# 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 with 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 arjafi♦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. – arjafi Sep 18 '14 at 10:37
PSA: Between the creation of this sandbox (in July 2012) and today (December 2015), technology has advanced. Something like StackEdit (or others, it's simply the only one I know) essentially solves all the limitations of this sandbox. You can have multiple concurrent drafts, you don't have to worry about polluting meta's front page, you can leave your draft untouched for days and expect it to still be there, you don't have to explicitly clear up your draft when you're done... Maybe someday we can get rid of this outdated crutch. – Najib Idrissi Dec 2 '15 at 14:07
@GraceNote Could you please also change to Community owned all of the posts in our Formatting Sandbox., including deleted posts too (I just got many pings when someone started using a deleted post). Thanks. – Bill Dubuque Dec 24 '15 at 19:50
@arjafi Please make that more clear in the sandbox itself, most users would think that would be a way of getting around such closing. – uoɥʇʎPʎzɐɹC Jun 21 at 23:40
There should be a rule to not vote on sandbox "answers" to keep the sandbox clean. Also, why are there only 17 answer slots? Why not an even number like 20 or 25? It would make the process more sleek. – uoɥʇʎPʎzɐɹC Jun 21 at 23:43

The triangle's vertices are:
$O(0\mid 0);\;A(15\mid 0);\;B(8.4\mid 11.2)$

The equations of $OB\quad$ and $AB\quad$ are, respectively,
$4x=3y\quad$ and $56x+33y=840\quad$

The intersections of the line $y=H\quad$ with
$OB\quad$ and $AB\quad$ are, respectively,
$M\left(\frac{3H}4\left\vert\right. H\right)\quad$ and $N\left(\frac{840-33H}{56}\mid H\right)$

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$$\begin{array}\\ P((x,y)=1) \\ = P(x \le n, y \le n | (x,y)=1) \\ +P(x \gt n, y \le n | (x,y)=1) +P(x \le n, y \gt n | (x,y)=1) +P(x \gt n, y \gt n | (x,y)=1) \end{array}$$

$$\begin{array}\\ &P(x \gt n, y \le n | (x,y)=1) +P(x \le n, y \gt n | (x,y)=1) +P(x \gt n, y \gt n | (x,y)=1) \\ \le &P(x \gt n, y \le n ) +P(x \le n, y \gt n ) +P(x \gt n, y \gt n ) \\ \le & 2 P(x \gt n) P( x \le n ) + P( x \gt n)^2 \\ = & P(x \gt n) (2 P( x \le n ) + P( x \gt n) ) \\ = &( 1 -P( x \le n ) ) ( 1 + P( x \le n )) \\ = & 1 -P( x \le n )^2 \\ = & 1 - (\sum_{i=1}^n2^{-i})^2 \\ = & 1 - (1 - 2^{-n})^2 \\ \lt & 2^{-n+1} \end{array}$$

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This answer is available to anyone.

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This answer 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 for anyone to use.

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Accelerated convergence series for $\zeta(4)$ using creative telescoping?

Can the $\zeta(4)$ accelerated convergence series proved e.g. in Corollaire 5.3 of Henri Cohen's Généralisation d'une Construction de R. Apéry $$\zeta(4):=\sum_{n=1}^{\infty }\frac{1}{n^{4}}=\frac{36}{17}\sum_{1}^{\infty } \frac{1}{n^{4}\binom{2n}{n}}\tag{1}$$ be obtained by similar techniques to those used in §1 of Alfred van der Poorten's Some wonderful formulae... Footnotes to Apery’s proof of the irrationality of $\zeta( 3)$ to derive the following accelerated convergence series for $\zeta(3),\zeta(2)$? \begin{equation*} \zeta (2):=\sum_{n=1}^{\infty }\frac{1}{n^{2}}=3\sum_{1}^{\infty } \frac{1}{n^{2}\binom{2n}{n}},\tag{2} \end{equation*} \begin{equation*} \zeta (3):=\sum_{n=1}^{\infty }\frac{1}{n^{3}}=\frac{5}{2}\sum_{1}^{\infty } \frac{(-1)^{n-1}}{n^{3}\binom{2n}{n}}\tag{3}. \end{equation*}

For instance $(3)$ can be obtained by letting $N\rightarrow \infty$ in \begin{equation*} \sum_{n=1}^{N}\frac{1}{n^{3}}-2\sum_{n=1}^{N}\frac{\left( -1\right) ^{n-1}}{n^{3}\binom{2n}{n}}=\sum_{k=1}^{N}\frac{(-1)^{k}}{2k^{3}\binom{N+k}{k}\binom{N}{k}}-\sum_{k=1}^{N}\frac{(-1)^{k}}{2k^{3}\binom{2k}{k}}\tag{4}. \end{equation*}

This last equality can be explained as follows:

