THEOREM 1 Taylor Series Expansion

If \(f(x)\) is represented by a power series centered at \(a\) in an interval \(|x-a|< R\) with \(R>0\), then that power series is the Taylor series \[ \boxed{\bbox[#FAF8ED,5pt]{T(x) = \sum\limits_{n=0}^\infty \frac{f^{(n)}(a)}{n!}(x-a)^n}} \]

EXAMPLE 1

Find the Taylor series for \(f(x)=x^{-3}\) centered at \(a=1\).

Solution The derivatives of \(f(x)\) are \(f'(x)=-3x^{-4}\), \(f''(x)=(-3)(-4)x^{-5}\), and in general, \[ f^{(n)}(x) = (-1)^n(3)(4)\cdots(n+2) x^{-3-n} \]

Note that \((3)(4)\cdots(n+2) =\tfrac12 (n+2)!\). Therefore, \[f^{(n)}(1) = (-1)^n\frac12(n+2)!\] Noting that \((n+2)! = (n+2)(n+1)n!\), we write the coefficients of the Taylor series as: \[ a_n = \frac{f^{(n)}(1)}{n!}=\frac{(-1)^n\frac12(n+2)!}{n!} = (-1)^n\frac{(n+2)(n+1)}2 \]

The Taylor series for \(f(x)=x^{-3}\) centered at \(a=1\) is \[ \begin{align*} T(x) &= 1-3(x-1)+6(x-1)^2-10(x-1)^3+\cdots \\&= \sum\limits_{n=0}^\infty\,(-1)^n\frac{(n+2)(n+1)}2(x-1)^n \end{align*} \]

See Exercise 92 for an example where a Taylor series \(T(x)\) converges but does not converge to \(f(x)\).

Reminder

\(f(x)\) is called “infinitely differentiable” if \(f^{(n)}(x)\) exists for all \(n\).

THEOREM 2

Let \(I =(a-R, a+R)\), where \(R>0\). Suppose there exists \(K>0\) such that all derivatives of \(f\) are bounded by \(K\) on \(I\): \[ \big|f^{(k)}(x)\big|\le K\qquad \textrm{for all}\quad k\geq 0\quad \textrm{and}\quad x\in I \] Then \(f(x)\) is represented by its Taylor series in \(I\): \[ f(x) = \sum_{n=0}^\infty \frac{f^{(n)}(a)}{n!} (x-a)^n \qquad\textrm{for all}\quad x\in I \]

Proof

According to the Error Bound for Taylor polynomials (Theorem 2 in Section 8.4), \[ |R_k(x)|=|f(x)-T_k(x)| \le K\frac{|x-a|^{k+1}}{(k+1)!} \] If \(x \in I\), then \(|x-a|<R\) and \[ |R_k(x)| \le K\frac{R^{k+1}}{(k+1)!} \] We showed in Example 9 of Section 10.1 that \(R^k/k!\) tends to zero as \(k\to\infty\). Therefore, \(\lim\limits_{k\to\infty} R_k(x)=0\) for all \(x\in (a-R, a+R)\), as required.

Taylor expansions were studied throughout the seventeenth and eighteenth centuries by Gregory, Leibniz, Newton, Maclaurin, Taylor, Euler, and others. These developments were anticipated by the great Hindu mathematician Madhava (c. 1340–1425), who discovered the expansions of sine and cosine and many other results two centuries earlier.

EXAMPLE 2 Expansions of Sine and Cosine

Show that the following Maclaurin expansions are valid for all \(x\). \[ \boxed{\bbox[#FAF8ED,5pt]{\sin x = \sum\limits_{n=0}^{\infty} (-1)^n\frac{x^{2n+1}}{(2n+1)!} = x- \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots}} \]

\[ \boxed{\bbox[#FAF8ED,5pt]{\cos x = \sum\limits_{n=0}^{\infty} (-1)^n\frac{x^{2n}}{(2n)!} = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \cdots}} \]

Solution Recall that the derivatives of \(f(x) = \sin x\) and their values at \(x=0\) form a repeating pattern of period 4:

In other words, the even derivatives are zero and the odd derivatives alternate in sign: \(f^{(2n+1)}(0) = (-1)^n\). Therefore, the nonzero Taylor coefficients for \(\sin x\) are \[a_{2n+1} = \frac{(-1)^n}{(2n+1)!}\] For \(f(x)=\cos x\), the situation is reversed. The odd derivatives are zero and the even derivatives alternate in sign: \(f^{(2n)}(0)=(-1)^n\cos 0 = (-1)^n\). Therefore the nonzero Taylor coefficients for \(\cos x\) are \({a_{2n} = {(-1)^n}/{(2n)!}}\).

