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badreferences Group Title

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

  • 2 years ago
  • 2 years ago

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  1. badreferences Group Title
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    @KingGeorge @UnkleRhaukus @TuringTest @Zarkon @bahrom7893

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

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

    • 2 years ago
  4. badreferences Group Title
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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?

    • 2 years ago
  5. hartnn Group Title
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    1/2 ?

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

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

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

    • 2 years ago
  9. badreferences Group Title
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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.

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

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

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

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

    • 2 years ago
  14. badreferences Group Title
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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.

    • 2 years ago
  15. sauravshakya Group Title
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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.

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

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

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

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

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

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

    • 2 years ago
  22. badreferences Group Title
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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.

    • 2 years ago
  23. badreferences Group Title
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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.

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

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

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

    • 2 years ago
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