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Let $g:R→R$ be a twice differentiable function satisfying $g(0)=1,g'(0)=0$ and $g''(x)−g(x)=0$ for all $x \in \mathbb{R}$

Fix $x \in R$. Show that there exists $M>0$ such that for all natural numbers $n$ and all $\theta$ from 0 to 1 $|g^{(n)}(\theta x)|≤M$

Also, find the coefficients of the Taylor expansion of $g$ about $0$, and prove that this expansion converges to $g(x)$ for all $x \in \mathbb{R}$.

I have proved that $g(x)$ has derivatives of all orders, not sure how to use this to proceed with the above question.

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  • $\begingroup$ Do you require that $M$ depends on $x$ or not? $\endgroup$
    – ntt
    Apr 9, 2017 at 16:01

1 Answer 1

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Some hints: Given non-zero $x$ let $[0;x]$ denote the segment between zero and $x$. The constant $$ M_x = \max\left\{ \max\{|g(t)|: t\in [0;x]\} \ , \ \max\{|g'(t)|: t\in [0;x]\} \right\}$$ should do. For the Taylor expansion use e.g. a Lagrange remainder estimate and the above constant.

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  • $\begingroup$ So I've found the taylor coefficients to be 1, when n is even and 0 when n is odd, would I now need to show the the lagrange remainder tends to zero as n tends to infinity? $\endgroup$
    – KaiReed
    Apr 9, 2017 at 20:49
  • $\begingroup$ One has $g^{(2n)}(0)=1$ so for the $2n$-Taylor coff you get $1/(2n)!$. For fixed $x$ you indeed have to show that the Lagrange remainder converges to zero with $n$ (still for fixed $x$). $\endgroup$
    – H. H. Rugh
    Apr 9, 2017 at 21:56

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