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I understand that $f^{2}(x)=f(f(x))$, and that $f^{3}(x)=f(f(f(x)))$, and so on and so on.

I have been given the result $$(f^n)'(x_0)=f^{'}(x_0)f^{'}(x_1)f^{'}(x_2)f^{'}(x_3)...f^{'}(x_{n-1})=\prod^{n-1}_{i=0}f^{'}(x_i)$$ during my learning about the Lyapunov exponent for one-dimensional maps. I understand how the middle expression can be written as the the end expression, but I'm unsure how to expand the initial expression to arrive at the middle expression. Could someone explain this expansion please? I understand it has something to do with the chain rule but I can't wrap my head around it.

Alp Uzman
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Nemon27
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    This is just the chain rule, expanded to deeper levels of composition $(f^n)'=f'(f^{n-1})(f^{n-1})'=f'(f^{n-1})f'(f^{n-2})(f^{n-2})'=...=f'(f^{n-1})f'(f^{n-2})···f'$. – Lutz Lehmann Jul 21 '21 at 12:03
  • @LutzLehmann Could you explain this further please? – Nemon27 Jul 21 '21 at 15:41
  • It really just simply is Alexandre Chain's rule $(f\circ g)'=(f'\circ g)·g'$ applied to $g=f^{n-1}$ and then continued recursively. – Lutz Lehmann Jul 21 '21 at 15:49

2 Answers2

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I believe the confusion is due to notation. Let's focus on the case when $f:X\to X$ is differentiable, where $X$ is the closed interval or the circle, as the OP is interested in one-dimensional dynamics. (Though note that the formula works (at least) for a differentiable map on an arbitrary differentiable manifold, when derivatives are interpreted as differential operators. Also note that traditionally the theory of Lyapunov exponents require a bit more than the existence of $f'$ (e.g. that it be $log$-integrable for systems whose time parameter is one dimensional).)

If $x\in X$, we have its orbit under $f$, and this orbit is parameterized by powers (w/r/t composition) of $f$. Thus we have a sequence defined recursively by

$$x_\bullet: \mathbb{Z}_{\geq0}\to X, x_0=x, x_{n+1}=f(x_n).$$

The common notation for this is $x_n=f^n(x)$.

Next recall the chain rule: $(f\circ g)'(x)=f'(g(x)) g'(x)$; more succinctly $(f\circ g)'=f'\circ g \,\cdot\, g'$, where $\cdot$ denotes pointwise multiplication. To get a better idea for the derivative of the composition of many maps, let's also write down the case of three functions:

\begin{align*} (f\circ g\circ h)' = (f \circ (g\circ h))' = f'\circ (g\circ h) \,\cdot\, (g\circ h)' = f'\circ (g\circ h) \,\cdot\, g'\circ h \,\cdot\, h' \end{align*}

In particular if we have $f$ composed with itself three times this gives

$$(f^3)'(x_0)=f'(f^2(x_0))f'(f(x_0))f'(x_0) = f'(x_2) f'(x_1) f'(x_0).$$

Now in one dimension derivatives are numbers and in particular they commute, whence we get $(f^3)'(x_0)=f'(x_0)f'(x_1)f'(x_2)$ (Though in my opinion it is better to not get used to this ordering).

I hope this much is sufficient for the OP to get the general formula by induction.

In fact, there is nothing special about iterating one fixed map (neither for this particular formula nor the theory of Lyapunov exponents); one can consider instead a nonautonomous system which is given by a sequence of maps $f_\bullet:\mathbb{Z}_{\geq0}\to \{\text{differentiable functions from } X \text{ to } X\}$ and define the orbit of $x$ like so:

$$x_\bullet: \mathbb{Z}_{\geq0}\to X, x_0=x, x_{n+1}=f_n(x_n).$$

As another exercise it is also lots of fun to write the formula for $(f^n)^{(k)}= \dfrac{d^k (f^n)}{dx^k}$ (or indeed $\left(\left(\left(f^{n_1}\right)^{(k_1)}\right)^{n_2}\right)^{(k_2)}$).

