# Proof or disproof a problem about critical points

There is a question im trying to solve, but im not sure im doing the right thing.

"Let $p(x) = \sum_{j=0}^na_jx^j$ a polynomial without multiple roots, then all critical points of $f(x,y)=cosh(x)+y\cdot p(x)$ are saddle points."

My task is to proof or disproof this statement.

I tried this:

Write $p(x)=(x-r_1)(x-r_2)\ldots(x-r_n)$, where each $r_i$, $i=1\ldots n$, are the roots of $p(x)$. Considering $v=(v_1,v_2)\in\mathbb{R^2}-\{(0,0)\}$ and the Hessian form $$H(x,y)v^2 = \frac{\partial^2f}{\partial x\partial x}v_1v_1+2\frac{\partial^2f}{\partial x\partial y}v_1v_2+\frac{\partial^2f}{\partial y\partial y}v_2v_2$$

if $H$ is indefinite for all critical points, then the statement is true, otherwise is false.

So, computing the partial derivates and the critical points i found $$H(r_i,y_i)v^2=\big(cosh(r_i)+y_i(2a_2+\ldots+n(n-1)a_nr_i^{n-2})\big)v_1^2+(a_1+2a_2+\ldots+na_nr_i^{n-1})v_1v_2$$

where the points $(r_i,y_i)$, $i=1\ldots n$, are the critical points($r_i$ is a root of $p(x)$).

I think i can prove that the statement is true by choosing $v_1,v_2$ properly and analyzing the signals, but im not sure how and im not even sure if this is right. Any help is welcome, thanks!

-

It is easy to see that we have exactly $\deg p$ critical points: if $r_1$, $\dots$, $r_n$ are the roots, they are the points $(r_i,-\frac{\sinh r_i}{p'(r_i)})$. The Hessian looks like $\begin{pmatrix}\text{something} & p'(x) \\ p'(x) & 0\end{pmatrix}$ and it obviously has negative determinant at each point $(x,y)$ where $p'(x)\neq0$, like the critical points. The critical points are, therefore, saddle points.
... where $p'(x) \ne 0$, you mean. – Robert Israel Sep 14 '12 at 19:45
I have a question. While reading about the Second partial derivative test, they say that if the determinant at $(x_0,y_0)$ is $<0$ then $(x_0,y_0)$ is a saddle point, but in this case the determinant is zero at the critical points. I dont get how did you conclude that they are saddle points. – Integral Sep 14 '12 at 20:44
The determinant is not zero at the critical points: as I wrote, if $(x,y)$ is a cvritical point, the determinant of the Hessian matrix there is $-p'(x)^2$, which is a strictly negative number because $x$ is a simple root of $p$ and therefore $p'(x)\neq0$. – Mariano Suárez-Alvarez Sep 14 '12 at 20:46
I confused the roots of $p(x)$ and $p'(x)$, my fault. Sorry. And thanks. – Integral Sep 14 '12 at 20:49