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closed as off-topic by Arthur Fischer Sep 18 '14 at 10:37

  • This question does not appear to be about Mathematics Stack Exchange or the software that powers the Stack Exchange network within the scope defined in the help center.
If this question can be reworded to fit the rules in the help center, please edit the question.

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I have added a [sandbox] tag to allow people ignore it more easily (via software support of ignoring tags), and since it seems that we have two sandboxes now, a tag may seem a bit more in place here. –  Asaf Karagila Jul 18 '12 at 8:35
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(+1) For thinking outside the (sand)box. –  cardinal Jul 18 '12 at 19:40
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At the suggestion of the moderators, I have gone and changed the associated owners of all the answers here to the Community user. This way, the original owners will not receive excess pings for each time another user uses the draft space for their work. Enjoy! –  Grace Note Oct 5 '12 at 14:45
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To prevent crashes I've found the "Bookmarks to disable/enable MathJax", provided in here, pretty useful. –  leo Dec 17 '12 at 18:03
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This "sandbox" is being closed to prevent the creation of new answers. To start a draft, simply edit one of the existing free answers. –  Arthur Fischer Sep 18 '14 at 10:37

17 Answers 17

HINT, not yet a complete answer.

Hermite polynomials: $$H_n(x)=(-1)^n e^{x^2}\frac{d^n}{dx^n}e^{-x^2}$$

so integral becomes:

$$ I_{m,n,k}=(-1)^{m+n}\int_{-\infty}^\infty H_m(x) H_n(x) x^k e^{-x^2} \mathrm{d}x $$

Particular case $k=0$:

$$ I_{m,n,0}=(-1)^{m+n}\int_{-\infty}^\infty H_m(x) H_n(x) e^{-x^2} \mathrm{d}x = (-1)^{m+n}\sqrt\pi 2^nn!\delta_{mn}$$

EDIT: The general case

Remembering that:

$$x^k = \frac{k!}{2^k} \sum_{j=0}^{\lfloor \tfrac{k}{2} \rfloor} \frac{1}{j!(k-2j)! } ~H_{k-2j}(x)~$$

and replacing in the general integral, we find:

$$I_{m,n,k}=(-1)^{m+n}\int_{-\infty}^\infty H_m(x) H_n(x) e^{-x^2} \frac{k!}{2^k} \sum_{j=0}^{\lfloor \tfrac{k}{2} \rfloor} \frac{1}{j!(k-2j)! } ~H_{k-2j}(x)~ \mathrm{d}x =$$ $$=\sum_{j=0}^{\lfloor \tfrac{k}{2} \rfloor} \frac{1}{j!(k-2j)! } \frac{k!}{2^k} (-1)^{m+n}\int_{-\infty}^\infty H_m(x) H_n(x) H_{k-2j} e^{-x^2} (x)~ \mathrm{d}x =$$

Now we can make use of the following integral:

$$\int_{-\infty}^\infty H_m(x) H_n(x) H_l(x) e^{-x^2} dx = \frac{2^{\frac{m+n+l}{2}}l!m!n!\sqrt\pi}{\left(\frac{m+l-n}{2}\right)!\left(\frac{n+l-m}{2}\right)!\left(\frac{m+n-l}{2}\right)!}$$ when $\frac{m+n+l}{2}$ is integer and $m+n\ge l$ and $m+l \ge n$ and $l+n\ge m$ ; Zero otherwise.

Therefore:

$$I_{m,n,k}=\sum_{j=0}^{\lfloor \tfrac{k}{2} \rfloor} \frac{1}{j!(k-2j)! } \frac{k!}{2^k} (-1)^{m+n} \frac{2^{\frac{m+n+k-2j}{2}}(k-2j)!m!n!\sqrt\pi}{\left(\frac{m+k-2j-n}{2}\right)!\left(\frac{n+k-2j-m}{2}\right)!\left(\frac{m+n-k+2j}{2}\right)!}=$$

