y don u use binomial approximation for (1-x^{2n})^{-1/2}
3 Answers
Tush Watts
·2009-10-07 05:04:34
Ans) 0 < x <1/2
x2n < x2
-x2n > -x2
(1-x2n ) >(1 -x2)
Now, I > √(1-x2n ) > √(1 -x2)
Therefore, I < 1/√(1-x2n ) < 1 / √(1 -x2)
< <
Therefore, (1/2) < I < sin -1 (1/2)
(1/2) < I < ∩/6
or 0.5 < I < 0.52