Section 2.2 An example of a Taylor series that converges
Consider the function \(f(x) = e^x\text{.}\) We know that \(f^{(n)}(x) = e^x\) for all \(n\geq 0\text{.}\) Therefore, the Taylor series generated by \(f\) at \(x=0\) is given by:
\begin{equation*}
\sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!} x^n = \sum_{n=0}^{\infty} \frac{1}{n!} x^n
\end{equation*}
We will show that this series converges to \(f(x) = e^x\) for all \(x\in \R\text{.}\) To do this, we will use the remainder theorem. We have:
\begin{equation*}
R_n(x) = \frac{f^{(n+1)}(c)}{(n+1)!} x^{n+1} = \frac{e^c}{(n+1)!} x^{n+1}
\end{equation*}
First, we consider the case when \(x>0\text{.}\) As in Remark 2.4, we do not need the exact value of \(c\text{,}\) only an upper bound for \(|f^{(n+1)}(c)| = e^c\text{.}\) Since \(c\) is between \(0\) and \(x\text{,}\) we have \(0 \leq c \leq x\text{,}\) so \(M = e^x\) works; see Figure 2.6. Therefore, we can bound the remainder term as follows:
\begin{equation*}
R_n(x) \leq \frac{e^x}{(n+1)!} x^{n+1}
\end{equation*}
Now we can compute the limit of the remainder term as \(n\to \infty\text{,}\) keeping in mind that \(x\) is fixed and only \(n\) is changing:
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Why is \(\lim_{n\to\infty} \frac{x^{n+1}}{(n+1)!} = 0\) for every fixed \(x\text{?}\) Write the quotient as a product of \(n+1\) factors, \(\frac{|x|^{n+1}}{(n+1)!} = \frac{|x|}{1}\cdot\frac{|x|}{2}\cdot\frac{|x|}{3}\cdots\frac{|x|}{n+1}\text{.}\) The numerators never change, but the denominators keep growing, so eventually every new factor is small. Precisely, fix an integer \(N\) with \(N \geq 2|x|\text{.}\) The first \(N\) factors contribute the fixed number \(C = \frac{|x|^{N}}{N!}\text{,}\) and each factor after that satisfies \(\frac{|x|}{k} \leq \frac{1}{2}\) for \(k > N\text{.}\) Hence for \(n+1 > N\) we get \(0 \leq \frac{|x|^{n+1}}{(n+1)!} \leq C \left(\frac{1}{2}\right)^{n+1-N}\text{,}\) and the right-hand side tends to \(0\) as \(n\to\infty\text{,}\) so the squeeze theorem gives the limit \(0\text{.}\) In short, the factorial in the denominator eventually outgrows any fixed power in the numerator, no matter how large \(|x|\) is.
\begin{equation*}
\lim_{n\to \infty} |R_n(x)| \leq \lim_{n\to \infty} \frac{e^x}{(n+1)!} x^{n+1}
=e^x \lim_{n\to \infty} \frac{x^{n+1}}{(n+1)!} = 0.
\end{equation*}
This shows that the Taylor series converges to \(f(x) = e^x\) for all \(x>0\text{.}\)
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The second case is when \(x < 0\text{.}\) In this case, since \(c\) is between \(x\) and \(0\text{,}\) we have \(x \leq c \leq 0\text{,}\) so this time \(M = e^0 = 1\) is an upper bound for \(e^c\text{;}\) see Figure 2.6 again. Hence, we can bound the remainder term as follows:
\begin{equation*}
|R_n(x)| \leq \frac{1}{(n+1)!} |x|^{n+1}.
\end{equation*}
Now we can compute the limit of the remainder term as \(n\to \infty\text{:}\)
\begin{equation*}
\lim_{n\to \infty} |R_n(x)| \leq \lim_{n\to \infty} \frac{1}{(n+1)!} |x|^{n+1} = 0
\end{equation*}
This shows that the Taylor series converges to \(f(x) = e^x\) for all \(x < 0\text{.}\) Therefore, we conclude that the Taylor series converges to \(f(x) = e^x\) for all \(x\in \R\text{.}\)
The animation below illustrates this convergence geometrically. As the degree \(n\) increases, the Taylor polynomials \(p_n(x) = \sum_{k=0}^{n} \frac{x^k}{k!}\) hug the graph of \(f(x) = e^x\) over a wider and wider interval, matching the fact that the remainder \(R_n(x)\to 0\) for every \(x\in\mathbb{R}\text{.}\)
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