I'm trying to solve the following recurrence relation for $\alpha_j$, for which Mathematica is not helpful.

$$ \lambda\alpha_j + (j+1)\alpha_{j+1} = \sum_{\mu = 0}^{j}\frac{\mu(j-1)!}{\mu!(j-\mu)!}\alpha_{\mu} $$

Don't be deterred by the inclusion of the extra parameter $\lambda$. Any value of $\lambda$ is interesting for my purposes, so if somebody were to show me a solution even for a particular $\lambda$, I would be ecstatic :). I am by no means expecting a solution for arbitrary $\lambda$. $\alpha_0$ will clearly be left hanging, so don't worry about that, either.

I can do the first few $\alpha_j$ out by hand, so it seems solvable, but I'm wondering if there is an analytic method that I could use for future like problems also.

In short, anything helps! Thank you!


The sum's coefficient is just: $$ \frac{\mu (j - 1)!}{\mu! (j - \mu)!} = \frac{\mu}{j} \binom{j}{\mu} $$ thus your recurrence is: $$ (j + 1) a_{j + 1} + \lambda a_j = \frac{1}{j} \sum_\mu \binom{j}{\mu} \mu a_\mu $$ This suggests the change of variables $u_j = j a_j$: $$ j u_{j + 1} + \lambda u_j = \sum_\mu \binom{j}{\mu} u_\mu $$ and the binomial convolution hints at exponential generating functions: $$ U(z) = \sum_{j \ge 0} a_j \frac{z^j}{j!} $$ Multiplying by $z^j /j!$, adding over $j \ge 0$: \begin{align} z \sum_{j \ge 0} u_{j + 1} \frac{z^{j - 1}}{(j - 1)!} + \lambda U(z) &= e^z U(z) \\ z \sum_{j \ge 0} u_{j + 2} \frac{z^j}{j!} + \lambda U(z) &= e^z U(z) \\ z U''(z) + (\lambda - e^z) U(z) &= 0 \end{align} This doesn't seem to have a nice solution.

  • $\begingroup$ This was extremely helpful. Thanks very much for your work. $\endgroup$ – Nicola Jul 2 '14 at 21:52

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