Home / AP Calculus AB and BC: Chapter 9 -Infinite Sequences and Series : 9.9 -Taylor Series and Maclaurin Series Study Notes

AP Calculus AB and BC: Chapter 9 -Infinite Sequences and Series : 9.9 -Taylor Series and Maclaurin Series Study Notes

9.9 Taylor Series and Maclaurin Series

Taylor Series and Maclaurin Series

If a function $f$ has derivatives of all orders at $x=c$, then the series

$
\begin{aligned}
& \sum_{n=0}^{\infty} \frac{f^{(n)}(c)}{n !}(x-c)^n \\
& =f(c)+f^{\prime}(c)(x-c)+\frac{f^{\prime \prime}(c)}{2 !}(x-c)^2+\frac{f^{\prime \prime \prime}(c)}{3 !}(x-c)^3+\cdots+\frac{f^{(n)}(c)}{n !}(x-c)^n+\cdots
\end{aligned}
$

is called the Taylor series for $\boldsymbol{f}(\boldsymbol{x})$ at $c$. Moreover, if $c=0$, then the series is called the Maclaurin series for $\boldsymbol{f}$.

Example 1

  • Find the Maclaurin series for the function $f(x)=\ln (1+x)$.
▶️Answer/Explanation

Solution 

$\begin{array}{ll}f(x)=\ln (1+x) & f(0)=0 \\ f^{\prime}(x)=\frac{1}{1+x} & f^{\prime}(0)=1 \\ f^{\prime \prime}(x)=-\frac{1}{(1+x)^2} & f^{\prime \prime}(0)=-1 \\ f^{\prime \prime \prime}(x)=\frac{2}{(1+x)^3} & f^{\prime \prime \prime}(0)=2 \\ f^{(4)}(x)=-\frac{6}{(1+x)^4} & f^{(4)}(0)=-6\end{array}$

$\begin{aligned} & \ln (1+x)=f(0)+f^{\prime}(0) x+\frac{f^{\prime \prime}(0)}{2 !} x^2+\frac{f^{\prime \prime \prime}(0)}{3 !} x^3+\cdots+\frac{f^{(n)}(0)}{n !} x^n+\cdots \\ & =0+1 \cdot x+\frac{-1}{2 !} x^2+\frac{2}{3 !} x^3+\frac{-6}{4 !} x^4+\cdots+\frac{(-1)^{n-1}(n-1) !}{n !} x^n+\cdots \cdot \\ & =x-\frac{1}{2} x^2+\frac{1}{3} x^3-\frac{1}{4} x^4+\cdots \cdot+\frac{(-1)^{n-1}}{n} x^n+\cdots \cdot\end{aligned}$

Example 2

  • Let $f$ be a function having derivatives of all orders. The fourth degree Taylor polynomial for $f$ about $x=1$ is given
▶️Answer/Explanation

solution

$
T(x)=4+3(x-1)-6(x-1)^2+7(x-1)^3-4(x-1)^4 \text {. }
$

Find $f(1), f^{\prime}(1), f^{\prime \prime}(1), f^{\prime \prime \prime}(1)$ and $f^{(4)}(1)$.

$\begin{array}{ll}f(1)=T(1)=4 & f^{\prime}(1)=3 \\ \frac{f^{\prime \prime}(1)}{2 !}=-6 \Rightarrow f^{\prime \prime}(1)=-12 & \frac{f^{\prime \prime \prime}(1)}{3 !}=7 \Rightarrow f^{\prime \prime \prime}(1)=42 \\ \frac{f^{(4)}(1)}{4 !}=-4 \Rightarrow f^{(4)}(1)=-96 & \end{array}$

Direct computation of the Taylor or Maclaurin coefficients is usually a tedious procedure. The easiest way to find a Taylor or Maclaurin series is to develop a power series from a list of elementary functions. From the list of power series for elementary functions, you can develop power series for other functions by the operations of addition, subtraction, multiplication, division, differentiation, integration, or composition with known power series.

Power Series for Elementary Functions

Function                                                                                                             Interval of Convergence
$
\begin{aligned}
& \frac{1}{x}=1-(x-1)+(x-1)^2-(x-1)^3+\cdots \cdot+(-1)^n(x-1)^n+\cdots \cdot 0<x<2 \\
& \frac{1}{1-x}=1+x+x^2+x^3+\cdots+x^n+\cdots \cdot \quad-1<x<1 \\
& \ln x=(x-1)-\frac{(x-1)^2}{2}+\frac{(x-1)^3}{3}-\cdots+\frac{(-1)^{n-1}(x-1)^n}{n}+\cdots \cdot 0<x \leq 2 \\
& e^x=1+x+\frac{x^2}{2 !}+\frac{x^3}{3 !}+\cdots+\frac{x^n}{n !}+\cdots \cdot \quad-\infty<x<\infty \\
& \sin x=x-\frac{x^3}{3 !}+\frac{x^5}{5 !}-\frac{x^7}{7 !}+\cdots+\frac{(-1)^n x^{2 n+1}}{(2 n+1) !}+\cdots \cdot \quad-\infty<x<\infty \\
& \cos x=1-\frac{x^2}{2 !}+\frac{x^4}{4 !}-\frac{x^6}{6 !}+\cdots+\frac{(-1)^n x^{2 n}}{(2 n) !}+\cdots \cdot \quad-\infty<x<\infty \\
& \tan ^{-1} x=x-\frac{x^3}{3}+\frac{x^5}{5}-\frac{x^7}{7}+\cdots+\frac{(-1)^n x^{2 n+1}}{2 n+1}+\cdots \cdot \quad-1 \leq x \leq 1 \\
&
\end{aligned}
$

Multiplication of Power Series
Power series can be multiplied the way we multiply polynomials. We usually find only the first few terms because the calculations for the later terms become tedious and the initial terms are the most important ones.

