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- #1

- Thread starter dwsmith
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- Thread starter
- #1

- Feb 13, 2012

- 1,704

Writing in explicit form...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?

$\displaystyle \prod_{n=0}^{\infty} (1+z^{2^{n}})= (1+z)\ (1+z^{2})\ (1+z^{4})...\ (1+z^{2^{N}})$

... 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

$\chi$ $\sigma$

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- #3

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

$\chi$ $\sigma$

- Feb 13, 2012

- 1,704

Let's proceed 'step by step'...There is no odd power. With the partial sum, we would have odd powers too.

$(1+z)$ 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

$\chi$ $\sigma$

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- #5

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

$\chi$ $\sigma$