The discussion centers on the mathematical transformation of the product $\prod_{n=0}^N(1+z^{2^n})$ into the sum $\sum_{n=0}^{2^{N+1}-1}z^n$. Participants clarify that the product generates all powers of z from 0 to $2^{N+1}-1$, confirming that each coefficient is 1. The absence of odd powers in the product is highlighted, emphasizing the need for a comprehensive understanding of how these transformations occur through multiplication of terms like $(1+z)$ and $(1+z^2)$.
PREREQUISITESMathematicians, students of algebra, and anyone interested in combinatorial mathematics or the properties of power series and infinite products.
dwsmith said:I don't understand the bounds of the partial sum.
$$
\prod_{n=0}^N(1+z^{2^n}) =\sum_{n=0}^{2^{N+1}-1}z^n = \frac{1-z^{2^{N+1}}}{1-z}
$$
How does it go from N in the product to $2^{N+1}-1$ in the sum?
chisigma said:Writing in explicit form...
$\displaystyle \prod_{n=0}^{\infty} (1+z^{2^{n}})= (1+z)\ (1+z^{2})\ (1+z^{4})...\ (1+z^{2^{N}})$
... it is easy to observe that any power of z from 0 to $2^{N+1}-1$, each with coefficient 1, is present...
Kind regards
$\chi$ $\sigma$
dwsmith said:There is no odd power. With the partial sum, we would have odd powers too.
chisigma said:Let's proceed 'step by step'...
$(1+z)$ and we have even and odd powers...
$(1+z)\ (1+z^{2})= 1 + z + z^{2}+z^{3}$ and we have even an odd powers...
$(1+z)\ (1+z^{2})\ (1+z^{4})= 1+z+z^{2}+z^{3}+z^{4}+z^{5}+z^{6}+z^{7}$ and we have even an odd powers...
Shall I continue?...
Kind regards
$\chi$ $\sigma$