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Define a sequence of polynomials in the following way:

$P_m(t)=\frac {1} {m!}\cdot t\cdot (t-1)\cdot...\cdot (t-m+1) $.

(Where $P_0(t)=1$).

I'm trying to prove the following identity:

$\frac d {dt} P_{m+1}(t) = \sum_{k=0}^{m} \frac {(-1)^{m-k}} {m-k+1} \cdot P_k(t)$

Induction on $m$ doesn't seem to work here.

I've observed that if $f(x)=x^t$, then $P_m(t)=\frac {f^{(m)}(1)} {m!} $, but it didn't lead me anywhere.

Any ideas how to prove that?

user333618
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    Where did this question originate? Often the context gives the best hints... – darij grinberg Apr 21 '16 at 23:09
  • Actually, it originated from an ODE: The identity I wrote proves that $(P_{0}(t),P_{1}(t),...,P_{n}(t))$ is a solution of a linear ODE system. The (m+1)-th equation in the system is derived from the above identity. I managed to solve the system in another way (matrix exponentiation), which implies this identity, but i'm looking for a simpler, straightforward proof. – user333618 Apr 21 '16 at 23:48
  • Ah, nice. I had posted this question long ago on a linear algebra problem set (not as an exercise but as a remark after the solution); happy to hear that it comes up somewhere else. Hint 1: Notice that $P_m\left(t\right) = \dbinom{t}{m}$. Hint 2: Differentiate the equality $\left(1+x\right)^t = \exp\left(t\log\left(1+x\right)\right) \in \mathbb{Q}\left[\left[x,t\right]\right]$ with respect to $x$. – darij grinberg Apr 21 '16 at 23:51

1 Answers1

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Generating functions to the rescue! Your observation and Taylor series show us that $$ x^t = \sum_{m=0}^\infty P_m(t) (x-1)^m; $$ this converges uniformly for $\frac12\le x\le\frac32$ say and for $t$ in any bounded range. Therefore we may differentiate in $t$ term-by-term: \begin{align*} \sum_{m=0}^\infty \frac d{dt}P_m(t) (x-1)^m &= \frac d{dt} x^t \\ &= x^t \log x \\ &= \bigg( \sum_{m=0}^\infty P_m(t) (x-1)^m \bigg) \bigg( \sum_{m=1}^\infty \frac{(-1)^{m-1}}{m} (x-1)^m \bigg). \end{align*} Your identity then follows by comparing the coefficients of $(x-1)^{m+1}$ on both sides.

Greg Martin
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