# Sandbox for long or complicated edits. [duplicate]

This sandbox is intended for a collective collaboration to edit posts which would require a lot of effort and a lot of time, in particular, $$\LaTeX$$ifying images. It serves to localize to one thread the front-page "bumps" caused by edits, so that they may be easily ignored and mainly to facilitate that procedure. (Since if everyone attributed only a small amount of time, we would then together finish them without loosing too much time) Also, it helps to guard against losing longly-composed edits due to system crashes.

At the top of each post, type a link to the post you wanna edit followed with what should be edited in the post itself (directly copied from the code of the post), followed by a small space. $$\text{(\star)}$$

Proper Use of the Sandbox

Look for a post which indicates that it is free and then edit it to your heart's content. If you want to edit a post that is not in the list of answers, create a new one with the requirements mentioned in $$\text{(\star)}$$.

• If you didn't understand the purpose of this sandbox well, see the answers below. Thanks. – user93957 Dec 14 '13 at 15:14
• I wiki-fied this entire question, and so removed your first two rules (since they are obsolete by virtue of all posts being CW). – Willie Wong Dec 16 '13 at 9:39
• Could the four downvoters explain their act? – user93957 Dec 20 '13 at 11:47
• @Adobe Are you aware that such a sandbox already exists? The point of having a unique sandbox is that it minimizes the noise on the front page. Multiple sandboxes increase the noise, since they both can be simultaneously on the front page. As such, I recommend deleting this one to avoid this problem. – Gone Jan 6 '14 at 17:06
• @BillDubuque I don't that the sandbox you linked to is similar to this one, since one is about answers and the other about edits, one is personal (only the one who wants to post an answer use it) the other is non-personal. – user93957 Jan 6 '14 at 18:37
• @Aðøbe When I created the other sandbox the intent was that it would be used for all such purposes (except for TeX testing, for which there is already a sandbox, which is better not reused for this because of the TeX torture tests there) I don't understand why you think that the original sandbox does not suffice. Could you please elaborate. – Gone Jan 6 '14 at 18:41
• @BillDubuque I didn't know the intent of the first sandbox since I though it was only for answers. If it is the case, then it is more convenient to delete this sandbox. However, is it permissible to post like this answer: meta.math.stackexchange.com/a/12055/93957 in that sandbox? – user93957 Jan 6 '14 at 18:47
• @Aðøbe Having been away I was not sure if there might have been other motivations for a new sandbox. We can amend the other sandbox description to specifically state any protocols for specialized uses, e.g. maybe adding a banner to an answer stating any special purpose of the answer, e.g. inviting collaboration, or whatever other specific purpose it is intended to be used for. Does that sound ok to you? I mentioned similar things when I created the sandbox, but maybe it was in a comment that never made it into the thread. – Gone Jan 6 '14 at 18:54
• @BillDubuque "Does that sound ok to you?" - Yes, perfect. – user93957 Jan 6 '14 at 18:56
• @Aðøbe Ok, great. Please be sure that no one is actively using this box before closing it. – Gone Jan 6 '14 at 19:03
• @BillDubuque Sure. ${}{}$ – user93957 Jan 6 '14 at 19:05
• @Adobe Hmm, closing it doesn't stop people from using it. It seems we need mod intervention to lock it. Since you are the author, it probably makes more sense for you to flag with such a request. – Gone Jan 10 '14 at 19:03
• @BillDubuque Done. ${}{}{}{}$ – user93957 Jan 10 '14 at 19:07
• @Aðøbe Do you think we should add some remarks to the original sandbox to make it clearer that its intent is universal? (except for LaTeX torture tests) – Gone Jan 10 '14 at 19:13
• @BillDubuque Of course. – user93957 Jan 10 '14 at 19:15

• Let's try to compute it. First, we can take advantage of the conditional independence of the $z_i$'s given the cluster centers, $$p(x_{1:n})=\int_{\mu_1: \text{Which symbol here?}}$$

Could someone please check this? I don't understand what that thing that looks like $+y$ is supposed to be about, and I'm a little confused about some of the variables.

I tried to find the general integrals of the given P.D.E in the yellow box.

And I found $c_1$. But I cannot find another one, say $c_2$.

