# Function composition notation

1. Dec 26, 2014

### bjshnog

I've looked around on the internet a bunch, for a standard way to write an arbitrary number of nested functions (eg. ${f_1}\circ{f_2}\circ\cdots\circ{f_n}$) without ellipses and with a second input variable (eg. the i in $f_i$), but never found anything. If anyone does this, what is the standard way to write that kind of thing? To use messy ellipses?

I more or less want to be able to generalize things like arbitrary sums, products and continued fractions (as well as arbitrary functions).

Last edited: Dec 26, 2014
2. Dec 26, 2014

### Staff: Mentor

Nearest thing I could find is:

http://homepages.inf.ed.ac.uk/stg/NOTES/node34.html

Where they use exponential notation on the function name.

For something more complex you might have to invent your notation like the summation or product functional operators,

3. Dec 26, 2014

### bjshnog

I see. I have to say the main reason was because I went to the effort to invent something that I didn't think anyone had bothered doing in the past so that I could experiment with it, and I didn't want to be unoriginal. I posted about it here about 3 years ago, but I was bad at organization and posting back then, so let's just ignore that, heh heh. I guess the main thing was that I didn't put all this effort into something that was already done before (as far as I know).

Below is the current definition of my notation (though it might be necessary for me to add limits in there for infinites, if it isn't implied by the structure). $a$ is the lower limit and $b$ is the upper limit, where $i$ is the alternative input variable (in most "large" operators, you would see $i=a$ at the bottom, but that implies a starting point, and this notation needs to be able to go infinitely high ($\cdots\circ{f_1}\circ{f_0}$) and infinitely deep (${f_{-1}}\circ{f_{-2}}\circ\cdots$)). The higher the value of $i$, the more functions are nested within it, and the lower the value, the more functions it is nested in. The square brackets are where you place the function you're quasi-iterating, and the round brackets are where you put your independent variable (which always goes on the deepest-nested function, as we all know). $z$ is the important bit. $z$ is the symbol that represents what part of the function in the square brackets is replaced by the deeper nested function ($f_{i-1}$).

$$\Omega_{a}^{b}[f_{i}(z)]_{z}^{i}(x)= \begin{cases} {\Omega_{0}^{\infty}[f_{i}(z)]_{z}^{i}}\circ{\Omega_{-\infty}^{-1}[f_{i}(z)]_{z}^{i}} & {-\infty}={a},{b}=\infty \\ {f_{b}}\circ{\Omega_{a}^{b-1}[f_{i}(z)]_{z}^{i}} & {-\infty}\neq{a}<{b}\neq{\infty} \\ {\Omega_{a+1}^{b}[f_{i}(z)]_{z}^{i}}\circ{f_{a}}(x) & {\infty}\neq{a}<{b} \\ {f_{a}(x)} & {a}={b} \\ {x} & {a}>{b} \end{cases}$$

Note: I didn't put $(x)$ after the two operators that go infinitely deep ($\Omega_{-\infty}^{b}$), because, well... it goes infinitely deep, so you would never get to the value and you'd calculate the result via a limit, as a direct composition of an infinite number of functions. However, I'm now questioning whether I should put it there, since if you did set a value to the independent variable and calculated the limit with that in each step, you might get a totally different result depending on the type of function you're quasi-iterating.
Example #1: $\Omega_{1}^{\infty}[z+\frac{1}{2^i}]_{z}^{i}(-1)=1$
Example #2: $\lim_{b\rightarrow\infty}\Omega_{1}^{b}[z+\frac{1}{2^i}]_{z}^{i}(-1)=0$
The limit goes straight to 1 in the first example, but if you took into account the very final step (which is infinitely embedded in the functions), you would get 0. I don't think there's any mathematical "rule" that says what to do when you "reach" the final function if it's infinitely deep. If I make it a part of the definition, then it would look more like this:

$$\Omega_{a}^{b}[f_{i}(z)]_{z}^{i}(x)= \begin{cases} {x} & {a}>{b} \\ {f_{a}(x)} & {a}={b} \\ {f_{b}}\circ{\Omega_{a}^{b-1}[f_{i}(z)]_{z}^{i}}(x) & {-\infty}\neq{a}<{b}\neq{\infty} \\ \lim_{b\rightarrow{\infty}}[{\Omega_{a}^{b}[f_{i}(z)]_{z}^{i}}(x)] & {-\infty}\neq{a}<{b}={\infty} \\ \lim_{a\rightarrow-\infty}[{\Omega_{a}^{b}[f_{i}(z)]_{z}^{i}}(x)] & {-\infty}={a}<{b}\neq{\infty} \\ \lim_{b\rightarrow{\infty}}[\lim_{a\rightarrow{-\infty}}[{\Omega_{a}^{b}[f_{i}(z)]_{z}^{i}}(x)]] & {-\infty}={a},{b}=\infty \end{cases}$$

Can I have input, if anyone's interested?

By the way, here are some common expressions generalized by this notation:
$e$ expressed as a continued fraction: $e=2+\Omega_{-\infty}^{0}\left[\frac{1}{1+(1-2i/3)\cdot[i\in\mathbb{Z}_3]+z}\right]_{z}^{i}(0)$
Capital sigma notation: $\Sigma_{i=a}^{b}{x_i}=\Omega_{a}^{b}[z+{x_i}]_{z}^{i}(0)$
Capital pi notation: $\Pi_{i=a}^{b}{x_i}=\Omega_{a}^{b}[z\cdot{x_i}]_{z}^{i}(1)$

Last edited: Dec 26, 2014
4. Dec 26, 2014

### bjshnog

I thought about it a bit and it would be impossible to write an endless string of functions and not have a $z$ sitting around in the result. It must use a limit, as in the second definition.

Here's Graham's number. $G=\Omega_{1}^{64}[{3}\uparrow^{z}{3}]_{z}^{i}(4)$