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# 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 $$\displaystyle p_{2n}(0) = \dfrac{(2n)!}{n!}$$ for all positive integral $$n$$.

Bonus 1: Prove this without induction.

Bonus 2: Prove that for $$p_{n+1} = mx^{m-1}p_{n}+p_{n}'$$ and $$p_{0} = 1$$,

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

Note by Jake Lai
11 months, 2 weeks ago

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Let $e^{2xt + t^2} = \sum_{n = 0}^\infty \frac{q_n(x) t^n}{n!}.$ Taking the derivative with respect to $$x$$, we get $2te^{2xt + t^2} = \sum_{n = 0}^\infty \frac{q_n'(x) t^n}{n!}.$ Taking the derivative with respect to $$t$$, we get $(2x + 2t) e^{2xt + t^2} = \sum_{n = 1}^\infty \frac{q_n(x) t^{n - 1}}{(n - 1)!} = \sum_{n = 0}^\infty \frac{q_{n + 1}(x) t^n}{n!}.$ But $(2x + 2t) e^{2xt + t^2} = 2xe^{2xt + t^2} + 2te^{2xt + t^2} = \sum_{n = 0}^\infty \frac{2xq_n(x) t^n}{n!} + \sum_{n = 0}^\infty \frac{q_n'(x) t^n}{n!}.$ Comparing the coefficients of $$t^n/n!$$, we get $q_{n + 1} (x) = 2xq_n(x) + q_n'(x).$

Furthermore, expanding $$e^{2xt + t^2}$$, we find $$q_0 (x) = 1$$. Hence, the sequences $$(p_n)$$ and $$(q_n)$$ are identical.

Setting $$x = 0$$ in the first equation, we get $\sum_{n = 0}^\infty \frac{p_n(0) t^n}{n!} = e^{t^2} = \sum_{m = 0}^\infty \frac{t^{2m}}{m!}.$ Therefore, for $$n = 2m$$, $p_{2m}(0) = \frac{(2m)!}{m!}.$ · 10 months, 1 week ago

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