(i) Area AFDB = `int_0^h 1/{1+x} = int_0^h 1 - x + x^2 - x^3 + ... = h - h^2/2 + h^3/3 - h^4/4 + ....`
(i) Area AFdb = `int_0^h 1/{1-x} = int_0^h 1 + x + x^2 + x^3 + ... = h + h^2/2 + h^3/3 + h^4/4 ....+ h^k/k + ... `
These allow the estimation of the sum and difference of the two areas:
Total Area dbDB = ` 2h + 2h^3/3 + 2 h^5/5 + 2 h^7/7 + ....`
Now Newton uses the first eight terms with h = .1 (and .2) to estimate the hyperbolic log of .9, 1.1 (.8 , and 1.2).
h | 0.1 | .2 | .01 |
2h | 0.2 | 0.4 | 0.02 |
2h3/3 | 0.000666666666666 | 0.00533333333333 | 0.00000066666666667 |
2h5/5 | 0.000004 | 0.000128 | 0.00000000004 |
2h7/7 | 0.0000000285714286 | 0.00000365714285714 | 0.00000000000000286 |
2h9/9 | 0.0000000002222222 | 0.00000011377777778 | 2.22222222222e-19 |
2h11/11 | 0.0000000000018182 | 0.00000000372363636 | |
2h13/13 | 0.0000000000000154 | 0.00000000012603077 | |
2h15/15 | 0.0000000000000001 | 0.00000000000436907 | |
Sum of Areas | 0.200670695462151 | 0.405465108108002 | 0.020000666706670 |
h | 0.1 | .2 | .01 |
h2 | 0.01 | 0.04 | 0.0001 |
h4/2 | 0.00005 | 0.0008 | 0.000000005 |
h6/3 | 0.0000003333333333 | 0.00002133333333 | 0.000000000000333 |
h8/4 | 0.0000000025 | 0.00000064 | 2.50000000000e-17 |
h10/5 | 0.00000000002 | 0.00000002048 | |
h12/6 | 0.0000000000001667 | 0.00000000068267 | |
h14/7 | 0.0000000000000014 | 0.00000000002341 | |
Diff'ce of Areas | 0.010050335853501 | 0.040821994519406 | 0.000100005000333 |
Thus ln(1.1)= 1/2 ( 0.2006706954621511- 0.0100503358535014)
= 0.0953101798043248
while ln(.9) = -(1/2)( 0.2006706954621511 + 0.0100503358535014)
= -0.105360516578263 .
And ln(1.2)= 1/2 (0.405465108108002 -0.040821994519406 )
= 0.18232155576939546 (from Newton)
while ln(.8) = -(1/2)(0.405465108108002 +0.040821994519406 )
= -0.2231435513142097 (from Newton) .
Newton can also find ln(1.01) and ln(.99) by
using
h=.01
from the same tables.
Then Newton finds other logarithms with great accuracy using these.
For Example:
`ln(2) = ln ( 1.44/.72) = ln( 1.2/0.8 1.2/.9) = 2 ln(1.2) - [ln(.9) + ln(.8)] ~~ 2(0.18232155576939546) +0.105360516578263+ 0.2231435513142097`