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badreferences

  • 2 years ago

Difficult question. Find the value of\[\sqrt{1-\sqrt{\frac{17}{16}-\sqrt{1-\sqrt{\frac{17}{16}-\sqrt{1-\cdots}}}}}\]

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  1. badreferences
    • 2 years ago
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    @KingGeorge @UnkleRhaukus @TuringTest @Zarkon @bahrom7893

  2. badreferences
    • 2 years ago
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    I have the solution, by the way. This is like a challenge. :)

  3. hartnn
    • 2 years ago
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    y^2 = 1-sqrt{17/16-y} (y^2-1)^2 = 17/16-y solve...

  4. badreferences
    • 2 years ago
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    But \(y\) has one definite value as bounded by the equation before taking the square of both sides. So, which one is it?

  5. hartnn
    • 2 years ago
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    1/2 ?

  6. badreferences
    • 2 years ago
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    That's a good heuristic guess. It's correct. Why?

  7. UnkleRhaukus
    • 2 years ago
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    \[x=\sqrt{1-\sqrt{\frac{17}{16}-\sqrt{1-\sqrt{\frac{17}{16}-\sqrt{1-\cdots}}}}}\] \[1-x^2={\sqrt{\frac{17}{16}-\sqrt{1-\sqrt{\frac{17}{16}-\sqrt{1-\cdots}}}}}\] \[\left((1-x^2)^2-\frac{17}{16}\right)^2={{\sqrt{1-\sqrt{\frac{17}{16}-\sqrt{1-\cdots}}}}}\] \[\left((1-x^2)^2-\frac{17}{16}\right)^2=x\]

  8. UnkleRhaukus
    • 2 years ago
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    am i on the right track?

  9. badreferences
    • 2 years ago
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    A polynomial equation that doesn't rigorously determine which of the \(x\)'s are correct. But yes, you and @hartnn are on the right track. From here it's about iteration behavior at certain values.

  10. UnkleRhaukus
    • 2 years ago
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    i think i accidentally dropped a negative sign

  11. badreferences
    • 2 years ago
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    Yes, you did. Haha, I didn't notice. I just saw the polynomials and went, "k"

  12. badreferences
    • 2 years ago
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    Not that it makes a difference. A mild algebraic rearrangement leads to the same problem.

  13. UnkleRhaukus
    • 2 years ago
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    \[\left(\frac{17}{16}-(1-x^2)^2\right)^2=x\]\[\left(\frac{1}{16}+2x^2-x^4\right)^2=x\]\[\frac1{16^2}+\frac{x^2}4-\frac{31x^4}{8}-4x^6+x^8=x\]\[\frac1{256}-x+\frac{x^2}4-\frac{31x^4}{8}-4x^6+x^8=0\]

  14. badreferences
    • 2 years ago
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    Ahaha, that's some unnecessary complexity, but ostensibly it seems right. The difficulty is in showing which one of the \(x\)'s it is, as bounded by the one value \(x\) in the original equation.

  15. sauravshakya
    • 2 years ago
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    cant we just substitute the value in y^2 = 1-sqrt{17/16-y} and see if it is TRUE OR NOT.

  16. sauravshakya
    • 2 years ago
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    As squaring gives some unnecessary roots.

  17. UnkleRhaukus
    • 2 years ago
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    \[1-256x+64{x^2}-992x^4-1024x^6+256x^8=0\] \[1-2^8x+2^6{x^2}-31\times2^5x^4-2^{10}x^6+2^8x^8=0\] \[1-2^8x+2^4{(2x)^2}-62\times(2x)^4-2^{4}(2x)^6+(2x)^8=0\]

  18. sauravshakya
    • 2 years ago
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    |dw:1349505055344:dw|

  19. sauravshakya
    • 2 years ago
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    Only 1/2 satisfies that loop

  20. badreferences
    • 2 years ago
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    That's correct, and answers the question, but I feel like I'm missing a technical detail here.

  21. sauravshakya
    • 2 years ago
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    |dw:1349505240174:dw|

  22. badreferences
    • 2 years ago
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    Hmm, you used the same solution as I did, but for some reason my solution was corrected to include rates of change around \(x=0.5\). Haha, I'm still trying to catch why.

  23. badreferences
    • 2 years ago
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    I think the idea is that asserting\[y=\sqrt{1-\sqrt{\frac{17}{16}-y}}\]assumes that \(y\) has the innermost radical as \(1\), and iteratively doing this will result in a technical approximation. So we had to determine that the values converge around \(x=0.5\) using straightforward slopes.

  24. badreferences
    • 2 years ago
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    But whatever, if there was subtlety here, I'm still missing it. You got the answer.

  25. UnkleRhaukus
    • 2 years ago
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    \[y=\sqrt{1-\sqrt{\frac{17}{16}-y}}\]\[1-y^2=\sqrt{\frac{17}{16}-y}\]\[(1-y^2)^2={\frac{17}{16}-y}\]\[y+(1-y^2)^2-{\frac{17}{16}}=0\]\[y+(1-2y^2+y^4)-{\frac{17}{16}}=0\]\[y-2y^2+y^4-{\frac{1}{16}}=0\]\[16y-32y^2-16y^4-1=0\]\[2^3(2y)-2^3(2y)^2-(2y)^4-1=0\]

  26. UnkleRhaukus
    • 2 years ago
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    \[2^3(2y)-2^3(2y)^2-(2y)^4-(2y)^0=0\]

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