# User:Tohline/Appendix/Ramblings/PowerSeriesExpressions

(diff) ← Older revision | Current revision (diff) | Newer revision → (diff)

# Approximate Power-Series Expressions

## Broadly Used Mathematical Expressions (shown here without proof)

### Binomial

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

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,

 $~(a+b)^{n}$ $~=$ $~ a^{n}+\binom{n}{1}a^{n-1}b+\binom{n}{2}a^{n-2}b^{2}+\dots+\binom{n}{n-1}ab^{n-1}+b^{n},$

where, for nonnegative integer values of $~k$ and $~n$ and $~k \le n$, the notation,

 $~\binom{n}{k}$ $~=$ $~ \frac{n!}{(n-k)!k!}=\binom{n}{n-k}.$

Our Example:  Setting $~a = 1$ gives,

 $~(1+b)^{n}$ $~=$ $~ 1+\binom{n}{1}b+\binom{n}{2}b^{2}+\binom{n}{3}b^{3}+\binom{n}{4}b^{4}+\dots$ $~=$ $~ 1+\frac{n!}{(n-1)!}~b + \frac{n!}{(n-2)! 2!}~b^{2} + \frac{n!}{(n-3)! 3!}~b^{3} + \frac{n!}{(n-4)! 4!}~b^{4} + \dots$ $~=$ $~ 1+ nb + \biggl[ \frac{n(n-1)}{2!}\biggr] b^{2} + \biggl[ \frac{n(n-1)(n-2)}{3!} \biggr] b^{3} + \biggl[ \frac{n(n-1)(n-2)(n-3)}{4!} \biggr] b^{4} + \dots$

Note, for example, that,

 $~(1+x)^{-1}$ $~=$ $~1 - x +x^2 - x^3 + x^4 - x^5 + \cdots \, ;$ $~(1+x)^{-2}$ $~=$ $~1 - 2x + 3x^2 - 4x^3 + 5x^4 - 6x^5 + \cdots \, ;$ $~(1+x)^{-3}$ $~=$ $~1 - 3x + \biggl[ \frac{3\cdot 4 }{ 2} \biggr]x^2 - \biggl[ \frac{ 3\cdot 4 \cdot 5}{ 2\cdot 3} \biggr]x^3 + \biggl[ \frac{3\cdot 4 \cdot 5 \cdot 6 }{2\cdot 3 \cdot 4 } \biggr]x^4 - \biggl[ \frac{3\cdot 4 \cdot 5 \cdot 6 \cdot 7}{2\cdot 3 \cdot 4 \cdot 5 } \biggr]x^5 + \cdots$ $~=$ $~1 - 3x + 6x^2 - 10x^3 + 15x^4 - 21x^5 + \cdots \, .$

### Exponential

 $~e^x$ $~=$ $~ 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \cdots$

## Expressions with Astrophysical Relevance

### Polytropic Lane-Emden Function

We seek a power-series expression for the polytropic, Lane-Emden function, $~\Theta_\mathrm{H}(\xi)$ — expanded about the coordinate center, $~\xi = 0$ — that approximately satisfies the Lane-Emden equation,

 $~\frac{1}{\xi^2} \frac{d}{d\xi}\biggl( \xi^2 \frac{d\Theta_H}{d\xi} \biggr) = - \Theta_H^n$

A general power-series should be of the form,

 $~\Theta_H$ $~=$ $~ \theta_0 + a\xi + b\xi^2 + c\xi^3 + d\xi^4 + e\xi^5 + f\xi^6 + g\xi^7 + h\xi^8 + \cdots$

First derivative:

 $~\frac{d\Theta_H}{d\xi}$ $~=$ $~ a + 2b\xi + 3c\xi^2 + 4d\xi^3 + 5e\xi^4 + 6f\xi^5 + 7g\xi^6 + 8h\xi^7 + \cdots$

Left-hand-side of Lane-Emden equation:

 $~\frac{1}{\xi^2} \frac{d}{d\xi}\biggl( \xi^2 \frac{d\Theta_H}{d\xi} \biggr)$ $~=$ $~ \frac{2a}{\xi} + 2\cdot 3b + 2^2\cdot 3c\xi + 2^2\cdot 5d\xi^2 + 2\cdot 3\cdot 5e\xi^3 + 2\cdot 3\cdot 7f\xi^4 + 2^3\cdot 7g\xi^5 + 2^3\cdot 3^2h\xi^6 + \cdots$

