The discussion revolves around the bounds of the partial sum in the context of the product representation of a series involving powers of \( z \). Participants are exploring the transition from the index \( N \) in the product to the upper limit \( 2^{N+1}-1 \) in the sum, examining the implications of this transformation.
Participants generally agree on the observation that the product leads to a comprehensive representation of powers of \( z \), but there is no consensus on the implications of odd powers or the necessity of including them in the partial sum. The discussion remains unresolved regarding the clarity of the transition from \( N \) to \( 2^{N+1}-1 \).
Some assumptions about the nature of the powers and their coefficients may not be fully articulated, and the dependence on the definitions of even and odd powers is implicit in the discussion.
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
$\chi$ $\sigma$