# Math Help - taylor series

1. ## taylor series

I need to prove that if f is defined for abs(x) < R and if there exists a constant B such that abs(f*(x)) <= B (where f*(x) denotes the n derivatives of f) for all abs(x) < R and n in the natural numbers, then the Taylor Series expansion,
summation from n = 0 to infinity of [f*(0)/(n!)]*x^n converges to f(x) for abs(x) < R.

I did the following work: I have a defined function f(x). Since abs(f*(x)) <= B, it follows that abs(R(x)) <= (abs(x)^n)/n! for n in N and x in R where (R(x)) denotes the sequence of remainders. I know that lim as n goes to infinity of (R(x)) = 0 for each x in R. Because the sequence of remainders converges to 0 for each x in some interval {x: abs(x-c) < R}, I can write that f(x) = summation from n = 0 to infinity of [f*(c)/(n!)]*(x-c)^n. Evaluating at c = 0, the Taylor expansion becomes the summation from n = 0 to infinity of [f*(0)/(n!)]*x^n.
I use the Cauchy-Hadamard Theorem, which states that if R is the radius of convergence of the power series sum(a_n*x^n), then the series is absolutely convergent if abs(x) < R. Since abs(x) < R, I know that the above Taylor series expansion does converge. But how do I show it converges to f(x)?

I need to prove that if f is defined for abs(x) < R and if there exists a constant B such that abs(f*(x)) <= B (where f*(x) denotes the n derivatives of f) for all abs(x) < R and n in the natural numbers, then the Taylor Series expansion,
summation from n = 0 to infinity of [f*(0)/(n!)]*x^n converges to f(x) for abs(x) < R.
Let $x\in (-R,R)$ where $R>0$ and $x\not = 0$ (at $x=0$ the series converges at there is nothing to prove). Let $T_n(x) = \sum_{k=0}^n \frac{f^{(k)}(0)}{k!}x^k$ and let $R_n(x) = f(x) - T_n (x)$ then by Lagrange remainder we have that $\left| R_n(x) \right| = \left| \frac{f^{(n+1)}(y)}{(n+1)!} x^{n+1} \right|$ where $y$ is between $0$ and $x$, this means that $|R_n(x)| \leq \frac{B R^{n+1}}{(n+1)!}$ but $\frac{R^{n+1}}{(n+1)!} \to 0$.

3. Where does this show that it converges to f(x)?

BTW thanks for the help.