Difference between revisions of "User:Tohline/Appendix/Ramblings/PowerSeriesExpressions"

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</div>
</div>


See also:
<table border="1" align="center" cellpadding="8" width="70%">
* [http://mathworld.wolfram.com/BinomialTheorem.html Wolfram's presentation]
<tr>
 
  <th align="center" bgcolor="yellow">
===Exponential===
LaTeX mathematical expressions cut-and-pasted directly from
<div align="center">
<br />
NIST's ''Digital Library of Mathematical Functions''
  </th>
</tr>
<tr>
  <td align="left">
As a primary point of reference, note that according to [http://dlmf.nist.gov/1.2 &sect;1.2 of NIST's ''Digital Library of Mathematical Functions''], the binomial theorem states that,
<table border="0" cellpadding="5" align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~e^x</math>
<math>~(a+b)^{n}</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 47: Line 53:
   <td align="left">
   <td align="left">
<math>~
<math>~
1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \cdots
a^{n}+\binom{n}{1}a^{n-1}b+\binom{n}{2}a^{n-2}b^{2}+\dots+\binom{n}{n-1}ab^{n-1}+b^{n},
</math>
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</table>
</div>
where, for nonnegative integer values of <math>~k</math> and <math>~n</math> and <math>~k \le n</math>, the notation,
 


==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> &#8212; expanded about the coordinate center, <math>~\xi = 0</math> &#8212; that approximately satisfies the Lane-Emden equation,
<div align="center">
{{ User:Tohline/Math/EQ_SSLaneEmden01 }}
</div>
A general power-series should be of the form,
<div align="center">
<table border="0" cellpadding="5" align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\Theta_H</math>
<math>~\binom{n}{k}</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 75: Line 71:
   <td align="left">
   <td align="left">
<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
\frac{n!}{(n-k)!k!}=\binom{n}{n-k}.
</math>
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</table>
</div>


First derivative:
----
<div align="center">
 
'''Our Example:''' &nbsp;Setting <math>~a = 1</math> gives,
<table border="0" cellpadding="5" align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\frac{d\Theta_H}{d\xi}</math>
<math>~(1+b)^{n}</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 95: Line 91:
   <td align="left">
   <td align="left">
<math>~
<math>~
a + 2b\xi + 3c\xi^2 + 4d\xi^3 + 5e\xi^4 + 6f\xi^5 + 7g\xi^6 + 8h\xi^7 + \cdots
1+\binom{n}{1}b+\binom{n}{2}b^{2}+\binom{n}{3}b^{3}+\binom{n}{4}b^{4}+\dots
</math>
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>


Left-hand-side of Lane-Emden equation:
<tr>
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
   <td align="right">
   <td align="right">
<math>~\frac{1}{\xi^2} \frac{d}{d\xi}\biggl( \xi^2 \frac{d\Theta_H}{d\xi} \biggr)</math>
&nbsp;
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 115: Line 105:
   <td align="left">
   <td align="left">
<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
1+\frac{n!}{(n-1)!}~b + \frac{n!}{(n-2)! 2!}~b^{2+ \frac{n!}{(n-3)! 3!}~b^{3} + \frac{n!}{(n-4)! 4!}~b^{4} + \dots
</math>
</math>
   </td>
   </td>
</tr>
</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>
<tr>
   <td align="right">
   <td align="right">
<math>~\Theta_H^n</math>
&nbsp;
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 135: Line 119:
   <td align="left">
   <td align="left">
<math>~
<math>~
1 ~+ ~nF + \biggl[\frac{n(n-1)}{2!}\biggr]F^2
1+ nb + \biggl[ \frac{n(n-1)}{2!}\biggr] b^{2+ \biggl[ \frac{n(n-1)(n-2)}{3!} \biggr] b^{3} + \biggl[ \frac{n(n-1)(n-2)(n-3)}{4!} \biggr] b^{4} + \dots
~+~ \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>
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</table>
</div>
 
where,
  </td>
<div align="center">
</tr>
</table>
 
 
Note, for example, that,
<table border="0" cellpadding="5" align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~F</math>
<math>~(1+x)^{-1}</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~\equiv</math>
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~1 - x +x^2 - x^3 + x^4 - x^5 + \cdots \, ;</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
   </td>
</math>
   </td>
</tr>
</tr>


<tr>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>~(1+x)^{-2}</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
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   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~1 - 2x + 3x^2 - 4x^3 + 5x^4 - 6x^5 + \cdots \, ;</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>
   </td>
</tr>
</tr>
</table>
</div>
<font color="red">First approximation</font>:  &nbsp;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>
<tr>
   <td align="right">
   <td align="right">
<math>~\Theta_H^n</math>
<math>~(1+x)^{-3}</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~\approx</math>
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~1 - 3x + \biggl[ \frac{3\cdot 4 }{ 2} \biggr]x^2 - \biggl[ \frac{ 3\cdot 4 \cdot 5}{ 2\cdot 3} \biggr]x^3
1 ~+ ~nF + \biggl[\frac{n(n-1)}{2!}\biggr]F^2
+ \biggl[ \frac{3\cdot 4 \cdot 5 \cdot 6 }{2\cdot 3 \cdot 4 } \biggr]x^4 - \biggl[ \frac{3\cdot 4 \cdot 5 \cdot 6 \cdot 7}{2\cdot 3 \cdot 4 \cdot 5 } \biggr]x^5 + \cdots </math>
</math>
   </td>
   </td>
</tr>
</tr>
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   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~\approx</math>
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~1 - 3x + 6x^2 - 10x^3 + 15x^4 - 21x^5 + \cdots \, ;</math>
1 ~+ ~n(a\xi+b\xi^2) + \biggl[\frac{n(n-1)}{2!}\biggr]a^2\xi^2
</math>
   </td>
   </td>
</tr>
</tr>
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<tr>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>~(1+x)^{-4}</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~\approx</math>
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~1 - 4x + \biggl[ \frac{4\cdot 5 }{ 2} \biggr]x^2 - \biggl[ \frac{ 4\cdot 5 \cdot 6}{ 2\cdot 3} \biggr]x^3
1 ~+~na\xi + ~\biggl[n b + \frac{n(n-1)a^2}{2}\biggr]\xi^2\, .
+ \biggl[ \frac{4\cdot 5 \cdot 6 \cdot 7 }{2\cdot 3 \cdot 4 } \biggr]x^4 - \biggl[ \frac{4\cdot 5 \cdot 6 \cdot 7 \cdot 8}{2\cdot 3 \cdot 4 \cdot 5 } \biggr]x^5 + \cdots </math>
</math>
   </td>
   </td>
</tr>
</tr>
</table>
 
</div>
<tr>
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">
   <td align="right">
<math>~\xi^{-1}:</math>
&nbsp;
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~2a</math>
<math>~=</math>
  </td>
  <td align="center">
<math>~0</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\Rightarrow ~~~a=0</math>
<math>~1 - 4x + 10 x^2 - 20x^3
+ 35x^4 - 56x^5 + \cdots \, .</math>
   </td>
   </td>
</tr>
</tr>
</table>
See also:
* [http://mathworld.wolfram.com/BinomialTheorem.html Wolfram's presentation]
===Exponential===
<div align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\xi^{0}:</math>
<math>~e^x</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~2\cdot 3 b</math>
<math>~=</math>
  </td>
  <td align="center">
<math>~-1</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\Rightarrow ~~~b=- \frac{1}{6}</math>
<math>~
1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \cdots
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
==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> &#8212; expanded about the coordinate center, <math>~\xi = 0</math> &#8212; that approximately satisfies the Lane-Emden equation,
<div align="center">
{{ User:Tohline/Math/EQ_SSLaneEmden01 }}
</div>
A general power-series should be of the form,
<div align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\xi^{1}:</math>
<math>~\Theta_H</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~2^2\cdot 3 c</math>
<math>~=</math>
  </td>
  <td align="center">
<math>~-na</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\Rightarrow ~~~c=0</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>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
First derivative:
<div align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\xi^{2}:</math>
<math>~\frac{d\Theta_H}{d\xi}</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~2^2\cdot 5 d</math>
<math>~=</math>
  </td>
  <td align="center">
<math>~-\biggl[n b + \frac{n(n-1)a^2}{2}\biggr]</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\Rightarrow ~~~d=+\frac{n}{120}</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>
   </td>
   </td>
</tr>
</tr>
</table>
</table>
</div>
</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.
Left-hand-side of Lane-Emden equation:
* 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">
<div align="center">
<table border="0" cellpadding="5" align="center">
<table border="0" cellpadding="5" align="center">
Line 306: Line 287:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~f</math>
<math>~\frac{1}{\xi^2} \frac{d}{d\xi}\biggl( \xi^2 \frac{d\Theta_H}{d\xi} \biggr)</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 312: Line 293:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~- \frac{n}{378}\biggl(\frac{n}{5}-\frac{1}{8}  \biggr) \, ;</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>
   </td>
   </td>
</tr>
</tr>
</table>
</div>


<tr>
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">
   <td align="right">
<math>~h</math>
<math>~\Theta_H^n</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
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   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\frac{n(122n^2 -183n + 70)}{3265920} \, .</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>
   </td>
   </td>
</tr>
</tr>
</table>
</table>
</div>
</div>
 
where,
 
<div align="center">
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><td align="center">
<table border="0" cellpadding="5" align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\theta</math>
<math>~F</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=</math>
<math>~\equiv</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<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
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>
   </td>
   </td>
</tr>
</tr>
</table>
</td></tr></table>
</div>


===Isothermal Lane-Emden Function===
<tr>
 
   <td align="right">
<!-- As we have discussed in [[User:Tohline/SSC/Structure/IsothermalSphere#Governing_Relations|a separate chapter]], the 2<sup>nd</sup>-order ODE that governs the radial density distribution in an isothermal sphere is,
&nbsp;
<div align="center" id="Chandrasekhar">
<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\psi}{d\xi}\biggr)</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 367: Line 349:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~e^{-\psi} \, .</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>
   </td>
   </td>
</tr>
</tr>
</table>
</table>
</div>
</div>
-->
<font color="red">First approximation</font>&nbsp;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,
 
Here we seek a power-series expression for the isothermal, Lane-Emden function &#8212; expanded about the coordinate center &#8212; that approximately satisfies the [[User:Tohline/SSC/Structure/IsothermalSphere#Chandrasekhar|isothermal Lane-Emden equation]]; making the variable substitution (sorry for the unnecessary complication!), <math>~\psi(\xi) \leftrightarrow w(r)</math>, the governing ODE is,
<div align="center">
<div align="center">
<table border="0" cellpadding="5" align="center">
<table border="0" cellpadding="5" align="center">
Line 380: Line 362:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\frac{d^2w}{dr^2} +\frac{2}{r} \frac{d w}{dr}
<math>~\Theta_H^n</math>
</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=</math>
<math>~\approx</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~e^{-w} \, . </math>
<math>~
1 ~+ ~nF + \biggl[\frac{n(n-1)}{2!}\biggr]F^2
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
A general power-series should be of the form,
<div align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~w</math>
&nbsp;
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=</math>
<math>~\approx</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~
w_0 + ar + br^2 + cr^3 + dr^4 + er^5 + fr^6 + gr^7 + hr^8 +\cdots
1 ~+ ~n(a\xi+b\xi^2) + \biggl[\frac{n(n-1)}{2!}\biggr]a^2\xi^2
</math>
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
Derivatives:
<div align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\frac{dw}{dr}</math>
&nbsp;
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=</math>
<math>~\approx</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~
a + 2br + 3cr^2 + 4dr^3 + 5er^4 + 6fr^5 + 7gr^6 + 8hr^7 +\cdots \, ;
1 ~+~na\xi + ~\biggl[n b + \frac{n(n-1)a^2}{2}\biggr]\xi^2\, .
</math>
</math>
   </td>
   </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>


