Sandbox for drafts of long, complex posts [closed]

This sandbox is intended for saving drafts of long, complex posts, especially posts whose composition takes a long time. It serves to localize to one thread the front-page "bumps" caused by edits to drafts of such posts, so that they may be easily ignored. Also, it helps to guard against losing longly-composed posts due to system crashes.

When you are happy with your draft here, you may simply copy the code and paste it to the desired location.

Proper Use of the Sandbox

1. Do not post a new answer! We wish all the answers on this page to be owned by the Community user (so that only a non-sentient bot is informed of edits to these answers). Posting a new answer will make you the owner, meaning that you will be notified whenever another user makes an edit to that answer.

The sandbox has been closed to prevent the creation of new answers. There are more than enough existing answers for users to edit over, and this will greatly reduce the frequency at which we request that the answers be disassociated from specific users.

2. Do not delete answers! Deleting seems like a reasonable option, but there are no "hard deletions" on Stack Exchange, and users will sufficient privileges will still see your supposedly deleted postings. Deleted answers will be undeleted and cleared for the use of others.

3. Do look for an answer which indicates that it is free and then edit it to your heart's content. If none appears available, take over the one that has been left unchanged the longest (which will appear at the bottom of the page if you order answers by "activity").

4. Do not expect your draft to remain untouched for days. There are no guarantees that your draft will be the latest revision if you return days later. While users will try not to step over others' toes, it may happen that an unfinished draft is edited out. Your draft will, however, still exist as a revision of the answer it was made in. If your drafting is expected to take place over a longer period of time, either

• take note of the URL of the answer provided by clicking the share button, or
• save a copy of your draft locally (or even "in the cloud").
5. Do clear your draft when you are finished. This includes removing all $\LaTeX$ from your answers. Replacing all code with a simple statement like

This answer is free for anyone to use

is sufficient. Periodically users may go through and free up answer slots that have not been edited in, say, over one month. But you can aid in the smooth running of this sandbox by clearing away your drafts when you are finished with them.

6. Do not "claim" multiple answers concurrently. Since this post is closed, the answers are a limited resource. If you really must compose several long, complex posts at the same time, you can still use a single answer, separating the different drafts using Markup: horizontal rules (---) and/or headings (# Header 1 #) are natural choices.

7. Do not create new such sandboxes. The point of having a unique such sandbox is that it minimizes the noise on the front page when the sandbox is edited. If there were multiple sandboxes they will frequently occupy numerous front page slots, pushing other topics off the front page, and increasing noise.

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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.

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

This answer is free to be used.

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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.

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 $$r(t) = \sin^2{t}[ i] + \cos^2{t}[j] + 1/(2t)[k]$$

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

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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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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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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This answer is free to be used now

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This space is free for anyone to use.

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This answer is free for anyone to use!

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This answer is free to be used.

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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 $i'j=\mathbf1_Z$.

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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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