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 one year ago
Compute
\[\huge{\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{\sqrt{2}^\ldots}}}}}\]
 one year ago
Compute \[\huge{\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{\sqrt{2}^\ldots}}}}}\]

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SithsAndGiggles
 one year ago
Best ResponseYou've already chosen the best response.0I'm not sure your expression makes any sense. Is the continued exponentiation finite or not?

SithsAndGiggles
 one year ago
Best ResponseYou've already chosen the best response.0Ah, thanks for fixing it

klimenkov
 one year ago
Best ResponseYou've already chosen the best response.0Hope, you like it now. What will it be?

SithsAndGiggles
 one year ago
Best ResponseYou've already chosen the best response.0I'm pretty sure it's infinite. \[\left(\left(\left(\sqrt2^\sqrt2\right)^\sqrt2\right)^\sqrt2\right)^{\ldots}\\ \left(\left(\left( \left(2^\frac{1}{2}\right) ^\sqrt2\right)^\sqrt2\right)^\sqrt2\right)^{\ldots}\\ \left(\left( \left(2^\frac{\sqrt2}{2}\right) ^\sqrt2\right)^\sqrt2\right)^{\ldots}\\ \left( \left(2^\frac{2}{2}\right)^\sqrt2\right)^{\ldots}\\ \left(2^\sqrt2\right)^{\ldots}\\ \left(\left(2^\sqrt2\right)^\sqrt2\right)^{\ldots}\\ \left(\left(\left(2^2\right)^\sqrt2\right)^\sqrt2\right)^{\ldots}\\ \left(\left(4^\sqrt2\right)^\sqrt2\right)^{\ldots}\\ \left(4^2\right)^{\ldots}\] Continuing in this way, the base keeps getting squared.

klimenkov
 one year ago
Best ResponseYou've already chosen the best response.0The order is wrong. Compare: \[2^{2^3}=2^8=256,\quad (2^2)^3=4^3=64.\]

shubhamsrg
 one year ago
Best ResponseYou've already chosen the best response.2why don't let whole of it = x ?

shubhamsrg
 one year ago
Best ResponseYou've already chosen the best response.2then if, you observe properly, you'll be reduced to [ sqrt(2) ]^x = x

klimenkov
 one year ago
Best ResponseYou've already chosen the best response.0Nice, what will it be?

shubhamsrg
 one year ago
Best ResponseYou've already chosen the best response.2hmm, we get stuck ther! _

shubhamsrg
 one year ago
Best ResponseYou've already chosen the best response.22^(x/2) = x => 2^x = x^2 symmetrically, there is a solution , see that ?

hartnn
 one year ago
Best ResponseYou've already chosen the best response.0integer solutions, 2 and 4...

shubhamsrg
 one year ago
Best ResponseYou've already chosen the best response.2which one will it be ?

klimenkov
 one year ago
Best ResponseYou've already chosen the best response.0Now tell me, what is the answer: 2 or 4?

hartnn
 one year ago
Best ResponseYou've already chosen the best response.0because while squaring, we attached an extra root, x=4

klimenkov
 one year ago
Best ResponseYou've already chosen the best response.0Hm... \(\sqrt 2^4=4\). Why do you call it "extra root"?

hartnn
 one year ago
Best ResponseYou've already chosen the best response.0:O .. let me think more...

shubhamsrg
 one year ago
Best ResponseYou've already chosen the best response.22 also fits in though.

shubhamsrg
 one year ago
Best ResponseYou've already chosen the best response.2on the calculator, it approaches 2, I am not too sure how can we prove that ?

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.0Let \(\epsilon > 0\)

klimenkov
 one year ago
Best ResponseYou've already chosen the best response.0A medal will be given only if you prove this.

shubhamsrg
 one year ago
Best ResponseYou've already chosen the best response.2arey to hell with the medal !

klimenkov
 one year ago
Best ResponseYou've already chosen the best response.0Can somebody say what means "arey to hell" ?

shubhamsrg
 one year ago
Best ResponseYou've already chosen the best response.2ignore "arey" I meant To Hell with the medal.

klimenkov
 one year ago
Best ResponseYou've already chosen the best response.0Ok. I found an interesting proof to this. Hope, you will find another interesting proof.

shubhamsrg
 one year ago
Best ResponseYou've already chosen the best response.2unable to prove, what proof you got ?

shubhamsrg
 one year ago
Best ResponseYou've already chosen the best response.2I mean I am unable to prove*

klimenkov
 one year ago
Best ResponseYou've already chosen the best response.0dw:1362220984668:dw Hope, you understand.

shubhamsrg
 one year ago
Best ResponseYou've already chosen the best response.2how can you say thats not gonna intersect at x=4 ?

shubhamsrg
 one year ago
Best ResponseYou've already chosen the best response.2And that seems sort of forced proof.

shubhamsrg
 one year ago
Best ResponseYou've already chosen the best response.2@experimentX @mukushla

experimentX
 one year ago
Best ResponseYou've already chosen the best response.1dw:1362247620602:dwdw:1362247717003:dw dw:1362247538096:dw Hence the above value only converges to Sqrt 2

experimentX
 one year ago
Best ResponseYou've already chosen the best response.1*The last figure is first figure, the first figure is second figure and second figure is last figure.

shubhamsrg
 one year ago
Best ResponseYou've already chosen the best response.2Awesome! thanks sire.

Dodo1
 one year ago
Best ResponseYou've already chosen the best response.0thats actually look really awsome!
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