I have been trying to get my head around how Roger Cotes first discovered Euler Formula.

I knew how Euler did it, but I wanted a new perspective, especially from someone who discovered it earlier.

Cotes used geometry to explain his discovery, but it was very terse (or at least to my preference). I tried to draw the diagram after his explanation, but I still can't understand how he suddenly conclude that $\ln{\frac{EX+XC\sqrt{-1}}{CE}}=\text{arc length}\cdot\sqrt{-1}$

He used many old keywords such as "measurement of ratio" to represent natural log. How is natural log a "measurement of ratio", and why does taking "measurement of ratio" of $\cos{\theta}+i\sin{\theta}$ equate to arc length of circle intuitively?

  • 1
    $\begingroup$ For equidistant points on a logarithmic scale, neighboring points are in identical ratio to each other. $\endgroup$
    – njuffa
    Commented May 2, 2021 at 23:14
  • $\begingroup$ You may find the my answer to HSM question 13004 "Why did John Napier use logos 'ratio', to coin 'logarithm'?" helpful. $\endgroup$
    – Somos
    Commented May 7, 2021 at 0:24

3 Answers 3


This answer is now mostly complete, and it's really long. I've made substantial changes since originally posting, but now it's pretty much in its final state. Finding the right interpretation of Cotes' work proved much more difficult, and required a lot more translation, than I had expected! A translation exists in Roger Cotes - Natural Philosopher by Cook and Gowing, but I didn't have access to it. I might continue the translation, in which case I'll post a link to it later.

My reference is:

  • Cotes, Roger. Logometria. Philosophical transactions [of the Royal Society of London], N. 338, vol. 29 (1714), pp. 31-32. Link.

This proof is pretty involved, mostly because of the large "knowledge debt" from how the practice of mathematics has changed since Cotes' time. He was working with the surface area of an ellipsoid of revolution. For that part, he includes two diagrams, and they are reproduced with the text.

Background: Cotes' Definition of Measure of a Ratio and its Modulus

I translated some passages from the beginning of Logometria that might be relevant. These are just rough translations, maybe I'll revisit later.

In this article we treat Measures of Ratios. These Measures are quantities of any sort, whose magnitudes are proportional with magnitudes of ratios. Thus in a given System, the measure of the same ratio is the same, of a doubled ratio is double, of a tripled ratio is triple, of a halved ratio is half, of a sesquiplicate ratio [in ratio of 3:2] is sesquialtera: hence whatever ratios are increased or diminished by composition or resolution, the measure is likewise increased or diminished. A ratio of equals has no magnitude, for addition or detraction do not lead to change; ratios which are called major and minor of inequality have opposite affections of their magnitudes, since they always make opposites in composition and resolution: therefore if we imagine the rational measure having a major term to the minor as positive, the rational measure having the minor term to the major will be negative, the rational measure between equal terms will of course be of zero magnitude. Further, different Systems of measures arise, just as the determined proportion is expressed by different ways and that it is unalterable between rational magnitudes. Hence it is clear, that the Systems can be expressed by an infinite number, by decreasing or increasing all measures of whatever given System in the same given proportion, or [otherwise for quantities by applying measures of different kinds]. In such varieties however some confusion necessarily arises, unless it is well agreed to what System of reference each of the measures which the conversation touches upon is established. One can best prepare a remedy for this issue if the given measure of some ratio, which is considered most suitable, is taken as the Modulus against which in all Systems the measures of remaining ratios are continuously weighed. This being done, immediately all Systems will be determined from this Modulus : as from these measures it will consist of those which will be homogeneous to the Modulus, and which will have greater or lesser magnitudes than it, the larger or smaller it is, such that the invariance of the magnitudes of ratios to be measured is preserved by the ratio between the same measures. It will therefore be clear in the following that a certain ratio is given, between the double and triple ratios, but approaching nearer to the triple, which by our proposition may be judged most appropriate not without reason, since the very nature of the thing suggests its use and seems to insist upon it in a way by no uncertain indication. This I will call the Modular Ratio, so named by virtue of its application; the way, however, that it may be more accurately defined, will be shown further below, as now a word on Logarithms must be added.

