möbius function on poset Let $P$ be the poset of all subsets of $\{1,2,\ldots, n\}$ with av even number of elements, ordered by inclusion. There is a recursive formula for the Möbius function on a poset:
$$ \mu(x,y) = -\sum_{x\leq z < y} \mu(x,z) ~~~~ \textrm{when} ~x<y$$
and
$$\mu(x,x)=1$$
Is it possible to find a explicit formula in the case of this particular poset? For the Boolean lattice (all subsets of $\{1,2,\ldots, n\}$) it is $\mu(X,Y)=(-1)^{|Y\setminus X|}$, but the property of the subsets having an even number of elements makes it more complicated. 
 A: It is easy to see that $\mu(S,T)$ depends only on $\lvert T \setminus S \rvert$. Hence, $\mu(S,T) = \hat{\mu}(\lvert T \setminus S \rvert)$ for some function $\hat{\mu}$ defined on the non-negative even integers. We will show that
\begin{equation*}
\sum_{\substack{k=0\\\text{$k$ even}}}^{+\infty} \hat{\mu}(k) \frac{t^k}{k!} = \frac{1}{\cosh(t)} \:.
\end{equation*}
Indeed,
\begin{equation*}
\cosh(t) \sum_{\substack{k=0\\\text{$k$ even}}}^{+\infty} \hat{\mu}(k) \frac{t^k}{k!} = \sum_{\substack{j=0\\\text{$j$ even}}}^{+\infty} \frac{t^j}{j!} \sum_{\substack{k=0\\\text{$k$ even}}}^{+\infty} \hat{\mu}(k) \frac{t^k}{k!} = \sum_{\substack{j=0\\\text{$j$ even}}}^{+\infty} \sum_{\substack{k=0\\\text{$k$ even}}}^{+\infty} \hat{\mu}(k) \frac{t^{j+k}}{j!k!} \:.
\end{equation*}
And thus
\begin{equation*}
\cosh(t) \sum_{\substack{k=0\\\text{$k$ even}}}^{+\infty} \hat{\mu}(k) \frac{t^k}{k!} = \sum_{\substack{r=0\\\text{$r$ even}}}^{+\infty} \sum_{\substack{k=0\\\text{$k$ even}}}^{r} \hat{\mu}(k) \frac{t^{r}}{k!(r-k)!} = \sum_{\substack{r=0\\\text{$r$ even}}}^{+\infty} \frac{t^r}{r!}\sum_{\substack{k=0\\\text{$k$ even}}}^{r} \! \binom{r}{k} \: \hat{\mu}(k) = 1 \:.
\end{equation*}
We conclude with the Taylor expansion of $\frac{1}{\cosh(t)}$ involving the Euler's numbers.
