In this subsection, $I = ]{-1,1}[$ and $w(x) = \frac{1}{\sqrt{1-x^2}}$.
Show that, for every integer $k \in \mathbb{N}$, the function $x \mapsto x^k w(x)$ is integrable on $I$.

In this subsection, $I = ]{-1,1}[$ and $w(x) = \frac{1}{\sqrt{1-x^2}}$. For every integer $n \in \mathbb{N}$, consider the function $Q_n : \left|\,\begin{array}{ccl} [-1,1] & \rightarrow & \mathbb{R} \\ x & \mapsto & \cos(n \arccos(x)) \end{array}\right.$.
Calculate $Q_0$, $Q_1$ and, for all $n \in \mathbb{N}$, express simply $Q_{n+2}$ in terms of $Q_{n+1}$ and $Q_n$.

In this subsection, $I = ]{-1,1}[$ and $w(x) = \frac{1}{\sqrt{1-x^2}}$. For every integer $n \in \mathbb{N}$, consider the function $Q_n(x) = \cos(n \arccos(x))$ on $[-1,1]$.
Deduce that, for all $n \in \mathbb{N}$, $Q_n$ is polynomial and determine its degree and leading coefficient.

In this subsection, $I = ]{-1,1}[$ and $w(x) = \frac{1}{\sqrt{1-x^2}}$. For every integer $n \in \mathbb{N}$, let $Q_n$ denote the polynomial of $\mathbb{R}[X]$ that coincides with $x \mapsto \cos(n \arccos(x))$ on $[-1,1]$. Let $(p_n)_{n \in \mathbb{N}}$ be the sequence of orthogonal polynomials associated with the weight $w$.
Show that $$\begin{cases} p_0 = Q_0 \\ \forall n \in \mathbb{N}^*, \quad p_n = \dfrac{1}{2^{n-1}} Q_n \end{cases}$$

In this subsection, $I = ]{-1,1}[$ and $w(x) = \frac{1}{\sqrt{1-x^2}}$. The orthogonal polynomials associated with $w$ satisfy $p_n = \frac{1}{2^{n-1}} Q_n$ for $n \geqslant 1$, where $Q_n(x) = \cos(n \arccos(x))$.
For $n \in \mathbb{N}$, explicitly determine the points $(x_j)_{0 \leqslant j \leqslant n}$ of $I$ such that the quadrature formula $I_n(f) = \sum_{j=0}^n \lambda_j f(x_j)$ has maximal order.

Using the expansion $\frac{z}{\mathrm{e}^z - 1} = \sum_{n=0}^{+\infty} \frac{b_n}{n!} z^n$ (valid for $0 < |z| < r$), by performing a Cauchy product, show that $b_0 = 1$ and, for all integer $n \geqslant 2$, $$\sum_{p=0}^{n-1} \binom{n}{p} b_p = 0.$$

Using the relation $b_0 = 1$ and $\sum_{p=0}^{n-1} \binom{n}{p} b_p = 0$ for all integer $n \geqslant 2$, deduce the value of $b_1, b_2, b_3$ and $b_4$.

Using the expansion $\frac{z}{\mathrm{e}^z - 1} = \sum_{n=0}^{+\infty} \frac{b_n}{n!} z^n$ and a parity argument, show that $b_{2p+1} = 0$ for all integer $p \geqslant 1$.

The polynomials $B_m$ are defined by $$\forall m \in \mathbb{N}, \quad B_m(x) = \sum_{k=0}^m \binom{m}{k} b_k x^{m-k}.$$
Determine $B_0, B_1, B_2$ and $B_3$.

The polynomials $B_m$ are defined by $$\forall m \in \mathbb{N}, \quad B_m(x) = \sum_{k=0}^m \binom{m}{k} b_k x^{m-k}.$$
Show that, for all integer $m \geqslant 2$, $B_m(1) = b_m$, then that, for all integer $m \geqslant 1$, $B_m' = m B_{m-1}$.

