Prove that the series $ \sum_{n=0}^\infty \frac{(-1)^n \cdot x^{2n}}{(2n)!} $ represents $ \cos x $ for all values of $ x $ guys.
The question is as stated in the title: prove that the series $ \sum_{n=0}^\infty \frac{(-1)^n \cdot x^{2n}}{(2n)!} $ represents $\cos x $ for all values of $ x $
My doubt is quite theoretical:
I did this exercise the following way:
$ \cos x = \frac{d}{dx} \sin x \therefore \cos x = \frac{d}{dx} \sum_{n=0}^\infty \frac{(-1)^n \cdot x^{2n+1}}{(2n+1)!} $
My doubt is right here. According to the book, when we take the derivative here we get $ \sum_{n=0}^\infty \frac{(-1)^n \cdot x^{2n}}{(2n)!} $, but shouldn't we get $ \sum_{n=1}^\infty \frac{(-1)^n \cdot x^{2n}}{(2n)!} $ ?
When my teacher was teaching how to differentiate series, he said that when you did, you always had to add one to the index. 
Am I missing something here?
Thanks in advance.
Pedro.
 A: That "rule" is deceptive; I will use an example to demystify it.
Note that formally we have
$D\sum_{k=1}^{n}x^{k} = D(x + x^{2} + \cdots + x^{n}) = 1 + 2x + \cdots + nx^{n-1} = \sum_{k=1}^{n}kx^{k-1};$
no need to apply the rule here. But 
$D\sum_{k=0}^{n}x^{k} = D(1 + x + \cdots + x^{n}) = 1 + \cdots + nx^{n-1} = \sum_{k=1}^{n}kx^{k-1}$;
it does appear like "we add one to the index"!
Can you see now a why?
A: I think what your teacher is reffering to is that the derivative of a constant is zero, and if the first term of the series is a constant it follows that the first term of the derivative of the series is 0, so you can just start the series at $n = 1$. What your teacher said is correct IF the first term of the series is a constant. For example if you take the derivative of the cosine series then it is ok. The first term of the sine series is $x$ which is not a constant with respect to $x$. 
A: 
Expand $\sin(x)$ using Taylor series as follows:
$$\sin(x)=\sum_{k=0}^\infty\frac{(-1)^k}{(2k+1)!}x^{2k+1}=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\ldots$$
Given the above information you can write $\cos(x)$ as follows:
$$\begin{align}
\cos(x)&=\frac{\mathrm d}{\mathrm dx}\sin(x)\\
&=\frac{\mathrm d}{\mathrm dx}\sum_{k=0}^\infty\frac{(-1)^k}{(2k+1)!}x^{2k+1}\\
&=\frac{\mathrm d}{\mathrm dx}\left(x-\frac{x^3}{3!}+\frac{x^5}{5!}-\ldots\right)\\
&=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\ldots\\
&=\sum_{k=0}^\infty\frac{(-1)^kx^{2k}}{(2k)!}
\end{align}$$

Source: Taylor Series Expansion - UBC - CA.
