OBJECTIVES

When you finish this section, you should be able to:

1. Express a function as a Taylor series or a Maclaurin series (p. 613)
2. Determine the convergence of a Taylor/Maclaurin series (p. 614)
3. Find Taylor/Maclaurin expansions (p. 616)
4. Work with a binomial series (p. 619)

ORIGINS

Colin Maclaurin (1698–1746) was a Scottish mathematician. An orphan, he and his brother were raised by an uncle. Maclaurin was 11 years old when he entered the University of Glasgow, and within a year he had taught himself Euclid’s elements (geometry). Maclaurin was 19 when he was appointed professor of Mathematics at the University of Aberdeen, Scotland. Maclaurin was an avid supporter of Newton’s mathematics. In response to a claim that Newton’s calculus lacked rigor, Maclaurin wrote the first systematic treatise on Newton’s methods. In it he used Taylor series centered about \(0,\) which are now known as Maclaurin series.

THEOREM Taylor Series; Maclaurin Series

Suppose \(f\) is a function that has derivatives of all orders on the open interval \((c-R,c+R),\) \(R>0.\) If \(f\) can be represented by the power series \(\sum\limits_{k\,=\,0}^{\infty }a_{k}\,(x-c)^{k},\) whose radius of convergence is \(R\), then \[\bbox[5px, border:1px solid black, #F9F7ED]{\bbox[#FAF8ED,4pt]{ \begin{array}{@{\hskip0pc}rcl} f(x)&=&f(c)+f^{\prime} (c)(x-c)+\frac{f^{\prime \prime} (c) }{2!}(x-c) ^{2}+\cdots +\frac{f^{(n) }(c)\,}{n!}(x-c)^{n}+\cdots \\[11pt] &=&\sum\limits_{k\,=\,0}^{\infty }\frac{f^{(k) }(c) }{k!}(x-c) ^{k} \hspace{120pt} (2)\\ \\ \end{array}}} \]

for all numbers \(x\) in the open interval \((c-R,\,c+R)\). A power series that has the form of equation (2) is called a Taylor series of the function \(f\).

When \(c=0\), the Taylor series representation of a function \(f\) \[\bbox[5px, border:1px solid black, #F9F7ED]{\bbox[#FAF8ED,5pt]{ f(x)=f(0)+f^{\prime} (0)\,x+\frac{f^{\prime \prime}(0) }{2!}x^{2}+\cdots +\frac{f^{(n)}(0)}{n!}x^{n}+\cdots =\sum\limits_{k\,=\,0}^{\infty }\frac{f^{(k) }(0) }{k!}x^{k} }} \]

is called a Maclaurin series.

Expressing a Function as a Maclaurin Series

Assuming that \(f(x)=e^{x}\) can be represented by a power series in \(x\), find its Maclaurin series.

Solution To express a function \(f\) as a Maclaurin series, we begin by evaluating \(f\) and its derivatives at \(0\). \[ \begin{array}{rl@{\qquad}rll} f(x) & =e^{x} & f(0) &=1 \\ f^{\prime} (x)& =e^{x} & f^{\prime} (0) & =1 \\ f^{\prime \prime} (x) & =e^{x} & f^{\prime \prime} (0) &=1 \\ & \vdots & & \vdots \end{array} \]

Then we use the definition of a Maclaurin series.

NOW WORK

Problem 3.

THEOREM Taylor’s Formula with Remainder

Let \(f\) be a function whose first \(n+1\) derivatives are continuous on an open interval \(I\) containing the number \(c\). Then for every \(x\) in \(I\), there is a number \(u\) between \(x\) and \(c\) for which \[ f(x) =f(c) +f^\prime (c) (x-c) +\frac{f^{\prime \prime} (c) }{2!}(x-c) ^{2}+ \cdots +\,\frac{f^{(n) }(c) }{n!}(x-c) ^{n}+R_{n}(x) \]

where \[ R_{n}(x) =\frac{f^{(n+1)}(u)}{(n+1) !}(x-c) ^{n+1} \]

is the remainder after \(n\) terms.

