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I am currently studying the CMB power spectrum from a numerical approach (easier than the analytical approach). In a Mathematica notebook that I am following, they work with spherical Bessel functions in order to free stream the multipole solution of the fluid equations in Fourier space. I understand the analytical implementation of the Bessel functions in the formula, but in the Mathematica code, they approximate these functions in a way which I have not been able to derive for myself or find online. The approximation of the Bessel function is \begin{equation} l^2 j_l^2(xl) = \frac{1}{2x\sqrt{x^2-1}}. \end{equation} I also have to use the derivative of the Bessel function which they approximate as \begin{equation} l^2 j_l'^2(xl) = \frac{1}{2x\sqrt{x^2-1}}\frac{x^2-1}{x^2}. \end{equation}

This approximations are done in the limit where both $x$ and $xl$ are large. I would very much appreciate anyone bringing some insights the derive these approximations!

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    $\begingroup$ I have never seen either form. It is definitely not applicable to the $x\to0$ limit, which means that it is presumably designed for some form of large-$x$ situation. The standard formula is this one, and it is oscillatory. If both $x$ and $l$ are large, then the oscillations could be seen as averaging out and it could be acceptable to substitute $\sin^2 \to \frac12$. But in that regime the square root thing could equally well be substituted as $\sqrt{x^2-1} \mapsto x$, so who knows what they're doing. $\endgroup$
    – Emilio Pisanty
    Commented Jun 10, 2022 at 10:12
  • $\begingroup$ ... and, speaking of which: who is "they"? what notebook is this, and who made it? what papers does it relate to? $\endgroup$
    – Emilio Pisanty
    Commented Jun 10, 2022 at 10:12
  • $\begingroup$ Thank you very much for your comments. This mathematica notebook is from Daniel Baumann and it is the one that he suggests as simple exercise in equation 3.3.70lecture notes Baumann.@EmilioPisanty $\endgroup$
    – god_operator
    Commented Jun 10, 2022 at 10:34
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    $\begingroup$ but what of $\ell$? It's not enough to know that $x$ is small: we also need to know if $\ell$ is also small or if it is large... what matters here is the size of $x\ell$, not the size of $x$ alone. $\endgroup$
    – ZeroTheHero
    Commented Jun 10, 2022 at 15:00
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    $\begingroup$ I'm too lazy to work through it, but it appears to be a stationary phase approximation to the Bessel integral $J_m(z) = 1/(2\pi i^m) \int_0^{2\pi} d\phi e^{ix\cos\phi-im\phi}$, with $\ell j_\ell(x\ell) = \sqrt{\frac{\pi}{2x\ell}}J_{\ell+1/2}(x\ell)$. The stationary phase points are at $\sin\phi = \frac{2\ell+1}{2\ell x}$, or $\cos(\phi) = \pm\sqrt{x^2-(2\ell+1)^2/(2\ell)^2}/x \simeq \pm\frac{ \sqrt{x^2-1}}{x}$, so it looks like you might get a result proportional to $1/(x\sqrt{x^2-1})$ in the limit $x\ell$ and $\ell$ both large. $\endgroup$
    – user200143
    Commented Jun 12, 2022 at 1:25

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This is too long for a comment so I wrote this answer.

I looked in the obvious place, G. N. Watson, "Treatise on the Theory of Bessel Functions", (Cambridge University Press,Cambridge,1980), second edition, in section 8.12 he gives an expansion first derived by Meissel for large order and $x$ times the order large. Watson then discusses the stationary phase approximation in section 8.2.

Watson gives the Meissel series, where he says this dominant term had been derived by L. Lorenz earlier, \begin{equation} J_\nu(x) \simeq \sqrt{\frac{2}{\pi\sqrt{x^2-\nu^2}}} \cos\left (Q_\nu(x)-\frac{1}{4}\pi\right) \end{equation} \begin{equation} Q_\nu(x) = \sqrt{x^2-\nu^2}-\frac{1}{2}\nu\pi+\nu\arcsin(\nu/x) \end{equation} If I substitute, $j_\ell (x) = \sqrt{\frac{\pi}{2x}} J_{\ell+1/2}(x)$, \begin{equation} j_\ell(\ell x) \simeq \sqrt{\frac{\pi}{2x\ell}} \sqrt{\frac{2}{\pi\sqrt{\ell^2 x^2-(\ell+\frac{1}{2})^2}}} \cos\left (Q_{\ell+1/2}(\ell x)-\frac{1}{4}\pi\right) \,. \end{equation} Simplifying, squaring, and multiplying by $\ell^2$, \begin{equation} \ell^2 j_\ell(\ell x) \simeq \frac{1}{x\sqrt{x^2-\left (\frac{2\ell+1}{2\ell} \right)^2}} \cos^2\left (Q_{\ell+1/2}(\ell x)-\frac{1}{4}\pi\right ) \,. \end{equation} If, as suggested by Emilio Pisanty in the first comment, that the cosine squared is approximated by its average, $\frac{1}{2}$, in the sums or integrals you are doing, and you approximate $\left (\frac{2\ell+1}{2\ell} \right)^2 \simeq 1$, you get your result.

In the stationary phase approximation, it looks to me like the two stationary phase points give integrals that give the $\frac{1}{x\sqrt{x^2-\left (\frac{2\ell+1}{2\ell} \right)^2}}$ factor, and their phases give the cosine term, but as I said, I'm too lazy to spend the time to carefully calculate and check that term.

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