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

$e^x$

$=$

$1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+\frac{x^5}{5!}+\frac{x^6}{6!}\cdots$

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

See also:

• Equation (62) from §5 in Chapter IV of C67.

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.

Theta-Function 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

Derivation

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.

For Later Use

We have just determined that,

$e^{-w}$

$\approx$

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

where,

$a=c=e=g$	$=$	$0,$
$b$	$=$	$+\frac{1}{2\cdot 3},$
$d$	$=$	$-\frac{1}{2^3\cdot 3\cdot 5},$
$f$	$=$	$+\frac{1}{2\cdot 3^3\cdot 5\cdot 7}.$

That is to say,

$e^{-w}$	$\approx$	$1-\frac{r^2}{6}+r^4(\frac{1}{2^3\cdot 3^2}+\frac{1}{2^3\cdot 3\cdot 5})+r^6(-\frac{1}{2^4\cdot 3^2\cdot 5}-\frac{1}{2^4\cdot 3^4}-\frac{1}{2\cdot 3^3\cdot 5\cdot 7})$
	$\approx$	$1-\frac{r^2}{6}+r^4(\frac{5}{2^3\cdot 3^2\cdot 5}+\frac{3}{2^3\cdot 3^2\cdot 5})+r^6(-\frac{3^2\cdot 7}{2^4\cdot 3^4\cdot 5\cdot 7}-\frac{5\cdot 7}{2^4\cdot 3^4\cdot 5\cdot 7}-\frac{2^3\cdot 3}{2^4\cdot 3^4\cdot 5\cdot 7})$
	$\approx$	$1-\frac{r^2}{6}+r^4\left(\frac{1}{3^2\cdot 5}\right)-r^6\left(\frac{61}{2^3\cdot 3^4\cdot 5\cdot 7}\right).$

Therefore, also,

$e^{+w}=(e^{-w}{)}^{-1}$	$\approx$	$(1+ϵ{)}^{-1}$
	$\approx$	$1-ϵ+{ϵ}^2-{ϵ}^3+{ϵ}^4-{ϵ}^5,$

where,

$ϵ$	$\equiv$	$-\frac{r^2}{6}+r^4\left(\frac{1}{3^2\cdot 5}\right)-r^6\left(\frac{61}{2^3\cdot 3^4\cdot 5\cdot 7}\right);$
${ϵ}^2$	$=$	$[-\frac{r^2}{6}+r^4(\frac{1}{3^2\cdot 5})-r^6(\frac{61}{2^3\cdot 3^4\cdot 5\cdot 7}\left)\right]\cdot [-\frac{r^2}{6}+r^4(\frac{1}{3^2\cdot 5})-r^6(\frac{61}{2^3\cdot 3^4\cdot 5\cdot 7}\left)\right]$
	$\approx$	$-\frac{r^2}{6}\cdot [-\frac{r^2}{6}+r^4(\frac{1}{3^2\cdot 5}\left)\right]+r^4\left(\frac{1}{3^2\cdot 5}\right)\cdot [-\frac{r^2}{6}]$
	$\approx$	$\frac{r^4}{2^2\cdot 3^2}-r^6\left(\frac{1}{3^3\cdot 5}\right);$
${ϵ}^3$	$\approx$	$-\frac{r^2}{2\cdot 3}\cdot \frac{r^4}{2^2\cdot 3^2}=-\frac{r^6}{2^3\cdot 3^3}.$

