- #1

- 221

- 0

\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}

- Thread starter ClamShell
- Start date

- #1

- 221

- 0

\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

Mentallic

Homework Helper

- 3,798

- 94

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 2Anybody 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}

\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}

- #3

- 648

- 18

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

- #4

- 221

- 0

Much better, thanks. BTW, you're the first person I've shown it toThat'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}

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.

- #5

- 221

- 0

And listen to this...unity, the pride of the rationals, is an infiniteThat'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}

product of irrationals in this context.

- #6

- 221

- 0

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

- #7

- 221

- 0

\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}

- #8

- 221

- 0

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.

- #9

- 221

- 0

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.

- #10

- 221

- 0

wonderful telegram from Publisher's Clearing House......

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