Logarithm of Grassmann numbers What is $\log \theta$ where $\theta$ is a Grassmann number such that $\theta^2 = 0$. How does one then look at its logarithm? Does it even make sense?
 A: I don't think we can assign a meaningful value to $\log(\theta)$. We can, however, find $\log(1+\theta)$ as follows.
Since $\theta^2=0$ then we have $\theta^k=0 \space \forall k \ge 2$. So
$\displaystyle\exp(\theta) = \sum_{k=0}^\infty \frac {\theta^k}{k!} = 1 + \theta$
and since $\log$ is the inverse of $\exp$ we have
$\log(1 + \theta) = \log(\exp(\theta)) = \theta$
A: *

*In general, if $f: \Omega \subseteq \mathbb{C}\to \mathbb{C}$ is a complex analytic function, then we can define the value
$$ f(z)~:=~ \sum_{n\in\mathbb{N}_0} \frac{f^{(n)}(z_B)}{n!}z_S^n\tag{1.1.6}$$ to a supernumber $$z~=~\underbrace{z_B}_{\text{body}}+\underbrace{z_S}_{\text{soul}}\tag{1.1.2}$$
if the body
$$z_B~\in~\Omega$$
belongs to the open neighborhood $\Omega\subseteq \mathbb{C}$.


*The problem with OP's example $\log\theta$ is that the body of the Grassmann number $\theta$ is $0$, but $f=\log$ is not defined in $0\notin \Omega$, so we cannot apply definition (1.1.6).
References:

*

*Bryce DeWitt, Supermanifolds, Cambridge Univ. Press, 1992.

