Timeline for Bound of Partial Sum of Partition Generating Function
Current License: CC BY-SA 3.0
14 events
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| Jun 25, 2012 at 3:34 | history | edited | Gilyoung Cheong | CC BY-SA 3.0 |
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| Jun 25, 2012 at 3:26 | comment | added | Gilyoung Cheong | For the second estimation, I think it's either I made a horrendous mistake or the value $m$ happens in a weird way. Thanks for clarifying things. Using truncated sum sounds like a good idea! Thanks a ton. | |
| Jun 25, 2012 at 3:23 | vote | accept | Gilyoung Cheong | ||
| Jun 25, 2012 at 2:03 | comment | added | Douglas Zare | The asymptotic formula is good asymptotically, but of course the lower degree terms dominate and getting the $1/q$ coefficient exactly right gives you a better approximation than only approximating it as $1.87.../q$. The approximation $q/(q-1)$ has the wrong coefficient of $1/q^2$, $1$ instead of $2$. Try approximating the truncated sum as $\sum_{j=0}^\infty p(j)/q^j - \sum_{j=k+1}^\infty f(j)/q^j$ or just drop the second series and approximate it by $\sum_{j=0}^\infty p(j)/q^j$. This a lot more accurate than being off by about $1/q^2$ except perhaps for very low values like $k=2$. | |
| Jun 25, 2012 at 1:07 | comment | added | Gilyoung Cheong | Also, this estimation seems better than using asymptotic formula in this situation. For example, we have $\sum_{j=0}^{3}p(j)/97^{j} = 1.01052512783878$; $97/96 = 1.01041666666667$; $1 + \sum_{j=1}^{3}f(j)/97^{j} = 1.01964017022542$, where $f(j) = $\exp(\pi \sqrt{2j/3})/(4\sqrt{3}j)$. | |
| Jun 25, 2012 at 0:23 | history | edited | Gilyoung Cheong | CC BY-SA 3.0 |
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| Jun 25, 2012 at 0:10 | answer | added | Douglas Zare | timeline score: 3 | |
| Jun 24, 2012 at 23:45 | comment | added | Gilyoung Cheong | I am perfectly fine with any argument with all $x$ in any interval near $0$ since that will be useful for most of the finite fields that I am interested in. | |
| Jun 24, 2012 at 23:42 | comment | added | Gilyoung Cheong | Moreover, it looks like as $x$ become close to zero the conjecture seems a lot better. I have only investigated when $x = 1/N$ where $N > 1$. But the property is probably what's happening over a continuum (subset of $\mathbb{R}$), and I also suspect if there is, any argument must be easier in continuous or differentiable $x$ than discrete $x = n$. | |
| Jun 24, 2012 at 23:35 | comment | added | Gilyoung Cheong | @Gerry Myerson: Thanks for pointing out. In my head it was looking like the left-hand side of $q/(q-1) = 1/(1-q^{-1})$. | |
| Jun 24, 2012 at 23:34 | history | edited | Gilyoung Cheong | CC BY-SA 3.0 |
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| Jun 24, 2012 at 23:33 | comment | added | Gilyoung Cheong | You are right. I should have written it more carefully. Let me correct it. | |
| Jun 24, 2012 at 22:53 | comment | added | Gerry Myerson | $x/(x-1)$ is going to be negative for the kind of $x$ you ask about, so that seems like an odd conjecture. | |
| Jun 24, 2012 at 22:36 | history | asked | Gilyoung Cheong | CC BY-SA 3.0 |