Vector-Jacobian Product

[pharringtonp19]

In the Vector-Jacobain product section of The Jax for the Impatient tutorial, the authors explain how to efficiently compute the gradient of a function composed of two functions.

They note that the outer function, \phi, is an element of the dual space of R^m. I am curious about why they make this note. What changes if we don't impose that \phi is an element of the dual space?

[andsteing]

@BertrandRdp who was originally writing this tutorial.

[BertrandRdp]

Hi! I don't remember exactly what I was thinking, but at the time I was reflecting on why you would be interested in computing those quantities (from a mathematical perspective). However rereading now seems that as I was thinking about the space where the gradient lives (dual space), and I somehow mixed the original function and its gradient.

Hence \phi is a scalar valued function (not a linear form), its differential is a linear form (hence the dual space). In the case where \phi is a linear form, the only thing that changes is that the gradient is a constant vector, which is just a special case.

I'll fix this, thanks for catching this :)

[andsteing]

Thanks for fixing this in #1442, Bertrand!

[pharringtonp19]

@BertrandRdp Thanks for the explanation!