k^2 \Psi$$
Combining them into an equation:
$$\nabla^2 \Psi + \partial_t^2 \Psi = -k^2 \Psi - c^2 k^2 \Psi$$
Multiplying the second time derivative with $-1/c^2$ we get:
$$\nabla^2 \Psi - \frac{1}{c^2}\partial_t^2 \Psi = -k^2 \Psi + k^2 \Psi = 0$$
Reordering we just arrive at the wave equation:
$$\partial_t^2 \Psi = c^2 \nabla^2 \Psi$$
And the same thing can be done to obtain the Schrödinger or Klein-Gordon equations, from their respective dispersion relations. |
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Now showing items 1-10 of 26
Production of light nuclei and anti-nuclei in $pp$ and Pb-Pb collisions at energies available at the CERN Large Hadron Collider
(American Physical Society, 2016-02)
The production of (anti-)deuteron and (anti-)$^{3}$He nuclei in Pb-Pb collisions at $\sqrt{s_{\rm NN}}$ = 2.76 TeV has been studied using the ALICE detector at the LHC. The spectra exhibit a significant hardening with ...
Forward-central two-particle correlations in p-Pb collisions at $\sqrt{s_{\rm NN}}$ = 5.02 TeV
(Elsevier, 2016-02)
Two-particle angular correlations between trigger particles in the forward pseudorapidity range ($2.5 < |\eta| < 4.0$) and associated particles in the central range ($|\eta| < 1.0$) are measured with the ALICE detector in ...
Measurement of D-meson production versus multiplicity in p-Pb collisions at $\sqrt{s_{\rm NN}}=5.02$ TeV
(Springer, 2016-08)
The measurement of prompt D-meson production as a function of multiplicity in p–Pb collisions at $\sqrt{s_{\rm NN}}=5.02$ TeV with the ALICE detector at the LHC is reported. D$^0$, D$^+$ and D$^{*+}$ mesons are reconstructed ...
Measurement of electrons from heavy-flavour hadron decays in p–Pb collisions at $\sqrt{s_{\rm NN}} = 5.02$ TeV
(Elsevier, 2016-03)
The production of electrons from heavy-flavour hadron decays was measured as a function of transverse momentum ($p_{\rm T}$) in minimum-bias p–Pb collisions at $\sqrt{s_{\rm NN}} = 5.02$ TeV with ALICE at the LHC for $0.5 ...
Direct photon production in Pb-Pb collisions at $\sqrt{s_{NN}}$=2.76 TeV
(Elsevier, 2016-03)
Direct photon production at mid-rapidity in Pb-Pb collisions at $\sqrt{s_{\rm NN}} = 2.76$ TeV was studied in the transverse momentum range $0.9 < p_{\rm T} < 14$ GeV/$c$. Photons were detected via conversions in the ALICE ...
Multi-strange baryon production in p-Pb collisions at $\sqrt{s_\mathbf{NN}}=5.02$ TeV
(Elsevier, 2016-07)
The multi-strange baryon yields in Pb--Pb collisions have been shown to exhibit an enhancement relative to pp reactions. In this work, $\Xi$ and $\Omega$ production rates have been measured with the ALICE experiment as a ...
$^{3}_{\Lambda}\mathrm H$ and $^{3}_{\bar{\Lambda}} \overline{\mathrm H}$ production in Pb-Pb collisions at $\sqrt{s_{\rm NN}}$ = 2.76 TeV
(Elsevier, 2016-03)
The production of the hypertriton nuclei $^{3}_{\Lambda}\mathrm H$ and $^{3}_{\bar{\Lambda}} \overline{\mathrm H}$ has been measured for the first time in Pb-Pb collisions at $\sqrt{s_{\rm NN}}$ = 2.76 TeV with the ALICE ...
Multiplicity dependence of charged pion, kaon, and (anti)proton production at large transverse momentum in p-Pb collisions at $\sqrt{s_{\rm NN}}$= 5.02 TeV
(Elsevier, 2016-09)
The production of charged pions, kaons and (anti)protons has been measured at mid-rapidity ($-0.5 < y < 0$) in p-Pb collisions at $\sqrt{s_{\rm NN}} = 5.02$ TeV using the ALICE detector at the LHC. Exploiting particle ...
Jet-like correlations with neutral pion triggers in pp and central Pb–Pb collisions at 2.76 TeV
(Elsevier, 2016-12)
We present measurements of two-particle correlations with neutral pion trigger particles of transverse momenta $8 < p_{\mathrm{T}}^{\rm trig} < 16 \mathrm{GeV}/c$ and associated charged particles of $0.5 < p_{\mathrm{T}}^{\rm ...
Centrality dependence of charged jet production in p-Pb collisions at $\sqrt{s_\mathrm{NN}}$ = 5.02 TeV
(Springer, 2016-05)
Measurements of charged jet production as a function of centrality are presented for p-Pb collisions recorded at $\sqrt{s_{\rm NN}} = 5.02$ TeV with the ALICE detector. Centrality classes are determined via the energy ... |
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What was your best troll you did in math?
Note by Sharky Kesa 5 years, 1 month ago
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sinxn=six \frac{sin x}{n} = six nsinx=six
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MIND = BLOWN!
LOL..So true! I have another troll with me posed by my friends about trig. but unfortunately that's in a vernacular (Tamil)
Why don't you post it? There must be some Tamilians on Brilliant. As for me, I could have understood it if it was in Kannada or Malayalam, but Tamil is too alien to me!!
@Shabarish Ch – SURE...I'll add it in soon! :)
See this.
Cody Johnson\Huge{\color{#D61F06}{\textbf{Cody }}} \Huge{\color{#D61F06}{\textbf{Johnson}}}Cody Johnson
The King of troll in everything !
@Cody Johnson
By the way, here is something that proves the importance of google ! Google Google-the awesome song
@My brilliantian friends...@Finn Hulse , @Sharky Kesa , @Satvik Golechha , @Krishna Ar, @Kartik Sharma , @Ronak Agarwal
Yeah! I can't live without Google\large \color{#3D99F6}{G}\color{#D61F06}{o}\color{#EC7300}{o}\color{#3D99F6}{g}\color{#20A900}{l}\color{#D61F06}{e}Google
@Satvik Golechha – Hey, this is off-topic but I have recently seen that the position of the horse in your photo is changed. Why so?
@Kartik Sharma – Are you sure? It looks same to me.
@Satvik Golechha – Hey, this question is for the Google - What is the significance of this color difference in its spelling?
@Kartik Sharma – I believe it is more artistic. But you can google that.
@Sharky Kesa – Yes, I googled it and found this -
https://www.quora.com/What-is-the-significance-of-Googles-logo-colors
@Satvik Golechha – What a perfect coloring !!! ;)
What a song.
Yeah, I have heard it, it's very addicting(I mean that high-pitched, annoying but attractive voice) and of course the lyrics !! :) :P
@Kartik Sharma – This song is the best troll!! :P
LOLOLOLOLOL!!!
Best song ever. :D
@Finn Hulse – Twice replied for same thing, why?
@Aditya Raut – Oh, haha, I forgot that I wrote the first one. I wrote them at like 1:00 in the morning, so I wasn't exactly in tip-top shape. :P
When I learnt Binomial Theorem
kartik, 18 downvotes on 2 comments
Of course it's Trollathon!
The best kind of troll is Trollathon. No disputes on that !
There's a problem that I posted called called Troller. Look it up. It's easy :P
But it took me pains to compose that poem for the problem
Can I get as much downvotes as possible?
With pleasure.
As you wish
5!=5+4+3+2+1
How can 5! be 5+4+3+2+1???
I believe that's the troll we would have done.
Why did I get 3 down votes for that comment??
Some people downvote just for fun, but that your comment was a WORTH downvote, who'll ask search where ? You use brilliant means you have internet and you have google ! The king of all searches, the google ! Why need to ask where to search ?
And btw, why my comment got down voted when I asked what means troll?? What is bad in asking questions??
I searched it on internet but the meaning of troll is a person who starts argument. :D
When I asked my friend what is this - 89801∑n=0∞(4n)!(n!)4×26390n+11033964n\frac{\sqrt{8}}{9801} \sum_{n=0}^{\infty}\frac{(4n)!}{(n!)^4}\times\frac{26390n + 1103}{396^{4n}} 98018∑n=0∞(n!)4(4n)!×3964n26390n+1103
and
153360640320∑n=0∞(−1)n(6n)!n!3(3n)!×13591409+545140134n6403203n\frac{1}{53360 \sqrt{640320}} \sum_{n=0}^\infty (-1)^n \frac{(6n)!}{n!^3(3n)!} \times \frac{13591409 + 545140134n}{640320^{3n}} 533606403201∑n=0∞(−1)nn!3(3n)!(6n)!×6403203n13591409+545140134n
:P
What's Troll about this -_-
Troll on my friend because he must have died taking the value of it. But after all, he asked me what's the answer and then, you know what is it.
Why should I get 5 downvotes??
@Kartik Sharma – Lo and behold you have 8 downvotes!!!!
@Krishna Ar – 5 downvotes on the above comment too.. hahahaha I am loving these downvotes, really. It is pleasure for me that I have 13 downvotes on 2 comments!!!
@Kartik Sharma – I will upvote every comment I even remotely like.
@Sharky Kesa – oh thanks, I have my first upvote but now I am used to these downvotes. Did you get the troll?
@Kartik Sharma – Yeah.
I believe first one is equals to the ramanujan formula for finding π\piπ. Second one is chudnovsky formula for finding π\piπ. Both are equal to 1π\frac{1}{\pi}π1.
Sorry, @Ronak Agarwal , although it is correct in the eyes of all those (8+5) downvotes but not in the eyes of a troll. I am sorry for all of them who didn't get my point/troll and just hit the downvote button.
*Actually, the troll here is that these 2 are nothing but mathematical expressions only. I asked "what is this" and not "what is the value of it or what is it equal to" *
So good luck trolling and I would love if I break the world record for downvotes on this comment! ;) :P
What means troll??
Search it up.
Where??
@Anuj Shikarkhane – Dude you sure u r from this century? ; )
@Pankaj Joshi – Haha. :D
@Pankaj Joshi – Off course!
@Anuj Shikarkhane – Get back on course.
@Joshua Ong – Nice one!
Problem Loading...
Note Loading...
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Consider a $2\times 2$ block matrix and a linear system of equations associated to it:
\begin{equation} \begin{pmatrix} - A & B \\ B^t & C \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} \phi \\ \psi \end{pmatrix} \end{equation}
Assume that $A$ and $C$ are symmetric, and symmetric positive semi-definite, and that $A$ is even invertible. One can construct the Schur complement system $(C + B^t A^{-1} B ) y = \psi - B^t A^{-1} \phi$ where $S = C + B^t A^{-1} B$ is symmetric positive-definite, by assumption.
How do you precondition such a system in general? I have noted that preconditioners for $S$ are derived from block matrix preconditioners for the original block matrix. Is there a general consensus how such a block matrix preconditioner looks like? |
Hessian Valuations Speaker(s):Monika Ludwig (Technische Universität Wien) Location:MSRI: Simons Auditorium Tags/Keywords
Valuation
convex function
intrinsic volume
Primary Mathematics Subject Classification Secondary Mathematics Subject ClassificationNo Secondary AMS MSC 11-Ludwig
Different approaches to introduce intrinsic volumes and more generally mixed volumes for convex and log-concave functions were proposed by Bobkov, Colesanti and Fragal\` a, by Rotem and Milman and by Alesker. They all turn out to be valuations on the corresponding spaces. Here a new class of continuous valuations on the space of convex functions on ${\mathbb R}^n$ is introduced. On smooth convex functions, they are defined for $i=0,\dots,n$ by \begin{equation*} u\mapsto \int_{{\mathbb R}^n} \zeta(u(x),x,\nabla u(x))\,[{{D}^2} u(x)]_i\,d x \end{equation*} where $\zeta\in C({\mathbb R}\times{\mathbb R}^n\times{\mathbb R}^n)$ and $[{{D}^2} u]_i$ is the $i$-th elementary symmetric function of the eigenvalues of the Hessian matrix, ${{D}^2} u$, of $u$. Under suitable assumptions on $\zeta$, these valuations are shown to be invariant under translations and rotations on convex and coercive functions. Ultimately, a complete classification of continuous and rigid motion invariant valuations on this space of functions is the aim of this approach. The connection to Hadwiger's theorem will be discussed. The results presented in this talk are joint with Andrea Colesanti (University of Florence) and Fabian Mussnig (Technische Universit\"at Wien).
Ludwig Notes Download 11-Ludwig
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Given a Hopf-algebra $H$ (over a commutative ring), it is a classical fact that its category of (left) modules is monoidal, even if $H$ is not commutative. Given two left modules $M$ and $N$, we can form a new left module structure on $M\otimes N$ via the structure map $$H\otimes M\otimes N\overset{\Delta\otimes 1\otimes 1}\to H\otimes H\otimes M\otimes N\overset{1\otimes\tau\otimes 1}\to H\otimes M\otimes H \otimes N\to M\otimes N.$$
When working in a derived setting (let's assume $H$ is an object in a symmetric monoidal quasicategory $\mathscr{C}$), things can be slightly more complicated, and we should probably have $H$ with monoidal structure and comonoidal structure given by some operads like the little $n$-disk operads $\mathbb{E}_n.$ It's basically formal when working in quasicategories to say that $H$ is an $\mathbb{E}_n$-algebra with a compatible $\mathbb{E}_m$-coalgebra structure, making it into an $\mathbb{E}_n/\mathbb{E}_m$-bialgebra in $\mathscr{C}$. We just say that $H$ is an $\mathbb{E}_n$-algebra object in the quasicategory of $\mathbb{E}_m$-coalgebra objects in $\mathscr{C}$.
It's known that, in general, given an $\mathbb{E}_n$-algebra, the category of left modules over it is $\mathbb{E}_{n-1}$-monoidal. This is why, for instance, left modules over a noncommutative ring (i.e. an $\mathbb{E}_1$-algebra) are not monoidal at all. So my question is, to what extent can we perform the above trick in a "derived" way? Obviously it does not suffice to simply write down the structure map, since we need a whole lot of coherent data to write down a module structure now, but is there some other way to do it?
A good example would be, I think, the example of an $n$-fold loop space $X$. Any space, via the diagonal map, is an $\mathbb{E}_\infty$-coalgebra. In fact there's an equivalence of quasicategories $CoAlg_{\mathbb{E}_\infty}(Top)\simeq Top$. So an $n$-fold loop space is definitely an $\mathbb{E}_n$-algebra in $\mathbb{E}_\infty$-coalgebras in $Top$. So, is the category of modules in $Top$ over $X$ somehow "more monoidal than it should be?" In general, how well does this type of thing work? |
I have some hints for you and I suggest you work from there. I might have made errors as well. If you need more help or find an error, leave a comment.
First we write the probability mass function for a having the result $N$ on a single die:
$$prob(N_i)=1/6,\, \text{with }N=1,2,3,4,5,6$$
This is a Discrete Uniform Distribution. All results are equally likely since the die is fair. Now consider the result of summing the faces of 25 rolls of the die and taking the average. The average $S$ is given by the sum of $N_i$ with $i=1,...25$ over 25 where the $N_i$ are i.i.d. distributed as above.
$$S= \frac{N_1 + N_2 + ... +N_{25}}{25}$$
So we ask what is $prob(S>4)$.
For that you first need the probability distribution of $S$. Here is where the hint with the Central Limit Theorem comes in handy. It tells us that we can approximate the density of $prob(S)$ can be approximated as Gaussian with
$$prob(S) \sim \mathcal{N}(\mu,\sigma^2/25)$$
where $\mu$ is the expected value of the uniform distribution from above and $\sigma^2$ is its variance. You can calculate them using the information on this Wikipedia page.
EDITIf you do not want to use the Gaussian approximation, have a look at the Bates Distribution as suggested by @wolfies.
Do you think you can take it from here? |
In physics,
coherence length is the propagation distance over which a coherent wave (e.g. an electromagnetic wave) maintains a specified degree of coherence. Wave interference is strong when the paths taken by all of the interfering waves differ by less than the coherence length. A wave with a longer coherence length is closer to a perfect sinusoidal wave. Coherence length is important in holography and telecommunications engineering.
This article focuses on the coherence of classical electromagnetic fields. In quantum mechanics, there is a mathematically analogous concept of the quantum coherence length of a wave function.
Formulas
In radio-band systems, the coherence length is approximated by
L={c \over n\, \Delta f},
where
c is the speed of light in a vacuum, n is the refractive index of the medium, and \Delta f is the bandwidth of the source.
In optical communications, the coherence length L is given by
[1] L={2 \ln(2) \over \pi n} {\lambda^2 \over \Delta\lambda},
where \lambda is the central wavelength of the source, n is the refractive index of the medium, and \Delta\lambda is the spectral width of the source. If the source has a Gaussian spectrum with FWHM spectral width \Delta\lambda, then a path offset of ±L will reduce the fringe visibility to 50%.
Coherence length is usually applied to the optical regime.
The expression above is a frequently used approximation. Due to ambiguities in the definition of spectral width of a source, however, the following definition of coherence length has been suggested:
The coherence length can be measured using a Michelson interferometer and is the optical path length difference of a self-interfering laser beam which corresponds to a 1/e=37\% fringe visibility,
[2] where the fringe visibility is defined as V = {I_\max - I_\min \over I_\max + I_\min} ,\,
where I is the fringe intensity.
In long-distance transmission systems, the coherence length may be reduced by propagation factors such as dispersion, scattering, and diffraction.
Lasers
Multimode helium–neon lasers have a typical coherence length of 20 cm, while the coherence length of singlemode ones can exceed 100 m. Semiconductor lasers reach some 100 m. Singlemode fiber lasers with linewidths of a few kHz can have coherence lengths exceeding 100 km. Similar coherence lengths can be reached with optical frequency combs due to the narrow linewidth of each tooth. Non-zero visibility is present only for short intervals of pulses repeated after cavity length distances up to this long coherence length.
See also References
^ Drexler, Fujimoto (2008). Optical Coherence Tomography. Springer Berlin Heidelberg. ^ Ackermann, Gerhard K. (2007). Holography: A Practical Approach. Wiley-VCH.
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Before I start, I'll admit that I've criticized the question based on its improbability; however, I've been persuaded otherwise. I'm going to try to do the calculations based on completely different formulas than I think have been used; I hope you'll stay with me as I work it out.
Let's imagine that Lucifer becomes a main-sequence star - in fact, let's call it a low-mass red dwarf. Main-sequence stars follow the mass-luminosity relation:
$$\frac{L}{L_\odot} = \left(\frac{M}{M_\odot}\right)^a$$
Where $L$ and $M$ are the star's luminosity and mass, and $L_\odot$ and $M_\odot$ and the luminosity and mass of the Sun. For stars with $M < 0.43M_\odot$, $a$ takes the value of 2.3. Now we can plug in Jupiter's mass ($1.8986 \times 10 ^{27}$ kg) into the formula, as well as the Sun's mass ($1.98855 \times 10 ^ {30}$ kg) and luminosity ($3.846 \times 10 ^ {26}$ watts), and we get
$$\frac{L}{3.846 \times 10 ^ {26}} = \left(\frac{1.8986 \times 10 ^ {27}}{1.98855 \times 10 ^ {30}}\right)^{2.3}$$
This becomes $$L = \left(\frac{1.8986 \times 10 ^ {27}}{1.98855 \times 10 ^ {30}}\right)^{2.3} \times 3.846 \times 10 ^ {26}$$
which then becomes
$$L = 4.35 \times 10 ^ {19}$$ watts.
Now we can work out the apparent brightness of Lucifer, as seen from Earth. For that, we need the formula
$$m = m_\odot - 2.5 \log \left(\frac {L}{L_\odot}\left(\frac {d_\odot}{d}\right) ^ 2\right)$$
where $m$ is the apparent magnitude of the star, $m_\odot$ is the apparent magnitude of the Sun, $d_\odot$ is the distance to the Sun, and $d$ is the distance to the star. Now, $m = -26.73$ and $d(s)$ is 1 (in astronomical units). $d$ varies. Jupiter is about 5.2 AU from the Sun, so at its closest distance to Earth, it would be ~4.2 AU away. We plug these numbers into the formula, and find
$$m = -6.25$$
which is a lot less brighter than the Sun. Now, when Jupiter is farthest away from the Sun, it is ~6.2 AU away. We plug
that into the formula, and find
$$m = -5.40$$
which is dimmer still - although, of course, Jupiter would be completely blocked by the Sun. Still, for finding the apparent magnitude of Jupiter at some distance from Earth, we can change the above formula to
$$m = -26.73 - 2.5 \log \left(\frac {4.35 \times 10 ^ {19}}{3.846 \times 10 6 {26}}\left(\frac {1}{d}\right) ^ 2\right)$$
By comparison, the Moon can have an average apparent magnitude of -12.74 at full moon - much brighter than Lucifer. The apparent magnitude of both bodies can, of course, change - Jupiter by transits of its moon, for example - but these are the optimal values.
While the above calculations really don't answer most parts of your question, I hope it helps a bit. And please, correct me if I made a mistake somewhere. LaTeX is by no means my native language, and I could have gotten something wrong.
I hope this helps.
Edit
The combined brightness of Lucifer and the Sun would depend on the angle of the Sun's rays and Lucifer's rays. Remember how we have different seasons because of the tilt of the Earth's axis? Well, the added heat would have to do with the tilt of Earth's and Lucifer's axes relative to one another. I can't give you a numerical result, but I can add that I hope it wouldn't be too much hotter than it is now, as I'm writing this!
Second Edit
Like I said in a comment somewhere on this page, the mass-luminosity relation really only works for main-sequence stars. If Lucifer was not on the main sequence. . . Well, then none of my calculations would be right. |
An
antichain in a DAG $(V, E)$ is a subset $A \subseteq V$ of vertices that are pairwise unreachable, namely, there are no $v \neq v' \in A$ such that $v$ is reachable from $v'$ in $E$. From Dilworth's theorem in partial order theory, it is known that if DAG has no antichain of size $k \in \mathbb{N}$, then it can be decomposed in a union of at most $k-1$ disjoint chains, i.e., directed paths.
Now, I am interested in
labeled DAGs, i.e., DAGs where each vertex $v$ carries a label $\lambda(v)$ in some fixed finite set $\Sigma$ of labels. Given an antichain $A \subseteq V$, I can define its labeled size as the minimal number of occurrences of the labels of $\Sigma$ in $A$, namely, $\min_{a \in \Sigma} |\{v \in A \mid \lambda(v) = a\}|$. Is there an analogue of Dilworth's theorem in this context? In other words, if I assume that a DAG has no antichain of labeled size $k \in \mathbb{N}$, what can I assume about its structure? Can I decompose it in some special way? I am already puzzled by the case of $\Sigma = \{a, b\}$, but also interested in the case of a general finite label set.
To visualize this for $\Sigma = \{a, b\}$, saying that $G$ has no antichain of labeled size $k$ means that there is no antichain containing at least $k$ vertices labeled $a$ and $k$ vertices labeled $b$; there can be arbitrarily large antichains but they have to contain only $a$ elements or only $b$ elements, up to $k-1$ exceptions at most. It seems that disallowing large antichains should enforce that the DAG essentially "alternates" between parts of large width for $a$-labeled vertices, and large width for $b$-labeled vertices, but I have not been able to formalize this intuition. (Of course, a suitable structural characterization must talk about the labels of vertices in addition to the shape of the DAG, because already for $k \geq 1$ and on $\{a, b\}$ the condition is satisfied by completely arbitrary DAGs whenever all vertices carry the same label.) |
Arkiv för Matematik Ark. Mat. Volume 38, Number 1 (2000), 21-36. Vector-valued Hardy inequalities and B-convexity Abstract
Inequalities of the form$\sum\nolimits_{k = 0}^\infty {|\hat f(m_k )|/(k + 1) \leqslant C||f||_1 } $ for all
f∈ H 1, where { m } are special subsequences of natural numbers, are investigated in the vector-valued setting. It is proved that Hardy's inequality and the generalized Hardy inequality are equivalent for vector valued Hardy spaces defined in terms ff atoms and that they actually characterize k B-convexity. It is also shown that for 1< q<∞ and 0<α<∞ the space X=H(1, q,γa) consisting of analytic functions on the unit disc such that $\int_0^1 {(1 - r)^{q\alpha - 1} M_1^q (f,r) dr< \infty } $ satisfies the previous inequality for vector valued functions in H 1( X), defined as the space of X-valued Bochner integrable functions on the torus whose negative Fourier coefficients vanish, for the case { m }={2 k k} but not for { m }={ k k } for any α ∈ N. a Note
The author has been partially supported by the Spanish DGICYT, Proyecto PB95-0291.
Article information Source Ark. Mat., Volume 38, Number 1 (2000), 21-36. Dates Received: 23 September 1998 First available in Project Euclid: 31 January 2017 Permanent link to this document https://projecteuclid.org/euclid.afm/1485898663 Digital Object Identifier doi:10.1007/BF02384487 Mathematical Reviews number (MathSciNet) MR1749355 Zentralblatt MATH identifier 1028.42016 Rights 2000 © Institut Mittag-Leffler Citation
Blasco, Oscar. Vector-valued Hardy inequalities and B -convexity. Ark. Mat. 38 (2000), no. 1, 21--36. doi:10.1007/BF02384487. https://projecteuclid.org/euclid.afm/1485898663 |
Here is a summary of what is known about approximability of $k$-CSP over a domain of size $q$:
The best known approximation algorithms for the problem give an $\Omega(q \max(k, \log q)/q^k)$ approximation [MM14 and MNT16]. For $k = \Omega(q)$, there is a matching hardness of $O(kq/q^k)$ by Håstad (UGC-hardness) and Chan (NP-hardness) [Chan13]. For $k$ between $c \log q / \log \log q$ and $q$, the best known NP-hardness result is $O(q^2/q^k)$ (follows from [Chan13]). For $2 < k < c \log q / \log \log q$, the best known UGC-hardness is $2^{O(k \log k)} q \, (\log q)^{k/2}/q^k$ by Manurangsi et al. [MNT16]. For $k=2$, the best NP-hardness is $O(\log q/\sqrt{q})$ by Chan [Chan13], the best UGC-hardness is $O(\log q/q^k)$ by Khot et al. [KKMO07]
[Chan13] Siu On Chan. Approximation resistance from pairwise independent subgroups. In Proceedings of the Symposium on Theory of Computing, pages 447–456, 2013.
[KKMO07] Subhash Khot, Guy Kindler, Elchanan Mossel, and Ryan O’Donnell. Optimal inapproxima bility results for MAX-CUT and other 2-variable CSPs? SIAM J. Comput., 37(1):319–357, 2007.
[MM14] Konstantin Makarychev and Yury Makarychev. Approximation algorithm for non-Boolean Max $k$-CSP. Theory of Computing, 10(13):341–358, 2014.
[MNT16] Pasin Manurangsi, Preetum Nakkiran, and Luca Trevisan. Near-optimal UGC-hardness of approximating Max $k$-CSP$_r$ . In Proceedings of the Workshop on Approximation Algorithms for Combinatorial Optimization Problems (to appear), 2016. |
What happens to all of the electrons and protons in the material of a neutron star?
Could there ever be an electron star or a proton star?
If a dense, spherical star were made of uniformly charged matter, there'd be an attractive gravitational force and a repulsive electrical force. These would balance for a very small net charge: $$ dF = \frac1{r^2}\left( - GM_\text{inside} dm + \frac1{4\pi\epsilon_0}Q_\text{inside} dq \right) $$ which balances if $$ \frac{dq}{dm} = \frac{Q_\text{inside}}{M_\text{inside}} = \sqrt{G\cdot 4\pi\epsilon_0} \approx 10^{-18} \frac{e}{\mathrm{GeV}/c^2}. $$ This is approximately one extra fundamental charge per $10^{18}$ nucleons, or a million extra charges per mole — not much. Any more charge than this and the star would be unbound and fly apart.
What actually happens is that the protons and electrons undergo electron capture to produce neutrons and electron-type neutrinos.
The other answers cover your second question well enough, but there's some detail still missing on the first one - what happens to the protons and electrons in a star when it collapses into a neutron star. The basic answer is simple: they
become those neutrons.
The reason that this happens is that, as it turns out, an {electron, proton} pair is sort of interchangeable with a neutron, or at least it's interchangeable given enough energy. The "more natural" version of the reaction, in fact, goes the other way: on its own, a neutron will in fact decay into a proton, emitting an electron in the process to keep the charge balance happy.
$$ n\to p^++e^-+\bar{\nu}_e $$
This is the most basic example of beta decay, and it has a half-life of about 15 minutes, which is fairly fast for a weak-force reaction.
The $\bar{\nu}_e$ thing is an antineutrino, which needs to be emitted to keep the lepton number constant. It has no mass, but it does carry energy, so what happens is that the neutron can turn into a proton and thereby lose a bit of mass, which becomes enough energy to materialize the electron and the antineutrino and accelerate them to keV energies.
Now, one of the cool things about particle physics is that it is all essentially time-reversible, which means that you can run any reaction in reverse. In this case, you can do something like $$p^++e^-\to n+\nu_e$$ if you have enough energy around to run it.
In any given star, you will have both reactions happening with some probability. You will have some amount of free neutrons sticking around, and these will decay into proton-electron pairs, but you will also have a lot of protons and electrons around in an energetic environment, so if two of them crash with enough energy, they will coalesce into a neutron for a bit.
The key word, though, is "enough" energy, and in a normal star the thermal energy - say, ~1keV for the 16 MK at the sun's core - is not enough to provide a significant fraction of the proton-electron collisions the ~780keV they need to produce a neutron. Nevertheless, in any thermal environment there will be some bits of the system that fluctuate to energies $E$ bigger than the thermal energy $k_B T$, with probability $e^{-E/k_B T}$. In this case, this gives a rough estimate that $e^{-1.35/780}\approx 0.1\%$ of proton-electron collisions produce a neutron, which is small but not completely negligible.
So much for normal stars in equilibrium. To make a neutron star, you need something else to break this equation, and this turns out to be an immense amount of pressure: the electron is essentially pushed into the proton by the surrounding plasma. Once nuclear fusion ceases to have fuel, the temperature can no longer keep up with the pressure and, at fairly constant temperatures*, the pressure rises to huge levels.
The reason the pressure changes the game is that the electron capture reaction significantly reduces the volume occupied by the system, which means that the environment performs work on the system by pushing it in, in exactly the same way that a piston performs work on a gas that's inside a box. It is this extra work, performed over a tiny volume by an absolutely humongous pressure, that provides the sizable >780 keV of energy required to make the electron capture reaction favourable.
* Or something like it. Experts, please correct me if I'm wrong.
The inner force of gravitation is so strong than outward pressure that the electron is forced inside the nucleus and fuses with the proton so become a neutral particle similar to neutron. In a sense , we can tell that the nuclues contains only neutron and thus called neutron star.
Addon to the present answers. They so far neglect the strong interaction, which keeps the known atom cores together, working "against" the mutual electric repulsion of the protons. But even $^2$He is not stable. Since the gravitational force is significantly weaker as the electromagnetic, proton stars are (as far as I know) not possible.
The answer to the main question is
no. The repulsive force due to "like charges" is orders of magnitude larger than the attractive gravitational force, so it would be impossible to form a star. In the case of "opposite charges" the situation is now reversed, the opposite charges attract and Helium atoms are created. If enough of them are created, the attractive gravitational force increases until it is larger than the force separating the protons and the electrons and they "fuse," creating neutrons. This continues until all the electrons (or protons) are gone, thus a neutron star would be created.
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The Annals of Probability Ann. Probab. Volume 27, Number 3 (1999), 1283-1303. Integrated Brownian Motion, Conditioned to be Positive Abstract
We study the two-dimensional process of integrated Brownian motion and Brownian motion, where integrated Brownian motion is conditioned to be positive. The transition density of this process is derived from the asymptotic behavior of hitting times of the unconditioned process. Explicit expressions for the transition density in terms of confluent hypergeometric functions are derived, and it is shown how our results on the hitting time distributions imply previous results of Isozaki-Watanabe and Goldman. The conditioned process is characterized by a system of stochastic differential equations (SDEs) for which we prove an existence and unicity result. Some sample path properties are derived from the SDEs and it is shown that $t \to t^{9/10}$ is a “critical curve” for the conditioned process in the sense that the expected time that the integral part of the conditioned process spends below any curve $t \to t^{\alpha}$ is finite for $\alpha < 9 /10$ and infinite for $\alpha \geq 9/10$.
Article information Source Ann. Probab., Volume 27, Number 3 (1999), 1283-1303. Dates First available in Project Euclid: 29 May 2002 Permanent link to this document https://projecteuclid.org/euclid.aop/1022677447 Digital Object Identifier doi:10.1214/aop/1022677447 Mathematical Reviews number (MathSciNet) MR1733148 Zentralblatt MATH identifier 0983.60078 Subjects Primary: 60J65: Brownian motion [See also 58J65] 60G40: Stopping times; optimal stopping problems; gambling theory [See also 62L15, 91A60] 60J25: Continuous-time Markov processes on general state spaces Citation
Groeneboom, Piet; Jongbloed, Geurt; Wellner, Jon A. Integrated Brownian Motion, Conditioned to be Positive. Ann. Probab. 27 (1999), no. 3, 1283--1303. doi:10.1214/aop/1022677447. https://projecteuclid.org/euclid.aop/1022677447 |
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Kaon femtoscopy in Pb-Pb collisions at $\sqrt{s_{\rm{NN}}}$ = 2.76 TeV
(Elsevier, 2017-12-21)
We present the results of three-dimensional femtoscopic analyses for charged and neutral kaons recorded by ALICE in Pb-Pb collisions at $\sqrt{s_{\rm{NN}}}$ = 2.76 TeV. Femtoscopy is used to measure the space-time ...
Anomalous evolution of the near-side jet peak shape in Pb-Pb collisions at $\sqrt{s_{\mathrm{NN}}}$ = 2.76 TeV
(American Physical Society, 2017-09-08)
The measurement of two-particle angular correlations is a powerful tool to study jet quenching in a $p_{\mathrm{T}}$ region inaccessible by direct jet identification. In these measurements pseudorapidity ($\Delta\eta$) and ...
Online data compression in the ALICE O$^2$ facility
(IOP, 2017)
The ALICE Collaboration and the ALICE O2 project have carried out detailed studies for a new online computing facility planned to be deployed for Run 3 of the Large Hadron Collider (LHC) at CERN. Some of the main aspects ...
Evolution of the longitudinal and azimuthal structure of the near-side peak in Pb–Pb collisions at $\sqrt{s_{\rm NN}}=2.76$ TeV
(American Physical Society, 2017-09-08)
In two-particle angular correlation measurements, jets give rise to a near-side peak, formed by particles associated to a higher $p_{\mathrm{T}}$ trigger particle. Measurements of these correlations as a function of ...
J/$\psi$ elliptic flow in Pb-Pb collisions at $\sqrt{s_{\rm NN}}$ = 5.02 TeV
(American Physical Society, 2017-12-15)
We report a precise measurement of the J/$\psi$ elliptic flow in Pb-Pb collisions at $\sqrt{s_{\rm NN}} = 5.02$ TeV with the ALICE detector at the LHC. The J/$\psi$ mesons are reconstructed at mid-rapidity ($|y| < 0.9$) ...
Enhanced production of multi-strange hadrons in high-multiplicity proton-proton collisions
(Nature Publishing Group, 2017)
At sufficiently high temperature and energy density, nuclear matter undergoes a transition to a phase in which quarks and gluons are not confined: the quark–gluon plasma (QGP)1. Such an exotic state of strongly interacting ...
K$^{*}(892)^{0}$ and $\phi(1020)$ meson production at high transverse momentum in pp and Pb-Pb collisions at $\sqrt{s_\mathrm{NN}}$ = 2.76 TeV
(American Physical Society, 2017-06)
The production of K$^{*}(892)^{0}$ and $\phi(1020)$ mesons in proton-proton (pp) and lead-lead (Pb-Pb) collisions at $\sqrt{s_\mathrm{NN}} =$ 2.76 TeV has been analyzed using a high luminosity data sample accumulated in ...
Production of $\Sigma(1385)^{\pm}$ and $\Xi(1530)^{0}$ in p-Pb collisions at $\sqrt{s_{\rm NN}} = 5.02$ TeV
(Springer, 2017-06)
The transverse momentum distributions of the strange and double-strange hyperon resonances ($\Sigma(1385)^{\pm}$, $\Xi(1530)^{0}$) produced in p-Pb collisions at $\sqrt{s_{\rm NN}} = 5.02$ TeV were measured in the rapidity ...
Charged–particle multiplicities in proton–proton collisions at $\sqrt{s}=$ 0.9 to 8 TeV, with ALICE at the LHC
(Springer, 2017-01)
The ALICE Collaboration has carried out a detailed study of pseudorapidity densities and multiplicity distributions of primary charged particles produced in proton-proton collisions, at $\sqrt{s} =$ 0.9, 2.36, 2.76, 7 and ...
Energy dependence of forward-rapidity J/$\psi$ and $\psi(2S)$ production in pp collisions at the LHC
(Springer, 2017-06)
We present ALICE results on transverse momentum ($p_{\rm T}$) and rapidity ($y$) differential production cross sections, mean transverse momentum and mean transverse momentum square of inclusive J/$\psi$ and $\psi(2S)$ at ... |
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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 ...
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 ...
Elliptic flow of muons from heavy-flavour hadron decays at forward rapidity in Pb-Pb collisions at $\sqrt{s_{\rm NN}}=2.76$ TeV
(Elsevier, 2016-02)
The elliptic flow, $v_{2}$, of muons from heavy-flavour hadron decays at forward rapidity ($2.5 < y < 4$) is measured in Pb--Pb collisions at $\sqrt{s_{\rm NN}}$ = 2.76 TeV with the ALICE detector at the LHC. The scalar ...
Centrality dependence of the pseudorapidity density distribution for charged particles in Pb-Pb collisions at $\sqrt{s_{NN}}$ = 2.76 TeV
(Elsevier, 2013-11)
We present the first wide-range measurement of the charged-particle pseudorapidity density distribution, for different centralities (the 0-5%, 5-10%, 10-20%, and 20-30% most central events) in Pb-Pb collisions at $\sqrt{s_{NN}}$ ...
Beauty production in pp collisions at √s=2.76 TeV measured via semi-electronic decays
(Elsevier, 2014-11)
The ALICE Collaboration at the LHC reports measurement of the inclusive production cross section of electrons from semi-leptonic decays of beauty hadrons with rapidity |y|<0.8 and transverse momentum 1<pT<10 GeV/c, in pp ... |
By Murray Bourne, 22 Mar 2007
Last week was "Pi Day".
Google trends shows there is a periodic peak in search trends for pi on that day each year.
My post on pi was one of them.
There was a surge in 2013, after the movie "Life of Pi" was released.
See the 1 Comment below.
Posted in Mathematics category - 22 Mar 2007 [Permalink]
Tags: Technology
Yeh - quite an explosion of interest.
But it seems that people are not interested in math on weekends...
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LLVM optimizes power sums, such as
int sum(int count) { int result = 0; for (int j = 0; j < count; ++j) result += j*j; return result; }to code that calculates the result without a loop (godbolt) sum(int): test edi, edi jle .LBB0_1 lea eax, [rdi - 1] lea ecx, [rdi - 2] imul rcx, rax lea eax, [rdi - 3] imul rax, rcx shr rax imul eax, eax, 1431655766 add eax, edi shr rcx lea ecx, [rcx + 2*rcx] lea eax, [rax + rcx] add eax, -1 ret .LBB0_1: xor eax, eax retIt handles more complex cases too (godbolt) – that is, the optimization is not just a simple pattern matcher. This post will show how the optimization is done. Loop analysis – scalar evolution
There are many cases where compilers need to track how values are updated within a loop. For example, the loop vectorizer needs to check that the pointers are moved to the adjacent element in the next iteration, and check that no other pointer indexing may alias the range we are vectorizing.
