Appendix/Ramblings/PowerSeriesExpressions

Approximate Power-Series Expressions

Broadly Used Mathematical Expressions (shown here without proof)

Binomial

$(1\pm x{)}^n$

$=$

$1\pm nx+\left[\frac{n(n-1)}{2!}\right]x^2\pm \left[\frac{n(n-1)(n-2)}{3!}\right]x^3+\left[\frac{n(n-1)(n-2)(n-3)}{4!}\right]x^4\pm \cdots$      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, $(a+b{)}^n$ $=$ $a^n+(\genfrac{}{}{0ex}{}{n}{1})a^{n-1}b+(\genfrac{}{}{0ex}{}{n}{2})a^{n-2}{b}^2+\dots +(\genfrac{}{}{0ex}{}{n}{n-1})a{b}^{n-1}+b^n,$ where, for nonnegative integer values of $k$ and $n$ and $k\le n$, the notation, $(\genfrac{}{}{0ex}{}{n}{k})$ $=$ $\frac{n!}{(n-k)!k!}=(\genfrac{}{}{0ex}{}{n}{n-k}).$ Our Example:  Setting $a=1$ gives, $(1+b{)}^n$ $=$ $1+(\genfrac{}{}{0ex}{}{n}{1})b+(\genfrac{}{}{0ex}{}{n}{2})b^2+(\genfrac{}{}{0ex}{}{n}{3})b^3+(\genfrac{}{}{0ex}{}{n}{4})b^4+\dots$   $=$ $1+\frac{n!}{(n-1)!}b+\frac{n!}{(n-2)!2!}{b}^2+\frac{n!}{(n-3)!3!}{b}^3+\frac{n!}{(n-4)!4!}{b}^4+\dots$   $=$ $1+nb+\left[\frac{n(n-1)}{2!}\right]b^2+\left[\frac{n(n-1)(n-2)}{3!}\right]b^3+\left[\frac{n(n-1)(n-2)(n-3)}{4!}\right]b^4+\dots$

Note, for example, that,

$(1+x{)}^{-1}$	$=$	$1-x+x^2-x^3+x^4-x^5+\cdots ;$
$(1+x{)}^{-2}$	$=$	$1-2x+3x^2-4x^3+5x^4-6x^5+\cdots ;$
$(1+x{)}^{-3}$	$=$	$1-3x+\left[\frac{3\cdot 4}{2}\right]x^2-\left[\frac{3\cdot 4\cdot 5}{2\cdot 3}\right]x^3+\left[\frac{3\cdot 4\cdot 5\cdot 6}{2\cdot 3\cdot 4}\right]x^4-\left[\frac{3\cdot 4\cdot 5\cdot 6\cdot 7}{2\cdot 3\cdot 4\cdot 5}\right]x^5+\cdots$
	$=$	$1-3x+6x^2-10x^3+15x^4-21x^5+\cdots ;$
$(1+x{)}^{-4}$	$=$	$1-4x+\left[\frac{4\cdot 5}{2}\right]x^2-\left[\frac{4\cdot 5\cdot 6}{2\cdot 3}\right]x^3+\left[\frac{4\cdot 5\cdot 6\cdot 7}{2\cdot 3\cdot 4}\right]x^4-\left[\frac{4\cdot 5\cdot 6\cdot 7\cdot 8}{2\cdot 3\cdot 4\cdot 5}\right]x^5+\cdots$
	$=$	$1-4x+10x^2-20x^3+35x^4-56x^5+\cdots .$

See also:

• Wolfram's presentation

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

Power-Series Derivation

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

$\frac{1}{{\xi }^2}\frac{d}{d\xi }\left({\xi }^2\frac{d{\mathrm{\Theta }}_H}{d\xi }\right)=-{\mathrm{\Theta }}_H^n$

A general power-series should be of the form,

${\mathrm{\Theta }}_H$

$=$

${\theta }_0+a\xi +b{\xi }^2+c{\xi }^3+d{\xi }^4+e{\xi }^5+f{\xi }^6+g{\xi }^7+h{\xi }^8+\cdots$

First derivative:

$\frac{d{\mathrm{\Theta }}_H}{d\xi }$

$=$

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

Left-hand-side of Lane-Emden equation:

$\frac{1}{{\xi }^2}\frac{d}{d\xi }\left({\xi }^2\frac{d{\mathrm{\Theta }}_H}{d\xi }\right)$

$=$

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

Right-hand-side of Lane-Emden equation (adopt the normalization, ${\theta }_0=1$, then use the binomial theorem recursively):

${\mathrm{\Theta }}_H^n$

$=$

$1+nF+\left[\frac{n(n-1)}{2!}\right]F^2+\left[\frac{n(n-1)(n-2)}{3!}\right]F^3+\left[\frac{n(n-1)(n-2)(n-3)}{4!}\right]F^4+\cdots$

where,

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

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,

${\mathrm{\Theta }}_H^n$	$\approx$	$1+nF+\left[\frac{n(n-1)}{2!}\right]F^2$
	$\approx$	$1+n(a\xi +b{\xi }^2)+\left[\frac{n(n-1)}{2!}\right]a^2{\xi }^2$
	$\approx$	$1+na\xi +[nb+\frac{n(n-1)a^2}{2}]{\xi }^2.$

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
${\xi }^{-1}:$	$2a$	$0$	$\Rightarrow a=0$
${\xi }^0:$	$2\cdot 3b$	$-1$	$\Rightarrow b=-\frac{1}{6}$
${\xi }^1:$	$2^2\cdot 3c$	$-na$	$\Rightarrow c=0$
${\xi }^2:$	$2^2\cdot 5d$	$-[nb+\frac{n(n-1)a^2}{2}]$	$\Rightarrow d=+\frac{n}{120}$

By including higher and higher order terms in the series expansion for ${\mathrm{\Theta }}_H$, 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,

$f$	$=$	$-\frac{n}{378}(\frac{n}{5}-\frac{1}{8});$
$h$	$=$	$\frac{n(122n^2-183n+70)}{3265920}.$

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

For Spherically Symmetric Configurations
${\mathrm{\Theta }}_H$ $=$ $1-\frac{{\xi }^2}{6}+\frac{n}{120}{\xi }^4-\frac{n}{378}(\frac{n}{5}-\frac{1}{8}){\xi }^6+\left[\frac{n(122n^2-183n+70)}{3265920}\right]{\xi }^8+\cdots$   $=$ $1-\frac{{\xi }^2}{3!}+\left(\frac{n}{5!}\right){\xi }^4-\frac{n}{7!}\left(\frac{8n-5}{3}\right){\xi }^6+\frac{n}{9!}\left[\frac{(122n^2-183n+70)}{9}\right]{\xi }^8+\cdots$

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.

Examples

When $n=0$, all of the terms in the power-series expression higher than quadratic go to zero. As a result, we see that

${\theta }_{n=0}$

$=$

$1-\frac{{\xi }^2}{6}.$

---

When $n=1$, the first few terms in the power-series expression are,

${\theta }_{n=1}$	$=$	$1-\frac{{\xi }^2}{6}+\frac{{\xi }^4}{120}-\frac{{\xi }^6}{5040}+\frac{{\xi }^8}{362880}+\cdots$
	$=$	$1-\frac{{\xi }^2}{3!}+\frac{{\xi }^4}{5!}-\frac{{\xi }^6}{7!}+\frac{{\xi }^8}{9!}+\cdots$

which is consistent with the power-series expression for $(\sin \xi )/\xi$.

---

When $n=5$, the first few terms in the power-series expression are,

${\theta }_{n=5}$	$=$	$1-\frac{{\xi }^2}{3!}+\left(\frac{n}{5!}\right){\xi }^4-\frac{n}{7!}\left(\frac{8n-5}{3}\right){\xi }^6+\frac{n}{9!}\left[\frac{(122n^2-183n+70)}{9}\right]{\xi }^8+\cdots$
	$=$	$1-\frac{{\xi }^2}{3!}+\frac{{\xi }^4}{4!}-\frac{{\xi }^6}{6!}\left(\frac{5^2}{3}\right)+\frac{{\xi }^8}{8!}[\frac{5^2}{3}\cdot \frac{7^2}{3}]+\cdots$

This should be compared with the known analytic solution, which is

$({\mathrm{\Theta }}_H{)}_{n=5}$

$=$

$[1+\frac{{\xi }^2}{3}{]}^{-1/2}.$

From the binomial theorem, we start with,

$(1+b{)}^m$

$=$

$1+mb+\left[\frac{m(m-1)}{2!}\right]b^2+\left[\frac{m(m-1)(m-2)}{3!}\right]b^3+\left[\frac{m(m-1)(m-2)(m-3)}{4!}\right]b^4+\dots$

then set $b={\xi }^2/3$ and $m=-1/2$ to obtain:

${\theta }_{n=5}$	$=$	$1+m\left(\frac{{\xi }^2}{3}\right)+\frac{1}{2!}\left[m\right(m-1\left)\right](\frac{{\xi }^2}{3}{)}^2+\frac{1}{3!}[m(m-1)(m-2)\left]\right(\frac{{\xi }^2}{3}{)}^3+\frac{1}{4!}\left[m\right(m-1\left)\right(m-2\left)\right(m-3\left)\right](\frac{{\xi }^2}{3}{)}^4+\dots$
	$=$	$1+(-\frac{1}{2})\left(\frac{{\xi }^2}{3}\right)+\frac{1}{2!}[-\frac{1}{2}(-\frac{1}{2}-1\left)\right]\left(\frac{{\xi }^4}{3^2}\right)+\frac{1}{3!}[-\frac{1}{2}(-\frac{1}{2}-1\left)\right(-\frac{1}{2}-2\left)\right]\left(\frac{{\xi }^6}{3^3}\right)+\frac{1}{4!}[-\frac{1}{2}(-\frac{1}{2}-1\left)\right(-\frac{1}{2}-2\left)\right(-\frac{1}{2}-3\left)\right]\left(\frac{{\xi }^8}{3^4}\right)+\dots$
	$=$	$1-\left(\frac{{\xi }^2}{2\cdot 3}\right)+\frac{1}{2!}\left[\frac{3}{4}\right]\left(\frac{{\xi }^4}{3^2}\right)+\frac{1}{3!}[-\frac{15}{8}]\left(\frac{{\xi }^6}{3^3}\right)+\frac{1}{4!}\left[\right(\frac{105}{16}\left)\right]\left(\frac{{\xi }^8}{3^4}\right)+\dots$
	$=$	$1-\frac{{\xi }^2}{3!}+\frac{{\xi }^4}{4!}-\frac{{\xi }^6}{6!}\left(\frac{5^2}{3}\right)+\frac{{\xi }^8}{8!}\left(\frac{5^2\cdot 7^2}{3^2}\right)+\dots$

QED

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,

$\frac{d^{2}w}{d{r}^2}+\frac{2}{r}\frac{dw}{dr}$

$=$

$e^{-w}.$

A general power-series should be of the form,

$w$

$=$

$w_0+ar+b{r}^2+c{r}^3+d{r}^4+e{r}^5+f{r}^6+g{r}^7+h{r}^8+\cdots$

Derivatives:

$\frac{dw}{dr}$	$=$	$a+2br+3c{r}^2+4d{r}^3+5e{r}^4+6f{r}^5+7g{r}^6+8h{r}^7+\cdots ;$
$\frac{d^{2}w}{d{r}^2}$	$=$	$2b+2\cdot 3cr+2^2\cdot 3d{r}^2+2^2\cdot 5e{r}^3+2\cdot 3\cdot 5f{r}^4+2\cdot 3\cdot 7g{r}^5+2^3\cdot 7h{r}^6+\cdots .$