<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\frac{d^2w}{dr^2}</math>
<math>~\xi^{-1}:</math>
  </td>
  <td align="center">
<math>~2a</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=</math>
<math>~0</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~\Rightarrow ~~~a=0</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>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
Put together, then, the left-hand-side of the isothermal Lane-Emden equation becomes:
<div align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\frac{d^2w}{dr^2} +\frac{2}{r} \frac{d w}{dr} </math>
<math>~\xi^{0}:</math>
  </td>
  <td align="center">
<math>~2\cdot 3 b</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=</math>
<math>~-1</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~\Rightarrow ~~~b=- \frac{1}{6}</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>
   </td>
   </td>
</tr>
</tr>
Line 468: Line 445:
<tr>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>~\xi^{1}:</math>
  </td>
  <td align="center">
<math>~2^2\cdot 3 c</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=</math>
<math>~-na</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<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)
<math>~\Rightarrow ~~~c=0</math>
+ 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
  </td>
</math>
</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>
   </td>
</tr>
</tr>
</table>
</table>
</div>
</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:


Drawing on the [[#Exponential|above power-series expression for an exponential function]], and adopting the convention that <math>~w_0 = 0</math>, the right-hand-side becomes,
* 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">
<div align="center">
<table border="0" cellpadding="5" align="center">
<table border="0" cellpadding="5" align="center">
Line 488: Line 485:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~e^{-w}</math>
<math>~f</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 494: Line 491:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~- \frac{n}{378}\biggl(\frac{n}{5}-\frac{1}{8} \biggr) \, ;</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>
   </td>
   </td>
</tr>
</tr>
Line 502: Line 497:
<tr>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>~h</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 508: Line 503:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~\frac{n(122n^2 -183n + 70)}{3265920} \, .</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>
   </td>
   </td>
</tr>
</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>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>~\theta</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
&nbsp;
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<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]
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
\times \biggl[1 - dr^4\biggr] \times \biggl[1 - er^5\biggr]\times \biggl[1 - fr^6\biggr]
</math>
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</td></tr></table>
</div>
NOTE:  &nbsp;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''.
===Isothermal Lane-Emden Function===
<!-- As we have discussed in [[User:Tohline/SSC/Structure/IsothermalSphere#Governing_Relations|a separate chapter]], the 2<sup>nd</sup>-order ODE that governs the radial density distribution in an isothermal sphere is,
<div align="center" id="Chandrasekhar">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>~\frac{1}{\xi^2}\frac{d}{d\xi}\biggl( \xi^2 \frac{d\psi}{d\xi}\biggr)</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~\approx</math>
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~e^{-\psi} \, .</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]
  </td>
\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]  
</tr>
\times \biggl[1 - dr^4 - er^5 - fr^6\biggr]
</table>
</math>
</div>
  </td>
-->
</tr>
 
Here we seek a power-series expression for the isothermal, Lane-Emden function &#8212; expanded about the coordinate center &#8212; that approximately satisfies the [[User:Tohline/SSC/Structure/IsothermalSphere#Chandrasekhar|isothermal Lane-Emden equation]]; making the variable substitution (sorry for the unnecessary complication!), <math>~\psi(\xi) \leftrightarrow w(r)</math>, the governing ODE is,
<div align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>~\frac{d^2w}{dr^2} +\frac{2}{r} \frac{d w}{dr}
</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~\approx</math>
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\biggl\{
<math>~e^{-w} \, . </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]
- dr^4 \biggl[ 1 -ar + \frac{a^2r^2}{2} \biggr] - er^5 \biggl[ 1 -ar \biggr] - fr^6
\biggr\}
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
A general power-series should be of the form,
<div align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>~w</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
&nbsp;
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<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]
w_0 + ar + br^2 + cr^3 + dr^4 + er^5 + fr^6 + gr^7 + hr^8 +\cdots
</math>
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
Derivatives:
<div align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>~\frac{dw}{dr}</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~\approx</math>
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\biggl[
<math>~
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}
a + 2br + 3cr^2 + 4dr^3 + 5er^4 + 6fr^5 + 7gr^6 + 8hr^7 +\cdots \, ;
- dr^+ adr^5 - \frac{a^2d r^6}{2}  - er^5  +  aer^6  - fr^6
\biggr]
</math>
</math>
   </td>
   </td>
Line 593: Line 616:
<tr>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>~\frac{d^2w}{dr^2}</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
&nbsp;
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<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]
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>
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
Put together, then, the left-hand-side of the isothermal Lane-Emden equation becomes:
<div align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>~\frac{d^2w}{dr^2} +\frac{2}{r} \frac{d w}{dr} </math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~\approx</math>
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\biggl[
<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)
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
+ r^6 \biggl(\frac{a^6}{30\cdot 24} - \frac{a^2d}{2}   +  ae  - f \biggr)
+ \frac{2}{r}\biggl[ a + 2br + 3cr^2 + 4dr^3 + 5er^4 + 6fr^5 + 7gr^6 + 8hr^\biggr] + \cdots
\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>
   </td>
   </td>
Line 627: Line 654:
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~\approx</math>
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<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)
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^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
+ r^6 \biggl(\frac{a^6}{30\cdot 24} - \frac{a^2d}{2}  + ae  - f \biggr)
</math>
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
Drawing on the [[#Exponential|above power-series expression for an exponential function]], and adopting the convention that <math>~w_0 = 0</math>, the right-hand-side becomes,
<div align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>~e^{-w}</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
&nbsp;
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<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]
<math>~
-cr^3 \biggl[ 1 -ar + \frac{a^2r^2}{2} - \frac{a^3r^3}{6} \biggr]
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
+ \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>
</math>
   </td>
   </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>~r</math>.  Beginning with the highest order terms, we initially 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>


<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~r^{-1}:</math>
&nbsp;
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~2a</math>
<math>~=</math>
  </td>
  <td align="center">
<math>~0</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\Rightarrow ~~~a=0</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>
   </td>
   </td>
</tr>
</tr>
Line 683: Line 699:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~r^{0}:</math>
&nbsp;
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~6b</math>
&nbsp;
  </td>
  <td align="center">
<math>~1</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\Rightarrow ~~~b = + \frac{1}{6}</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>
   </td>
   </td>
</tr>
</tr>
Line 698: Line 714:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~r^{1}:</math>
&nbsp;
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~2^2\cdot 3c</math>
<math>~\approx</math>
  </td>
  <td align="center">
<math>~-a</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\Rightarrow ~~~c = -\frac{a}{2^2\cdot 3} =0</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>
   </td>
   </td>
</tr>
</tr>
Line 713: Line 730:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~r^{2}:</math>
&nbsp;
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~(2^2\cdot 3d + 2^3d)</math>
<math>~\approx</math>
  </td>
  <td align="center">
<math>~\frac{a^2}{2} - b</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\Rightarrow ~~~d = \frac{1}{20}\biggl( \frac{a^2}{2} - b \biggr) = - \frac{1}{120}</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>
   </td>
   </td>
</tr>
</tr>
</table>
</div>


With this initial set of coefficient values in hand, we can rewrite (and significantly simplify) our approximate expression for the RHS, namely,
<tr>
 
  <td align="right">
<div align="center">
&nbsp;
<table border="0" cellpadding="5" align="center">
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<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>
  </td>
</tr>


<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~e^{-w}</math>
&nbsp;
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 741: Line 766:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~\biggl[
1 -d r^4 -e r^5 -f r^6  
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}  
-br^2 ( 1 -d r^4 ) + \frac{b^2r^4}{2} - \frac{b^3r^6}{6}
- dr^4  + adr^5 - \frac{a^2d r^6}{2}  - er^5  +  aer^6  - fr^6
\biggr]
</math>
</math>
   </td>
   </td>
Line 753: Line 779:
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=</math>
&nbsp;
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~
1 -br^2+ r^4 \biggl(\frac{b^2}{2}  -d \biggr) -e r^5   
\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]
+r^6\biggl( bd - \frac{b^3}{6} -f \biggr) \, .
</math>
</math>
   </td>
   </td>
</tr>
</table>
</div>
Continuing, then, with equating terms with like powers on both sides of the equation, 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>


<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~r^{3}:</math>
&nbsp;
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~30e</math>
<math>~\approx</math>
  </td>
  <td align="center">
<math>~0</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\Rightarrow ~~~e=0</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>
   </td>
   </td>
</tr>
</tr>
Line 792: Line 807:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~r^{4}:</math>
&nbsp;
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~(2\cdot 3\cdot 5 f + 2^2\cdot 3f)</math>
<math>~\approx</math>
  </td>
  <td align="center">
<math>~\biggl(\frac{b^2}{2}  -d \biggr) </math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<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>~
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>
   </td>
   </td>
</tr>
</tr>
Line 807: Line 822:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~r^{5}:</math>
&nbsp;
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~(2\cdot 3\cdot 7 g+ 2\cdot 7g)</math>
&nbsp;
  </td>
  <td align="center">
<math>~-e</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\Rightarrow ~~~g = 0</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>
   </td>
   </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>~r</math>.  Beginning with the highest order terms, we initially 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>


<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~r^{6}:</math>
<math>~r^{-1}:</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~(2^3 \cdot 7 h + 2^4 h)</math>
<math>~2a</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~\biggl( bd - \frac{b^3}{6} -f \biggr)</math>
<math>~0</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\Rightarrow ~~~
<math>~\Rightarrow ~~~a=0</math>
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>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
Result:
<div align="center" id="IsothermalLaneEmden">
<table border="1" width="80%" cellpadding="8" align="center"><tr><td align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~w(r)
<math>~r^{0}:</math>
</math>
  </td>
  <td align="center">
<math>~6b</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=</math>
<math>~1</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\frac{r^2}{6} - \frac{r^4}{120} + \frac{r^6}{1890} - \frac{61 r^8}{1,632,960} + \cdots \, .</math>
<math>~\Rightarrow ~~~b = + \frac{1}{6}</math>
   </td>
   </td>
</tr>
</tr>
</table>
</td></tr></table>
</div>


 
<tr>
See also:
  <td align="right">
* Equation (377) from &sect;22 in Chapter IV of [[User:Tohline/Appendix/References#C67|C67]]).
<math>~r^{1}:</math>
 
