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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

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    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.

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  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").

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    • take note of the URL of the answer provided by clicking the share button, or
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  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

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  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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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

Read more about locked posts here.

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    $\begingroup$ 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. $\endgroup$ – Asaf Karagila Jul 18 '12 at 8:35
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    $\begingroup$ (+1) For thinking outside the (sand)box. $\endgroup$ – cardinal Jul 18 '12 at 19:40
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    $\begingroup$ 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! $\endgroup$ – Grace Note Oct 5 '12 at 14:45
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    $\begingroup$ To prevent crashes I've found the "Bookmarks to disable/enable MathJax", provided in here, pretty useful. $\endgroup$ – leo Dec 17 '12 at 18:03
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    $\begingroup$ 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. $\endgroup$ – Najib Idrissi Dec 2 '15 at 14:07

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My post needs more work, and I don't have the time for it today. Leaving this to be used by anyone.

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This slot is as vacant as the expression of someone who's been staring at the same line of a proof for three hours.

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

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This space free for whoever wants it

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This slot is open for anyone to use.

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My first ansatz, which gives at best an asymptotic power series (with convergence radius zero!), seemed to be rough and likely much suboptimal. But for completeness I'll give the method here, also because it provides that nice initial guess $a_\text{init}$ which is very near at the true value.

The ansatz employs essentially the Bernoulli-numbers, but not as in the classic Bernoulli-polynomials but in some "transposed" manner.

The individual $\log(1+1/a_k)$ - epressions have a mercator series, such that we can reformulate the original sum $S_d(a,N)$ as a double sum $$ \begin{array}{rll} S_d(a,N) &=& 1/a - 1/2/a^2 + 1/3/ a^3 - 1/4/ a^4 + \cdots - \cdots \\ &+& 1/(a+d) - 1/2/(a+d)^2 + 1/3/ (a+d)^3 - 1/4/ (a+d)^4 + \cdots - \cdots \\ &+& 1/(a+2d) - 1/2/(a+2d)^2 + 1/3/ (a+2d)^3 - 1/4/ (a+2d)^4 + \cdots - \cdots \\ &...& \\ &+& 1/(a+(N-1)d) - 1/2/(a+(N-1)d)^2 + 1/3/ (a+(N-1)d)^3 - 1/4/ (a+(N-1)d)^4 + \cdots - \cdots \\ \end{array}$$ Summing this vertically gives the column-sums $$ \begin{array}{rll} C_{d,1}(a,N) &=& 1/a + 1/(a+d)+ 1/(a+2d) + ... + 1/(a+(N-1)d) \\ C_{d,2}(a,N) &=&\frac12 ( 1/a^2 + 1/(a+d)^2+ 1/(a+2d)^2 + ... + 1/(a+(N-1)d)^2 )\\ C_{d,3}(a,N) &=&\frac13 ( 1/a^3 + 1/(a+d)^3+ 1/(a+2d)^3 + ... + 1/(a+(N-1)d)^3 )\\ &...& \end{array}$$ Such sums are essentially Hurwitz-Zeta sums, and can be expressed as power-series with zeta at negative integers as coefficients.
For easiness of handling and memorizing I've the "ZETA"-matrix in my toolbox, which provides that coefficients for the application to power series.
The ZETA-matrix looks like
image
and has the composition of Zeta-values
image2 Reading it horizontally one observes the coefficients for the integrals of the Bernoulli-polynomials which are configured to allow the summing-of-like-powers.
However, reading vertically, they give the coeffients for power series to determine the sums-of-reciprocals-of-like-powers. However, that power series have convergence-radius zero, but because of alternating signs we can use them as asymptotic power series, truncated at some meaningful index.

For instance, the power series taken from the second,third, fourth and so on column, and calculated as $$ f_1(a) = a z_{0,1} + a^2 z_{1,1} + a^3 z_{2,1} + ... \to \zeta(2,a) \\ f_2(a) = a z_{0,2} + a^2 z_{1,2} + a^3 z_{2,2} + ... \to \zeta(3,a) \\ ... $$ where $z_{r,c}$ indicates the element of the ZETA-matrix from row $r$ and column $c$ (index based at $0$).
That allows immediately to determine the sums of reciprocals from $a$ to $a+(N-1)$ by $$ s_1(a) = f_1(a) - f_1(a+N) = 1/a^2 + 1/(a+1)^2+... + 1/(a+(N-1))^2 \\ s_2(a) = f_2(a) - f_2(a+N) = 1/a^3 + 1/(a+1)^3+... + 1/(a+(N-1))^3 \\ ... $$

For the first column we need a correction factor which comes out to be $$ f_0(a) = a z_{0,0} + a^2 z_{1,0} + a^3 z_{2,0} + ... \to \\ \log(1/a)+\left(\sum_{k=1}^{a-1} \frac1k \right) -\gamma $$
such that $$ s_0(a) = f_0(a) - f_0(a+N) = 1/a + 1/(a+1)+... + 1/(a+(N-1)) \\ + \log(1/a) - \log(1/(a+(N-1)) $$

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Unused, available for anyone.$ $

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

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This entry is free for anyone to commandeer since Oct. 27th, 2018

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This answer is being edited right now.

