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Take Derivative of Quadratic Form but instead do it with a 3rd order tensor in the middle, which maps to a vector. That is, $x \in \mathbb{R}^n$, $A \in \mathbb{R}^{n \times n \times n}$, where we are interested in a reduction along the lines of (apologies if I used the wrong notation) $$ c^i = x_j \, A^{ijk} \,x_k \equiv Q(x, A) $$

So the question then is to find (1) $\frac{\partial Q(x, A)}{\partial x}$ and (2) $\frac{\partial Q(x, A)}{\partial A}$?

jlperla
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1 Answers1

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You're identifying $x$ with $x_\flat =\langle x,\cdot\rangle$. We have $A:(\Bbb R^n)^*\times (\Bbb R^n)^*\times (\Bbb R^n)^*\to \Bbb R$ and $Q:\Bbb R^n \to (\Bbb R^n)^*$ given by $$Q(x) = A(\cdot,x_\flat,x_\flat).$$Then $DQ(x):\Bbb R^n \to (\Bbb R^n)^*$ is given by $$DQ(x)(h) = A(\cdot, h_\flat,x_\flat)+A(\cdot, x_\flat,h_\flat)$$since $A$ is trilinear and $\flat:\Bbb R^n \to (\Bbb R^n)^*$ is linear. I have carried out all the possible isomorphisms because you committed the crime of writing the components of a vector with lower indices, against Einstein's convention (which dictates that $x\in (\Bbb R^n)^*$ in your case).

Ivo Terek
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  • Thanks for your rapid response. And I am happy to change the notation if you think it would make things more clear. – jlperla Sep 03 '20 at 20:37
  • In Euclidean space, if the components of $x$ are $x^i$ and the components of $x_\flat$ are $x_i$, then we have the numerical equality $x^i = x_i$. So you're not exactly in danger now. Usually people working in applied math and computational methods, thinking only of tensors as arrays of numbers, don't care about this distinction. But if you want to do calculus on manifolds, deal with tensors as multilinear maps and so on, this makes a difference. – Ivo Terek Sep 03 '20 at 20:39
  • Amazing. I have to say that I am not aware of the notation you are using for the answer, but since we are mostly interested in defining a reverse auto-differentation rule, it seems like what you are describing might become relatively easy to implement? – jlperla Sep 03 '20 at 20:41
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    Also for the edited question: if $Q:\Bbb R^n \times (\Bbb R^n)^{\otimes 3}\to \Bbb R$ instead, $D_xQ(x,A)$ was computed above, while $D_AQ(x,A) = Q(x,\cdot)$, since $Q(x,A)$ is linear in the variable $A$. – Ivo Terek Sep 03 '20 at 20:42
  • @Ivo Terek do you mind explaining the notation $Q(x)=A(\cdot,x_b,x_b)$ (you probably remember i am trying to learn tensor analysis)? I am asking since I would think $Q(x)=A^{ijk}x_jx_k$ should be equally good. Does $\cdot$ represent index $i$? – Ted Black Oct 06 '23 at 10:40
  • $Q(x)^i = A^{ijk}x_jx_k$. – Ivo Terek Oct 06 '23 at 15:10