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In class today I was learning about the old factorization of $x^2-1$ to be $(x+1)(x-1)$.
I was thinking that $x-1$ could be factored to $(\sqrt[2]{x}+1)(\sqrt[2]{x}-1)$.
And in turn, $\sqrt[2]{x}-1$ could then be factored into $(\sqrt[4]{x}+1)(\sqrt[4]{x}-1)$.
I thought that possibly this process could continue forever, with the same solution of $x^2-1$.
What I wanted to ask is whether or not $\prod\limits_{n=1}^{\infty}(\sqrt[2^n]{x}+1)$ would truly be equal to $x-1$.
Clever thought, but no, unfortunately. For $x = 0$, this product is $1 \cdot 1 \cdot 1 \cdot \cdots = 1 \neq -1$. For $x > 0$, notice that $x^{1/2^n}$ goes to $1$ as $n$ goes to $\infty$; so for $n$ sufficiently large, $x^{1/2^n} + 1$ is at least (for example) $1.5$. So the infinite product $\prod_n(x^{1/2^n} + 1)$ is at least $c\prod_n1.5^n$, where $c$ is some positive constant. But this second product doesn't converge ($1.5 \cdot 1.5 \cdot 1.5 \cdot \cdots = \infty$, speaking loosely). So the first product has to be infinite, too. And for $x < 0$, $x^{1/2}+1$ doesn't even make sense (do we mean $i+1$ or $-i+1$?) so the product doesn't work there either.
If you're not familiar with the notation, $\prod_nf(n)$ just means "the infinite product of $f(n)$ as $n$ goes to $\infty$.
However, you can do this:
$\begin{array}\\ x-1 &=(x^{1/2}+1)(x^{1/2}-1)\\ &=(x^{1/2}+1)(x^{1/4}+1)(x^{1/4}-1)\\ &=(x^{1/2}+1)(x^{1/4}+1)(x^{1/8}+1)(x^{1/8}-1)\\ &...\\ &=(x^{1/2}+1)(x^{1/4}+1)...(x^{1/2^m}+1)(x^{1/2^m}-1)\\ &=\left(\prod_{k=1}^m(x^{1/2^k}+1)\right)(x^{1/2^m}-1)\\ \end{array} $
As $m$ gets large, $x^{1/2^m}-1 =e^{\ln(x)/2^m}-1 \approx (1+\ln(x)/2^m)-1 =\ln(x)/2^m $ so that $x-1 \approx \dfrac{\ln(x)}{2^m}\prod_{k=1}^m(x^{1/2^k}+1) =\ln(x)\prod_{k=1}^m\dfrac{x^{1/2^k}+1}{2} $.
I don't know how useful this is, but I enjoy playing around like this.