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

 $~\Theta_H^n$ $~=$ $~ 1 ~+ ~nF + \biggl[\frac{n(n-1)}{2!}\biggr]F^2 ~+~ \biggl[\frac{n(n-1)(n-2)}{3!}\biggr]F^3 + \biggl[\frac{n(n-1)(n-2)(n-3)}{4!}\biggr]F^4 ~~+ ~~ \cdots$

where,

 $~F$ $~\equiv$ $~ a\xi + b\xi^2 + c\xi^3 + d\xi^4 + e\xi^5 + f\xi^6 + g\xi^7 + h\xi^8 + \cdots$ $~=$ $~ a\xi\biggl[1 + \frac{b}{a}\xi + \frac{c}{a}\xi^2 + \frac{d}{a}\xi^3 + \frac{e}{a}\xi^4 + \frac{f}{a}\xi^5 + \frac{g}{a}\xi^6 + \frac{h}{a}\xi^7 + \cdots\biggr] \, .$

First approximation:  Assume that $~e=f=g=h=0$, in which case the LHS contains terms only up through $~\xi^2$. This means that we must ignore all terms on the RHS that are of higher order than $~\xi^2$; that is,

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

Expressions for the various coefficients can now be determined by equating terms on the LHS and RHS that have like powers of $~\xi$. Remembering to include a negative sign on the RHS, we find:

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

By including higher and higher order terms in the series expansion for $~\Theta_H$, and proceeding along the same line of deductive reasoning, one finds:

• Expressions for the four coefficients, $~a, b, c, d$, remain unchanged.
• The coefficient is zero for all other terms that contain odd powers of $~\xi$; specifically, for example, $~e = g = 0$.
• The coefficients of $~\xi^6$ and $~\xi^8$ are, respectively,
 $~f$ $~=$ $~- \frac{n}{378}\biggl(\frac{n}{5}-\frac{1}{8} \biggr) \, ;$ $~h$ $~=$ $~\frac{n(122n^2 -183n + 70)}{3265920} \, .$

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

For Spherically Symmetric Configurations
 $~\theta$ $~=$ $~ 1 - \frac{\xi^2}{6} + \frac{n}{120} \xi^4 - \frac{n}{378} \biggl( \frac{n}{5} - \frac{1}{8} \biggr) \xi^6 + \biggl[ \frac{n(122n^2 -183n + 70)}{3265920} \biggr] \xi^8 + \cdots$

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!), $~\psi(\xi) \leftrightarrow w(r)$, the governing ODE is,

 $~\frac{d^2w}{dr^2} +\frac{2}{r} \frac{d w}{dr}$ $~=$ $~e^{-w} \, .$

A general power-series should be of the form,

 $~w$ $~=$ $~ w_0 + ar + br^2 + cr^3 + dr^4 + er^5 + fr^6 + gr^7 + hr^8 +\cdots$

Derivatives:

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

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

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

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

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

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

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

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

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

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

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

Result:

For Spherically Symmetric Configurations
 $~w(r)$ $~=$ $~\frac{r^2}{6} - \frac{r^4}{120} + \frac{r^6}{1890} - \frac{61 r^8}{1,632,960} + \cdots \, .$

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

NOTE:  For cylindrically symmetric, rather than spherically symmetric, configurations, an analytic expression for the function, $~w(r)$, 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,

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

where, $~\theta(\xi)$ is the polytropic Lane-Emden function describing the configuration's unperturbed radial density distribution, and $~\gamma$, $~\sigma_c^2$, and $~\alpha \equiv (3-4/\gamma)$ are constants. Here we seek a power-series expression for the displacement function, $~x(r)$, 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,

 $~\theta$ $~=$ $~ 1 - \frac{\xi^2}{6} + \frac{n}{120} \xi^4 - \frac{n}{378} \biggl( \frac{n}{5} - \frac{1}{8} \biggr) \xi^6 + \cdots$

Hence,

 $~-\frac{d\theta}{d\xi}$ $~\approx$ $~ \frac{1}{3} \biggl[ \xi - \frac{n}{10} \xi^3 + \frac{n}{21} \biggl( \frac{n}{5} - \frac{1}{8} \biggr) \xi^5 \biggr] \, .$

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

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

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

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

and,

 $~\frac{d^2x}{d\xi^2}$ $~=$ $~ 2b + 6c\xi + 12d\xi^2 + 20e\xi^3 + 30f\xi^4 + \cdots$

