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[SOLVED] bounds of a series

dwsmith

Well-known member
Feb 1, 2012
1,673
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

Well-known member
Feb 13, 2012
1,704
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

$\chi$ $\sigma$
 

dwsmith

Well-known member
Feb 1, 2012
1,673
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

Well-known member
Feb 13, 2012
1,704
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

$\chi$ $\sigma$
 

dwsmith

Well-known member
Feb 1, 2012
1,673
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.