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I know that if we have functions $f\colon \mathbb{R}^{3}\to \mathbb{R}$ and $\vec{g}\colon \mathbb{R}^{3}\to\mathbb{R}^{3}$ then we can write $$g_{1}f_{x} + g_{2}f_{y} + g_{3}f_{z} = \vec{g}\cdot\nabla f. $$ If we have another function $\vec{h}\colon \mathbb{R}^{3}\to\mathbb{R}^{3}$ is there any similar notation for rewriting $$g_{1}\nabla h_{1} + g_{2}\nabla h_{2} + g_{3}\nabla h_{3} \ ?$$

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  • $\begingroup$ Product rules have this term, if I remember correctly. $\endgroup$ – A---B Aug 30 '17 at 19:45
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Yes, you can write it in terms of the tensor product: $[ (\nabla \otimes h)\cdot \vec{g}]_j = \sum_{i=1}^3 (\partial_j h_i)g_i$.

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  • $\begingroup$ I should probably learn tensor notation... +1 $\endgroup$ – qbert Aug 30 '17 at 19:46
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Sure, you could write this as $$ g(\vec{x})^TDh $$ where $Dh$ is the total derivative of $h$, which will be the three by three matrix $$ \begin{bmatrix}\frac{\partial h_1}{\partial x}&\frac{\partial h_1}{\partial y}&\frac{\partial h_1}{\partial z}\\ \frac{\partial h_2}{\partial x}&\frac{\partial h_2}{\partial y}&\frac{\partial h_2}{\partial z}\\ \frac{\partial h_3}{\partial x}&\frac{\partial h_3}{\partial y}& \frac{\partial h_3}{\partial z} \end{bmatrix} $$ and $g(\vec{x})^T$ is the vector $$ \begin{bmatrix}g_1(\vec{x})&g_2(\vec{x})&g_3(\vec{x})\end{bmatrix} $$

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No, because the symbolic gradient operations mimic ordinary vector calculus, and there is nothing like a vector times three vectors giving a vector.

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