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#1 2008-10-16 09:25:23

mikau
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Registered: 2005-08-22
Posts: 1,504

second derivative in multivariable functions

my book lost me with this introduction of the second derivative:

to define an analogue for the second derivative, let us consider the following. Let


be differentiable on R[sup]n[/sup].
We know that for for fixed

that

is a linear transformation from R[sup]n[/sup] to R[sup]m[/sup].
This means that we have a function T:


where L(R[sup]n[/sup], R[sup]m[/sup]) denotes the space of linear transformations from R[sup]n[/sup] to R[sup]m[/sup]

Hence if we differentiate T at x[sub]0[/sub] again, we obtain a linear transformation:


from R[sup]n[/sup] to L(R[sup]n[/sup], R[sup]m[/sup])

And there you lost me...  I don't understand exactly how we are taking the derivative of a linear mapping, which to my understanding, is just a matrix thats a function of x[sub]0[/sub].

Particularly because of my following observation:
let f: R[sup]n[/sup] to R[sup]m[/sup].
Then the derivative of f is an m x n matrix such that when right multiplied by an element of the domain (an n x 1 vector) you obtain an element of the target set (an m x 1 vector).

Likewise, if T: R[sup]n[/sup] to M[sub] m x n[/sub] (meaning the space of mxn matrices, or equivalantly the space of linear maps from R[sup]n[/sup] to R[sup]m[/sup]), then the derivative of f should (by my understanding) be a matrix such that when right multiplied by an (n x 1) vector in the domain, gives you an m x n matrix. But obviously, right multiplication of any matrix by an n x 1 vector cannot produce an m x n matrix.

Hence, I lose grasp on the meaning of the derivative in this context. By my understanding, the derivative must produce a linear map from the departure space to the arrival space, but the linear map (appearantly) cannot be represented by a matrix, which simply can't be true!

Can someone clarify whats wrong with my reasoning here?

Last edited by mikau (2008-10-16 09:52:19)


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#2 2008-10-16 14:28:59

Ricky
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Registered: 2005-12-04
Posts: 3,791

Re: second derivative in multivariable functions

Edit: The derivative at a point is a linear transformation.  The derivative as a matrix of partials is not.

Remember that a linear map is just a special type of map, or rather, function.  As such, it can be differentiated like any other function.

A derivative is a matrix of partials, and if you plug in a point x_0, you get a matrix of real numbers.  It is the real numbers which is an element of L(R^n, R^m).  This is why the derivative T is a function from R^n to L(R^n, R^m).  We need to plug in a point from R^n to get a matrix of numbers, an element of L(R^n, R^m).  A matrix is basically the slope in multiple directions, and all this is saying is that the slope can change at each point.

When studying derivatives in higher dimensions, it makes sense to see what would happen in the 1-dimensional case.  Take f(x) = x^2.  The matrix of partials is simply [2x].  The value of this matrix at x_0 = 2 is [4].  The linear map this defines is:

f'(x) = [4][x] = 4x

It is at this point which I feel you might be confused.  Remember that when we have our linear map represented by a matrix A, it is a linear transformation that sends x to Ax.  In this case, A = [4], so we send x to 4x.

Now what is the derivative in this case?  Simply 4.  In general, the derivative of a linear transformation is the matrix which represents that transformation.  In the 1-dimensional case, the "slope" of a linear transformation will be a constant.  In the general case, the slope of a linear transformation will be a matrix, which is like a constant but in multiple dimensions.  That is to say that the derivative of a linear transformation at any point x is equal to the derivative at any other point, i.e. constant.


"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

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#3 2008-10-16 17:38:08

mikau
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Registered: 2005-08-22
Posts: 1,504

Re: second derivative in multivariable functions

Thanks for the response, Ricky. Its pretty late right now and i don't have the time or energy to read it now, but I'll check it tomorrow. Much appreciated!


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#4 2008-10-18 06:08:02

mikau
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Registered: 2005-08-22
Posts: 1,504

Re: second derivative in multivariable functions

as far as I can see, what you did was take a function:

f(x) = x^2,

took the derivative mapping:

f'(x) = [2x]

evaluated it at 2,

f'(2) = 4,

and constructed a new mapping from this:

g(x) = 4x

and took the derivative of this to get 4 again.
This seems, in no way, analogous to second, or even first derivative. I chose g(x), but you wrote f'(x) = 4x after saying f(x) = x^2, and that just makes me want to scream.

Furthermore, I'd have said the second derivative of f(x) = x^2 was 2, not 4.

"Remember that a linear map is just a special type of map, or rather, function.  As such, it can be differentiated like any other function."
You see, the problem is i see T as a function that creates a linear map, not a linear map itself, and hence the notion of derivative escapes me. T, to my understanding, maps to a matrix, but to make a linear mapping (call it g), we need to grab the resulting matrix at a particular point, and say g(x) = Ax.

moreover, since the derivative of a linear map is just the linear map itself, it seems by this reasoning that the 2nd, 3rd, 4th, n'th derivative, with n > 1, is just A.

"A derivative is a matrix of partials, and if you plug in a point x_0, you get a matrix of real numbers.  It is the real numbers which is an element of L(R^n, R^m).  This is why the derivative T is a function from R^n to L(R^n, R^m)."

see, this is the part i don't get. I thought it said T was the function that produces an element of L(R^n, R^m). T maps x to  D[sub]x[/sub](f), so we plug in a value of x, and we get a mapping in L(R^n, R^m). Makes sense, but i don't see why the derivative of T would be a function from R^n to  L(R^n, R^m).


