Prove Taylor series converges to $f$.

Solution 1:

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 $\left| {x - a} \right| < R$

Solution 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.