# Taylor Expansion of Power Series

Suppose that $\space f:[0,1]\rightarrow \mathbb{R}$ is real analytic and that its power series expansion is:
$\\ f(x)=\sum\limits_{n=0}^\infty a_nx^n$

Prove that there exists an $x_0\epsilon (0,x)$ such that

$f(x) = a_0 + a_1x + \frac{1}{2!}f''(x_0)x^2$

For my approach:

• The $f(x)$ given looks very close to the taylor series, so I assume that $f''(x_0)$ will be the remainder of terms from the rest of the taylor series
• I used mean value theorem to get $f''(x_0)=\frac{f'(x)-f'(0)}{x-0}$ which led to the $f''(x_0)$ term to be close to the taylor terms once multiplied by 1/2! and $x^2$, but there is still remaining coefficients: $f''(x_0)=2a_2+3a_3x+...$ which when put back into the f(x), leaves coefficients on the taylor expansion that shouldn't be there
• In your title: of real analytic what? – Mariano Suárez-Álvarez May 15 '13 at 6:26
• sorry latex is still new to me, fixed it – Vitale Watre May 15 '13 at 6:31
• You shouldn't start by manipulating the expression $f''(x_0)$, since we have no idea what $x_0$ should be. I like the idea of using the mean value theorem somehow, though. Perhaps apply it to the function $(f(x)-a_0)/x$? – Greg Martin May 15 '13 at 7:18
• This is the mean value theorem, generalized to a higher derivative. (See Taylor's theorem with Lagrange's form of the error.) – copper.hat May 15 '13 at 7:25
• I wonder: the given power series for $\,f\,$ is around the point $\,z=0\,$...How can this be possible if the function's defined on $\,[0,1]\,$ ? Don't we have to require definition in a neighbourhood of some point for analicity there? – DonAntonio May 15 '13 at 13:21