Logarithms are Numerical measures of ratios: they are typically so distributed in a Canon, so that each number increasing by the natural order, and set in continuous series, is ascribed to the Logarithm; in fact the use of the number itself is not commonly observed, but rather the use of the ratio which the number has to Unity. From this it is easy to find the Logarithm of a ratio denoted by whatever terms.

I had a difficult time understanding what Cotes meant here, although certain things stand out (e.g. the "certain ratio, between double and triple ratios, approaching nearer to triple" is $e$, and hence it gives the natural log when used as the base of a logarithm). But from his Proposition 1, the subsequent Corollaries, and Scholium 1, I was able to verify that:

  • The measure of a ratio, say the ratio $p:q$, is its logarithm, $\log \frac{p}{q}$, to some unspecified base which must be agreed upon. For a logarithm, he says that it is usually calculated using $y=\frac{p}{q}$, and calculating $\log \frac{y}{1}$, i.e. $\log y$. His actual definition is more abstract: "Measures of ratio are quantities of any sort, whose magnitudes are analogous to the magnitudes of ratios. In a given system, the measure of the same ratio is the same, etc..." But later the definition becomes more clear and is calculated explicitly (Schol. 1).
  • The modulus of the measure is related to the logarithm base. More properly, the modulus is the measure of ratio of a particular reference ratio, so that (I suppose) the reference measure is 1 for that ratio. This should be more clear with an example. Suppose we want to know the measure of ratio of quantities $x,y,z,...$, i.e. the values $\log x, \log y, \log z,...$ with no base specified. If we have calculated these in base $a$, then it follows that the measure of ratio $a$ should be 1, i.e. $\log_a a$ as expected. Hence all other measures of ratio are determined by their relationship to the modulus $a$.
  • In Proposition 1, Cotes calls $M$ the modulus of the system. This seems to conflict with what I'm saying, since it's actually the inverse of the measure of ratio, and referred to another measure of ratio ($e$). So there's probably still a misunderstanding in my description above!
  • Calculating measures of ratios in a different base $b$ can always be done with $$ \log_a x = M \log_b x $$ where of course $M=\frac{1}{\log_b a}$ (from the change of base formula). Similarly, every other value, e.g. $\log_a y,\log_a z$, etc, is obtained in base $b$ by multiplying by $M$. If $a$ is the modular ratio (later revealed to be the base $e$) which is readily calculated, then logs to every other base can be found with a suitable modulus $M$ in this way. This is the way log tables are used.
  • Practically speaking, if we wish to find the measure of ratio of $x$ to some base, i.e. $\log_a x$ with base $a$, then we can simply calculate $\ln x$, and multiply it by $M$, this value being given by $$ M=\frac{1}{\ln a}. $$

Surface Area of an Ellipsoid of Revolution Part 1.

I've translated the details from pp.31-32:

enter image description here

...To this may be added the surfaces generated by an Ellipse. Let $ANB$ be an Ellipse described about the center $C$, with vertices $A$ and $B$, focus $F$, principal semiaxis $CB$, conjugate semiaxis $CA$; and to some point $X$ on the axis $CA$ let $XN$ be the ordinate, which the Ellipse crosses at $N$. In the axis $CB$, is taken $CE$ to $CA$ as $CA$ is to $CF$, and $EX$ is joined. Then $KL$ is set such that it be to $XC$ as $XE$ is to $CE$, and $LM$ that which is the measure of ratio between $EX+XC$ and $CE$ to the Modulus $CE$: and the surface generated from the arc $BN$ rotating about the axis $CX$, will be to the semidiameter $CB$ described, as of the lines $KL$ and $LM$ combined as $KM$, is to the semidiameter $CB$. For this latter construction to exist, it is necessary that the semiaxis $CA$ around which the revolution was made, be lesser than the other semiaxis $CB$; otherwise the quantity $\frac{CAq}{\sqrt{ CBq-CAq }}$ of the modulus $CE$ becomes impossible, and the construction by Logometry (which in these cases often arises) turns itself into Trigonometry, as follows. enter image description here

