# A cool identity

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}

Mentallic
Homework Helper
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}
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}

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}
I have yet to play with infinite products and if it works then yes indeed, very super-cool!

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}
Much better, thanks. BTW, you're the first person I've shown it to
who did not doubt it, at first.

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.

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}
And listen to this...unity, the pride of the rationals, is an infinite
product of irrationals in this context.

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}
And yes, I do have what I think is a rigorous proof of this.

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}

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.

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.

Well, that's all I have to say...I'll be waiting for that
wonderful telegram from Publisher's Clearing House......