# Maclaurin Series from sin(x) to cos(x) using derivative

I understand how to find the MacLaurin series for $$\sin(x)$$. $$\sum_{n=1}^\infty \frac{x^{2n-1}\cdot\!(-1)^{n-1}}{(2n-1)!}$$ Now I am trying to find the MacLaurin series for $$\cos(x)$$ by taking the derivative of the above sum with respect to $$x$$. Using power rule, I got the following series: $$\cos(x) = \sum_{n=1}^\infty \frac{x^{2n-2}\cdot\!\mathrm{(-1)}^{n-1}}{(2n-2)!}$$

However, the MacLaurin series is: $$\cos(x) = \sum_{n=0}^\infty \frac{x^{2n}\cdot\!\mathrm{(-1)}^{n}}{(2n)!}$$ How are these two $$\cos(x)$$ MacLaurin series equal? What makes the second one more correct than the series I got by taking the derivative of the $$\sin(x)$$ series.

A sort of related question: if you choose to start at $$n=1$$ vs $$n=0$$, how would you change the terms of the $$\sin(x)$$ Maclaurin series?

• The two series are the same. They only differ by "re-indexing". Shifting the start from $n=1$ to $n=0$ just replaces all the $n$'s by $(n+1)$ in the formula. – Nick Mar 6 at 21:34

You got twice the same series. Both$$\sum_{n=0}^\infty \frac{x^{2n}\cdot\!\mathrm{(-1)}^{n}}{(2n)!}\text{ and }\sum_{n=1}^\infty \frac{x ^{2n-2}\cdot\!\mathrm{(-1)}^{n-1}}{(2n-2)!}$$are equal to$$1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+\cdots$$
To reindex, replace $$n$$ with $$n+1$$ in the summand of the first series and lower the range of $$n$$ by $$1$$ (so the new initial index will now be $$0$$ instead of $$1$$).
More generally, if the index $$n$$ runs from $$n=a$$ to $$n=b$$, then replace $$n$$ with $$n+K$$ in the summand and change the index limits to run from $$n=a-K$$ to $$n=b-K$$. You are essentially adding and subtracting $$K$$ from each value of $$n$$, so it is unchanged in the end. Symbolically: $$\sum_{n=a}^b t_n =\sum_{n=a-K}^{b-K}t_{n+K}$$