If #f(x)=(1+x)(1+x^2)(1+x^4)(1+x^8)...oo# then #f(1/2)=#?

1 Answer
Jul 21, 2018

Answer:

The answer is #=2#

Explanation:

The series is

#f(x)=Pi_(k=0)^oo(1+x^(2^k))#

Let's start with #2# terms

#(1+x)(1+x^2)=1+x+x^2+x^3#

And with #3# terms

#(1+x)(1+x^2)(1+x^4)=1+x+x^2+x^3+x^4+x^5+x^6+x^7#

Therefore,

#f(x)=Pi_(k=0)^oo(1+x^(2^k))=1+x+x^2+x^3+x^4+x^5+x^6+.......................x^(oo)#

This is a geometric series

And

#f(x)=1/(1-x)# iif #|x|<1#

Therefore,

#f(1/2)=1/(1-1/2)=2#