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I'm having trouble understanding how this expression:

$$\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2+\dots}}}} \cdot \left(\frac{\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\dots+\sqrt2}}}}}{\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\dots+\sqrt2}}}}}\right)=$$

got to this one:

$$\frac2{\sqrt2\cdot\sqrt{2+\sqrt2}\cdot\sqrt{2+\sqrt{2+\sqrt{2}}}\cdot\dots}$$

Martin Sleziak
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  • For some basic information about writing math on SO network sites supporting MathJax see e.g. here, here, here and here. (I have edited your post, if you are satisfied with the result, you can remove the images and keep the only the TeX formulas.) – Martin Sleziak Nov 17 '13 at 06:53
  • This question is related to yours (although it does not answer your question): http://math.stackexchange.com/questions/186652/how-can-one-solve-for-x-x-sqrt2-sqrt2-sqrt2-cdots – Martin Sleziak Nov 17 '13 at 07:17
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    Is it intentional that the numerator and denominator of the fraction in parentheses are the same? –  Nov 17 '13 at 08:31
  • BTW it would be nice to include the source where this is comming from. – Martin Sleziak Nov 17 '13 at 10:47

1 Answers1

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Now that I am reading this correctly I think I can make some sense of the claim.

Let $x=\sqrt{2+\sqrt{2+\sqrt{2+\cdots}}}$ and $ y=\sqrt{2-\sqrt{2+\sqrt{2+\cdots}}}$ The "equation" you ask about is $$y=y\frac{x}{x}=\frac{2}{\sqrt{2}\cdot\sqrt{2+\sqrt{2}}\cdot\sqrt{2+\sqrt{2+\sqrt{2}}}\cdot\ \cdots} $$ Since the righthand side has a finite numerator and denominator an infinite product (of factors at least $\sqrt{2}$) this says $$y=0.$$ That does make sense, at least informally, since $x$ (if it makes any sense) is positive and satisfies $x=\sqrt{2+x}$ meaning $x^2=2+x$ and $x=2.$ Then, indeed, $$y=\sqrt{2-x}=\sqrt{0}=0. $$ But why that particular way of writing $0?$

The presumed meaning of the expression for $x$ is

$x=\lim_{n \to \infty}x_n$ where $x_0=0$, $x_1=\sqrt{2}$, $ x_2=\sqrt{2+\sqrt{2}}$ and in general $x_{n+1}=\sqrt{2+x_n}.$

Then we might also define

$y=\lim_{n \to \infty}y_n$ where $y_1=\sqrt{2}$, $y_2=\sqrt{2-\sqrt{2}}$, $y_3=\sqrt{2-\sqrt{2+\sqrt{2}}}$ and in general $y_{n+1}=\sqrt{2-x_n}.$

So the thing to be explained is $$y=y\frac{x}{x}\stackrel{?}{=}\frac{2}{\prod_{j=1}^{\infty}x_j} .\tag{*}$$

The sense I make of this is as the limit (in a somewhat informal sense) of $$y_n=y_n\frac{x_n}{x_n}=\frac{2}{\prod_{j=1}^nx_j}. \tag{**}$$ This can be shown by induction.

The base case is $y_1=\frac{2}{x_1}$ i.e. $$\sqrt{2}=\sqrt{2}\frac{x_1}{x_1}=\sqrt{2}\frac{\sqrt{2}}{x_1}=\frac{2}{x_1} $$

then the needed induction step is $y_{n+1}=\frac{y_{n}}{x_{n+1}}$ and indeed:

$$y_{n+1}=y_{n+1}\frac{x_{n+1}}{x_{n+1}}=\sqrt{2-x_{n}}\frac{\sqrt{2+x_{n}}}{x_{n+1}}=\frac{\sqrt{4-x_{n}^2}}{x_{n+1}}=\frac{y_{n}}{x_{n+1}}. $$

The last step makes use of $x_{n}^2=2+x_{n-1}$ so $\sqrt{4-x_{n}^2}=\sqrt{2-x_{n-1}}=y_{n}$

So, given $(**)$, I can make sense of $$y=\lim_{n \to \infty}y_n=\lim_{n \to \infty}y_n\frac{x_n}{x_n}=\lim_{n \to \infty}\frac{2}{\prod_{j=1}^nx_j}.$$ But, since my proof of $(**)$ is by induction, I do not see how one can, in one step, say $$y\frac{x}{x}=\frac{2}{\prod_{j=1}^{\infty}x_j} .$$

One way to more formally justify the value $x=2$ is to show $2-\frac{1}{2^{n-1}} \le x_n \lt 2$.

Finally, it does not directly relate to the specific question you ask, but see this question for a proof that $$y_n=2\sin{\frac{\pi}{2^{n+1}}}$$ so that $$ \frac{\pi}{2^n}-\frac{\left(\pi/2^n\right)^3}{24} \lt y_n \lt \frac{\pi}{2^n} $$

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Aaron Meyerowitz
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