IRL_ABooklet_2023

Define a sequence $(p_i{)}_{i\ge 0}$ by letting $p_0=p$ and $p_i=f^{(i)}(p)$ for $i\ge 1$. Putting $x=p_n$ in the given equation, $f^{(2)}(x)=sf(x)+tx$, gives the following linear recurrence $$p_{n+2}=s{p}_{n+1}+t{p}_n.(25)$$ Applying $f^{(i)}$ for $i\ge 1$ to the condition $f^{(n+m)}(p)=f^{(m)}(p)$ implies $p_{n+k}=p_k$ for all $k\ge m$, i.e. the sequence $(p_i{)}_{i\ge 0}$ is periodic. We wish to prove $p_0=p_1$. The characteristic polynomial of the linear recurrence (25) is $X^2-sX-t$, with discriminant $s^2+4t$.

Case 1: $s^2+4t=0$. In this case, the characteristic polynomial has a double root, say $r_0$. Since $r_0^2=-t\ne 0$, we have $r_0\ne 0$. The sequence then has the form $p_k=(\alpha +k\beta )r_0^k$ for $k\ge 0$. The condition $p_{n+k}=p_k$ for all $k\ge m$ then translates into $$(\alpha +(n+k)\beta )r_0^{n+k}=(\alpha +k\beta )r_0^k\text{for }k\ge m,$$ which could be rewritten as $k\beta (r_0^n-1)=\alpha -(\alpha +n\beta )r_0^n$. Since this holds true for at least two positive values of $k$, we must have $\beta (r_0^n-1)=0$, i.e. $\beta =0$ or $r_0^n=1$. If $r_0^n=1$, the previous equation implies $n\beta =0$, hence $\beta =0$. If $\beta =0$, the previous equation implies $\alpha (r_0^n-1)=0$. If $\alpha =0$, we obtain $p_i=0$ for all $i\ge 0$, in particular $p_0=p_1$. Otherwise, we need to have $r_0^n=1$. We can conclude that, unless $\alpha =\beta =0$, we need to have $\beta =0$ and $r_0^n=1$. This gives $p_k=\alpha r_0^k$ for all $k\ge 0$, in particular $p_0=\alpha \ge 0$ and $p_1=\alpha r_0\ge 0$. Therefore $r_0$ is a non-negative real number (we continue to assume $\alpha \ne 0$). Since $r_0^n=1$ this implies $r_0=1$, and so $p_k=\alpha$ for all $k\ge 0$.

Case 2: $s^2+4t\ne 0$. In this case, the characteristic polynomial has two distinct roots $r_1\ne r_2$. They are non-zero, because $r_1{r}_2=-t\ne 0$. The sequence then has the form $p_k=\alpha r_1^k+\beta r_2^k$ for $k\ge 0$, and the condition $p_{n+k}=p_k$ for all $k\ge m$ translates into $\alpha r_1^{n+k}+\beta r_2^{n+k}=\alpha r_1^k+\beta r_2^k$ for $k\ge m$. This can be rewritten as $$r_1^k\alpha (r_1^n-1)+r_2^k\beta (r_2^n-1)=0.(26)$$ If $\alpha =\beta =0$, we clearly have $p_0=p_1=0$. From now on, we assume $(\alpha ,\beta )\ne (0,0)$. There are three cases: $r_1^n\ne 1$, $r_2^n\ne 1$, and $r_1^n=r_2^n=1$. However, symmetry between $r_1$ and $r_2$ covers the second case once we prove the first one.

Case 2A: $r_1^n\ne 1$. We can rewrite (26) as $$\alpha {\left(\frac{r_1}{r_2}\right)}^k=-\frac{\beta (r_2^n-1)}{r_1^n-1}\text{for }k\ge m$$ and obtain $\alpha =0$. Otherwise the non-zero LHS would give different values for different $k\ge m$, because $r_1\ne r_2$ in Case 2. With $\alpha =0$, we get $p_0=\beta >0$ and $p_1=\beta r_2\ge 0$. In particular, $r_2\ge 0$ is a real number; in fact, $r_2>0$ since $r_2$ is non-zero. On the other hand, (26) with $\alpha =0$ and $\beta \ne 0$ implies $r_2^n=1$, hence $r_2=1$ and $p_0=p_1$, as required.

