User:Tohline/Appendix/Ramblings/PowerSeriesExpressions

Approximate Power-Series Expressions

Broadly Used Mathematical Expressions (shown here without proof)

Binomial

<math>~(1 \pm x)^n</math>

<math>~=</math>

<math>~ 1 ~\pm ~nx + \biggl[\frac{n(n-1)}{2!}\biggr]x^2 ~\pm~ \biggl[\frac{n(n-1)(n-2)}{3!}\biggr]x^3 + \biggl[\frac{n(n-1)(n-2)(n-3)}{4!}\biggr]x^4 ~~\pm ~~ \cdots </math>      for <math>~(x^2 < 1)</math>

See also:

• Wolfram's presentation

Exponential

<math>~e^x</math>

<math>~=</math>

<math>~ 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \cdots </math>

Expressions with Astrophysical Relevance

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,

<math>~\frac{1}{\xi^2} \frac{d}{d\xi}\biggl( \xi^2 \frac{d\Theta_H}{d\xi} \biggr) = - \Theta_H^n</math>

A general power-series should be of the form,

<math>~\Theta_H</math>

<math>~=</math>

<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>

First derivative:

<math>~\frac{d\Theta_H}{d\xi}</math>

<math>~=</math>

<math>~ a + 2b\xi + 3c\xi^2 + 4d\xi^3 + 5e\xi^4 + 6f\xi^5 + 7g\xi^6 + 8h\xi^7 + \cdots </math>

Left-hand-side of Lane-Emden equation:

<math>~\frac{1}{\xi^2} \frac{d}{d\xi}\biggl( \xi^2 \frac{d\Theta_H}{d\xi} \biggr)</math>

<math>~=</math>

<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>

Right-hand-side of Lane-Emden equation (adopt the normalization, <math>~\theta_0=1</math>, then use the binomial theorem recursively):

<math>~\Theta_H^n</math>

<math>~=</math>

<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>

where,

<math>~F</math>	<math>~\equiv</math>	<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>
	<math>~=</math>	<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>

First approximation:  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,

<math>~\Theta_H^n</math>	<math>~\approx</math>	<math>~ 1 ~+ ~nF + \biggl[\frac{n(n-1)}{2!}\biggr]F^2 </math>
	<math>~\approx</math>	<math>~ 1 ~+ ~n(a\xi+b\xi^2) + \biggl[\frac{n(n-1)}{2!}\biggr]a^2\xi^2 </math>
	<math>~\approx</math>	<math>~ 1 ~+~na\xi + ~\biggl[n b + \frac{n(n-1)a^2}{2}\biggr]\xi^2\, . </math>

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:

Term	LHS	RHS	Implication
<math>~\xi^{-1}:</math>	<math>~2a</math>	<math>~0</math>	<math>~\Rightarrow ~~~a=0</math>
<math>~\xi^{0}:</math>	<math>~2\cdot 3 b</math>	<math>~-1</math>	<math>~\Rightarrow ~~~b=- \frac{1}{6}</math>
<math>~\xi^{1}:</math>	<math>~2^2\cdot 3 c</math>	<math>~-na</math>	<math>~\Rightarrow ~~~c=0</math>
<math>~\xi^{2}:</math>	<math>~2^2\cdot 5 d</math>	<math>~-\biggl[n b + \frac{n(n-1)a^2}{2}\biggr]</math>	<math>~\Rightarrow ~~~d=+\frac{n}{120}</math>

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,

<math>~f</math>	<math>~=</math>	<math>~- \frac{n}{378}\biggl(\frac{n}{5}-\frac{1}{8} \biggr) \, ;</math>
<math>~h</math>	<math>~=</math>	<math>~\frac{n(122n^2 -183n + 70)}{3265920} \, .</math>

In summary, the desired, approximate power-series expression for the polytropic Lane-Emden function is:

<math>~\theta</math> <math>~=</math> <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>

Isothermal Lane-Emden Function

Here we seek a power-series expression for the isothermal, Lane-Emden function — expanded about the coordinate center — that approximately satisfies the isothermal Lane-Emden equation; making the variable substitution (sorry for the unnecessary complication!), <math>~\psi(\xi) \leftrightarrow w(r)</math>, the governing ODE is,

<math>~\frac{d^2w}{dr^2} +\frac{2}{r} \frac{d w}{dr} </math>

<math>~=</math>

<math>~e^{-w} \, . </math>

A general power-series should be of the form,

<math>~w</math>

<math>~=</math>

<math>~ w_0 + ar + br^2 + cr^3 + dr^4 + er^5 + fr^6 + gr^7 + hr^8 +\cdots </math>

Derivatives:

<math>~\frac{dw}{dr}</math>	<math>~=</math>	<math>~ a + 2br + 3cr^2 + 4dr^3 + 5er^4 + 6fr^5 + 7gr^6 + 8hr^7 +\cdots \, ; </math>
<math>~\frac{d^2w}{dr^2}</math>	<math>~=</math>	<math>~ 2b + 2\cdot 3cr + 2^2\cdot 3dr^2 + 2^2\cdot 5er^3 + 2\cdot 3 \cdot 5fr^4 + 2\cdot 3 \cdot 7gr^5 + 2^3\cdot 7hr^6 +\cdots \, . </math>

Put together, then, the left-hand-side of the isothermal Lane-Emden equation becomes:

