Matematyka.pl

\(\displaystyle{
a = 6378.137\\
f = \frac{21.385}{6378.137}
}\)

a to długość promienia równikowego
natomiast f to spłaszczenie
Nie jest to do końca prawdą ale przyjmijmy że są to wartości stałe
Dokładny model tej bryły to geoida a rozpatrywana tutaj bryła to tylko model przybliżony

Przyjmijmy że równanie elipsy jest następujące

\(\displaystyle{
\left(\frac{x}{a}\right)^{2}+\left( \frac{y}{b}\right)^{2} = 1\\
}\)

Zastosujmy parametryzację
\(\displaystyle{
\begin{cases} \frac{x}{a} = \cos{\left( t\right) } \\ \frac{y}{b} = \sin{\left( t\right) } \end{cases} \\
\begin{cases} x = a\cos{\left( t\right) } \\ y = b\sin{\left( t\right) } \end{cases} \\
}\)

Wstawmy to do wzoru na długość krzywej

\(\displaystyle{ \int\limits_{\theta_{1}}^{\theta_{2}}{ \sqrt{\left(x'\left( t\right) \right)^2+\left(y'\left( t\right) \right)^2} \mbox{d}t}\\
\int\limits_{\theta_{1}}^{\theta_{2}}{ \sqrt{\left(-a\sin{\left( t\right) }\right)^2+\left( b\cos{\left( t\right) }\right)^2 }\mbox{d}t}\\
\int\limits_{\theta_{1}}^{\theta_{2}}{ \sqrt{a^2\sin^2{\left( t\right) }+b^2\cos^{2}{\left( t\right) }} \mbox{d}t}\\
\int\limits_{\theta_{1}}^{\theta_{2}}{ \sqrt{a^2\sin^2{\left( t\right) }+b^2 -b^2\sin^{2}{\left( t\right) }} \mbox{d}t}\\
\int\limits_{\theta_{1}}^{\theta_{2}}{ \sqrt{b^2+\left(a^2-b^2\right)\sin^2{\left( t\right) } } \mbox{d}t}\\
\int\limits_{\theta_{1}}^{\theta_{2}}{ b\sqrt{1+\frac{\left(a^2-b^2\right)}{b^2}\sin^2{\left( t\right) } } \mbox{d}t}\\
}\)

Niech \(\displaystyle{ \varepsilon^2 = \frac{a^2 - b^2}{b^2}}\)


\(\displaystyle{ b\int\limits_{\theta_{1}}^{\theta_{2}}{ \sqrt{1+\varepsilon^2 \sin^{2}{\left( t\right) }} \mbox{d}t}\\
b\int\limits_{\theta_{1}}^{\theta_{2}}{ \sqrt{1+\left(\varepsilon \sin{\left( t\right) }\right)^2} \mbox{d}t}\\
}\)

Rozwińmy funkcję podcałkową w szereg z dwumianu Newtona

\(\displaystyle{
b\int\limits_{\theta_{1}}^{\theta_{2}}{\left(\sum\limits_{n=0}^{ \infty }{\frac{1}{2} \choose n}\varepsilon^{2n}\sin^{2n}{\left( t\right) } \right) \mbox{d}t}\\
b\sum\limits_{n=0}^{ \infty }{ {\frac{1}{2} \choose n}\varepsilon^{2n}\int\limits_{\theta_{1}}^{\theta_{1}}\sin^{2n}{\left(t \right) } \mbox{d}t}\\
}\)

Rozpiszmy symbol Newtona

\(\displaystyle{
{\frac{1}{2} \choose n} = \frac{\frac{1}{2}\left( \frac{1}{2} - 1\right)\cdot\ldots\cdot\left( \frac{1}{2}-\left( n-1\right) \right) }{n!}\\
{\frac{1}{2} \choose n} = \frac{\left( -1\right)^{n-1} }{2^{n}} \cdot \frac{1 \cdot 1 \cdot 3 \cdot \ldots\cdot\left( 2n-3\right) }{n!}\\
{\frac{1}{2} \choose n} = \frac{\left( -1\right)^{n}}{2^{n}}\frac{1 \cdot 3 \cdot \ldots\cdot\left( 2n-3\right) \cdot \left( 2n-1\right) }{\left( 1-2n\right)n! }\\
{\frac{1}{2} \choose n} = \frac{\left( -1\right)^{n}}{2^{n}}\frac{1 \cdot 3 \cdot \ldots\cdot\left( 2n-3\right) \cdot \left( 2n-1\right) \cdot 2\cdot 4 \cdot\ldots\cdot 2n}{\left( 1-2n\right)n! \cdot 2\cdot 2 \cdot 2 \cdot\ldots\cdot 2 \cdot n}\\
{\frac{1}{2} \choose n} = \frac{\left( -1\right)^{n} }{2^{2n}} \cdot \frac{1}{1-2n} \cdot \frac{\left(2n \right)! }{n! \cdot n!} \\
{\frac{1}{2} \choose n} = \frac{\left( -1\right)^{n} }{2^{2n}} \cdot \frac{1}{1-2n} \cdot {2n \choose n}\\
}\)

