9.4: The Binomial Theorem and Applications of Taylor Series

Learning Objectives

• Write the terms of the binomial series.
• Recognize the Taylor series expansions of common functions.
• Recognize and apply techniques to find the Taylor series for a function.
• Use Taylor series to evaluate non-elementary integrals.

In the preceding section, we defined Taylor series and showed how to find the Taylor series for several common functions by explicitly calculating the coefficients of the Taylor polynomials. We also showed how to use those Taylor series to derive Taylor series for other functions. We then showed how power series can be used to evaluate integrals when the antiderivative of the integrand cannot be expressed in terms of elementary functions. We now venture into the final discussion surrounding Taylor series - expansions of powers of expressions involving two terms.

The Binomial Series

Our first goal in this section is to determine the Maclaurin series for the function \( f(x)=(1+x)^r\) for all real numbers \( r\). The Maclaurin series for this function is known as the binomial series. We begin by considering the simplest case: \( r\) is a non-negative integer. We recall that, for \( r=0,\,1,\,2,\,3,\,4,\;f(x)=(1+x)^r\) can be written as\[\begin{array}{rcl}
f(x) & = & (1+x)^0=1, \\[6pt]
f(x) & = & (1+x)^1=1+x, \\[6pt]
f(x) & = & (1+x)^2=1+2x+x^2, \\[6pt]
f(x) & = & (1+x)^3=1+3x+3x^2+x^3, \\[6pt]
f(x) & = & (1+x)^4=1+4x+6x^2+4x^3+x^4. \\[6pt]
\end{array} \nonumber \]The expressions on the right-hand side are known as binomial expansions and the coefficients are known as binomial coefficients. More generally, for any nonnegative integer \( r\), the binomial coefficient of \( x^n\) in the binomial expansion of \( (1+x)^r\) is given by\[\binom{r}{n}=\dfrac{r!}{n!(r−n)!}\label{eq6.6} \]and\[\begin{array}{rcl}
f(x) & = & (1+x)^r \\[6pt]
& = & \displaystyle \binom{r}{0}+\binom{r}{1}x+\binom{r}{2}x^2+\binom{r}{3}x^3+ \cdots +\binom{r}{r-1}x^{r−1}+\binom{r}{r}x^r \\[6pt]
& = & \displaystyle \sum_{n=0}^r\binom{r}{n}x^n. \\[6pt]
\end{array} \label{eq6.7} \]For example, using this formula for \( r=5\), we see that\[ \begin{array}{rcl}
f(x) & = & (1+x)^5 \\[6pt]
& = & \displaystyle \binom{5}{0}1+\binom{5}{1}x+\binom{5}{2}x^2+\binom{5}{3}x^3+\binom{5}{4}x^4+\binom{5}{5}x^5 \\[6pt]
& = & \dfrac{5!}{0!5!}1+\dfrac{5!}{1!4!}x+\dfrac{5!}{2!3!}x^2+\dfrac{5!}{3!2!}x^3+\dfrac{5!}{4!1!}x^4+\dfrac{5!}{5!0!}x^5 \\[6pt]
& = & 1+5x+10x^2+10x^3+5x^4+x^5. \\[6pt]
