Timeline for Approximate sum by an integral: valid or not?
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
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| Jul 14, 2017 at 15:37 | comment | added | fedja | @becko The problem with the endpoint is that you'll have to start the contour right there and won't be able to bypass the maximum on the ascent and the descent like I did. That will result in the usual Laplace asymptotics along the imaginary axis as the main term, and that one is just a series in inverse powers. | |
| Jul 14, 2017 at 15:30 | comment | added | valle | Uhm, I suspected it should be similar to Laplace's saddle-point method, where if the max lies on an end-point, you need to include a factor of $1/2$ in the estimate... But I did not think it would affect convergence .... | |
| Jul 14, 2017 at 15:26 | comment | added | fedja | Yep. Unfortunately, when the maximum is at an endpoint, the exponential decay of $R_n$ is out of question more often than not (say, for $f(x)=-x^2$ and $g(x)=1$ you still have $c=1$ but the convergence is no better than the typical convergence of the left Riemann sum to the integral ($R_n=O(1/n)$ or something like that). That's why I didn't try to address this case in the hope that the OP doesn't really need it. | |
| Jul 14, 2017 at 15:04 | comment | added | M. Dus | I've thought about your question a little bit. Fedja's answer is really interesting, but in the case where the maximum is not in the interior, you do have to take a constant $c$. I tried a little bit to play with some functions. I took $f$ increasing and $g=1$ and I think that $S_n$ is equivalent to $\frac{\mathrm{e}^{f'(1)}}{\mathrm{e}^{f'(1)}-1}\frac{1}{n}\mathrm{e}^{nf(1)}$, while $I_n$ is equivalent to $\frac{1}{f'(1)}\frac{1}{n}\mathrm{e}^{nf(1)}$, but I don't have a formal proof. That should give us the $c$. | |
| Jul 14, 2017 at 8:33 | comment | added | M. Dus | No you're right, that was a mistake. I thought about it and the constant $c$ should be related to the derivative of $f$, somehow... though it is not easy to find the right one. I'm sorry for my last post anyway. | |
| Jul 13, 2017 at 16:15 | comment | added | valle | $(1/n)\mathrm e^{n\mathrm e} / \mathrm e^n \rightarrow \infty$, so I dont your understand your proof. Can you add some steps? Maybe add it into your answer. If you keep coming up with counterexamples, eventually we'll hit on something true :P | |
| Jul 13, 2017 at 16:06 | comment | added | M. Dus | Anyway, here is a counter-example (with a non-polynomial function): Take $f(x)=\mathrm{e}^{x}=g(x)$. Then, $S_n=\frac{1}{n}\sum\mathrm{e}^{n\mathrm{e}^{k/n}}\mathrm{e}^{k/n}$, so $S_n\geq \mathrm{e}^n$. However, $I_n=\frac{1}{n}(\mathrm{e}^{n\mathrm{e}}-\mathrm{e}^n)$, so that $I_n/S_n$. Then, even with a constant $c$, you get $(S_n-cI_n)/S_n$ converges to $1$. | |
| Jul 13, 2017 at 16:02 | comment | added | M. Dus | Hum, I think that should be true for polynomials, but I'm not sure. | |
| Jul 13, 2017 at 15:28 | comment | added | valle | You are correct. I modified the question. $I_n$ needs to be multiplied by a constant so that it approaches $S_n$ (I think). Please don't delete this answer because it is a good example. And sorry for modifying the question. | |
| Jul 13, 2017 at 15:06 | history | answered | M. Dus | CC BY-SA 3.0 |