## klimenkov 2 years ago Compute $\huge{\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{\sqrt{2}^\ldots}}}}}$

1. SithsAndGiggles

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

2. klimenkov

It is infinite.

3. SithsAndGiggles

Ah, thanks for fixing it

4. klimenkov

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

5. borgore420

faliure

6. SithsAndGiggles

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.

7. klimenkov

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

8. shubhamsrg

why don't let whole of it = x ?

9. shubhamsrg

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

10. klimenkov

Nice, what will it be?

11. shubhamsrg

hmm, we get stuck ther! -_-

12. shubhamsrg

there*

13. shubhamsrg

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

14. hartnn

integer solutions, 2 and 4...

15. shubhamsrg

which one will it be ?

16. klimenkov

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

17. hartnn

i think 2

18. klimenkov

Why not 4?

19. hartnn

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

20. klimenkov

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

21. hartnn

:O .. let me think more...

22. shubhamsrg

2 also fits in though.

23. shubhamsrg

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

24. ParthKohli

Let $$\epsilon > 0$$

25. klimenkov

A medal will be given only if you prove this.

26. shubhamsrg

-_-

27. shubhamsrg

arey to hell with the medal !

28. hartnn

lol

29. ParthKohli

LOL!!

30. klimenkov

Can somebody say what means "arey to hell" ?

31. shubhamsrg

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

32. klimenkov

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

33. shubhamsrg

unable to prove, what proof you got ?

34. shubhamsrg

I mean I am unable to prove*

35. klimenkov

|dw:1362220984668:dw| Hope, you understand.

36. shubhamsrg

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

37. shubhamsrg

And that seems sort of forced proof.

38. shubhamsrg

@experimentX @mukushla

39. experimentX

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

40. experimentX

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

41. shubhamsrg

Awesome! thanks sire.

42. Dodo1

thats actually look really awsome!