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

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

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

$e^{-w}$	$=$	$e^0\cdot e^{-ar}\cdot e^{-b{r}^2}\cdot e^{-c{r}^3}\cdot e^{-d{r}^4}\cdot e^{-e{r}^5}\cdot e^{-f{r}^6}\cdot e^{-g{r}^7}\cdot e^{-h{r}^8}\cdots$
	$=$	$[1-ar+\frac{a^2{r}^2}{2!}-\frac{a^3{r}^3}{3!}+\frac{a^4{r}^4}{4!}-\frac{a^5{r}^5}{5!}+\frac{a^6{r}^6}{6!}+\cdots ]$
		$\times [1-b{r}^2+\frac{b^2{r}^4}{2!}-\frac{b^3{r}^6}{3!}+\cdots ]\times [1-c{r}^3+\frac{c^2{r}^6}{2!}+\cdots ]\times [1-d{r}^4]\times [1-e{r}^5]\times [1-f{r}^6]$
	$\approx$	$[1-ar+\frac{a^2{r}^2}{2}-\frac{a^3{r}^3}{6}+\frac{a^4{r}^4}{24}-\frac{a^5{r}^5}{5\cdot 24}+\frac{a^6{r}^6}{30\cdot 24}]\times [1-c{r}^3+\frac{c^2{r}^6}{2}-b{r}^2+bc{r}^5+\frac{b^2{r}^4}{2}-\frac{b^3{r}^6}{6}]\times [1-d{r}^4-e{r}^5-f{r}^6]$
	$\approx$	$\left\{\right[1-ar+\frac{a^2{r}^2}{2}-\frac{a^3{r}^3}{6}+\frac{a^4{r}^4}{24}-\frac{a^5{r}^5}{5\cdot 24}+\frac{a^6{r}^6}{30\cdot 24}]-d{r}^4[1-ar+\frac{a^2{r}^2}{2}]-e{r}^5[1-ar]-f{r}^6\}$
		$\times [1-b{r}^2-c{r}^3+\frac{b^2{r}^4}{2}+bc{r}^5+r^6(\frac{c^2}{2}-\frac{b^3}{6}\left)\right]$
	$\approx$	$[1-ar+\frac{a^2{r}^2}{2}-\frac{a^3{r}^3}{6}+\frac{a^4{r}^4}{24}-\frac{a^5{r}^5}{5\cdot 24}+\frac{a^6{r}^6}{30\cdot 24}-d{r}^4+ad{r}^5-\frac{a^{2}d{r}^6}{2}-e{r}^5+ae{r}^6-f{r}^6]$
		$\times [1-b{r}^2-c{r}^3+\frac{b^2{r}^4}{2}+bc{r}^5+r^6(\frac{c^2}{2}-\frac{b^3}{6}\left)\right]$
	$\approx$	$[1-ar+\frac{a^2{r}^2}{2}-\frac{a^3{r}^3}{6}+r^4(\frac{a^4}{24}-d)+r^5(ad-e-\frac{a^5}{5\cdot 24})+r^6(\frac{a^6}{30\cdot 24}-\frac{a^{2}d}{2}+ae-f\left)\right]\times [1-b{r}^2-c{r}^3+\frac{b^2{r}^4}{2}+bc{r}^5+r^6(\frac{c^2}{2}-\frac{b^3}{6}\left)\right]$
	$\approx$	$1-ar+\frac{a^2{r}^2}{2}-\frac{a^3{r}^3}{6}+r^4(\frac{a^4}{24}-d)+r^5(ad-e-\frac{a^5}{5\cdot 24})+r^6(\frac{a^6}{30\cdot 24}-\frac{a^{2}d}{2}+ae-f)$
		$-b{r}^2[1-ar+\frac{a^2{r}^2}{2}-\frac{a^3{r}^3}{6}+r^4(\frac{a^4}{24}-d\left)\right]-c{r}^3[1-ar+\frac{a^2{r}^2}{2}-\frac{a^3{r}^3}{6}]+\frac{b^2{r}^4}{2}[1-ar+\frac{a^2{r}^2}{2}]+bc{r}^5[1-ar]+r^6(\frac{c^2}{2}-\frac{b^3}{6})$

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
$r^{-1}:$	$2a$	$0$	$\Rightarrow a=0$
$r^0:$	$6b$	$1$	$\Rightarrow b=+\frac{1}{6}$
$r^1:$	$2^2\cdot 3c$	$-a$	$\Rightarrow c=-\frac{a}{2^2\cdot 3}=0$
$r^2:$	$(2^2\cdot 3d+2^{3}d)$	$\frac{a^2}{2}-b$	$\Rightarrow d=\frac{1}{20}(\frac{a^2}{2}-b)=-\frac{1}{120}$

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

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

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

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

Result:

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

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, $w(r)$, 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,

$0$	$=$	$\frac{d^{2}x}{d{\xi }^2}+[\frac{4}{\xi }-\frac{(n+1)}{\theta }(-\frac{d\theta }{d\xi }\left)\right]\frac{dx}{d\xi }+\frac{(n+1)}{\theta }[\frac{{\sigma }_c^2}{6\gamma }-\frac{\alpha }{\xi }(-\frac{d\theta }{d\xi }\left)\right]x$
	$=$	$\theta \frac{d^{2}x}{d{\xi }^2}+[4\theta -(n+1)\xi (-\frac{d\theta }{d\xi }\left)\right]\frac{1}{\xi }\frac{dx}{d\xi }+\frac{(n+1)}{6}[\frac{{\sigma }_c^2}{\gamma }-\frac{6\alpha }{\xi }(-\frac{d\theta }{d\xi }\left)\right]x,$

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,

$\theta$

$=$

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

Hence,

$-\frac{d\theta }{d\xi }$

$\approx$

$\frac{1}{3}[\xi -\frac{n}{10}{\xi }^3+\frac{n}{21}(\frac{n}{5}-\frac{1}{8}\left){\xi }^5\right].$

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

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

where, $\mathfrak{𝔉}\equiv ({\sigma }_c^2/\gamma -2\alpha )$ and, for present purposes, we have kept terms in the series no higher than ${\xi }^4$.

This is a derivation check ... $\left[6\theta \right]\frac{d^{2}x}{d{\xi }^2}+\left\{24\right[\theta ]-6(n+1)\xi [-\frac{d\theta }{d\xi }\left]\right\}\frac{1}{\xi }\frac{dx}{d\xi }$ $=$ $-(n+1)\{\frac{{\sigma }_c^2}{\gamma }-\frac{6\alpha }{\xi }[-\frac{d\theta }{d\xi }\left]\right\}x$ $6[1-\frac{{\xi }^2}{6}+\frac{n}{120}{\xi }^4-\frac{n}{378}(\frac{n}{5}-\frac{1}{8}){\xi }^6+\cdots ]\frac{d^{2}x}{d{\xi }^2}+\left\{24\right[1-\frac{{\xi }^2}{6}+\frac{n}{120}{\xi }^4-\frac{n}{378}(\frac{n}{5}-\frac{1}{8}){\xi }^6+\cdots ]-2(n+1)\xi [\xi -\frac{n}{10}{\xi }^3+\frac{n}{21}(\frac{n}{5}-\frac{1}{8}){\xi }^5\left]\right\}\frac{1}{\xi }\frac{dx}{d\xi }$ $=$ $-(n+1)\{\frac{{\sigma }_c^2}{\gamma }-\frac{2\alpha }{\xi }[\xi -\frac{n}{10}{\xi }^3+\frac{n}{21}(\frac{n}{5}-\frac{1}{8}){\xi }^5\left]\right\}x$ $[6-{\xi }^2+\frac{n}{20}{\xi }^4-\frac{n}{63}(\frac{n}{5}-\frac{1}{8}\left){\xi }^6\right]\frac{d^{2}x}{d{\xi }^2}+\{12-(n+3){\xi }^2+[\frac{n(n+2)}{10}]{\xi }^4-[\frac{n(n+19)}{7\cdot 27}\left]\right(\frac{8n-5}{40}\left){\xi }^6\right\}\frac{2}{\xi }\frac{dx}{d\xi }$ $=$ $-(n+1)\{\mathfrak{𝔉}+\frac{n\alpha }{5}{\xi }^2-\frac{2n\alpha }{21}(\frac{8n-5}{40}\left){\xi }^4\right\}x$ Try $\cdots$ $x$ $=$ $1+a\xi +b{\xi }^2+c{\xi }^3+d{\xi }^4+e{\xi }^5+f{\xi }^6\cdots$ $\frac{dx}{d\xi }$ $=$ $a+2b\xi +3c{\xi }^2+4d{\xi }^3+5e{\xi }^4+6f{\xi }^5\cdots$ $\Rightarrow \frac{2}{\xi }\cdot \frac{dx}{d\xi }$ $=$ $\frac{2a}{\xi }+4b+6c\xi +8d{\xi }^2+10e{\xi }^3+12f{\xi }^4\cdots$ $\frac{d^{2}x}{d{\xi }^2}$ $=$ $2b+6c\xi +12d{\xi }^2+20e{\xi }^3+30f{\xi }^4+\cdots$ in which case, $0$ $\approx$ $[6-{\xi }^2+\frac{n}{20}{\xi }^4-\frac{n}{63}(\frac{n}{5}-\frac{1}{8}\left){\xi }^6\right][2b+6c\xi +12d{\xi }^2+20e{\xi }^3+30f{\xi }^4]$     $+\{12-(n+3){\xi }^2+[\frac{n(n+2)}{10}]{\xi }^4-[\frac{n(n+19)}{7\cdot 27}\left]\right(\frac{8n-5}{40}\left){\xi }^6\right\}[\frac{2a}{\xi }+4b+6c\xi +8d{\xi }^2+10e{\xi }^3+12f{\xi }^4]$     $+(n+1)\{\mathfrak{𝔉}+\frac{n\alpha }{5}{\xi }^2-\frac{2n\alpha }{21}(\frac{8n-5}{40}\left){\xi }^4\right\}[1+a\xi +b{\xi }^2+c{\xi }^3+d{\xi }^4+e{\xi }^5+f{\xi }^6]$ ${\xi }^{-1}:$ $0=24a$ $\Rightarrow a=0$ ${\xi }^0:$ $0=12b+48b+(n+1)\mathfrak{𝔉}$ $\Rightarrow b=-\frac{(n+1)\mathfrak{𝔉}}{60}$ ${\xi }^1:$ $0=36c+72c+{\xcancel{a}}^0(n+1)\mathfrak{𝔉}$ $\Rightarrow c=0$ ${\xi }^2:$ $0=72d-2b+96d-4b(n+3)+b(n+1)\mathfrak{𝔉}+\frac{n(n+1)\alpha }{5}=168d+b[-14-4n+(n+1)\mathfrak{𝔉}]+\frac{n(n+1)\alpha }{5}$ $\Rightarrow d=-(n+1)\left\{\frac{12n\alpha +\mathfrak{𝔉}[(4n+14)-(n+1)\mathfrak{𝔉}]}{10080}\right\}$

Displacement Finite at Center

Let's adopt a power-series expression for the displacement function of a form that is finite at the center of the configuration, namely,

$x$	$=$	$1+a\xi +b{\xi }^2+c{\xi }^3+d{\xi }^4+e{\xi }^5+f{\xi }^6\cdots$
$\frac{dx}{d\xi }$	$=$	$a+2b\xi +3c{\xi }^2+4d{\xi }^3+5e{\xi }^4+6f{\xi }^5\cdots$
$\Rightarrow \frac{1}{\xi }\frac{dx}{d\xi }$	$=$	$\frac{a}{\xi }+2b+3c\xi +4d{\xi }^2+5e{\xi }^3+6f{\xi }^4+\cdots$
$\frac{d^{2}x}{d{\xi }^2}$	$=$	$2b+6c\xi +12d{\xi }^2+20e{\xi }^3+30f{\xi }^4+\cdots$

Substituting these expressions into the LAWE gives,

$(6-{\xi }^2+\frac{n}{20}{\xi }^4)(2b+6c\xi +12d{\xi }^2+20e{\xi }^3+30f{\xi }^4)+[12-(n+3){\xi }^2+\frac{n(n+2)}{10}{\xi }^4](\frac{2a}{\xi }+4b+6c\xi +8d{\xi }^2+10e{\xi }^3+12f{\xi }^4)$

$\approx$

$-(n+1)[\mathfrak{𝔉}+\frac{n\alpha }{5}{\xi }^2-\frac{2n\alpha }{21}(\frac{n}{5}-\frac{1}{8}\left){\xi }^4\right](1+a\xi +b{\xi }^2+c{\xi }^3+d{\xi }^4)$

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
${\xi }^{-1}:$	$24a$	$0$	$\Rightarrow a=0$
${\xi }^0:$	$(12b+48b)$	$-(n+1)\mathfrak{𝔉}$	$\Rightarrow b=-\frac{(n+1)\mathfrak{𝔉}}{60}$
${\xi }^1:$	$[36c+72c-2a(n+3)]$	$-a(n+1)\mathfrak{𝔉}$	$\Rightarrow 108c=2a(n+3)-a(n+1)\mathfrak{𝔉}\Rightarrow c=0$
${\xi }^2:$	$[72d-2b+96d-4b(n+3)]$	$[-b(n+1)\mathfrak{𝔉}-\frac{n(n+1)\alpha }{5}]$	$\Rightarrow d=-(n+1)\left\{\frac{12n\alpha +\mathfrak{𝔉}[(4n+14)-(n+1)\mathfrak{𝔉}]}{10080}\right\}$
NOTE:  On 9/10/2026, we realized that the first term in the numerator of the expression for the coefficient, $d$, should be $12n\alpha$ instead of just $n\alpha$.