<math>~\frac{d^2w}{dr^2} +\frac{2}{r} \frac{d w}{dr} </math>	<math>~=</math>	<math>~ 2b + 2\cdot 3cr + 2^2\cdot 3dr^2 + 2^2\cdot 5er^3 + 2\cdot 3 \cdot 5fr^4 + 2\cdot 3 \cdot 7gr^5 + 2^3\cdot 7hr^6 + \frac{2}{r}\biggl[ a + 2br + 3cr^2 + 4dr^3 + 5er^4 + 6fr^5 + 7gr^6 + 8hr^7 \biggr] + \cdots </math>
	<math>~=</math>	<math>~\frac{2a}{r} + r^0(6b) + r^1(2^2\cdot 3c) + r^2(2^2\cdot 3d + 2^3d) + r^3(2^2\cdot 5e + 2\cdot 5e) + r^4(2\cdot 3\cdot 5 f + 2^2\cdot 3f) + r^5(2\cdot 3\cdot 7 g+ 2\cdot 7g) + r^6(2^3 \cdot 7 h + 2^4 h) + \cdots </math>

Drawing on the above power-series expression for an exponential function, and adopting the convention that <math>~w_0 = 0</math>, the right-hand-side becomes,

<math>~e^{-w}</math>	<math>~=</math>	<math>~ e^{0}\cdot e^{-ar} \cdot e^{-br^2} \cdot e^{-cr^3} \cdot e^{-dr^4} \cdot e^{-er^5} \cdot e^{-fr^6} \cdot e^{-gr^7} \cdot e^{-hr^8} \cdots </math>
	<math>~=</math>	<math>~ \biggl[ 1 -ar + \frac{a^2r^2}{2!} - \frac{a^3r^3}{3!} + \frac{a^4r^4}{4!} - \frac{a^5r^5}{5!} + \frac{a^6r^6}{6!} + \cdots \biggr] </math>
		<math>~ \times \biggl[ 1 -br^2 + \frac{b^2r^4}{2!} - \frac{b^3r^6}{3!} + \cdots \biggr] \times \biggl[ 1 -cr^3 + \frac{c^2r^6}{2!} + \cdots \biggr] \times \biggl[1 - dr^4\biggr] \times \biggl[1 - er^5\biggr]\times \biggl[1 - fr^6\biggr] </math>
	<math>~\approx</math>	<math>~ \biggl[ 1 -ar + \frac{a^2r^2}{2} - \frac{a^3r^3}{6} + \frac{a^4r^4}{24} - \frac{a^5r^5}{5\cdot 24} + \frac{a^6r^6}{30\cdot 24} \biggr] \times \biggl[ 1 -cr^3 + \frac{c^2r^6}{2} -br^2 + bcr^5 + \frac{b^2r^4}{2} - \frac{b^3r^6}{6} \biggr] \times \biggl[1 - dr^4 - er^5 - fr^6\biggr] </math>
	<math>~\approx</math>	<math>~\biggl\{ \biggl[ 1 -ar + \frac{a^2r^2}{2} - \frac{a^3r^3}{6} + \frac{a^4r^4}{24} - \frac{a^5r^5}{5\cdot 24} + \frac{a^6r^6}{30\cdot 24} \biggr] - dr^4 \biggl[ 1 -ar + \frac{a^2r^2}{2} \biggr] - er^5 \biggl[ 1 -ar \biggr] - fr^6 \biggr\} </math>
		<math>~ \times \biggl[ 1 -br^2 -cr^3 + \frac{b^2r^4}{2} + bcr^5 + r^6\biggl(\frac{c^2}{2}- \frac{b^3}{6}\biggr) \biggr] </math>
	<math>~\approx</math>	<math>~\biggl[ 1 -ar + \frac{a^2r^2}{2} - \frac{a^3r^3}{6} + \frac{a^4r^4}{24} - \frac{a^5r^5}{5\cdot 24} + \frac{a^6r^6}{30\cdot 24} - dr^4 + adr^5 - \frac{a^2d r^6}{2} - er^5 + aer^6 - fr^6 \biggr] </math>
		<math>~ \times \biggl[ 1 -br^2 -cr^3 + \frac{b^2r^4}{2} + bcr^5 + r^6\biggl(\frac{c^2}{2}- \frac{b^3}{6}\biggr) \biggr] </math>
	<math>~\approx</math>	<math>~\biggl[ 1 -ar + \frac{a^2r^2}{2} - \frac{a^3r^3}{6} + r^4\biggl(\frac{a^4}{24} - d\biggr) + r^5\biggl(ad - e-\frac{a^5}{5\cdot 24}\biggr) + r^6 \biggl(\frac{a^6}{30\cdot 24} - \frac{a^2d}{2} + ae - f \biggr) \biggr] \times \biggl[ 1 -br^2 -cr^3 + \frac{b^2r^4}{2} + bcr^5 + r^6\biggl(\frac{c^2}{2}- \frac{b^3}{6}\biggr) \biggr] </math>
	<math>~\approx</math>	<math>~ 1 -ar + \frac{a^2r^2}{2} - \frac{a^3r^3}{6} + r^4\biggl(\frac{a^4}{24} - d\biggr) + r^5\biggl(ad - e-\frac{a^5}{5\cdot 24}\biggr) + r^6 \biggl(\frac{a^6}{30\cdot 24} - \frac{a^2d}{2} + ae - f \biggr) </math>
		<math>~-br^2\biggl[ 1 -ar + \frac{a^2r^2}{2} - \frac{a^3r^3}{6} + r^4\biggl(\frac{a^4}{24} - d\biggr) \biggr] -cr^3 \biggl[ 1 -ar + \frac{a^2r^2}{2} - \frac{a^3r^3}{6} \biggr] + \frac{b^2r^4}{2}\biggl[ 1 -ar + \frac{a^2r^2}{2} \biggr] + bcr^5\biggl[1 -ar \biggr] + r^6\biggl(\frac{c^2}{2}- \frac{b^3}{6}\biggr) </math>

Expressions for the various coefficients can now be determined by equating terms on the LHS and RHS that have like powers of <math>~r</math>. Beginning with the highest order terms, we initially find,

