An interesting radical I came up with and I'd like to know your approach.
Each term is of the form: $\frac{2^{2^r}}{2^{2^{r}+2^{r-1}+...+2+1}} = \frac{1}{2^{2^{r-1}+...+2+1}}$
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1I think the pattern is in the difference of powers $0,1,3,7$ subtract from previous term we have a GP $1-0=1,3-1=2,7-3=4$ – Archis Welankar Jun 18 '17 at 12:40
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Hint:$$\sqrt{1+\sqrt{\frac{1}{2^1}+\sqrt{\frac{1}{2^3}+\sqrt{\frac{1}{2^7}+\sqrt{\frac{1}{2^{15}}+…}}}}}=\ \sqrt{1+\sqrt{\frac{1}{2^1}+\frac{1}{2}\sqrt{\frac{1}{2^{3-2}}+\frac{1}{4}\sqrt{\frac{1}{2^{7-4}}+\frac{1}{16}\sqrt{\frac{1}{2^{15-8}}+…}}}}}=\ \sqrt{1+\frac{1}{\sqrt2}\sqrt{\frac{1}{1}+\frac{1}{1}\sqrt{\frac{1}{2^{3-2}}+\frac{1}{1}\sqrt{\frac{1}{2^{7-4}}+\frac{1}{1}\sqrt{\frac{1}{2^{15-8}}+…}}}}}=\ \sqrt{1+\frac{1}{\sqrt2}\sqrt{1+\sqrt{\frac{1}{2^1}+\sqrt{\frac{1}{2^3}+\sqrt{\frac{1}{2^7}+\sqrt{\frac{1}{2^{15}}+…}}}}}}=\$$ – Khosrotash Jun 18 '17 at 13:04
2 Answers
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If you multiply the expression by $1/\sqrt{2}$ and push it inside each successive radical, you get the second radical. So you have $x = \sqrt{1+x/\sqrt{2}}.$ When you solve for $x$, you get $x=\sqrt{2}.$
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Let the number be $x$. Then
\begin{align} \sqrt{2}(x^2-1)&=x\\ 16x^2-8\sqrt{2}x-16&=0\\ (4x-\sqrt{2})^2&=18\\ 4x-\sqrt{2}&=\pm3\sqrt{2}\\ 4x&=4\sqrt{2}\qquad (\text{as }x\text{ is positive})\\ x&=\sqrt{2} \end{align}
CY Aries
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