## value of sqrt(2sqrt(2sqrt(...))) (I'm told there's a trick?)

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testing
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### value of sqrt(2sqrt(2sqrt(...))) (I'm told there's a trick?)

I'm trying to find the value of sqrt(2sqrt(2sqrt(...))). I figure you start by setting this equal to "x". But then what?

I'm told there's a trick to this, so maybe that's why I'm stuck?

Edited to correct misunderstanding of original puzzle.

stapel_eliz
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I'm trying to find the value of sqrt(2sqrt(2sqrt(...))). I figure you start by setting this equal to "x"....
Yes, and then you use this equality in a helpful way.

. . . . .$\sqrt{2\sqrt{2\sqrt{2\sqrt{2\mbox{...}}}}}\, =\, x$

But then:

. . . . .$\sqrt{2\sqrt{2\sqrt{2\sqrt{2\mbox{...}}}}}\, =\, \sqrt{2\left(\sqrt{2\sqrt{2\sqrt{\sqrt{2\mbox{...}}}}}\right)}\, =\, \sqrt{2x}\, =\, x$

So:

. . . . .$2x\, =\, x^2$

. . . . .$0\, =\, x^2\, -\, 2x\, =\, x(x\, -\, 2)$

One solution is "obvious" (when you think about it). The other solution, the one they're probably looking for, is surprising!

testing
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Okay, I never would have thought of that.

Martingale
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### Re: value of sqrt(2sqrt(2sqrt(...))) (I'm told there's a trick?)

the trick where you get $\sqrt{2x}=x$ only works if we know that the original sequence of numbers converges. Therefore, before one uses this technique you should determine if it converges first then use the 'trick'. Otherwise you could get the wrong answer.

Also, there is only one solution to the problem in this thread...2 ... since the sequence of the sqrt of 2's is monotone increasing.