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Im looking for a recurrence formula of type: $$(\mu-\nu) x P_\mu^\nu(x) + P_{\mu-1}^\nu(x)=?, \quad \mu,\nu\in \mathbb R$$ where $P_\mu^\nu(x)$ is the Legendre function of the first kind (solution to the Legendre differential equation which is regular at the origin).

My goal is to rewrite the sum in one expression, i.e. $(\mu-\nu) x P_\mu^\nu(x) + P_{\mu-1}^\nu(x)= C P_\alpha^\beta(x) $

Any useful reference, I will be very grateful. Thank you in advance

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    $\begingroup$ why do you think there would exist such a recursion? $\endgroup$ May 8, 2020 at 14:53
  • $\begingroup$ I don't know, I just asked the question to know if it exists or not $\endgroup$
    – Z. Alfata
    May 8, 2020 at 14:58
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    $\begingroup$ Try this. people.math.sfu.ca/~cbm/aands/intro.htm#006. If it's not there then you have a problem on your hands. $\endgroup$ May 8, 2020 at 15:08
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    $\begingroup$ You can also check it in I.S. Gradshteyn and I.M. Ryzhik-Table of integrals, series, and products-Academic Press (2007) If it's not there, then it has little chances of being true. $\endgroup$
    – Tony419
    May 8, 2020 at 15:09
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    $\begingroup$ It's not really a well-posed question. Of course one can just go to Abramowitz/Stegun or to the NIST handbook and pick out two recurrence relations that contain the two terms on the l.h.s., shift everything else over to the r.h.s. and one has an answer to the OP's question as stated. The r.h.s. will probably be ugly and the answer not particularly useful. $\endgroup$ May 8, 2020 at 15:17

2 Answers 2

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Let's take relation 14.10.3 from the NIST Handbook, which, after renaming $\mu \leftrightarrow \nu $ and shifting the new $\mu \rightarrow \mu -2$ reads $$ (\mu -\nu ) P_{\mu }^{\nu } (x) - (2\mu -1)x P_{\mu -1}^{\nu } (x) + (\mu+\nu -1) P_{\mu -2}^{\nu } (x) =0 $$ We can thus isolate the desired l.h.s., $$ (\mu -\nu ) xP_{\mu }^{\nu } (x) + P_{\mu -1}^{\nu } (x) = ((2\mu -1)x^2 +1) P_{\mu -1}^{\nu } (x) - (\mu+\nu -1)x P_{\mu -2}^{\nu } (x) $$ The r.h.s. can be consolidated into an expression containing a single Legendre function by also invoking relation 14.10.4 from the NIST Handbook, which, after renaming $\mu \leftrightarrow \nu $ and shifting the new $\mu \rightarrow \mu -2$ reads $$ (1-x^2 ) \frac{d}{dx} P_{\mu -2}^{\nu } (x) = (\nu-\mu +1) P_{\mu -1}^{\nu } (x) + (\mu -1)x P_{\mu -2}^{\nu } (x) $$ Solving for $P_{\mu -1}^{\nu } (x)$ and inserting above, we end up with $$ (\mu -\nu ) xP_{\mu }^{\nu } (x) + P_{\mu -1}^{\nu } (x) = \left[ \frac{(2\mu -1)x^2 +1}{\nu -\mu +1} \left( (1-x^2 ) \frac{d}{dx} - (\mu -1)x \right) -(\mu +\nu -1)x \right] P_{\mu -2}^{\nu } (x) $$

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  • $\begingroup$ Thank you very much M. Engelhardt $\endgroup$
    – Z. Alfata
    May 10, 2020 at 12:43
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See equation 37 here. And now some more characters.

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  • $\begingroup$ The formula 37 is for the associated Legendre polynomials $P^n_m$ (i.e, $n,m$ are positive integers). In our case: $\mu, \nu\in \mathbb R$. $\endgroup$
    – Z. Alfata
    May 9, 2020 at 0:59
  • $\begingroup$ This formula is valid for any real $n$, $m$. $\endgroup$ May 9, 2020 at 19:05
  • $\begingroup$ Which is what i can take for $ \nu $ and $\mu$ to arrive at the formula $37$? $\endgroup$
    – Z. Alfata
    May 9, 2020 at 21:15
  • $\begingroup$ You can take two instances of this formula and construct your lhs, for example. The problem is that your question is not well formulated. Recursion in what? In $\mu $? In $\nu $? In both? Upwards? Downwards? Are derivatives allowed? What data can we assume to be known? Usually the point of a recursion is to be able to take a set of known instances and derive an unknown one. You haven't specified any of this, so the best we can do is point you to some general relations. $\endgroup$ May 9, 2020 at 21:41
  • $\begingroup$ my goal is to rewrite the sum in one expression, i.e. $(\mu-\nu+1) x P_\mu^\nu(x) + P_{\mu-1}^\nu(x)= C P_\alpha^\beta(x) $ $\endgroup$
    – Z. Alfata
    May 9, 2020 at 22:02

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