Timeline for Is there any numerical technique to sum x^(n^alpha), n=0,1,...?
Current License: CC BY-SA 2.5
8 events
| when toggle format | what | by | license | comment | |
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| Mar 18, 2010 at 7:40 | comment | added | Robin Chapman | Sorry Kevin, I ought to learn how to read :-( | |
| Mar 17, 2010 at 21:36 | comment | added | Kevin Buzzard | My definition of the theta function is non-standard Robin (I put the square in already) so I think I'm OK. Mea culpa for using the wrong definition, of course. I think the problem with editing comments is that math overflow is running on software written by someone else so you have to request changes to a higher god who is not always benevolent. If you feel strongly about it, ask at tea.mathoverflow.net and they'll explain the procedure. | |
| Mar 17, 2010 at 21:00 | comment | added | Robin Chapman | For $\theta(1/t)=t\theta(t)$ read $\theta(1/t)=\sqrt{t}\theta(t)$. (Not that this helps much with the original question). So again, when can we edit comments? | |
| Mar 17, 2010 at 19:59 | comment | added | Kevin Buzzard | [aargh slip above "reduce to the case $t\geq1$". When can we edit comments?? | |
| Mar 17, 2010 at 19:57 | comment | added | Kevin Buzzard | [...and if you want more accuracy then use more than one term when expanding $\theta(u)$!] | |
| Mar 17, 2010 at 19:55 | comment | added | Kevin Buzzard | As an explicit example, if $x=0.9$ and $\alpha=2$ then summing 1000 terms gives an answer of 3.230272... . But solving $e^{-t^2}=x$ gives t=0.183... so $u:=1/t=5.4605...$ and summing just one term for the series with $x=e^{-\Pi u^2}$ gives 1, correct to 40 or so dec places, so $\theta(t)$ is approximately $u$ (to about 40 dec places!) so the sum we're interested in is hence (1+u)/2. In summary, the answer with $x=0.9$ and $\alpha=2$ is very very close to $(1+u)/2$ with $u=(-\log(0.9)/\pi)^{-1/2}$. As $x$ gets closer to 1 this formula gets better and better. | |
| Mar 17, 2010 at 19:47 | comment | added | Kevin Buzzard | For $\alpha=2$ you can use the functional equation for the theta function to get a massive speed-up. Set $\theta(t)=\sum_ne^{-\pi n^2 t^2}$, where the sum is over all integers. This converges for $t>0$ real. Much less clear, but completely standard, is that $\theta(1/t)=t\theta(t)$. Using this trick you can reduce to the case $0<t\leq 1$ which corresponds to $0<x<0.0432...$ where the series is converging super-fast. | |
| Mar 17, 2010 at 15:29 | history | answered | Douglas Zare | CC BY-SA 2.5 |