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

$x(\xi )$ $=$ $1-\frac{(n+1)\mathfrak{𝔉}}{60}{\xi }^2-(n+1)\left\{\frac{12n\alpha +\mathfrak{𝔉}[(4n+14)-(n+1)\mathfrak{𝔉}]}{10080}\right\}{\xi }^4+\cdots$

Displacement Function for Isothermal LAWE

The LAWE for isothermal spheres may be written as,

$\frac{d^{2}x}{d{r}^2}+[4-r(\frac{dw}{dr}\left)\right]\frac{1}{r}\frac{dx}{dr}$

$=$

$-[\frac{{\sigma }_c^2}{6\gamma }-\frac{\alpha }{r}(\frac{dw}{dr}\left)\right]x,$

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,

$w(r)$

$=$

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

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,

$\frac{d^{2}x}{d{r}^2}+[4-(\frac{r^2}{3}-\frac{r^4}{30}+\frac{r^6}{315}\left)\right]\frac{1}{r}\frac{dx}{dr}$

$\approx$

$-\frac{1}{6}[\frac{{\sigma }_c^2}{\gamma }-2\alpha (1-\frac{r^2}{10}+\frac{r^4}{105}\left)\right]x.$

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

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

and,

$\frac{d^{2}x}{d{r}^2}$

$=$

$2b+6cr+12d{r}^2+\cdots$

Substituting these expressions into the LAWE gives,

$2b+6cr+12d{r}^2+[4-(\frac{r^2}{3}-\frac{r^4}{30}+\frac{r^6}{315}\left)\right][\frac{a}{r}+2b+3cr+4d{r}^2]$

$\approx$

$-\frac{1}{6}[\frac{{\sigma }_c^2}{\gamma }-2\alpha (1-\frac{r^2}{10}+\frac{r^4}{105}\left)\right](1+ar+b{r}^2+c{r}^3+d{r}^4).$

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

$2b+6cr+12d{r}^2+4[\frac{a}{r}+2b+3cr+4d{r}^2]-\frac{r^2}{3}[\frac{a}{r}+2b]$

$\approx$

$-\frac{\mathfrak{𝔉}}{6}(1+ar+b{r}^2)-\frac{\alpha }{3}\left(\frac{r^2}{10}\right)$

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

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

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

This is a derivation check ... As was discovered by Yabushita, the analytically prescribed isothermal displacement function is, ${\sigma }_c^2=0$  and   $x_Y=1-\left(\frac{1}{\xi e^{-\psi }}\right)\frac{d\psi }{d\xi }.$

Maclaurin Spheroid Index Symbols

In our accompanying discussion of the equilibrium properties of models along the Maclaurin spheroid sequence, we find the "Index Symbols" expressions,

$A_1$	$=$	$\frac{1}{e^2}[\frac{{\sin }^{-1}e}{e}-(1-e^2{)}^{1/2}](1-e^2{)}^{1/2},$
$A_3$	$=$	$\frac{2}{e^2}[(1-e^2{)}^{-1/2}-\frac{{\sin }^{-1}e}{e}](1-e^2{)}^{1/2},$

where,

$e$

$\equiv$

$[1-(\frac{c}{a}{)}^2{]}^{1/2}$       (always positive).

Our aim, here, is to derive a power-series expression for these two index symbols (a) in the case of nearly spherical configurations $(e\ll 1)$, and (b) in the case of an infinitesimally thin disk $(c/a\ll 1)$.

Nearly Spherical Configurations

On p. 457 of [CRC], we find that,

$\frac{{\sin }^{-1}e}{e}$

$=$

$1+\left[\frac{1}{2\cdot 3}\right]e^2+\left[\frac{1\cdot 3}{2\cdot 4\cdot 5}\right]e^4+\left[\frac{1\cdot 3\cdot 5}{2\cdot 4\cdot 6\cdot 7}\right]e^6+\left[\frac{1\cdot 3\cdot 5\cdot 7}{2\cdot 4\cdot 6\cdot 8\cdot 9}\right]e^8+\cdots$       for,       $(e^2<1,-\frac{\pi }{2}<{\sin }^{-1}e<\frac{\pi }{2}).$

Also, from the above binomial-theorem expression, we have,

$(1-e^2{)}^{1/2}$	$=$	$1-\frac{1}{2}{e}^2+\left[\frac{\frac{1}{2}(\frac{1}{2}-1)}{2!}\right]e^4-\left[\frac{\frac{1}{2}(\frac{1}{2}-1)(\frac{1}{2}-2)}{3!}\right]e^6+\left[\frac{\frac{1}{2}(\frac{1}{2}-1)(\frac{1}{2}-2)(\frac{1}{2}-3)}{4!}\right]e^8-\cdots$      for $(e^4<1)$
	$=$	$1-\left[\frac{1}{2}\right]e^2-\left[\frac{1}{2^3}\right]e^4-\left[\frac{1}{2^4}\right]e^6-\left[\frac{5}{2^7}\right]e^8-\cdots$

So we can write,

${{ℱ}}_1\equiv (1-e^2{)}^{1/2}[\frac{{\sin }^{-1}e}{e}]$	$=$	$\{1-[\frac{1}{2}]e^2-[\frac{1}{2^3}]e^4-[\frac{1}{2^4}]e^6-[\frac{5}{2^7}]e^8-\cdots \}\times \{1+[\frac{1}{2\cdot 3}]e^2+[\frac{1\cdot 3}{2\cdot 4\cdot 5}]e^4+[\frac{1\cdot 3\cdot 5}{2\cdot 4\cdot 6\cdot 7}]e^6+[\frac{1\cdot 3\cdot 5\cdot 7}{2\cdot 4\cdot 6\cdot 8\cdot 9}]e^8+\cdots \}$
	$=$	$\{1+[\frac{1}{2\cdot 3}]e^2+[\frac{1\cdot 3}{2\cdot 4\cdot 5}]e^4+[\frac{1\cdot 3\cdot 5}{2\cdot 4\cdot 6\cdot 7}]e^6+[\frac{1\cdot 3\cdot 5\cdot 7}{2\cdot 4\cdot 6\cdot 8\cdot 9}\left]e^8\right\}-\frac{e^2}{2}\{1+[\frac{1}{2\cdot 3}]e^2+[\frac{1\cdot 3}{2\cdot 4\cdot 5}]e^4+[\frac{1\cdot 3\cdot 5}{2\cdot 4\cdot 6\cdot 7}\left]e^6\right\}$
		$-\frac{e^4}{2^3}\{1+[\frac{1}{2\cdot 3}]e^2+[\frac{1\cdot 3}{2\cdot 4\cdot 5}\left]e^4\right\}-\frac{e^6}{2^4}\{1+[\frac{1}{2\cdot 3}\left]e^2\right\}-\left[\frac{5}{2^7}\right]e^8+{𝒪}(e^{10})$
	$=$	$1+e^2[\frac{1}{2\cdot 3}-\frac{1}{2}]+e^4[\frac{1\cdot 3}{2\cdot 4\cdot 5}-\frac{1}{2^2\cdot 3}-\frac{1}{2^3}]+e^6[\frac{1\cdot 3\cdot 5}{2\cdot 4\cdot 6\cdot 7}-\frac{1}{2}\cdot \frac{1\cdot 3}{2\cdot 4\cdot 5}-\frac{1}{2^3}\cdot \frac{1}{2\cdot 3}-\frac{1}{2^4}]$
		$+e^8[\frac{1\cdot 3\cdot 5\cdot 7}{2\cdot 4\cdot 6\cdot 8\cdot 9}-\frac{1}{2}\cdot \frac{1\cdot 3\cdot 5}{2\cdot 4\cdot 6\cdot 7}-\frac{1}{2^3}\cdot \frac{1\cdot 3}{2\cdot 4\cdot 5}-\frac{1}{2^4}\cdot \frac{1}{2\cdot 3}-\frac{5}{2^7}]+{𝒪}(e^{10})$
	$=$	$1-e^2\left[\frac{1}{3}\right]+e^4\left[\frac{3^2-2\cdot 5-3\cdot 5}{2^3\cdot 3\cdot 5}\right]+e^6\left[\frac{3\cdot 5^2-3^2\cdot 7-5\cdot 7-3\cdot 5\cdot 7}{2^4\cdot 3\cdot 5\cdot 7}\right]+e^8\left[\frac{5^2\cdot 7^2-2^2\cdot 3^2\cdot 5^2-2\cdot 3^3\cdot 7-2^2\cdot 3\cdot 5\cdot 7-3^2\cdot 5^2\cdot 7}{2^7\cdot 3^2\cdot 5\cdot 7}\right]+{𝒪}(e^{10})$
	$=$	$1-e^2\left[\frac{1}{3}\right]-e^4\left[\frac{2}{3\cdot 5}\right]-e^6\left[\frac{2^3}{3\cdot 5\cdot 7}\right]-e^8\left[\frac{2^4}{3^2\cdot 5\cdot 7}\right]+{𝒪}(e^{10})$

Hence,

$e^2{A}_1$	$=$	${{ℱ}}_1-(1-e^2)$
	$=$	$e^2\left[\frac{2}{3}\right]-e^4\left[\frac{2}{3\cdot 5}\right]-e^6\left[\frac{2^3}{3\cdot 5\cdot 7}\right]-e^8\left[\frac{2^4}{3^2\cdot 5\cdot 7}\right]+{𝒪}(e^{10})$
$\Rightarrow A_1$	$=$	$\frac{2}{3}-e^2\left[\frac{2}{3\cdot 5}\right]-e^4\left[\frac{2^3}{3\cdot 5\cdot 7}\right]-e^6\left[\frac{2^4}{3^2\cdot 5\cdot 7}\right]+{𝒪}(e^8).$

And,

$\left(\frac{e^2}{2}\right)A_3$	$=$	$1-{{ℱ}}_1$
	$=$	$e^2\left[\frac{1}{3}\right]+e^4\left[\frac{2}{3\cdot 5}\right]+e^6\left[\frac{2^3}{3\cdot 5\cdot 7}\right]+e^8\left[\frac{2^4}{3^2\cdot 5\cdot 7}\right]+{𝒪}(e^{10})$
$\Rightarrow A_3$	$=$	$\frac{2}{3}+e^2\left[\frac{2^2}{3\cdot 5}\right]+e^4\left[\frac{2^4}{3\cdot 5\cdot 7}\right]+e^6\left[\frac{2^5}{3^2\cdot 5\cdot 7}\right]+{𝒪}(e^8).$

This looks okay, in the sense that $(2A_1+A_3)=2$.

Infinitesimally Thin Axisymmetric Disk

As $e\to 1$ — that is, in the case of an infinitesimally thin, axisymmetric disk — the preferred small parameter is,

$\frac{c}{a}$

$=$

$(1-e^2{)}^{1/2}\ll 1.$

Recognizing as well that,

${\sin }^{-1}e$	$=$	${\cos }^{-1}(1-e^2{)}^{1/2}={\cos }^{-1}(\frac{c}{a})$
$\Rightarrow {{ℱ}}_2\equiv \frac{{\sin }^{-1}e}{e^3}\cdot (1-e^2{)}^{1/2}$	$=$	$\left(\frac{c}{a}\right)\left[{\cos }^{-1}\right(\frac{c}{a}\left)\right][1-\frac{c^2}{a^2}{]}^{-3/2},$

the expressions for the pair of relevant index symbols may be rewritten as,

$A_1$	$=$	${{ℱ}}_2-\left(\frac{c}{a}{)}^2\right(1-\frac{c^2}{a^2}{)}^{-1},$
$\frac{A_3}{2}$	$=$	$(1-\frac{c^2}{a^2}{)}^{-1}-{{ℱ}}_2.$

Pulling again from p. 457 of [CRC], we find that,

${\cos }^{-1}\left(\frac{c}{a}\right)$

$=$

$\frac{\pi }{2}-\frac{c}{a}\{1+[\frac{1}{2\cdot 3}\left]\right(\frac{c}{a}{)}^2+\left[\frac{1\cdot 3}{2\cdot 4\cdot 5}\right](\frac{c}{a}{)}^4+[\frac{1\cdot 3\cdot 5}{2\cdot 4\cdot 6\cdot 7}\left]\right(\frac{c}{a}{)}^6+\left[\frac{1\cdot 3\cdot 5\cdot 7}{2\cdot 4\cdot 6\cdot 8\cdot 9}\right](\frac{c}{a}{)}^8+\cdots \}$       for,       $(\frac{c^2}{a^2}<1,0<{\cos }^{-1}(\frac{c}{a})<\pi ).$

