Derivation of a 2D trigonometric function - the pseudo-elliptic cosine

For the surface normalisation factor of the pseudo-elliptic cone, the integral from equation 16.51b on page [*] is

$\displaystyle \int_{S}^{}$ dS = $\displaystyle \int_{{-\sigma_{\textrm{max}}}}^{{\sigma_{\textrm{max}}}}$$\displaystyle \int_{{-\pi}}^{{\pi}}$$\displaystyle \left(\vphantom{ 1 - \cos\theta_{\textrm{max}}}\right.$1 - cosθmax$\displaystyle \left.\vphantom{ 1 - \cos\theta_{\textrm{max}}}\right)$ dφ dσ. (16.41)

When combined with the pseudo-ellipse of equation 16.39, this becomes the intractable integral

$\displaystyle \int_{S}^{}$ dS = $\displaystyle \bigintsss_{{-\sigma_{\textrm{max}}}}^{{\sigma_{\textrm{max}}}}$$\displaystyle \bigintsss_{{-\pi}}^{{\pi}}$$\displaystyle \left(\vphantom{ 1 - \cos \left( \frac{1}{\sqrt{\frac{\cos^2\phi}{\theta_x ^2} + \frac{\sin^2\phi}{\theta_y ^2}}} \right) }\right.$1 - cos$\displaystyle \left(\vphantom{ \frac{1}{\sqrt{\frac{\cos^2\phi}{\theta_x ^2} + \frac{\sin^2\phi}{\theta_y ^2}}} }\right.$$\displaystyle {\frac{{1}}{{\sqrt{\frac{\cos^2\phi}{\theta_x ^2} + \frac{\sin^2\phi}{\theta_y ^2}}}}}$$\displaystyle \left.\vphantom{ \frac{1}{\sqrt{\frac{\cos^2\phi}{\theta_x ^2} + \frac{\sin^2\phi}{\theta_y ^2}}} }\right)$$\displaystyle \left.\vphantom{ 1 - \cos \left( \frac{1}{\sqrt{\frac{\cos^2\phi}{\theta_x ^2} + \frac{\sin^2\phi}{\theta_y ^2}}} \right) }\right)$ dφ dσ, (16.42)

Instead the cosine series expansion will be used

cosθmax = 1 - $\displaystyle {\frac{{{\theta_{\textrm{max}}}^2}}{{2!}}}$ + $\displaystyle {\frac{{{\theta_{\textrm{max}}}^4}}{{4!}}}$ - $\displaystyle {\frac{{{\theta_{\textrm{max}}}^6}}{{6!}}}$ + $\displaystyle {\frac{{{\theta_{\textrm{max}}}^8}}{{8!}}}$ - $\displaystyle {\frac{{{\theta_{\textrm{max}}}^{10}}}{{10!}}}$ + ... , (16.43)
  = $\displaystyle \sum_{{n=0}}^{\infty}$$\displaystyle {\frac{{(-1)^n}}{{(2n)!}}}$θmax2n. (16.44)

Integrating each element of the sum over the φ parameter and using the assumption that θx, θy≥ 0 gives

