Taylor Series

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Calculus

Cegep/1

Cegep/2

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186 words

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2 minutes

Power series that estimates a function in terms of its derivatives at a single point
Extension of linear approximation

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

[!abstract]+ Taylor polynomial
nth partial sum of a Taylor series

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

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

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,

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

Contributors

Ajitani Hifumi

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