Both GCC and LLVM does this in the same way in their scalar evolution passes, where each variable at iteration \(i\) (we start enumerating iterations from \(0\)) is represented as a function \(f_0(i)\) defined as a linear recurrence of the form
\[f_j(i) = \begin{cases} \phi_j & \text{if $i = 0$} \\ f_j(i-1) \odot_{j+1} f_{j+1}(i-1) & \text{if $i > 0$} \end{cases}\] where \(\odot\in\{+,*\}\). \begin{cases} 0 & \text{if $i = 0$} \\ f(i-1) + 1 & \text{if $i > 0$} \end{cases}\] \begin{cases} 2\phantom{f_0(i-1) + f_1(i-1)} & \text{if $i = 0$} \\ f_2(i-1) + 6 & \text{if $i > 0$} \end{cases}\\ f_1(i) & = \begin{cases} k-1 & \text{if $i = 0$} \\ f_1(i-1) + f_2(i-1)\phantom{2} & \text{if $i > 0$} \end{cases}\\ f(i) = f_0(i) & = \begin{cases} 7 & \text{if $i = 0$} \\ f_0(i-1) + f_1(i-1)\phantom{2} & \text{if $i > 0$} \end{cases}\end{align}\] One optimization we can see directly from these functions is that the value can be calculated by just three additions within the loop \begin{cases} 2\phantom{f_0(i-1) + f_1(i-1)} & \text{if $i = 0$} \\ f_2(i-1) + 6 & \text{if $i > 0$} \end{cases} \phantom{xx}\text{is written as $\{2,+,6\}$}\\ f_1(i) & = \begin{cases} k-1 & \text{if $i = 0$} \\ f_1(i-1) + f_2(i-1)\phantom{2} & \text{if $i > 0$} \end{cases} \phantom{xx}\text{is written as $\{k-1,+,f_2\}$}\\ f(i) = f_0(i) & = \begin{cases} 7 & \text{if $i = 0$} \\ f_0(i-1) + f_1(i-1)\phantom{2} & \text{if $i > 0$} \end{cases} \phantom{xx}\text{is written as $\{7,+,f_1\}$}\end{align}\] These can be chained, so \(f(i)\) can be written as a \{\phi_0, +, \phi_1\} + \{\psi_0, +, \psi_1\} & \phantom{xx} \Rightarrow \phantom{xx} \{\phi_0 + \psi_0, +, \phi_1 + \psi_1\} \\ \{\phi_0, +, \phi_1\}* \{\psi_0, +, \psi_1\} & \phantom{xx} \Rightarrow \phantom{xx} \{\phi_0 * \psi_0, +, \psi_1 * \{\phi_0, +, \phi_1\} + \phi_1 * \{\psi_0, +, \psi_1\} + \phi_1*\psi_1\} \\ \{\phi_0, +, \phi_1,+,0\} & \phantom{xx} \Rightarrow \phantom{xx} \{\phi_0, +, \phi_1\}\end{align}\] So for the loop from the & = \{0 * 0, +, 1 * \{0, +,1\} + 1 * \{0, +, 1\} + 1*1\} \\ & = \{0, +, 1,+,2\}\end{align}\] Similar calculations for
\[f_j(i) =
\begin{cases}
\phi_j & \text{if $i = 0$} \\
f_j(i-1) \odot_{j+1} f_{j+1}(i-1) & \text{if $i > 0$}
\end{cases}\] where \(\odot\in\{+,*\}\).
Example 1The simplest case is a loop such as void foo(int m, int *p) { for (int j = 0; j < m; j++) *p++ = j; }The loop writes \(0\) to
*p++ in the first iteration, \(1\) in the second, etc. So we can express the value written at iteration \(i\) as\[f(i) =
\begin{cases}
0 & \text{if $i = 0$} \\
f(i-1) + 1 & \text{if $i > 0$}
\end{cases}\]
Example 2Polynomials in the iteration variable can also be expressed in this form. void foo(int m, int k, int *p) { for (int j = 0; < m; j++) *p++ = j*j*j - 2*j*j + k*j + 7; }We will see below how to build the functions, but the result for the value stored in this loop is \[\begin{align}f_2(i) & =
\begin{cases}
2\phantom{f_0(i-1) + f_1(i-1)} & \text{if $i = 0$} \\
f_2(i-1) + 6 & \text{if $i > 0$}
\end{cases}\\
f_1(i) & =
\begin{cases}
k-1 & \text{if $i = 0$} \\
f_1(i-1) + f_2(i-1)\phantom{2} & \text{if $i > 0$}
\end{cases}\\
f(i) = f_0(i) & =
\begin{cases}
7 & \text{if $i = 0$} \\
f_0(i-1) + f_1(i-1)\phantom{2} & \text{if $i > 0$}
\end{cases}\end{align}\] One optimization we can see directly from these functions is that the value can be calculated by just three additions within the loop
void foo(int m, int k, int *p) { int t0 = 7; int t1 = k-1; int t2 = 2; for (int j = 0; j < m; j++) { *p++ = t0; t0 = t0 + t1; t1 = t1 + t2; t2 = t2 + 6; } }which is a useful optimization for architectures were multiplication is expensive. This kind of code is, however, uncommon, so most compilers do not do this optimzation – but they usually do this for simpler cases, such as void foo(int m, int k, int *p) { for (int j = 0; < m; j++) *p++ = k*j + 7; }as constructs of the form
k*j+7 are common in address calculations.
Chains of recurrencesIt is cumbersome to write the recursive functions all the time, so the functions are usually written in the form \(\{\phi_j, \odot_{j+1}, f_{j+1}\}\). For example \[\begin{align}f_2(i) & =
\begin{cases}
2\phantom{f_0(i-1) + f_1(i-1)} & \text{if $i = 0$} \\
f_2(i-1) + 6 & \text{if $i > 0$}
\end{cases} \phantom{xx}\text{is written as $\{2,+,6\}$}\\
f_1(i) & =
\begin{cases}
k-1 & \text{if $i = 0$} \\
f_1(i-1) + f_2(i-1)\phantom{2} & \text{if $i > 0$}
\end{cases} \phantom{xx}\text{is written as $\{k-1,+,f_2\}$}\\
f(i) = f_0(i) & =
\begin{cases}
7 & \text{if $i = 0$} \\
f_0(i-1) + f_1(i-1)\phantom{2} & \text{if $i > 0$}
\end{cases} \phantom{xx}\text{is written as $\{7,+,f_1\}$}\end{align}\] These can be chained, so \(f(i)\) can be written as a
chain of recurrences(CR) \(\{7,+,\{k-1,+,\{2,+,6\}\}\}\). The inner curly braces are redundant, so the CR is usually written as a tuple \(\{7,+,k-1,+,2,+,6\}\). Building the chains of recurrencesThe chains of recurrences are built by iterating over the code and calculating the CR for the result of each operation (or marking it as unknown if we cannot handle the operation), using simplification rules such as\[\begin{align}c * \{\phi_0, +, \phi_1\} & \phantom{xx} \Rightarrow \phantom{xx} \{c * \phi_0, +, c * \phi_1\} \\
\{\phi_0, +, \phi_1\} + \{\psi_0, +, \psi_1\} & \phantom{xx} \Rightarrow \phantom{xx} \{\phi_0 + \psi_0, +, \phi_1 + \psi_1\} \\
\{\phi_0, +, \phi_1\}* \{\psi_0, +, \psi_1\} & \phantom{xx} \Rightarrow \phantom{xx} \{\phi_0 * \psi_0, +, \psi_1 * \{\phi_0, +, \phi_1\} + \phi_1 * \{\psi_0, +, \psi_1\} + \phi_1*\psi_1\} \\
\{\phi_0, +, \phi_1,+,0\} & \phantom{xx} \Rightarrow \phantom{xx} \{\phi_0, +, \phi_1\}\end{align}\] So for the loop from the
sum function
for (int j = 0; j < count; ++j) result += j*j;we start with
j which we know has the CR \(\{0,+,1\}\) per Example 1. This is then used as
j*j when calculating
result, so we calculate the CR for
j*j using the simplification rules as \[\begin{align}j*j& = \{0,+,1\} * \{0,+,1\} \\
& = \{0 * 0, +, 1 * \{0, +,1\} + 1 * \{0, +, 1\} + 1*1\} \\
& = \{0, +, 1,+,2\}\end{align}\] Similar calculations for
result gives us the CR \(\{0,+,0,+,1,+,2\}\) for the value at the beginning of the iteration, and \(\{0,+,1,+,3,+,2\}\) after adding
j*j.
Doing the optimization
The optimization is done during induction variable simplification, and LLVM has transformed the function to a form more convenient for analysis and optimization
One nice property of the chains of recurrences is that it is easy to calculate the value at a specific iteration – if we have a CR \(\{\phi_0,+,\phi_1,+,\ldots,+,\phi_n\}\), then the value at iteration \(i\) can be calculated as \[\begin{align}f(i) & = \sum_{j=0}^{n}\phi_j{i \choose j} \\ & = \phi_0 + \phi_1i + \phi_2{i(i-1)\over 2!} + \ldots + \phi_n{i(i-1)\cdots(i-n+1)\over n!}\end{align}\] Inserting the values for the CR \(\{0,+,1,+,3,+,2\}\) describing One problem with this optimization is that it is hard for developers to make the compiler generate a loop for this if they know that the majority of values used in reality are small enough for the loop to be the fastest option. GCC does, therefore, not replace the final value of a loop if the expression is expensive int sum(int count) { int result = 0; if (count > 0) { int j = 0; do { result = result + j*j; ++j; } while (j < count); } return result; }or as it looks like in the LLVM IR define i32 @sum(i32) { %2 = icmp sgt i32 %0, 0 br i1 %2, label %3, label %6 ; <label>:3: br label %8 ; <label>:4: %5 = phi i32 [ %12, %8 ] br label %6 ; <label>:6: %7 = phi i32 [ 0, %1 ], [ %5, %4 ] ret i32 %7 ; <label>:8: %9 = phi i32 [ %13, %8 ], [ 0, %3 ] ; {0,+,1} %10 = phi i32 [ %12, %8 ], [ 0, %3 ] ; {0,+,0,+,1,+,2} %11 = mul nsw i32 %9, %9 ; {0,+,1,+,2} %12 = add nuw nsw i32 %11, %10 ; {0,+,1,+,3,+,2} %13 = add nuw nsw i32 %9, 1 ; {1,+,1} %14 = icmp slt i32 %13, %0 br i1 %14, label %8, label %4 }The compiler can see that the function returns
0 if
count <= 0, otherwise it returns the value of
result at loop iteration
count-1.
One nice property of the chains of recurrences is that it is easy to calculate the value at a specific iteration – if we have a CR \(\{\phi_0,+,\phi_1,+,\ldots,+,\phi_n\}\), then the value at iteration \(i\) can be calculated as \[\begin{align}f(i) & = \sum_{j=0}^{n}\phi_j{i \choose j} \\ & = \phi_0 + \phi_1i + \phi_2{i(i-1)\over 2!} + \ldots + \phi_n{i(i-1)\cdots(i-n+1)\over n!}\end{align}\] Inserting the values for the CR \(\{0,+,1,+,3,+,2\}\) describing
result gives us \[f(i) = i + {3i(i-1)\over 2} + {i(i-1)(i-2) \over 3}\] The compiler now only need to insert code that calculates this with \(i =\)
count-1, after the loop
result = count-1 + 3*(count-1)*(count-2)/2 + (count-1)*(count-2)(count-3)/3;but it need to take some care to calculate in the correct precision (as temporary values may not fit in 32-bit integers). Integer division is slow, so it is also doing tricks to replace the divisions with multiplication and shift instructions. The result is the LLVM IR %4 = add i32 %0, -1 %5 = zext i32 %4 to i33 %6 = add i32 %0, -2 %7 = zext i32 %6 to i33 %8 = mul i33 %5, %7 %9 = add i32 %0, -3 %10 = zext i32 %9 to i33 %11 = mul i33 %8, %10 %12 = lshr i33 %11, 1 %13 = trunc i33 %12 to i32 %14 = mul i32 %13, 1431655766 %15 = add i32 %14, %0 %16 = lshr i33 %8, 1 %17 = trunc i33 %16 to i32 %18 = mul i32 %17, 3 %19 = add i32 %15, %18 %20 = add i32 %19, -1Inserting this makes the loop become dead, so it is later removed by dead code elimination, and we eventually end up with the code sum(int): test edi, edi jle .LBB0_1 lea eax, [rdi - 1] lea ecx, [rdi - 2] imul rcx, rax lea eax, [rdi - 3] imul rax, rcx shr rax imul eax, eax, 1431655766 add eax, edi shr rcx lea ecx, [rcx + 2*rcx] lea eax, [rax + rcx] add eax, -1 ret .LBB0_1: xor eax, eax ret PerformanceThis optimization is not always profitable. For example, int sum(int count) { int result = 0; for (int j = 0; j < count; ++j) result += j*j*j*j*j*j; return result; }is calculated by three 32-bit multiplications and one addition within the loop, while the optimized version needs six 64-bit multiplications, five 32-bit multiplications, and a slew of other instructions (godbolt), so the optimized version is slower for small values of
count.I benchmarked on my PC, and
count must be larger than
5 for the optimized version to be faster than the loop. Smaller CPUs with, for example, more expensive 64-bit multiplication, will need a higher
count for the optimization to help. And CPUs not having instructions for 64-bit multiplication (godbolt) will need a
muchhigher
count.
One problem with this optimization is that it is hard for developers to make the compiler generate a loop for this if they know that the majority of values used in reality are small enough for the loop to be the fastest option. GCC does, therefore, not replace the final value of a loop if the expression is expensive
/* Do not emit expensive expressions. The rationale is that when someone writes a code like while (n > 45) n -= 45; he probably knows that n is not large, and does not want it to be turned into n %= 45. */ || expression_expensive_p (def))So GCC not doing this optimization is a feature – not a bug. Further readingsChains of recurrences: Olaf Bachmann, Paul S. Wang, Eugene V. Zima. “Chains of recurrences – a method to expedite the evaluation of closed-form functions” Eugene V. Zima. “On computational properties of chains of recurrences” Robert A. van Engelen. “Symbolic Evaluation of Chains of Recurrences for Loop Optimization” Robert A. van Engelen. “Efficient Symbolic Analysis for Optimizing Compilers” Torbjörn Granlund, Peter L. Montgomery. “Division by Invariant Integers using Multiplication” Updated: The original post incorrectly called the power sums “geometric sums”. |
The Krylov-Bogolioubov theorem is a fundamental result in the ergodic theory of dynamical systems which is typically stated as follows: if $T$ is a continuous transformation of a nonempty compact metric space $X$, then there exists a Borel probability measure $\mu$ on $X$ which is invariant under $T$ in the sense that $\mu(A)=\mu(T^{-1}A)$ for all Borel sets $A \subseteq X$, or equivalently $\int f d\mu = \int (f \circ T)d\mu$ for every continuous function $f \colon X \to \mathbb{R}$. This question is concerned with proofs of that theorem.
The most popular proof of the Krylov-Bogolioubov theorem operates as follows. Let $\mathcal{M}$ denote the set of all Borel probability measures on $X$, and equip $\mathcal{M}$ with the coarsest topology such that for every continuous $f \colon X \to \mathbb{R}$, the function from $\mathcal{M}$ to $\mathbb{R}$ defined by $\mu \mapsto \int f d\mu$ is continuous. In this topology $\mathcal{M}$ is compact and metrisable, and a sequence $(\mu_n)$ of elements of $\mathcal{M}$ converges to a limit $\mu$ if and only if $\int fd\mu_n \to \int fd\mu$ for every continuous $f \colon X \to \mathbb{R}$. Now let $x \in X$ be arbitrary, and define a sequence of elements of $\mathcal{M}$ by $\mu_n:=(1/n)\sum_{i=0}^{n-1}\delta_{T^ix}$, where $\delta_z$ denotes the Dirac probability measure concentrated at $z$. Using the sequential compactness of $\mathcal{M}$ we may extract an accumulation point $\mu$ which is invariant under $T$ by an easy calculation. This proof, together with minor variations thereupon, is fairly ubiquitous in ergodic theory textbooks. In the answers to this question, Vaughn Climenhaga notes the following alternative proof: the map taking the measure $\mu$ to the measure $T_*\mu$ defined by $(T_*\mu)(A):=\mu(T^{-1}A)$ is a continuous transformation of the compact convex set $\mathcal{M}$, and hence has a fixed point by the Schauder-Tychonoff theorem. A couple of years ago I thought of another proof, given below. The first part of this question is: has the following proof ever been published?
This third proof is as follows. Clearly it suffices to show that there exists a
finite Borel measure on $X$ which is invariant under $T$, since we may normalise this measure to produce a probability measure. By the Hahn decomposition theorem it follows that it suffices to find a nonzero finite signed measure on $X$ which is invariant under $T$. By the Riesz representation theorem for measures this is equivalent to the statement that there exists a nonzero continuous linear functional $L \colon C(X) \to \mathbb{R}$ such that $L(f \circ T)=f$ for all continuous functions $f \colon X \to \mathbb{R}$. Let $B(X)$ be the closed subspace of $C(X)$ which is equal to the closure of the set of all continuous functions which take the form $g \circ T - g$ for some continuous $g$. Clearly a continuous linear functional $L \colon C(X) \to \mathbb{R}$ satisfies $L(f \circ T)=f$ if and only if it vanishes on $B(X)$, so to construct an invariant measure it suffices to show that the dual of $C(X)/B(X)$ is nontrivial. A consequence of the Hahn-Banach theorem is that the dual of $C(X)/B(X)$ is nontrivial as long as $C(X)/B(X)$ is itself nontrivial, so to prove the theorem it is sufficient to show that the complement of $B(X)$ in $C(X)$ is nonempty. But the constant function $h(x):=1$ is not in $B(X)$ because if $|(g \circ T - g)(x) - 1|<1/2$ for all $x \in X$, then $g(Tx)>g(x)+1/2$ for all $x \in X$ and hence $g(T^nx)>g(x)+n/2$ for all $n \geq 1$ and $x \in X$. This is impossible since $g$ is continuous and $X$ is compact. We conclude that $B(X)$ is a proper Banach subspace of $C(X)$ and the desired functional exists.
The second part of the question deals with the Krylov-Bogolioubov theorem for measures invariant under amenable groups of transformations. Let $\Gamma = \{T_\gamma\}$ be a countable amenable group of continuous transformations of the compact metric space $X$. We shall say that $\mu \in \mathcal{M}$ is invariant under $\Gamma$ if $(T_\gamma)_*\mu = \mu$ for all $T_\gamma$. I believe that by using Følner sequences one may generalise the first proof of the Krylov-Bogolioubov theorem to show that every such amenable group has an invariant Borel probability measure. I seem to recall that the second proof also generalises to this scenario (in Glasner's book, perhaps?). Despite a certain amount of thought I have not been able to see how the third proof might generalise to this situation, even when $\Gamma$ is generated by just two commuting elements. So, the second part of this question is: can anyone see how the third proof generalises to the amenable case? |
This question is largely about definitions of PCA/FA, so opinions might differ. My opinion is that PCA+varimax should not be called either PCA or FA, bur rather explicitly referred to e.g. as "varimax-rotated PCA".
I should add that this is quite a confusing topic. In this answer I want to explain what a rotation actually
is; this will require some mathematics. A casual reader can skip directly to the illustration. Only then we can discuss whether PCA+rotation should or should not be called "PCA".
One reference is Jolliffe's book "Principal Component Analysis", section 11.1 "Rotation of Principal Components", but I find it could be clearer.
Let $\mathbf X$ be a $n \times p$ data matrix which we assume is centered. PCA amounts (see my answer here) to a singular-value decomposition: $\mathbf X=\mathbf{USV}^\top$. There are two equivalent but complimentary views on this decomposition: a more PCA-style "projection" view and a more FA-style "latent variables" view.
According to the PCA-style view, we found a bunch of orthogonal directions $\mathbf V$ (these are eigenvectors of the covariance matrix, also called "principal directions" or "axes"), and "principal components" $\mathbf{US}$ (also called principal component "scores") are projections of the data on these directions. Principal components are uncorrelated, the first one has maximally possible variance, etc. We can write: $$\mathbf X = \mathbf{US}\cdot \mathbf V^\top = \text{Scores} \cdot \text{Principal directions}.$$
According to the FA-style view, we found some uncorrelated unit-variance "latent factors" that give rise to the observed variables via "loadings". Indeed, $\widetilde{\mathbf U}=\sqrt{n-1}\mathbf{U}$ are standardized principal components (uncorrelated and with unit variance), and if we define loadings as $\mathbf L = \mathbf{VS}/\sqrt{n-1}$, then $$\mathbf X= \sqrt{n-1}\mathbf{U}\cdot (\mathbf{VS}/\sqrt{n-1})^\top =\widetilde{\mathbf U}\cdot \mathbf L^\top = \text{Standardized scores} \cdot \text{Loadings}.$$ (Note that $\mathbf{S}^\top=\mathbf{S}$.) Both views are equivalent. Note that loadings are eigenvectors scaled by the respective eigenvalues ($\mathbf{S}/\sqrt{n-1}$ are eigenvalues of the covariance matrix).
(I should add in brackets that PCA$\ne$FA; FA explicitly aims at finding latent factors that are linearly mapped to the observed variables via loadings; it is more flexible than PCA and yields different loadings. That is why I prefer to call the above "FA-style view on PCA" and not FA, even though some people take it to be one of FA methods.)
Now, what does a rotation do? E.g. an orthogonal rotation, such as varimax. First, it considers only $k<p$ components, i.e.: $$\mathbf X \approx \mathbf U_k \mathbf S_k \mathbf V_k^\top = \widetilde{\mathbf U}_k \mathbf L^\top_k.$$ Then it takes a square orthogonal $k \times k$ matrix $\mathbf T$, and plugs $\mathbf T\mathbf T^\top=\mathbf I$ into this decomposition: $$\mathbf X \approx \mathbf U_k \mathbf S_k \mathbf V_k^\top = \mathbf U_k \mathbf T \mathbf T^\top \mathbf S_k \mathbf V_k^\top = \widetilde{\mathbf U}_\mathrm{rot} \mathbf L^\top_\mathrm{rot},$$ where rotated loadings are given by $\mathbf L_\mathrm{rot} = \mathbf L_k \mathbf T$, and rotated standardized scores are given by $\widetilde{\mathbf U}_\mathrm{rot} = \widetilde{\mathbf U}_k \mathbf T$. (The purpose of this is to find $\mathbf T$ such that $\mathbf L_\mathrm{rot}$ became as close to being sparse as possible, to facilitate its interpretation.)
Note that what is rotated are: (1) standardized scores, (2) loadings. But not the raw scores and not the principal directions! So the rotation happens in the latent space, not in the original space. This is absolutely crucial.
From the FA-style point of view, nothing much happened. (A) The latent factors are still uncorrelated and standardized. (B) They are still mapped to the observed variables via (rotated) loadings. (C) The amount of variance captured by each component/factor is given by the sum of squared values of the corresponding loadings column in $\mathbf L_\mathrm{rot}$. (D) Geometrically, loadings still span the same $k$-dimensional subspace in $\mathbb R^p$ (the subspace spanned by the first $k$ PCA eigenvectors). (E) The approximation to $\mathbf X$ and the reconstruction error did not change at all. (F) The covariance matrix is still approximated equally well:$$\boldsymbol \Sigma \approx \mathbf L_k\mathbf L_k^\top = \mathbf L_\mathrm{rot}\mathbf L_\mathrm{rot}^\top.$$
But the PCA-style point of view has practically collapsed. Rotated loadings do not correspond to orthogonal directions/axes in $\mathbb R^p$ anymore, i.e. columns of $\mathbf L_\mathrm{rot}$ are not orthogonal! Worse, if you [orthogonally] project the data onto the directions given by the rotated loadings, you will get correlated (!) projections and will not be able to recover the scores. [Instead, to compute the standardized scores after rotation, one needs to multiply the data matrix with the
pseudo-inverse of loadings $\widetilde{\mathbf U}_\mathrm{rot} = \mathbf X (\mathbf L_\mathrm{rot}^+)^\top$. Alternatively, one can simply rotate the original standardized scores with the rotation matrix: $\widetilde{\mathbf U}_\mathrm{rot} = \widetilde{\mathbf U} \mathbf T$.] Also, the rotated components do not successively capture the maximal amount of variance: the variance gets redistributed among the components (even though all $k$ rotated components capture exactly as much variance as all $k$ original principal components).
Here is an illustration. The data is a 2D ellipse stretched along the main diagonal. First principal direction is the main diagonal, the second one is orthogonal to it. PCA loading vectors (eigenvectors scaled by the eigenvalues) are shown in red -- pointing in both directions and also stretched by a constant factor for visibility. Then I applied an orthogonal rotation by $30^\circ$ to the loadings. Resulting loading vectors are shown in magenta. Note how they are not orthogonal (!).
An FA-style intuition here is as follows: imagine a "latent space" where points fill a small circle (come from a 2D Gaussian with unit variances). These distribution of points is then
stretched along the PCA loadings (red) to become the data ellipse that we see on this figure. However, the same distribution of points can be rotated and then stretched along the rotated PCA loadings (magenta) to become the same data ellipse.
[To actually
see that an orthogonal rotation of loadings is a rotation, one needs to look at a PCA biplot; there the vectors/rays corresponding to original variables will simply rotate.]
Let us summarize. After an orthogonal rotation (such as varimax), the "rotated-principal" axes are not orthogonal, and orthogonal projections on them do not make sense. So one should rather drop this whole axes/projections point of view. It would be weird to still call it PCA (which is all about projections with maximal variance etc.).
From FA-style point of view, we simply rotated our (standardized and uncorrelated) latent factors, which is a valid operation. There are no "projections" in FA; instead, latent factors generate the observed variables via loadings. This logic is still preserved. However, we started with principal components, which are not actually factors (as PCA is not the same as FA). So it would be weird to call it FA as well.
Instead of debating whether one "should" rather call it PCA or FA, I would suggest to be meticulous in specifying the exact used procedure: "PCA followed by a varimax rotation".
Postscriptum. It is possible to consider an alternative rotation procedure, where $\mathbf{TT}^\top$ is inserted between $\mathbf{US}$ and $\mathbf V^\top$. This would rotate raw scores and eigenvectors (instead of standardized scores and loadings). The biggest problem with this approach is that after such a "rotation", scores will not be uncorrelated anymore, which is pretty fatal for PCA. One can do it, but it is not how rotations are usually being understood and applied. |
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Kaon femtoscopy in Pb-Pb collisions at $\sqrt{s_{\rm{NN}}}$ = 2.76 TeV
(Elsevier, 2017-12-21)
We present the results of three-dimensional femtoscopic analyses for charged and neutral kaons recorded by ALICE in Pb-Pb collisions at $\sqrt{s_{\rm{NN}}}$ = 2.76 TeV. Femtoscopy is used to measure the space-time ...
J/$\psi$ elliptic flow in Pb-Pb collisions at $\sqrt{s_{\rm NN}}$ = 5.02 TeV
(American Physical Society, 2017-12-15)
We report a precise measurement of the J/$\psi$ elliptic flow in Pb-Pb collisions at $\sqrt{s_{\rm NN}} = 5.02$ TeV with the ALICE detector at the LHC. The J/$\psi$ mesons are reconstructed at mid-rapidity ($|y| < 0.9$) ...
Highlights of experimental results from ALICE
(Elsevier, 2017-11)
Highlights of recent results from the ALICE collaboration are presented. The collision systems investigated are Pb–Pb, p–Pb, and pp, and results from studies of bulk particle production, azimuthal correlations, open and ...
Event activity-dependence of jet production in p-Pb collisions at $\sqrt{s_{\rm NN}} = 5.02$ TeV measured with semi-inclusive hadron+jet correlations by ALICE
(Elsevier, 2017-11)
We report measurement of the semi-inclusive distribution of charged-particle jets recoiling from a high transverse momentum ($p_{\rm T}$) hadron trigger, for p-Pb collisions at $\sqrt{s_{\rm NN}} = 5.02$ TeV, in p-Pb events ...
System-size dependence of the charged-particle pseudorapidity density at $\sqrt {s_{NN}}$ = 5.02 TeV with ALICE
(Elsevier, 2017-11)
We present the charged-particle pseudorapidity density in pp, p–Pb, and Pb–Pb collisions at sNN=5.02 TeV over a broad pseudorapidity range. The distributions are determined using the same experimental apparatus and ...
Photoproduction of heavy vector mesons in ultra-peripheral Pb–Pb collisions
(Elsevier, 2017-11)
Ultra-peripheral Pb-Pb collisions, in which the two nuclei pass close to each other, but at an impact parameter greater than the sum of their radii, provide information about the initial state of nuclei. In particular, ...
Measurement of $J/\psi$ production as a function of event multiplicity in pp collisions at $\sqrt{s} = 13\,\mathrm{TeV}$ with ALICE
(Elsevier, 2017-11)
The availability at the LHC of the largest collision energy in pp collisions allows a significant advance in the measurement of $J/\psi$ production as function of event multiplicity. The interesting relative increase ...
Multiplicity dependence of jet-like two-particle correlations in pp collisions at $\sqrt s$ =7 and 13 TeV with ALICE
(Elsevier, 2017-11)
Two-particle correlations in relative azimuthal angle (Δ ϕ ) and pseudorapidity (Δ η ) have been used to study heavy-ion collision dynamics, including medium-induced jet modification. Further investigations also showed the ...
Electroweak boson production in p–Pb and Pb–Pb collisions at $\sqrt{s_\mathrm{NN}}=5.02$ TeV with ALICE
(Elsevier, 2017-11)
W and Z bosons are massive weakly-interacting particles, insensitive to the strong interaction. They provide therefore a medium-blind probe of the initial state of the heavy-ion collisions. The final results for the W and ...
Investigating the Role of Coherence Effects on Jet Quenching in Pb-Pb Collisions at $\sqrt{s_{NN}} =2.76$ TeV using Jet Substructure
(Elsevier, 2017-11)
We report measurements of two jet shapes, the ratio of 2-Subjettiness to 1-Subjettiness ($\it{\tau_{2}}/\it{\tau_{1}}$) and the opening angle between the two axes of the 2-Subjettiness jet shape, which is obtained by ... |
We say that a multilinear polynomial $P(x_1,\ldots,x_n)$ in $n$ commuting variables over $\mathbb{R}$ has
zero trace if$$ \frac{d}{dt} P(t,\ldots,t) = 0. $$Equivalently,$$ \left(\sum_{i=1}^n \frac{\partial}{\partial x_i}\right) P = 0. $$(Suggestions for a better name are welcome.)
Such functions come up, for example, when considering the Johnson association scheme: every function on $\binom{[n]}{k} = \{(x_1,\ldots,x_n\}) \in \{0,1\}^n : \sum_i x_i = k\}$ can be represented as a zero trace multilinear polynomial of degree at most $k$ (assuming $k \leq n/2$); see Bannai and Ito,
Association schemes, Proposition III.2.7.
For $d \leq n/2$, the dimension of the linear space of all zero trace multilinear polynomials of degree at most $d$ is $\binom{n}{d}$.
Are there known orthogonal bases for the linear span of all zero-trace multilinear polynomials in $n$ variables of degree at most $n/2$?
Orthogonality is with respect to the uniform measure on $\binom{[n]}{k}$, though the orthogonal basis proposed below is orthogonal with respect to
any measure which is permutation-invariant.
Is the following orthogonal basis known?
Here is a conjectured orthogonal basis which can be gleaned from Young's orthogonal representation of $S_n$. Given two sequences $A = a_1,\ldots,a_d$ and $B = b_1,\ldots,b_d$ of distinct numbers in $[n]$, we say that $A < B$ if $A$ and $B$ are disjoint and $a_i < b_i$ for all $i$. We say that a sequence $B$ is a
top set if $B$ is increasing and there exists a sequence $A$ smaller than $B$. It turns out that there are $\binom{n}{d} - \binom{n}{d-1}$ top sets of size $d$, a fact mentioned by Frankl and Graham (this can be proved using Bertrand's ballot problem). For each top set $B$ of size $d$ there is a basis vector$$ \sum_{A < B} \prod_{i=1}^d (x_{a_i} - x_{b_i}). $$For example, for $d=1$ the $n-1$ basis vectors are$$ \sum_{i=1}^{b-1} (x_i - x_b), \quad 2 \leq b \leq n. $$ |
I watched a YouTube movie about Reaction Engines development of the Sabre engine which main feature is cooling intake air. However, I cannot find an overview of the temperature of intake air at different airspeed levels. I am looking for an overview based on speed of sound levels and the increase on intake air temperatures.
Thermodynamics is your friend here. Compression in a ram intake is ideally isentropic (meaning entropy stays constant, so the whole process is reversible). In reality, some viscous losses cannot be avoided but are in the low percentage range in a well designed intake.
The temperature $T$ in isentropic compression from a state 1 to a state 2 can be calculated from the pressure $p$ and the ratio of specific heats $\kappa$: $$T_2 = T_1\cdot\left(\frac{p_2}{p_1}\right)^{\frac{\kappa-1}{\kappa}}$$
$\kappa$ is 1.4 for air, so if the saber engine flies in air 250 K cold (-23.15°C) and a pressure of 500 mbar, and the intake compresses this air by a factor of 10 to 5,000 mbar, the temperature will ideally rise to 483 K (210°C). In reality, this might well be 220°C or 230°C due to viscous heating of the air in the process.
How the temperature ratio depends on flight Mach number $Ma$ (again, ideally) can be seen from this plot, based on this way to calculate the pressure rise: |
Given a field $F$, can you necessarily construct a field extension $E \supset F$ such that $\operatorname{Gal}(E/F) = S_n\,$?
We will prove that there exists a finite Galois extension $K/\mathbb{Q}$ such that $S_n$ = $Gal(K/\mathbb{Q})$ for every integer $n \geq 1$.We will follow mostly van der Waerden's book on algebra.You can also see his proof on Milne's course note on Galois theory.However, Milne refers to his book for a crucial theorem(
Proposition 1 below) whose proof uses multivariate polynomials.Instead, we will use elementary commutative algebra to prove this theorem. Notations
We denote by |S| the number of elements of a finite set S.
Let $K$ be a field. We denote by $K^*$ the multiplicative group of $K$.
Let $\tau$ = $(i_1, ..., i_m)$ be a cycle in $S_n$. The set {$i_1, ..., i_m$} is called the support of $\tau$. Let $\sigma \in S_n$. Let $\sigma$ = $\tau_1\dots\tau_r$, where each $\tau_i$ is a cycle of length $m_i$ and they have mutually disjoint supports. Then we say $\sigma$ is of type [$m_1, ..., m_r$].
Definition 1Let $F$ be a field.Let $f(X)$ be a non-constant polynomial of degree n in $F[X]$.Let $K/F$ be a splitting field of $f(X)$.Suppose $f(X)$ has distinct $n$ roots in $K$.Then $f(X)$ is called separable.Since the splitting fields of $f(X)$ over $F$ are isomorphic to each other,this definition does not depend on a choice of a splitting field of $f(X)$. Definition 2Let $F$ be a finite field.Let $|F| = q$.Let $K/F$ be a finite extension of $F$.Let $\sigma$ be a map: $K \rightarrow K$ defined by $\sigma(x) = x^q$ for each $x \in K$.$\sigma$ is an automorphism of $K/F$.This is called the Frobenius automorphism of $K/F$. Definition 3Let $G$ be a permutation group on a set $X$.Let $G'$ be a permutation group on a set $X'$.Let $f:X \rightarrow X'$ be a bijective map.Let $\lambda:G \rightarrow G'$ be an isomophism.Suppose $f(gx) = \lambda(g).f(x)$ for any $g \in G$ and any $x \in X$.Then G and G' are said to be isomorphic as permutation groups. Lemma 1Let $F$ be a field.Let $f(X)$ be a separable polynomial of degree $n$ in $F[X]$.Let $K/F$ be a splitting field of $f(X)$.Let $G = Gal(K/F)$.Let $S$ be the set of roots of $f(X)$ in K.Then $G$ acts transitively on $S$ if and only if $f(X)$ is irreducible in $F[X]$.
Proof: If $f(X)$ is irreducible, clearly $G$ acts transitively on $S$.
Conversely, suppose $f(X)$ is not irreducible.Let $f(X) = g(X)h(X)$, where $g$ and $h$ are non-constant polynomials in $F[X]$.Let $T$ be the set of roots of $g(X)$ in $K$.Since $G$ acts on $T$ and $S \neq T$, $S$ is not transitive.
QED Lemma 2Let $F$ be a field.Let $f(X)$ be a separable polynomial in F[X].Let $f(X) = f_1(X)...f_r(X)$, where $f_1(X), ..., f_r(X)$ are distinct irreducible polynomials in $F[X]$.Let $K/F$ be a splitting field of $f(X)$.Let $G = Gal(K/F)$.Let $S$ be the set of roots of $f(X)$ in $K$.Let $S_i$ be the set of roots of $f_i(X)$ in $K$ for each $i$.Then $S = \cup S_i$ is a disjoint union and each $S_i$ is a $G$-orbit.
Proof: This follows immediately from Lemma 1.
Lemma 3Let $F$ be a finite field.Let $K/F$ be a finite extension of $F$.Then $K/F$ is a Galois extension and $Gal(K/F)$ is a cyclic group generated by the Frobenius automorphism $\sigma$.
Proof: Let $|F| = q$. Let $n = (K : F)$. Since $|K^*| = q^n - 1$, $x^{q^n - 1} = 1$ for each $x \in K^*$. Hence $x^{q^n} = x$ for each $x \in K$. Hence $\sigma^n = 1$.
Let m be an integer such that $1 \leq m < n$.Since the polynomial $X^{q^m} - X$ has at most $q^m$ roots in $K$, $\sigma^m \neq 1$.Hence $\sigma$ generates a subgroup $G$ of order n of $Aut(K/F)$.Since $n = (K : F)$, $G = Aut(K/F)$.Since $|Aut(K/F)| = n$, $K/F$ is a Galois extension.
QED Lemma 4Let $F$ be a finite field.Let $f(X)$ be an irreducible polynomial of degree $n$ in $F[X]$.Let $K/F$ be a splitting field of $f(X)$.Let $\sigma$ be the Frobenius automorphism of $K/F$.Then $Gal(K/F)$ is a cyclic group of order $n$ generated by $\sigma$.
Proof:Let $\alpha$ be a root of $f(X)$ in $K$.By Lemma 3, $F(\alpha)/F$ is a Galois extension. Hence $K = F(\alpha)/F$
By Lemma 3, $Gal(F(\alpha)/F)$ is a cyclic group of order $n$ generated by $\sigma$. QED Lemma 5Let $F$ be a finite field.Let $f(X)$ be an irreducible polynomial of degree $n$ in $F[X]$.Let $K/F$ be a splitting field of $f(X)$.Let $G = Gal(K/F)$.Let $\sigma$ be the Frobenius automorphism of $K/F$.Let $S$ be the set of roots of $f(X)$.We regard $G$ as a permutation group on $S$.Then $\sigma$ is an $n$-cycle.
Proof: This follows immediately from Lemma 4.
Lemma 6Let $F$ be a finite field.Let $f(X)$ be a separable polynomial in F[X].Let $f(X) = f_1(X)...f_r(X)$, where $f_1(X), ..., f_r(X)$ are distinct irreducible polynomials in $F[X]$.Let $m_i$ = deg $f_i(X)$ for each $i$.Let $K/F$ be a splitting field of $f(X)$.Let $G = Gal(K/F)$.Let $\sigma$ be the Frobenius automorphism of $K/F$.Let $S$ be the set of roots of $f(X)$.We regard $G$ as a permutation group on $S$.Then $\sigma$ is a permutation of type $[m_1, ..., m_r]$.
Proof: This follows immediately from Lemma 2, Lemma 3 and Lemma 5.
Lemma 7$S_n$ is generated by $(k, n)、k = 1、．．．、n - 1$.
Proof:Let $(a, b)$ be a transpose on {$1, ..., n$}.If $a \neq n$ and $b \neq n$, then $(a, b) = (a, n)(b, n)(a, n)$.Since $S_n$ is generated by transposes, we are done.
QED Lemma 8Let $G$ be a transitive permutation group on a finite set $X$.Let $n = |X|$.Suppose $G$ contains a transpose and a $(n-1)$-cycle.Then $G$ is a symmetric group on X.
Proof:Without loss of generality, we can assume that $X$ = {$1, ..., n$} and $G$ containsa cycle $\tau$ = $(1, ..., n-1)$ and transpose $(i, j)$.Since $G$ acts transitively on $X$, there exists $\sigma \in G$ such that $\sigma(j)$ = $n$.Let $k$ = $\sigma(i)$.Then $\sigma(i, j)\sigma^{-1}$ = $(k, n) \in G$.Taking conjugates of $(k, n)$ by powers of $\tau$, we get $(m, n), m = 1, ..., n - 1$. Hence, by Lemma 7, $G = S_n$.
QED Lemma 9Let $F$ be a finite field.Let $n \geq 1$ be an integer.Then there exists an irreducible polynomial of degree $n$ in $F[X]$.
