We use the notation $R$ introduced in part I and $V_n(z) = U_{n+1}(z,-1)$. Let $z \in \mathbb{C}$ such that $z^2 \neq 1$, with $r$, $s$, $t$, $h$ as defined in II.C.1. What can be said about the radius of convergence of the power series $Z \mapsto \sum_{n=0}^{+\infty} V_n(z) Z^n$? We denote $g_z$ its sum.

We use the notation $V_n(z) = U_{n+1}(z,-1)$ and $g_z$ the sum of the power series $\sum_{n=0}^{+\infty} V_n(z) Z^n$. When it makes sense, calculate $\left(1 - 2zZ + Z^2\right) g_z(Z)$.

Let $z \in \mathbb{C}$. We denote $\Omega_z$ the set of points in the plane with complex affixe $Z$ such that $|Z(Z-2z)| < 1$, and $V_n(z) = U_{n+1}(z,-1)$. Show that there exists a non-empty open disk $\Delta$ with center $O$ included in $\Omega_z$ such that $$\forall Z \in \Delta, \quad \frac{1}{1 - 2zZ + Z^2} = \sum_{n=0}^{+\infty} V_n(z) Z^n = \sum_{p=0}^{+\infty} \left(Z^p(2z - Z)^p\right)$$

Using the results of II.C.6, recover the relation $$V_n(z) = \sum_{j=0}^{\lfloor n/2 \rfloor} \binom{n-j}{j} (2z)^{n-2j} (-1)^j$$

We assume $\alpha = 1$ and use the notation $V_n(z) = U_{n+1}(z,-1)$. Let $t \in ]0,\pi[$. Show that the function $$H_t : x \mapsto \frac{1}{1 - 2x\cos(t) + x^2}$$ is expandable as a power series on $]-1,1[$.

We assume $\alpha = 1$ and use the notation $V_n(z) = U_{n+1}(z,-1)$. Using the expansion of $H_t$ as a power series, deduce that $$\forall n \in \mathbb{N},\, \forall t \in ]0,\pi[, \quad V_n(\cos t) = \frac{\sin((n+1)t)}{\sin t}$$

We consider a function $f$ of class $\mathcal{C}^2$ on $[0,1]$ taking values in $[0,1]$ such that $f'$ and $f''$ take non-negative values. We assume $f(1)=1$, $f'(0)<1$ and $f''(1)>0$. We consider the recurrent sequence $(u_n)_{n\in\mathbb{N}}$ defined by $u_0=0$ and, for all $n\in\mathbb{N}$, $u_{n+1}=f(u_n)$.
Show that the sequence $(u_n)_{n\in\mathbb{N}}$ is increasing, then that it is convergent. We denote its limit by $l$.

We consider a function $f$ of class $\mathcal{C}^2$ on $[0,1]$ taking values in $[0,1]$ such that $f'$ and $f''$ take non-negative values. We assume $f(1)=1$, $f'(0)<1$ and $f''(1)>0$. We consider the recurrent sequence $(u_n)_{n\in\mathbb{N}}$ defined by $u_0=0$ and, for all $n\in\mathbb{N}$, $u_{n+1}=f(u_n)$.
Show that the equation $f(x)=x$ admits a smallest solution. In what follows, we denote it by $x_f$.

We consider a function $f$ of class $\mathcal{C}^2$ on $[0,1]$ taking values in $[0,1]$ such that $f'$ and $f''$ take non-negative values. We assume $f(1)=1$, $f'(0)<1$ and $f''(1)>0$. We consider the recurrent sequence $(u_n)_{n\in\mathbb{N}}$ defined by $u_0=0$ and, for all $n\in\mathbb{N}$, $u_{n+1}=f(u_n)$. We denote by $x_f$ the smallest solution of $f(x)=x$ and by $l$ the limit of $(u_n)$.
Show that $l=x_f$.

We consider a function $f$ of class $\mathcal{C}^2$ on $[0,1]$ taking values in $[0,1]$ such that $f'$ and $f''$ take non-negative values. We assume $f(1)=1$, $f'(0)<1$ and $f''(1)>0$. We set $m=f'(1)$. We denote by $x_f$ the smallest solution of $f(x)=x$.
We assume $m>1$. Show that $x_f\in[0,1[$.