Simplifying ( we have employed the fact that if $m+n-k$ is an even integer also $\pm m \pm n \pm k$ and $\pm m \pm n \pm k \pm 2j$ are even integers) we can rewrite:

$$I_{m,n,k} = \begin{cases}2^{\frac{m+n-k}{2}}(-1)^{m+n}m!n!k! \sqrt\pi \sum_{j=0}^{\lfloor \tfrac{k}{2} \rfloor} \frac{2^{-j}[m+n+\ge k-2j][m+k-2j \ge n][k-2j+n\ge m] }{j!\left(\frac{m+k-2j-n}{2}\right)!\left(\frac{n+k-2j-m}{2}\right)!\left(\frac{m+n-k+2j}{2}\right)!} & \text{when $\frac{m+n+k}{2}$ is integer} \\ 0 & \text{ otherwise} \end{cases}$$

where the Iverson convention (see "Concrete Mathematics" Graham,Knuth,Patshnik, and http://en.wikipedia.org/wiki/Iverson_bracket ) has been employed:

$$[statement]=\begin{cases}1 & \text{when statement is true} \\ 0 & \text{otherwise}\end{cases}$$

SUM SIMPLIFICATION : WORK-IN-PROGRESS We want to study the integer solutions of system: $$\begin{cases}m+n\ge k-2j \\k-2j+m \ge n \\k-2j+n \ge m \\ j \ge 0 \\j \le \lfloor k/2 \rfloor \\ m+n+k=2p \end{cases}$$

where $p$ is an integer.

ADDENDUM (Alternative solution) At Wolfram:

http://mathworld.wolfram.com/HermitePolynomial.html

Equation (52) gives another solution of aboveseen integral. I think this alternative solution has a complexity comparable to that I have proposed.

http://mathworld.wolfram.com/images/equations/HermitePolynomial/NumberedEquation14.gif

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The problem says, Find the rate of change of $$(x,y,z) = x/z + y/z$$ with respect to t along the curve $$r(t) = \sin^2{t}[ i] + \cos^2{t}[j] + 1/(2t)[k]$$

The answer is apparently

$$(z/z^2)(2\sin{t}\cos{t}) - (z/z^2)(2\sin{t}\cos{t}) + (-x-y/z^2)(-2/4t^2)$$

i get everyting except where the $$(z/z^2)$$ comes from. should the partial derivative of x and y just be z?

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Question: How is $x,y,z$ done?

Motivation: I have been working on a problem in field $\alpha$ in regards to $\beta$ and I have come across a problem $\zeta$


What I have tried: Blahblahblah....[working,methods]$$$$$$$$$$$$


How do I fix this?

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Given: Points $A(-2\mid-2)$ and $B(1\mid2)$.
Goal: Find point $C(x\mid y)$ such that
(1) $\overline{AC}=\overline{BC}$,
(2) $C$ is north of the x-axis, and
(3) Triangle $ABC$ has an area of 10 units.
Solution: The perpendicular-bisector of $AB$ is $6x+8y+3=0$. Any point on this line, together with the base $AB$ can be the vertex of an isosceles triangle. The degenerate conic (two parallel lines) $(4x-3y+2)^2=400$ is the locus of all points that, together with the base $AB$, form triangles of 10 units area. This locus intersects the perpendicular-bisector at two points. One of these is south of the x-axis and can therefore be disregarded. The one we want is $C(-3.7\mid2.4)$.

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\newpage \section{Central extensions} \subsection*{General terminologies:} An injective group homomorphism is called a monomorphism, and a surjective group homomorphism is called an epimorphism. These are denoted by the mapping symbol $\hookrightarrow$ and $\twoheadrightarrow$ respectively. This definition holds in the categories $\text{FinGrp},\text{AbGrp},\text{Grp}$ \subsection*{Definition of an exact sequence:} A sequence of homomorphisms, is called exact if we have:

$$1\overset{\phi_1}{\to} N \overset{\phi_2}{\to} G \overset{\phi_3}{\to} H \overset{\phi_4}{\to} 1$$ where we have the following:

$$\text{Im}(\phi_1) =\ker(\phi_2)$$ $$\text{Im}(\phi_2) =\ker(\phi_3)$$ $$\text{Im}(\phi_3) =\ker(\phi_4)$$

\subsection*{Quick Example of a short exact sequence}

$$0\hookrightarrow \Bbb Z \overset{\times n}{\hookrightarrow} \mathbb Z \twoheadrightarrow \mathbb Z/n \twoheadrightarrow 0$$

$$\text{Im}(\phi_1) = 0 = \ker(\phi_2)$$ $$\text{Im}(\phi_2) = n\mathbb Z = \ker(\phi_3)$$ $$\text{Im}(\phi_3) = \mathbb{Z}/n = \ker(\phi_4)$$

