Suppose that the function f is infinitely differentiable on $(-1,1)$ and there are constants $A>0$ and $B>0$ such that

Question

Suppose that the function f is infinitely differentiable on $(-1,1)$ and there are constants $A>0$ and $B>0$ such that $|f^{(n)}(x)|\le A \frac{n!}{B^{n}}$ for all $x\in (-1,1)$ and all $n\in \mathbb{N} $. Prove that there exists $\delta>0$ such that $f(x)=\sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!}x^n $ for $x\in(-\delta,\delta)$.

I guess I have to use Taylor's theorem in this question.

Answer 1

It suffices to show that $R_n(x)=\frac{f^{n+1}(c)}{(n+1)!}x^{n}\to 0$ as $n\to \infty $, where without loss of generality we may assume $0<x<1$ so that $0<c<x$.

We have $\left | \frac{f^{n+1}(c)}{(n+1)!}x^{n} \right |\le A \frac{x^n}{B^{n}}\cdot \frac{1}{n+1}\to 0$ as $n\to \infty$ as soon as we take $x<\delta$ where $\delta$ is so small that $\delta /B<1$.

Answer 2

You're trying to prove that $f$ is analytic in $(-\delta,\delta)$, which by definition means that $$ R_n(x)=f(x)-\sum_{k=0}^n\frac{f^{(k)}(0)}{k!}x^k\to0\hspace{.4cm}\text{ as }\hspace{.4cm}n\to\infty $$ for $x\in(-\delta,\delta)$. We know that $|f^{(n)}(x)|\leq A\frac{n!}{B^n}$ for $x\in I=[-1+\varepsilon,1-\varepsilon]$, where $0<\varepsilon<1$ is any constant, so by Taylor's Inequality $$ |R_n(x)|\leq\frac{An!}{(n+1)!B^n}|x|^{n+1}=\frac{AB}{n+1}\left|\frac{x}{B}\right|^{n+1},\hspace{.3cm}x\in I. $$ For fixed $x\in I$ this sequence converges to $0$ if $|\frac{x}{B}|<1$. As $|x|<1$, if $|B|\geq1$ then we are done. If, on the contrary, $|B|<1$ then we can assure $|\frac{x}{B}|<1$ for $|x|<|B|$. Either case we can take $\delta=\min(\varepsilon,B)$.

[[Notice that if we know $|B|\geq1$ then by the arbitrarity of $\varepsilon$ we actually get that $f$ is analytic in $(-1,1)$.]]