We consider a function $f$ of class $\mathcal{C}^2$ on $[0,1]$ taking values in $[0,1]$ such that $f'$ and $f''$ take non-negative values. We assume $f(1)=1$, $f'(0)<1$ and $f''(1)>0$. We set $m=f'(1)$. We consider the recurrent sequence $(u_n)_{n\in\mathbb{N}}$ defined by $u_0=0$ and, for all $n\in\mathbb{N}$, $u_{n+1}=f(u_n)$. We denote by $x_f$ the smallest solution of $f(x)=x$.
We now assume $m\leqslant 1$. Show that $x_f=1$ and that for all $n\in\mathbb{N}$, $u_n\neq 1$.

We consider a function $f$ of class $\mathcal{C}^2$ on $[0,1]$ taking values in $[0,1]$ such that $f'$ and $f''$ take non-negative values. We assume $f(1)=1$, $f'(0)<1$ and $f''(1)>0$. We set $m=f'(1)$. We consider the recurrent sequence $(u_n)_{n\in\mathbb{N}}$ defined by $u_0=0$ and, for all $n\in\mathbb{N}$, $u_{n+1}=f(u_n)$.
In this question, we assume $m=1$. We set, for $n\in\mathbb{N}$, $\varepsilon_n=1-u_n$. Show that $\lim_{n\rightarrow+\infty}\left(\frac{1}{\varepsilon_{n+1}}-\frac{1}{\varepsilon_n}\right)=\frac{f''(1)}{2}$.

We consider a function $f$ of class $\mathcal{C}^2$ on $[0,1]$ taking values in $[0,1]$ such that $f'$ and $f''$ take non-negative values. We assume $f(1)=1$, $f'(0)<1$ and $f''(1)>0$. We set $m=f'(1)$. We consider the recurrent sequence $(u_n)_{n\in\mathbb{N}}$ defined by $u_0=0$ and, for all $n\in\mathbb{N}$, $u_{n+1}=f(u_n)$.
In this question, we assume $m=1$. We set, for $n\in\mathbb{N}$, $\varepsilon_n=1-u_n$. Deduce that, as $n$ tends to infinity, $1-u_n\sim\frac{2}{f''(1)n}$.
One may use Cesaro's lemma: if $(a_n)_{n\in\mathbb{N}}$ is a sequence of real numbers converging to $l$ and if we set, for $n\in\mathbb{N}^*$, $b_n=\frac{1}{n}(a_1+\cdots+a_n)$, then the sequence $(b_n)_{n\geqslant 1}$ converges to $l$.

We consider a function $f$ of class $\mathcal{C}^2$ on $[0,1]$ taking values in $[0,1]$ such that $f'$ and $f''$ take non-negative values. We assume $f(1)=1$, $f'(0)<1$ and $f''(1)>0$. We set $m=f'(1)$. We consider the recurrent sequence $(u_n)_{n\in\mathbb{N}}$ defined by $u_0=0$ and, for all $n\in\mathbb{N}$, $u_{n+1}=f(u_n)$.
We now assume $m<1$ and we set again, for $n\in\mathbb{N}$, $\varepsilon_n=1-u_n$.
Show that the series with general term $\varepsilon_n$ is absolutely convergent and deduce the convergence of the series with general term $\ln\left(\frac{m^{-(n+1)}\varepsilon_{n+1}}{m^{-n}\varepsilon_n}\right)$.

We consider a function $f$ of class $\mathcal{C}^2$ on $[0,1]$ taking values in $[0,1]$ such that $f'$ and $f''$ take non-negative values. We assume $f(1)=1$, $f'(0)<1$ and $f''(1)>0$. We set $m=f'(1)$. We consider the recurrent sequence $(u_n)_{n\in\mathbb{N}}$ defined by $u_0=0$ and, for all $n\in\mathbb{N}$, $u_{n+1}=f(u_n)$.
We now assume $m<1$ and we set again, for $n\in\mathbb{N}$, $\varepsilon_n=1-u_n$.
Deduce that there exists $c>0$ such that, as $n$ tends to infinity, $1-u_n\sim cm^n$.

