Consider the sequence of polynomials $(P_n)_{n\geqslant0}$ uniquely defined by the recursion $$(P_n)'=\sum_{k=0}^{n-1}\frac{P_k}{n-k+1},$$ valid for every $n\geqslant0$, with the initial conditions $P_0=1$ and $P_n(0)=0$ for every $n\geqslant1$.

The polynomials $P_n$ appear in the context of a recent MSE question asking about the expansion of the function $(1+\epsilon)^{x/\epsilon}$ as a series in $\epsilon$, more precisely, one has $$(1+\epsilon)^{x/\epsilon}=e^x\sum_{n=0}^\infty(-1)^nP_n(x)\epsilon^n.$$ Each $P_n$ is a polynomial of degree $n$, with nonnegative rational coefficients, whose terms of higher and lower degrees can be readily computed.

The question, admittedly a little vague, is:

What else is known about the family $(P_n)$? To begin with, is this a well-known family of polynomials, with a given name?

Edit: As mentioned in a comment below, differentiating and rearranging, one gets the pair of identities, valid for every $n\geqslant0$, $$xP'_n(x)+nP_n(x)=x\sum_{k=0}^{n-1}P_k(x)$$ and $$nP_n(x)=x\sum_{k=0}^{n-1}\left(1-\frac{1}{n-k+1}\right)P_k(x).$$

  • $\begingroup$ $xP_n'(x)+nP_n(x)=x\sum_{k=0}^{n-1}P_k(x)$; $P_n(x)=\frac{x}{n}\sum_{k=0}^{n-1}\left(\frac{n-k}{n-k+1}\right)P_k(x)$. $\endgroup$ – vnd May 27 '16 at 13:50
  • $\begingroup$ Differentiating wrt $\epsilon$ and equating coefficients as before, one obtains $P_n(x)$. Generates the same set of polynomials if $P_0=1$. $\endgroup$ – vnd May 27 '16 at 15:25
  • $\begingroup$ The relations certainly hold. The steps between differentiating the equation $$(1+\epsilon)^{x/\epsilon}=e^x\sum_{n=0}^\infty(-1)^nP_n(x)\epsilon^n$$ wrt $\epsilon$ till equating coefficients of equal powers of $\epsilon$ from both sides to arrive at the form of $P_n(x)$ (Second equation in my first comment) are fairly routine. The equation $xP_n'(x)+nP_n(x)=x\sum_{k=0}^{n-1}P_k(x)$ follows from the form of $P_n$ and that of $P_n'$ (mentioned in the problem statement). $\endgroup$ – vnd May 27 '16 at 19:46
  • $\begingroup$ @vnd Right. $ $ $\endgroup$ – Did May 30 '16 at 13:21
  • $\begingroup$ How does $P_0=1$ and $P_n(0)=0$ for $n\geq 0$? Shouldn't this be only for $n>0$? $\endgroup$ – vrugtehagel Mar 21 '17 at 8:41

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