# analytic function

• Apr 18th 2010, 12:13 PM
Archi
analytic function
let $\displaystyle g:[0,1] \rightarrow R$ be a continuous function. Let $\displaystyle \epsilon >0$. Prove that there is a real analytic function $\displaystyle h: [0,1] \rightarrow R$ such that $\displaystyle |g(x)-h(x)| < \epsilon$ for all $\displaystyle x \in [0,1]$.
• Apr 18th 2010, 12:43 PM
Opalg
Quote:

Originally Posted by Archi
let $\displaystyle g:[0,1] \rightarrow R$ be a continuous function. Let $\displaystyle \epsilon >0$. Prove that there is a real analytic function $\displaystyle h: [0,1] \rightarrow R$ such that $\displaystyle |g(x)-h(x)| < \epsilon$ for all $\displaystyle x \in [0,1]$.

You can even find a polynomial that will do this (that is the Weierstrass approximation theorem). A concrete way to construct a polynomial that approximates a given continuous function to any desired degree of accuracy is to use Bernstein polynomials.
• Apr 18th 2010, 04:28 PM
Archi
can i just say it is proven by weierstrass approximation theorem? or do i have to actually show the proof ? i am not sure how to even start. whould u guide me?