Compute Taylor series at $x = 0$ of $e^{1-\cos{x}}$ in summation notation? I am doing an exercise from textbook which asks to compute Taylor series at  $x = 0$ of $e^{1-\cos{x}}$ in summation notation?
My intuition was to expand function $e^x = \sum^{\infty}_{k=0} \frac{x^k}{k!}$ and $\cos{x} = \sum^{\infty}_{k=0}(-1)^k\frac{x^{2k}}{(2k)!}$ at ${x = 0}$, so I got the result: 
$\sum^{\infty}_{k=0}\frac{1 - \sum^{\infty}_{t=0}(-1)^t\frac{x^{2t}}{(2t)!}}{k!}$. 
The result is a form of 2 summations, thus I am not too sure whether it is correct. Could anyone please help to verify or give some comments?
Thanks for VVejalla's comment, the double summation result I got should be: $\sum^{\infty}_{k=0}\frac{(1 - \sum^{\infty}_{t=0}(-1)^t\frac{x^{2t}}{(2t)!})^k}{k!}$
 A: Let $f(x) =e^{1-\cos x} $ so that $$f'(x) =f(x) \sin x\tag{1}$$ Let us assume that $f(x) $ can be represented as a power series in terms of even powers of $x$ (because $f$ is even) so that $$f(x) =\sum_{n=0}^{\infty} a_nx^{2n}\tag{2}$$ Differentiating term by term we get $$f'(x)=\sum_{n=1}^{\infty} 2na_nx^{2n-1}=\sum_{n=0}^{\infty} 2(n+1)a_{n+1}x^{2n+1}$$ Putting the series for $f'$ and $f$ in $(1)$ we get $$\sum_{n=0}^{\infty} 2(n+1)a_{n+1}x^{2n+1}=\sum_{n=0}^{\infty} a_nx^{2n}\cdot\sum_{n=0}^{\infty} (-1)^n\frac{x^{2n+1}}{(2n+1)!}$$ Cancelling $x$ from both sides we get $$\sum_{n=0}^{\infty} 2(n+1)a_{n+1}x^{2n}=\sum_{n=0}^{\infty} a_nx^{2n}\cdot\sum_{n=0}^{\infty} (-1)^n\frac{x^{2n}}{(2n+1)!}$$ Comparing coefficients of $x^{2n}$ on both sides we get $$2(n+1)a_{n+1}=\sum_{k=0}^{n}(-1)^{n-k}\frac{a_k}{(2n-2k+1)!}$$ Starting with $a_0=1$ (as $f(0)=1$) we get $$2a_1=\frac{a_0}{1!}$$ so that $a_1=1/2$. Similarly $$4a_2=-\frac{a_0}{3!}+\frac{a_1}{1!}$$ ie $a_2=1/12$ and $$6a_3=\frac{a_0}{5!}-\frac{a_1}{3!}+a_2$$ ie $a_3=1/720$.
Thus $$f(x) =1+\frac{x^2}{2}+\frac{x^4}{12}+\frac{x^6}{720}+\dots$$
A: This seems to be quite difficult but I think that I found something which is a least amazing.
Consider the expansion of
$$e^{1-\cosh (y)}=\sum_{n=0}^\infty \frac{a_n}{(2n)!} \,y^{2n}$$ The first $a_n$'s make the sequence
$$\{1,-1,2,-1,-43,254,4157,-70981,-1310398,40933619,1087746617\}$$ which is sequence $A260884$ in OEIS. According to the documentation, they are given by
$$a_n=\sum _{k=0}^{2 n} \text{BellY}[2 n,k,\left[\text{Range}[2 n] \bmod 2\right]-1]\qquad \text{with} \qquad a_0=1$$
Now, make $y=i\,x$ to make
$$e^{1-\cosh (y)-1}=e^{1-\cos (x)}$$
A: This is the problem of computing the moments from the cumulants, which is the inverse problem as the one usually encountered. 
