Showing $p_{n+1}(x)=(x-2n-1)p_n(x)-n^2p_{n-1}(x)$ for $p_n(x)=(-1)^nn!\sum_{k=0}^{n}\binom{n}{k}\frac{(-x)^k}{k!}$

Question

Let $n \in \mathbb N_0$ and

$$p_n(x)=(-1)^nn!\sum_{k=0}^{n}\binom{n}{k}\frac{(-x)^k}{k!}$$

I need to show that $p_0(x)=1, p_1(x)=x-1$ and $p_{n+1}(x)=(x-2n-1)p_n(x)-n^2p_{n-1}(x)$.

$p_0(x)=1$ and $p_1(x)=x-1$ are easy to show but I struggle to show $p_{n+1}(x)=(x-2n-1)p_n(x)-n^2p_{n-1}(x)$

Answer 1

We obtain for $n\geq 1$:

\begin{align*} \color{blue}{p_{n+1}(x)}&=(-1)^{n+1}(n+1)!\sum_{k=0}^{n+1}\binom{n+1}{k}\frac{(-x)^k}{k!}\\ &=(-1)^{n+1}(n+1)!\left(1+\sum_{k=1}^{n}\left[\binom{n}{k}+\binom{n}{k-1}\right]\frac{(-x)^k}{k!}+\frac{(-x)^{n+1}}{(n+1)!}\right)\tag{1}\\ &=(-1)^{n+1}(n+1)!\sum_{k=0}^n\binom{n}{k}\frac{(-x)^k}{k!}\\ &\qquad+(-1)^{n+1}(n+1)!\sum_{k=1}^n\binom{n}{k-1}\frac{(-x)^k}{k!}+x^{n+1}\tag{2}\\ &=-(n+1)p_n(x)+(-1)^{n}n!x\sum_{k=1}^n\binom{n+1}{k}\frac{(-x)^{k-1}}{(k-1)!}+x^{n+1}\tag{3}\\ &=-(n+1)p_n(x)+(-1)^{n}n!x\sum_{k=1}^n\left[\binom{n}{k}+\binom{n}{k-1}\right]\frac{(-x)^{k-1}}{(k-1)!}+x^{n+1}\tag{4}\\ &=-(n+1)p_n(x)+(-1)^{n}n!nx\sum_{k=1}^n\binom{n-1}{k-1}\frac{(-x)^{k-1}}{k!}\\ &\qquad+(-1)^{n}n!x\sum_{k=0}^{n-1}\binom{n}{k}\frac{(-x)^{k}}{k!}+x^{n+1}\tag{5}\\ &=(x-n-1)p_n(x)+(-1)^{n}n!nx\sum_{k=0}^{n-1}\binom{n-1}{k}\frac{(-x)^{k}}{(k+1)!}\tag{6}\\ &=(x-n-1)p_n(x)-nx^n+(-1)^{n}n!nx\sum_{k=0}^{n-2}\binom{n-1}{k}\frac{(-x)^{k}}{(k+1)!}\tag{7}\\ &=(x-n-1)p_n(x)-nx^n+(-1)^{n}n!nx\sum_{k=1}^{n-1}\binom{n-1}{k-1}\frac{(-x)^{k-1}}{k!}\tag{8}\\ &=(x-n-1)p_n(x)-nx^n\\ &\qquad-(-1)^{n}n!n\sum_{k=0}^{n-1}\left[\binom{n}{k}-\binom{n-1}{k}\right]\frac{(-x)^{k}}{k!}\tag{9}\\ &=(x-n-1)p_n(x)-n(-1)^nn!\sum_{k=0}^n\binom{n}{k}\frac{(-x)^k}{k!}\\ &\qquad-n^2(-1)^{n-1}(n-1)!\sum_{k=0}^{n-1}\binom{n-1}{k}\frac{(-x)^k}{k!}\tag{10}\\ &\,\,\color{blue}{=(x-2n-1)p_n(x)-n^2p_{n-1}(x)} \end{align*}

and the claim follows.

Comment:

• In (1) we separate the first and last term of the sum and apply the binomial identity $\binom{p}{q}=\binom{p-1}{q}+\binom{p-1}{q-1}$.

• In (2) we split the sum, merge $1$ into the left-hand sum and multiply out.

