Question

If $f\left(x\right)=(1+x)(1+x^2)(1+x^4)(1+x^8)...\infty$ then $f\left(\frac{1}{2}\right)=$?

Answer 1

The answer is $=2$

The series is

$f\left(x\right)={\Pi }_{k=0}^{\infty }(1+x^{2k})$

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,

$f\left(x\right)={\Pi }_{k=0}^{\infty }(1+x^{2^k})=1+x+x^2+x^3+x^4+x^5+x^6+............x^{\infty }$

This is a geometric series.

And

$f\left(x\right)=\frac{1}{1-x}$ if $|x|<1$

therefore,

$f\left(\frac{1}{2}\right)=\frac{1}{1-\frac{1}{2}}=2$