Goodness, I'm confounded..

Last edited by mikau (2008-10-18 06:32:07)


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#5 2008-10-18 14:38:05

Ricky
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Registered: 2005-12-04
Posts: 3,791

Re: second derivative in multivariable functions

Ok, I think I got confused by:

And there you lost me...  I don't understand exactly how we are taking the derivative of a linear mapping...

Remember that T is not a linear mapping.  I thought you were talking about taking a derivative of the linear mapping at a point.

Hence if we differentiate T at x0 again, we obtain a linear transformation:

from Rn to L(Rn, Rm)

This is not correct.  T is a function from R^n to R^(n*m).  Thus, taking a derivative we get a function from R^n into L(Rn, R(n*m))

Take the 2 dimensional example, f(x,y) = z.  The first derivative will consist of two partials, fx and fy in a matrix with a single row.  The 2nd derivative will consist of four partials, fxx, fxy, fyx, fyy, in a 2x2 matrix.


"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

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#6 2008-10-27 07:33:51

mikau
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Registered: 2005-08-22
Posts: 1,504

Re: second derivative in multivariable functions

Thanks, Ricky. I didn't have time to look at this till just now. Trying to spend time actually understanding the underlying theory in this course has just been holding me back, as my teachers unpublished textbook is so riddled with error, I don't have time to clarify every detail and keep up with the pace of the course. It really makes me sad.. topics being presented for the first time really need to be straight in order to make sense. You can't learn this way...

i 'll have to buy a published textbook on vector calc and teach myself over the winter i guess. Seems to be the only proper way to learn anything these days..

But I will do my best to understand what you just explained, thank you for that.

Last edited by mikau (2008-10-27 07:34:53)


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#7 2008-10-27 07:46:34

Ricky
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Registered: 2005-12-04
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Re: second derivative in multivariable functions

i 'll have to buy a published textbook on vector calc and teach myself over the winter i guess.

Remember that Vector Calculus is typically different than higher dimensional analysis.  While they would give you the definition of higher order derivatives in a vector calc book, the problems would be way different.


"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

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#8 2008-10-27 08:22:38

mikau
Member
Registered: 2005-08-22
Posts: 1,504

Re: second derivative in multivariable functions

Incidentally, remember how i asked you what the difference between Vector and Multivariable Calculus was?? well, for some reason my school (attempts) to teach both in one course. So maybe the problems wouldn't be so different? I don't know.

Anyway, so from the above discussion, we have the derivative of T would be an (m*n)x(n) matrix, which i presume we could multiply by an nx1 vector (the point in question) to get an (m*n)x1 vector, which techinically represents an m x n matrix ( i guess?). And we could multiply this by an nx1 vector again to obtain a vector in R^m. Do these operations mean anything?

I ask because my book points out that we can right multiply D[sub]x0[/sub](T) = D[sup]2[/sup][sub]x0[/sub](f) first by x1 in R^n to obtain an element of L(R^n, R^m), and then right multiply again by x2 in R^n, to obtain an element in R^m, and thus construct a bilinear mapping B:

     R[sup]n[/sup] x R[sup]n[/sup] -> L(R[sup]n[/sup], R[sup]m[/sup])
B: (x1, x2) -> D[sup]2[/sup][sub]x0[/sub](f)(x1,x2)

whats does this mapping represent? and um... shouldn't B go to R^m? neutral

Last edited by mikau (2008-10-27 08:53:17)


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#9 2008-10-27 17:58:23

Ricky
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Registered: 2005-12-04
Posts: 3,791

Re: second derivative in multivariable functions

Anyway, so from the above discussion, we have the derivative of T would be an (m*n)x(n) matrix, which i presume we could multiply by an nx1 vector (the point in question) to get an (m*n)x1 vector, which techinically represents an m x n matrix ( i guess?). And we could multiply this by an nx1 vector again to obtain a vector in R^m. Do these operations mean anything?

It would be the derivative of T at x0 is an (m*n)x(n) matrix.  This represents our linear transformation from R^n to R^m x R^n.  Multiplying by a vector in R^n, we are applying this map to the vector, winding up in R^m x R^n.  Remembering that in the first derivative case, our transformation sent things so that when we had h sufficiently small, our linear transformation would send h to be close to the value of f(x0+h) - f(x0).  Here, our f is really T.  So what we are doing by plugging in values to the linear transformation from R^n to R^mxR^n is really the same thing.  When h is sufficiently small, it represents the difference between T(x0 + h) and T(x0).  This is why we get an R^m x R^n matrix, because the change between T(x0 + h) and T(x0) is of course the difference in each component of the matrix that represents our first derivative.

whats does this mapping represent?

I'm not certain it represents anything that would be from a calculus standpoint, intuitive.

and um... shouldn't B go to R^m?

Indeed it does, and that's exactly what you wrote.

D2x0(f)(x1,x2)

This means to multiply D2x0(f) by x1, and then by x2.  In general, we can represent an binary operation as a mapping, and that's all they did.  In this case, we are thinking of D2x0(f) as:

D2x0(f) : R^n x R^n -> R^m by D2x0(f)(x1,x2) = (D2x0(f).(x1)).(x2)


"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

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