Find the Taylor series (about x=0) of the function $cos (x)$. Suppose we wish to find a polynomial which approximates to the function to within $\epsilon$ ($>0$) throughout the interval $[-k,k]$. Show that such a polynomial exists.

I understand how to find the Taylor series, which is $1-\frac{x^2}{2!}+\frac{x^4}{4!}-...\pm\frac{x^n}{n!}$.

I also know that the difference between the original function and the polynomial is $\frac{x_0^{n+1}}{(n+1)!}f^{n+1}(c)$ for some $c$ between $0$ and $x_0$. I'm just not sure how to use this fact to show that such a polynomial exists on the interval $[-k,k]$.

  • $\begingroup$ You will want to use the fact that the absolute value of any derivative of $f$ is bounded by 1. Can you take it from there? $\endgroup$ Apr 7 '17 at 0:16

For $f(x)=\cos x$, the $(n+1)$th derivative $f^{(n+1)}(x)$ is $\pm \cos x$ or $\pm \sin x$, so for all $x$ we have $|f^{(n+1)}(x)|\leq 1.$

So for all $x_0\in [-k,k]$ the remainder $R(x_0,n)=x_0^{n+1}f^{(n+1)}(c)/(n+1)!$ satisfies $$|R(x_0,n)|\leq |x_0|^{n+1}/(n+1)!\leq k^{n+1}/(n+1)!\;.$$ Now $k^n/n!\to 0$ as $n\to \infty.$ So for a given $\epsilon >0,$ if we take $n$ large enough that $k^{n+1}/(n+1)!<\epsilon,$ then $|R(x_0,n)|<\epsilon$ for all $x_0\in [-k,k]$.


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