Let $ANB$ be an ellipse described about the center $C$, with vertices $A$ & $B$, focus $F$, principal semiaxis $CA$, conjugate semiaxis $CB$; and to some point $X$ on the axis $CA$ let $XN$ be the ordinate, which the Ellipse crosses at $N$. The line $CE$ is inscribed from the angle $CXN$, such that it is to $CA$ as $CA$ is to $CF$. Then let $KL$ be taken such that it is to $XC$ as $XE$ is to $CE$, and let $LM$ be the measure of the angle $XEC$ to the Modulus $CE$, i.e. the measure of that equal to the arc whose sine is $XC$, to the radius $CE$: and the surface generated from the arc $BN$ by revolving around the axis $CX$, will be to the semidiameter $CB$ of the circle described, as that of the lines $KL$ & $LM$ combined as $KM$ are, to the semidiameter $CB$. One can obtain the area of this surface by Logometry, but only by the inexplicable [imaginary numbers]. Since if any arc of the quadrant of a circle, described with radius $CE$, has the sine $CX$ and the complementary sine to the quadrant $XE$: taking the radius $CE$ as the Modulus, the arc will be a measure of the ratio between $EX+CX\sqrt{-1}$ & $CE$ multiplied by $\sqrt{-1}$. Here I leave the more diligent examinations to others, who would find the work valuable.

Such is the text leading to this result. The first part of this describes the problem of finding the surface area of an ellipsoid of revolution, which leads to the latter part, the crux of the problem, describing the sum of cosine and the product of the sine and the imaginary unit. This is what we're interested in, but we'll need to understand the context.

Surface Area Part 2 (Cajori)

How Cajori describes the above in History of the Exponential and Logarithmic Concepts, Amer. Math. Monthly v20 no. 2 (Feb 1913), pp. 35-47, as linked in the other answer, is as follows:

Let $a$ be the horizontal radius, $b$ be the vertical radius, $\sqrt{a^2-b^2}$ be the horizontal distance from the center to the focus; if $\theta$ is the angle $XEC$, then using $l=CE$, the value of $y$ is $y=l\sin\theta$, and the surface area of the revolved arc of the ellipse $b^2 x^2 + a^2 y^2 = a^2 b^2$ up to height $y$ becomes $$ \begin{align} S&= a\pi \left\{ y\sqrt{ 1+\frac{a^{2}-b^{2}}{b^4} y^2 } + \frac{b^{2}}{\sqrt{ a^{2}-b^{2} }} \log \left( y\sqrt{ \frac{a^{2}-b^{2}}{b^4} }+\sqrt{ 1+\frac{a^2-b^{2}}{b^4}y^{2} } \right) \right\} , \\ S&=a\pi \left\{ y\sqrt{ 1+\frac{a^{2}-b^{2}}{b^4}y^{2} }+\frac{b^{2}}{\sqrt{ b^{2}-a^{2} }}\varphi \right\}, \end{align} $$ where $\sin \theta = y\sqrt{ \frac{b^{2}-a^{2}}{b^4} }=iy\sqrt{ \frac{a^{2}-b^{2}}{b^4} }$, $\cos \theta = \sqrt{ 1+\frac{a^{2}-b^{2}}{b^4}y^2 }$. The imaginary unit appears when flipping the sign under the root. Substituting these in, the surface area becomes $$ \begin{align} S=&a\pi \left( \cos \theta + \frac{b^{2}}{f} \log \left( i\sin \theta +\cos \theta \right) \right) \\ S= & a\pi \left( \cos \theta + \frac{b^{2}}{if}\theta \right) \end{align} $$ and comparing like terms, $$ \log(i\sin \theta+\cos \theta)=-i \theta. $$ Note that Cajori was missing the negative sign on the right, but Cotes had it, as seen from his having $i$ on the left side. There's evidently a mistake here somewhere, because the correct result would not have a negative on the right-hand side.

Surface Area Part 3 (adapted from Cotes)

Let's go through Cotes' work, adapting it to modern notation. Cotes uses his own propositions and corollaries to draw conclusions with strictly geometric arguments, something we don't have the luxury of doing, so it will probably seem complex compared to his short passage above. But the development is the same, and here it is self-contained.

