How would I go about showing that the function defined by f(x)=exp(-1/x^2) for x<>0 and f(x)=0 for x=0 has derivatives of all orders and the value of all the derivatives at x=0 is 0?

 

It seem obvious that f is infinitely many times differentiable for x not equal to 0, but I don't know how I would write down a proof. Taylor series come to mind, but nothing in the book deals with that, so there should be another way.

 

I've shown f'(0)=0 by writing down the limit and using L'Hospital. But how would I show it for higher order derivatives without explicitly calculating the derivatives and evaluating the limits?

Would induction work? I've thought of letting g(x)=-1/x^2, so f(x)=exp(g(x)) and:

f'=g'e^g

f''=(g'+g'')e^g

f'''=(g'+2g''+g''')e^g

'etc'

whatever etcetera means. I explicitly find the relation using induction and then use induction to calculate the limits for all orders, but doesn't seem to go anywhere.

Anyone know of a better way?

One way to do it is to simply note that it is always e^(-1/x^2)*(a0 + a1/x^2 + ... + an/x^n) for some finite n (you can actually find what n is). In that case, there clearly is only 1 x at which this is discontinuous.

 

Another way is to try to use extend that function to the entire complex plane. If you can create what is known as a holomorphic function over the entire plane, then the function over the real line must be infinitely continuous.

=Uncool-