Taylor Series Representation of $e^{1-\cos(x)}$

Question

Hello I was wondering how to simplify this Taylor Series

$$ e^{1-\cos(x)} =\sum_{k=0}^\infty\frac{(1-\cos(x))}{k!} ^k\ $$

to where I can write out the first couple of terms which are $ 1+\frac{x^2}{2}+\frac{x^4}{12}+\frac{x^6}{720}$

Answer 1

For the full power series you could use Faà di Bruno's formula.

However, to get the first couple of terms, you just have to compose them as truncated polynomials as is done in this example.

Answer 2

One way to see this is to note that $f(x)=e^{1-\cos x}$ is analytic and has a power series expansion, and that it is an even function. So $f(x)=\sum_{n=0}^\infty a_nx^{2n}$. Noting that $f'(x)=\sin(x)f(x)$ we conclude that $$ \left(\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)!}x^{2n+1}\right)\left(\sum_{n=0}^\infty a_nx^{2n}\right)=\sum_{n=0}^\infty 2(n+1) a_{n+1}x^{2n+1} $$ Equating the coefficients of $x^{2n+1}$ and noting that $f(0)=1$, we conclude that the sequence $(a_n)_{n\geq0}$ is defined by the recursion : $$ a_0=1,\quad\hbox{and for $n\geq0$},\quad a_{n+1}=\frac{1}{2(n+1)}\sum_{k=0}^n\frac{(-1)^k}{(2k+1)!}a_{n-k}. $$ For example this gives $$f(x)=1+\frac{x^2}{2}+\frac{x^4}{12}+\frac{x^6}{720}-\frac{43 x^8}{40320}-\frac{127 x^{10}}{1814400}+\mathcal{O}\left(x^{12}\right).$$