Term	LHS	RHS	Implication
<math>~r^{-1}:</math>	<math>~2a</math>	<math>~0</math>	<math>~\Rightarrow ~~~a=0</math>
<math>~r^{0}:</math>	<math>~6b</math>	<math>~1</math>	<math>~\Rightarrow ~~~b = + \frac{1}{6}</math>
<math>~r^{1}:</math>	<math>~2^2\cdot 3c</math>	<math>~-a</math>	<math>~\Rightarrow ~~~c = -\frac{a}{2^2\cdot 3} =0</math>
<math>~r^{2}:</math>	<math>~(2^2\cdot 3d + 2^3d)</math>	<math>~\frac{a^2}{2} - b</math>	<math>~\Rightarrow ~~~d = \frac{1}{20}\biggl( \frac{a^2}{2} - b \biggr) = - \frac{1}{120}</math>

With this initial set of coefficient values in hand, we can rewrite (and significantly simplify) our approximate expression for the RHS, namely,

<math>~e^{-w}</math>	<math>~\approx</math>	<math>~ 1 -d r^4 -e r^5 -f r^6 -br^2 ( 1 -d r^4 ) + \frac{b^2r^4}{2} - \frac{b^3r^6}{6} </math>
	<math>~=</math>	<math>~ 1 -br^2+ r^4 \biggl(\frac{b^2}{2} -d \biggr) -e r^5 +r^6\biggl( bd - \frac{b^3}{6} -f \biggr) \, . </math>

Continuing, then, with equating terms with like powers on both sides of the equation, we find,

Term	LHS	RHS	Implication
<math>~r^{3}:</math>	<math>~30e</math>	<math>~0</math>	<math>~\Rightarrow ~~~e=0</math>
<math>~r^{4}:</math>	<math>~(2\cdot 3\cdot 5 f + 2^2\cdot 3f)</math>	<math>~\biggl(\frac{b^2}{2} -d \biggr) </math>	<math>~\Rightarrow ~~~f = \frac{1}{2\cdot 3\cdot 7}\biggl(\frac{1}{2^3\cdot 3^2}+\frac{1}{2^3\cdot 3 \cdot 5}\biggr) = \frac{1}{2\cdot 3^3\cdot 5 \cdot 7}</math>
<math>~r^{5}:</math>	<math>~(2\cdot 3\cdot 7 g+ 2\cdot 7g)</math>	<math>~-e</math>	<math>~\Rightarrow ~~~g = 0</math>
<math>~r^{6}:</math>	<math>~(2^3 \cdot 7 h + 2^4 h)</math>	<math>~\biggl( bd - \frac{b^3}{6} -f \biggr)</math>	<math>~\Rightarrow ~~~ h = -\frac{1}{2^3\cdot 3^2}\biggl( \frac{1}{2^4\cdot 3^2 \cdot 5} + \frac{1}{2^4\cdot 3^4} + \frac{1}{2\cdot 3^3\cdot 5\cdot 7}\biggr) = -\frac{61}{2^{6} \cdot 3^6\cdot 5\cdot 7} </math>

Result:

<math>~w(r) </math> <math>~=</math> <math>~\frac{r^2}{6} - \frac{r^4}{120} + \frac{r^6}{1890} - \frac{61 r^8}{1,632,960} + \cdots \, .</math>

See also:

• Equation (377) from §22 in Chapter IV of C67).

Displacement Function for Polytropic LAWE

The LAWE for polytropic spheres may be written as,

<math>~0 </math>	<math>~=</math>	<math>~\frac{d^2x}{d\xi^2} + \biggl[\frac{4}{\xi} - \frac{(n+1)}{\theta} \biggl(- \frac{d\theta}{d\xi}\biggr)\biggr] \frac{dx}{d\xi} + \frac{(n+1)}{\theta}\biggl[\frac{\sigma_c^2}{6\gamma } - \frac{\alpha}{\xi} \biggl(- \frac{d\theta}{d\xi}\biggr)\biggr] x </math>
	<math>~=</math>	<math>~\theta \frac{d^2x}{d\xi^2} + \biggl[4\theta - (n+1)\xi \biggl(- \frac{d\theta}{d\xi}\biggr)\biggr] \frac{1}{\xi}\frac{dx}{d\xi} + \frac{(n+1)}{6} \biggl[\frac{\sigma_c^2}{\gamma } - \frac{6\alpha}{\xi} \biggl(- \frac{d\theta}{d\xi}\biggr)\biggr] x \, ,</math>

where, <math>~\theta(\xi)</math> is the polytropic Lane-Emden function describing the configuration's unperturbed radial density distribution, and <math>~\gamma</math>, <math>~\sigma_c^2</math>, and <math>~\alpha \equiv (3-4/\gamma)</math> are constants. Here we seek a power-series expression for the displacement function, <math>~x(r)</math>, expanded about the center of the configuration, that approximately satisfies this LAWE.

First we note that, near the center, an accurate power-series expression for the polytropic Lane-Emden function is,

<math>~\theta</math>

<math>~=</math>

<math>~ 1 - \frac{\xi^2}{6} + \frac{n}{120} \xi^4 - \frac{n}{378} \biggl( \frac{n}{5} - \frac{1}{8} \biggr) \xi^6 + \cdots </math>

Hence,

<math>~-\frac{d\theta}{d\xi}</math>

<math>~\approx</math>

<math>~ \frac{1}{3} \biggl[ \xi - \frac{n}{10} \xi^3 + \frac{n}{21} \biggl( \frac{n}{5} - \frac{1}{8} \biggr) \xi^5 \biggr] \, .</math>