We can apply Theorem 2 with \(K=1\) and any value of \(R\) because both sine and cosine satisfy \(|f^{(n)}(x)|\le 1\) for all \(x\) and \(n\). The conclusion is that the Taylor series converges to \(f(x)\) for \(\left|x\right|<R\). Since \(R\) is arbitrary, the Taylor expansions hold for all \(x\).

EXAMPLE 3 Taylor Expansion of \(f(x) = e^x\) at \(x=a\)

Find the Taylor series \(T(x)\) of \(f(x) = e^x\) at \(x=a\).

Solution We have \(f^{(n)}(a)=e^a\) for all \(x\), and thus \[ T(x) = \sum_{n=0}^\infty \frac{e^a}{n!} (x-a)^n \] Because \(e^x\) is increasing for all \(R>0\) we have \(|f^{(k)}(x)|\le e^{a+R}\) for \(x \in (a-R, a+R)\). Applying Theorem 2 with \(K=e^{a+R}\), we conclude that \(T(x)\) converges to \(f(x)\) for all \(x \in (a-R, a+R)\). Since \(R\) is arbitrary, the Taylor expansion holds for all \(x\). For \(a=0\), we obtain the standard Maclaurin series \[\boxed{\bbox[#FAF8ED,5pt]{e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots}} \]

EXAMPLE 4

Find the Maclaurin series for \(f(x) = x^2e^x\).

Solution Multiply the known Maclaurin series for \(e^x\) by \(x^2\). \[ \begin{eqnarray*} x^2e^x & = & x^2\left(1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \frac{x^5}{5!} + \cdots\right)\\ &=& x^2 +x^3+ \frac{x^4}{2!} + \frac{x^5}{3!} + \frac{x^6}{4!} +\frac{x^7}{5!}+\cdots = \sum_{n = 2}^{\infty} \frac{x^{n}}{(n-2)!} \end{eqnarray*} \]

In Example 4, we can also write the Maclaurin series as \[ \sum_{n = 0}^{\infty} \frac{ x^{n+2} } { n! }\]

EXAMPLE 5 Substitution

Find the Maclaurin series for \(e^{-x^2}\).

Solution Substitute \(-x^2\) in the Maclaurin series for \(e^x\). \[ \begin{equation*} e^{-x^2}= \sum_{n=0}^\infty \frac{(-x^2)}{n!}^n = \sum_{n=0}^\infty \frac{(-1)^nx^{2n}}{n!}= 1 - x^2 + \frac{x^4}{2!} - \frac{x^6}{3!} + \frac{x^8}{4!} - \cdots\tag{2} \end{equation*} \] The Taylor expansion of \(e^x\) is valid for all \(x\), so this expansion is also valid for all \(x\).

EXAMPLE 6 Integration

Find the Maclaurin series for \(f(x) = \ln (1 + x)\).

Solution We integrate the geometric series with common ratio \(-x\) (valid for \(\left|x\right|<1\)): \[ \begin{align*} \frac1{1+x}&= 1-x+x^2-x^3+\cdots \\ \ln(1+x)&=\int\frac{dx}{1+x}= x-\frac{x^2}2+\frac{x^3}3-\frac{x^4}4+\cdots= \sum_{n=1}^{\infty} (-1)^{n-1} \frac{x^n}{n} \end{align*} \] The constant of integration on the right is zero because \(\ln(1+x) = 0\) for \(x=0\). This expansion is valid for \(\left|x\right|<1\). It also holds for \(x=1\) (see Exercise 84.)

EXAMPLE 7 Multiplying Taylor Series

Write out the terms up to degree five in the Maclaurin series for \(f(x) = e^x \cos x\).

