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Homework Help: Nested Radicals

  1. Dec 12, 2011 #1
    We can see that if

    [itex]u=\sqrt{x+\sqrt{x+\sqrt{x+\dots}}}[/itex]

    then [itex]u^2=x+u[/itex]

    so [itex]u^2-u-x=0[/itex]

    This has solution

    [itex]\left( u-\frac{1}{2} \right)^2 -\frac{1}{4}-x=0 \Rightarrow u=\frac{1}{2} \pm \sqrt{x + \frac{1}{4}}[/itex]

    This means that [itex]u \in \mathbb{R} \forall x \geq \frac{1}{4}[/itex]

    In other words [itex]\sqrt{ -\frac{1}{8} + \sqrt{ - \frac{1}{8} + \sqrt{-\frac{1}{8} + \dots}}}[/itex] is real.

    This is clearly true according to the above formula. However, I cannot get my head around it - to me it seems like it must be imaginary! Can anyone give an explanation of why this is turning out to be real?
     
  2. jcsd
  3. Dec 12, 2011 #2

    ehild

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    Writing in the opposite way, it will not look imaginary:

    [tex]\sqrt {\dots \sqrt{ -\frac{1}{8} + \sqrt{ - \frac{1}{8} + \sqrt{-\frac{1}{8} + v}}}}[/tex]

    with v=1/4 for example.

    Edit: I changed the last term to v>0 instead of 1/4,
    ehild
     
    Last edited: Dec 12, 2011
  4. Dec 12, 2011 #3
    i dont see where you got the 1/4 from?
     
  5. Dec 12, 2011 #4

    ehild

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    Sorry, it was an example to show that the -1/8 terms do not lead necessarily to imaginary result.

    ehild
     
    Last edited: Dec 12, 2011
  6. Dec 12, 2011 #5
    ok. but surely the original expression will have a negative under the first root giving an imaginary result and then the second root will be the root of an imaginary number giving a complex result. All subsequent roots will involve roots of complex numbers and will give complex results won't they?
     
  7. Dec 12, 2011 #6

    Office_Shredder

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    The first thing you do is take the square root o
    [tex] -1/8+\sqrt{-1/8+\sqrt{-1/8+....}}[/tex]

    As long as the set of nested square roots is bigger than -1/8 then we're OK. But how can we check if this nested set of square roots is OK? Well, the first thing we do is take the square root of
    [tex]-1/8+\sqrt{-1/8+\sqrt{-1/8+....}}[/tex]

    So it's not at all obvious whether these things are positive or negative.
     
  8. Dec 12, 2011 #7

    SammyS

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    Shouldn't that say: [itex]u \in \mathbb{R} \forall x \geq -\frac{1}{4}[/itex] rather than [itex]x \geq \frac{1}{4}\ ?[/itex]
     
  9. Dec 13, 2011 #8

    ehild

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    When you write an expression you need to say if you mean it for real or for complex numbers.
    Defining the expression for complex x with the use of the complex square root, the expression does not have a unique value for any x. It can corresponds to a set of complex numbers.

    In case we require that x and u are real, and the expression has a defined value, there are some constraints. The square of a real number is non-negative real, the square-root function is defined only for non-negative numbers, and the square root of a number is non-negative.
    The condition that the value of the nested radicals exist and real, implies that u=√(x+u). From this, it follows that u and x must satisfy the equation u2=x+u, or
    (u-1/2)2=x+1/4, that is x≥-1/4, and u=1/2±√(x+1/4).
    For x>0 only the + sign is valid as u ≥0. But there can be two values for u if -1/4<x<0. So the expression of the nestled radicals is not single-valued for x<0. The value of the nestled radicals does not exist.

    So the statement that the value of the nestled radicals must be complex for a certain x, say for x=-1/8 does not hold, but we can say that it is undefined. But in case x>0 the expression returns a single real number u=1/2+√(x+1/4)

    You can consider the nestled radicals as a sequence given by an+1=√(an+x), for x≥-1/4 and a0 given. If a0=u=1/2±√(x+1/4) all elements of the sequence are equal.

    ehild
     
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