Contenuto cancellato Contenuto aggiunto
|
|
|
:<math>
\arcsec z = \arccos\left(\frac{1}{z}\right) = \frac {\pi} {2} - (\left[z^{-1} + \left( \frac {1} {2} \right) \frac {z^{-3}} {3} + \left( \frac {1 \cdot 3} {2 \cdot 4} \right) \frac {z^{-5}} {5} + \left( \frac{1 \cdot 3 \cdot 5} {2 \cdot 4 \cdot 6 } \right) \frac{z^{-7}} {7} + \cdots )\right]= \frac {\pi} {2} - \sum_{n=0}^\infty \left( \frac {(2n)!} {2^{2n}(n!)^2} \right) \frac {z^{-(2n+1)}} {(2n+1)}
\ , \quad \left| z \right| > 1
</math>
:<math>
&\arccot z = \frac {\pi} {2} - \arctan z = & \frac {\pi} {2} - ( z - \frac {z^3} {3} +\frac {z^5} {5} -\frac {z^7} {7} +\cdots ) = \ frac {\ pi} {2} - \ sum_{n=0}^\ infty \frac {(-1)^n z^{2n+1}} {2n+1}▼\begin{matrix}
\arccot z & = & \frac {\pi} {2} - \arctan z \\ \\
▲& = & \frac {\pi} {2} - ( z - \frac {z^3} {3} +\frac {z^5} {5} -\frac {z^7} {7} +\cdots ) \\ \\
& = & \frac {\pi} {2} - \sum_{n=0}^\infty \frac {(-1)^n z^{2n+1}} {2n+1}
\end{matrix}
\ , \quad \left| z \right| < 1
</math>
|
