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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
1 year, 7 months 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 \(n = 2m\), \[p_{2m}(0) = \frac{(2m)!}{m!}.\]

Jon Haussmann - 1 year, 6 months ago

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