• In (3) we write the left-hand sum in terms of $p_n$ and apply the binomial identity $\binom{p+1}{q+1}=\frac{p+1}{q+1}\binom{p-1}{q-1}$ to the right-hand sum.

• In (4) we apply the binomial identity as in (1).

• In (5) we spit the sum and apply the binomial identity as in (3) to the left-hand sum. We also shift the index of the right-hand sum to start with $k=0$.

• In (6) we merge $x^{n+1}$ into the right-hand sum of (5) and observe the right-hand sum is $xp_n(x)$. We also shift the index of the left-hand sum of (5) to start with $k=0$.

• In (7) we separate the summand with $k=n-1$.

• In (8) we shift the index to start with $k=1$.

• In (9) we apply the binomial identity as in (1) in the form $\binom{p-1}{q-1}=\binom{p}{q}-\binom{p-1}{q}$. We also start the sum with $k=0$ instead of $k=1$ without changing anything since we add $0$ only.

• In (10) we separate the sums, merge $-nx^n$ into the left-hand sum and write them conveniently as preparation for the last step.

Answer 2

Let $$f(t,x)=\sum_{n=0}^\infty\frac{p_n(x)}{n!}t^n.$$ So $$f(t,x)=\sum_{n=0}^\infty(-1)^nt^n\sum_{k=0}^\infty(-1)^k\binom{n}{k}\frac{x^k}{k!}.$$ Therefore, $$f(t,x)=\sum_{k=0}^\infty(-1)^k\frac{x^k}{k!}\sum_{n=0}^\infty(-1)^n\binom{n}{k}t^n.$$ Hence $$f(t,x)=\sum_{k=0}^\infty\frac{x^kt^k}{k!}\sum_{n=k}^\infty(-1)^{n-k}\binom{n}{n-k}t^{n-k}.$$ That is $$f(t,x)=\sum_{k=0}^\infty\frac{(xt)^k}{k!}\sum_{m=0}^\infty(-1)^m\binom{m+k}{m}t^m.$$ As $$\sum_{m=0}^\infty(-1)^m\binom{m+k}{m}t^m=\sum_{m=0}^\infty\binom{-k-1}{m}t^m=(1+t)^{-k-1},$$ we obtain $$f(t,x)=\sum_{k=0}^\infty\frac{(xt)^k}{k!}(1+t)^{-k-1}=\frac{1}{1+t}\sum_{k=0}^\infty\frac{1}{k!}\left(\frac{xt}{1+t}\right)^k.$$ Hence $$f(t,x)=\frac{1}{1+t}\exp\left(\frac{xt}{1+t}\right).$$

This means $$(1+t)\frac{\partial }{\partial t}\big((1+t)f(t,x)\big)=x f(t,x).\tag{1}$$ However, defining $p_{-1}(x)=0$, we get $$(1+t)f(t,x)=\sum_{n=0}^\infty\frac{p_n(x)+np_{n-1}(x)}{n!}t^n$$ so $$\frac{\partial }{\partial t}\big((1+t)f(t,x)\big)=\sum_{n=1}^\infty\frac{p_n(x)+np_{n-1}(x)}{(n-1)!}t^{n-1}=\sum_{n=0}^\infty\frac{p_{n+1}(x)+(n+1)p_n(x)}{n!}t^n.$$ Hence $$(1+t)\frac{\partial }{\partial t}\big((1+t)f(t,x)\big)=\sum_{n=0}^\infty\frac{\big(p_{n+1}(x)+(n+1)p_n(x)\big)+n\big(p_n(x)+np_{n-1}(x)\big)}{n!}t^n,$$ or $$(1+t)\frac{\partial }{\partial t}\big((1+t)f(t,x)\big)=\sum_{n=0}^\infty\frac{p_{n+1}(x)+(2n+1)p_n(x)+n^2p_{n-1}(x)}{n!}t^n.\tag{2}$$ Note also that $$xf(t,x)=\sum_{n=0}^\infty\frac{xp_n(x)}{n!}t^n.\tag{3}$$ From $(1)$, $(2)$, and $(3)$, we have $$\sum_{n=0}^\infty\frac{xp_n(x)}{n!}t^n=\sum_{n=0}^\infty\frac{p_{n+1}(x)+(2n+1)p_n(x)+n^2p_{n-1}(x)}{n!}t^n.$$ This is equivalent to the given recursion.