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Square roots

  1. Aug 19, 2005 #1
    Hi, I came across this puzzle, see if you can solve it :smile: :

    [tex]\sqrt{1 + \sqrt{1 + 2\sqrt{1 + 3\sqrt{...}}}} = ?[/tex]
  2. jcsd
  3. Aug 19, 2005 #2

    -- AI
    P.S -> .dettimbus steg egassem taht os dedda texT
  4. Aug 19, 2005 #3


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    For all natural n:

    [tex]n = \sqrt{1 + (n-1)\sqrt{1 + n\sqrt{1 + (n + 1)\sqrt{\dots }}}}[/tex]

    For n = 1, we see the right side is:

    [tex]\sqrt{1 + 0\sqrt{\dots }} = \sqrt{1} = 1[/tex]

    Note that for now, we're assuming that what we're looking at is even a number, and doesn't go to infinity. Next, assume that:

    [tex]k = \sqrt{1 + (k-1)\sqrt{1 + k\sqrt{1 + (k + 1)\sqrt{\dots }}}}[/tex]

    for some natural k. Then:

    [tex]k+1 = \frac{(k - 1)(k + 1)}{(k - 1)}[/tex]
    [tex]= \frac{k^2 - 1}{k - 1}[/tex]

    [tex]= \frac{\left (\sqrt{1 + (k-1)\sqrt{1 + k\sqrt{1 + (k + 1)\sqrt{\dots }}}}\right )^2 - 1}{k - 1}[/tex]

    [tex]= \frac{(k - 1)\sqrt{1 + k\sqrt{1 + (k + 1)\sqrt{\dots }}}}{k - 1}[/tex]

    [tex]= \sqrt{1 + k\sqrt{1 + (k + 1)\sqrt{\dots }}}[/tex]​
    as required.

    From this, TenaliRaman's answer follows. It remains to prove, however, that the stuff really is a number and doesn't "diverge". Any ideas as to how to prove that?
    Last edited: Aug 19, 2005
  5. Aug 19, 2005 #4
    Do i prove this with series? Don't have my Calculus book with me... I got to the part of general representation, the last part of k-expression...
  6. Aug 19, 2005 #5
    There is a bit easier way to go abt the above,
    Let f(x) = x+1
    = sqrt((x+1)^2)
    = sqrt(x^2 + 2x + 1)
    = sqrt(1 + x (x+2))
    = sqrt(1 + xf(x+1))

    Recursively applying the above gives,
    f(x) = sqrt(1+x*sqrt(1+(x+1)*sqrt(1+(x+2)*sqrt(......))) = x + 1
    place x = 1 and et voila!

    -- AI
  7. Aug 19, 2005 #6
    ahhh, i see... how did you see that?
  8. Aug 19, 2005 #7
    Its one of those *aha* moments. Personally, these sort of "functional equations" are good fun to play around with (i have played with this one, quite a while back). Just remember the trick i used above and try to come up with some more radicals just like those.

    For example,
    f(x) = x + a = sqrt(x^2+2ax+a^2) = sqrt(a^2+x*f(x+a))
    Recursively apply the above and get a new oh_so_cool radical expression, fit in a value for a and set some value for x and give it to your friends to torture them for a while.

    Note in all of the above, its important to analyse that the infinite recursion is valid and that the infinite recursion actually converges. This aspect ofcourse i have pushed under the rug for now, since its a puzzle after all :tongue2: .

    There is a paper by T Vijayaraghavan (IIRC) which discussed convergence issues of such radicals, which could probably be the starting point of such analysis.

    -- AI
  9. Aug 19, 2005 #8
    cool, thanks! yeah...i guess if you do a lot of these, you kind of develop an intuition.
    have a link? or a reference?
  10. Aug 19, 2005 #9


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    How would you check that the recursion is valid? EDIT: actually, since each step in the recursion produces the same number, why would the infinite recursion be any different? If it is different, what we have to do to check convergence?
    Last edited: Aug 19, 2005
  11. Aug 21, 2005 #10
    When i said recursion being valid, i meant, making sure that the final number doesnt change by infinite application of recursion.

    umm, let me try and give an example,
    Let f(x) = x
    f(x) = x = (x+1) - 1
    f(x) = f(x+1)-1
    .....-3-2-1 = f(x) = x ,for any x
    I guess this example is pretty lame but i hope it drives home the point.

    I am not sure whether there is a convergence check (the paper i mentioned above, discusses convergence of infinite radicals IIRC and not exactly recursion, and to be honest, when i had read that paper, i couldnt understand a penny out of it, :uhh: cmon i am just 21 and an engineer at that so give me a break :biggrin:) .

    However, i guess one could model the convergence condition as that in the convergence of a series,
    Let R(k,f(x)) denote the value of recursive function f(x) at depth k.
    Then a possible convergence condition would be,
    R(1,f(x)) = R(2,f(x)) = R(k,f(x)) = R(k+1,f(x))

    -- AI
  12. Aug 21, 2005 #11
    If i had, i would have definitely linked to it in the original post. I dont even remember where i had read it, it was quite a long time back. Hell i was a toddler in maths when i had read that, i was just able to quote that paper because i had read that in reference to Srinivasa Ramanujan, who btw was the brains behind such infinite radicals.

    -- AI
  13. Aug 21, 2005 #12


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    Why not apply the usual condition for convergence of a sequence? :tongue2:

    I.E. does this limit exist?

    \lim_{k \rightarrow \infty} R(k, f)
  14. Aug 22, 2005 #13
    :eek: Indeed!
    Damn, my post looks stupid now! :frown:

    -- AI
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