Therefore, near the center of the configuration, the LAWE may be written as,

<math>~6~\theta \frac{d^2x}{d\xi^2} + \biggl\{ 12~\theta - (n+1)\xi \biggl[ \xi - \frac{n}{10} \xi^3 + \frac{n}{21} \biggl( \frac{n}{5} - \frac{1}{8} \biggr) \xi^5 \biggr] \biggr\} \frac{2}{\xi}\frac{dx}{d\xi}</math>	<math>~\approx</math>	<math>~ - (n+1) \biggl\{ \frac{\sigma_c^2}{\gamma } - \frac{2\alpha}{\xi} \biggl[ \xi - \frac{n}{10} \xi^3 + \frac{n}{21} \biggl( \frac{n}{5} - \frac{1}{8} \biggr) \xi^5 \biggr] \biggr\} x </math>
<math>\Rightarrow~~~ ~6\biggl[ 1 - \frac{\xi^2}{6} + \frac{n}{120} \xi^4 \biggr] \frac{d^2x}{d\xi^2} + \biggl\{ 12 \biggl[ 1 - \frac{\xi^2}{6} + \frac{n}{120} \xi^4 \biggr] - (n+1)\biggl[ \xi^2 - \frac{n}{10} \xi^4 \biggr] \biggr\} \frac{2}{\xi}\frac{dx}{d\xi}</math>	<math>~\approx</math>	<math>~ - (n+1) \biggl\{ \mathfrak{F} + 2\alpha \biggl[ \frac{n}{10} \xi^2 - \frac{n}{21} \biggl( \frac{n}{5} - \frac{1}{8} \biggr) \xi^4 \biggr] \biggr\} x </math>
<math>\Rightarrow~~~ ~\biggl( 6 - \xi^2 + \frac{n}{20} \xi^4 \biggr) \frac{d^2x}{d\xi^2} + \biggl[ 12 - (n+3)\xi^2 + \frac{n(n+2)}{10} \xi^4 \biggr] \frac{2}{\xi}\frac{dx}{d\xi}</math>	<math>~\approx</math>	<math>~ - (n+1) \biggl[ \mathfrak{F} + \frac{n\alpha}{5} \xi^2 - \frac{2n\alpha}{21} \biggl( \frac{n}{5} - \frac{1}{8} \biggr) \xi^4 \biggr] x \, ,</math>

where, for present purposes, we have kept terms in the series no higher than <math>~\xi^4</math>. Let's now adopt a power-series expression for the displacement function of the form,

<math>~x</math>	<math>~=</math>	<math>~ 1 + a\xi + b\xi^2 + c\xi^3 + d\xi^4 + e\xi^5 + f\xi^6\cdots </math>
<math>~\Rightarrow ~~~ \frac{1}{\xi}\frac{dx}{d\xi}</math>	<math>~=</math>	<math>~ \frac{a}{\xi} + 2b + 3 c\xi + 4d\xi^2 + 5e\xi^3 + 6f\xi^4 +\cdots </math>

and,

<math>~\frac{d^2x}{d\xi^2}</math>

<math>~=</math>

<math>~ 2b + 6c\xi + 12d\xi^2 + 20e\xi^3 + 30f\xi^4 + \cdots </math>

Substituting these expressions into the LAWE gives,

<math>~\biggl( 6 - \xi^2 + \frac{n}{20} \xi^4 \biggr) \biggl( 2b + 6c\xi + 12d\xi^2 + 20e\xi^3 + 30f\xi^4 \biggr) + \biggl[ 12 - (n+3)\xi^2 + \frac{n(n+2)}{10} \xi^4 \biggr] \biggl( \frac{2a}{\xi} + 4b + 6 c\xi + 8d\xi^2 + 10e\xi^3 + 12f\xi^4 \biggr)</math>

<math>~\approx</math>

<math>~ - (n+1) \biggl[ \mathfrak{F} + \frac{n\alpha}{5} \xi^2 - \frac{2n\alpha}{21} \biggl( \frac{n}{5} - \frac{1}{8} \biggr) \xi^4 \biggr] \biggl( 1 + a\xi + b\xi^2 + c\xi^3 + d\xi^4 \biggr)</math>

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>.

Term	LHS	RHS	Implication
<math>~\xi^{-1}:</math>	<math>~24a</math>	<math>~0</math>	<math>~\Rightarrow ~~~a=0</math>
<math>~\xi^{0}:</math>	<math>~(12b + 48b)</math>	<math>~-(n+1)\mathfrak{F}</math>	<math>~\Rightarrow ~~~b = - \frac{(n+1)\mathfrak{F}}{60}</math>
<math>~\xi^{1}:</math>	<math>~[36c+72c-2a(n+3)]</math>	<math>~-a(n+1)\mathfrak{F}</math>	<math>~\Rightarrow ~~~108c = 2a(n+3)-a(n+1)\mathfrak{F} \Rightarrow~~c=0</math>
<math>~\xi^{2}:</math>	<math>~[72d-2b+96d-4b(n+3)]</math>	<math>~\biggl[-b(n+1)\mathfrak{F}-\frac{n(n+1)\alpha}{5}\biggr]</math>	<math>~\Rightarrow ~~~d = - (n+1)\biggl\{ \frac{n\alpha +\mathfrak{F}[(4n+14)-(n+1)\mathfrak{F} ]}{10080} \biggr\}</math>

In summary, the desired, approximate power-series expression for the polytropic displacement function is:

<math>~x(\xi)</math> <math>~=</math> <math>~ 1 - \frac{(n+1)\mathfrak{F}}{60} \xi^2- (n+1)\biggl\{ \frac{n\alpha +\mathfrak{F}[(4n+14)-(n+1)\mathfrak{F} ]}{10080} \biggr\} \xi^4 + \cdots </math>

Displacement Function for Isothermal LAWE

The LAWE for isothermal spheres may be written as,

<math>~\frac{d^2 x}{dr^2} + \biggl[4 - r \biggl(\frac{dw }{dr}\biggr) \biggr] \frac{1}{r}\frac{dx}{dr}</math>

<math>~=</math>

<math>~ - \biggl[ \frac{\sigma_c^2}{6\gamma} - \frac{\alpha}{r} \biggl(\frac{dw }{dr}\biggr)\biggr] x \, , </math>

where, <math>~w(r)</math> is the isothermal Lane-Emden function describing the configuration's unperturbed radial density distribution, and <math>~\gamma</math>, <math>~\sigma_c^2</math>, and <math>~\alpha \equiv (3-4/\gamma)</math> are constants. Here we seek a power-series expression for the displacement function, <math>~x(r)</math>, expanded about the center of the configuration, that approximately satisfies this LAWE.

