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klimenkov
 3 years ago
Compute
\[\huge{\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{\sqrt{2}^\ldots}}}}}\]
klimenkov
 3 years ago
Compute \[\huge{\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{\sqrt{2}^\ldots}}}}}\]

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anonymous
 3 years 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?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Ah, thanks for fixing it

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

anonymous
 3 years 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
 3 years 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
 3 years ago
Best ResponseYou've already chosen the best response.2why don't let whole of it = x ?

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

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

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

shubhamsrg
 3 years 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
 3 years ago
Best ResponseYou've already chosen the best response.0integer solutions, 2 and 4...

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

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

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

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

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

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

shubhamsrg
 3 years 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
 3 years ago
Best ResponseYou've already chosen the best response.0Let \(\epsilon > 0\)

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

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

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

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

klimenkov
 3 years 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
 3 years ago
Best ResponseYou've already chosen the best response.2unable to prove, what proof you got ?

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

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

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

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

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

experimentX
 3 years 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
 3 years 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
 3 years ago
Best ResponseYou've already chosen the best response.2Awesome! thanks sire.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0thats actually look really awsome!
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