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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 \(x>0\text{.}\) Since \(c\) is between \(0\) and \(x\text{,}\) we have \(0 \leq c \leq x\text{.}\) 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 \(n\to \infty\text{:}\)
\begin{equation*} \lim_{n\to \infty} |R_n(x)| \leq \lim_{n\to \infty} \frac{e^x}{(n+1)!} x^{n+1} = 0 \end{equation*}
This shows that the Taylor series converges to \(f(x) = e^x\) for all \(x>0\text{.}\)
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Figure 2.6. Schematic graph of \(f(x) = e^x\) and its Taylor polynomials \(p_1, \dots, p_4\) about \(0\text{.}\) For a positive \(x\) the remainder uses some \(c\) with \(0 < c < x\text{,}\) giving \(e^c < e^x\text{;}\) for a negative \(x\) it uses some \(c\) with \(x < c < 0\text{,}\) giving \(e^c < 1\text{.}\)
The second case is when \(x < 0\text{.}\) In this case, since \(c\) is between \(x\) and \(0\text{,}\) we have \(x \leq c \leq 0\text{.}\) 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 \(x\in\mathbb{R}\text{.}\)
Figure 2.7. The Taylor polynomials \(p_1,\dots,p_8\) of \(e^x\) about \(x=0\) converging to \(y=e^x\text{,}\) followed by the remainder-theorem argument.
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