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$\text{Given}\quad y = \sqrt{k\sqrt{k\sqrt{k \sqrt{k\sqrt{\dots}}}}}\quad \text{ where }\,k\geq 0,\;\;\text{find the value of }\,y.$
I have no idea on how to solve problems like this.

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2 Answers 2

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$$y=\sqrt{k\cdot\sqrt{k\cdot\sqrt{\dots}}}=k^{1/2+1/4+1/8+...}=k^{\frac{1}{2}\frac{1}{1-1/2}}=k^1=k$$

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  • $\begingroup$ That's a nice approach too. I hadn't thought of it that way. $\endgroup$
    – Ragnar
    Dec 30, 2013 at 14:28
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We know that $$ y=\sqrt{k\cdot\sqrt{k\cdot\sqrt{\dots}}} $$ We can square the equation and divide by $k$: $$ \frac{y^2}k=\sqrt{k\cdot\sqrt{\dots}} $$ But the right hand side is just $y$ again, so we have $$ \frac{y^2}k=y $$ Solving this gives $y=0$ or $y=k$.

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    $\begingroup$ This answer assumes that $y$ exists. One also needs to show that the sequence of iterated square roots converge. $\endgroup$ Aug 15, 2014 at 12:46

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