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I have the following spherical harmonic $$Y_3^1(\upsilon, \varphi) = (-1)^1\sqrt {\frac{(2\times 3 +1)}{4\pi}\frac{(3-1)!}{(3+1)!}}P_3^1(\cos(\upsilon))e^{i\varphi}$$ Which simplifies down to $$-\sqrt{\frac{7}{48\pi}}P_3^1(\cos(\upsilon))e^{i\varphi}$$ Inputting the values for $P_3^1(\cos(\upsilon))$ we obtain the following equation $$P_3^1(\cos(\upsilon)) = \frac{(-1)^1}{2^3\times3!}(1-\cos^2(\upsilon))^\frac{1}{2}\frac{d^{3+1}(\cos^2(\upsilon)-1)}{d\cos^4(\upsilon)}$$ Which simplifies down to $$P_3^1(\cos(\upsilon)) = -\frac{\sin(\upsilon)}{48}\frac{d^4(-\sin^2(\upsilon))^3}{d\cos^4(\upsilon)}$$

This seems unfeasible to solve, does anyone know where I went wrong/improvements that can be made to make it easier?

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  • $\begingroup$ There is a typo in the first line, it should be $(-1)^m$. Another typo is in the third equations where you should take $\frac{d^4 (\cos^2(\nu) -1)^3}{d \cos^4 \nu}$, it is this one that gives the incorrect result. It is better to formulate it as $\frac{d^4 (x^2-1)^3}{d x^4 }$ and evaluate before substitution of $x=\cos(\nu)$, although strictly speaking not wrong. $\endgroup$ – Ronald Blaak Aug 1 '17 at 14:10
  • $\begingroup$ So, whilst looking horrific, this is still (sort of) correct. $\endgroup$ – CooperCape Aug 1 '17 at 14:14
  • $\begingroup$ Yes, once you get used to the notation and know how to interpret this correctly, it works just fine. $\endgroup$ – Ronald Blaak Aug 1 '17 at 14:15
  • $\begingroup$ Okay, thanks a bunch :) $\endgroup$ – CooperCape Aug 1 '17 at 14:15

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