Notation for infinite iteration

[Ssnow]

TL;DR

I want to know what are the possible notations for the infinite self-iteraction of a function

Hi Physics Forum, I want to ask if there is an "appropriate" notation for the infinite self-iteraction of an analytic function $f(x)$, that is $f(f(f(...)))$. For example I know $f^{(+\infty)}(x)$ can be a way, but there is an operator notation as for the infinite sum $\sum_{k=1}^{+\infty}f_{n}$ ...
Thank you,
Ssnow

 

[fresh_42]

Summary: I want to know what are the possible notations for the infinite self-iteraction of a function

Hi Physics Forum, I want to ask if there is an "appropriate" notation for the infinite self-iteraction of an analytic function $f(x)$, that is $f(f(f(...)))$. For example I know $f^{(+\infty)}(x)$ can be a way, but there is an operator notation as for the infinite sum $\sum_{k=1}^{+\infty}f_{n}$ ...
Thank you,
Ssnow

The sum is usually not abbreviated, except you simply call it $S.$ The iterations have to be defined in order to avoid confusion. You already used the most common usage: $f^{(n+1)}(x):=f^{(n)}(f(x)).$ Since $f^{(n+1)}(x)$ is often used as $f^{(n+1)}(x)=\dfrac{d}{dx}f^{(n)}(x),$ it requires an explicit definition. Hence, after defining it, you have
$$
S(x):=\sum_{k=1}^\infty f^{(k)}(x)
$$

 

[FactChecker]

I do not think there is a notation whose meaning would be commonly recognized by mathematicians. I recommend that you clearly define one of your own where you need it.
UPDATE: I think that @pasmith's notation, $\bigcirc_{n=1}^\infty f$, in post #4 serves the purpose very well and might be a standard that I was not familiar with.

 

[pasmith]

Summary: I want to know what are the possible notations for the infinite self-iteraction of a function

Hi Physics Forum, I want to ask if there is an "appropriate" notation for the infinite self-iteraction of an analytic function $f(x)$, that is $f(f(f(...)))$. For example I know $f^{(+\infty)}(x)$ can be a way, but there is an operator notation as for the infinite sum $\sum_{k=1}^{+\infty}f_{n}$ ...
Thank you,
Ssnow

\DeclareMathOperator{bigcomp}{\bigcirc}
\begin{split}
\bigcomp_{n=1}^N f_n &= f_N \circ f_{N - 1} \circ \cdots \circ f_2 \circ f_1  \\
\bigcomp_{n=1}^\infty f &= f \circ f \circ f \circ \cdots \end{split}

<br /> \DeclareMathOperator{bigcomp}{\bigcirc}<br /> \begin{split}<br /> \bigcomp_{n=1}^N f_n &amp;= f_N \circ f_{N_-1} \circ \cdots \circ f_2 \circ f_1 \\<br /> \bigcomp_{n=1}^\infty f &amp;= f \circ f \circ f \circ \cdots \end{split}<br /> (It seemed logical to apply f_1 first and f_N last.)

There may be a way to make LaTeX put the limits above and below the \bigcomp in displayed formulas, but \limits on its own won't do it.

It is, however, incredibly easy for the limit to not exist, as for example if f : [0,1] \to [0,1] : x \mapsto 4x(1-x) and x_0 falls on one of the many unstable periodic orbits of that map, or falls on the chaotic attractor.

 

[Ssnow]

I think the notation of @pasmith is appropriate,
thank you!
Ssnow

 

[jgill]

\mathop L\limits_{k = 1}^n {g_k}(z) = {g_n} \circ {g_{n - 1}} \circ \cdots \circ {g_1}(z)
{G_n}(z) = \mathop R\limits_{k = 1}^n {g_k}(z)\mathop L\limits_{k = 1}^\infty {g_k}(z) = \mathop {\lim }\limits_{n \to \infty } {G_n}(z)
and
\mathop R\limits_{k = 1}^n {f_k}(z) = {f_1} \circ {f_2} \circ \cdots \circ {f_n}(z), {F_n}(z) = \mathop R\limits_{k = 1}^n {f_k}(z)
\mathop R\limits_{k = 1}^\infty {f_k}(z) = \mathop {\lim }\limits_{n \to \infty } {F_n}(z)
See the Wikipedia article on infinite compositions of analytic functions. There is a fairly well developed elementary theory of infinite compositions of complex functions, even functions in Banach spaces.