# 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 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 at 10:37

This answer is free for anyone to use

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This answer is free for anyone to use

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@Committingtoaname , sorry, forgot this was here. Is not going so well, can't find a way of clearly proving an important argument. –  Edvin Orlov Oct 11 at 3:46

This answer is free for anyone to use

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Definition A: Given $b\in\Bbb R$ we define a map $m\mapsto b^m$ from $\Bbb Z^{\ge 0}$ to $\Bbb R$ recursively by $b^0:=1,b^{m+1}:=b^m\cdot b.$

Observe that this map is increasing for $b>1,$ decreasing for $0<b<1,$ and constant for $b=1.$

Claim 1: For $b>1,$ the map from Definition A is unbounded above.

Proof: Given $m\in\Bbb Z^{>0},$ let $a_m:=b^m-b^{m-1},$ so that each $a_m>0.$ It is readily shown by induction on $m$ that for all $m\in\Bbb Z^{>0},$ we have $a_{m+1}=a_m\cdot b$ and $$b^m=1+\sum_{k=1}^m a_k,$$ so in particular, $$b^m\ge 1+\sum_{k=1}^m a_1=1+ma_1>ma_1.$$ Take any $y\in\Bbb R.$ By the Archimedean Property of Reals, since $a_1>0,$ then there is some $m\in\Bbb Z^{>0}$ such that $ma_1>y,$ and so $b^m>y.$ Since this holds for all $y\in\Bbb R,$ then the map is unbounded above, as desired. $\Box$

Claim 2: Given $m\in\Bbb Z^{\ge0},$ the map $\Bbb R^{>0}\to\Bbb R^{>0}$ given by $b\mapsto b^m$ is non-decreasing, and is increasing for $m\ne 0$.

Proof: It is readily seen that the map $b\mapsto b^0=1$ is nondecreasing, and that the map $b\mapsto b^1=b$ is increasing. Suppose for some $m>0$ that the map $b\mapsto b^m$ is increasing, and take $b,c\in\Bbb R$ with $0<b<c.$ We know that $b^m<c^m$ by inductive hypothesis. Since $0<b$ and $b^m<c^m,$ then $b^{m+1}:=b^m\cdot b<c^m\cdot b.$ Since $b<c$ and $0<c^m,$ then $c^m\cdot b<c^m\cdot c=:c^{m+1},$ and so $b^{m+1}<c^{m+1},$ as desired. $\Box$

Definition B: Given $c\in\Bbb R^{\ge1}$ and $n\in\Bbb Z^{>0},$ we define $c^{\frac1n}:=\sup\{b\in\Bbb R:b^n\le c\}.$ Note in particular that $1^n=1\le c,$ so $\{b\in\Bbb R:b^n\le c\}$ is non-empty, and by Claim 1 is bounded above, so that $c^{\frac1n}$ is well-defined for $n\in\Bbb Z^{\ge 2}$. Furthermore, observing that $b^1=b$ for any $b\in\Bbb R,$ we have $$c^{\frac11}:=\sup\{b\in\Bbb R:b^1\le c\}=\sup\{b\in\Bbb R:b\le c\}=c=c^1,$$ so that Definitions A and B agree, and $c^{\frac1n}$ is well-defined for all $n\in\Bbb Z^{>0}.$

Claim 3: Given $c\in\Bbb R^{\ge 1}$ and $n\in\Bbb Z^{>0},$ $x=c^{\frac1n}$ is a positive real solution to $x^n=c.$ Moreover, it is the unique such solution by Claim 2.

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@cameronbuie How did this go? –  Committing to a challenge Oct 22 at 0:43
@Committingtoachallenge: I've not yet finished. Would you like me to tag you once I've done so? –  Cameron Buie Oct 22 at 2:14
@CameronBuie Yes :), what is it for? –  Committing to a challenge Oct 22 at 2:16

This answer is free to be used

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This answer is free for anyone to use.

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This is a free answer for the use of anyone

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This answer is free for anyone to use

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@committingtoaname: no, and I'd actually forgotten this was here. For one, I realized that there were some errors in what is here; moreover, achielle hui's comments in that question both shows the location I was heading and why it's hard to proceed further. So I abandoned this direct route. –  Semiclassical Oct 8 at 2:04

This answer is free for anyone to use

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This answer is free for anyone to use

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System

$$\dot{x}_1 = 3x_1 + x_2$$ $$\dot{x}_2 = 4x_1 + 3x_2 + u$$

where $|u|\leq 1$

First we will convert this to maxtrix form and compare eigenvalues:

$\def\b{\begin{pmatrix}}\def\e{\end{pmatrix}}\def\d{\dot{x}}\b\d_1 \\ \d_2\e=\b3&1\\4&3\e\b x_1\\x_2 \e+ \b 0 \\u\e$

So we can clearly see that we have eigenvalues calculated from $(3-\lambda)^2 -4=0$

$\lambda^2 -6\lambda +5=0=(\lambda-5)(\lambda-1), \lambda=1,5$

Hence we have a repulsive node.