\(\displaystyle{ b\sum\limits_{n=0}^{\infty}{\left( -1\right)^{n} \cdot \left( \frac{\varepsilon}{2} \right)^{2n} \cdot {2n \choose n} \cdot \frac{1}{1-2n} \cdot \int\limits_{\theta_{1}}^{\theta_{2}}{\sin^{2n}{\left( t\right) }\mbox{d}t} } }\)

\(\displaystyle{
\int\limits_{\theta_{1}}^{\theta_{2}}{\sin^{2n}{\left( t\right) }\mbox{d}t}=\int\limits_{\theta_{1}}^{\theta_{2}}{\sin{\left( t\right) } \cdot \sin^{2n-1}{\left( t\right) }\mbox{d}t}\\
\int\limits_{\theta_{1}}^{\theta_{2}}{\sin^{2n}{\left( t\right) }\mbox{d}t}= -\cos{\left( t\right) }\sin^{2n-1}{\left( t\right) }\biggl|_{\theta_{1}}^{\theta_{2}} - \int\limits_{\theta_{1}}^{\theta_{2}}{\left( -\cos{\left( t\right)}\right)\left( 2n-1\right) \sin^{2n-2}{\left( t\right) } \cos{\left( t\right) }\mbox{d}t}\\
\int\limits_{\theta_{1}}^{\theta_{2}}{\sin^{2n}{\left( t\right) }\mbox{d}t}= -\cos{\left( t\right) }\sin^{2n-1}{\left( t\right) }\biggl|_{\theta_{1}}^{\theta_{2}} + \left( 2n-1\right) \int\limits_{\theta_{1}}^{\theta_{2}}{\sin^{2n-2}{\left( t\right) }\cos^{2}{\left( t\right) }\mbox{d}t}\\
\int\limits_{\theta_{1}}^{\theta_{2}}{\sin^{2n}{\left( t\right) }\mbox{d}t}= -\cos{\left( t\right) }\sin^{2n-1}{\left( t\right) }\biggl|_{\theta_{1}}^{\theta_{2}} + \left( 2n-1\right) \int\limits_{\theta_{1}}^{\theta_{2}}{\sin^{2n-2}{\left( t\right) }\left( 1-\sin^{2}{\left( t\right) }\right) \mbox{d}t}\\
\int\limits_{\theta_{1}}^{\theta_{2}}{\sin^{2n}{\left( t\right) }\mbox{d}t}= -\cos{\left( t\right) }\sin^{2n-1}{\left( t\right) }\biggl|_{\theta_{1}}^{\theta_{2}} +\left( 2n-1\right)\int\limits_{\theta_{1}}^{\theta_{2}}{\sin^{2n-2}{\left( t\right) }\mbox{d}t} -\left( 2n-1\right)\int\limits_{\theta_{1}}^{\theta_{2}}{\sin^{2n}{\left( t\right) }\mbox{d}t}\\
2n\int\limits_{\theta_{1}}^{\theta_{2}}{\sin^{2n}{\left( t\right) }\mbox{d}t}= -\cos{\left( t\right) }\sin^{2n-1}{\left( t\right) }\biggl|_{\theta_{1}}^{\theta_{2}} +\left( 2n-1\right)\int\limits_{\theta_{1}}^{\theta_{2}}{\sin^{2n-2}{\left( t\right) }\mbox{d}t}\\
\int\limits_{\theta_{1}}^{\theta_{2}}{\sin^{2n}{\left( t\right) }\mbox{d}t}= -\frac{1}{2n}\cos{\left( t\right) }\sin^{2n-1}{\left( t\right) }\biggl|_{\theta_{1}}^{\theta_{2}} + \frac{2n-1}{2n}\int\limits_{\theta_{1}}^{\theta_{2}}{\sin^{2n-2}{\left( t\right) }\mbox{d}t}\\
\int\limits_{\theta_{1}}^{\theta_{2}}{\sin^{2n}{\left( t\right) }\mbox{d}t}= -\frac{1}{2n}\cos{\left( \theta_{2}\right) }\sin^{2n-1}{\left( \theta_{2}\right) }+\frac{1}{2n}\cos{\left( \theta_{1}\right) }\sin^{2n-1}{\left( \theta_{1}\right) } + \frac{2n-1}{2n}\int\limits_{\theta_{1}}^{\theta_{2}}{\sin^{2n-2}{\left( t\right) }\mbox{d}t}\\
}\)

Mamy wzór rekurencyjny na tę całkę

\(\displaystyle{
\begin{cases} I_{0} = \theta_{2} - \theta_{1} \\ I_{n+1} =-\frac{1}{2n+2}\cos{\left( \theta_{2}\right) }\sin^{2n+1}{\left( \theta_{2}\right) }+\frac{1}{2n+2}\cos{\left( \theta_{1}\right) }\sin^{2n+1}{\left( \theta_{1}\right) } + \frac{2n+1}{2n+2}I_{n} \end{cases}
}\)


Czy dałoby się tę całkę wykorzystać do policzenia długości łuku gdy mamy podane współrzędne punktów

