The Purplemath Forums

by **testing** on Sun Mar 22, 2009 4:15 pm

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.

**Sponsor**

by **stapel_eliz** on Mon Mar 23, 2009 5:37 pm

testing wrote: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. :wink:

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

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!

by **testing** on Mon Mar 23, 2009 6:24 pm

Okay, I never would have thought of that. :shock:

by **Majstor** on Tue Mar 24, 2009 10:48 am

The answer will be two. I don't know how to put it mathmatically, but I'm sure it's two.

sqrt(2sqrt(2sqrt(...) keeps getting closer to 2, so when you continue to infinity it will equal two.

by **Martingale** on Wed Apr 01, 2009 11:52 am

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.

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

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

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

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