LAGNIAPPE: According to the above binomial-theorem expression, we find for $(c^2/a^2<1)$, $\frac{1}{e}=[1-\frac{c^2}{a^2}{]}^{-1/2}$ $=$ $1+\frac{1}{2}(\frac{c}{a}{)}^2+[\frac{-\frac{1}{2}(-\frac{1}{2}-1)}{2!}\left]\right(\frac{c}{a}{)}^4-\left[\frac{-\frac{1}{2}(-\frac{1}{2}-1)(-\frac{1}{2}-2)}{3!}\right](\frac{c}{a}{)}^6+[\frac{-\frac{1}{2}(-\frac{1}{2}-1)(-\frac{1}{2}-2)(-\frac{1}{2}-3)}{4!}\left]\right(\frac{c}{a}{)}^8-\cdots$   $=$ $1+\frac{1}{2}(\frac{c}{a}{)}^2+[\frac{3}{2^2\cdot 2!}\left]\right(\frac{c}{a}{)}^4+\left[\frac{3\cdot 5}{2^3\cdot 3!}\right](\frac{c}{a}{)}^6+[\frac{3\cdot 5\cdot 7}{2^4\cdot 4!}\left]\right(\frac{c}{a}{)}^8+\cdots$   $=$ $1+\frac{1}{2}(\frac{c}{a}{)}^2+[\frac{3}{2^3}\left]\right(\frac{c}{a}{)}^4+\left[\frac{5}{2^4}\right](\frac{c}{a}{)}^6+[\frac{5\cdot 7}{2^7}\left]\right(\frac{c}{a}{)}^8+\cdots$ Hence, $\frac{{\sin }^{-1}e}{e}={\cos }^{-1}\left(\frac{c}{a}\right)[1-\frac{c^2}{a^2}{]}^{-1/2}$ $=$ $\frac{\pi }{2}\{1+\frac{1}{2}(\frac{c}{a}{)}^2+\left[\frac{3}{2^3}\right](\frac{c}{a}{)}^4+[\frac{5}{2^4}\left]\right(\frac{c}{a}{)}^6+\left[\frac{5\cdot 7}{2^7}\right](\frac{c}{a}{)}^8+\cdots \}$     $-\frac{c}{a}\{1+[\frac{1}{2\cdot 3}\left]\right(\frac{c}{a}{)}^2+\left[\frac{1\cdot 3}{2\cdot 4\cdot 5}\right](\frac{c}{a}{)}^4+[\frac{1\cdot 3\cdot 5}{2\cdot 4\cdot 6\cdot 7}\left]\right(\frac{c}{a}{)}^6+\left[\frac{1\cdot 3\cdot 5\cdot 7}{2\cdot 4\cdot 6\cdot 8\cdot 9}\right](\frac{c}{a}{)}^8+\cdots \}$     $\times \{1+\frac{1}{2}(\frac{c}{a}{)}^2+\left[\frac{3}{2^3}\right](\frac{c}{a}{)}^4+[\frac{5}{2^4}\left]\right(\frac{c}{a}{)}^6+\left[\frac{5\cdot 7}{2^7}\right](\frac{c}{a}{)}^8+\cdots \}$   $=$ $\frac{\pi }{2}\{1+[\frac{1}{2}\left]\right(\frac{c}{a}{)}^2+\left[\frac{3}{2^3}\right](\frac{c}{a}{)}^4+[\frac{5}{2^4}\left]\right(\frac{c}{a}{)}^6+\left[\frac{5\cdot 7}{2^7}\right]\left(\frac{c}{a}{)}^8\right\}$     $-\left\{\right(\frac{c}{a})+[\frac{1}{2\cdot 3}\left]\right(\frac{c}{a}{)}^3+\left[\frac{1\cdot 3}{2\cdot 4\cdot 5}\right](\frac{c}{a}{)}^5+[\frac{1\cdot 3\cdot 5}{2\cdot 4\cdot 6\cdot 7}\left]\right(\frac{c}{a}{)}^7+\left[\frac{1\cdot 3\cdot 5\cdot 7}{2\cdot 4\cdot 6\cdot 8\cdot 9}\right]\left(\frac{c}{a}{)}^9\right\}$     $-\frac{1}{2}\left\{\right(\frac{c}{a}{)}^3+\left[\frac{1}{2\cdot 3}\right](\frac{c}{a}{)}^5+[\frac{1\cdot 3}{2\cdot 4\cdot 5}\left]\right(\frac{c}{a}{)}^7+\left[\frac{1\cdot 3\cdot 5}{2\cdot 4\cdot 6\cdot 7}\right]\left(\frac{c}{a}{)}^9\right\}$     $-\left[\frac{3}{2^3}\right]\left\{\right(\frac{c}{a}{)}^5+\left[\frac{1}{2\cdot 3}\right](\frac{c}{a}{)}^7+[\frac{1\cdot 3}{2\cdot 4\cdot 5}\left]\right(\frac{c}{a}{)}^9\}-[\frac{5}{2^4}\left]\right\{(\frac{c}{a}{)}^7+[\frac{1}{2\cdot 3}\left]\right(\frac{c}{a}{)}^9\}-[\frac{5\cdot 7}{2^7}\left]\right(\frac{c}{a}{)}^9+{𝒪}\left(\frac{c^{10}}{a^{10}}\right)$   $=$ $\frac{\pi }{2}+\frac{\pi }{2}\left[\frac{1}{2}\right](\frac{c}{a}{)}^2+\frac{\pi }{2}[\frac{3}{2^3}\left]\right(\frac{c}{a}{)}^4+\frac{\pi }{2}\left[\frac{5}{2^4}\right](\frac{c}{a}{)}^6+\frac{\pi }{2}[\frac{5\cdot 7}{2^7}\left]\right(\frac{c}{a}{)}^8$     $-\left(\frac{c}{a}\right)-\left[\frac{1}{2\cdot 3}\right](\frac{c}{a}{)}^3-[\frac{1\cdot 3}{2\cdot 4\cdot 5}\left]\right(\frac{c}{a}{)}^5-\left[\frac{1\cdot 3\cdot 5}{2\cdot 4\cdot 6\cdot 7}\right](\frac{c}{a}{)}^7-[\frac{1\cdot 3\cdot 5\cdot 7}{2\cdot 4\cdot 6\cdot 8\cdot 9}\left]\right(\frac{c}{a}{)}^9$     $-\frac{1}{2}(\frac{c}{a}{)}^3-[\frac{1}{2^2\cdot 3}\left]\right(\frac{c}{a}{)}^5-\left[\frac{1\cdot 3}{2^2\cdot 4\cdot 5}\right](\frac{c}{a}{)}^7-[\frac{1\cdot 3\cdot 5}{2^2\cdot 4\cdot 6\cdot 7}\left]\right(\frac{c}{a}{)}^9$     $-\left[\frac{3}{2^3}\right](\frac{c}{a}{)}^5-[\frac{3}{2^3}\left]\right[\frac{1}{2\cdot 3}\left]\right(\frac{c}{a}{)}^7-\left[\frac{3}{2^3}\right]\left[\frac{1\cdot 3}{2\cdot 4\cdot 5}\right](\frac{c}{a}{)}^9-[\frac{5}{2^4}\left]\right(\frac{c}{a}{)}^7-\left[\frac{5}{2^4}\right]\left[\frac{1}{2\cdot 3}\right](\frac{c}{a}{)}^9-[\frac{5\cdot 7}{2^7}\left]\right(\frac{c}{a}{)}^9+{𝒪}\left(\frac{c^{10}}{a^{10}}\right).$ (continue expression simplification) $\frac{{\sin }^{-1}e}{e}={\cos }^{-1}\left(\frac{c}{a}\right)[1-\frac{c^2}{a^2}{]}^{-1/2}$ $=$ $\frac{\pi }{2}+\frac{\pi }{2}\left[\frac{1}{2}\right](\frac{c}{a}{)}^2+\frac{\pi }{2}[\frac{3}{2^3}\left]\right(\frac{c}{a}{)}^4+\frac{\pi }{2}\left[\frac{5}{2^4}\right](\frac{c}{a}{)}^6+\frac{\pi }{2}[\frac{5\cdot 7}{2^7}\left]\right(\frac{c}{a}{)}^8$     $-\left(\frac{c}{a}\right)-\left[\frac{1}{2\cdot 3}\right](\frac{c}{a}{)}^3-[\frac{3}{2^3\cdot 5}\left]\right(\frac{c}{a}{)}^5-\left[\frac{5}{2^4\cdot 7}\right](\frac{c}{a}{)}^7-[\frac{5\cdot 7}{2^7\cdot 3^2}\left]\right(\frac{c}{a}{)}^9$     $-\frac{1}{2}(\frac{c}{a}{)}^3-[\frac{1}{2^2\cdot 3}\left]\right(\frac{c}{a}{)}^5-\left[\frac{3}{2^4\cdot 5}\right](\frac{c}{a}{)}^7-[\frac{5}{2^5\cdot 7}\left]\right(\frac{c}{a}{)}^9$     $-\left[\frac{3}{2^3}\right](\frac{c}{a}{)}^5-[\frac{1}{2^4}\left]\right(\frac{c}{a}{)}^7-\left[\frac{3^2}{2^6\cdot 5}\right](\frac{c}{a}{)}^9-[\frac{5}{2^4}\left]\right(\frac{c}{a}{)}^7-\left[\frac{5}{2^5\cdot 3}\right](\frac{c}{a}{)}^9-[\frac{5\cdot 7}{2^7}\left]\right(\frac{c}{a}{)}^9+{𝒪}\left(\frac{c^{10}}{a^{10}}\right)$   $=$ $\frac{\pi }{2}+\frac{\pi }{2}\left[\frac{1}{2}\right](\frac{c}{a}{)}^2+\frac{\pi }{2}[\frac{3}{2^3}\left]\right(\frac{c}{a}{)}^4+\frac{\pi }{2}\left[\frac{5}{2^4}\right](\frac{c}{a}{)}^6+\frac{\pi }{2}[\frac{5\cdot 7}{2^7}\left]\right(\frac{c}{a}{)}^8$     $-\left(\frac{c}{a}\right)-\left\{\right[\frac{1}{2\cdot 3}]+\frac{1}{2}\}(\frac{c}{a}{)}^3-\{\left[\frac{3}{2^3\cdot 5}\right]+\left[\frac{1}{2^2\cdot 3}\right]+\left[\frac{3}{2^3}\right]\left\}\right(\frac{c}{a}{)}^5-\left\{\right[\frac{5}{2^4\cdot 7}]+[\frac{3}{2^4\cdot 5}]+[\frac{1}{2^4}]+[\frac{5}{2^4}\left]\right\}(\frac{c}{a}{)}^7$     $-\left\{\right[\frac{5\cdot 7}{2^7\cdot 3^2}]+[\frac{5}{2^5\cdot 7}]+[\frac{3^2}{2^6\cdot 5}]+[\frac{5}{2^5\cdot 3}]+[\frac{5\cdot 7}{2^7}\left]\right\}(\frac{c}{a}{)}^9+{𝒪}(\frac{c^{10}}{a^{10}})$   $=$ $\frac{\pi }{2}+\frac{\pi }{2}\left[\frac{1}{2}\right](\frac{c}{a}{)}^2+\frac{\pi }{2}[\frac{3}{2^3}\left]\right(\frac{c}{a}{)}^4+\frac{\pi }{2}\left[\frac{5}{2^4}\right](\frac{c}{a}{)}^6+\frac{\pi }{2}[\frac{5\cdot 7}{2^7}\left]\right(\frac{c}{a}{)}^8$     $-\left(\frac{c}{a}\right)-\left[\frac{2}{3}\right](\frac{c}{a}{)}^3-[\frac{2^3}{3\cdot 5}\left]\right(\frac{c}{a}{)}^5-\left[\frac{2^4}{5\cdot 7}\right](\frac{c}{a}{)}^7-[\frac{2^7}{3^2\cdot 5\cdot 7}\left]\right(\frac{c}{a}{)}^9+{𝒪}\left(\frac{c^{10}}{a^{10}}\right).$

Referring again to the above binomial-theorem expression, we find for $(c^2/a^2<1)$,

