Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $f$ be an infinitely differentiable function on an interval $I$. If $a \in I$ and there are positive constants $C$, $R$ such that for every $x$ in a neighborhood of $a$ and every $k$ it holds that

$|f^{(k)}(x)| \leq C \frac{k!}{R^k}$

then prove that the Taylor series of $f$ about $a$ converges to $f(x)$.

I think a good approach would be to estimate the error term. I'm not sure how to proceed exactly though. Thoughts?

share|improve this question
1  
The way I stated it is correct. –  Tony D. Mar 8 '12 at 0:27

2 Answers 2

Your approach is not only correct, it's the only possible way to proceed.

Express the error term for the kth polynomial and take it's limit. How far it converges depends very heavily on the constant called $R$, but it's straightforward computation.

share|improve this answer

I'm almost sure what you need for convergence is:

$$|f^{(n)}(x)|<R^n$$

In such a case you would have the following:

$${R_n}\left( x \right) = \int\limits_a^x {\frac{{{{\left( {x - t} \right)}^n}}}{{n!}}{f^{\left( {n + 1} \right)}}\left( t \right)dt} $$

Set $$t = x + \left( {a - x} \right)u$$

$${R_n}\left( x \right) = \frac{{{{\left( {x - a} \right)}^{n + 1}}}}{{n!}}\int\limits_0^1 {{u^n}{f^{\left( {n + 1} \right)}}\left[ {x + \left( {a - x} \right)u} \right]du} $$

Then

$$\eqalign{ & 0 \leqslant \left| {{R_n}\left( x \right)} \right| \leqslant \frac{{{{\left| {x - a} \right|}^{n + 1}}}}{{n!}}{R^{n + 1}}\int\limits_0^1 {{u^n}du} \cr & 0 \leqslant \left| {{R_n}\left( x \right)} \right| \leqslant \frac{{{{\left| {x - a} \right|}^{n + 1}}}}{{\left( {n + 1} \right)!}}{R^{n + 1}} \cr} $$

And for $n \to \infty$ we have that $|R_n(x)| \to 0$

Here's my pick on your condition. If

$${f^{\left( {n + 1} \right)}}\left( x \right) \leqslant C\frac{{\left( {n + 1} \right)!}}{{{R^{n + 1}}}}$$

The you'd have

$$\eqalign{ & 0 \leqslant \left| {{R_n}\left( x \right)} \right| \leqslant C \frac{{{{\left| {x - a} \right|}^{n + 1}}}}{{n!}}\frac{{\left( {n + 1} \right)!}}{{{R^{n + 1}}}}\int\limits_0^1 {{u^n}du} \cr & 0 \leqslant \left| {{R_n}\left( x \right)} \right| \leqslant C{\left( {\frac{{\left| {x - a} \right|}}{R}} \right)^{n + 1}} \cr} $$

And the limit would be $0$ if $\left| {x - a} \right| < R$

share|improve this answer
    
The second part, beginning with "Here's my pick", is the actual answer. –  Thursday Jul 9 at 22:46
    
@Thisismuchhealthier. Agreed. –  Pedro Tamaroff Jul 9 at 23:12

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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