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 20598
Nonlinear objectives, nonlinear constraints, non-convex objective, non-convex feasible region.
4
votes
Maximize $f(0)+\cdots+f(n-1)$ subject to $f(x)f(y) + f(x+y) \leq 1$
Gerhard Paseman's solution is correct, and shows that $f \equiv 1/\phi$ is the unique maximizer of $d_n$ for each $n\geq 0$. Just reproducing here in as few words as I can, for the purpose of succinct …
- 9,720
10
votes
Elementary inhomogeneous inequality for three non-negative reals
Write $x = 1-X$, $y=1-Y$, $z=1-Z$. Then the inequality reduces to
$$2XYZ \leq X^2 + Y^2 + Z^2$$
for $X, Y, Z \leq 1$. If $X, Y, Z < 0$ then the inequality is trivial, since LHS < 0. Otherwise suppose …
- 9,720