Zareon Posted November 28, 2007 Share Posted November 28, 2007 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? Link to comment Share on other sites More sharing options...
uncool Posted November 29, 2007 Share Posted November 29, 2007 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- Link to comment Share on other sites More sharing options...
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