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

  1. Nov 5, 2005 #1

    benorin

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    I hope you have fun with these...

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

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

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

    [tex]\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[/tex]

    Prove it in two ways.

    Suppose a function [tex]f(z)[/tex] is equal to its own logarithmic derivative, namely, suppose that

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

    Prove that [itex]f(z)=\frac{1}{C-z}[/itex] 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):

    [tex] \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[/tex]

    We have derived what new series by virtue of the fact that the log of a product is the sum of the logs?
     
    Last edited: Nov 5, 2005
  2. jcsd
  3. Nov 5, 2005 #2

    benorin

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    Would a hint help? two?

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

    [tex]\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[/tex]

    Prove it in two ways.

    Hint #1: Convert binary to decimal.

    Hint #2: Algebra I students know this factoring trick: use it in reverse.
     
  4. Nov 8, 2005 #3

    benorin

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    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:

    [tex](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) [/tex]
    [tex]= \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) [/tex]
    [tex]= \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)[/tex]
    [tex] =\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) [/tex]
    [tex]=\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)[/tex]
    [tex] =\lim_{M \rightarrow \infty} \left( 1- z^{2^{M+1}}\right)=1,\forall z\in\mathbb{C}\mbox{ such that }\left| z\right|<1[/tex]

    QED.
     
    Last edited: Nov 8, 2005
  5. Nov 11, 2005 #4

    benorin

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    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:
    [tex](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) [/tex]
    [tex]= \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) [/tex]
    [tex]= \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)[/tex]
    [tex] =\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) [/tex]
    [tex]=\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)[/tex]
    [tex] =\lim_{M \rightarrow \infty} \left( 1- z^{2^{M+1}}\right)=1,\forall z\in\mathbb{C}\mbox{ such that }\left| z\right|<1[/tex]
    QED.
     
  6. Nov 11, 2005 #5

    benorin

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    It's separable

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

    Our ODE, namely [itex]\frac{f^{\prime}(z)}{f(z)} = f(z)[/itex], is separable:

    [tex]\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[/tex]

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

    To check,

    [tex]\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)[/tex].
     
    Last edited: Nov 11, 2005
  7. Nov 11, 2005 #6

    benorin

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    Strange series transformation of the geometric Series

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

    Put C=1 above, and recall that:

    [tex]\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[/tex].

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

    [tex] \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}[/tex]

    Someone help? :confused: [itex]\rightarrow[/itex] Should this hold [tex]\forall z\in\mathbb{C}\mbox{ such that }\left| z\right|<1[/tex], or just [tex]\forall z\in\mathbb{R}\mbox{ such that }\left| z\right|<1[/tex]?

    -Ben
     
    Last edited: Nov 11, 2005
  8. Nov 11, 2005 #7

    benorin

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    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:

    [tex]\prod_{k=0}^{\infty} \left( 1+ z^{2^{k}}\right) = \frac{1}{1-z}[/tex]

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

    [tex]\prod_{k=0}^{\infty} \left( 1+ z^{2^{k}}\right) = \sum_{k=0}^{\infty} z^k[/tex]

    but we know that is just the geometric series, hence

    [tex]\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[/tex]

    as required.

    Is this the last post on this thread? :cry: You decide.
     
  9. Nov 11, 2005 #8
    here's what euler did

    [tex]\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}}....[/tex]

    i guess you'd have to make it rigourous since euler was the sloppiest mathematcian of all time but that might help anyway.
     
    Last edited: Nov 11, 2005
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