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let $f:\mathbb R\to \mathbb R$,and such

$$f(f(x))=\dfrac{x+1}{x+2}$$

Find the $f(x)$

My try

I found $f(x)=\dfrac{1}{x+1}$ because when $f(x)=\dfrac{1}{x+1}$,then $$f(f(x))=f\left(\dfrac{1}{x+1}\right)=\dfrac{1}{\dfrac{1}{x+1}+1}=\dfrac{x+1}{x+2}$$ so $f(x)=\dfrac{1}{x+1}$ such this condition,But $f(x)$ Have other form? Thank you

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1  
$\frac{1}{x+1}$ is not defined on all of ${\mathbb R}$ but only for $x\neq -1$. –  Ewan Delanoy Nov 8 '13 at 14:47

7 Answers 7

up vote 22 down vote accepted
+50

You can easily check that if we define

$$f_{A} = \frac{ax+b}{cx+d} \quad \text{for} \quad A = \begin{pmatrix}a & b \\ c & d \end{pmatrix}, $$

then $f_{A} \circ f_{B} = f_{AB}$. Thus any matrix $A$ satisfying

$$ A^{2} = k \begin{pmatrix} 1 & 1 \\ 1 & 2 \end{pmatrix}, \quad \text{for some} \ k \neq 0 $$

gives rise to a solution. Mathematica yields two different solutions

$$ f(x) = \frac{1}{x+1} \quad \text{and} \quad f(x) = \frac{2x+1}{x+3}, $$

but I'm not sure if other solutions exist.

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maybe have many solution? But I can't –  china math Nov 3 '13 at 8:19
    
I consider sometimes,and I think only have two solutions,But also I can't prove –  china math Nov 3 '13 at 9:18
    
How prove have this problem have only this two solution? –  user94270 Nov 5 '13 at 17:42
5  
@nanchangjan: there is also the "mixed" solution $f(x)=\frac{1}{x+1}$ over $\mathbb{Q}[\alpha]$ and $f(x)=\frac{2x+1}{x+3}$ over $\mathbb{R}\setminus\mathbb{Q}[\alpha]$, with $\alpha$ algebraic number of degree $2$ over $\mathbb{Q}$, so the number of solutions is infinite, for sure. –  Jack D'Aurizio Nov 10 '13 at 1:40
    
Nice solution sos440. –  juantheron Nov 11 '13 at 2:53

Suppose $f(x)=\frac{ax+b}{x+c}$. Then $$ f(f(x))= \frac{(a^2+b)x+b(x+c)}{(a+c)x + b+c^2}=\frac{x+1}{x+2} $$ yields the equations $$ \frac{a^2+b}{a+c}=1, \frac{b(a+c)}{a+c}=b=1, \frac{b+c^2}{a+c}=2, \frac{a^2+1}{a+c}=1$$ from which we find $$c=a^2-a+1$$ and hence $$a^4-2a^3+a^2-2a=a(a^3-2a^2+a-2)=0$$ This last equation has but two real solutions, $a=0$ and $a=2$, corresponding to the two solutions given by sos440: $$ f(x)=\frac{1}{x+1} \mbox{ and } f(x)=\frac{2x+1}{x+3}.$$

Added: The last equation has $a=\pm i$ as roots, but these yield the degenerate functions $f(x)=i$ and $f(x)=-i$ which do not yield the required $f(f(x))$.

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Ok, but this proof that the only solutions of the form $$f(x)=\frac{ax+b}{x+c}$$ are those which you gave. How do you prove that any solution is of this form? –  leo Nov 5 '13 at 22:15
    
@leo, it happens that there are no solutions of form $ax+c$, that is, there are none where the matrix has $0$ in the lower left entry. –  Lubin Nov 6 '13 at 2:18
    
@Lubin yes, I see. My problem then is, how do we now that the solutions are among linear transformations? Why not other kind of functions? –  leo Nov 6 '13 at 2:22
    
@leo, look at my answer, I found another function which is not a linear fractional transforms. –  i707107 Nov 6 '13 at 2:30
    
@leo, I agree with the above comment completely. If you go beyond the restriction of asking only for fractional-linear transformations, the field opens up tremendously. The question then is whether this restriction is a reasonable one. –  Lubin Nov 6 '13 at 19:38

As sos440 mentioned already, linear fractional transforms correspond to matrices.

Using that, we can simplify the problem by using conjugation.

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

Here, we use also linear fractional transform $g$.

The RHS of the equation above corresponds to a matrix that is conjugate to $$\pmatrix{{1}&{1}\\{1}&{2}}$$

This matrix is diagonalizable with distinct real positive eigenvalues $\lambda_1, \lambda_2$. Let $k= \lambda_1/\lambda_2 >0$.