THEOREM Convergence of a Taylor Series

If a function \(f\) has derivatives of all orders in an open interval \(I=\) \((c-R,c+R),\) \(R>0,\) centered at \(c,\) and if \[ \lim\limits_{n\rightarrow \infty }R_{n}(x)=0 \]

for all numbers \(x\) in \(I\), then \[\bbox[5px, border:1px solid black, #F9F7ED]{\bbox[#FAF8ED,5pt]{ \begin{array}{ll} f(x)=\sum\limits_{k\,=\,0}^{\infty }\frac{f^{(k)}(c)}{ k!}\,(x-c)^{k}&= f(c)+f^{\prime} (c)(x-c)+\frac{f^{\prime\prime} (c)}{2!} (x-c)^{2}\\ &\quad+\cdots +\frac{f^{(n)}(c)}{n!}(x-c)^{n}+\cdots \end{array}}} \]

for all numbers \(x\) in \(I\).

Determining the Convergence of a Maclaurin Series

Show that \(1+x+\frac{x^{2}}{2!}+\frac{x^{3}}{3!}+\cdots +\frac{x^{n}}{n!}+\cdots\) converges to \(e^{x}\) for every number \(x\). That is, prove that \[\bbox[5px, border:1px solid black, #F9F7ED]{\bbox[#FAF8ED,5pt]{ e^{x}=1+\frac{x}{1!}+\frac{x^{2}}{2!}+\frac{x^{3}}{3!} +\cdots +\frac{x^{n}}{n!}+\cdots =\sum\limits_{k\,=\,0}^{\infty }\frac{ x^{k}}{k!}}} \]

for all real numbers.

Solution To prove that \(1+\frac{x}{1!}+\frac{x^{2}}{2!}+\frac{x^{3}}{3!}+\cdots =e^{x}\) for every number \(x\), we need to show that \(\lim\limits_{n\rightarrow \infty }R_{n}(x)=0\). Since \(f^{\,(n+1)}(x)=e^{x}\), we have \[ R_{n}(x)=\frac{f^{(n+1)}(u) x^{n+1}}{(n+1) !}=\frac{e^{u}x^{n+1}}{(n+1)!} \]

where \(u\) is between \(0\) and \(x.\) To show that \(\lim\limits_{n\rightarrow\infty }R_{n}(x) = 0\), we consider two cases: \(x>0\) and \(x<0.\)

Finding the Maclaurin Expansion for \(f(x) = \sin x\)

1. Find the Maclaurin expansion for \(f(x)=\sin x.\)
2. Show that it converges to \(\sin x\) for all numbers \(x.\)

Solution (a) The value of \(f\) and its derivatives at \(0\) are \[ \begin{array}{rl@{\qquad}rll} f(x) &=\sin x & f(0) &=0 \\ f^{\prime} (x) &=\cos x & f^\prime (0) &=1 \\ f^{\prime \prime} (x)&=-\sin x & f^{\prime \prime} (0) &=0 \\ f^{\prime \prime \prime}(x) &=-\cos x & f^{\prime \prime \prime} (0) &=-1 \end{array} \]

Higher-order derivatives follow this same pattern, so if \(f(x) =\sin x\) can be represented by a power series in \(x,\) then \[ \begin{eqnarray*} \sin x &=& f(0) +\frac{f^{\prime} (0) }{1!}x+\frac{ f^{\prime \prime} (0) }{2!}x^{2}+\frac{f^{\prime \prime \prime} (0) }{3!}x^{3}+\cdots\,=x-\frac{x^{3}}{3!}+\frac{x^{5}}{5!}- \frac{x^{7}}{7!}+\cdots\\[4pt] &=& \sum\limits_{k\,=\,0}^{\infty }\,(-1) ^{k}\frac{x^{2k+1}}{(2k+1) !} \end{eqnarray*} \]

