# 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

TRISECTION OF A CLOCKFACE BY ITS CLOCKHANDS by Donn S. Miller (dsm5442@gmail.com) At the outset, I shall disappoint everyone by baldly stating that the hands of a theoretical clock will never trisect their clockface. Having said that, however, I shall develop an exact solution for the closest approach possible to a trisection. (Actually, there are two symmetrical solutions, with the vertical diameter of the clockface as their axis of symmetry.) For the purpose of this discussion, I shall always describe the positions of the three clockhands exactly using rational numbers in the domain $[0, 1)$, where the upper vertical radius is $0$ and the position of a hand representing the fraction of a complete revolution of the hand clockwise (how appropriate!) from $0$. For example, the position of the hourhand, minutehand and secondhand — we shall always discuss the clockhands in this same order — at 8:00:00 would be, respectively, $\frac23$, $0$ and $0$. As another example, at 8:00:07, the corresponding positions would be presented as, respectively, $\frac{28807}{43200}$, $\frac7{3600}$ and $\frac7{60}$, because all the hands on our theoretical clock are moving constantly and smoothly, not with the herky-jerky motion of a mechanical clock. The angular velocities of the three hands are proportional, respectively, to 1, 12 and 720; hence, in the 7 elapsed seconds between 8:00:00 and 8:00:07, all the hands have traversed angles in those proportions. Such exactitude is necessary for an exact answer to this interesting problem. It should be noted that if the exact position (as described above) of the hourhand is known, the exact positions of the minutehand and the secondhand are knowable; also, if the exact position of the minutehand is known, the exact position of the secondhand is knowable. The angle between two clockhands is calculated, of course, by subtracting the position fraction of one from the position fraction of the other. If the result of any calculation falls outside the domain [0, 1), it can easily be adjusted by adding or subtracting an appropriate integer. The method I shall use will now be explained. First of all, note that we are striving for an ideal clockhand configuration in which all interhand angles are $\frac13$. Suppose we designate the clockhands as $A$, $B$ and $C$. We shall be intercepted only in the valid configurations in which the angle $A-B$ is $\frac13$; there are more than one. There are two conceptions of $C$ that we shall deal with: An ideal $C$ $[C_i]$ which, taken in conjunction with $A$ and $B$, would comprise an exact trisection (but which, unfortunately, never occurs); and a real $C$ $[C_r]$ which is actually attainable but which does not result in an exact trisection. There are numerous hand configurations in which the $\left[A-B=\frac13\right]$ criterion is met; the one we seek is the one in which the difference between $C_r$ and $C_i$ is a minimum. $$h-m=\frac13$$ $h=\frac8{33}+x$
$m=\frac{10}{11}+12·x$
$s_r=\frac6{11}+720·x$
$h-m=\frac13-11·x$
$m-s=\frac4{11}-708·x$
$s-h=\frac{10}{33}+719·x$
$\left(\frac13-11·x-\frac13\right)^2 +\left(\frac4{11}-708·x-\frac13\right)^2 +\left(\frac{10}{33}+719·x-\frac13\right)^2$ $(-11·x)^2+\left(\frac1{33}-708·x\right)^2+\left(-\frac1{33}+719·x\right)^2$
$(363·x)^2+(23364·x-1)^2+(23727·x-1)^2$
$1108978794·x^2-94182·x+2=y$
$2217957588·x-94182=\frac{dy}{dx}=0$
$x=\frac{94182}{2217957588}=\frac{1427}{33605418}$
$h=\frac8{33}+\frac{1427}{33605418}=\frac{8148195}{33605418}=\frac{246915}{1018346}$
2. secondhand trails minutehand by 1/3
3. hourhand trails secondhand by 1/3

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1169384
$\begin{matrix} a=&x\land y\\ b=&\lnot a\\ c=&x\lor b\\ d=&c\lor z\\ e=&\lnot x\\ f=&\lnot y\\ g=&e\land f\\ h=&g\land z\\ i=&a\lor h\\ j=&\lnot z\\ k=&y\land j\\ l=&i\lor k\\ m=&d\land l\\ \end{matrix}$

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This answer is free for anyone to use!

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rollback concurrent claim of multiple answers. –  achille hui yesterday


Definition 5.1.1: The function $\norm{\cdot}\colon V \to \R$ is a norm if for all $x,y\in V$ and all $c\in\F$,

(1) $\norm{x}\geq 0$

(1a) $\norm{x}=0\iff x=0$

(2) $\norm{cx}=|c|\norm{x}$

(3)$\norm{x+y}\leq \norm{x}+\norm{y}$

If $\norm{\cdot}$ only satisfies (1),(2),(3), then $\norm{\cdot}$ is a seminorm.

Lemma 5.1.2: If $\norm{\cdot}$ is a seminorm, then $\big|\norm{x}-\norm{y}\big|\leq\norm{x-y}$ for all $x,y\in V$.