\subsection*{Definition of an extension:} An extension of a group $G$ by a group $A$ is given by an exact sequence of group homomorphisms:

$$1\hookrightarrow A \hookrightarrow E \twoheadrightarrow G \twoheadrightarrow 1$$

This extension is called central if $A$ is abelian and the image of the monomorphism from $A\hookrightarrow E$ is in the center of $E$. \newpage \subsection*{Example of a central extension: Discrete Heisenberg group}

The Heisenburg group is also a lie group. It is not only a group but with topology inherited from $\Bbb R^3$ it becomes a smooth manifold with applications in Fourier analysis, quantum mechanics and other subjects. We will consider the discrete Heisenburg group, for simplicity. $$$$ Why do we care about central extensions? Finding group extensions is hard. Split extensions are the easiest to classify, these are semi-direct products, and general extensions are not much better than impossible to classify. This Heisenburg group is an example of one that doesn't split.

$$H_3(\mathbb{Z})$$

$$H_3(\mathbb{Z})= \begin{bmatrix}1&x&z\\0&1&y\\0&0&1 \end{bmatrix}|a,b,c\in \mathbb Z$$

[don't need to say]: We have the relations $z=xyx^{-1}y^{-1},xz=zx,yz=zy$

[don't need to say]: It has generators $$x=\begin{bmatrix}1&1&0\\0&1&0\\0&0&1\end{bmatrix},y=\begin{bmatrix}1&0&0\\0&1&1\\0&0&1\end{bmatrix},z=\begin{bmatrix}1&0&1\\0&1&0\\0&0&1\end{bmatrix}$$

The center of this is: $\begin{bmatrix}1&0&z\\0&1&0\\0&0&1\end{bmatrix}$

and we have a central extension from the short exact sequence: $$I\hookrightarrow N = Z(H_3(\Bbb Z)) \hookrightarrow H_3{\Bbb Z} \twoheadrightarrow \Bbb H_3(\Bbb Z)/N \twoheadrightarrow I$$

Since $N$ is the center, this is a central extension as long as we can show that the sequence is exact:

What we are working with in disguise: $$1\hookrightarrow \Bbb Z \hookrightarrow H_3{\Bbb Z} \twoheadrightarrow \Bbb Z^2 \twoheadrightarrow 1$$

$$\text{Im}(\phi_1)=I=\ker(\phi_2)$$ $$\text{Im}(\phi_2)=N=\ker(\phi_3)$$ $$\text{Im}(\phi_3)=H_3(\mathbb Z)/N=\ker(\phi_4)$$

The explicit maps are:

$$\phi_2: (0,0,z)\mapsto \begin{bmatrix}1&0&z\\0&1&0\\0&0&1\end{bmatrix},\phi_3: \begin{bmatrix}1&x&z\\0&1&y\\0&0&1\end{bmatrix}\mapsto (x,y,0)$$

[don't need to say:] $1=(0,0,0)$ [don't need to say:] $H/N$ is not a subgroup, it is a quotient group

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for t:=1 to 10 $$\theta_{\mathrm{new}}:=\theta-\alpha \frac{1}{1000}\sum \limits_{i=1}^{1000} \left(h(x^{(i)},\theta)-y^{(i)}\right)x^{(i)}$$

for t:=1 to 10 $$\begin{array}{lcl} \\ \theta_{\mathrm{new}}&:=&\theta\\ \theta_{\mathrm{new}}&:=&\theta_{\mathrm{new}}-\alpha \frac{1}{1000}\left(h(x^{(1)},\theta)-y^{(1)}\right)x^{(1)}\\ \theta_{\mathrm{new}}&:=&\theta_{\mathrm{new}}-\alpha \frac{1}{1000}\left(h(x^{(2)},\theta)-y^{(2)}\right)x^{(2)}\\ &\vdots& \\ \theta_{\mathrm{new}}&:=&\theta_{\mathrm{new}}-\alpha \frac{1}{1000}\left(h(x^{(1000)},\theta)-y^{(1000)}\right)x^{(1000)} \\ \theta&:=&\theta_{\mathrm{new}} \end{array}$$

for t:=1 to 10 $$\begin{array}{lcl} \\ \theta&:=&\theta-\alpha \left(h(x^{(1)},\theta)-y^{(1)}\right)x^{(1)}\\ \theta&:=&\theta-\alpha \left(h(x^{(2)},\theta)-y^{(2)}\right)x^{(2)}\\ &\vdots& \\ \theta&:=&\theta-\alpha \left(h(x^{(1000)},\theta)-y^{(1000)}\right)x^{(1000)} \\ \end{array}$$

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A presheaf on a category $J$ is a functor $J^{\text{op}}\to\mathbf{Set}$. The category of presheaves on $J$, denoted $\hat J$, with natural transformations as arrows is an example of functor category, and every category $J$ embeds into $\hat J=\mathbf{Set}^{J^{\text{op}}}$ via the Yoneda embedding $Y:x\mapsto J(-,x)$.