\end{array} \nonumber \]We now consider the case when the exponent \(r\) is any real number, not necessarily a non-negative integer. If \(r\) is not a nonnegative integer, then \(f(x)=(1+x)^r\) cannot be written as a finite polynomial. However, we can find a power series for \(f\). Specifically, we look for the Maclaurin series for \(f\). To do this, we find the derivatives of \(f\) and evaluate them at \(x=0\).\[ \begin{array}{rclcrcl}
f(x) & = & (1+x)^r & \implies & f(0) & = & 1 \\[6pt]
f^{\prime}(x) & = & r(1+x)^{r−1} & \implies & f^{\prime}(0) & = & r \\[6pt]
f^{\prime\prime}(x) & = & r(r−1)(1+x)^{r−2} & \implies & f^{\prime\prime}(0) & = & r(r−1) \\[6pt]
f^{\prime\prime\prime}(x) & = & r(r−1)(r−2)(1+x)^{r−3} & \implies & f^{\prime\prime\prime}(0) & = & r(r−1)(r−2) \\[6pt]
f^{(n)}(x) & = & r(r−1)(r−2) \cdots (r−n+1)(1+x)^{r−n} & \implies & f^{(n)}(0) & = & r(r−1)(r−2) \cdots (r−n+1) \\[6pt]
\end{array} \nonumber \]We conclude that the coefficients in the binomial series are given by\[\dfrac{f^{(n)}(0)}{n!}=\dfrac{r(r−1)(r−2) \cdots (r−n+1)}{n!}.\label{eq6.8} \]We note that if \(r\) is a non-negative integer, then the \((r+1)^{\text{st}}\) derivative \( f^{(r+1)}\) is the zero function, and the series terminates. In addition, if \( r\) is a non-negative integer, then the coefficients in Equation \ref{eq6.8} agree with those from Equation \ref{eq6.6}, and the formula for the binomial series agrees with Equation \ref{eq6.7} for the finite binomial expansion. More generally, to denote the binomial coefficients for any real number \( r\), we define\[\binom{r}{n}=\dfrac{(r−1)(r−2) \cdots (r−n+1)}{n!}. \nonumber \]With this notation, we can write the binomial series for \( (1+x)^r\) as\[\sum_{n=0}^ \infty \binom{r}{n}x^n=1+rx+\dfrac{r(r−1)}{2!}x^2+ \cdots +\dfrac{r(r−1) \cdots (r−n+1)}{n!}x^n+ \cdots . \label{bin1} \]We now need to determine the interval of convergence for the binomial series Equation \ref{bin1}. To do so, we apply the Ratio Test. Consequently, we consider\[\begin{array}{rcl}
\dfrac{|a_{n+1}|}{|a_n|} & = & \dfrac{|r(r−1)(r−2) \cdots (r−n)|x||^{n+1}}{(n+1)!} \cdot \dfrac{n}{|r(r−1)(r−2) \cdots (r−n+1)||x|^n} \\[6pt]
& = & \dfrac{|r−n||x|}{|n+1|}. \\[6pt]
\end{array} \nonumber \]Since\[\lim_{n \to \infty }\dfrac{|a_{n+1}|}{|a_n|} = |x| \lt 1 \nonumber \]if and only if \( |x| \lt 1\), we conclude that the interval of convergence for the binomial series is \( (−1,1)\). The behavior at the endpoints depends on \( r\). It can be shown that, for \( r \geq 0\), the series converges at both endpoints; for \( −1 \lt r \lt 0\), the series converges at \( x=1\) and diverges at \( x=−1\); and for \( r \lt −1\), the series diverges at both endpoints. The binomial series does converge to \( (1+x)^r\) in \( (−1,1)\) for all real numbers \( r\), but proving this fact by showing that the remainder \( R_n(x) \to 0\) is difficult.

