Calc 2 prove the series is itself

[brushman]

Homework Statement

A function defined by a power series $\sum_{n=0}^\infty \(a_{n}x^n$ with a radius of convergence $c > 0$ has a Maclaurin series that converges to the function at every point of $(-c, c)$. Show this by showing that the Maclaurin series generated by $f(x) = \sum_{n=0}^\infty \(a_{n}x^n$ is the series $\sum_{n=0}^\infty \(a_{n}x^n$ itself.

An immediate consequence of this is that series like

$$x\sin(x) = x^2 - \frac{x^4}{3!} + \frac{x^6}{5!} - \cdot \cdot \cdot$$

and

$$x^{2}e^{x} = x^2 + x^3 + \frac{x^4}{2!} + \cdot \cdot \cdot$$

obtained by multiplying Maclaurin series by powers of x, as well as series obtained by integration and differentiation of convergent power series, are themselves the Maclaurin series generated by the functions they represent.

The Attempt at a Solution

So I don't even know what to do. It seems like the Maclaurin series already equals itself by definition.

 

[micromass]

The MacLaurin series of a function f is defined as

$$\sum_{n=0}^{+\infty}{\frac{f^{(n)}(0)x^n}{n!}}$$

So given a function $$f(x)=\sum_{n=0}^{+\infty}{a_nx^n}$$, you'll need to prove that $$a_n=\frac{f^{(n)}(0)}{n!}$$.