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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 ├Âlk 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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