It's time to summarize this in a formal definition.

Definition: Binomial Series

For any real number \( r\), the Maclaurin series for \( f(x)=(1+x)^r\) is called the binomial series. It converges to \( f\) for \( |x| \lt 1\), and we write\[(1+x)^r=\sum_{n=0}^ \infty \binom{r}{n}x^n=1+rx+\dfrac{r(r−1)}{2!}x^2+ \cdots +r\dfrac{(r−1) \cdots (r−n+1)}{n!}x^n+ \cdots \nonumber \]for \( |x| \lt 1\).

We can use this definition to find the binomial series for \( f(x)=\sqrt{1+x}\) and use the series to approximate \( \sqrt{1.5}\).

Example \(\PageIndex{1}\): Finding Binomial Series

1. Find the binomial series for \( f(x)=\sqrt{1+x}\).
2. Use the third-order Maclaurin polynomial \( p_3(x)\) to estimate \( \sqrt{1.5}\). Use Taylor's Theorem to bound the error. Use graphing technology to compare the graphs of \( f\) and \( p_3\).

Solutions

1. Here \( r=\frac{1}{2}\). Using the definition for the binomial series, we obtain\[ \begin{array}{rcl}
   \sqrt{1+x} & = & 1+\dfrac{1}{2}x+\dfrac{(1/2)(−1/2)}{2!}x^2+\dfrac{(1/2)(−1/2)(−3/2)}{3!}x^3+ \cdots \\[6pt]
   & = & 1+\dfrac{1}{2}x−\dfrac{1}{2!}\dfrac{1}{2^2}x^2+\dfrac{1}{3!}\dfrac{1 \cdot 3}{2^3}x^3− \cdots +\dfrac{(−1)^{n+1}}{n!}\dfrac{1 \cdot 3 \cdot 5 \cdots (2n−3)}{2^n}x^n+ \cdots \\[6pt]
   & = & 1+ \displaystyle \sum_{n=1}^ \infty \dfrac{(−1)^{n+1}}{n!}\dfrac{1 \cdot 3 \cdot 5 \cdots (2n−3)}{2^n}x^n. \\[6pt]
   \end{array} \nonumber \]
2. From the result in part a., the third-order Maclaurin polynomial is\[ p_3(x)=1+\dfrac{1}{2}x−\dfrac{1}{8}x^2+\dfrac{1}{16}x^3. \nonumber \]Therefore,\[ \sqrt{1.5}=\sqrt{1+0.5} \approx 1+\dfrac{1}{2}(0.5)−\dfrac{1}{8}(0.5)^2+\dfrac{1}{16}(0.5)^3 \approx 1.2266. \nonumber \]From Taylor’s Theorem, the error satisfies\[ R_3(0.5)=\dfrac{f^{(4)}(c)}{4!}(0.5)^4 \nonumber \]for some \( c\) between \( 0\) and \( 0.5\). Since \( f^{(4)}(x)=−\frac{15}{2^4(1+x)^{7/2}}\), and the maximum value of \( |f^{(4)}(x)|\) on the interval \( (0,0.5)\) occurs at \( x=0\), we have\[ |R_3(0.5)| \leq \dfrac{15}{4!2^4}(0.5)^4 \approx 0.00244. \nonumber \]The function and the Maclaurin polynomial \( p_3\) are graphed in Figure \(\PageIndex{1}\).

   
   Figure \(\PageIndex{1}\): The third-order Maclaurin polynomial \( p_3(x)\) provides a good approximation for \( f(x)=\sqrt{1+x}\) for \( x\) near zero.

Checkpoint \(\PageIndex{1}\)

Find the binomial series for \( f(x)=\frac{1}{(1+x)^2}\).

Answer

\(\displaystyle \sum_{n=0}^ \infty (−1)^n(n+1)x^n\)

A Summary of Common Functions Expressed as Taylor Series

As in Section 4.3, we summarize the common Maclaurin series we have derived up to this point.

Table \(\PageIndex{1}\): Maclaurin Series for Common Functions

Function	Maclaurin Series	Interval of Convergence
\( f(x)=\frac{1}{1−x}\)	\(\displaystyle \sum_{n=0}^ \infty x^n\)	\( −1<x<1\)
\( f(x)=e^x\)	\(\displaystyle \sum_{n=0}^ \infty \frac{x^n}{n!}\)	\( − \infty <x< \infty \)
\( f(x)=\sin x\)	\(\displaystyle \sum_{n=0}^ \infty (−1)^n\frac{x^{2n+1}}{(2n+1)!}\)	\( − \infty <x< \infty \)
\( f(x)=\cos x\)	\(\displaystyle \sum_{n=0}^ \infty (−1)^n\frac{x^{2n}}{(2n)!}\)	\( − \infty <x< \infty \)
\( f(x)=\ln(1+x)\)	\(\displaystyle \sum_{n=1}^ \infty (−1)^{n-1}\frac{x^n}{n}\)	\( −1<x \leq 1\)
\( f(x)=\tan^{−1}x\)	\(\displaystyle \sum_{n=0}^ \infty (−1)^n\frac{x^{2n+1}}{2n+1}\)	\( −1 \leq x \leq 1\)
\( f(x)=(1+x)^r\)	\(\displaystyle \sum_{n=0}^ \infty \binom{r}{n}x^n\)	\( −1<x<1\)

