| (1−x2)−1/2+1 | (1−x2)−1/2−1 | |||
[ | ]−1/2 +[ | ]−1/2 dla x=2k1/2(1+k)−1i k>1 | ||
| 2 | 2 |
| 2√k | 4k | |||
x= | wiec x2= | |||
| 1+k | (1+k)2 |
| 4k | (1+k)2−4k | 1−2k+k2 | ||||
(1−( | )−1/2=[ | ]−1/2=[ | ]−1/2 | |||
| (1+k)2 | (1+k)2 | (1+k)2 |
| (1−k)2 | ||
=[ | ]−1/2 | |
| (1+k)2 |
| 1 | (k+1)2 | (1+k)2 | ||||||||||
=( | )1/2=( | )1/2=√ | = | |||||||||
| (1−k)2 | (1−k)2 |
| |1+k| | |1+k| | 1+k | ||||
a przy warunku k>1 | = | |||||
| |1−k| | |1−k| | k−1 |
| 1+k | ||
Wiec (1−x2)−1/2= | ||
| k−1 |
| 1+k | 1 | 1+k+k−1 | 1 | |||||
[( | +1)* | )]−1/2=[( | )* | ]−1/2=[(U{k}{ | ||||
| k−1 | 2 | k−1 | 2 |
| k−1 | √k−1 | |||
k−1})]−1/2=( | )1/2= | |||
| k | √k |
| 1+k | 1 | 1+k−(k−1) | 1 | 1 | ||||||
[( | −1)* | )]−1/2=[( | )* | ]−1/2=( | )−1/2 | |||||
| k−1 | 2 | k−1 | 2 | k−1 |
| √k−1 | 1 | ||
+√k−1=√k−1( | +1) dla k>1 | ||
| √k | √k |
To super .