I am looking for a way to relate the numbers in the tables of Napier Mirifici Logarithmorum Canonis Descriptio with the number $e$. For example, as Napier used the number $\left(1-\frac{1}{10^7}\right)$ to calculate the logarithms. His successive calculations might have generated a number close to $\left(1-\frac{1}{10^7}\right)^{10^7}$. Therefore, the tables might have contained a number close to $3,678,794$, which is equal to the integer part of $10^7\times\left(1-\frac{1}{10^7}\right)^{10^7}$, as $\left(1-\frac{1}{10^7}\right)^{10^7}$ gets close to $ \frac{1}{e}$ and Napier multiplies the result by $10^7$.

However, we cannot use this reasoning directly. Is there a way of relating a number in the table with the number $e$? Maybe taking the $\arcsin$ of a number in the table or doing another kind of calculation?

My goal is to make the findings of Napier closer to a modern reader and to clarify the evolution of the ideas that brought the definition of the number $e$.


1 Answer 1


The entries of the tables (as seen, for example, here by following the link to the "spreadsheet tables") give an angle $\theta$ (in degrees and minutes), then $\sin\theta$, then $-\ln\sin\theta$.

If $-\ln\sin\theta = 1$, then $\sin\theta = \frac1e$, and $\theta = \arcsin \frac1e$, which is between $21^\circ 35'$ and $21^\circ 36'$ (closer to the former). Even without anachronistically using a calculator to compute that, we can find this entry by looking through the $-\ln\sin\theta$ column for a value close to $1$. And we find (on page 44 of the PDF) a table with the following fragment:

\begin{array}{cccc} \text{Degree} \\ 21 \\ \\ \text{Minute} & \text{Sine} & \text{Logarithm} & \cdots \\ \vdots & \vdots & \vdots \\ 35 & 3678541 & 10000690 & \cdots \\ 36 & 3681246 & 9993339 & \cdots \\ \vdots & \vdots & \vdots \end{array}

This tells us that the value of $\sin\theta$ which gives $-\ln\sin\theta = 1$ (the value which we eventually identify as $\frac1e$) is between $0.3678541$ and $0.3681246$.

  • $\begingroup$ Beautiful. Following your steps, I interpolated to find sintheta that implies the log equal to 1 (angle 21 35.09'). From the actual table, I used 10000685 and 9993335 for the logs. I got sintheta=3678793/1e7, which implies etil=1/sintheta=2.718283. Very close to a modern approximation of e given by 2.718282. Thanks! $\endgroup$
    – Andre
    Oct 4, 2023 at 20:14

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