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I have the following recurrence equation: $$(\mu\ n + \nu) f_{n} + J\Phi^{*} \sqrt{n+1}f_{n+1} + J\Phi\ \sqrt{n}f_{n-1} = 0$$ for complex numbers $f_{n}$ where $n = 0,1,2,3,...,\infty$ and complex $\Phi$ and real $\mu, \nu, J$. Is there a way to find a general expression for $f_{n}$ in terms of $f_0$?

A few first terms: $$n = 0:\ \ \ \nu\ f_{0} + J\Phi^{*}f_{1} = 0 \rightarrow f_{1} = -\frac{\nu}{J\Phi^{*}}f_{0}$$ $$n = 1:\ \ \ (\mu + \nu)\ f_{1} + J\Phi^{*}\sqrt{2} f_{2} + J\Phi f_{0} = 0 \rightarrow f_{2} = -\frac{\Phi}{\Phi^{*}\sqrt{2}}f_{0} - \frac{\mu + \nu}{J\sqrt{2}\Phi^{*}}f_{1}$$

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  • $\begingroup$ Since you wrote that this comes from mathematical physics, and out of sheer curiosity, where does this problem come from? $\endgroup$
    – Amir Sagiv
    Commented Apr 14, 2016 at 15:23
  • $\begingroup$ @AmirSagiv This is equation for ground state of the Gutzwiller bosonic wavefunction when you deal with hopping hamiltonian - cold atoms in optical lattice. $\endgroup$
    – WoofDoggy
    Commented Apr 14, 2016 at 19:03

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Multiply the equation with $x^n/sqrt(n!)$. Define a generating function $g(x)$ with coefficients $g_n:=f_n/sqrt(n!)$.

You will get a homogeneous differential equation for $g(x)$.

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    $\begingroup$ The answer above is an answer to the question, because: It's simple to create the differential equation, which can be easily solved. I don't think, that it is necessary to explain elementary transformations here. @Nex_Friedrich: To develop the result for g(x) into a Taylor series gives the result for g_n and therefore for f_n . $\endgroup$
    – user90369
    Commented Apr 15, 2016 at 7:49

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