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Prove that $$\underbrace{\sqrt{k\sqrt{k\sqrt{k\sqrt{\cdots\sqrt{k}}}}}}_{n\text { times}}=k^{1-1/2^n}$$ How do I derive this formula?

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    $\begingroup$ Glibly: induction. $\endgroup$ – Semiclassical Sep 4 '14 at 13:31
  • $\begingroup$ Try to rewrite this expession using $k^{1/2}$. $\endgroup$ – Marc Bogaerts Sep 4 '14 at 13:31
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    $\begingroup$ Take logarithms (you could try base $k$) $\endgroup$ – Mark Bennet Sep 4 '14 at 13:35
  • $\begingroup$ what an interesting question! $\endgroup$ – hypergeometric Sep 4 '14 at 14:36
  • $\begingroup$ if k = 0 left side is 0 right side is not defined. $\endgroup$ – djechlin Sep 4 '14 at 16:17
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$$\underbrace{\sqrt{k\sqrt{k\sqrt{k\sqrt{\cdots\sqrt{k}}}}}}_{n\text { times}} = \sqrt{k}\times\sqrt[4]{k}\times \ldots \times \sqrt[2^n]{k} = k^{1/2}\times k^{1/4}\times \ldots \times k^{1/2^n} = \\ = k^{\Large\frac{1}{2} + \frac{1}{4} + \ldots + \frac{1}{2^n}} = k^{\Large1-\frac{1}{2^n}}$$

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Base case: $\sqrt{k}=k^{1/2}=k^{1-1/2^1}$.

Induction step: let $a_n$ be the LHS when there are $n$ square root signs. If $a_n=k^{1-1/2^n}$, then $$ a^2_{n+1}=ka_n=kk^{1-1/2^n}=k^{2-1/2^n}\implies a_{n+1}=k^{1-1/2^{n+1}}. $$

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Take the logarithm (as already suggested in comments): $$ \frac{1}{2}\left(\ln k + \frac{1}{2}\left(\ln k + \frac{1}{2}\left( \ln k + \ldots \right) \right)\right)$$ $$ =\left(\frac{1}{2}+ \frac{1}{2^2} + \frac{1}{2^3} + \ldots + \frac{1}{2^n}\right)\ln k$$ $$ =\left(1 - \frac{1}{2^n}\right) \ln k.$$

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The first $k$ is powered by $\frac{1}{2}$, the second by $\frac{1}{2^2}$ the $n$-therm is then $k^{\frac{1}{2^n}}$ then the expression is equal to:

$$k^{\frac{1}{2^1}}\cdot k^{\frac{1}{2^2}}\cdot \dots k^{\frac{1}{2^n}}=k^{\sum_{i=1}^n \frac{1}{2^i}}$$

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