# Maclaurin Polynomial

• Dec 8th 2010, 06:23 AM
hmmmm
Maclaurin Polynomial
How do we find the maclaurin polynomial for $f(x)=\left\{\begin{array}{cc}\frac{cos(x)-1}{x},&\mbox{ if }x\not= 0\\0, & \mbox{ if } x=0\end{array}\right,$
Its not the maclaurin polynomial part that i dont understand its how i differentiate the pairwise function and find $f(0),f'(0),f''(0)....$
Thanks for any help
• Dec 8th 2010, 06:50 AM
TheEmptySet
Quote:

Originally Posted by hmmmm
How do we find the maclaurin polynomial for $f(x)=\left\{\begin{array}{cc}\frac{cos(x)-1}{x},&\mbox{ if }x\not= 0\\0, & \mbox{ if } x=0\end{array}\right,$
Its not the maclaurin polynomial part that i dont understand its how i differentiate the pairwise function and find $f(0),f'(0),f''(0)....$
Thanks for any help

Remember that

$\displaystyle \cos(x)=\sum_{n=0}^{\infty}\frac{(-1)^nx^{2n}}{n!}$

This gives that

$\displaystyle \cos(x)-1=\sum_{n=1}^{\infty}\frac{(-1)^nx^{2n}}{n!}$

and finally that

$\displaystyle \frac{\cos(x)-1}{x}=\sum_{n=1}^{\infty}\frac{(-1)^nx^{2n-1}}{n!}$

Now what is the value of this series at 0? How does it compare with the limit

$\displaystyle \lim_{x \to 0}\frac{\cos(x)-1}{x}$?
• Dec 8th 2010, 07:03 AM
hmmmm
sorry im still a bit confused with the pairwise part when x=0, I understand what you have do to get the maclaurin expansion when x is not 0 but what happens to the x=0 bit?
• Dec 8th 2010, 07:18 AM
TheEmptySet
Quote:

Originally Posted by hmmmm
sorry im still a bit confused with the pairwise part when x=0, I understand what you have do to get the maclaurin expansion when x is not 0 but what happens to the x=0 bit?

What I was trying to point out is that

$\displaystyle \lim_{x \to 0}f(x)=\lim_{x \to 0}\frac{\cos(x)-1}{x}=0$

So the function f is actually continuous moreover it actually infinity differentiable there. They key idea is that the function can be extended to be defined at 0.

The 2nd point is the series should sum to 0 at 0 right. So when we evaluate the series at zero we get 0. So the series does not care that the function was defined piecewise because of the continuous(analytic) extension. Maybe a plot of the function would help.Attachment 20024
• Dec 8th 2010, 07:28 AM
hmmmm
ok Im pretty sure i understand what your getting at thanks very much sorry about that