I am working with the expression
$$\det\big{|}f(-kx), f(-(k-1)x),\cdots,f(0),\cdots, f((k-1)x), f(kx), g(x)\big{|},$$
where $f,g\colon\mathbb{R}\mapsto \mathbb{R}^{2k+2}$, and want to use the Taylor series of $f$ and $g$ to expand the $\det$ expression in terms of $x$ and then extract the $x^i$ coefficients. However, I cannot find a way to make the determinant function expand linearly and anti-symmetrically, with the understanding that $f$ and $g$ are vector functions and hence cannot be taken out of the determinant.
A potential method would be to define a completely new function $\textrm{newdet}$, with an unspecified number of arguments, which is both antisymmetric and linear over $\mathbb{R}$ in each argument. Indeed, I am not interested in the actual computation of the determinant, simply the expansion, so this would be sufficient. To make $\textrm{newdet}$ antisymmetric, I can use this or this, yeilding
newdet[a__] := Signature[{a}] (newdet @@ Sort@{a}) /; ! OrderedQ[{a}];
newdet[a__] := 0 /; ! Unequal[a];
but how do I make it linear (i.e. distributive over addition and factor real constants out) in each argument over the reals? I have found several ways to make it linear in one or two variables (such as this or this), but I don't know how to extend this to an arbitrator number of inputs. Is there a way (using TensorProduct or WedgeProduct, perhaps) to address both of these issues at once?
Det[m] == HodgeDual[TensorWedge @@ m]
. $\endgroup$Simplify[HodgeDual[TensorWedge @@ {v, w}] + HodgeDual[TensorWedge @@ {w, v}]]
will yieldHodgeDual[v\[TensorWedge]w] + HodgeDual[w\[TensorWedge]v]
instead of $0$ $\endgroup$TensorReduce[]
with assumptions is supposed to be able to handle that simple case, but it doesn't... hmm... $\endgroup$