TITLE: Finding the Jacobian of Taylor series of a vector valued function
QUESTION [0 upvotes]: This question is related to one I asked over on physics stackexchange, but at this point it is purely mathematical in nature and I thought it would make sense to ask here. As the title suggests, I want to find an expression for the elements of a Jacobian, where the vector valued function in question is expressed as a Taylor series. More precisely I want to show that:
\begin{equation}
\tag{1}
\bigg(\frac{d \mathbf{P}^{(n)}}{d \mathbf{E}}\bigg)_{ij} = n \chi^{(n)}_{i jk_1 ... k_{n-1}} \prod_{i\in I_{n-2}} E_i
\end{equation}
where $P^{(n)}$ is the n-th order term in:
\begin{equation}\tag{2}
P_i = \chi_{ij}^{(1)} E_i+ \chi_{ijk}^{(2)} E_i E_j+ \chi_{ijkl}^{(3)} 
E_i E_j E_k + ...
\end{equation}
and $I_n$ is a set containing the indices of the n-th order tensor. The tensor $\chi^{(n)}$ has $n+1$ Cartesian indices, and is symmetric with respect to permutation of these indices. I know how to show this for say the 2nd order term:
\begin{equation}\tag{3}
\bigg(\frac{d \mathbf{P}^{(2)}}{d \mathbf{E}}\bigg)_{ij} = 
2\delta_{jk} \delta_{jm} \chi^{(2)}_{imk}E_k+
(1-\delta_{jm}) \delta_{jk} \chi^{(2)}_{imk}E_l +
(1-\delta_{jk}) \delta_{im} \chi^{(2)}_{imk}E_k
\end{equation}
In the first term, because $m=k$, there's a factor 2 when taking the derivative. In the second term I can simply relabel the indices, and then use the permutation symmetry to cancel all the terms with two delta functions, leaving:
\begin{equation}\tag{4}
2\delta_{jm}\chi_{imk}^{(2)} E_k = 2\chi_{ijk}^{(2)} E_k
\end{equation}
Similarly for the 3rd order term:
\begin{equation}\tag{5}
\begin{gathered}
\bigg(\frac{d \mathbf{P}^{(3)}}{d \mathbf{E}}\bigg)_{ij} = 
3\delta_{jm}\delta_{jk}\delta_{jl} \chi_{imkl}^{(3)}+
2(1-\delta_{jm})\delta_{jk}\delta_{jl} \chi_{imkl}^{(3)} E_m E_k\\+
2(1-\delta_{jk})\delta_{jm}\delta_{jl} \chi_{imkl}^{(3)} E_m E_k+
2(1-\delta_{jl})\delta_{jk}\delta_{jm} \chi_{imkl}^{(3)} E_l E_k\\+
(1-\delta_{jm})(1-\delta_{jk})\delta_{jl} \chi_{imkl}^{(3)} E_k+
(1-\delta_{jl})(1-\delta_{jm})\delta_{jk} \chi_{imkl}^{(3)} E_k\\+
(1-\delta_{jl})(1-\delta_{jk})\delta_{jm} \chi_{imkl}^{(3)} E_l
\end{gathered}
\end{equation}
All terms with more than one delta-function cancel, leaving:
\begin{equation}\tag{6}
3\delta_{jm}\chi_{imkl}^{(2)} E_kE_l = 3\chi_{ijkl}^{(2)} E_kE_l
\end{equation}
from the last three terms in the long sum above. Now, I don't know how to show this for the general case, or even a good way how to express it. I could define something like:
$$\tag{7} \delta_{ij}^{\mu} = \begin{cases} 
      \delta_{ij} & \mu = 1 \\
      (1-\delta_{ij}) & \mu = 0
   \end{cases}
$$
and then I should take the product of  $n-1$ such functions:
\begin{equation}\tag{8}
\prod_{j\in I_{n-1}}  \delta_{ij}^{\mu_{j}}
\end{equation}
but summed over every combination of $\mu_{j}$, so something like:
\begin{equation}\tag{9}
\sum_{\mu_1=0}^{1}...\sum_{\mu_{n-1}=0}^{1} \prod_{j\in I_{n-1}}  \delta_{ij}^{\mu_{j}} = \prod_{j=0}^{n-1} \sum_{\mu_j = 0}^{1} \prod_{j\in I_{n-2}}  \delta_{ij}^{\mu_{j}}
\end{equation}
Each product of delta functions should also be multiplied by the sum of $\mu_j$ to account for the factor from the derivative:
\begin{equation}\tag{10}
\prod_{j=0}^{n-1} \sum_{\mu_j = 0}^{1} \sum_{j=1}^{n-1}\mu_{j}\prod_{j\in I_{n-1}}  \delta_{ij}^{\mu_{j}}
\end{equation}
In total:
\begin{equation}\tag{11}
\bigg(\frac{d \mathbf{P}^{(n)}}{d \mathbf{E}}\bigg)_{ij} = \prod_{j=0}^{n-1} \sum_{\mu_j = 0}^{1} \sum_{j=1}^{n-1}\mu_{j}\prod_{j\in I_{n-1}}  \delta_{ij}^{\mu_{j}}
\chi^{(n)}_{i_1 ... i_n} \prod_{i\in I_{n-2}} E_i
\end{equation}
but I'm not convinced this is correct, and I also don't know how to usefully manipulate that expression at all. I'm sure there's some better way of approaching this problem that I'm not seeing.

REPLY [0 votes]: Nonlinear optics! One of my favorite subjects.
Anyway consider the following gradient:
$$\frac{\partial (\prod_{i\in I_n} E_i)}{\partial E_k} = \sum_{j\in I_n} \frac{\partial E_j}{\partial E_k} \prod_{i\neq j}{E_i} = \sum_{j\in I_n} \delta_{jk} \prod_{i\neq j} E_i = n\prod_{i\in I_{n-1}} E_i $$
where the last equality holds by the symmetry of the arguments (permutation of the indices). Given that $\chi^{(n)}$ shares that same symmetry (and is constant w.r.t. $\mathbf{E}$), I believe the relation you want follows.
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