Note that your category $\cal A$ can be identified with $\hat{\Bbb N}$ since $\Bbb N=\Bbb N^{\text{op}}$ and every endomorphism determines a functor from the monoid $\Bbb N$ to $\mathbf{Set}$.

A nice property of a $\hat J$ for small $J$ is that it is Cartesian closed. Given any presheaves $B$ and $C$, if there exists an exponential $C^B$, then by Yoneda's lemma $$ C^B(x)=\text{Nat}(J(-,x),C^B-) = \hat J(Y(x),C^B)\cong \hat J(Y(x)×B,C) $$ and the arrow $C^B(x)\to C^B(y)$ is the function precomposing with $Yf×1_B$. So we only need to check that this presheaf is really the exponential. So we define $ε:C^B×B\Rightarrow C$ as $$ \begin{align} ε_x:\hat J(J(-,x)×B,C)×B(x) &\to C(x) \\ (\sigma,b)\qquad &\mapsto \sigma_x(1_x,b) \end{align} $$ It is easy to check that $ε$ is natural.

Note how similar this is to the construction in your case. Setting $J=\Bbb N, B=α,$ and $C=β$, and noting that $Y(\bullet)$ is the functor sending every $k\in \Bbb N$ to the addition "$...+k$", we get $β^α(\bullet)=\mathcal A(s×α,β)$. Precomposing with $s^k×1_α$ is the same as causing $h$ to be applied to $(n+k,a)$ instead of $(n,a)$. And the transformation $ε_x$ sends a map $h$ and an element $a\in α(\bullet)$ to $h(0,a)$.

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Let $M$ be a monoid. A homomorphism $\varphi:M\to N$ recognizes $L\subset M$, if $L=\varphi^{-1}(\varphi(L))$. A monoid $N$ recognizes $L \subset M$, if there exists a homomorphism $\varphi:M\to N$ that recognizes $L$. A subset of $M$ is recognizable, if it is recognized by a finite monoid. The family $\operatorname{REC}(M)$ of recognizable sets over $M$ is defined as the set of recognizable subsets of $M$.

A similar definition can be found in "J.-E. Pin, Mathematical Foundations of Automata Theory", and already "S. Eilenberg, Automata, Languages, and Machines" defines recognizable sets.

A deterministic $M$-automaton $\mathcal A=(Q,\cdot,q_0,F)$ consists of a set of states $Q$, a transition function $\cdot:Q\times M\to Q$, an initial state $q_0\in Q$, and a set of final states $F\subset Q$. For $q\in Q$ and $u,v\in M$, the transition function has to satisfy $$\begin{array}{rcl}(q\cdot u)\cdot v &=& q \cdot (uv)\\ q\cdot 1 &=& q\end{array}$$ The subset of $M$ accepted by $\mathcal A$ is $$L(\mathcal A) = \{u\in M|q_0\cdot u\in F\}$$

If $Q$ is a finite set, then $\mathcal A$ is said to have finitely many states.

I found this definition in chapter 7 Automatentheorie of the German text book "Volker Diekert, Manfred Kufleitner, Gerhard Rosenberger: Diskrete algebraische Methoden: Arithmetik, Kryptographie, Automaten und Gruppen”. There we also find the following theorem:

Theorem 7.10 Let $L\subset M$. The following statements are equivalent:

  1. $L$ is recognizable, i.e. $L\in \operatorname{REC}(M)$.
  2. $L$ gets accepted by a deterministic $M$-automaton with finitely many states.
  3. The "minimal automaton $\mathcal A_L$" has finitely many states.
  4. The "syntactic monoid $\operatorname{Synt}(L)$" is finite.

I have the impression that this definition of deterministic $M$-automata is unsuitable for characterizing deterministic finite state transducers as deterministic $\Sigma^*\times \Sigma^*$-automata.

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