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 nonlinear-optimization
Search options answers only
not deleted
user 4312
Nonlinear objectives, nonlinear constraints, non-convex objective, non-convex feasible region.
37
votes
Accepted
How to prove $e^x\left|\int_x^{x+1}\sin(e^t) \,\mathrm d t\right|\le 1.4$?
Integrate by parts:
\begin{align}
\int_x^{x+1}\sin(e^t)dt
& =\int_x^{x+1}e^{-t}d(-\cos(e^t)) \\
& =e^{-x}\cos e^x-e^{-x-1}\cos e^{x+1}-\int_x^{x+1}e^{-t}\cos e^{t}dt\\
& =e^{-x}\cos e^x-e^{-x-1}\cos e …
- 99k
4
votes
Accepted
Proving an infinite norm minimization problem has finite support (non-convex p-norms)
If $p=1$, $N=1$ and $a_1=(1/2,2/3,3/4,4/5,\ldots)$, the infimum equals 1 and is not achieved on a finitely supported vector (moreover, it is not achieved at all).
However if $0<p<1$ and the minimize …
- 99k
2
votes
Accepted
Maximizing the distance sum of some points inside a circle
Let 0 be the centre of your circle of radius $r$.
If $a=2$, we expand the brackets and write $\sum_{i,j} (p_i-p_j)^2=2n\sum p_i^2-2(\sum p_i)^2\leqslant 2nr^2$, with equality if and only if all points …
- 99k
8
votes
Accepted
Elementary inhomogeneous inequality for three non-negative reals
Denote $x^2=a^3,y^2=b^3,z^2=c^3$. By AM-GM we have $1+2xyz=1+(abc)^{3/2}+(abc)^{3/2}\geqslant 3\sqrt[3]{1\cdot (abc)^{3/2}\cdot (abc)^{3/2}}=3abc$, so LHS is not less then $$a^3+b^3+c^3+3abc\geqslant …
- 99k
5
votes
Accepted
Bound on the sum of arguments
Let's prove that
$$
\arg \frac{1-zf(s-u)}{1-zf(s+u)}< \pi/2- \arg(1-z\bar{z}f(2u)).
$$
Then summing this up with an analogous inequality
$$
\arg \frac{1-zf(t+u)}{1-zf(t-u)}< \pi/2- \arg(1-z\bar{z}f(- …
- 99k