Proof:Let $|F| = q$.Let $K/F$ be a splitting field of the polynomial $X^{q^n} - X$ in $F[X]$.Let $S$ be the set of roots of $X^{q^n} - X$ in $K$.It's easy to see that $S$ is a subfield of $K$ containing $F$.Hence $S = K$.Since $X^{q^n} - X$ is separable, $|S| = q^n$.Hence $(K : F) = n$.Since $K^*$ is a cyclic group, $K^*$ has a generator $\alpha$.Let $f(X)$ be the minimal polynomial of $\alpha$ over $F$.Since $K = F(\alpha)$, the degree of $f(X)$ is $n$.
QED Lemma 10Let $f(X) \in \mathbb{Z}[X]$ be a monic polynomial.Let $p$ be a prime number.Suppose $f(X)$ mod $p$ is separable in $\mathbb{Z}/p\mathbb{Z}[X]$.Then $f(X)$ is separable in $\mathbb{Q}$.
Proof:Suppose $f(X)$ is not separable in $\mathbb{Q}$.Since $\mathbb{Q}$ is perfect, there exists a monic irreducible $g(X) \in \mathbb{Z}[X]$ such that $f(X)$ is divisible by $g(X)^2$. Then $f(X)$ (mod $p$) is divisible by $g(X)^2$ (mod $p$). This is a contradiction.
QED Proposition 1Let $A$ be an integrally closed domain and let $P$ be a prime ideal of $A$.Let $K$ be the field of fractions of A.Let $\tilde{K}$ be the field of fractions of $A/P$.Let $f(X) ∈ A[X]$ be a monic polynomial without multiple roots.Let $\tilde{f}(X) \in (A/P)[X]$ be the reduction of $f(X)$ mod $P$.Suppose $\tilde{f}(X)$ is also wihout multiple roots.Let $L$ be the splitting field of $f(X)$ over $K$.Let $G$ be the Galois group of $L/K$.Let S be the set of roots of $f(X)$ in $L$.We regard $G$ as a permutation group on $S$.Let $\tilde{L}$ be the splitting field of $\tilde{f}(X)$ over $\tilde{K}$.Let $\tilde{G}$ be the Galois group of $\tilde{L}/\tilde{K}$.Let $\tilde{S}$ be the set of roots of $\tilde{f}(X)$ in $\tilde{L}$.We regard $\tilde{G}$ as a permutation group on $\tilde{S}$.
Then there exists a subgroup $H$ of $G$ such that $H$ and $\tilde{G}$ are isomorphic as permutation groups.
Proof: See my answer here.
CorollaryLet $f(X) \in \mathbb{Z}[X]$ be a monic polynomial of degree $m$.Let p be a prime number.Suppose $f(X)$ mod $p$ is separable in $\mathbb{Z}/p\mathbb{Z}[X]$.Suppose $f \equiv f_1...f_r$ (mod $p$), where each $f_i$ is monic and irreducible of degree $m_i$ in $\mathbb{Z}/p\mathbb{Z}[X]$.Let $K/\mathbb{Q}$ be a splitting field of $f(X)$.Let $M$ be the set of roots of $f(X)$.$G = Gal(K/\mathbb{Q})$ can be regarded as a permutation group on $M$.Then $G$ contains an element of type [$m_1, ..., m_r$].
Proof:By Lemma 10, $f(X)$ is separable in $\mathbb{Q}[X]$.Let $F_p$ = $\mathbb{Z}/p\mathbb{Z}[X]$.Let $\tilde{f}(X) \in F_p[X]$ be the reduction of $f(X)$ mod $p$.Let $\tilde{K}/F_p$ be a splitting field of $\tilde{f}(X)$.Let $\tilde{G}$ be the Galois group of $\tilde{K}/F_p$.Let $\tau$ be the Frobenius automorphism of $\tilde{K}/F_p$.Let $\tilde{M}$ be the set of roots of $\tilde{f}(X)$.We regard $\tilde{G}$ as a permutation group on $\tilde{M}$.By Lemma 6, $\tau$ is a permutation of type $[m_1, ..., m_r]$.Hence the assertion follows by Proposition 1.
QED TheoremThere exists a finite Galois extension $K/\mathbb{Q}$ such that $S_n$ = $Gal(K/\mathbb{Q})$ for every integer $n \geq 1$.
Proof(van der Waerden): By Lemma 9, we can find the following irreducible polynomials.
Let $f_1$ be a monic irreducible polynomial of degree $n$ in $\mathbb{Z}/2\mathbb{Z}[X]$.
Let $g_0$ be a monic polynomial of degree 1 in $\mathbb{Z}/3\mathbb{Z}[X]$. Let $g_1$ be a monic irreducible polynomial of degree $n - 1$ in $\mathbb{Z}/3\mathbb{Z}[X]$. Let $f_2 = g_0g_1$. If $n - 1 = 1$, we choose $g_1$ such that $g_0 \ne g_1$. Hence $f_2$ is separable.
Let $h_0$ be a monic irreducible polynomial of degree 2 in $\mathbb{Z}/5\mathbb{Z}[X]$. If $n - 2$ is odd, Let $h_1$ be a monic irreducible polynomial of degree $n - 2$ in $\mathbb{Z}/5\mathbb{Z}[X]$. Let $f_3 = h_0h_1$. Since $h_0 \neq h_1$, $f_3$ is separable.
If $n - 2$ is even, $n - 2 = 1 + a$ for some odd integer $a$. Let $h_1$ and $h_2$ be monic irreducible polynomials of degree $1$ and $a$ respectively in $\mathbb{Z}/5\mathbb{Z}[X]$. Let $f_3 = h_0h_1h_2$. If a = 1, we choose $h_2$ such that $h_1 \ne h_2$. Hence $f_3$ is separable.
Let $f = -15f_1 + 10f_2 + 6f_3$. Since each of $f_1, f_2, f_3$ is a monic of degree $n$, f is a monic of degree $n$.
Then,
$f \equiv f_1$ (mod 2)
$f \equiv f_2$ (mod 3)
$f \equiv f_3$ (mod 5)
Since $f \equiv f_1$ (mod 2), $f$ is irreducible. Let $K/\mathbb{Q}$ be the splitting field of $f$. Let $G = Gal(K/\mathbb{Q})$. Let $M$ be the set of roots in $K$. We regard $G$ as a permutation group on $M$.
Since $f$ is irreducible, $G$ acts transitively on $M$.
Since $f \equiv f_2$ (mod 3), $G$ contains a $(n-1)$-cycle by Corollary of Proposition 1.
Similarly, since $f \equiv f_3$ (mod 5), $G$ contains a permutation $\tau$ of type [$2, a$] or [$2, 1, a$], where $a$ is odd. Then $\tau^a$ is a transpose. Hence $G$ contains a transpose.
Hence, $G$ is a symmetric group on $M$ by Lemma 8.
QED
There is no general solution to your question. It depends on the base field $F$. I will show that your problem is solved affirmatively when $F$ is a field of rational functions over any field.
Let $k$ be a field. Let $K$ = $k(X_1, ..., X_n)$ be the field of rational functions over k. Each element of $S_n$ acts on $K$ as an $k$-automorphism. Hence $S_n$ can be regarded as a subgroup of Aut($K/k$). Let $F$ be the fixed field by $S_n$. By Artin's theorem, $K/F$ is a Galois extension and its Galois group is $S_n$. Let $s_1, ..., s_n$ be the elementary symmetric functions of $X_1, ..., X_n$. Then $F = k(s_1, ..., s_n)$. It can be easily proved that $s_1, ..., s_n$ are algebraically independent over $k$. Hence $F$ can be regarded as the field of rational functions of $n$ variables.
Therefore we get the following proposition.
PropositionLet $k$ be a field.Let $F = k(x_1, ..., x_n)$ be the field of rational functoins of $n$ variables over $k$.There exists a Galois extension $E$ of $F$ such that $S_n = Gal(E/F)$. |
For which values of $z$ does $$\sum_{n=1}^\infty \frac{\tan(nz)}{n^2}$$ converge? For which values of $z$ is the limiting function analytic?
One can show, as in this answer, that $$\left|\frac{e^{inz}-e^{-inz}}{e^{inz}+e^{-inz}}\right|$$ is bounded as $n\to \infty$, so long as $\text{Im}(z)\neq 0$. But the article above really does not discuss the case $\text{Im}(z)=0$, although it thinks it does. It doesn't deal with the poles at $\frac{(2k+1)\pi}{2}$, which can make some of the terms of the series undefined.
If $\text{Im}(z)=0$, obviously the estimate $$\left| \frac{e^{inz}-e^{-inz}}{e^{inz}+e^{-inz}} \right|\leq \frac{1+e^{2ny}}{|1-e^{2ny}|} $$
does not work. (Here $y=\text{Im}(z)$.) For $x\in \mathbb{R}$ of the form $j^2\frac{(2k+1)\pi}{2}$, there will be undefined terms.
Suppose there are no undefined terms. What can we say then about convergence? And in what way can we describe these singularities of the limiting function, corresponding to $x$ with undefined terms? Perhaps these points are not even isolated... |
rencsee wrote:
What is the remainder when the positive integer x is divided by 9?
(1) x + 34 is a multiple of 18 (2) x is a multiple of 11
\(\left\{ \matrix{
\,\,1\, \le \,x = 9Q + r\,\,,\,\,Q \ge 0\,\,{\mathop{\rm int}} \hfill \cr
\,\,0 \le r \le 8\,\,{\mathop{\rm int}} \,\,\,\,\, \Rightarrow \,\,\, - 2\,\, \le r - 2\,\, \le \,\,6\,\,\,\left( * \right) \hfill \cr} \right.\)
\(? = r\)
\(\left( 1 \right)\,\,\,\frac{{x + 34}}{{18}} = \operatorname{int} \,\,\,\, \Rightarrow \,\,\,\,\operatorname{int} = \,\frac{{9Q + r + 36 - 2}}{{18}} = \frac{{9Q + \left( {r - 2} \right)}}{{18}}\,\,\,\,\mathop \Rightarrow \limits^{\left( * \right)} \,\,\,\,\left\{ \begin{gathered}
\,Q\,\,{\text{even}} \hfill \\
\,r - 2 = 0\,\,\,\,\,\, \Rightarrow \,\,\,\,\,? = 2 \hfill \\
\end{gathered} \right.\)
\(\left( 2 \right)\,\,\,\frac{x}{{11}} = \operatorname{int} \,\,\,\left\{ \begin{gathered}
\,{\text{Take}}\,\,x = 11\,\,\,\,\left[ {Q = 1,\,\,r = 2} \right]\,\,\,\,\, \Rightarrow \,\,\,\,{\text{?}}\,\,{\text{ = }}\,\,{\text{2}}\,\, \hfill \\
\,{\text{Take}}\,\,x = 22\,\,\,\,\left[ {Q = 2,\,\,r = 4} \right]\,\,\,\,\, \Rightarrow \,\,\,\,{\text{?}}\,\,{\text{ = }}\,\,{\text{4}}\,\, \hfill \\
\end{gathered} \right.\,\,\)
This solution follows the notations and rationale taught in the GMATH method.
Regards,
Fabio.
_________________
Fabio Skilnik :: GMATH
method creator (Math for the GMAT)
Our high-level "quant" preparation starts here: https://gmath.net |
Let $\mathcal{C}$ be a small category and $\Sigma$ a collection of morphisms in $\mathcal{C}$. Denote by $F_\Sigma:\mathcal{C} \to \mathcal{C}[\Sigma^{-1}]$ the usual quotient functor from $\mathcal{C}$ to its localization about $\Sigma$.
Are there conditions on $\Sigma$ which guarantee that $F_\Sigma$ induces a homotopy equivalence $B\mathcal{C} \sim B\mathcal{C}[\Sigma^{-1}]$ of classifying spaces?
For example: if $\mathcal{C}$ consists of two objects with a single arrow from one to the other, then localization about that single arrow preserves homotopy type of classifying spaces: everything is contractible before and after localization. On the other hand, see this paper for a counter-example to the conjecture that the group completion of a monoid has the same classifying space as that monoid. Clearly, we can't just shamelessly start inverting arrows all over the place without destroying homotopy type.
One always has Quillen's Theorem A: if the under categories $F_\Sigma \downarrow c$ are all contractible, then $BF_\Sigma$ is a homotopy-equivalence of classifying spaces. So, one possible answer would highlight those conditions on $\Sigma$ which magically give contractible over/under categories. Is there a known result that does the trick? |
Existing algorithms for solving ODEs handle functions $\frac{dy}{dt} = f(y, t)$, where $y \in \mathbb R^n$. But in many physical systems, the differential equation is autonomous, so $\frac{dy}{dt} = f(y)$, $y \in \mathbb R^n$, with the $t$ left out. With this simplifying assumption, what improvements can be seen in existing numerical methods? For example, if $n=1$, the problem turns into $t = \int \frac{dy}{f(y)}$ and we turn to a wholly different class of algorithms for integrating one-dimensional integrals. For $n>1$, the maximum possible improvement is reducing the dimension of $y$ by 1, because the time-dependent case can be simulated by appending $t$ to $y$, changing the domain of $y$ from $\mathbb R^n$ to $\mathbb R^{n+1}$.
I would say one significant improvement is that in the scope of time-stepping approaches, where you propagate $y_n \rightarrow y_{n+1}=\mathcal{U}(y_n)$ using a solution map $\mathcal{U}$, you can determine the propagator (or at least parts of it) once and then re-use it at every time step.
For example, in the linear case you would have $\partial_t y = A y$, where $A$ is a matrix. The solution operator $\mathcal{U}(y) = \exp(A \Delta t)y$ consists mainly of a matrix exponential. For autonomous systems, this costly matrix exponential evaluation is required only once for the complete propagation -- in contrast to a time-dependent system, where you have to perform this evaluation at every time step.
For nonlinear systems it is not that easy, but depending on the algorithm certain costly evaluations can be re-used. |
There's several issues here.
(1) Is a given incidence correspondence actually a closed variety?
(2) What are explicit equations for the correspondence in the product of the relevant spaces?
(3) What are geometric properties of the incidence correspondence?
Most often, questions (1) and (3) are studied and little attention is paid to (2).
For (1), things can be generalized a fair bit. For instance, suppose $X$ is a variety, $H$ is an ample divisor, and $P$, $Q$ are two Hilbert polynomials with respect to the ample divisor $X$. Then there are (projective) Hilbert schemes $Hilb_P(X)$ and $Hilb_Q(X)$ parameterizing closed subschemes of $X$ with Hilbert polynomials $P$ and $Q$. Then there are a couple different natural incidence correspondences, for example
$\{(Z,Z'):Z\subset Z'\} \subset Hilb_P(X)\times Hilb_Q(X)$
$\{(Z,Z'):Z\cap Z'\neq \emptyset\}\subset Hilb_P(X)\times Hilb_Q(X),$
and it is easy to verify that these conditions are closed (although the correct scheme structure may be less clear). One can instead restrict attention to a closed subvariety of the Hilbert scheme (so as to not use all the components of the Hilbert scheme, for instance, in case the geometric objects you care about are not entirely determined by their Hilbert polynomials). It is also easy to generalize to cases with more factors. These types of constructions mean that arguments for the closedness of incidence correspondences are almost never written down, as anything reasonable that you can write down will be closed so long as the families of objects under consideration are themselves closed.
In practice, (2) is rarely of any theoretical interest, unless these are very special varieties. Perhaps there is a large algebraic group acting and the ideal can be studied via representation theory, or perhaps the variety has very small dimension or is otherwise very simple, in which case some information might be learned from the ideal.
Regarding (3), geometric properties of an incidence correspondence are only very rarely (roughly in the same cases as discussed for (2)) determined by studying the defining equations. Most often, the reason we study an incidence correspondence $\Sigma \subset X\times Y$ is because we have some question about one of the projection maps, say $\Sigma \to X$; for instance, we may wonder if it is dominant. We then ideally answer this by studying the other projection, which ideally is easier to understand. Likewise, properties like dimension and irreducibility are hopefully easily understood by studying the projections, and smoothness can sometimes be analyzed as well (although smoothness is often not so important in basic applications). |
№ 9
All Issues Petrenko S. M.
Ukr. Mat. Zh. - 2016. - 68, № 1. - pp. 64-77
The notion of conflict system is introduced in terms of couples of probability measures. We construct several models of conflict systems and show that every trajectory with initial state given by a couple of measures $\mu, \nu$ converges to an equilibrium state specified by the normalized components $\mu_+, \nu_+$ of the classical Hahn – Jordan decomposition of the signed measure $\omega = \mu - \nu$.
Ukr. Mat. Zh. - 2011. - 63, № 3. - pp. 369-384
A continuous infinite systems of point particles with strong superstable interaction are considered in the framework of classical statistical mechanics. The family of approximated correlation functions is determined in such a way that they take into account only those configurations of particles in the space $\mathbb{R}^d$ which, for a given partition of $\mathbb{R}^d$ into nonintersecting hypercubes with a volume $a^d$, contain no more than one particle in every cube. We prove that so defined approximations of correlation functions pointwise converge to the proper correlation functions of the initial system if the parameter of approximation a tends to zero for any positive values of an inverse temperature $\beta$ and a fugacity $z$. This result is obtained for both two-body and many-body interaction potentials. |
The Rudin-Shapiro sequence (also known as the Golay-Rudin-Shapiro sequence) is defined as follows.
Let $a_n = \sum \epsilon_i\epsilon_{i+1}$ where $\epsilon_1,\epsilon_2,\dots$ are the digits in the binary expansion of $n$. $WS(n)$, the $n$th term of the Rudin Shapiro sequence is defined by $$WS(n)=(-1)^{a_n}.$$
Question:
Prove that $$\sum_{i=0}^n WS(i) \mu (i) = o(n).$$
Here, $\mu(n)$ is the Möbius function.
Motivation
This question continues a one-year old question walsh-fourier-transform-of-the-mobius-function . The two parts of the old question on "Mobius randomness" was settled by Green and by Bourgain, respectively. This question represent the simplest case, which is quite important in its own right, where some new idea/method may be needed.
Motivation (2)
Under the translation 0 --> 1, 1 --> -1, the "Walsh-Fourier" functions can be considered as (all) linear functions over Z/2Z. It tuned out that proving Mobius randomness for a few of them suffices to deduce Mobious randomness for $AC^0$ functions. This was the second part of our old question that was proved by Green. Bourgain showed Mobius randomness for all Walsh functions (namely all linear functions over Z/2Z.) What about low degree polynomials instead of linear polynomials? The Rudin-Shapiro sequence represent a very simple example of quadratic polynomial.
If we can extend the results to polynomials over Z/2Z of degree at most polylog (n) this will imply by a result of Razborov Mobius randomness for $AC^0(2)$ circuits. (This is interesting also under GRH). |
trying to determine if the series is conditionally convergent or divergent. $$\sum_{n = 1}^\infty \frac{2^{n^{2}}}{n!}$$ with n! i tried the ratio test on the series $$\frac{2^{(n+1)^{2}}}{(n+1)!} * \frac{n!}{2^{n^{2}}} = \frac{2^{2n+1}}{(n+1)} $$ which is > 1 as $n\to \infty$ and is overall divergent ? not sure if I am on the right track.
Well, applying the ratio test:
$$\lim_{\text{n}\to\infty}\space\left|\frac{\frac{2^{\left(\text{n}+1\right)^2}}{\left(\text{n}+1\right)!}}{\frac{2^{\text{n}^2}}{\text{n}!}}\right|=\lim_{\text{n}\to\infty}\space\left|\frac{2^{2\text{n}+1}}{\text{n}+1}\right|\space\to\space\infty\tag1$$
Note that ${2^{n^2} \over n!} = { (2^n)^n \over n!} \ge { (2^n)^n \over n^n}= ({2^n \over n})^n \ge 1$. Hence $\sum_n {2^{n^2} \over n!} \ge N$ for all $N$. |
Just for fun, here's a purely abstract way of seeing neither of these can happen:
The functors you wrote down both preserve weak equivalences, so are morphisms of relative categories, and induce adjunctions of the associated $\left(\infty,1\right)$-categories, which are $\infty$-groupoids and $1$-groupoids (which is actually a $(2,1)$-category) respectively. The adjunction is precisely the one exhibiting $1$-groupoids as a reflective subcategory of $\infty$-groupoids. The full and faithful inclusion is modeled by $N$. The $\left(\infty,1\right)$-category of $\infty$-groupoids is freely generated by the (any) contractible $\infty$-groupoid (i.e. the one point space), (i.e. it's the free colimit cocompletion of the terminal category). Since the $\left(2,1\right)$-category of groupoids is cocomplete, and contains the terminal groupoid, if the inclusion preserved colimits, it would have to be essentially surjective:
If $X$ is any $\infty$-groupoid, you can write $X$ as the colimit of the terminal functor $X \to \infty\mbox{-}\mathbf{Gpd}$ (the constant functor with value the point) (this is basically just the Yoneda Lemma for $\infty$-categories), and this functor (obviously) factors through the inclusion $\mathbf{Gpd} \hookrightarrow \infty\mbox{-}\mathbf{Gpd}$.
If the left adjoint $\tau_1$ of this inclusion (which is modeled by $\Pi_1$) preserved even
finite limits, then this would exhibit the $(2,1)$-category of groupoids as a left-exact localization of $\infty\mbox{-}\mathbf{Gpd}=\mathbf{Psh}_\infty\left(*\right),$ i.e. it would exhibit $\mathbf{Gpd}$ as an $\infty$-topos. This means there would exist a unique Grothendieck topology $J$ on the terminal category such that $\mathbf{Gpd}$ sat somewhere between $\mathbf{Sh}_\infty\left(*,J\right)$ and its hypercompletion $\mathbf{HSh}_\infty\left(*,J\right).$ But, there is a unique Grothendieck topology on the terminal category (the trivial one!), so we'd have $$\mathbf{HSh}_\infty\left(*,J\right)=\mathbf{Sh}_\infty\left(*,J\right)=\mathbf{Psh}_\infty\left(*\right).$$ So we'd have to have $\mathbf{Gpd}\simeq \mathbf{Psh}_\infty\left(*\right)=\infty\mbox{-}\mathbf{Gpd}$ which is clearly nonsense. |
User:Jan A. Sanders/An introduction to Lie algebra cohomology/Lecture 2 Central extensions
The definition of projective is important when \(R\) is not \(\mathbb{R}\ ,\) \(\mathbb{C}\) or in general a field of characteristic zero, or when the dimension of the Lie algebra is not finite, otherwise it can be skipped.
definition - projective
A module \(\mathfrak{g}\) is
projective if for every surjective morphism of modules \(\alpha:\mathfrak{k}\rightarrow\mathfrak{g}\rightarrow 0\) and every morphism \(\beta:\mathfrak{h}\rightarrow\mathfrak{g}\ ,\) there exists a morphism \(\gamma:\mathfrak{h}\rightarrow\mathfrak{k}\) such that \(\beta=\alpha\gamma\ .\)
In particular, if \(\mathfrak{g}\) is projective and \(\alpha\) surjective, there exists a \(\gamma:\mathfrak{g}\rightarrow\mathfrak{k}\) such that \(id_{\mathfrak{g}}=\alpha\gamma\ ,\)that is, \(\gamma\) is a
section of \(\alpha\ .\)
When \(\mathfrak{g}\) is
free, that is, \(\mathfrak{g}\) has an \(R\)-basis \(\{g_\iota\}_{\iota\in I}\ ,\)one can find for each \(g_\iota\) a preimage \(k_\iota\) under \(\alpha\ .\)
If \(\beta(h)=\sum_{\iota\in I} b_\iota g_\iota, b_\iota\in R\ ,\) (finite sum, since \(\{g_\iota\}_{\iota\in I}\) is a basis) one defines \(\gamma(h)=\sum_{\iota\in I} b_\iota k_\iota\ .\)
When \(\mathrm{rank}_R \mathfrak{g}=\#I<\infty\) this can be done explicitly.
If \(R\) is a field with characteristic zero, it suffices to require \(\mathrm{dim}_R \mathfrak{g}<\infty\) in order to conclude to the existence of sections.
Consider the exact sequence of Lie algebras with Lie algebra morphisms
\[ \iota \qquad \quad \ p\]
\[\tag{1} 0 \quad \rightarrow \quad \mathfrak{a} \quad \rightarrow \quad \mathfrak{k} \quad \rightarrow \quad \mathfrak{g} \quad \rightarrow 0 \ .\]
problem statement
If \(\mathfrak{g}\) is projective as an \(R\)-module, is there a section of \(p\) which is also a Lie algebra morphism, that is, is it also projective as a Lie algebra?
To answer this question, first assume \(\mathfrak{a}\) to be abelian, that is, \([x,y]=0\) for all \(x,y\in\mathfrak{a}\ .\)
Let \(\sigma_1\) be a section, but not necessarily a Lie algebra morphism (Exists, since \(\mathfrak{g}\) is projective).
Define \(\sigma_2\in C_{\wedge}^2(\mathfrak{g},\mathfrak{k})\ ,\) (antisymmetric 2-forms) by \[\tag{2} \sigma_2(x,y)=[\sigma_1(x),\sigma_1(y)]_{\mathfrak{k}}-\sigma_1([x,y]_{\mathfrak{g}}).\]
One has \[\tag{3} p \sigma_2(x,y)=p[\sigma_1(x),\sigma_1(y)]_{\mathfrak{k}}-p\sigma_1([x,y]_{\mathfrak{g}})=[p\sigma_1(x),p\sigma_1(y)]_{\mathfrak{g}}-[x,y]_{\mathfrak{g}}=0. \]
Since \(\sigma_2(x,y)\in\ker p=\mathrm{im\ }\iota\ ,\) one can define \(a_2\in C_{\wedge}^2(\mathfrak{g},\mathfrak{a})\) by \[\tag{4} \iota(a_2(x,y))=\sigma_2(x,y). \]
Now define \(d_1:\mathfrak{g}\rightarrow End(\mathfrak{a})\) by \[\tag{5} \iota(d_1(x)a)=[\sigma_1(x),\iota(a)]_{\mathfrak{k}}\]
This is well-defined since, for instance, \(p[\sigma_1(x),\iota(a)]_{\mathfrak{k}}=[p\sigma_1(x),p\iota(a)]_{\mathfrak{g}}=0\ .\)
Moreover, \(d_1\) is a representation, since \[\iota(d_1([x,y]_{\mathfrak{g}})a) = [\sigma_1([x,y]),\iota(a)]_{\mathfrak{k}}\ :\]
\[ = -[\sigma_2(x,y),\iota(a)]_{\mathfrak{k}} +[[\sigma_1(x),\sigma_1(y)]_{\mathfrak{k}},\iota(a)]_{\mathfrak{k}}\ :\] \[ = -[\iota a_2(x,y),\iota(a)]_{\mathfrak{k}} +[\sigma_1(x),[\sigma_1(y),\iota(a)]_{\mathfrak{k}}]_{\mathfrak{k}}-[\sigma_1(y),[\sigma_1(x),\iota(a)]_{\mathfrak{k}}]_{\mathfrak{k}}\ :\] \[ =- \iota[a_2(x,y),a]_{\mathfrak{a}}+[\sigma_1(x),\iota(d_1(y)a)]_{\mathfrak{k}}-[\sigma_1(y),\iota(d_1(x)a)]_{\mathfrak{k}}\ :\] \[ = \iota(d_1(x)d^{(0)}(y)a) -\iota(d_1(y)d_1(x)a)\]
or \(d_1([x,y]_{\mathfrak{g}})=[d_1(x),d_1(y)]_{\mathfrak{k}}\ .\)
In the sequel we drop the subscript of the bracket.
Suppose now there exists a \(\delta\sigma_1\in C^1(\mathfrak{g},\mathfrak{k})\) such that \(\sigma_1+\delta\sigma_1\) is a Lie algebra homomorphism and a section of \(p\ .\) Then \[ x=p(\sigma_1+\delta\sigma_1)(x)=p\sigma_1(x)+p \delta\sigma_1(x)=x+p \delta\sigma_1(x), \] implying that \(\delta\sigma_1 (x)\in\ker(p)=\mathrm{im\ }\iota\ .\) Define \(a_1\in C^1(\mathfrak{g},\mathfrak{a})\) by \(\iota a_1(x)=\delta\sigma_1 (x)\ .\) Then (by assumption!) \[ 0 =(\sigma_1+\delta\sigma_1)([x,y])-[(\sigma_1+\delta\sigma_1)(x),(\sigma_1+\delta\sigma_1)(y)]\ :\] \[ =\sigma_1([x,y])+\delta\sigma_1([x,y])-[\sigma_1(x),\sigma_1(y)]-[\delta\sigma_1(x),\sigma_1(y)]-[\sigma_1(x),\delta\sigma_1(y)]-[\delta\sigma_1(x),\delta\sigma_1(y)]\ :\]
\[ = -\sigma_2(x,y) +\delta\sigma_1([x,y])-[\iota a_1(x),\sigma_1(y)]-[\sigma_1(x),\iota a_1(y)]-[\iota a_1(x),\iota a_1(y)]\ :\] \[ = -\iota(a_2(x,y)+(d_1(x)a_1(y)-d_1(y)a_1(x)-a_1([x,y])-[a_1(x),a_1(y)])\ :\] \[= -\iota((a_2+d^1 a_1)(x,y)).\] This implies that the existence of such a \(\delta\sigma_1\) is equivalent to \(a_2\in\mathrm{im\ }d^1\ .\)
On the other hand, if it would turn out that \(d^2 a_2\neq 0\ ,\) this would be a definite obstruction of the existence of such a \(\delta\sigma_1\ .\)
We can, however, rule out the last possibility. \[ \iota d^2 a_2(x,y,z)= \iota d_1(x)a_2(y,z) -\iota d_1(y)a_2(x,z) +\iota d_1(z)a_2(x,y)-\iota a_2([x,y],z)-\iota a_2(y,[x,z])+\iota a_2(x,[y,z])\ :\]
\[=-[\sigma_1(x),\sigma_2(y,z)]-[\sigma_1(y),\sigma_2(x,z)]-[\sigma_2(x,y),\sigma_1(z)]-\sigma_2([x,y],z)-\sigma_2(y,[x,z])+\sigma_2(x,[y,z])\ :\] \[=[\sigma(x),\sigma([y,z])-[\sigma(y),\sigma(z)]]-[\sigma(y),\sigma([x,z])-[\sigma(x),\sigma(z)]]-[\sigma([x,y])-[\sigma(x),\sigma(y)],\sigma(z)]\ :\] \[ -\sigma_1([[x,y],z])+[\sigma_1([x,y]),\sigma_1(z)]-\sigma_1([y,[x,z]])+[\sigma_1(y),\sigma_1([x,z]]+\sigma_1([x,[y,z]])-[\sigma_1(x),\sigma_1([y,z])]\ :\] \[= -[\sigma(x),[\sigma(y),\sigma(z)]]+[\sigma(y),[\sigma(x),\sigma(z)]] +[[\sigma(x),\sigma(y)],\sigma(z)] -\sigma_1([[x,y],z]+[y,[x,z]]-[x,[y,z]]) \ :\] \[=0.\]
It follows that \(d^2 a_2=0\ .\) |
The state of a qubit, or of a collection of qubits, is given by a state vector. However, most people would say that the
output of a quantum computer is in fact the result of a measurement, which transforms the state into one which indicates a single outcome.
There are many ways to approach this, but the simplest one to describe mathematically is
complete measurement in the standard basis. In this case, you transform a quantum state on $n$ qubits into an $n$-bit string.
Suppose that you have a state $|\psi\rangle \in \mathbb C^{2^n}$ on $n$ qubits, given by $$ |\psi\rangle \;=\; \sum_{x \in \{0,1\}^n} \!u_x \;|x_1 x_2 \cdots x_n \rangle \;, \quad \text{such that } \sum\limits_{x\in\{0,1\}^n} \!|u_x|^2 = 1 .$$
In order to get a classical output from a quantum computer, one thing you can do is to perform a complete measurement in the standard basis. What this means is that we collapse the state to some single output vector $y \in \{0,1\}^n$. We get different bit-strings $y$ with different probabilities, governed by the formula$$ \Pr\nolimits_\psi(y) \;=\; \Bigl| \langle y | \psi \rangle \Bigr|^2\;=\; \left| \sum_{x \in \{0,1\}^n} \!u_x \; \langle y_1 y_2 \cdots y_n | x_1 x_2 \cdots x_n \rangle \right|^2\;=\; \left| u_y \right|^2 \;.$$The outcome is then just a random output string $y \in \{0,1\}^n$, where the probabilities are given by the non-negative real numbers $|u_y|^2$. Afterwards, you can compute with the measurement result $y \in \{0,1\}^n$ however you like, with a classical computer.
It's possible to describe performing a measurement only on a single bit, or with respect to bases other than the standard basis, but that's not really necessary to answer your question; measuring all of the qubits is certainly something you can do, and it's enough to get classical outputs from quantum computations. |
Yes, and no. There is a "categorically comprehensive" reason for this trace map to exist, but not necessarily to construct it. And to prove that this reason is valid, one does require categorical machinery. The elevator speech answer is:
THH is an algebra in some symmetric monoidal category. K is the unit in this symmetric monoidal category, so THH receives a unique algebra map from K theory.
(If you are looking for a less theoretical reason, and want some explicit constructions, you might take a look at Kantorovitz-Miller, "An explicit description of the Dennis trace map.")
Everything I write below, I learned from Blumberg-Gepner-Tabuada, "Uniqueness of the multiplicative cyclotomic trace."
First, we note that both THH and K define functors
$\infty Cat^{perf} \to Spectra$.
On the righthand side is the category of spectra. (Take any model you like, so long as it's not the
homotopy category of the model. You can take Lurie's oo-categorical model, or symmetric spectra if you like.)
On the lefthand side is the category of perfect stable oo-categories. Roughly, these are the categories that look like modules over some ring spectrum. A different way you might describe this category is as follows: The category of spectrally enriched categories, localized with respect to Morita equivalence.
Note that both categories--$\infty Cat^{perf}$ and $Spectra$--have a symmetric monoidal structure. The latter has the usual smash product, while the former has the tensor product of stable $\infty$-categories. This is given by the cocompletion of the following naive tensor product: Given two categories $A$ and $B$, the objects of $A \otimes^{naive} B$ are pairs of objects $(a,b)$, and the hom spectrum between $(a,b)$ and $(a',b')$ is given by $hom(a,a') \wedge hom(b,b')$.
Moreover, we note that both THH and K satisfy the following properties:
They are lax monoidal. (In fact, THH is symmetric monoidal.) This means that we have specified natural maps $K(A) \otimes K(B) \to K(A \otimes B)$, but these need not be equivalences.
They are localizing: If we have a short exact sequence of categories $A \to B \to C$, we have a cofibration sequence of spectra $K(A) \to K(B) \to K(C)$, and likewise for THH. The proof of this for THH can be found in Blumberg-Mandell ("Localization theorems in topological Hochschild homology and topological cyclic homology").
Now, consider the category of all functors $\infty Cat^{perf} \to Spectra$ satisfying (2). One can construct a symmetric monoidal structure on this category. And it turns out that any functor further satisfying (1) can be made into an $E_\infty$ algebra in this category, and that K theory is in fact the unit! Since THH satsifies (1) and (2), the corresponding algebra for THH receives a unique algebra map from K theory.
When $A$ is an $E_\infty$ ring, then $THH(A)$ is an $E_\infty$ ring as well; by the algebra map from $K$ to $THH$, one obtains an $E_\infty$ ring map $K(A) \to THH(A)$.
More generally, if $A$ is an $E_n$-algebra, then $K(A) = K(AMod)$ is an $E_{n-1}$ ring. This is because the category of $A$-modules can be given an $E_{n-1}$-structure, and $K$ theory is lax monoidal. Moreover, you can also prove that $THH(A)$ has an $E_{n-1}$ structure as well (you can see this also using factorization homology, for instance). The fact that there is an algebra map $K \to THH$ implies that one also obtains an $E_{n-1}$-algebra map $K(A) \to THH(A)$. |
For the heat equation
\begin{equation} u_t(t,x) = \nu u_{xx}(t,x) \end{equation}
for $x \in [0,1]$ with boundary conditions $u(t,0) = u(t,1) = 0$ and initial value $u(0,x) = u_0(x)$ it is easy to show that the "energy" defined as
\begin{equation} E(t) = \frac{1}{2} \int_{0}^{1} u^2(t,x)~dx \end{equation}
decays over time, that is $E(t) \leq E(0)$. I wonder if there is any physical interpretation of the quantity $E$?
In terms of units, $u$ is temperature in Kelvin while the thermal diffusivity $\nu$ has units $m^2/s$ and is composed from \begin{equation} \nu = \frac{k}{\rho c_p} \end{equation} where $k$ is the thermal conductivity in $W/(m*K)$, $\rho$ the density in $kg/m^3$ and $c_p$ the specific heat capacity in $J/(kg*K)$.
I figured out that $u$ is related to the internal energy in Joule per volume via \begin{equation} I = c_p \rho u. \end{equation}
But what is the interpretation of $E$, which is related to $u^2$? |
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Now showing items 1-10 of 167
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 ...
Kaon femtoscopy in Pb-Pb collisions at $\sqrt{s_{\rm{NN}}}$ = 2.76 TeV
(Elsevier, 2017-12-21)
We present the results of three-dimensional femtoscopic analyses for charged and neutral kaons recorded by ALICE in Pb-Pb collisions at $\sqrt{s_{\rm{NN}}}$ = 2.76 TeV. Femtoscopy is used to measure the space-time ...
Anomalous evolution of the near-side jet peak shape in Pb-Pb collisions at $\sqrt{s_{\mathrm{NN}}}$ = 2.76 TeV
(American Physical Society, 2017-09-08)
The measurement of two-particle angular correlations is a powerful tool to study jet quenching in a $p_{\mathrm{T}}$ region inaccessible by direct jet identification. In these measurements pseudorapidity ($\Delta\eta$) and ...
Online data compression in the ALICE O$^2$ facility
(IOP, 2017)
The ALICE Collaboration and the ALICE O2 project have carried out detailed studies for a new online computing facility planned to be deployed for Run 3 of the Large Hadron Collider (LHC) at CERN. Some of the main aspects ...
Evolution of the longitudinal and azimuthal structure of the near-side peak in Pb–Pb collisions at $\sqrt{s_{\rm NN}}=2.76$ TeV
(American Physical Society, 2017-09-08)
In two-particle angular correlation measurements, jets give rise to a near-side peak, formed by particles associated to a higher $p_{\mathrm{T}}$ trigger particle. Measurements of these correlations as a function of ...
J/$\psi$ elliptic flow in Pb-Pb collisions at $\sqrt{s_{\rm NN}}$ = 5.02 TeV
(American Physical Society, 2017-12-15)
We report a precise measurement of the J/$\psi$ elliptic flow in Pb-Pb collisions at $\sqrt{s_{\rm NN}} = 5.02$ TeV with the ALICE detector at the LHC. The J/$\psi$ mesons are reconstructed at mid-rapidity ($|y| < 0.9$) ...
