Let $f$ be a real valued continuous function on $\left[−1, 1\right]$ such that $f(x) = f(−x)$ for all $x \in \left[−1, 1\right]$.
Show that for every $\epsilon > 0$ there is a polynomial $p\left(x\right)$ with rational coefficients such that for every $x \in \left[−1, 1\right]$, $$|f\left(x\right) − p\left(x^2\right)| < \epsilon$$.
I was trying something like this
$$f(x)=f(0)+\frac{f^{'}(0)}{1!}x+\frac{f^{''}(0)}{2!}x^2+\frac{f^{'''}(0)}{3!}x^3+\frac{f^{''''}(0)}{4!}x^4+\dots \dots$$
$$f(-x)=f(0)-\frac{f^{'}(0)}{1!}x+\frac{f^{''}(0)}{2!}x^2-\frac{f^{'''}(0)}{3!}x^3+\frac{f^{''''}(0)}{4!}x^4-\dots \dots$$
adding these two we get $$f(x)+f(-x)=2f(0)+2\frac{f^{''}(0)}{2!}x^2+2\frac{f^{''''}(0)}{4!}x^4+2\frac{f^{''''''}(0)}{6!}x^6+\dots \dots$$
Because $f(x)=f(-x)$ we have $$2f(x)=2f(0)+2\frac{f^{''}(0)}{2!}x^2+2\frac{f^{''''}(0)}{4!}x^4+2\frac{f^{''''''}(0)}{6!}x^6+\dots \dots$$
Hence $$f(x)=f(0)+2\frac{f^{''}(0)}{2!}x^2+\frac{f^{''''}(0)}{4!}x^4+\frac{f^{''''''}(0)}{6!}x^6+\dots \dots$$
Thus taking $$p(x)=f(x^{\frac12})=f(0)+\frac{f^{''}(0)}{2!}x+\frac{f^{''''}(0)}{4!}x^2+2\frac{f^{''''''}(0)}{6!}x^3+\dots \dots$$
This gives me $$|f\left(x\right) − p\left(x^2\right)|=0$$
But I want a polynomial $p(x)$ which should of finite degree. So I have trouble with this part
Can Someone write the answer of this question formally. I am not sure about presenting my answer.
Your approach is not valid because $f$ need not admit a Taylor series expansion. $f(\sqrt {|x|})$ is a continuous function. By Wierstrass Theorem there is a polynomial $p$ such that $|f(\sqrt {|x|})-p(x)| < \epsilon$ for all $x$. Replacing $x$ by $x^{2}$ we get $|f(x)-p(x^{2})| < \epsilon$ for all $x \in [0,1]$. Now use the fact that $f(-x)=f(x)$ to conclude that the inequality holds for $x \in [-1,0] $ also.