Hence,

$e^{+w}$	$\approx$	$1-ϵ+{ϵ}^2-{ϵ}^3+{\xcancel{{ϵ}^4}}^0-{\xcancel{{ϵ}^5}}^0$
	$\approx$	$1-[-\frac{r^2}{6}+r^4(\frac{1}{3^2\cdot 5})-r^6(\frac{61}{2^3\cdot 3^4\cdot 5\cdot 7}\left)\right]+[\frac{r^4}{2^2\cdot 3^2}-r^6(\frac{1}{3^3\cdot 5}\left)\right]-[-\frac{r^6}{2^3\cdot 3^3}]$
	$\approx$	$1+\frac{r^2}{6}+[\frac{1}{2^2\cdot 3^2}-\frac{1}{3^2\cdot 5}]r^4+\left[\right(\frac{61}{2^3\cdot 3^4\cdot 5\cdot 7})-(\frac{1}{3^3\cdot 5})+\frac{1}{2^3\cdot 3^3}]r^6$
	$\approx$	$1+\frac{r^2}{6}+\left[\frac{1}{2^2\cdot 3^2\cdot 5}\right]r^4+[\frac{61}{2^3\cdot 3^4\cdot 5\cdot 7}-\frac{2^3\cdot 3\cdot 7}{2^3\cdot 3^4\cdot 5\cdot 7}+\frac{3\cdot 5\cdot 7}{2^3\cdot 3^4\cdot 5\cdot 7}]r^6$
	$\approx$	$1+\frac{r^2}{6}+\left[\frac{1}{2^2\cdot 3^2\cdot 5}\right]r^4-\left[\frac{1}{2^2\cdot 3^4\cdot 5\cdot 7}\right]r^6.$
	$\approx$	$1+\frac{r^2}{6}+\left[\frac{1}{180}\right]r^4-\left[\frac{1}{11,340}\right]r^6.$

We also have shown that,

$w(r)$	$=$	$\frac{r^2}{6}-\frac{r^4}{120}+\frac{r^6}{1890}-\frac{61r^8}{1,632,960}+\cdots .$
$\Rightarrow \frac{1}{r}\frac{dw}{dr}$	$\approx$	$\frac{1}{3}-\frac{4r^2}{120}+\frac{6r^4}{1890}-\frac{8\cdot 61r^6}{1,632,960}.$

Therefore, together we have,

$-\left[\frac{1}{r}\frac{dw}{dr}\right]e^{+w}$

$\approx$

$\{-\frac{1}{3}+\frac{r^2}{2\cdot 3\cdot 5}-\frac{r^4}{3^2\cdot 5\cdot 7}+\frac{61r^6}{204,120}\}\times \{1+\frac{r^2}{2\cdot 3}+\frac{r^4}{2^2\cdot 3^2\cdot 5}-\frac{r^6}{2^2\cdot 3^4\cdot 5\cdot 7}\}$

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:

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

Polytropic Displacement-Function Examples

We will only consider situations in which ${\sigma }_c^2=0$ and $\gamma =(n+1)/n$, in which case, $\alpha =(3-n)/(n+1)$ and $\mathfrak{𝔉}=2(n-3)/(n+1)$.

When $n=5$ $\Rightarrow \alpha =-1/3$ and $\mathfrak{𝔉}=2/3\dots$ we have,

$x(\xi )$	$=$	$1-(\frac{6}{60}\cdot \frac{2}{3}){\xi }^2-(6)\left\{\frac{-20+(2/3)[(34)-4]}{2^5\cdot 3^2\cdot 5\cdot 7}\right\}{\xi }^4+\cdots$
	$=$	$1-\left(\frac{1}{15}\right){\xi }^2-0+\cdots$

Compare this result with the expression for $x_P{|}_{n=5}$ found in an an accompanying "related thoughts" paragraph.

---

When $n=3$ $\Rightarrow \alpha =0$ and $\mathfrak{𝔉}=0\dots$ we have,

$x(\xi )$	$=$	$1-\left(0\right){\xi }^2-(6)\left\{\frac{0}{2^5\cdot 3^2\cdot 5\cdot 7}\right\}{\xi }^4+\cdots$
	$=$	$1+\cdots$

Compare this result with the expression for $x_P{|}_{n=3}$ found in an an accompanying "related thoughts" paragraph.

