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A cool identity

  1. Feb 25, 2014 #1
    Anybody think this identity is not super-cool?

    \begin{eqnarray}
    \frac {1} {1-x} =
    (1+x) \prod_{n=1} ^{\infty}
    [ \frac {(1+x^{2^n})} {(1-x^{2^n})} ]^{2^{-n}} , {\;}
    for {\;} 0{\le}x<1
    \nonumber
    \end{eqnarray}
     
  2. jcsd
  3. Feb 25, 2014 #2

    Mentallic

    User Avatar
    Homework Helper

    That's one I've never seen before. Handy tip, if you need larger brackets, type \left[ ... \right] and latex will make them a suitable size. I was first going to respond saying that the identity can't be right because it looked as though 2-n was a factor and not an exponent due to the small brackets.

    \begin{eqnarray}
    \frac {1} {1-x} =
    (1+x) \prod_{n=1} ^{\infty}
    \left( \frac {1+x^{2^n}} {1-x^{2^n}} \right)^{2^{-n}} , {\;}
    \text{for} {\;} 0\leq x<1
    \nonumber
    \end{eqnarray}
     
  4. Feb 25, 2014 #3
    I have yet to play with infinite products and if it works then yes indeed, very super-cool!
     
  5. Feb 25, 2014 #4
    Much better, thanks. BTW, you're the first person I've shown it to
    who did not doubt it, at first.
    What's your secret?

    Please notice that substituting principle powers of of x, and then taking
    a principle root yields an infinite number of infinite product identities
    and all that changes is the first value of the counter "n". Has many
    other interesting properties too; like it really likes its logarithm taken.
     
  6. Feb 25, 2014 #5
    And listen to this...unity, the pride of the rationals, is an infinite
    product of irrationals in this context.
     
  7. Feb 25, 2014 #6
    And yes, I do have what I think is a rigorous proof of this.
     
  8. Feb 25, 2014 #7
    Euclid helped....

    \textbf{I. Lemma.}

    \begin{eqnarray}
    1-x^{2^N} =
    (1-x) \prod_{n=0}^{N-1}
    (1+x^{2^n}), {\;} N=1,2,...
    \nonumber
    \end{eqnarray}


    \underline{\textit{Proof}}\\

    \indent a) at N = 1,

    \begin{eqnarray}
    1-x^2 =
    (1-x) (1+x)
    \nonumber
    \end{eqnarray}

    \indent b) Inductively, the lemma is true for N = 1,...,M

    \begin{eqnarray}
    1-x^{2^{M+1}} =
    1-(x^{2^M})^2
    \nonumber
    \end{eqnarray}

    \begin{eqnarray}
    =
    (1-x^{2^M}) (1+x^{2^M})
    \nonumber
    \end{eqnarray}

    \begin{eqnarray}
    =
    (1-x) \prod_{n=0} ^{M-1} (1+x^{2^n}) (1+x^{2^M})
    \nonumber
    \end{eqnarray}

    \begin{eqnarray}
    =
    (1-x) \prod_{n=0} ^{M} (1+x^{2^n})
    \nonumber
    \end{eqnarray}

    \textbf{II. Lemma.}

    \begin{eqnarray}
    \frac {1} {1-x} =
    \prod_{n=0} ^{\infty} (1+x^{2^n}), {\;}
    for {\;} 0{\le}x<1
    \nonumber
    \end{eqnarray}

    \underline{\textit{Proof}}

    \begin{eqnarray}
    1-x^{2^{N}} =
    (1-x) \prod_{n=0} ^{N-1} (1+x^{2^n})
    \nonumber
    \end{eqnarray}

    or,

    \begin{eqnarray}
    1 =
    (1-x) \prod_{n=0} ^{\infty} (1+x^{2^n})
    \nonumber
    \end{eqnarray}

    \newpage

    \textbf{III. Lemma.}

    \begin{eqnarray}
    \prod_{n=m} ^{\infty} (1+x^{2^n}) =
    \frac{1} {1-x^{2^m}}
    \nonumber
    \end{eqnarray}

    \underline{\textit{Proof}}

    \begin{eqnarray}
    \prod_{n=m} ^{\infty} (1+x^{2^n}) =
    \prod_{n=0} ^{\infty} (1+x^{2^n})
    [\prod_{n=0}^{m-1} (1+x^{2^n})]^{-1}
    \nonumber
    \end{eqnarray}

    \begin{eqnarray}
    =
    \frac {1} {1-x} [\frac{1-x^{2^m}} {1-x}]^{-1}
    \nonumber
    \end{eqnarray}

    \begin{eqnarray}
    =
    \frac {1} {1-x^{2^m}}
    \nonumber
    \end{eqnarray}

    \textbf{IV. Lemma.}

    \begin{eqnarray}
    \sum_{n=0} ^{m} y^n =
    \frac {1-y^{m+1}} {1-y}
    \nonumber
    \end{eqnarray}

    \underline{\textit{Proof}}

    \begin{eqnarray}
    \sum_{n=0} ^{m} y^n =
    1+ \sum_{n=1} ^{m} y^n
    \nonumber
    \end{eqnarray}

    \begin{eqnarray}
    =
    1+ y \sum_{n=0} ^{m} y^n
    -y^{m+1}
    \nonumber
    \end{eqnarray}

    \begin{eqnarray}
    (1-y)\sum_{n=0} ^{m} y^n =
    1-y^{m+1}
    \nonumber
    \end{eqnarray}

    \begin{eqnarray}
    \sum_{n=0} ^{m} y^n =
    \frac{1-y^{m+1}} {1-y}
    \nonumber
    \end{eqnarray}

