# If $\;x\in\left[1,\infty \right)$, then $x=\prod_{r=0}^\infty\left(1+\left(\frac{x-1}{x}\right)^{2^r}\right)$

Let $$\; t\in\mathbb{R} \;$$ and let $$\; \alpha,\beta\in(t,t+1) \;$$ with $$\; \alpha\not=\beta \;$$ and consider $$\;\displaystyle\frac{1-\alpha+\beta}{1+\alpha-\beta}$$.

My goal was to "simplify" $$\; \displaystyle\frac{1-\alpha+\beta}{1+\alpha-\beta} \;$$ as the product of terms. In the process of attempting this, I found that $$\frac{1-\alpha+\beta}{1+\alpha-\beta} = \left(1-\alpha+\beta\right)^2 \cdot\prod_{r=1}^\infty\left(1+\left(\alpha-\beta\right)^{2^r}\right)$$

Through a bit of algebraic reworking, I came to the more interesting conclusion that for any $$x\in\left[1,\infty\right)$$, $$x=\prod_{r=0}^\infty\left(1+\left(\frac{x-1}{x}\right)^{\displaystyle2^r}\right)$$

I'm not sure whether my proof is airtight. I think that (if true) this is a really interesting property. If anyone recognizes it or something similar, please let me know. I tried searching for it but came up with nothing. If anyone would check my proof and/or provide one of their own that would be splendid.

My proof:

$$\frac{1-\alpha+\beta}{1+\alpha-\beta}=\frac{\left(1-\alpha+\beta\right)^2}{\left(1+\alpha-\beta\right) \left(1-\alpha+\beta\right)}=\frac{\left(1-\alpha+\beta\right)^2}{1-\epsilon}$$

where

$$\epsilon=1-\left(1+\alpha-\beta\right)\left(1-\alpha+\beta\right)=\alpha^2+\beta^2-2\alpha\beta=\left(\alpha-\beta\right)^2$$ 

Since $$\; \alpha - \beta \in(-1,1)\text{,}\;$$ we know that $$1> \epsilon >0$$.

\begin{align} \frac{\left(1-\alpha+\beta\right)^2}{1-\epsilon}&=\frac{\left(1-\alpha+\beta\right)^2\left(1+\epsilon\right)}{1-\epsilon^2} \\ \\ &= \frac{\left(1-\alpha+\beta\right)^2\left(1+\epsilon\right)\left(1+\epsilon^2\right)}{1-\epsilon^4} \\ \\ &= \frac{\left(1-\alpha+\beta\right)^2\left(1+\epsilon\right)\left(1+\epsilon^2\right)\left(1+\epsilon^4\right)}{1-\epsilon^8} \\ \\ &= \cdots = \frac{\left(1-\alpha+\beta\right)^2\left(1+\epsilon\right)\left(1+\epsilon^2\right)\left(1+\epsilon^4\right)\cdots\left(1+\epsilon^{2^{n-1}}\right)\left(1+\epsilon^{2^n}\right)}{1-\epsilon^{2^{n+1}}}\end{align}

As $$n\to\infty$$, $$\displaystyle\epsilon^{2^{n+1}}\to0$$, and we have

\begin{align} \frac{1-\alpha+\beta}{1+\alpha-\beta}&=\left(1-\alpha+\beta\right)^2\left(1+\epsilon\right)\left(1+\epsilon^2\right)\left(1+\epsilon^4\right)\cdots \\ \\ &=\left(1-\alpha+\beta\right)^2 \cdot\prod_{r=0}^\infty\Bigr(1+\epsilon^{2^r}\Bigr) \\ \\ &=\left(1-\alpha+\beta\right)^2 \cdot\prod_{r=0}^\infty \Bigr(1+\left(\left(\alpha-\beta\right)^2\right)^{2^r}\Bigr) \\ \\ &= \left(1-\alpha+\beta\right)^2 \cdot\prod_{r=1}^\infty \Bigr(1+ \left(\alpha-\beta\right)^{2^r}\Bigr) \end{align}  Dividing both sides of the equation above by $$\left(1-\alpha+\beta\right)^2 \text{,}\;$$ (a nonzero term), we arrive at

$$\frac{1}{\left(1+\alpha-\beta\right)\left(1-\alpha+\beta\right)} = \frac{1}{\left(1+(\alpha-\beta)\right)\left(1-(\alpha-\beta)\right)} = \prod_{r=1}^\infty \Bigr(1+ \left(\alpha-\beta\right)^{2^r}\Bigr)$$ 

$$\alpha - \beta\;$$ is an arbitrary value in $$\left(-1,1\right) \text{,} \;$$ so for any $$\gamma\in \left(-1,1\right)$$, we have

$$\frac{1}{\left(1+\gamma\right)\left(1-\gamma\right)} = \frac{1}{1-\gamma^2} = \prod_{r=1}^\infty \Bigr(1+ \gamma^{2^r}\Bigr)$$ 

Now let $$\;x\in\mathbb{R}\;$$ such that $$\; x=\frac{1}{1-\gamma^2} \;$$ with $$\gamma \in \left(-1,1\right)\text{.}\;$$ Then $$x=\frac{1}{1-\gamma^2}\iff \gamma=\pm \sqrt{\frac{x-1}{x}}$$ 

We can see here that $$\; x \;$$ is necessarily an element of $$\;\left[1,\infty\right)\;$$because $$\; x\in\left(-\infty,0\right)\Longrightarrow \gamma>1\;$$ (a contradiction), and $$\; x\in\left[0,1\right)\Longrightarrow \gamma\not\in\mathbb{R}\;$$ (a contradiction). Note that $$\; \frac{x-1}{x}\geq 0 \;$$ for any $$\;x\in\left[1,\infty\right)$$.

Thus, if $$\; x\in\left[1,\infty\right)\text{,}\;$$ then $$x=\prod_{r=1}^\infty\left(1+\left(\sqrt{\frac{x-1}{x}}\right)^{\displaystyle2^r}\right)=\prod_{r=0}^\infty\left(1+\left(\frac{x-1}{x}\right)^{\displaystyle2^r}\right)$$

Example ($$x=7$$): Wolfram Alpha Example 1

We have a telescoping product,$$\prod_{r\ge0}\left(1+y^{2^r}\right)=\prod_{r\ge0}\frac{1-y^{2^{r+1}}}{1-y^{2^r}}=\frac{1}{1-y}$$for $$|y|<1$$. Now take $$y=1-\frac1x$$ for $$x\ge1$$.