How to show that $\cos(x)=\sum\limits_{n=0}^\infty (-1)^n \frac {x^{2n}}{ (2n)!} $

Question

We know that cos(x) is infinitely differentiable and the Lagrange remainder $\rightarrow 0$ for all $x$, so the Taylor series indeed produces the function. We also know that
$\cos^{(4k)}(x)=\cos(x)$ at $x=0$ its $1$
$\cos^{(4k+1)}(x)=-\sin(x)$ at $x=0$ its $0$
$\cos^{(4k+2)}(x)=-\cos(x)$ at $x=0$ its $-1$
$\cos^{(4k+3)}(x)=\sin(x)$ at $x=0$ its $0$

Now the question is how do i get that sum from these informations?

Answer 1

It is the Taylor expansion around $0$ of the cosine function. With $f(x) = \cos(x)$ you just proved that

$\dfrac{d^{2n}}{dx^{2n}}f(x) = (-1)^n$ and $\dfrac{d^{2n+1}}{dx^{2n+1}}f(x) = 0$

Just introduce this in the Taylor expansion definition and you will get your sum.

Answer 2

By Taylor's expansion (with $f^n(a) = \left.\frac{d^n}{dx^n}f(x)\right|_{x=a}$) $$ f(0+x) = \sum_{n=0}^{\infty} \frac{x^nf^n(0)}{n!} $$ $\cos(x)$ becomes: \begin{align*} \cos(x) &= \sum_{n=0}^{\infty} \left(\frac{x^{4n}\cdot1}{(4n)!}+\frac{x^{4n+1}\cdot 0}{(4n+1)!}+\frac{x^{4n+2}\cdot(-1)}{(4n+2)!}+\frac{x^{4n+3}\cdot 0}{(4n+3)!} \right) \\ &= \sum_{n=0}^{\infty} \left(\frac{x^{4n}\cdot 1}{(4n)!}+\frac{x^{4n+2}\cdot(-1)}{(4n+2)!}\right)\\ &= \sum_{n=0}^{\infty} \left(\frac{x^{2(2n)}\cdot(-1)^{2n}}{(2(2n))!}+\frac{x^{2(2n+1)}\cdot (-1)^{(2n+1)}}{(2(2n+1))!}\right) \end{align*} Now, put $2n=m$, to get $$ \cos(x) = \sum_{m=0,2,4}^{\infty} \left(\frac{x^{2m}.(-1)^{m}}{(2m)!}+\frac{x^{2(m+1)}.(-1)^{(m+1)}}{(2(m+1))!}\right) $$ Now, the first term uses even powers for $-1$ and second term uses odd. So, this can be rewritten as the required form.

PS: It might be easier to prove using the expansion of $e^{jx}$ and collecting the real part of that.