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How were logs and roots calculated before calculators?

  1. Oct 24, 2005 #1


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    I'm not sure whether this is the correct forum, so I apologise if it's in the incorrect forum.

    Anyway, when studying A level maths a few years ago, we came across a technique for calculating roots that my teacher claimed was used before calculators were invented. I can't remember the actual technique used, but I remember thinking at the time that it was particularly clever. He also claimed that before calculators, mathematicians were forced to invent other, similarly clever techniques to work out their logs and roots etc.

    So, what were these techniques? How exactly were the logs and roots calculated before modern computing machines?
  2. jcsd
  3. Oct 24, 2005 #2
  4. Oct 24, 2005 #3

    Math Is Hard

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    How is it possible that BobG hasn't shown up in this thread yet? :biggrin:
  5. Oct 25, 2005 #4
    There is an algorithm to calculate square roots to any required degree of accuracy with merely pen and paper alone. I learnt it from a book published in about 1920 aimed at 16 year olds when I was about 10; the book is also no longer in print. Anyway, my point is, I can't really remember it, but if someone really wanted to, I could half-remember and half-work it out. Or maybe someone else knows it too.
  6. Oct 25, 2005 #5


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    Suppose you want the squareroot of 55. It must lie between 7 and 8.
    Try 7.5. [tex]\frac{55}{7.5} = 7.333[/tex]
    Now, calculate the mean:
    [tex]\frac{7.5 + 7.333}{2} = 7.41666[/tex]
    and use it in the new iteration:
    [tex]\frac{55}{7.41666} = 7.41573[/tex]
    [tex]\frac{7.4166 + 7.41573}{2} = 7.416198468[/tex]
    Using the calculator of Windows, you get for the squareroot of 55 the value 7.416198487...
  7. Oct 25, 2005 #6
    This "divide-and-average" method can be derived using the more general http://planetmath.org/encyclopedia/NewtonsMethod.html [Broken] of finding zeroes of nonlinear functions.
    Last edited by a moderator: May 2, 2017
  8. Oct 25, 2005 #7
    These algorithms are slow; a faster one realises that
    [tex](10a+b)^2 = 100a^2 + 20ab + b^2[/tex]
    and works out the digits of the square root based on pairs of digits of the square, by finding a [itex]b[/itex] such that [itex]b(20a+b)[/itex] matches whatever is left (having subtracted the first term already).
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