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    Re: First sine table (Ptolemy)
    From: J Cora
    Date: 2009 Jan 26, 07:34 -0800

    I prepare to be corrected here but I recall reading that
    mathematicians studying continued fractions were the
    first to find finite sums for infinite series, perhaps
    this is how James Gregory worked out arctan x.
    
    I was curious how napier and briggs worked out the
    values and found some information.  I have a detailed
    example and a simpler version which I am attaching
    The calculator I used is called pari/gp and is free sw.
    
    
    
    On Wed, Jan 21, 2009 at 6:33 PM, James N Wilson  wrote:
    >
    > Were infinite series used in the initial creation of tables? I find in my
    > ancient college handbook listings of series for just about everything,
    > including the trigonometric functions.
    >
    > Jim Wilson
    > ____________________________________________________________
    > Click here for free information on how to reduce your debt by filing for bankruptcy.
     
    >
    > >
    >
    
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    After reading how Napier and Briggs calculated some values for the first log tables, 
    I decided to follow the procedure and post a summary.
    
    The method I used was explained on mathforum.org in these two posts but I decided 
    to limit the calculations to 7 decimal places to make it easier to follow.
    
    http://mathforum.org/library/drmath/view/51432.html
    http://mathforum.org/library/drmath/view/55600.html
    
    
    
      01.   let x = 8th root of ten, which will give us a
            number close to 1.01
    
                x = 10^2^-8  
            x = sqrt(sqrt(sqrt(sqrt(sqrt(sqrt(sqrt(sqrt(10))))))))
                   1    2    3    4    5    6    7    8
    
            from a 4 function calculator take sqrt(10) 8 times
    
            sqrt01         sqrt(10)          =   3.1622776   
            sqrt02         sqrt(3.1622776)   =   1.7782793
            sqrt03         sqrt(1.7782793)   =   1.3335213
            sqrt04         sqrt(1.3335213)   =   1.1547819
            sqrt05         sqrt(1.1547819)   =   1.0746077
            sqrt06         sqrt(1.0746077)   =   1.0366328
            sqrt07         sqrt(1.0366328)   =   1.0181516
            sqrt08         sqrt(1.0181516)   =   1.0090349
    
    
      02.   log(10^2^-8) = 2^-8*log(10) = 2^-8
    
            2^-8 is the same as 1/2^8  or 1/256
    
            log(10^2^-8) = log(1.0090349) = 1/256
            from a 4 function calculator    1/256  =  0.0039062
    
    
      03.   assume that log is continous and can be approximated
            by a straight line near 1 in the interval from
            1 to 1.009... and 1.01
            what the formula says below is that the slope of the lhs
            is approximately equal to the slope of the rhs
           
    
            log(x2)  -  log(1)          log(x1)  -  log(1)
            --------------------  =~    ---------------------
                x2   -   1                  x1   -   1
    
    
            log(1.01)  -  log(1)          log(10^2^-8)  -  log(1)
            --------------------    =~    -----------------------
                1.01   -  1                  1.0090349  -  1
    
    
                         x2 - 1
            log(x2)  =~  ------ * log(x1)          since log(1) = 0
                         x1 - 1
    
    
                            1.01 - 1
            log(1.01)  =~  --------------   *  0.0039062     
                            1.0090349 - 1
    
    
                           0.01
            log(1.01)  =~  --------- * 0.0039602  =  0.0043235
                           0.0090349
    
            or in pari/gp terms
            log(1.01)  =~  (1/100)/( 10^2^-8 - 1) * 1/256
    
    
      04.   Here the rules for exponents, and correspondingly
            logarithms are used to find log(2)
    
            log(2^10/10^3)  =  log(1024/1000)  =  log(1.024)
             
            The lhs becomes  10*log(2) - 3*log(10)  and since
            log(10) = 1    is    10*log(2) -3
             
            log(1.024)  =  10*log(2) - 3
    
    
      05.   A linear approximation is applied for log(1.024)
    
    
            log(1.024) - log(1)      log(1.01) - log(1)
            -------------------  =~  -------------------
                1.024  -  1              1.01  -  1
    
    
                             0.024                   0.024 
            log(1.024)  =~  ------- * log(1.01)  =  ------- * 0.0043235
                             0.01                    0.01
    
             
            log(1.024)   =~  2.4 * 0.004325    =    0.0103764
    
            log(1.024)   =   10*log(2) - 3     =    0.0103764
            
            10*log(2)    =   3.013764
    
            log(2)       =   0.3010376
    
    
      06.   Use pari/gp to look at the accuracy of the estimate
    
            log(2)/log(10) - (3 + 2.4*(1/100)/( 10^2^-8 -1) * 1/256)/10
    
            The log base 10 value for log(2) to 7 dec places
    
            0.3010299       versus the above approximation       0.3010376
               good to 5 decimal places.
    
    
    
    
    

       
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