Scamuzzig2

2022-10-02

How did Napier rounded his logarithms?
How did Napier round his logarithms? Wikipedia says:
By repeated subtractions Napier calculated $\left(1-{10}^{-7}{\right)}^{L}$ for $L$ ranging from 1 to 100. The result for $L=100$ is approximately $0.99999=1-{10}^{-5}$. Napier then calculated the products of these numbers with ${10}^{7}\left(1-{10}^{-5}{\right)}^{L}$ from 1 to 50, and did similarly with $0.9998\approx \left(1-{10}^{-5}{\right)}^{20}$ and $0.9\approx 0.99520$
What was his rule for rounding? May one say that he was computing in the ring
$\mathbb{Z}\left[x\right]/\left(10x-1,{x}^{8}\right)$?

### Answer & Explanation

Derick Ortiz

This is not "Napier's rule" but to round $\left(1-{10}^{-7}{\right)}^{L}$ you can use the binomial theorem and only keep the first (or two if $L$ is large) elements in the expansion :
$\left(1-{10}^{-7}{\right)}^{L}\approx 1-L\cdot {10}^{-7}+\frac{{L}^{2}}{2}\cdot {10}^{-14}$
Hence only using one term: $\left(1-{10}^{-7}{\right)}^{100}\approx 1-100\cdot {10}^{-7}=1-{10}^{-5}=0.99999$
This also explains $\left(1-{10}^{-5}{\right)}^{20}\approx 1-20\cdot 0.00001=1-0.0002=0.9998$
Generally, taking only the first term is enough unless $L$ is very large.
Note that what you are doing is calculating $\left(\left(1-{10}^{-7}{\right)}^{100}{\right)}^{20}=\left(1-{10}^{-7}{\right)}^{2000}\approx 0.9998$ always using the same rule.

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