# taylor series power question..

• Jan 13th 2010, 10:11 AM
transgalactic
taylor series power question..
when i develop the series of a cosine i have a (-1) member
i wanted to represent the series as a sum
so i need to take only the odd members so the power of -1 is 2k+1 i got
but the solution says that the power of -1 is equal $\displaystyle (-1)^{k-1}$

is it the same??
why they have such an expression
(they use n istead of k)
http://i45.tinypic.com/6sszue.jpg

how they got the power?
• Jan 13th 2010, 10:38 AM
parkhid
In Summation We can use the Merge Role :

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

So We Can Merge Like Below :

$\displaystyle Cosx = \sum_{n=1}^{n=\infty} {(-1)}^{(n-1)}{\frac {x^{2(n-1)}}{2(n-1)!}}$
• Jan 13th 2010, 12:10 PM
transgalactic
in the original expression had (-1)^(2n+1)
which represent the odd number members

while you write as the original
(-1)^(n)

its not the same
• Jan 13th 2010, 01:54 PM
parkhid
We Can Play By Sigma Indexes .

I think the $\displaystyle (-1)^n$ is easier to understand