Consider an ellipse, where $a$ is the horizontal radius, $b$ is the vertical radius. Thus $a=CB,b=CA$. Let the angle $\theta=\angle XEC$. This all remains true in the two following cases:

  1. $a>b$, so $CB=a$ is the principal axis, $CA=b$ is the conjugate
  2. $a<b$, so $CA=b$ is the principal axis, $CB=a$ is the conjugate

enter image description here

Case 1. The drawing above shows the first case. We'll consider this one to begin. Let $x=CE$ be the distance from center to $E$, $y=CX$ be the height of a section of the ellipse, $f=CF$ be the horizontal distance from center to focus, $l=XE$ be the hypotenuse of a right triangle $XEC$, $k_{1}=KL$ be the first auxiliary length, $k_{2}=LM$ be the second. Let $S$ be the surface area of the ellipsoid of revolution up to height $y$ in the drawing above. Cotes gives the following results: $$ \begin{align} \text{I. }& CE:CA :: CA:CF\\ \text{II. }& KL:XC :: XE:CE\\ \text{III. }& LM = CE\ln \left( \frac{EX+XC}{CE} \right) \\ \text{IV. }& S:CB :: KM:CB \end{align} $$ Where $KM=KL+LM$. With the variables given previously, and in modern notation, these are $$ \begin{align} \text{I. }& \frac{x}{b}=\frac{b}{f}\\ \text{II. }& \frac{k_{1}}{y}=\frac{l}{x}\\ \text{III. }& k_{2} = x\ln \left( \frac{l+y}{x} \right) \\ \text{IV. }& \frac{S}{a} = \frac{k_{1}+k_{2}}{a}. \end{align} $$ We can deduce that $$ \begin{align} x=& \frac{b^{2}}{f}, \\ k_{1} = & \frac{yl}{x}, \\ f =& \sqrt{ a^{2}-b^{2} } \end{align} $$ and finally the surface area of the section of the ellipsoid of revolution would be $$ S=k_{1}+k_{2}=\frac{yl}{x}+x\log \left( \frac{l}{x}+\frac{y}{x} \right) . $$ The variables $l,x,y$ are legs of the right triangle, and taking $y$ as given, without using trigonometry, we get the others, $$ \begin{align} x= & \frac{b^{2}}{\sqrt{ a^{2}-b^{2} }}, \\[1em] l= & \sqrt{ x^{2}+y^{2} }. \end{align} $$ Note, however, the denominator of $x$ (the focus) becomes imaginary if $b>a$. This is where imaginary numbers are introduced into the problem. We'll consider that case on its own later.

We can obtain Cajori's expression for $S$ by first rearranging the expression for $k_{1}$ as $$ k_{1}=\frac{yl}{x}=\frac{y}{x}\sqrt{ x^{2}+y^{2} }=y\sqrt{ 1+\frac{y^{2}}{x^{2}} }, $$ then substituting $x=\frac{b^{2}}{f}$, and $f=\sqrt{ a^{2}-b^{2} }$ $$ k_{1}=y\sqrt{ 1+\frac{f^{2}}{b^{4}}y^{2} }=y\sqrt{ 1+\frac{a^{2}-b^{2}}{b^{4}}y^{2} }. $$ Next, we rearrange the expression for $k_{2}$ as follows, using a similar expansion: $$ k_{2}=x\ln \left( \frac{y}{x}+\frac{l}{x} \right) =x\ln \left( \frac{y}{x}+ \sqrt{ 1+\frac{a^{2}-b^{2}}{b^{4}}y^{2} } \right) $$ And now comes a tricky bit. To get Cajori's expression, we make the following substitution for $x$: $$ x=\frac{b^{2}}{\sqrt{ a^{2}-b^{2} }}=\sqrt{ \frac{b^{4}}{a^{2}-b^{2}} } $$ But this is only valid if $a^{2}-b^{2}>0$. This sort of manipulation does not hold when the denominator is negative. This being noted, and substituting $x=\frac{b^{2}}{\sqrt{ a^{2}-b^{2} }}$ out front, we can see $$ k_{2}= \frac{b^{2}}{\sqrt{ a^{2}-b^{2} }} \ln \left( y \sqrt{ \frac{a^{2}-b^{2}}{b^{4}} }+\sqrt{ 1+\frac{a^{2}-b^{2}}{b^{4}}y^{2} } \right) , $$ and hence we can obtain for the surface area $S$ the expression $$ S=k_{1}+k_{2}=y\sqrt{ 1+\frac{a^{2}-b^{2}}{b^{4}}y^{2} } + \frac{b^{2}}{\sqrt{ a^{2}-b^{2} }} \ln \left( y \sqrt{ \frac{a^{2}-b^{2}}{b^{4}} }+\sqrt{ 1+\frac{a^{2}-b^{2}}{b^{4}}y^{2} } \right) $$ which is the same as Cajori's expression, but missing the coefficient $a \pi$ (from integration?).