Enhanced production of multi-strange hadrons in high-multiplicity proton-proton collisions
(Nature Publishing Group, 2017)
At sufficiently high temperature and energy density, nuclear matter undergoes a transition to a phase in which quarks and gluons are not confined: the quark–gluon plasma (QGP)1. Such an exotic state of strongly interacting ... |
2018-09-11 04:29
Proprieties of FBK UFSDs after neutron and proton irradiation up to $6*10^{15}$ neq/cm$^2$ / Mazza, S.M. (UC, Santa Cruz, Inst. Part. Phys.) ; Estrada, E. (UC, Santa Cruz, Inst. Part. Phys.) ; Galloway, Z. (UC, Santa Cruz, Inst. Part. Phys.) ; Gee, C. (UC, Santa Cruz, Inst. Part. Phys.) ; Goto, A. (UC, Santa Cruz, Inst. Part. Phys.) ; Luce, Z. (UC, Santa Cruz, Inst. Part. Phys.) ; McKinney-Martinez, F. (UC, Santa Cruz, Inst. Part. Phys.) ; Rodriguez, R. (UC, Santa Cruz, Inst. Part. Phys.) ; Sadrozinski, H.F.-W. (UC, Santa Cruz, Inst. Part. Phys.) ; Seiden, A. (UC, Santa Cruz, Inst. Part. Phys.) et al. The properties of 60-{\mu}m thick Ultra-Fast Silicon Detectors (UFSD) detectors manufactured by Fondazione Bruno Kessler (FBK), Trento (Italy) were tested before and after irradiation with minimum ionizing particles (MIPs) from a 90Sr \b{eta}-source . [...] arXiv:1804.05449. - 13 p. Preprint - Full text დეტალური ჩანაწერი - მსგავსი ჩანაწერები 2018-08-25 06:58
Charge-collection efficiency of heavily irradiated silicon diodes operated with an increased free-carrier concentration and under forward bias / Mandić, I (Ljubljana U. ; Stefan Inst., Ljubljana) ; Cindro, V (Ljubljana U. ; Stefan Inst., Ljubljana) ; Kramberger, G (Ljubljana U. ; Stefan Inst., Ljubljana) ; Mikuž, M (Ljubljana U. ; Stefan Inst., Ljubljana) ; Zavrtanik, M (Ljubljana U. ; Stefan Inst., Ljubljana) The charge-collection efficiency of Si pad diodes irradiated with neutrons up to $8 \times 10^{15} \ \rm{n} \ cm^{-2}$ was measured using a $^{90}$Sr source at temperatures from -180 to -30°C. The measurements were made with diodes under forward and reverse bias. [...] 2004 - 12 p. - Published in : Nucl. Instrum. Methods Phys. Res., A 533 (2004) 442-453 დეტალური ჩანაწერი - მსგავსი ჩანაწერები 2018-08-23 11:31 დეტალური ჩანაწერი - მსგავსი ჩანაწერები 2018-08-23 11:31
Effect of electron injection on defect reactions in irradiated silicon containing boron, carbon, and oxygen / Makarenko, L F (Belarus State U.) ; Lastovskii, S B (Minsk, Inst. Phys.) ; Yakushevich, H S (Minsk, Inst. Phys.) ; Moll, M (CERN) ; Pintilie, I (Bucharest, Nat. Inst. Mat. Sci.) Comparative studies employing Deep Level Transient Spectroscopy and C-V measurements have been performed on recombination-enhanced reactions between defects of interstitial type in boron doped silicon diodes irradiated with alpha-particles. It has been shown that self-interstitial related defects which are immobile even at room temperatures can be activated by very low forward currents at liquid nitrogen temperatures. [...] 2018 - 7 p. - Published in : J. Appl. Phys. 123 (2018) 161576 დეტალური ჩანაწერი - მსგავსი ჩანაწერები 2018-08-23 11:31 დეტალური ჩანაწერი - მსგავსი ჩანაწერები 2018-08-23 11:31
Characterization of magnetic Czochralski silicon radiation detectors / Pellegrini, G (Barcelona, Inst. Microelectron.) ; Rafí, J M (Barcelona, Inst. Microelectron.) ; Ullán, M (Barcelona, Inst. Microelectron.) ; Lozano, M (Barcelona, Inst. Microelectron.) ; Fleta, C (Barcelona, Inst. Microelectron.) ; Campabadal, F (Barcelona, Inst. Microelectron.) Silicon wafers grown by the Magnetic Czochralski (MCZ) method have been processed in form of pad diodes at Instituto de Microelectrònica de Barcelona (IMB-CNM) facilities. The n-type MCZ wafers were manufactured by Okmetic OYJ and they have a nominal resistivity of $1 \rm{k} \Omega cm$. [...] 2005 - 9 p. - Published in : Nucl. Instrum. Methods Phys. Res., A 548 (2005) 355-363 დეტალური ჩანაწერი - მსგავსი ჩანაწერები 2018-08-23 11:31
Silicon detectors: From radiation hard devices operating beyond LHC conditions to characterization of primary fourfold coordinated vacancy defects / Lazanu, I (Bucharest U.) ; Lazanu, S (Bucharest, Nat. Inst. Mat. Sci.) The physics potential at future hadron colliders as LHC and its upgrades in energy and luminosity Super-LHC and Very-LHC respectively, as well as the requirements for detectors in the conditions of possible scenarios for radiation environments are discussed in this contribution.Silicon detectors will be used extensively in experiments at these new facilities where they will be exposed to high fluences of fast hadrons. The principal obstacle to long-time operation arises from bulk displacement damage in silicon, which acts as an irreversible process in the in the material and conduces to the increase of the leakage current of the detector, decreases the satisfactory Signal/Noise ratio, and increases the effective carrier concentration. [...] 2005 - 9 p. - Published in : Rom. Rep. Phys.: 57 (2005) , no. 3, pp. 342-348 External link: RORPE დეტალური ჩანაწერი - მსგავსი ჩანაწერები 2018-08-22 06:27
Numerical simulation of radiation damage effects in p-type and n-type FZ silicon detectors / Petasecca, M (Perugia U. ; INFN, Perugia) ; Moscatelli, F (Perugia U. ; INFN, Perugia ; IMM, Bologna) ; Passeri, D (Perugia U. ; INFN, Perugia) ; Pignatel, G U (Perugia U. ; INFN, Perugia) In the framework of the CERN-RD50 Collaboration, the adoption of p-type substrates has been proposed as a suitable mean to improve the radiation hardness of silicon detectors up to fluencies of $1 \times 10^{16} \rm{n}/cm^2$. In this work two numerical simulation models will be presented for p-type and n-type silicon detectors, respectively. [...] 2006 - 6 p. - Published in : IEEE Trans. Nucl. Sci. 53 (2006) 2971-2976 დეტალური ჩანაწერი - მსგავსი ჩანაწერები 2018-08-22 06:27
Technology development of p-type microstrip detectors with radiation hard p-spray isolation / Pellegrini, G (Barcelona, Inst. Microelectron.) ; Fleta, C (Barcelona, Inst. Microelectron.) ; Campabadal, F (Barcelona, Inst. Microelectron.) ; Díez, S (Barcelona, Inst. Microelectron.) ; Lozano, M (Barcelona, Inst. Microelectron.) ; Rafí, J M (Barcelona, Inst. Microelectron.) ; Ullán, M (Barcelona, Inst. Microelectron.) A technology for the fabrication of p-type microstrip silicon radiation detectors using p-spray implant isolation has been developed at CNM-IMB. The p-spray isolation has been optimized in order to withstand a gamma irradiation dose up to 50 Mrad (Si), which represents the ionization radiation dose expected in the middle region of the SCT-Atlas detector of the future Super-LHC during 10 years of operation. [...] 2006 - 6 p. - Published in : Nucl. Instrum. Methods Phys. Res., A 566 (2006) 360-365 დეტალური ჩანაწერი - მსგავსი ჩანაწერები 2018-08-22 06:27
Defect characterization in silicon particle detectors irradiated with Li ions / Scaringella, M (INFN, Florence ; U. Florence (main)) ; Menichelli, D (INFN, Florence ; U. Florence (main)) ; Candelori, A (INFN, Padua ; Padua U.) ; Rando, R (INFN, Padua ; Padua U.) ; Bruzzi, M (INFN, Florence ; U. Florence (main)) High Energy Physics experiments at future very high luminosity colliders will require ultra radiation-hard silicon detectors that can withstand fast hadron fluences up to $10^{16}$ cm$^{-2}$. In order to test the detectors radiation hardness in this fluence range, long irradiation times are required at the currently available proton irradiation facilities. [...] 2006 - 6 p. - Published in : IEEE Trans. Nucl. Sci. 53 (2006) 589-594 დეტალური ჩანაწერი - მსგავსი ჩანაწერები |
Without a lot of expensive equipment the best you can do is make a rough estimate. By incorrectly assuming that the light that you see describes the black body temperature of the object, by assuming that the system is in thermal equilibrium with itself and that the temperature within the system is spatially and temporally uniform, you can use Planck's law to calculate the spectrum.
Planck's law describes the spectral density of electromagnetic radiation emitted by a black body in thermal equilibrium at a given temperature $T$.
The spectral radiance of a body, $B_ν$, describes the amount of energy it gives off as radiation of different frequencies. It is measured in terms of the power emitted per unit area of the body, per unit solid angle that the radiation is measured over, per unit frequency. Planck showed that the spectral radiance of a body for frequency $ν$ at absolute temperature $T$ is given by:
$$ B_{\nu }(\nu ,T)={\frac {2h\nu ^{3}}{c^{2}}}{\frac {1}{e^{\frac {h\nu }{k_{\mathrm {B} }T}}-1}}$$
where $k_B$ is the Boltzmann constant, $h$ is the Planck constant, and $c$ is the speed of light in the medium, whether material or vacuum. The spectral radiance can also be measured per unit wavelength $\lambda$ instead of per unit frequency. In this case, it is given by:
$$B_{\lambda }(\lambda ,T)={\frac {2hc^{2}}{\lambda ^{5}}}{\frac {1}{e^{\frac {hc}{\lambda k_{\mathrm {B} }T}}-1}}.$$
The law may also be expressed in other terms, such as the number of photons emitted at a certain wavelength, or the energy density in a volume of radiation. The SI units of $B_ν$ are $W·sr^{−1}·m^{−2}·Hz^{−1}$, while those of $B_\lambda$ are $W·sr^{−1}·m^{−3}$.
The "Planck Calculator for Infrared Remote Sensing" can be used to make some calculations:
Temperature T(K) 10,000
Center Wavelength μm 11.0
Spectral Radiance (W/m^2-sr-μm) 5292.388800699331
With a temperature of 10K the center frequency is 11 $\mu m$.
That's the oversimplified (inaccurate) layperson's calculation. More accurate calculations are made by allowing for everything from: relative velocity, the composition of the object's atmosphere (and our own), the effect of distance, even other elements, upon the spectral distribution (short list).
Consequently, after correcting for light travel time effects the response of the Balmer and optical helium lines should generally be strongest during low continuum luminosity states. Responsivity that depends upon photon flux and continuum state may explain a number of outstanding problems currently under investigation in broad line variability studies of these and other emission lines. [Abridged]
Once calculations are made for the blackbody curve in accordance with Planck's law it is necessary to transpose that to Rayleigh–Jeans law which is applicable at large wavelengths (low frequencies) but strongly disagrees at short wavelengths (high frequencies) where Planck's law is more useful.
See Wikipedia's "Comparison [of Rayleigh–Jeans law] to Planck's law".
Further reading:
If you want to include the effect of quantum gravity in your calculations a recent paper is: "Black body radiation in a model universe with largeextra dimensions and quantum gravity effects" 8 Jan 2018, by Nozari, Anvari, and Sefiedgar
Abstract: "We analyze the problem of black body radiation in the presence of quantum gravity effects encoded in modified dispersion relations and in a model universe with large extra dimensions. In this context, modified form of Planck distribution, Jeans number, equipartition theorem, spectral energy density, Stefan-Boltzmann law and Wien's law are found and the corresponding results are interpreted. Then, entropy and specific heat of black body radiation are obtained in this setup. Finally, the modified form of Debye law and Dulong-Petit law are investigated in this framework.".
This article from the PrimaLuce Lab's website: "Radio astronomy at school with SPIDER radio telescopes" explains backyard radio astronomy.
Their 3M diameter SPIDER 500A's H142-One receiver has a 1.42 GHz superheterodyne type radiometer/ spectrometer, double conversion (type UP/DOWN) with 50 MHz bandwidth (RF=1.395MHz-1.445MHz) and 14-bit analog to digital converter. It has a spectrometer with 1024 channels (each 61 KHz) and is able to record many radio sources in the Universe, with a theoretical flow of at least 5 Jy. Price available upon request.
Their SPIDER 230C Compact Radio Telescope measures 2.3 meters and is priced at U$16K, from what I can determine from reading the literature it features the same receiver and software; dish diameter and mount capacity being the only difference, along with beam width and sensitivity.
The mesh on their dishes is good up to 5 GHz but the feedhorn normally supplied is for 1.42 GHz, I imagine that they are willing to do some amount of customization.
An explanation of the atmospheric window can be found here: "The Effects of Earth's Upper Atmosphere on Radio Signals", this image is from NASA's website:
Question: From literature, I assume 3 cm is 10GHz, which should fall under either synchroton or free-free emission. But how do I know exactly?
3cm is 10 GHz and can be received with a 10 GHz receiver. The difference between synchrotron and free-free is explained on NRAO's webpages Free–Free Radiation and Synchrotron Radiation.
Acceleration by a magnetic field produces magnetobremsstrahlung, the German word for “magnetic braking radiation.” The character of magnetobremsstrahlung depends on the speeds of the electrons, so these somewhat different types of radiation are given specific names. Gyro radiation comes from electrons whose velocities are much smaller than the speed of light: $v≪c$. Mildly relativistic electrons whose kinetic energies are comparable with their rest mass $m_ec^2$ emit cyclotron radiation, and ultrarelativistic electrons (kinetic energies $≫m_ec^2$) produce synchrotron radiation.
Acceleration by an electric field accounts for free–free radiation, the resulting emission is called free–free radiation because the electron is free both before and after the interaction; it is not captured by the ion. If the ionized interstellar cloud is reasonably dense, the electrons and ions interact often enough that they quickly come into local thermodynamic equilibrium (LTE) at some common temperature, so free–free radiation is usually thermal emission. |
Many believe that $\mathsf{BPP} = \mathsf{P} \subseteq \mathsf{NP}$. However we only know that $\mathsf{BPP}$ is in the second level of polynomial hierarchy, i.e. $\mathsf{BPP}\subseteq \Sigma^ \mathsf{P}_2 \cap \Pi^ \mathsf{P}_2$. A step towards showing $\mathsf{BPP} = \mathsf{P}$ is to first bring it down to the first level of the polynomial hierarchy, i.e. $\mathsf{BPP} \subseteq \mathsf{NP}$.
The containment would mean that nondeterminism is at least as powerful as randomness for polynomial time.
It also means that if for a problem we can find the answers using efficient (polynomial time) randomized algorithms then we can verify the answers efficiently (in polynomial time) .
Are there any known interesting consequences for $\mathsf{BPP} \subseteq \mathsf{NP}$?
Are there any reasons to believe that proving $\mathsf{BPP} \subseteq \mathsf{NP}$ is out of reach right now (e.g. barriers or other arguments)? |
To evaluate detection performance, we plot the
miss-rate $mr(c) = \frac{fn(c)}{tp(c) + fn(c)}$ against the number of false positives per image $fppi(c)=\frac{fp(c)}{\text{#img}}$ in log-log plots. $tp(c)$ is the number of true positives, $fp(c)$ is the number of false positives, and $fn(c)$ is the number of false negatives, all for a given confidence value $c$ such that only detections are taken into account with a confidence value greater or equal than $c$. As commonly applied in object detection evaluation the confidence threshold $c$ is used as a control variable. By decreasing $c$, more detections are taken into account for evaluation resulting in more possible true or false positives, and possible less false negatives. We define the log average miss-rate (LAMR) as shown, where the 9 fppi reference points are equally spaced in the log space: $\DeclareMathOperator*{\argmax}{argmax}LAMR = \exp\left(\frac{1}{9}\sum\limits_f \log\left(mr(\argmax\limits_{fppi\left(c\right)\leq f} fppi\left(c\right))\right)\right)$
For each fppi reference point the corresponding mr value is used. In the absence of a miss-rate value for a given f the highest existent
fppi value is used as new reference point. This definition enables LAMR to be applied as a single detection performance indicator at image level. At each image the set of all detections is compared to the groundtruth annotations by utilizing a greedy matching algorithm. An object is considered as detected (true positive) if the Intersection over Union (IoU) of the detection and groundtruth bounding box exceeds a pre-defined threshold. Due to the high non-rigidness of pedestrians we follow the common choice of an IoU threshold of 0.5. Since no multiple matches are allowed for one ground-truth annotation, in the case of multiple matches the detection with the largest score is selected, whereas all other matching detections are considered false positives. After the matching is performed, all non matched ground-truth annotations and detections, count as false negatives and false positives, respectively.
Neighboring classes and ignore regions are used during evaluation. Neighboring classes involve entities that are semantically similar, for example bicycle and moped riders. Some applications might require their precise distinction (
enforce) whereas others might not ( ignore). In the latter case, during matching correct/false detections are not credited/penalized. If not stated otherwise, neighboring classes are ignored in the evaluation. In addition to ignored neighboring classes all persons annotations with the tags behind glass or sitting-lying are treated as ignore regions. Further, as mentioned in Section 3.2, EuroCity Persons Dataset Publication, ignore regions are used for cases where no precise bounding box annotation is possible (either because the objects are too small or because there are too many objects in close proximity which renders the instance based labeling infeasible). Since there is no precise information about the number or the location of objects in the ignore region, all unmatched detections which share an intersection of more than $0.5$ with these regions are not considered as false positives.
Note that submissions with provided publication link and/or code will get priorized in below list (COMING SOON).
Method User LAMR (reasonable) LAMR (small) LAMR (occluded) LAMR (all) External data used▲ Publication URL Publication code Submitted on YOLOv3_640 HUI_Tsinghua-Daim... 0.273 0.564 0.623 0.456 no no 2019-05-17 04:56:27 View Faster R-CNN ECP Team 0.101 0.196 0.381 0.251 ImageNet yes no 2019-04-01 17:06:33 View YOLOv3 ECP Team 0.097 0.186 0.401 0.242 ImageNet yes no 2019-04-01 17:08:05 View R-FCN (with OHEM) ECP Team 0.163 0.245 0.507 0.330 ImageNet yes no 2019-04-01 17:10:03 View SSD ECP Team 0.131 0.235 0.460 0.296 ImageNet yes no 2019-04-02 13:56:14 View HRNet Hongsong Wang 0.061 0.138 0.287 0.183 ImageNet no no 2019-08-05 17:11:04 View
Method User LAMR (reasonable) LAMR (small) LAMR (occluded) LAMR (all) External data used▲ Publication URL Publication code Submitted on Faster R-CNN ECP Team 0.201 0.359 0.701 0.358 ImageNet yes no 2019-05-02 10:10:01 View FasterRCNN with M... Qihua Cheng 0.150 0.253 0.653 0.295 ImageNet no no 2019-07-08 08:48:13 View HRNet Hongsong Wang 0.079 0.156 0.265 0.153 ImageNet no no 2019-08-05 17:11:04 View |
Given that $p$ is a Mersenne prime, it is easy to check that ${\rm gcd}(p+1,p^{2}+1) = {\rm gcd}(p+1,p^{4}+1)= {\rm gcd}(p^{2}+1,p^{4}+1) = 2$, and we recall that $p+1$ is a power of $2$. Let $G$ be a solvable group of order $p^{4}(p+1)(p^{2}+1)(p^{4}+1)$. Suppose that $G$ has an irreducible character of degree $p^{4}$. Then by results (eg of Brauer), we know that $O_{p}(G) = 1$. Let us examine the structure of $O_{p^{\prime}}(G)$.
Since $O_{p}(G) = 1$, by the well-known centralizer Lemma of Hall-Higman, no element of order $p$ in $G$ acts trivially on $O_{p^{\prime}}(G)$, so that $O_{p^{\prime}}(G)$ admits a group of (outer) automorphisms $P$ of order $p^{4}$.
We claim though that every $P$-invariant section $M$ of $O_{p^{\prime}}(G)$ of odd order is centralized by $P$. By standard results on coprime automorphisms, it suffices to assume that $M$ is an elementary $q$-group, and is minimal $P$-invariant, but not centralized by $P$. Then $|M|$ divides one (and only one) of $\frac{p^{2}+1}{2}$ or $\frac{p^{4}+1}{2}$, and $|M| \equiv 1$ (mod $p$) ( note for the purposes of dealing with other odd primes that for any odd prime $p$, $\frac{p+1}{2}$ clearly has no divisor (greater than $1$) which is congruent to $1$ (mod $p$)).
Suppose that $|M|$ divides $p^{2}+1$. Then $\frac{p^{2}+1}{|M|} \equiv 1$ (mod $p$), while $|M| \neq p^{2}+1$, since $p$ is Mersenne. Then we have the contradiction $p^{2}+1 = |M|\frac{p^{2}+1}{|M|} \geq (p+1)^{2}.$ Hence $|M|$ divides $p^{4}+1$.
By Y. Cor's answer to your previous question MO208645, we know that no divisor of $p^{4}+1$ other than $1$ and $p^{4}+1$ is congruent to $1$ (mod $p$). But we can't have $|M| = p^{4}+1$ since $|M|$ is a prime power and $p^{4}+1 \equiv 2$ (Mod $8$). This is a contradiction.
We conclude that for $P \in {\rm Syl}_{p}(G)$, it must be the case that $[O_{p^{\prime}}(G),P]$ is a $2$-group, say $X$. By standard results on coprime automorphisms, $[P,X] = X$ and $P$ acts faithfully on $X$.
Now a Sylow $2$-subgroup of $G$ has order $2^{n+2}$. Now $n$ is the smallest positive integer with $2^{n} \equiv 1$ (mod $p$). Now let $L = PX$. Then $|L|$ divides $2^{n+2}p^{4}$ and $[L:N_{L}(P)]$ is a power of $2$, and is congruent to $1$ (mod $p$), but is greater than $1$.
It follows that $[L:N_{L}(P)] = 2^{n}$ if $p > 3$, and that $[L:N_{L}(P)] \in \{2^{n},2^{n+2} \}$ if $p = 3$. But then it is clear that $X$ admits no group of automorphisms of order $p^{4}$, which is a contradiction.
An alternative way to conclude is to take a minimal $P$-invariant $M$ section of $X$with $[P,M] \neq 1$. Then we have $|M| \geq 2^{n}$, and for every choice of $p$ (even $p=3$), there can be at most two such sections of $X$, each of which admits an automorphism of order $p$ but no group of automorphisms of order $p^{2}$. Hence in no case can $X$ admit a faiithful group of automorphisms of order $p^{4}$.
What has actually been proved is that when $p$ is a Mersenne prime, a $p$-solvable finite group $G$ of order $p^{4}(p+1)(p^{2}+1)(p^{4}+1)$ has a non-identity normal $p$-subgroup, so certainly has no $p$-block of defect zero.
( I don't think that the fact that $p$ is a Mersenne prime is essential here, I think the argument can be modified to work for any odd prime $p$. In fact, I believe the argument really shows that if $p$ is an odd prime, then a $p$-solvable group of order $p^{k}(p+1)(p^{2}+1)(p^{4}+1)$ has a non-identity normal $p$-subgroup if $k >1$ when $p \neq 3$ and if $k >2$ when $p =3$). |
I've known about Noether's theorem for some time and reading some things about it recently I've realised I haven't completely understood it. In that case I've been trying to understand a more rigorous approach to it as presented in Spivak's Mechanics book.
There Spivak first derives a corollary from the derivation of the Euler-Lagrange's equation for the extremal of the functional
$$J(f) = \int_a^b F(f(t),f'(t),t)dt,$$
for $f : [a,b]\to \mathbb{R}^n$. This corollary is what he calls the boundary term corollary which states that if $f$ is a critical point of $J$ and if $\alpha : (-\epsilon,\epsilon)\times [a,b]\to \mathbb{R}^n$ is a variation of $f$ then if we set $\bar{\alpha}(u)=\alpha(u, \cdot)$ we have
$$\dfrac{dJ(\bar{\alpha}(u))}{du}\bigg|_{u=0}=\sum_{i=1}^n \dfrac{\partial \alpha^i}{\partial u}(0,t)\dfrac{\partial F}{\partial y^i}(f(t),f'(t),t)\bigg|_a^b.$$
Then, if $M$ is the configuration manifold of a system and if $\phi : (-\epsilon,\epsilon)\times M\to M$ is a one-parameter family of diffeomorphisms defining $\phi_s = \phi(s,\cdot)$ and $\Phi_s = (\phi_s)_{\ast}$ Spivak defines that $\phi_s$ preserves $L : TM\to \mathbb{R}$ if for all $v\in TM$ we have $L(\Phi_s (v)) = L(v)$.
He also defines $W = \partial \phi/\partial s$ and defines the quantity
$$\Phi_c(t) = \lim_{h\to 0}\dfrac{L(c'(t)+hW(c(t)))-L(c'(t))}{h}$$
After that Noether's theorem is stated and proven like this
If the $\phi_s$ preserve $L$, then $\Phi_c$ is constant along any solution $c$ of Lagrange's equations for $L$
PROOF: Since the $\phi_s$ preserve $L$, each of the curves
$$c_s(t)=\phi_s(c(t))=\phi(s,c(t)),$$
is also a solution of Lagrange's equations for $L$, and thus an extremal for $\int_a^bL(c(t),c'(t),t)dt$ for all $a,b$ in the interval under consideration.
The boundary term corollary then says that for all such $a$ and $b$ we have
$$0 = \sum_{i=1}^n \dfrac{\partial q^i}{\partial x}\dfrac{\partial L}{\partial \dot{q}^i}\bigg|_a^b,$$
so that
$$\sum_{i=1}^n \dfrac{\partial q^i}{\partial x}\dfrac{\partial L}{\partial \dot{q}^i}$$
is constant.
Now there are a couple of points. In summary I can't understand what is going on here. My doubts are
That quantity $\Phi_c$ never entered the proof and I really don't see a connetion between the quantity proven constant and $\Phi_c$ at first.
How the boundary term corollary applies here? Are we considering the variation $\alpha:(-\epsilon,\epsilon)\times [a,b]\to M$ given by $\alpha(u,t)=c_u(t)$? But this variation might not keep endpoints fixed. What is the meaning of this?
Why we need each $c_s$ to be a solution of the equations? On the boundary term corollary we have that quantity is zero even if the variation is not yet a solution.
In summary what is the idea behind the proof of Noether's theorem as stated above? What is really going on? |
Consider the sphere of unit radius centered at $(0,0,1)$,and the cone of equation $z^2=x^2+y^2.$ Find the volume above the cone and inside the sphere.
The equation of the sphere is $$1 \leq x^2+y^2+(z-1)^2=x^2+y^2+z^2-2z+1$$
Using spherical coordinates this gives $$1 \leq \rho ^2\sin^2\varphi \cos^2\theta + \rho^2\sin^2\varphi sin^2\theta + \rho^2 \cos^2\varphi -2\rho \sin\varphi \sin\theta +1 $$
$$=\rho^2 -2\rho\sin\varphi\sin\theta+1$$
The solution says I should come to $\rho \leq 2\cos\varphi$. I don't understand where they have got this from though, could someone explain? |
Suppose we have a $k$-dimensional subspace $V$ in $\mathbb{R}^n$ given by a basis $\{v_1,\cdots,v_k\in \mathbb{R}^n\}$, find an index set $I\subset [n]$ with $|I|=m$ where $k\le m\le n$, such that $$\max_{v\in V}\frac{||v||_2}{||v_I||_2}$$ is minimum, where $v_I\in \mathbb{R}^m$ is the projection of $v$ to the coordinates indexed by $I$.
My question is:
Is this problem NP-hard? If $k$ is a constant, $m$ can be unbounded, is this problem still NP-hard? If it is NP-hard, what is the best (or any) approximate ratio we can achieve?
Any references in the literature will also be welcome. |
$$ \min \delta CVaR - (1-\delta) \sum_i^{n} \mu_i x_i \\ \sum x_i = \sum x^{old}_i \\ Losses(s) = \sum x_i - \sum_i^{n} (R(s,i))x_i \\ VaRDev(s) = Losses(s) - VaR \\ CVaR = VaR + \frac{\sum_s^{} p_s VaRDev(s)}{1-\alpha} $$
The Markowitz style Mean-cVaR portfolio is stated above. It Appears in
by Zenios. $x^{old}$ is exciting portfolio and now you want to optimize and create a new portfolio $x$. $s$ is the different scenarios. Practical financial optimization
But what if transaction costs existing so you should pay $c_{variable}$ % of how much you trade an asset and each trade have a minimum cost as well.
How can we add transaction cost to this model? |
Let $\Delta ABC$ be our triangle and $KLMN$ be our rectangle.
There are three cases:
1) $KN\subset AC$;
2) $KN\subset AB$ and
3) $KN\subset BC$.
In the first case let $L\in AB$ and $M\in BC$.
By the Heron's formula $$S_{\Delta ABC}=\frac{1}{4}\sqrt{2(5^28^2+5^2(\sqrt{41})^2+8^2(\sqrt{41})^2)-5^4-(\sqrt{41})^4-8^4}=16.$$Let $BD$ be an altitude of $\Delta ABC$.
Thus, $$\frac{8BD}{2}=16,$$ which gives $$BD=4.$$Now, let $MN=x$ and since $\Delta LBM\sim\Delta ABC,$ we obtain$$\frac{4-x}{4}=\frac{ML}{8},$$ which gives$$ML=2(4-x).$$Thus, by AM-GM $$S_{KLMN}=2x(4-x)\leq2\left(\frac{x+4-x}{2}\right)^2=8.$$The equality occurs for $x=2$, which says that $8$ is a maximal value in this case.
By the same way we can consider another cases and choose the maximal value. |
Superfunctions Contents Covers
The back cover suggests short abstract of the Book and few notes about the Author.
About the topic
Assume some given holomorphic function \(T\). The superfunction is holomorphic solution F of equation \(T(F(z))=F(z+1)\)
The Abel function (or abelfunction) is the inverse of superfunction, \(G=F^{-1}\)
The abelfunction is solution of the Abel equation \(G(T(z))=G(z)+1\)
As the superfunction \(F\) and the abelfunction \(G=F^{-1}\) are established, the \(n\)th iterate of transfer function \(T\) can be expressed as follows: \(T^n(z)=F(n+G(z))\)
This expression allows to evaluate the non-integer iterates. The number n of iterate can be real or even complex. In particular, for integer \(n\), the iterates have the common meaning: \(T^{-1}\) is inverse function of \(T\),
\(T^0(z)=z\), \(T^1(z)=T(z)\), \(T^2(z)=T(T(z)) \), \(T^3(z)=T(T(T(z))) \), and so on. The group property holds: \(T^m(T^n(z))=T^{m+n}(z)\)
The special notation is used in through the book; the number of iterate is indicated as superscript. For example, In these notations, \(\sin^2(z)=\sin(\sin(z))\), but never \(\sin(z)^2\).
This notation is borrowed from the Quantum mechanics, where \(P^2(\psi)=P(P(\psi))\), but never \(P(\psi)^2\). About the Book
Tools for evaluation of superfunctions, abelfunctions and non-integer iterates of holomorphic functions are collected. For a giver transfer function T, the superfunction is solution F of the transfer equation F(z+1)=T(F(z)) . The abelfuction is inverse of F. In particular, thesuperfunctions of factorial, exponent, sin; the holomorphic extensions of the logistic sequence and of the Ackermann functions are suggested. from ackermanns, the tetration (mainly to the base b>1) and pentation (to base e) are presented. The efficient algorithm for the evaluation of superfunctions and abelfunctions are described. The graphics and complex maps are plotted. The possible applications are discussed. Superfunctions significantly extend the set of functions that can be used in scientific research and technical design. Generators of figures are loaded to the site TORI, http://mizugadro.mydns.jp for the free downloading. With these generators, the Readers can reproduce (and modify) the figures from the Book. The Book is intended to be applied and popular. I try to avoid the complicated formulas, but some basic knowledge of the complex arithmetics, Cauchi integral and the principles of the asymptotical analysis should help at the reading.
About the Author
Dmitrii Kouznetsov
Graduated from the Physics Department of theMoscow State University (1980). Work: USSR, Mexico, USA, Japan.
Century 20: Proven the quantum stability of the optical soliton, suggested the low bound of the quantum noise of nonlinear amplifier, indicated the limit of the single mode approximation in the quantum optics. Century 21: Theorem about boundary behaviour of modes of Dirichlet laplacian, Theory of ridged atomic mirrors, formalism of superfunctions, TORI axioms. Summary
The summary suggests main notations used in the Book:
\(T\)\( ~ ~ ~ ~ ~\) Transfer function
\(F\big(G(z)\big)=z\) \(~ ~ ~ ~ ~\) Identity function
\(T^n(z)=F\big(n+G(z)\big)\) \(~ ~ ~\) \(n\)th iterate
\(\displaystyle F(z)=\frac{1}{2\pi \mathrm i} \oint \frac{F(t) \, \mathrm d t}{t-z}\) \(~ ~ ~\) Cauchi integral
\(\mathrm{tet}_b(z\!+\!1)=b^{\mathrm{tet}_b(z)}\) \(~ ~ ~\) tetration to base \(b\)
\(\mathrm{tet}_b(0)=1\) \(~, ~ ~\) \( \mathrm{tet}_b\big(\mathrm{ate}_b(z)\big)=z\)
\(\mathrm{ate}_b(b^z)=\mathrm{ate}_b(z)+1\) \(~ ~\) arctetration to base \(b\)
\(\exp_b^{~n}(z)=\mathrm{tet}_b\big(n+\mathrm{ate}_b(z)\big)\) \(~ ~\) \(n\)th iterate of function \(~\) \(z\!\mapsto\! b^z\)
\(\displaystyle \mathrm{Tania}^{\prime}(z)=\frac{\mathrm{Tania}(z)}{\mathrm{Tania}(z)\!+\!1}\) \(~ ~\) Tania function,\(~\) \(\mathrm{Tania}(0)\!=\!1\)
\(\displaystyle \mathrm{Doya}(z)=\mathrm{Tania}\big(1\!+\!\mathrm{ArcTania}(z)\big)\) \(~ ~\) Doya function
\(\displaystyle \mathrm{Shoka}(z)=z+\ln(\mathrm e^{-z}\!+\!\mathrm e \!-\! 1)\) \(~\) Shoka function
\(\displaystyle \mathrm{Keller}(z)=\mathrm{Shoka}\big(1\!+\!\mathrm{ArcShoka}(z)\big)\) \(~ ~\) Keller function
\(\displaystyle \mathrm{tra}(z)=z+\exp(z)\) \(~ ~ ~\) Trappmann function
\(\displaystyle \mathrm{zex}(z)=z\,\exp(z)\) \(~ ~ ~ ~\) Zex function
\(\displaystyle \mathrm{Nem}_q(z)=z+z^3+qz^4\) \(~ ~ ~ ~\) Nemtsov function
Recent advance
Most of results presented in the book, are published in scientific journals; the links (without numbers) are supplied at the bottom.
After the appearance of the first version of the Book, certain advances are observed about evaluation of tetrationof complex argument; the new algorithm is suggested, that seems to be mode efficient, than the Cauchi integral described in the Book.
[3] [4] [5] [6] References https://www.morebooks.de/store/ru/book/Суперфункции/isbn/978-3-659-56202-0Дмитрий Кузнецов. Суперфунцкии.ISBN-13: 978-3-659-56202-0ISBN-10: 3659562025EAN: 9783659562020 http://www.ils.uec.ac.jp/~dima/BOOK/202.pdf http://mizugadro.mydns.jp/BOOK/202.pdf http://www.ils.uec.ac.jp/~dima/BOOK/443.pdf (a little bit out of date) http://mizugadro.mydns.jp/BOOK/444.pdf D.Kouznetov. Superfunctions. 2018. http://journal.kkms.org/index.php/kjm/article/view/428 William Paulsen. Finding the natural solution to f(f(x))=exp(x). Korean J. Math. Vol 24, No 1 (2016) pp.81-106. https://link.springer.com/article/10.1007/s10444-017-9524-1 William Paulsen, Samuel Cowgill. Solving F(z + 1) = b F(z) in the complex plane. Advances in Computational Mathematics, December 2017, Volume 43, Issue 6, pp 1261–1282 https://search.proquest.com/openview/cb7af40083915e275005ffca4bfd4685/1?pq-origsite=gscholar&cbl=18750&diss=y Cowgill, Samuel. Exploring Tetration in the Complex Plane. Arkansas State University, ProQuest Dissertations Publishing, 2017. 10263680. https://link.springer.com/article/10.1007/s10444-018-9615-7 William Paulsen. Tetration for complex bases. Advances in Computational Mathematics, 2018.06.02.
The book combines the main results from the following publications:
http://www.ams.org/mcom/2009-78-267/S0025-5718-09-02188-7/home.html
http://mizugadro.mydns.jp/PAPERS/2009analuxpRepri.pdf D.Kouznetsov. Analytic solution of F(z+1)=exp(F(z)) in complex z-plane. Mathematics of Computation 78 (2009), 1647-1670.
http://www.jointmathematicsmeetings.org/journals/mcom/2010-79-271/S0025-5718-10-02342-2/S0025-5718-10-02342-2.pdf
http://www.ams.org/journals/mcom/2010-79-271/S0025-5718-10-02342-2/home.html http://eretrandre.org/rb/files/Kouznetsov2009_215.pdf http://www.ils.uec.ac.jp/~dima/PAPERS/2010q2.pdf http://mizugadro.mydns.jp/PAPERS/2010q2.pdf D.Kouznetsov, H.Trappmann. Portrait of the four regular super-exponentials to base sqrt(2). Mathematics of Computation, 2010, v.79, p.1727-1756.
http://www.springerlink.com/content/qt31671237421111/fulltext.pdf?page=1
http://mizugadro.mydns.jp/PAPERS/2010superfae.pdf D.Kouznetsov, H.Trappmann. Superfunctions and square root of factorial. Moscow University Physics Bulletin, 2010, v.65, No.1, p.6-12.
http://www.ils.uec.ac.jp/~dima/PAPERS/2010vladie.pdf
http://mizugadro.mydns.jp/PAPERS/2010vladie.pdf D.Kouznetsov. Tetration as special function. Vladikavkaz Mathematical Journal, 2010, v.12, issue 2, p.31-45.
http://www.springerlink.com/content/qt31671237421111/fulltext.pdf?page=1
D.Kouznetsov, H.Trappmann. Superfunctions and square root of factorial. Moscow University Physics Bulletin, 2010, v.65, No.1, p.6-12
http://mizugadro.mydns.jp/t/index.php/Place_of_science_in_the_human_knowledge D.Kouznetsov. Place of science and physics in human knowledge. English translation from http://ufn.ru/tribune/trib120111 Russian Physics:Uspekhi, v.191, Tribune, p.1-9 (2010)
http://www.ils.uec.ac.jp/~dima/PAPERS/2010logistie.pdf http://mizugadro.mydns.jp/PAPERS/2010logistie.pdf D.Kouznetsov. Continual generalisation of the Logistic sequence. Moscow State University Physics Bulletin, 3 (2010) No.2, стр.23-30.
http://www.ams.org/journals/mcom/0000-000-00/S0025-5718-2012-02590-7/S0025-5718-2012-02590-7.pdf
http://www.ils.uec.ac.jp/~dima/PAPERS/2012e1eMcom2590.pdf http://mizugadro.mydns.jp/PAPERS/2012e1eMcom2590.pdf H.Trappmann, D.Kouznetsov. Computation of the Two Regular Super-Exponentials to base exp(1/e). Mathematics of Computation, 2012, 81, February 8. p.2207-2227.
http://www.ils.uec.ac.jp/~dima/PAPERS/2012or.pdf
http://mizugadro.mydns.jp/PAPERS/2012or.pdf Dmitrii Kouznetsov. Superfunctions for optical amplifiers. Optical Review, July 2013, Volume 20, Issue 4, pp 321-326.
http://www.scirp.org/journal/PaperInformation.aspx?PaperID=36560
http://mizugadro.mydns.jp/PAPERS/2013jmp.pdf D.Kouznetsov. TORI axioms and the applications in physics. Journal of Modern Physics, 2013, v.4, p.1151-1156.
http://www.ingentaconnect.com/content/asp/asl/2013/00000019/00000003/art00071 http://mizugadro.mydns.jp/PAPERS/2012thaiSuper.pdf D.Kouznetsov. Recovery of Properties of a Material from Transfer Function of a Bulk Sample (Theory). Advanced Science Letters, Volume 19, Number 3, March 2013, pp. 1035-1038(4).
http://link.springer.com/article/10.1007/s10043-013-0058-6 D.Kouznetsov. Superfunctions for amplifiers. Optical Review, July 2013, Volume 20, Issue 4, pp 321-326.
http://www.m-hikari.com/ams/ams-2013/ams-129-132-2013/kouznetsovAMS129-132-2013.pdf http://mizugadro.mydns.jp/PAPERS/2013hikari.pdf D.Kouznetsov. Entire function with logarithmic asymptotic. Applied Mathematical Sciences, 2013, v.7, No.131, p.6527-6541.
