Numerical evaluation of an infinite sum

Question

I am trying to evaluate numerically : \begin{equation} G = \frac{-1}{4\pi}\sum_{l=0}^{\infty}\frac{2l+1}{\frac{l(l+1)}{R^2}+\frac{1}{L_d^2}}P_l(\cos(\gamma)) \end{equation} Where $P_l$ is the $l_{th}$ Legendre Polynomial, $R = 6371000, Ld = 1000000$ and $\cos(\gamma)\in [-1,1) $. I know that the series converges for any $\cos(\gamma) \neq 1$.

I did a very simple code on fortran90 that calculates the sum, but i really dont know how much terms to sum, is there any tolerance or relative errors i can include in my code?

Answer 1

Not a complete answer, just some thoughts:

$$\frac{2\ell+1}{\frac{\ell(\ell+1)}{R^2} + \frac1{L_d^2}} \leq \max(R^2,L_d^2) \cdot \ell^{-1}$$

Further, there is the asymptotic formula

$$ P_\ell(\cos \gamma) = J_0(\ell\gamma) + \mathcal O(\ell^{-1}) $$

where $J_\nu$ denotes the Bessel functions of the first kind. From what I found, the constant in the error term seems to be related to the error term in Stirling's formula for factorials, for which estimates are known. Unfortunately, I cannot provide a good reference for that.

Also, something is known about the error term in the asymptotic form of $J_0$:

$$ \biggl|J_0(\ell \gamma) - \sqrt{\frac2{\pi \ell \gamma}}\cos(\ell \gamma-\tfrac14)\biggr| \leq \frac14 \cdot \Bigl(\frac2 \pi\Bigr)^{3/2}\cdot (\ell \gamma)^{-3/2}. $$ This follows from Theorem 10 in https://arxiv.org/pdf/1107.2007.pdf.

It should be possible to get a numerical estimate on the error of the partial sums in your problem by putting these pieces together. I guess it will be some work though. Sorry that I can't provide much more detail. I'm not an expert in approximation theory.

Answer 2

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Lets $\ds{r}$ and $\ds{\bar{r}}$ the roots of $\ds{x\pars{x + 1}/R^{2} + 1/L_{d}^{2} = 0}$ for the given values of $\ds{R = 6371000}$ and $\ds{L_{d} = 1000000}$.

Note that $\ds{r = -\,{1 \over 2} + {\root{40339641} \over 1000}\,\ic \approx -\,{1 \over 2} + 6.3514\,\ic}$.

Then, \begin{align} G & \equiv -\,{1 \over 4\pi}\sum_{\ell = 0}^{\infty}{2\ell + 1 \over \ell\pars{\ell + 1}/R^{2} + 1/L_{d}^{2}}\,\mrm{P}_{\ell}\pars{\cos\pars{\gamma}} = -\,{1 \over 4\pi}\sum_{\ell = 0}^{\infty}{2\ell + 1 \over \pars{\ell - r}\pars{\ell - \bar{r}}}\,\mrm{P}_{\ell}\pars{\cos\pars{\gamma}} \\[5mm] & = -\,{1 \over 4\pi}\sum_{\ell = 0}^{\infty}\pars{% {2\ell + 1 \over \ell - r} + {2\ell + 1 \over \ell -\bar{r}}} {1 \over r - \bar{r}}\,\mrm{P}_{\ell}\pars{\cos\pars{\gamma}} \\[5mm] & = -\,{1 \over 4\pi\,\Im\pars{r}}\,\Im\sum_{\ell = 0}^{\infty} {2\ell + 1 \over \ell - r}\,\mrm{P}_{\ell}\pars{\cos\pars{\gamma}} = -\,{1 \over 4\pi\,\Im\pars{r}}\,\Im\bracks{\pars{2r + 1}% \sum_{\ell = 0}^{\infty}{\mrm{P}_{\ell}\pars{\cos\pars{\gamma}} \over \ell - r}} \\[5mm] & = -\,{1 \over 4\pi\,\Im\pars{r}}\,\Im\bracks{\pars{2r + 1}% \sum_{\ell = 0}^{\infty}\mrm{P}_{\ell}\pars{\cos\pars{\gamma}} \int_{0}^{1}x^{\ell - r - 1}\,\dd x} \\[5mm] & = -\,{1 \over 4\pi\,\Im\pars{r}}\,\Im\bracks{\pars{2r + 1}% \int_{0}^{1}x^{-r - 1} \sum_{\ell = 0}^{\infty}\mrm{P}_{\ell}\pars{\cos\pars{\gamma}}x^{\ell}\,\dd x} \\[5mm] & = -\,{1 \over 4\pi\,\Im\pars{r}}\,\Im\bracks{\pars{2r + 1}\int_{0}^{1}{x^{-r - 1} \over \root{1 - 2x\cos\pars{\gamma} + x^{2}}}\,\dd x} \end{align}

It turns out that $\ds{\Re\pars{r} = -1/2}$ such that

\begin{align} G & \equiv -\,{1 \over 4\pi}\sum_{\ell = 0}^{\infty}{2\ell + 1 \over \ell\pars{\ell + 1}/R^{2} + 1/L_{d}^{2}}\,\mrm{P}_{\ell}\pars{\cos\pars{\gamma}} = -\,{1 \over 2\pi}\,\Re\bracks{\int_{0}^{1}{x^{-1/2 - \Im\pars{r}\,\ic} \over \root{1 - 2x\cos\pars{\gamma} + x^{2}}}\,\dd x} \\[5mm] & = -\,{1 \over 2\pi}\,\int_{0}^{1}{\cos\pars{\Im\pars{r}\ln\pars{x}} \over \root{1 - 2x\cos\pars{\gamma} + x^{2}}}\,{\dd x \over \root{x}} = \bbx{-\,{1 \over \pi}\,\int_{0}^{1}{\cos\pars{2\,\Im\pars{r}\ln\pars{x}} \over \root{1 - 2x^{2}\cos\pars{\gamma} + x^{4}}}\,\dd x} \end{align}

Now, you can try some quadrature !!!.

Moreover, \begin{align} G & \equiv -\,{1 \over 4\pi}\sum_{\ell = 0}^{\infty}{2\ell + 1 \over \ell\pars{\ell + 1}/R^{2} + 1/L_{d}^{2}}\,\mrm{P}_{\ell}\pars{\cos\pars{\gamma}} \\[5mm] & = {1 \over \pi}\int_{0}^{1}\mrm{f}\pars{x,\gamma}\,\dd x - {1 \over \pi}\,\ \underbrace{% \int_{0}^{1}\cos\pars{2\,\Im\pars{r}\ln\pars{x}}\,\dd x} _{\ds{1 \over 1 + 4\bracks{\Im\pars{r}}^{2}}} \\[5mm] & \mbox{where}\quad \mrm{f}\pars{x,\gamma} = \left\{\begin{array}{ll} \ds{\cos\pars{2\,\Im\pars{r}\ln\pars{x}}\bracks{% 1 - {1 \over \root{1 - 2x^{2}\cos\pars{\gamma} + x^{4}}}}\,,} & \ds{0 < x \leq 1} \\[2mm] \ds{0\,,} & \mbox{otherwise} \end{array}\right. \end{align}

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The following picture depicts $\ds{G}$ as a function of $\ds{\gamma \in \pars{-4\pi,4\pi}}$. The integration was performed with a $\ds{20}$-Points Trapezoidal Rule (TR). There isn't a 'visible' improvement when we increase the number of points. Even with a $\ds{10}$-points TR the picture is very similar. I hope I didn't do any mistake !!!.