2
$\begingroup$

I am trying to solve the following problem:

Let $f: \mathbb R^2\rightarrow\mathbb R$ be a function defined by $$ f(x,y) = x^{2n} + y^{2n} - nx^2 + 2nxy - ny^2, $$ where $n$ is a natural number greater than 1. Decide whether $f$ has a (global) minimum. Also find all the points at which $f$ attains its local maxima and local minima.

I calculated the partial derivatives $f_x = 2nx^{2n-1}-2nx+2ny$, $f_{xx} = 2n(2n-1)x^{2n-2}-2n$, $f_{xy} = 2n$, and so on. But I got an equation system $f_x = f_y = 0$ of degree three, which I had hard times solving it. I found out that $(x,y) = (0,0)$ is one of its solutions, but at that point the the determinant of the Hessian is zero, from which I could not conclude whether it was a local extremum.

I was not sure about how to prove that the minimum value of $f$ exists, either.

I would be most grateful if you could help me solve this problem.,

$\endgroup$
  • $\begingroup$ @Amzoti Done. I also added a condition for $n$, which I forgot to write about. $\endgroup$ – Semo Ponume Aug 25 '13 at 6:55
2
$\begingroup$

The point $(0,0)$ is a saddle point for $n>0$.

To see this, look at the function along the lines $y=x$ and $y=-x$: $$y=\ \ x\quad \ \to \quad f(t) = 2 x^{2 n}\quad\quad\quad\quad\quad\ \ \ $$ $$y=-x\quad \ \to \quad f(t) = (-x)^{2 n} - 4 n x^2 + x^{2 n}$$

Now take the second derivative on the line $y=-x$ to get $-8n<0$, and take the $2n$-th derivative on the line $y=x$ to get $2(2n!)>0$, thus showing this point is a saddle point and not a local maximum or minimum.

To find the minima, note that $f_x =0, f_y=0$ leads to: $$2 n x^{2 n-1} = -2 n y^{2 n-1} \quad \to \quad y = -x$$ Using the fact that $2n-1$ is odd. Now plug this into $f_x=0$ to get:

$$x\to \pm2^{-1/(2-2 n)}, \ y=-x$$ You can verify these are indeed minima.

In the graphs below:

$f(x,y)$ along the lines $y=x$ and $y=-x$ for $n=3$.

enter image description here

$\endgroup$
  • $\begingroup$ Thank you for your helpful explanation and figures. How did you find the directions in which $f$ behaves like this? Also, I would appreciate if you could help me with the other extrema than (0,0)? (I cannot vote up because my reputation is low.) $\endgroup$ – Semo Ponume Aug 25 '13 at 7:10
  • $\begingroup$ @SemoPonume - you can always vote up, as well as accept answers. I found the directions by noting the symmetry in xy, and I added the formula for the minima. $\endgroup$ – nbubis Aug 25 '13 at 7:54
  • 1
    $\begingroup$ Silly me! It seems that math.se does not allow users of reputation less than 15 to vote up. I accepted your answer, though. If somebody vote up to my comments I can vote up your answer. $\endgroup$ – Semo Ponume Aug 25 '13 at 10:02
  • 1
    $\begingroup$ @user64494 I do not see any problem with it. Could you please elaborate on that? $\endgroup$ – Semo Ponume Aug 25 '13 at 11:04
  • 1
    $\begingroup$ @nnubis: You are right. $\endgroup$ – user64494 Aug 25 '13 at 11:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.