A function $$F(n)=f(F(n+1))$$ is called $n$ times, from $n=b$ to $n=a$, where $b>a$, with the purpose of acquiring $F(a)$. Is there an elegant, mathematical way of depicting this?

To illustrate the iterations:

Loop 0: F(b)=0

Loop 1: F(b-1)=f(F(b))

Loop 2: F(b-2)=f(F(b-1))


Loop n: F(a)=f(F(a-1))

  • $\begingroup$ Welcome to Math.SE :) People will be more receptive to your questions if you show what effort you have made and where exactly you are stuck. If you truly cannot get started, knowing the techniques you are familiar with helps people pose an answer at a level you will hopefully understand $\endgroup$ – lioness99a Jan 25 '19 at 15:09
  • $\begingroup$ Usually, we use indices - it's a (recursive) sequence, after all. It's a bit unusual to have it defined backwards (at first sight, it implies implicit definition that requires inversion of $f$), but it doesn't change the notation: $F_n=f(F_{n+1})$. $\endgroup$ – orion Jan 25 '19 at 15:33

Mathematically, you were almost there:

\begin{align} F(n) &= f(F(n+1)) & n \in \mathbb Z, \ a \leq n < b, \\ F(b) &= 0. \end{align}

That's all a mathematician would need to see. The value of $F(a)$ is completely defined by this, although this requires the reader to actually think about how to produce the value of $F(a)$ given the value of $F(b).$

If you want to encode explicit step-by-step instructions, you can write the algorithm in the form of a computer program. If you want that to be a mathematical expression, you could try to learn denotational semantics, but that probably wouldn't actually make anything better.

  • $\begingroup$ Thanks! Could this also be denoted as: $$F(a)=\int_{b}^{a} F(f(n-1))dn$$ or $$F(a)=f^{b-a}(F(b))$$ $\endgroup$ – cheesus Jan 26 '19 at 9:34
  • $\begingroup$ I would not use the integral notation since this is a finite number of discrete steps, but the $f^{b-a}$ notation was suggested in the other answer and it works. $\endgroup$ – David K Jan 26 '19 at 10:26

A fairly standard way of writing the $k$th iterate of a function $f$ is $f^k$. Because this is sometimes confused with the $k$th power of the function, especially when we write things like $$ \sin^2 x + \cos^2 x = 1, $$ where exponentiation really is what's intended, some folks write

$$f^{\circ k}$$

to indicate the $k$th iterate.

Read more here.


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