(b) To prove that the series actually converges to \(\sin x\) for all \(x\), we need to show that the remainders \(R_{2n+1}(x)\) and \(R_{2n}(x)\) approach zero. For \(R_{2n+1}(x)\) we have \[ R_{2n+1}(x)=(-1)^{n+1}\frac{f^{(2n+2)}(u)\,x^{2n+2}}{(2n+2)!} \]

where \(u\) is between \(0\) and \(x\). Since \(\vert f^{(2n+2)}(u)\vert = \vert \sin u \vert \leq 1\) for every number \(u\), then \[ 0\leq \vert R\,_{2n+1}(x)\vert =\frac{\left\vert f^{(2n+2)}(u)\right\vert }{(2n+2)!}\vert x\vert ^{2n+2}\leq \frac{ \vert x\vert ^{2n+2}}{(2n+2)!} \]

By the Ratio Test, the series \(\sum\limits_{k\,=\,0}^{\infty }\frac{|x|^{2k+2} }{( 2k+2) !}\) converges for all \(x\), so \[ \lim_{n\,\dot{\rightarrow}\infty }\frac{|x|^{2n+2}}{(2n+2)!}=0 \]

By the Squeeze Theorem, \(\lim\limits_{n\rightarrow \infty }\)\(\vert R_{2n+1}(x)\vert =0\) and therefore \(\lim\limits_{n\rightarrow \infty }\) \(R_{2n+1}(x)=0\) for all \(x.\)

A similar argument holds for the remainder \(\left\vert R_{2n}(x) \right\vert =\frac{\left\vert \cos u\right\vert x^{2n+1}}{(2n+1) !}.\)

We conclude that the series \(\sum\limits_{k\,=\,0}^{\infty }\,(-1) ^{k}\frac{x^{2k+1}}{(2k+1) !}\) converges to \(\sin x\) for all \(x.\) That is, \[\bbox[5px, border:1px solid black, #F9F7ED]{\bbox[#FAF8ED,5pt]{ \sin x=x-\frac{x^{3}}{3!}+\frac{x^{5}}{5!} -\cdots +(-1)^{n}\frac{x^{2n+1}}{(2n+1)!}+\cdots =\sum\limits_{k\,=\,0}^{\infty }\,(-1) ^{k}\frac{x^{2k+1}}{(2k+1)!} }} \]

Finding the Maclaurin Expansion for \(f(x) =\cos x\)

Find the Maclaurin expansion for \(f(x) =\cos x.\)

Solution We apply the differentiation property of power series to the Maclaurin expansion for \(\sin x.\) \[ \begin{eqnarray*} \frac{d}{dx}\sin x &=& \frac{d}{dx}\!\left( x-\frac{x^{3}}{3!}+\frac{x^{5}}{5!} -\cdots +(-1)^{n}\frac{x^{2n+1}}{(2n+1)!}+\cdots \right) \\[5pt] &=&\frac{d}{dx}\sum\limits_{k\,=\,0}^{\infty }\,(-1) ^{k}\frac{x^{2k+1}}{ (2k+1) !} \end{eqnarray*} \]

Then \[\bbox[5px, border:1px solid black, #F9F7ED]{\bbox[#FAF8ED,5pt]{ \cos x=1-\frac{x^{2}}{2!}+\frac{x^{4}}{4!}-\cdots +(-1)^{n}\frac{x^{2n}}{(2n)!}+\cdots =\sum\limits_{k\,=\,0}^{\infty }\,(-1) ^{k}\frac{x^{2k}}{(2k) !}}} \]

for all numbers \(x\).

NOTE

The Maclaurin expansion for \(f(x)=\cos x\) could also have been found by integrating the Maclaurin expansion for \(f(x)=\sin x\).