  </td>
===Displacement Function for Polytropic LAWE===
   <td align="center">
The [[User:Tohline/SSC/Stability/Polytropes#Adiabatic_.28Polytropic.29_Wave_Equation|LAWE for polytropic spheres]] may be written as,
<math>~2^2\cdot 3c</math>
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
   <td align="right">
<math>~0 </math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=</math>
<math>~-a</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<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} +
<math>~\Rightarrow ~~~c = -\frac{a}{2^2\cdot 3} =0</math>
\frac{(n+1)}{\theta}\biggl[\frac{\sigma_c^2}{6\gamma } -
\frac{\alpha}{\xi} \biggl(- \frac{d\theta}{d\xi}\biggr)\biggr]  x </math>
   </td>
   </td>
</tr>
</tr>
Line 888: Line 896:
<tr>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>~r^{2}:</math>
  </td>
  <td align="center">
<math>~(2^2\cdot 3d + 2^3d)</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=</math>
<math>~\frac{a^2}{2} - b</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<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} +
<math>~\Rightarrow ~~~d = \frac{1}{20}\biggl( \frac{a^2}{2} - b \biggr) = - \frac{1}{120}</math>
\frac{(n+1)}{6} \biggl[\frac{\sigma_c^2}{\gamma } -  
\frac{6\alpha}{\xi} \biggl(- \frac{d\theta}{d\xi}\biggr)\biggr]  x \, ,</math>
   </td>
   </td>
</tr>
</tr>
</table>
</table>
</div>
</div>
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 [[#PolytropicLaneEmden|power-series expression for the polytropic Lane-Emden function]] is,  
With this initial set of coefficient values in hand, we can rewrite (and significantly simplify) our approximate expression for the RHS, namely,
 
<div align="center">
<table border="0" cellpadding="5" align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\theta</math>
<math>~e^{-w}</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=</math>
<math>~\approx</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<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
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>
   </td>
   </td>
</tr>
</tr>
</table>
Hence,
<div align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~-\frac{d\theta}{d\xi}</math>
&nbsp;
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~\approx</math>
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<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]
1 -br^2+ r^4 \biggl(\frac{b^2}{2} -d \biggr) -e r^
\, .</math>
+r^6\biggl( bd - \frac{b^3}{6} -f \biggr) \, .
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</table>
</div>
</div>
Therefore, near the center of the configuration, the LAWE may be written as,
 
Continuing, then, with equating terms with like powers on both sides of the equation, we find,
<div align="center">
<div align="center">
<table border="0" cellpadding="5" 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>
<tr>
   <td align="right">
   <td align="right">
<math>~6~\theta \frac{d^2x}{d\xi^2} + \biggl\{ 12~\theta
<math>~r^{3}:</math>
- (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>
  </td>
  <td align="center">
<math>~30e</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~\approx</math>
<math>~0</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~ -
<math>~\Rightarrow ~~~e=0</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>
   </td>
   </td>
</tr>
</tr>
Line 961: Line 975:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>\Rightarrow~~~ ~6\biggl[ 1 - \frac{\xi^2}{6} + \frac{n}{120} \xi^4 \biggr] \frac{d^2x}{d\xi^2}  
<math>~r^{4}:</math>
+ \biggl\{ 12 \biggl[ 1 - \frac{\xi^2}{6} + \frac{n}{120} \xi^4 \biggr]
  </td>
- (n+1)\biggl[ \xi^2 - \frac{n}{10} \xi^4 \biggr] \biggr\} \frac{2}{\xi}\frac{dx}{d\xi}</math>
  <td align="center">
<math>~(2\cdot 3\cdot 5 f + 2^2\cdot 3f)</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~\approx</math>
<math>~\biggl(\frac{b^2}{2}  -d \biggr) </math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<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>
(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>
   </td>
   </td>
</tr>
</tr>
Line 977: Line 990:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>\Rightarrow~~~ ~\biggl( 6 - \xi^2 + \frac{n}{20} \xi^4  \biggr) \frac{d^2x}{d\xi^2}
<math>~r^{5}:</math>
+ \biggl[ 12 - (n+3)\xi^2 + \frac{n(n+2)}{10} \xi^4  \biggr] \frac{2}{\xi}\frac{dx}{d\xi}</math>
  </td>
  <td align="center">
<math>~(2\cdot 3\cdot 7 g+ 2\cdot 7g)</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~\approx</math>
<math>~-e</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~ -
<math>~\Rightarrow ~~~g = 0</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>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
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,
<div align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~x</math>
<math>~r^{6}:</math>
  </td>
  <td align="center">
<math>~(2^3 \cdot 7 h + 2^4 h)</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=</math>
<math>~\biggl( bd - \frac{b^3}{6} -f \biggr)</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~\Rightarrow ~~~
1 + a\xi + b\xi^2 + c\xi^3 + d\xi^4 + e\xi^5 + f\xi^6\cdots
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>
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
Result:
<div align="center" id="IsothermalLaneEmden">
<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>
<tr>
   <td align="right">
   <td align="right">
<math>~\Rightarrow ~~~ \frac{1}{\xi}\frac{dx}{d\xi}</math>
<math>~w(r)
</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 1,017: Line 1,041:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~\frac{r^2}{6} - \frac{r^4}{120} + \frac{r^6}{1890} - \frac{61 r^8}{1,632,960} + \cdots \, .</math>
\frac{a}{\xi} + 2b + 3 c\xi + 4d\xi^2 + 5e\xi^3 + 6f\xi^4 +\cdots
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</table>
</td></tr></table>
</div>
</div>
and,
 
 
See also:
* Equation (377) from &sect;22 in Chapter IV of [[User:Tohline/Appendix/References#C67|C67]]).
 
 
NOTE:  &nbsp;For cylindrically symmetric, rather than spherically symmetric, configurations, an analytic expression for the function, <math>~w(r)</math>, is presented as equation (56) in a paper by [http://adsabs.harvard.edu/abs/1964ApJ...140.1056O J. P. Ostriker (1964, ApJ, 140, 1056)] titled, ''The Equilibrium of Polytropic and Isothermal Cylinders''.
 
===Displacement Function for Polytropic LAWE===
The [[User:Tohline/SSC/Stability/Polytropes#Adiabatic_.28Polytropic.29_Wave_Equation|LAWE for polytropic spheres]] may be written as,
<div align="center">
<div align="center">
<table border="0" cellpadding="5" align="center">
<table border="0" cellpadding="5" align="center">
Line 1,030: Line 1,062:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\frac{d^2x}{d\xi^2}</math>
<math>~0 </math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 1,036: Line 1,068:
   </td>
   </td>
   <td align="left">
   <td align="left">
<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} +
2b + 6c\xi + 12d\xi^2 + 20e\xi^3 + 30f\xi^4 + \cdots
\frac{(n+1)}{\theta}\biggl[\frac{\sigma_c^2}{6\gamma } -
</math>
\frac{\alpha}{\xi} \biggl(- \frac{d\theta}{d\xi}\biggr)\biggr]  x </math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
Substituting these expressions into the LAWE gives,
<div align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
<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)
&nbsp;
+ \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>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~\approx</math>
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<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} +
(n+1) \biggl[ \mathfrak{F}
\frac{(n+1)}{6} \biggl[\frac{\sigma_c^2}{\gamma } -  
+ \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>
\frac{6\alpha}{\xi} \biggl(- \frac{d\theta}{d\xi}\biggr)\biggr]  x \, ,</math>
   </td>
   </td>
</tr>
</tr>
</table>
</table>
</div>
</div>
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.


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>. 
First we note that, near the center, an accurate [[#PolytropicLaneEmden|power-series expression for the polytropic Lane-Emden function]] is,
<div align="center">
<table border="0" cellpadding="5" 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>
<tr>
   <td align="right">
   <td align="right">
<math>~\xi^{-1}:</math>
<math>~\theta</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~24a</math>
<math>~=</math>
  </td>
  <td align="center">
<math>~0</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\Rightarrow ~~~a=0</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>
   </td>
   </td>
</tr>
</tr>
</table>
Hence,
<div align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\xi^{0}:</math>
<math>~-\frac{d\theta}{d\xi}</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~(12b + 48b)</math>
<math>~\approx</math>
  </td>
  <td align="center">
<math>~-(n+1)\mathfrak{F}</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\Rightarrow ~~~b = - \frac{(n+1)\mathfrak{F}}{60}</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>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
Therefore, near the center of the configuration, the LAWE may be written as,
<div align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\xi^{1}:</math>
<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>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~[36c+72c-2a(n+3)]</math>
<math>~\approx</math>
  </td>
  <td align="center">
<math>~-a(n+1)\mathfrak{F}</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\Rightarrow ~~~108c = 2a(n+3)-a(n+1)\mathfrak{F} \Rightarrow~~c=0</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>
   </td>
   </td>
</tr>
</tr>
Line 1,122: Line 1,149:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\xi^{2}:</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>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~[72d-2b+96d-4b(n+3)]</math>
<math>~\approx</math>
  </td>
  <td align="center">
<math>~\biggl[-b(n+1)\mathfrak{F}-\frac{n(n+1)\alpha}{5}\biggr]</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\Rightarrow ~~~d = - (n+1)\biggl\{ \frac{n\alpha +\mathfrak{F}[(4n+14)-(n+1)\mathfrak{F} ]}{10080} \biggr\}</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>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
In summary, the desired, approximate power-series expression for the polytropic displacement function is:
<div align="center" id="PolytropicDisplacement">
<table border="1" width="80%" cellpadding="8" align="center"><tr><td align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~x(\xi)</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>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=</math>
<math>~\approx</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<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
(n+1) \biggl[ \mathfrak{F}
</math>
+ \frac{n\alpha}{5} \xi^2 - \frac{2n\alpha}{21} \biggl( \frac{n}{5} - \frac{1}{8} \biggr) \xi^4 \biggr]  x \, ,</math>
   </td>
   </td>
</tr>
</tr>
</table>
</table>
</td></tr></table>
</div>
</div>
 
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,
===Displacement Function for Isothermal LAWE===
The [[User:Tohline/SSC/Stability/Isothermal#Taff_and_Van_Horn_.281974.29|LAWE for isothermal spheres]] may be written as,
<div align="center">
<div align="center">
<table border="0" cellpadding="5" align="center">
<table border="0" cellpadding="5" align="center">
Line 1,166: Line 1,185:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\frac{d^2 x}{dr^2} + \biggl[4 - r \biggl(\frac{dw }{dr}\biggr) \biggr] \frac{1}{r}\frac{dx}{dr}</math>
<math>~x</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 1,172: Line 1,191:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~
- \biggl[ \frac{\sigma_c^2}{6\gamma}  - \frac{\alpha}{r} \biggl(\frac{dw }{dr}\biggr)\biggr]  x  \, ,
1 + a\xi + b\xi^2 + c\xi^3 + d\xi^4 + e\xi^5 + f\xi^6\cdots
</math>
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
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 [[#Isothermal_Lane-Emden_Function|power-series expression for the isothermal Lane-Emden function]] is,
<div align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~w(r)
<math>~\Rightarrow ~~~ \frac{1}{\xi}\frac{dx}{d\xi}</math>
</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 1,194: Line 1,205:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\frac{r^2}{6} - \frac{r^4}{120} + \frac{r^6}{1890} - \frac{61 r^8}{1,632,960} + \cdots \, .</math>
<math>~
\frac{a}{\xi} + 2b + 3 c\xi + 4d\xi^2 + 5e\xi^3 + 6f\xi^4 +\cdots
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</table>
</div>
</div>
Hence,
and,
<div align="center">
<div align="center">
<table border="0" cellpadding="5" align="center">
<table border="0" cellpadding="5" align="center">
Line 1,205: Line 1,218:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\frac{dw}{dr}</math>
<math>~\frac{d^2x}{d\xi^2}</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~\approx</math>
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\frac{r}{3} - \frac{r^3}{30} + \frac{r^5}{315} \, .</math>
<math>~
2b + 6c\xi + 12d\xi^2 + 20e\xi^3 + 30f\xi^4 + \cdots
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</table>
</div>
</div>
Therefore, near the center of the configuration, the LAWE may be written as,
Substituting these expressions into the LAWE gives,
 
