[Math] Maclaurin series of cos(x) using sin(x)

calculustaylor expansion

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

Best Answer

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.

Related Question