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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?

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  • $\begingroup$ 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. $\endgroup$ – Nick Mar 6 at 21:34
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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$$1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+\cdots$$

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These are the same series, just indexed differently. Write out the first few terms of each and you will see that.

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: $$\sum_{n=a}^b t_n =\sum_{n=a-K}^{b-K}t_{n+K}$$

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