Sandbox for drafts of long, complex posts

Question

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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   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.

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Answer 1

This answer is free for anyone to use.

Answer 2

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

Answer 3

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

Answer 4

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

Answer 5

This space free for whoever wants it

Answer 6

This answer is available for anyone to use.

Answer 7

This slot is open for anyone to use.

Answer 8

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

and has the composition of Zeta-values
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 $$ s_0(a) = f_0(a) - f_0(a+N) = 1/a + 1/(a+1)+... + 1/(a+(N-1)) \\ + \log(1/a) - \log(1/(a+(N-1)) $$

Answer 9

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Answer 10

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Answer 11

Unused, available for anyone.$ $

Answer 12

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

Answer 13

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

Answer 14

This entry is available for everyone.

Answer 15

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 $$f(x)=\frac{1}{(x-1)}\cdot\frac{1}{(x^2-1)}\cdot\frac{1}{(x^3-1)}=\sum_{i=0}^\infty x^i \cdot\sum_{k=0}^\infty x^{2k} \cdot\sum_{j=0}^\infty x^{3j}$$

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

Answer 16

This entry is free for anyone to use.

Answer 17

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.