Finding the Maclaurin Expansion for \(f(x) =\cosh x\)

Find the Maclaurin expansion for \(f(x) =\cosh x.\)

Solution Since \[ \cosh x=\frac{e^{x}+e^{-x}}{2} \]

its Maclaurin expansion can be found by adding corresponding terms of the Maclaurin expansions for \(e^{x}\) and \(e^{-x}\) and then dividing by 2. The result is \[ \begin{eqnarray*} \hspace{-10pc} \cosh x &=&\frac{e^{x}+e^{-x}}{2}=\frac{\left( 1+x+\frac{x^{2}}{2!}+ \frac{x^{3}}{3!}+\cdots +\frac{x^{n}}{n!}+\cdots \right) +\left( 1-x+ \frac{x^{2}}{2!}-\frac{x^{3}}{3!}+\cdots +(-1)^{n}\frac{x^{n}}{n!}+\cdots \right) }{2} \\[10pt] \hspace{-10pc} &=&\frac{1+1}{2}+\frac{x-x}{2}+\frac{x^{2}+x^{2}}{2\cdot 2!}+\frac{ x^{3}-x^{3}}{2\cdot 3!}+\cdots+\frac{x^{n}+(-1) ^{n}x^{n}}{ 2\cdot n!}+\cdots \end{eqnarray*} \]

\[\bbox[5px, border:1px solid black, #F9F7ED]{\bbox[#FAF8ED,5pt]{ \cosh x=1+\frac{x^{2}}{2!}+\frac{x^{4}}{4!}+\frac{x^{6}}{6!}+\cdots + \frac{x^{2n}}{(2n)!}+\cdots =\sum\limits_{k\,=\,0}^{\infty }\,\frac{x^{2k}}{(2k) !}}} \]

for all \(x.\)

NOW WORK

Problem 27.

Finding the Maclaurin Expansion for \(f(x) =e^{x}\cos x\)

Find the first five terms of the Maclaurin expansion for \(f(x) = e^{x}\cos x.\)

Solution The Maclaurin expansion for \(f(x) =\) \(e^{x}\cos x\) is obtained by multiplying the Maclaurin expansion for \(e^{x}\) by the Maclaurin expansion for \(\cos x\). That is, \[ e^{x}\cos x=\left( 1+x+\frac{x^{2}}{2!}+\frac{x^{3}}{3!}+\frac{x^{4}}{4!}+ \frac{x^{5}}{5!}+\cdots \right) \left( 1-\frac{x^{2}}{2!}+\frac{x^{4}}{4!} -\cdots \right) \]

Then \[ \begin{eqnarray*} e^{x}\cos x &=& 1 \!\left( 1-\frac{x^{2}}{2!}+\frac{x^{4}}{4!}-\cdots\right) +x\!\left( 1-\frac{x^{2}}{2!}+\frac{x^{4}}{4!}-\cdots\right) \\[5pt] && +\, \frac{x^{2}}{2!}\left( 1-\frac{x^{2}}{2!}+\frac{x^{4}}{4!}-\cdots\right) +\frac{x^{3}}{3!}\left( 1-\frac{x^{2}}{2!}+\frac{x^{4}}{4!}-\cdots\right) \\[5pt] && +\, \frac{x^{4}}{4!}\left(1-\frac{x^{2}}{2!}+\frac{x^{4}}{4!}-\cdots \right)+\frac{x^{5}}{5!} \left(1-\frac{x^{2}}{2!}+\frac{x^{4}}{4!}-\cdots\right)+\cdots \\[5pt] &=& \left( 1-\frac{x^{2}}{2}+\frac{x^{4}}{24}\right) +\left( x-\frac{x^{3}}{ 2}+\frac{x^{5}}{24}\right) +\left( \frac{x^{2}}{2}-\frac{x^{4}}{4}\right) +\left( \frac{x^{3}}{6}-\frac{x^{5}}{12}\right) \\[5pt] && +\, \frac{x^{4}}{24}+\frac{x^{5}}{120}+\cdots \\[5pt] &=& 1+x+\left( -\frac{1}{2}+\frac{1}{2}\right) x^{2}+\left( -\frac{1}{2}+ \frac{1}{6}\right) x^{3}+\left( \frac{1}{24}-\frac{1}{4}+\frac{1}{24}\right) x^{4}\\[5pt] &&+\,\left( \frac{1}{24}-\frac{1}{12}+\frac{1}{120}\right) x^{5}+\cdots \\[5pt] &=& 1+x-\frac{1}{3}x^{3}-\frac{1}{6}x^{4}-\frac{1}{30}x^{5}+\cdots \end{eqnarray*} \]

NOW WORK

Problem 29.