<div align="center">
<div align="center">
<table border="0" cellpadding="5" align="center">
<table border="0" cellpadding="5" align="center">
Line 1,222: Line 1,238:
<tr>
<tr>
   <td align="right">
   <td align="right">
<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>~\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>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 1,228: Line 1,245:
   </td>
   </td>
   <td align="left">
   <td align="left">
<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 \, .
(n+1) \biggl[ \mathfrak{F}  
</math>
+ \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>
   </td>
   </td>
</tr>
</tr>
Line 1,236: Line 1,253:
</div>
</div>


Let's now adopt a power-series expression for the displacement function of the form,
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>. 
<div align="center">
<div align="center">
<table border="0" cellpadding="5" 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>
<tr>
   <td align="right">
   <td align="right">
<math>~x</math>
<math>~\xi^{-1}:</math>
  </td>
  <td align="center">
<math>~24a</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=</math>
<math>~0</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~\Rightarrow ~~~a=0</math>
1 + ar + br^2 + cr^3 + dr^4 + \cdots
</math>
   </td>
   </td>
</tr>
</tr>
Line 1,256: Line 1,280:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\Rightarrow ~~~ \frac{1}{r}\frac{dx}{dr}</math>
<math>~\xi^{0}:</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=</math>
<math>~(12b + 48b)</math>
  </td>
  <td align="center">
<math>~-(n+1)\mathfrak{F}</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~\Rightarrow ~~~b = - \frac{(n+1)\mathfrak{F}}{60}</math>
\frac{a}{r} + 2b + 3 cr + 4dr^2 + \cdots
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
and,
<div align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\frac{d^2x}{dr^2}</math>
<math>~\xi^{1}:</math>
  </td>
  <td align="center">
<math>~[36c+72c-2a(n+3)]</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=</math>
<math>~-a(n+1)\mathfrak{F}</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~\Rightarrow ~~~108c = 2a(n+3)-a(n+1)\mathfrak{F} \Rightarrow~~c=0</math>
2b + 6cr + 12dr^2 + \cdots
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
Substituting these expressions into the LAWE gives,
<div align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
<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>~\xi^{2}:</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~\approx</math>
<math>~[72d-2b+96d-4b(n+3)]</math>
   </td>
   </td>
   <td align="left">
   <td align="center">
<math>~
<math>~\biggl[-b(n+1)\mathfrak{F}-\frac{n(n+1)\alpha}{5}\biggr]</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) \, .
  </td>
</math>
  <td align="left">
<math>~\Rightarrow ~~~d = - (n+1)\biggl\{ \frac{n\alpha +\mathfrak{F}[(4n+14)-(n+1)\mathfrak{F} ]}{10080} \biggr\}</math>
   </td>
   </td>
</tr>
</tr>
</table>
</table>
</div>
</div>
Keeping terms only up through <math>~r^2</math> leads to the following simplification:
 
<div align="center">
In summary, the desired, approximate power-series expression for the polytropic displacement function is:
<div align="center" id="PolytropicDisplacement">
<table border="1" width="80%" cellpadding="8" align="center"><tr><td align="center">
<table border="0" cellpadding="5" align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~
<math>~x(\xi)</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>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~\approx</math>
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~
- \frac{\mathfrak{F} }{6}  \biggl( 1 + ar + br^2 \biggr)
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
- \frac{\alpha}{3} \biggl(\frac{r^2}{10} \biggr)
</math>
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</table>
</td></tr></table>
</div>
</div>
where,
 
===Displacement Function for Isothermal LAWE===
The [[User:Tohline/SSC/Stability/Isothermal#Taff_and_Van_Horn_.281974.29|LAWE for isothermal spheres]] may be written as,
<div align="center">
<div align="center">
<math>~\mathfrak{F} \equiv \frac{\sigma_c^2}{\gamma} - 2\alpha \, .</math>
<table border="0" cellpadding="5" align="center">
</div>
 
Finally, balancing terms of like powers on both sides of the equation leads us to conclude the following:
 
<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>
<tr>
   <td align="right">
   <td align="right">
<math>~r^{-1}:</math>
<math>~\frac{d^2 x}{dr^2} + \biggl[4 - r \biggl(\frac{dw }{dr}\biggr) \biggr] \frac{1}{r}\frac{dx}{dr}</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~4a</math>
<math>~=</math>
  </td>
  <td align="center">
<math>~0</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\Rightarrow ~~~a = 0 </math>
<math>~
- \biggl[ \frac{\sigma_c^2}{6\gamma}  - \frac{\alpha}{r} \biggl(\frac{dw }{dr}\biggr)\biggr]  x  \, ,
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
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 [[#Isothermal_Lane-Emden_Function|power-series expression for the isothermal Lane-Emden function]] is,
<div align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~r^{0}:</math>
<math>~w(r)
</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~2b + 8b</math>
<math>~=</math>
  </td>
  <td align="center">
<math>~- \frac{\mathfrak{F}}{6}</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\Rightarrow ~~~b = - \frac{\mathfrak{F}}{60}</math>
<math>~\frac{r^2}{6} - \frac{r^4}{120} + \frac{r^6}{1890} - \frac{61 r^8}{1,632,960} + \cdots \, .</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
Hence,
<div align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~r^{1}:</math>
<math>~\frac{dw}{dr}</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~6c + 12c - \frac{a}{3}</math>
<math>~\approx</math>
  </td>
  <td align="center">
<math>~-\frac{a\mathfrak{F}}{6}</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\Rightarrow ~~~c=0</math>
<math>~\frac{r}{3} - \frac{r^3}{30} + \frac{r^5}{315} \, .</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
Therefore, near the center of the configuration, the LAWE may be written as,
<div align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~r^{2}:</math>
<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>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~12d + 16d - \frac{2b}{3}</math>
<math>~\approx</math>
  </td>
  <td align="center">
<math>~-\frac{\mathfrak{F}b}{6} - \frac{\alpha}{30}</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\Rightarrow ~~~
<math>~
28d = \frac{1}{30}\biggl[ 5b (4- \mathfrak{F} ) - \alpha \biggr] ~
- \frac{1}{6} \biggl[ \frac{\sigma_c^2}{\gamma} - 2\alpha \biggl(1 - \frac{r^2}{10} + \frac{r^4}{105}\biggr) \biggr] x \, .
\Rightarrow~
d = \frac{1}{10080}\biggl[ \mathfrak{F}(\mathfrak{F} -4) - 12\alpha \biggr]
</math>
</math>
   </td>
   </td>
Line 1,413: Line 1,424:
</div>
</div>


In summary, the desired, approximate power-series expression for the isothermal displacement function is:
Let's now adopt a power-series expression for the displacement function of the form,
<div align="center" id="IsothermalDisplacement">
<div align="center">
<table border="1" width="80%" cellpadding="8" align="center"><tr><td align="center">
<table border="0" cellpadding="5" align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~x(r)</math>
<math>~x</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 1,427: Line 1,437:
   <td align="left">
   <td align="left">
<math>~
<math>~
1 - \frac{\mathfrak{F}}{60} r^2 + \frac{1}{10080}\biggl[ \mathfrak{F}(\mathfrak{F} -4) - 12\alpha \biggr] r^4 + \cdots
1 + ar + br^2 + cr^3 + dr^4 + \cdots
</math>
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</td></tr></table>
</div>
==Taylor Series (Type 1)==
First:
<div align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~f_0</math>
<math>~\Rightarrow ~~~ \frac{1}{r}\frac{dx}{dr}</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 1,450: Line 1,451:
   <td align="left">
   <td align="left">
<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)
\frac{a}{r} + 2b + 3 cr + 4dr^2 + \cdots
</math>
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
and,
<div align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>~\frac{d^2x}{dr^2}</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 1,464: Line 1,470:
   <td align="left">
   <td align="left">
<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)
2b + 6cr + 12dr^2 + \cdots
</math>
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
Substituting these expressions into the LAWE gives,
<div align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\Rightarrow~~~
<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>
- \frac{3^2}{2} (\Delta)^2 f^{''}_3
</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=</math>
<math>~\approx</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<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)
- \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>
</math>
   </td>
   </td>
Line 1,486: Line 1,495:
</table>
</table>
</div>
</div>
 
Keeping terms only up through <math>~r^2</math> leads to the following simplification:
Second:
<div align="center">
<div align="center">
<table border="0" cellpadding="5" align="center">
<table border="0" cellpadding="5" align="center">
Line 1,493: Line 1,501:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~f_1</math>
<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>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=</math>
<math>~\approx</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<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)
- \frac{\mathfrak{F} }{6} \biggl( 1 + ar + br^2 \biggr)  
- \frac{\alpha}{3} \biggl(\frac{r^2}{10} \biggr)  
</math>
</math>
   </td>
   </td>
</tr>
</table>
</div>
where,
<div align="center">
<math>~\mathfrak{F} \equiv \frac{\sigma_c^2}{\gamma} - 2\alpha \, .</math>
</div>
Finally, balancing terms of like powers on both sides of the equation leads us to conclude the following:
<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>


<tr>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>~r^{-1}:</math>
  </td>
  <td align="center">
<math>~4a</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=</math>
<math>~0</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~\Rightarrow ~~~a = 0 </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>
   </td>
   </td>
</tr>
</tr>
Line 1,521: Line 1,552:
<tr>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>~r^{0}:</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=</math>
<math>~2b + 8b</math>
  </td>
  <td align="center">
<math>~- \frac{\mathfrak{F}}{6}</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~\Rightarrow ~~~b = - \frac{\mathfrak{F}}{60}</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>
   </td>
   </td>
</tr>
</tr>
Line 1,536: Line 1,567:
<tr>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>~r^{1}:</math>
  </td>
  <td align="center">
<math>~6c + 12c - \frac{a}{3}</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
&nbsp;
<math>~-\frac{a\mathfrak{F}}{6}</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~\Rightarrow ~~~c=0</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>
   </td>
   </td>
</tr>
</tr>
Line 1,550: Line 1,582:
<tr>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>~r^{2}:</math>
  </td>
  <td align="center">
<math>~12d + 16d - \frac{2b}{3}</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=</math>
<math>~-\frac{\mathfrak{F}b}{6} - \frac{\alpha}{30}</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~\Rightarrow ~~~
\biggl[\frac{2^2}{3^2} \biggr] f_0
28d = \frac{1}{30}\biggl[ 5b (4- \mathfrak{F} ) - \alpha \biggr] ~
+ f_3 \biggl[1-\frac{2^2}{3^2} \biggr]  
\Rightarrow~
+ \biggl[ \frac{2^2}{3^2} (3\Delta) - 2(\Delta) \biggr] f_3^'  
d = \frac{1}{10080}\biggl[ \mathfrak{F}(\mathfrak{F} -4) - 12\alpha \biggr]
+ \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>
  </td>
   </td>
</tr>
</tr>
</table>
 