The Fabius function is a continuous function $F:\mathbb R\to[0, 1]$ uniquely determined by the following conditions:

  • $F(0)=0$,
  • $F(1)=1$, and
  • $F'(x)=2\,F(2{\tiny\text{ }}x)\,$ for all $x\in\mathbb R.$

It is well known that the Fabius function assumes rational values at dyadic rational arguments, and several algorithms/formulas to compute those values have been published (usually making use of some auxiliary rational sequences defined by bulky recurrence relations).

Now, the rumors are these values can be calculated using the following nice formula: $$\large\!\!\!\!\!\!F\!\left({\normalsize\frac m{2^n}}\right) = \frac1{2^{n^2}\,\left(\frac12;{\tiny\text{ }}\frac12\right)_n}\,\sum _{k=0}^n \frac{\binom n k_{1/2}}{2^{{\tiny\text{ }}k\,(k-1)}\,(n+k)!}\,\sum _{r=1}^{2^k{\tiny\text{ }}m}\,(-1)^{s_2\left(2^k{\tiny\text{ }}m-r\right)}\,\left(\tfrac12-r\right)^{n+k}{\small,}$$ where $m,\,n$ are non-negative integers, $\left(a;{\tiny\text{ }}q\right)_n$ is the q-Pochhammer symbol, ${\binom n k}_q$ is the q-binomial coefficient, and $s_2(r)$ is the sum of binary digits of $r$.

Unfortunately, I could not find a way to prove it so far.

$$\begin{align} \int_0^1&f(x)\,dx =\sum_{n=1}^\infty\sum_{m=1}^{2^n-1}\frac{(-1)^{m+1}}{2^n}f\left(\frac m{2^n}\right)\\ \int_0^1&f(x)\,dx = f(0)-\sum_{n=1}^\infty\sum_{m=0}^{2^n-1}\frac{(-1)^m}{2^n}f\left(\frac m{2^n}\right)\\ \int_0^\infty&f(x)\,dx = \sum_{m=1}^{\infty } f(m)-\sum _{n=1}^{\infty} \sum _{m=1}^{\infty } \frac{(-1)^m}{2^n}f\left(\frac{m}{2^n}\right) \end{align}$$

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Function $f$ is defined as $$f(x)=\frac{1}{(x-1)^3(x^3+2x^2+x+1)}\tag 1$$ Note that $$x^3+2x^2+x+1=(x+1)(x^2+x+1)\tag 2$$ and so $$f(x)=\frac{1}{(x-1)^3(x+1)(x^2+x+1)}\tag 3$$ and also $$f(x)=\frac{1}{(x-1)}\cdot\frac{1}{(x^2-1)}\cdot\frac{1}{(x^3-1)}=\sum_{i=0}^\infty x^i \cdot\sum_{k=0}^\infty x^{2k} \cdot\sum_{j=0}^\infty x^{3j}$$

$$\frac{1}{(x-1)^3(x+1)(x^2+x+1}=\frac{A_1}{(x-1)}+\frac{A_2}{(x-1)^1}+\frac{A_3}{(x-1)^3}+\frac{B}{(x+1)}+\frac{Cx+D}{(x^2+x+1)}$$ $$1=A_1(x-1)^2(x+1)(x^2+x+1)+A_2(x-1)(x+1)(x^2+x+1)+A_3(x+1)(x^2+x+1)+B(x-1)^3(x^2+x+1)+(Cx+D)(x-1)^3(x+1)$$

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  • $\begingroup$ Oh god, I still remember having tests that required computing these king of generating functions by hand. If result was correct, the method was verified, otherwise no points. $\endgroup$ – enedil Feb 11 at 21:53
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It's been a long time since I last did this stuff, but nobody better seems to have turned up, so I'll have a go.

I'll give the box $\square \phi$ its usual meaning here: "when I translate $\phi$ into the formal system using Gödel numbers, then there is a collection of mechanical transformations inside the formal system that together form an internal-system-proof of $\phi$'s quotation". I will also refer to the formal system in question in the first person, with the pronoun "I".

Then the formula $\square (\square \phi \to \phi) \to \square \phi$ should be read as:

If there is a number which encodes a proof that "having a number which encodes a proof of $\phi$ means I can actually prove $\phi$", then there is a number which encodes a proof of $\phi$.

Or, a bit more loosely,

If I, internally, can prove that my proof of $\phi$ respects the logic in which I operate (so my proof of $\phi$ means that in my surrounding logic, $\phi$ is actually true), then in fact there is a proof of $\phi$ that I will be satisfied with.

This is more concretely understood through its universal quantification, Löb's theorem (which is true in any system rich enough to support Peano arithmetic):

If I can prove within myself that my own proof-checking process respects the semantics of the logic in which I operate, then the logic realises that actually I can satisfy myself of any statement at all.

Specifically, $\square (\square \phi \to \phi)$ means "I contain a proof that my proofs are semantically valid in the surrounding logic", while $\square \square \phi \to \square \phi$ means "it's true in the surrounding logic that if I want to prove $\phi$, it's enough to prove provability-of-$\phi$".

The two statements were very unlikely to be the same: one is talking about proof within the system, and one is talking about facts of the surrounding logic.

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