    \newpage

    \underline{\textit{Corollary}}

    \begin{eqnarray}
    1-2^{-m} =
    [\sum_{i=0} ^{m-1} 2^{i}]2^{-m}
    \nonumber
    \end{eqnarray}

    \textbf{V. Lemma.}

    \begin{eqnarray}
    \prod_{m=1} ^{\infty} (1+x^{2^m})^{1-2^{-m}} =
    \prod_{m=1} ^{\infty}
    \prod_{n=m} ^{\infty} (1+x^{2^n})^{ 2^{-m}}
    \nonumber
    \end{eqnarray}

    \underline{\textit{Proof}}

    \begin{eqnarray}
    \prod_{m=1} ^{\infty} (1+x^{2^m})^{1-2^{-m}} =
    \prod_{m=1} ^{\infty} (1+x^{2^m})
    ^{(\sum_{i=0} ^{m-1}2^i)2^{-m}}
    \nonumber
    \end{eqnarray}

    \begin{eqnarray}
    =
    \prod_{m=1} ^{\infty} (1+x^{2^n})
    ^{(\sum_{i=0} ^{n-1}2^{i-n})}
    \nonumber
    \end{eqnarray}

    \begin{eqnarray}
    =
    \lim_{N \to {\infty}}\prod_{n=1} ^{N}
    \prod_{i=1} ^n (1+x^{2^n}) ^{2^{-i}}
    \nonumber
    \end{eqnarray}

    \begin{eqnarray}
    =
    \lim_{N \to {\infty}}
    %\prod_{\frac {1 \le n \le N} {1 \le i \le n}}
    \prod_{\substack{1 \le n \le N\\{1 \le i \le n}}}
    (1+x^{2^n})^{2^{-i}}
    \nonumber
    \end{eqnarray}




    \begin{eqnarray}
    =
    \lim_{N \to {\infty}}
    %\prod_{\frac {1 \le n \le N} {1 \le i \le n}}
    \prod_{\substack{i \le n \le N\\{1 \le i \le N}}}
    (1+x^{2^n})^{2^{-i}}
    \nonumber
    \end{eqnarray}


    \begin{eqnarray}
    =
    \prod_{m=1} ^{\infty}
    \prod_{n=m} ^{\infty} (1+x^{2^n})^{ 2^{-m}}
    \nonumber
    \end{eqnarray}

    \textbf{VI. Theorem.}

    \begin{eqnarray}
    \frac {1} {1-x} =
    (1+x) \prod_{n=1} ^{\infty}
    [ \frac {(1+x^{2^n})} {(1-x^{2^n})} ]^{2^{-n}} , {\;}
    for {\;} 0{\le}x<1
    \nonumber
    \end{eqnarray}

    \underline{\textit{Proof}}

    \begin{eqnarray}
    \frac {1} {1-x} =
    \prod_{n=0} ^{\infty}
    (1+x^{2^n})
    \nonumber
    \end{eqnarray}

    \begin{eqnarray}
    =
    (1+x) \prod_{n=1} ^{\infty}
    (1+x^{2^n}) ^{2^{-n}} (1+x^{2^n}) ^{1-2^{-n}}
    \nonumber
    \end{eqnarray}

    \begin{eqnarray}
    =
    [(1+x) \prod_{n=1} ^{\infty}
    (1+x^{2^n}) ^{2^{-n}}] \;
    [\prod_{n=1} ^{\infty}
    (1+x^{2^n}) ^{1-2^{-n}}]
    \nonumber
    \end{eqnarray}

    \begin{eqnarray}
    =
    [(1+x) \prod_{n=1} ^{\infty}
    (1+x^{2^n}) ^{2^{-n}}] \;
    [\prod_{n=1} ^{\infty}
    \prod_{i=n} ^{\infty}
    (1+x^{2^i}) ^{2^{-n}}]
    \nonumber
    \end{eqnarray}

    \begin{eqnarray}
    =
    [(1+x) \prod_{n=1} ^{\infty}
    (1+x^{2^n}) ^{2^{-n}}] \;
    [\prod_{n=1} ^{\infty}
    (\frac {1} {1-x^{2^n}}) ^{2^{-n}}]
    \nonumber
    \end{eqnarray}

    \begin{eqnarray}
    =
    (1+x) \prod_{n=1} ^{\infty}
    [ \frac {(1+x^{2^n})} {(1-x^{2^n})} ]^{2^{-n}} , {\;}
    for {\;} 0{\le}x<1
    \nonumber
    \end{eqnarray}
     
  9. Feb 25, 2014 #8
    I didn't write the proof, a good internet friend, Dr. David R. Fischer,
    wrote it after I showed him a "down and dirty" bifurcation of roots
    pyramid "proof". The key is Lemma II, a standard infinite product
    found in many texts concerning infinite products.
     
  10. Feb 26, 2014 #9
    I'm actually surprised that no one has "naively" asked...
    whether this infinite product is "unique"? I think it would
    have been a very good question (mainly because I've
    pondered that question). The answer is that out of the
    infinity of infinities of infinite products equal to "1/(1-x)",
    the super-cool-identity has a "uniqueness" in that the
    dimensions/units are right. That is, if "x" has a unit,
    then the answer is in that unit, from the "(1+x)" factor;
    and in all the other factors the units are cancelled out.
    Actual proof of this "uniqueness" escapes me though;
    perhaps some smart gal and/or guy might solve it for
    me some day.
     
  11. Feb 26, 2014 #10
    Well, that's all I have to say...I'll be waiting for that
    wonderful telegram from Publisher's Clearing House......
     
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