What is wrong with my derivation of the general formula of the power series for $\cos^{2}(x)$ I have arduously obtained the series for $\cos^2(x)$ by using long multiplication method (multiplying twice the Taylor series of $\cos(x)$ which is $\sum_{n=0}^{+\infty} \frac{(-1)^nx^{2n}}{(2n)!}$
$$\cos^2(x)=1-x^2+\dfrac{1}{3}x^4-\dfrac{2}{45}x^6+\dfrac{1}{315}x^8+...$$
I wish to manipulate with the sigma notation to reduce the cost of labour, so I square the above general formula for $\cos(x)$:
$$\cos^2(x)=\sum_{n=1}^{+\infty} \left(\frac{(-1)^nx^{2n}}{(2n!)}\right)^2=\sum_{n=1}^{+\infty} \dfrac{(-1)^{2n}x^{4n}}{[(2n)!]^2}$$
But this is plain wrong, the formula that I obtain is not the general formula for the series of $\cos^2(x)$.
So what is wrong with my manipulation with the sigma notation of this series. The power of $x$ and $-1$ is too high, it should be $2n$. Also, is my denominator correct?
How do you obtain the general formulae for this series?
 A: When you square a summation in general, the result is not equal to the sum of the squares of each term of the summation, as you have assumed.
I will demonstrate this with a couple of examples. Firstly, it's clear that
$(a+b)^2=a^2+b^2+2ab$. But according to what you wanted to do, then $(a+b)^2=a^2+b^2$, which is clearly wrong.
Similarly, consider the value of
$$\left(\sum_{r=1}^n 1\right)^2$$
This is clearly $(n)^2=n^2$, but according to what you're saying the value should be equal to
$$\sum_{r=1}^n 1^2=\sum_{r=1}^n 1=n$$
which again is clearly wrong.
Secondly, it's important to realize that $(n!)^2$ is not equal to $n^2!$. It cannot be simplified further I'm afraid; $(n!)^2$ is the best you can do. Again, verify this with some examples, eg $n=4$.
The best strategy to find the power series for your series is to utilise the double angle formulae, as Olivia Oloa has noted.
Firstly, recall that
$$\cos2x=2\cos^2x-1\implies\cos^2x=\frac{1}{2}(1+\cos2x)$$
Now, we know that
$$\cos2x=\sum_{n=0}^\infty (-1)^n\frac{(2x)^{2n}}{(2n)!}$$
Hence,
$$\cos^2x=\frac{1}{2}+\frac{1}{2}\sum_{n=0}^\infty (-1)^n\frac{(2x)^{2n}}{(2n)!}$$
If you have any questions please don't hesitate to ask.
A: Since your error was already explained in other answers let us see what one obtains if the series multiplication in your example is applied correctly. We have:
$$\begin{align}\cos^2 x&=\left[\sum_{i\ge0}(-1)^i\frac{x^{2i}}{(2i)!}\right]^2\\
&=\sum_{j\ge0}\sum_{i\ge0}(-1)^{i+j}\frac{x^{2i+2j}}{(2i)!(2j)!}\\
&=\sum_{k\ge0}(-1)^k \frac{x^{2k}}{(2k)!}\sum_{i\ge0}\frac{(2k)!}{(2i)!(2k-2i)!}\\
&=\frac12+\frac12\sum_{k\ge0}(-1)^k\frac{(2x)^{2k}}{(2k)!}\\
&=\frac{1+\cos2x}2,
\end{align}$$
where the proof of
$$
\sum_{i=0}^k\frac{(2k)!}{(2i)!(2k-2i)!}
=\begin{cases}1,& k=0,\\
2^{2k-1},& k>0
\end{cases}
$$
is left to you.
