Maclaurin Series

Tags

Calculus

Cegep/1

Cegep/2

Word count

402 words

Reading time

3 minutes

Taylor series where $a=0$

$$f(x)=\sum _{n=0}^{\mathrm{\infty }}\frac{f^{(n)}(0)}{n!}{x}^n$$

Examples

Find the Maclaurin series of the following functions:

$\cos x$

$$\begin{array}{rl}\cos x& =\frac{1}{0!}{x}^0+0+\frac{-1}{2!}{x}^2+0+\frac{1}{4!}{x}^4+0+\dots \\ & =1-\frac{1}{2!}{x}^2+\frac{1}{4!}{x}^4+\dots \\ & =\sum _{n=0}^{\mathrm{\infty }}\frac{(-1{)}^n{x}^{2n}}{(2n)!}\end{array}$$

Knowing that $\arctan x=\sum _{n=0}^{\mathrm{\infty }}\frac{(-1{)}^n{x}^{2n+1}}{2n+1}$,

a. Find a Maclaurin series for $\arctan x^2$.

$$\begin{array}{rl}\arctan x& =\sum _{n=0}^{\mathrm{\infty }}\frac{(-1{)}^n{x}^{2n+1}}{2n+1}\\ ⟹\arctan x^2& =\sum _{n=0}^{\mathrm{\infty }}\frac{(-1{)}^n{x}^{4n+2}}{2n+1}\\ & =\sum _{n=0}^{\mathrm{\infty }}\frac{\frac{(-1{)}^n{x}^{3n+2}n!}{2n+1}}{n!}{x}^n\end{array}$$

b. Find a series expansion for $F(x)={\int }_0^x\arctan t^2\mathrm{d}t$.

$$\begin{array}{rl}F(x)& ={\int }_0^x\sum _{n=0}^{\mathrm{\infty }}\frac{\frac{(-1{)}^n{t}^{3n+2}n!}{2n+1}}{n!}{t}^n\mathrm{d}t\\ & =\sum _{n=0}^{\mathrm{\infty }}{\int }_0^x\frac{\frac{(-1{)}^n{t}^{3n+2}n!}{2n+1}}{n!}{t}^n\mathrm{d}t\\ & =\sum _{n=0}^{\mathrm{\infty }}\left(\frac{(-1{)}^n}{2n+1}{\int }_0^x{t}^{4n+2}\mathrm{d}t\right)\\ & =\sum _{n=0}^{\mathrm{\infty }}\left(\frac{(-1{)}^n}{2n+1}\frac{x^{4n+3}}{4n+3}\right)\\ & =\sum _{n=0}^{\mathrm{\infty }}\frac{(-1{)}^n{x}^{4n+3}}{(2n+1)(4n+3)}\\ & =\frac{x^3}{1\cdot 3}-\frac{x^7}{3\cdot 7}+\frac{x^{11}}{5\cdot 11}-\dots \end{array}$$

Use the degree 5 Maclaurin polynomial of $f(x)=ln(1-x)$ to approximate $\ln 2$.

n	0	1	2	3	4	5
$f^{(n)}(x)$	$ln(1-x)$	$-\frac{1}{1-x}$	$-\frac{1}{(1-x{)}^2}$	$-\frac{2}{(1-x{)}^3}$	$-\frac{2\cdot 3}{(1-x{)}^4}$	$-\frac{2\cdot 3\cdot 4}{(1-x{)}^5}$
$f^{(n)}(0)$	$0$	$-1$	$-1$	$-2$	$-6$	$-24$

So,

$$\begin{array}{rl}{P}_5(x)& =\frac{f^{(0)}(0)}{0!}{x}^0+\frac{f^{(1)}(0)}{1!}{x}^1+\frac{f^{(2)}0}{2!}{x}^2+\frac{f^{(3)}(0)}{3!}{x}^3+\frac{f^{(4)}(0)}{4!}{x}^4+\frac{f^{(5)}(0)}{5!}{x}^5\\ & =-x-\frac{x^2}{2!}-\frac{2x^3}{3!}-\frac{6x^4}{4!}-\frac{24x^5}{5!}\end{array}$$

Note that $\ln 2=f(-1)$, then

$$\begin{array}{rl}f(-1)& \approx P_5(-1)\\ & =-(-1)-\frac{(-1{)}^2}{2!}-\frac{2(-1{)}^3}{3!}-\frac{6(-1{)}^4}{4!}-\frac{24(-1{)}^5}{5!}\\ & =\frac{47}{60}\\ & =0.78\bar{3}\end{array}$$

Contributors

Ajitani Hifumi

Changelog

Last edited about 2 hours ago

View full history

7a6541d-feat: feet