Finding Max and Min Extremes of a Function with Second Derivatives Equal to Zero

In summary, the conversation discusses finding the max and min of a function with multiple variables. The speaker suggests differentiating the function with respect to x and y, setting them equal to zero, and then using the determinant of the hessian matrix to determine further calculations. The speaker also mentions that switching to a different coordinate system may provide better insight into the behavior of the function.
  • #1
NODARman
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TL;DR Summary
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What should I do when the f(x, y) function's second derivatives or Δ=AC-B² is zero? When the function is f(x) then we can differentiate it until it won't be a zero, but if z = some x and y then can I just continue this process to find what max and min (extremes) it has?

What I've done is:
Differentiated z=f(x, y) by partial derivatives with respect to x and y;
Made them equal to zero in the system of numbers;
Where I got x=0, y=0;
Differentiated function by the second partial derivatives;
Used Δ = AC-B² where if Δ=0 then it needs more calculations.
I got Δ=0.

What should I do now? Differentiate again (third partial) as we do to function f(x) with only one variable x until it won't be equal to zero?
 
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  • #2
What the determinant of the hessian matrix of [itex]f[/itex] vanishing tells you is that is has a zero eigenvalue with a corresponding eigenvector [itex]\mathbf{v}[/itex]. To determine the behaviour of [itex]f[/itex] in this direction, you should look at [tex]
\frac{d^3}{dt^3}f(\mathbf{x}_0 + t\mathbf{v})[/tex] and higher derivatives until you find one which does not vanish. If both eigenvalues are zero then the hessian is zero, and you need to look at the cubic terms to see what is going on.

Switching to a different coordinate system may give better insight into what is going on, for example https://www.wolframalpha.com/input?i=(x^2+y^2)*(1+-+(x^2+y^2)) where in polar coordinates it is easy to find that [itex]r^2(1-r^2)[/itex] has a ring of global maxima at [itex]r = 1/\sqrt{2}[/itex] and a local minimum at [itex]r = 0[/itex].
 
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Related to Finding Max and Min Extremes of a Function with Second Derivatives Equal to Zero

1. What is the purpose of the function min and max?

The function min and max is used to find the minimum and maximum values in a set of numbers or a range of values. It can be used to compare values and determine the smallest or largest value.

2. How do I use the function min and max in my code?

To use the function min and max, you can pass in a set of numbers or a range of values as parameters. For example, min(5, 10, 2) would return 2 as the smallest value, while max(5, 10, 2) would return 10 as the largest value.

3. Can I use the function min and max on non-numeric values?

Yes, the function min and max can be used on non-numeric values such as strings. However, it will compare the values based on their ASCII codes, so it may not always give the expected result.

4. Is there a limit to the number of parameters I can pass into the function min and max?

No, there is no limit to the number of parameters you can pass into the function min and max. You can pass in as many values as you need to compare and the function will return the smallest or largest value.

5. Are there any other functions that are similar to min and max?

Yes, there are other functions that are similar to min and max such as Math.min() and Math.max(). These functions are part of the Math object in JavaScript and can be used to find the minimum and maximum values in a set of numbers.

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