Substituting these expressions into the LAWE gives,

 $~\biggl( 6 - \xi^2 + \frac{n}{20} \xi^4 \biggr) \biggl( 2b + 6c\xi + 12d\xi^2 + 20e\xi^3 + 30f\xi^4 \biggr) + \biggl[ 12 - (n+3)\xi^2 + \frac{n(n+2)}{10} \xi^4 \biggr] \biggl( \frac{2a}{\xi} + 4b + 6 c\xi + 8d\xi^2 + 10e\xi^3 + 12f\xi^4 \biggr)$ $~\approx$ $~ - (n+1) \biggl[ \mathfrak{F} + \frac{n\alpha}{5} \xi^2 - \frac{2n\alpha}{21} \biggl( \frac{n}{5} - \frac{1}{8} \biggr) \xi^4 \biggr] \biggl( 1 + a\xi + b\xi^2 + c\xi^3 + d\xi^4 \biggr)$

Expressions for the various coefficients can now be determined by equating terms on the LHS and RHS that have like powers of $~\xi$.

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

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

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

### Displacement Function for Isothermal LAWE

The LAWE for isothermal spheres may be written as,

 $~\frac{d^2 x}{dr^2} + \biggl[4 - r \biggl(\frac{dw }{dr}\biggr) \biggr] \frac{1}{r}\frac{dx}{dr}$ $~=$ $~ - \biggl[ \frac{\sigma_c^2}{6\gamma} - \frac{\alpha}{r} \biggl(\frac{dw }{dr}\biggr)\biggr] x \, ,$

where, $~w(r)$ is the isothermal Lane-Emden function describing the configuration's unperturbed radial density distribution, and $~\gamma$, $~\sigma_c^2$, and $~\alpha \equiv (3-4/\gamma)$ are constants. Here we seek a power-series expression for the displacement function, $~x(r)$, 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,

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

Hence,

 $~\frac{dw}{dr}$ $~\approx$ $~\frac{r}{3} - \frac{r^3}{30} + \frac{r^5}{315} \, .$

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

 $~\frac{d^2 x}{dr^2} + \biggl[4 - \biggl(\frac{r^2}{3} - \frac{r^4}{30} + \frac{r^6}{315}\biggr) \biggr] \frac{1}{r}\frac{dx}{dr}$ $~\approx$ $~ - \frac{1}{6} \biggl[ \frac{\sigma_c^2}{\gamma} - 2\alpha \biggl(1 - \frac{r^2}{10} + \frac{r^4}{105}\biggr) \biggr] x \, .$

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

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

and,

 $~\frac{d^2x}{dr^2}$ $~=$ $~ 2b + 6cr + 12dr^2 + \cdots$

Substituting these expressions into the LAWE gives,

 $~2b + 6cr + 12dr^2 + \biggl[4 - \biggl(\frac{r^2}{3} - \frac{r^4}{30} + \frac{r^6}{315}\biggr) \biggr] \biggl[ \frac{a}{r} + 2b + 3 cr + 4dr^2 \biggr]$ $~\approx$ $~ - \frac{1}{6} \biggl[ \frac{\sigma_c^2}{\gamma} - 2\alpha \biggl(1 - \frac{r^2}{10} + \frac{r^4}{105}\biggr) \biggr] \biggl( 1 + ar + br^2 + cr^3 + dr^4 \biggr) \, .$

Keeping terms only up through $~r^2$ leads to the following simplification:

 $~ 2b + 6cr + 12dr^2 + 4 \biggl[ \frac{a}{r} + 2b + 3 cr + 4dr^2 \biggr] - \frac{r^2}{3} \biggl[ \frac{a}{r} + 2b \biggr]$ $~\approx$ $~ - \frac{\mathfrak{F} }{6} \biggl( 1 + ar + br^2 \biggr) - \frac{\alpha}{3} \biggl(\frac{r^2}{10} \biggr)$

where,

$~\mathfrak{F} \equiv \frac{\sigma_c^2}{\gamma} - 2\alpha \, .$

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

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

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

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

## Taylor Series (Hunter77)

### First (Unsuccessful) Try

First:

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

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

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

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

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

Second:

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

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

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

Third:

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

And, finally:

