262f(x)= (x-1) + (1+x2 + x4 + ....)
= (x-1) + (1-x2)-1
but f(f(x)) is getting too complicated .
looking forward to a simpler method
\hspace{-16}$Let $f(x)=x+x^2+x^4+x^8+x^{16}+x^{32}+...........................$\\\\ Then find Coeff. of $x^{10}$ in $f(f(x))$
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8 Answers
1057f(x)=x1-x2
f(f(x)=x(1-x2)(1+x4-3x2)-1
now i think it can be done!!!!!
262
71@Ketan/Aditya Isn't your formula for GP Sum valid only when |x| < 1
There isn't any mention of that. However,I think it has to be assumed.
1057@vivek if |x| > 1 then f(x) will tend to infinity......f(x) wont be defined den....
although my method was incorrect... :P
262sorry, my earlier post is incorrect.
directly plugging in f(f(x)) and finding the coeff of x10 , i am getting 40
71Isn't that complicated Aditya? Please post the solution.
262f(f(x) = (x+x2+x4+..) + (x+x2+x4+..)2 + (x+x2+x4+..)4 + (x+x2+x4+..)8 + ...
i)coeff of x10→ (x+x2+x4+..) = 0
ii)coeff of x10→ (x+x2+x4+..)2
= coeff of x8→ (1+x1+x3+x7+..)2 = 2!1!1! (7,1) = 2
iii)coeff of x10→ (x+x2+x4+..)4
=coeff of x6→ (1+x1+x3+..)4 = 4!2!2! (3,3,0,0) + 4!3!1! (3,1,1,1) =6+4=10
iv)coeff of x10→ (x+x2+x4+..)8
= coeff of x2→ (1+x1+x3+..)8 = 8!2!6! (1,1,0,0,0,0,0,0) = 28
hence total = 28 +10 +2 =40
