Timeline for Ellipsoidal harmonics - A Series expansion for Lame functions of the second kind
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
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| Jul 28, 2016 at 6:06 | comment | added | Carlo Beenakker | You have assumed that $\rho$ is much larger than one, so the appropriate expansion of $E(\rho)$ is in powers of $1/\rho$, not in powers of $\rho$. | |
| Jul 27, 2016 at 9:30 | comment | added | ina | How do i proceed with $(2n+1)E(\rho)E(\mu)E(\nu)I_n(\rho)$ ? Do I just take the first few terms and multiplicate with the $I_n(\rho)$ that i ve found? ie $(2n+1)(a_0\rho^n+a_1\rho^{n-2}+a_2\rho^{n-4}+\cdots)(a_0\mu^n+a_1 \mu^{n-2}+a_2\mu^{n-4}+\cdots)(a_0\nu^n+a_1\nu^{n-2}+a_2\nu^{n-4}+\cdots)$ | |
| Jul 21, 2016 at 15:16 | history | bounty ended | ina | ||
| Jul 19, 2016 at 11:09 | comment | added | Carlo Beenakker | no binomial expansion is needed; I just took $K(t)$, substituted the series expansion for $E_n(t)$ and made a Taylor expansion of the whole thing; the first three terms are the ones I listed. | |
| Jul 19, 2016 at 8:58 | comment | added | ina | Thank you very much! Just to be sure: you found the coefficients $c_0, c_1,c_2,\dots$ by equating $\left[\frac{t^{2n}/h_2^{2n}}{(a_0+a_1t^2+a_2t^4)^2 }\sum_{n=0}^{+\infty}\sum_{k=0}^n(-1)^n\binom{-\frac{1}{2}}{n-k}\binom{-\frac{1}{2}}{k}\frac{h_3^{2k}}{h_2^{2k+1}}t^{2n}\right]=c_0(t/h_2)^{2n}+c_1(t/h_2)^{2n+2}+c_2(t/h_2)^{2n+4} $ ? | |
| Jul 19, 2016 at 7:33 | history | edited | Carlo Beenakker | CC BY-SA 3.0 |
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| Jul 19, 2016 at 7:12 | history | edited | Carlo Beenakker | CC BY-SA 3.0 |
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| Jul 18, 2016 at 21:13 | history | edited | Carlo Beenakker | CC BY-SA 3.0 |
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| Jul 18, 2016 at 20:57 | history | answered | Carlo Beenakker | CC BY-SA 3.0 |