2. Notice that $$X_{n,k}=D_{n,k-1}-D_{n,k}.\tag{5}$$ Hence \begin{eqnarray*} \sum_{k=1}^{n-1}\frac{X_{n,k}}{n} &=&\sum_{k=1}^{n-1}\left( \frac{D_{n,k-1}}{ n}-\frac{D_{n,k}}{n}\right) =\frac{D_{n,0}}{n}-\frac{D_{n,n-1}}{n} \\ &=&\frac{1}{n^{3}}-2\frac{\left( -1\right) ^{n-1}}{n^{3}\binom{2n}{n}} \\ \frac{D_{n,0}}{n} &=&\frac{1}{n^{3}},\qquad \frac{D_{n,n-1}}{n}=2\frac{ \left( -1\right) ^{n-1}}{n^{3}\binom{2n}{n}} \end{eqnarray*}
3. Sum over $k$, $1\leq k\leq n-1$ \begin{equation*} \sum_{k=1}^{n-1}X_{n,k}=\sum_{k=1}^{n-1}\left( D_{n,k-1}-D_{n,k}\right) =D_{n,0}-D_{n,n-1}. \end{equation*}
4. Now, sum over $n$, $1\leq n\leq N$, and noticing that \begin{equation*} \frac{X_{n,k}}{n}=E_{n,k}-E_{n-1,k},\qquad E_{n,k}=\frac{(-1)^{k}}{2k^{3}\binom{n+k}{k}\binom{n}{k}},\tag{6} \end{equation*} get \begin{equation*} \sum_{k=1}^{N-1}\sum_{n=k+1}^{N}\frac{X_{n,k}}{n}=\sum_{k=1}^{N-1} \sum_{n=k+1}^{N}\left( E_{n,k}-E_{n-1,k}\right) =\sum_{k=1}^{N}\left( E_{N,k}-E_{k,k}\right). \end{equation*}
5. Thus \begin{eqnarray*} \sum_{n=1}^{N}\sum_{k=1}^{n-1}\frac{X_{n,k}}{n} &=&\sum_{n=1}^{N}\frac{1}{ n^{3}}-2\sum_{n=1}^{N}\frac{\left( -1\right) ^{n-1}}{n^{3}\binom{2n}{n}},\tag{7} \end{eqnarray*} and \begin{eqnarray*} \sum_{n=1}^{N}\sum_{k=1}^{n-1}\frac{X_{n,k}}{n} &=&\sum_{k=1}^{N}E_{N,k}-\sum_{k=1}^{N}E_{k,k} \\ &=&\sum_{k=1}^{N}\frac{(-1)^{k}}{2k^{3}\binom{n+k}{k}\binom{n}{k}} -\sum_{k=1}^{N}\frac{(-1)^{k}}{2k^{3}\binom{2k}{k}}.\tag{8} \end{eqnarray*}
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This answer 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 available to anyone.

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This answer is available to anyone.

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Disconnected Cases

As noted in the Question, disconnected examples will amount to vertex transitive 6-regular graphs on $d$ nodes where $d\ge 7$ is a divisor of $42$. Thus $d=7,14,21$ are the possibilities, and small enough that all have been exhaustively cataloged by McKay and Royle (see Comment under Question).

The files there are (Unix) ASCII format/gzipped. However after extracting the files, one needs a utility showg to convert them to human-readable form.

Case $d=7$

The only $6$-regular graph on $7$ vertices is the complete one $K_7$. This may be as good a place as any to point out that this achieves $\binom{6}{2}=15$ edges among the neighbors of any fixed vertex $v$ (a clique), and thus the maximum count of such edges.

To do: insert picture

Case $d=14$

Among the $51$ vertex transitive connected graphs on $14$ vertices, there are eight which are $6$-regular. Of these, one has no edges between neighbors of $v$, two have $3$ edges there, four have $6$ such edges, and one has $9$.

Case $d=21$

Among the $235$ vertex transitive connected graphs on $21$ vertices, there are twenty-eight which are $6$-regular. Of these, some have a number of edges between neighbors of $v$ which is not a multiple of three.

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This answer is free for anyone to use.

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Suggestion of a post serving for duplicate posts, see here for details. If needed, we can also discuss improvements in chat.

This thread is intended to help with closure of duplicates. It should serve for situations where a first attempt at duplicate closure was not successful. The creation of this thread was decided here.

Some basic rules:

• Do not use this thread for other types of closures, it is only for duplicates.
• Only use this thread if the usual means did not work. In particular, you should have cast already a vote or flag, the question already has been through the close votes review queue, and it was still not marked as duplicate.1
• Keep in mind that very short answers are automatically turned into comments. So your post should probably contain at least some brief explanation in addition to links to the duplicates. (Maybe short discussion why they are duplicates, if it is not clear that they are indeed duplicates. Or a brief explanation which of the two posts is more suitable as the duplicate target.)
• If it turns out that discussion grows long and that it is contentious whether the particular question is indeed a duplicate, the question might deserve a separate post on meta.