A może by tak rozwiązać to równanie rekurencyjne czynnikiem sumacyjnym
Wtedy dostalibyśmy tę całkę w postaci podwójnej sumy

\(\displaystyle{
\begin{cases} I_{0} = \theta_{2} - \theta_{1} \\ I_{n} = \frac{2n-1}{2n}I_{n-1}+\frac{1}{2n}\left(\cos{\left( \theta_{1}\right) }\sin^{2n-1}{\left( \theta_{1}\right) } - \cos{\left( \theta_{2}\right) }\sin^{2n-1}{\left( \theta_{2}\right) }\right) \end{cases} \\
\begin{cases} I_{0} = \theta_{2} - \theta_{1} \\ 2n I_{n} = \left( 2n-1\right)I_{n-1} +\cos{\left( \theta_{1}\right) }\sin^{2n-1}{\left( \theta_{1}\right) } - \cos{\left( \theta_{2}\right) }\sin^{2n-1}{\left( \theta_{2}\right) }\end{cases}
}\)

Teraz \(\displaystyle{ a_{n} = 2n \qquad b_{n} = 2n-1 \qquad c_{n} = \cos{\left( \theta_{1}\right) }\sin^{2n-1}{\left( \theta_{1}\right) } - \cos{\left( \theta_{2}\right) }\sin^{2n-1}{\left( \theta_{2}\right) }}\)

Niech \(\displaystyle{ s_{n} = \frac{2n-2}{2n-1}s_{n-1}}\)

Jeżeli pomnożymy równanie rekurencyjne przez \(\displaystyle{ s_{n}}\)
to otrzymamy \(\displaystyle{ 2ns_{n}I_{n} = \frac{2n-2 }{2n-1}s_{n-1} \cdot \left( 2n-1\right)I_{n-1}+s_{n}\left(\cos{\left( \theta_{1}\right) }\sin^{2n-1}{\left( \theta_{1}\right) } - \cos{\left( \theta_{2}\right) }\sin^{2n-1}{\left( \theta_{2}\right) } \right) \\
2ns_{n}I_{n} = \left( 2n-2\right)s_{n-1}I_{n-1} +s_{n}\left(\cos{\left( \theta_{1}\right) }\sin^{2n-1}{\left( \theta_{1}\right) } - \cos{\left( \theta_{2}\right) }\sin^{2n-1}{\left( \theta_{2}\right) } \right) \\
U_{n} = 2ns_{n}I_{n}\\
U_{n} = U_{n-1}+ s_{n}\left(\cos{\left( \theta_{1}\right) }\sin^{2n-1}{\left( \theta_{1}\right) } - \cos{\left( \theta_{2}\right) }\sin^{2n-1}{\left( \theta_{2}\right) } \right) \\
U_{n} = U_{0} + \sum\limits_{k=1}^{n}{s_{k}\left(\cos{\left( \theta_{1}\right) }\sin^{2k-1}{\left( \theta_{1}\right) } - \cos{\left( \theta_{2}\right) }\sin^{2k-1}{\left( \theta_{2}\right) } \right)}\\
2ns_{n}I_{n} = \left( s_{1}I_{0} + \sum\limits_{k=1}^{n}{s_{k}\left(\cos{\left( \theta_{1}\right) }\sin^{2k-1}{\left( \theta_{1}\right) } - \cos{\left( \theta_{2}\right) }\sin^{2k-1}{\left( \theta_{2}\right) } \right)}\right) \\
I_{n} = \frac{1}{2ns_{n}} \cdot \left( s_{1}\left( \theta_{2} - \theta_{1}\right)+ \sum\limits_{k=1}^{n}{s_{k}\left(\cos{\left( \theta_{1}\right) }\sin^{2k-1}{\left( \theta_{1}\right) } - \cos{\left( \theta_{2}\right) }\sin^{2k-1}{\left( \theta_{2}\right) } \right)}\right) \\
}\)

\(\displaystyle{ s_{n} = \frac{2n-2}{2n-1}s_{n-1}\\
}\)

Czynnik sumacyjny można zapisać w postaci iloczynu

\(\displaystyle{
s_{n} = s_{1} \cdot \prod\limits_{j = 1}^{n-1}{\frac{2n-2j}{2n-2j + 1}}
}\)

Co po odwróceniu kolejności czynników mamy

\(\displaystyle{ s_{n} = s_{1} \cdot \prod\limits_{j = 1}^{n-1}{\frac{2j}{2j+1}} }\)

Po rozpisaniu tego iloczynu powinniśmy otrzymać

\(\displaystyle{
I_{n} = \frac{1}{2^{2n}}\cdot {2n \choose n} \cdot \left( \theta_{2}-\theta_{1} + \sum\limits_{k=1}^{n}{ \frac{\left( 2^{k} \cdot k!\right)^2 }{2k \cdot \left( 2k\right)! } \cdot \left( \cos{\left( \theta_{1}\right) }\sin^{2k-1}{\left( \theta_{1}\right) } - \cos{\left( \theta_{2}\right) }\sin^{2k-1}{\left( \theta_{2}\right) }\right)}\right)
}\)