Definition 5.1.3: $\ps{\cdot}{\cdot}\colon V \times V \to \F$ is an inner product if for all $x,y,z\in V$ and all $c \in \F$

(1) $\ps{x}{x}\geq 0$

(1a) $\ps{x}{x}=0\iff x=0$

(2) $\ps{x+y}{z} =\ps{x}{z}+\ps{y}{z}$

(3) $\ps{cx}{y}=c\ps{x}{y}$

(4) $\ps{x}{y}=\overline{\ps{y}{x}}$

Axioms (2),(3),(4) imply that $\ps{\cdot}{\cdot}$ is a sesquilinear function. If $\ps{\cdot}{\cdot}$ satisfies (1),(2),(3),(4) then it is a semi-inner product.

Theorem 5.1.4 & 5,1,8: (Cauchy-Schwartz) Let $\ps{\cdot}{\cdot}$ be a semi inner product on $V$, then $$|\ps{x}{y}|^2\leq\ps{x}{x}\ps{y}{y} \qquad \forall x,y\in V.$$ Moreover, $\norm{x}:=\ps{x}{x}^{1/2}$ is a semi-norm on $V$. If $\ps{\cdot}{\cdot}$ is an inner product, then we have equality above if and only if $x=\alpha y$ for some $\alpha \in \F$ and $\norm{x}$ is a norm.

Remark: The characterization for equality in Cauchy-Schwartz fails in general for semi-inner product.

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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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$$(E^o)^c=\overline{(E^c)}$$

Proving left is a subset of right.

$$x\in(E^o)^c\implies x\in((E^c)' \cup E^c)$$

Proof:

$$x\in(E^o)^c\implies x\in((E^c)' \cup E^c)$$

Proving right is a subset of left.

$$x\in((E^c)' \cup E^c)\implies x\in(E^o)^c$$

Proof:

$$E^o \subset E \implies E^c\subset (E^o)^c$$

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This answer is free for anyone to use

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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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Summary of chapter 1 (Symmetries) of Galois Theory (J. Rotman, second edition)

Definition: For $x,y \in \Bbb R^2$, let us denote by $\|x-y\|$ the distance between $x$ and $y$. A linear transformation $\sigma : \Bbb R^2 \to \Bbb R^2$ is called orthogonal if $\|\sigma(x)-\sigma(y)\|=\|x-y\|$ for every $x,y \in \Bbb R^2$. The set of all orthogonal transformation of $\Bbb R^2$ is denoted by $O(2,\Bbb R)$.

Remark: Every $\sigma \in O(2,\Bbb R)$ is bijective and $\sigma^{-1}\in O(2,\Bbb R)$. $O(2,\Bbb R)$ is a group under composition called the real orthogonal group.

Lemma 1: Let $\sigma \in O(2,\Bbb R),x,y,z \in \Bbb R^2$ and $x' = \sigma(x),y'=\sigma(y),z'=\sigma(z)$. Then $\angle xyz = \angle x'y'z'$, where $\angle abc$ is the angle in $b$ formed by $a,b,c \in \Bbb R^2$.

Definition: Let $F$ a figure in the plane, its symmetry group is defined by $\Sigma(F):=\{\sigma \in O(2,\Bbb R)\mid \sigma(F)=F\}$.

Lemma 2: If $P$ is a polygon and $\operatorname{Vert}(P)$ denotes its vertices, then $\sigma\big(\operatorname{Vert}(P)\big)=\operatorname{Vert}(P)$ for all $\sigma \in \Sigma(P)$.

Theorem 3: If $P$ is a polygon with $n$ vertices, then $\Sigma(P)$ is isomorphic to a subgroup of the symmetric group $S_n$.

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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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This answer is free for anyone to use.

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$$\sum_{k=1}^K y^{(i)}_k \log ((h_\Theta (x^{(i)}))_k) + (1 - y^{(i)}_k)\log (1 - (h_\Theta(x^{(i)}))_k)$$

$$\sum_{\{k|y_k=1\}} \log ((h_\Theta (x))_k) + \sum_{\{k|y_k=0\}} \log (1-(h_\Theta (x))_k)$$

$$h_\theta(x)=\left(h_{\theta,1}(x),\ldots,h_{\theta,K}(x)\right)$$

$$h_\theta(x)=\left((h_{\theta}(x))_1,\ldots,(h_{\theta}(x))_K\right)$$

$$\sum_{k=1}^K y^{(i)}_k \log (h_{\Theta,k} (x^{(i)})) + (1 - y^{(i)}_k)\log (1 - h_{\Theta,k}(x^{(i)}))$$

$$\sum_{\{k|y_k=1\}} \log (h_{\Theta,k} (x)) + \sum_{\{k|y_k=0\}} \log (1-h_{\Theta,k} (x))$$

$$\sum_{\{k|y_k=1\}} \mathrm{penalty} (h_{\Theta,k} (x),1) + \sum_{\{k|y_k=0\}} \mathrm{penalty} (h_{\Theta,k} (x),0)$$

the index $^{(i)}$

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