User:官翁/Draft for a general formula for the meridian arc length

Generalized series
The above series, to eighth order in eccentricity or fourth order in third flattening, are adequate for most practical applications. Each can be written quite generally. For example, Kazushige Kawase (2009) derived following general formula.:

\begin{align} m(\phi)&=a(1-n)^2(1+n)\sum_{j=0}^\infty\left(\prod_{k=1}^j\delta_k\right)^2\left\{\phi+\sum_{l=1}^{2j}\frac{\sin 2l\phi}{l}\prod_{m=1}^l\delta_{j+(-1)^m\lfloor m/2\rfloor}^{(-1)^m}\right\}\\ &=\frac{a}{1+n}\sum_{j=0}^\infty\left(\prod_{k=1}^j\varepsilon_k\right)^2\left\{\phi+\sum_{l=1}^{2j}\left(\frac{1}{l}-4l\right)\sin 2l\phi\prod_{m=1}^l\varepsilon_{j+(-1)^m\lfloor m/2\rfloor}^{(-1)^m}\right\} \end{align} $$ where
 * $$\delta_i=-\frac{n}{2i}-n,\quad\varepsilon_i=\frac{3n}{2i}-n.

$$

The first result is for Bessel's series. This is derived by regarding the integrand of $$m(\phi)\,\!$$ as a generating function of Gegenbauer polynomials $$C_i^{3/2}(\cos 2\phi)$$ and expanding it to series of trigonometric functions in accordance with well known formula: (See for instance NIST library)

C_i^{\lambda}(\cos\theta)=\sum _{\ell=0}^{i}\frac{\left(\lambda\right)_{\ell}\left(\lambda\right)_{i-\ell}}{\ell!\;(i-\ell)!}\cos\left((i-2\ell)\theta\right). $$

That is, the integrand becomes

\frac{1}{(1+2n\cos 2\phi+n^2)^{3/2}}=\sum_{i=0}^{\infty}(-n)^i\sum _{\ell=0}^{i}\frac{\left(3/2\right)_{\ell}\left(3/2\right)_{i-\ell}}{\ell!\;(i-\ell)!}\cos\left((2i-4\ell)\phi\right). $$

Integrating term by term and rearranging the indices of the summation and product give the final result of the first general formula.

The second result is for Helmert's series. This is derived by firstly taking account of the well known relation between a special case of an incomplete elliptic integral of the third kind and that of the second kind with the same arguments as follows:

\big(1-e^2\big)\,\Pi(\phi,e,e)=E\big(\phi,e\big)+\frac{d^2}{d\phi^2}E\big(\phi,e\big). $$

Substituting $$e^2 = \frac{4n}{(1+n)^2}$$, the right side of above equation becomes

\frac{1}{1+n}\left[\int_{0}^{\phi}\sqrt{1+2n\cos 2\phi+n^2}d\phi+\frac{d^2}{d\phi^2}\int_{0}^{\phi}\sqrt{1+2n\cos 2\phi+n^2}d\phi\right]. $$

Since the integrand of the first term of above corresponds to a generating function of $$C_i^{-1/2}(\cos 2\phi)$$, the first term is identical with the first result with substituting $$\delta_i\,\!$$ to $$\varepsilon_i\,\!$$ analogically. The contribution from the second term remains only trigonometric terms with the coefficient which is multiplied $$-(2l)^2=-4l^2\,\!$$ as a result of the second derivative of $$\sin 2l\phi\,\!$$. Consolidating the first and second terms gives the final result of the second general formula.

Truncating the summation of the second result at $$j=2\,\!$$ gives Helmert's approximation.