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I was going through backpropagation in time for RNN, in the deep learning book of Joshua Bengio et.al. (deep learning book , section 10.2.1 ).

Given a network as:

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the book tells that the backpropagation of the error gradient respect with the state h(t) is given by:

enter image description here

So, why is the matrix of the D(h_(t+1))/D(h_t) diagonal? How is its solution obtained?

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I'll provide a sketch of the derivation. Omitting the bias term (since anyways we take derivatives later), the recursion looks like: $$\mathbf{h_{t+1}}=tanh(\mathbf{ U x_{t}+W h_{t}}) $$ where the $tanh$ is taken elementwise.

Now, since $\mathbf{h_{t}}$ and $\mathbf{h_{t+1}}$ are vectors, the derivative $\frac{\partial \mathbf{h_{t+1}}}{\partial \mathbf{h_{t}}}$ is a Jacobian.

  1. Notice that if $y=tanh(x), dy/dx=1-tanh^2(x)=1-y^2$.
  2. Let's see how a single element of the Jacobian looks like. Assume that the hidden layers are of dimension $n$ Now $h_{t+1, i}=tanh(\sum_{k=1} ^{n} w_{ik}h_{t,k}+ g(x))$. Here $g(x)$ stands for some function of x.

  3. Hence $\frac{\partial h_{t+1, i}}{\partial h_{t, j}}$ = $1-h_{t+1,i}^{2}w_{i,j}$. This corresponds to element in position $(i,j)$ of the Jacobian (from #1 above).

  4. This multiplier $1-h_{t+1,i}^{2}$ applies to each element of the $i^{th}$ row of the Jacobian.
  5. From basic linear algebra you can show that pre-multiplying a matrix $A$ by a diagonal matrix $diag(b_{i,i})$ is equivalent to scaling up row $i$ of matrix $A$ by $b_{i,i}$

Hence the term $diag(1-\mathbf{h_{t+1}^{2}})$ occurs as a premultiplier to the matrix $\mathbf{W}$

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  • $\begingroup$ Could you please share how the derivation is performed? I can not understand form where come term W(i, j) ? if dy/dx=1-y^2 it needs to be just 1-h^2(t+1, i) I do not understand how the W(i,j) appears at the point 3. $\endgroup$ – Petar Georgiev Feb 13 '18 at 9:08

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