Search Results
| Search type | Search syntax |
|---|---|
| Tags | [tag] |
| Exact | "words here" |
| Author |
user:1234 user:me (yours) |
| Score |
score:3 (3+) score:0 (none) |
| Answers |
answers:3 (3+) answers:0 (none) isaccepted:yes hasaccepted:no inquestion:1234 |
| Views | views:250 |
| Code | code:"if (foo != bar)" |
| Sections |
title:apples body:"apples oranges" |
| URL | url:"*.example.com" |
| Saves | in:saves |
| Status |
closed:yes duplicate:no migrated:no wiki:no |
| Types |
is:question is:answer |
| Exclude |
-[tag] -apples |
| For more details on advanced search visit our help page | |
Results tagged with convolution
Search options not deleted
user 62193
3
votes
1
answer
403
views
Fast computation of convolution integral of a gaussian function
Given a convolution integral
$$
g(y) =\int_a^b\varphi(y-x)f(x)dx=\int_{-\infty}^{+\infty}\varphi(y-x)f(x)\mathbb{I}_{[a,b]}(x)dx
$$
where
$\varphi(x)= \frac{1}{\sqrt{2\pi}}\exp{\left(-\frac{x^2}{2}\right … Perhaps there exists a clever method using
the property that $\varphi$ is a Gaussian function and/or
$g(y)$ is a convolution of two function $\phi$ and $f\cdot \mathbb{I}_{[a,b]}$. …
- 250