Grassmann variable Taylor expansion I have a problem concerning Taylor expansion of functions of Grassmann variables. Let us consider a general function $ f(\eta_1,\eta_2)$. Then the Taylor expansion around $\eta=0$ should be: $$ f(\eta_1,\eta_2)=f(0,0)+\left. \frac{\partial f}{\partial \eta_1 }\right|_0\eta_1+\left. \frac{\partial f}{\partial \eta_2 }\right|_0\eta_2+\frac{1}{2}\left. \frac{\partial^2 f}{\partial \eta_1 \partial\eta_2}\right|_0\eta_2\eta_1 +\frac{1}{2}\left. \frac{\partial^2 f}{\partial \eta_2 \partial\eta_1}\right|_0\eta_1\eta_2\qquad (1)  $$
Then since Grassmann numbers and derivatives anti-commute, I can write: $$ \frac{\partial^2f}{\partial\eta_2\partial\eta_1}=-\frac{\partial^2f}{\partial\eta_1\partial\eta_2},\qquad\eta_1\eta_2=-\eta_2\eta_1$$Then the expansion of $f(\eta_1,\eta_2)$ becomes: $$ f(\eta_1,\eta_2)=f(0,0)+\left. \frac{\partial f}{\partial \eta_1 }\right|_0\eta_1+\left. \frac{\partial f}{\partial \eta_2 }\right|_0\eta_2+\frac{1}{2}\left. \frac{\partial^2 f}{\partial \eta_1 \partial\eta_2}\right|_0\eta_2\eta_1 +(-1)^2\frac{1}{2}\left. \frac{\partial^2 f}{\partial \eta_1 \partial\eta_2}\right|_0\eta_2\eta_1\qquad (2)$$That means: $$ f(\eta_1,\eta_2)=f(0,0)+\left. \frac{\partial f}{\partial \eta_1 }\right|_0\eta_1+\left. \frac{\partial f}{\partial \eta_2 }\right|_0\eta_2+\left. \frac{\partial^2 f}{\partial \eta_1 \partial\eta_2}\right|_0\eta_2\eta_1$$
My problem arises when treating functions like $e^{\eta_1+\eta_2}$; I have in fact, using expression 1: $$ \frac{1}{2}\left. \frac{\partial^2 e^{\eta_1+\eta_2}}{\partial \eta_1 \partial\eta_2}\right|_0\eta_2\eta_1 +\frac{1}{2}\left. \frac{\partial^2 e^{\eta_1+\eta_2}}{\partial \eta_2 \partial\eta_1}\right|_0\eta_1\eta_2=\frac{1}{2}\eta_1\eta_2+\frac{1}{2}\eta_2\eta_1=0 $$ Since singularly those derivatives should give 1. But if I use (2) I only have: $$ \left. \frac{\partial^2e^{\eta_1 +\eta_2}}{\partial\eta_1\eta_2}\right|_0\eta_1\eta_2=\eta_1\eta_2 $$Where am I wrong?
 A: Your last two equations are not correct. Basically, you seem to use
$$\partial_1\partial_2 e^{\eta_1+\eta_2} \stackrel{?}{=} \partial_1e^{\eta_1} \partial_2e^{\eta_2}\,,$$
but that's wrong. The easiest way to see it is that $\chi=\eta_1+\eta_2$ is again a Grassmann variable, so
$$e^\chi=1+\chi+\frac{1}{2}\chi^2 + \dotsm=1+\chi\,,$$
and so 
$$\partial_1\partial_2 e^{\eta_1+\eta_2}=\partial_1\partial_2 \left(1+\eta_1+\eta_2\right)\,.$$
Hence, your double-derivative term vanishes.
Note that $e^\eta=1+\eta$ in particular implies that $\partial e^\eta/\partial\eta\neq e^\eta$, and that $e^\eta$ does not have definite Grassmann parity, so you have to take care e.g. when applying the Leibniz rule.
A: Note that your second equation does not hold in general, e.g. for $f(\eta_1,\eta_2) = \eta_3 \eta_1 \eta_2$  you get 
$$
\partial_{\eta_1}\partial_{\eta_2}f = -\eta_3 = \partial_{\eta_2}\partial_{\eta_1}f.
$$
The error in your example is the last equation. 
$$
\partial_{\eta_2} e^{\eta_1+\eta_2} = 1
$$
Hence this also yields zero.
