# Analytic Functions

• Nov 15th 2009, 01:14 PM
Analytic Functions
How would I prove that the sum and product of analytic functions are analytic?

With an analytic function being defined as...
A function $\displaystyle f$ is (real) analytic on an open set D in the real line if for any $\displaystyle x_0$ in D one can write...
$\displaystyle f(x) = \sum_{n=0}^\infty a_n \left( x-x_0 \right)^n = a_0 + a_1 (x-x_0) + a_2 (x-x_0)^2 + a_3 (x-x_0)^3 + \cdots$
in which the coefficients $\displaystyle a_0$, $\displaystyle a_1$, ... are real numbers and the series is convergent to f for x in a neighborhood of $\displaystyle x_0$.

Thinking something along the lines of grouping the x terms together, then can find epsilon so that it converges after the nth term..? This along the right lines?
• Nov 15th 2009, 03:24 PM
Enrique2
You can use that $\displaystyle f$ is analytic on $\displaystyle D\subseteq\mathbb{R}$ if and only if there exists an open set $\displaystyle G\subseteq C$ containing $\displaystyle D$ such that
there exists $\displaystyle g$ holomorphic on $\displaystyle G$ extending $\displaystyle f$. Then the assertion follows from the fact that sums and products of holomorphic functions are holomorphic.

Also you can use that $\displaystyle f$ is analytic on $\displaystyle D\subseteq\mathbb{R}$ if and only if for each compact subset
$\displaystyle K$ of $\displaystyle D$ there exists $\displaystyle M>0$ such that $\displaystyle \left|\frac{f^{(n)}(x)}{k!}\right|\leq M^k$ for each $\displaystyle k\in\mathbb{N}, x\in K$. Here the sum is almost immediate but for the product you will need Leibnitz rule.

If you want a direct proof for the product, puff, let say using Cauchy Series could be done, but anyway you will need the fact that every analytic expansion converges absolutely, which follows from the characterization above.