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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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