We fix an integer $n \in \mathbb{N}^*$ and consider a function $g : [0,n] \rightarrow \mathbb{R}$ of class $\mathcal{C}^\infty$. The polynomial $B_1$ is as defined in the sequence $(B_m)$.
Show that $$\int_0^n g(x)\,\mathrm{d}x = \sum_{k=0}^{n-1} \frac{g(k) + g(k+1)}{2} - \int_0^n B_1(x - \lfloor x \rfloor) g'(x)\,\mathrm{d}x.$$

We fix an integer $n \in \mathbb{N}^*$ and consider a function $g : [0,n] \rightarrow \mathbb{R}$ of class $\mathcal{C}^\infty$. The polynomials $B_m$ are defined by $B_m(x) = \sum_{k=0}^m \binom{m}{k} b_k x^{m-k}$.
Deduce that for all integer $m \geqslant 2$, $$\int_0^n g(x)\,\mathrm{d}x = \sum_{k=0}^{n-1} \frac{g(k)+g(k+1)}{2} + \sum_{p=2}^m \frac{(-1)^{p-1} b_p}{p!}\left(g^{(p-1)}(n) - g^{(p-1)}(0)\right) + \frac{(-1)^m}{m!} \int_0^n B_m(x - \lfloor x \rfloor) g^{(m)}(x)\,\mathrm{d}x.$$

We consider a function $f : [a,b] \rightarrow \mathbb{R}$ of class $\mathcal{C}^\infty$ and the trapezoidal method $$T_n(f) = h \sum_{i=0}^{n-1} \frac{f(a_i) + f(a_{i+1})}{2},$$ where $h = \frac{b-a}{n}$ and $\forall i \in \llbracket 0, n-1 \rrbracket, a_i = a + ih$.
Using the result of question 38, show that, for all integer $m \geqslant 1$, $$\int_a^b f(x)\,\mathrm{d}x = T_n(f) - \sum_{p=1}^m \frac{\gamma_{2p}}{n^{2p}} + \rho_{2m}(n)$$ where the coefficients $\gamma_{2p}$ are given by $$\gamma_{2p} = \frac{(b-a)^{2p} b_{2p}}{(2p)!}\left(f^{(2p-1)}(b) - f^{(2p-1)}(a)\right)$$ and $\rho_{2m}(n)$ is a remainder integral satisfying the bound $$|\rho_{2m}(n)| \leqslant \frac{C_{2m}}{n^{2m}}$$ where $C_{2m}$ is a constant to be determined depending only on $m$, $a$ and $b$.

For $x \in \mathbb{R} \backslash \mathbb{Z}$, justify that the series defining $g(x)$ is convergent, where $$g(x) = \frac{1}{x} + \sum_{n=1}^{+\infty} \left(\frac{1}{x+n} + \frac{1}{x-n}\right)$$

Let $f(x) = \pi \operatorname{cotan}(\pi x)$, $g(x) = \frac{1}{x} + \sum_{n=1}^{+\infty}\left(\frac{1}{x+n} + \frac{1}{x-n}\right)$, and $D = f - g$. Show that the functions $g$ and $D$ are odd.

Let $f(x) = \pi \operatorname{cotan}(\pi x)$, $g(x) = \frac{1}{x} + \sum_{n=1}^{+\infty}\left(\frac{1}{x+n} + \frac{1}{x-n}\right)$, and $D = f - g$. Show that the functions $g$ and $D$ are periodic with period 1.

Let $f(x) = \pi \operatorname{cotan}(\pi x)$, $g(x) = \frac{1}{x} + \sum_{n=1}^{+\infty}\left(\frac{1}{x+n} + \frac{1}{x-n}\right)$, and $D = f - g$. Show that the functions $g$ and $D$ are continuous on $\mathbb{R} \backslash \mathbb{Z}$.

Let $D = f - g$ where $f(x) = \pi \operatorname{cotan}(\pi x)$ and $g(x) = \frac{1}{x} + \sum_{n=1}^{+\infty}\left(\frac{1}{x+n} + \frac{1}{x-n}\right)$, and let $\widetilde{D}$ be its continuous extension to $\mathbb{R}$. Deduce that the function $\widetilde{D}$ is zero on $\mathbb{R}$, then that: $$\forall x \in \mathbb{R} \backslash \mathbb{Z}, \quad \pi x \operatorname{cotan}(\pi x) = 1 + 2\sum_{n=1}^{+\infty} \frac{x^2}{x^2 - n^2}$$

For $(n,N) \in \mathbf{N} \times \mathbf{N}^*$, we denote by $P_{n,N}$ the set of lists $(a_1, \ldots, a_N) \in \mathbf{N}^N$ such that $\sum_{k=1}^{N} k a_k = n$. If this set is finite, we denote by $p_{n,N}$ its cardinality.
Let $n \in \mathbf{N}$. Show that $P_{n,N}$ is included in $[0,n]^N$ and non-empty for all $N \in \mathbf{N}^*$, that the sequence $(p_{n,N})_{N \geq 1}$ is increasing and that it is constant from rank $\max(n,1)$ onwards.

