MATH 152 Chapter 10.7

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Taylor Series

Goal: every differentiable function can be written as an infinite sum.

Maclaurin Series center is at 0: ${\begin{aligned}f\left(x\right)&=\sum _{n=0}^{\infty }c_{n}x^{n}=c_{0}+c_{1}x+c_{2}x^{2}+c_{3}x^{3}+c_{4}x^{4}+\dots \\f(0)&=c_{0}\\f'\left(x\right)&=0+c_{1}+2c_{2}x+3c_{3}x^{2}+4c_{4}x^{3}+...\\f'\left(0\right)&=c_{1}\\f''\left(x\right)&=0+0+2c_{2}+6c_{3}x+16c_{4}x^{2}+\dots \\f''\left(0\right)&=2c_{2}\end{aligned}}$

Therefore, the equation to find any coefficient in a Maclaurin Series is:

$c_{n}={\frac {f^{\left(n\right)}\left(0\right)}{n!}}$

Where $f^{\left(n\right)}\left(a\right)$ is the nth derivative at center.

More generally, if our center is at a, then

${\begin{aligned}c_{n}&={\frac {f^{\left(n\right)}\left(a\right)}{n!}}\\f\left(x\right)&=\sum _{n=0}^{\infty }{{\frac {f^{\left(n\right)}\left(a\right)}{n!}}\,\left(x-a\right)^{n}}\end{aligned}}$

Vocabulary Terms

• $T_{N}$ is the $N$th partial sum of the Taylor series of $f$ (also known as the $n$th degree Taylor Polynomial of $f$ if $n={\mbox{deg}}\,x$)
• $R_{N}$ is the remainder; the sum of the rest of the terms in the series
  • $R_{N}=\left|f\left(x\right)-T_{N}\right|$; note the similarity to $\left|s-s_{N}\right|$

Example

Find the Maclaurin series for $f\left(x\right)=\sin {x}$ and explain why the series converges to $\sin {x}$

$\sin {x}=\sum _{n=0}^{\infty }{{\frac {f^{\left(n\right)}\left(a\right)}{n!}}\,x^{n}}$

Take derivatives and find patterns:

$n$	$f^{\left(n\right)}\left(x\right)$	$f^{\left(n\right)}\left(a\right)$	${\frac {f^{\left(n\right)}\left(a\right)}{n!}}\,x^{n}$
0	$f\left(x\right)=\sin {x}$	0	0
1	$f'\left(x\right)=\cos {x}$	1	${\frac {1}{1!}}\,x^{1}$
2	$f''\left(x\right)=-\sin {x}$	0	0
3	$f'''\left(x\right)=-\cos {x}$	-1	$-{\frac {1}{3!}}\,x^{3}$
4	$f^{\left(4\right)}\left(x\right)=\sin {x}$	0	0
5	$f^{\left(5\right)}\left(x\right)=\cos {x}$	1	${\frac {1}{5!}}\,x^{5}$
etc...

The pattern for odd terms (because even terms are 0): $f^{\left(2n+1\right)}\left(0\right)=\left(-1\right)^{n}$

And therefore, the sum is: $\sin {x}=\sum _{n=0}^{\infty }{\left(-1\right)^{n}\,{\frac {x^{2n+1}}{\left(2n+1\right)!}}}$

Remainder Formula

Upper-bound calculation on the error for a Taylor Series (not important now; will be covered later)

$R_{N}\left(x\right)\leq {\frac {M}{\left(N+1\right)!}}\,\left|x-a\right|^{N+1}$

Example

Find the Taylor series for $f\left(x\right)=e^{x}$ centered at $x=2$. {{note|the derivative of $e^{x}$ is always $e^{x}$ ${\begin{aligned}e^{x}&=\sum _{n=0}^{\infty }{{\frac {f^{\left(n\right)}\left(2\right)}{n!}}\,\left(x-2\right)^{n}}\\e^{x}&=\sum _{n=0}^{\infty }{{\frac {e^{2}}{n!}}\,\left(x-2\right)^{n}}\end{aligned}}$

Important Maclaurin Series to Know

$e^{x}=\sum _{n=0}^{\infty }{\frac {x^{n}}{n!}}$	$\sin {x}=\sum _{n=0}^{\infty }{\frac {\left(-1\right)^{n}x^{2n+1}}{\left(2n+1\right)!}}$
${\frac {1}{1-x}}=\sum _{n=0}^{\infty }x^{n}$	$\cos {x}=\sum _{n=0}^{\infty }{\frac {\left(-1\right)^{n}x^{2n}}{\left(2n\right)!}}$
	$\arctan {x}=\sum _{n=0}^{\infty }{\frac {\left(-1\right)^{n}x^{2n+1}}{2n+1}}$

Friday, November 12, 2010

Manipulation of Taylor Series

Given $\textstyle {\sin {t}=\sum _{n=0}^{\infty }{\left(-1\right)^{n}\,{\frac {t^{2n+1}}{\left(2n+1\right)!}}}}$, we can find $\textstyle {\int {{\frac {\sin {t}}{t}}\,dt}}$:

${\begin{aligned}\sin {t}&=\sum _{n=0}^{\infty }{\left(-1\right)^{n}\,{\frac {t^{2n+1}}{\left(2n+1\right)!}}}\\{\frac {\sin {t}}{t}}&=\sum _{n=0}^{\infty }{\left(-1\right)^{n}\,{\frac {t^{2n}}{\left(2n+1\right)!}}}\\\int {{\frac {\sin {t}}{t}}\,dt}&=\sum _{n=0}^{\infty }{\left(-1\right)^{n}\,{\frac {t^{2n+1}}{\left(2n+1\right)\,\left(2n+1\right)!}}}\end{aligned}}$

Interesting Results

$\textstyle {\sum _{n=1}^{\infty }{\left(-1\right)^{n}\,{\frac {\left(n-1\right)^{n}}{n}}}}$ . . .

${\begin{aligned}\ln {2}&=\sum _{n=1}^{\infty }{\left(-1\right)^{n+1}\,{\frac {\left(2-1\right)^{n}}{n}}}\\&=\sum _{n=1}^{\infty }{\left(-1\right)^{n+1}\,{\frac {1}{n}}}\\&=1-{\frac {1}{2}}+{\frac {1}{3}}-{\frac {1}{4}}+\dots &{\mbox{(alt. harmonic series)}}\end{aligned}}$

${\begin{aligned}e^{ix}&=1+{\frac {ix}{1!}}+{\frac {i^{2}x^{2}}{2!}}+{\frac {i^{3}x^{3}}{3!}}+{\frac {i^{4}x^{4}}{4!}}+\dots \\e^{ix}&=1+{\frac {ix}{1!}}-{\frac {x^{2}}{2!}}-{\frac {ix^{3}}{3!}}+{\frac {x^{4}}{4!}}+{\frac {ix^{5}}{5!}}+\dots \\e^{ix}&=\cos {x}+i\sin {x}\\\left({\mbox{let}}\,x=\pi \right)\quad e^{i\pi }&=-1+i\left(0\right)=-1\\e^{i\pi }+1&=0\end{aligned}}$