$[1-\frac{c^2}{a^2}{]}^{-3/2}$	$=$	$1+\frac{3}{2}(\frac{c}{a}{)}^2+[\frac{-\frac{3}{2}(-\frac{3}{2}-1)}{2!}\left]\right(\frac{c}{a}{)}^4-\left[\frac{-\frac{3}{2}(-\frac{3}{2}-1)(-\frac{3}{2}-2)}{3!}\right](\frac{c}{a}{)}^6+[\frac{-\frac{3}{2}(-\frac{3}{2}-1)(-\frac{3}{2}-2)(-\frac{3}{2}-3)}{4!}\left]\right(\frac{c}{a}{)}^8-\cdots$
	$=$	$1+\frac{3}{2}(\frac{c}{a}{)}^2+[\frac{\frac{3}{2}(\frac{3}{2}+1)}{2!}\left]\right(\frac{c}{a}{)}^4+\left[\frac{\frac{3}{2}(\frac{3}{2}+1)(\frac{3}{2}+2)}{3!}\right](\frac{c}{a}{)}^6+[\frac{\frac{3}{2}(\frac{3}{2}+1)(\frac{3}{2}+2)(\frac{3}{2}+3)}{4!}\left]\right(\frac{c}{a}{)}^8+\cdots$
	$=$	$1+\frac{3}{2}(\frac{c}{a}{)}^2+[\frac{3\cdot 5}{2^2\cdot 2!}\left]\right(\frac{c}{a}{)}^4+\left[\frac{3\cdot 5\cdot 7}{2^3\cdot 3!}\right](\frac{c}{a}{)}^6+[\frac{3\cdot 5\cdot 7\cdot 9}{2^4\cdot 4!}\left]\right(\frac{c}{a}{)}^8+\cdots$
	$=$	$1+\frac{3}{2}(\frac{c}{a}{)}^2+[\frac{3\cdot 5}{2^3}\left]\right(\frac{c}{a}{)}^4+\left[\frac{5\cdot 7}{2^4}\right](\frac{c}{a}{)}^6+[\frac{5\cdot 7\cdot 9}{2^7}\left]\right(\frac{c}{a}{)}^8+\cdots$

We therefore can write,

${{ℱ}}_2$	$=$	$\left\{\frac{\pi }{2}\right(\frac{c}{a})-(\frac{c}{a}{)}^2-\left[\frac{1}{2\cdot 3}\right](\frac{c}{a}{)}^4-[\frac{1\cdot 3}{2\cdot 4\cdot 5}\left]\right(\frac{c}{a}{)}^6-\left[\frac{1\cdot 3\cdot 5}{2\cdot 4\cdot 6\cdot 7}\right](\frac{c}{a}{)}^8-[\frac{1\cdot 3\cdot 5\cdot 7}{2\cdot 4\cdot 6\cdot 8\cdot 9}\left]\right(\frac{c}{a}{)}^{10}-\cdots \}$
		$\times \{1+\frac{3}{2}(\frac{c}{a}{)}^2+\left[\frac{3\cdot 5}{2^3}\right](\frac{c}{a}{)}^4+[\frac{5\cdot 7}{2^4}\left]\right(\frac{c}{a}{)}^6+\left[\frac{5\cdot 7\cdot 9}{2^7}\right](\frac{c}{a}{)}^8+\cdots \}$
	$=$	$\frac{\pi }{2}\left(\frac{c}{a}\right)-(\frac{c}{a}{)}^2-[\frac{1}{2\cdot 3}\left]\right(\frac{c}{a}{)}^4-\left[\frac{1\cdot 3}{2\cdot 4\cdot 5}\right](\frac{c}{a}{)}^6-[\frac{1\cdot 3\cdot 5}{2\cdot 4\cdot 6\cdot 7}\left]\right(\frac{c}{a}{)}^8-\left[\frac{1\cdot 3\cdot 5\cdot 7}{2\cdot 4\cdot 6\cdot 8\cdot 9}\right](\frac{c}{a}{)}^{10}$
		$+\frac{3}{2}\left(\frac{c}{a}{)}^2\right\{\frac{\pi }{2}\left(\frac{c}{a}\right)-(\frac{c}{a}{)}^2-[\frac{1}{2\cdot 3}\left]\right(\frac{c}{a}{)}^4-\left[\frac{1\cdot 3}{2\cdot 4\cdot 5}\right](\frac{c}{a}{)}^6-[\frac{1\cdot 3\cdot 5}{2\cdot 4\cdot 6\cdot 7}\left]\right(\frac{c}{a}{)}^8\}$
		$+\left[\frac{3\cdot 5}{2^3}\right]\left(\frac{c}{a}{)}^4\right\{\frac{\pi }{2}\left(\frac{c}{a}\right)-(\frac{c}{a}{)}^2-[\frac{1}{2\cdot 3}\left]\right(\frac{c}{a}{)}^4-\left[\frac{1\cdot 3}{2\cdot 4\cdot 5}\right]\left(\frac{c}{a}{)}^6\right\}$
		$+\left[\frac{5\cdot 7}{2^4}\right]\left(\frac{c}{a}{)}^6\right\{\frac{\pi }{2}\left(\frac{c}{a}\right)-(\frac{c}{a}{)}^2-[\frac{1}{2\cdot 3}\left]\right(\frac{c}{a}{)}^4\}+[\frac{5\cdot 7\cdot 9}{2^7}\left]\right(\frac{c}{a}{)}^8\left\{\frac{\pi }{2}\right(\frac{c}{a})-(\frac{c}{a}{)}^2\}+{𝒪}(\frac{c^{10}}{a^{10}})$
	$=$	$\frac{\pi }{2}\left(\frac{c}{a}\right)-(\frac{c}{a}{)}^2-[\frac{1}{2\cdot 3}\left]\right(\frac{c}{a}{)}^4-\left[\frac{1\cdot 3}{2\cdot 4\cdot 5}\right](\frac{c}{a}{)}^6-[\frac{1\cdot 3\cdot 5}{2\cdot 4\cdot 6\cdot 7}\left]\right(\frac{c}{a}{)}^8+\pi \left[\frac{3}{2^2}\right](\frac{c}{a}{)}^3-[\frac{3}{2}\left]\right(\frac{c}{a}{)}^4-\left[\frac{1}{2^2}\right](\frac{c}{a}{)}^6-[\frac{3^2}{2^4\cdot 5}\left]\right(\frac{c}{a}{)}^8$
		$+\pi \left[\frac{3\cdot 5}{2^4}\right](\frac{c}{a}{)}^5-[\frac{3\cdot 5}{2^3}\left]\right(\frac{c}{a}{)}^6-\left[\frac{5}{2^4}\right](\frac{c}{a}{)}^8+\pi [\frac{5\cdot 7}{2^5}\left]\right(\frac{c}{a}{)}^7-\left[\frac{5\cdot 7}{2^4}\right](\frac{c}{a}{)}^8+\pi [\frac{5\cdot 7\cdot 9}{2^8}\left]\right(\frac{c}{a}{)}^9+{𝒪}\left(\frac{c^{10}}{a^{10}}\right)$
	$=$	$\frac{\pi }{2}\left(\frac{c}{a}\right)-(\frac{c}{a}{)}^2+\pi [\frac{3}{2^2}\left]\right(\frac{c}{a}{)}^3-[\frac{1}{2\cdot 3}+\frac{3}{2}](\frac{c}{a}{)}^4+\pi [\frac{3\cdot 5}{2^4}\left]\right(\frac{c}{a}{)}^5-[\frac{1\cdot 3}{2\cdot 4\cdot 5}+\frac{1}{2^2}+\frac{3\cdot 5}{2^3}](\frac{c}{a}{)}^6$
		$+\pi \left[\frac{5\cdot 7}{2^5}\right](\frac{c}{a}{)}^7-[\frac{1\cdot 3\cdot 5}{2\cdot 4\cdot 6\cdot 7}+\frac{3^2}{2^4\cdot 5}+\frac{5}{2^4}+\frac{5\cdot 7}{2^4}\left]\right(\frac{c}{a}{)}^8+\pi \left[\frac{5\cdot 7\cdot 9}{2^8}\right](\frac{c}{a}{)}^9+{𝒪}(\frac{c^{10}}{a^{10}})$
	$=$	$\frac{\pi }{2}\left(\frac{c}{a}\right)-(\frac{c}{a}{)}^2+\pi [\frac{3}{2^2}\left]\right(\frac{c}{a}{)}^3-\left[\frac{5}{3}\right](\frac{c}{a}{)}^4+\pi [\frac{3\cdot 5}{2^4}\left]\right(\frac{c}{a}{)}^5-\left[\frac{11}{5}\right](\frac{c}{a}{)}^6+\pi [\frac{5\cdot 7}{2^5}\left]\right(\frac{c}{a}{)}^7-\left[\frac{93}{5\cdot 7}\right](\frac{c}{a}{)}^8+\pi [\frac{5\cdot 7\cdot 9}{2^8}\left]\right(\frac{c}{a}{)}^9+{𝒪}\left(\frac{c^{10}}{a^{10}}\right).$

Once again from the binomial theorem,

$(1-\frac{c^2}{a^2}{)}^{-1}$

$=$

$1+(\frac{c}{a}{)}^2+(\frac{c}{a}{)}^4+(\frac{c}{a}{)}^6+(\frac{c}{a}{)}^8+(\frac{c}{a}{)}^{10}+\cdots$

which gives us,

$A_1={{ℱ}}_2-\left(\frac{c}{a}{)}^2\right(1-\frac{c^2}{a^2}{)}^{-1}$	$=$	$\frac{\pi }{2}\left(\frac{c}{a}\right)-(\frac{c}{a}{)}^2+\pi [\frac{3}{2^2}\left]\right(\frac{c}{a}{)}^3-\left[\frac{5}{3}\right](\frac{c}{a}{)}^4+\pi [\frac{3\cdot 5}{2^4}\left]\right(\frac{c}{a}{)}^5-\left[\frac{11}{5}\right](\frac{c}{a}{)}^6+\pi [\frac{5\cdot 7}{2^5}\left]\right(\frac{c}{a}{)}^7-\left[\frac{93}{5\cdot 7}\right](\frac{c}{a}{)}^8+\pi [\frac{3^2\cdot 5\cdot 7}{2^8}\left]\right(\frac{c}{a}{)}^9$
		$-\left\{\right(\frac{c}{a}{)}^2+(\frac{c}{a}{)}^4+(\frac{c}{a}{)}^6+\left(\frac{c}{a}{)}^8\right\}+{𝒪}\left(\frac{c^{10}}{a^{10}}\right)$
	$=$	$\frac{\pi }{2}\left(\frac{c}{a}\right)-2(\frac{c}{a}{)}^2+\pi [\frac{3}{2^2}\left]\right(\frac{c}{a}{)}^3-\left[\frac{8}{3}\right](\frac{c}{a}{)}^4+\pi [\frac{3\cdot 5}{2^4}\left]\right(\frac{c}{a}{)}^5-\left[\frac{2^4}{5}\right](\frac{c}{a}{)}^6+\pi [\frac{5\cdot 7}{2^5}\left]\right(\frac{c}{a}{)}^7-\left[\frac{2^7}{5\cdot 7}\right](\frac{c}{a}{)}^8+\pi [\frac{3^2\cdot 5\cdot 7}{2^8}\left]\right(\frac{c}{a}{)}^9+{𝒪}\left(\frac{c^{10}}{a^{10}}\right).$

And,

$\frac{A_3}{2}=(1-\frac{c^2}{a^2}{)}^{-1}-{{ℱ}}_2$	$=$	$\{1+(\frac{c}{a}{)}^2+(\frac{c}{a}{)}^4+(\frac{c}{a}{)}^6+(\frac{c}{a}{)}^8+(\frac{c}{a}{)}^{10}+\cdots \}$
		$-\left\{\frac{\pi }{2}\right(\frac{c}{a})-(\frac{c}{a}{)}^2+\pi \left[\frac{3}{2^2}\right](\frac{c}{a}{)}^3-[\frac{5}{3}\left]\right(\frac{c}{a}{)}^4+\pi \left[\frac{3\cdot 5}{2^4}\right](\frac{c}{a}{)}^5-[\frac{11}{5}\left]\right(\frac{c}{a}{)}^6+\pi \left[\frac{5\cdot 7}{2^5}\right](\frac{c}{a}{)}^7-[\frac{93}{5\cdot 7}\left]\right(\frac{c}{a}{)}^8+\pi \left[\frac{3^2\cdot 5\cdot 7}{2^8}\right]\left(\frac{c}{a}{)}^9\right\}+{𝒪}\left(\frac{c^{10}}{a^{10}}\right)$
	$=$	$1-\frac{\pi }{2}\left(\frac{c}{a}\right)+2(\frac{c}{a}{)}^2-\pi [\frac{3}{2^2}\left]\right(\frac{c}{a}{)}^3+\left[\frac{8}{3}\right](\frac{c}{a}{)}^4-\pi [\frac{3\cdot 5}{2^4}\left]\right(\frac{c}{a}{)}^5+\left[\frac{2^4}{5}\right](\frac{c}{a}{)}^6-\pi [\frac{5\cdot 7}{2^5}\left]\right(\frac{c}{a}{)}^7+\left[\frac{2^7}{5\cdot 7}\right](\frac{c}{a}{)}^8-\pi [\frac{3^2\cdot 5\cdot 7}{2^8}\left]\right(\frac{c}{a}{)}^9\}+{𝒪}(\frac{c^{10}}{a^{10}}).$

Notice that, to the highest order retained in these expressions, we find as expected that, $(A_1+A_3/2)=1$.