Please help me finding $c_2$. Thank you.

$$\color{#F3D159}{\boxed{\displaystyle\,\,\color{black}{px(z-2y^2)=(z-9y)(z-y^2-2x^3)}\,\,}}$$ \begin{gather*} P=x(x-2y^2)\\ Q=+y(z-y^2-2x^3)\\ R=z(z-y^2-2x^3) \end{gather*} $$\frac{\mathrm dx}{x(z-2y^2)}=\frac{\mathrm dy}{+y(z-y^2-2x^3)}=\frac{\mathrm dz}{z(z-y^2-2x^3)}$$ $$\frac{\mathrm dy} y = \frac {\mathrm dz} z \implies \ln y = \ln z + \ln c_1$$ $$\frac y z = c_1$$

This space is free. Anyone can use it.

• @GEdgar what are you doing?? – user93957 Dec 18 '13 at 20:37
• I found one that says FREE and used it. I guess it wasn't FREE after all? – GEdgar Dec 18 '13 at 21:20
• FREE in this context doesn't mean anyone could use it rather it means anyone could edit it. – user93957 Dec 18 '13 at 21:25
• @GEdgar, if it were free free, all content would have been deleted. The terminology is non-optimal. "Free" currently means something akin to "not checked out" in a revision control system. – dfeuer Dec 19 '13 at 6:25
• @GEdgar By the way here is a free to use post: meta.math.stackexchange.com/a/12054/93957 – user93957 Dec 19 '13 at 12:30
• @dfeuer: What? Confusing terminology defined by a mathematician? That's a first! – GEdgar Dec 19 '13 at 15:26
• @KarlKronenfeld Now that you edited that post here, you can directly edit it in the original post (math.stackexchange.com/questions/509831/…). – user93957 Dec 22 '13 at 12:03
• @Adobe Do you recommend replacing the image? – user714630 Dec 22 '13 at 12:11
• @KarlKronenfeld Yes in the original post. (math.stackexchange.com/questions/509831/…) – user93957 Dec 22 '13 at 12:12

This space is free for rent...

This space is free for use. $\text{}\text{}$

This space is free for use. ${}{}{}{{}{}{}}{}{}$

• Many of these seem to be quite old and should not be edited just for the sake of removing images. For example math.stackexchange.com/questions/548511/… is from November – Carl Mummert Dec 14 '13 at 22:37
• @CarlMummert, it's from November of this year. That's pretty recent in my book. Personally, I think images of text are bad enough to justify editing them away in posts of any age. – dfeuer Dec 15 '13 at 0:33
• @dfeuer: is there really any significant benefit, particularly when the question is already answered? – Carl Mummert Dec 15 '13 at 0:44
• @CarlMummert, in my opinion, the ability of people who have disabilities preventing them from accessing images to use this site is very important. That includes being able not only to ask questions and get answers but also to read others' questions, vote, perform moderation functions, etc. Obviously, the SE platform has some unfortunate inherent limitations (e.g., not supporting xyjax or any function graphing system), but that does not free us from the responsibility to do what we can within those constraints. – dfeuer Dec 15 '13 at 0:57
• @CarlMummert See this topic: Should I edit a question/answer every-time I see an image in it? meta.math.stackexchange.com/questions/11696/… – user93957 Dec 15 '13 at 10:56

Clairuts' Equation - FREE

$\displaystyle\left(x\dfrac{\mathrm dy}{\mathrm dx}-y\right)^2=a\left(1+\left(\dfrac{\mathrm dy}{\mathrm dx}\right)^2\right)(x^2+y^2)^{3/2}$

$\displaystyle \begin{cases}x=\rho \cos(\theta) \\ y=\rho \sin(\theta)\end{cases}\,\to\,\begin{cases}\mathrm dx=\mathrm d\rho\cos(\theta)-\rho\sin(\theta)\mathrm d\theta \\ \mathrm dy=\mathrm d\rho\sin(\theta)+\rho\cos(\theta)\mathrm d\theta\end{cases}$

$\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{\dfrac{\mathrm d\rho}{\mathrm d\theta}\sin(\theta)+\rho\cos(\theta)}{\dfrac{\mathrm d\rho}{\mathrm d\theta}\cos(\theta)-\rho\sin(\theta)}$

$\left( \rho \cos(\theta) \dfrac{\dfrac{\mathrm d\rho}{\mathrm d\theta}\sin(\theta)+\rho\cos(\theta)}{\dfrac{\mathrm d\rho}{\mathrm d\theta}\cos(\theta)-\rho\sin(\theta)}-\rho\sin(\theta) \right)^2=a\left[1+\left(\dfrac{\dfrac{\mathrm d\rho}{\mathrm d\theta}\sin(\theta)+\rho\cos(\theta)}{\dfrac{\mathrm d\rho}{\mathrm d\theta}\cos(\theta)-\rho\sin(\theta)}\right)^2\right]\rho^3$

This space is free for rent...