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kghedriu (2333)

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Olaaa!! Perrrfect answer. 423  [532 rates]

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say f(x)= e^(sin-1x) ....it is e to the power of sin inverse x...
 
now... f(1/2)= e^pi/6 and f(-1/2)= e^-pi/6
cubing this....we get e^pi/2 or e^-pi/2 in RHS.....
 
by expansions...
 
f(x)= 1+x+ x^2/2! +2x^3/3!+ 5x^4/4!+.......
 
     ={1 +t 2k+1}+ {x+t 2k+2}  where k runs frm 1 to infinity and mod x less dan 1..
 
and t2k+1= (1)(2^2+1)(4^2+1).....[(2k-1)^2+1]x^2k/(2k)!
 
and
t 2k+2 = (1^2+1)(3^2+1)........x^2k+1/(2k+1)!
now...use these formulas to calculate 85 terms in each with each term rounded to 54 decimal places..this processs gives  fifty two decimal places of e^pi/6
 
simly calc for e^-pi/6
 
den multiply to get e ^-pi/3
calc e^-pi/3
multiply to get e^-pi/2 which is LHS and i^i is our RHS
 
this is the only way to do this...
it was given in 1922 by UHLER....
 
finally the value of i^i is as follows...
 
0.2078795763
 this reply: 12 points  (with Olaaa!! Perrrfect answer.   in 3 votes )   [?]
 
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