 $~f_4$ $~=$ $~ f_3 + (\Delta) f_3^' + \frac{1}{2} ( \Delta)^2 f^{''}_3 + \frac{1}{6} (\Delta)^3 f_3^{'''} + \frac{1}{24}(\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)$ $~=$ $~ f_3 + (\Delta) f_3^' + \mathcal{O}(\Delta^5)$ $~ + \frac{1}{2} \biggl\{ - f_0 + \biggl[\frac{3^3}{2^2}\biggr] f_1 - 3^3 f_2 + \biggl[\frac{5\cdot 17}{2^2} \biggr] f_3 + \biggl[- \frac{3^2\cdot 5}{2}\biggr] (\Delta) f_3^' \biggr\} \biggl[ - \frac{2}{3^2} \biggr]$ $~ + \frac{1}{6} \biggl\{ \biggl[-\frac{2}{3} \biggr] f_0 + \biggl[4\biggr] f_1 + \biggl[- 2\cdot 5 \biggr] f_2 + \biggl[\frac{2^2\cdot 5}{3}\biggr] f_3 + \biggl[- \frac{2\cdot 11}{3}\biggr] (\Delta) f_3^' \biggr\} \biggl[ -\frac{3}{2} \biggr]$ $~ + \frac{1}{24}\biggl\{ \biggl[-\frac{1}{3^2} \biggr]f_0 + \biggl[\frac{1}{2} \biggr] f_1 - f_2 + \biggl[ \frac{11}{2\cdot 3^2} \biggr] f_3 + \biggl[- \frac{2}{3}\biggr] (\Delta) f_3^' \biggr\} \biggl[ -2^2\cdot 3 \biggr]$ $~=$ $~ f_3 + (\Delta) f_3^' + \mathcal{O}(\Delta^5)$ $~ + \biggl\{ \biggl[ \frac{1}{3^2} \biggr] f_0 + \biggl[- \frac{3}{2^2}\biggr] f_1 +\biggl[ 3 \biggr] f_2 + \biggl[- \frac{5\cdot 17}{2^2\cdot 3^2} \biggr] f_3 + \biggl[\frac{5}{2}\biggr] (\Delta) f_3^' \biggr\}$ $~ + \biggl\{ \biggl[\frac{1}{2\cdot 3} \biggr] f_0 + \biggl[-1 \biggr] f_1 + \biggl[ \frac{5}{2} \biggr] f_2 + \biggl[- \frac{5}{3}\biggr] f_3 + \biggl[\frac{11}{2\cdot 3}\biggr] (\Delta) f_3^' \biggr\}$ $~ + \biggl\{ \biggl[\frac{1}{2\cdot 3^2} \biggr]f_0 + \biggl[- \frac{1}{2^2} \biggr] f_1 + \biggl[ \frac{1}{2} \biggr] f_2 + \biggl[ -\frac{11}{2^2 \cdot 3^2} \biggr] f_3 + \biggl[\frac{1}{3}\biggr] (\Delta) f_3^' \biggr\}$ $~=$ $~ \biggl[ \frac{1}{3^2} + \frac{1}{2\cdot 3} + \frac{1}{2\cdot 3^2} \biggr] f_0 + \biggl[- \frac{3}{2^2} - 1 - \frac{1}{2^2} \biggr] f_1 +\biggl[ 3 + \frac{5}{2} + \frac{1}{2} \biggr] f_2 + \mathcal{O}(\Delta^5)$ $~ + \biggl[1 - \frac{5\cdot 17}{2^2\cdot 3^2} - \frac{5}{3} - \frac{11}{2^2 \cdot 3^2} \biggr] f_3 + \biggl[1 + \frac{5}{2} + \frac{11}{2\cdot 3} + \frac{1}{3} \biggr] (\Delta) f_3^'$ $~=$ $~ \biggl[ \frac{1}{3} \biggr] f_0 + \biggl[- 2\biggr] f_1 +\biggl[ 6 \biggr] f_2 + \biggl[- \frac{10}{3} \biggr] f_3 + \biggl[\frac{17}{3} \biggr] (\Delta) f_3^' + \mathcal{O}(\Delta^5)$

Result:

Definitely WRONG!

