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===Polytropic Lane-Emden Function=== We seek a power-series expression for the polytropic, Lane-Emden function, <math>~\Theta_\mathrm{H}(\xi)</math> — expanded about the coordinate center, <math>~\xi = 0</math> — that approximately satisfies the Lane-Emden equation, <div align="center"> {{ Math/EQ_SSLaneEmden01 }} </div> A general power-series should be of the form, <div align="center"> <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\Theta_H</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \theta_0 + a\xi + b\xi^2 + c\xi^3 + d\xi^4 + e\xi^5 + f\xi^6 + g\xi^7 + h\xi^8 + \cdots </math> </td> </tr> </table> </div> First derivative: <div align="center"> <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\frac{d\Theta_H}{d\xi}</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ a + 2b\xi + 3c\xi^2 + 4d\xi^3 + 5e\xi^4 + 6f\xi^5 + 7g\xi^6 + 8h\xi^7 + \cdots </math> </td> </tr> </table> </div> Left-hand-side of Lane-Emden equation: <div align="center"> <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\frac{1}{\xi^2} \frac{d}{d\xi}\biggl( \xi^2 \frac{d\Theta_H}{d\xi} \biggr)</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \frac{2a}{\xi} + 2\cdot 3b + 2^2\cdot 3c\xi + 2^2\cdot 5d\xi^2 + 2\cdot 3\cdot 5e\xi^3 + 2\cdot 3\cdot 7f\xi^4 + 2^3\cdot 7g\xi^5 + 2^3\cdot 3^2h\xi^6 + \cdots </math> </td> </tr> </table> </div> Right-hand-side of Lane-Emden equation (adopt the normalization, <math>~\theta_0=1</math>, then use the [[#Binomial|binomial theorem]] recursively): <div align="center"> <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\Theta_H^n</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ 1 ~+ ~nF + \biggl[\frac{n(n-1)}{2!}\biggr]F^2 ~+~ \biggl[\frac{n(n-1)(n-2)}{3!}\biggr]F^3 + \biggl[\frac{n(n-1)(n-2)(n-3)}{4!}\biggr]F^4 ~~+ ~~ \cdots </math> </td> </tr> </table> </div> where, <div align="center"> <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~F</math> </td> <td align="center"> <math>~\equiv</math> </td> <td align="left"> <math>~ a\xi + b\xi^2 + c\xi^3 + d\xi^4 + e\xi^5 + f\xi^6 + g\xi^7 + h\xi^8 + \cdots </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ a\xi\biggl[1 + \frac{b}{a}\xi + \frac{c}{a}\xi^2 + \frac{d}{a}\xi^3 + \frac{e}{a}\xi^4 + \frac{f}{a}\xi^5 + \frac{g}{a}\xi^6 + \frac{h}{a}\xi^7 + \cdots\biggr] \, . </math> </td> </tr> </table> </div> <font color="red">First approximation</font>: Assume that <math>~e=f=g=h=0</math>, in which case the LHS contains terms only up through <math>~\xi^2</math>. This means that we must ignore all terms on the RHS that are of higher order than <math>~\xi^2</math>; that is, <div align="center"> <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\Theta_H^n</math> </td> <td align="center"> <math>~\approx</math> </td> <td align="left"> <math>~ 1 ~+ ~nF + \biggl[\frac{n(n-1)}{2!}\biggr]F^2 </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~\approx</math> </td> <td align="left"> <math>~ 1 ~+ ~n(a\xi+b\xi^2) + \biggl[\frac{n(n-1)}{2!}\biggr]a^2\xi^2 </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~\approx</math> </td> <td align="left"> <math>~ 1 ~+~na\xi + ~\biggl[n b + \frac{n(n-1)a^2}{2}\biggr]\xi^2\, . </math> </td> </tr> </table> </div> Expressions for the various coefficients can now be determined by equating terms on the LHS and RHS that have like powers of <math>~\xi</math>. Remembering to include a negative sign on the RHS, we find: <div align="center"> <table border="1" cellpadding="5" align="center"> <tr> <td align="center">Term</td> <td align="center">LHS</td> <td align="center">RHS</td> <td align="center">Implication</td> </tr> <tr> <td align="right"> <math>~\xi^{-1}:</math> </td> <td align="center"> <math>~2a</math> </td> <td align="center"> <math>~0</math> </td> <td align="left"> <math>~\Rightarrow ~~~a=0</math> </td> </tr> <tr> <td align="right"> <math>~\xi^{0}:</math> </td> <td align="center"> <math>~2\cdot 3 b</math> </td> <td align="center"> <math>~-1</math> </td> <td align="left"> <math>~\Rightarrow ~~~b=- \frac{1}{6}</math> </td> </tr> <tr> <td align="right"> <math>~\xi^{1}:</math> </td> <td align="center"> <math>~2^2\cdot 3 c</math> </td> <td align="center"> <math>~-na</math> </td> <td align="left"> <math>~\Rightarrow ~~~c=0</math> </td> </tr> <tr> <td align="right"> <math>~\xi^{2}:</math> </td> <td align="center"> <math>~2^2\cdot 5 d</math> </td> <td align="center"> <math>~-\biggl[n b + \frac{n(n-1)a^2}{2}\biggr]</math> </td> <td align="left"> <math>~\Rightarrow ~~~d=+\frac{n}{120}</math> </td> </tr> </table> </div> By including higher and higher order terms in the series expansion for <math>~\Theta_H</math>, and proceeding along the same line of deductive reasoning, one finds: * Expressions for the four coefficients, <math>~a, b, c, d</math>, remain unchanged. * The coefficient is zero for all other terms that contain ''odd'' powers of <math>~\xi</math>; specifically, for example, <math>~e = g = 0</math>. * The coefficients of <math>~\xi^6</math> and <math>~\xi^8</math> are, respectively, <div align="center"> <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~f</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~- \frac{n}{378}\biggl(\frac{n}{5}-\frac{1}{8} \biggr) \, ;</math> </td> </tr> <tr> <td align="right"> <math>~h</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\frac{n(122n^2 -183n + 70)}{3265920} \, .</math> </td> </tr> </table> </div> In summary, the desired, approximate power-series expression for the polytropic Lane-Emden function is: <div align="center" id="PolytropicLaneEmden"> <table border="1" width="80%" cellpadding="8" align="center"> <tr><th align="center">For Spherically Symmetric Configurations</th></tr> <tr><td align="center"> <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\theta</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ 1 - \frac{\xi^2}{6} + \frac{n}{120} \xi^4 - \frac{n}{378} \biggl( \frac{n}{5} - \frac{1}{8} \biggr) \xi^6 + \biggl[ \frac{n(122n^2 -183n + 70)}{3265920} \biggr] \xi^8 + \cdots </math> </td> </tr> </table> </td></tr></table> </div> NOTE: For cylindrically symmetric, rather than spherically symmetric, configurations, the analogous power-series expression appears as equation (15) in the article by [http://adsabs.harvard.edu/abs/1964ApJ...140.1056O J. P. Ostriker (1964, ApJ, 140, 1056)] titled, ''The Equilibrium of Polytropic and Isothermal Cylinders''.
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