Derivative

Taylor Series

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Cegep1

Mathematics

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306 words

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2 minutes

Series of polynomials used to approximate functions
The Taylor polynomial of degree $n$ at $x=a$ is

$$P_n(x)=\sum _{i=0}^n\frac{f^{(i)}(a)}{i!}(x-a{)}^i$$

Extension of linear approximation

$a$ is called the **center of approximation**.

[!abstract] Maclaurin Polynomial
Taylor Polynomial where $a=0$

Examples

Estimate $f(x)=\arctan x$ at $x=0$.

n	0	1	2	3
$f^{(n)}(x)$	$\arctan x$	$\frac{1}{1+x^2}$	$-\frac{2x}{(1+x^2{)}^2}$	$\frac{2(3x^2-1)}{(1+x^2{)}^3}$
$f^{(n)}(1)$	$\frac{\pi }{4}$	$\frac{1}{2}$	$-\frac{1}{2}$	$\frac{1}{2}$

So,

$$\begin{array}{rl}{P}_3(x)& =\frac{f^{(0)}(1)}{0!}(x-1{)}^0+\frac{f^{(1)}(1)}{1!}(x-1{)}^1+\frac{f^{(2)}(1)}{2!}(x-1{)}^2+\frac{f^{(3)}(1)}{3!}(x-1{)}^3\\ & =\frac{\pi }{4}+\frac{1}{2}(x-1)-\frac{1}{4}(x-1{)}^2+\frac{1}{12}(x-1{)}^3\\ ⟹P_3(0)& =\frac{\pi }{4}+\frac{1}{2}(0-1)-\frac{1}{4}(0-1{)}^2+\frac{1}{12}(0-1{)}^3\\ & =\frac{\pi }{4}-\frac{1}{2}-\frac{1}{4}-\frac{1}{12}\\ & =\frac{3\pi -10}{12}\end{array}$$

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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

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