Keywords
Abel function, Book, Doya function, Iteration, Keller function, Maple and tea, LambertW, Shoka function, Superfunction, SuperFactorial, SuSin, SuTra, SuZex, Tania function, Tetration, Trappmann function, Tetration, Zex function, |
User:Jan A. Sanders/An introduction to Lie algebra cohomology/Lecture 8 The trace and Killing form
Let \( R\) be \(\mathbb{C}\) and \(\dim_\mathbb{C}\mathfrak{a}<\infty\ .\) Then define \(K_\mathfrak{a}\in C^2(\mathfrak{g},\mathbb{C})\) by\[ K_\mathfrak{a}(x,y)=\mathrm{tr}(d_1(x) d_1(y))\]In the case \(\mathfrak{a}=\mathfrak{g}\) and \(d_1=\mathrm{ad}\ ,\) this is called the
Killing form.In general, one calls \(K_\mathfrak{a}\) the trace form. example - of a trace form
Let \(\mathfrak{g}=\mathfrak{sl}_2\) and \(\mathfrak{a}=\R^2\ ,\) with the standard representation (see Lecture 1). Then \[ K_{\R^2}(M,M)=0, \quad K_{\R^2}(M,N)=1,\quad K_{\R^2}(M,H)=0,\] \[ K_{\R^2}(N,M)=1, \quad K_{\R^2}(N,N)=0,\quad K_{\R^2}(N,H)=0,\] \[ K_{\R^2}(H,M)=0, \quad K_{\R^2}(H,N)=0,\quad K_{\R^2}(H,H)=2.\]
proposition - trace form symmetric
\(K_\mathfrak{a}\) is symmetric.
proof
This follows from \(\mathrm{tr}(AB)=\mathrm{tr}(BA)\).\(\square\)
proposition - trace form invariant
\( K_\mathfrak{a} \) is \(\mathfrak{g}\)-invariant, that is, \(K_\mathfrak{a}\in C^2(\mathfrak{g},\mathbb{C})^\mathfrak{g}\ .\)
proof
Given the trivial action of \(\mathfrak{g}\) on \(\mathbb{C}\ ,\) one has \[ d_1^{2}(x)K_\mathfrak{a}(y,z)=-K_\mathfrak{a}([x,y],z)-K_\mathfrak{a}(y,[x,z])\ :\] \[=-\mathrm{tr}(d_1([x,y]) d_1(z))-\mathrm{tr}(d_1(y) d_1([x,z]))\ :\] \[=-\mathrm{tr}(d_1(x) d_1(y) d_1(z))+\mathrm{tr}(d_1(y) d_1(x) d_1(z)) -\mathrm{tr}(d_1(y) d_1(x)d_1(z))+\mathrm{tr}(d_1(y) d_1(z)d_1(x))\ :\] \[=0\]
proposition - \(d^2 K_\mathfrak{a} \) antisymmetric
\[d^2 K_\mathfrak{a}\in C_{\wedge}^3(\mathfrak{g},\mathbb{C})\]
proof
From the \(\mathfrak{g}\)-invariance it follows that \[d^2 K_\mathfrak{a}(x,y,z)=K_\mathfrak{a}(x,[y,z])\] Furthermore, \[K_\mathfrak{a}(x,[z,y])=-K_\mathfrak{a}(x,[y,z])\] and \[K_\mathfrak{a}(z,[x,y])=-K_\mathfrak{a}(z,[y,x])=K_\mathfrak{a}([y,z],x)=K_\mathfrak{a}(x,[y,z])\]\(\square\)
corollary - nontrivial third cohomology
Let \(\mathfrak{g}\) be a Lie algebra. Then \[[d^2 K_\mathfrak{a}]\in H_{\wedge}^3(\mathfrak{g},\mathbb{C})\] Observe that this class is not trivial, since \(K_\mathfrak{a}\) is symmetric, not antisymmetric.
musical maps
Let \(\mathfrak{g}^\star=C^1(\mathfrak{g},\mathbb{C})\) and define \( \flat: \mathfrak{g}\rightarrow \mathfrak{g}^\star\) by \[ \flat(x)(y)=K_\mathfrak{a}(x,y)\]
proposition
\[ \flat\in Hom_\mathfrak{g}(\mathfrak{g},\mathfrak{g}^\star)\]
proof
\[ \flat([x,y])(z)=K_\mathfrak{a}([x,y],z)\ :\] \[=-K_\mathfrak{a}(y,[x,z])\ :\] \[=-\flat(y)([x,z])\ :\] \[=d_1^{1}(x)\flat(y)(z)\] or \( \flat([x,y])=d_1^{1}(x)\flat(y)\quad \square\ .\)
Define \[ \sharp:\mathfrak{g}^\star\rightarrow \mathfrak{g}\] by \[ K_\mathfrak{a}(\sharp(c_1),y)=c_1(y)\] Then \[ K_\mathfrak{a}(x,y)=\flat(x)(y)=K_\mathfrak{a}(\sharp(\flat(x)),y)\ ,\] or \(x-\sharp(\flat(x))\in \ker K_\mathfrak{a}\ .\)
proposition
\(\ker K_\mathfrak{a}\) is an ideal.
proof
Let \(y\in\ker K_\mathfrak{a}\ ,\) that is \(K_\mathfrak{a}(x,y)=0\) for all \(x\in\mathfrak{g}\ .\)
Then it follows from the invariance of \(K_\mathfrak{a}\) that \[ K_\mathfrak{a}([y,x],z)+K_\mathfrak{a}(y,[x,z])=0\] and therefore \(K_\mathfrak{a}([y,x],z)=0\) for all \(z\in\mathfrak{g}\ .\)
This shows that \([\mathfrak{g},\ker K_\mathfrak{a}]\subset \ker K_\mathfrak{a}\ .\)
The statement that \([\ker K_\mathfrak{a},\mathfrak{g}]\subset \ker K_\mathfrak{a}\) follows by a symmetry argument.
definition
A Lie algebra \(\mathfrak{g}\) is called
simple if \([\mathfrak{g},\mathfrak{g}]\neq 0\)and \(\mathfrak{g}\) contains no ideals besides \(0\) and itself. proposition - simple Lie algebra
If \(\mathfrak{g}\) is
simple, then \(\mathfrak{g}=[\mathfrak{g},\mathfrak{g}]\ .\) proof
\([\mathfrak{g},\mathfrak{g}]\neq 0\) is an ideal, so it must equal \([\mathfrak{g},\mathfrak{g}]=\mathfrak{g}\ .\)
proposition
If \( K_\mathfrak{a} \) is nonzero, and \(\mathfrak{g}\) is simple, then \(\flat\) is injective.
proof
Let \( x\in\ker\flat\ .\)
Then \( 0=\flat(x)(y)=K_\mathfrak{a}(x,y)\) for all \(y\in\mathfrak{g}\ ,\) that is, \(x\in \ker K_\mathfrak{a}\ .\)
But \( \ker K_\mathfrak{a}\) must be zero, so \(x=0\ .\)
proposition
Let \(\mathfrak{h}\) be an ideal in \(\mathfrak{g}\ .\) Define \[\mathfrak{h}^\perp=\{x\in\mathfrak{g}|K_\mathfrak{g}(x,y)=0 \quad \forall y\in \mathfrak{h}\}\] Then \(\mathfrak{h}^\perp\) is an ideal in \(\mathfrak{g}\ .\)
proof
Let \(g\in\mathfrak{g}\ ,\) \( h\in\mathfrak{h}\) and \(k\in\mathfrak{h}^\perp\ .\) Then \[K_\mathfrak{g}([g,k],h)=-K_\mathfrak{g}(k,[g,h])=0\] This shows that \([\mathfrak{g},\mathfrak{h}^\perp]\subset \mathfrak{h}^\perp\) and similarly \([\mathfrak{h}^\perp,\mathfrak{g}]\subset \mathfrak{h}^\perp\ .\)
definition - derived series, solvable
One defines a series of ideals of \(\mathfrak{g}\ ,\) the
derived series, as follows.\[\mathfrak{g}^{(0)}=\mathfrak{g}\]\[\mathfrak{g}^{(i+1)}=[\mathfrak{g}^{(i)},\mathfrak{g}^{(i)}]\]If, for some \(n\in\mathbb{N}\ ,\) \(\mathfrak{g}^{(n)}=0\) then \(\mathfrak{g}\) is called solvable. well defined
\( \mathfrak{g}^{(0)}\) is an ideal in \(\mathfrak{g}\ .\) Suppose that \(\mathfrak{g}^{(i)}\) is an ideal for \( i=0,\dots,n\ .\) Then \[ [\mathfrak{g},\mathfrak{g}^{(n+1)}]=[\mathfrak{g},[\mathfrak{g}^{(n)},\mathfrak{g}^{(n)}]]\ :\] \[ \subset [[\mathfrak{g},\mathfrak{g}^{(n)}],\mathfrak{g}^{(n)}]+[\mathfrak{g}^{(n)},[\mathfrak{g},\mathfrak{g}^{(n)}]]\ :\] \[\subset [\mathfrak{g}^{(n)},\mathfrak{g}^{(n)}]=\mathfrak{g}^{(n+1)}\] The inclusion \( [\mathfrak{g}^{(n+1)},\mathfrak{g}]\subset \mathfrak{g}^{(n+1)}\) follows in a similar way. By induction it follows that all the \( g^{(i)}\)'s are ideals in \(\mathfrak{g}\)
corollary
For \(i\leq j \ ,\) \(\mathfrak{g}^{(j)}\) is an ideal in \(\mathfrak{g}^{(i)}\ .\)
remark
If \(\mathfrak{g}\) is
solvable (that is, \(\mathfrak{g}^{(n)}=0\) for some \(n\)), then it contains an abelian ideal (namely \(\mathfrak{g}^{(n-1)}\)). proposition - solvable
If \(\mathfrak{g}\) is
solvable, then all its subalgebras and homomorphic images are. proof
Let \( \mathfrak{h}\) be a subalgebra.
Then \( \mathfrak{h}^{(0)}\subset\mathfrak{g}^{(0)}\ .\)
Assume \(\mathfrak{h}^{(i)}\subset\mathfrak{g}^{(i)}\ .\)
Then \[ \mathfrak{h}^{(i+1)}=[\mathfrak{h}^{(i)},\mathfrak{h}^{(i)}]\subset [\mathfrak{g}^{(i)},\mathfrak{g}^{(i)}]=\mathfrak{g}^{(i+1)}\] and the statement is proved by induction.
Similarly, let \(\phi:\mathfrak{g}\rightarrow \mathfrak{h}\) be surjective, and assume \(\phi:\mathfrak{g}^{(i)}\rightarrow \mathfrak{h}^{(i)}\) to be surjective.
Then \[\phi(\mathfrak{g}^{(i+1)})=\phi([\mathfrak{g}^{(i)},\mathfrak{g}^{(i)}])=[\phi(\mathfrak{g}^{(i)}),\phi(\mathfrak{g}^{(i)})]= [\mathfrak{h}^{(i)},\mathfrak{h}^{(i)}]=\mathfrak{h}^{(i+1)}\]
proposition - solvable quotient
If \(\mathfrak{h}\) is a solvable ideal such that \(\mathfrak{g}/\mathfrak{h}\) is solvable, then \(\mathfrak{g}\) is solvable.
proof
Say \((\mathfrak{g}/\mathfrak{h})^{(n)}=0\ .\)
Let \(\pi:\mathfrak{g}\rightarrow \mathfrak{g}/\mathfrak{h}\) be the canonical projection.
Then \(\pi(\mathfrak{g}^{(n)})=(\mathfrak{g}/\mathfrak{h})^{(n)}=0\) or \(\mathfrak{g}^{(n)}\subset \mathfrak{h}\ .\)
Since \(\mathfrak{h}^{(m)}=0\ ,\) \(\mathfrak{g}^{(n+m)}=(\mathfrak{g}^{(n)})^{(m)}\subset \mathfrak{h}^{(m)}=0\ ,\) implying the statement.
proposition
If \(\mathfrak{h}, \mathfrak{k}\) are solvable ideals of \(\mathfrak{g}\ ,\) then so is \(\mathfrak{h}+\mathfrak{k}\ .\)
proof
One has \[ (\mathfrak{h}+ \mathfrak{k})/\mathfrak{k}\equiv \mathfrak{h}/(\mathfrak{h}\cap \mathfrak{k}) \] Since \(\mathfrak{h}\) is solvable, so is \(\mathfrak{h}/(\mathfrak{h}\cap \mathfrak{k}) \ .\) But this implies that \(\mathfrak{h}+\mathfrak{k}\ ,\) since \(\mathfrak{k}\) is solvable.
proposition - radical
If \(\dim \mathfrak{g}<\infty\ ,\) there exists a unique maximal solvable ideal in \(\mathfrak{g}\ ,\) the
radical of \(\mathfrak{g}\ ,\)denoted by \( \mathrm{Rad\ }\mathfrak{g}\ .\) proof
Let \(\mathfrak{s}\) be a maximal solvable ideal in \(\mathfrak{g}\ .\)
Suppose \(\mathfrak{h}\) is another solvable ideal.
Then \(\mathfrak{s}+\mathfrak{h}\supset \mathfrak{s}\) is solvable, and by the maximality, \(\mathfrak{s}+\mathfrak{h}= \mathfrak{s}\ ,\) that is, \(\mathfrak{h}\subset\mathfrak{s}\ .\)
definition - semisimple
A Lie algebra \(\mathfrak{g}\) is called
semisimple if \(\mathrm{Rad\ }\mathfrak{g}=0\ .\) proposition - simple implies semisimple
If \(\mathfrak{g}\) is
simple, it is semisimple proof
For a
simple Leibniz algebra the derived series is stationary, that is, \(\mathfrak{g}^{(i)}=\mathfrak{g}\) for all \(i\in\mathbb{N}\ .\)
The only other possible ideal is \(0\ ,\) so this must be \(\mathrm{Rad\ }\mathfrak{g}\ .\)
proposition
\(\mathfrak{g}/\mathrm{Rad\ }\mathfrak{g}\) is
semisimple. proof
Let \([\mathfrak{h}]\) be a nonzero solvable ideal in \(\mathfrak{g}/\mathrm{Rad\ }\mathfrak{g}\ .\)
Then \(\mathfrak{h}+\mathrm{Rad\ }\mathfrak{g}\) strictly contains \(\mathrm{Rad\ }\mathfrak{g}\ ,\) which is in contradiction with its maximality.
Thus \(\mathfrak{h}\subset \mathrm{Rad\ }\mathfrak{g}\ ,\) that is, \([\mathfrak{h}]\) is the zero ideal in \(\mathfrak{g}/\mathrm{Rad\ }\mathfrak{g}\ .\)
proposition
If \(\ker K_\mathfrak{g}=0\ ,\) then \(\mathfrak{g}\) is
semisimple. proof
Let \(\mathfrak{h}\) be an abelian ideal of \(\mathfrak{g}\ .\) Take \(h\in\mathfrak{h}, g\in\mathfrak{g}\ .\) Then \( ad(h)ad(g)\) maps \( \mathfrak{g}\) to \(\mathfrak{h}\ .\) Thus \( (ad(h)ad(g))^2=0\ .\) This implies that \[ K_\mathfrak{g}(h,g)=\mathrm{tr}(ad(h)ad(g))=0\] In other words, \(\mathfrak{h}\subset\ker K_\mathfrak{g}=0\ .\) If there are no abelian ideals, then there are no solvable ideals besides \(0\ ,\) that is, \(\mathfrak{g}\) is semisimple.
theorem - common eigenvector
Let \(\mathfrak{g}\) be a solvable subalgebra of \(\mathfrak{gl}(\mathfrak{a})\ ,\) \(\dim\mathfrak{a}<\infty\ .\) If \(\mathfrak{a}\neq 0\ ,\) then \(\mathfrak{a}\) contains a common eigenvector for all endomorphisms in \(\mathfrak{g}\ .\)
proof
Induction on \(\dim\mathfrak{g}\ .\) Since \(\mathfrak{g}\) is solvable, it properly contains \(\mathfrak{g}^{(1)}=[\mathfrak{g},\mathfrak{g}]\ ,\) otherwise \(\mathfrak{g}^{(i)}=\mathfrak{g}\) for \( i\in\mathbb{N}\ .\)
Since \(\mathfrak{g}/[\mathfrak{g},\mathfrak{g}]\) is abelian, subspaces are ideals.
Take a subspace of codimension one.
Then the inverse image \(\mathfrak{h}\) in \(\mathfrak{g}\) is an ideal of codimension one which includes \([\mathfrak{g},\mathfrak{g}]\ .\)
\( \mathfrak{h}\) is solvable, and by the induction assumption there exists a vector \(a\in\mathfrak{a}\) such that \(a \) is an eigenvector for each \(h\in\mathfrak{h}\ ,\) that is, \[ h a=\lambda(h)a,\quad\lambda\in C^1(\mathfrak{h},\mathbb{C})\] (the exceptional case here is when \(\dim \mathfrak{h}=0\ .\)
In that case, \(\mathfrak{g}\) onedimensional and abelian, so one takes an eigenvector of a generator of \(\mathfrak{g}\)). Let \[\mathcal{W}=\{a\in\mathfrak{a}|x a=\lambda(x)a \quad \forall x\in \mathfrak{h}\}\] Now for \(x\in\mathfrak{g}\) and \(y\in\mathfrak{h}\) one finds \[ y x w=x y w-[x,y] w=\lambda(y) x w-\lambda([x,y])w\ .\] If one can prove that \(\lambda([x,y])=0\) then \(\mathcal{W}\) is invariant under the action of \(\mathfrak{g}\ .\)
Fix \(x\in \mathfrak{g}\ ,\) \(w\in\mathcal{W}\ .\)
Let \( n>0 \) be the smallest integer such that \(w, xw, \dots, x^n w\) are linearly dependent.
Let \(\mathcal{W}_0=0\) and \(\mathcal{W}_i\) be the subspace of \(\mathfrak{a}\) spanned by \(w, xw,\dots, x^{i-1} w\ .\)
It follows that \(\dim\mathcal{W}_n=n\) and \(W_{n+i}=W_n, i\geq 0\ .\)
Each \(\mathcal{W}_i\) is invariant under \(y\in\mathfrak{h}\ .\)
The matrix of \(y\) is upper triangular with eigenvalue \(\lambda(y)\) on the diagonal.
This implies \(\mathrm{tr}_{\mathcal{W}_i}(y)=i\lambda(y)\ .\)
Since \([x,y]\in\mathfrak{h}\ ,\) one also has \[\mathrm{tr}_{\mathcal{W}_n}([x,y])=i\lambda([x,y])\] Both \(x\) and \(y\) leave \(\mathcal{W}_n\) invariant, so the trace of \([x,y]\) must be zero.
Thus \(n\lambda([x,y])=0\ .\)
This shows that \(\mathcal{W}\) is invariant under the action of \(\mathfrak{g}\ .\)
Write \(\mathfrak{g}=\mathfrak{h}+\mathbb{C} z\ .\) Let \(w_0 \in\mathcal{W}\) be an eigenvector of \(z\) (acting on \(\mathcal{W}\)).
Then \(w_0\) is a common eigenvector of \(\mathfrak{g}\ .\)
definition - flag
Let \(\mathfrak{a}\) be a finite dimensional vectorspace (\(\dim\mathfrak{a}=n\)). A
flag is a chain of subspaces \[0=\mathfrak{a}_0\subset\mathfrak{a}_1\subset\dots\subset\mathfrak{a}_n=\mathfrak{a},\quad \dim\mathfrak{a}_i=i\]If \(x\in\mathrm{End}(\mathfrak{a})\ ,\) one says that \( x\) leaves the flag invariant if \(x \mathfrak{a}_i\subset \mathfrak{a}_i\) for \(i=1,\dots,n\ .\) theorem (Lie)
Let \(\mathfrak{g}\) be a solvable subalgebra of \(\mathfrak{gl}(\mathfrak{a}), \dim\mathfrak{a}=n<\infty\ .\) Then \( \mathfrak{g}\) leaves a flag in \(\mathfrak{a}\) invariant.
proof
It follows from the proof above that there exists a codimension one \( \mathfrak{g}\)-invariant subspace.
Let that be \(\mathfrak{a}_{n-1}\ .\)
Repeat the argument starting with \(\mathfrak{a}_{n-1}\) instead of \(\mathfrak{a}_{n}\) and use induction.
lemma - flag of ideals
Let \( \mathfrak{g}\) be solvable. Then there exists a flag of ideals \[ 0=\mathfrak{g}_0\subset\mathfrak{g}_1\subset\dots\subset\mathfrak{g}_n=\mathfrak{g},\quad \dim\mathfrak{g}_i=i\]
proof
Let \(d_1:\mathfrak{g}\rightarrow \mathfrak{gl}(\mathfrak{a})\) be a finite dimensional representation of \(\mathfrak{g}\ .\)
Then \(d_1(\mathfrak{g})\) is solvable, and stabilizes a flag in \(\mathfrak{a}\ .\)
Take \(\mathfrak{a}=\mathfrak{g}\) and \(d_1=\mathrm{ad}\ ,\) then the \(\mathfrak{g}_i\) are ideals (since they are \(\mathfrak{g}\)-invariant) and they obey the flag condition.
lemma
Let \(\mathfrak{g}\) be solvable. Then \(x\in\mathfrak{g}^{(1)}\) implies that \( \mathrm{ad}_\mathfrak{g}(x)\) is nilpotent.
proof
From the flag of ideals construct a basis. relative to this basis the matrix of \(\mathrm{ad}_\mathfrak{g}(y), y\in \mathfrak{g}\ ,\) is upper triangular.
Thus the matrix of \(\mathrm{ad}_\mathfrak{g}(x), x\in \mathfrak{g}^{(1)}\) is strictly upper triangular, and therefore nilpotent.
remark
In the next lecture it is shown that this implies that \(\mathfrak{g}^{(1)}\) is
nilpotent (to be defined). |
2019-10-14 17:21
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Unfoldings
James Murdock (2006), Scholarpedia, 1(12):1904. doi:10.4249/scholarpedia.1904 revision #91898 [link to/cite this article]
Contents Introduction
A mathematical model of a real-world problem is usually based onidealized assumptions. If the model proves inadequate, it can beimproved by adding small terms that were neglected at first. A model obtained by adding small parametersto a given system is called an
unfolding of the originalsystem. (One may picture the various behaviors of the expanded system as being hidden or "folded up" when the parameters are set to zero.)
For instance, a pair of coupled oscillators that are first modelled as a conservative system in exact resonance might be improved by adding three small parameters representing damping in each oscillator and detuning of the resonance. Perturbation methods may then be used to obtain approximate solutions expanded in the small parameters, and bifurcation analysis may be used to determine the qualitative changes in the behavior of the system in a neighborhood of the original model.
The mathematical theory of unfolding originated in the theory of singularities of mappings and in catastrophe theory. (For an introduction from this point of view, see Bruce and Giblin 1992.) In dynamical systems, unfolding means the attempt to exhibit all possible behaviors for systems close to a given original system (sometimes called the
organizing center of the unfolding) byadding a finite number \(k\) of small parameters \(\mu_1,\dots,\mu_k\ .\) The number \(k\) is called the codimension of the organizing center. In order to begin, it is necessary to specify some space of admissible systems (at least a topological space, usually asmooth manifold) and some equivalence relation on this spaceexpressing the idea that two equivalent systems "have the samebehavior". Under these conditions it makes sense to specify anoriginal system (or organizing center) and ask whether there existsa \(k\)-parameter family (for some \(k\)) of systems that intersects each equivalence class in a neighborhood of the organizing center. If so, the goal of the theory can be achieved. If not, the organizing center is said to have "infinite codimension". Unfoldings of Matrices
Many finite-dimensional linear systems can be represented by asquare matrix, whether it be the matrix of a linear transformationor of a linear system of differential equations \(\dot x = Ax\ .\) In either case, a natural equivalence relation is similarity. Suppose that \(A_0\) is a given \(n\times n\) matrix, taken as the organizingcenter. We wish to construct a family \(A(\mu_1,\dots,\mu_k)\) ofmatrices that depends continuously (or better, smoothly) on\(\mu_1,\dots,\mu_k\ ,\) reduces to \(A_0\) when \(\mu_1=\cdots=\mu_k=0\ ,\) and intersects each similarity class near \(A_0\ .\) We may assume \(A_0\) is in Jordan normal form, but it cannot always be the case that\(A(\mu_1,\dots,\mu_k)\) will be in Jordan form for all\(\mu_1,\dots,\mu_k\) near zero, because the Jordan form of a matrix does not (always) depend continuously on the matrix. Since thesimilarity class \(M\) of \(A_0\) is a smooth submanifold of \(\R^{n^2}\) (the space of \(n\times n\) matrices), we require that \(A(\mu_1,\dots,\mu_k)\ ,\) for \(\mu_1,\dots,\mu_k\) near zero, be a smoothly embedded submanifold transverse to \(M\ .\) Such an unfolding of \(A_0\) is called
versal (an abbreviation of transversal), and automatically intersects all similarity classes near \(A_0\) (even though these classes have various dimensions). The smallest possible number \(k\) of parameters will equal the codimension (in the usual manifold sense) of \(M\) in \({\mathbb R}^{n^2}\ ;\) this explains the use of "codimension" as defined above. A versal unfolding of this kind is called miniversal.
If \(n=2\) and \(A_0=\begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}\ ,\) the codimension is two and twodifferent miniversal unfoldings are \[\begin{pmatrix} \mu_2&1\\ \mu_1&\mu_2\end{pmatrix}\] and \(\begin{pmatrix} 0&1\\ \mu_1&\mu_2\end{pmatrix}\ .\) The first form is known as a
striped matrix. We may observe that this striped matrix commuteswith \(A_0^*\) (the adjoint or conjugate transpose of \( A_0 \)), and it is alwaystrue that a miniversal unfolding of a matrix can be found byobtaining the most general matrix that commutes with \( A_0^*\ .\) Thesecond form illustrates that the striped matrix is not the simplestchoice for the unfolding, if "simplest" is interpreted as "having the most zero entries". (The striped matrix has the advantagethat this unfolding is not only transverse to \(M\) but is orthogonal with respect to the inner product \(\langle P,Q\rangle = {\rm tr}\, PQ^*\ .\))These unfoldings (for matrices of any size) are due to Arnold andare explained in Arnold (1988, section 30), Wiggins (2003, section20.5), and Murdock (2003, chapter 3). Relation to Normal Forms
There is a close relationship between unfoldings and normal formideas. Any smooth one-parameter family \(A(\varepsilon)\) of matrices with \(A(0)=A_0\) can be embedded (up to similarity) into any miniversalunfolding \(A(\mu_1,\dots,\mu_k)\) of \(A_0\ ;\) that is, there is a smooth family of matrices \(T(\varepsilon)\) with \(T(0)=I\) such that\[T(\varepsilon)^{-1}A(\varepsilon)T(\varepsilon) = A(\mu_1(\varepsilon),\dots,\mu_k(\varepsilon))\ ,\]where the functions \(\mu_i(\varepsilon)\) are smooth. The form of theunfolding, as well as the power series expansions of \(\mu_i(\varepsilon)\) canbe computed by normal form methods. Writing \(U(\varepsilon)=I+\varepsilon U_1+\cdots\) and \(A(\varepsilon)=A_0+\varepsilon A_1 +\cdots\ ,\) and setting \(T(\varepsilon)^{-1}A(\varepsilon)T(\varepsilon)=B(\varepsilon)=A_0 + \varepsilon B_1 +\cdots\ ,\) one finds that\[L_{A_0} U_1 = A_1-B_1\ ,\]where \(L_P Q = QP-PQ\ .\) Similar homological equations exist at higher orders. Choosing a complement to the image of \(L_{A_0}\) (that is, choosing a
normal form style) fixes the form of theunfolding to which \(B(\varepsilon)\) belongs, and the rest of the computation determines the \(\mu_i(\varepsilon)\ .\) The striped matrix unfolding comes from the inner product normal form style \(\ker\Lambda_{A_0^*}\ ,\) and the second type of unfolding illustrated above comes from the simplified normal form style. Unfoldings of Dynamical Systems
For nonlinear dynamical systems, it is much more difficult todefine an appropriate space of systems and equivalence relationwith which to begin. Any suitable space of systems will beinfinite-dimensional, and under the most natural equivalencerelations (either topological equivalence or topological conjugacy), most systems turn out to have infinite codimension,so a versal unfolding is impossible. We must either restrictattention to those few (but important) systems that have finitecodimension with respect to topological equivalence, or else adopta coarser equivalence relation. One often-used equivalencerelation is
static equivalence, in which attention is limitedto the equilibrium solutions.
An unfolding of a dynamical system under static equivalence is one that exhibits all possible bifurcations of the equilibrium (rest) points, up to topological equivalence of the set of equilibria. It is easiest to localize the problem to the bifurcations of a single equilibrium point of the organizing center. Since no bifurcations take place in hyperbolic directions, it is enough to unfold the system on its center manifold. The various cases are classified by the eigenvalues of the Jacobian matrix (i.e., the linearized system at the equilibrium) on the imaginary axis.
A single zero eigenvalue
The simplest case is a system with a single zero eigenvalue at the equilibrium, leading to a center manifold of dimension one. Since the behavior of the system should be dominated by the lowest order term, one considers a (scalar) organizing center of the form \(\dot x = x^k\) (for \(k\) a positive integer). An unfolding under static equivalence is \[\dot x = \mu_1 +\mu_2x+\cdots+\mu_{k-1}x^{k-2} + x^k\ ,\] the interesting point being the absence of \(x^{k-1}\ .\) For instance,
the unfolding of \(\dot x = x^2\) is \(\dot x = \mu_1 + x^2\ ,\) which exhibits a saddle-node bifurcation as \(\mu_1\) is varied. The unfolding of \(\dot x = x^3\) is \(\dot x = \mu_1 + \mu_2x + x^3\ .\) If \(\mu_1=0\) this gives a pitchfork bifurcation as \(\mu_2\) is varied; \(\mu_1\) is an "imperfection parameter" that splits the pitchfork into a saddle-node bifurcation and a continuation curve (i.e., a curve of equilibria that does not bifurcate).
This sort of analysis is very close to the original use of unfoldings in singularity theory. For further information see section 6.3 of Murdock (2003), and for complete details of this approach see Golubitsky and Schaeffer (1985) and Golubitsky et al. (1988). (In the last references, one of the unfolding parameters is treated as the bifurcation parameter and is not counted in the codimension.)
A conjugate pair
The organizing center \[\begin{pmatrix} \dot x \\ \dot y\end{pmatrix} =\begin{pmatrix} 0 &-1\\1 &0\end{pmatrix} \begin{pmatrix} x\\y\end{pmatrix}+ \alpha(x^2+y^2) \begin{pmatrix} x\\y\end{pmatrix} + \beta(x^2+y^2)\begin{pmatrix} -x\\y\end{pmatrix}\ ,\] which is in (semisimple) normal form truncated at the quadratic terms and has a conjugate pair of eigenvalues \(\pm \imath\ ,\) takes the form \[\dot r = \alpha r^3\] \[\dot\theta = 1 + \beta r^2\ .\] If \(\alpha\ne 0\ ,\) an unfolding under local topological equivalence (but not topological conjugacy) is \[\dot r = \mu_1 r + \alpha r^3\] \[\dot\theta = 1 + \beta r^2\ .\] This exhibits an Andronov-Hopf bifurcation as \(\mu_1\) is varied.
A nonsemisimple double eigenvalue
For the case of a double zero eigenvalue with a nonsemisimple linear part, the organizing center is \[\begin{pmatrix} \dot x\\ \dot y\end{pmatrix} = \begin{pmatrix} 0&1\\0&0\end{pmatrix}\begin{pmatrix} x\\y\end{pmatrix} + \begin{pmatrix} 0\\ \alpha x^2+\beta xy\end{pmatrix}\ ,\] with quadratic term in (simplified) normal form. Assuming that \(\alpha\ne 0\ ,\) an unfolding is \[ \begin{pmatrix} \dot x\\ \dot y\end{pmatrix} = \begin{pmatrix} 0\\ \mu_1\end{pmatrix} + \begin{pmatrix} 0&1\\0&\mu_2\end{pmatrix}\begin{pmatrix} x\\ y\end{pmatrix} + \begin{pmatrix} 0\\ \alpha x^2+\beta xy\end{pmatrix}\ .\] It is remarkable that this can be proved to be an unfolding under topological equivalence. (The proof is difficult and uses one of or another of several "blowing-up" techniques.) For further discussion see Bogdanov-Takens bifurcation.
Comparing this unfolding to the matrix unfolding of \(\begin{pmatrix} 0&1\\0&0\end{pmatrix}\) given above, it is seen that the codimension is the same but that one unfolding parameter appears in the constant term rather than in the matrix. This phenomenon is typical, as can be seen using asymptotic unfoldings, sketched below.
Additional Examples
For additional examples of unfoldings presented in an elementary manner, see Kuznetsov (1998), Guckenheimer and Holmes (1986), and Wiggins (2003). For a detailed treatment of some unfoldings with respect to topological equivalence, proved via blowup techniques, see Dumortier et al. (1991).
Asymptotic Unfoldings
As in the case of matrices, unfoldings of dynamical systems can be approached from a normal form viewpoint. Beginning with an organizing center \[\dot x = Ax + a_1(x) + a_2(x)+\cdots\] in normal form (of some chosen style), consider an arbitrary one-parameter perturbation of the following form (where the degree of a term is the subscript plus \(1\)): \[\dot x = Ax + a_1(x) + a_2(x)+\cdots\ :\]
\[+\varepsilon(p + Bx + b_1(x) + b_2(x) + \cdots) + \cdots\ .\]
The final \(\cdots\) refer to higher powers of \(\varepsilon\ .\) Notice that the \(\varepsilon\) part contains a constant term \(p\ ,\) not present in the unperturbed system. Normal form methods can be applied to simplify \(p\ ,\) \(B\ ,\) \(b_i\ ,\) and so forth. Whatever coefficients cannot be eliminated become unfolding parameters expressed as functions of \(\varepsilon\ .\) Stopping the calculation at a finite degree in \(x\) gives an unfolding with finite codimension, but it is (usually) not a versal unfolding with respect to topological equivalence. Nevertheless, it is often possible to prove that the unfolding correctly exhibits specific features of the behavior. Under generic hypotheses on the quadratic terms \(a_2\ ,\) the number of unfolding parameters in the constant and linear terms (coming from \(p\) and \(B\)) always equals the codimension of the matrix unfolding of \(A\ ,\) explaining the remark in the last section. Asymptotic unfoldings have been used informally without a name for many years, and a number of them are computed by Elphick et al. (1992). A general treatment is given in section 6.4 of Murdock (2003). (The restriction to the simplified normal form style has since been removed, see Murdock and Malonza.) This approach to unfoldings makes the computation of unfoldings quite easy, as illustrated (in the references)by an example of codimension 14. (Deriving useful dynamical conclusions from unfoldings of high codimension is another matter altogether.)
References Arnold V. I. (1988) Geometrical Methods in the Theory of Ordinary Differential Equations.Springer, New York, second edition. Bruce J. W. and Giblin P. G. (1992) Curves and Singularities. Cambridge University Press, Cambridge, England, second edition. Dumortier, F. and Roussarie, R. and Sotomayor, J. (1991) Generic three-parameter families of planar vector elds, unfoldings of saddle, focus, and elliptic singularities with nilpotent linear parts. Lecture Notes in Mathematics, 1480:1-164. Elphick C., Tirapegui E., Brachet M. E., Coullet P., and Iooss G. (1987) A simple global characterization for normal forms of singular vector fields. Physica D, 29:95–127. Golubitsky M. and Schaeffer D.G. (1985) Singularities and Groups in Bifurcation Theory, volume 1. Springer, New York. Golubitsky M., Stewart I., and Schaeffer D.G. (1988) Singularities and Groups in Bifurcation Theory, volume 2. Springer, New York. Guckenheimer J. and Holmes P. (1986) Nonlinear Oscillations, Dynamical Systems, and Bifurcations of Vector Fields, Corrected second printing, Springer, N.Y. Kuznetsov Y.A. (1998) Elements of Applied Bifurcation Theory, Second edition, Springer, N.Y. Murdock J. (2003) Normal Forms and Unfoldings for Local Dynamical Systems. Springer, New York. Murdock J. and Malonza D.M. Improved computation of asymptotic unfoldings. In preparation. Wiggins S. (2003) Introduction to Applied Nonlinear Dynamical Systems and Chaos. Springer, New York, second edition. Internal references Yuri A. Kuznetsov (2006) Andronov-Hopf bifurcation. Scholarpedia, 1(10):1858. Jack Carr (2006) Center manifold. Scholarpedia, 1(12):1826. James Murdock (2006) Normal forms. Scholarpedia, 1(10):1902. Jeff Moehlis, Kresimir Josic, Eric T. Shea-Brown (2006) Periodic orbit. Scholarpedia, 1(7):1358. Yuri A. Kuznetsov (2006) Saddle-node bifurcation. Scholarpedia, 1(10):1859. |
Some of its key features are Non-orthogonal tight-binding with a minimal sp (d,f…) basis Self-consistent and spin polarized calculations Standard DFT-like properties including thermodynamic and electrical levels, vibrational modes/intensities/Raman cross-sections, … LDA+U like corrections for strongly correlated electrons Spin-orbit and non-collinear spin Electron transport simulation through nano-structures Excited state calculations (time dependent DFTB)
Key ideas
The equations for DFTB come from making an expansion around the density functional energy for a reference system. This is chosen to be a set of neutral atoms in a condensed phase environment. The non-self consistent hamiltonian, \(H^0\), is constructed from interactions between the atomic orbitals on pairs of atoms, while the double-counting terms are grouped together into a short range pairwise repulsive potential, \(E^{rep.}(R_{AB})\). DFTB then solves a set of Kohn-Sham like single particle equations for the quantum mechanical states of the system, and uses this to construct the single particle density matrix \(\rho\). The atomic orbitals on neighbouring atoms are not orthogonal, so the hamiltonian and energy are given by:
$$ \begin{aligned} H^0_{jk} c_{ik} =& \;\varepsilon_i S_{jk} c_{ik} \\ \rho_{jk} =& \sum\limits_{i} f(\varepsilon_i) c_{ij} c_{ik}\\ E =& \;Trace \left(\rho H^0\right) + \sum\limits_{ab} E^{rep.}\left(R_{ab}\right) \end{aligned} $$ where \(f(\varepsilon_i)\) is the occupation of state \(i\) which has a single particle energy \(\varepsilon_i\).
DFTB is usually applied in its self-consistent form (SCC-DFTB or DFTB2), where the effects of charge transfer (and spin polarization when relevant) are included in equations which must now be solved iteratively. Terms in the fluctuation of charge (and magnetisation) from the neutral reference system are included.
Some typical applications are shown below
In diamond crystals, a common defect is a complex between a missing “vacant” site and a nitrogen impurity. These VN centres have some remarkable properties.
Topological insulators are a new category of materials which are electrically insulating inside while being
guaranteed to be conductive on the surface. References
[Bibtex]
@Article{strathprints31166,author = {B. Aradi and B. Hourahine and Th. Frauenheim},title = {DFTB+, a sparse matrix-based implementation of the DFTB method},journal = {Journal of Physical Chemistry A},year = {2007},volume = {111},number = {26},pages = {5678--5684},month = {July},abstract = {A new Fortran 95 implementation of the DFTB (density functional-based tight binding) method has been developed, where the sparsity of the DFTB system of equations has been exploited. Conventional dense algebra is used only to evaluate the eigenproblems of the system and long-range Coulombic terms, but drop-in O(N) or O(N-2) modules are planned to replace the small code sections that these entail. The developed sparse storage structure is discussed in detail, and a short overview of other features of the new code is given.},doi = {10.1021/jp070186p},keywords = {electronic structur calculations, density functional theory, tight binding method, ewald sums, convergence, simulations, potentials, molecules, Physics, Physical and Theoretical Chemistry},url = {http://strathprints.strath.ac.uk/31166/}} |
I read the Devinatz-Hopkins-Smith proof of the nilpotence conjectures last year, and while I followed along sentence to sentence I don't think I understood much of the motivating reasons for why what they did was a sensible mode of proof. I intend to go back and figure some of these conceptual pieces out; this question is a step in that direction.
The proof of smash nilpotence is, on the face of it, accomplished by a sequence of interpolations. The original statement is:
(Smash nilpotence:) If a map off a finite spectrum induces the zero map on $MU$-homology, then the map is in fact smash nilpotent.
Without any hassle, they reduce this to the following statement:
(Special case of Hurewicz nilpotence:) Suppose $\alpha$ is an element of the homotopy $\pi_* R$ of an associative ring spectrum $R$ of finite type. If $\alpha$ is in the kernel of the map $\pi_* R \to MU_* R$ induced by smashing $R$ with the unit map $S \to MU$, then $\alpha$ is nilpotent.