Consider a (analytic) function $K(x)$ with $K(0)=0$
\begin{align}
M(x) &:=e^{K(x)}=\exp{\sum_{n=1} \kappa_n \frac{x^n}{n!} } \\
     &:= \sum_{n=0} \mu_n \frac{x^n}{n!}
\end{align}
with $\mu_0 =1$. Then one has the following recursion
$$
\mu_{n}=\kappa_{n}+\sum_{m=1}^{n-1}\left(\begin{array}{c}
n-1\\
m-1
\end{array}\right)\kappa_{m}\mu_{n-m}
$$
Indeed the identity holds with greater generality at the level of formal power series. In our case $K(x) = 1- \cos(x)$ so $\kappa_{2n} = -(-1)^n$ and $\kappa_{2n+1}=0$. Of course one also has that $\mu_{2n+1}=0$. 
With the above formula one gets, for the first few non-zero coefficients
$$
\mu_n = (1, 2, 1, -43, -254, 4157, 70981, -1310398, -40933619, 1087746617,\ldots)
$$
which agrees with Claude Leibovici's results (but here we have the correct signs). 
A: You can express the general coefficient in terms of the complete Bell polynomials $B_n$ in the following manner:
\begin{align*}
\exp (1 - \cos x) & = \exp \left( {\sum\limits_{n = 1}^\infty  {( - 1)^{n + 1} \frac{{x^{2n} }}{{(2n)!}}} } \right) \\ &= \sum\limits_{n = 0}^\infty  {\frac{1}{{n!}}B_n \!\left( {\frac{{1!}}{{2!}}, \ldots ,( - 1)^{n + 1} \frac{{n!}}{{(2n)!}}} \right)x^{2n} } .
\end{align*}
An explicit, though somewhat complicated, formula is as follows:
$$
B_n \!\left( {\frac{{1!}}{{2!}}, \ldots ,( - 1)^{n + 1} \frac{{n!}}{{(2n)!}}} \right) \\ = \sum\limits_{k = 1}^n {\sum\limits_{\substack{ j_1  + j_2  +  \cdots  + j_{n - k + 1}  = k \\ j_1  + 2j_2  +  \cdots  + (n - k + 1)j_{n - k + 1}  = n \\ j_1 ,j_2 , \ldots ,j_{n - k + 1}  \in \mathbb{Z}_{ \ge 0} }} {\frac{{( - 1)^{n + k} n!}}{{j_1 !j_2 ! \cdots j_{n - k + 1} !}}\prod\limits_{m = 1}^{n - k + 1} {\frac{1}{{(2m)!^{j_m } }}} } } .
$$
A: The higher order derivatives of $f(x)=\exp(1-\cos(x))$ have a nice structure.
$$
  \begin{align}
  f'(x) &= f(x)\sin(x) \nonumber\\
  f''(x) &= f'(x)\sin(x) + f(x)\cos(x) \nonumber\\
  f^{(3)}(x) &= f''(x)\sin(x) + 2f'\cos(x) - f(x)\sin(x) \nonumber\\
  f^{(4)}(x) &= f^{(3)}(x)\sin(x) + 3f''(x)\cos(x) - 3f'(x)\sin x - f(x)\cos(x) \nonumber\\
  f^{(5)} &= f^{(4)}\sin + 4f^{(3)}\cos - 6f''\sin - 4f'\cos + f\sin \nonumber\\
  \vdots
  \end{align}
$$
All odd-order derivatives are 0 at $x=0$.  You can see that the coefficients on the right hand side
are the binomial coefficients (with periodic signs $+,+,-,-$).  So the equation for the $2n$th order 
derivative of $f$ at 0 is:
$$
  f^{(2n)}(0) = \sum_{i=0}^{n-1}\binom{2n-1}{2i+1} f^{(2n-1-2i)}(0)(-1)^i.
$$
The values for $n=0,1,\dots 10$ are
$$
1, 1, 2, 1, -43, -254, 4157, 70981, -1310398, -40933619, 1087746617.
$$
