Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $X$ and $C$ be two sets, I have defined an iterated function on them $f: X \times C \rightarrow X$. What interests me is the iterations of $f$ on an initial value $x \in X$, and a sequence $(c_n)_{n \in \mathbb{N}} \in C$, for instance $f(f(f(x, c_0), c_1),c_2)$.

I am wondering if there is some conventional way to express these iterations, for example by adding superscript on $f$. What I need to write are:

1) iterations from $c_0$ to $c_{n_0}$ where $n_0 \in \mathbb{N}$ $$f(\ldots f(f(f(x, c_0), c_1), c_2) \ldots, c_{n_0})$$

2) iterations from $c_k$ to $c_{k+m}$: $$f(\ldots f(f(\ldots, c_k), c_{k+1}) \ldots, c_{k+m})$$

Could anyone help?

share|cite|improve this question
There's something off about your notation. $X\to C\to X$ means two two functions, but it looks like, since you write $f(x,c_0)$, that you really mean $f:X\times C\to X$. Alternatively, you could have meant: $f:X\to X^C$. Perhaps you meant $C\to X$ to mean the collection of functions from $C$ to $X$, but then you need to be specific with parentheses that you mean $f:X\to(C\to X)$. – Thomas Andrews Feb 19 '13 at 16:43
I agree w/ Thomas that there's an interpretation issue. However, $f:X \times C \to X$ is more likely what you intend, ie, $C$ is the parameter space. This interpretation is typical in dynamics. But $ f:X \to X^C$ is inconsistent with argument list $f(x,c_i)$ – alancalvitti Feb 19 '13 at 17:12
It is definitely $f: X \times C \rightarrow X$, i have amended the OP... – SoftTimur Feb 19 '13 at 19:00

I doubt that there is any such formula. Let $g_i(x)=f(x,c_i)$. Then you really just want $h_{0}(x)=x$ and $h_{n+1}(x)=g_{n+1}(h_n(x))$.

But also note that if there is a sequence of $g_i:X\to X$ we can let $C=\mathbb N$ and define $f(x,i)=g_i(x)$ and $c_i=i$. So given any sequence $g_i$, we can get this recursive sequence.

So any sequence that looks like:


will be of this form.

I don't think there is a a general notation for this sort of induction. Certainly, none that also includes $C$ and $c_i$ in the formulation, since they are essentially arbitrary.

share|cite|improve this answer

I don't know about conventional in mathematics, but in the symbolic math system Mathematica, your operation is called Fold (see the comments below your Q regarding proper functional notation however).

Fold[f, Subscript[x, 0], {Subscript[c, 1], Subscript[c, 2], Subscript[
  c, 3], Subscript[c, 4]}]


f[f[f[f[Subscript[x, 0], Subscript[c, 1]], Subscript[c, 2]], Subscript[c, 3]], Subscript[c, 4]]

This remains unevaluated since $f$ is undefined. Suppose you define:

f[x_, c_] := c + Sqrt[x] 

(Note f takes two arguments, the 2nd interpreted as parameter, and returns a real number hence the functional signature is $f:X \times C \to X$).

Then evaluating this gives:

enter image description here

The related function FoldList returns all the iterates, not just the last one.

I don't see a way to obtain an output sequence on $C$. It is an exogenous input to the dynamic, it is not determined by the dynamic.

You could in principle add a second equation $g: X \to C$ or even $h: X \times C \to C$ to determine $c_i$ at each iteration ($g$ just depends on the state, whereas $h$ depends on both the state and parameter values); either way this second equation would be separate from $f$.

share|cite|improve this answer
The sequence on $C$ is input, not an output. It is indeed a Fold... – SoftTimur Feb 19 '13 at 19:01
@SoftTimur, if you're satisfied w/ my answer, do you mind upvoting and accepting it? – alancalvitti Feb 20 '13 at 18:24

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.