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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 $10^7(1 − 10^{−5})^L$ for $L$ 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 $\mathbb{Z}[x]/(10x-1, x^8)$?

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  • $\begingroup$ That ring is the trivial ring. Specifically, $10x-1$ generates $10^8x^8-1$, which, together with $x^8$, generates $1$. $\endgroup$ – Dustan Levenstein May 18 '14 at 17:55
  • $\begingroup$ Or better yet, $x$ is a unit by $10x=1$, and therefore so is $x^8$. $\endgroup$ – Dustan Levenstein May 18 '14 at 17:56
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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 :

$$(1 - 10^{-7})^L \approx 1 - L\cdot10^{-7} + \frac{L^2}{2}\cdot 10^{-14}$$ 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 $\Big((1 - 10^{-7})^{100}\Big)^{20} = (1 - 10^{-7})^{2000} \approx 0.9998$ always using the same rule.

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