Find a polynomial $P_{n}\in{\mathbb{C}[X]}$ such as $S_{n}=\ker P_{n}(D)$ Find a polynomial $P_{n}\in{\mathbb{C}[X]}$ such as $S_{n}=\ker P_{n}(D)$
-With $S_{n}$ $(n\geq 1)$ all of the functions such as $y^{(n)}=y  $
(where $y^{(n)}$ is the nth derivative of $y$)
-D the endomorphism sending the functions $C^{\infty}$ on their derivatives   D : y -> y'
I already proved that D is an endomorphism and $S_{n}$ a vector space containing the functions $ke^x$
But my probleme is that i dont really understand the meaning of $P_{n}(D)$, and so dont see what $S_{n}=ker P_{n}(D)$ could be. It probably have to do with eigenvalues, eigenvectors...But really I dont see anything.
Thanks for your help
 A: Let $P_n = X^n - 1$.
Saying $x \in \ker(P_n(D))$ is precisely saying $P_n(D)(x) = 0$.
Then, we have $D^n(x) - Id(x) = x^{(n)} - x = 0$, which is exactly $x^{(n)} = x$.
Also, $P_n(D)$ is a polynomial evaluated at an endomorphism.
If $f$ is an endomorphism of a $\mathbb{K}$-vector space $E$ and $P = a_nX^n + \cdots + a_1X + a_0 \in \mathbb{K}[X]$ a polynomial, then $P(f)$ is the endomorphism $a_nf^n + \cdots + a_1f + a_0$ where $f^i$ is $f$ composed with itself $i$ times.
A: Let me just try to improve the formulation of the question, which is quite confused. Before talking about endomorphisms, one should state which vector space one is considering; apparently this is the space $\def\CC{\mathcal C^\infty}E=\CC(\Bbb R)$ of smooth functions on$~\Bbb R$ (with values in the field we are working over, which I presume is $\Bbb C$ or maybe $\Bbb R$).
Now the operation$~D$ of differentiation is an endomorphism of $E$ (because it is a linear operation, and also because differentiating a $\CC$-function results in another $\CC$-function; note that the latter would not have worked for any finite differentiability class). Composition is a (bilinear) operation on endomorphisms, so $D\circ D$ (differentiation twice) is another endomorphism, and so is the $k$-fold composition $D^k$ for any $k\in\Bbb N$; the latter is written $D^k:f\mapsto f^{(k)}$ on the level of individual vectors ($\CC$-functions) $f\in E$. (For $k=0$ one has $f^{(0)}=f$.)
Since the set of endomorphism is also a vector space, we can form (finite!) linear combinations of these $D^k$ for different values of $k$ with coefficients $c_k$ to get a quite general $d$-th order differential operator $\sum_{k=0}^dc_kD^k$ that maps $f\mapsto \sum_{k=0}^dc_kf^{(k)}$. It is natural to consider this in relation to the polynomial $P=\sum_{k=0}^dc_kX^k$, namely as the result of substituting $D$ for $X$ into $P$. I will then write $\sum_{k=0}^dc_kD^k=P[X:=D]$, which (since $X$ is the only candidate to substitute for) can be shortened to$~P[D]$ (but I refuse to write this as $P(D)$ as almost everybody does, since that would suggest that $P$ is used as a function, which it is not). Finally, like for any endomorphism, one can form $\ker(P[D])$ to denote the subspace $\{\,f\in E\mid P[D](f)=0\,\}=\{\,f\in E\mid \sum_{k=0}^dc_kf^{(k)}=0\,\}$ of vectors that are set to the zero vector by $P[D]$.
Now the question becomes for which polynomial$~P$ this kernel gives the set of solutions to the differential equation $f^{(n)}=f$. It should be immediately obvious that $P=X^n-1$ is the unique polynomial that satisfies the requirement.
