Scamuzzig2

2022-10-02

How did Napier rounded his logarithms?

How did Napier round his logarithms? Wikipedia says:

By repeated subtractions Napier calculated $(1-{10}^{-7}{)}^{L}$ 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 ${10}^{7}(1-{10}^{-5}{)}^{L}$ from 1 to 50, and did similarly with $0.9998\approx (1-{10}^{-5}{)}^{20}$ and $0.9\approx 0.99520$

What was his rule for rounding? May one say that he was computing in the ring

$\mathbb{Z}[x]/(10x-1,{x}^{8})$?

How did Napier round his logarithms? Wikipedia says:

By repeated subtractions Napier calculated $(1-{10}^{-7}{)}^{L}$ 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 ${10}^{7}(1-{10}^{-5}{)}^{L}$ from 1 to 50, and did similarly with $0.9998\approx (1-{10}^{-5}{)}^{20}$ and $0.9\approx 0.99520$

What was his rule for rounding? May one say that he was computing in the ring

$\mathbb{Z}[x]/(10x-1,{x}^{8})$?

Derick Ortiz

Beginner2022-10-03Added 11 answers

This is not "Napier's rule" but to round $(1-{10}^{-7}{)}^{L}$ you can use the binomial theorem and only keep the first (or two if $L$ is large) elements in the expansion :

$(1-{10}^{-7}{)}^{L}\approx 1-L\cdot {10}^{-7}+\frac{{L}^{2}}{2}\cdot {10}^{-14}$

Hence only using one term: $(1-{10}^{-7}{)}^{100}\approx 1-100\cdot {10}^{-7}=1-{10}^{-5}=0.99999$

This also explains $(1-{10}^{-5}{)}^{20}\approx 1-20\cdot 0.00001=1-0.0002=0.9998$

Generally, taking only the first term is enough unless $L$ is very large.

Note that what you are doing is calculating $(}(1-{10}^{-7}{)}^{100}{{\textstyle )}}^{20}=(1-{10}^{-7}{)}^{2000}\approx 0.9998$ always using the same rule.

$(1-{10}^{-7}{)}^{L}\approx 1-L\cdot {10}^{-7}+\frac{{L}^{2}}{2}\cdot {10}^{-14}$

Hence only using one term: $(1-{10}^{-7}{)}^{100}\approx 1-100\cdot {10}^{-7}=1-{10}^{-5}=0.99999$

This also explains $(1-{10}^{-5}{)}^{20}\approx 1-20\cdot 0.00001=1-0.0002=0.9998$

Generally, taking only the first term is enough unless $L$ is very large.

Note that what you are doing is calculating $(}(1-{10}^{-7}{)}^{100}{{\textstyle )}}^{20}=(1-{10}^{-7}{)}^{2000}\approx 0.9998$ always using the same rule.

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