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


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

  • 1
    $\begingroup$ $\frac{1}{x+1}$ is not defined on all of ${\mathbb R}$ but only for $x\neq -1$. $\endgroup$ Nov 8, 2013 at 14:47

6 Answers 6


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.

  • $\begingroup$ maybe have many solution? But I can't $\endgroup$
    – math110
    Nov 3, 2013 at 8:19
  • $\begingroup$ I consider sometimes,and I think only have two solutions,But also I can't prove $\endgroup$
    – math110
    Nov 3, 2013 at 9:18
  • $\begingroup$ How prove have this problem have only this two solution? $\endgroup$
    – user94270
    Nov 5, 2013 at 17:42
  • 5
    $\begingroup$ @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. $\endgroup$ Nov 10, 2013 at 1:40
  • $\begingroup$ Nice solution sos440. $\endgroup$
    – juantheron
    Nov 11, 2013 at 2:53

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.

  • $\begingroup$ What is the linear fractional transform $g$? Can you add it to your post? $\endgroup$
    – miracle173
    Nov 6, 2013 at 14:30

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 $f(f(x))$.

  • $\begingroup$ 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? $\endgroup$
    – leo
    Nov 5, 2013 at 22:15
  • $\begingroup$ @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. $\endgroup$
    – Lubin
    Nov 6, 2013 at 2:18
  • $\begingroup$ @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? $\endgroup$
    – leo
    Nov 6, 2013 at 2:22
  • $\begingroup$ @leo, look at my answer, I found another function which is not a linear fractional transforms. $\endgroup$ Nov 6, 2013 at 2:30
  • $\begingroup$ @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. $\endgroup$
    – Lubin
    Nov 6, 2013 at 19:38

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 $(z+1)/(z+2)$.


$f$ 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.


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 $n=1$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy