- #1
Dustinsfl
- 2,281
- 5
Show that the infinite product [itex]f(z) = \prod\limits_{n = 0}^{\infty}(1 + z^{2^n})[/itex] converges on the open disc [itex]D(0,1)[/itex] to the function [itex]1/(1 - z)[/itex]. Is this convergence uniform on compact subsets of the disc?
This should actually be done by the comparison test.
For [itex]|z| < 1[/itex], we have that
$$
\frac{1}{1-z}=\sum_{n=0}^{\infty}|z|^{2^n}\leq \sum_{n=0}^{\infty}|z|^n
$$
So now I need to show by partial products that there are no zeros in [itex]K\subset D(0,1)[/itex]. So I need to find an N such that forall n > N this holds. I am looking for some guidance on this piece.
$$
f(z) = \prod_{n=0}^{N}\left(1+z^{2^n}\right) \prod_{n=N+1}^{\infty}\left(1+z^{2^n}\right)
$$
We need to fix [itex]R\in\mathbb{R}^+[/itex]. Let [itex]N\in\mathbb{N}[/itex] such that [itex]|z_N|\leq 2R < |z_{N+1}[/itex] (is this correct-the inequalities?).
The first partial product is finite on [itex]D(0,1)[/itex] and the second partial product behaves well on [itex]D(0,1)[/itex].
What would be my choice of [itex]k_n[/itex] for this product?
This should actually be done by the comparison test.
For [itex]|z| < 1[/itex], we have that
$$
\frac{1}{1-z}=\sum_{n=0}^{\infty}|z|^{2^n}\leq \sum_{n=0}^{\infty}|z|^n
$$
So now I need to show by partial products that there are no zeros in [itex]K\subset D(0,1)[/itex]. So I need to find an N such that forall n > N this holds. I am looking for some guidance on this piece.
$$
f(z) = \prod_{n=0}^{N}\left(1+z^{2^n}\right) \prod_{n=N+1}^{\infty}\left(1+z^{2^n}\right)
$$
We need to fix [itex]R\in\mathbb{R}^+[/itex]. Let [itex]N\in\mathbb{N}[/itex] such that [itex]|z_N|\leq 2R < |z_{N+1}[/itex] (is this correct-the inequalities?).
The first partial product is finite on [itex]D(0,1)[/itex] and the second partial product behaves well on [itex]D(0,1)[/itex].
What would be my choice of [itex]k_n[/itex] for this product?
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