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klimenkov

  • 3 years ago

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

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  1. SithsAndGiggles
    • 3 years ago
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    I'm not sure your expression makes any sense. Is the continued exponentiation finite or not?

  2. klimenkov
    • 3 years ago
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    It is infinite.

  3. SithsAndGiggles
    • 3 years ago
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    Ah, thanks for fixing it

  4. klimenkov
    • 3 years ago
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    Hope, you like it now. What will it be?

  5. borgore420
    • 3 years ago
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    faliure

  6. SithsAndGiggles
    • 3 years ago
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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.

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

  8. shubhamsrg
    • 3 years ago
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    why don't let whole of it = x ?

  9. shubhamsrg
    • 3 years ago
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    then if, you observe properly, you'll be reduced to [ sqrt(2) ]^x = x

  10. klimenkov
    • 3 years ago
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    Nice, what will it be?

  11. shubhamsrg
    • 3 years ago
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    hmm, we get stuck ther! -_-

  12. shubhamsrg
    • 3 years ago
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    there*

  13. shubhamsrg
    • 3 years ago
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    2^(x/2) = x => 2^x = x^2 symmetrically, there is a solution , see that ?

  14. hartnn
    • 3 years ago
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    integer solutions, 2 and 4...

  15. shubhamsrg
    • 3 years ago
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    which one will it be ?

  16. klimenkov
    • 3 years ago
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    Now tell me, what is the answer: 2 or 4?

  17. hartnn
    • 3 years ago
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    i think 2

  18. klimenkov
    • 3 years ago
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    Why not 4?

  19. hartnn
    • 3 years ago
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    because while squaring, we attached an extra root, x=4

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

  21. hartnn
    • 3 years ago
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    :O .. let me think more...

  22. shubhamsrg
    • 3 years ago
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    2 also fits in though.

  23. shubhamsrg
    • 3 years ago
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    on the calculator, it approaches 2, I am not too sure how can we prove that ?

  24. ParthKohli
    • 3 years ago
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    Let \(\epsilon > 0\)

  25. klimenkov
    • 3 years ago
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    A medal will be given only if you prove this.

  26. shubhamsrg
    • 3 years ago
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    -_-

  27. shubhamsrg
    • 3 years ago
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    arey to hell with the medal !

  28. hartnn
    • 3 years ago
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    lol

  29. ParthKohli
    • 3 years ago
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    LOL!!

  30. klimenkov
    • 3 years ago
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    Can somebody say what means "arey to hell" ?

  31. shubhamsrg
    • 3 years ago
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    ignore "arey" I meant To Hell with the medal.

  32. klimenkov
    • 3 years ago
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    Ok. I found an interesting proof to this. Hope, you will find another interesting proof.

  33. shubhamsrg
    • 3 years ago
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    unable to prove, what proof you got ?

  34. shubhamsrg
    • 3 years ago
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    I mean I am unable to prove*

  35. klimenkov
    • 3 years ago
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    |dw:1362220984668:dw| Hope, you understand.

  36. shubhamsrg
    • 3 years ago
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    how can you say thats not gonna intersect at x=4 ?

  37. shubhamsrg
    • 3 years ago
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    And that seems sort of forced proof.

  38. shubhamsrg
    • 3 years ago
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    @experimentX @mukushla

  39. experimentX
    • 3 years ago
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    |dw:1362247620602:dw||dw:1362247717003:dw| |dw:1362247538096:dw| Hence the above value only converges to Sqrt 2

  40. experimentX
    • 3 years ago
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    *The last figure is first figure, the first figure is second figure and second figure is last figure.

  41. shubhamsrg
    • 3 years ago
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    Awesome! thanks sire.

  42. Dodo1
    • 3 years ago
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    thats actually look really awsome!

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