The matrix that corresponds to $g^{-1}$ is the matrix $X^{-1}=\pmatrix{{v_1} & {v_2}}$ where $v_1$, $v_2$ are eigenvectors corresponding to eigenvalues $\lambda_1$, $\lambda_2$ respectively. Thus, $g$ corresponds to the matrix $X=\pmatrix{{v_1} & {v_2}}^{-1}$. Then $$ X \pmatrix{{1}&{1}\\{1}&{2}}X^{-1}= \pmatrix{{\lambda_1}&{0}\\{0}&{\lambda_2}} $$ so that $$ g\left(\frac{g^{-1}(x)+1}{g^{-1}(x)+2} \right) \mbox{ corresponds to the diagonal matrix } \pmatrix{{\lambda_1}&{0}\\{0}&{\lambda_2}} $$

After conjugation, our problem becomes a simpler problem finding solution to $$ F\circ F (x) = kx $$ where $F=g\circ f\circ g^{-1}$.

This new problem looks similar to the following problem: $$G\circ G(x) = x$$

The solution to this one is not only $x$, $-x$, but also $$G(x)=\cases{{-\frac{1}{2}x \mbox{ if $x<0$}}\\{-2x \mbox{ if $x\geq 0$}}}$$

Similarly if we let $F$ as follows, $$F(x)=\cases{{-\frac{\sqrt{k}}{2}x \mbox{ if $x<0$}}\\{-2\sqrt{k}x \mbox{ if $x\geq 0$}}}$$

then $F\circ F (x) = kx$.

For this $F$, we recover $f$ by reversing the conjugation $f= g^{-1}\circ F\circ g$. This resulting function $f$ will not be one of the two functions given.

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What is the linear fractional transform $g$? Can you add it to your post? –  miracle173 Nov 6 '13 at 14:30

If one is looking only for solutions that are fractional-linear, $f(z)=(az+b)/(cz+d)$, then the two that @sos440 has found are the only ones, as suggested by the interesting response of @i707107.

Here’s a somewhat more elaborate response than that of @MatthewConroy, his answer really disposes of the question fully. The fractional linear transformations with real coefficients may be looked at as transformations of the real projective line, $\mathbb P^1(\mathbb R)$ and equally well as transformations of the complex line, $\mathbb P^1(\mathbb C)$. Those that are unequal to the identity fall into three classes, the elliptic, without fixed points on the real line, such as $(z+1)/(-z+1)$, which permutes the four points $\{\infty, -1, 0, 1\}$ cyclically; the parabolic, which have a single fixed point that in some sense is of multiplicity two, such as $z\mapsto z+1$, whose only fixed point is $\infty$; and the hyperbolic, with two fixed points.

Our given map $z\mapsto(z+1)/(z+2)$ is of this last type, and its fixed points are $\rho>0$ and $\rho'<0$, the two roots of $X^2+X-1$. As you see, they satisfy $\rho+\rho'=\rho\rho'=-1$, and it happens that $\rho$ is the reciprocal of the “Golden Ratio”, but that is just an accident here. Now, working in the group of complex fractional-linear transformations, we may conjugate $\rho'$ to $0$ and $\rho$ to $\infty$, and ask about the transformations that leave leave these two fixed. Of course these are all $z\mapsto\lambda z$, we’re just talking multiplication here. In particular, if we have a transformation of this type that takes a particular $\zeta$ to $\lambda\zeta$, then there are just two transformations that square to this, namely $z\mapsto\mu z$ for the two values of $\mu$ with $\mu^2=\lambda$.

By hand and with the help of a symbolic program, I conjugated $(\infty,0)$ to $(\rho,\rho')$, and the transformation that corresponds to $z\mapsto\lambda z$ is $$ \pmatrix{1+(1-\rho)\lambda&-\rho+\rho\lambda\\-\rho+\rho\lambda&1-\rho+\lambda } \>\colon\>z\mapsto\frac{\bigl(1+(1-\rho)\lambda\bigr)z-\rho+\rho\lambda } {(-\rho+\rho\lambda)z + 1-\rho+\lambda}\,. $$ You can check that $\lambda_-=-2-\rho$ gives $1/(z+1)$, $\lambda_+=2+\rho$ gives $(2z+1)/(z+3)$, and $\Lambda=\lambda_+^2$ gives $(z+1)/(z+2)$.

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$f$ is a bijection from $\Bbb R \setminus \{-2\}$ to $\Bbb R \setminus \{1\}$.

Strictly speaking, it cannot have a functional square root : If $f = g \circ g$, then $g(g(-5/3)) = f(-5/3) = -2$. If $g(-5/3) = -2$ then $g(-2)=-2$ and we get that $g\circ g$ is defined at $-2$ when it should not. So $g(-5/3) \neq -2$ and $f$ is defined at $g(-5/3)$ : $f(g(-5/3)) = g(-2)$. Again, if $f(g(-5/3)) = -2$ then $g(-2) = -2$, so $f$ is defined at $f(g(-5/3))$, and so $g$ has to be defined at $g(-2)$. In any case, $g(g(-2))$ exists while it shouldn't.