First we note that, near the center, an accurate power-series expression for the isothermal Lane-Emden function is,

<math>~w(r) </math>

<math>~=</math>

<math>~\frac{r^2}{6} - \frac{r^4}{120} + \frac{r^6}{1890} - \frac{61 r^8}{1,632,960} + \cdots \, .</math>

Hence,

<math>~\frac{dw}{dr}</math>

<math>~\approx</math>

<math>~\frac{r}{3} - \frac{r^3}{30} + \frac{r^5}{315} \, .</math>

Therefore, near the center of the configuration, the LAWE may be written as,

<math>~\frac{d^2 x}{dr^2} + \biggl[4 - \biggl(\frac{r^2}{3} - \frac{r^4}{30} + \frac{r^6}{315}\biggr) \biggr] \frac{1}{r}\frac{dx}{dr}</math>

<math>~\approx</math>

<math>~ - \frac{1}{6} \biggl[ \frac{\sigma_c^2}{\gamma} - 2\alpha \biggl(1 - \frac{r^2}{10} + \frac{r^4}{105}\biggr) \biggr] x \, . </math>

Let's now adopt a power-series expression for the displacement function of the form,

<math>~x</math>	<math>~=</math>	<math>~ 1 + ar + br^2 + cr^3 + dr^4 + \cdots </math>
<math>~\Rightarrow ~~~ \frac{1}{r}\frac{dx}{dr}</math>	<math>~=</math>	<math>~ \frac{a}{r} + 2b + 3 cr + 4dr^2 + \cdots </math>

and,

<math>~\frac{d^2x}{dr^2}</math>

<math>~=</math>

<math>~ 2b + 6cr + 12dr^2 + \cdots </math>

Substituting these expressions into the LAWE gives,

<math>~2b + 6cr + 12dr^2 + \biggl[4 - \biggl(\frac{r^2}{3} - \frac{r^4}{30} + \frac{r^6}{315}\biggr) \biggr] \biggl[ \frac{a}{r} + 2b + 3 cr + 4dr^2 \biggr] </math>

<math>~\approx</math>

<math>~ - \frac{1}{6} \biggl[ \frac{\sigma_c^2}{\gamma} - 2\alpha \biggl(1 - \frac{r^2}{10} + \frac{r^4}{105}\biggr) \biggr] \biggl( 1 + ar + br^2 + cr^3 + dr^4 \biggr) \, . </math>

Keeping terms only up through <math>~r^2</math> leads to the following simplification:

<math>~ 2b + 6cr + 12dr^2 + 4 \biggl[ \frac{a}{r} + 2b + 3 cr + 4dr^2 \biggr] - \frac{r^2}{3} \biggl[ \frac{a}{r} + 2b \biggr] </math>

<math>~\approx</math>

<math>~ - \frac{\mathfrak{F} }{6} \biggl( 1 + ar + br^2 \biggr) - \frac{\alpha}{3} \biggl(\frac{r^2}{10} \biggr) </math>

where,

<math>~\mathfrak{F} \equiv \frac{\sigma_c^2}{\gamma} - 2\alpha \, .</math>

Finally, balancing terms of like powers on both sides of the equation leads us to conclude the following:

Term	LHS	RHS	Implication
<math>~r^{-1}:</math>	<math>~4a</math>	<math>~0</math>	<math>~\Rightarrow ~~~a = 0 </math>
<math>~r^{0}:</math>	<math>~2b + 8b</math>	<math>~- \frac{\mathfrak{F}}{6}</math>	<math>~\Rightarrow ~~~b = - \frac{\mathfrak{F}}{60}</math>
<math>~r^{1}:</math>	<math>~6c + 12c - \frac{a}{3}</math>	<math>~-\frac{a\mathfrak{F}}{6}</math>	<math>~\Rightarrow ~~~c=0</math>
<math>~r^{2}:</math>	<math>~12d + 16d - \frac{2b}{3}</math>	<math>~-\frac{\mathfrak{F}b}{6} - \frac{\alpha}{30}</math>	<math>~\Rightarrow ~~~ 28d = \frac{1}{30}\biggl[ 5b (4- \mathfrak{F} ) - \alpha \biggr] ~ \Rightarrow~ d = \frac{1}{10080}\biggl[ \mathfrak{F}(\mathfrak{F} -4) - 12\alpha \biggr] </math>

In summary, the desired, approximate power-series expression for the isothermal displacement function is:

<math>~x(r)</math> <math>~=</math> <math>~ 1 - \frac{\mathfrak{F}}{60} r^2 + \frac{1}{10080}\biggl[ \mathfrak{F}(\mathfrak{F} -4) - 12\alpha \biggr] r^4 + \cdots </math>

Taylor Series (Type 1)

First:

<math>~f_0</math>	<math>~=</math>	<math>~ f_3 + (- 3\Delta) f_3^' + \frac{1}{2} (- 3\Delta)^2 f^{}_3 + \frac{1}{6} (- 3\Delta)^3 f_3^{'} + \frac{1}{24}(- 3\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5) </math>
	<math>~=</math>	<math>~ f_3 - (3\Delta) f_3^' + \frac{3^2}{2} (\Delta)^2 f^{}_3 - \frac{3^2}{2} (\Delta)^3 f_3^{'} + \frac{3^3}{2^3}(\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5) </math>
<math>~\Rightarrow~~~ - \frac{3^2}{2} (\Delta)^2 f^{}_3 </math>	<math>~=</math>	<math>~ f_3 - f_0 - (3\Delta) f_3^' - \frac{3^2}{2} (\Delta)^3 f_3^{} + \frac{3^3}{2^3}(\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5) </math>

Note that, replacing the <math>~(\Delta)^3 f_3^{}</math> term with the expression derived in the Second step, below, gives,

<math>~ - \frac{3^2}{2} (\Delta)^2 f^{}_3 </math>	<math>~=</math>	<math>~ f_3 - f_0 - (3\Delta) f_3^' + \frac{3^3}{2^3}(\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5) </math>
		<math>~ - \frac{3^2}{2} \biggl\{ \biggl[\frac{2^2}{3^2} \biggr] f_0 - f_1 + f_3 \biggl[\frac{5}{3^2} \biggr] + \biggl[- \frac{2}{3}\biggr] (\Delta) f_3^' + \biggl[- \frac{5}{6} \biggr] (\Delta)^4 f_3^{iv} \biggr\}\biggl[ -\frac{3}{2} \biggr] </math>
	<math>~=</math>	<math>~ f_3 - f_0 - 3 (\Delta) f_3^' + \frac{3^3}{2^3}(\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5) </math>
		<math>~ + \biggl\{ 3 f_0 + \biggl[- \frac{3^3}{2^2}\biggr] f_1 + \biggl[\frac{15}{2^2} \biggr] f_3 + \biggl[- \frac{3}{2}\biggr] (\Delta) f_3^' + \biggl[- \frac{3^2\cdot 5}{2^3} \biggr] (\Delta)^4 f_3^{iv} \biggr\} </math>
	<math>~=</math>	<math>~ 2f_0 + \biggl[- \frac{3^3}{2^2}\biggr] f_1 + \biggl[1 + \frac{15}{2^2} \biggr] f_3 + \biggl[-3 - \frac{3}{2}\biggr] (\Delta) f_3^' + \biggl[\frac{3^3}{2^3}- \frac{3^2\cdot 5}{2^3} \biggr] (\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5) </math>
	<math>~=</math>	<math>~ 2f_0 + \biggl[- \frac{3^3}{2^2}\biggr] f_1 + \biggl[\frac{19}{2^2} \biggr] f_3 + \biggl[- \frac{9}{2}\biggr] (\Delta) f_3^' + \biggl[- \frac{9}{4} \biggr] (\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5) </math>

Then, replacing the <math>~(\Delta)^4 f_3^{iv}</math> term with the expression derived in the Third step, below, gives,

<math>~ - \frac{3^2}{2} (\Delta)^2 f^{}_3 </math>	<math>~=</math>	<math>~ 2f_0 + \biggl[- \frac{3^3}{2^2}\biggr] f_1 + \biggl[\frac{19}{2^2} \biggr] f_3 + \biggl[- \frac{9}{2}\biggr] (\Delta) f_3^' + \mathcal{O}(\Delta^5) </math>
		<math>~ + \biggl[- \frac{9}{4} \biggr] \biggl\{ \biggl[-\frac{1}{3^2} \biggr]f_0 + \biggl[\frac{1}{2} \biggr] f_1 - f_2 + \biggl[ \frac{11}{2\cdot 3^2} \biggr] f_3 + \biggl[- \frac{2}{3}\biggr] (\Delta) f_3^' \biggr\}\biggl[- 2^2\cdot 3 \biggr] </math>
	<math>~=</math>	<math>~ 2f_0 + \biggl[- \frac{3^3}{2^2}\biggr] f_1 + \biggl[\frac{19}{2^2} \biggr] f_3 + \biggl[- \frac{9}{2}\biggr] (\Delta) f_3^' + \mathcal{O}(\Delta^5) </math>
		<math>~ + \biggl\{ \biggl[-3 \biggr]f_0 + \biggl[\frac{3^3 }{2} \biggr] f_1 - 3^3 f_2 + \biggl[ \frac{3 \cdot 11}{2} \biggr] f_3 + \biggl[- 2\cdot 3^2 \biggr] (\Delta) f_3^' \biggr\} </math>
	<math>~=</math>	<math>~ - f_0 + \biggl[\frac{3^3 }{2} - \frac{3^3}{2^2}\biggr] f_1 - 3^3 f_2 + \biggl[\frac{3 \cdot 11}{2} + \frac{19}{2^2} \biggr] f_3 + \biggl[- 2\cdot 3^2- \frac{9}{2}\biggr] (\Delta) f_3^' + \mathcal{O}(\Delta^5) </math>
	<math>~=</math>	<math>~ - f_0 + \biggl[\frac{3^3}{2^2}\biggr] f_1 - 3^3 f_2 + \biggl[\frac{5\cdot 17}{2^2} \biggr] f_3 + \biggl[- \frac{3^2\cdot 5}{2}\biggr] (\Delta) f_3^' + \mathcal{O}(\Delta^5) </math>

Second:

<math>~f_1</math>	<math>~=</math>	<math>~ f_3 + (- 2\Delta) f_3^' + \frac{1}{2} (- 2\Delta)^2 f^{}_3 + \frac{1}{6} (- 2\Delta)^3 f_3^{'} + \frac{1}{24}(- 2\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5) </math>
	<math>~=</math>	<math>~ f_3 - 2(\Delta) f_3^' + 2 (\Delta)^2 f^{}_3 - \frac{2^2}{3} (\Delta)^3 f_3^{'} + \frac{2}{3}(\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5) </math>
	<math>~=</math>	<math>~ f_3 - 2(\Delta) f_3^' - \frac{2^2}{3} (\Delta)^3 f_3^{} + \frac{2}{3}(\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5) </math>
		<math>~ - \biggl[\frac{2^2}{3^2} \biggr] \biggl[ f_3 - f_0 - (3\Delta) f_3^' - \frac{3^2}{2} (\Delta)^3 f_3^{} + \frac{3^3}{2^3}(\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5) \biggr] </math>
	<math>~=</math>	<math>~ \biggl[\frac{2^2}{3^2} \biggr] f_0 + f_3 \biggl[1-\frac{2^2}{3^2} \biggr] + \biggl[ \frac{2^2}{3^2} (3\Delta) - 2(\Delta) \biggr] f_3^' + \biggl[ \biggl(\frac{2^2}{3^2} \biggr) \frac{3^2}{2} (\Delta)^3 - \frac{2^2}{3} (\Delta)^3 \biggr] f_3^{} </math>
		<math>~ + \biggl[ \frac{2}{3}(\Delta)^4 - \biggl( \frac{2^2}{3^2} \biggr) \frac{3^3}{2^3}(\Delta)^4 \biggr] f_3^{iv} + \mathcal{O}(\Delta^5) </math>
	<math>~=</math>	<math>~ \biggl[\frac{2^2}{3^2} \biggr] f_0 + f_3 \biggl[\frac{5}{3^2} \biggr] + \biggl[- \frac{2}{3}\biggr] (\Delta) f_3^' + \biggl[ \frac{2}{3} \biggr] (\Delta)^3f_3^{} + \biggl[- \frac{5}{6} \biggr] (\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5) </math>
<math>~\Rightarrow~~~ - \biggl[ \frac{2}{3} \biggr] (\Delta)^3f_3^{} </math>	<math>~=</math>	<math>~ \biggl[\frac{2^2}{3^2} \biggr] f_0 - f_1 + f_3 \biggl[\frac{5}{3^2} \biggr] + \biggl[- \frac{2}{3}\biggr] (\Delta) f_3^' + \biggl[- \frac{5}{6} \biggr] (\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5) </math>

Now, replacing the <math>~(\Delta)^4 f_3^{iv}</math> term with the expression derived in the Third step, below, gives,

<math>~ - \biggl[ \frac{2}{3} \biggr] (\Delta)^3f_3^{} </math>	<math>~=</math>	<math>~ \biggl[\frac{2^2}{3^2} \biggr] f_0 - f_1 + f_3 \biggl[\frac{5}{3^2} \biggr] + \biggl[- \frac{2}{3}\biggr] (\Delta) f_3^' + \mathcal{O}(\Delta^5) </math>
		<math>~ + \biggl[- \frac{5}{6} \biggr] \biggl\{ \biggl[-\frac{1}{3^2} \biggr]f_0 + \biggl[\frac{1}{2} \biggr] f_1 - f_2 + \biggl[ \frac{11}{2\cdot 3^2} \biggr] f_3 + \biggl[- \frac{2}{3}\biggr] (\Delta) f_3^' \biggr\} \biggl[ -2^2\cdot 3\biggr] </math>
	<math>~=</math>	<math>~ \biggl[\frac{2^2}{3^2} \biggr] f_0 - f_1 + f_3 \biggl[\frac{5}{3^2} \biggr] + \biggl[- \frac{2}{3}\biggr] (\Delta) f_3^' + \mathcal{O}(\Delta^5) </math>
		<math>~ + \biggl\{ \biggl[-\frac{2\cdot 5 }{3^2} \biggr]f_0 + \biggl[5\biggr] f_1 + \biggl[- 2\cdot 5 \biggr] f_2 + \biggl[ \frac{5\cdot 11}{3^2} \biggr] f_3 + \biggl[- \frac{2^2\cdot 5}{3}\biggr] (\Delta) f_3^' \biggr\} </math>
	<math>~=</math>	<math>~ \biggl[\frac{2^2}{3^2} -\frac{2\cdot 5 }{3^2} \biggr] f_0 + \biggl[4\biggr] f_1 + \biggl[- 2\cdot 5 \biggr] f_2 + \biggl[\frac{5}{3^2} + \frac{5\cdot 11}{3^2}\biggr] f_3 + \biggl[- \frac{2^2\cdot 5}{3} - \frac{2}{3}\biggr] (\Delta) f_3^' + \mathcal{O}(\Delta^5) </math>
	<math>~=</math>	<math>~ \biggl[-\frac{2}{3} \biggr] f_0 + \biggl[4\biggr] f_1 + \biggl[- 2\cdot 5 \biggr] f_2 + \biggl[\frac{2^2\cdot 5}{3}\biggr] f_3 + \biggl[- \frac{2\cdot 11}{3}\biggr] (\Delta) f_3^' + \mathcal{O}(\Delta^5) </math>

Third:

