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Sequence of rationals that converge to irrational

  1. Mar 14, 2005 #1
    Find a sequence of rational numbers that converges to the square root of 2
     
  2. jcsd
  3. Mar 14, 2005 #2

    mathwonk

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    use newton's binomial theorem? i.e. (1+1)^(1/2) = 1 + (1/2) + (1/2)(-1/2)(1/2) +

    + (1/2)(-1/2)(-3/2)(1/2)(1/3)+......

    this might work.
     
  4. Mar 14, 2005 #3
    Or another of Newton's tricks:

    How do I approximate the positive root of [tex]x^2-2[/tex]?

    [tex]x_1 = 1[/tex]

    [tex]x_2 = x_1 - \frac{x_1^2 - 2}{2x_1}[/tex]

    [tex]. \ . \ .[/tex]

    [tex]x_n = x_{n-1} - \frac{x_{n-1}^2 - 2}{2x_{n-1}}[/tex]

    [tex]. \ . \ .[/tex]

    Obviously each term is rational and [tex]\{x_n\}[/tex] converges to [tex]\sqrt{2}[/tex].
     
    Last edited: Mar 14, 2005
  5. Mar 14, 2005 #4
    And converges quite quickly, I might add.
     
  6. Mar 15, 2005 #5

    HallsofIvy

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    Do you mean giving a general formula?

    If not, take the square root of 2 on a calculator:
    1.4142135623730950488016887242097

    So a sequence of rationals converging to square root of 2 is:
    1, 1.4, 1.41, 1.414, 1.4142, 1.41421, 1.414213, 1.4142135, 1.41421356, etc.
     
  7. Mar 15, 2005 #6

    mathwonk

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    the problem here is that a calculator has a bounded number of terms.
     
  8. Mar 15, 2005 #7

    shmoe

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    You could take the sequence of convergents of the continued fraction of sqrt(2).

    [tex]\sqrt{2}=1+\frac{1}{2+\frac{1}{2+\frac{1}{2+\ldots}}}[/tex]
     
  9. Mar 15, 2005 #8
    I think Data's use of Newton's Method is the best way to display the sequence that converges. Shmoe's definition of the square root of two is correct, but it isn't really written in a form that converges, although I'm sure shmoe could easily do that. I think good old Newton can help you best.
     
  10. Mar 16, 2005 #9

    shmoe

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    I'm not sure what you mean here. The notation I used is pretty standard, and the sequence of convergents does converge to [tex]\sqrt{2}[/tex].

    Maybe I should have defined what I meant by the convergents. I just mean the sequence begining

    [tex]1,1+\frac{1}{2}, 1+\frac{1}{2+\frac{1}{2}}, 1+\frac{1}{2+\frac{1}{2+\frac{1}{2}}},\ldots[/tex]

    Or do you want some proof that this continued fraction is actually [tex]\sqrt{2}[/tex]?

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

    Then repeat:

    [tex]\sqrt{2}=1+\frac{1}{2+\frac{1}{1+\sqrt{2}}}=1+\frac{1}{2+\frac{1}{2+\frac{1}{1+\sqrt{2}}}},\ldots[/tex]

    In this way you can easily see [tex]\sqrt{2}[/tex] is larger than all the even convergents and smaller than all the odd convergents (calling the first term the 0th). If you believe that they convegre to something (this is pretty standard) then that something has to be [tex]\sqrt{2}[/tex]
     
    Last edited: Mar 16, 2005
  11. Mar 16, 2005 #10
    I didn't see the "..." you had included. I just wanted to make sure it was understood that the sequence had to keep going in order to converge. It sounded worse than I meant it. I just liked the way Data wrote the convergence of Newton's Method, with "nth" notation. You are of course very correct with all you said... didn't mean to imply otherwise.
     
  12. Mar 16, 2005 #11

    shmoe

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    Oh I see :smile: The ... are quite small, I would have probably missed them if I didn't know I put it there.
     
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