---

When $n=0$ $\Rightarrow \alpha =+3$ and $\mathfrak{𝔉}=-6\dots$ we have,

$x(\xi )$

$=$

$1+\frac{1}{10}{\xi }^2+\frac{1}{84}{\xi }^4+\cdots$

Sterne (1937) derived analytic expressions for the eigenvectors that describe radial oscillations in uniform-density spheres. The fundamental (j = 0) mode exhibits homologous expansion/contraction, that is to say,

$x{|}_{\mathrm{S}\mathrm{t}\mathrm{e}\mathrm{r}\mathrm{n}\mathrm{e}}$

$=$

$1.$

But in his analysis, the accompanying value of $\mathfrak{𝔉}$ is zero; and the accompanying (square of the) dimensionless eigenfrequency is ${\sigma }_c^2=2(\gamma -4/3)$. We cannot adopt the requirement that $\gamma =(n+1)/n$ because $n=0$.

---

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 [modified leading term on 9/29/2026],

$x$	$=$	$x_0+ar+b{r}^2+c{r}^3+d{r}^4+\cdots$
$\Rightarrow \frac{dx}{dr}$	$=$	$a+2br+3c{r}^2+4d{r}^3+\cdots$
$\Rightarrow \frac{1}{r}\frac{dx}{dr}$	$=$	$\frac{a}{r}+2b+3cr+4d{r}^2+\cdots$

and,

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

$=$

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

Substituting these expressions into the LAWE gives,

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

$\approx$

$-\frac{1}{6}[\frac{{\sigma }_c^2}{\gamma }-2\alpha (1-\frac{r^2}{10}+\frac{r^4}{105}\left)\right](x_0+ar+b{r}^2+c{r}^3+d{r}^4).$

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

$2b+6cr+12d{r}^2+4[\frac{a}{r}+2b+3cr+4d{r}^2]-\frac{r^2}{3}[\frac{a}{r}+2b]$	$\approx$	$-\frac{1}{6}[\mathfrak{𝔉}+\frac{\alpha r^2}{5}-\frac{2\alpha r^4}{105}](x_0+ar+b{r}^2+c{r}^3+d{r}^4)$
	$\approx$	$-\frac{\mathfrak{𝔉}}{6}(x_0+ar+b{r}^2+c{r}^3+d{r}^4)-\frac{\alpha r^2}{30}(x_0+ar+b{r}^2+c{r}^3+d{r}^4)+\frac{\alpha r^4}{315}(x_0+ar+b{r}^2+c{r}^3+d{r}^4)$
	$\approx$	$-\frac{\mathfrak{𝔉}}{6}(x_0+ar+b{r}^2)-\frac{\alpha r^2}{30}\left(x_0\right).$

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
$r^{-1}:$	$4a$	$0$	$\Rightarrow a=0$
$r^0:$	$2b+8b$	$-\frac{\mathfrak{𝔉}{x}_0}{6}$	$\Rightarrow b=-\frac{\mathfrak{𝔉}{x}_0}{60}$
$r^1:$	$6c+12c-\frac{a}{3}$	$-\frac{a\mathfrak{𝔉}}{6}$	$\Rightarrow c=0$
$r^2:$	$12d+16d-\frac{2b}{3}$	$-\frac{\mathfrak{𝔉}b}{6}-\frac{\alpha x_0}{30}$	$\Rightarrow 28d=\frac{1}{30}[20b-5\mathfrak{𝔉}b-\alpha x_0]\Rightarrow d=\frac{x_0}{10080}[\mathfrak{𝔉}(\mathfrak{𝔉}-4)-12\alpha ]$

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

$\frac{x(r)}{x_0}$ $=$ $1-\frac{\mathfrak{𝔉}}{60}{r}^2+\frac{1}{10080}[\mathfrak{𝔉}(\mathfrak{𝔉}-4)-12\alpha ]r^4+\cdots$ Then, in the particular case where $\gamma =1$ and ${\sigma }_c^2=0$ — which implies that $\alpha =(3-4/\gamma )=-1$ and $\mathfrak{𝔉}=({\sigma }_c^2/\gamma -2\alpha )=+2$ — we have, $x_Y(\xi )$ $=$ $1-\frac{{\xi }^2}{30}+\frac{{\xi }^4}{1260}+\cdots \Rightarrow \frac{2}{3}{x}_Y=\frac{2}{3}-\frac{{\xi }^2}{45}+\frac{{\xi }^4}{1890}+\cdots$

