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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
> ____________________________________________________________

>
> >
>

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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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