Frequency (temporary)

${\omega }_0^2$	$=$	$2\pi G\rho [A_1-A_3(1-e^2)]$
$\Rightarrow {\mathrm{\Omega }}_{\mathrm{M}\mathrm{c}}^2\equiv \frac{{\omega }_0^2}{\pi G\rho }$	$=$	$2(3-2e^2)(1-e^2{)}^{1/2}\cdot \frac{{\sin }^{-1}e}{e^3}-\frac{6(1-e^2)}{e^2},$

Taylor Series (Hunter77)

First (Unsuccessful) Try

First:

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

Note that, replacing the $(\mathrm{\Delta }{)}^3{f}_3^{{\text{'}}^{″}}$ term with the expression derived in the Second step, below, gives,

$-\frac{3^2}{2}(\mathrm{\Delta }{)}^2{f}_3^{{\text{'}}^{\prime }}$	$=$	$f_3-f_0-(3\mathrm{\Delta })f_3^{\text{'}}+\frac{3^3}{2^3}(\mathrm{\Delta }{)}^4{f}_3^{iv}+{𝒪}({\mathrm{\Delta }}^5)$
		$-\frac{3^2}{2}\left\{\right[\frac{2^2}{3^2}]f_0-f_1+f_3[\frac{5}{3^2}]+[-\frac{2}{3}](\mathrm{\Delta })f_3^{\text{'}}+[-\frac{5}{6}](\mathrm{\Delta }{)}^4{f}_3^{iv}\left\}\right[-\frac{3}{2}]$
	$=$	$f_3-f_0-3(\mathrm{\Delta })f_3^{\text{'}}+\frac{3^3}{2^3}(\mathrm{\Delta }{)}^4{f}_3^{iv}+{𝒪}({\mathrm{\Delta }}^5)$
		$+\{3f_0+[-\frac{3^3}{2^2}]f_1+[\frac{15}{2^2}]f_3+[-\frac{3}{2}](\mathrm{\Delta })f_3^{\text{'}}+[-\frac{3^2\cdot 5}{2^3}](\mathrm{\Delta }{)}^4{f}_3^{iv}\}$
	$=$	$2f_0+[-\frac{3^3}{2^2}]f_1+[1+\frac{15}{2^2}]f_3+[-3-\frac{3}{2}](\mathrm{\Delta })f_3^{\text{'}}+[\frac{3^3}{2^3}-\frac{3^2\cdot 5}{2^3}](\mathrm{\Delta }{)}^4{f}_3^{iv}+{𝒪}({\mathrm{\Delta }}^5)$
	$=$	$2f_0+[-\frac{3^3}{2^2}]f_1+\left[\frac{19}{2^2}\right]f_3+[-\frac{9}{2}](\mathrm{\Delta })f_3^{\text{'}}+[-\frac{9}{4}](\mathrm{\Delta }{)}^4{f}_3^{iv}+{𝒪}({\mathrm{\Delta }}^5)$

Then, replacing the $(\mathrm{\Delta }{)}^4{f}_3^{iv}$ term with the expression derived in the Third step, below, gives,

$-\frac{3^2}{2}(\mathrm{\Delta }{)}^2{f}_3^{{\text{'}}^{\prime }}$	$=$	$2f_0+[-\frac{3^3}{2^2}]f_1+\left[\frac{19}{2^2}\right]f_3+[-\frac{9}{2}](\mathrm{\Delta })f_3^{\text{'}}+{𝒪}({\mathrm{\Delta }}^5)$
		$+[-\frac{9}{4}]\left\{\right[-\frac{1}{3^2}]f_0+[\frac{1}{2}]f_1-f_2+[\frac{11}{2\cdot 3^2}]f_3+[-\frac{2}{3}](\mathrm{\Delta })f_3^{\text{'}}\}[-2^2\cdot 3]$
	$=$	$2f_0+[-\frac{3^3}{2^2}]f_1+\left[\frac{19}{2^2}\right]f_3+[-\frac{9}{2}](\mathrm{\Delta })f_3^{\text{'}}+{𝒪}({\mathrm{\Delta }}^5)$
		$+\left\{\right[-3]f_0+[\frac{3^3}{2}]f_1-3^3{f}_2+[\frac{3\cdot 11}{2}]f_3+[-2\cdot 3^2](\mathrm{\Delta })f_3^{\text{'}}\}$
	$=$	$-f_0+[\frac{3^3}{2}-\frac{3^3}{2^2}]f_1-3^3{f}_2+[\frac{3\cdot 11}{2}+\frac{19}{2^2}]f_3+[-2\cdot 3^2-\frac{9}{2}](\mathrm{\Delta })f_3^{\text{'}}+{𝒪}({\mathrm{\Delta }}^5)$
	$=$	$-f_0+\left[\frac{3^3}{2^2}\right]f_1-3^3{f}_2+\left[\frac{5\cdot 17}{2^2}\right]f_3+[-\frac{3^2\cdot 5}{2}](\mathrm{\Delta })f_3^{\text{'}}+{𝒪}({\mathrm{\Delta }}^5)$

Second:

$f_1$	$=$	$f_3+(-2\mathrm{\Delta })f_3^{\text{'}}+\frac{1}{2}(-2\mathrm{\Delta }{)}^2{f}_3^{{\text{'}}^{\prime }}+\frac{1}{6}(-2\mathrm{\Delta }{)}^3{f}_3^{{\text{'}}^{″}}+\frac{1}{24}(-2\mathrm{\Delta }{)}^4{f}_3^{iv}+{𝒪}({\mathrm{\Delta }}^5)$
	$=$	$f_3-2(\mathrm{\Delta })f_3^{\text{'}}+2(\mathrm{\Delta }{)}^2{f}_3^{{\text{'}}^{\prime }}-\frac{2^2}{3}(\mathrm{\Delta }{)}^3{f}_3^{{\text{'}}^{″}}+\frac{2}{3}(\mathrm{\Delta }{)}^4{f}_3^{iv}+{𝒪}({\mathrm{\Delta }}^5)$
	$=$	$f_3-2(\mathrm{\Delta })f_3^{\text{'}}-\frac{2^2}{3}(\mathrm{\Delta }{)}^3{f}_3^{{\text{'}}^{″}}+\frac{2}{3}(\mathrm{\Delta }{)}^4{f}_3^{iv}+{𝒪}({\mathrm{\Delta }}^5)$
		$-\left[\frac{2^2}{3^2}\right][f_3-f_0-(3\mathrm{\Delta })f_3^{\text{'}}-\frac{3^2}{2}(\mathrm{\Delta }{)}^3{f}_3^{{\text{'}}^{″}}+\frac{3^3}{2^3}(\mathrm{\Delta }{)}^4{f}_3^{iv}+{𝒪}({\mathrm{\Delta }}^5)]$
	$=$	$\left[\frac{2^2}{3^2}\right]f_0+f_3[1-\frac{2^2}{3^2}]+[\frac{2^2}{3^2}(3\mathrm{\Delta })-2(\mathrm{\Delta })]f_3^{\text{'}}+\left[\right(\frac{2^2}{3^2})\frac{3^2}{2}(\mathrm{\Delta }{)}^3-\frac{2^2}{3}(\mathrm{\Delta }{)}^3]f_3^{{\text{'}}^{″}}$
		$+[\frac{2}{3}(\mathrm{\Delta }{)}^4-\left(\frac{2^2}{3^2}\right)\frac{3^3}{2^3}(\mathrm{\Delta }{)}^4]f_3^{iv}+{𝒪}({\mathrm{\Delta }}^5)$
	$=$	$\left[\frac{2^2}{3^2}\right]f_0+f_3\left[\frac{5}{3^2}\right]+[-\frac{2}{3}](\mathrm{\Delta })f_3^{\text{'}}+\left[\frac{2}{3}\right](\mathrm{\Delta }{)}^3{f}_3^{{\text{'}}^{″}}+[-\frac{5}{6}](\mathrm{\Delta }{)}^4{f}_3^{iv}+{𝒪}({\mathrm{\Delta }}^5)$
$\Rightarrow -\left[\frac{2}{3}\right](\mathrm{\Delta }{)}^3{f}_3^{{\text{'}}^{″}}$	$=$	$\left[\frac{2^2}{3^2}\right]f_0-f_1+f_3\left[\frac{5}{3^2}\right]+[-\frac{2}{3}](\mathrm{\Delta })f_3^{\text{'}}+[-\frac{5}{6}](\mathrm{\Delta }{)}^4{f}_3^{iv}+{𝒪}({\mathrm{\Delta }}^5)$

Now, replacing the $(\mathrm{\Delta }{)}^4{f}_3^{iv}$ term with the expression derived in the Third step, below, gives,

$-\left[\frac{2}{3}\right](\mathrm{\Delta }{)}^3{f}_3^{{\text{'}}^{″}}$	$=$	$\left[\frac{2^2}{3^2}\right]f_0-f_1+f_3\left[\frac{5}{3^2}\right]+[-\frac{2}{3}](\mathrm{\Delta })f_3^{\text{'}}+{𝒪}({\mathrm{\Delta }}^5)$
		$+[-\frac{5}{6}]\left\{\right[-\frac{1}{3^2}]f_0+[\frac{1}{2}]f_1-f_2+[\frac{11}{2\cdot 3^2}]f_3+[-\frac{2}{3}](\mathrm{\Delta })f_3^{\text{'}}\}[-2^2\cdot 3]$
	$=$	$\left[\frac{2^2}{3^2}\right]f_0-f_1+f_3\left[\frac{5}{3^2}\right]+[-\frac{2}{3}](\mathrm{\Delta })f_3^{\text{'}}+{𝒪}({\mathrm{\Delta }}^5)$
		$+\left\{\right[-\frac{2\cdot 5}{3^2}]f_0+[5]f_1+[-2\cdot 5]f_2+[\frac{5\cdot 11}{3^2}]f_3+[-\frac{2^2\cdot 5}{3}](\mathrm{\Delta })f_3^{\text{'}}\}$
	$=$	$[\frac{2^2}{3^2}-\frac{2\cdot 5}{3^2}]f_0+\left[4\right]f_1+[-2\cdot 5]f_2+[\frac{5}{3^2}+\frac{5\cdot 11}{3^2}]f_3+[-\frac{2^2\cdot 5}{3}-\frac{2}{3}](\mathrm{\Delta })f_3^{\text{'}}+{𝒪}({\mathrm{\Delta }}^5)$
	$=$	$[-\frac{2}{3}]f_0+\left[4\right]f_1+[-2\cdot 5]f_2+\left[\frac{2^2\cdot 5}{3}\right]f_3+[-\frac{2\cdot 11}{3}](\mathrm{\Delta })f_3^{\text{'}}+{𝒪}({\mathrm{\Delta }}^5)$

Third:

