Taylor Series

Section 5.6 Taylor Series

We have seen the usefulness of a power series representation of a function. In this section, we discuss a way of showing analyticity of a function. The key result is the following theorem.

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Theorem 5.72. Taylor’s Theorem.

Suppose \(f \colon [a,b] \to \mathbb{R}\) has continuous derivatives up to order \(n\) and \(f^{(n+1)}\) exists on \((a,b)\text{.}\) Let \(x_0\) be a point in \([a,b]\text{.}\) Then for any \(x \in [a,b]\) other than \(x_0\text{,}\)

\begin{equation} \begin{split} f(x) = f(x_0) & + \frac{f'(x_0)}{1!}(x-x_0) + \frac{f''(x_0)}{2!}(x-x_0)^2 + \cdots \\ & + \frac{f^{(n)}(x_0)}{n!}(x-x_0)^n + \frac{f^{(n+1)}(c)}{(n+1)!}(x-x_0)^{n+1} \end{split}\tag{5.7} \end{equation}

for some \(c\) in between \(x\) and \(x_0\text{.}\)

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Remark 5.73.

The right-hand side of (5.7) is the \(n\)th order Taylor expansion of \(f\) at \(x_0\). The polynomial
\begin{equation} \begin{split} f(x_0) & + \frac{f'(x_0)}{1!}(x-x_0) + \frac{f''(x_0)}{2!}(x-x_0)^2 +\\ & \cdots + \frac{f^{(n)}(x_0)}{n!}(x-x_0)^n \end{split}\tag{5.8} \end{equation}
is called the \(n\)th order Taylor polynomial of \(f\) at \(x_0\) . The term \(R_n(x) :=\frac{f^{(n+1)}(c)}{(n+1)!}(x-x_0)^{n+1}\) is the remainder term (in Lagrange form) of the \(n\)th order Taylor expansion .

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A function \(f\) is infinitely differentiable at \(x_0\) if \(f^{(n)}(x_0)\) exists for all \(n\text{.}\) For such a \(f\text{,}\) the Taylor series of \(f\) at \(x_0\) is the power series
\begin{equation*} \sum_{n=0}^{\infty} \frac{f^{(n)}(x_0)}{n!}(x-x_0)^n = f(x_0) + \frac{f'(x_0)}{1!}(x-x_0) + \frac{f''(x_0)}{2!}(x-x_0)^2 + \cdots \end{equation*}
A Maclaurin Series is simply a Taylor series with center \(0\)

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The Taylor series of \(f\) at \(x_0\) converges to \(f(x)\) if and only if \(R_n(x) \to 0\) as \(n \to \infty\text{.}\)

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Corollary 5.74.

Suppose \(f\) satisfies the conditions in Theorem 5.72 and there exists some \(B \gt 0\) such that for all \(n\) \(|f^{(n)}(c)| \le B\) for all \(c \in [a,b]\) then \(\lim_n R_n(x) = 0\) and hence the Taylor series of \(f\) at \(x_0\text{,}\) converges to \(f(x)\) for any \(x \in [a,b]\)

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Proof.

The statement is clearly true for \(x=x_0\text{.}\) Under the assumptions, for \(x \in [a,b]\) other than \(x_0\text{,}\)

\begin{equation*} |R_n(x)| = \left|\frac{f^{(n+1)}(c)}{(n+1)!}(x-x_0)^{n+1}\right| \le \frac{B}{(n+1)!}|x-x_0|^{n+1}. \end{equation*}

It follows from Proposition 5.20 that the last quantity tends to \(0\) as \(n \to \infty\text{.}\)

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Let us see some applications

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Example 5.75.

The exponential function \(f(x)=e^x\) has derivative of all orders on \(\mathbb{R}\) and \(f^{(n)}(x) = e^x\) for each \(n\text{.}\) So, for \(M \gt 0\text{,}\) \(|f^{(n)}(c)| = e^c \le e^M\) for each \(n\) and \(c \in [-M,M]\text{.}\) Therefore,

\begin{equation*} e^x = f(x) = \sum_{n=0}^{\infty} \frac{x^n}{n!} \end{equation*}

for all \(x \in [-M,M]\text{.}\) Since \(M \gt 0\) is arbitrary, this shows that \(e^x\) is analytic on \(\mathbb{R}\text{.}\)

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Example 5.76.

The function \(\sin(x)\) has derivative of all orders on \(\mathbb{R}\) and all of them are bounded by \(1\) on \(\mathbb{R}\text{.}\) Therefore, by Corollary 5.74, for all \(x \in \mathbb{R}\text{,}\)

\begin{equation*} \sin(x) = \sum_{n=0}^{\infty} (-1)^n \frac{x^{2n+1}}{(2n+1)!} \end{equation*}

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One important application of Taylor Series is that we can use it to estimate values of various functions.

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Example 5.77.

