Urgent: Prove Smoothness of f(x) with Infinite Derivatives

[ARozanski]

Homework Statement

[tex] f(x)=\left\{\begin{array}{cc}e^{-x^{-2}},& \mbox{ if } x!=0 \\ 0, \mbox{ if } x=0 \end{array}\right [/tex]
Prove the smoothness of f(x) - as in - prove it has infinite derivatives.
I was asked this question in a calculas class i have in university...It just so happens to be that this question was asked on this site last year =>
LINK: https://www.physicsforums.com/showthread.php?t=135116

The Attempt at a Solution

Now - looking at what the person said last year - they got to what i also got to - which was that after a few derivatives you get - [tex]\frac{c}{x^n}[/tex] where is c is some number

I did not however understand why they had to get to the point where they wanted to multiply the limit of the derivatives with the actual function f(x).

Please can someone point me in the right direction into solving this...i need it done by sunday - any help well be greatly appreciated

 

[chickendude]

It is pretty obvious that the function is smooth whenever [tex]x \neq 0 [/tex]

At x=0, we need to prove that [tex]f(0) = \lim_{x\to 0}f(x)[/tex] by the definition of smoothness

f(0)=0 by the def of f(x)

[tex]\lim_{x\to 0}e^{-x^{-2}}[/tex]
[tex]=\lim_{x\to 0}(1/e)^{\frac{1}{x^2}}[/tex]
[tex]=\lim_{x\to \infty}(1/e)^{x^2}[/tex]

since 1/e < 0, this limit is comparable to a geometric series which approaches 0
so the limit is 0, finishing the proof

 

[mathboy]

since 1/e < 0, this limit is comparable to a geometric series which approaches 0
so the limit is 0, finishing the proof

Well, that proves the first derivative exists only (and not even that the derivative is continuous). He wants to prove that all the higher order derivatives of f exist and are continuous.


All the higher derivatives of f is the sum of terms of the form [tex]\frac{c}{x^n}[/tex]f(x) (prove this by induction!). Then just prove that this term goes to zero as x goes to zero, and so all nth derivatives goes to zero.

 

[mathboy]

To prove that [tex]\frac{c}{x^n}[/tex]f(x) goes to zero as x goes to zero, use the substitution y= 1/x. Then use L'Hopital's rule n times.

It turns out that induction had to be used 3 times to finish the proof (at least for me). The most important of the 3 inductions is to prove that if the nth derivative of f at x=0 is 0, then so is the (n+1)th derivative of f at x=0. This inductive step proves 2 things: that the (n+1)th derivative exists, and that the nth derivative is continuous. Chickendude already demonstrated the case n=1. Carry out the inductive step and that will prove the smoothness of f.

 

[ARozanski]

Thanks a lot for all your help...I have finished the project and it looks alright.