klimenkov
  • klimenkov
Compute \[\huge{\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{\sqrt{2}^\ldots}}}}}\]
Calculus1
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
I'm not sure your expression makes any sense. Is the continued exponentiation finite or not?
klimenkov
  • klimenkov
It is infinite.
anonymous
  • anonymous
Ah, thanks for fixing it

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klimenkov
  • klimenkov
Hope, you like it now. What will it be?
anonymous
  • anonymous
faliure
anonymous
  • anonymous
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
  • klimenkov
The order is wrong. Compare: \[2^{2^3}=2^8=256,\quad (2^2)^3=4^3=64.\]
shubhamsrg
  • shubhamsrg
why don't let whole of it = x ?
shubhamsrg
  • shubhamsrg
then if, you observe properly, you'll be reduced to [ sqrt(2) ]^x = x
klimenkov
  • klimenkov
Nice, what will it be?
shubhamsrg
  • shubhamsrg
hmm, we get stuck ther! -_-
shubhamsrg
  • shubhamsrg
there*
shubhamsrg
  • shubhamsrg
2^(x/2) = x => 2^x = x^2 symmetrically, there is a solution , see that ?
hartnn
  • hartnn
integer solutions, 2 and 4...
shubhamsrg
  • shubhamsrg
which one will it be ?
klimenkov
  • klimenkov
Now tell me, what is the answer: 2 or 4?
hartnn
  • hartnn
i think 2
klimenkov
  • klimenkov
Why not 4?
hartnn
  • hartnn
because while squaring, we attached an extra root, x=4
klimenkov
  • klimenkov
Hm... \(\sqrt 2^4=4\). Why do you call it "extra root"?
hartnn
  • hartnn
:O .. let me think more...
shubhamsrg
  • shubhamsrg
2 also fits in though.
shubhamsrg
  • shubhamsrg
on the calculator, it approaches 2, I am not too sure how can we prove that ?
ParthKohli
  • ParthKohli
Let \(\epsilon > 0\)
klimenkov
  • klimenkov
A medal will be given only if you prove this.
shubhamsrg
  • shubhamsrg
-_-
shubhamsrg
  • shubhamsrg
arey to hell with the medal !
hartnn
  • hartnn
lol
ParthKohli
  • ParthKohli
LOL!!
klimenkov
  • klimenkov
Can somebody say what means "arey to hell" ?
shubhamsrg
  • shubhamsrg
ignore "arey" I meant To Hell with the medal.
klimenkov
  • klimenkov
Ok. I found an interesting proof to this. Hope, you will find another interesting proof.
shubhamsrg
  • shubhamsrg
unable to prove, what proof you got ?
shubhamsrg
  • shubhamsrg
I mean I am unable to prove*
klimenkov
  • klimenkov
|dw:1362220984668:dw| Hope, you understand.
shubhamsrg
  • shubhamsrg
how can you say thats not gonna intersect at x=4 ?
shubhamsrg
  • shubhamsrg
And that seems sort of forced proof.
shubhamsrg
  • shubhamsrg
@experimentX @mukushla
experimentX
  • experimentX
|dw:1362247620602:dw||dw:1362247717003:dw| |dw:1362247538096:dw| Hence the above value only converges to Sqrt 2
experimentX
  • experimentX
*The last figure is first figure, the first figure is second figure and second figure is last figure.
shubhamsrg
  • shubhamsrg
Awesome! thanks sire.
anonymous
  • anonymous
thats actually look really awsome!

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