# Maclaurin series question

• Apr 29th 2009, 09:59 AM
Kaitosan
Maclaurin series question
It seems like it's impossible to get the maclaurin series of sinx, cosx, and ln(x+1) to have a series index beginning at n=0. I'm confused because I think I was taught that taylor series supposedly start at n=0. ?
• Apr 29th 2009, 11:28 AM
Jameson
Quote:

Originally Posted by Kaitosan
It seems like it's impossible to get the maclaurin series of sinx, cosx, and ln(x+1) to have a series index beginning at n=0. I'm confused because I think I was taught that taylor series supposedly start at n=0. ?

I don't quite understand your confusion. By series index, do you mean the beginning value of the sum. Like for $\displaystyle \sum_{n=0}^{\infty}$ you mean the n=0 part? Please explain your confusion a little more.

Taylor series do always start with n=0 if my above assumption is true. Note that by the formula for generating every term in a Taylor series that when you apply the fact that 0!=1 and that $\displaystyle f^{0}(a)=f(a)$ (the 0th derivative of f(x) is f(x)) then you are simply left with f(a), which can be 0 for a lot of things.
• Apr 29th 2009, 11:45 AM
Kaitosan
Will you please show me the maclaurin series of cosx, sinx, and ln(x+1) with "n" beginning at zero? I can't find it...
• Apr 29th 2009, 11:55 AM
running-gag
The maclaurin series of f is $\displaystyle \sum_{n=0}^{+\infty} \frac{f^{(n)}(0)}{n!}\:x^n$

For $\displaystyle f(x) = \cos x$, $\displaystyle f^{(2n+1)}(0) = 0$ and $\displaystyle f^{(2n)}(0) = (-1)^n$

Therefore the maclaurin series of cos x is $\displaystyle \sum_{n=0}^{+\infty} \frac{(-1)^n}{(2n)!}\:x^{2n}$
• Apr 29th 2009, 12:05 PM
Jester
Quote:

Originally Posted by Kaitosan
Will you please show me the maclaurin series of cosx, sinx, and ln(x+1) with "n" beginning at zero? I can't find it...

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

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

$\displaystyle \ln x + 1 = \sum_{n=1}^\infty \frac{(-1)^{(n+1)}}{n}x^{n}$

and the only resaon that the last one starts at $\displaystyle n = 1$ is that when $\displaystyle n= 0$, the first term of the series is zero.
• Apr 29th 2009, 12:06 PM
Kaitosan
Thanks guys, I was just confused about few things.