 $~f_4$ $~=$ $~ \frac{1}{3} ~ f_0 - 2 f_1 + 6 f_2 - \frac{10}{3} ~ f_3 + \frac{17}{3} (\Delta) f_3^' + \mathcal{O}(\Delta^5) \, .$

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 $~(\Delta) f_3^'$, namely,

Somewhat Improved

 $~f_4$ $~=$ $~ \frac{1}{3} ~ f_0 - 2 f_1 + 6 f_2 - \frac{10}{3} ~ f_3 + 4 (\Delta) f_3^' + \mathcal{O}(\Delta^5) \, .$

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 $~f_4$ is the variable that we desire to evaluate, and the "known" quantities are:   $~f_3$, $~f_3^'$, $~f_2$, $~f_1$, and $~f_0$.

 $~f_4$ $~=$ $~ f_3 + (\Delta) f_3^' + \frac{1}{2} ( \Delta)^2 f^{''}_3 + \frac{1}{6} (\Delta)^3 f_3^{'''} + \frac{1}{24}(\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)$

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

 $~f_2$ $~=$ $~ f_3 + (- \Delta) f_3^' + \frac{1}{2} (- \Delta)^2 f^{''}_3 + \frac{1}{6} (- \Delta)^3 f_3^{'''} + \frac{1}{24}(- \Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)$ $~f_1$ $~=$ $~ f_3 + (- 2\Delta) f_3^' + \frac{1}{2} (- 2\Delta)^2 f^{''}_3 + \frac{1}{6} (- 2\Delta)^3 f_3^{'''} + \frac{1}{24}(- 2\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)$ $~f_0$ $~=$ $~ f_3 + (- 3\Delta) f_3^' + \frac{1}{2} (- 3\Delta)^2 f^{''}_3 + \frac{1}{6} (- 3\Delta)^3 f_3^{'''} + \frac{1}{24}(- 3\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)$

First:

 $~-\frac{1}{2} (- \Delta)^2 f^{''}_3$ $~=$ $~ f_3 + (- \Delta) f_3^' - f_2+ \frac{1}{6} (- \Delta)^3 f_3^{'''} + \frac{1}{24}(- \Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)$ $~\Rightarrow~~~ \frac{1}{2} (\Delta)^2 f^{''}_3$ $~=$ $~ - f_3 + (\Delta) f_3^' + f_2+ \frac{1}{6} (\Delta)^3 f_3^{'''} - \frac{1}{24}(\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)$ $~\Rightarrow~~~ f_4$ $~=$ $~ f_3 + (\Delta) f_3^' - f_3 + (\Delta) f_3^' + f_2 + \mathcal{O}(\Delta^3)$ $~=$ $~ f_2 + 2(\Delta) f_3^' + \mathcal{O}(\Delta^3)$
$~\mathcal{O}(\Delta^3)$
 $~ f_4$ $~=$ $~ f_2 + 2(\Delta) f_3^' + \mathcal{O}(\Delta^3)$

This expression works very well for a parabola.

Second:

 $~f_1$ $~=$ $~ f_3 + (- 2) \Delta f_3^' + 2 (\Delta)^2 f^{''}_3 + \biggl[- \frac{2^3}{6}\biggr] \Delta^3 f_3^{'''} + \biggl[ \frac{2^4}{2^3\cdot 3} \biggr] \Delta^4 f_3^{iv} + \mathcal{O}(\Delta^5)$ $~=$ $~ f_3 + (- 2) \Delta f_3^' + \biggl[- \frac{2^3}{6}\biggr] \Delta^3 f_3^{'''} + \biggl[ \frac{2^4}{2^3\cdot 3} \biggr] \Delta^4 f_3^{iv} + \mathcal{O}(\Delta^5)$ $~ + 2 \biggl\{ - f_3 + (\Delta) f_3^' + f_2+ \frac{1}{2\cdot 3} (\Delta)^3 f_3^{'''} - \frac{1}{2^3\cdot 3}(\Delta)^4 f_3^{iv} \biggr\} \biggl[ 2 \biggr]$ $~=$ $~ f_3\biggl[ 1 - 2^2\biggr] + (2^2 - 2) \Delta f_3^' + 2^2f_2 + \biggl[\frac{2}{3} - \frac{2^3}{6}\biggr] \Delta^3 f_3^{'''} + \biggl[ \frac{2^4}{2^3\cdot 3} - \frac{1}{2\cdot 3}\biggr] \Delta^4 f_3^{iv} + \mathcal{O}(\Delta^5)$ $~=$ $~ f_3\biggl[ -3\biggr] + (2) \Delta f_3^' + 2^2f_2 + \biggl[- \frac{2}{3}\biggr] \Delta^3 f_3^{'''} + \biggl[ \frac{1}{2} \biggr] \Delta^4 f_3^{iv} + \mathcal{O}(\Delta^5)$ $~\Rightarrow~~~ \biggl[\frac{2}{3}\biggr] \Delta^3 f_3^{'''}$ $~=$ $~ - f_1 + 2^2f_2 -3 f_3 + 2 \Delta f_3^' + \biggl[ \frac{1}{2} \biggr] \Delta^4 f_3^{iv} + \mathcal{O}(\Delta^5)$

