I know the Approximation Theorem of Weierstrass. I think one can apply it to my question but I don't see directly how.
Assume $f$ is a continuous function on the unit interval $[0,1]$ such that $f(0)=0$ and that $\varepsilon>0$ is arbitary. I want to show that there exists a polynomial of the form $P(x)=a_nx^n+\cdots+a_1x$ such that $|f(x)-P(x)|<\varepsilon$ for each $x\in [0,1]$. I thought that this is a corollary of thhe approxmiation Theorem, but I think one has to prove that $a_0$ (the absolute term of the polynomial) has to be Zero. Follows this from the fact that $\varepsilon$ was arbitrary Chosen (that we have a unifrom Approximation)?
If we can do that for arbitrary continuous functions we can do this in particular for $\sqrt{x}$ and for $x^2$. But this can be done more specific. I want to Show that for abitrary $\varepsilon>0$ and arbitrary $A>0$ the following holds: There exists a polynomial $Q$ such that $Q(x)=b_1x^4+b_2x^8+\cdots+b_mx^{4m}$ such that $|x^2-Q(x)|<\varepsilon$ for all $x\in [-A,A]$. My first thought was that this is true since $x^2$ is the forth power of $\sqrt{x}$. Is this true and how can I make the Argument precise?
Thanks a lot.
There is a polynomial $P^*$ such that $|f(x)-P(x)|\le\epsilon/2$ for all $x\in[0,1]$. Let $P(x)=P^*(x)-P^*(0)$. Then $P(0)=0$ and $$ |f(x)-P(x)|\le|f(x)-P^*(x)|+|P^*(x)-P(x)|\le\epsilon\quad\forall x\in[0,1]. $$