0

I am studying the second partial derivative test and it says when determinant of the hessian matrix is $0$ then there is no conclusion. However I saw on a website a method that tells you how to workout what kind of point is the one with zero determinant. Given an initial function $f$, it says to write a new function $f_1=f-c$, where $c=f(x_0,y_0)$. Then, the rule states what follows. If there is a neighborhood of $(x_0,y_0)$ where $f_1$ is negative in every point except $(x_0,y_0)$ then $(x_0,y_0)$ is a local maximum. If there is a neighborhood of $(x_0,y_0)$ where $f_1$ is positive in every point except$(x_0,y_0)$ then $(x_0,y_0)$ is a local minimum. Lastly, if in every neighborhood of $(x_0,y_0)$ $f_1$ has both positive and negative values then $(x_0,y_0)$ is a saddle point. However, I couldn't verify if the method is correct, is there a proof of this result?

Shaun Gagnier
  • 1
  • What is your definition of local maximum and local minimum? Because if you read through that, then read through this test, I'd be willing to bet they line up almost word for word. – Arthur Jul 23 '19 at 12:28
  • Actually it makes super sense like you phrased it. My problem is for example I apply it to z=x^2*y and I find (0,y) with y>0 local minima and (0,y) with y<0 as local maxima but softwares like wolfram dont give me that result – Shaun Gagnier Jul 23 '19 at 12:31

1 Answers1

1

No proof needed, since that's not a “rule”, it's the definition of what the words “(strict) local maximum” (etc.) mean!

Hans Lundmark
  • 53,395
  • Thanks, may I ask a question then? When applying this method to a function like z=x^2*y I find that all points (0,y) with y >0 are local minima while all points (0,y) with y<0 are local maxima. However when trying to compute local extrema with wolframalpha or other softwares the only thing I find is the saddle point (0,0) which I found but I don't understand why maxima and minima aren't coming out of the result. – Shaun Gagnier Jul 23 '19 at 12:29
  • 1
    This is not necessarily the definition. It's pretty darn close, but personally I haven't before seen any definition which first subtracts a constant value, then looks at the sign of the values in a neighborhood. It's basically the difference between saying $x=y$ and saying $x-y=0$, but still. One requires arithmetic and ordering on the codomain, the other only ordering. – Arthur Jul 23 '19 at 12:30
  • I don't know it seemed strange to me too, however I didn't find many descriptions on what to do when determinant is 0, but when applying it I'd say it works but why doesn't the software give me that result? – Shaun Gagnier Jul 23 '19 at 12:36
  • @Arthur: Fair enough. But to me, $f(x,y)>f(x_0,y_0)$ is so obviously the same thing as $f(x,y)-f(x_0,y_0)>0$ (for real-valued functions) that I didn't even notice the difference! – Hans Lundmark Jul 23 '19 at 12:53
  • @ShaunGagnier: You're right about $z=x^2 y$. WA might not be clever enough to figure this out, or it maybe it only searches for strict local extrema. – Hans Lundmark Jul 23 '19 at 12:58
  • To be precise, the definitions you give in the questions are for strict local extrema, if by “positive” you mean “greater than zero”. To just define “local minimum” one usually only requires the difference to be nonnegative in a neighbourhood of the point. – Hans Lundmark Jul 23 '19 at 13:00
  • Yes by positive I meant greater than zero. Not sure why WA doesn't find that so I just wanted to make sure – Shaun Gagnier Jul 23 '19 at 13:03