Let :
- $n \in \mathbb{N}$
- $\Phi(P)=[ (X^2-1)P']'$ for $P \in \mathbb{R}_n[X]$
- $m \leq n$
- $W_m(X)=(X^2-1)^m$
- $P_m(X)= k_m \frac{ d^m}{dX^m}[ W_m(X)]$ with $k_m$ such that $P_m(1)=1$
We want to express $\Phi(P_m)$ as a function of $P_m$, in fact that $P_m$ is an eigenvector of $\Phi$.
My attempt :
Let $f(X)= (X-1)^m$ and $g(X)=(X+1)^m$. $\forall 0 \leq k \leq m-1, ~ f^{(k)}(1) =0$
$\frac{ d ^ m }{ d X^m} (X^2-1) =\frac{ d ^ m }{ d X^m} (fg) = \sum_{k=0}^{m} \binom{m}{k} f^{(k)} g^{ ( m-k) }$ so
$\frac{ d ^ m }{ d X^m} (X^2-1)|_{X=1} = \binom{m}{m} f^{(m)}|_{X=1} g^{ ( 0) }|_{X=1} = 1 \times m ! \times 2^m $
but $P_m(1)= 1$ so $k_m= \frac{1}{2^m m!}$
We notice that $ (X^2-1) W_m'(X)= 2m XW_m(X)$
I derive $1$ time $\Phi(W_m)= (2m W_m + 2m X W_m')$, then $m$ times.
$ \begin{align*} \frac{ d^m }{ d X^m} \Phi(W_m) &= 2m \frac{ d^m}{ d X^m} W_m + 2m \frac{ d^m }{ d X^m} (X W_m) \\ &= \frac{2m}{k_m} P_m(X) + 2 m \sum_{k=0}^{m} \frac{ d^k }{ d X^k} W_m \frac{ d^{(m-k)} }{ d X^{(m-k)} } X \\ &= \frac{2m}{k_m} P_m(X) + 2 m \sum_{k=m-1}^{m} \binom{m}{k} \frac{ d^k }{ d X^k} W_m \frac{ d^{(m-k)} }{ d X^{(m-k)} } X \\ &= \frac{2m}{k_m} P_m(X) + 2 m ( W_m^{(m)} X + m W_m^{(m-1)} ) \\ &=\frac{2m}{k_m} P_m(X) + 2 m ( \frac{1}{k_m} P_m(X) X + m W_m^{(m-1)} ) \\ \end{align*} $