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badreferences

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

  • one year ago
  • one year ago

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

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

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

    • one year ago
  4. badreferences
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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?

    • one year ago
  5. hartnn
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    1/2 ?

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

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

    • one year ago
  8. UnkleRhaukus
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    am i on the right track?

    • one year ago
  9. badreferences
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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.

    • one year ago
  10. UnkleRhaukus
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    i think i accidentally dropped a negative sign

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

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

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

    • one year ago
  14. badreferences
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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.

    • one year ago
  15. sauravshakya
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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.

    • one year ago
  16. sauravshakya
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    As squaring gives some unnecessary roots.

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

    • one year ago
  18. sauravshakya
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    |dw:1349505055344:dw|

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

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

    • one year ago
  21. sauravshakya
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    |dw:1349505240174:dw|

    • one year ago
  22. badreferences
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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.

    • one year ago
  23. badreferences
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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.

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

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

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

    • one year ago
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