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Napier's original conceptualization of the logarithm was as a relationship between an arithmetic progression and a geometric progression; a point moving with zero acceleration and a point moving with negative acceleration. This is problem II in Book I of Maria Agnesi's Analyical Institutions. Agnesi uses ratios and proportions to describe the problem but shifts to fractions in her example solutions.

Does anybody have a historical reference to what might be called differential ratios? I'm thinking of ratios or proportions with a 'dt' term that were solved with non-arithmetic reasoning involving for example the mediant.

Attempts to find inchoate logarithms before Napier, for example in the work of Wallis, have as far as I know been fully discounted but the link between Napier's thinking and Agnesi's problem is so obvious one imagines that a ratio formulation of the logarithm might have existed.

Thanks for comments and insight.

Cheers, Scott

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As this is not a mathematical question, I suggest to wikify it. – Wadim Zudilin Jun 30 '10 at 14:44
@Wadim: I don't think that is a reason to wikify. If the question is off topic (I don't), it should be closed, not wikified. If the question is likely to generate a big list of answered, it should be wikified, but I don't think that's the case. There is a specific question in there, so an answer can be given. The vague-looking final paragraph notwithstanding, I think the question is fine as is. – Harald Hanche-Olsen Jun 30 '10 at 15:46
All right, Harald. "MathOverflow's primary goal is for users to ask and answer research level math questions." As I don't count this as a math question, I vote to close. – Wadim Zudilin Jun 30 '10 at 23:41
I started a meta discussion:… – Harald Hanche-Olsen Jul 1 '10 at 0:17
up vote 1 down vote accepted

The most famous historical reference to differential ratios that I know of is Euler's Institutiones calculi differentialis vol. 1, caput 3. Starting on p. 64 you can find a lot of differential ratios like

$\frac{dx + dx^2}{dx} = 1$


$a\sqrt{dx} + bdx = a\sqrt{dx}$.

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Thanks much, Rhett. – Scott Guthery Mar 25 '13 at 13:08
@Scott You are welcome Scott. It seems that we both have a strong interest in history. – Rhett Butler Mar 25 '13 at 15:33

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