Timeline for Approximate sum by an integral: valid or not?
Current License: CC BY-SA 3.0
15 events
| when toggle format | what | by | license | comment | |
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| Jul 21, 2017 at 12:41 | comment | added | valle | I am trying to solve this problem by approximating it with this one. | |
| Jul 16, 2017 at 11:41 | comment | added | fedja | @becko Yes, we can. Let's say $f$ attains its maximum at $0$ (and $0$ only). If $f'(0)=0$, then it is still $1$. Otherwise it is the same as for $f(x)=f'(0)x$, $g(x)=1$, as M.Dus heas already observed. However the convergence can be slow now. If not a secret, what are you really after with all that? | |
| Jul 16, 2017 at 8:13 | comment | added | valle | Can we compute the proportionality constant between the sum and the integral in general, even if the max of $f$ is at an endpoint? | |
| Jul 15, 2017 at 20:03 | comment | added | valle | True. Thanks! I'll get back to you on the endpoint issue. | |
| Jul 15, 2017 at 19:49 | comment | added | fedja | @becko A negative continuous function on a compact set is separated from $0$, isn't it? | |
| Jul 15, 2017 at 19:46 | comment | added | valle | Also, $f$ is not Gaussian. That was a mistake, sorry. Let me see if I can formulate the problem more better to see what kind of generalization we can do if the max occurs at the endpoint. I'll get back to you on that. | |
| Jul 15, 2017 at 19:42 | comment | added | valle | I know that $f<0$ on $(-2\delta , 2\delta)$, $x\ne0$, but nothing precludes $f$ from getting arbitrarily close to $0$ even if $x\ne0$. That's why I'm not sure about the bounding from $-2\delta$ to $-\Delta$. | |
| Jul 15, 2017 at 18:40 | comment | added | fedja | @becko We have a cutoff $\psi$, remember? So $-\infty$ is actually $-2\delta$ and we have assumed that $f<0$ on $(-2\delta,2\delta)$ except at $0$. If your $f$ and $g$ are both even, you still can pretend that you integrate between $-1$ and $1$ and have the exponential decay if you introduce the extra factor $1/2$ for the $0$ endpoint in your Riemann-type sum. If $g$ is not even, you may be out of luck; I have to check. | |
| Jul 15, 2017 at 11:47 | comment | added | valle | Turns out I now have a case where the max of $f$ occurs at an endpoint. If you need to know, my $f$ is a Gaussian, and the mean is now at $0$ (before it was between $0$ and $1$. Will I lose the exponential decay in this case? I assume the conclusion in this case does not depend on $g$, correct? | |
| Jul 15, 2017 at 11:41 | comment | added | valle | I do not understand how you bound the integrand in the new contour, from $-\infty$ to $-\Delta$, and from $\Delta$ to $\infty$. | |
| Jul 14, 2017 at 15:43 | comment | added | fedja | @becko Indeed. I corrected. I suspect there are more stupid misprints though :-) | |
| Jul 14, 2017 at 15:42 | history | edited | fedja | CC BY-SA 3.0 |
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| Jul 14, 2017 at 15:24 | comment | added | valle | Thanks! I have to digest this. Any chance of looking into the case where the max of $f$ occurs at $0$ or $1$? In this case we probably need a proportionality constant between the sum and the integral that is $\ne 1$. | |
| Jul 14, 2017 at 15:21 | history | edited | fedja | CC BY-SA 3.0 |
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| Jul 14, 2017 at 15:08 | history | answered | fedja | CC BY-SA 3.0 |