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Nested Radicals

  1. Oct 31, 2008 #1
    I been trying to solve some nested radicals. I've been able to do:

    Which is pretty cool since it equals to the "Golden Ratio" or [tex]\[\frac{\sqrt{5}+1}{2}\][/tex]

    But I can't seem to do the following:

    Using a calculator, it seems that this converges to [tex]$\sqrt{3}$[/tex]. Could anyone show me how? Thanks
  2. jcsd
  3. Oct 31, 2008 #2


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    If [tex]

    converged to [tex]\sqrt{3}[/tex]
    (I'm going to play a bit fast and loose with limits here)
    Then by squaring the nested radical you would have

    [tex]1+\sqrt{2+...} = 3[/tex]

    [tex]\sqrt{2+....} = 2[/tex]

    Then squaring both sides and subtracting again we get

    [tex]\sqrt{3+...} = 2[/tex]

    Repeat this to get

    [tex]\sqrt{4+....} = 1[/tex]

    Repeating one more time gives

    [tex]\sqrt{5+...} = -3[/tex]

    But that last equation looks unlikely
  4. Oct 31, 2008 #3
    I'm not sure exactly how you solve it, but with these kinds of problems, I start out getting rid of silly infinities.

    [tex]f(x) = \sqrt{x + f(x+1)}[/tex]

    Then, square both sides, and we've removed the radicals:

    [tex]f(x)^2 = x + f(x+1)[/tex]

    Rewrite it as a root-looking problem.

    [tex]f(x)^2 - f(x+1) - x = 0[/tex]

    And then, I don't know. Note that in the equation, f is *not* a specified function -- it's an unknown. It's something akin to differential equations, since we know various properties about the function (in this case, we know a recursive property of it), but we don't know *which* function it is yet. There might be a unique solution. There might be multiple solutions. There may not be a solution (if f diverges).

    But you don't necessarily need to know the actual function, f. It suffices to find the value of f(1).

    Anyway, good luck!

    Oh... also. How did you manage to plug it into your calculator? If you plugged in a finite expansion of the expression, you would have had to to make a guess for what f(k) would be for some integer k. If you assumed it to be 0, then that sequence may converge where the infinite one may not.
    Last edited: Oct 31, 2008
  5. Oct 31, 2008 #4


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    If x is a natural number you get

    [tex]f(x)^2 - x = f(x+1)[/tex] which is a recurrence relationship. So if I remembered anything about how to solve those, maybe you can find what shape f(x) is. But we don't even need that, since taking the limit of both sides as x goes to infinity does the trick. If the limit exists, call it F. Then

    [tex] f(x+1) \rightarrow F if x \rightarrow \infty[/tex]

    So the RHS of the equation is F. Then the left hand side's limit doesn't exist, since if lim(a+b) exists and lim(a) exists, lim(b) must exist too. (in this case a is f(x)2 and b is -x) Hence the function can't converge to a limit as x goes to infinity
  6. Oct 31, 2008 #5
    Ah, I get it now. I ignored the remainder term when I put the expansion into the calculator. Thanks!
  7. Oct 31, 2008 #6


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    The sequence
    definitely converges. Define the double sequence

    [tex]f(a, b) = \begin{cases}
    \sqrt{a + f(a+1, b-1)} & b > 0
    \0 & b = 0 \end{cases}[/tex]

    Then [itex]f(1, n)[/itex] yields the sequence presumably intended by the nested radicals. The value of the infinite nesting is then
    [tex]\lim_{n \rightarrow +\infty} f(1, n)[/tex]

    For [itex]a \geq 1[/itex] and [itex]b \geq 1[/itex], we have the overestimate:
    [tex]f(a, b) \leq a + 1[/tex]
    which can be proven by induction on b:

    [tex]f(a, 1) = \sqrt{a} \leq a + 1[/tex]
    [tex]f(a, b) = \sqrt{ a + f(a+1, b-1) }
    \leq \sqrt{a + a + 1 + 1} = \sqrt{2(a+1)} \leq a + 1[/tex]

    [itex]f(1, b)[/itex] is clearly an increasing function of b, and is bounded above by 2, and so must be a convergent sequence.
    Last edited: Oct 31, 2008
  8. Oct 31, 2008 #7
    Forgive my untrained brain, but I don't get the proof fully. So you used a new variable (b) in order to propagate the sequence to infinity, but I don't get how you can say that [tex]\sqrt{2(a+1)} \leq a + 1
    [/tex] when before it was said only that [tex]\sqrt{a} \leq a + 1
    [/tex]. Then shouldn't the statement be more like: [tex]\sqrt{2(a+1)} \leq 2a + 3
  9. Nov 1, 2008 #8
    [tex]\sqrt{2(a+1)} \leq a + 1[/tex] was not based on [tex]\sqrt{a} \leq a + 1[/tex], but meant to be proved on its own ("left to the reader") :)

    It is easier to see if you write it as [tex]\sqrt 2 \sqrt{a+1} \leq a + 1[/tex]; this is an equality for a = 1, and as a increases, [tex]\sqrt 2[/tex] is smaller than [tex]\sqrt{a+1}[/tex], so their product is smaller than [tex]\sqrt{a+1} \sqrt{a+1} = a + 1[/tex].
  10. Nov 1, 2008 #9
    Thanks, now I get it.

    But is there a way to find out where the sequence converges? We do know that [tex]\sqrt{3} < \lim_{n\rightarrow\infty} f(1,n) < 2 [/tex]. Where:

    [tex] f(a, b) = \begin{cases}
    \sqrt{a + f(a+1, b-1)} & b > 0
    \0 & b = 0 \end{cases}[/tex]
  11. Nov 1, 2008 #10


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