</div>
<tr>
 
   <td align="right">
In summary, the desired, approximate power-series expression for the isothermal displacement function is:
&nbsp;
<div align="center" id="IsothermalDisplacement">
<table border="1" width="80%" cellpadding="8" align="center"><tr><td align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~x(r)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
1 - \frac{\mathfrak{F}}{60} r^2 + \frac{1}{10080}\biggl[ \mathfrak{F}(\mathfrak{F} -4) - 12\alpha \biggr] r^4 + \cdots
</math>
  </td>
</tr>
</table>
</td></tr></table>
</div>
 
==Taylor Series (Hunter77)==
 
===First (Unsuccessful) Try===
 
First:
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~f_0</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<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>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<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>
  </td>
</tr>
 
<tr>
  <td align="right">
<math>~\Rightarrow~~~
- \frac{3^2}{2} (\Delta)^2 f^{''}_3
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<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>
  </td>
</tr>
</table>
</div>
 
Note that, replacing the <math>~(\Delta)^3 f_3^{'''}</math> term with the expression derived in the ''Second'' step, below, gives,
 
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~
- \frac{3^2}{2} (\Delta)^2 f^{''}_3
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
f_3 - f_0 - (3\Delta) f_3^' + \frac{3^3}{2^3}(\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<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>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
f_3 - f_0 - 3 (\Delta) f_3^' + \frac{3^3}{2^3}(\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<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>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<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>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<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>
  </td>
</tr>
</table>
</div>
 
Then, replacing the <math>~(\Delta)^4 f_3^{iv}</math> term with the expression derived in the ''Third'' step, below, gives,
 
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~
- \frac{3^2}{2} (\Delta)^2 f^{''}_3
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<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>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<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>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<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>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<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>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<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>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<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>
  </td>
</tr>
</table>
</div>
 
 
Second:
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~f_1</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<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>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<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>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<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>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<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>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<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>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<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>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<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>
  </td>
</tr>
 
<tr>
  <td align="right">
<math>~\Rightarrow~~~
- \biggl[ \frac{2}{3} \biggr] (\Delta)^3f_3^{'''}
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<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>
  </td>
</tr>
</table>
</div>
 
Now, replacing the <math>~(\Delta)^4 f_3^{iv}</math> term with the expression derived in the ''Third'' step, below, gives,
 
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~
- \biggl[ \frac{2}{3} \biggr] (\Delta)^3f_3^{'''}
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<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>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<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>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<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>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<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>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<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>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<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>
  </td>
</tr>
</table>
</div>
 
 
Third:
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~f_2</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<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>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<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>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<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>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<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>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<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>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<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>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<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>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<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>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<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>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<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>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<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>
  </td>
</tr>
 
<tr>
  <td align="right">
<math>~\Rightarrow~~~
- \biggl[\frac{1}{2^2\cdot 3} \biggr] (\Delta)^4 f_3^{iv}
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<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>
  </td>
</tr>
</table>
</div>
 
And, finally:
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~f_4</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<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>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
f_3 + (\Delta) f_3^' + \mathcal{O}(\Delta^5)
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<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>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<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>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<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>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
f_3 + (\Delta) f_3^' + \mathcal{O}(\Delta^5)
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ \biggl\{
\biggl[ \frac{1}{3^2}  \biggr] f_0
+ \biggl[- \frac{3}{2^2}\biggr] f_1
+\biggl[ 3 \biggr] f_2
+ \biggl[- \frac{5\cdot 17}{2^2\cdot 3^2} \biggr] f_3
+  \biggl[\frac{5}{2}\biggr] (\Delta) f_3^'
\biggr\}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ \biggl\{
\biggl[\frac{1}{2\cdot 3}  \biggr] f_0
+ \biggl[-1 \biggr] f_1
+ \biggl[ \frac{5}{2} \biggr] f_2
+ \biggl[- \frac{5}{3}\biggr] f_3
+  \biggl[\frac{11}{2\cdot 3}\biggr] (\Delta) f_3^'
\biggr\}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ \biggl\{
\biggl[\frac{1}{2\cdot 3^2}  \biggr]f_0
+ \biggl[- \frac{1}{2^2} \biggr] f_1
+ \biggl[ \frac{1}{2} \biggr] f_2
+ \biggl[ -\frac{11}{2^2 \cdot 3^2} \biggr] f_3
\biggl[\frac{1}{3}\biggr] (\Delta) f_3^'
\biggr\}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[ \frac{1}{3^2} + \frac{1}{2\cdot 3} + \frac{1}{2\cdot 3^2} \biggr] f_0 +
\biggl[- \frac{3}{2^2} - 1 - \frac{1}{2^2} \biggr] f_1
+\biggl[ 3 + \frac{5}{2} + \frac{1}{2} \biggr] f_2
+ \mathcal{O}(\Delta^5)
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ \biggl[1 - \frac{5\cdot 17}{2^2\cdot 3^2} - \frac{5}{3} - \frac{11}{2^2 \cdot 3^2} \biggr] f_3
+  \biggl[1 + \frac{5}{2} + \frac{11}{2\cdot 3} + \frac{1}{3} \biggr] (\Delta) f_3^'
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[ \frac{1}{3} \biggr] f_0 +
\biggl[- 2\biggr] f_1
+\biggl[ 6 \biggr] f_2
+ \biggl[- \frac{10}{3} \biggr] f_3  
\biggl[\frac{17}{3} \biggr] (\Delta) f_3^'
+ \mathcal{O}(\Delta^5)
</math>
  </td>
</tr>
</table>
</div>
 
Result:
 
<div align="center">
<table border="1" cellpadding="8" align="center">
<tr>
  <th align="center">
Definitely WRONG!
  </th>
</tr>
<tr><td>
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~f_4</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{1}{3} ~ f_0 - 2 f_1 + 6 f_2
- \frac{10}{3} ~ f_3
+  \frac{17}{3}  (\Delta) f_3^'
+ \mathcal{O}(\Delta^5) \, .
</math>
  </td>
</tr>
</table>
</td></tr>
</table>
</div>
 
When I used an Excel spreadsheet to test this out against a parabola, the integration quickly became wildly unstable, strongly suggesting that there is an error in the derivation.  My first attempt to uncover this error produced a new coefficient on the <math>~(\Delta) f_3^'</math>, namely,
 
<div align="center">
<table border="1" cellpadding="8" align="center">
<tr>
  <th align="center">
Somewhat Improved
  </th>
</tr>
<tr><td>
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~f_4</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{1}{3} ~ f_0 - 2 f_1 + 6 f_2
- \frac{10}{3} ~ f_3
+  4  (\Delta) f_3^'
+ \mathcal{O}(\Delta^5) \, .
</math>
  </td>
</tr>
</table>
</td></tr>
</table>
</div>
Although it showed improvement, this expression still blows up.  So I have not bothered to revise the original (definitely WRONG!) derivation.  Instead, let's start all over and approach it with a more gradual derivation.
 
===Second Try===
 
We will work from the following foundation expression in which <math>~f_4</math> is the variable that we desire to evaluate, and the "known" quantities are: &nbsp; <math>~f_3</math>, <math>~f_3^'</math>,  <math>~f_2</math>, <math>~f_1</math>, and <math>~f_0</math>.
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~f_4</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<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>
  </td>
</tr>
</table>
</div>
 
Let's use similar Taylor-series expansions for <math>~f_2</math>, <math>~f_3</math>, etc. in order to eliminate the <math>~f_3^{''}</math> term,  the <math>~f_3^{'''}</math> term, etc.
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~f_2</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<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>
  </td>
</tr>
 
<tr>
  <td align="right">
<math>~f_1</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<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>
  </td>
</tr>
 
<tr>
  <td align="right">
<math>~f_0</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<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>
  </td>
</tr>
</table>
</div>
 
----
 
First:
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~-\frac{1}{2} (- \Delta)^2 f^{''}_3 </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
f_3 + (- \Delta) f_3^' - f_2+ \frac{1}{6} (- \Delta)^3 f_3^{'''} + \frac{1}{24}(- \Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
<math>~\Rightarrow~~~
\frac{1}{2} (\Delta)^2 f^{''}_3
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
- f_3 + (\Delta) f_3^' + f_2+ \frac{1}{6} (\Delta)^3 f_3^{'''} - \frac{1}{24}(\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
<math>~\Rightarrow~~~
f_4
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
f_3 + (\Delta) f_3^' - f_3 + (\Delta) f_3^' + f_2 + \mathcal{O}(\Delta^3)
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
f_2 + 2(\Delta) f_3^' + \mathcal{O}(\Delta^3)
</math>
  </td>
</tr>
</table>
</div>
 
<div align="center">
<table border="1" cellpadding="8" align="center">
<tr><td align="center"><math>~\mathcal{O}(\Delta^3)</math></td></tr>
<tr><td align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~
f_4
</math>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
f_2 + 2(\Delta) f_3^' + \mathcal{O}(\Delta^3)
</math>
  </td>
</tr>
</table>
</td></tr></table>
</div>
 
This expression works very well for a parabola.
 