Solution We multiply the fifth-order Taylor polynomials of \(e^x\) and \(\cos x\) together, dropping the terms of degree greater than 5: \[ \left(1+x +\frac{x^2}2 + \frac{x^3}6 + \frac{x^4}{24}+\frac{x^5}{120} \right)\left(1 - \frac{x^2}{2} + \frac{x^4}{24} \right) \] Distributing the term on the left (and ignoring terms of degree greater than \(5\)), we obtain \[ \begin{eqnarray*} &\displaystyle\left(1+x +\frac{x^2}2 + \frac{x^3}6 + \frac{x^4}{24}+\frac{x^5}{120} \right) - \left(1+x +\frac{x^2}2 + \frac{x^3}6\right)\left(\frac{x^2}{2}\right)+ (1+x) \left(\frac{x^4}{24}\right)&\\ &= \underbrace{ 1+x-\frac{x^3}3-\frac{x^4}6 -\frac{x^5}{30}}_{\text{Retain terms of degree \(\leq 5\)}}{}& \end{eqnarray*} \] We conclude that the fifth Maclaurin polynomial for \(f(x) = e^x \cos x\) is \[ T_5(x) = 1+x-\frac{x^3}3-\frac{x^4}6 -\frac{x^5}{30} \]

EXAMPLE 8

Let \(J=\int_0^1 \sin(x^2)\,dx\).

1. Express \(J\) as an infinite series.
2. Determine \(J\) to within an error less than \(10^{-4}\).

Solution

1. The Maclaurin expansion for \(\sin x\) is valid for all \(x\), so we have \[ \sin x = \sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!}x^{2n+1}\quad \Rightarrow\quad \sin(x^2) = \sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!}x^{4n+2} \] We obtain an infinite series for \(J\) by integration: \[ \begin{align*} J&=\int_0^1\sin(x^2)\,dx = \sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!}\int_0^1 x^{4n+2}dx =\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!}\left(\frac1{4n+3}\right)\notag\\ &= \frac13-\frac1{42} +\frac1{1320}- \frac1{75{,}600} + \cdots\tag{3} \end{align*} \]
2. The infinite series for \(J\) is an alternating series with decreasing terms, so the sum of the first \(N\) terms is accurate to within an error that is less than the \((N+1)\)st term. The absolute value of the fourth term \(1/75{,}600\) is smaller than \(10^{-4}\) so we obtain the desired accuracy using the first three terms of the series for \(J\): \[ J \approx \frac13-\frac1{42} +\frac1{1320} \approx 0.31028 \] The error satisfies \[ \left|J- \left( \frac13-\frac1{42} +\frac1{1320}\right)\right|< \frac1{75{,}600} \approx1.3\times 10^{-5} \] The percentage error is less than 0.005% with just three terms.

When \(a\) is a whole number, \({\binom an}\) is zero for \(n>a\), and in this case, the binomial series breaks off at degree \(n\). The binomial series is an infinite series when \(a\) is not a whole number.

THEOREM 3 The Binomial Series

For any exponent \(a\) and for \(\left|x\right|<1\), \[ \begin{equation*}\boxed{\bbox[#FAF8ED,5pt]{ (1+x)^a = 1 + \frac{a}{1!}x + \frac{a(a-1)}{2!}x^2 + \frac{a(a-1)(a-2)}{3!}x^3 + \cdots + \binom an x^n + \cdots}} \end{equation*} \]

EXAMPLE 9

Find the terms through degree four in the Maclaurin expansion of \[ f(x) = (1 + x)^{4/3} \]

Solution The binomial coefficients \(\binom an\) for \(a = \tfrac{4}{3}\) for \(0<n<4\) are \[ 1, \qquad \frac{\frac{4}{3}}{1!} = \frac43, \qquad \frac{\frac{4}{3}\big(\frac{1}{3}\big)}{2!} = \frac29,\qquad \frac{\frac{4}{3}\big(\frac{1}{3}\big)\big(-\frac{2}{3}\big)}{3!} = -\frac4{81}, \qquad \frac{\frac{4}{3}\big(\frac{1}{3}\big)\big(-\frac{2}{3}\big)\big(-\frac{5}{3}\big)}{4!} = \frac{5}{243} \] Therefore, \( (1 + x)^{4/3} \approx 1 + \tfrac{4}{3}x + \tfrac{2}{9}x^2 - \tfrac{4}{81}x^3 + \tfrac{5}{243}x^4 + \cdots.\)

EXAMPLE 10

Find the Maclaurin series for \[f(x) = \frac1{\sqrt{1-x^2}}\]