Case 2B: $r_1^n=r_2^n=1$. In this case, $r_1$ and $r_2$ are complex numbers that satisfy $|r_1|=|r_2|=1$. We recall that $r_1{r}_2=-t$, hence we have $(-t{)}^n=r_1^n{r}_2^n=1$. Since $t$ is a real number, we need to have $t=\pm 1$. If $t=1$, then $r_2=-1/r_1=-{\bar{r}}_1$, which implies $s=r_1+r_2=r_1-{\bar{r}}_1=2\Im (r_1)i$. As $s$ is real, this is only possible if the imaginary part of $r_1$ is equal to zero, but then $s=0$, which was excluded. Hence, we must have $t=-1$, which implies $r_2=1/r_1={\bar{r}}_1$. If we let $r_1=u+iv$ with real numbers $u,v$, then $r_2=u-iv$ and $$\begin{gathered} p_0=\alpha +\beta \\ {p}_1=\alpha r_1+\beta r_2=(\alpha +\beta )u+(\alpha -\beta )vi. \end{gathered}$$ Since $p_1$ is real, we need to have $(\alpha -\beta )v=0$. If $v=0$, then $r_1=r_2$, which was excluded in Case 2. Hence $\alpha =\beta$ and $$p_k=\alpha (r_1^k+r_2^k)=\alpha (r_1^k+{\bar{r}}_1^k)=2\alpha \Re (r_1^k).$$ Because $r_1$ is an $n$-th root of unity, there exists a smallest positive integer $d$ for which $r_1^d=1$, and $r_1$ is a primitive $d$-th root of unity. As $r_1\ne r_2$ we have $d>1$. A complete list of $d$-th roots of unity then is $$r_1,r_1^2,r_1^3,\dots ,r_1^{d-1},r^d.$$ Since $p_0=\alpha >0$ and $p_k\ge 0$ for all $k\ge 0$, the real parts of all these $d$-th roots of unity need to be non-negative. However, for each $d\ge 2$ there is a $d$-th root of unity with negative real part. This can be seen explicitly by considering the $d$-th roots of unity in trigonometric form $${\zeta }_k=\cos \left(\frac{2k\pi }{d}\right)+i\sin \left(\frac{2k\pi }{d}\right)k=0,1,\dots ,d-1.$$ The real part of ${\zeta }_k$ is equal to $\cos (\frac{2k\pi }{d})$ and this is negative for $k=\lfloor \frac{d}{2}\rfloor$ if $d\ge 2$. We can conclude now that Case 2B is not possible, and we have shown in all other cases that $p_0=p_1$, i.e. $f(p)=p$.

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Alternative solution.

Since $x=(f^{(2)}(x)-sf(x))/t$, it is clear that $f$ is injective. This allows us to “peel off” layers of $f$ from the assumed equation $$f^{(n+m)}(p)=f^{(m)}(p)$$ to deduce that $$f^{(n)}(p)=p.(27)$$ Thus, we need to show that every periodic point is fixed. We assume that $n\in \mathbb{N}$ is minimal in (27). From now on, a subscripted variable will denote an iterate, so $x_k=f^{(k)}(x)$, $p_k=f^{(k)}(p)$, etc., for every $k\ge 0$; in particular, $x_0=x$ and $x_1=f(x)$. Inductively, we see that $p_{n+m}=p_m$ for all $m\in \mathbb{N}$. Putting $x=p_m$ for each $0\le m<n$ in $f^{(2)}(x)=sf(x)+tx$, and adding these equations, we get $$\sigma =(s+t)\sigma$$ where $\sigma :={\sum }_{i=0}^{n-1}{p}_i$. We conclude that either $\sigma =0$ or $s+t=1$. By minimality of $n$, the points $p_i$, $0\le i<n$, are distinct, so if $\sigma =0$, then $n=1$ and $p=0$. This proves the desired result except in the case $s+t=1$.

From now on, we assume $s+t=1$. For the sake of contradiction assume that the least period $n$ of $p$ strictly exceeds 1. We rewrite $f^{(2)}(x)=sf(x)+tx$ as $$f^{(2)}(x)=(1-t)f(x)+tx,x\in \mathbb{R}.(28)$$ This equation allows us to write all higher iterates in a similar form: $$x_{k+1}=s_k{x}_1+t_k{x}_0,x\in \mathbb{R},k\in \mathbb{N}.(29)$$ Now, $$x_{k+2}=s_k{x}_2+t_k{x}_1=(s_k(1-t)+t_k)x_1+s_{k}t{x}_0.$$ so we get the recurrence relations $$\begin{array}{l}{s}_{k+1}=s_k(1-t)+t_k\\ t_{k+1}=s_{k}t\end{array}\}(30)$$ Together with $s_1=1-t$, $t_1=t$, (30) defines $s_k$ and $t_k$ for all $k\in \mathbb{N}$. Note that $s_{k+1}+t_{k+1}=s_k+t_k$, so we have $s_k+t_k=1$ for all $k\in \mathbb{N}$. Let us also use (30) to write $t_{k+2}$ in terms of $s_k$ and $t_k$. Since $t_{k+2}=s_{k+1}t$, we obtain $$t_{k+2}=s_{k}t(1-t)+t_{k}t(31)$$ Applying the identity $x_k=(1-t_k)x_1+t_k{x}_0$ with $x=p$ and $k=n$, we deduce $$(1-t_n)p_0=(1-t_n)p_1,$$ giving a contradiction (since the least period of $p$ strictly exceeds 1) unless $t_n=1$. To finish the proof, we prove that $t_k$ never equals 1. We assume without loss of generality that $k>1$. We are given that $s,t\ne 0$. Since now $s=1-t$, we have $t\ne 0,1$. These omitted values break the line into three intervals and we consider each separately. Suppose first that $0<t<1$. In (30), $s_{k+1}$ and $t_{k+1}$ are linear combinations of $s_k$ and $t_k$, where the coefficients are positive. Since $s_1,t_1>0$, it follows inductively that $s_k,t_k>0$ for all $k\in \mathbb{N}$. Since $s_k+t_k=1$, we deduce that $t_k\ne 1$. Suppose next that $t<0$, and so $s_1=1-t>1$. Using (30), it follows by induction that $t_k<0$ and $s_k>1$ for all $k\in \mathbb{N}$. Finally, we consider $t>1$, and so $s_1=1-t<0$. By (30), we have $t_2<0$ and so $s_2>1$. It now follows inductively from (31) that $t_{2j}<0$ and $s_{2j}>1$ for all $j\in \mathbb{N}$. We deduce from (30) that $t_{2j+1}>1$, and so $s_{2j+1}<0$, for all $j\in \mathbb{N}$. This finishes the proof that $t_k$ never equals 1, and so we are done.