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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
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    (+1) For thinking outside the (sand)box. – cardinal Jul 18 '12 at 19:40
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    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
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    To prevent crashes I've found the "Bookmarks to disable/enable MathJax", provided in here, pretty useful. – leo Dec 17 '12 at 18:03
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    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 Maybe someday we can get rid of this outdated crutch.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

First let me rename things a bit, for instance the length-of-a-cycle is much in the focus of many formulas and often a given parameter, so I avoid "indexing-letters" like i or k for this. I use N for the length of a cycle, counted in the odd-steps only. Also the number of even-steps becomes the letter S because it is also the sum-of-exponents-of-2 involved in such formulae.
The transformation from one odd to the next odd number itself I write consequently $$ b = {3a+1\over2^A} \qquad \qquad \text{ $A$ taken to make $b$ odd}$$

With this there is a nice formula for an important property of cycles. Here I use the odd numbers to be transformed by the letter a but indexed from $1$ to $N$ $$ a_2 \cdot a_3 \cdot \ldots\cdot a_N \cdot a_1 = ({3a_1+1\over 2^{A_1}})\cdot({3a_2+1\over 2^{A_2}}) \cdot \ldots\cdot ({3a_N+1\over 2^{A_N}}) $$ Here the lhs is simply the product of all elements $a_k$ of an assumed cycle of length $N$ and sum-of-exponents $S$ and the rhs the same, only where the numbers are written in their transformed form.
We can rewrite this in the very practical formula: $$ 2^S = (3 + {1\over a_1})\cdot({3+1\over a_2}) \cdot \ldots\cdot (3+{1\over a_n}) $$ From this we can extract some properties of any assumed cycle given length $N$ and $S$.

For instance, we can immediately see, that the rhs gives a lower bound for the exponent $S$ in the perfect power of $2$ on the lhs: $$ rhs \gt (3+1/\infty)^N = 3^N \qquad \text{ assumed all $a_k$ are $\infty$ }$$ so $$ 2^S \gt 3^N \implies S \ge \lceil N\log_2(3) \rceil$$ On the other hand, if all $a_k=1$ then $$ rhs \gt (3+1/1)^N = 4^N \qquad \text{ assumed all $a_k=1$ }$$ so also $$ 2^S \le 4^N \implies S \le N \cdot 2 $$ Interestingly, we have immediately one solution for any $N$, namely the trivial cycle of all elements $a_k=1$.
Now, if there is another cycle and all elements $a_k \ge 5$ then $S$ must be smaller than $2 N$ and it seems that $$S=\lceil N \cdot \beta \rceil$$ is even the upper bound for $N>10$

$$m = \dfrac{10n^2+2n}{7n+1}$$

$$10n^2+(2-7m)n - m= 0$$

$$10\left(n-\dfrac{7m-2}{20}\right)^2 = \dfrac{(7m-2)^2}{40} + m$$

\begin{align} 7m-2 &\equiv 0 \pmod{20} \\ 7m &\equiv 2 \pmod{20} \\ m &= 20k+6 \\ \dfrac{7m - 2}{20} &= 7k + 2 \end{align}

$$10(n-7k-2)^2 = 49k^2+48k+10$$

\begin{align} 49k^2+48k+10 &\equiv 0 \pmod{10} \\ k^2+2k &\equiv 0 \pmod{10} \\ (k+1)^2 &\equiv 1 \pmod{10} \\ k+1 &\in \{1,9\} \pmod{10} \\ k &\in \{0, 8\} \pmod{10} \end{align}

$$k=10x+8 \implies (70x - n + 58)^2 = 490x^2 + 832x + 353$$

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.

You can commandeer this post after 1/10/2018

Whenever I'm faced with a big number (anything bigger than $10$ I guess!), I start with smaller numbers and go from there.

How about we start with $4$? Easy, there's a maximum of $1$ four-weak subsets (fws).

How about $5$? If we have $1$ edge, there's $4$ fws. With two (not necessarily non-adjacent) edges,

This entry is free for anyone to use since May 26th, 2018.

When we have removed fractions from a equation fo the third digree, 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 rfoot in integer numbers, but also that tha 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.

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}

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  • 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
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    @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

This answer is currently available for use.

free for anyone to use since April 18th, 2018.

The general idea is to interpolate the terms to get a function and then analyze its properties.


Let $\{a_n(x)\}$ be a sequence of once-differentiable functions.

Define the recurrence relation $$A_n(x)=a_n(x)^{A_{n+1}(x)}$$ (where often the '$(x)$' part will be omitted for simplicity.)

Then, we have $$A_n'=A_n\left(A'_{n+1}\ln a_n+A_{n+1}\frac{a_n'}{a_n}\right)$$


Let $$t_n=\frac1n-\frac1{n+1}$$ Let $$H(x)= \begin{cases} 1, &x<0 \\ \frac{\cos(\pi x)+1}2, &0\le x\le1\\ 0, &x>0 \end{cases} $$ Define $$a_n(x)=(t_n)^{t_{n+1}H(n-x)}$$

OP’s sequence of even terms thus becomes $$\{A_1(1),A_1(2),A_1(3),\cdots\}$$

Then, the limit of the OP's sequence for even terms (i.e. $\lim_{n\to\infty}a_{2n}$, not to be confused with the $a_n(x)$ in this answer) is $$A_1(\infty)\equiv \lim_{x\to\infty}A_1(x)$$

So our question would become

Does $\lim_{x\to\infty}A_1(x)$ exists?

Let's analyze the derivatives.

Firstly, $$a_n'=-t_{n+1}\ln(t_n)H'(n-x)a_n$$ So, $$A_n'=\overbrace{\cdots}^{\text{messy algebra}}=A_nb_n(A_{n+1}H'(n-x)-A'_{n+1}H(n-x))$$ where $b_n=\frac{\ln(n+1)+\ln(n)}{(n+1)(n+2)}$.

For $n<\lfloor x\rfloor$, $H'(n-x)=0$. Therefore, we can recursively write out $$A_1'=\left(\prod^{\lfloor x\rfloor}_{k=1}(-A_kb_k)\right) A'_{\lfloor x\rfloor+1}$$

With $$A'_{\lfloor x\rfloor+1}=A_{\lfloor x\rfloor+1}b_{\lfloor x\rfloor+1}(A_{\lfloor x\rfloor+2}H'(\lfloor x\rfloor+1-x)-\underbrace{A'_{\lfloor x\rfloor+2}H(\lfloor x\rfloor+1-x)}_{=0})$$ we can finally write out something neater $$A_1'=-\left(A_{\lfloor x\rfloor+2}\prod^{\lfloor x\rfloor+1}_{k=1}(-A_kb_k)\right)\frac{\sin\pi(x-\lfloor x\rfloor)}2$$

Recalling $b_n=\frac{\ln(n+1)+\ln(n)}{(n+1)(n+2)}$, and observing $A_k<1$ for all $k$, immediately $A_1'(\infty)=0$. Thus, we may argue that $A_1(\infty)$ exists, so as the 'even term limit' of OP's sequence.


The existence of 'odd term limit' is equivalent to the existence of of $A_1(\infty)$, with $t_n=\frac1{n+1}-\frac1{n+2}$ and the same $a_n(x)$ as above.

The change of definition in $t_n$ would only lead to the change of $b_n$: $$b_n=\frac{\ln(n+2)+\ln(n+1)}{(n+2)(n+3)}$$ But we can still obtain $A_1'(\infty)=0$. Thus, the 'odd term limit' exists as well.

The following formula gives values of the Fabius function at dyadic rationals for $0\le\frac n{2^m}\le1$: $$F\left(\frac n{2^m}\right)=\frac12-\frac{1}{2^{m^2+1} \, \left(\frac12;\frac12\right)_m} \, \sum_{\mu=0}^m \frac{\binom m\mu_{\!1/2}}{2^{\mu(\mu-1)} \, (m+\mu)!} \, \sum_{\lambda=0}^{2^{m+\mu}-1}(-1)^{s_2(\lambda)} \, \frac{\left(\lambda-2^\mu n+\tfrac12\right)^{m+\mu+1}}{\left|\lambda-2^\mu n+\tfrac12\right|},$$ where $$(q;\,q)_m = \prod_{k=1}^m (1-q^k), \quad \binom m \mu_{\!q} = \frac{(q;\,q)_m}{(q;\,q)_\mu \, (q;\,q)_{m-\mu}}$$ are the $q$-Pochhammer symbol and the $q$-binomial coefficient, and $s_2(\lambda)$ is the sum of digits of $\lambda$ in base-2. Note that $(-1)^{s_2(\lambda)}=t_\lambda$ is just the signed Thue–Morse sequence, satisfying the recurrence $t_0=1,\,\,t_\lambda=(-1)^\lambda\,t_{\lfloor\lambda/2\rfloor}$.

This answer is now open for your use.

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.

Records for the analytic soul of the holonomic approximation theorem.

Suppose $V \subset \Bbb R^2$ is a bounded domain and $h_t : \Bbb R^2 \to \Bbb R^2$ a diffeotopy for $t \in (0, 1]$ where $h_1 = \text{id}$ and $\|h_{1/n}|_V - \text{id}\|_{C^0} \to 0$. Call $U_n = h_{1/n}(V)$. Suppose $f_n : \mathcal{O}(U_n) \to \Bbb R$ and $f : \Bbb R^2 \to \Bbb R$ are smooth functions, where $\mathcal{O}(A)$ denotes some unspecified neighborhood of $A$. Assume that $\|f|_{U_n} - f_n\|_{C^0} \to 0$.

Is it true that $\int_{U_n} f_n \to \int_V f$?

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.

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