23. Taylor Series

e. Taylor Remainder

Taylor Remainder Inequality
If $a$ is in an open interval $I$ and $M \ge |f^{(k+1)}(c)|$ for all $c$ in the interval $I$, then the Taylor remainder of degree $k$ centered at $x=a$ satisfies $|R_k f(x)| \le \dfrac{M}{(k+1)!}|x-a|^{k+1}$ for all $x$ in the interval $I$. The quantity on the right is called the Taylor bound on the remainder.

3. Convergence Proofs

To prove the Taylor series for $f(x)$ converges to $f(x)$, it suffices to prove the limit of the Taylor remainders is $0$: $\lim_{k\rightarrow\infty}R_k f(x)=0$

There are a couple of technical points which need to be discussed before we start the proof. We need to distinguish between a function $f(x)$ and the sum of its Taylor series which we denote by $Tf(x)$. We know that the Taylor series converges to $Tf(x)$ but we do not yet know that it converges to $f(x)$. Thus, the Taylor series is: $Tf(x)=\sum_{n=0}^\infty \dfrac{f^{(n)}(a)}{n!}(x-a)^n$ Further, if the Taylor series is approximated by the $k^\text{th}$ Taylor polynomial $T_k f(x)=\sum_{n=0}^k \dfrac{f^{(n)}(a)}{n!}(x-a)^n$ then the error in the approximation is $E_k=Tf(x)-T_k f(x) =\sum_{n=k+1}^\infty \dfrac{f^{(n)}(a)}{n!}(x-a)^n$ This needs to be distinguished from the Taylor remainder which is defined to be: $R_k f(x)=f(x)-T_k f(x)$ using $f(x)$ instead of $Tf(x)$. It is $R_k f(x)$ which appears in the Lemma as well as the Taylor Remainder Formula and the Taylor Remainder Inequality.

To prove the Taylor series for $f(x)$ converges to $f(x)$, we need to show the partial sums (the Taylor polynomials) converge to $f(x)$: $\lim_{k\rightarrow\infty}T_k f(x)=f(x)$ Subtracting each side from $f(x)$, this is equivalent to: $\lim_{k\rightarrow\infty}\left[f(x)-T_k f(x)\right]=0$ or $\lim_{k\rightarrow\infty}R_k f(x)=0$

Prove that the Maclaurin series for $f(x)=\sin x$ converges to $\sin x$ for all real numbers $x$.

The Taylor series is: $Tf(x)=\sum_{n=0}^\infty \dfrac{(-1)^n}{(2n+1)!}x^{2n+1}$ and the $k^\text{th}$ partial sum is the Taylor polynomial of degree $2k+1$: $T_{2k+1}f(x)=\sum_{n=0}^k \dfrac{(-1)^n}{(2n+1)!}x^{2n+1}$ and the Taylor remainder is: $R_{2k+1}f(x)=\sin x-\sum_{n=0}^k \dfrac{(-1)^n}{(2n+1)!}x^{2n+1}$ To show the remainders converge to $0$, we use the Taylor bound on the remainder. For $f(x)=\sin x$, the $(2k+2)^\text{nd}$ derivative is either $\sin x$ or $-\sin x$. In either of these cases, $|f^{(2k+2)}(x)| \le 1$. So we take $M=1$. Thus, $|R_{2k+1}f(x)| \le \dfrac{M}{(2k+2)!}|x|^{2k+2} =\dfrac{|x|^{2k+2}}{(2k+2)!}$ and so $\lim_{k\rightarrow\infty}|R_{2k+1}f(x)| \le \lim_{k\rightarrow\infty}\dfrac{|x|^{2k+2}}{(2k+2)!}=0$ Why?

This shows $\displaystyle \lim_{k\rightarrow\infty}R_{2k+1}f(x)=0$, and hence, the Taylor series for $\sin x$ converges to $\sin x$.