Q on Second partial derivative test for functions of n variables

In summary, the steps to find local maxima/minima of a function are: 1) find all the stationary points 2) form the Hessian matrix and calculate the determinants 3) have the following cases: i) if Di > 0 for i = 1 to n then P is definitely a local minimum point ii) if Di*(-1)^i > 0 for i = 1 to n then P is definitely a local maximum point iii) if Dn = 0 this test cannot help us determine whether the point is a local minimum or maximum iv) in ALL other cases (for example Di = 0 for i other than n or Di with sign other than what i and ii
  • #1
D_Tr
45
4
Hi, I would like to confirm that I have understood this correctly.

The steps to find local maxima/minima of a function f(x1, ... , xn) are:

1) We find all the stationary points.
2) We form the Hessian matrix and calculate the determinants D1, D2... Dn for a stationary point P we want to check.
3) We have the following cases:

i) if Di > 0 for i = 1 to n then P is definitely a local minimum point
ii) if Di*(-1)^i > 0 for i = 1 to n then P is definitely a local maximum point
iii) if Dn = 0 this test cannot help us determine whether the point is a local minimum or maximum
iv) in ALL other cases (for example Di = 0 for i other than n or Di with sign other than what i and ii indicate) we definitely have a saddle point.

Are iii and iv correct? More specifically, I would like a clarification on what exactly happens when we have one or more zero Di.

Thanks in advance for your time.
 
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  • #2
what you have as the conditions is right. When the determinant is zero, the test fails. This is to say that we might have a maximum, minimum or saddle but the test will not help us identify such a critical point. You might have to plot the function or see some values to be able to identify critical points. My old calculus book had this question which you might want to do and see what happens when the test fails:
Consider
[tex] f(x,y)=6xy^{2}-2x^{3}-3y^{4} [/tex]

You can plot the function in Mathematica, Matlab, Maple or any programs and see the behavior. Another way to be able to see this is to see the proof for why these conditions come about and see why the test fails. I hope this was clear.
 
  • #3
Thanks! I wanted a confirmation that I got the conditions right. I will study the example you gave; I just plotted it with wolfram alpha and it seems interesting!
 

FAQ: Q on Second partial derivative test for functions of n variables

1. What is the second partial derivative test for functions of n variables?

The second partial derivative test is a method used to determine the nature of a critical point (maximum, minimum, or saddle point) for a function with multiple variables. It involves calculating the second partial derivatives of the function and evaluating them at the critical point.

2. How do you determine the critical points for a function of n variables?

To determine the critical points for a function of n variables, you need to find the values of the variables that make all the first partial derivatives equal to zero. This will give you a set of equations, which can be solved to find the critical points.

3. What is the significance of the second partial derivative test?

The second partial derivative test is significant because it allows us to determine the nature of a critical point without having to graph the function. This is especially useful for functions with a large number of variables, where graphing is not feasible.

4. Are there any limitations to the second partial derivative test?

Yes, there are limitations to the second partial derivative test. It can only be used to determine the nature of critical points for functions of n variables that have continuous second partial derivatives. It also cannot be used to determine the global maximum or minimum of a function.

5. Can the second partial derivative test be extended to functions with more than n variables?

Yes, the second partial derivative test can be extended to functions with more than n variables. However, the calculations become more complex and the test may not always be applicable. In such cases, other methods, such as the Hessian matrix, may be used to determine the nature of critical points.

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