This is a derivation check ...      (Numerical values in parentheses assume $\xi =0.056$, or 0.107, or 0.322) As was discovered by Yabushita, the analytically prescribed isothermal displacement function is, ${\sigma }_c^2=0$  and   $x_Y=1-\left(\frac{1}{\xi e^{-\psi }}\right)\frac{d\psi }{d\xi }.$ Given that, $\psi (\xi )$ $=$ $\frac{{\xi }^2}{6}-\frac{{\xi }^4}{120}+\frac{{\xi }^6}{1890}-\frac{61{\xi }^8}{1,632,960}+\cdots$      (5.22585 × 10-4, or 1.907075 × 10-03, or 0.017191666)$\Rightarrow e^{\psi }=$(1.000522, or 1.001908895, or 1.017340293) $\Rightarrow \frac{1}{\xi }\cdot \frac{d\psi }{d\xi }$ $=$ $\frac{1}{3}-\frac{{\xi }^2}{30}+\frac{6{\xi }^4}{1890}-\frac{8\cdot 61{\xi }^6}{1,632,960}+\cdots$      (3.33229× 10-1, or 0.332952116, or 0.329910995) $\Rightarrow x_Y=$(1 -  0.465010   =   0.34990)     $(1-x_Y)=$ (xxx, or 0.333587080, or 0.335631748) and that, ${\psi }^2$ $=$ $[\frac{{\xi }^2}{6}-\frac{{\xi }^4}{120}+\frac{{\xi }^6}{1890}-\frac{61{\xi }^8}{1,632,960}]\times [\frac{{\xi }^2}{6}-\frac{{\xi }^4}{120}+\frac{{\xi }^6}{1890}-\frac{61{\xi }^8}{1,632,960}]+{𝒪}({\xi }^{10})$   $=$ $\left[\frac{{\xi }^2}{6}\right]\times [\frac{{\xi }^2}{6}-\frac{{\xi }^4}{120}+\frac{{\xi }^6}{1890}]+[-\frac{{\xi }^4}{120}]\times [\frac{{\xi }^2}{6}-\frac{{\xi }^4}{120}]+\left[\frac{{\xi }^6}{1890}\right]\times \left[\frac{{\xi }^2}{6}\right]+{𝒪}({\xi }^{10})$   $=$ $[\frac{{\xi }^4}{6\cdot 6}-\frac{{\xi }^6}{6\cdot 120}+\frac{{\xi }^8}{6\cdot 1890}]+[-\frac{{\xi }^6}{6\cdot 120}+\frac{{\xi }^8}{120\cdot 120}]+\left[\frac{{\xi }^8}{6\cdot 1890}\right]+{𝒪}({\xi }^{10})$   $=$ $\frac{{\xi }^4}{2^2\cdot 3^2}-\frac{{\xi }^6}{2^4\cdot 3^2\cdot 5}+\frac{{\xi }^8}{2^2\cdot 3^4\cdot 5\cdot 7}-\frac{{\xi }^6}{2^4\cdot 3^2\cdot 5}+\frac{{\xi }^8}{2^6\cdot 3^2\cdot 5^2}+\frac{{\xi }^8}{2^2\cdot 3^4\cdot 5\cdot 7}+{𝒪}({\xi }^{10})$   $=$ $\frac{{\xi }^4}{2^2\cdot 3^2}-\frac{{\xi }^6}{2^3\cdot 3^2\cdot 5}+\frac{{\xi }^8}{2\cdot 3^4\cdot 5\cdot 7}+\frac{{\xi }^8}{2^6\cdot 3^2\cdot 5^2}+{𝒪}({\xi }^{10})$   $=$ $\frac{{\xi }^4}{2^2\cdot 3^2}-\frac{{\xi }^6}{2^3\cdot 3^2\cdot 5}+\frac{{\xi }^8}{2^6\cdot 3^4\cdot 5^2\cdot 7}[2^5\cdot 5+3^2\cdot 7]+{𝒪}({\xi }^{10})$      (2.73095 × 10-07, or 3.636931 × 10-06, or 2.955536 × 10-04   vs. from above  2.73095 × 10-7, or 3.636935 × 10-06 , or 2.955533 × 10-04) and that, ${\psi }^3$ $=$ $\{\frac{{\xi }^4}{2^2\cdot 3^2}-\frac{{\xi }^6}{2^3\cdot 3^2\cdot 5}+\frac{{\xi }^8}{2^6\cdot 3^4\cdot 5^2\cdot 7}[2^5\cdot 5+3^2\cdot 7\left]\right\}\times [\frac{{\xi }^2}{6}-\frac{{\xi }^4}{120}+\frac{{\xi }^6}{1890}-\frac{61{\xi }^8}{1,632,960}+\cdots ]+{𝒪}({\xi }^{10})$   $=$ $\frac{{\xi }^6}{2^3\cdot 3^3}-\frac{{\xi }^8}{2^2\cdot 3^2\cdot 120}-\left[\frac{{\xi }^8}{2^4\cdot 3^3\cdot 5}\right]+{𝒪}({\xi }^{10})$   $=$ $\frac{{\xi }^6}{2^3\cdot 3^3}-\frac{{\xi }^8}{2^5\cdot 3^3\cdot 5}-\left[\frac{{\xi }^8}{2^4\cdot 3^3\cdot 5}\right]+{𝒪}({\xi }^{10})$   $=$ $\frac{{\xi }^6}{2^3\cdot 3^3}-\frac{{\xi }^8}{2^5\cdot 3^2\cdot 5}+{𝒪}({\xi }^{10})$      (1.42715 × 10-10  , or 6.935893 × 10-09, or 5.080120 × 10-06   vs. from above  1.42715 × 10-10  , or 6.935909 × 10-09, or 5.081054 × 10-06 ) and that, ${\psi }^4$ $=$ $\frac{{\xi }^8}{2^4\cdot 3^4}+{𝒪}({\xi }^{10})$      (7.46276 × 10-14  , or 1.325760 × 10-11, or 8.917476 × 10-08 vs. from above  7.45808 × 10-14  , or 1.322730 × 10-11, or 8.735179 × 10-08) we appreciate that the exponential function gives, $e^{\psi }$ $=$ $1+\psi +\frac{{\psi }^2}{2!}+\frac{{\psi }^3}{3!}+\frac{{\psi }^4}{4!}+\cdots$      (1 + 5.227213 × 10-4    , or 1+1.908894 × 10-03, or 1.017340 vs. from above   1.000522 × 100  , or 1+1.908895 × 10-03, or 1.017340293)   $=$ $1+[\frac{{\xi }^2}{6}-\frac{{\xi }^4}{120}+\frac{{\xi }^6}{1890}-\frac{61{\xi }^8}{1,632,960}]+\frac{1}{2!}[\frac{{\xi }^4}{2^2\cdot 3^2}-\frac{{\xi }^6}{2^3\cdot 3^2\cdot 5}+\frac{{\xi }^8}{2^6\cdot 3^4\cdot 5^2\cdot 7}(2^5\cdot 5+3^2\cdot 7\left)\right]$     $+\frac{1}{3!}[\frac{{\xi }^6}{2^3\cdot 3^3}-\frac{{\xi }^8}{2^5\cdot 3^2\cdot 5}]+\frac{1}{4!}\left[\frac{{\xi }^8}{2^4\cdot 3^4}\right]+{𝒪}({\xi }^{10})$   $\approx$ $1+[\frac{{\xi }^2}{6}-\frac{{\xi }^4}{120}+\frac{{\xi }^6}{1890}]+\frac{1}{2!}[\frac{{\xi }^4}{2^2\cdot 3^2}-\frac{{\xi }^6}{2^3\cdot 3^2\cdot 5}]+\frac{1}{3!