For $n \geq 1$, define the following operator $M_n$ on the ring of all polynomials with real coefficients. $$M_n P(x) = nP(x)^2 - x \int_0^x (P'(t))^2 \, \mathrm{d}t$$ Monomials $x^k$ are mapped to $n - \frac{k^2}{2k-1} x^{2k}$, so this operator doubles the degree. For Hermite polynomials $H_n$ (probabilist's form), one has the identity $$M_n H_n(x) = \sum_{k=1}^n (n-k)! \binom{n}{k}^2 \frac{k(k-1)}{2k-1} H_{2k},$$ which is the ordinary multiplication formula where the $k$th summand gets the "weight" $\frac{k(k-1)}{2k-1}$.

I encountered this while working around diffusive Markov generators (the operators $M_n$ are related to the generator of the Ornstein-Uhlenbeck semigroup) and can't make sense of this identity in my context.

My question is rather general:

Has anybody already encountered an operator of the form of $M_n$ somewhere? What about the numbers $\frac{k(k-1)}{2k-1}?$ The first ones are $\frac{2}{3}$, $\frac{6}{5}$, $\frac{12}{7}$, $\frac{20}{9}$, $\frac{30}{11}$, $\frac{42}{13}$. Does this ring a bell?

I'm thankful for any hint!

  • $\begingroup$ Why does this preserve the degree? Why doesn't it double it? $\endgroup$
    – Will Sawin
    Aug 27, 2014 at 0:45
  • $\begingroup$ @WillSawin You're of course correct, the degree is doubled. This was a leftover from my original notes (I originally considered it as a bilinear operator). I've corrected my question, thanks for pointing it out! $\endgroup$ Aug 27, 2014 at 9:45


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