analytic function

**Archi** · April 18th 2010, 12:13 PM

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

**Opalg** · April 18th 2010, 12:43 PM

Originally Posted by Archi

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

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.

**Archi** · April 18th 2010, 04:28 PM

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?

Thread: analytic function

LinkBack

Thread Tools

Display

analytic function

Similar Math Help Forum Discussions

[SOLVED] Analytic Function

analytic function

Analytic Function

Analytic function

Analytic function

Search Tags