# Differentiability Proof

• Apr 20th 2011, 03:10 AM
OXF
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
• Apr 20th 2011, 03:59 AM
girdav
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.
• Apr 20th 2011, 04:32 AM
OXF
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 (k-1)th, (k-2)th, ... etc exist also), then apply the defintion of the derivative? I don't see how it would work?
• Apr 20th 2011, 04:40 AM
girdav
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\$.
• Apr 20th 2011, 04:47 AM
OXF
is it valid to say that as the k-th 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).