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I understand how to get the proper maclaurin series representation for $\cos x$, but I'm having trouble understanding the following part conceptually:

I get $\cos x$ as $\sum_{n=0}^\infty (-1)^n\frac{x^{2n}}{2n!}$ but,

Can the maclaurin series of $\cos x$ also be $\sum_{n=0}^\infty (-1)^n\frac{x^n}{n!}$?

I'm confused because even though the odd powers of this functions are going to $0$, wouldn't it still be valid to include them in our maclaurin series? Furthermore, why do we omit terms if they are $0$?

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Maclaurin series are unique when they exists and the second one is $$1-x+\frac{x^2}{2}\mp \dots$$ which is $\exp(-x)$. But $$\cos(x)=1-\frac{x^2}{2} +\frac{x^4}{4!} \mp \dots$$ For sure we allow terms when they are zero, but here those would it make much more complicate to write it in a closed form, thats why we don't write them.

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The answer to the first question is no. The latter series is that for $e^{-x}$. The inclusion of only even terms is a consequence of $\cos{x}$ being an even function. That said, sure you can include zero terms in a sum, but do you go around saying that $5$ is really $5+0$? There may be instances where it makes sense to do so, but under normal circumstances, no.

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