We showed previously in this chapter how power series can be differentiated term-by-term to create a new power series. In Example \(\PageIndex{2}\), we differentiate the binomial series for \( \sqrt{1+x}\) term by term to find the binomial series for \( \frac{1}{\sqrt{1+x}}\). Note that we could construct the binomial series for \( \frac{1}{\sqrt{1+x}}\) directly from the definition, but differentiating the binomial series for \( \sqrt{1+x}\) is an easier calculation.

Example \(\PageIndex{2}\): Differentiating a Series to Find a New Series

Use the binomial series for \( \sqrt{1+x}\) to find the binomial series for \( \frac{1}{\sqrt{1+x}}\).

Solution

The two functions are related by\[ \dfrac{d}{dx}\sqrt{1+x}=\dfrac{1}{2\sqrt{1+x}}, \nonumber \]so the binomial series for \( \frac{1}{\sqrt{1+x}}\) is given by\[ \dfrac{1}{\sqrt{1+x}}=2\dfrac{d}{dx}\sqrt{1+x}=1+\sum_{n=1}^ \infty \dfrac{(−1)^n}{n!}\dfrac{1 \cdot 3 \cdot 5 \cdots (2n−1)}{2^n}x^n. \nonumber \]

Checkpoint \(\PageIndex{2}\)

Find the binomial series for \( f(x)=\frac{1}{(1+x)^{3/2}}\)

Answer

\(\displaystyle \sum_{n=1}^ \infty \frac{(−1)^n}{n!}\frac{1 \cdot 3 \cdot 5 \cdots (2n−1)}{2^n}x^n\)

In this example, we differentiated a known Taylor series to construct a Taylor series for another function. The ability to differentiate power series term-by-term makes them a powerful tool for solving differential equations.

Applications Involving Taylor Series

Probability

Taylor and Maclaurin series are used frequently when encountering non-elementary integrals. For example, the integral \(\displaystyle \int e^{−x^2}\,dx\) arises often in probability theory. Specifically, it is used when studying data sets that are normally distributed, meaning the data values lie under a bell-shaped curve. For example, if a set of data values is normally distributed with mean \( \mu \) and standard deviation \( \sigma \), then the probability that a randomly chosen value lies between \( x=a\) and \( x=b\) is given by\[\dfrac{1}{ \sigma \sqrt{2 \pi }}\int ^b_ae^{−(x− \mu )^2/(2 \sigma ^2)}\, dx.\label{probeq} \](See Figure \(\PageIndex{2}\).)

Figure \(\PageIndex{2}\): If data values are normally distributed with mean \( \mu \) and standard deviation \( \sigma \), the probability that a randomly selected data value is between \(a\) and \(b\) is the area under the curve \( y=\frac{1}{ \sigma \sqrt{2 \pi }}e^{−(x− \mu )^2/(2 \sigma ^2)}\) between \( x=a\) and \( x=b\).

To simplify this integral, we typically let \( z=\frac{x− \mu }{ \sigma }\). This quantity \(z\) is the data value's \(z\) score. With this simplification, integral Equation \ref{probeq} becomes\[\dfrac{1}{\sqrt{2 \pi }}\int ^{(b− \mu )/ \sigma }_{(a− \mu )/ \sigma }e^{−z^2/2}\,dz. \nonumber \]In Example \(\PageIndex{3}\), we show how we can use this integral in calculating probabilities.