Now we want to find the eigenvectors:

For $\def\l{\lambda}\l=5. \b-2 &1\\4&-2\e=\b 0\\0\e\implies \underset{\sim}{v}_1=\b1\\2\e$

For $\l=1. \b2 & 1\\4&2\e =\b 0 \\ 0\e\implies \underset{\sim}{v}_2=\b1\\-2\e$

Thus we get the general solution: $A\b1\\2\e e^{5t} + B \b 1 \\-2 \e e^t$

Hamiltonian:$$H= -1 + \psi_1(3x_1+x_2) + \psi_2(4x_1+3x_2 +u)$$ $$= -1 + \psi_1(3x_1+x_2) + \psi_2(4x_1+3x_2) +\psi_2u$$

$$\underset{\sim}{\dot{\psi}}=-A^T\underset{\sim}{\psi}=\b3&4\\1&3\e\underset{\sim}{\psi}$$

$$\begin{vmatrix} 3 -\lambda & 4 \\ 1 & 3-\lambda\end{vmatrix}=(3-\lambda)^2 -4 = 0$$$$\lambda^2 - 6\lambda + 5=(\lambda-1)(\lambda-5)=0, \lambda = 1,5$$

$\lambda = 1, \begin{bmatrix}2 & 4|\,0 \\ 1 & 2|\,0 \end{bmatrix}$ Eigen-vector for $\lambda=1$ is $\b -2\\1 \e$

$\lambda = 5, \begin{bmatrix}-2 & 4|\,0 \\ 1 & -2|\,0 \end{bmatrix}$ Eigen-vector for $\lambda=1$ is $\b 2\\1 \e$

$\psi$

$H$ is linear in $u$, $|u|\leq 1$, hence we have $H$ max at $u^*=sgn(\psi_2) = \pm 1$

We now need equilibrium points for $u^*=1$ $$3x_1+x_2=0$$ $$4x_1+3x_2+1=0$$ $$x_1 = \frac15,x_2 = -\frac35$$

$$\underset{\sim}{x}=A\b1\\2\e e^{5t} + B \b 1 \\-2 \e e^t + \b \frac15 \\-\frac35 \e$$

And for $u^*=-1$

$$3x_1+x_2=0$$ $$4x_1+3x_2-1=0$$ $$x_1 = -\frac15,x_2 = \frac35$$ $$\underset{\sim}{x}=A\b1\\2\e e^{5t} + B \b 1 \\-2 \e e^t + \b -\frac15 \\\frac35 \e$$

Slope of these curves at $(0,0)$ is $\frac{\frac{dx_2}{dt}}{\frac{dx_1}{dt}}=\frac{dx_2}{dx_1}=\frac{5A+B}{10A-2B}$

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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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I have trouble with a proof of the following fact

Let $(F,U,\eta,\varepsilon):M\to N$ be a Quillen adjunction, i.e. an adjunction between model categories such that $F$ preserves cofibrations and acyclic cofibrations or, equivalently, $U$ preserves fibrations and acyclic fibrations. Then the total left derived functor $\Bbb LF:HoM\to HoN$ and the total right derived functor $\Bbb RU:HoN→HoM$ exist and form an adjoint pair.

The proof goes as follows

Let $\gamma:M→HoM$ and $\delta:N→HoN$ be the localizations. Since $F$ maps acyclic cofibrations to weak equivalences, $δF$ sends acyclic cofibrations between cofibrant objects to iso's, which implies that it has a left derived functor $\Bbb LF$. Similarly, since $U$ preserves acyclic fibrations, the right derived functor $\Bbb RU$ exists.

So the functors between the homotopy categories exist. We want to obtain an isomorphism $HoN(LFX,Y)\to HoM(X,RUY)$

Assume $X$ is cofibrant and $Y$ is fibrant. From $U$ preserving acyclic fibrations, it follows easily that if $f\simeq g:FX\to Y$, then $Uf\simeq_l Ug$. Since $U$ preserves fibrations and terminal objects, $UY$ is fibrant, hence $Uf\circ\eta\simeq Ug\circ\eta:X\to UY$. This gives a bijection $$\pi N(FX,Y)\cong\pi M(X,UY)$$

We generalize to arbitrary $X$ and $Y$:

On objects we have $\Bbb LF=FQ$ and $\Bbb RU=UR$. We have bijections: $$HoN(\Bbb LFX,Y)\cong HoN(FQX,RY)\cong HoM(QX,URY)\cong HoM(X,\Bbb RUY)$$ The first bijection is induced by $r:Y\to RY$, the last one by $q:QX\to X$. For the middle bijection, note that there is a bijection $$\pi(X,Y)\to HoM(X,Y)\\ [f]\mapsto[RQf]$$ if $X$ is cofibrant and $Y$ is fibrant.

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This answer is free for anyone to use

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