Computing the Jacobian $J_F$ with $F = h \circ f$

Asked Feb 17 '21 at 04:31

Active Feb 17 '21 at 12:40

Viewed 24 times

Let

$$ f: \mathbb{R}^l \rightarrow{} \mathbb{R}^m\\[.7ex] h: \mathbb{R}^m \rightarrow{} \mathbb{R}^o$$

and let $$F = h \circ f \quad (F : \mathbb{R}^l \rightarrow{} \mathbb{R}^o)$$

I want to compute the Jacobian using Forward mode accumulation in one path.

I do understand the automatic differentiation way of working in the forward mode for simple cases.

So I think based on my knowledge if I want to compute $J_F$ using as many paths as I want I could do $\ (l\times m)$ paths. Since I am constrained with only one path, I start to get confused. I know I can do something with the initialization but it's very confused in my mind. Could you please help me to understand how to implement the forward mode just in one path using the Jacobian of $f$ and $h$ along the way?

Thanks

edited Feb 17 '21 at 12:40

MarianD

1,493
2
8
17

asked Feb 17 '21 at 04:31

lalaland

3

$$\frac{\text dx}{\text dz}=\frac{\text dx}{\text dy}\frac{\text dy}{\text dz}$$ – Xi'an Feb 17 '21 at 08:46
1

https://math.stackexchange.com/q/2648147/14893 – Xi'an Feb 17 '21 at 08:46
That's the chain rule, but you can implement it in reverse or forward mode. I was looking at the forward mode and was a bit confused? – lalaland Feb 17 '21 at 14:28

Computing the Jacobian $J_F$ with $F = h \circ f$

0 Answers0