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 "wiki locked" 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.

locked by user642796 Jan 12 '17 at 6:11

This question's answers are a collaborative effort: if you see something that can be improved, just edit the answer to improve it! No additional answers can be added here

  • 8
    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
  • 21
    (+1) For thinking outside the (sand)box. – cardinal Jul 18 '12 at 19:40
  • 14
    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
  • 3
    To prevent crashes I've found the "Bookmarks to disable/enable MathJax", provided in here, pretty useful. – leo Dec 17 '12 at 18:03
  • 3
    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

17 Answers 17

this box is free for everyone to use

A possibly interesting, further generalization becomes obvious, when we insert $\zeta()$-values for the Bernoulli-numbers $B(n)$ . Since $B(n)/n \underset{n>0}= (-1)^{n-1 } \zeta(1-n)$ the second identity becomes $$ \sum_{k=1}^\infty \zeta(1-2k) = \gamma + \zeta(1-1) + B(0) = \gamma +1/2 - 1 = \gamma-1/2 \\ \sum_{k=1}^\infty \zeta(1-2k)2k = \zeta(2) + (\zeta(1-1)\cdot 1) - B(0)\cdot 0) \\ $$ (... to be corrected... and continued...)

Please allow this draft to remain for at least a week after it was last edited, or until I release it

Let $F$ be a field, and $p(x) = a_0 + a_1x + \cdots a^nx^n$ a polynomial over $F$. We can define the formal derivative of $p(x)$ to be the polynomial $a_1 + 2a_2x + \cdots na_nx^{n-1}$. The formal derivative clearly mimics the standard differentiation operator from calculus for polynomials over an arbitrary field, and satisfies many of the usual properties of derivatives - in particular linearity and the Leibniz law $(p(x)q(x))' = p'(x)q(x) + p(x)q'(x)$. The fact that it coincides with the differentiation operator for subfields of $\mathbb{C}$ immediately shows that they hold for that special case, but for arbitrary fields we have to verify them formally, which is nevertheless quite simple.

I would also like to say that the formal derivative also satisfies the chain rule $(p \circ q)'(x) = p'(q(x))q'(x)$, where the composition of two polynomials is defined formally as the polynomial obtained by substituting $q(x)$ for the formal variable $x$ in $p(x)$ and then expanding and collecting like terms; this definition is because abstract polynomials are not the same as polynomial functions.

This answer is available for anyone to use.

When we have removed fractions from a equation of the third degree, according to the manner which has been explained, and none of the divisors of the last term are found to be a root of the equation, it is a certain proof, not only that the equation has no root in integer numbers, but also that the fractional root cannot exist, which may be proved as follows.

Let here be given the equation $x^3-ax^2+bx-c=0$, in which, $a,b,c$ express integer numbers. If we suppose, for example, $x = \dfrac 23$, we shall have $\dfrac{27}{8} - \dfrac 94a + \dfrac 32 b - c = 0$. Now here, the first term alone has $8$ for the denominator; the others being either integer numbers, or numbers divided by $4$ or $2$, and therefore cannot make $0$ with the first term. The same thing happens with every other fraction.

This entry is available for anyone to commandeer since Oct. 17th, 2018.

It might be easier to use a coordinate system which is adapted to $\Sigma$. Let

$$\Phi : V\to \Sigma, $$

where $(x_1, x_2) \in V$, be a local parametrization of $\Sigma$. Then

$$\Psi (x_1, x_2, z) = \Phi(x_1, x_2) + z \vec n (\Phi(x_1, x_2))$$

is a local parametrization of an open set $U\subset \mathbb R^3$ containing $\Phi (V)$.

I will prove the identity $Dp = P_\Sigma - \rho \mathcal H$ only when you consider points on $\Sigma$ (that is, $z=0$ situation)

Let $X$ be a vector on $U$. Write $X = X_\Sigma + Y$, there $X_\Sigma$ is tangent to $\Sigma$. Then

$$D_{X_\Sigma} p = X- (\nabla \rho \cdot X) \vec n (p(x)) - \rho(x) \nabla_X \vec n (p(x)).$$

Since $\nabla \rho = \vec n$, the second term is

$$-(\nabla \rho \cdot X) \vec n(p(x))= -(X\cdot \vec n) \vec n = -Y.$$

Thus the first two terms gives $X - Y = P_\Sigma X$. For the remaining term, note that from the local coordinates, it is of the form

$$\vec n(p(x) ) = \vec n( \Phi(x_1, x_2)).$$

This implies

$$\nabla_Y \vec n (\Phi(x_1, x_2))$$

since it has no dependence on $z$, and

$$\nabla_{X_\Sigma}

Olympiad Inequality $\sum\limits_{cyc} \frac{x^4}{8x^3+5y^3} \geqslant \frac{x+y+z}{13}$

$x,y,z >0$, prove $$\frac{x^4}{8x^3+5y^3}+\frac{y^4}{8y^3+5z^3}+\frac{z^4}{8z^3+5x^3} \geqslant \frac{x+y+z}{13}$$

Note: there is equality when $x = y = z$, so assume wlog $x > y$ and $x > z$.

