I'm stuck trying to calculate $2C(x)^2 = C(2x)+1$
with $C(x) =\sum_{n=0}^{\infty} (-1)^n \frac {x^{2n}} {(2n)!}$ (Cosine Power series)
So I've seen proofs for this identity before, but never using the Cauchy product formula.
So this is what I've tried:
Plugging in the power series we get:
$2\left(\sum_{n=0}^{\infty} (-1)^n \frac {x^{2n}} {(2n)!}\right)^2$
$=$ $2$$\left(\sum_{n=0}^\infty\sum_{k=0}^n (-1)^{n-k}\frac{x^{2(n-k)}}{(2(n-k))!}(-1)^k\frac{x^{2k}}{(2k)!}\right)$
$=$ $2\left(\sum_{n=0}^\infty\sum_{k=0}^n(-1)^n\binom{2n}{2k}\frac{x^{2n}}{(2n)!}\right)\\$ I hope this is correct at least.
For the other side of the equation I'm actually not quite sure how to write down.
It seems to me that this is going to be quite a long equation in general and I'm not able to calculate this further. Is it possible that I'm missing something or how can I proceed further to prove this? (Using the Cauchy product formula)
By definition, $$C(x) =\sum_{n=0}^{\infty} (-1)^n \frac {x^{2n}} {(2n)!}$$ The series is absolutely convergent, so you can consider the Cauchy product with itself, which gives you :
$$C(x)^2 = \left(\sum_{n=0}^{\infty} (-1)^n \frac {x^{2n}} {(2n)!} \right)\left(\sum_{n=0}^{\infty} (-1)^n \frac {x^{2n}} {(2n)!} \right)=\sum_{n=0}^{\infty}\sum_{k=0}^n (-1)^k \frac{x^{2k}}{(2k)!}(-1)^{n-k} \frac{x^{2(n-k)}}{(2(n-k))!}$$
which simplifies to give : $$C(x)^2 =\sum_{n=0}^{\infty}\sum_{k=0}^n (-1)^n \frac{x^{2n}}{(2n)!}{2n \choose 2k}= \sum_{n=0}^{\infty} (-1)^n \left(\sum_{k=0}^n{2n \choose 2k}\right)\frac{x^{2n}}{(2n)!}$$
By developping $(1+i)^{2n}$, you can see that for all $n \geq 1$, one has $$\sum_{k=0}^n{2n \choose 2k}=2^{2n-1}$$
Plugging this in the expression, you get $$C(x)^2 = 1 +\sum_{n=1}^{\infty} (-1)^n \left(\sum_{k=0}^n{2n \choose 2k}\right)\frac{x^{2n}}{(2n)!} = 1 + \sum_{n=1}^{\infty} (-1)^n 2^{2n-1}\frac{x^{2n}}{(2n)!}$$
So $$2C(x)^2 = 2 + \sum_{n=1}^{\infty} (-1)^n 2^{2n}\frac{x^{2n}}{(2n)!} = 2 +\sum_{n=1}^{\infty} (-1)^n \frac{(2x)^{2n}}{(2n)!}$$
Finally, reincorporing the term for $n=0$, you get $$2C(x)^2 = 1 + \sum_{n=0}^{\infty} (-1)^n \frac{(2x)^{2n}}{(2n)!} = 1+ C(2x)$$