----
 
Second:
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~f_1</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
f_3 + (- 2) \Delta f_3^' + 2 (\Delta)^2 f^{''}_3 + \biggl[- \frac{2^3}{6}\biggr] \Delta^3 f_3^{'''} + \biggl[ \frac{2^4}{2^3\cdot 3} \biggr] \Delta^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
f_3 + (- 2) \Delta f_3^' + \biggl[- \frac{2^3}{6}\biggr]  \Delta^3 f_3^{'''} + \biggl[ \frac{2^4}{2^3\cdot 3} \biggr] \Delta^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ 2 \biggl\{
- f_3 + (\Delta) f_3^' + f_2+ \frac{1}{2\cdot 3} (\Delta)^3 f_3^{'''} - \frac{1}{2^3\cdot 3}(\Delta)^4 f_3^{iv} 
\biggr\} \biggl[ 2 \biggr]
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
f_3\biggl[ 1 - 2^2\biggr] + (2^2 - 2) \Delta f_3^' + 2^2f_2
+  \biggl[\frac{2}{3} - \frac{2^3}{6}\biggr]  \Delta^3 f_3^{'''}
+ \biggl[ \frac{2^4}{2^3\cdot 3} - \frac{1}{2\cdot 3}\biggr] \Delta^4 f_3^{iv}
+ \mathcal{O}(\Delta^5)
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
f_3\biggl[ -3\biggr] + (2) \Delta f_3^' + 2^2f_2
+ \biggl[- \frac{2}{3}\biggr]  \Delta^3 f_3^{'''}
+ \biggl[ \frac{1}{2} \biggr] \Delta^4 f_3^{iv}
+ \mathcal{O}(\Delta^5)
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
<math>~\Rightarrow~~~
\biggl[\frac{2}{3}\biggr]  \Delta^3 f_3^{'''}
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
- f_1 + 2^2f_2 -3 f_3 + 2 \Delta f_3^'
+ \biggl[ \frac{1}{2} \biggr] \Delta^4 f_3^{iv}
+ \mathcal{O}(\Delta^5)
</math>
  </td>
</tr>
</table>
</div>
 
This also allows us to improve the expression for the <math>~f_3^{''}</math> term, as initially derived in the "First" subsection, above.  Namely,
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~
\frac{1}{2} (\Delta)^2 f^{''}_3
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
f_2 - f_3 + (\Delta) f_3^' - \frac{1}{24}(\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ \frac{1}{6} \biggl\{
- f_1 + 2^2f_2 -3 f_3 + 2 \Delta f_3^'
+ \biggl[ \frac{1}{2} \biggr] \Delta^4 f_3^{iv}
\biggr\} \biggl[ \frac{3}{2} \biggr]
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
- \frac{1}{4} f_1 + 2f_2 + \biggl[ - \frac{7}{4} \biggr] f_3 + \biggl[ \frac{3}{2} \biggr] (\Delta) f_3^'
+ \biggl[\frac{1}{2^2\cdot 3} \biggr](\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
  </td>
</tr>
</table>
</div>
 
Hence, an improved expression for <math>~f_4</math> is,
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~f_4</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
f_3 + (\Delta) f_3^' + \mathcal{O}(\Delta^4)
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ \biggl\{
- \frac{1}{4} f_1 + 2f_2 + \biggl[ - \frac{7}{4} \biggr] f_3 + \biggl[ \frac{3}{2} \biggr] (\Delta) f_3^'
\biggr\}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ \frac{1}{6} \biggl\{
- f_1 + 2^2f_2 -3 f_3 + 2 \Delta f_3^'
\biggr\} \biggl[ \frac{3}{2} \biggr]
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
- \frac{1}{2} f_1 + 3f_2 - \frac{3}{2} f_3 + 3(\Delta) f_3^' + \mathcal{O}(\Delta^4)
</math>
  </td>
</tr>
</table>
</div>
 
 
<div align="center">
<table border="1" cellpadding="8" align="center">
<tr><td align="center"><math>~\mathcal{O}(\Delta^4)</math></td></tr>
<tr><td align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~
f_4
</math>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
- \frac{1}{2} f_1 + 3f_2 - \frac{3}{2} f_3 + 3(\Delta) f_3^' + \mathcal{O}(\Delta^4)
</math>
  </td>
</tr>
</table>
</td></tr></table>
</div>
 
 
----
 
Third:
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~f_0</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<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>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
f_3 + \biggl[ - 3 \biggr] (\Delta) f_3^' + \biggl[ \frac{3^3}{2^3} \biggr] (\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ 3^2 \biggl\{
- \frac{1}{4} f_1 + 2f_2 + \biggl[ - \frac{7}{4} \biggr] f_3 + \biggl[ \frac{3}{2} \biggr] (\Delta) f_3^'
+ \biggl[\frac{1}{2^2\cdot 3} \biggr](\Delta)^4 f_3^{iv}
\biggr\}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ \biggl[-\frac{3^3}{2^2} \biggr] \biggl\{
- f_1 + 2^2f_2 -3 f_3 + 2 \Delta f_3^'
+ \biggl[ \frac{1}{2} \biggr] \Delta^4 f_3^{iv}
\biggr\}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
f_3 + \biggl[ - 3 \biggr] (\Delta) f_3^'  + \biggl[ \frac{3^3}{2^3} \biggr] (\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ \biggl\{
\biggl[- \frac{3^2 }{4} \biggr]  f_1 + \biggl[ 2\cdot 3^2 \biggr]f_2 + \biggl[ - \frac{3^2 \cdot 7}{4} \biggr] f_3 + \biggl[ \frac{3^3}{2} \biggr] (\Delta) f_3^'
+ \biggl[\frac{3}{2^2} \biggr](\Delta)^4 f_3^{iv}
\biggr\}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+  \biggl\{
\biggl[\frac{3^3}{2^2} \biggr] f_1
+ \biggl[-3^3 \biggr] f_2
+ \biggl[\frac{3^4}{2^2} \biggr]f_3
+ \biggl[- \frac{3^3}{2} \biggr] \Delta f_3^'
+ \biggl[- \frac{3^3}{2^3} \biggr] \Delta^4 f_3^{iv}
\biggr\}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[\frac{3^3}{2^2} - \frac{3^2 }{4} \biggr]  f_1 + \biggl[ 2\cdot 3^2 -3^3\biggr]f_2 + \biggl[ 1+ \frac{3^4}{2^2} - \frac{3^2 \cdot 7}{4} \biggr] f_3
+ \biggl[ \frac{3^3}{2} - \frac{3^3}{2} -3\biggr] (\Delta) f_3^'
+ \biggl[\frac{3^3}{2^3}  + \frac{3}{2^2} - \frac{3^3}{2^3}\biggr](\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[\frac{3^2}{2} \biggr]  f_1 + \biggl[ - 3^2 \biggr]f_2 + \biggl[ \frac{11}{2} \biggr] f_3
+ \biggl[ -3\biggr] (\Delta) f_3^'
+ \biggl[\frac{3}{2^2} \biggr](\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
<math>~\Rightarrow ~~~
- \biggl[\frac{3}{2^2} \biggr](\Delta)^4 f_3^{iv}
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
- f_0 + \biggl[\frac{3^2}{2} \biggr]  f_1 + \biggl[ - 3^2 \biggr]f_2 + \biggl[ \frac{11}{2} \biggr] f_3
+ \biggl[ -3\biggr] (\Delta) f_3^'
+ \mathcal{O}(\Delta^5)
</math>
  </td>
</tr>
</table>
</div>
 
Hence,
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~
\frac{1}{2} (\Delta)^2 f^{''}_3
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
- \frac{1}{4} f_1 + 2f_2 + \biggl[ - \frac{7}{4} \biggr] f_3 + \biggl[ \frac{3}{2} \biggr] (\Delta) f_3^'
+ \mathcal{O}(\Delta^5)
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ \biggl[\frac{1}{2^2\cdot 3} \biggr]\biggl\{
- f_0 + \biggl[\frac{3^2}{2}\biggr]  f_1 + \biggl[ - 3^2 \biggr]f_2 + \biggl[ \frac{11}{2} \biggr] f_3
+ \biggl[ -3\biggr] (\Delta) f_3^'
\biggr\} \biggl[ - \frac{2^2}{3}  \biggr]
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
- \frac{1}{4} f_1 + 2f_2 + \biggl[ - \frac{7}{4} \biggr] f_3 + \biggl[ \frac{3}{2} \biggr] (\Delta) f_3^'
+ \mathcal{O}(\Delta^5)
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ \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{1}{3} \biggr] (\Delta) f_3^'
\biggr\}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[\frac{1}{3^2} \biggr] f_0 + \biggl[- \frac{1}{2} - \frac{1}{4} \biggr] f_1
+ 3 f_2 + \biggl[- \frac{11}{2 \cdot 3^2} - \frac{7}{4} \biggr] f_3
+ \biggl[\frac{1}{3} +  \frac{3}{2} \biggr] (\Delta) f_3^'
+ \mathcal{O}(\Delta^5)
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[\frac{1}{3^2} \biggr] f_0 + \biggl[- \frac{3}{4} \biggr] f_1
+ 3 f_2 + \biggl[- \frac{5\cdot 17}{2^2\cdot 3^2} \biggr] f_3
+ \biggl[\frac{11}{2\cdot 3} \biggr] (\Delta) f_3^'
+ \mathcal{O}(\Delta^5)
</math>
  </td>
</tr>
</table>
</div>
 
And,
 
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~
\biggl[\frac{2}{3}\biggr]  \Delta^3 f_3^{'''}
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
- f_1 + 2^2f_2 -3 f_3 + 2 \Delta f_3^'
+ \mathcal{O}(\Delta^5)
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ \biggl[ \frac{1}{2} \biggr] \biggl\{
- f_0 + \biggl[\frac{3^2}{2} \biggr]  f_1 + \biggl[ - 3^2 \biggr]f_2 + \biggl[ \frac{11}{2} \biggr] f_3
+ \biggl[ -3\biggr] (\Delta) f_3^'
\biggr\} \biggl[ - \frac{2^2}{3}  \biggr]
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
- f_1 + 2^2f_2 -3 f_3 + 2 \Delta f_3^'
+ \mathcal{O}(\Delta^5)
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ \biggl\{
\biggl[ \frac{2}{3}  \biggr] f_0 + \biggl[- 3 \biggr]  f_1
+ \biggl[ 2\cdot 3 \biggr] f_2 + \biggl[- \frac{11}{3}  \biggr] f_3
+ \biggl[ 2 \biggr] (\Delta) f_3^'
\biggr\}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<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[ 4 \biggr] (\Delta) f_3^'
+ \mathcal{O}(\Delta^5)
</math>
  </td>
</tr>
</table>
</div>
 
----
 
Finally, then:
 
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~f_4</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<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>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
f_3 + (\Delta) f_3^' + \mathcal{O}(\Delta^5)
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ \frac{1}{2} \biggl\{
\biggl[\frac{1}{3^2} \biggr] f_0 + \biggl[- \frac{3}{4} \biggr] f_1
+ 3 f_2 + \biggl[- \frac{5\cdot 17}{2^2\cdot 3^2} \biggr] f_3
+ \biggl[\frac{11}{2\cdot 3} \biggr] (\Delta) f_3^'
\biggr\}\biggl[ 2 \biggr]
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+  \frac{1}{2\cdot 3} \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[ 4 \biggr] (\Delta) f_3^'
\biggr\}\biggl[ \frac{3}{2} \biggr]
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ \frac{1}{2^3\cdot 3} \biggl\{
- f_0 + \biggl[\frac{3^2}{2} \biggr]  f_1 + \biggl[ - 3^2 \biggr]f_2 + \biggl[ \frac{11}{2} \biggr] f_3
+ \biggl[ -3\biggr] (\Delta) f_3^'
\biggr\}\biggl[- \frac{2^2}{3}  \biggr]
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
f_3 + (\Delta) f_3^' + \mathcal{O}(\Delta^5)
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ \biggl\{
\biggl[\frac{1}{3^2} \biggr] f_0 + \biggl[- \frac{3}{4} \biggr] f_1
+ 3 f_2 + \biggl[- \frac{5\cdot 17}{2^2\cdot 3^2} \biggr] f_3
+ \biggl[\frac{11}{2\cdot 3} \biggr] (\Delta) f_3^'  
\biggr\}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+  \frac{1}{2^2} \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[ 4 \biggr] (\Delta) f_3^'  
\biggr\}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ \frac{1}{2\cdot 3^2} \biggl\{
f_0 + \biggl[ - \frac{3^2}{2} \biggr]  f_1 + \biggl[ 3^2 \biggr]f_2 + \biggl[- \frac{11}{2} \biggr] f_3
+ \biggl[ 3\biggr] (\Delta) f_3^'
\biggr\}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
f_3 + (\Delta) f_3^' + \mathcal{O}(\Delta^5)
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ \biggl\{
\biggl[\frac{1}{3^2} \biggr] f_0 + \biggl[- \frac{3}{4} \biggr] f_1
+ 3 f_2 + \biggl[- \frac{5\cdot 17}{2^2\cdot 3^2} \biggr] f_3
+ \biggl[\frac{11}{2\cdot 3} \biggr] (\Delta) f_3^'
\biggr\}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+  \biggl\{
\biggl[ \frac{1}{2\cdot 3}  \biggr] f_0 + \biggl[- 1 \biggr]  f_1
+ \biggl[ \frac{5}{2} \biggr] f_2 + \biggl[- \frac{5}{3}  \biggr] f_3
+ \biggl[ 1 \biggr] (\Delta) f_3^'
\biggr\}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
\biggl\{
\biggl[ \frac{1}{2\cdot 3^2} \biggr] f_0 + \biggl[ - \frac{1}{2^2} \biggr]  f_1 + \biggl[ \frac{1}{2} \biggr]f_2 + \biggl[- \frac{11}{2^2\cdot 3^2} \biggr] f_3
+ \biggl[ \frac{1}{2\cdot 3} \biggr] (\Delta) f_3^'
\biggr\}
</math>
   </td>
</tr>
 
