A Geometric Product (a series of exercises for the curious)

[benorin]

I hope you have fun with these...

OK, so you know the geometric series, right? It goes like this:

$$\sum_{k=0}^{\infty} z^k = \frac{1}{1-z},\forall z\in\mathbb{C}\mbox{ such that }\left| z\right|<1$$

How about this one? Call it, say, the geometric product:

$$\prod_{k=0}^{\infty} \left( 1+ z^{2^{k}}\right) = \frac{1}{1-z},\forall z\in\mathbb{C}\mbox{ such that }\left| z\right|<1$$

Prove it in two ways.

Suppose a function $$f(z)$$ is equal to its own logarithmic derivative, namely, suppose that

$$\exists f(z)\mbox{ such that } \frac{d}{dz}\ln\left( f(z)\right) = \frac{f^{\prime}(z)}{f(z)} = f(z)$$

Prove that $f(z)=\frac{1}{C-z}$ where C is a constant, is a family of solutions to this differential equation.

That being known, and taking C=1 above, we have (hand-waving the convergence details):

$$\frac{d}{dz}\ln\left[ \prod_{k=0}^{\infty} \left( 1+ z^{2^{k}}\right) \right] = \frac{1}{1-z},\forall z\in\mathbb{C}\mbox{ such that }\left| z\right|<1$$

We have derived what new series by virtue of the fact that the log of a product is the sum of the logs?

 

[benorin]

Would a hint help? two?

How about this one? Call it, say, the geometric product:

$$\prod_{k=0}^{\infty} \left( 1+ z^{2^{k}}\right) = \frac{1}{1-z},\forall z\in\mathbb{C}\mbox{ such that }\left| z\right|<1$$

Prove it in two ways.

Hint #1: Convert binary to decimal.

Hint #2: Algebra I students know this factoring trick: use it in reverse.

 

[benorin]

An answer

Prove it via Hint #2: Algebra I students know this factoring trick: use it in reverse. Said factoring trick is the difference of squares.
Proof:

$$(1-z)\prod_{k=0}^{\infty} \left( 1+ z^{2^{k}}\right) = \lim_{M \rightarrow \infty} (1-z)\prod_{k=0}^{M} \left( 1+ z^{2^{k}}\right)$$
$$= \lim_{M \rightarrow \infty} (1-z)(1+z)(1+z^2)(1+z^4)(1+z^8)(1+z^16)\cdot\cdot\cdot \left( 1+ z^{2^{M}}\right)$$
$$= \lim_{M \rightarrow \infty} (1-z^2)(1+z^2)(1+z^4)(1+z^8)(1+z^16)\cdot\cdot\cdot \left( 1+ z^{2^{M}}\right)$$
$$=\lim_{M \rightarrow \infty} (1-z^4)(1+z^4)(1+z^8)(1+z^16)\cdot\cdot\cdot \left( 1+ z^{2^{M}}\right)$$
$$=\lim_{M \rightarrow \infty} (1-z^8)(1+z^8)(1+z^16)\cdot\cdot\cdot \left( 1+ z^{2^{M}}\right)=\cdot\cdot\cdot =\lim_{M \rightarrow \infty} \left( 1- z^{2^{M}}\right) \left( 1+ z^{2^{M}}\right)$$
$$=\lim_{M \rightarrow \infty} \left( 1- z^{2^{M+1}}\right)=1,\forall z\in\mathbb{C}\mbox{ such that }\left| z\right|<1$$

QED.

 

[benorin]

Sorry, typo

Prove it via Hint #2: Algebra I students know this factoring trick: use it in reverse. Said factoring trick is the difference of squares.
Proof:
$$(1-z)\prod_{k=0}^{\infty} \left( 1+ z^{2^{k}}\right) = \lim_{M \rightarrow \infty} (1-z)\prod_{k=0}^{M} \left( 1+ z^{2^{k}}\right)$$
$$= \lim_{M \rightarrow \infty} (1-z)(1+z)(1+z^2)(1+z^4)(1+z^8)(1+z^{16})\cdot\cdot\cdot \left( 1+ z^{2^{M}}\right)$$
$$= \lim_{M \rightarrow \infty} (1-z^2)(1+z^2)(1+z^4)(1+z^8)(1+z^{16})\cdot\cdot\cdot \left( 1+ z^{2^{M}}\right)$$
$$=\lim_{M \rightarrow \infty} (1-z^4)(1+z^4)(1+z^8)(1+z^{16})\cdot\cdot\cdot \left( 1+ z^{2^{M}}\right)$$
$$=\lim_{M \rightarrow \infty} (1-z^8)(1+z^8)(1+z^{16})\cdot\cdot\cdot \left( 1+ z^{2^{M}}\right)=\cdot\cdot\cdot =\lim_{M \rightarrow \infty} \left( 1- z^{2^{M}}\right) \left( 1+ z^{2^{M}}\right)$$
$$=\lim_{M \rightarrow \infty} \left( 1- z^{2^{M+1}}\right)=1,\forall z\in\mathbb{C}\mbox{ such that }\left| z\right|<1$$
QED.

