23. Taylor Series

e. Taylor Remainder

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 \[ \lim_{k\rightarrow\infty}|R_{2k+1}f(x)| \le \lim_{k\rightarrow\infty}\dfrac{|x|^{2k+2}}{(2k+2)!}=0 \]

We'll show \(\displaystyle \lim_{n\rightarrow\infty}\dfrac{|x|^n}{n!}=0\).
Pick an integer \(N>2|x|\). Then \(|x| \lt \dfrac{N}{2}\).
Now if \(p>N\), then \(\dfrac{1}{p} \lt \dfrac{1}{N}\) and \(\dfrac{|x|}{ p}<\dfrac{N}{2N}=\dfrac{1}{2}\).
We now compute (with \(n>N\)) \[\begin{aligned} \lim_{n\rightarrow\infty}\dfrac{|x|^n}{n!} &=\lim_{n\rightarrow\infty}\dfrac{|x|}{1}\dfrac{|x|}{2} \cdots\dfrac{|x|}{N}\dfrac{|x|}{N+1}\cdots\dfrac{|x|}{n} \\ &=\dfrac{|x|^N}{N!}\lim_{n\rightarrow\infty}\dfrac{|x|}{N+1}\cdots\dfrac{|x|}{n} \\ &\lt \dfrac{|x|^N}{N!}\lim_{n\rightarrow\infty}\left(\dfrac{1}{2}\right)^{n-N}=0 \end{aligned}\]

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

Prove that the Maclaurin series for \(f(x)=e^{2x}\) converges to \( e^{2x}\) for all real numbers \(x\).

Let \(N\) be a number \(\gt |x|\). Find the Taylor bound on the remainder for the interval \((-N,N)\)

To show \(\displaystyle \lim_{k\rightarrow\infty}R_k f(x)=0\), we use the Taylor bound on the remainder. Let \(N\) be a number \(\gt |x|\). Consider the interval \((-N,N)\). For \(f(x)=e^{2x}\), we have \(f^{(k+1)}(x)=2^{k+1}e^{2x}\). So, \(|f^{(k+1)}(x)| \le 2^{k+1}e^{2N}\). So we take \(M=2^{k+1}e^{2N}\). Thus, \[ |R_k f(x)| \le \dfrac{M}{(k+1)!}|x|^{k+1} =\dfrac{e^{2N}|2x|^{k+1}}{(k+1)!} \] and so \[ \lim_{k\rightarrow\infty}|R_k f(x)| \le e^{2N}\lim_{k\rightarrow\infty}\dfrac{|2x|^{k+1}}{(k+1)!}=0 \]

23. Taylor Series

e. Taylor Remainder

3. Convergence Proofs

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