Solution First, let’s find the coefficients in the binomial series for \((1+x)^{-1/2}\): \[ 1, \qquad \frac{-\frac12}{1!}=-\frac12, \qquad \frac{-\frac12\big(-\frac32\big)}{1\cdot2} = \frac{1\cdot 3}{2\cdot 4},\qquad \frac{-\frac12\big(-\frac32\big)\big(-\frac52\big)}{1\cdot2\cdot3} = -\frac{1\cdot 3\cdot 5}{2\cdot 4\cdot 6} \] The general pattern is \[ \binom{-\frac12}{n} = \frac{-\frac12 \big(-\frac32\big)\big(-\frac52\big)\cdots\big(-\frac{2n-1}{2}\big)}{1\cdot2\cdot3\cdots n}= (-1)^n \frac{1\cdot3\cdot5\cdots(2n-1)}{2\cdot4\cdot6\cdot2n} \]

Thus, the following binomial expansion is valid for \(\left|x\right|<1\): \[ \frac1{\sqrt{1+x}} = 1 + \sum_{n=1}^\infty (-1)^n\frac{1\cdot 3\cdot 5 \cdots (2n-1)}{2\cdot 4\cdot 6 \cdots (2n)} x^n = 1 - \frac12 x + \frac{1\cdot 3}{2\cdot 4}x^2 - \cdots \] If \(\left|x\right| < 1\), then \(\left|x\right|^2 < 1\), and we can substitute \(-x^2\) for \(x\) to obtain \[ \begin{equation*} \frac{1}{\sqrt{1-x^2}} = 1 + \sum_{n=1}^\infty \frac{1\cdot 3\cdot 5 \cdots (2n-1)}{2\cdot 4\cdot 6 \cdots 2n} x^{2n} = 1 + \frac12x^2+\frac{1\cdot 3 }{2\cdot 4}x^4+\cdots\tag{4} \end{equation*} \]

EXAMPLE 11 Elliptic Function

Find the Maclaurin series for \(E(k)\) and estimate \(E(k)\) for \(k=\sin\tfrac{\pi}{6}\).

Solution Substitute \(x = k\sin t\) in the Taylor expansion (4): \[ \frac1{\sqrt{1-k^2\sin^2t}} = 1 + \frac12k^2~\sin^2t+\frac{1\cdot 3}{2\cdot 4}k^4~\sin^4t+ \frac{1\cdot 3\cdot 5}{2\cdot 4\cdot 6}k^6~\sin^6t+ \cdots \] This expansion is valid because \(\left|k\right|<1\) and hence \(\left|x\right| = \left|k\sin t\right| < 1\). Thus \(E(k)\) is equal to \[ \int_0^{\pi/2}\frac{dt}{\sqrt{1-k^2\sin^2~t}} = \int_0^{\pi/2} dt + \sum_{n=1}^\infty \frac{1\cdot 3 \cdots (2n-1)}{2\cdot 4 \cdot (2n)}\left(\int_0^{\pi/2}\sin^{2n}~t\,dt\right)k^{2n} \] According to Exercise 78 in Section 7.2, \[ \int_0^{\pi/2}\sin^{2n}~t\,dt = \left(\frac{1\cdot 3 \cdots (2n-1)}{2\cdot 4 \cdot (2n)}\right)\frac{\pi}2 \] This yields \[ E(k) = \frac{\pi}{2} + \frac{\pi}2\sum_{n=1}^\infty \left(\frac{1\cdot 3 \cdots (2n-1)^2}{2\cdot 4 \cdots (2n)}\right)^2k^{2n} \]

We approximate \(E(k)\) for \(k =\sin\big(\frac{\pi}6\big)=\frac12\) using the first five terms: \begin{align*} E\left(\frac12\right) &\approx \frac{\pi}2\left(1+\left(\frac12\right)^2\left(\frac12\right)^2+\left(\frac{1\cdot 3}{2\cdot 4}\right)^2\left(\frac12\right)^4\right.\\ &\qquad\quad\left. +\left(\frac{1\cdot 3\cdot 5}{2\cdot 4\cdot 6}\right)^2\left(\frac12\right)^6 +\left(\frac{1\cdot 3\cdot 5\cdot 7}{2\cdot 4\cdot 6\cdot 8}\right)^2\left(\frac12\right)^8 \right)\\ &\approx1.68517 \end{align*} The value given by a computer algebra system to seven places is \(E\big(\frac12\big) \approx 1.6856325\).