To address this question, they interpolate between the sphere spectrum and $MU$ in two ways. First, they produce a sequence of spectra $X(n)$, each given by the Thom spectrum associated to the composite $\Omega SU(n) \to \Omega SU \to BU$, sort of a restriction of the Bott map. The colimit of the $X(n)$ is $MU$, and hence $X(N)_* \alpha$ is zero for some sufficiently large $N$, where $\alpha$ is considered as a map $S^t \to R$. Since $X(1)$ is the sphere spectrum, the new goal is to show that nilpotence in $X(n+1)$-homology forces nilpotence in $X(n)$-homology for any $n$, then to walk down from $X(N)$ to the sphere spectrum. To move between $X(n+1)$ and $X(n)$, they interpolate between
these spectra by pulling $X(n+1)$ apart using the filtered James construction; this results in an increasing sequence of $X(n)$-module spectra $F_{n, k}$ satisfying $F_{n, 0} \simeq X(n)$ and converging to $X(n+1)$ in the limit.
The rest of the argument falls into two pieces:
1) If $\alpha: S^t \to R$ is nilpotent in $X(n+1)$-homology then it induces the zero map in $F_{n,p^k-1}$-homology for some large $k$ --- that is, our argument continues somewhere in the approximating tower between $X(n+1)$ and $X(n)$.
2) If it induces the zero map in $F_{n, p^k-1}$-homology it also induces the zero map in $F_{n, p^{k-1}-1}$-homology.
To prove part 1, they investigate the $X(n+1)$-based Adams spectral sequence $$\mathrm{Ext}^{*, *}_{X(n+1)_* X(n+1)}(X(n+1)_*, X(n+1)_* F_{n, p^k-1} \wedge R) \Rightarrow (F_{n, p^k-1})_* R.$$ The key is the existence of vanishing lines in these spectral sequences, where the slope of the vanishing line can be made small by making $k$ large. In order to establish these vanishing lines, they perform a sequence of approximations, finishing with spectral sequences with the following $E_2$ terms: $$\mathrm{Ext}^{*, *}_{\mathbb{F}_p[b_n]}(\mathbb{F}_p, \mathbb{F}_p\{1, \ldots, b_n^{p^k-1}\}),$$ $$\mathrm{Ext}^{*, *}_{\mathbb{F}_p[b_n^{p^k}]}(\mathbb{F}_p, \mathbb{F}_p).$$
My question is:Is there a geometric interpretation for the stacks associated to the Hopf algebroids above? ( --or any of the other Hopf algebroids involved which I haven't listed.) Or: what's the geometric content of Part II of D-H-S?
For instance, it's well-known that there's a spectral sequence computing the homotopy groups of real K-theory whose input corresponds to the moduli stack of quadratics and translations. This stack is supposed to parametrize the available multiplicative groups over some non-algebraically closed field, which provides some geometric insight into the problem. I'd like to know if there's some kind of geometry that corresponds to the controlling stacks that sit at the bottom of the D-H-S argument.
Note: Of course, a positive answer to this question as phrased might not mean much. The process that D-H-S uses to reduce to this much smaller Ext calculation is an extremely lossy one with the very clear intention of just getting at the existence of a vanishing line. The geometry of these bottom stacks may have very little to say about the geometry of the stacks we started with. |
Say I am measuring a quantity $x$ in physical system whose true value is approximately sinusoidal in time. I have an instrument to sample this quantity, for which the manufacturer gives an accuracy spec. Can I use this accuracy spec $(\Delta x?)$ to compute the uncertainty of the sample mean $\bar{x}$ of the quantity based on a finite amount of samples, $N$?
My guess is that the naive calculation would be
$$ \Delta{\bar{x}} = \frac{\Delta{x}}{\sqrt{N}}, $$
but this would assume the samples are all statistically independent. Can I use this assumption, or should $N$ be based on the number of periods sampled, or something else? |
This question is from HRW's Fundamentals of Physics. It goes:
A tall, cylindrical chimney falls over when its base is ruptured. Treat the chimney as a thin rod of length $55.0\;m$. At the instant it makes an angle of $35.0^\circ$ with the vertical as it falls, what is the radial acceleration of the top?
What I did is this:
The energy of the top-most point at the chimney must've been conserved. Therefore: $$K_0 + U_0 = K + U$$ $$\Rightarrow mgl = \frac{1}{2}mv^2 + mgl\cos{35^\circ}$$ $$\Rightarrow \frac{v^2}{l} = 2g(1 - \cos{35^\circ})$$
Now $\frac{v^2}{l}$ is the radial acceleration, considering $l$ to be the length of the chimney, in other words the radius of the circle which is the path that the top of the chimney takes. Calculating that expression gives us $\frac{v^2}{l} = a_r = 3.54\;m/s^2$. However the correct answer according to the book is $a_r = 5.32\;m/s^2$.
Where did I make a wrong assumption? |
Let $a_n>0$ and $b_n>0$ be two strictly declining sequences such that the series $$\sum_{n=1}^\infty \frac{a_n}{b_n}$$ is convergent. For $\sigma>0$ define $$f^N(\sigma) = \sum_{n=1}^N \frac{a_n}{b_n + \sigma/N}$$ Is it generally true that $\lim_{N \to \infty} f^N(\sigma)$ is independent of $\sigma$ or are there counterexamples?
Remarks:
The answer is trivially true if $\sum \frac{a_n}{b_n^2}$ is convergent as well. In this case $$\left|\frac{d}{d\sigma} f^N(\sigma)\right| = \frac{1}{N}\sum_{n=1}^{N} \frac{a_n}{(b_n+\sigma/N)^2} \leq \frac{1}{N}\sum_{n=1}^N \frac{a_n}{b_n^2} \to 0$$ More interesting is the case of divergent $\sum \frac{a_n}{b_n^2}$, e.g. $a_n = c^{-2n}$ and $b_n = c^{-n}$, or $a_n = 1/n^4$ and $b_n = 1/n^2$. In both these cases $$ \frac{d}{d\sigma} \left.f^N(\sigma)\right|_{\sigma=0} \to 1, $$ but from playing around with Maple and Mathematica I have the suspicion that $\frac{d}{d{\sigma}}f^N(\sigma)$ converges to $0$ for every $\sigma>0$, i.e. $f^N(\sigma)$ becomes non-differentiable in the limit. If that is true it would still allow for the possibility of $f^N(\sigma)$ converging pointwise to a constant. Eventually I am interested in the case $a_n = n^2I_n(K)^2$ and $b_n=I_n(K)$, where $I_n(K)$ is the modified Bessel function of the first kind. It might be related to the Stolz-Cesaro theorem, but I can't figure out how.
Any help or pointer to relevant literature is very much appreciated! |
I imagine you'll get many answers to your question- here is my humble attempt.
From the Mathematical point of view, the first thing I always remember is to think of equations and formulas as part of the sentence- they are not separate objects that stand alone on the page From the typesetting point of view, it's always good to try and allow LaTeX to do the heavy lifting and tedious tasks for you. When I say 'tedious' tasks, I mean things such as automatic enumerations of environments cross referencing that is updated automatically (after 2 compilations) pagination- allow your figures and tables to
float, and try to avoid manually specifying page breaks
When it comes to Mathematical typesetting, the first package you should explore is the
amsmath package- once you're comfortable with it, and perhaps need additional enhancements, you can study the
mathtools package, which supplements it.
Of course, one final detail is to keep your code as tidy as possible so that it can be read easily by you (and perhaps others) in the future.
I've included a very simple sample document below that I hope might get you started- happy TeXing!
\documentclass{article}
\usepackage[left=3cm,right=3cm,top=0cm,bottom=2cm]{geometry} % page settings
\usepackage{amsmath} % provides many mathematical environments & tools
\setlength{\parindent}{0mm}
\begin{document}
\title{MTH 251: Week 2 lab write up}
\author{C. M. Hughes}
\date{\today}
\maketitle
\subsection*{Lab activity 1.2.4}
Find the difference quotient of $f(x)$ when $f(x)=x^3$.
We proceed as demonstrated in the lab manual; assuming that $h\ne 0$
we have
\begin{align*}
\frac{f(x+h)-f(x)}{h} & = \frac{(x+h)^3-x^3}{h} \\
& = \frac{x^3+3x^2h+3xh^2+h^3 - x^3}{h}\\
& = \frac{3x^2h+2xh^2+h^3}{h}\\
& = \frac{h(3x^2+2xh+h^2)}{h}\\
& = 3x^2+2xh+h^2
\end{align*}
\subsection*{Lab activity 2.3.4}
Use the definition of the derivative to find $f'(x)$ when $f(x)=x^{\frac{1}{4}}$.
Using the definition of the derivative, we have
\begin{align*}
f'(x) &= \lim_{h\rightarrow 0}\frac{(x+h)^{1/4}-x^{1/4}}{h} \\
&= \lim_{h\rightarrow 0}\frac{(x+h)^{1/4}-x^{1/4}}{h}\cdot \frac{((x+h)^{1/4}+x^{1/4})((x+h)^{1/2}+x^{1/2})}{((x+h)^{1/4}+x^{1/4})((x+h)^{1/2}+x^{1/2})}\\
&= \lim_{h\rightarrow 0}\frac{(x+h)-x}{h((x+h)^{1/4}+x^{1/4})((x+h)^{1/2}+x^{1/2})} \\
&= \lim_{h\rightarrow 0}\frac{1}{((x+h)^{1/4}+x^{1/4})((x+h)^{1/2}+x^{1/2})} \\
&= \frac{1}{(x^{1/4}+x^{1/4})(x^{1/2}+x^{1/2})} \\
&= \frac{1}{(2x^{1/4})(2x^{1/2})} \\
&= \frac{1}{4x^{3/4}} \\
&= \frac{1}{4}x^{-3/4}
\end{align*}
Note: the key observation here is that
\begin{align*}
a^4-b^4 &= (a^2-b^2)(a^2+b^2) \\
&= (a-b)(a+b)(a^2+b^2),
\end{align*}
with
\[
a = (x+h)^{1/4}, \qquad b = x^{1/4},
\]
which allowed us to rationalize the denominator.
\end{document} |
1,062 6
Quick question, what is the approach to this problem?
Keep in mind I am supposed to use the Fundamental Theorem of Line Integrals.
[tex]\int_{C} 2ydx + 2xdy [/tex]
Where C is the line segment from (0,0) to (4,4).
Unless I am missing something I need to make that into the form of [tex]\vec{F} \cdot \vec_d{R}[/tex] to use the F.T. of L.I., but I have no clue where to start there. Thanks.
Keep in mind I am supposed to use the Fundamental Theorem of Line Integrals.
[tex]\int_{C} 2ydx + 2xdy [/tex]
Where C is the line segment from (0,0) to (4,4).
Unless I am missing something I need to make that into the form of [tex]\vec{F} \cdot \vec_d{R}[/tex] to use the F.T. of L.I., but I have no clue where to start there. Thanks. |
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Now showing items 1-10 of 18
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 ...
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 ...
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 ...
Multi-strange baryon production at mid-rapidity in Pb-Pb collisions at $\sqrt{s_{\rm NN}}$ = 2.76 TeV
(Elsevier, 2014-01)
The production of ${\rm\Xi}^-$ and ${\rm\Omega}^-$ baryons and their anti-particles in Pb-Pb collisions at $\sqrt{s_{\rm NN}}$ = 2.76 TeV has been measured using the ALICE detector. The transverse momentum spectra at ...
Measurement of charged jet suppression in Pb-Pb collisions at √sNN = 2.76 TeV
(Springer, 2014-03)
A measurement of the transverse momentum spectra of jets in Pb–Pb collisions at √sNN = 2.76TeV is reported. Jets are reconstructed from charged particles using the anti-kT jet algorithm with jet resolution parameters R ...
Two- and three-pion quantum statistics correlations in Pb-Pb collisions at √sNN = 2.76 TeV at the CERN Large Hadron Collider
(American Physical Society, 2014-02-26)
Correlations induced by quantum statistics are sensitive to the spatiotemporal extent as well as dynamics of particle-emitting sources in heavy-ion collisions. In addition, such correlations can be used to search for the ...
Exclusive J /ψ photoproduction off protons in ultraperipheral p-Pb collisions at √sNN = 5.02TeV
(American Physical Society, 2014-12-05)
We present the first measurement at the LHC of exclusive J/ψ photoproduction off protons, in ultraperipheral proton-lead collisions at √sNN=5.02 TeV. Events are selected with a dimuon pair produced either in the rapidity ... |
Linear codes and ciphers¶ Codes¶
A linear code of length \(n\) is a finite dimensional subspace of \(GF(q)^n\). Sage can compute with linear error-correcting codes to a limited extent. It basically has some wrappers to GAP and GUAVA commands. GUAVA 2.8 is not included with Sage 4.0’s install of GAP but can be installed as an optional package.
Sage can compute Hamming codes
sage: C = codes.HammingCode(GF(3), 3)sage: C[13, 10] Hamming Code over GF(3)sage: C.minimum_distance()3sage: C.generator_matrix()[1 0 0 0 0 0 0 0 0 0 1 2 0][0 1 0 0 0 0 0 0 0 0 0 1 2][0 0 1 0 0 0 0 0 0 0 1 0 2][0 0 0 1 0 0 0 0 0 0 1 1 1][0 0 0 0 1 0 0 0 0 0 1 1 2][0 0 0 0 0 1 0 0 0 0 2 0 2][0 0 0 0 0 0 1 0 0 0 1 2 1][0 0 0 0 0 0 0 1 0 0 2 1 1][0 0 0 0 0 0 0 0 1 0 2 2 0][0 0 0 0 0 0 0 0 0 1 0 1 1]
the four Golay codes
sage: C = codes.GolayCode(GF(3))sage: C[12, 6, 6] Extended Golay code over GF(3)sage: C.minimum_distance()6sage: C.generator_matrix()[1 0 0 0 0 0 2 0 1 2 1 2][0 1 0 0 0 0 1 2 2 2 1 0][0 0 1 0 0 0 1 1 1 0 1 1][0 0 0 1 0 0 1 1 0 2 2 2][0 0 0 0 1 0 2 1 2 2 0 1][0 0 0 0 0 1 0 2 1 2 2 1]
as well as binary Reed-Muller codes, quadratic residue codes, quasi-quadratic residue codes, “random” linear codes, and a code generated by a matrix of full rank (using, as usual, the rows as the basis).
For a given code, \(C\), Sage can return a generator matrix, a check matrix, and the dual code:
sage: C = codes.HammingCode(GF(2), 3)sage: Cperp = C.dual_code()sage: C; Cperp[7, 4] Hamming Code over GF(2)[7, 3] linear code over GF(2)sage: C.generator_matrix() [1 0 0 0 0 1 1] [0 1 0 0 1 0 1] [0 0 1 0 1 1 0] [0 0 0 1 1 1 1]sage: C.parity_check_matrix() [1 0 1 0 1 0 1] [0 1 1 0 0 1 1] [0 0 0 1 1 1 1]sage: C.dual_code()[7, 3] linear code over GF(2)sage: C = codes.HammingCode(GF(4,'a'), 3)sage: C.dual_code()[21, 3] linear code over GF(4)
For \(C\) and a vector \(v\in GF(q)^n\), Sage can try to decode \(v\) (i.e., find the codeword \(c\in C\) closest to \(v\) in the Hamming metric) using syndrome decoding. As of yet, no special decoding methods have been implemented.
sage: C = codes.HammingCode(GF(2), 3)sage: MS = MatrixSpace(GF(2),1,7)sage: F = GF(2); a = F.gen()sage: v = vector([a,a,F(0),a,a,F(0),a])sage: c = C.decode_to_code(v, "Syndrome"); c(1, 1, 0, 1, 0, 0, 1)sage: c in CTrue
To plot the (histogram of) the weight distribution of a code, one can use the matplotlib package included with Sage:
sage: C = codes.HammingCode(GF(2), 4)sage: C[15, 11] Hamming Code over GF(2)sage: w = C.weight_distribution(); w [1, 0, 0, 35, 105, 168, 280, 435, 435, 280, 168, 105, 35, 0, 0, 1]sage: J = range(len(w))sage: W = IndexedSequence([ZZ(w[i]) for i in J],J)sage: P = W.plot_histogram()
Now type
show(P) to view this.
There are several coding theory functions we are skipping entirely.Please see the reference manual or the file
coding/linear_codes.py for examples.
Sage can also compute algebraic-geometric codes, called AG codes,via the Singular interface § sec:agcodes. One may also use the AGcodes implemented in GUAVA via the Sage interface to GAP
gap_console(). See the GUAVA manual for more details. {GUAVA}
Ciphers¶ LFSRs¶
A special type of stream cipher is implemented in Sage, namely, a linear feedback shift register (LFSR) sequence defined over a finite field. Stream ciphers have been used for a long time as a source of pseudo-random number generators. {linear feedback shift register}
S. Golomb {G} gives a list of three statistical properties a sequence of numbers \({\bf a}=\{a_n\}_{n=1}^\infty\), \(a_n\in \{0,1\}\), should display to be considered “random”. Define the autocorrelation of \({\bf a}\) to be
In the case where \(a\) is periodic with period \(P\) then this reduces to
Assume \(a\) is periodic with period \(P\).
balance: \(|\sum_{n=1}^P(-1)^{a_n}|\leq 1\).
low autocorrelation:\[\begin{split}C(k)= \left\{ \begin{array}{cc} 1,& k=0,\\ \epsilon, & k\not= 0. \end{array} \right.\end{split}\]
(For sequences satisfying these first two properties, it is known that \(\epsilon=-1/P\) must hold.)
proportional runs property: In each period, half the runs have length \(1\), one-fourth have length \(2\), etc. Moveover, there are as many runs of \(1\)’s as there are of \(0\)’s.
A sequence satisfying these properties will be called pseudo-random. {pseudo-random}
A general feedback shift register is a map \(f:{\bf F}_q^d\rightarrow {\bf F}_q^d\) of the form
where \(C:{\bf F}_q^d\rightarrow {\bf F}_q\) is a given function. When \(C\) is of the form
for some given constants \(c_i\in {\bf F}_q\), the map is called a linear feedback shift register (LFSR). The sequence of coefficients \(c_i\) is called the key and the polynomial
is sometimes called the connection polynomial.
Example: Over \(GF(2)\), if \([c_0,c_1,c_2,c_3]=[1,0,0,1]\) then \(C(x) = 1 + x + x^4\),
The LFSR sequence is then
The sequence of \(0,1\)’s is periodic with period\(P=2^4-1=15\) and satisfies Golomb’s three randomnessconditions. However, this sequence of period 15 can be “cracked”(i.e., a procedure to reproduce \(g(x)\)) by knowing only 8terms! This is the function of the Berlekamp-Massey algorithm {M},implemented as
lfsr_connection_polynomial (which produces thereverse of
berlekamp_massey).
sage: F = GF(2)sage: o = F(0)sage: l = F(1)sage: key = [l,o,o,l]; fill = [l,l,o,l]; n = 20sage: s = lfsr_sequence(key,fill,n); s[1, 1, 0, 1, 0, 1, 1, 0, 0, 1, 0, 0, 0, 1, 1, 1, 1, 0, 1, 0]sage: lfsr_autocorrelation(s,15,7)4/15sage: lfsr_autocorrelation(s,15,0)8/15sage: lfsr_connection_polynomial(s)x^4 + x + 1sage: from sage.matrix.berlekamp_massey import berlekamp_masseysage: berlekamp_massey(s)x^4 + x^3 + 1
Classical ciphers¶
has a type for cryptosystems (created by David Kohel, who also wrote the examples below), implementing classical cryptosystems. The general interface is as follows:
sage: S = AlphabeticStrings()sage: SFree alphabetic string monoid on A-Zsage: E = SubstitutionCryptosystem(S)sage: ESubstitution cryptosystem on Free alphabetic string monoid on A-Zsage: K = S([ 25-i for i in range(26) ])sage: e = E(K)sage: m = S("THECATINTHEHAT")sage: e(m)GSVXZGRMGSVSZG
Here’s another example:
sage: S = AlphabeticStrings()sage: E = TranspositionCryptosystem(S,15);sage: m = S("THECATANDTHEHAT")sage: G = E.key_space()sage: GSymmetric group of order 15! as a permutation groupsage: g = G([ 3, 2, 1, 6, 5, 4, 9, 8, 7, 12, 11, 10, 15, 14, 13 ])sage: e = E(g)sage: e(m)EHTTACDNAEHTTAH
The idea is that a cryptosystem is a map\(E: KS \to \text{Hom}_\text{Set}(MS,CS)\) where\(KS\), \(MS\), and \(CS\) are the key space,plaintext (or message) space, and ciphertext space, respectively.\(E\) is presumed to be injective, so
e.key() returns thepre-image key. |
If $A$ is a ($\mathbb Z$-)graded algebra, we can define its Hochschild cohomology in the usual way, via the standard complex of $A$-bimodules:
$$C_*(A)=\cdots\rightarrow A\otimes A\otimes A\rightarrow A\otimes A.$$
Since $A$ is graded, then so is the envelopìng algebra $A^{op}\otimes A$, therefore there is a ($\mathbb Z$-)graded $Hom$ in the category of $A$-bimodules, that we can use to define the Hochschild cohomology complex,
$$C^{*,\bullet}(A)=Hom_{A^{op}\otimes A}^{\bullet}(C_*,A).$$
Therefore Hochschild cohomology is bigraded $HH^{*,\bullet}(A)$ (I'm just interested in $A$ as a bimodule of coefficients).
Following some computations of mine, the $(1,0)$-cochain $\delta$ which multiplies each homogenoeus element in $A$ by its degree
$$\delta(x)=|x|\cdot x$$
seems to be a cocycle, and its cohomology class
$$\{\delta\}\in HH^{1,0}(A)$$
is surprisingly (in my opinion) non-trivial since it can somehow detect the degree of any other cohomology class $y\in HH^{p,q}(A)$, $|y|=q$, via the Lie bracket $$[\{\delta\},y]=|y|\cdot y.$$ That formula actually holds at the cochain level. It's turned out to play an important role in a current project and I wonder whether anybody has seen this class, or anything like that before. Is there maybe any mistake in what I say?
Notice that it is a very particular phenomenon of the graded setting. In the ungraded case $\delta$ is trivial already as a cochain.
EDIT: The degree of $y$ is $q$, not $p$ (which is what was written before). |
Suppose I had a ray of unpolarized light, and I was sitting inside the beam and looking at the electric fields oscillating, then , if I am looking at a point how would the oscillations look like? I cannot seem to understand it.
I guess you have this image in mind:
This hods just for a single photon, or elementary packet of light. An unpolarized beam of light contains a bunch of photons with many different polarizations. The total electric field will randomly jump all around, still the interactions with matter typically involve a single photon at a time. This means that if you ride the beam, you are allowed to look at the photons one by one, each time seeing a different, but consistent with itself, oscillating field.
They will look all over the place. If you take a particular photon, you will see it in a particular polarisation state, but it will have nothing to do with the photon next to it, or the one before.
Actually, getting a perfectly unpolarised source of light is quite difficult. One of the best cheap options are the sodium discharge lamps, used extensively in undergrad optics labs. They work by passing an electric current through some sodium vapour, so that each individual atom gets excited, and then releases the energy in light. They are very hot, so the state when the light is emitted is quite random; but still you will see a faint polarisation in the direction of the discharge. That is, if you measure the intensity of the light through a
very good polariser at 90º, you will see a small difference.
A full classical answer to your question is actually rather subtle and tricky, but its summary is pretty much as in Davidmh and DarioP's answer, both good answers: the electric field vector's direction is changing wildly. The classical description of the nature of depolarised light is bound up with decoherence and partially coherence, a topic which Born and Wolf in "Principles of Optics" give a whole chapter to. The following is roughly analogous to Born and Wolf's desctipyion: if the transverse (normal to propagation) plane is the $x,y$ plane, then we represent the electric field at a point as:
$$\mathbf{E} = \left(\begin{array}{cc}E_x(t) \cos(\omega t + \phi_x(t))\\E_y(t) \cos(\omega t + \phi_y(t))\end{array}\right)\tag{1}$$
where $\omega$ is the centre frequency and the phases $\phi_x(t)$, $\phi_y(t)$ and envelopes $E_x(t)$, $E_y(t)$ are stochastic processes, which can be as complicated as you like. The formulas I cite above just assume that:
$E_x$, $E_y$ and $\phi$ behave like independent random variables, and They vary with time swiftly compared to your observation interval (the time interval whereover you gather light in a sensor to come up with an "intensity" measurement) but not so swiftly that the light's spectrum broadened so much that we cannot still think of the light as roughly monochromatic.
Although it cannot really be "visualised", I actually find the quantum description of depolarised light is a great deal clearer and easier: certainly it involves a great deal less of the heavy duty stochastic process theory expounded on in Born and Wolf. We think of a lone photon; it propagates following Maxwell's equations, which are the first quantised propagation equations for the photon (only here the $\vec{E},\,\vec{B}$ represent a quantum state, not measurable electric and magnetic fields). Then we think of an ensemble of pure quantum states, exactly analogous to classical, fully polarised light states. A partially depolarised state is nothing more than a
classical mixture of pure polarisation states: what this means is illustrated by the Wigner's Friend Thought Experiment. For a large ensemble of photons, you simply do any calculations you need for all the different pure polarisation states present in the classical mixture, then sum the squared amplitudes weighted by the classical probabilities. The density matrix formalism is invaluable for such calculations, and I talk about these calcualtions for light in this answer here. |
I am dealing with a highly nonlinear system of two PDEs. I already have a code to solve the system in case of Dirichlet boundary conditions. The explicit system is:
$$ \begin{eqnarray*} \partial_{t}u & = & D_{11}\partial_{x}^{2}u+\partial_{x}D_{11}\cdot\partial_{x}u+D_{12}\partial_{x}^{2}v+\partial_{x}D_{12}\cdot\partial_{x}v\\ \partial_{t}v & = & D_{21}\partial_{x}^{2}u+\partial_{x}D_{21}\cdot\partial_{x}u+D_{22}\partial_{x}^{2}v+\partial_{x}D_{22}\cdot\partial_{x}v \end{eqnarray*}$$
with boundary conditions:
$$u(0,t)=0 \ (\partial_x v)(0,t) = 0$$
And the discretized form of the scheme are:
$$\begin{eqnarray*} -D_{21}(x_{j})u_{n+1,j-1}-D_{22}(x_{j})v_{n+1,j-1}\\ {}[\mu+2D_{11}(x_{j})]u_{n+1,j}+2D_{12}(x_{j})v_{n+1,j}\\ -D_{11}(x_{j})u_{n+1,j+1}-D_{12}(x_{j})v_{n+1,j+1} & = & \frac{1}{4}[4\mu-D_{11}(x_{j+1})+D_{11}(x_{j-1})]u_{n,j}\\ & & -[D_{12}(x_{j+1})-D_{12}(x_{j})]v_{n,j}\\ & & +\frac{1}{4}[D_{11}(x_{j+1})-D_{11}(x_{j-1})]u_{n,j+1}\\ & & +\frac{1}{4}[D_{12}(x_{j+1})-D_{12}(x_{j-1})]v_{n,j+1} \end{eqnarray*}$$
$$ \begin{eqnarray*} -D_{21}(x_{j})u_{n+1,j-1}-D_{22}(x_{j})v_{n+1,j-1}\\ 2D_{21}(x_{j})u_{n+1,j}+[\mu+2D_{22}(x_{j})]v_{n+1,j}\\ -D_{21}(x_{j})u_{n+1,j+1}-D_{22}(x_{j})v_{n+1,j+1} & = & \frac{1}{4}[4\mu-D_{21}(x_{j+1})+D_{21}(x_{j-1})]u_{n,j}\\ & & -[D_{22}(x_{j+1})-D_{22}(x_{j})]v_{n,j}\\ & & +\frac{1}{4}[D_{21}(x_{j+1})-D_{21}(x_{j-1})]u_{n,j+1}\\ & & +\frac{1}{4}[D_{22}(x_{j+1})-D_{22}(x_{j-1})]v_{n,j+1} \end{eqnarray*} $$
That is, I am solving a linear system of the form:
$$A\boldsymbol{x}=\boldsymbol{b}$$
where $A$ has $2J\times 2J$ entries, according to the structure:
$$\left(\begin{array}{cccccccccc} \mbox{1st equation}\rightarrow & | & c_{u}^{j} & c_{v}^{j} & c_{u}^{j+1} & c_{v}^{j+1} & 0 & & ... & 0\\ \mbox{2nd equation}\rightarrow & | & d_{u}^{j} & d_{v}^{j} & d_{u}^{j+1} & d_{v}^{j+1} & 0 & & ... & 0\\ \vdots & | & . & . & . & . & . & . & . & .\\ \mbox{1st}\rightarrow & | & 0 & 0 & c_{u}^{j-1} & c_{v}^{j-1} & c_{u}^{j} & c_{v}^{j} & c_{u}^{j+1} & c_{v}^{j+1}\\ \mbox{2nd }\rightarrow & | & 0 & 0 & d_{u}^{j-1} & d_{v}^{j-1} & d_{u}^{j} & d_{v}^{j} & d_{u}^{j+1} & d_{v}^{j+1}\vdots \end{array}\right)$$
where the $c$'s and $d$'s are the the coefficients corresponding to the scheme above.
Now, I am trying to implement the Neumann boundary condition (the one on $v$), using the ghost point method as discussed, for instance, here for a simpler equation.
Thus, I would like to change my code in order to add some extra terms to the array, but I can't work out what to put in them. Also, in the respective components of the matrix $A$. Because when I write the condition:
$$\frac{u_1-u_{-1}}{(\Delta x)^2}=0$$ (and same for $v$)
I need to eliminate $u_{-1}$ and $v_{-1}$, using also the third and fourth equations I wrote in this post. If I inverted for $u_{-1}$ and $v_{-1}$ (assuming I know how to), I do not see how I could write down a code which works for any kind of scheme I have, without putting the explicit inversion in the matrix. |
In a paper by Joos and Zeh, Z Phys B 59 (1985) 223, they say:This 'coming into being of classical properties' appears related to what Heisenberg may have meant by his famous remark [7]: 'Die "Bahn" entsteht erst dadurch, dass wir sie beobachten.'Google Translate says this means something ...
@EmilioPisanty Tough call. It's technical language, so you wouldn't expect every German speaker to be able to provide a correct interpretation—it calls for someone who know how German is used in talking about quantum mechanics.
Litmus are a London-based space rock band formed in 2000 by Martin (bass guitar/vocals), Simon (guitar/vocals) and Ben (drums), joined the following year by Andy Thompson (keyboards, 2001–2007) and Anton (synths). Matt Thompson joined on synth (2002–2004), while Marek replaced Ben in 2003. Oli Mayne (keyboards) joined in 2008, then left in 2010, along with Anton. As of November 2012 the line-up is Martin Litmus (bass/vocals), Simon Fiddler (guitar/vocals), Marek Bublik (drums) and James Hodkinson (keyboards/effects). They are influenced by mid-1970s Hawkwind and Black Sabbath, amongst others.They...
@JohnRennie Well, they repeatedly stressed their model is "trust work time" where there are no fixed hours you have to be there, but unless the rest of my team are night owls like I am I will have to adapt ;)
I think u can get a rough estimate, COVFEFE is 7 characters, probability of a 7-character length string being exactly that is $(1/26)^7\approx 1.2\times 10^{-10}$ so I guess you would have to type approx a billion characters to start getting a good chance that COVFEFE appears.
@ooolb Consider the hyperbolic space $H^n$ with the standard metric. Compute $$\inf\left\{\left(\int u^{2n/(n-2)}\right)^{-(n-2)/n}\left(4\frac{n-1}{n-2}\int|\nabla u|^2+\int Ru^2\right): u\in C^\infty_c\setminus\{0\}, u\ge0\right\}$$
@BalarkaSen sorry if you were in our discord you would know
@ooolb It's unlikely to be $-\infty$ since $H^n$ has bounded geometry so Sobolev embedding works as expected. Construct a metric that blows up near infinity (incomplete is probably necessary) so that the inf is in fact $-\infty$.
@Sid Eating glamorous and expensive food on a regular basis and not as a necessity would mean you're embracing consumer fetish and capitalism, yes. That doesn't inherently prevent you from being a communism, but it does have an ironic implication.
@Sid Eh. I think there's plenty of room between "I think capitalism is a detrimental regime and think we could be better" and "I hate capitalism and will never go near anything associated with it", yet the former is still conceivably communist.
Then we can end up with people arguing is favor "Communism" who distance themselves from, say the USSR and red China, and people who arguing in favor of "Capitalism" who distance themselves from, say the US and the Europe Union.
since I come from a rock n' roll background, the first thing is that I prefer a tonal continuity. I don't like beats as much as I like a riff or something atmospheric (that's mostly why I don't like a lot of rap)
I think I liked Madvillany because it had nonstandard rhyming styles and Madlib's composition
Why is the graviton spin 2, beyond hand-waiving, sense is, you do the gravitational waves thing of reducing $R_{00} = 0$ to $g^{\mu \nu} g_{\rho \sigma,\mu \nu} = 0$ for a weak gravitational field in harmonic coordinates, with solution $g_{\mu \nu} = \varepsilon_{\mu \nu} e^{ikx} + \varepsilon_{\mu \nu}^* e^{-ikx}$, then magic? |
Geometry and Topology Seminar Contents Fall 2016
date speaker title host(s) September 9 Bing Wang (UW Madison) "The extension problem of the mean curvature flow" (Local) September 16 Ben Weinkove (Northwestern University) "Gauduchon metrics with prescribed volume form" Lu Wang September 23 Jiyuan Han (UW Madison) "Deformation theory of scalar-flat ALE Kahler surfaces" (Local) September 30 October 7 Yu Li (UW Madison) "Ricci flow on asymptotically Euclidean manifolds" (Local) October 14 Sean Howe (University of Chicago) "Representation stability and hypersurface sections" Melanie Matchett Wood October 21 Nan Li (CUNY) "Quantitative estimates on the singular Sets of Alexandrov spaces" Lu Wang October 28 Ronan Conlon(Florida International University) "New examples of gradient expanding K\"ahler-Ricci solitons" Bing Wang November 4 Jonathan Zhu (Harvard University) "Entropy and self-shrinkers of the mean curvature flow" Lu Wang November 11 Canceled. November 18 Caglar Uyanik (Illinois) "Geometry and dynamics of free group automorphisms" Kent Thanksgiving Recess December 2 Peyman Morteza (UW Madison) "TBA" (Local) December 9 Yu Zeng(University of Rochester) "TBA" Bing Wang December 16 Spring 2017
date speaker title host(s) Jan 20 Carmen Rovi (University of Indiana Bloomington) "TBA" Maxim Jan 27 Feb 3 Feb 10 Feb 17 Feb 24 March 3 March 10 March 17 March 24 Spring Break March 31 April 7 April 14 April 21 April 28 Bena Tshishiku (Harvard) "TBA" Dymarz Fall Abstracts Ronan Conlon New examples of gradient expanding K\"ahler-Ricci solitons
A complete K\"ahler metric $g$ on a K\"ahler manifold $M$ is a \emph{gradient expanding K\"ahler-Ricci soliton} if there exists a smooth real-valued function $f:M\to\mathbb{R}$ with $\nabla^{g}f$ holomorphic such that $\operatorname{Ric}(g)-\operatorname{Hess}(f)+g=0$. I will present new examples of such metrics on the total space of certain holomorphic vector bundles. This is joint work with Alix Deruelle (Universit\'e Paris-Sud).
Jiyuan Han Deformation theory of scalar-flat ALE Kahler surfaces
We prove a Kuranishi-type theorem for deformations of complex structures on ALE Kahler surfaces. This is used to prove that for any scalar-flat Kahler ALE surfaces, all small deformations of complex structure also admit scalar-flat Kahler ALE metrics. A local moduli space of scalar-flat Kahler ALE metrics is then constructed, which is shown to be universal up to small diffeomorphisms (that is, diffeomorphisms which are close to the identity in a suitable sense). A formula for the dimension of the local moduli space is proved in the case of a scalar-flat Kahler ALE surface which deforms to a minimal resolution of \C^2/\Gamma, where \Gamma is a finite subgroup of U(2) without complex reflections. This is a joint work with Jeff Viaclovsky.
Sean Howe Representation stability and hypersurface sections
We give stability results for the cohomology of natural local systems on spaces of smooth hypersurface sections as the degree goes to \infty. These results give new geometric examples of a weak version of representation stability for symmetric, symplectic, and orthogonal groups. The stabilization occurs in point-counting and in the Grothendieck ring of Hodge structures, and we give explicit formulas for the limits using a probabilistic interpretation. These results have natural geometric analogs -- for example, we show that the "average" smooth hypersurface in \mathbb{P}^n is \mathbb{P}^{n-1}!
Nan Li Quantitative estimates on the singular sets of Alexandrov spaces
The definition of quantitative singular sets was initiated by Cheeger and Naber. They proved some volume estimates on such singular sets in non-collapsed manifolds with lower Ricci curvature bounds and their limit spaces. On the quantitative singular sets in Alexandrov spaces, we obtain stronger estimates in a collapsing fashion. We also show that the (k,\epsilon)-singular sets are k-rectifiable and such structure is sharp in some sense. This is a joint work with Aaron Naber.
Yu Li
In this talk, we prove that if an asymptotically Euclidean (AE) manifold with nonnegative scalar curvature has long time existence of Ricci flow, it converges to the Euclidean space in the strong sense. By convergence, the mass will drop to zero as time tends to infinity. Moreover, in three dimensional case, we use Ricci flow with surgery to give an independent proof of positive mass theorem. A classification of diffeomorphism types is also given for all AE 3-manifolds with nonnegative scalar curvature.
Gaven Marin TBA Peyman Morteza TBA Caglar Uyanik Geometry and dynamics of free group automorphisms
A common theme in geometric group theory is to obtain structural results about infinite groups by analyzing their action on metric spaces. In this talk, I will focus on two geometrically significant groups; mapping class groups and outer automorphism groups of free groups.We will describe a particular instance of how the dynamics and geometry of their actions on various spaces provide deeper information about the groups.
Bing Wang The extension problem of the mean curvature flow
We show that the mean curvature blows up at the first finite singular time for a closed smooth embedded mean curvature flow in R^3. A key ingredient of the proof is to show a two-sided pseudo-locality property of the mean curvature flow, whenever the mean curvature is bounded. This is a joint work with Haozhao Li.
Ben Weinkove Gauduchon metrics with prescribed volume form
Every compact complex manifold admits a Gauduchon metric in each conformal class of Hermitian metrics. In 1984 Gauduchon conjectured that one can prescribe the volume form of such a metric. I will discuss the proof of this conjecture, which amounts to solving a nonlinear Monge-Ampere type equation. This is a joint work with Gabor Szekelyhidi and Valentino Tosatti.
Jonathan Zhu Entropy and self-shrinkers of the mean curvature flow
The Colding-Minicozzi entropy is an important tool for understanding the mean curvature flow (MCF), and is a measure of the complexity of a submanifold. Together with Ilmanen and White, they conjectured that the round sphere minimises entropy amongst all closed hypersurfaces. We will review the basics of MCF and their theory of generic MCF, then describe the resolution of the above conjecture, due to J. Bernstein and L. Wang for dimensions up to six and recently claimed by the speaker for all remaining dimensions. A key ingredient in the latter is the classification of entropy-stable self-shrinkers that may have a small singular set.
Yu Zeng Short time existence of the Calabi flow with rough initial dataCalabi flow was introduced by Calabi back in 1950’s as a geometric flow approach to the existence of extremal metrics. Analytically it is a fourth order nonlinear parabolic equation on the Kaehler potentials which deforms the Kaehler potential along its scalar curvature. In this talk, we will show that the Calabi flow admits short time solution for any continuous initial Kaehler metric. This is a joint work with Weiyong He. Spring Abstracts Bena Tshishiku
"TBA"
Archive of past Geometry seminars
2015-2016: Geometry_and_Topology_Seminar_2015-2016
2014-2015: Geometry_and_Topology_Seminar_2014-2015 2013-2014: Geometry_and_Topology_Seminar_2013-2014 2012-2013: Geometry_and_Topology_Seminar_2012-2013 2011-2012: Geometry_and_Topology_Seminar_2011-2012 2010: Fall-2010-Geometry-Topology |
To evaluate detection performance, we plot the
miss-rate $mr(c) = \frac{fn(c)}{tp(c) + fn(c)}$ against the number of false positives per image $fppi(c)=\frac{fp(c)}{\text{#img}}$ in log-log plots. $tp(c)$ is the number of true positives, $fp(c)$ is the number of false positives, and $fn(c)$ is the number of false negatives, all for a given confidence value $c$ such that only detections are taken into account with a confidence value greater or equal than $c$. As commonly applied in object detection evaluation the confidence threshold $c$ is used as a control variable. By decreasing $c$, more detections are taken into account for evaluation resulting in more possible true or false positives, and possible less false negatives. We define the log average miss-rate (LAMR) as shown, where the 9 fppi reference points are equally spaced in the log space: $\DeclareMathOperator*{\argmax}{argmax}LAMR = \exp\left(\frac{1}{9}\sum\limits_f \log\left(mr(\argmax\limits_{fppi\left(c\right)\leq f} fppi\left(c\right))\right)\right)$
For each fppi reference point the corresponding mr value is used. In the absence of a miss-rate value for a given f the highest existent
fppi value is used as new reference point. This definition enables LAMR to be applied as a single detection performance indicator at image level. At each image the set of all detections is compared to the groundtruth annotations by utilizing a greedy matching algorithm. An object is considered as detected (true positive) if the Intersection over Union (IoU) of the detection and groundtruth bounding box exceeds a pre-defined threshold. Due to the high non-rigidness of pedestrians we follow the common choice of an IoU threshold of 0.5. Since no multiple matches are allowed for one ground-truth annotation, in the case of multiple matches the detection with the largest score is selected, whereas all other matching detections are considered false positives. After the matching is performed, all non matched ground-truth annotations and detections, count as false negatives and false positives, respectively.