Finding the Taylor Expansion for \(f(x) =\cos x\) about \(\frac{\pi }{2}\)

Find the Taylor expansion for \(f(x) =\cos x\) about \(\frac{\pi}{2}.\)

Solution To express \(f(x) =\cos x\) as a Taylor expansion about \(\frac{\pi}{2}\), we evaluate \(f\) and its derivatives at \(\frac{\pi}{2}.\) \[ \begin{array}{rl@{\qquad}rll} f(x) &=\cos x & f\!\!\left(\frac{\pi }{2}\right)& =0 \\ f^\prime (x) &=-\sin x & f^\prime \!\!\left( \frac{\pi }{2}\right) &=-1 \\ f^{\prime \prime} (x) &=-\cos x & f^{\prime \prime} \!\!\left( \frac{\pi }{2}\right) &=0 \\ f^{\prime \prime \prime} (x) &=\sin x & f^{\prime \prime \prime} \!\!\left( \frac{\pi }{2}\right) &=1 \end{array} \]

For derivatives of odd order, \(f^{(2n+1) }\left( \frac{\pi }{2}\right) =(-1) ^{n+1}.\) For derivatives of even order, \(f^{(2n) }\left( \frac{\pi }{2}\right) =0.\) The Taylor expansion for \(f(x) =\cos x\) about \(\frac{\pi }{2}\) is

\[ \begin{eqnarray*} f(x) &=&\cos x=f\!\!\left( \frac{\pi }{2}\right) +f^\prime \!\!\left( \frac{\pi }{2}\right) \left( x-\frac{\pi }{2}\right) +\frac{f^{\prime \prime} \!\!\left( \frac{\pi }{2}\right) }{2!}\left( x-\frac{\pi }{2}\right) ^{2}\\[6pt] &&+\, \frac{f^{\prime \prime \prime} \!\!\left( \frac{\pi }{2}\right) }{3!}\left( x-\frac{\pi }{2}\right) ^{3}+\cdots \\[6pt] &=&-\left( x-\frac{\pi }{2}\right) +\frac{1}{3!}\left( x-\frac{\pi }{2} \right) ^{3}-\frac{1}{5!}\left( x-\frac{\pi }{2}\right) ^{5}+\text{ }\cdots \\[6pt] &=& \sum\limits_{k\,=\,0}^{\infty }\frac{(-1) ^{k+1}}{(2k+1) !}\left( x-\frac{\pi }{2}\right) ^{2k+1} \end{eqnarray*} \]

The radius of convergence is \(\infty ;\) the interval of convergence is \((-\infty ,\infty).\)

NOW WORK

Problem 15.

NEED TO REVIEW?

The Binomial Theorem is discussed in Appendix A.5, p. A-43.

Finding the Maclaurin Expansion for \(f(x) =(1+x) ^{m}\)

Find the Maclaurin series for \(f(x) =(1+x) ^{m},\) where \(m\) is any real number.

Solution We begin by finding the derivatives of \(f\) at \(0\): \[ \begin{array}{@{\hspace{-8pc}}ll@{\qquad}rll} f(x)&=&(1+x)^{m} && f(0)&=&1 \\ f^\prime (x)&=&m(1+x)^{m-1} && f^{\prime} (0)&=&m \\ f^{\prime \prime} (x)&=&m(m-1)(1+x)^{m-2} && f^{\prime \prime} (0)&=&m(m-1) \\ \vdots & & & & & \vdots & & & &\\ f^{(n)}(x)&=&m(m-1)(m-2)\cdots (m-n+1)(1+x)^{m-n} && f^{(n)}(0) &=& m(m-1)(m-2)\,\cdots \,(m-n+1) \end{array} \]