<tr>
   <td align="right">
&nbsp;
   </td>
   </td>
   <td align="center">
   <td align="center">
&nbsp;
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<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}
\biggl[\frac{1}{3^2} + \frac{1}{2\cdot 3} + \frac{1}{2\cdot 3^2} \biggr] f_0
+ \biggl[- 1  - \frac{3}{4} - \frac{1}{2^2} \biggr] f_1
+ \biggl[ 3 + \frac{5}{2} + \frac{1}{2} \biggr] f_2
+ \biggl[1 - \frac{5\cdot 17}{2^2\cdot 3^2} - \frac{5}{3} - \frac{11}{2^2\cdot 3^2} \biggr] f_3
+ \biggl[2 + \frac{11}{2\cdot 3} + \frac{1}{2\cdot 3} \biggr] (\Delta) f_3^'
+ \mathcal{O}(\Delta^5)
+ \mathcal{O}(\Delta^5)
</math>
</math>
Line 1,589: Line 3,691:
   <td align="left">
   <td align="left">
<math>~
<math>~
\biggl[\frac{2^2}{3^2} \biggr] f_0
\biggl[\frac{1}{3} \biggr] f_0  
+ f_3 \biggl[\frac{5}{3^2} \biggr]  
+ \biggl[- 2\biggr] f_1
+ \biggl[- \frac{2}{3}\biggr] (\Delta) f_3^'
+ \biggl[ 6 \biggr] f_2
+ \biggl[ \frac{2}{3} \biggr] (\Delta)^3f_3^{'''}
+ \biggl[- \frac{2\cdot 5}{3} \biggr] f_3
+ \biggl[- \frac{5}{6} \biggr] (\Delta)^4 f_3^{iv}
+ \biggl[4 \biggr] (\Delta) f_3^'
+ \mathcal{O}(\Delta^5)
+ \mathcal{O}(\Delta^5)
</math>
</math>
Line 1,601: Line 3,703:
</div>
</div>


Third:
 
<div align="center">
<div align="center">
<table border="1" cellpadding="8" align="center">
<tr><td align="center"><math>~\mathcal{O}(\Delta^5)</math></td></tr>
<tr><td align="center">
<table border="0" cellpadding="5" align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~f_2</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<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)
f_4
</math>
</math>
  </td>
</tr>
</table>
</div>
And, finally:
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~f_4</math>
  </td>
   <td align="center">
   <td align="center">
<math>~=</math>
<math>~=</math>
Line 1,634: Line 3,720:
   <td align="left">
   <td align="left">
<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)
\frac{1}{3} f_0 - 2 f_1 + 6 f_2 - \frac{2\cdot 5}{3} f_3 + (\Delta) f_3^' + \mathcal{O}(\Delta^5)
</math>
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</table>
</td></tr></table>
</div>
</div>




{{LSU_HBook_footer}}
{{LSU_HBook_footer}}

Latest revision as of 20:29, 11 July 2020

Approximate Power-Series Expressions

Whitworth's (1981) Isothermal Free-Energy Surface
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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>

LaTeX mathematical expressions cut-and-pasted directly from
NIST's Digital Library of Mathematical Functions

As a primary point of reference, note that according to §1.2 of NIST's Digital Library of Mathematical Functions, the binomial theorem states that,

<math>~(a+b)^{n}</math>

<math>~=</math>

<math>~ a^{n}+\binom{n}{1}a^{n-1}b+\binom{n}{2}a^{n-2}b^{2}+\dots+\binom{n}{n-1}ab^{n-1}+b^{n}, </math>

where, for nonnegative integer values of <math>~k</math> and <math>~n</math> and <math>~k \le n</math>, the notation,

<math>~\binom{n}{k}</math>

<math>~=</math>

<math>~ \frac{n!}{(n-k)!k!}=\binom{n}{n-k}. </math>


Our Example:  Setting <math>~a = 1</math> gives,

<math>~(1+b)^{n}</math>

<math>~=</math>

<math>~ 1+\binom{n}{1}b+\binom{n}{2}b^{2}+\binom{n}{3}b^{3}+\binom{n}{4}b^{4}+\dots </math>

 

<math>~=</math>

<math>~ 1+\frac{n!}{(n-1)!}~b + \frac{n!}{(n-2)! 2!}~b^{2} + \frac{n!}{(n-3)! 3!}~b^{3} + \frac{n!}{(n-4)! 4!}~b^{4} + \dots </math>

 

<math>~=</math>

<math>~ 1+ nb + \biggl[ \frac{n(n-1)}{2!}\biggr] b^{2} + \biggl[ \frac{n(n-1)(n-2)}{3!} \biggr] b^{3} + \biggl[ \frac{n(n-1)(n-2)(n-3)}{4!} \biggr] b^{4} + \dots </math>


Note, for example, that,

<math>~(1+x)^{-1}</math>

<math>~=</math>

<math>~1 - x +x^2 - x^3 + x^4 - x^5 + \cdots \, ;</math>

<math>~(1+x)^{-2}</math>

<math>~=</math>

<math>~1 - 2x + 3x^2 - 4x^3 + 5x^4 - 6x^5 + \cdots \, ;</math>

<math>~(1+x)^{-3}</math>

<math>~=</math>

<math>~1 - 3x + \biggl[ \frac{3\cdot 4 }{ 2} \biggr]x^2 - \biggl[ \frac{ 3\cdot 4 \cdot 5}{ 2\cdot 3} \biggr]x^3 + \biggl[ \frac{3\cdot 4 \cdot 5 \cdot 6 }{2\cdot 3 \cdot 4 } \biggr]x^4 - \biggl[ \frac{3\cdot 4 \cdot 5 \cdot 6 \cdot 7}{2\cdot 3 \cdot 4 \cdot 5 } \biggr]x^5 + \cdots </math>

 

<math>~=</math>

<math>~1 - 3x + 6x^2 - 10x^3 + 15x^4 - 21x^5 + \cdots \, ;</math>

<math>~(1+x)^{-4}</math>

<math>~=</math>

<math>~1 - 4x + \biggl[ \frac{4\cdot 5 }{ 2} \biggr]x^2 - \biggl[ \frac{ 4\cdot 5 \cdot 6}{ 2\cdot 3} \biggr]x^3 + \biggl[ \frac{4\cdot 5 \cdot 6 \cdot 7 }{2\cdot 3 \cdot 4 } \biggr]x^4 - \biggl[ \frac{4\cdot 5 \cdot 6 \cdot 7 \cdot 8}{2\cdot 3 \cdot 4 \cdot 5 } \biggr]x^5 + \cdots </math>

 

<math>~=</math>

<math>~1 - 4x + 10 x^2 - 20x^3 + 35x^4 - 56x^5 + \cdots \, .</math>


See also:

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,

LSU Key.png

<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:

For Spherically Symmetric Configurations

<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>

NOTE:  For cylindrically symmetric, rather than spherically symmetric, configurations, the analogous power-series expression appears as equation (15) in the article by J. P. Ostriker (1964, ApJ, 140, 1056) titled, The Equilibrium of Polytropic and Isothermal Cylinders.

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:

For Spherically Symmetric Configurations

<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).


NOTE:  For cylindrically symmetric, rather than spherically symmetric, configurations, an analytic expression for the function, <math>~w(r)</math>, is presented as equation (56) in a paper by J. P. Ostriker (1964, ApJ, 140, 1056) titled, The Equilibrium of Polytropic and Isothermal Cylinders.

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 (Hunter77)

First (Unsuccessful) Try

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>

 

<math>~=</math>

<math>~ f_3 + (\Delta) f_3^' + \mathcal{O}(\Delta^5) </math>

 

 

<math>~ + \biggl\{ \biggl[ \frac{1}{3^2} \biggr] f_0 + \biggl[- \frac{3}{2^2}\biggr] f_1 +\biggl[ 3 \biggr] f_2 + \biggl[- \frac{5\cdot 17}{2^2\cdot 3^2} \biggr] f_3 + \biggl[\frac{5}{2}\biggr] (\Delta) f_3^' \biggr\} </math>

 

 

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

 

 

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

 

<math>~=</math>

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

 

 

<math>~ + \biggl[1 - \frac{5\cdot 17}{2^2\cdot 3^2} - \frac{5}{3} - \frac{11}{2^2 \cdot 3^2} \biggr] f_3 + \biggl[1 + \frac{5}{2} + \frac{11}{2\cdot 3} + \frac{1}{3} \biggr] (\Delta) f_3^' </math>

 

<math>~=</math>

<math>~ \biggl[ \frac{1}{3} \biggr] f_0 + \biggl[- 2\biggr] f_1 +\biggl[ 6 \biggr] f_2 + \biggl[- \frac{10}{3} \biggr] f_3 + \biggl[\frac{17}{3} \biggr] (\Delta) f_3^' + \mathcal{O}(\Delta^5) </math>

Result:

Definitely WRONG!

<math>~f_4</math>

<math>~=</math>

<math>~ \frac{1}{3} ~ f_0 - 2 f_1 + 6 f_2 - \frac{10}{3} ~ f_3 + \frac{17}{3} (\Delta) f_3^' + \mathcal{O}(\Delta^5) \, . </math>

When I used an Excel spreadsheet to test this out against a parabola, the integration quickly became wildly unstable, strongly suggesting that there is an error in the derivation. My first attempt to uncover this error produced a new coefficient on the <math>~(\Delta) f_3^'</math>, namely,

Somewhat Improved

<math>~f_4</math>

<math>~=</math>

<math>~ \frac{1}{3} ~ f_0 - 2 f_1 + 6 f_2 - \frac{10}{3} ~ f_3 + 4 (\Delta) f_3^' + \mathcal{O}(\Delta^5) \, . </math>

Although it showed improvement, this expression still blows up. So I have not bothered to revise the original (definitely WRONG!) derivation. Instead, let's start all over and approach it with a more gradual derivation.

