Maclaurin series of cos(x) using sin(x) I know Maclaurin series for sin(x) is $$\sum_{n>0}(-1)^n\frac{x^{(2n+1)}}{(2n+1)!}$$
so I  can say cos(x)=  $$\sum_{n>0}\frac{d}{dx}(-1)^n\frac{x^{(2n+1)}}{(2n+1)!}$$
$$=\sum_{n>1}(-1)^n\frac{x^{(2n)}}{(2n)!} $$
but my text says that $$cos(x)=\sum_{n>0}(-1)^n\frac{x^{(2n)}}{(2n)!}$$
Can somebody explain what I did wrong?
Thank you
 A: Just to move this from the comments to an answer (and to respond to the followup), when we differentiate the series
$\sin{x} = \sum_{n \geq 0} (-1)^n \frac{x^{(2n+1)}}{(2n+1!)} = x - \frac{x^3}{3} + 
\dots $
we get 
$\cos{x} = \sum_{n \geq 0} (-1)^n \frac{x^{(2n)}}{(2n!)} = 1 - \frac{x^2}{2} + 
\dots $
We don't have to increase the lower bounds because there is no constant term in the series for $\sin{x}$ that vanishes. If you had a series that looked like
$1 + x^2 + x^4 + \dots = \sum_{n>0} x^{2n-2}$
we could chose to increase the bounds because the first term vanishes when we differentiate, and we get
$0 + 2x + 4x^3 + \dots = \sum_{n>1}(2n-2)x^{2n-3}$
though note you could leave the bounds as they were and get the same series. In general, I recommed writing out the first few terms in the series so you can get a better idea of what happens as you manipulate it.
A: This is all wrong in the index. If you write
$$
\sum_{n>0}(-1)^n\frac{x^{2n+1}}{(2n+1)!}
$$
then the first term in that series is for $n=1$, that is $-\dfrac{x^3}{3!}$. What you intended to use is
$$
\sin x = \sum_{n\ge0}(-1)^n\frac{x^{2n+1}}{(2n+1)!}
$$
where then the formal derivative correctly results in
$$
\cos x = \sum_{n\ge0}(-1)^n\frac{x^{2n}}{(2n)!}
$$
where the first term for $n=0$ is correctly $\dfrac{x^0}{0!}=1$.
