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

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

klimenkov Group TitleBest ResponseYou've already chosen the best response.0
It is infinite.
 one year ago

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

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

borgore420 Group TitleBest ResponseYou've already chosen the best response.0
faliure
 one year ago

SithsAndGiggles Group TitleBest ResponseYou've already chosen the best response.0
I'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.
 one year ago

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

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

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

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

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

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

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

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

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

klimenkov Group TitleBest ResponseYou've already chosen the best response.0
Why not 4?
 one year ago

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

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

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

shubhamsrg Group TitleBest ResponseYou've already chosen the best response.2
2 also fits in though.
 one year ago

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

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

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

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

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

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

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

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

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

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

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

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

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

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

experimentX Group TitleBest 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.
 one year ago

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

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