Problem from Baby Rudin Chapter 8

[Poopsilon]

The problem I am having trouble with is Exercise #1 in Chapter 8 of Rudin's Principles of Mathematical Analysis. For those of you who don't own the book the problem states:

Define f(x) = { exp(-1/x^2) for x≠0.
......{ 0 for x=0.

Prove that f(x) has derivatives of all orders at x=0 and that the nth derivative is equal to zero for all n=1,2,3,... .



There is a theorem in the book concerning power-series which are convergent in some interval having derivatives of all orders in this interval. Clearly exp(-1/x^2) has a convergent power-series expansion for the intervals (-R,0) and (0,R), but I'm a bit confused on how to approach proving statements about the derivative at zero.

Now for both f(x) = exp(-1/x^2) and the first derivative f'(x) = (2/x)exp(-1/x^2) go to 0 as x->0 (from both directions). So since f(x)=0 for x=0 does this mean I can 'stitch in' 0 into the original function and conclude that the derivative is 0 at f(x)=0? If this is sufficient what would be a good way to generalize this to all n=1,2,3,...? Thanks.

 

[Dick]

If f(x)=exp(-1/x^2) then f'(x) is NOT equal to (2/x)*exp(-1/x^2). And even if it were it's not clear that goes to zero. Why don't you warm up by trying to find a correct expression for the first derivative and show it goes to zero as x->0?

 

[Poopsilon]

oops, ok so x^3 on the bottom not x. It still goes to zero like before.

 

[Dick]

oops, ok so x^3 on the bottom not x. It still goes to zero like before.

Sure it does. But why? If you can figure that out it might be clearer how to extend the result.

 

[Poopsilon]

I'm not sure I understand. You want me to go back to it's power-series expansion and do a term by term differentiation kinda thing? I'm still just applying the chain rule term by term and the power-series still doesn't converge for x=0.. Or are you asking me to apply the formal definition of the derivative?

 

[Dick]

I'm not sure I understand. You want me to go back to it's power-series expansion and do a term by term differentiation kinda thing? I'm still just applying the chain rule term by term and the power-series still doesn't converge for x=0.. Or are you asking me to apply the formal definition of the derivative?

The power series expansion is baloney. The first derivative is like exp(-1/x^2)/x^3. I would take the log and use l'Hopital.

 

[Poopsilon]

I'm already aware both first derivative and the original function go to zero as x->0, what I want to know is what these facts do for me as far as finding the derivative of all orders of f at 0. Is there a theorem in the book somewhere or something? I don't know what you mean by baloney, whether you mean its a dead end or whether my interpretation of it is incorrect, could you be a bit more explanatory in your responses?

 

[Dick]

I'm already aware both first derivative and the original function go to zero as x->0, what I want to know is what these facts do for me as far as finding the derivative of all orders of f at 0. Is there a theorem in the book somewhere or something? I don't know what you mean by baloney, whether you mean its a dead end or whether my interpretation of it is incorrect, could you be a bit more explanatory in your responses?

I mean the power series expansion of the function doesn't exist. If all of the derivatives are zero, then the power series expansion doesn't converge to the function, obviously. Since the power series expansion is zero and the function is not zero around x=0. That's what I mean by baloney. It doesn't work. And while being 'aware' the the first derivative goes to zero as x->0, that doesn't tell my why. If 'awareness' is enough then why don't you just tell me that you are 'aware' that the nth derivative goes to zero also, and then the problem is solved. Do you see what I'm getting at? You haven't told me why the FIRST derivative goes to zero yet. It's a 0/0 limit, isn't it?

 

[Poopsilon]

Showing that the first derivative goes to zero as x->0 was never the issue for me, its clear when looking at it and when I went to prove it formally it would have been then that I searched for a way to show it rigorously. But at this stage I am still mapping out the broad strokes of the proof and what I am confused about is dealing with the piece-wise aspect of the function. More specifically, is showing that the first derivative goes to zero as x->0 enough to conclude that the derivative actually exists at 0, and is equal to zero, as a result of the fact that f(x)=0 for x=0? Maybe this was treated in the chapters on continuity and differentiation, and if it was then I have forgotten and I apologize, but upon looking back at those two chapters nothing jumps out at me.

Your final remark about the problem being solved suggests to me that it is sufficient, and if that is the case, then maybe you could point me in the direction of why.

 

[Dick]

Showing that the first derivative goes to zero as x->0 was never the issue for me, its clear when looking at it and when I went to prove it formally it would have been then that I searched for a way to show it rigorously. But at this stage I am still mapping out the broad strokes of the proof and what I am confused about is dealing with the piece-wise aspect of the function. More specifically, is showing that the first derivative goes to zero as x->0 enough to conclude that the derivative actually exists at 0, and is equal to zero, as a result of the fact that f(x)=0 for x=0? Maybe this was treated in the chapters on continuity and differentiation, and if it was then I have forgotten and I apologize, but upon looking back at those two chapters nothing jumps out at me.

Your final remark about the problem being solved suggests to me that it is sufficient, and if that is the case, then maybe you could point me in the direction of why.

My final comment was ironic. If it's 'clear' to you the first derivative is zero, that's not a proof it's zero. Any more than saying 'it's clear to me all of the derivatives are zero'. Aren't we looking for a proof here? Not just an 'it's clear to me', statement? WHY is it zero? Just the first derivative for a starter, ok?

 

[Poopsilon]

I never said that it was clear to me that the first derivative is zero. I said its clear to me that it goes to zero as x->0, and by 'goes to' maybe I should be more precise and say 'approaches'. My issue, which I feel like I am making clear but apparently I am not, is with what this says about the possible existence of the derivative AT x=0 and about what its value might be at that point.

 

[Dick]

I never said that it was clear to me that the first derivative is zero. I said its clear to me that it goes to zero as x->0, and by 'goes to' maybe I should be more precise and say 'approaches'. My issue, which I feel like I am making clear but apparently I am not, is with what this says about the possible existence of the derivative AT x=0 and about what its value might be at that point.

You can form the difference quotient for f'(0) directly as (f(x)-f(0))/(x-0)=exp(-1/x^2)/x. If you show that limit is 0 and you also show f'(x)=2*exp(-1/x^2)/x^3 also approaches 0 then you've shown you have a continuous derivative both around x=0 and AT x=0. Is that what you are asking?

 

[Poopsilon]

Ah yes, that is what I was looking for, thank you. I will see if I can generalize to the nth derivative from here.