# Taylor series power question

1. Jan 13, 2010

### nhrock3

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 (-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?

2. Jan 13, 2010

### Staff: Mentor

You want the signs of your Taylor's series to alternate, right? (-1)k - 1 gives you that sign alternation. If you had (-1)2k + 1, the sign would always be negative, since you have odd powers of -1.

3. Jan 14, 2010

### nhrock3

you are correct
how to get this expression?

4. Jan 14, 2010

### Staff: Mentor

What exactly are you asking? Are you asked to find the Taylor's series for cos z at z = 2? There is a standard technique for finding the coefficients of this series.

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