MHB Partial Sum Bounds: Explaining the Shift from N to $2^{N+1}-1$

[Dustinsfl]

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]

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?

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

$\chi$ $\sigma$

 

[Dustinsfl]

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

$\chi$ $\sigma$

There is no odd power. With the partial sum, we would have odd powers too.

 

[chisigma]

There is no odd power. With the partial sum, we would have odd powers too.

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

$\chi$ $\sigma$

 

[Dustinsfl]

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

$\chi$ $\sigma$

Ok, I wasn't exactly thinking of multiplying out the products.