$f_2$	$=$	$f_3+(-\mathrm{\Delta })f_3^{\text{'}}+\frac{1}{2}(-\mathrm{\Delta }{)}^2{f}_3^{{\text{'}}^{\prime }}+\frac{1}{6}(-\mathrm{\Delta }{)}^3{f}_3^{{\text{'}}^{″}}+\frac{1}{24}(-\mathrm{\Delta }{)}^4{f}_3^{iv}+{𝒪}({\mathrm{\Delta }}^5)$
	$=$	$f_3+[-1](\mathrm{\Delta })f_3^{\text{'}}+\left[\frac{1}{2}\right](\mathrm{\Delta }{)}^2{f}_3^{{\text{'}}^{\prime }}+[-\frac{1}{2\cdot 3}](\mathrm{\Delta }{)}^3{f}_3^{{\text{'}}^{″}}+\left[\frac{1}{2^3\cdot 3}\right](\mathrm{\Delta }{)}^4{f}_3^{iv}+{𝒪}({\mathrm{\Delta }}^5)$
	$=$	$f_3+[-1](\mathrm{\Delta })f_3^{\text{'}}+\left[\frac{1}{2^3\cdot 3}\right](\mathrm{\Delta }{)}^4{f}_3^{iv}+{𝒪}({\mathrm{\Delta }}^5)$
		$+\left[\frac{1}{2}\right]\{2f_0+[-\frac{3^3}{2^2}]f_1+[\frac{19}{2^2}]f_3+[-\frac{9}{2}](\mathrm{\Delta })f_3^{\text{'}}+[-\frac{9}{4}](\mathrm{\Delta }{)}^4{f}_3^{iv}\left\}\right[-\frac{2}{3^2}]$
		$+[-\frac{1}{2\cdot 3}]\left\{\right[\frac{2^2}{3^2}]f_0-f_1+f_3[\frac{5}{3^2}]+[-\frac{2}{3}](\mathrm{\Delta })f_3^{\text{'}}+[-\frac{5}{6}](\mathrm{\Delta }{)}^4{f}_3^{iv}\left\}\right[-\frac{3}{2}]$
	$=$	$f_3+[-1](\mathrm{\Delta })f_3^{\text{'}}+\left[\frac{1}{2^3\cdot 3}\right](\mathrm{\Delta }{)}^4{f}_3^{iv}+{𝒪}({\mathrm{\Delta }}^5)$
		$+\left\{\right[-\frac{2}{3^2}]f_0+[\frac{3}{2^2}]f_1+[-\frac{19}{2^2\cdot 3^2}]f_3+[\frac{1}{2}](\mathrm{\Delta })f_3^{\text{'}}+[\frac{1}{4}](\mathrm{\Delta }{)}^4{f}_3^{iv}\}$
		$+\left\{\right[\frac{1}{3^2}]f_0+[-\frac{1}{2^2}]f_1+f_3[\frac{5}{2^2\cdot 3^2}]+[-\frac{1}{2\cdot 3}](\mathrm{\Delta })f_3^{\text{'}}+[-\frac{5}{2^3\cdot 3}](\mathrm{\Delta }{)}^4{f}_3^{iv}\}$
	$=$	$f_3+[-1](\mathrm{\Delta })f_3^{\text{'}}+\left[\frac{1}{2^3\cdot 3}\right](\mathrm{\Delta }{)}^4{f}_3^{iv}+{𝒪}({\mathrm{\Delta }}^5)$
		$+\left\{\right[\frac{1}{3^2}-\frac{2}{3^2}]f_0+[\frac{3}{2^2}-\frac{1}{2^2}]f_1+[\frac{5}{2^2\cdot 3^2}-\frac{19}{2^2\cdot 3^2}]f_3+[\frac{1}{2}-\frac{1}{2\cdot 3}](\mathrm{\Delta })f_3^{\text{'}}+[\frac{1}{4}-\frac{5}{2^3\cdot 3}](\mathrm{\Delta }{)}^4{f}_3^{iv}\}$
	$=$	$[-\frac{1}{3^2}]f_0+\left[\frac{1}{2}\right]f_1+\left[\frac{11}{2\cdot 3^2}\right]f_3+[-\frac{2}{3}](\mathrm{\Delta })f_3^{\text{'}}+\left[\frac{1}{2^2\cdot 3}\right](\mathrm{\Delta }{)}^4{f}_3^{iv}+{𝒪}({\mathrm{\Delta }}^5)$
$\Rightarrow -\left[\frac{1}{2^2\cdot 3}\right](\mathrm{\Delta }{)}^4{f}_3^{iv}$	$=$	$[-\frac{1}{3^2}]f_0+\left[\frac{1}{2}\right]f_1-f_2+\left[\frac{11}{2\cdot 3^2}\right]f_3+[-\frac{2}{3}](\mathrm{\Delta })f_3^{\text{'}}+{𝒪}({\mathrm{\Delta }}^5)$

And, finally:

$f_4$	$=$	$f_3+(\mathrm{\Delta })f_3^{\text{'}}+\frac{1}{2}(\mathrm{\Delta }{)}^2{f}_3^{{\text{'}}^{\prime }}+\frac{1}{6}(\mathrm{\Delta }{)}^3{f}_3^{{\text{'}}^{″}}+\frac{1}{24}(\mathrm{\Delta }{)}^4{f}_3^{iv}+{𝒪}({\mathrm{\Delta }}^5)$
	$=$	$f_3+(\mathrm{\Delta })f_3^{\text{'}}+{𝒪}({\mathrm{\Delta }}^5)$
		$+\frac{1}{2}\{-f_0+[\frac{3^3}{2^2}]f_1-3^3{f}_2+[\frac{5\cdot 17}{2^2}]f_3+[-\frac{3^2\cdot 5}{2}](\mathrm{\Delta })f_3^{\text{'}}\}[-\frac{2}{3^2}]$
		$+\frac{1}{6}\left\{\right[-\frac{2}{3}]f_0+[4]f_1+[-2\cdot 5]f_2+[\frac{2^2\cdot 5}{3}]f_3+[-\frac{2\cdot 11}{3}](\mathrm{\Delta })f_3^{\text{'}}\}[-\frac{3}{2}]$
		$+\frac{1}{24}\left\{\right[-\frac{1}{3^2}]f_0+[\frac{1}{2}]f_1-f_2+[\frac{11}{2\cdot 3^2}]f_3+[-\frac{2}{3}](\mathrm{\Delta })f_3^{\text{'}}\}[-2^2\cdot 3]$
	$=$	$f_3+(\mathrm{\Delta })f_3^{\text{'}}+{𝒪}({\mathrm{\Delta }}^5)$
		$+\left\{\right[\frac{1}{3^2}]f_0+[-\frac{3}{2^2}]f_1+[3]f_2+[-\frac{5\cdot 17}{2^2\cdot 3^2}]f_3+[\frac{5}{2}](\mathrm{\Delta })f_3^{\text{'}}\}$
		$+\left\{\right[\frac{1}{2\cdot 3}]f_0+[-1]f_1+[\frac{5}{2}]f_2+[-\frac{5}{3}]f_3+[\frac{11}{2\cdot 3}](\mathrm{\Delta })f_3^{\text{'}}\}$
		$+\left\{\right[\frac{1}{2\cdot 3^2}]f_0+[-\frac{1}{2^2}]f_1+[\frac{1}{2}]f_2+[-\frac{11}{2^2\cdot 3^2}]f_3+[\frac{1}{3}](\mathrm{\Delta })f_3^{\text{'}}\}$
	$=$	$[\frac{1}{3^2}+\frac{1}{2\cdot 3}+\frac{1}{2\cdot 3^2}]f_0+[-\frac{3}{2^2}-1-\frac{1}{2^2}]f_1+[3+\frac{5}{2}+\frac{1}{2}]f_2+{𝒪}({\mathrm{\Delta }}^5)$
		$+[1-\frac{5\cdot 17}{2^2\cdot 3^2}-\frac{5}{3}-\frac{11}{2^2\cdot 3^2}]f_3+[1+\frac{5}{2}+\frac{11}{2\cdot 3}+\frac{1}{3}](\mathrm{\Delta })f_3^{\text{'}}$
	$=$	$\left[\frac{1}{3}\right]f_0+[-2]f_1+\left[6\right]f_2+[-\frac{10}{3}]f_3+\left[\frac{17}{3}\right](\mathrm{\Delta })f_3^{\text{'}}+{𝒪}({\mathrm{\Delta }}^5)$

Result:

Definitely WRONG!
$f_4$ $=$ $\frac{1}{3}{f}_0-2f_1+6f_2-\frac{10}{3}{f}_3+\frac{17}{3}(\mathrm{\Delta })f_3^{\text{'}}+{𝒪}({\mathrm{\Delta }}^5).$

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 $(\mathrm{\Delta })f_3^{\text{'}}$, namely,

Somewhat Improved
$f_4$ $=$ $\frac{1}{3}{f}_0-2f_1+6f_2-\frac{10}{3}{f}_3+4(\mathrm{\Delta })f_3^{\text{'}}+{𝒪}({\mathrm{\Delta }}^5).$

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:   $f_3$, $f_3^{\text{'}}$, $f_2$, $f_1$, and $f_0$.

$f_4$

$=$

$f_3+(\mathrm{\Delta })f_3^{\text{'}}+\frac{1}{2}(\mathrm{\Delta }{)}^2{f}_3^{{\text{'}}^{\prime }}+\frac{1}{6}(\mathrm{\Delta }{)}^3{f}_3^{{\text{'}}^{″}}+\frac{1}{24}(\mathrm{\Delta }{)}^4{f}_3^{iv}+{𝒪}({\mathrm{\Delta }}^5)$

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

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

---

First:

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

${𝒪}({\mathrm{\Delta }}^3)$

$f_4$ $=$ $f_2+2(\mathrm{\Delta })f_3^{\text{'}}+{𝒪}({\mathrm{\Delta }}^3)$

This expression works very well for a parabola.

---

Second:

$f_1$	$=$	$f_3+(-2)\mathrm{\Delta }{f}_3^{\text{'}}+2(\mathrm{\Delta }{)}^2{f}_3^{{\text{'}}^{\prime }}+[-\frac{2^3}{6}]{\mathrm{\Delta }}^3{f}_3^{{\text{'}}^{″}}+[\frac{2^4}{2^3\cdot 3}]{\mathrm{\Delta }}^4{f}_3^{iv}+{𝒪}({\mathrm{\Delta }}^5)$
	$=$	$f_3+(-2)\mathrm{\Delta }{f}_3^{\text{'}}+[-\frac{2^3}{6}]{\mathrm{\Delta }}^3{f}_3^{{\text{'}}^{″}}+\left[\frac{2^4}{2^3\cdot 3}\right]{\mathrm{\Delta }}^4{f}_3^{iv}+{𝒪}({\mathrm{\Delta }}^5)$
		$+2\{-f_3+(\mathrm{\Delta })f_3^{\text{'}}+f_2+\frac{1}{2\cdot 3}(\mathrm{\Delta }{)}^3{f}_3^{{\text{'}}^{″}}-\frac{1}{2^3\cdot 3}(\mathrm{\Delta }{)}^4{f}_3^{iv}\}\left[2\right]$
	$=$	$f_3[1-2^2]+(2^2-2)\mathrm{\Delta }{f}_3^{\text{'}}+2^2{f}_2+[\frac{2}{3}-\frac{2^3}{6}]{\mathrm{\Delta }}^3{f}_3^{{\text{'}}^{″}}+[\frac{2^4}{2^3\cdot 3}-\frac{1}{2\cdot 3}]{\mathrm{\Delta }}^4{f}_3^{iv}+{𝒪}({\mathrm{\Delta }}^5)$
	$=$	$f_3[-3]+(2)\mathrm{\Delta }{f}_3^{\text{'}}+2^2{f}_2+[-\frac{2}{3}]{\mathrm{\Delta }}^3{f}_3^{{\text{'}}^{″}}+\left[\frac{1}{2}\right]{\mathrm{\Delta }}^4{f}_3^{iv}+{𝒪}({\mathrm{\Delta }}^5)$
$\Rightarrow \left[\frac{2}{3}\right]{\mathrm{\Delta }}^3{f}_3^{{\text{'}}^{″}}$	$=$	$-f_1+2^2{f}_2-3f_3+2\mathrm{\Delta }{f}_3^{\text{'}}+\left[\frac{1}{2}\right]{\mathrm{\Delta }}^4{f}_3^{iv}+{𝒪}({\mathrm{\Delta }}^5)$

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

$\frac{1}{2}(\mathrm{\Delta }{)}^2{f}_3^{{\text{'}}^{\prime }}$	$=$	$f_2-f_3+(\mathrm{\Delta })f_3^{\text{'}}-\frac{1}{24}(\mathrm{\Delta }{)}^4{f}_3^{iv}+{𝒪}({\mathrm{\Delta }}^5)$
		$+\frac{1}{6}\{-f_1+2^2{f}_2-3f_3+2\mathrm{\Delta }{f}_3^{\text{'}}+[\frac{1}{2}\left]{\mathrm{\Delta }}^4{f}_3^{iv}\right\}\left[\frac{3}{2}\right]$
	$=$	$-\frac{1}{4}{f}_1+2f_2+[-\frac{7}{4}]f_3+\left[\frac{3}{2}\right](\mathrm{\Delta })f_3^{\text{'}}+\left[\frac{1}{2^2\cdot 3}\right](\mathrm{\Delta }{)}^4{f}_3^{iv}+{𝒪}({\mathrm{\Delta }}^5)$

