Derivative in a recursion

Define a sequence of polynomials in \(x\) \(\lbrace p_{n} \rbrace\) by the recursion \(p_{n+1} = 2xp_{n}+p_{n}'\), with \(p_{0} = 1\).

Prove that p2n(0)=(2n)!n!\displaystyle p_{2n}(0) = \dfrac{(2n)!}{n!} for all positive integral nn.

Bonus 1: Prove this without induction.

Bonus 2: Prove that for pn+1=mxm−1pn+pn′p_{n+1} = mx^{m-1}p_{n}+p_{n}' and p0=1p_{0} = 1,

pmn(0)=(mn)!n!  .p_{mn}(0) = \dfrac{(mn)!}{n!} \; .

#Calculus

Note by Jake Lai
5 years, 2 months ago

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Comments

Let e2xt+t2=∑n=0∞qn(x)tnn!.e^{2xt + t^2} = \sum_{n = 0}^\infty \frac{q_n(x) t^n}{n!}. Taking the derivative with respect to xx, we get 2te2xt+t2=∑n=0∞qn′(x)tnn!.2te^{2xt + t^2} = \sum_{n = 0}^\infty \frac{q_n'(x) t^n}{n!}. Taking the derivative with respect to tt, we get (2x+2t)e2xt+t2=∑n=1∞qn(x)tn−1(n−1)!=∑n=0∞qn+1(x)tnn!.(2x + 2t) e^{2xt + t^2} = \sum_{n = 1}^\infty \frac{q_n(x) t^{n - 1}}{(n - 1)!} = \sum_{n = 0}^\infty \frac{q_{n + 1}(x) t^n}{n!}. But (2x+2t)e2xt+t2=2xe2xt+t2+2te2xt+t2=∑n=0∞2xqn(x)tnn!+∑n=0∞qn′(x)tnn!.(2x + 2t) e^{2xt + t^2} = 2xe^{2xt + t^2} + 2te^{2xt + t^2} = \sum_{n = 0}^\infty \frac{2xq_n(x) t^n}{n!} + \sum_{n = 0}^\infty \frac{q_n'(x) t^n}{n!}. Comparing the coefficients of tn/n!t^n/n!, we get qn+1(x)=2xqn(x)+qn′(x).q_{n + 1} (x) = 2xq_n(x) + q_n'(x).

Furthermore, expanding e2xt+t2e^{2xt + t^2}, we find q0(x)=1q_0 (x) = 1. Hence, the sequences (pn)(p_n) and (qn)(q_n) are identical.

Setting x=0x = 0 in the first equation, we get ∑n=0∞pn(0)tnn!=et2=∑m=0∞t2mm!.\sum_{n = 0}^\infty \frac{p_n(0) t^n}{n!} = e^{t^2} = \sum_{m = 0}^\infty \frac{t^{2m}}{m!}. Therefore, for n=2mn = 2m, p2m(0)=(2m)!m!.p_{2m}(0) = \frac{(2m)!}{m!}.

Jon Haussmann - 5 years ago
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