Let $\left(u_{n}\right)_{n \geqslant 0}$ be a real sequence such that: $\forall (m,n) \in \mathbb{N}^{2}, \quad u_{m+n} \geqslant u_{m} + u_{n}$. We suppose that the set $\left\{\frac{u_{n}}{n}, n \in \mathbb{N}^{*}\right\}$ is bounded above and we denote by $s$ its supremum.
Let $m$, $q$ and $r$ be elements of $\mathbb{N}$. We set $n = mq + r$. Compare the two real numbers $u_{n}$ and $qu_{m} + u_{r}$ and show that $u_{n} - ns \geqslant q\left(u_{m} - ms\right) + u_{r} - rs$.

Let $\left(u_{n}\right)_{n \geqslant 0}$ be a real sequence such that: $\forall (m,n) \in \mathbb{N}^{2}, \quad u_{m+n} \geqslant u_{m} + u_{n}$. We suppose that the set $\left\{\frac{u_{n}}{n}, n \in \mathbb{N}^{*}\right\}$ is bounded above and we denote by $s$ its supremum.
We fix $m$ in $\mathbb{N}^{*}$ and $\varepsilon$ in $\mathbb{R}^{+*}$. Using the Euclidean division of $n$ by $m$, show that there exists an integer $N$ such that for all $n > N$, $$\frac{u_{n}}{n} \geqslant \frac{u_{m}}{m} - \varepsilon$$

Let $\left(u_{n}\right)_{n \geqslant 0}$ be a real sequence such that: $\forall (m,n) \in \mathbb{N}^{2}, \quad u_{m+n} \geqslant u_{m} + u_{n}$. We suppose that the set $\left\{\frac{u_{n}}{n}, n \in \mathbb{N}^{*}\right\}$ is bounded above and we denote by $s$ its supremum.
Show $\lim_{n \rightarrow \infty} \frac{u_{n}}{n} = s$.

Let $f$ be a function with real values, defined and continuous on $\mathbb{R}^{+}$, and admitting a finite limit at $+\infty$.
a) Show that $f$ is bounded on $\mathbb{R}^{+}$.
b) Deduce that the function $g$ defined on $\mathbb{R}^{+}$ by $\forall t \in \mathbb{R}^{+}, g(t)=t e^{\gamma t}$ where $\gamma$ is a strictly negative real, is bounded on $\mathbb{R}^{+}$.

Verify that a periodic sequence is bounded.

What can be said about 1-periodic sequences?

Verify that, if $\left( z _ { k } \right)$ is $p$-periodic, then $\forall n \in \mathbb { N } , \forall k \in \mathbb { N } , z _ { n + k p } = z _ { n }$.

What can be said about sequences that are both periodic and convergent?

In this subsection II.A, $a$ is a nonzero real number. We denote by Sol(II.1) the set of complex sequences $\left( z _ { k } \right) _ { k \in \mathbb { N } }$ satisfying the recurrence relation $$\forall k \in \mathbb { N } ^ { * } , \quad z _ { k + 1 } + a z _ { k } + z _ { k - 1 } = 0$$ Give the general form of sequences belonging to $\operatorname { Sol } ($ II.1 $)$ as a function of the complex roots $r _ { 1 }$ and $r _ { 2 }$ of the equation $r ^ { 2 } + a r + 1 = 0$. What are $r _ { 1 } + r _ { 2 }$ and $r _ { 1 } r _ { 2 }$?

In this subsection II.A, $a$ is a nonzero real number. We denote by Sol(II.1) the set of complex sequences $\left( z _ { k } \right) _ { k \in \mathbb { N } }$ satisfying the recurrence relation $$\forall k \in \mathbb { N } ^ { * } , \quad z _ { k + 1 } + a z _ { k } + z _ { k - 1 } = 0$$ Show that if $| a | > 2$, the zero sequence is the only periodic solution of (II.1).