# $f(x)=f(x+1)$ and $f(-1/x)=f(x)$

Is there any function $f$ which would satisfy $f(x)=f(x+1)$ and $f(-1/x)=f(x)$ for every $x$ or at least positive $x$? For the widest possible domains of $x$?

If I could turn this functional equation into differential equations, I could use some approximate analytic method to get the solution.

In a more general case, is the a function $g$ so $f \left( \frac{ax+b}{cx+d} \right) = g(x)$?

For real $a$, $b$, $c$ and $d$?

-
I think you want to ask: "Find all the function satisfy..." instead of asking "is there any function which satisfy..." since there are many functions satisfying the functional equation you stated, e.g. constant functions. – Paul Dec 16 '11 at 23:22
@Jose Garcia: You would want to impose additional conditions on $f$ relevant to your needs. For example, continuity everywhere is too strong, for then only the constant functions work. – André Nicolas Dec 17 '11 at 0:12
What you are asking for is a modular form of weight $0$. You may want to look here (en.wikipedia.org/wiki/Modular_form) for more details. – user17762 Jun 26 '12 at 21:18

The general solution of $f(x)=f(x+1)$, according to http://eqworld.ipmnet.ru/en/solutions/fe/fe1101.pdf, should be $f(x)=\Theta(x)$, where $\Theta(x)$ is an arbitrary periodic function with unit period.

The general solution of $f(-\frac{1}{x})=f(x)$, according to http://eqworld.ipmnet.ru/en/solutions/fe/fe1120.pdf, should be $f(x)=\Phi(x,-\frac{1}{x})$, where $\Phi(x,-\frac{1}{x})$ is any symmetric function of $x$ and $-\frac{1}{x}$.

So the function $f(x)$ that satisfly $f(x)=f(x+1)$ and $f(-\frac{1}{x})=f(x)$ is the intersection of the above two general solutions.

-
Is it really interesting ? – Lierre Jun 24 '12 at 9:09
(-1) "The general solution of $f(−1/x)=f(x)$, according to [...], should be $f(x)=\Phi(x,−1/x)$, where $\Phi(x)$ is any symmetric function of $x$ and $−1/x$." -- This is as empty as a statement can be. – TMM Jun 26 '12 at 21:27
I am not sure why this was accepted by the OP. While the answer is correct, it is vacuous. – Eric Naslund Jul 1 '12 at 20:36
Captain obvious nervously smoking in aside – Norbert Jul 4 '12 at 0:32

I'm not sure if this is of interest, but if, instead of real $x$, you consider $z$ in the complex upper half plane, then the two linear fractional transformations $$z\to z+1,\quad z\to -1/z$$ generate the modular group. I.e., writing the linear fractional transformations as matrices, they generate $\text{PSL}_2(\mathbb Z)$. The classical $j$ invariant http://en.wikipedia.org/wiki/J-invariant is an example of a function invariant under the modular group

-
This neat property of the Klein invariant also makes for a good numerical algorithm for evaluating it... – J. M. Dec 17 '11 at 5:52

A very simple solution to these functional equations is any constant function of the form $f(x)=c$ for some constant $c$. Let f(x)=3. Then $f(x+1)=f(x)=f(-1/x)=3$.

-
May I ask why this was downvoted? If this answer does not respond to the question appropriately, I will remove it. – analysisj Dec 16 '11 at 23:42
Your answer is correct, I can not understand why anyone would have downvoted it. – Arjang Dec 16 '11 at 23:59

(Edit)

You have

$$\frac{-1}{\frac{-1}{\frac{-1}{x}+1}+1}=x-1,$$

so

$$f(x)=f\left(\frac{-1}{x}\right)=f\left(\frac{-1}{x}+1\right)=\dots =f\left(\frac{-1}{\frac{-1}{\frac{-1}{x}+1}+1}\right)=f(x-1),$$

and so you can even explicitly shift to the left. One can actually construct any continued fraction and so you see that the function takes the same value on all rational values (even for only a finite number of operations for each number).

Due to $f(x)=f(x\pm 1)$ you have a grid with gap distance $1$ at the right of any starting value. There the function always take the same value, i.e. $f(n)=f(0)$ for all integers. Now consider $n=-2$, then $f(1/2)=f(-2)=f(0-2)=f(0),$ and consequently also $f(1/2+n)=f(0)$. So the grid really only has distance $1/2$. We can go on and collect more point with that value. Look at numbers with bigger absolute values, like $-3:\ f(1/3)=f(-3)=f(0)$. Obviously $f(1/n)=f(0)$ for any $n$. And so in general we have $f(1/n+m)=f(0)$ for all integers $n$ and $m$. Therefore also $f(1/(1/n+m))=f(0)$, and so $f(1/(1/n+m)+k)=f(0)$ and so on.

Sidenote: The first condition is obviously periodicity. And regarding the other condition, notice that whenever you have an operation $g$ with $g(g(x))=x$, like it is the case for $g(x)=-\frac{1}{x}$ or in fact all real functions that can be mirrored w.r.t. the $45°$ axis, then for any function $f$, the function $$\hat{f}(x):=f(x)+f(g(x))$$ fulfills $$\hat{f}(g(x))=f(g(x))+f(g(g(x)))=f(g(x))+f(x)=\hat{f}(x)$$

Actually, via the identity

$$f(x)=\frac{1}{2}\left(f(x)+f(g(x))\right)+\frac{1}{2}\left(f(x)-f(g(x))\right),$$

all functions have a part which fullfills the relation, except the ones which fulfill the anti-relation $f(g(x))=-f(x)$ for which that part is zero.

-
(+1) I like the first part of your answer. – TMM Jun 26 '12 at 21:30

As stopple noted, the transformations $z \to z+1$ and $z \to -1/z$ generate the modular group. For any two rationals $r$ and $s$ there is a transformation in this group that takes $r$ to $s$, and thus $f(r) = f(s)$. In particular, if $f$ is continuous that says $f$ is constant.

-