Permalink Submitted by Peter L. Griffiths on July 23, 2016

Napier recognised that the formula for the hyperbola 1/(1+x) = 1-x +x^2 - x^3....
could be converted into measuring the area under the hyperbola by integration, similar
to the conversion of the circumference of a circle 2PIr intothe area of the circle PIr^2.
S(1/1+x) = x - x^2/2 + x^3/3 - x^4/4.....
Multiplying both sides by 1/2 gives
1/2S(1/1+x) = x/2 - x^2/4 + x^3/6 -x^3/8.....
Napier very cleverly recognised that S(1/1+x) could be represented log(1 + x),
so that 1/2S(1/[1+x]) could be represented as 1/2 log(1 +x) or log (1 +x)^1/2.
In this way with x = 1, log2 = 1 - 1/2 + 1/3 - 1/4....= 0.693148
and log 2^1/2 = 1/2 - 1/4 + 1/6 -1/8.......= 0.346574.
Napier mentions amounts close to 0.693148 and 0.346574 in paragraphs 46-53
of the Constructio.

## Napier's Hyperbolic Logarithms

Napier recognised that the formula for the hyperbola 1/(1+x) = 1-x +x^2 - x^3....

could be converted into measuring the area under the hyperbola by integration, similar

to the conversion of the circumference of a circle 2PIr intothe area of the circle PIr^2.

S(1/1+x) = x - x^2/2 + x^3/3 - x^4/4.....

Multiplying both sides by 1/2 gives

1/2S(1/1+x) = x/2 - x^2/4 + x^3/6 -x^3/8.....

Napier very cleverly recognised that S(1/1+x) could be represented log(1 + x),

so that 1/2S(1/[1+x]) could be represented as 1/2 log(1 +x) or log (1 +x)^1/2.

In this way with x = 1, log2 = 1 - 1/2 + 1/3 - 1/4....= 0.693148

and log 2^1/2 = 1/2 - 1/4 + 1/6 -1/8.......= 0.346574.

Napier mentions amounts close to 0.693148 and 0.346574 in paragraphs 46-53

of the Constructio.