Example \(\PageIndex{3}\): Using Maclaurin Series to Approximate a Probability

Suppose a set of standardized test scores are normally distributed with mean \( \mu =100\) and standard deviation \( \sigma =50\). Use Equation \ref{probeq} and the first six terms in the Maclaurin series for \( e^{−x^2/2}\) to approximate the probability that a randomly selected test score is between \( x=100\) and \( x=200\). Use the Alternating Series Test to determine how accurate your approximation is.

Solution

Since \( \mu =100\), \(\sigma =50\), and we are trying to determine the area under the curve from \( a=100\) to \( b=200\), integral Equation \ref{probeq} becomes\[ \dfrac{1}{\sqrt{2 \pi }}\int ^2_0e^{−z^2/2}\,dz.\nonumber \]The Maclaurin series for \( e^{−x^2/2}\) is given by\[ \begin{array}{rcl}
e^{−x^2/2} & = & \sum_{n=0}^ \infty \dfrac{\left(−\dfrac{x^2}{2}\right)^n}{n!} \\[6pt]
& = & 1−\dfrac{x^2}{2^1 \cdot 1!}+\dfrac{x^4}{2^2 \cdot 2!}−\dfrac{x^6}{2^3 \cdot 3!}+ \cdots +(−1)^n\dfrac{x^{2n}}{2^n \cdot n}!+ \cdots \\[6pt]
& = & \sum_{n=0}^ \infty (−1)^n\dfrac{x^{2n}}{2^n \cdot n!}. \\[6pt]
\end{array} \nonumber \]Therefore,\[\begin{array}{rcl}
\dfrac{1}{\sqrt{2 \pi }}\int e^{−z^2/2}\,dz & = & \dfrac{1}{\sqrt{2 \pi }}\int \left(1−\dfrac{z^2}{2^1 \cdot 1!}+\dfrac{z^4}{2^2 \cdot 2!}−\dfrac{z^6}{2^3 \cdot 3!}+ \cdots +(−1)^n\dfrac{z^{2n}}{2^n \cdot n!}+ \cdots \right) \, dz \\[6pt]
& = & \dfrac{1}{\sqrt{2 \pi }}\left(C+z−\dfrac{z^3}{3 \cdot 2^1 \cdot 1!}+\dfrac{z^5}{5 \cdot 2^2 \cdot 2!}−\dfrac{z^7}{7 \cdot 2^3 \cdot 3!}+ \cdots +(−1)^n\dfrac{z^{2n+1}}{(2n+1)2^n \cdot n!}+ \cdots \right) \\[6pt]
\end{array} \nonumber \]\[\dfrac{1}{\sqrt{2 \pi }}\int ^2_0e^{−z^2/2}\,dz=\dfrac{1}{\sqrt{2 \pi }}\left(2−\dfrac{8}{6}+\dfrac{32}{40}−\dfrac{128}{336}+\dfrac{512}{3456}−\dfrac{2^{11}}{11 \cdot 2^5 \cdot 5!}+ \cdots \right)\nonumber \]Using the first five terms, we estimate that the probability is approximately 0.4922. By the Alternating Series Test, we see that this estimate is accurate to within\[ \dfrac{1}{\sqrt{2 \pi }}\dfrac{2^{13}}{13 \cdot 2^6 \cdot 6!} \approx 0.00546.\nonumber \]

Analysis

The probability that a data value is within two standard deviations of the mean is approximately \( 95\%\). Here, we calculated the probability that a data value is between the mean and two standard deviations above the mean, so the estimate should be around \( 47.5\%\). The estimate, combined with the bound on the accuracy, falls within this range.

Checkpoint \(\PageIndex{3}\)

Use the first five terms of the Maclaurin series for \( e^{−x^2/2}\) to estimate the probability that a randomly selected test score is between \( 100\) and \( 150\). Use the Alternating Series Test to determine the accuracy of this estimate.