A more low-key way to draw attention to a possible duplicate is this chat room, which has been created for similar purposes. For example, if just one vote is missing you might consider posting there instead.

1 You can find review for a particular post using this SEDE query. If the close votes are less than week old, it is possible that they are not in the last data dump. A more laborious alternative is to look in close votes review history. Or, just wait 24 hours after your vote; then it is very likely the post has passed review.

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This answer is free for anyone to use.

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This answer is free for anyone to use.

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## $$AP = \lambda BP$$

Notationally, $XY$ will represent the length of the segment $\overline{XY}$ and $XY^2$ means $(XY)^2$. $X-Y-Z$ means that point $Y$ is on the segment $\overline{XZ}$. We say that $Y$ is between $X$ and $Z$. This is true if and only if $XY + YZ = XZ$.

An expression like $A-B-C-D-E$ would then indicate how the points $B$, $C$, and $D$ are positioned on the segment $\overline{AE}$.

We need to show that $\Gamma_{AB}(\lambda) = \{ P \in \mathbb R^2: AP = \lambda BP\}$ is a circle.

$\Gamma_{AB}(1) = \left\{ \dfrac 12(A+B) \right\}$

Since $AP = \lambda BP \iff BP = \dfrac 1\lambda AP$, then we can assume that points $A$ and $B$ have been chosen so that $\lambda \gt 1$.

LEMMA $1$: For $\lambda > 1$, there are two points, $P_L$ and $P_R$, on the line $\overleftrightarrow{AB}$ such that $AP = \lambda BP$. They can be described by

• $A-P_L-B-P_R$.
• $AP_L = \dfrac{1}{\lambda+1}AB$.
• $AP_R = \dfrac{1}{\lambda - 1}AB$.

DEFINITION $2$: Let $\Gamma$ be the circle with diameter $\overline{P_LP_R}$.

We will show that $\Gamma_{AB}(\lambda) = \Gamma$.

We will assume that the following theorem is already known.

THEOREM $3$: $P \not \in \{X, Y\}$ is on the circle with diameter $\overline{XY}$ if and only if $\angle XPY$ is a right angle.

Let's say that a solution, $P$, to $AP = \lambda BP$ that isn't on the line $\overleftrightarrow{AB}$ is a non trivial solution.

then $\Gamma_{AB}(\lambda) = \Gamma$ will follow if we can prove the following theorem.

THEOREM $4$: $P$ is a non trivial solution to $PA = \lambda PB$ if and only if $\angle P_LPP_R$ is a right angle.

PROOF. First we show that $AP = \lambda BP \implies \angle P_LPP_R$ is a right angle.

We know that $AP = \lambda BP$ and $AP_L = \lambda BP_L$. Hence $\dfrac{AP}{AP_L} = \dfrac{BP}{BP_L}$. This implies that ray $\overrightarrow{PP_L}$ bisects $\angle APB$. So let $m\angle APP_L = m\angle BPP_L = \theta$.

Let $D$ be the point on $\overline{AP}$ such that $PA::PD = \lambda::1$. Then

\begin{align} \lambda = \dfrac{PA}{PD} = \dfrac{PA}{PB} &\implies PB = PD \\ &\implies \triangle PCB \sim \triangle PCD \\ &\implies m \angle DBP = \theta' = \dfrac{\pi}{2} - \theta \end{align}

Also \begin{align} \lambda = \dfrac{PA}{PD} = \dfrac{P_RA}{P_RB} &\implies \triangle ADB \sim \triangle APP_R &\text{By SAS for similar triangles.}\\ &\implies \overline{BD} \parallel \overline{P_RP} \\ &\implies m\angle BPP_R = m\angle DBP = \theta'\\ &\implies m\angle P_LPB + m\angle BPP_R = \dfrac{\pi}{2} \\ &\implies \angle P_LPP_R \text{ is a right triangle.} \end{align}

Finally, we show that $\angle P_LPP_R$ is a right angle $\implies AP = \lambda BP$.

Consider the abelian group $\mathcal G = \sum_\limits{i=1}^A \mathbb Z/p_i^{a_i}\mathbb Z$, where all of the $p_i$ are pairwise prime. Let $\overline{h} \in \mathcal G$ and let $\mathcal H = \langle \overline h \rangle$ be the cyclic subgroup of $\mathcal G$ generated by $\overline h$. Let $P = \prod_\limits{i-1}^A p_i^{a_i}$ and let $\widehat{\mathcal G} = \mathcal Z_P$. By the Chinese Remainder Theorem, there exists an isomorphism $f:\mathcal G \to \widehat{\mathcal G}$. Let $f(\mathcal H) = \widehat{\mathcal H}$.