For $(n,N) \in \mathbf{N} \times \mathbf{N}^*$, we denote by $P_{n,N}$ the set of lists $(a_1, \ldots, a_N) \in \mathbf{N}^N$ such that $\sum_{k=1}^{N} k a_k = n$. If this set is finite, we denote by $p_{n,N}$ its cardinality. We denote by $p_n$ the final value of $(p_{n,N})_{N \geq 1}$.
Let $N \in \mathbf{N}^*$. Give a sequence $(a_{n,N})_{n \in \mathbf{N}}$ such that $$\forall z \in D, \frac{1}{1-z^N} = \sum_{n=0}^{+\infty} a_{n,N} z^n$$ Deduce, by induction, the formula $$\forall N \in \mathbf{N}^*, \forall z \in D, \prod_{k=1}^{N} \frac{1}{1-z^k} = \sum_{n=0}^{+\infty} p_{n,N} z^n$$

For $(n,N) \in \mathbf{N} \times \mathbf{N}^*$, we denote by $P_{n,N}$ the set of lists $(a_1, \ldots, a_N) \in \mathbf{N}^N$ such that $\sum_{k=1}^{N} k a_k = n$. If this set is finite, we denote by $p_{n,N}$ its cardinality. We denote by $p_n$ the final value of $(p_{n,N})_{N \geq 1}$.
We fix $\ell \in \mathbf{N}$ and $x \in [0,1[$. Using the result of the previous question, establish the bound $\sum_{n=0}^{\ell} p_n x^n \leq P(x)$. Deduce the radius of convergence of the power series $\sum_n p_n z^n$.

For $(n,N) \in \mathbf{N} \times \mathbf{N}^*$, we denote by $P_{n,N}$ the set of lists $(a_1, \ldots, a_N) \in \mathbf{N}^N$ such that $\sum_{k=1}^{N} k a_k = n$. If this set is finite, we denote by $p_{n,N}$ its cardinality. We denote by $p_n$ the final value of $(p_{n,N})_{N \geq 1}$.
Let $z \in D$. By examining the difference $\sum_{n=0}^{+\infty} p_n z^n - \sum_{n=0}^{+\infty} p_{n,N} z^n$, prove that $$P(z) = \sum_{n=0}^{+\infty} p_n z^n$$

Let $n \in \mathbf{N}$. Show that for all real $t > 0$, $$p_n = \frac{e^{nt} P(e^{-t})}{2\pi} \int_{-\pi}^{\pi} e^{-in\theta} \frac{P(e^{-t}e^{i\theta})}{P(e^{-t})} \mathrm{d}\theta \tag{1}$$

Let $x \in [0,1[$ and $\theta \in \mathbf{R}$. Using the function $L$, show that $$\left|\frac{1-x}{1-xe^{i\theta}}\right| \leq \exp(-(1-\cos\theta)x)$$ Deduce that for all $x \in [0,1]$ and all real $\theta$, $$\left|\frac{P(xe^{i\theta})}{P(x)}\right| \leq \exp\left(-\frac{1}{1-x} + \operatorname{Re}\left(\frac{1}{1-xe^{i\theta}}\right)\right)$$

Let $x \in [0,1[$ and $\theta$ a real. Show that $$\frac{1}{1-x} - \operatorname{Re}\left(\frac{1}{1-xe^{i\theta}}\right) \geq \frac{x(1-\cos\theta)}{(1-x)\left((1-x)^2 + 2x(1-\cos\theta)\right)}.$$ Deduce that if $x \geq \frac{1}{2}$ then $$\left|\frac{P(xe^{i\theta})}{P(x)}\right| \leq \exp\left(-\frac{1-\cos\theta}{6(1-x)^3}\right) \quad \text{or} \quad \left|\frac{P(xe^{i\theta})}{P(x)}\right| \leq \exp\left(-\frac{1}{3(1-x)}\right).$$ For this last result, distinguish two cases according to the relative values of $x(1-\cos\theta)$ and $(1-x)^2$.