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

 $~ \frac{1}{2} (\Delta)^2 f^{''}_3$ $~=$ $~ f_2 - f_3 + (\Delta) f_3^' - \frac{1}{24}(\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)$ $~ + \frac{1}{6} \biggl\{ - f_1 + 2^2f_2 -3 f_3 + 2 \Delta f_3^' + \biggl[ \frac{1}{2} \biggr] \Delta^4 f_3^{iv} \biggr\} \biggl[ \frac{3}{2} \biggr]$ $~=$ $~ - \frac{1}{4} f_1 + 2f_2 + \biggl[ - \frac{7}{4} \biggr] f_3 + \biggl[ \frac{3}{2} \biggr] (\Delta) f_3^' + \biggl[\frac{1}{2^2\cdot 3} \biggr](\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)$

Hence, an improved expression for $~f_4$ is,

 $~f_4$ $~=$ $~ f_3 + (\Delta) f_3^' + \mathcal{O}(\Delta^4)$ $~ + \biggl\{ - \frac{1}{4} f_1 + 2f_2 + \biggl[ - \frac{7}{4} \biggr] f_3 + \biggl[ \frac{3}{2} \biggr] (\Delta) f_3^' \biggr\}$ $~ + \frac{1}{6} \biggl\{ - f_1 + 2^2f_2 -3 f_3 + 2 \Delta f_3^' \biggr\} \biggl[ \frac{3}{2} \biggr]$ $~=$ $~ - \frac{1}{2} f_1 + 3f_2 - \frac{3}{2} f_3 + 3(\Delta) f_3^' + \mathcal{O}(\Delta^4)$

$~\mathcal{O}(\Delta^4)$
 $~ f_4$ $~=$ $~ - \frac{1}{2} f_1 + 3f_2 - \frac{3}{2} f_3 + 3(\Delta) f_3^' + \mathcal{O}(\Delta^4)$

Third:

 $~f_0$ $~=$ $~ f_3 + (- 3\Delta) f_3^' + \frac{1}{2} (- 3\Delta)^2 f^{''}_3 + \frac{1}{6} (- 3\Delta)^3 f_3^{'''} + \frac{1}{24}(- 3\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)$ $~=$ $~ f_3 + \biggl[ - 3 \biggr] (\Delta) f_3^' + \biggl[ \frac{3^3}{2^3} \biggr] (\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)$ $~ + 3^2 \biggl\{ - \frac{1}{4} f_1 + 2f_2 + \biggl[ - \frac{7}{4} \biggr] f_3 + \biggl[ \frac{3}{2} \biggr] (\Delta) f_3^' + \biggl[\frac{1}{2^2\cdot 3} \biggr](\Delta)^4 f_3^{iv} \biggr\}$ $~ + \biggl[-\frac{3^3}{2^2} \biggr] \biggl\{ - f_1 + 2^2f_2 -3 f_3 + 2 \Delta f_3^' + \biggl[ \frac{1}{2} \biggr] \Delta^4 f_3^{iv} \biggr\}$ $~=$ $~ f_3 + \biggl[ - 3 \biggr] (\Delta) f_3^' + \biggl[ \frac{3^3}{2^3} \biggr] (\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)$ $~ + \biggl\{ \biggl[- \frac{3^2 }{4} \biggr] f_1 + \biggl[ 2\cdot 3^2 \biggr]f_2 + \biggl[ - \frac{3^2 \cdot 7}{4} \biggr] f_3 + \biggl[ \frac{3^3}{2} \biggr] (\Delta) f_3^' + \biggl[\frac{3}{2^2} \biggr](\Delta)^4 f_3^{iv} \biggr\}$ $~ + \biggl\{ \biggl[\frac{3^3}{2^2} \biggr] f_1 + \biggl[-3^3 \biggr] f_2 + \biggl[\frac{3^4}{2^2} \biggr]f_3 + \biggl[- \frac{3^3}{2} \biggr] \Delta f_3^' + \biggl[- \frac{3^3}{2^3} \biggr] \Delta^4 f_3^{iv} \biggr\}$ $~=$ $~ \biggl[\frac{3^3}{2^2} - \frac{3^2 }{4} \biggr] f_1 + \biggl[ 2\cdot 3^2 -3^3\biggr]f_2 + \biggl[ 1+ \frac{3^4}{2^2} - \frac{3^2 \cdot 7}{4} \biggr] f_3 + \biggl[ \frac{3^3}{2} - \frac{3^3}{2} -3\biggr] (\Delta) f_3^' + \biggl[\frac{3^3}{2^3} + \frac{3}{2^2} - \frac{3^3}{2^3}\biggr](\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)$ $~=$ $~ \biggl[\frac{3^2}{2} \biggr] f_1 + \biggl[ - 3^2 \biggr]f_2 + \biggl[ \frac{11}{2} \biggr] f_3 + \biggl[ -3\biggr] (\Delta) f_3^' + \biggl[\frac{3}{2^2} \biggr](\Delta)^4 f_3^{iv} + \mathcal{O}(\Delta^5)$ $~\Rightarrow ~~~ - \biggl[\frac{3}{2^2} \biggr](\Delta)^4 f_3^{iv}$ $~=$ $~ - f_0 + \biggl[\frac{3^2}{2} \biggr] f_1 + \biggl[ - 3^2 \biggr]f_2 + \biggl[ \frac{11}{2} \biggr] f_3 + \biggl[ -3\biggr] (\Delta) f_3^' + \mathcal{O}(\Delta^5)$