Neighboring classes and ignore regions are used during evaluation. Neighboring classes involve entities that are semantically similar, for example bicycle and moped riders. Some applications might require their precise distinction (
enforce) whereas others might not ( ignore). In the latter case, during matching correct/false detections are not credited/penalized. If not stated otherwise, neighboring classes are ignored in the evaluation. In addition to ignored neighboring classes all persons annotations with the tags behind glass or sitting-lying are treated as ignore regions. Further, as mentioned in Section 3.2, EuroCity Persons Dataset Publication, ignore regions are used for cases where no precise bounding box annotation is possible (either because the objects are too small or because there are too many objects in close proximity which renders the instance based labeling infeasible). Since there is no precise information about the number or the location of objects in the ignore region, all unmatched detections which share an intersection of more than $0.5$ with these regions are not considered as false positives.
Note that submissions with provided publication link and/or code will get priorized in below list (COMING SOON).
Method User LAMR (reasonable) LAMR (small) LAMR (occluded) LAMR (all)▲ External data used Publication URL Publication code Submitted on HRNet Hongsong Wang 0.061 0.138 0.287 0.183 ImageNet no no 2019-08-05 17:11:04 View YOLOv3 ECP Team 0.097 0.186 0.401 0.242 ImageNet yes no 2019-04-01 17:08:05 View Faster R-CNN ECP Team 0.101 0.196 0.381 0.251 ImageNet yes no 2019-04-01 17:06:33 View SSD ECP Team 0.131 0.235 0.460 0.296 ImageNet yes no 2019-04-02 13:56:14 View R-FCN (with OHEM) ECP Team 0.163 0.245 0.507 0.330 ImageNet yes no 2019-04-01 17:10:03 View YOLOv3_640 HUI_Tsinghua-Daim... 0.273 0.564 0.623 0.456 no no 2019-05-17 04:56:27 View
Method User LAMR (reasonable) LAMR (small) LAMR (occluded) LAMR (all)▲ External data used Publication URL Publication code Submitted on HRNet Hongsong Wang 0.079 0.156 0.265 0.153 ImageNet no no 2019-08-05 17:11:04 View FasterRCNN with M... Qihua Cheng 0.150 0.253 0.653 0.295 ImageNet no no 2019-07-08 08:48:13 View Faster R-CNN ECP Team 0.201 0.359 0.701 0.358 ImageNet yes no 2019-05-02 10:10:01 View |
How can we prove that the inverse of an upper (lower) triangular matrix is upper (lower) triangular?
Another method is as follows. An invertible upper triangular matrix has the form $A=D(I+N)$ where $D$ is diagonal (with the same diagonal entries as $A$) and $N$ is upper triangular with zero diagonal. Then $N^n=0$ where $A$ is $n$ by $n$. Both $D$ and $I+N$ have upper triangular inverses: $D^{-1}$ is diagonal, and $(I+N)^{-1}=I-N+N^2-\cdots +(-1)^{n-1}N^{n-1}$. So $A^{-1}=(I+N)^{-1}D^{-1}$ is upper triangular.
Personally, I prefer arguments which are more geometric to arguments rooted in matrix algebra. With that in mind, here is a proof.
First, two observations on the geometric meaning of an upper triangular invertible linear map.
Define $S_k = {\rm span} (e_1, \ldots, e_k)$, where $e_i$ the standard basis vectors. Clearly, the linear map $T$ is upper triangular if and only if $T S_k \subset S_k$.
If $T$ is in addition invertible, we must have the stronger relation $T S_k = S_k$.
Indeed, if $T S_k$ was a strict subset of $S_k$, then $Te_1, \ldots, Te_k$ are $k$ vectors in a space of dimension strictly less than $k$, so they must be dependent: $\sum_i \alpha_i Te_i=0$ for some $\alpha_i$ not all zero. This implies that $T$ sends the
nonzerovector $\sum_i \alpha_i e_i$ to zero, so $T$ is not invertible.
With these two observations in place, the proof proceeds as follows. Take any $s \in S_k$. Since $TS_k=S_k$ there exists some $s' \in S_k$ with $Ts'=s$ or $T^{-1}s = s'$. In other words, $T^{-1} s$ lies in $S_k$, so $T^{-1}$ is upper triangular.
I'll add nothing to alext87 answer, or J.M. comments. Just "display" them. :-)
Remeber that you can compute the inverse of a matrix by reducing it to row echelon form and solving the simultaneous systems of linear equations $ (A \vert I)$, where $A$ is the matrix you want to invert and $I$ the unit matrix. When you have finished the process, you'll get a matrix like $(I\vert A^{-1})$ and the matrix on the right, yes!, is the inverse of $A$. (Why?)
In your case, half of the work is already done:
$$ \begin{pmatrix} a^1_1 & a^1_2 & \cdots & a^1_{n-1} & a^1_n & 1 & 0 & \cdots & 0 & 0 \\\ & a^2_2 & \cdots & a^2_{n-1} & a^2_n & & 1 & \cdots & 0 & 0 \\\ & & \ddots & \vdots & \vdots & & & \ddots & \vdots & \vdots \\\ & & & a^{n-1}_{n-1} & a^{n-1}_n & & & & 1 & 0 \\\ & & & & a^n_n & & & & & 1 \end{pmatrix} $$
Now, what happens when you do back substitution starting with $a^n_n$ and then continuing with $a^{n-1}_{n-1}$...?
You can prove by induction.
Suppose $A$ is upper triangular. It is easy to show that this holds for any $2\times 2$ matrix. (In fact, $A^{-1}=\left[\begin{array}{cc} a & b\\ 0 & d \end{array}\right]^{-1} =\frac{1}{ad}\left[\begin{array}{cc} d & -b\\ 0 & a \end{array}\right]$. )
Suppose the result holds for any $n\times n$ upper triangular matrix. Let $A=\left[\begin{array}{cc} A_{1} & a_{2}\\ 0 & x \end{array}\right]$, $B=\left[\begin{array}{cc} B_{1} & b_{2}\\ b_{3}^{T} & y \end{array}\right]$ be any $(n+1)\times (n+1)$ upper triangular matrix and its inverse. (Mind that $a_2$, $b_2$, $b_3$ are $n\times 1$ vectors, $x$, $y$ are scalars.) Then $AB=BA=I_{n+1}$ implies that $$ \left[\begin{array}{cc} A_{1} & a_{2}\\ 0 & x \end{array}\right] \left[\begin{array}{cc} B_{1} & b_{2}\\ b_{3}^{T} & y \end{array}\right]= \left[\begin{array}{cc} B_{1} & b_{2}\\ b_{3}^{T} & y \end{array}\right] \left[\begin{array}{cc} A_{1} & a_{2}\\ 0 & x \end{array}\right] =I_{n+1}, $$
From the upper left corner of the second multiplication, we have $B_1 A_1 = I_n$. Hence $B_1$ is upper triangular from our hypothesis. From the lower left block of the multiplication , we know that $b_3=0$. ($x\ne 0$ since $A$ is invertible.) Therefore $B=\left[\begin{array}{cc} B_{1} & b_{2}\\ 0 & y \end{array}\right]$ is also upper triangular.
Another proof is by contradiction. Let $A = [a_{ij}]$ be an upper triangular matrix of size $N$. Assume $B = A^{-1} = [b_{ij}]$ is not upper triangular. Thus there exists an entry $b_{ij} \neq 0$ for $i > j$. Let $b_{ik}$ be the element with the smallest $k$ in row $i$ such that $b_{ik} \neq 0$ and $ i > k$. Consider the product $C = B A$. The element $c_{ik}$ of matrix C is off-diagonal ($i > k$) and computed as $$ c_{ik} = \sum b_{ij}a_{jk} = b_{i1}a_{1k} + b_{i2}a_{2k} + \dots + b_{ik}a_{kk} + \dots + b_{iN}a_{Nk} $$
Since $b_{ik}$ is the first non-zero element in its row, all the terms to the left of $b_{ik}a_{kk}$ vanish. Since A is upper triangular (given), all the terms to the right of $b_{ik}a_{kk}$ vanish. Since $A$ is invertible, all its diagonal elements are non-zero. Thus $c_{ik} = b_{ik}a_{kk} \neq 0$. However, since $C$ is the identity matrix and $c_{ik}$ is off diagonal, this is a contradiction! Thus, $B = A^{-1}$ is upper triangular.
Same applies to lower triangular matrix by noticing that $(A^T)^{-1} = (A^{-1})^T$
Suppose that $U$ is upper. The $i$th column $x_i$ of the inverse is given by $Ux_i=e_i$ where $e_i$ is the $i$th unit vector. By backward subsitution you can see that $(x_i)_j=0$ for $i+1\leq j\leq n$. I.e all the entries in the $i$th column of the inverse below the diagonal are zero. This is true for all $i$ and hence the inverse $U^{-1}=[x_1|\ldots|x_n]$ is upper triangular.
The same thing works for lower triangular using forward subsitution. |
One can explore things visually/experimentally and make good discoveries, Theory is needed to justify them. Here, however, are more experiments and speculation. This is somewhat the same thing looked at in different ways, but I think each adds something.
We see that if the (real) coefficients of $p(x)=\sum_0^n a_ix^i$ are drawn randomly from a positive interval then we can normalize to get the same roots with coefficients from $[1-\delta,1+\delta]$ and if $n$ is
large enough and $\delta$ small enough relative to each other (whatever that means) the roots will be near the unit circle and almost equally distributed.
Optional example for illustration and checking: The polynomial $\sum_0^{299}z^n$ has roots $z_m=r_me^{i\theta_m}$ for $r_m=1$ and $\theta_m=\frac{2\pi m}{300}$ $1 \le m \le 299.$ I generated a single random polynomial $f(x)=1+\sum_1^{298} a_n x^x+x^{299}$ with the $a_i$ random and uniformly selected from $[0.8,1.2].$ The $299$ roots (actually, half of them) are shown below. Much can be seen but specifically: The roots, in order of increasing argument, are $r_me^{i \theta_m}$ where in all cases $0.978 \lt r_m \lt 1.036$ and $300|\theta_m-{2\pi m}| \lt 0.78.$ Here are the most extreme deviations in argument (in the upper half):$[123, -.7783]$, $[73, .7107]$, $[61, -.7036]$, $[100, -.5640]$,$[56, .5493]$, $[67, -.5482]$, $[102, -.5382]$, $[72, .5213]$,$[117, .5156]$, $[97, .5099]$, $[43, .4866]$, $[86, .4827]$,$[87, -.4749]$
Before going on:
better to say the unit circle except the neighborhood of $1$. Though we could fix that by multiplying through by $x-1$ and discussing polynomials $x^{n+1}+\sum_1^{n}a_ix^i+a_0$ with $a_0$ as before but the $a_i$ in $[-2\delta,2\delta]$ (but denser near $0$.) Then the roots really are almost equally distributed. maybe it is better to look at complex coefficients with magnitude in $[1-\delta,1+\delta]$ or $[0,1]$ or $\{{0,1\}}$ or real coefficients of that form or from one of the sets $\{{-1,1\}},\{{0,1\}},\{{-1,0,1\}}.$
Actually these all relate to each other. Of course
small perturbations of coefficients should move roots only a bit. But does that explain how little these moved?
If we are allowed to cook the coefficients to really move a root it seems best to move $-1$ by picking $n$ even and make the coefficients alternately $1-\delta$ and $1+\delta.$Then $-1$ is no longer a root, $p(-1)=(n+2)\delta.$ But $p'(-1) =\frac{n+2}{2} -\frac{n^2+n}{2}\delta$ is so large that we shouldn't have to go far. In fact calculation shows that the root is roughly $-(1+2\delta).$ More precisely, exactly $-(1+2\delta+2\delta^2\cdots)=-(1+\frac{1+\delta}{1-\delta})$ and the other roots seem unchanged. Of course (in hindsight) we should just factor to see that $$p(x)=((1-\delta)x+(1+\delta))\frac{x^n-1}{x^2-1}.$$
So a root can move more than $2 \delta.$ Is that tight? Seems reasonable, but I'm not going to check. We saw only a fraction of that above. For random coefficients in our model $p(-1)=\sum(a_i-1)(-1)^i$, the sum of $n+1$ values uniformly drawn from $[-\delta,\delta]$ so not that large in relation to $p'(-1)$ . So an actual root is likely not far away at all. As far as it goes, that reasoning is valid for any of the other roots of $\sum x^i$. The root $-1$ is special, but only because we are using real positive coefficients. Arbitrary coefficient near the unit circle should resolve that.
If the cyclotomic roots don't move "much" then they can't end up "too close" together. But perhaps it is also good to just consider if roots (want to be) separated from each other. We could try to add a new root very near an old one or move two roots until they touch or are near. I'll leave it to you to check that, if the other roots are fixed, we get coefficients almost as large as $2.$ Can one do better moving all the roots?
What is the effect of each coefficient? If we change just one interior coefficient $a_k$ to something extremely huge, then there will be seen to be about $n-k$ "big" roots near equally spaced on the circle of radius $a_k^{1/(n-k)}$ and about $k$ "small" roots near the circle of radius $a_k^{-1/k}.$ We can see why. Also, for $k$ not too near either extreme, and $a_k$ merely kind of huge, we would still have all those roots quite close to the unit circle. It is far from obvious how that carries over allowing all the coefficients moving a moderate amount (randomly). Perhaps it could be said that, as long as things like $(\frac{a_{n-k}}{a_n})^{1/k}$ and $(\frac{a_k}{a_0})^{-1/k}$ are all near $1$, there should be many pullings and pushings, none very large, that usually cancel out. That is very vague and unsupported but it works for me as a motivation. With the right deliberate choices we could shove a particular root or small set of roots. That actually seems like the case above for $x=-1.$
Here is an idea in perhaps more detail than it deserves. If we trace $q(x)=\frac{x^{n+1}-1}{x-1}$ as $x$ moves around the unit circle, we get a path not too hard to describe that touches the origin at the roots and then goes somewhat far away until coming back for the next one. When we perturb the coefficients and look at the position $p(x)$ on the new path, it will differ from the position on the old path by $\sum (a_i-1)x^i$ where these coefficients are random and distributed in $[-\delta,\delta]$ so the $x$ that were (near) roots of $q(x)$ will (usually) be near roots of $p(x)$ and those with $|q(x)|$ not that close to $0$ will? have the same true. It is possible to cook things to get a big deviation in one or a few places but unlikely to hapen by random. |
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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 ...
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 ...
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 ...
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 ...
Multi-strange baryon production at mid-rapidity in Pb-Pb collisions at $\sqrt{s_{\rm NN}}$ = 2.76 TeV
(Elsevier, 2014-01)
The production of ${\rm\Xi}^-$ and ${\rm\Omega}^-$ baryons and their anti-particles in Pb-Pb collisions at $\sqrt{s_{\rm NN}}$ = 2.76 TeV has been measured using the ALICE detector. The transverse momentum spectra at ...
Measurement of charged jet suppression in Pb-Pb collisions at √sNN = 2.76 TeV
(Springer, 2014-03)
A measurement of the transverse momentum spectra of jets in Pb–Pb collisions at √sNN = 2.76TeV is reported. Jets are reconstructed from charged particles using the anti-kT jet algorithm with jet resolution parameters R ...
Two- and three-pion quantum statistics correlations in Pb-Pb collisions at √sNN = 2.76 TeV at the CERN Large Hadron Collider
(American Physical Society, 2014-02-26)
Correlations induced by quantum statistics are sensitive to the spatiotemporal extent as well as dynamics of particle-emitting sources in heavy-ion collisions. In addition, such correlations can be used to search for the ...
Exclusive J /ψ photoproduction off protons in ultraperipheral p-Pb collisions at √sNN = 5.02TeV
(American Physical Society, 2014-12-05)
We present the first measurement at the LHC of exclusive J/ψ photoproduction off protons, in ultraperipheral proton-lead collisions at √sNN=5.02 TeV. Events are selected with a dimuon pair produced either in the rapidity ...
Measurement of quarkonium production at forward rapidity in pp collisions at √s=7 TeV
(Springer, 2014-08)
The inclusive production cross sections at forward rapidity of J/ψ , ψ(2S) , Υ (1S) and Υ (2S) are measured in pp collisions at s√=7 TeV with the ALICE detector at the LHC. The analysis is based on a data sample corresponding ... |
Difference between revisions of "Dev:Bline Speed"
m (→Bline's parameter's "speed")
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== Bline's parameter's "speed" ==
== Bline's parameter's "speed" ==
−
If you have played enough with Blines, "
+
If you have played enough with Blines, "Vertex" and "Tangent" converts you perhaps have noticed that a change in the "Amount" parameter doesn't always step forward/backwards the same amount. For example, adding 0.1 to "Amount" doesn't move a "Bline Vertex" by the same distance all the time.
−
Near the bline's
+
Near the bline's (or near the curved parts) you'll notice that evenly spaced "Amount" values are either compressed together or expanded away from each other. Trying to make an object follow a bline will lead to the object changing speeds as it goes along it.
−
The problem lies in how Blines are defined and how a position in the Bline changes as "Amount" parameter changes. I'll refer to the rate of change as the Bline's "speed".
+
The problem lies in how Blines are defined and how a position in the Bline changes as "Amount" parameter changes. I'll refer to the rate of change as the Bline's "speed".
== Why does "speed" changes? ==
== Why does "speed" changes? ==
Revision as of 23:42, 15 February 2008 Contents Bline's parameter's "speed"
If you have played enough with Blines, "BLine Vertex" and "BLine Tangent" converts you perhaps have noticed that a change in the "Amount" parameter doesn't always step forward/backwards the same amount. For example, adding 0.1 to "Amount" doesn't move a "Bline Vertex" by the same distance all the time.
Near the bline's vertices (or near the curved parts) you'll notice that evenly spaced "Amount" values are either compressed together or expanded away from each other. Trying to make an object follow a bline will lead to the object changing speeds as it goes along it.
The problem lies in how Blines are defined and how a position in the Bline changes as the "Amount" parameter changes. I'll refer to the rate of change as the Bline's "speed".
Why does "speed" changes?
Firstly, a Synfig Bline is composed of several bezier sections. Each segment is a cubic bezier line. This sections are joined back to back, allowing for arbitrarily complex shapes. All the properties that for a single section, also hold true for any number of sections. So I'm gonna focus on Blines with a single section, in other words, Blines with only two vertexes.
A Bline with a single section reduces to a Cubic Bezier defined like this:
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle \mathbf{B}(t)=(1-t)^3\mathbf{P}_0+3t(1-t)^2\mathbf{P}_1+3t^2(1-t)\mathbf{P}_2+t^3\mathbf{P}_3 \mbox{ , } t \in [0,1].}
This equation describes the shape of the curve. As the
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle t\,\!} parameter increases from zero up to one, the point defined by the equation moves from the Bezier's start towards it's end. The rate of the motion to as Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle t\,\!} increases describes the curve's "speed". Taking the derivative of this equation yields the "speed": Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle \frac{d\mathbf{B}(t)}{dt}= (1-t)^2 [ 3 ( \mathbf{P}_1 - \mathbf{P}_0 ) ] + 2t(1-t) [ 3 ( \mathbf{P}_2 - \mathbf{P}_1 ) ] + t^2 [ 3 ( \mathbf{P}_3 - \mathbf{P}_2 ) ] \mbox{ , }t \in [0,1]. }
You may have noticed that this equation is equivalent to a Quadratic Bezier. This tells us that the "speed" can and does change as the
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle t\,\!} parameter changes. Adjusting a Bline's "speed"
Our objective is now to compensate the derivative to achieve a desired "speed". We cannot change the control points to the curve without changing it's shape. The only other thing we can change is the parameter
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle t\,\!}. Therefore, we define a function Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle g(t)\,\!} so that: Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle \frac{d\mathbf{B}(g(t))}{dt}=\boldsymbol{s}(t)\,\,}
Where
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle \boldsymbol{s}(t)} is a vector Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle (s_x(t),s_y(t))\,\!} that defines the desired speed as a function of Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle t\,\!}. The curve needs to move in a whole range of directions as the curve describes its shape. Our objective is only to control its magnitude. This magnitude condition can be expressed as: Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle s_x^2(t)+s_y^2(t)=s_{mag}^2(t)}
Where
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle s_{mag}(t)\,\!} is a function defining the desired "speed" magnitude.
We can expand our first equation a bit:
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle \frac{d\mathbf{B}(g(t))}{dt}=\frac{d\mathbf{B}(g(t))}{d(g(t))}\,\frac{dg(t)}{dt}=\boldsymbol{s}(t)\,\,}
Expanding the equation like this lets us use the original Bline's derivative definition, by replacing
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle t\,\!} with Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle g(t)\,\!}. Next we replace the x and y components into the magnitude condition equation: Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle \Bigg[\frac{dB_x(g(t))}{d(g(t))}\,\frac{dg(t)}{dt}\Bigg]^2+\Bigg[\frac{dB_y(g(t))}{d(g(t))}\,\frac{dg(t)}{dt}\Bigg]^2=s_{mag}^2(t)}
Rearranging we obtain an ordinary non-linear differential equation:
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle \frac{dg(t)}{dt}=\frac{s_{mag}(t)}{\sqrt{\Big[\frac{dB_x(g(t))}{d(g(t))}\Big]^2+\Big[\frac{dB_y(g(t))}{d(g(t))}\Big]^2}}}
Solving this equation yields a function
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle g(t)\,\!} such that the curve's "speed" is dictated by the function Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle s_{mag}(t)\,\!}. Solving the equation
All that is left is to solve the equation. It is quite complex and as I said before, the differential equation that we got is non-linear. This makes it hard to find
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle g(t)\,\!} in a clear formula.
But even such a complex equation is easy to solve numerically. Keeping in mind that what we want is simply the value of
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle g(t)\,\!} so we can plug it into "Amount". A numerical solution of the equation gives us just that, the value of Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle g(t)\,\!} at certain intervals.
The Runge-Kutta method serves this purpose quite well, and it's quite simple also. All we need is to evaluate the derivative of the function that we need to find, and feed the values into the Runge-Kutta method.
Let's try a simple case, constant speed. If
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle s_{mag}(t)\,\!} is a constant value, then it would need to be equal to the Bline's length, so that as the Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle t\,\!} goes from 0.0 up to 1.0, the curve moves from the start to the end. Too little speed and the curve won't reach the end when Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle t\,\!} reaches 1.0. Too much and the curve will go past the end when Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle t\,\!} reaches 1.0.
Conveniently, this method also allows to find a Bline's length. If we assume
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle s_{mag}(t)=1\,\!} then the curve will reach it's end when Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle t=LENGTH\,\!}, where LENGTH is the Bline's length. |
Please note: this question was posted first (September 4) in math.stackeschange.com and then (September 16) in stats.stackeschange.com. It got no answers in neither of those sites.
Let the regularized incomplete beta and gamma functions be defined as usual: \begin{equation} I_p(z,w) = \frac a {B(z,w)} \int_0^p t^{z-1} (1-t)^{w-1} \,\mathrm dt, \end{equation} \begin{equation} \gamma(r,t) = \frac 1 {\Gamma(r)} \int_0^t s^{r-1} e^{-s} \,\mathrm ds. \end{equation}
Let $r \in \mathbb N$, and consider $a, b > 0$. As is well known, $I_p(r,a/p-b) \rightarrow \gamma(r,a)$ as $p \rightarrow 0$.
I have been able to prove that the difference between the incomplete beta function and its limit,\begin{equation}I_p(r,a/p-b) - \gamma(r,a),\end{equation}is positive for all $p \in (0,1)$ if $a$ is sufficiently larger than $r$, and negative if $a$ is sufficiently smaller than $r$. For example, if you have access to Matlab you can check that (note that Matlab's
gammainc uses reverse order for its arguments)
p = 1e-3; r = 5; a = 10; b = 5; betainc(p,r,a/p-b) - gammainc(a,r)
is positive, whereas
p = 1e-3; r = 5; a = 2; b = 5; betainc(p,r,a/p-b) - gammainc(a,r)
is negative.
I'd like to know the widest range of values of $a$ for which that difference is assured to be positive/negative.
So, my question is: Is there a known theorem that gives sufficient conditions on $a, b, r$ that assure that $I_p(r,a/p-b) - \gamma(r,a)$ is positive/negative? Even if you don't know any such theorem, any pointer in the right direction would be appreciated.
Alternate formulation
Since the incomplete gamma is related to the distribution function of a Poisson random variable, and the incomplete beta is related to that of a binomial random variable, the question can also be posed from a statistical point of view, as follows.
Let $B(n,p,r)$ denote the binomial distribution function (DF) with parameters $n \in \mathbb N$ and $p \in (0,1)$ evaluated at $r \in \{0,1,\ldots,n\}$: \begin{equation} B(n,p,r) = \sum_{i=0}^r \binom{n}{i} p^i (1-p)^{n-i}, \end{equation} and let $F(\nu,r)$ denote the Poisson DF with parameter $a \in \mathbb R^+$ evaluated at $r \in \{0,1,2,\ldots\}$: \begin{equation} F(a,r) = e^{-a} \sum_{i=0}^r \frac{a^i}{i!}. \end{equation}
Consider $p \rightarrow 0$, and let $n$ be defined as $\lceil a/p-d \rceil$, where $d$ is a constant of the order of $1$. Since $np \rightarrow a$, the function $B(n,p,r)$ converges to $F(a,r)$ for all $r$, as is well known.
With the above definition for $n$, it is interesting to know the values of $a$ for which \begin{equation} B(n,p,r) < F(a,r) \quad \forall p \in (0,1), \end{equation} and similarly those for which \begin{equation} B(n,p,r) > F(a,r) \quad \forall p \in (0,1). \end{equation}
I have been able to prove that the first inequality (in either of the two formulations) holds for $a$ sufficiently larger than $r$; more specifically, for $a$ greater than a certain bound $g(r)$, with $g(r)>r$. Similarly, the second inequality holds for $a$ sufficiently smaller than $r$, i.e. for $a$ lower than a certain bound $h(r)$, with $h(r)<r$. (The expressions of the bounds $g(r)$ and $h(r)$ are irrelevant here. I will provide the details to anyone interested.) However, numerical results suggest that those inequalities hold for less stringent bounds, that is, for $a$ closer to $r$ than I can prove.
So, I'd like to know if there is some theorem or result which establishes under which conditions each inequality holds (for all $p$); that is, when the binomial DF is guaranteed to be above/below its limiting Poisson DF. If such theorem doesn't exist, any idea or pointer in the right direction would be appreciated. |
OP mistakenly believes the relationship between these two functions is due to the number of samples (i.e. single vs all). However, the actual difference is simply how we select our training labels.
In the case of binary classification we may assign the labels $y=\pm1$ or $y=0,1$.
As it has already been stated, the logistic function $\sigma(z)$ is a good choice since it has the form of a probability, i.e. $\sigma(-z)=1-\sigma(z)$ and $\sigma(z)\in (0,1)$ as $z\rightarrow \pm \infty$. If we pick the labels $y=0,1$ we may assign
\begin{equation}\begin{aligned}\mathbb{P}(y=1|z) & =\sigma(z)=\frac{1}{1+e^{-z}}\\\mathbb{P}(y=0|z) & =1-\sigma(z)=\frac{1}{1+e^{z}}\\\end{aligned}\end{equation}
which can be written more compactly as $\mathbb{P}(y|z) =\sigma(z)^y(1-\sigma(z))^{1-y}$.
It is easier to maximize the log-likelihood. Maximizing the log-likelihood is the same as minimizing the negative log-likelihood. For $m$ samples $\{x_i,y_i\}$, after taking the natural logarithm and some simplification, we will find out:
\begin{equation}\begin{aligned}l(z)=-\log\big(\prod_i^m\mathbb{P}(y_i|z_i)\big)=-\sum_i^m\log\big(\mathbb{P}(y_i|z_i)\big)=\sum_i^m-y_iz_i+\log(1+e^{z_i})\end{aligned}\end{equation}
Full derivation and additional information can be found on this jupyter notebook. On the other hand, we may have instead used the labels $y=\pm 1$. It is pretty obvious then that we can assign
\begin{equation}\mathbb{P}(y|z)=\sigma(yz). \end{equation}
It is also obvious that $\mathbb{P}(y=0|z)=\mathbb{P}(y=-1|z)=\sigma(-z)$. Following the same steps as before we minimize in this case the loss function
\begin{equation}\begin{aligned}L(z)=-\log\big(\prod_j^m\mathbb{P}(y_j|z_j)\big)=-\sum_j^m\log\big(\mathbb{P}(y_j|z_j)\big)=\sum_j^m\log(1+e^{-yz_j})\end{aligned}\end{equation}
Where the last step follows after we take the reciprocal which is induced by the negative sign. While we should not equate these two forms, given that in each form $y$ takes different values, nevertheless these two are equivalent:
\begin{equation}\begin{aligned}-y_iz_i+\log(1+e^{z_i})\equiv \log(1+e^{-yz_j})\end{aligned}\end{equation}
The case $y_i=1$ is trivial to show. If $y_i \neq 1$, then $y_i=0$ on the left hand side and $y_i=-1$ on the right hand side.
While there may be fundamental reasons as to why we have two different forms (see Why there are two different logistic loss formulation / notations?), one reason to choose the former is for practical considerations. In the former we can use the property $\partial \sigma(z) / \partial z=\sigma(z)(1-\sigma(z))$ to trivially calculate $\nabla l(z)$ and $\nabla^2l(z)$, both of which are needed for convergence analysis (i.e. to determine the convexity of the loss function by calculating the Hessian). |
Series
Series refers to the infinite sum, that is defined as limit of sum, is it converges, and as way to get the asymptotic approximations, if diverge.
Example 1
Sometimes, the results of application of the Series operator are difficult to interpret. One example squared result of function HankelH0 of square root of parameter of the expansion is copypasted below.
Series[HankelH1[0, Sqrt[x] ]^2 Pi I Sqrt[x]/2, {x, Infinity, 2}]
does \( ~ e^{2 i \sqrt{x}} \left(1-\frac{1}{4} i \sqrt{\frac{1}{x}}-\frac{5}{32 x}+\frac{21}{128} i \left(\frac{1}{x}\right)^{3/2}+\frac{381}{16384 x^2}-\frac{675 i \left(\frac{1}{x}\right)^{5/2}}{65536}+O\left(\left(\frac{1}{x}\right)^3\right)\right) \)
while
Series[HankelH1[0, Sqrt[x] ]^2 Pi I Sqrt[x]/2, {x, Infinity, 1}]
does \( ~ e^{2 i \sqrt{x}} \left(1-\frac{1}{4} i \sqrt{\frac{1}{x}}-\frac{5}{32 x}+\frac{9}{512} i \left(\frac{1}{x}\right)^{3/2}+O\left(\left(\frac{1}{x }\right)^2\right)\right) \)
In the last case, coefficient with \( ~ e^{2 i \sqrt{x}} \left(\frac{1}{x}\right)^{3/2}~\) is just wrong. (The same refers to coefficients at higher terms in the first series.)
In the examples above, in order to see the bug, both, square root of argument and second power of value of function seem to be important. Perhaps this bug refers only to the Bessel function; attempts to reproduce similar inconsistencies with other functions are not successful.
Example 2
Kori[z_]=BesselJ[0,BesselJZero[0,1] Sqrt[z]]/(1-z)
naga[p_] = Assuming[{Im[p] == 0}, Integrate[Kori[z]^2 Exp[I p z], {z, 0, Infinity}]]
N[Assuming[p > 0, Series[naga[p], {p, 0, 2}]],20]]
gives
\(1.00000000000000000000000 +(0.\times 10^{-72}+1.00000000000000000000000 i)p -(0.50000000000000000000000 +0.\times 10^{-72}+i)p+O[p]^3\)
instead of expected
Integrate::idiv: "Integral of ((6+6\I\p\z-3\p^2\z^2)\BesselJ[0,Sqrt[z] <<1>>]^2)/(-1+z)^2 does not converge on {0,\[Infinity]}. "
References |
As per Ohms law , V = IR OR I = V/ R
So if I set R to 0 , then V should be equal to I ?
However when I connect 9 Volt battery poles to each other with multimeter in between the reading says 0.2 A.
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As per Ohms law , V = IR OR I = V/ R
So if I set R to 0 , then V should be equal to I ?
However when I connect 9 Volt battery poles to each other with multimeter in between the reading says 0.2 A.
Fortunately for you, batteries have internal resistance. If they hadn't then you would have either blown your meter's current limit fuse or destroyed your meter.
Never short-circuit a battery or power supply with an ammeter. Always connect it in series with the load.
You can model your battery as an "ideal" 9 V battery with a series resistance. From you measurement you can calculate that the battery's internal resistance. \$ R = \frac {V}{I} = \frac {9}{0.2} = 45 \ \Omega \$.
From the comments:
So does it mean if I short circuit 220 V (in house voltage), then reading should be 220 amps, since now there is no battery hence no internal resistance?
If there is no internal resistance on your 220 V supply then \$ I = \frac {V}{R} = \frac {220}{0} = \infty \ \text A \$.
Never short out a supply like that. Explosive amounts of energy can be released. (Look up "arc flash" on YouTube.) If you try it with your multimeter the results could be lethal.
First, as already said- don’t ever attempt to deliberately short any power source unless you are trying to obtain a UL rating for a device- you run a high risk of fire or (perhaps minor) explosion! I once worked for a power supply manufacturer and was assigned to do this sort of work one week- the supply was behind a blast shield for this reason.
Second, the ammeter itself has resistance in addition to the wires connecting it as well as the battery itself. There’s no way in a practical circuit to achieve zero resistance however close you might get and your current reading would only be 9A if the total resistance was exactly 1 ohm.
For internal resistance of typical battery types, see this article (the table therein shows a 9V zinc carbon battery is typically 35 ohms): |
I am learning EE, and about complex frequencies, but what is its physical meaning? What is it used for? Why is it? And only happen in the laplace transform?
closed as too broad by gented, The Photon, ZeroTheHero, Jon Custer, SRS Mar 19 at 17:15
Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
It is hard to know what you mean without any examples, but typically complex frequency means either loss or gain, depending on the sign of the imaginary part and the convention you are using.
For example, I could try to find solutions to:
$\ddot{x}+\gamma \dot{x}+ \omega_0^2 x = 0$, where $\gamma,\omega_0 \in \mathbb{R}$
by trying $x=\exp\left(i\omega t\right)$, I would find that such are valid provided $-\omega^2+i\gamma\omega+\omega_0^2=0$, which will necessitate for $\omega$ to have an imaginary part. So I would end up with
$x=\exp\left(i\Re\left(\omega\right)t\right)\exp\left(-\Im\left(\omega\right)t\right)$, so now you can see that depending on the sign of the imaginary part, $x$ will either grow (gain) or decay (loss).
Note that you can perfectly easily put complex frequency into $\sin/\cos$, e.g.
$\sin\left(\left(\omega+i\omega'\right)t\right)=\sin\left(\omega t\right)\cosh\left(\omega ' t\right)+i\cos\left(\omega t\right)\sinh\left(\omega't\right)$ |
Let $C_1$ be our $\color{blue}{\text{blue circle}}$, and $C_2$ our $\color{darkorange}{\text{orange circle}}$ with the following coordinates. Note that $C_2$ lies at $(0,R)$ which is an important detail for this problem.
Our two circles' equations are
$\begin{eqnarray}C_1&:& x^2+y^2 &= R^2 \tag{1}\\C_2&:& (x-R)^2+y^2 &= r^2 \tag{2}\end{eqnarray}$
Combining $(1)$ and $(2)$ and solving for $x$ we get
$$(x-R)^2+(R^2-x^2) = r^2 \\ x_1 = \dfrac{2R^2-r^2}{2R}$$
So what did we just find? We found the $x$-coordinate at which the two circles intersect.Using the formula to find an area for a circular segment. Let $R'$ be a radius, and $d$ a distance from a circle's centre to a chord.
For our case, our chord would be a vertical line going through the intersection points of our circles.
$$A(R',d) = R'^2\cdot\cos^{-1}\left(\frac{d}{R'}\right) - d\sqrt{R'^2-d^2}$$
Using this, and plugging in respectively value for our circles we get the following
$$A(R,x_1) = R^2\cdot\cos^{-1}\left(\frac{x_1}{R}\right) - x_1\sqrt{R^2-x_1^2}$$
So where are we now? We've calculated the shaded area below, and we need to find $d$ so that we can do the same for the other circular segment.
Well, we know that the distance between our circles is $R$. So why not put $d = R-x_1$. Hence,
$$A(r,d) = r^2\cdot\cos^{-1}\left(\frac{d}{r}\right) - d\sqrt{r^2-d^2}$$
This would mean our $A_{total} = A(R,x_1) + A(r,d)$. Writing this out becomes tedious, but using Mathematica gives,
$$A_{total} = r^2\cos^{-1}\left(\frac{r^2}{2R\cdot r}\right) + R^2\cos^{-1}\left(\frac{2R^2-r^2}{2R^2}\right) - \frac{1}{2}\sqrt{r^2\cdot 2R(2R+r)}$$
Now, using our $A_{total}$ we can find values for $r$ by specifying $R$ and the area ratio you want. I.e.
If $R = 2$ and you want the area between the circles to equal half of $C_1$'s area, you would put:
$$R = 2: A_{total} = \frac{1}{2}\pi\cdot R^2 = 2\pi$$
Let Mathematica calculate $A_{total}$ numerically after specifying $R$, and you will find it gives you two solutions. Use the
positive numerical solution and that's the radius you were looking for.
NB! By trying different values for $R$ and dividing the result $r$ with $R$, respectively, you'll find $\frac{R}{r} \approx 1.15872847 \ldots$. (OEIS A133731)
And as such, for this particular problem, the radius $r$ is always near $1.16\cdot R$ length units. |
That post has a more generalized form of a metric I occasionally see $d(x,y) = \frac{|x-y|}{1+|x-y|}$. When would using this metric be useful exactly? It occasionally comes up when I study analysis, but I don't know why, I don't know what people use it for or what benefit it could ever bring over the standard metric. I've merely only seen it as an example of a metric in books or sites, but if so many sources mention it, then it's very unlikely that it's useless.
It's a metric that is bounded above by $1$, while maintaining the same topology. This means that bounded metrics are just as powerful as general metrics (which is arguably interesting in itself).
More concretely, there's a commonly used construction for turning a countable product of metric spaces into a metric space itself. Specifically, if we have spaces $(X_n, d_n)$ where $n \in \Bbb{N}$ and $d_n$ is bounded uniformly (e.g. $d_n \le 1$ for all $n$), then $\prod_n X_n$ is a metric space with the metric $$d(x, y) = \sum_{n=1}^\infty \frac{d_n(x_n, y_n)}{2^n}.$$ Boundedness is important to guarantee convergence. This function is a metric, and it proves that a countable product of metrisable spaces are metrisable. This, in turn, is used to prove a bunch of interesting metrisability theorems. Coming from a functional analysis background, one consequence I'm partial to is the metrisability of the weak topology of a separable normed linear space when restricted to the unit ball. From this, we get the handy Eberlein-Smulian theorem.
Of course, this is just one field's use of this metric!
One advantage of $\rho$ is that $\rho\le1$ regardless of the size of $d$. Suppose you're fitting a model to data, penalising the model for each data point's distance from the model's predictions, with the aim of parameter estimation. If $d$ is unbounded, a sum of $d$ penalisations is very sensitive to outliers, especially if large $d$ values aren't all that improbable (they're not always Gaussian). By contrast, $\rho$ gives at most a penalty of $1$ to any one data point, so the sensitivity to outliers is reduced.
