Can all functions be theoretically iterated succesfully?

Question

I'm trying to solve the Collatz conjecture, and am having some trouble designing a function that divides a number by two until it's odd.

Here is what I've thought of.

We know $a\mod b = \arctan(\tan(\frac{a\pi}{b}-\frac{\pi}2))\frac{b}\pi + \frac{b}2 $

Thus,

$$f(x) = \frac{x}2 + \frac{x}2(x \mod 2) = \frac{x}2 + \frac{x}2(arctan(\tan(\frac{x\pi}{2}-\frac{\pi}2))\frac{2}\pi + \frac{2}2) = \frac{x}2 + \frac{x}2(arctan(\tan(\frac{x\pi}{2}-\frac{\pi}2))\frac{2}\pi +1)$$

What would this do? This function would return half the number, if $ x \mod 2 = 0$, else, return the whole number, because it's already odd.

The thing is, if I want to make sure any number is odd, I'd have to iterate this infinite times. I know some functions are easily iterated, but can all functions be iterated in theory? I know it may be hard, but the whole conjecture is tough by itself, so I don't really care.

---

One extra question, notice

$$a\mod b = \arctan(\tan(\frac{a\pi}{b}-\frac{\pi}2))\frac{b}\pi + \frac{b}2 = (\frac{a\pi}{b}-\frac{\pi}2)\frac{b}\pi + \frac{b}2$$

Is this simplification correct?

EDIT: The simplification collapses to $ a \neq a \mod b$, so that's out of the question.

Answer 1

If you insist on using trig functions you can do the following. Consider the sequence of functions $$ \begin{aligned} f_0(x)&=1,\\ f_1(x)&=\frac12(1+\cos\pi x),\\ f_2(x)&=\frac14(1+\cos\frac{\pi x}2+\cos\pi x+\cos\frac{3\pi x}2),\\ \vdots&\qquad\\ f_k(x)&=\frac1{2^k}\sum_{j=0}^{2^k-1}\cos\frac{j\pi x}{2^{k-1}},\\ \vdots&\qquad \end{aligned} $$ At integer points they have the properties:

• $f_0(n)=1$ for all $n\in\Bbb{N}$,
• $f_1(n)=1$ when $2\mid n$ and $f_1(n)=0$ at other integer points,
• $f_2(n)=1$ when $4\mid n$ and $f_2(n)=0$ at other integer points,
• $f_k(n)=1$ when $2^k\mid n$ and $f_k(n)=0$ at other integer points.

Therefore the series $$ F(n):=f_0(n)-\frac12f_1(n)-\frac14f_2(n)-\cdots=1-\sum_{k=1}^\infty 2^{-k}f_k(n) $$ has the value $F(n)=2^{-\nu_2(n)}$ at any integer $n$.

Consequently $$ x F(x) $$ divides an integer $x$ by all the powers of two that still produce an integer.

Unless I made a mistake the series $F(x)$ converges uniformly everywhere, so it is a continuous function. Here's a plot of the sum of the first five terms, call it $S_4(x)$, in the interval $x\in[0,16]$.

Here's the corresponding plot of $x S_4(x)$.

I doubt you can approach Collatz in this way, but who am I to forbid anyone to think outside the box :)

Answer 2

Let $f(x) = b^x $ with $b=\sqrt{2}$ then $f(1)$,$f(f(1))$, ..., $f°^h(1)$ are all well defined and computable and converging to $c=2$.
We can even use the analytial result that the limit of infinite iteration is equal to $2$ and can therefor derive from that any further (appropriate) corrolary, lemma or theorem in all generality (well, for bases $b$ with appropriate values) ...

Another example: let $g(x) = x/2 $ then we can iterate $g(1)$,$g(g(1))$ ... $g°^h(1)$ and even can say an analytical result for the sum of all iterates : this is the geometric series and evaluates to $1+ g(1) + g°^2(1) + g°^3(1)... = 2$