OBJECTIVES

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

1. Find a Taylor Polynomial (p. 240)

ORIGINS

The Taylor polynomial is named after the English mathematician Brook Taylor (1685–1731). Taylor grew up in an affluent but strict home. He was home-schooled in the arts and the classics until he attended Cambridge University in 1703, where he studied mathematics. He was an accomplished musician and painter as well as a mathematician.

DEFINITION Taylor Polynomial

If a function $f$ and its first $n$ derivatives are defined on an interval containing the number $x_{0}$, then $$\bbox[5px, border:1px solid black, #F9F7ED]{P_{n}( x) =f( x_{0}) +f^\prime ( x_{0}) ( x-x_{0}) +\dfrac{f^{\prime \prime} ( x_{0}) }{2!}( x-x_{0}) ^{2}+\cdots+\dfrac{f^{( n) }( x_{0}) }{\,n!}( x-x_{0}) ^{n}}$$ is called the $n$th Taylor Polynomial for $f$ at $x_{0}$.

NOTE

The sum of the first two terms of a Taylor Polynomial is the linear approximation of $f$ at $x$. That is, ${P}_{1}({x}) = L({x})$.

Find the Taylor Polynomial $P_{3}( x)$ for $f( x) =\sqrt{x}$ at $1.$

Solution The first three derivatives of $f(x) =\sqrt{x}$ are $$\begin{eqnarray*} f^\prime ( x) &=&\dfrac{1}{2\sqrt{x}}\qquad\qquad f^{\prime \prime} ( x) =\dfrac{d}{dx}\left( \dfrac{x^{-1/2}}{2}\right) =-\dfrac{1}{4x^{3/2}} \\ f^{\prime \prime \prime} ( x) &=&\dfrac{d}{dx}\left( -\dfrac{x^{-3/2}}{4}\right) =\dfrac{3}{8x^{5/2}} \end{eqnarray*}$$

Figure 17

Then $$f(1) =1\qquad f^\prime (1) =\dfrac{1}{2}\qquad f^{\prime \prime} (1) =-\dfrac{1}{4}\qquad f^{\prime \prime \prime} (1) =\dfrac{3}{8}$$

The Taylor Polynomial $P_{3}( x)$ for $f( x) =\sqrt{x}$ at $1$ is $$\begin{eqnarray*} P_{3}( x) &=&f(1) +f^\prime (1) ( x-1) +\dfrac{f^{\prime \prime} (1) }{2!}( x-1) ^{2}+\dfrac{f^{\prime \prime \prime} (1) }{3!}( x-1)^{3}\\ &=&1+\dfrac{x-1}{2}-\dfrac{( x-1) ^{2}}{8}+\dfrac{( x-1) ^{3}}{16} \end{eqnarray*}$$

The graphs of the function $f$ and the Taylor Polynomial $P_{3}$ are shown in Figure 17. Notice how $P_{3}( x) \approx f( x)$ for values of $x$ near $1$.

NOW WORK

Problem 11.

Find the Taylor Polynomial $P_{7}( x)$ for $f( x) =\sin x$ at $0$.

Solution The derivatives of $f( x) =\sin x$ at $0$ are $$\begin{array}{rcl@{\qquad}rcl@{\qquad}rcl@{\qquad}rcl} f( x) &=&\sin x & f^\prime ( x) &=&\cos x & f^{\prime \prime} ( x) &=&-\!\sin x & f^{\prime \prime \prime} ( x) &=&-\!\cos x \\ f( 0) &=&0 & f^\prime ( 0) &=&1 & f^{\prime \prime} ( 0) &=&0 & f^{\prime \prime \prime} ( 0) &=&-1 \\ f^{( 4) }( x) &=&\sin x & f^{(5) }( x) &=&\cos x & f^{( 6)}( x) &=&-\!\sin x & f^{( 7) }( x) &=&-\!\cos x \\ f^{( 4) }( 0) &=&0 & f^{(5) }( 0) &=&1 & f^{( 6) }( 0) &=&0 & f^{( 7) }( 0) &=&-1 \end{array}$$

The Taylor Polynomial $P_{7}( x)$ for $\sin x$ at $0$ is $$P_{7}( x) =f( 0) +f^\prime ( 0) x+\dfrac{f^{\prime \prime} ( 0) }{2!}x^{2}+\dfrac{f^{\prime\prime \prime} ( 0) }{3!}x^{3}+\cdots +\dfrac{f^{( 7) }( 0) }{7!}x^{7}=x-\dfrac{x^{3}}{3!}+\dfrac{x^{5}}{5!}-\dfrac{x^{7}}{7!}$$

Figure 18 shows the graph of $P_{7}$ superimposed on the graph of $y=\sin x$.

NOW WORK

Problem 13.

Find the Taylor Polynomial $P_{n}( x)$ for $f( x) =e^{x}$ at $0$.

Solution The derivatives of $f( x) =e^{x}$ are $$\begin{array}{rcl@{\qquad}rcl@{\qquad}rcl@{\qquad}rcl@{ }c@{ }rcl} f( x) = e^{x} & f^\prime ( x) = e^{x} & f^{\prime \prime} ( x) = e^{x} & f^{\prime \prime \prime} ( x) = e^{x} & \ldots & f^{( n) }( x) = e^{x} \\ f( 0) = 1 & f^\prime ( 0) = 1 & f^{\prime \prime} ( 0) = 1 & f^{\prime \prime \prime} ( 0) = 1 & \ldots & f^{( n)}( 0) = 1 \end{array}$$

The Taylor Polynomial $P_{n}( x)$ at $0$ is $$\begin{eqnarray*} P_{n}( x) &=&f( 0) +f^\prime ( 0) x+\dfrac{ f^{\prime \prime} ( 0) }{2!}x^{2}+\dfrac{f^{\prime \prime\prime} ( 0) }{3!}x^{3}+\cdots +\dfrac{f^{( n) }( 0) }{n!}x^{n}\\ &=&1+x+\dfrac{x^{2}}{2!}+\dfrac{x^{3}}{3!}+\cdots +\dfrac{ x^{n}}{n!} \end{eqnarray*}$$

NOTE

In Chapter 8 we investigate the relationship between a function $f$ and its Taylor Polynomials.