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mikau

Member

Registered: 2005-08-22

Posts: 1,504

derivatives in multiple variables

I'm having trouble understanding this proof in my book. The book is written by my teacher and free to copy and distribute, so i just scanned the portion I'm stuck on.

The proof is to show that the linear map which we call the derivative is unique.

Here are the relevant portions from the book (sorry for the low quality):

if those pics don't look right, try these URLS:
http://i21.photobucket.com/albums/b299/mikau16/derivative0.jpg
http://i21.photobucket.com/albums/b299/mikau16/derivative2.jpg

What throws me is the part I put question marks around, when they suddenly plug v into the linear map, and claim that the difference between D[sub]a[/sub](f)(v) - L(v) = etc...
I have no idea where that observation came from! No idea whatsoever.

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A logarithm is just a misspelled algorithm.

Ricky

Moderator

Registered: 2005-12-04

Posts: 3,791

Re: derivatives in multiple variables

There has to be a typo.  First off, Baby Rudin uses the same proof but he has:

for just after it says "by the triangle inequality".  Furthermore, rewrite the thing you have question marks around by canceling terms and taking a limit on both sides shows that there is no need for the triangle inequality.  Finally, no where is linearity of the derviative explicitly used.  That's typically how the proof goes.  Prove it for small h, then use linearity to extend it to all vectors.

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"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..."

mikau

Member

Registered: 2005-08-22

Posts: 1,504

Re: derivatives in multiple variables

I think i follow you...

You're providing various reasons why the proof doesn't make sense, if i'm not mistaken. (though I don't think I fully understand them all)

If you could show me, or link me too a correct version of this proof, that would be nice!

Last edited by mikau (2008-09-24 02:09:25)

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A logarithm is just a misspelled algorithm.

Ricky

Moderator

Registered: 2005-12-04

Posts: 3,791

Re: derivatives in multiple variables

Stolen from Baby Rudin, with a bit more explanation.

Let A_1 and A_2 suffice as derivatives in the definition of "differentiable", and let B = A_1 - A_2.  Then we have:

Applying the norm (and subsequently, triangle inequality), we get:

This shows that:

Where we divide both sides by |h| and then use the definition "differentiable" for both A_1 and A_2.  Fix h to be any nonzero vector, then we have that:

Since again, that would make th go to 0.  But the derivative is linear and constants can be "factored out" of norms, so:

Shows that this is actually independent of t, and therefore, Bh = 0, making B = A_1 - A_2 = 0, so A_1 = A_2.

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"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..."

mikau

Member

Registered: 2005-08-22

Posts: 1,504

Re: derivatives in multiple variables

That was awesome! And easy to understand!

My only problem is where you used the triangle inequality. Shouldn't there be a plus in between the norms on the right side, and not a minus?

Last edited by mikau (2008-09-24 11:07:09)

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A logarithm is just a misspelled algorithm.

mikau

Member

Registered: 2005-08-22

Posts: 1,504

Re: derivatives in multiple variables

I think I may have a weak understanding of limits thats causing me to struggle with derivative mappings as a whole. 

Let f : R[sup]n[/sup] -> R, n ≥ 1, f(x) = ||x|| be the usual norm in R[sup]n[/sup]

prove that:

using simple algebra (particularly be multiplying above and below by the conjugate of the numerator), I can show:

now I can see the idea is that the denominator approaches 2||x|| but in order to complete the proof, we need to kill the h's in the denominator (by setting the equal to zero) and leave h in place in the numerator, giving us the form:

which is of the form:
L(h) + o(||h||), with L being a linear mapping. However, we took h to zero in the denominator, and left it in place in the numerator. This makes me uncomfortable and I'm not sure if this is even legal. I seem to remember a quotient rule for limits in real numbers, but i don't know if this is legal for dot products of vectors in R[sup]n[/sup]

Last edited by mikau (2008-09-25 09:33:00)

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A logarithm is just a misspelled algorithm.

Ricky

Moderator

Registered: 2005-12-04

Posts: 3,791

Re: derivatives in multiple variables

My only problem is where you used the triangle inequality. Shouldn't there be a plus in between the norms on the right side, and not a minus?

Indeed, it should.  But notice how that doesn't cause a problem in the proof since 0 + 0 = 0.

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"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..."

Ricky

Moderator

Registered: 2005-12-04

Posts: 3,791

Re: derivatives in multiple variables

You're right that you can't just have "some" h's go to 0.

L(h) + o(||h||), with L being a linear mapping.

Not just a linear mapping, but L must be the derivative.  Put this in, and then your result (the limit) should pop right out.

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"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..."

mikau

Member

Registered: 2005-08-22

Posts: 1,504

Re: derivatives in multiple variables

I'm not sure I follow you. As far as I can tell, i don't have a mapping L to put in the limit until I bring the h's in the denominator to 0.

Oh, and i was kind of wondering, can we use the quotient rule for limits with vectors?

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A logarithm is just a misspelled algorithm.

Ricky

Moderator

Registered: 2005-12-04

Posts: 3,791

Re: derivatives in multiple variables

I'm not sure I follow you. As far as I can tell, i don't have a mapping L to put in the limit until I bring the h's in the denominator to 0.

That's what you want to show, so plug it in then prove that the linear mapping you're given is the derivative by having the limit go to 0.

Oh, and i was kind of wondering, can we use the quotient rule for limits with vectors?

I'm not entirely sure what you mean.  Are you referring to the quotient rule for derivatives and applying it somehow to limits?

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"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..."

mikau

Member

Registered: 2005-08-22

Posts: 1,504

Re: derivatives in multiple variables

Perhaps I'm using the wrong name, but I seem to remember, as one of the limit laws:

I'm just not sure if we can apply this as a general rule, particularly, in the context of vectors.

That's what you want to show, so plug it in then prove that the linear mapping you're given is the derivative by having the limit go to 0.

you mean plug in the linear mapping they presented me with in the first place and show it works, without deriving it?

well that would work, certainly. However the solution in the answer key of my textbook, obtains it from scratch by the same process I did. But it does what made me uncomfortable, taking the limit in the denominator, and not the numerator. Thats why I wondered if they were applying some sort of quotient rule for limits.

Last edited by mikau (2008-09-26 01:38:28)

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A logarithm is just a misspelled algorithm.

Ricky

Moderator

Registered: 2005-12-04

Posts: 3,791

Re: derivatives in multiple variables

Remember that when we talk about limits of vectors, we talk about limits of each of the components.  Thus, a whole lot of properties carry over from single dimensional analysis, including this one.  So yes, that step is valid, though written rather oddly.

you mean plug in the linear mapping they presented me with in the first place and show it works, without deriving it?

That is typically how these proofs go.  You're making it much more hard on yourself trying to derive a given.

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"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..."

mikau

Member

Registered: 2005-08-22

Posts: 1,504

Re: derivatives in multiple variables

So yes, that step is valid, though written rather oddly.

how would you write it? (a non odd fashion)

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A logarithm is just a misspelled algorithm.