Proving kth Derivative of f:R->R is 0 at 0

[Nusc]

Define f:R->R by fx = 0 if x<=0 and e^-1/x if x>0

Show that f is c^inf and that all the derivatives of f at 0 vanish; that is, f^(k)0=0 for every k.


After taking first and secondderivatives we just apply l'hostpials rule to show that the derivatives vanish.

My problem is for the kth derivative.

We know after taking the derivative we get some polynomail say p(x) so

f^k(x) = p(x) * e^-1/x


What do I do now?

 

[matt grime]

After you differentiate e^-{1/x} some number of times you only ever end up with e^-{1/x}*(some type of function of x), work out what 'type' of function I mean, and then show that for any function of that type the limit is zero as x tends to zero.

 

[Nusc]

I assume that by that type of function you mean the polynomial I spoke of. Do I need to contruct a lemma?

 

[matt grime]

You need to prove something. What you call it is up to you.

 

[Nusc]

Show that If g: R -> R is analytic in a nbhrd of 0 then gx = sum g^k0 x^k/k! for all x in the symmetric int. with center 0. then f, as defined above, cannot be analytic at 0

Well it is assumed that all derivatives vanish at 0 therefore no power series centered at 0 converges to f on a nbhrd of 0

Is that enough justifcation?