Answer

The estimate is approximately \( 0.3414\). This estimate is accurate to within \( 0.0000094\).

Elliptical Integrals

Another application in which a non-elementary integral arises involves the period of a pendulum. The integral is\[\int ^{ \pi /2}_0\dfrac{d \theta }{\sqrt{1−k^2\sin^2 \theta }}.\nonumber \]An integral of this form is known as an elliptic integral of the first kind. Elliptic integrals originally arose when trying to calculate the arc length of an ellipse. We now show how to use power series to approximate this integral.

Example \( \PageIndex{4}\): Period of a Pendulum

The period of a pendulum is the time it takes for a pendulum to make one complete back-and-forth swing. For a pendulum with length \( L\) that makes a maximum angle \( \theta _{max}\) with the vertical, its period \( T\) is given by\[ T=4\sqrt{\dfrac{L}{g}}\int ^{ \pi /2}_0\dfrac{d \theta }{\sqrt{1−k^2\sin^2 \theta }}\nonumber \]where \( g\) is the acceleration due to gravity and \( k=\sin\left(\dfrac{ \theta _{max}}{2}\right)\) (see Figure \(\PageIndex{3}\)). (We note that this formula for the period arises from a non-linearized model of a pendulum. In some cases, for simplification, a linearized model is used, and \(\sin \theta \) is approximated by \( \theta \).)

Figure \(\PageIndex{3}\): This pendulum has length \( L\) and makes a maximum angle \( \theta _{max}\) with the vertical.

Use the binomial series\[ \dfrac{1}{\sqrt{1+x}}=1+\sum_{n=1}^ \infty \dfrac{(−1)^n}{n!}\dfrac{1 \cdot 3 \cdot 5 \cdots (2n−1)}{2^n}x^n\nonumber \]to estimate the period of this pendulum. Specifically, approximate the period of the pendulum if

1. you use only the first term in the binomial series, and
2. you use the first two terms in the binomial series.

Solution

We use the binomial series, replacing x with \( −k^2\sin^2 \theta\). Then we can write the period as\[ T=4\sqrt{\dfrac{L}{g}}\int ^{ \pi /2}_0\left(1+\dfrac{1}{2}k^2\sin^2 \theta +\dfrac{1 \cdot 3}{2!2^2}k^4\sin^4 \theta + \cdots \right)\,d \theta .\nonumber \]

a. Using just the first term in the integrand, the first-order estimate is\[ T \approx 4\sqrt{\dfrac{L}{g}}\int ^{ \pi /2}_0\,d \theta =2 \pi \sqrt{\dfrac{L}{g}}.\nonumber \]If \( \theta _{max}\) is small, then \( k=\sin\left(\frac{ \theta _{max}}{2}\right)\) is small. This is a good estimate when \( k\) is small. To justify this claim, consider\[ \int ^{ \pi /2}_0\left(1+\dfrac{1}{2}k^2\sin^2 \theta +\dfrac{1 \cdot 3}{2!2^2}k^4\sin^4 \theta + \cdots \right)\,d \theta .\nonumber \]Since \( |\sin x| \leq 1\), this integral is bounded by\[ \int ^{ \pi /2}_0\left(\dfrac{1}{2}k^2+\dfrac{1.3}{2!2^2}k^4+ \cdots \right)\,d \theta \;<\;\dfrac{ \pi }{2}\left(\dfrac{1}{2}k^2+\dfrac{1 \cdot 3}{2!2^2}k^4+ \cdots \right).\nonumber \]Furthermore, it can be shown that each coefficient on the right-hand side is less than \( 1\) and, therefore, that this expression is bounded by\[ \dfrac{ \pi k^2}{2}(1+k^2+k^4+ \cdots )=\dfrac{ \pi k^2}{2} \cdot \dfrac{1}{1−k^2}, \nonumber \]which is small for \( k\) small.

