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

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions.

A community for students.

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

Calculus1
See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions

I'm not sure your expression makes any sense. Is the continued exponentiation finite or not?
It is infinite.
Ah, thanks for fixing it

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

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

Not the answer you are looking for?

Search for more explanations.

Ask your own question