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(Serbia) Just squaring won't work

  1. Jul 26, 2014 #1
    1. The problem statement, all variables and given/known data
    Find the real solution [itex] x = \sqrt{2+\sqrt{2-\sqrt{2+x}}} [/itex]

    2. Relevant equations
    Algebric equations

    3. The attempt at a solution
    I tried to find some kind of recurrence by expanding as follows:
    x = [itex]\sqrt{2+\sqrt{2-\sqrt{2+x}}} [/itex]
    [itex] 2 - x = 2 -\sqrt{2+\sqrt{2-\sqrt{2+x}}} [/itex]
    [itex]\sqrt{2-x} = \sqrt{2-\sqrt{2+\sqrt{2-\sqrt{2+x}}}} [/itex]
    then
    [itex]2+\sqrt{2-x} = 2 + \sqrt{2-\sqrt{2+\sqrt{2-\sqrt{2+x}}}} [/itex]
    [itex]\sqrt{2+\sqrt{2-x}} = \sqrt{2 + \sqrt{2-\sqrt{2+\sqrt{2-\sqrt{2+x}}}}} [/itex]
    but I couldn't identify any, so I tried some substitutions and no solution yet :p
    How should I do this?
    Note: no calculators allowed
     
  2. jcsd
  3. Jul 26, 2014 #2

    phinds

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    Just out of curiosity, what does Serbia have to do with this?
     
  4. Jul 26, 2014 #3
    It is a Serbian olympic question. I thought indicating its origin would help finding a solution though
     
  5. Jul 26, 2014 #4

    Ray Vickson

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    You can analyze some general properties of the solution, but getting the solution itself is challenging. Let ##f(x) = \sqrt{2+\sqrt{2-\sqrt{2+x}}}##, so the equation is ##x = f(x)##. In order for all square roots defining f(x) to be real, we need ##2 \geq \sqrt{x+2}##, hence ##x \leq 2##. Also, ##f(x) \geq \sqrt{2}## (because ##f(x) = \sqrt{2 + \text{something} \geq 0}##) , so ##x \geq \sqrt{2}##. Note that ##f(\sqrt{2}) > \sqrt{2}## and ##f(2) = \sqrt{2} < 2## so there IS a root between ##\sqrt{2}## and ##2##.

    Getting an exact root can be done, but I don't see how without using a computer algebra package. Squaring DOES work, but it introduces 7 extraneous roots that are not part of the original problem.
    [tex] x = f(x) \: \Longrightarrow x^2-2 = \sqrt{2 - \sqrt{2+x}} \: \Longrightarrow
    (x^2-2)^2 = 2 - \sqrt{2+x} \: \longrightarrow x+2 = (2 - (x^2-2)^2)^2[/tex]
    This is a polynomial equation of degree 8. Packages like Maple or Mathematica can find all the roots, but they are complicated. Only one of the roots lies in the region ##[\sqrt{2},2]##.
     
    Last edited: Jul 26, 2014
  6. Jul 26, 2014 #5
    I have been at this problem for about half an hour now, and couldn't deduce much beyond what you said, Ray. But I am curious to know how you deduced there is only one root in the region [itex][\sqrt 2,2][/itex]. I think there is at least one other real root(assuming the others are complex), which may be in that range, right?
     
  7. Jul 27, 2014 #6

    SammyS

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    You can determine that any real root must be in that interval by considering the following.

    You have ##\ \sqrt{2+x} \ ## and ##\ \sqrt{2-\sqrt{2+x}\,}\,. \ ## Right?
     
  8. Jul 27, 2014 #7
    Dude, I have spent like 2 hours in it and I am really considering to explode this book now. The worst part is this question is at the easy exercises page of the chapter
     
    Last edited: Jul 27, 2014
  9. Jul 27, 2014 #8
    #%$&(@&*$, I got it! just take x = 2cos(y)
     
    Last edited: Jul 27, 2014
  10. Jul 27, 2014 #9
    What a relief! But this exercise is at factorizations chapter, I wonder if there is a solution by factoring this monster
     
  11. Jul 27, 2014 #10

    Mentallic

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    Would you mind sharing your solution? I'm not seeing it.
     
  12. Jul 27, 2014 #11
    Ok
    [itex] x = \sqrt{2+\sqrt{2-\sqrt{2+2cosy}}} [/itex]
    note that
    2 + 2cosy = 2(1 + cosy)
    and cosy = 2cos²(y/2) -1
    so 2 + 2cosy = 4cos²(y/2)
    Now, we have
    [itex] x = \sqrt{2+\sqrt{2-2cos(y/2)}} [/itex]
    analogously 2 - 2 cos(y/2) = 4sin(y/4)
    [itex] x = \sqrt{2+2sin(y/4)} [/itex]
    Finally, x = 2cos(y/8) <=> 2cos(y) = 2 cos(y/8)
    y = 2pi/9
    x = 2cos(2pi/9)

    xD
     
  13. Jul 27, 2014 #12

    Ray Vickson

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    By finding all the roots and evaluating them numerically. Alternatively, plot f(x) and x in the range [√2,2]. All eight roots of the polynomial are real, but 7 of them are extraneous to the original problem. All work done in Maple.
     
  14. Jul 27, 2014 #13
    My point exactly. Which is why I was confused when Ray said only one real root lies in that interval, I believe there at least need to be two.

    I tried that earlier in my attempt. The substitution does yield a 'neat' looking equation, but I don't see how to reach a numerical answer for x without the use of calculators/tables. Could you please share how?
     
  15. Jul 27, 2014 #14

    Ray Vickson

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    The function ##f(x)## is strictly decreasing on the interval ##[\sqrt{2},2]## (from the derivative), while the function ##x## is strictly increasing on that interval. There is only one root.
     
  16. Jul 27, 2014 #15
    That makes sense, thanks Ray.

    As a side note, I should start using one of these softwares soon :tongue2:

    The expressions of x in the two lines aren't equal. That, or I'm doing some trigonometric errors that are embarrassingly silly. :redface:
     
  17. Jul 27, 2014 #16

    ehild

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    After repeatedly squaring, you can simplify out a first-order factor and get an odd -number order polynomial equal to zero.

    ehild
     
  18. Jul 27, 2014 #17

    Mentallic

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    At the step

    [tex]2\cos y = \sqrt{2+2\sin\frac{y}{4}}[/tex]

    squaring and then using the double angle identity

    [tex]\cos^2\alpha = \frac{1}{2}\left(1+\cos{2\alpha}\right)[/tex]

    yields

    [tex]\cos{2y} = \sin{\frac{y}{4}}[/tex]

    And I can't proceed.
     
  19. Jul 27, 2014 #18

    ehild

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    [tex]\cos{2y} = \sin{(\pi/2\pm2y +2k\pi)}=\sin{\frac{y}{4}}[/tex]


    ehild
     
  20. Jul 27, 2014 #19

    Mentallic

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    Of course! :blushing:
     
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