Hence, an improved expression for $f_4$ is,

$f_4$	$=$	$f_3+(\mathrm{\Delta })f_3^{\text{'}}+{𝒪}({\mathrm{\Delta }}^4)$
		$+\{-\frac{1}{4}{f}_1+2f_2+[-\frac{7}{4}]f_3+[\frac{3}{2}](\mathrm{\Delta })f_3^{\text{'}}\}$
		$+\frac{1}{6}\{-f_1+2^2{f}_2-3f_3+2\mathrm{\Delta }{f}_3^{\text{'}}\}\left[\frac{3}{2}\right]$
	$=$	$-\frac{1}{2}{f}_1+3f_2-\frac{3}{2}{f}_3+3(\mathrm{\Delta })f_3^{\text{'}}+{𝒪}({\mathrm{\Delta }}^4)$

${𝒪}({\mathrm{\Delta }}^4)$

$f_4$ $=$ $-\frac{1}{2}{f}_1+3f_2-\frac{3}{2}{f}_3+3(\mathrm{\Delta })f_3^{\text{'}}+{𝒪}({\mathrm{\Delta }}^4)$

---

Third:

$f_0$	$=$	$f_3+(-3\mathrm{\Delta })f_3^{\text{'}}+\frac{1}{2}(-3\mathrm{\Delta }{)}^2{f}_3^{{\text{'}}^{\prime }}+\frac{1}{6}(-3\mathrm{\Delta }{)}^3{f}_3^{{\text{'}}^{″}}+\frac{1}{24}(-3\mathrm{\Delta }{)}^4{f}_3^{iv}+{𝒪}({\mathrm{\Delta }}^5)$
	$=$	$f_3+[-3](\mathrm{\Delta })f_3^{\text{'}}+\left[\frac{3^3}{2^3}\right](\mathrm{\Delta }{)}^4{f}_3^{iv}+{𝒪}({\mathrm{\Delta }}^5)$
		$+3^2\{-\frac{1}{4}{f}_1+2f_2+[-\frac{7}{4}]f_3+[\frac{3}{2}](\mathrm{\Delta })f_3^{\text{'}}+[\frac{1}{2^2\cdot 3}](\mathrm{\Delta }{)}^4{f}_3^{iv}\}$
		$+[-\frac{3^3}{2^2}]\{-f_1+2^2{f}_2-3f_3+2\mathrm{\Delta }{f}_3^{\text{'}}+[\frac{1}{2}\left]{\mathrm{\Delta }}^4{f}_3^{iv}\right\}$
	$=$	$f_3+[-3](\mathrm{\Delta })f_3^{\text{'}}+\left[\frac{3^3}{2^3}\right](\mathrm{\Delta }{)}^4{f}_3^{iv}+{𝒪}({\mathrm{\Delta }}^5)$
		$+\left\{\right[-\frac{3^2}{4}]f_1+[2\cdot 3^2]f_2+[-\frac{3^2\cdot 7}{4}]f_3+[\frac{3^3}{2}](\mathrm{\Delta })f_3^{\text{'}}+[\frac{3}{2^2}](\mathrm{\Delta }{)}^4{f}_3^{iv}\}$
		$+\left\{\right[\frac{3^3}{2^2}]f_1+[-3^3]f_2+[\frac{3^4}{2^2}]f_3+[-\frac{3^3}{2}]\mathrm{\Delta }{f}_3^{\text{'}}+[-\frac{3^3}{2^3}\left]{\mathrm{\Delta }}^4{f}_3^{iv}\right\}$
	$=$	$[\frac{3^3}{2^2}-\frac{3^2}{4}]f_1+[2\cdot 3^2-3^3]f_2+[1+\frac{3^4}{2^2}-\frac{3^2\cdot 7}{4}]f_3+[\frac{3^3}{2}-\frac{3^3}{2}-3](\mathrm{\Delta })f_3^{\text{'}}+[\frac{3^3}{2^3}+\frac{3}{2^2}-\frac{3^3}{2^3}](\mathrm{\Delta }{)}^4{f}_3^{iv}+{𝒪}({\mathrm{\Delta }}^5)$
	$=$	$\left[\frac{3^2}{2}\right]f_1+[-3^2]f_2+\left[\frac{11}{2}\right]f_3+[-3](\mathrm{\Delta })f_3^{\text{'}}+\left[\frac{3}{2^2}\right](\mathrm{\Delta }{)}^4{f}_3^{iv}+{𝒪}({\mathrm{\Delta }}^5)$
$\Rightarrow -\left[\frac{3}{2^2}\right](\mathrm{\Delta }{)}^4{f}_3^{iv}$	$=$	$-f_0+\left[\frac{3^2}{2}\right]f_1+[-3^2]f_2+\left[\frac{11}{2}\right]f_3+[-3](\mathrm{\Delta })f_3^{\text{'}}+{𝒪}({\mathrm{\Delta }}^5)$

Hence,

$\frac{1}{2}(\mathrm{\Delta }{)}^2{f}_3^{{\text{'}}^{\prime }}$	$=$	$-\frac{1}{4}{f}_1+2f_2+[-\frac{7}{4}]f_3+\left[\frac{3}{2}\right](\mathrm{\Delta })f_3^{\text{'}}+{𝒪}({\mathrm{\Delta }}^5)$
		$+\left[\frac{1}{2^2\cdot 3}\right]\{-f_0+[\frac{3^2}{2}]f_1+[-3^2]f_2+[\frac{11}{2}]f_3+[-3](\mathrm{\Delta })f_3^{\text{'}}\}[-\frac{2^2}{3}]$
	$=$	$-\frac{1}{4}{f}_1+2f_2+[-\frac{7}{4}]f_3+\left[\frac{3}{2}\right](\mathrm{\Delta })f_3^{\text{'}}+{𝒪}({\mathrm{\Delta }}^5)$
		$+\left\{\right[\frac{1}{3^2}]f_0+[-\frac{1}{2}]f_1+f_2+[-\frac{11}{2\cdot 3^2}]f_3+[\frac{1}{3}](\mathrm{\Delta })f_3^{\text{'}}\}$
	$=$	$\left[\frac{1}{3^2}\right]f_0+[-\frac{1}{2}-\frac{1}{4}]f_1+3f_2+[-\frac{11}{2\cdot 3^2}-\frac{7}{4}]f_3+[\frac{1}{3}+\frac{3}{2}](\mathrm{\Delta })f_3^{\text{'}}+{𝒪}({\mathrm{\Delta }}^5)$
	$=$	$\left[\frac{1}{3^2}\right]f_0+[-\frac{3}{4}]f_1+3f_2+[-\frac{5\cdot 17}{2^2\cdot 3^2}]f_3+\left[\frac{11}{2\cdot 3}\right](\mathrm{\Delta })f_3^{\text{'}}+{𝒪}({\mathrm{\Delta }}^5)$

And,

$\left[\frac{2}{3}\right]{\mathrm{\Delta }}^3{f}_3^{{\text{'}}^{″}}$	$=$	$-f_1+2^2{f}_2-3f_3+2\mathrm{\Delta }{f}_3^{\text{'}}+{𝒪}({\mathrm{\Delta }}^5)$
		$+\left[\frac{1}{2}\right]\{-f_0+[\frac{3^2}{2}]f_1+[-3^2]f_2+[\frac{11}{2}]f_3+[-3](\mathrm{\Delta })f_3^{\text{'}}\}[-\frac{2^2}{3}]$
	$=$	$-f_1+2^2{f}_2-3f_3+2\mathrm{\Delta }{f}_3^{\text{'}}+{𝒪}({\mathrm{\Delta }}^5)$
		$+\left\{\right[\frac{2}{3}]f_0+[-3]f_1+[2\cdot 3]f_2+[-\frac{11}{3}]f_3+[2](\mathrm{\Delta })f_3^{\text{'}}\}$
	$=$	$\left[\frac{2}{3}\right]f_0+[-4]f_1+[2\cdot 5]f_2+[-\frac{2^2\cdot 5}{3}]f_3+\left[4\right](\mathrm{\Delta })f_3^{\text{'}}+{𝒪}({\mathrm{\Delta }}^5)$

---

Finally, then:

$f_4$	$=$	$f_3+(\mathrm{\Delta })f_3^{\text{'}}+\frac{1}{2}(\mathrm{\Delta }{)}^2{f}_3^{{\text{'}}^{\prime }}+\frac{1}{6}(\mathrm{\Delta }{)}^3{f}_3^{{\text{'}}^{″}}+\frac{1}{24}(\mathrm{\Delta }{)}^4{f}_3^{iv}+{𝒪}({\mathrm{\Delta }}^5)$
	$=$	$f_3+(\mathrm{\Delta })f_3^{\text{'}}+{𝒪}({\mathrm{\Delta }}^5)$
		$+\frac{1}{2}\left\{\right[\frac{1}{3^2}]f_0+[-\frac{3}{4}]f_1+3f_2+[-\frac{5\cdot 17}{2^2\cdot 3^2}]f_3+[\frac{11}{2\cdot 3}](\mathrm{\Delta })f_3^{\text{'}}\}\left[2\right]$
		$+\frac{1}{2\cdot 3}\left\{\right[\frac{2}{3}]f_0+[-4]f_1+[2\cdot 5]f_2+[-\frac{2^2\cdot 5}{3}]f_3+[4](\mathrm{\Delta })f_3^{\text{'}}\}\left[\frac{3}{2}\right]$
		$+\frac{1}{2^3\cdot 3}\{-f_0+[\frac{3^2}{2}]f_1+[-3^2]f_2+[\frac{11}{2}]f_3+[-3](\mathrm{\Delta })f_3^{\text{'}}\}[-\frac{2^2}{3}]$
	$=$	$f_3+(\mathrm{\Delta })f_3^{\text{'}}+{𝒪}({\mathrm{\Delta }}^5)$
		$+\left\{\right[\frac{1}{3^2}]f_0+[-\frac{3}{4}]f_1+3f_2+[-\frac{5\cdot 17}{2^2\cdot 3^2}]f_3+[\frac{11}{2\cdot 3}](\mathrm{\Delta })f_3^{\text{'}}\}$
		$+\frac{1}{2^2}\left\{\right[\frac{2}{3}]f_0+[-4]f_1+[2\cdot 5]f_2+[-\frac{2^2\cdot 5}{3}]f_3+[4](\mathrm{\Delta })f_3^{\text{'}}\}$
		$+\frac{1}{2\cdot 3^2}\{f_0+[-\frac{3^2}{2}]f_1+[3^2]f_2+[-\frac{11}{2}]f_3+[3](\mathrm{\Delta })f_3^{\text{'}}\}$
	$=$	$f_3+(\mathrm{\Delta })f_3^{\text{'}}+{𝒪}({\mathrm{\Delta }}^5)$
		$+\left\{\right[\frac{1}{3^2}]f_0+[-\frac{3}{4}]f_1+3f_2+[-\frac{5\cdot 17}{2^2\cdot 3^2}]f_3+[\frac{11}{2\cdot 3}](\mathrm{\Delta })f_3^{\text{'}}\}$
		$+\left\{\right[\frac{1}{2\cdot 3}]f_0+[-1]f_1+[\frac{5}{2}]f_2+[-\frac{5}{3}]f_3+[1](\mathrm{\Delta })f_3^{\text{'}}\}$
		$\left\{\right[\frac{1}{2\cdot 3^2}]f_0+[-\frac{1}{2^2}]f_1+[\frac{1}{2}]f_2+[-\frac{11}{2^2\cdot 3^2}]f_3+[\frac{1}{2\cdot 3}](\mathrm{\Delta })f_3^{\text{'}}\}$
	$=$	$[\frac{1}{3^2}+\frac{1}{2\cdot 3}+\frac{1}{2\cdot 3^2}]f_0+[-1-\frac{3}{4}-\frac{1}{2^2}]f_1+[3+\frac{5}{2}+\frac{1}{2}]f_2+[1-\frac{5\cdot 17}{2^2\cdot 3^2}-\frac{5}{3}-\frac{11}{2^2\cdot 3^2}]f_3+[2+\frac{11}{2\cdot 3}+\frac{1}{2\cdot 3}](\mathrm{\Delta })f_3^{\text{'}}+{𝒪}({\mathrm{\Delta }}^5)$
	$=$	$\left[\frac{1}{3}\right]f_0+[-2]f_1+\left[6\right]f_2+[-\frac{2\cdot 5}{3}]f_3+\left[4\right](\mathrm{\Delta })f_3^{\text{'}}+{𝒪}({\mathrm{\Delta }}^5)$

${𝒪}({\mathrm{\Delta }}^5)$

$f_4$ $=$ $\frac{1}{3}{f}_0-2f_1+6f_2-\frac{2\cdot 5}{3}{f}_3+4(\mathrm{\Delta })f_3^{\text{'}}+{𝒪}({\mathrm{\Delta }}^5)$