matrix expression for $e^{iA}Be^{iA}$ in terms of anticommutators? I'm familiar with the expression 
$$e^{-iA}Be^{iA} = \sum_{n=0}^{\infty} \frac{i^n}{n!}[..[B,A],\dots A]_{n \; \rm times}$$
for square matrices $A$ and $B$ and was wondering if equivalently
$$e^{iA}Be^{iA} = \sum_{n=0}^{\infty} \frac{i^n}{n!}\{..\{B,A\},\dots A\}_{n \; \rm times},$$
or something similar?
 A: Let $A,B$ be fixed square matrices of the same size, and define $F(t) = e^{itA}Be^{itA}.$ We then have
$$\begin{align}
F'(t) &= e^{itA}iABe^{itA} + e^{itA}BiAe^{itA} = e^{itA}(iAB+BiA)e^{itA} = e^{itA}i\{A,B\}e^{itA} \\
F''(t) &= e^{itA}iA(iAB+BiA)e^{itA} + e^{itA}(iAB+BiA)iAe^{itA} = e^{itA}i^2\{A,\{A,B\}\}e^{itA} \\
F'''(t) &= \cdots = i^3e^{itA}\{A,\{A,\{A,B\}\}\}e^{itA} \\
&\vdots \\
F^{(k)}(t) &= i^k e^{itA}(\{A,\bullet\}^kB)e^{itA}
\end{align}$$
so
$$
F(t) 
= \sum_{k=0}^{\infty} \frac{t^k}{k!} f^{(k)}(0)
= \sum_{k=0}^{\infty} \frac{t^k}{k!} i^k (\{A,\bullet\}^kB)
$$
and
$$
e^{iA}Be^{iA} = F(1) = \sum_{k=0}^{\infty} \frac{1}{k!} i^k (\{A,\bullet\}^kB).
$$
A: Nice question and nice answers. There is a generalization that admits an even simpler proof that I think it's worth mentioning. 
For (square) matrices $X,Y,B$,  let me define
$$
B(t):=e^{tX} B e^{tY}.
$$
Differentiating with respect to $t$ we see that $B(t)$ satisfies the following differential equation:
\begin{align}
\frac{dB}{dt}(t) &= XB(t) + B(t) Y \\
 &= \mathcal{K}_{X,Y}(B)& = XZ+ZY, \tag{1}
\end{align}
having defined the linear (super) operator
$$
\mathcal{K}_{X,Y}(Z) = XZ+ZY
$$
Now the solution of $(1)$ with initial condition $B(0)=B$ is  $
B(t) =  e^{t \mathcal{K}_{X,Y}} B$. We have then
$$
B(t) = e^{tX} B e^{tY} = e^{t \mathcal{K}_{X,Y}} B =\sum_{n=0}^\infty \frac{(t\mathcal{K}_{X,Y})^n}{n!} B .
$$
Set $Y=\pm X$ to obtain your formulas with commutators/anticommutators. 
A: Define $L_A(B):=AB$ and $R_A(B):=BA$ and write $\mathrm{ad}_A(B):=[A,B]=(L_A-R_A)B$. 
Then use the binomial theorem to expand $(\mathrm{ad}_A)^n$ (since $L_A,R_A$ commute):
$$ \begin{array}{lcl} \displaystyle \sum_{n=0}^{\infty} \frac{(i\,\mathrm{ad}_A)^n}{n!}B & = & \displaystyle \sum_{n=0}^{\infty} \frac{i^n}{n!}(L_A-R_A)^nB \\[5pt] & = & \displaystyle \sum_{n=0}^{\infty}\frac{i^n}{n!} \left[ \sum_{k+\ell=n} \binom{n}{k,\ell} L_A^k (-R_A)^{\ell} \right] B \\[8pt] & = & \displaystyle \sum_{k=0}^{\infty}\sum_{\ell=0}^{\infty}\frac{(iL_A)^k}{k!}\frac{(-iR_A)^{\ell}}{\ell!}B \\[8pt] & = & \displaystyle \left[\sum_{k=0}^{\infty}\frac{(iA)^k}{k!}\right]B\left[\sum_{\ell=0}^{\infty} \frac{(-iA)^\ell}{\ell!}\right] \\[6pt] & = & =\exp(iA)B\exp(-iA). \end{array} $$
The same thing works for $\{A,B\}$, just get rid of the minus sign in $L_A-R_A$.