Subtract $\frac{x + y + z}{13}$ from both sides:

$$\frac{x^4}{8x^3+5y^3} - \frac{x}{13}+\frac{y^4}{8y^3+5z^3} - \frac{y}{13}+\frac{z^4}{8z^3+5x^3} - \frac{z}{13} \geqslant 0 $$ $$\frac{13x^4 - x(8x^3+5y^3)}{13(8x^3+5y^3)} +\frac{13y^4 - y (8y^3+5z^3)}{13(8y^3+5z^3)}+\frac{13z^4 - z(8z^3+5x^3)}{13(8z^3+5x^3)} \geqslant 0 $$

$$\frac{5}{13}\left(\frac{x^4 - xy^3}{8x^3+5y^3} +\frac{y^4 - yz^3}{8y^3+5z^3}+\frac{z^4 - zx^3}{8z^3+5x^3}\right) \geqslant 0 $$ $$x(x-y)\frac{x^2 + xy + y^2}{8x^3+5y^3} +y(y-z)\frac{y^2 + yz + z^2}{8y^3+5z^3}+z(z-x)\frac{z^2 + zx + x^2}{8z^3+5x^3} \geqslant 0 $$

Note, the only negative quantities are the $(z - x)$ and possibly $(y - z)$.

This answer is available for anyone to use.

This answer is available for anyone.

This answer is available for anyone to use.

This answer is available for use by anyone.

  • I am surprised that for the fraction $\frac{dy}{dx}$ the community did not (properly) use \dd y and \dd x to make $\frac{\text{d}y}{\text{d}x}$. Note the slight difference? – user477343 Oct 14 '17 at 23:07
  • 1
    @user477343 test $\dd x$ and $\frac{\dd}{\dd x}f(x)$ are not working for me... – zwim Nov 7 '17 at 10:37
  • @zwim yes they are not working for me for some reason as well. Just use \text{...} instead, I guess, where we can put d in {...} – user477343 Nov 10 '17 at 15:36
  • @zwim I worked it out (after nearly a year, hahah). It is just a new command! – user477343 Sep 19 at 6:59

This entry is available for anyone to commandeer since Oct. 13th, 2018.

my wiki edit to category-theory tag was rejected. The reason for rejection was

This edit copies a significant amount of content from an external source. Generic descriptions such as encyclopedia articles and ad copy do not provide useful guidance; try creating something useful to this community specifically, and be sure to attribute the original author. See: How to reference material written by others.

But in fact it was not copied, I composed the proposed text myself.

Here is my proposed edit (look in history of this question to see them side-by-side):


excerpt:

Category theory is the study of categories, functors, and natural transformations. Categories are directed graphs with a composition law. Most mathematical structures can form vertices (objects) of a category with homomorphisms as arrows, so some constructions, such as products, common across different subjects can be given unified descriptions. Functors are homomorphisms of categories, and natural transformations are arrows between functors.


Category theory]1 is the study of categories, functors, and natural transformations. Categories are directed graphs containing vertices and arrows (also called objects and morphisms) with a composition law on compatible arrows. They simultaneously generalize monoids and posets.

Most mathematical structures form the objects of a category with homomorphisms as arrows, such as the category of sets and functions between them, the category of groups and homomorphisms, or the category of topological spaces and continuous maps.

Some constructions common across many fields of mathematics, such as products, pullbacks, limits, sums, pushouts, colimits, universal properties, Galois connections, adjunctions, monads, and Kan extensions are most naturally phrased in category theoretic language.

Functors are homomorphisms between categories. Canonical constructions like cohomology or the ring of functions on a space, which transform geometric data into algebraic, are naturally viewed as functors. Natural transformations are families of arrows which furnish a comparison between functors. An invertible natural transformation is called a natural isomorphism.

Category theory allows one to draw a distinction between equality and isomorphism, and encourages one to keep track of isomorphisms between objects, rather than identifying isomorphic objects.

Generalizations of category theory include enriched category, where the arrows constitute objects in another category, and higher category theory, where there are also arrows between arrows. A specialization is topos theory; a topos is a category with enough structure to provide an alternative to set theory.

This entry is available for anyone to use since Jun 2nd, 2018.

This entry is available for anyone to commandeer since Oct. 17th, 2018.

I record here the draft of my answer to deleted question Exterior product with coefficients from finite fields..


There are several statements which are true for symmetric and alternating powers of vector spaces (or modules) over fields (or rings) of characteristic zero, which fail over positive characteristic, including... (link to https://math.stackexchange.com/a/674299/16490).

Exterior algebra, Symmetric algebra, Divided power algebra in Characteristic $p>0$.

Discussion of forms as subspaces of tensor algebra versus quotient. link to https://en.wikipedia.org/wiki/Ε-quadratic_form

For example, over characteristic two, alternating bilinear forms $B(v,v)=0$ are distinct from skew-symmetric bilinear forms $B(u,v)+B(v,u)=0.$ Alternating implies skew-symmetry, as we can see by looking at

$$B(u+v,u+v)=0=B(u,u)+B(u,v)+B(v,u)+B(v,v)=B(u,v)+B(v,u).$$

Of course over a ring where $2$ is a unit, this implication is reversible.

For example, over $\mathbb{F}_2$, given a 2-dimensional vector space $V=\langle e_1,e_2\rangle,$ there are two alternating forms, $B_0=0$, and $B_1(e_1,e_2)=1.$ But there are eight antisymmetric forms, because $B(e_i,e_j)$ may take any value for $i\leq j.$

In general $\Lambda^2 V$ is supposed to be a representing object for multilinear maps satisfying . But over characteristic 2, there

Give a module or vector space $V$ over a field or ring $K$, the exterior power $\Lambda^2 V$ is properly defined as $$\Lambda^2 V=\frac{V\otimes V}{(u\otimes v+v\otimes u)},$$ here $(u\otimes v+v\otimes u)$ is the vector space spanned by all vectors of the stated form.

You must log in to answer this question.