Before embarking on the detailed calculation of his logarithms, according to paragraphs 46-53 of the Constructio,
Napier seemed to be able to apply the hyperbola and its expansion y = 1/(1+x) = 1 -x + x^2 - x^3 ....also
its integration for computing the area S1/(1+x) = x - x^2/2 + x^3/3 ...Napier could have obtained this integration idea from
the relationship between Pir^2 and 2Pir. If 1 +x =2 we have the formula for log2 from the integrated hyperbola
1 - 1/2 + 1/3 -1/4 ..... = log 2= 0.69. The log symbol seems to be negative integration.
Before embarking on the detailed calculation of his logarithms, according to paragraphs 46-53 of the Constructio,
Napier seemed to be able to apply the hyperbola and its expansion y = 1/(1+x) = 1 -x + x^2 - x^3 ....also
its integration for computing the area S1/(1+x) = x - x^2/2 + x^3/3 ...Napier could have obtained this integration idea from
the relationship between Pir^2 and 2Pir. If 1 +x =2 we have the formula for log2 from the integrated hyperbola
1 - 1/2 + 1/3 -1/4 ..... = log 2= 0.69. The log symbol seems to be negative integration.