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Archimedes sqrt(3)

  1. May 25, 2009 #1
    Hi guys,

    I am a 1st year Engineering/Maths student. I have recently started reading "journey through genius"; which so far is great! After reading chapter 3 on Archimedes i was very curious about how he approximated a lower and upper bound for sqrt(3). I looked at the translated proof in my "great western books - vol. 11", however, there is no mention of how he arrived at such figures 265/153 < sqrt(3) < 1351/780.

    After searching the net, i realise there is no way to know exactly how he came to these numbers. However, i did come across an interesting article at http://www.mathpages.com/home/kmath038.htm [Broken]

    This article suggests starting with (5/3) as the estimate. I found by using 5/3 and the bisecting method?
    a_{n + 1} = (1/2)(a_n + (N/a_n)), with (a_0)^2 = (5/3)^2 < 3 = N
    the first 3 numbers are
    5/3, 26/15, 1351/780, ...
    However this method seems to skip the lower bound of 265/153.

    In the above mentioned article it states "if we imagine that their first estimate for the square root of 3 was 5/3, perhaps based on the fact that 5^2 = 25 is close to 3(3^2) = 27. From here it isn't hard to see that if x is a bound on the square root of 3, then
    (5x+9)/(3x+5) is a closer bound on the opposite side."

    The last sentence is not "easy for me to see"; I am wandering if anybody can help me understand how (5x+9)/(3x+5) was derived? It seems related to a_{n + 1} = (1/2)(5/3) + (9/5)?

    If you use the article's formula and x_1 = 5/3, you get 26/15; allow x_2 = 26/15, you get 265/153; and finally, allow x_3 = 265/153, you get 1351/780. Interesting results!

    Thanks for your help,

    Brendan
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. May 25, 2009 #2

    Hurkyl

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    They explain in the very next line how it is "easy to see" -- you're asking a different question: "how would I go about looking for such a thing?"

    I'm not sure how that paper came up with it, but I can speculate. First, try solving the equation
    x = (5x + 9) / (3x + 5)​

    Done that? Good. Now what if you reverse the steps you just performed to solve that equation? You would come up with an iterative procedure just like this one! There are choices involved so you might not have produced (5x+9)/(3x+5), but you would have produced something that works.


    (A more systematic approach would probably involve continued fractions)
     
    Last edited: May 25, 2009
  4. May 25, 2009 #3

    Hurkyl

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    Oh, that formula just comes from what they described earlier in the article. If you look at that formula for s they mention, and grind through the algebra, you get:

    If N is your "first" approximation to sqrt(A), and x is a later approximation to sqrt(A), then the next approximation to use is (Nx + A) / (x + N).
     
  5. May 25, 2009 #4
    Thank you very much Hurkyl,

    this was very helpful. I think i have it right...

    (1/2)(N + A/N) = (N^2 + A)/(2N) = (Nx + A)/(x + N) for x=N. And if |x^2 - A| < |N^2 -A|, then (Nx + A)/(x + N) is a closer approximation then (N^2 + A)/(2N).

    Now i am trying to see why (Nx + A)/(x + N) alternates (with each new x : x /= N) between upper and lower bounds? Is there a proof or way of showing why this, or is it coincidence?

    EDIT: I know why this seems to happen, but can't prove it. If x > sqrt(3) then the change in denominator is more than the change in the numerator, and the opposite for x < sqrt(3).

    Also, although i cannot perform the large calculations, (1/2)(N + A/N) seems to go from 5/3 (lower bound), to 25/15 (upper bound) and then converges to sqrt(3) from the upper side, or at the least, it randomly alternates?

    Thanks again,

    Brendan
     
    Last edited: May 25, 2009
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