Understanding the Second Derivative Test

[theBEAST]

Homework Statement

So fx is how much f changes when you change x. Thus fxx is the rate of change of fx, or geometrically how fast the functions slope is changing. The same can be said for fy and fyy. But what about fxy and fyx? Could someone please explain to me what they mean?

I want to understand what it means to understand the second derivative test:
http://dl.dropbox.com/u/64325990/MATH%20253/Capture.PNG

I am not sure what the meaning of the term [fxy(a,b)]^2 is. I can't seem to visualize what's going on when you take the second derivative test and the textbook doesn't have a proof.

 

[Mark44]

Homework Statement

So fx is how much f changes when you change x. Thus fxx is the rate of change of fx, or geometrically how fast the functions slope is changing. The same can be said for fy and fyy. But what about fxy and fyx? Could someone please explain to me what they mean?

f_xy is the rate of change of f_x relative to a change in y.
f_xy is also written as
$$ \frac{\partial}{\partial y} \frac{\partial f}{\partial x}$$

Note that the two kinds of notation are a little confusing, as the order of x and y is reversed in the two kinds of notation.

I want to understand what it means to understand the second derivative test:
http://dl.dropbox.com/u/64325990/MATH%20253/Capture.PNG

I am not sure what the meaning of the term [fxy(a,b)]^2 is. I can't seem to visualize what's going on when you take the second derivative test and the textbook doesn't have a proof.

For [f_xy(a, b)]²,
1. take the partial of f with respect to x
2. take the partial of f_x with respect to y
3. evaluate the result of step 2 at the point (a, b).
4. square the result of step 3.

For example, if f(x, y) = 2x³y², and we need to evaluate it at (1, 1),
f_x = 6x²y² and f_xy = 12x²y.
f_xy(1, 1) = 12*1*1 = 12

Squaring that result gives you 144.

 

[theBEAST]

f_xy is the rate of change of f_x relative to a change in y.
f_xy is also written as
$$ \frac{\partial}{\partial y} \frac{\partial f}{\partial x}$$

Note that the two kinds of notation are a little confusing, as the order of x and y is reversed in the two kinds of notation.


For [f_xy(a, b)]²,
1. take the partial of f with respect to x
2. take the partial of f_x with respect to y
3. evaluate the result of step 2 at the point (a, b).
4. square the result of step 3.

For example, if f(x, y) = 2x³y², and we need to evaluate it at (1, 1),
f_x = 6x²y² and f_xy = 12x²y.
f_xy(1, 1) = 12*1*1 = 12

Squaring that result gives you 144.

Thank you but I what I really don't understand is what fxy looks like on a graph. What is it's visual representation? If I can understand that then I think I can answer other questions like:

Why you would subtract [f_xy(a, b)]², why not add it.

 

[xaos]

with the partial with respect to x, you are able to extract the flat structure of the function f but only in the x direction. so you're transforming the function f into another function df/dx, and you find its measures the flatness of f but only in the direction of x. now look at df/dx and we want to find the flat structure of this function but now in the y direction. so we create another function from this one that measures the flatness of f in the direction of x and the flatness of df/dx in the direction of y. this function does not necessarily measure flatness in other directions, in fact a derivative may not even exist in other directions.

sorry my cumbersome attempt to describe it.

 

[Mark44]

Thank you but I what I really don't understand is what fxy looks like on a graph. What is it's visual representation?

The graph of z = f(x, y) requires three dimensions: two for the domain, and one for the range.

Using my example of f(x, y) = 2x³y², can you get a graph of that function?
If so, you should also be able to graph f_x(x, y) = 6x²y² and f_xy(x, y) = 12x²y.

If I can understand that then I think I can answer other questions like:

Why you would subtract [f_xy(a, b)]², why not add it.

If I recall correctly, this is directly related to what is known as the Hessian matrix (http://en.wikipedia.org/wiki/Hessian_matrix), whose deteriminant consists of the second partials. I don't remember why this plays a role in categorizing critical points, but it does.

 

[theBEAST]

Alright it is starting to make sense now, thanks everyone.

Does anyone know how to complete the square for this? It looks so complicated :S
http://dl.dropbox.com/u/64325990/MATH%20253/Capture1.PNG

 

[Ray Vickson]

Thank you but I what I really don't understand is what fxy looks like on a graph. What is it's visual representation?

The graph of z = f(x, y) requires three dimensions: two for the domain, and one for the range.

Using my example of f(x, y) = 2x³y², can you get a graph of that function?
If so, you should also be able to graph f_x(x, y) = 6x²y² and f_xy(x, y) = 12x²y.

If I recall correctly, this is directly related to what is known as the Hessian matrix (http://en.wikipedia.org/wiki/Hessian_matrix), whose deteriminant consists of the second partials. I don't remember why this plays a role in categorizing critical points, but it does.

We want to know if the Hessian is positive definite, negative definite, or indefinite. In the first case the point is a strict local min, in the second case a strict local max, and in the third case a saddle point. The determinant is the product of the Hessian's eigenvalues, so if it is > 0 both eigenvalues have the same sign. THAT is why it plays a role.

RGV

 

[HallsofIvy]

In a more abstract, but more general, sense, the "derivative" of a function, f, from R^m to Rⁿ is the linear function, Lf, from R^m to Rⁿ that best approximates f ("best approximates" can be made precise through limit calculations). A linear function from Rⁿ to R^m can be written as an n by m matrix.

So the first derivative of f, from R³ to R is a "3 by 1" matrix or vector- the gradient vector, in fact. And since the first derivative is from R³ to R³, the second derivative is a linear transformation from R³ to R³- which, of course, can be represented by a 3 by 3 matrix- the "Hessian" that Ray Vickerson mentions:
$$\begin{bmatrix}\frac{\partial^2 f}{\partial x^2} & \frac{\partial^2 f}{\partial x\partial y} \\ \frac{\partial^2 f}{\partial y\partial x} & \frac{\partial^2 f}{\partial y^2}\end{bmatrix}$$

Since the "mixed" derivatives are equal, that is a symmetric matrix and there exist a "basis" (coordinate system) in which it is diagonal. That is, there exist x', y' such that the second derivative is
$$\begin{bmatrix}\frac{\partial^2 f}{\partial x'^2} & 0 \\ 0 & \frac{\partial^2 f}{\partial y'^2}\end{bmatrix}$$

And, since the first derivatives are 0 at a critical point, we can write f as
$$f(x,y)= f(x;_0, y'_0)+ \frac{\partial^2 f}{\partial x'^2)(x'- x'_0)^2+ \frac{\partial^2 f}{\partial y'^2}(y- y_0)^2$$
plus higher power terms. If those two derivatives are both positive (at $(x_0, y_0)$) then it f is increases in a neighborhood and so f has a minimum there. If they are both negative, f has a maximum there. If one is positive and the other negative, f has a saddle point.

Note that the determinant of that matrix is
$$\frac{\partial^2f}{\partial x'^2}\frac{\partial^2f}{\partial y'^2}$$
which is positive if they both have the same sign (so increasing or decreasing) and negative if they have opposite signs (so saddle point). Finally, the determinant of a matrix is independent of the change of coordinates so that it is sufficient to check
$$\frac{\partial^2f}{\partial x^2}\frac{\partial^2f}{\partial y^2}$$
without having to actually make that change of variable.

Unfortunately, that doesn't work for dimensions higher than 3. The sign of the determinant, the sign of the product of the diagonal elements, does not tell us much about the sign of the individual terms.