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Old 2006-05-29, 13:41   #1
Damian
 
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Default Multivariable differentiability

Sorry if this is a bit off-topic.
I searched the web for the proof that if a multivariable function has partial derivatives, and they are continuous, then it is differentiable. (and f is called a C1 function)
Any help will be appreciated.
Damian.
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Old 2006-05-30, 04:06   #2
Orgasmic Troll
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p. 219 of Rudin, Theorem 9.21
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Old 2006-05-30, 12:53   #3
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Thanks for the help. I don't have Rudin's book but neitherway I found the proof here http://www.maths.mq.edu.au/~wchen/ln...er/mva02-d.pdf

Now I have another question related to the same theorem.
Suppose the function:

f(x,y) ={ 1 (if x=0 or y=0)
{ 0 (otherwise)

the partial derivatives at the origin are
df/dx = 0 and df/dy = 0

so they exist and are continious.
However the function is not continious in (0,0), so it can't be differentiable there. (there can't be a tangent plane)

What did I misundertood? Because I think the theorem says that if the partial derivatives exists and they are continious then the function is differentiable there.
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Old 2006-05-30, 18:26   #4
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Quote:
Originally Posted by Damian
Suppose the function:

f(x,y) ={ 1 (if x=0 or y=0)
{ 0 (otherwise)

the partial derivatives at the origin are
df/dx = 0 and df/dy = 0

so they exist and are continious.
However the function is not continious in (0,0), so it can't be differentiable there. (there can't be a tangent plane)

What did I misundertood? Because I think the theorem says that if the partial derivatives exists and they are continious then the function is differentiable there.
That function is not differentiable at the origin - just use the definition of derivative in the sense of the limit of the divided difference delta(f)/delta(coordinate) - that limit does not exist at (0,0) in this case.

Also, it makes no sense to speak of a function being continuous at a single point - continuity only makes sense in regions, i.e. neighborhoods of a point.
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Old 2006-05-30, 19:03   #5
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Quote:
Originally Posted by ewmayer
Also, it makes no sense to speak of a function being continuous at a single point - continuity only makes sense in regions, i.e. neighborhoods of a point.
Although we usually think of it this way, continuity is classically defined at a point. The classic counter-intuitive case is
f(x) = 1/q if x is rational p/q in lowest form,
f(x)= 0 if x is irrational

This function is continuous at irrational points.

(f(x) is continuous at x' if, for every epsilon > 0 there exists a delta such that
for every x such that |x-x'|<delta, |f(x)-f(x')|<epsilon)

William
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Old 2006-06-01, 13:15   #6
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Thanks for the help. I've been thinking about it and I think I understand the flaw in the statement I posted above now.
The theorem says the functions partial derivatives have to be continious in the -neighborhood- of a point, in order to be diferentiable there, wich isn't the case in the f(x,y) I formulated above in the origin, so it does't say nothing about this function been differentiable in the origin.

Damian.
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Old 2006-06-01, 13:20   #7
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Another question: Does anybody knows who was the mathemathician that first proved that theorem? It sounds strange to me that the theorem hasn't got a name asociated with it.
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Old 2006-06-02, 13:40   #8
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Has anybody one example of a function that is differentiable at a point, but is not a class C1 function at that point?
Is that even possible?
Thanks,
Damian.
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Old 2006-06-02, 17:29   #9
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Quote:
Originally Posted by Damian
Has anybody one example of a function that is differentiable at a point, but is not a class C1 function at that point?
Is that even possible?
No - to begin with a function can in some sense never be "differentiable at a point", i.e. without referring to the function's values in a neighborhood of the point. A function is only differentiable at a point x0 if the limit of the divided difference [f(x)-f(x0)]/[x-x0] exists at that point, which implies that the limit exists and is the same, irrespective of the direction from which one approaches x0 in taking the limit. (I.e. from the left or the right if one restricts oneself to the real axis, or from an arbitrary direction if one generalizes to the complex plane.) For example, the function f(x) = |x| is continuous everywhere and the above limit has a well-defined value for a given direction of approach to x0 = 0 (-1 when approaching from the left, +1 from the right), but because the limit depends on the direction of approach the derivative does not in fact exist at x0 = 0. Since "C1" is equivalent to "derivative exists", if the function is not differentiable at a given point, it cannot be C1 there, and vice versa.

There are eminently readable online links to all of this stuff, you know - for example:

http://mathworld.wolfram.com/Derivative.html

http://mathworld.wolfram.com/C-kFunction.html
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Old 2006-06-02, 18:45   #10
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Thanks for your reply.
The {Since "C1" is equivalent to "derivative exists"} was what I needed to know

So what do you think, is it ok to say: a function is differentiable in the neighborhood of a point A, if and only if its partial derivatives are continious in the neighborhood of A?

In other words, is it a neccesary and sufficient condition for a function to have partial derivatives continious in an interval, to be differentiable there?
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Old 2006-06-02, 19:28   #11
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k-differentiable (k = 0,1,2,...) in an interval/region means derivatives (or partial derivatives, if we're in 2 or more dimensions) of order <= k exist in that interval/region.
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