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I'm encountering this strange behave of the mathjax-system with this question .
Everything is fine if I refresh this page in my browser (Firefox) ; but when I start editing the text then the preview stops rendering mathjax, and also after saving the page will be shown without mathjax-rendering. After refreshing the page everything looks ok again. (I've restarted the browser, but the behave did not improve, so I think I've treated mathjax to run into a bug...


[update]: I've copied the text of my question in an answer box so you need not move to the main-site to test the behave

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    $\begingroup$ Very strange, I have the same problem on Chrome. $\endgroup$ – Najib Idrissi Jan 8 '15 at 17:59
  • $\begingroup$ You should probably mention that here. $\endgroup$ – Daniel Fischer Jan 8 '15 at 18:43
  • $\begingroup$ @Daniel : done ${ }$ $\endgroup$ – Gottfried Helms Jan 8 '15 at 18:47
  • $\begingroup$ This happens to me quite a lot when I am making an answer with a lot tex. $\endgroup$ – dustin Mar 10 '15 at 19:37
  • $\begingroup$ Same on Internet Explorer. It must be a site problem. $\endgroup$ – Zach466920 Apr 9 '15 at 21:39
  • $\begingroup$ Having the same issue on Internet Explorer. It seems like once a certain length is reached, it stops updating. I've had to cut at least one answer short because I could no longer see what I was doing. $\endgroup$ – Mike Aug 15 '15 at 20:43
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For the opportunity of testing I include here just the original text of my question. Perhaps try to edit/update this "answer" to isolate the point from where the bug can be reproduced

I'm ölk considering some family of functions whose coefficients of their power series occur in the columns of the following matrix A
$ \qquad $ picture

The second of that functions is $$ f_1(x) = - \sum_{k=1}^\infty \zeta(1-k) x^k $$ By evaluating it numerically for various $x$ using Noerlund sums I arrived at the guessed closed-form expression for it as $$ f_1(x) = \log(1/x) - \psi(1/x) $$

  • Sanity-check: if I feed the following in W/A $$ \text{series [ log(t) - digamma(t) ]} $$ I get the same series, however in the form for some leading coefficients from which I conclude $$ [f_1(1/t) =] \quad \sum_{k=1}^\infty -\zeta(1-k) \frac 1{t^k} \qquad \qquad \text{ expansion at } t=\infty$$ so it appears that the guessed closed-form is correct.


I have now the next function $$ f_2(x) = \sum_{k=2}^\infty a_{2,k} x^k $$ where the compositions of the $a_{2,k}$ is of $\zeta()$-values form the above, but with some pattern. On searching for a closed form for this I'm stuck.
To understand the pattern in the $a_{2,k}$ I've found a very elegant solution, but I can't do the final step. It involves Borel- or Laplace-transformation and -reversion.

The Borel-transformation of the series representation of $f_1(x)$ is $$ F_1(x) = - \sum_{k=1}^\infty \zeta(1-k) {x^k \over k! } $$ and this has a closed form $$ F_1(x) = \log( {\exp(x)-1\over x}) $$ Now the "elegant moment" is, that the Borel-transformation of the series representation of $f_2(x)$ $$ F_2(x) = \sum_{k=2}^\infty a_{2,k} \frac {x^k}{k!} $$ has simply the closed form $$ F_2(x) = (F_1(x))^2 $$ and moreover, for all following $k>2$ it is $$ F_k(x) = (F_1(x))^k $$

The problem is now that I don't know how to find from this the closed form for $f_2(x)$ (or $f_k(x)$ for higher $k$. W/A was of no help here.

Q1: What is the closed form for $f_2(x)$ and that of the following members of that family of functions?


Additional remark: I'm not experienced with Borel- or Laplace-transform. But while Pari/GP has a function "serlaplace()" and this gives the expected transformation from $F_k(x)$ to $f_k(x)$ and I could experiment with this a bit, W/A gives another thing when called "laplacetransform()" , actually it gave $f_1(x)/x$ instead $f_1(x)$. So question 2:

Q2: What is the correct expression for the series-forward&backward transformations: Borel? Laplace?

P.s.: I'm not good with tagging of questions like this. Please don't hesitate to improve the tagging if not optimal

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  • $\begingroup$ I've removed the \tag{} -function and the \small - directives, but that alone does not cure the problem $\endgroup$ – Gottfried Helms Jan 8 '15 at 19:10

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