
Differentiability Proof
Let g:R>R be a twice differentiable function satisfying g(0)=g'(0)=1 and g''(x)=g(x)=0 for all x in R.
(i) Prove g has derivatives of all orders.
(ii)Let x>0. Show that there exists a constant M>0 such that g^n(Ax)<=M for all n in N and A in (0,1).
Any help would be appreciated; not really sure where to start with this one.
OXF

I guess you meant $\displaystyle g''(x)+g(x)=0$. You can try to show the result by induction on the order of the derivative.

Yeah, I meant g''(x)+g(x)=0.
How would I go about an inductive proof?
Would you assume the kth derivative exists (therefore we know the (k1)th, (k2)th, ... etc exist also), then apply the defintion of the derivative? I don't see how it would work?

If we assume that the $\displaystyle k$th derivative of $\displaystyle g$ exists and is continuous, then so is the $\displaystyle k$th derivative of $\displaystyle g''$, which is the $\displaystyle (k+2)$th derivative of $\displaystyle g$.

is it valid to say that as the kth derivative exists, so does the (k+2)th derivative? I thought the rule stated would only apply to g and g'', rather than g^(k) and g^(k+2).