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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.
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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
(+1) For thinking outside the (sand)box. –  cardinal Jul 18 '12 at 19:40
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
To prevent crashes I've found the "Bookmarks to disable/enable MathJax", provided in here, pretty useful. –  leo Dec 17 '12 at 18:03
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

This answer is free to be used.


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.


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

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 ) 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:

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


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?


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 $C(-3.7\mid2.4)$.


This answer is free to be used


This answer is free to be used


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?


This answer is free to be used now


This answer is free to be used.


This answer is free to be used.


The Stone-Čech compactification $\beta X$ of a space $X$ is defined as follows:
Let $C(X)$ be the set of all continuous maps from $X$ to the unit interval $I$. Then $\prod_{\varphi\in C(X)} I$ is a compact Hausdorff space. If $X$ is a Tychonoff space, then the map $i_X:x\mapsto(\varphi(x))_{\varphi\in C(X)}$ is an embedding, and the closure $\beta X=\overline{i_X(X)}$ is a compact Hausdorff space.
Now given any compact Hausdorff space $K$ and a map $f:X\to K$, there is a map $$f^{**}:\textstyle\prod_{\varphi\in C(X)}I\to\prod_{\psi\in C(K)}I\\ (y_\varphi)_\varphi\mapsto (y_{\psi f})_\psi$$ Since $f^{**}i_X(x)=f^{**}(\varphi(x))_\varphi=(\psi f(x))_\psi=i_Kf(x)$, this means $f^{**}$ restricts to a map $i(X)\to i(K)$, and thus to a map $\overline{i(X)}\to i(K)$ since $i(K)$ is closed. Since maps to Hausdorff spaces are completely determined by their values on a dense subspace, this restriction $\beta X\to i(K)$ is the only map $g$ satisfying $gi_X=i_Kf$, hence $f'=i^{-1}f^{**}:\beta X\to K$ is the unique map such that $f'i=f$.

Now assume $(Z,i)$ is a compactification such that any map $f:X\to K$ to a compact Hausdorff space $K$ has an extension $Z\to K$. Since $i_X:X\to\beta X$ is such a map, there is $j:Z\to \beta X$ such that $ji=i_X$. On the other hand, there is a unique $i':\beta X\to Z$ with $i'i_X=i$. Then $ji'$ is the unique map on $\beta X$ whose composition with $i_X$ equals $i_X$, so $ji'=\mathbf 1_{\beta X}$. Since $i'ji=i$, it follows that $i'j=\mathbf1_Z$.


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