That function is a metric because $f(x)=\frac{x}{1+x}$ is a monotone increasing function on $(0,\infty)$ so if you consider point $x,y,z$ if your space, then
$d(x,y)\leq d(x,z)+d(z,y)$
Thus
$f(d(x,y))\leq f(d(x,z)+d(z,y))= \frac{d(x,z)+d(z,y)}{1+d(x,z)+d(z,y)}= $
$\frac{d(x,z)}{1+d(x,z)+d(z,y)}+ \frac{d(z,y)}{1+d(x,z)+d(z,y)}\leq $
$\leq \frac{d(x,z)}{1+d(x,z)}+ \frac{d(z,y)}{1+d(z,y)}$
So $d’(x,y)=\frac{d(x,y)}{1+d(x,y)}$ is a metric
Why it is useful use this metric? Because this metric is always limited, in fact
$d’(x,y)< 1$
This is useful to prove, for example, that a countable product of metric space is also a metric space.
In fact you can observe that $d$ and $d’$ induce the same Topology on the space, so if you consider a countable family of metric spaces $\{(X_n,d_n)\}_n$ then
$D(x,y):=\sum_{n=1}^\infty \frac{d_n’(x_n,y_n)}{2^n}$ is a metric of $\prod_{n}X_n$ that induces the product Topology on this space.
I know this thing: if $(X, d)$ supports a measure $\mu$, then convergence in measure, for a sequence of functions, is the same as convergence with respect to the (integral of the) metric you cite. That is: on $\mathrm{Meas}(X)$, define the metric $$ d_\mu(f, g) = \int \frac{d(f(x), g(x))} {1+d(f(x), g(x))} d\mu(x) $$ Then a sequence of measurable functions converges in measure iff it converges with respect to the metric $d_\mu$.
This is stated in Tao's book on measure theory, if I recall correctly. |
I've studied that the spin space of an electron is a two-dimensions Hilbert space. A possible representation of this space can be constructed defining:$$\chi_+ = \begin{pmatrix}1 \\ 0 \end{pmatrix} \quad \chi_- = \begin{pmatrix}0 \\ 1 \end{pmatrix} $$
With this base the spin operators became: $$ \hat{S}_n = \frac{\hbar}{2} \sigma_n $$
where $\sigma_n$ is the nth Pauli matrix. If we take the system of two isolated electrons we have a four-dimension Hilbert space and a possible base for this space is: $$ \chi_+(1)\chi_+(2) \quad \chi_+(1)\chi_-(2) \quad \chi_-(1)\chi_+(2) \quad \chi_-(1)\chi_-(2) $$
(I'm ignoring indistinguishability and symmetrization principles because I think are not fundamentals in this question)
During lectures my professor didn't talked about possible representation of this space and what type of product is the one between the two spinors in the base we constructed.
I made some research and I've found out that this product is the tensor product between vector spaces which became the kronecker product in case of finite-dimension space. I applied this product to the base vectors and I've found this representation for the composite system: $$ \chi_+(1)\chi_+(2) = \begin{pmatrix}1\\0\\0\\0 \end{pmatrix} \quad \chi_+(1)\chi_-(2) = \begin{pmatrix}0\\1\\0\\0 \end{pmatrix} \quad \chi_-(1)\chi_+(2) = \begin{pmatrix}0\\0\\1\\0 \end{pmatrix} \quad \chi_-(1)\chi_-(2) = \begin{pmatrix}0\\0\\0\\1 \end{pmatrix} $$
which seemed to me pretty reasonable. Now I was trying to find a representation for the spin operators in this space but applying the kroneker product to the $\hat{S}_z$ operators in the two dimension space with itself: $$ \hat{S}_z \otimes \hat{S}_z = \frac{\hbar^2}{4} \begin{pmatrix}1 &0 &0 &0 \\ 0 &-1 &0 &0\\0& 0 &-1 &0\\ 0 &0 &0 &1\end{pmatrix} $$
But the correct operator should be: $$ \hat{S}_z = \hbar \begin{pmatrix}1 &0 &0 &0 \\ 0 &0 &0 &0\\0& 0 &0 &0\\ 0 &0 &0 &-1\end{pmatrix} $$
What am I doing wrong? How can I correctly build the representation of a composite system starting from the one of a single particle space? |
Liouville Numbers are Irrational Theorem Proof
Let $x$ be a Liouville number.
$x = \dfrac a b$
with $a, b \in \Z$ and $b > 0$.
By definition of a Liouville number, for all $n \in \N$, there exist $p,q \in \Z$ (which may depend on $n$) with $q > 1$ such that:
$0 < \left \lvert x - \dfrac p q \right \rvert < \dfrac 1 {q^n}$
Let $n$ be a positive integer such that $2^{n - 1} > b$.
Let $p$ and $q$ be any integers with $q > 1$.
We have:
$\left \lvert x - \dfrac p q \right \rvert = \dfrac {\left \lvert a q - b p \right \rvert} {b q}$
If $\left \lvert a q - b p \right \rvert = 0$, this would violate the first inequality.
If $\left \lvert a q - b p \right \rvert \ne 0$, then:
\(\displaystyle \left \lvert x - \frac p q \right \rvert\) \(\ge\) \(\displaystyle \frac 1 {b q}\) as $\left \lvert a q - b p \right \rvert$ is a positive integer \(\displaystyle \) \(>\) \(\displaystyle \frac 1 {2^{n - 1} q}\) by our choice of $n$ \(\displaystyle \) \(\ge\) \(\displaystyle \frac 1 {q^n}\) as $q > 1$ by definition
which would violate the second inequality.
Therefore, in any case, if $n$ is sufficiently large, there cannot exist integers $p$ and $q$ with $q > 1$ satisfying the two inequalities.
This is a contradiction.
Thus $x$ is irrational.
$\blacksquare$ |
I have some doubts related to some of the optical flow implementations.
How to compute angular error for optical flow algorithms? I know the formula for computing, $\theta=\arccos(\vec{v}_c\cdot\vec{v}_e)$
I have the computed flow field $\vec{v}_c$. How to get the exact flow field $\vec{v}_e$, ie the actual movement of pixels between frames ?
I am trying to implement Nagel's Paper for optical flow computation based on the discussion given in here. What I don't get is how to compute the weight matrix $W$. $W$ is given as
$W=(I_x^2+I_y^2+2\delta)^{-1}\begin{pmatrix}I_y^2+\delta &-I_xI_y\\ -I_xI_y & I_x^2+\delta \end{pmatrix}$
Assuming that the image dimensions are $m\times n$ the first matrix on the RHS has dimension $m\times n$, but the second block matrix is of dimension $2m\times 2n$. What am I doing wrong ? |
Superfluids passing an obstacle and vortex nucleation
1.
Courant Institute of Mathematical Sciences, 251 Mercer Street, New York, NY. 10012, USA
2.
Department of Mathematics, University of British Columbia, Vancouver BC V6T 1Z2, Canada
$ \epsilon^2 \Delta u+ u(1-|u|^2) = 0 \ \mbox{in} \ {\mathbb R}^d \backslash \Omega, \ \ \frac{\partial u}{\partial \nu} = 0 \ \mbox{on}\ \partial \Omega $
$ \Omega $
$ {\mathbb R}^d $
$ d\geq 2 $
$ \epsilon>0 $
$ u = \rho_\epsilon (x) e^{i \frac{\Phi_\epsilon}{\epsilon}} $
$ \rho_\epsilon (x) \to 1-|\nabla \Phi^\delta(x)|^2, \Phi_\epsilon (x) \to \Phi^\delta (x) $
$ \Phi^\delta (x) $
$ \nabla ( (1-|\nabla \Phi|^2)\nabla \Phi ) = 0 \ \mbox{in} \ {\mathbb R}^d \backslash \Omega, \ \frac{\partial \Phi}{\partial \nu} = 0 \ \mbox{on} \ \partial \Omega, \ \nabla \Phi (x) \to \delta \vec{e}_d \ \mbox{as} \ |x| \to +\infty $
$ |\delta | <\delta_{*} $
$ d = 2 $
$ |\nabla \Phi^\delta (x)|^2 $ Mathematics Subject Classification:35J25, 35B25, 35B40, 35Q35. Citation:Fanghua Lin, Juncheng Wei. Superfluids passing an obstacle and vortex nucleation. Discrete & Continuous Dynamical Systems - A, 2019, 39 (12) : 6801-6824. doi: 10.3934/dcds.2019232
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The conormal derivative problem for elliptic equations with BMO coefficients on Reifenberg flat domains,
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M. del Pino, M. Kowalczyk and J. Wei,
Entire solutions of the Allen-Cahn equation and complete embedded minimal surfaces of finite total curvature,
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M. del Pino, P. Felmer and M. Kowalczyk,
Minimality and nondegeneracy of degree-one Ginzburg-Landau vortex as a Hardy's type inequality,
[18]
Q. Du, J. Wei and C. Zhao,
Vortex solutions of the high-$\kappa$ high-field Ginzburg-Landau model with an applied current,
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Motions in a Bose condensate. Ⅲ. The structure and effective masses of charged and uncharged impurities,
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Gross-Pitaevskii dynamics of Bose-Einstein condensates and superfluid turbulence,
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Nonexistence of supersonic traveling waves for nonlinear Schrödinger equations with nonzero conditions at infinity,
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Boundary layers and emitted excitations in nonlinear Schröinger superflow past a disk,
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show all references
References:
[1] [2] [3] [4] [5] [6] [7] [8] [9]
L. Bers,
[10]
S. Byun and L. Wang,
The conormal derivative problem for elliptic equations with BMO coefficients on Reifenberg flat domains,
[11] [12] [13] [14]
M. del Pino, M. Kowalczyk and J. Wei,
Entire solutions of the Allen-Cahn equation and complete embedded minimal surfaces of finite total curvature,
[15] [16] [17]
M. del Pino, P. Felmer and M. Kowalczyk,
Minimality and nondegeneracy of degree-one Ginzburg-Landau vortex as a Hardy's type inequality,
[18]
Q. Du, J. Wei and C. Zhao,
Vortex solutions of the high-$\kappa$ high-field Ginzburg-Landau model with an applied current,
[19] [20] [21] [22] [23] [24] [25]
J. Grant and P. H. Roberts,
Motions in a Bose condensate. Ⅲ. The structure and effective masses of charged and uncharged impurities,
[26] [27]
M. Abid, C. Huepe, S. Metens, C. Nore, C. T. Pham, L. S. Tuckerman and M. E. Brachet,
Gross-Pitaevskii dynamics of Bose-Einstein condensates and superfluid turbulence,
[28] [29] [30] [31] [32] [33] [34] [35]
Y. Liu and J. Wei, Adler-Moser polynomials and traveling waves solutions of Gross-Pitaevskii, preprint.Google Scholar
[36] [37] [38]
M. Maris,
Nonexistence of supersonic traveling waves for nonlinear Schrödinger equations with nonzero conditions at infinity,
[39] [40]
C.-T. Pham, C. Nore and M. E. Brachet,
Boundary layers and emitted excitations in nonlinear Schröinger superflow past a disk,
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O. Rey and J. Wei,
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Traveling waves for some nonlocal 1D Gross–Pitaevskii equations with nonzero conditions at infinity.
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Asymptotic behaviors of ground states for a modified Gross-Pitaevskii equation.
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Gap solitons for the repulsive Gross-Pitaevskii equation with periodic potential: Coding and method for computation.
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Self-trapping and Josephson tunneling solutions to the nonlinear Schrödinger / Gross-Pitaevskii equation.
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Constraint minimizers of perturbed gross-pitaevskii energy functionals in $\mathbb{R}^N$.
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Another way to see that the difficulty there is the each appearance of "provable" in quotes refers to the formal Pvbl relation, which only resembles the real provability relation in certain respects.
In particular, there are consistent theories $T$ that extend PA such that $T \vdash \text{Pvbl}_T(\phi)$ for
every formula $\phi$. But of course $T$ does not actually prove every formula $\phi$, because $T$ is consistent.
In the argument in the question, there are several steps that mix real provability with formalized provability. Let's assume that by "false" statement you mean a statement disprovable in the theory.
For example, step 2 begins with the fact that, if $P$ is false, then $P$ is not provable, so by contraposition, if $P$ is provable then $P$ is true. But this does not mean the theory can prove "if $P$ is provable then $P$ is true", which is the formula $\text{Pvbl}(P) \to P$. We cannot expect the theory to prove $\lnot P \to \lnot \text{Pvbl}(P)$, either, because as above there are consistent theories in which $\text{Pvbl}(P)$ is provable for all $P$.
The way to avoid this sort of thing is to be more explicit about writing Pvbl for the formalized provability relation, rather than using "provable" in quotes, which tends to cause confusion.
In particular, Löb's theorem tells us that if $T \vdash \text{Pvbl}(P) \to P$ then $T \vdash P$. That is stronger than the claim that if $T \vdash \text{Pvbl}(P) \to P$ then $T \vdash \text{Pvbl}(P)$. The distinction between these is not nearly as clear when "provable" is used for both "$\vdash$" and "Pvbl". |
You are mixing up two different quantities (entropy of reaction, and molar entropies of the two gases involved in the reaction) and
either your text book is in error, or you have misread it somehow. For that reaction, as written, the formation of ozone from oxygen,$\Delta G$ is positive and $\Delta S$ is negative (the opposite of what you wrote in your question). It is unfavourable on both enthalpic and entropic grounds. Ozone is thermodynamically unstable with respect to oxygen under standard conditions.
Your experience, regarding the sign of $\Delta S$ and its relation to the change in the number of molecules in a gas-phase reaction, is by-and-large a good guide, because the translational contribution to the entropy typically dominates over the rotational and vibrational contributions in a gas. This reaction is consistent with this rule of thumb.
The relevant thermodynamic quantities are the standard Gibbs free energy, enthalpy, and entropy, of formation of ozone, since the oxygen is in its standard state. Taking the figures from this copy of CRC we have$$\Delta_{\text{f}} G^\circ (\text{O}_3) = 163.2 \, \text{kJ}\,\text{mol}^{-1}\quad\Delta_{\text{f}} H^\circ (\text{O}_3) = 142.7 \, \text{kJ}\,\text{mol}^{-1}$$and for the standard entropies$$S^\circ (\text{O}_3) = 238.9 \, \text{J}\,\text{K}^{-1}\text{mol}^{-1}\quadS^\circ (\text{O}_2) = 205.2 \, \text{J}\,\text{K}^{-1}\text{mol}^{-1}$$So indeed the standard molar entropy of ozone is a little higher than that of oxygen (as the comments say, if you want to explain this precisely you need to take into account internal degrees of freedom, as well as properly calculating the translational contributions), but this does not lead to a positive entropy of reaction. To calculate this, i.e. the entropy of formation of ozone, we need$$\Delta_{\text{f}} S^\circ (\text{O}_3) = S^\circ (\text{O}_3) - \tfrac{3}{2} S^\circ (\text{O}_2) = -68.9 \, \text{J}\,\text{K}^{-1}\text{mol}^{-1}$$remembering that all these standard quantities of formation are per mole of ozone. Equivalently you can calculate this number from $[\Delta_{\text{f}} H^\circ (\text{O}_3)-\Delta_{\text{f}} G^\circ (\text{O}_3)]/T$ giving the same answer.
So the reaction has a negative $\Delta S$ and a positive $\Delta G$.
[EDIT following OP comment]
This means that, in the oxygen-ozone equilibrium, both forward and backward reactions are happening, at the same rates,but the rate constants are such that the equilibrium concentration (or partial pressure) of O$_3$ is much lower than that of O$_2$.
This is not how things happen in the stratosphere, nor in most methods for preparing ozone. Generally, these involve a multi-step process, in whichthe first step is the dissociation of O$_2$ into two oxygen atoms,by absorbing a photon of UV light, or through an electrical discharge (for example). Wikipedia provides a reasonable summary of the Ozone-oxygen cycle, or Chapman cycle, with links to further resources, while various production methods are described on the Ozone page. The above equilibrium thermodynamics relations cannot be applied to processes like this, because of the injection of energy.
Instead, in the case of the Chapman cycle, a steady-state kinetic scheme can be set up for creation and destruction of O$_3$, and O atoms, where some of the rate constants depend on the intensity of UV light. The net result is a steady (small) population of O$_3$, and O atoms, accompanied by the conversion of energy from UV light into heat which is dissipated in the atmosphere. In real life, the situation is more complicated still, because of the roles of other species and reactions: books such as RP Wayne's
Chemistry of Atmospheres give a lot more detail.
[2nd EDIT following further OP comments]
For the equilibrium between oxygen and ozone, the standard Gibbs free energy change for the reaction $\Delta_r G^\circ$ may be related to the equilibrium constant $K$ (see e.g. here)$$\Delta_r G^\circ = -RT \ln K$$where $K$ is expressed (for ideal gases) in terms of the partial pressuresrelative to the standard pressure $p^\circ$ (1 bar)$$K = \frac{(p_{\text{O}_3}/p^\circ)^2}{(p_{\text{O}_2}/p^\circ)^3}$$For large positive $\Delta_r G^\circ$, $K$ is extremely small,so the equilibrium partial pressure of ozone is extremely small,if we are given that the partial pressure of oxygen has a reasonable value.Indeed, if we take (for the reaction as written, producing two moles of ozone)$$\Delta_r G^\circ=2\times\Delta_{\text{f}} G^\circ (\text{O}_3) = 326.4 \, \text{kJ}\,\text{mol}^{-1}$$then $K=\exp(-\Delta_r G^\circ/RT)\approx 6\times10^{-58}$;if the partial pressure of oxygen is $\approx 0.21$ bar(as in the atmosphere around us),then the equilibrium partial pressure of ozone is $p_{\text{O}_3}\approx 2.4\times 10^{-30}$ bar.
It is wrong to conclude that the reaction producing ozone "cannot happen"because the free energy change is positive.$\Delta_r G^\circ$ is simply the free energy change associated withthe
complete conversion of reactants into products.In reality, an equilibrium is always established somewhere in between.In this case, it is very far over to one side, but not 100%.If $\Delta_r G^\circ$ were not so large (but still positive) theequilibrium would not be so far in favour of reactants.Thermodynamically, you can think of the position of equilibriumas being determined by both $\Delta_r G^\circ$ and the entropy of mixing of the reactants with the products:this will always result in an equilibrium position that is neither 100% reactantsnor 100% products.
It is also possible to give a dynamical interpretation of chemical equilibriain terms of forward and reverse reaction rates,but I emphasize that this argument only applies, at least in this simplified form, to one-step chemical reactions. The oxygen-ozone equilibrium certainly involves multiple steps(nobody is pretending that three molecules of oxygen miraculously collidetogether at once). To see why this more complicated mechanismdoes not affect the argument above about the equilibrium constant, see How is it that the equilibrium constant does not depend on the mechanism? and you might also find Transition state and free energy useful. So consider the dynamical equilibrium$$\text{A} + \text{B} \leftrightharpoons \text{C} + \text{D}$$and assume that the forward and reverse reactions are simple one-step processesobeying the rate equations$$\text{Rate}(\rightarrow) = k_\rightarrow [\text{A}][\text{B}],\qquad\text{and}\qquad\text{Rate}(\leftarrow) = k_\leftarrow [\text{C}][\text{D}].$$At equilibrium
both forward and reverse reactions are happening at the same rate because otherwise the concentrations would be changing.We can write$$\text{Rate}(\rightarrow) = \text{Rate}(\leftarrow)\qquad\Rightarrow\qquadK = \frac{[\text{C}][\text{D}]}{[\text{A}][\text{B}]} = \frac{k_\rightarrow}{k_\leftarrow}$$So, if $\Delta_r G^\circ$ is large and positive, $K\ll 1$,and $k_\leftarrow \gg k_\rightarrow$.The rate constant for the reverse reaction is much larger than that for theforward reaction.However, both reactions are still happening, and the rates of these reactionsare equal.
As I said, you can't apply this argument directly to a multi-step reaction scheme, but the two links I gave above show how the same ideas apply in such a case.Also, most of the above material is covered in standard physical chemistry texts. |
For an undirected graph $G$, let us denote $\gamma(G)$ the chromatic number of $G$. Let us define for positive integer $n$, the language $$ L_n = \{\langle G\rangle : \gamma(G) = n\}. $$ This is the language of undirected graphs with chromatic number $n$. I think $L_n \in \mathrm{P}^{\mathrm{NP}}$, for any $n$.
My argument is the following. We can look at the language, $$ \mathrm{CHROMNUM} = \{\langle G, k\rangle : \gamma(G) \leq k\}. $$ It is clear that $\mathrm{CHROMNUM} \in \mathrm{NP}$, since we can verify an instance just by displaying an at most $k$-coloring of $G$.
Thus, consider oracle Turing machine $M^{\mathrm{CHROMNUM}}_n$, which on input $G$, simply checks that $\langle G, n \rangle \in \mathrm{CHROMNUM}$ and that $\langle G, k \rangle \not \in \mathrm{CHROMNUM}$ for all $k \leq n$.
I've just read about $\mathrm{P}^{\mathrm{NP}}$, so I wanted to be sure that this kind of argument is correct. |
Day Date Topic Text Exercise 50 Thurs Nov 15 Radian Measure pg 320 1aceg, 2aceg, 3bc, 4bc, 5, 7ab, 8ab, 9ac, 11, 12, 13
Challenge: 10, 16*
*the answer for 16 should be about 86.81 radians per second
51 Fri Nov 16 PD DAY 52 Mon Nov 19 Radian Measure and Angles on the Plane pg 330 2ab, 3, 5acdf, 6cdef, 7ad, 11, 13, 15, 16 53 Tues Nov 20 Exploring Graphs of Primary Trigonometric Functions pg 333
(A – L)
Supplementary Discovery Activity
pg 336
2c, 3, 5
54 Wed Nov 21 Transformations of Trigonometric Functions pg 343 1ad, 4bc, 5ac, 6c, 7bc, 8c d , 9, 10 , 11, 12, 14a graph, do not sketch! 55 Thurs Nov 22 Homework Takeup and Formative Quiz Complete any outstanding homework 56 Fri Nov 23 Exploring Graphs of Reciprocal Trigonometric Functions pg 353 1, 2, 3, 7*
* for 7 graph on the interval -2π ≤ x ≤ 2π
Also, graph, not sketches
57 Mon Nov 26 Modelling with Trigonometric Functions pg 360 1, 3, 5, 7, 8, 10 , 11% Answers in the text are incorrect. Change them to:
Challenge: 13
10a) $n(t)=3.7cos(\frac{2\pi}{365}(t-172))+12$
10b) 9.2 hours Also, for #10a change the instruction” from nth” day to “t-th day”
% remember Example 2?
Also, the answer for 11 is incorrect. The “d” value is not 116; it is approx. 102
58 Tues Nov 27 Review Day 1 IS YOUR CALCULATOR IN RADIAN MODE?
Enrichment Day
pg 376
1 - 4, 5b, 6, 8, 11d, 12, 13, 14, 16
, 19** a d-value of about 0.3 is acceptable
** sub in $h=\pm 0.01$ immediately
Note final answer corrections:
#6 a) $tan\theta=\pm \frac{12}{5}$
#6 c) $\theta=2.0\text{ or }4.3$
#19d) approx. -144
pg 378
Self-Test
Note: do not use a calculator for #2.
In #6 answers may vary.
For #8a, use only a cosine function.
Note final answer corrections:
#1: y = tan x is also a function that is possible!
#3: y = 94.9
59 Wed Nov 28 Review Day 2 60 Thurs Nov 29 Unit Test 6 |
How great would it be if you were able to perfectly characterize melodies in mathematical terms? You could populate an abstract "melodyspace" with your own melodies and find similar or totally different ones.
Lets reduce the complexity right away and consider only pitch and note length. These two quantities offer plenty possibilities to characterize melodies. Later we could expand our analysis by considering meter, dynamics in general, harmonic context (scale) and probably many more aspects.
I think it is sufficient to consider only a relative pitch and note length. You could shift a whole melody by some semitones and it would still be the same.
What is a note in our system?
We consider only two aspects. This results in a 2-tuple, e.g. $\left( \mathrm{A4}, \frac{1}{4} \right)$, of some pitch value, here A4 (see scientific notation), and a length, 1/4th.
But this is an absolute measure. We could look a little to the side and shift the focus to the gap
betweenthe notes - lets talk in terms of intervals! The pitch is represented as some stepsize, e.g. $+5$ for a perfect fourth (see interval), and could be limited to meaningful values. No stepsize should lead to a lower note than C0 or a higher note than G9 (highest possible MIDI note). Interestingly this restricts the ability to transpose the melody. The melody C0 G9 C0 G9 can't be shifted at all - in contrast to something like C4 E4 C4 E4.
It would be also a good idea to normalize the note length to a whole note and limit it to values between $\frac{1}{32}$ and $1$.
Finally this would lead to a 2-tuple of the form $\left( \pm n, \frac{1}{m} \right)$, with $n,m \in \mathbb{N}$ and some arbitrary meaningful restrictions for $n$ and $m$.
What is a melody in our system?
A melody would be just a sequence of notes. Something like
$$\left(\mathrm{A4}, \frac{1}{2} \right),\left( +5, \frac{1}{2} \right), \left( +5, \frac{1}{2} \right), \left( -7, 1 \right) $$
Which corresponds to the sequence A4, D5, G5, C5, if you start with a note A4.
To underline the generic nature it could be a good idea to define a special tuple for the first note. Something like $\left( n', \frac{1}{m} \right)$, with $n' \in \left[ n_{\mathrm{lowest}}, n_{\mathrm{highest}}\right]$ and $m$ as above. $\left[ n_{\mathrm{lowest}}, n_{\mathrm{highest}}\right]$ is the interval of possible starting notes, which has to be computed according to the limit of C0 as the lowest and G9 as the highest note.
Creating the melodyspace:
What quantities can we think of to characterize such a melody?
There are quantities which do not rely on the order of the involved intervals. These are for example a mean-interval + the according variance, the amount of dissonant/consonant intervals, the length of the melody, the mean note-length + according variance, etc.
Other quantities could be dependent on the order. I don't know if this is important later, but it surely is interesting!
And what the heck is this melodyspace I am talking of?
If you consider every of the above quantities as an (independent) direction in a multidimensional space, then a melody is characterized by a vector in this space.
For example you could imagine that we want to characterize a melody only by its total length and its tendency to be "rising" or "falling". (0 would mean something like 100% falling, e.g. the melody C9 C8 C7 C6, and 1 would mean something like 100% rising, like A6 A7 A8 A9)
In this case the melodyspace would only be a two dimensional graph with the length on the x-axis and the tendency on the y-axis.
Both example melodies, C9 C8 C7 C6 and A6 A7 A8 A9, would have a $x$ value of $4$, but the former would be at $y = 0$ and the later would be at $y = 1$
I have no clue at all if this leads to something useful, but I surely will implement this model in some programming language and look where it leads to. |
As you know, the peak AC input signal for delivering any particular power into any particular speaker load is:
$$V_\text{P}=\sqrt{2\:R\:P}$$
In your case, with \$1\:\text{W}\$ and \$R_\text{SPKR}=4\:\Omega\$, this works out to \$V_\text{P}=2.83\:\text{V}\$ and \$I_\text{P}=710\:\text{mA}\$.
Also, your capacitor needs to have a low impedance relative to the speaker, so let's say about \$400\:\text{m}\Omega\$ for the capacitor. If you decide that \$f=200\:\text{Hz}\$ is the lowest frequency you want to handle, this suggests a capacitance of \$2\:\text{mF}\$. A bit larger than you show. Let's also assume you set the capacitance, appropriately.
In general, you want the BJT to be operating with a collector current dynamic range of less than 5:1. (That's about \$42\:\text{mV}\$ variation of \$V_\text{BE}\$.) This means that your minimum emitter current should be \$I_{\text{E}_\text{MIN}}\ge \frac{1}{2}\left[I_\text{P}+\frac{V_\text{P}}{R_\text{E}}\right]=\frac{1}{2}\frac{V_\text{P}}{R_\text{E}\vert\vert R_\text{SPKR}}\$. In your circuit's case, this means \$I_{\text{E}_\text{MIN}}\ge 495\:\text{mA}\$. Whatever you decide for \$I_{\text{E}_\text{MIN}}\$ is your choice, though. That's just a recommendation.
Once you have that value, and I'm picking \$I_{\text{E}_\text{MIN}}=500\:\text{mA}\$, you can work out that the quiescent emitter voltage needs to be \$V_{\text{E}_\text{Q}}\ge R_\text{E}\left(I_\text{P}+I_{\text{E}_\text{MIN}}\right)+V_\text{P}\$. In your case, this means \$V_{\text{E}_\text{Q}}\ge 14.83\:\text{V}\$. I'm picking \$V_{\text{E}_\text{Q}}= 15\:\text{V}\$. (Your BJT's base will have to be driven with a quiescent voltage that is one \$V_\text{BE}\$ higher.)
(You could skip the steps shown above and simply compute \$V_{\text{E}_\text{Q}}\ge \frac{3}{2}V_\text{P}\left(1+\frac{R_\text{E}}{R_\text{SPKR}}\right)\$ and get the same answer. Note that the larger the ratio of \$\frac{R_\text{E}}{R_\text{SPKR}}\$ the higher the quiescent emitter voltage needs to be.)
The BJT's collector voltage should be at least \$2\:\text{V}\$ higher than the quiescent voltage plus the peak. So that tells you the rail you'll need, at minimum.
Just did a quick run on the basis of above analysis, except that I didn't bother working out the \$V_\text{BE}\$ for the BJT. I just set the DC bias for the input signal as shown. Close enough.
Spice reports the power in the speaker at \$1.0043\:\text{W}\$ and the RMS voltage across the speaker as \$V_\text{SPKR}=2.0043\:\text{V}_\text{RMS}\$. Not bad for shooting from the hip, so to speak.
Sadly, Spice also reports that the power for the BJT here is about \$3.9\:\text{W}\$ and for \$R_\text{E}\$ is \$20.7\:\text{W}\$. This ignores all the driver stuff that would normally have to back up this output section and would only increase the power dissipation. But in round numbers this means burning off about \$25\:\text{W}\$ in a circuit that will deliver \$1\:\text{W}\$ to the speaker. (Which is a good thing, I suppose, if you own a lot of stock with your local power company. You'll get some of your cost back in dividends.)
My question is: is there a process for determining a value for R2 that
can meet the objective of driving 1 watt into the speaker, or is this
just a waste of time?
You can work this backwards. Given that your \$V_\text{CC}=12\:\text{V}\$, then find that \$R_\text{E}\le 2.5\:\Omega\$. (From: \$R_\text{E}\le R_\text{SPKR}\left(\frac{2}{3}\frac{V_\text{CC}-V_\text{P}-2\:\text{V}}{V_\text{P}}-1\right)\$.)
FINAL NOTES
Here is an LTspice schematic that can be used to program up and test out any particular crazy-minded class-A amplifier of this kind. I've used a particular BJT in it that can handle some current -- but feel free to change it. Other than that, the specs are hopefully clear. Make some changes and run it.
Aside from setting up the desired power and load values, only one of the two -- either \$R_\text{E}\$ or else V_\text{CC}\$ -- can be set, as the other one is then determined by that choice. So set exactly one of them, not both. The other must be set to zero.
The .ASC file follows here to save time having to write up the above schematic, by hand.
Version 4
SHEET 1 880 680
WIRE 96 80 96 48
WIRE 32 128 -32 128
WIRE -32 160 -32 128
WIRE 96 192 96 176
WIRE 176 192 96 192
WIRE 288 192 240 192
WIRE 96 224 96 192
WIRE 288 224 288 192
WIRE -32 272 -32 240
WIRE 96 336 96 304
WIRE 288 336 288 304
FLAG 96 336 0
FLAG 288 336 0
FLAG 96 48 Vcc
FLAG -32 272 0
FLAG 288 192 Vspkr
SYMBOL npn2 32 80 R0
SYMATTR InstName Q1
SYMATTR Value D44H11
SYMBOL res 80 208 R0
SYMATTR InstName RE
SYMATTR Value {REV}
SYMBOL cap 240 176 R90
WINDOW 0 0 32 VBottom 2
WINDOW 3 32 32 VTop 2
SYMATTR InstName C1
SYMATTR Value {CV}
SYMBOL res 272 208 R0
SYMATTR InstName Rspkr
SYMATTR Value {LOAD}
SYMBOL voltage -32 144 M0
WINDOW 123 0 0 Left 2
WINDOW 39 0 0 Left 2
SYMATTR InstName V1
SYMATTR Value SINE({VQ+.7} {VPEAK} 1k)
TEXT 152 32 Left 2 !V99 Vcc 0 {VCCV}
TEXT 152 56 Left 2 !.tran 0 .1 0 100n
TEXT -392 -112 Left 2 !* INPUT SPECIFICATIONS\n.param POWER={1}\n.param LOAD={4}\n.param FMIN={200}
TEXT -392 376 Left 2 !* INTERMEDIATE CALCULATIONS\n.param VPEAK={sqrt(2*POWER*LOAD)}\n.param VQ=if({RE+.5},{3/2*VPEAK*(1+RE/LOAD)},{VCC-VPEAK-2})\n.param REV={LOAD*(2/3*VQ/VPEAK-1)}\n.param VCCV={VQ+VPEAK+2}\n.param CV={5/pi/FMIN/LOAD}
TEXT 48 -112 Left 2 !* SET ONLY ONE OF THESE TO NON-ZERO VALUE\n.param RE={0}\n.param VCC={20}
TEXT -232 -184 Left 2 ;CRAZY-MINDED CLASS-A OUTPUT STAGE AMPLIFIER |
I suspect your text is taking $$\hat x=x\times \;,\qquad \text{ and } \qquad \hat p_x=\frac {\hbar}i\frac d{dx},$$as postulates (it only holds in the Schrödinger Picture and with the Position Representation). And what it is saying is that it expects you to take any other observable $O$ and write it as a function of $t$, $x$, $p_x$, etcetera and replace every $x$ with a $\hat x$, and every $p_x$ with a $\hat p_x,$ etcetera to get an operator $\hat O$.
Obviously the order in which you write your observable $O$ as a function of $x$ and $p_x$ and such matters because you can get different operators $\hat O$ for different choices. And sometimes the operator you get won't even be self-adjoint or even hermitian. So really it only works for some observables and have to do it in a very specific way. And most very elementary texts don't want to get into it and think its OK to give a false sense of generality. Then you can feel good about yourself and imagine that you know how to make operators for anything you want and can get back to focusing on the next section of the book.
If you want to derive the results $$\hat x=x\times \;,\qquad \text{ and } \qquad \hat p_x=\frac {\hbar}i\frac d{dx},$$ then you have to make some other assumptions. But since it only holds in some pictures and some representations, maybe it isn't that important.
Then you asked a second question. Why is the eigenvalue of the operator corresponding to an observable the value of the observable? Your other result only held in the Schrödinger Picture and the Position Representation . Well, this result (about eigenvalues) doesn't hold for weak measurements. The whole idea is that a strong measurement, when repeated yields the same result. An eigenvector is very nice for that because before and after acting on an eigenvector you have (up to a scalar multiple) the same thing. So as long as you have different eigenvectors behave differently from each other but identically each time, then you can get that desirable (defining?) feature of a strong measurement. But many measurements don't send each eigenvector to someplace different. And they can't totally separate them because any dynamical development has to be linear, so if you have a whole 2d space of eigenvectors one you've sent two eigenvectors $v$ and $w$ that aren't both on the same line somewhere, everything else in the plane spanned by the two vectors is determined where it has to go, so ones that start out close to $v$ of those have to end up close to where $v$ went. These continuously close together nonzero eigenvectors have the same eigenvalues. So you are forced to some (some) eigenvectors with the same eigenvalue close together. One way to set up a strong measurement is to have it treat each eigenvector differently based solely on the different value of the eigenvalue. Then all the ones in the same eigenspace get operated on in the same way. Problem solved.
So one option is to just say that a measurement separates eigenvectors with different eigenvalues. For instance if a device deflects one eigenvector left and another one right then you can use multiple copies of the device and see that once deflected left the result is then always deflected left by the same machines. This reliability across reproduction at different times and places is a nice thing, and you can call that reliability the measurement outcome.
So in that sense it doesn't need a value per se. However if you actually described your particular machine to saw how quickly or how far it deflected something then you can relate it to a numerical eigenvalue. You could even have many detectors in a row and associate the value with the position of the detector that goes off. So there are senses where the numerical values make empirical sense.
Finally, if you assign the different outcomes the numerical value of the eigenvalue, then you can compute the expectation value of the operator (multiply each outcome by the frequency it happened and add them all up to get a sample average and compare that to a theoretical population average). if you do that, then the theoretical population average can be compared at the inner product of the original vector with the vector after if is acted on by the operator divided by the inner product of the vector with itself (that gives the eigenvalue for each eigenvector and a weighted average for the eigenvalues for a general vector so the relative weight of the linear combination is now related to the frequency weighing in the weighted average).
Obviously you can assign whatever labels you want to your outcomes, but if you assign the eigenvalues, then this simple computation "
vector inner producted with the result after operator acts on vector divided by vector inner producted with itself" gives you the thing you wanted. |
I) One problem is that the momentum operator $\hat{p}$ is an unbounded operator, which means that it is only defined on a domain $D(\hat{p}) \subsetneq {\cal H}$ of the Hilbert space ${\cal H}=L^2(\mathbb{R})$.
When we apply the differentiation operator $\hat{p}=\frac{\hbar}{i}\frac{d}{dx}$ to OP's wave function
$$\tag{1} \psi(x)~=~A(a-x)\theta(a-|x|), \qquad A>0, $$
we get a term proportional to a distribution, cf. Stephen Powell's answer, so that the image $\hat{p}\psi\notin{\cal H}$ is outside the Hilbert space ${\cal H}=L^2(\mathbb{R})$ of square integrable functions. Stated a bit more precise, the issue is that OP's wave function $\psi \notin D(\hat{p}) $ is
not in the domain $D(\hat{p})$ of the momentum operator $\hat{p}$, so that the calculation does not make mathematical sense, cf. answer & below comments by Valter Moretti.
II) Nevertheless, apart from standard arguments about expectation values of operators, we can perform various non-rigorous heuristic explicit calculations that indicate that the average momentum $\langle p \rangle$ should be interpreted as zero:
$$\tag{2} \langle p \rangle~=~\frac{\hbar }{i}\int_{\mathbb{R}} \! dx ~\psi(x)\psi^{\prime}(x)~=~\frac{\hbar }{2i}\int_{\mathbb{R}} \! dx \frac{d}{dx}\psi(x)^2~=~\frac{\hbar }{2i}\left[\psi(x)^2 \right]^{x=\infty}_{x=-\infty}~=~0,\qquad $$
or
$$\langle p \rangle~=~\frac{\hbar }{i}\int_{\mathbb{R}} \! dx ~\psi(x)\psi^{\prime}(x)~\stackrel{(1)}{=}~\frac{\hbar A^2}{i}\int_{\mathbb{R}} \! dx~ (a-x)\theta(a-|x|)\frac{d}{dx}\left\{ (a-x)\theta(a-|x|)\right\}$$$$~\stackrel{\begin{matrix}\text{Leibniz'}\\ \text{rule}\end{matrix}}{=}~\frac{\hbar A^2}{i}\int_{\mathbb{R}} \! dx~ (a-x)\theta(a-|x|)\left\{-\theta(a-|x|) - (a-x){\rm sgn}(x) \delta(a-|x|) \right\} $$ $$~=~\frac{\hbar A^2}{i}\left\{\int_{-a}^a \! dx~ (x-a) - \sum_{x=\pm a} (a-x)^2\theta(a-|x|){\rm sgn}(x) \right\} $$$$\tag{3}~=~\frac{\hbar A^2}{i}\left\{ -2a^2+4a^2 \theta(0)-0\right\}~=~0. $$
In the last equality we assigned the value $\theta(0)=\frac{1}{2}$ to the Heaviside step function $\theta$, cf. e.g. this Phys.SE post.
III) It becomes more ill-defined to try to calculate $$\tag{4}\langle p^2 \rangle~=~\left(\frac{\hbar }{i}\right)^2\int_{\mathbb{R}} \! dx ~\psi^{\prime}(x)^2~=~\infty,$$ partly because a square of the Dirac delta distribution is ill-defined, cf. e.g. this Phys.SE post. At least the value $\infty$ in eq. (4) doesn't conflict with the Heisenberg uncertainty relations! |
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