466IF X<0 ,
THEN S=1-X;
- Himanshu Giria NO SORRY
IT WILL B 1+X ONLY
Sayan Sinha My solution works that way, right??? :)
Sayantan Hazra Simple check : At x = -1, acc. to u, S = 0. But is it? 
Find the summation
s=x+x2+x4+x8.......till infinity
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UP 0 DOWN 0 1 6
6 Answers
337 s=x+x2+x4+x8.......∞
=>s= x(1+x2+x4+x8...∞)
=>s=x(1+s)
=>s=x1-x
Shaswata Roy Check again.You made a mistake.Upvote·0· Reply ·2013-05-24 11:01:31- Sayan Sinha In the second step, it should have been s=x(1+x+x^2+x^4...inf)
- Shaswata Roy You Sure...It's still wrong.
- Sayan Sinha I know the it should have been x^3..x^5 and so on. So, how about taking x^2 common?
- Sayan Sinha Nope. Doesn't work...
466IF X<1,
THEN S=1+X;
IF X≥1 ,
THEN S=∞
Asish Mahapatra A simple check : at x=0, S = 0 :)
466206Hint: Concentrate on 2nd,3rd,4th terms leaving aside the 1st term.... :)
2305This is a Lacunary function [See-"http://en.wikipedia.org/wiki/Lacunary_function"].
It does not have a closed form.
