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Dec 29, 2011 at 10:53 vote accept asd
Dec 28, 2011 at 19:26 comment added Robert Israel If $f'(p) \ne 0$, it's not a saddle point. The definition of a saddle point requires a stationary point.
Dec 28, 2011 at 14:46 comment added asd Is it possible $f'(z)\neq0$ and $f(z)\neq0$, but $z$ is saddle point of the above type.
Dec 28, 2011 at 14:42 comment added asd @Robert Israel: Yes, it is. but I was meaning that in one direction similar to $x^2$ and in the other direction similar to $x^3$. For example: $f(z)=cos(z)$ in $z=0$ has that condition, $|f|$ in one direction is like $x^2$ and in the other is like $(−x^2)$
Dec 27, 2011 at 20:33 history answered Robert Israel CC BY-SA 3.0