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



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SithsAndGiggles
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I'm not sure your expression makes any sense. Is the continued exponentiation finite or not?

klimenkov
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It is infinite.

SithsAndGiggles
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Ah, thanks for fixing it

klimenkov
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Hope, you like it now. What will it be?

borgore420
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faliure

SithsAndGiggles
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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.

klimenkov
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The order is wrong. Compare:
\[2^{2^3}=2^8=256,\quad (2^2)^3=4^3=64.\]

shubhamsrg
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why don't let whole of it = x ?

shubhamsrg
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then if, you observe properly, you'll be reduced to
[ sqrt(2) ]^x = x

klimenkov
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Nice, what will it be?

shubhamsrg
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hmm, we get stuck ther! _

shubhamsrg
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there*

shubhamsrg
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2^(x/2) = x
=> 2^x = x^2
symmetrically, there is a solution , see that ?

hartnn
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integer solutions, 2 and 4...

shubhamsrg
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which one will it be ?

klimenkov
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Now tell me, what is the answer: 2 or 4?

hartnn
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i think 2

klimenkov
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Why not 4?

hartnn
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because while squaring, we attached an extra root, x=4

klimenkov
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Hm... \(\sqrt 2^4=4\). Why do you call it "extra root"?

hartnn
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:O .. let me think more...

shubhamsrg
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2 also fits in though.

shubhamsrg
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on the calculator, it approaches 2, I am not too sure how can we prove that ?

ParthKohli
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Let \(\epsilon > 0\)

klimenkov
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A medal will be given only if you prove this.

shubhamsrg
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_

shubhamsrg
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arey to hell with the medal !

hartnn
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lol

ParthKohli
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LOL!!

klimenkov
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Can somebody say what means "arey to hell" ?

shubhamsrg
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ignore "arey"
I meant To Hell with the medal.

klimenkov
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Ok. I found an interesting proof to this. Hope, you will find another interesting proof.

shubhamsrg
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unable to prove, what proof you got ?

shubhamsrg
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I mean I am unable to prove*

klimenkov
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dw:1362220984668:dw
Hope, you understand.

shubhamsrg
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how can you say thats not gonna intersect at x=4 ?

shubhamsrg
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And that seems sort of forced proof.

shubhamsrg
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@experimentX @mukushla

experimentX
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dw:1362247620602:dwdw:1362247717003:dw
dw:1362247538096:dw
Hence the above value only converges to Sqrt 2

experimentX
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*The last figure is first figure, the first figure is second figure and second figure is last figure.

shubhamsrg
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Awesome!
thanks sire.

Dodo1
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thats actually look really awsome!