Second Try

We will work from the following foundation expression in which <math>~f_4</math> is the variable that we desire to evaluate, and the "known" quantities are:   <math>~f_3</math>, <math>~f_3^'</math>, <math>~f_2</math>, <math>~f_1</math>, and <math>~f_0</math>.

<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>

Let's use similar Taylor-series expansions for <math>~f_2</math>, <math>~f_3</math>, etc. in order to eliminate the <math>~f_3^{}</math> term, the <math>~f_3^{'}</math> term, etc.

<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>~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>~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>


First:

<math>~-\frac{1}{2} (- \Delta)^2 f^{}_3 </math>

<math>~=</math>

<math>~ f_3 + (- \Delta) f_3^' - f_2+ \frac{1}{6} (- \Delta)^3 f_3^{} + \frac{1}{24}(- \Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5) </math>

<math>~\Rightarrow~~~ \frac{1}{2} (\Delta)^2 f^{}_3 </math>

<math>~=</math>

<math>~ - f_3 + (\Delta) f_3^' + f_2+ \frac{1}{6} (\Delta)^3 f_3^{} - \frac{1}{24}(\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5) </math>

<math>~\Rightarrow~~~ f_4 </math>

<math>~=</math>

<math>~ f_3 + (\Delta) f_3^' - f_3 + (\Delta) f_3^' + f_2 + \mathcal{O}(\Delta^3) </math>

 

<math>~=</math>

<math>~ f_2 + 2(\Delta) f_3^' + \mathcal{O}(\Delta^3) </math>

<math>~\mathcal{O}(\Delta^3)</math>

<math>~ f_4 </math>

<math>~=</math>

<math>~ f_2 + 2(\Delta) f_3^' + \mathcal{O}(\Delta^3) </math>

This expression works very well for a parabola.


Second:

<math>~f_1</math>

<math>~=</math>

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

 

<math>~=</math>

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

 

 

<math>~ + 2 \biggl\{ - f_3 + (\Delta) f_3^' + f_2+ \frac{1}{2\cdot 3} (\Delta)^3 f_3^{} - \frac{1}{2^3\cdot 3}(\Delta)^4 f_3^{iv} \biggr\} \biggl[ 2 \biggr] </math>

 

<math>~=</math>

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

 

<math>~=</math>

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

<math>~\Rightarrow~~~ \biggl[\frac{2}{3}\biggr] \Delta^3 f_3^{} </math>

<math>~=</math>

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

This also allows us to improve the expression for the <math>~f_3^{}</math> term, as initially derived in the "First" subsection, above. Namely,

<math>~ \frac{1}{2} (\Delta)^2 f^{}_3 </math>

<math>~=</math>

<math>~ f_2 - f_3 + (\Delta) f_3^' - \frac{1}{24}(\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5) </math>

 

 

<math>~ + \frac{1}{6} \biggl\{ - f_1 + 2^2f_2 -3 f_3 + 2 \Delta f_3^' + \biggl[ \frac{1}{2} \biggr] \Delta^4 f_3^{iv} \biggr\} \biggl[ \frac{3}{2} \biggr] </math>

 

<math>~=</math>

<math>~ - \frac{1}{4} f_1 + 2f_2 + \biggl[ - \frac{7}{4} \biggr] f_3 + \biggl[ \frac{3}{2} \biggr] (\Delta) f_3^' + \biggl[\frac{1}{2^2\cdot 3} \biggr](\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5) </math>

Hence, an improved expression for <math>~f_4</math> is,

<math>~f_4</math>

<math>~=</math>

<math>~ f_3 + (\Delta) f_3^' + \mathcal{O}(\Delta^4) </math>

 

 

<math>~ + \biggl\{ - \frac{1}{4} f_1 + 2f_2 + \biggl[ - \frac{7}{4} \biggr] f_3 + \biggl[ \frac{3}{2} \biggr] (\Delta) f_3^' \biggr\} </math>

 

 

<math>~ + \frac{1}{6} \biggl\{ - f_1 + 2^2f_2 -3 f_3 + 2 \Delta f_3^' \biggr\} \biggl[ \frac{3}{2} \biggr] </math>

 

<math>~=</math>

<math>~ - \frac{1}{2} f_1 + 3f_2 - \frac{3}{2} f_3 + 3(\Delta) f_3^' + \mathcal{O}(\Delta^4) </math>


<math>~\mathcal{O}(\Delta^4)</math>

<math>~ f_4 </math>

<math>~=</math>

<math>~ - \frac{1}{2} f_1 + 3f_2 - \frac{3}{2} f_3 + 3(\Delta) f_3^' + \mathcal{O}(\Delta^4) </math>



Third:

<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 + \biggl[ - 3 \biggr] (\Delta) f_3^' + \biggl[ \frac{3^3}{2^3} \biggr] (\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5) </math>

 

 

<math>~ + 3^2 \biggl\{ - \frac{1}{4} f_1 + 2f_2 + \biggl[ - \frac{7}{4} \biggr] f_3 + \biggl[ \frac{3}{2} \biggr] (\Delta) f_3^' + \biggl[\frac{1}{2^2\cdot 3} \biggr](\Delta)^4 f_3^{iv} \biggr\} </math>

 

 

<math>~ + \biggl[-\frac{3^3}{2^2} \biggr] \biggl\{ - f_1 + 2^2f_2 -3 f_3 + 2 \Delta f_3^' + \biggl[ \frac{1}{2} \biggr] \Delta^4 f_3^{iv} \biggr\} </math>

 

<math>~=</math>

<math>~ f_3 + \biggl[ - 3 \biggr] (\Delta) f_3^' + \biggl[ \frac{3^3}{2^3} \biggr] (\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5) </math>

 

 

<math>~ + \biggl\{ \biggl[- \frac{3^2 }{4} \biggr] f_1 + \biggl[ 2\cdot 3^2 \biggr]f_2 + \biggl[ - \frac{3^2 \cdot 7}{4} \biggr] f_3 + \biggl[ \frac{3^3}{2} \biggr] (\Delta) f_3^' + \biggl[\frac{3}{2^2} \biggr](\Delta)^4 f_3^{iv} \biggr\} </math>

 

 

<math>~ + \biggl\{ \biggl[\frac{3^3}{2^2} \biggr] f_1 + \biggl[-3^3 \biggr] f_2 + \biggl[\frac{3^4}{2^2} \biggr]f_3 + \biggl[- \frac{3^3}{2} \biggr] \Delta f_3^' + \biggl[- \frac{3^3}{2^3} \biggr] \Delta^4 f_3^{iv} \biggr\} </math>

 

<math>~=</math>

<math>~ \biggl[\frac{3^3}{2^2} - \frac{3^2 }{4} \biggr] f_1 + \biggl[ 2\cdot 3^2 -3^3\biggr]f_2 + \biggl[ 1+ \frac{3^4}{2^2} - \frac{3^2 \cdot 7}{4} \biggr] f_3 + \biggl[ \frac{3^3}{2} - \frac{3^3}{2} -3\biggr] (\Delta) f_3^' + \biggl[\frac{3^3}{2^3} + \frac{3}{2^2} - \frac{3^3}{2^3}\biggr](\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5) </math>

 

<math>~=</math>

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

<math>~\Rightarrow ~~~ - \biggl[\frac{3}{2^2} \biggr](\Delta)^4 f_3^{iv} </math>

<math>~=</math>

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

Hence,

<math>~ \frac{1}{2} (\Delta)^2 f^{}_3 </math>

<math>~=</math>

<math>~ - \frac{1}{4} f_1 + 2f_2 + \biggl[ - \frac{7}{4} \biggr] f_3 + \biggl[ \frac{3}{2} \biggr] (\Delta) f_3^' + \mathcal{O}(\Delta^5) </math>

 

 

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

 

<math>~=</math>

<math>~ - \frac{1}{4} f_1 + 2f_2 + \biggl[ - \frac{7}{4} \biggr] f_3 + \biggl[ \frac{3}{2} \biggr] (\Delta) f_3^' + \mathcal{O}(\Delta^5) </math>

 

 

<math>~ + \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{1}{3} \biggr] (\Delta) f_3^' \biggr\} </math>

 

<math>~=</math>

<math>~ \biggl[\frac{1}{3^2} \biggr] f_0 + \biggl[- \frac{1}{2} - \frac{1}{4} \biggr] f_1 + 3 f_2 + \biggl[- \frac{11}{2 \cdot 3^2} - \frac{7}{4} \biggr] f_3 + \biggl[\frac{1}{3} + \frac{3}{2} \biggr] (\Delta) f_3^' + \mathcal{O}(\Delta^5) </math>

 

<math>~=</math>

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

And,

<math>~ \biggl[\frac{2}{3}\biggr] \Delta^3 f_3^{} </math>

<math>~=</math>

<math>~ - f_1 + 2^2f_2 -3 f_3 + 2 \Delta f_3^' + \mathcal{O}(\Delta^5) </math>

 

 

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

 

<math>~=</math>

<math>~ - f_1 + 2^2f_2 -3 f_3 + 2 \Delta f_3^' + \mathcal{O}(\Delta^5) </math>

 

 

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


Finally, then:

<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\{ \biggl[\frac{1}{3^2} \biggr] f_0 + \biggl[- \frac{3}{4} \biggr] f_1 + 3 f_2 + \biggl[- \frac{5\cdot 17}{2^2\cdot 3^2} \biggr] f_3 + \biggl[\frac{11}{2\cdot 3} \biggr] (\Delta) f_3^' \biggr\}\biggl[ 2 \biggr] </math>

 

 

<math>~ + \frac{1}{2\cdot 3} \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[ 4 \biggr] (\Delta) f_3^' \biggr\}\biggl[ \frac{3}{2} \biggr] </math>

 

 

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

 

<math>~=</math>

<math>~ f_3 + (\Delta) f_3^' + \mathcal{O}(\Delta^5) </math>

 

 

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

 

 

<math>~ + \frac{1}{2^2} \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[ 4 \biggr] (\Delta) f_3^' \biggr\} </math>

 

 

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

 

<math>~=</math>

<math>~ f_3 + (\Delta) f_3^' + \mathcal{O}(\Delta^5) </math>

 

 

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

 

 

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

 

 

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

 

<math>~=</math>

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

 

<math>~=</math>

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


<math>~\mathcal{O}(\Delta^5)</math>

<math>~ f_4 </math>

<math>~=</math>

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


Whitworth's (1981) Isothermal Free-Energy Surface

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Recommended citation:   Tohline, Joel E. (2021), The Structure, Stability, & Dynamics of Self-Gravitating Fluids, a (MediaWiki-based) Vistrails.org publication, https://www.vistrails.org/index.php/User:Tohline/citation