b. For larger values of \( \theta _{max}\), we can approximate \( T\) by using more terms in the integrand. By using the first two terms in the integral, we arrive at the estimate\[ T \approx 4\sqrt{\dfrac{L}{g}}\int ^{ \pi /2}_0\left(1+\dfrac{1}{2}k^2\sin^2 \theta \right)\,d \theta =2 \pi \sqrt{\dfrac{L}{g}}\left(1+\dfrac{k^2}{4}\right).\nonumber \]

Recentering Polynomials

One of the most useful applications of Taylor series occurs in Differential Equations. A polynomial centered at \( 0 \) can always be recentered at another point using Taylor series. To showcase how this is done, let\[ P(x) = a_n x^n + a_{n - 1}x^{n - 1} + \cdots + a_1 x + a_0. \nonumber \]By computing the derivatives of \( P \) at \( a \), we get the recentered polynomial\[ P(x) = \sum_{n = 0}^{\infty} \dfrac{P^{(n)}(a)}{n!}(x - a)^n = P(a) + P^{\prime}(a)(x - a) + \dfrac{P^{\prime\prime}(a)}{2!}(x - a)^2 + \cdots . \nonumber \]The usefulness of recentering polynomials cannot be overstated; however, to describe the reasons at this point is moot (it suffices to say that it is important). Let's see how this works in practice.

Example \(\PageIndex{5}\)

Recenter the polynomial at \( a = -3 \).\[ P(x) = x^6 - x^4 + x^3 + x - 2. \nonumber \]

Solution

We begin by making a table of derivatives and their values at \( a = -3 \).\[ \begin{array}{rclcrcl}
& & & & P(-3) & = & 616 \\[6pt]
P^{\prime}(x) & = & 6x^5 - 4x^3 + 3x^2 + 1 & \implies & P^{\prime}(-3) & = & -1322 \\[6pt]
P^{\prime\prime}(x) & = & 30x^4 - 12x^2 + 6x & \implies & P^{\prime\prime}(-3) & = & 2304 \\[6pt]
P^{\prime\prime\prime}(x) & = & 120x^3 - 24x + 6 & \implies & P^{\prime\prime\prime}(-3) & = & -3162 \\[6pt]
P^{(4)}(x) & = & 360x^2 - 24 & \implies & P^{(4)}(-3) & = & 3216 \\[6pt]
P^{(5)}(x) & = & 720x & \implies & P^{(5)}(-3) & = & -2160 \\[6pt]
P^{(6)}(x) & = & 720 & \implies & P^{(6)}(-3) & = & 720 \\[6pt]
P^{(n)}(x) & = & 0 & \text{for} & n & \geq & 7 \\[6pt]
\end{array} \nonumber \]Therefore,\[ \begin{array}{rcl}
P(x) & = & \displaystyle \sum_{n = 0}^{\infty} \dfrac{P^{(n)}}{n!} (x - a)^n \\[6pt]
& = & 616 - 1322(x + 3) + \dfrac{2304}{2!}(x + 3)^2 - \dfrac{3162}{3!}(x + 3)^3 + \dfrac{3216}{4!}(x + 3)^4 - \dfrac{2160}{5!}(x + 3)^5 + \dfrac{720}{6!}(x + 3)^6 \\[6pt]
& = & 616 - 1322(x + 3) + 1152(x + 3)^2 - 527(x + 3)^3 + 134(x + 3)^4 - 18(x + 3)^5 + (x + 3)^6 \\[6pt]
\end{array} \nonumber \]

Since \( P \) is a polynomial with a finite number of terms, the radius of convergence is always \( \infty \).

Exercise \(\PageIndex{5}\)

Recenter\[ P(x) = x^5 + 2x^3 + x \nonumber \]at \( a = 2 \).

