Back Prop Algorithm - What remains constant in derivatives

The following are a set of notes on the back-prop algorithm. I happen to work in dynamical systems where one needs to compute the stability of periodic orbits. One could argue a dynamical system is essentially a rule for iteration such as the one below.

\[x_{n+1} = f(x_n)\]

Interestingly, there are sytems which come back to the same point after some iterations. Lets say the following is true,

\[x_{n+4} = x_{n} = f(f(f(f(x_n)))) = f^4(x_n)\]

One computes the stability of a discrete system as above by computing its derivative. If the derivative is less than 1, then the system is stable else unstable.

So to compute the stability of the \(f^4(.)\) function one needs to find the following:

\[\frac{d}{dx_n}(f^4(x_n)) = \frac{d}{dx_n}(x_{n+4}) = \frac{dx_{n+1}}{dx_n}.\frac{dx_{n+2}}{dx_{n+1}}.\frac{dx_{n+3}}{dx_{n+2}}.\frac{dx_{n+4}}{dx_{n+3}}\]

This is essentially the backpropagation update for the function \(f^4(.)\) with respect to the previous layer outputs \(x_n,x_{n+1}..\). While this is obvious for many, how this helps me is to understand what is kept constant when one takes the layer’s derivative.

That is as follows,

\[\frac{d}{dx_n}(x_{n+4}) = \frac{dx_{n+1}}{dx_n}.\frac{dx_{n+2}}{dx_{n+1}}.\frac{dx_{n+3}}{dx_{n+2}}.\frac{dx_{n+4}}{dx_{n+3}} = f'(x_n).f'(x_{n+1}).f'(x_{n+2}).f'(x_{n+3})\]

Therefore, instead of taking the derivative of the overall function \(f^4(.)\), we could take the derivative of the function \(f(.)\) and just multiply.

Consequently, in general, one can just use \(f'(.)\) and apply the function on the \(n\) iterates of the function and get the derivative \(n^{th}\) application of the function. The interesting aspect is what is kept constant when one applies the function is clear here. Cheers.

Reference: Ott, E. (2002). Chaos in Dynamical Systems (2nd ed.). Cambridge: Cambridge University Press. doi:10.1017/CBO9780511803260


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