}\left[\frac{{\xi }^6}{2^3\cdot 3^3}\right]$   $\approx$ $1+[\frac{{\xi }^2}{6}-\frac{{\xi }^4}{120}+\frac{{\xi }^6}{1890}]+[\frac{{\xi }^4}{2^3\cdot 3^2}-\frac{{\xi }^6}{2^4\cdot 3^2\cdot 5}]+\left[\frac{{\xi }^6}{2^4\cdot 3^4}\right]$   $\approx$ $1+\frac{{\xi }^2}{6}+{\xi }^4[\frac{1}{2^3\cdot 3^2}-\frac{1}{2^3\cdot 3\cdot 5}]+{\xi }^6[\frac{1}{2\cdot 3^3\cdot 5\cdot 7}-\frac{1}{2^4\cdot 3^2\cdot 5}+\frac{1}{2^4\cdot 3^4}]$   $\approx$ $1+\frac{{\xi }^2}{6}+{\xi }^4\left[\frac{1}{2^2\cdot 3^2\cdot 5}\right]+{\xi }^6[\frac{2^3\cdot 3}{2^4\cdot 3^4\cdot 5\cdot 7}-\frac{3^2\cdot 7}{2^4\cdot 3^4\cdot 5\cdot 7}+\frac{5\cdot 7}{2^4\cdot 3^4\cdot 5\cdot 7}]$   $\approx$ $1+\frac{{\xi }^2}{6}+{\xi }^4\left[\frac{1}{180}\right]+{\xi }^6\left[\frac{-4}{45360}\right]$      (1 + 5.22721 × 10-4, or 1 + 1.908894 × 10-03, or 1.017340) As a result, the approximate expression for $x_Y$ is (also see above), $x_Y-1=-\left[\frac{1}{\xi }\frac{d\psi }{d\xi }\right]e^{\psi }$ $\approx$ $\stackrel{-0.33295212or-0.329910995}{\overbrace{[-\frac{1}{3}+\frac{{\xi }^2}{2\cdot 3\cdot 5}-\frac{{\xi }^4}{3^2\cdot 5\cdot 7}+\frac{61{\xi }^6}{2^3\cdot 3^6\cdot 5\cdot 7}]}}\times \underset{1.001908167or1.017340}{\underbrace{\{1+\frac{{\xi }^2}{6}+{\xi }^4[\frac{1}{2^2\cdot 3^2\cdot 5}]-{\xi }^6[\frac{1}{2^2\cdot 3^4\cdot 5\cdot 7}\left]\right\}}}\approx$   -0.333587444, or -0.3356316   $\approx$ $-\left[\frac{1}{3}\right]\times \{1+\frac{{\xi }^2}{6}+{\xi }^4[\frac{1}{2^2\cdot 3^2\cdot 5}]-{\xi }^6[\frac{1}{2^2\cdot 3^4\cdot 5\cdot 7}\left]\right\}$     $+\left[\frac{{\xi }^2}{2\cdot 3\cdot 5}\right]\times \{1+\frac{{\xi }^2}{6}+{\xi }^4[\frac{1}{2^2\cdot 3^2\cdot 5}\left]\right\}$     $-\left[\frac{{\xi }^4}{3^2\cdot 5\cdot 7}\right]\times \{1+\frac{{\xi }^2}{6}\}+\left[\frac{61{\xi }^6}{2^3\cdot 3^6\cdot 5\cdot 7}\right]$   $\approx$ $-\frac{1}{3}-\frac{{\xi }^2}{2\cdot 3^2}-{\xi }^4\left[\frac{1}{2^2\cdot 3^3\cdot 5}\right]+{\xi }^6\left[\frac{1}{2^2\cdot 3^5\cdot 5\cdot 7}\right]$     $+\left[\frac{{\xi }^2}{2\cdot 3\cdot 5}\right]+\left[\frac{{\xi }^2}{2\cdot 3\cdot 