Hence,

 $~ \frac{1}{2} (\Delta)^2 f^{''}_3$ $~=$ $~ - \frac{1}{4} f_1 + 2f_2 + \biggl[ - \frac{7}{4} \biggr] f_3 + \biggl[ \frac{3}{2} \biggr] (\Delta) f_3^' + \mathcal{O}(\Delta^5)$ $~ + \biggl[\frac{1}{2^2\cdot 3} \biggr]\biggl\{ - f_0 + \biggl[\frac{3^2}{2}\biggr] f_1 + \biggl[ - 3^2 \biggr]f_2 + \biggl[ \frac{11}{2} \biggr] f_3 + \biggl[ -3\biggr] (\Delta) f_3^' \biggr\} \biggl[ - \frac{2^2}{3} \biggr]$ $~=$ $~ - \frac{1}{4} f_1 + 2f_2 + \biggl[ - \frac{7}{4} \biggr] f_3 + \biggl[ \frac{3}{2} \biggr] (\Delta) f_3^' + \mathcal{O}(\Delta^5)$ $~ + \biggl\{ \biggl[\frac{1}{3^2} \biggr] f_0 + \biggl[- \frac{1}{2} \biggr] f_1 + f_2 + \biggl[- \frac{11}{2 \cdot 3^2} \biggr] f_3 + \biggl[\frac{1}{3} \biggr] (\Delta) f_3^' \biggr\}$ $~=$ $~ \biggl[\frac{1}{3^2} \biggr] f_0 + \biggl[- \frac{1}{2} - \frac{1}{4} \biggr] f_1 + 3 f_2 + \biggl[- \frac{11}{2 \cdot 3^2} - \frac{7}{4} \biggr] f_3 + \biggl[\frac{1}{3} + \frac{3}{2} \biggr] (\Delta) f_3^' + \mathcal{O}(\Delta^5)$ $~=$ $~ \biggl[\frac{1}{3^2} \biggr] f_0 + \biggl[- \frac{3}{4} \biggr] f_1 + 3 f_2 + \biggl[- \frac{5\cdot 17}{2^2\cdot 3^2} \biggr] f_3 + \biggl[\frac{11}{2\cdot 3} \biggr] (\Delta) f_3^' + \mathcal{O}(\Delta^5)$

And,

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

Finally, then:

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

$~\mathcal{O}(\Delta^5)$
 $~ f_4$ $~=$ $~ \frac{1}{3} f_0 - 2 f_1 + 6 f_2 - \frac{2\cdot 5}{3} f_3 + 4 (\Delta) f_3^' + \mathcal{O}(\Delta^5)$

 © 2014 - 2020 by Joel E. Tohline |   H_Book Home   |   YouTube   | Appendices: | Equations | Variables | References | Ramblings | Images | myphys.lsu | ADS |