The Maclaurin expansion for \(f\) is \[ \begin{eqnarray*} (1+x)^{m} &=&1+m\,x+\frac{m(m-1)}{2!}x^{2}+\cdots \\[4pt] && +\, \frac{m(m-1)(m-2)\,\cdots \,(m-n+1)}{n!}x^{n}+\cdots \\[4pt] &=&{m\choose 0}+{m\choose 1}x+{m\choose 2}x^{2}+\cdots +{m\choose n}x^{n}+\cdots =\sum_{k\,=\,0}^{\infty }{m\choose k}\,x^{k} \end{eqnarray*} \]

where \[\bbox[5px, border:1px solid black, #F9F7ED]{\bbox[#FAF8ED,5pt]{ {m\choose 0} =1 \quad\hbox{and}\quad {m\choose k} =\frac{m (m-1) (m-2) \,\cdots\, (m-k+1) }{k!} }} \]

THEOREM Convergence of a Binomial Series

The binomial series \[ (1+x)^{m}=\sum_{k\,=\,0}^{\infty }{m\choose k}x^{k} \] converges

• for all \(x\) if \(m\) is a nonnegative integer. (In this case, there are only \(m+1\) nonzero terms.)
• on the open interval \((-1,1)\) if \(m\leq -1\).
• on the half-open interval \((-1,\,1]\) if \(-1<m<0\).
• on the closed interval \([-1,\,1]\) if \(m>0\), but \(m\) is not an integer.

Using a Binomial Series

Represent the function \(f(x) =\sqrt{x+1}\) as a Maclaurin series, and find its interval of convergence.

Solution Write \(\sqrt{x+1}=(1+x) ^{1/2}\) and use the binomial series with \(m=\frac{1}{2}\). The result is \[ \begin{eqnarray*} (1+x)^{1/2} &=& 1+\frac{1}{2}x+\frac{\left( \frac{1}{2}\right) \left( -\frac{1 }{2}\right) }{2!}\,x^{2}+\frac{\frac{1}{2}\left( -\frac{1}{2}\right) \left( -\frac{3}{2}\right) }{3!}x^{3}+\cdots \\[12pt] &=& 1+\frac{1}{2}x-\frac{1}{8} x^{2}+\frac{1}{16}x^{3} - \cdots \end{eqnarray*} \]

Since \(m=\frac{1}{2}>0\) and \(m\) is not an integer, the series converges on the closed interval \([-1,\,1]\).

NOW WORK

Problem 37.

Using a Binomial Series

Represent the function \(f(x)=\sin ^{-1}x\) by a Maclaurin series.

Solution Recall that \[ \sin ^{-1}x=\int_{0}^{x}\frac{dt}{\sqrt{1-t^{2}}}=\int_{0}^{x}(1-t^{2})^{-1/2}dt \]

We write the integrand as a binomial series, with \(x=-t^{2}\) and \(m=-\frac{1}{2}\). \[ \begin{eqnarray*} (1-t^{2})^{-1/2} &=&1+\left( -\frac{1}{2}\right) (-t^{2}) + \frac{\left( -\frac{1}{2}\right) \left( -\frac{3}{2}\right) }{2!}(-t^{2}) ^{2}\\[8pt] && +\frac{\left( -\frac{1}{2}\right) \left( -\frac{3}{2} \right) \left( -\frac{5}{2}\right) }{3!}(-t^{2}) ^{3}+\cdots \\[8pt] &=&1+\frac{1}{2}t^{2}+\frac{3}{8}\,t^{4}+\frac{5}{16}t^{6}+\cdots \end{eqnarray*} \]

Now we use the integration property of a power series to obtain \[ \begin{eqnarray*} \int_{0}^{x}\frac{dt}{\sqrt{1-t^{2}}} &=&\int_{0}^{x}\left( 1+\frac{1}{2} t^{2}+\frac{3}{8}\,t^{4}+\frac{5}{16}t^{6}+\cdots \right)\,dt \\ \sin ^{-1}x &=& x+\left( \frac{1}{2}\right) \left( \frac{x^{3}}{3}\right) +\left( \frac{3}{8}\right) \left( \frac{x^{5}}{5}\right) +\left( \frac{5}{16} \right) \left( \frac{x^{7}}{7}\right) +\cdots \\ &=& x+\frac{x^{3}}{6}+\frac{3}{40}x^{5}+\frac{5}{112}x^{7}+\cdots \end{eqnarray*} \]