5}\right]\frac{{\xi }^2}{6}+\left[\frac{{\xi }^2}{2\cdot 3\cdot 5}\right]\left[\frac{{\xi }^4}{2^2\cdot 3^2\cdot 5}\right]$     $-\left[\frac{{\xi }^4}{3^2\cdot 5\cdot 7}\right]-\left[\frac{{\xi }^6}{2\cdot 3^3\cdot 5\cdot 7}\right]+\left[\frac{61{\xi }^6}{2^3\cdot 3^5\cdot 5\cdot 7}\right]$   $\approx$ $-\frac{1}{3}+{\xi }^2[-\frac{1}{2\cdot 3^2}]+{\xi }^4[-\frac{1}{2^2\cdot 3^3\cdot 5}]+{\xi }^6\left[\frac{1}{2^2\cdot 3^5\cdot 5\cdot 7}\right]+{\xi }^2\left[\frac{1}{2\cdot 3\cdot 5}\right]+{\xi }^4\left[\frac{1}{2^2\cdot 3^2\cdot 5}\right]+{\xi }^6\left[\frac{1}{2^3\cdot 3^3\cdot 5^2}\right]+{\xi }^4[-\frac{1}{3^2\cdot 5\cdot 7}]+{\xi }^6[-\frac{1}{2\cdot 3^3\cdot 5\cdot 7}]+{\xi }^6\left[\frac{61}{2^3\cdot 3^5\cdot 5\cdot 7}\right]$   $\approx$ $-\frac{1}{3}+{\xi }^2[-\frac{1}{2\cdot 3^2}+\frac{1}{2\cdot 3\cdot 5}]+{\xi }^4[-\frac{1}{2^2\cdot 3^3\cdot 5}+\frac{1}{2^2\cdot 3^2\cdot 5}-\frac{1}{3^2\cdot 5\cdot 7}]+{\xi }^6[\frac{1}{2^2\cdot 3^5\cdot 5\cdot 7}+\frac{1}{2^3\cdot 3^3\cdot 5^2}+\frac{61}{2^3\cdot 3^5\cdot 5\cdot 7}]$   $\approx$ $-\frac{1}{3}+{\xi }^2[-\frac{5}{2\cdot 3^2\cdot 5}+\frac{3}{2\cdot 3^2\cdot 5}]+{\xi }^4[-\frac{7}{2^2\cdot 3^3\cdot 5\cdot 7}+\frac{3\cdot 7}{2^2\cdot 3^3\cdot 5\cdot 7}-\frac{2^2\cdot 3}{2^2\cdot 3^3\cdot 5\cdot 7}]+{\xi }^6[\frac{2\cdot 5}{2^3\cdot 3^5\cdot 5^2\cdot 7}+\frac{3^2\cdot 7}{2^3\cdot 3^5\cdot 5^2}+\frac{5\cdot 61}{2^3\cdot 3^5\cdot 5^2\cdot 7}]$   $\approx$ $-\frac{1}{3}-{\xi }^2\left[\frac{1}{3^2\cdot 5}\right]+{\xi }^4\left[\frac{1}{2\cdot 3^3\cdot 5\cdot 7}\right]+{\xi }^6\left[\frac{1}{2^2\cdot 3^2\cdot 5^2}\right]$   $\approx$ $-\frac{1}{3}-\left[\frac{{\xi }^2}{45}\right]+\left[\frac{{\xi }^4}{1890}\right]+\left[\frac{{\xi }^6}{900}\right]$   (-0.333587685, or -0.3356304) $\Rightarrow x_Y$ $\approx$ $\frac{2}{3}-\left[\frac{{\xi }^2}{45}\right]+\left[\frac{{\xi }^4}{1890}\right]+\left[\frac{{\xi }^6}{900}\right]$ YES!!

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:

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

${𝒪}({\mathrm{\Delta }}^5)$

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