$\displaystyle \int_{{-\pi}}^{{\pi}}$$\displaystyle {\frac{{{\theta_{\textrm{max}}}^2}}{{2!}}}$ dφ = πθxθy,    
$\displaystyle \int_{{-\pi}}^{{\pi}}$$\displaystyle {\frac{{{\theta_{\textrm{max}}}^4}}{{4!}}}$ dφ = $\displaystyle {\frac{{\pi \theta_x \theta_y }}{{24}}}$$\displaystyle \left(\vphantom{ \theta_x ^2 + \theta_y ^2 }\right.$θx2 + θy2$\displaystyle \left.\vphantom{ \theta_x ^2 + \theta_y ^2 }\right)$,    
$\displaystyle \int_{{-\pi}}^{{\pi}}$$\displaystyle {\frac{{{\theta_{\textrm{max}}}^6}}{{6!}}}$ dφ = $\displaystyle {\frac{{\pi \theta_x \theta_y }}{{2880}}}$$\displaystyle \left(\vphantom{ 3\theta_x ^4 + 2\theta_x ^2\theta_y ^2 + 3\theta_y ^4 }\right.$3θx4 +2θx2θy2 +3θy4$\displaystyle \left.\vphantom{ 3\theta_x ^4 + 2\theta_x ^2\theta_y ^2 + 3\theta_y ^4 }\right)$,    
$\displaystyle \int_{{-\pi}}^{{\pi}}$$\displaystyle {\frac{{{\theta_{\textrm{max}}}^8}}{{8!}}}$ dφ = $\displaystyle {\frac{{\pi \theta_x \theta_y }}{{322560}}}$$\displaystyle \left(\vphantom{ 5\theta_x ^6 + 3\theta_x ^4\theta_y ^2 + 3\theta_x ^2\theta_y ^4 + 5\theta_y ^6 }\right.$5θx6 +3θx4θy2 +3θx2θy4 +5θy6$\displaystyle \left.\vphantom{ 5\theta_x ^6 + 3\theta_x ^4\theta_y ^2 + 3\theta_x ^2\theta_y ^4 + 5\theta_y ^6 }\right)$,    
$\displaystyle \int_{{-\pi}}^{{\pi}}$$\displaystyle {\frac{{{\theta_{\textrm{max}}}^{10}}}{{10!}}}$ dφ = $\displaystyle {\frac{{\pi \theta_x \theta_y }}{{232243200}}}$$\displaystyle \left(\vphantom{ 35\theta_x ^8 + 20\theta_x ^6\theta_y ^2 + 18\theta_x ^4\theta_y ^4 + 20x^2\theta_y ^6 + 35\theta_y ^8 }\right.$35θx8 +20θx6θy2 +18θx4θy4 +20x2θy6 +35θy8$\displaystyle \left.\vphantom{ 35\theta_x ^8 + 20\theta_x ^6\theta_y ^2 + 18\theta_x ^4\theta_y ^4 + 20x^2\theta_y ^6 + 35\theta_y ^8 }\right)$,    
... (16.45)

Therefore a new two dimension trigonometric function, the pseudo-elliptic cosine, can be defined as

\begin{subequations}\begin{align}
\pec (\theta_x , \theta_y )
&= \int_{-\pi}^{\p...
...ac{(-1)^n}{4^n(2n+2)!} f_n(\theta_x , \theta_y ) .
\end{align}\end{subequations}

Let

a = θx2,    
b = θy2, (16.47)

then the first of the fn(θx, θy) equations are

f0 = 1,    
f1 = 2a + 2b,    
f2 = 6a2 +4ab + 6b2,    
f3 = 20a3 +12a2b + 12ab2 +20b3,    
f4 = 70a4 +40a3b + 36a2b2 +40ab3 +70b4,    
f5 = 252a5 +140a4b + 120a3b2 +120a2b3 +140ab4 +252b5,    
f6 = 924a6 +504a5b + 420a4b2 +400a3b3 +420a2b4 +504ab5 +924b6,    
f7 = 3432a7 +1848a6b + 1512a5b2 +1400a4b3 +1400a3b4 +1512a2b5 +1848ab6 +3432b7,    
... (16.48)

Or

f0 = 1,    
f1 = 2(a + b),    
f2 = 6(a2 + b2) + 4ab,    
f3 = 20(a3 + b3) + 12(a2b + ab2),    
f4 = 70(a4 + b4) + 40(a3b + ab3) + 36a2b2,    
f5 = 252(a5 + b5) + 140(a4b + ab4) + 120(a3b2 + a2b3),    
f6 = 924(a6 + b6) + 504(a5b + ab5) + 420(a4b2 + a2b4) + 400a3b3,    
f7 = 3432(a7 + b7) + 1848(a6b + ab6) + 1512(a5b2 + a2b5) + 1400(a4b3 + a3b4),    
f8 = 12870(a8 + b8) + 6864(a7b + ab7) + 5544(a6b2 + a2b6) + 5040(a5b3 + a3b5) + 4900a4b4,    
f9 = 48620(a9 + b9) + 25740(a8b + ab8) + 20592(a7b2 + a2b7) + 18480(a6b3 + a3b6) + 17640(a5b4 + a4b5),    
f10 = 184756(a10 + b10) + 97240(a9b + ab9) + 77220(a8b2 + a2b8) + 68640(a7b3 + a3b7) + 64680(a6b4 + a4b6) + 63504a5b5,    
... (16.49)

This series expansion up to n = 10 is sufficient for writing a fast and accurate pec function implementation in computer code. The numerical representation of this function is shown in figure 16.14.

Figure 16.14: The pseudo-ellipse cosine 2D trigonometric function. This is the surface area on a unit sphere bounded by the pseudo-elliptic cone.
 \includegraphics[width=.5\textwidth,bb=150 150 650 650]{images/pec_y.eps} \includegraphics[width=.5\textwidth,bb=80 80 730 740]{images/pec_diag.eps}  

The relax user manual (PDF), created 2020-08-26.