Alp Uzman
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    Hi, thanks for your answer, the middle section in particular was very helpful. Would you mind taking a look at the answer I've given and let me know if the method I've used there is sound? I'd still be happy to accept your answer so you receive some reputation, I'd just like to know if I managed it on my own (I was busy typing out my answer as you posted yours). – Nemon27 Jul 21 '21 at 16:41
  • It seems about right. I would suggest using $(f^n)'$ instead of $f^{(n)^'}$ (I had made an edit in your question accordingly), as I think the latter is more ambiguous ($f^{(n)}$ stands for the $n$th derivative of $f$ traditionally). – Alp Uzman Jul 21 '21 at 17:03
  • Another comment about notation would be that I would suggest using the indices to be in accordance with the powers of $f$ (as in my answer). According to your notation the orbit of $x=x_0$ becomes $x_0 =x_n\stackrel{f}{\to} x_{n-1} \stackrel{f}{\to} x_{n-2} \stackrel{f}{\to} \cdots \stackrel{f}{\to} x_1 \stackrel{f}{\to} \cdots$, whereas w/r/t the notation I presented above this orbit is $x_0 \stackrel{f}{\to} x_1 \stackrel{f}{\to} x_2 \stackrel{f}{\to} \cdots \stackrel{f}{\to} x_{n-1}\stackrel{f}{\to} x_n \stackrel{f}{\to} x_{n+1}\stackrel{f}{\to} \cdots$. – Alp Uzman Jul 21 '21 at 17:06
  • I find the latter more convenient, but it's a matter of taste of course. The caveat is that w/r/t the notation I presented above to get the chain rule formula one needs to use the commutativity of derivatives. The benefit is that $x_n$ w/r/t my notation immediately means where $x=x_0$ is at time $n$. – Alp Uzman Jul 21 '21 at 17:09
  • I should also note that typically when one writes products of matrices the product goes from right to left, that is, $\prod_{i=0}^n A_i = A_n A_{n-1} \cdots A_1 A_0$. – Alp Uzman Jul 21 '21 at 17:13
  • A caveat of your ordering is that it might make one think that $x$ is a periodic point of $f$, whereas the rule in question holds without this assumption. – Alp Uzman Jul 21 '21 at 17:22
  • Thank you so much for all of your help! However, I think I've seen an issue with my answer - continuing $n$ times means the last term will be $(f^{n-n})'(x_0)=(f^{0})'(x_0)$. Is this possible? Should I instead be doing it $n-1$ times? If I do then the terms will only include $x_1$ through to $x_{n-1}$ which is one short for my final consolidation with Pi notation? – Nemon27 Jul 21 '21 at 18:15
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    I do think your answer is correct up to notational logistics. I had alluded to the problem you mention in my last comment: according to your notation the general rule for indices is $f^{n-i}(x_0)=x_i$, so in particular for $i=n$ this gives $x_0=f^0(x_0)=x_n$ ( or alternatively for $i=0$ this gives $f^n(x_0)=x_0$), that is, $x_0$ is a periodic point of $f$. If you insist on using your notation you could restrict $i$ to be in ${1,2,...,n-1}$. – Alp Uzman Jul 21 '21 at 19:07
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    Following your example of $(f^{3})(x_0)$ I wrote it all out on paper for $(f^{4})(x_0)$ and I can see the pattern, so while not a rigorous proof I can do it and understand it now, thanks so much for your help. – Nemon27 Jul 21 '21 at 19:22
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Jumping off of the back of a comment from LutzLehmann I think I now understand it. First we take $$f^{(n)'}(x_0)=f'(f^{n-1}(x_0))f^{(n-1)'}(x_0),$$ and then define $f^{n-1}(x_0)=x_1$ for $$f^{(n)'}(x_0)=f'(x_1)f^{(n-1)'}(x_0).$$ Then we compute $$f^{(n-1)'}(x_0)=f'(f^{n-2}(x_0))f^{(n-2)'}(x_0),$$ and define $f^{n-2}(x_0)=x_2$. Inserting this into the expression for $f^{(n)'}(x_0)$ reads $$f^{(n)'}(x_0)=f'(x_1)f'(x_2)f^{(n-2)'}(x_0).$$ Continuing this process $n$ times yields $$f^{(n)'}(x_0)=f'(x_1)f'(x_2)f'(x_3)...f'(x_n)$$ which can be written as $$\prod^{n}_{i=1}f^{'}(x_i),$$ which is equivalent to $$\prod^{n-1}_{i=0}f^{'}(x_i).$$ Am I correct?

Nemon27
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