Examples

Maclaurin series of e to the x

What is the Maclaurin series of $f(x)=e^x$?

Solution

Using ${\displaystyle \frac{d}{dx}{e}^x=e^x}$ repeatedly, we see that $f^{(n)}(x)=e^x$ for all $n$.

So $f^{(n)}(0)=e^0=1$ for all $n$. Therefore $a_n={\displaystyle \frac{1}{n!}}$ for all $n$ by the Derivative-Coefficient Identity:

$$e^x=1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+\cdots =\sum _{n=0}^{\infty }\frac{x^n}{n!}$$

Maclaurin series of cos x

Find the Maclaurin series representation of $\cos x$.

Solution

Use the Derivative-Coefficient Identity to solve for the coefficients:

$$a_n=\frac{f^{(n)}(0)}{n!}$$

$n$	$f^{(n)}(x)$	$f^{(n)}(0)$	$a_n$
0	$\cos x$	$1$	$1$
1	$-\sin x$	$0$	$0$
2	$-\cos x$	$-1$	$-1/2$
3	$\sin x$	$0$	$0$
4	$\cos x$	$1$	$1/24$
5	$-\sin x$	$0$	$0$
$⋮$	$⋮$	$⋮$	$⋮$

By studying this pattern, we find the series:

$$\cos x=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+\cdots =\sum _{n=0}^{\infty }{(-1)}^n\frac{x^{2n}}{(2n)!}$$

Maclaurin series from other Maclaurin series

(a) Find the Maclaurin series of $\sin x$ using the Maclaurin series of $\cos x$.

(b) Find the Maclaurin series of $f(x)=x^2{e}^{-5x}$ using the Maclaurin series of $e^x$.

(c) Using (b), find the value of $f^{(22)}(0)$.

Solution

(a)

Remember that $\frac{d}{dx}\cos x=-\sin x$. Let us differentiate the cosine series by terms:

$$\begin{array}{c}1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+\cdots {\gg \gg }^{\frac{d}{dx}}0-2\frac{x^1}{2!}+4\frac{x^3}{4!}-6\frac{x^5}{6!}+\cdots \\ \\ \gg \gg -\frac{x^1}{1!}+\frac{x^3}{3!}-\frac{x^5}{5!}-\cdots \end{array}$$

Take negative to get:

$$\sin x=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\cdots$$

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(b)

$$e^u=1+\frac{u^1}{1!}+\frac{u^2}{2!}+\frac{u^3}{3!}+\cdots$$

Set $u=-5x$:

$$e^{-5x}=1+\frac{(-5x)}{1!}+\frac{{(-5x)}^2}{2!}+\frac{{(-5x)}^3}{3!}+\cdots =\sum _{n=0}^{\infty }{(-1)}^n\frac{5^n}{n!}{x}^n$$

Multiply all terms by $x^2$:

$$\begin{array}{cc}{x}^2{e}^{-5x}\gg \gg & x^2(1+\frac{(-5x)}{1!}+\frac{{(-5x)}^2}{2!}+\frac{{(-5x)}^3}{3!}+\cdots )\\ & \\ \gg \gg & x^2-5x^3+\frac{25}{2}{x}^4-\frac{125}{3!}{x}^5+\cdots \\ & \\ \gg \gg & \sum _{n=0}^{\infty }{(-1)}^n\frac{5^n}{n!}{x}^{n+2}\end{array}$$

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(c)

For any series:

$$f(x)=a_0+a_{1}x+a_2{x}^2+a_3{x}^3+\cdots$$

we have:

$$f^{(n)}(0)=n!\cdot a_n$$

We can use this to compute $a_{22}$. From the series formula:

$$\sum _{n=0}^{\infty }{(-1)}^n\frac{5^n}{n!}{x}^{n+2}$$

we see that:

$$a_{n+2}={(-1)}^n\frac{5^n}{n!}$$

Power, NOT term number

The coefficient with $a_{n+2}$ corresponds to the term having $x^{n+2}$, not necessarily the ${(n+2)}^{\text{th}}$ term of the series.

Therefore:

$$\begin{array}{c}{a}_{22}={(-1)}^{20}\frac{5^{20}}{20!}\gg \gg 5^{20}\frac{1}{20!}\\ \\ f^{(22)}(0)=22!\cdot a_{22}\gg \gg 5^{20}\cdot \frac{22!}{20!}\gg \gg 5^{20}\cdot 22\cdot 21\end{array}$$

Computing a Taylor series

Find the first five terms of the Taylor series of $f(x)=\sqrt{x+1}$ centered at $c=3$.

Solution

A Taylor series is just a Maclaurin series centered at a nonzero number.

General format of a Taylor series:

$$f(x)=a_0+a_1(x-c)+a_2{(x-c)}^2+a_3{(x-c)}^3+\cdots$$

The coefficients satisfy $f^{(n)}(c)=n!a_n$.

Find the coefficients by computing the derivatives and evaluating at $x=3$:

$$\begin{array}{cccc}f(x)& ={(x+1)}^{1/2},& f(3)& =2\\ f^{\prime }(x)& =\frac{1}{2}{(x+1)}^{-1/2},& f^{\prime }(3)& =\frac{1}{4}\\ f^{\prime \prime }(x)& =-\frac{1}{4}{(x+1)}^{-3/2},& f^{\prime \prime }(3)& =-\frac{1}{32}\\ f^{\prime \prime \prime }(x)& =\frac{3}{8}{(x+1)}^{-5/2},& f^{\prime \prime \prime }(3)& =\frac{3}{256}\\ f^{(4)}(x)& =-\frac{15}{16}{(x+1)}^{-7/2},& f^{(4)}(3)& =-\frac{15}{2048}\end{array}$$

The first terms of the series:

$$\begin{array}{c}f(x)=\sqrt{x+1}\\ \\ =2+\frac{1}{4}(x-3)-\frac{1}{64}{(x-3)}^2+\frac{1}{512}{(x-3)}^3-\frac{5}{\mathrm{16,384}}{(x-3)}^4+\cdots \end{array}$$