Yes, My Method Can Close In on Logarithms

Assorted retorts from yahoo boards and elsewhere : Medieval Quadrivium and Logarithms (quora) · Φιλολoγικά/Philologica : Expressing Logarithms in Points of Royal Feet · Previous Work on Logarithms · Yes, My Method Can Close In on Logarithms

My previous work has in a big extent been done on paper, here I did use paper to arrive at 10^(5/17) and 10^(7/23) as approximations, and from here on, I use an online calculator.

10^(5/17) = 1.968 419 447 29

10^(7/23) = 2.015 337 685 94

But why 10^(5/17) and 10^(7/23) in the first place? Well, they mean, 10^5 is approximately 2^17, 10^7 is approximately 2^23.

Check : 10^5 = 100 000, 2^17 = 131 072
131 072 is closer to 100 000 than its half 65 536 is.

Check : 10^7 = 10 000 000, 2^23 = 8 388 608
8 388 608 is closer to 10 000 000 than its double 16 777 216 is.

And 10^(5/17) ~ 2 is just a rewriting of 10^5 ~ 2^17.

Now, as seen above, 10^(5/17) and 10^(7/23) are on opposite directions in their deviation from two. This means, their middle ought to deviate less from two. But the denominators must be united, we multiply 17*23 and get 391, and we multiply both numerator and denominator with 17 in the case of the original denominator 23 and vice versa. If 10^(5/17) = 10^(115/391) and 10^(7/23) = 10^(119/391), we take the scale from 115/391 to 119/391:

10^(115/391) = 10^(5/17) = 1.968 419 447 29
10^(116/391) = 1.980 045 598 8
10^(117/391) = 1.991 740 418 29
10^(118/391) = 2.003 504 311 35
10^(119/391) = 10^(7/23) = 2.015 337 685 94

The two relevant ones were 10^(117/391) and 10^(118/391), we add 117+118 to get 235, and this can so neatly get doubled into 470, we have now gone to four times the original mean numerator (117.5), so we need to go to four times the original denominator, 391 * 4 = 1564, as the below denominator will be. From 470/1564, we get away one step on each side, to get a scale 469/1564 to 471/1564. Let's see where it gets us:

10^(469/1564) = 1.994 674 900 01
10^(470/1564) = 1.997 613 705 18
10^(471/1564) = 2.000 556 840 17

The relevant ones are 10^(470/1564) and 10^(471/1564), we add up the numerators and double the denominator:

10^(941/3128) = 1.999 084 731 05

We double that and then scale around it, until we get past the 2:

[10^(1881/6256) ] = unnecessary, since even next one was below 2, see next
10^(1882/6256) = 10^(941/3128) = 1.999 084 731 05
10^(1883/6256) = 1.999 820 650 15
10^(1884/6256) = 10^(471/1564) = 2.000 556 840 17

No, it couldn't stay at 10^(1883/6256), we had to go on to 10^(1884/6256) ...

(1883+1884)*2 = 7534
6256*4 = 25 024

After I set below scale, I start getting the calculations from the top, the latter two are unnecessary:

10^(7532/25 024) = 10^(1883/6256) =1.999 820 650 15
10^(7533/25 024) = 2.00000467226
[10^(7534/25 024)]
[10^(7535/25 024)]

(7532+7533)*2 = 30 130
25 024 * 4 = 100 096

10^(30 129/100 096) = 1.999 866 654 09
10^(30 130/100 096) = 1.999 912 659 09
10^(30 131/100 096) = 1.999 958 665 14
10^(30 132/100 096) = 2.000 004 672 26

By now, we are pretty close on the actual logarithm of two, either fraction, so let's see the decimal fractions worked out, and then let's see the decimals for the logarithm:

30 131/100 096 = 0.301 021 019 82
30 132/100 096 = 0.301 031 010 23

(10)log(2) = 0.301 029 995 66

So, yeah, the method can get values for logarithms, and the point in this is not my competing in any way with others who have calculated the logarithms long before me, it's proving that my definition of logarithms is a good definition. You cannot actually get any number raised to an actual irrational power, and when you raise a number to a fraction, this is algebraic reshuffling for raising both numbers to powers that are actual numbers. Logarithms don't prove that irrational powers exist, and therefore irrational numbers exist, the logarithms, they are the irrational limes for converging fractions, and fractional powers are algebraic shortcuts.

Hans Georg Lundahl
Paris
Day after Annunciation
26.III.2022