TITLE: Can we use infinite series to multiply finite polynomial series?
QUESTION [5 upvotes]: I was trying to figure out the expansion of the product:
$$Q= (1+x)(1+x^2)(1+x^4)..(1+x^{2^n})$$
Multiplying and dividing by $1-x$
$$ Q =  \frac{1}{1-x}  [ 1-x^{2^{n+1}}]$$
Expanding the denominator as geometric series:
$$ Q =  - [ x^{2^{n+1}} - 1 ]   [ \sum_{j=0}^{\infty} x^j]  = - [ \sum_{j=0}^{\infty} x^{2^{n+1}+j} - x^{j}]$$
Now,
$$ \sum_{j=0}^{\infty} x^j =  \sum_{j=0}^{2^{n+1}}  x^j  + \sum_{j=2^{n+1}}^{\infty} x^{j} = \sum_{j=0}^{2^{n+1}} x^j + \sum_{j=0}^{\infty} x^{2^n+j}$$
Using the above on $Q$,
$$ Q = \sum_{j=0}^{2^{n+1}} x^j$$
Or,
$$  (1+x)(1+x^2)(1+x^4)..(1+x^{2^n})=\sum_{j=0}^{2^{n+1}} x^j$$

My question: $ \frac{1}{1-x} = \sum_{j=0}^{\infty} x^j$ only holds if $ |x|<1$ so this chain of logical implications? But it is weird to me because I had a finite polynomial end ended in a finite polynomial so I am pretty convinced that my final polynomial is actually the expansion of first.. so, how do I justify this gives me the equation which holds for all $ x \in R$

REPLY [2 votes]: Actually, you do not need to introduce the $1/(1-x)$ factor, since the equality between the two polynomials
can be established in a direct elegant way.
Consider the development of
$$
\eqalign{
  & \left( {1 + x} \right)\left( {1 + x^{\,2} } \right)\left( {1 + x^{\,4} } \right) =   \cr 
  &  = \left( {x^{\,0}  + x} \right)\left( {x^{\,0}  + x^{\,2} } \right)\left( {x^{\,0}  + x^{\,4} } \right) =   \cr 
  &  = x^{\,0}  \cdot x^{\,0}  \cdot x^{\,0}  +   \cr 
  &  + x^{\,1}  \cdot x^{\,0}  \cdot x^{\,0}  +   \cr 
  &  + x^{\,0}  \cdot x^{\,2}  \cdot x^{\,0}  +   \cr 
  &  + x^{\,1}  \cdot x^{\,2}  \cdot x^{\,0}  +   \cr 
  &  + x^{\,0}  \cdot x^{\,0}  \cdot x^{\,4}  +   \cr 
  &  +  \cdots  +   \cr 
  &  + x^{\,1}  \cdot x^{\,2}  \cdot x^{\,4}  \cr} 
$$
That, is from each of the three binomials you take (weight 1) or don't take (weight 0)
each of the exponents $1,2,4$ , thereby obtaining $2^3$ triplets where the exponents
are the $3$ -digit binary representation of the numbers $0$ to $7$.
Since the binary representation of a natural number is unique, then ...
Refer to this interesting paper.