Answer

\( P(x) = 50 + 105(x - 2) + 92(x - 2)^2 + 42(x - 2)^3 + 10(x - 2)^4 + (x - 2)^5 \).

Building Data Models

Our final application involves data given at a single point. If we know the quantity of something at a given time, how quickly that quantity changes at that moment, how quickly that change changes, and so on, we can use Taylor series to build a polynomial model of the underlying dataset. Let's transition that language into Mathematics.

If we know \( P(a) \), \( P^{\prime}(a) \), \( P^{\prime\prime}(a) \), and so on, then we can build a model of the data (that works locally) using the Taylor series\[ P(x) = \sum_{n = 0}^{\infty} \dfrac{P^{(n)}(a)}{n!}(x - a)^n = P(a) + P^{\prime}(a)(x - a) + P^{\prime\prime}(a)(x - a)^2 + \cdots .\nonumber \]The fact that we will only have a finite amount of information will keep this polynomial restricted to a finite number of terms. Moreover, this limitation means that our model will only be predictive in "real-world" situations for a short time; however, this still can be very useful.

Example \(\PageIndex{6}\)

The price of a certain stock is currently (at this very moment) at $23.10. If, based on computer models, we know this price is increasing at a rate of $12.93 per day, this rate is increasing at a rate of $0.50 per day (squared), and the rate of this is decreasing at $0.24 per day (cubed), build a model that can be used to predict the price of the stock in 5 minutes.

Solution

We are given a single data point and three of its derivatives. That is, if we let \( p(t) \) be the price of the stock at time \( t \), we were given\[ \begin{array}{rcl}
p(0) & = & 23.10 \\[6pt]
p^{\prime}(0) & = & 12.93 \\[6pt]
p^{\prime\prime}(0) & = & 0.50 \\[6pt]
p^{\prime\prime\prime}(0) & = & -0.24 \\[6pt]
\end{array} \nonumber \]Therefore, we can build a model for the price using a Taylor polynomial.\[ P_3(t) = \sum_{n = 0}^{3} \dfrac{p^{(n)}(0)}{n!}(t - 0)^n = 23.10 + 12.93t + \dfrac{0.50}{2!}t^2 - \dfrac{0.24}{3!}t^3 = 23.10 + 12.93t + 0.25t^2 - 0.04t^3 \nonumber \]Since \( t \) is in terms of days, the predicted value of the stock in 5 minutes using this model is\[ P_3\left( \dfrac{5}{1440} \right) \approx 23.14\nonumber \]Thus, we can expect the stock to rise by $0.04 in the next five minutes.

To be honest, Example \( \PageIndex{ 6 } \) might be a little too immersed in fantasy to be realistic; however, the impact of being able to predict future values given "dense" information at a single point (a value and the values of several derivatives at that point) should imply many uses.

The applications of Taylor series in this section are intended to highlight their importance. In general, Taylor series are helpful because they allow us to represent known functions using polynomials, thus providing a tool for approximating function values and estimating complicated integrals. In addition, they will enable us to define new functions as power series, thus providing us with a powerful tool for solving differential equations.

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Key Concepts

• The binomial series is the Maclaurin series for \( f(x)=(1+x)^r\). It converges for \( |x|<1\).
• Taylor series for functions can often be derived by algebraic operations with a known Taylor series or by differentiating or integrating a known Taylor series.
• Power series can be used to solve differential equations.
• Taylor series can be used to help approximate integrals that cannot be evaluated by other means.

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Glossary

binomial series

the Maclaurin series for \( f(x)=(1+x)^r\); it is given by \(\displaystyle (1+x)^r=\sum_{n=0}^ \infty \binom{r}{n}x^n=1+rx+\frac{r(r−1)}{2!}x^2+ \cdots +\frac{r(r−1) \cdots (r−n+1)}{n!}x^n+ \cdots \) for \( |x|<1\)

non-elementary integral

an integral for which the antiderivative of the integrand cannot be expressed as an elementary function