Let $z \in D$. Show that the function $\Psi : t \mapsto ( 1 - t z ) e ^ { L ( t z ) }$ is constant on $[ 0,1 ]$, and deduce that
$$\exp ( L ( z ) ) = \frac { 1 } { 1 - z }$$

For $( n , N ) \in \mathbf { N } \times \mathbf { N } ^ { * }$, we denote by $P _ { n , N }$ the set of lists $\left( a _ { 1 } , \ldots , a _ { N } \right) \in \mathbf { N } ^ { N }$ 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 $\left( p _ { n , N } \right) _ { N \geq 1 }$.
Let $N \in \mathbf { N } ^ { * }$. Give a sequence $\left( a _ { n , N } \right) _ { n \in \mathbf { N } }$ 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
$$\forall N \in \mathbf { N } ^ { * } , \forall z \in D , \prod _ { k = 1 } ^ { N } \frac { 1 } { 1 - z ^ { k } } = \sum _ { n = 0 } ^ { + \infty } p _ { n , N } z ^ { n }$$

Let $z \in D$. Show that the function $\Psi : t \mapsto (1-tz)e^{L(tz)}$ is constant on $[0,1]$, and deduce that $$\exp(L(z)) = \frac{1}{1-z}$$

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 $$\forall N \in \mathbf{N}^*, \forall z \in D, \prod_{k=1}^{N} \frac{1}{1-z^k} = \sum_{n=0}^{+\infty} p_{n,N} z^n$$

Show, for $r > 0$, that $$r < \rho(f) \Rightarrow \exists a > 0 \text{ such that } f \prec \frac{a}{r - z} \Rightarrow r \leqslant \rho(\hat{f})$$ deduce in particular that $\rho(\hat{f}) = \rho(f)$.

Show that $\widehat{f \cdot g} \prec \hat{f} \cdot \hat{g}$, deduce that $\rho(f \cdot g) \geqslant \min(\rho(f), \rho(g))$.

Let $f$ and $g$ be power series, with $g \in O_1$. Show that $\widehat{f \circ g} \prec \hat{f} \circ \hat{g}$. Deduce that, if $f$ and $g$ have strictly positive radius of convergence, then $\rho(f \circ g) > 0$.

Let $f$ and $g$ be power series, with $g \in O_1$. For all $z$ satisfying $|z| < \rho(\hat{f} \circ \hat{g})$, show that the series $f$ converges at $g(z)$ and that $f \circ g(z) = f(g(z))$.

Let $f, g$ and $h$ be power series, with $g, h \in O_1$, show that $(f \circ g) \circ h = f \circ (g \circ h)$.

We consider a power series $f = \lambda z + F, F \in O_2, \lambda = (f)_1 \neq 0$. Show that there exists a unique series $h \in O_1$ such that $h \circ f = I$, and that $(h)_1 = 1/\lambda$.

We consider a power series $f = \lambda z + F, F \in O_2, \lambda = (f)_1 \neq 0$. Show that there exists a unique series $g \in O_1$ such that $f \circ g = I$.

We consider the special case of a series of the form $f = I + F, F \in O_2$. We denote $G := f^\dagger - I, G \in O_2$. Show that $[G]_{d+1} + F \circ (I + [G]_d) \in O_{d+2}$ for all $d \geqslant 1$ (the notation $[f]_d$ is defined in the introduction to the subject).

Let $\left(a_k\right)_{k \in \mathbb{N}}$ be a sequence of elements of $\mathbb{K}$. For every polynomial $p \in \mathbb{K}[X]$, show that the expression $$\sum_{k=0}^{+\infty} a_k D^k p$$ makes sense and defines a polynomial of $\mathbb{K}[X]$.

Show that, for every sequence $\left(a_k\right)_{k \in \mathbb{N}}$ of elements of $\mathbb{K}$, $\sum_{k=0}^{+\infty} a_k D^k$ is a shift-invariant endomorphism.

Let $\left(a_k\right)_{k \in \mathbb{N}}$ and $\left(b_k\right)_{k \in \mathbb{N}}$ be sequences of elements of $\mathbb{K}$ such that $\sum_{k=0}^{+\infty} a_k D^k = \sum_{k=0}^{+\infty} b_k D^k$.
Show that, for all $k \in \mathbb{N}$, $a_k = b_k$.

For every $n \in \mathbb{N}$, define the polynomial $q_n = \frac{X^n}{n!}$. Let $T$ be an endomorphism of $\mathbb{K}[X]$.
Show that $T$ is a shift-invariant endomorphism if, and only if, $$T = \sum_{k=0}^{+\infty} \left(T q_k\right)(0) D^k$$

Let $T$ be a delta endomorphism of $\mathbb{K}[X]$.
Show that there exists a unique shift-invariant and invertible endomorphism $U$ such that $T = D \circ U$. Specify $U$ in the case $T = D$, then in the case $T = L$.

Let $T$ be a delta endomorphism of $\mathbb{K}[X]$.
For every polynomial $p \in \mathbb{K}[X]$ non-zero, verify that $\deg(Tp) = \deg(p) - 1$. Deduce $\ker(T)$ and the spectrum of $T$.

Let $T$ be a delta endomorphism of $\mathbb{K}[X]$. For $n \in \mathbb{N}$, let $T_n$ denote the restriction of $T$ to $\mathbb{K}_n[X]$.
Determine $\operatorname{Im}(T_n)$ in terms of $n \in \mathbb{N}$ and deduce that $T$ is surjective.

We wish to show that, for every delta endomorphism $Q$, there exists a unique sequence of polynomials $(q_n)_{n \in \mathbb{N}}$ of $\mathbb{K}[X]$ such that:

• $q_0 = 1$;
• $\forall n \in \mathbb{N}, \deg(q_n) = n$;
• $\forall n \in \mathbb{N}^*, q_n(0) = 0$;
• $\forall n \in \mathbb{N}^*, Q q_n = q_{n-1}$.

Let $Q$ be a delta endomorphism. Show the existence and uniqueness of the sequence $(q_n)_{n \in \mathbb{N}}$ of polynomials associated with $Q$.

Let $(q_n)_{n \in \mathbb{N}}$ be a sequence of polynomials of $\mathbb{K}[X]$ such that $\forall n \in \mathbb{N}, \deg(q_n) = n$ and $$\forall (x,y) \in \mathbb{K}^2, \quad q_n(x+y) = \sum_{k=0}^n q_k(x) q_{n-k}(y)$$
Show that there exists a unique delta endomorphism $Q$ for which $(q_n)_{n \in \mathbb{N}}$ is the associated sequence of polynomials.

Let $Q$ be a delta endomorphism, let $(q_n)_{n \in \mathbb{N}}$ be the sequence of polynomials associated with $Q$, and let $n$ be a natural number.
Show that the family $(q_0, q_1, \ldots, q_n)$ is a basis of $\mathbb{K}_n[X]$.

For $Q = D$, verify that $$\forall n \in \mathbb{N}, \quad q_n = \frac{X^n}{n!}$$

For $Q = E_1 - I$, verify that $$\forall n \in \mathbb{N}^*, \quad q_n = \frac{X(X-1)\cdots(X-n+1)}{n!}$$

Let $Q$ be a delta endomorphism with associated polynomial sequence $(q_n)_{n \in \mathbb{N}}$.
Prove that, for all $p \in \mathbb{K}[X]$, the expression $\sum_{k=0}^{+\infty} (Q^k p)(0) q_k$ makes sense and defines a polynomial of $\mathbb{K}[X]$, then that $$p = \sum_{k=0}^{+\infty} (Q^k p)(0) q_k$$