Let $s(x)=\sum_{n=0}^\infty \frac1{(2n+1)!}x^{2n+1}$, $c(x)=\sum_{n=0}^\infty \frac1{(2n)!}x^{2n}$ prove that: $c(x)^2-s(x)^2=1$

Question

Let $$s(x)=\sum_{n=0}^\infty \frac1{(2n+1)!}x^{2n+1}$$ $$c(x)=\sum_{n=0}^\infty \frac1{(2n)!}x^{2n}$$ prove that $$c(x)^2-s(x)^2=1$$

I know that the following series are representations of the cosh and sinh hyperbolic functions but if I was to properly prove this how would I go about it?

I have tried expressing both s(x) and c(x) as exponential functions but do not really know how to construct it.

Answer 1

Differentiate $c^2-s^2=f$ to get $f'=2cc'-2ss'=0$ since $s=c',s'=c$, so $f$ is constant. The constant is $1$ by evaluating at $0$.

Answer 2

The approach by Pedro Tamaroff is simplest one for the question at hand. However since you already know (in your mind) that these series represent $\cosh x$ and $\sinh x$ it is better to link them with $\exp(x)$.

Let $$F(x) = c(s) + s(x) = \sum_{n = 0}^{\infty}\frac{x^{n}}{n!}$$ then clearly $$c(x) - s(x) = \sum_{n = 0}^{\infty}(-1)^{n}\frac{x^{n}}{n!} = F(-x)$$ Now using binomial theorem for positive integral index and the law of multiplication of infinite series it is easy to prove that $F(x + y) = F(x)F(y)$ for all real (even complex) values of $x, y$.

Then $$c^{2}(x) - s^{2}(x) = (c(x) + s(x))(c(x) - s(x)) = F(x)F(-x) = F(0) = 1$$