<math>~f_2</math>	<math>~=</math>	<math>~ f_3 + (- \Delta) f_3^' + \frac{1}{2} (- \Delta)^2 f^{}_3 + \frac{1}{6} (- \Delta)^3 f_3^{'} + \frac{1}{24}(- \Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5) </math>
	<math>~=</math>	<math>~ f_3 + \biggl[ -1 \biggr](\Delta) f_3^' + \biggl[ \frac{1}{2} \biggr] (\Delta)^2 f^{}_3 + \biggl[ - \frac{1}{2\cdot 3} \biggr] (\Delta)^3 f_3^{'} + \biggl[ \frac{1}{2^3\cdot 3} \biggr] (\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5) </math>
	<math>~=</math>	<math>~ f_3 + \biggl[ -1 \biggr](\Delta) f_3^' + \biggl[ \frac{1}{2^3\cdot 3} \biggr] (\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5) </math>
		<math>~ + \biggl[ \frac{1}{2} \biggr] \biggl\{ 2f_0 + \biggl[- \frac{3^3}{2^2}\biggr] f_1 + \biggl[\frac{19}{2^2} \biggr] f_3 + \biggl[- \frac{9}{2}\biggr] (\Delta) f_3^' + \biggl[- \frac{9}{4} \biggr] (\Delta)^4 f_3^{iv} \biggr\} \biggl[-\frac{2}{3^2}\biggr] </math>
		<math>~ + \biggl[ - \frac{1}{2\cdot 3} \biggr] \biggl\{ \biggl[\frac{2^2}{3^2} \biggr] f_0 - f_1 + f_3 \biggl[\frac{5}{3^2} \biggr] + \biggl[- \frac{2}{3}\biggr] (\Delta) f_3^' + \biggl[- \frac{5}{6} \biggr] (\Delta)^4 f_3^{iv} \biggr\} \biggl[-\frac{3}{2}\biggr] </math>
	<math>~=</math>	<math>~ f_3 + \biggl[ -1 \biggr](\Delta) f_3^' + \biggl[ \frac{1}{2^3\cdot 3} \biggr] (\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5) </math>
		<math>~ + \biggl\{ \biggl[ -\frac{2}{3^2} \biggr]f_0 + \biggl[\frac{3}{2^2}\biggr] f_1 + \biggl[-\frac{19}{2^2\cdot 3^2} \biggr] f_3 + \biggl[\frac{1}{2}\biggr] (\Delta) f_3^' + \biggl[\frac{1}{4} \biggr] (\Delta)^4 f_3^{iv} \biggr\} </math>
		<math>~ + \biggl\{ \biggl[\frac{1}{3^2} \biggr] f_0 + \biggl[- \frac{1}{2^2} \biggr] f_1 + f_3 \biggl[\frac{5}{2^2\cdot 3^2} \biggr] + \biggl[- \frac{1}{2\cdot 3}\biggr] (\Delta) f_3^' + \biggl[- \frac{5}{2^3\cdot 3} \biggr] (\Delta)^4 f_3^{iv} \biggr\} </math>
	<math>~=</math>	<math>~ f_3 + \biggl[ -1 \biggr](\Delta) f_3^' + \biggl[ \frac{1}{2^3\cdot 3} \biggr] (\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5) </math>
		<math>~ + \biggl\{ \biggl[\frac{1}{3^2} -\frac{2}{3^2} \biggr]f_0 + \biggl[\frac{3}{2^2}- \frac{1}{2^2} \biggr] f_1 + \biggl[\frac{5}{2^2\cdot 3^2} -\frac{19}{2^2\cdot 3^2} \biggr] f_3 + \biggl[\frac{1}{2}- \frac{1}{2\cdot 3}\biggr] (\Delta) f_3^' + \biggl[\frac{1}{4} - \frac{5}{2^3\cdot 3} \biggr] (\Delta)^4 f_3^{iv} \biggr\} </math>
	<math>~=</math>	<math>~ \biggl[-\frac{1}{3^2} \biggr]f_0 + \biggl[\frac{1}{2} \biggr] f_1 + \biggl[ \frac{11}{2\cdot 3^2} \biggr] f_3 + \biggl[- \frac{2}{3}\biggr] (\Delta) f_3^' + \biggl[\frac{1}{2^2\cdot 3} \biggr] (\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5) </math>
<math>~\Rightarrow~~~ - \biggl[\frac{1}{2^2\cdot 3} \biggr] (\Delta)^4 f_3^{iv} </math>	<math>~=</math>	<math>~ \biggl[-\frac{1}{3^2} \biggr]f_0 + \biggl[\frac{1}{2} \biggr] f_1 - f_2 + \biggl[ \frac{11}{2\cdot 3^2} \biggr] f_3 + \biggl[- \frac{2}{3}\biggr] (\Delta) f_3^' + \mathcal{O}(\Delta^5) </math>

And, finally:

<math>~f_4</math>	<math>~=</math>	<math>~ f_3 + (\Delta) f_3^' + \frac{1}{2} ( \Delta)^2 f^{}_3 + \frac{1}{6} (\Delta)^3 f_3^{'} + \frac{1}{24}(\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5) </math>
	<math>~=</math>	<math>~ f_3 + (\Delta) f_3^' + \mathcal{O}(\Delta^5) </math>
		<math>~ + \frac{1}{2} \biggl\{ - f_0 + \biggl[\frac{3^3}{2^2}\biggr] f_1 - 3^3 f_2 + \biggl[\frac{5\cdot 17}{2^2} \biggr] f_3 + \biggl[- \frac{3^2\cdot 5}{2}\biggr] (\Delta) f_3^' \biggr\} \biggl[ - \frac{2}{3^2} \biggr] </math>
		<math>~ + \frac{1}{6} \biggl\{ \biggl[-\frac{2}{3} \biggr] f_0 + \biggl[4\biggr] f_1 + \biggl[- 2\cdot 5 \biggr] f_2 + \biggl[\frac{2^2\cdot 5}{3}\biggr] f_3 + \biggl[- \frac{2\cdot 11}{3}\biggr] (\Delta) f_3^' \biggr\} \biggl[ -\frac{3}{2} \biggr] </math>
		<math>~ + \frac{1}{24}\biggl\{ \biggl[-\frac{1}{3^2} \biggr]f_0 + \biggl[\frac{1}{2} \biggr] f_1 - f_2 + \biggl[ \frac{11}{2\cdot 3^2} \biggr] f_3 + \biggl[- \frac{2}{3}\biggr] (\Delta) f_3^' \biggr\} \biggl[ -2^2\cdot 3 \biggr] </math>

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