2
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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 $\delta /B<1$.

$\endgroup$
0
$\begingroup$

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 $(-1,1)$.]]

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.