Using Taylor’s theorem, we can find a range of values of \(x\) near \(x_0\) at which the function \(f(x)\) is approximated by the value at \(x\) of its n-th order Taylor polynomial at \(x_0\) to within a certain \(E \gt 0\text{.}\)

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For example, let’s approximate \(\cos(x)\) by \(P_5(x)\) its 5th order Taylor polynomial at \(0\text{.}\) According to Taylor’s theorem, there exist some \(c\) between \(x\) and \(0\) such that

\begin{equation*} |\cos(x)-P_5(x)| = |R_5(x)| = \left|\frac{\cos^{(6)}(c)}{6!}x^6 \right| \le \frac{1}{6!}|x|^6 \end{equation*}

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So, if we want to make sure that the error of this approximation is within, say \(E = 0.000748\text{,}\) then it suffices to make sure

\begin{equation*} \frac{|x|^6}{6!} \lt 0.000748. \end{equation*}

Solving the inequality, yields, \(|x| \le (0.000748*6!)^{1/6} \approx 0.902\text{.}\) This guarantees, for example, at \(x=1/2\) the value

\begin{equation*} P_5(1/2) = 1- \frac{1}{2!}\left(\frac{1}{2}\right)^2 + \frac{1}{4!}\left(\frac{1}{2}\right)^4 = \frac{337}{384} \approx 0.877604167. \end{equation*}

is within \(0.000748\) from \(\cos(1/2)\text{.}\) Pretty good approximation that one can compute by hand.

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Taylor series (expansion) can be used in computing limits.

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Example 5.78.

To compute the limit

\begin{equation*} \lim_{x \to 0} \frac{\ln(1-x) + x + \frac{x^2}{2}}{4x^3}, \end{equation*}

first recall that for \(x\) near \(0\text{,}\)

\begin{equation*} \ln(1+x) = \sum_{n=0}^{\infty} (-1)^n \frac{x^{n+1}}{n+1}. \end{equation*}

Thus, for \(x\) near \(0\text{,}\)

\begin{align*} \ln(1-x) & = \ln(1+(-x)) = \sum_{n=0}^{\infty} (-1)^n \frac{(-1)^{n+1}x^{n+1}}{n+1}\\ & = -\sum_{n=0}^{\infty} \frac{x^{n+1}}{n+1} = -x - \frac{x^2}{2} - \frac{x^3}{3} - \sum_{n=3}^{\infty} \frac{x^{n+1}}{n+1}. \end{align*}

And so,

\begin{align*} \frac{\ln(1-x) + x + \frac{x^2}{2}}{4x^3} & = -\frac{1}{4x^3}\frac{x^3}{3} - \frac{1}{4x^3}\sum_{n=3}^{\infty} \frac{x^{n+1}}{n+1}\\ & = -\frac{1}{12} - \frac{1}{4x^3}\left( \frac{x^4}{4} + \frac{x^5}{5} + \cdots\right)\\ & = -\frac{1}{12} - \frac{1}{4}\left( \frac{x}{4} + \frac{x^2}{5} + \cdots\right) \to -\frac{1}{12} \end{align*}

as \(x \to 0.\)

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The binomial series is the Taylor series of the binomial function \((1+x)^{\alpha}, (\alpha \in \mathbb{R})\) about \(0\text{.}\) One verifies directly that the binomial series is

\begin{equation} \sum_{n=0}^{\infty} \binom{\alpha}{n} x^n\tag{5.9} \end{equation}

where \(\binom{\alpha}{n} = \frac{\alpha(\alpha-1) \cdots (\alpha-n+1)}{n!}\)

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Here we state without proof the analyticity of \((1+x)^{\alpha}\) about 0 and the interval of convergence of the binomial series. We proof part of it in Appendix D. For a proof of the full statement, read this set of notes.

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Theorem 5.79.

The interval of convergence \(I\) of the binomial series is

\begin{equation*} \begin{cases} [-1,1] & \mbox{if } \alpha \gt 0 \\ (-1,1] & \mbox{if } -1 \lt \alpha \lt 0 \\ (-1,1) & \mbox{if } \alpha \le -1. \end{cases} \end{equation*}

On \(I\) we have

\begin{equation*} (1+x)^{\alpha} = \sum_{n=0}^{\infty} \binom{\alpha}{n} x^n. \end{equation*}

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Example 5.80.

Let us apply the binomial series to find the Taylor series expansion of \(f(x)=x^{\alpha}\) at \(x=3\text{.}\) We are going to express \(x^{\alpha}\) as a power series of the form \(\sum a_n(x-3)^n\text{,}\) so we write \(x^{\alpha}\) as \((3+(x-3))^{\alpha} = 3^{\alpha}(1+(x-3)/3)^{\alpha}\text{.}\) Expand the power as a binomial series and we get

\begin{equation*} x^{\alpha} = 3^{\alpha}\sum_{n=0}^{\infty} \binom{\alpha}{n}\frac{1}{3^n}(x-3)^n \end{equation*}

So \(\ds a_n = \binom{\alpha}{n}\frac{3^{\alpha}}{3^n}\text{.}\) Since

\begin{equation*} \left|\frac{a_{n+1}}{a_n}\right| = \frac{1}{3}\left|\frac{\alpha}{n+1}-1\right| \to \frac{1}{3}, \end{equation*}

therefore, the radius of convergence of the Taylor series is \(3\) and the convergence of the Taylor series the end points \(x=0,6\) depends on \(\alpha\) as in Theorem 5.79. For example, if \(\alpha > 0\text{,}\) the series converges at both end points and hence the interval of convergence is \([0,6]\text{.}\)

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