Example 4

  • Find the Maclaurin series for the function $f(x)=\cos x^2$.
▶️Answer/Explanation

Solution

$\begin{aligned} g(x) & =\cos x=1-\frac{x^2}{2 !}+\frac{x^4}{4 !}-\frac{x^6}{6 !}+\cdots+\frac{(-1)^n x^{2 n}}{(2 n) !}+\cdots \\ f(x) & =\cos x^2 \\ & =g\left(x^2\right) \\ & =1-\frac{\left(x^2\right)^2}{2 !}+\frac{\left(x^2\right)^4}{4 !}-\frac{\left(x^2\right)^6}{6 !}+\cdots+\frac{(-1)^n\left(x^2\right)^{2 n}}{(2 n) !}+\cdots \\ & =1-\frac{x^4}{2 !}+\frac{x^8}{4 !}-\frac{x^{12}}{6 !}+\cdots+\frac{(-1)^n x^{4 n}}{(2 n) !}+\cdots\end{aligned}$

Example 5

  • Find the Maclaurin series for the function $f(x)=x^2 e^x-x^2$
▶️Answer/Explanation

Solution

Use the series $e^x$.

$
e^x=1+x+\frac{x^2}{2 !}+\frac{x^3}{3 !}+\cdots+\frac{x^n}{n !}+.
$

Multiply $e^x$ by $x^2$.
$
\begin{aligned}
x^2 e^x & =x^2\left(1+x+\frac{x^2}{2 !}+\frac{x^3}{3 !}+\cdots+\frac{x^n}{n !}+\cdots\right) \\
& =x^2+x^3+\frac{x^4}{2 !}+\frac{x^5}{3 !}+\cdots \cdot+\frac{x^{n+2}}{n !}+\cdots \cdot
\end{aligned}
$

Subtract $x^2$ from each side.
$
\begin{aligned}
& x^2 e^x-x^2 \\
& =\left(x^2+x^3+\frac{x^4}{2 !}+\frac{x^5}{3 !}+\cdots+\frac{x^{n+2}}{n !}+\cdots\right)-x^2 \\
& =x^3+\frac{x^4}{2 !}+\frac{x^5}{3 !}+\cdots+\frac{x^{n+2}}{n !}+\cdots \cdot
\end{aligned}
$

Example 6

  • Find the first three nonzero terms in the Maclaurin series for $e^x \cos x$.
▶️Answer/Explanation

Solution 

Use the power series for $e^x$ and $\cos x$ in the table.

Use the power series for $e^x$ and $\cos x$ in the table.
$
\begin{aligned}
e^x \cos x= & \left(1+x+\frac{x^2}{2 !}+\frac{x^3}{3 !}+\cdots\right)\left(1-\frac{x^2}{2 !}+\frac{x^4}{4 !}-\frac{x^6}{6 !}+\cdots\right) \\
= & \left(1+x+\frac{x^2}{2 !}+\frac{x^3}{3 !}+\cdots\right)(1)+\left(1+x+\frac{x^2}{2 !}+\frac{x^3}{3 !}+\cdots\right)\left(-\frac{x^2}{2 !}\right)+\cdots \\
= & 1+x+\frac{x^2}{2 !}+\frac{x^3}{3 !}+\cdots \\
& \quad-\frac{x^2}{2 !}-\frac{x^3}{2 !}-\cdots \\
= & 1+x-\frac{x^3}{3}+\cdots
\end{aligned}
$

Example7

  • A series expansion of $\frac{\arctan x}{x}$ is

(A) $1-\frac{x}{3}+\frac{x^3}{5}-\frac{x^5}{7}+\cdots$

(B) $1-\frac{x^2}{3}+\frac{x^4}{5}-\frac{x^6}{7}+\cdots$

(C) $1-\frac{x^2}{2 !}+\frac{x^4}{4 !}-\frac{x^6}{6 !}+\cdots$

(D) $x-\frac{x^3}{3}+\frac{x^4}{5}-\frac{x^6}{7}+\cdots$

▶️Answer/Explanation

Ans:B

Example 8

  • The coefficient of $x^3$ in the Taylor series for $e^{-2 x}$ about $x=0$ is

(A) $-\frac{4}{3}$

(B) $-\frac{2}{3}$

(C) $-\frac{1}{3}$

(D) $\frac{4}{3}$

▶️Answer/Explanation

Ans:A

Example9

  • A function $f$ has a Maclaurin series given by $-\frac{x^4}{3 !}+\frac{x^6}{5 !}-\frac{x^8}{7 !}+\cdots+\frac{(-1)^n x^{2 n+2}}{(2 n+1) !}+\cdots$. Which of the following is an expression for $f(x)$ ?

(A) $x^3 e^x-x^2$

(B) $x \ln x-x^2$

(C) $\tan ^{-1} x-x$

(D) $x \sin x-x^2$

▶️Answer/Explanation

Ans:D

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