1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    User Avatar
    Homework Helper
    Gold Member

    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

    User Avatar
    Homework Helper
    Gold Member

    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

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    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

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    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

    User Avatar
    Homework Helper
    Gold Member

    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
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Nested Radicals
  1. Nested Quantifier (Replies: 10)

  2. Nested quantifiers (Replies: 3)

Loading...