A series expansion of $\cos(x)$ is
$$ \cos(x)=\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n)!}x^{2n} $$
I want to get the series expansion of $\cos^2(x)$, but using two different method I get different result
First Method : Using Trigonometric Identities
$$ \cos^2(x)=\frac{1+\cos(2x)}{2} $$
Then subtituting the series $$ \cos(2x)=\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n)!}(2x)^{2n} \\ \cos^2(x)=\frac{1}{2}+\frac{1}{2}\sum_{n=0}^{\infty}2^{2n}\frac{(-1)^n}{(2n)!}x^{2n} $$
Second Method : Squaring The Series
By directly squaring the series $$ \cos^2(x)=[\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n)!}x^{2n}]^2 $$
I get
$$ \cos^2(x)=(\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n)!}x^{2n})(\sum_{m=0}^{\infty}\frac{(-1)^m}{(2m)!}x^{2m})\\ \cos^2(x)=\sum_{n=0}^{\infty}\sum_{m=0}^{\infty}\frac{(-1)^{n+m}}{(2n)!(2m)!}x^{2(m+n)} $$
Which is different result, the only way I can think off to make this the same result as the first method above is to somehow show that
$$ \sum_{n=m=0}^{\infty}\frac{(-1)^{n+m}}{(2n)!(2m)!}x^{2(m+n)}=\frac{1}{2}\sum_{n=0}^{\infty}2^{2n}\frac{(-1)^n}{(2n)!}x^{2n} $$
and
$$ \sum_{n=0}^{\infty}\sum_{m\neq n}^{\infty}\frac{(-1)^{n+m}}{(2n)!(2m)!}x^{2(m+n)}=\frac{1}{2} $$
Can I use the second method ? If so how to show the double summation lead back to result in the first method (especially the second one) ?
In your second method, make a partition according to the value of $m+n$, you'll get $$ \cos^2(x)=\sum_{k=0}^{+\infty}\sum_{\substack{n,m\geqslant 0 \\ n+m=k}}\frac{(-1)^k}{(2n)!(2m)!}x^{2k}=\sum_{k=0}^{+\infty}(-1)^k\frac{x^{2k}}{(2k)!}\sum_{n=0}^k\binom{2k}{2n}$$ But $$\sum_{n=0}^k\binom{2k}{2n}=\sum_{n=0}^{2k}\frac{1+(-1)^n}{2}\binom{2k}{n}=\frac{2^{2k}+0^{2k}}{2}$$ Therefore the above sum is $2^{2k-1}$ when $k\geqslant 1$ and $1$ when $k=0$, which means that $$ \cos^2(x)=1+\frac{1}{2}\sum_{k=1}^{+\infty}2^{2k}\frac{(-1)^k}{(2k)!}x^{2k} $$ and you get back your first sum !