# NavList:

## A Community Devoted to the Preservation and Practice of Celestial Navigation and Other Methods of Traditional Wayfinding

**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 Wilsonwrote: > > 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. > > > > --~--~---------~--~----~------------~-------~--~----~ Navigation List archive: www.fer3.com/arc To post, email NavList@fer3.com To unsubscribe, email NavList-unsubscribe@fer3.com -~----------~----~----~----~------~----~------~--~--- 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.