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Question :

What is the value of $$\sqrt{11\sqrt{11\sqrt{11...4\,\text{times}}}}$$

I did it by solving square root one by one. $$\sqrt{11\sqrt{11\sqrt{11\times11^\frac{1}{2}}}}$$ $$\sqrt{11\sqrt{11\sqrt{11^\frac{3}{2}}}}$$ $$\sqrt{11\sqrt{11\times{11^\frac{3}{4}}}}$$ $$\sqrt{11\sqrt{11^\frac{7}{4}}}$$ $$\sqrt{11\times{11^\frac{7}{8}}}$$ $$\sqrt{11^\frac{15}{8}}$$ $$11^\frac{15}{16}$$

Is there any other way to solve this?

I don't want the complete solution, just tell me the approach.

ShBh
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Wolgwang
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    I think your method is more or less optimal. for a (minor) variant: Rewrite in exponential notation. There's one $11$ to which the square root is applied exactly once, so that's a factor of $11^{1/2}$. Then there's one $11$ to which it is applied exactly twice, so that's a factor of $11^{1/4}$. Continue this way. – lulu Sep 06 '20 at 14:01
  • @lulu I didn't get you. – Wolgwang Sep 06 '20 at 14:02
  • there was a typo in what I wrote, now corrected. In the end, you should get $11$ raised to the power $\frac 12+\frac 14+\frac 18+\frac 1{16}=\frac {15}{16}$. This method makes it easy to see how you'd handle $1000$ iterates, instead of just $4$. – lulu Sep 06 '20 at 14:04
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    Oh! It is a sum of geometric progression. – Wolgwang Sep 06 '20 at 14:08
  • Exactly. So you should easily be able to work out the result for $n-$ iterates. But, really, this is the same as your method...maybe written down in a way that makes the pattern easier to spot. – lulu Sep 06 '20 at 14:11
  • But why this works? Can you give me proof? – Wolgwang Sep 06 '20 at 14:11
  • My comment is the proof. Or, if that seems too vague for you, prove it by induction. Let $a_n$ be the expression with $n$ iterates. Remark that $a_{n+1}=11^{1/2}\times \sqrt {a_n}$. – lulu Sep 06 '20 at 14:13
  • Why do you think "it seems too vague to me"? – Wolgwang Sep 06 '20 at 14:14
  • Well, you asked for a proof where I'd have said that my first comment already was a proof. I made a guess as to why you were dissatisfied with my initial argument. In any case, you should be able to make my comment into a proper proof, or use the inductive method I sketched. – lulu Sep 06 '20 at 14:16
  • I just love this method(GP). Thanks a lot . – Wolgwang Sep 06 '20 at 14:17
  • Glad to have helped. – lulu Sep 06 '20 at 14:17

2 Answers2

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If you solve the square roots one by one, but from outer one to inner one, you see it is the sum of geometric sequence.

Ali Ashja'
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There is a faster way to solve it, actually:

$\sqrt{11·\sqrt{11·\sqrt{11·\sqrt{11·\color{blue}{11}}}}} = \sqrt{11·\sqrt{11·\sqrt{11·11}}} = \sqrt{11·\sqrt{11·11}} = \sqrt{11·11} = 11$

So clearly the extra factor is just $\sqrt{\sqrt{\sqrt{\sqrt{\color{blue}{11}}}}}$, since square-root commutes with multiplication on non-negative reals (which implies that in any expression comprising just multiplication and square-roots the number of square-roots in effect is just the number of 'radical lines' above it).

user21820
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