To make the question interesting we plug the hole and consider $f$ as a bijection from $X = \Bbb R \cup \{\infty \}$ to itself. $f$ has two fixpoints $0 < x_1 < x_2$, any iterate of $f$ is still an homography so still has only those two same fixpoints. This means that you can partition $X' = X \setminus \{x_1,x_2\}$ into sequences of the form $\{f^k(x), k \in \Bbb Z\}$

If you don't need $g$ to be continuous, there are lots of ways to do define a functional square root. Given any point $x \in X'$, you can choose $g(x)$ to be any point $y \in X'$ as long as $y \notin \{f^k(x), k \in \Bbb Z\}$. This choice will determine $g$ at all the $f^k(x)$ and $f^k(y)$ for $k \in \Bbb Z$. Repeat this until you have defined $g$ on all of $X'$. Finally define either $g(x_i)= x_i$ or $g(x_i) = x_{3-i}$.

If you want $g$ to be continuous, there are two cases to consider. $X'$ has two connected components, $(x_1;x_2)$ and $(x_2; \infty) \cup \{\infty\}\cup\{- \infty ; x_1\}$, which should just be called $(x_2 ; x_1)$. Those two components are stable by $f$. It is useful to put an order on those components such that $f$ is increasing. On the first component it is the standard order, but on the other one, we have $y < \infty < z$ if $y > x_2$ and $z < x_1$.

If $g$ switches the connected components $(x_1;x_2)$ and $(x_2 ; x_1)$ of $X'$, $g$ will be determined by its image on an interval of the form $[x ; f(x))$ where $x \in X'$ : you have to send $x$ to some $y$ (on the other component), send $f(x)$ to $f(y)$, and in-between, $g$ can be any strictly increasing (for the orders defined above) continuous function.

If $g$ doesn't switch them, we have to define $g$ separately on each component. To do that, pick an $x \in X'$, choose some $g(x) \in (x ; f(x))$. Then we must have $g(g(x)) = f(x)$, and we can pick any strictly increasing continuous function on $[x ; g(x)]$. This determines $g$ completely on the component of $x$. do the same on the other component and we are done.

In both cases, for $g$ to be continuous, we must have $g(x_i) = x_i$ if $x_i$ is one of the two fixpoints.

Also, those constructions can give you functional square roots that are smooth on $X'$ (I(m not sure what happens at the ficpoints) and are still not one of the two homographies that you have found.

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This is not an answer, but it does not fit into the comment space

There's a link between the function $f(x)$ and the Fibonacci numbers which is quite interesting.

We know that $$f(f(x)) = \frac{x+1}{x+2}$$ If we substitute $x$ with $f(f(x))$ again we get $$f(f(f(f(x)))) = \frac{5x+8}{8x+13}$$
If we repeat this process again, we get
$$f(f(f(f(f(f(x)))))) = \frac{13x+21}{21x+34}$$ If fact, if we apply the function $n$ times, where $n$ is even we obtain,
$$f(f(f(...f(x)...))) = \frac{F_{n-1}x+F_{n}}{F_{n}x+F_{n+1}}$$ where $F_n$ is the $n$th Fibonacci number

Notice that if we assume that the relation above holds for odd $n$ (i.e we apply $f(x)$ an odd number of times), and we compose this with itself (i.e we get $f(f(f(...f(x)...)))$ $2n$ times) we get exactly the value of $f(x)$ composed $2n$ times. This again, we can easily verify.
Since the relation above holds for odd $n$ we see that a possible function is $$\frac{1}{x+1}$$ when we substitute $n=1$

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Assume that $f(x)$ has an inverse such that $ f(f^{-1}(x))=x $

And we have

$f(f(x))=\dfrac{x+1}{x+2}$

So, (in general)

$f(x)=f(f(f^{-1}(x)))=\dfrac{f^{-1}(x)+1}{f^{-1}(x)+2}$

Or, $f$ satisfy the following

$ 2f(x)-f^{-1}(x)+f(x)f^{-1}(x)=1 , $ **((Not helpful!))

$ f(f(x))=\dfrac{x+1}{x+2} $

For example,

$ f(x)=\dfrac{1}{x+1} \implies f^{-1}(x)=\dfrac{1-x}{x}$

$2f(x)-f^{-1}(x)+f(x)f^{-1}(x)=\dfrac{2}{x+1}-\dfrac{1-x}{x}+\dfrac{1-x}{x(x+1)}=1$ **((Not helpful!))

so, $f(x)=\dfrac{f^{-1}(x)+1}{f^{-1}(x)+2}=\dfrac{\dfrac{1-x}{x}+1}{\dfrac{1-x}{x}+2}=\dfrac{1}{x+1}$ and $f(f(x))=\dfrac{x+1}{x+2}$

Now, The question is

Is $f$ unique?


Correction: Please don't consider this an answer!

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