 

[benorin]

It's separable

Suppose a function $$f(z)$$ is equal to its own logarithmic derivative, namely, suppose that
$$\exists f(z)\mbox{ such that } \frac{d}{dz}\ln\left( f(z)\right) = \frac{f^{\prime}(z)}{f(z)} = f(z)$$

Our ODE, namely $\frac{f^{\prime}(z)}{f(z)} = f(z)$, is separable:

$$\frac{1}{f}\frac{df}{dz}=f \Rightarrow \frac{1}{f^{2}}df = dz \Rightarrow \int \frac{1}{f^{2}}df = \int dz \Rightarrow -\frac{1}{f} + C = z$$

and hence $f(z)=\frac{1}{C-z}$, where C is a constant, is a family of solutions to this ODE.

To check,

$$\frac{d}{dz}\ln\left( f(z)\right) = \frac{d}{dz}\ln\left( \frac{1}{C-z}\right) = \frac{\frac{d}{dz} \left( \frac{1}{C-z}\right)}{\frac{1}{C-z}} = \frac{1}{C-z} = f(z)$$.

 

[benorin]

Strange series transformation of the geometric Series

It has been shown in the above post that $f(z)=\frac{1}{C-z}$, where C is a constant, is a family of solutions to $\frac{d}{dz}\ln\left( f(z)\right) = f(z)$.

Put C=1 above, and recall that:

$$\prod_{k=0}^{\infty} \left( 1+ z^{2^{k}}\right) = \frac{1}{1-z},\forall z\in\mathbb{C}\mbox{ such that }\left| z\right|<1$$.

We have (hand-waving the convergence details, I know that the following works $\forall z\in\mathbb{R}\mbox{ such that }\left| z\right|<1$, but I don't understand the complex branch cut structure of $\ln (z)$ well-enough to say for $z\in\mathbb{C}$) :

$$\frac{d}{dz}\ln\left[ \prod_{k=0}^{\infty} \left( 1+ z^{2^{k}}\right) \right] = \frac{d}{dz} \sum_{k=0}^{\infty} \ln \left( 1+ z^{2^{k}}\right) = \sum_{k=0}^{\infty} \frac{\frac{d}{dz} \left( 1+ z^{2^{k}}\right)}{ 1+ z^{2^{k}}} = \sum_{k=0}^{\infty} \frac{ 2^{k} z^{2^{k}-1}}{ 1+ z^{2^{k}}} = \frac{1}{1-z}$$

Someone help? $\rightarrow$ Should this hold $$\forall z\in\mathbb{C}\mbox{ such that }\left| z\right|<1$$, or just $$\forall z\in\mathbb{R}\mbox{ such that }\left| z\right|<1$$?

-Ben

 

[benorin]

Last post on this thread?

Prove it via Hint #1: Convert binary to decimal: think exponents.
Note that expanding (that is multiplying out) the left-hand side of:

$$\prod_{k=0}^{\infty} \left( 1+ z^{2^{k}}\right) = \frac{1}{1-z}$$

each term in the resulting sum is a result of multiplying some combination of $$\left\{ 1, z, z^2, z^4, z^8, z^{16},...,z^{2^{k}},... \right\}$$, and hence, for the term $$a_{k} z^{b_{k}}$$ in the resulting sum, $$b_{k}$$ is some sum of unique powers of two, and $$a_{k}$$ is the number of ways to sum such unique powers of two, and by that I do mean $$a_{k}=1, \forall k\in\mathbb{N}$$ and since each decimal integer represents exactly one integer in binary, we have $$b_{k}=k, \forall k\in\mathbb{N}$$, namley

$$\prod_{k=0}^{\infty} \left( 1+ z^{2^{k}}\right) = \sum_{k=0}^{\infty} z^k$$

but we know that is just the geometric series, hence

$$\prod_{k=0}^{\infty} \left( 1+ z^{2^{k}}\right) = \frac{1}{1-z},\forall z\in\mathbb{C}\mbox{ such that }\left| z\right|<1$$

as required.

Is this the last post on this thread? You decide.

 

[fourier jr]

here's what euler did

$$\frac{1}{1-z} = \frac{1-z^{2}}{1-z}\times\frac{1-z^{4}}{1-z^{2}}\times\frac{1-z^{8}}{1-z^{4}}\times\frac{1-z^{16}}{1-z^{8}}\times\frac{1-z^{32}}{1-z^{16}}...$$

i guess you'd have to make it rigourous since euler was the sloppiest mathematcian of all time but that might help anyway.