Case 2. For case 2, many quantities are the same, but some are notably different; we'll prime these, as $x'$. Here's the drawing for the second case:

enter image description here

Note that the nature of the triangle $XEC$ (and the angle $\angle XEC$) has changed compared to case 1 - now $CE$ is the hypotenuse, and $XC, XE$ are the legs. Let $f'=CF$ be the vertical distance from center to focus, $y=CX$ is the sine of angle $\angle XEC$ as in case 1, $x'=EX$ is the cosine of $\angle XEC$ (compare with case 1), and $l'=CE$ is the hypotenuse of $\angle XEC$. We'll take $\theta=\angle XEC$ to be in radians, so that $$ \begin{align} x'= & XE=l'\cos \theta, \\ y= & XC=l'\sin \theta. \end{align} $$ As in case 1, Cotes gives the following: $$ \begin{align} \text{i. }& CE:CA :: CA:CF\\ \text{ii. }& KL:XC::XE:CE\\ \text{iii. }& LM=\text{arc length of }\angle XEC\text{ with radius CE}\\ \text{iv. }& S:CB::KM:CB \end{align} $$ The only difference from before is the third result. Now similar to before, but with different meanings for $x'$ and $l'$, we have $$ \begin{align} \text{i. }& \frac{l'}{b}=\frac{b}{f'}\\ \text{ii. }& \frac{k_{1}}{y}=\frac{x'}{l'}\\ \text{iii. }& k_{2} = l' \theta \\ \text{iv. }& \frac{S}{a} = \frac{k_{1}+k_{2}}{a}. \end{align} $$ Thus we can obtain $$ \begin{align} f'= & \sqrt{ b^{2}-a^{2} } \\ l'= & \frac{b^{2}}{f'} \\[.5em] k_{1}= & \frac{yx'}{l'} \\[.5em] S= & k_{1}+k_{2}; \end{align} $$ where we flipped the sign of $f'$ because it is now the focus in the vertical direction. Again we repeat Cajori's development by first rearranging $k_{1}$. In fact, by definition of $\cos \theta$, we have $$ k_{1}=y \frac{x'}{l'} = y\cos \theta; $$ but we can use the fact that $x'^{2}=l'^{2}-y^{2}$ to also obtain $$ \frac{x'}{l'} = \frac{\sqrt{ l'^{2}-y^{2} }}{l'}=\sqrt{ 1-\frac{y^{2}}{l'^{2}} } $$ then substituting $l'=\frac{b^{2}}{f'}$, and $f'=b^{2}-a^{2}$, $$ \frac{x'}{l'}=\sqrt{ 1- \frac{f'^{2}}{b^{4}} y^{2} }=\sqrt{ 1- \frac{b^{2}-a^{2}}{b^{4}}y^{2} }=\sqrt{ 1+\frac{a^{2}-b^{2}}{b^{4}}y^{2} }. $$ Hence $$ k_{1}=y \frac{x'}{l'} = y\sqrt{ 1+\frac{a^{2}-b^{2}}{b^{4}}y^{2} } $$ which is identical to case 1.

Rearranging $k_{2}$ (arc length $l'\ \theta$ for $\theta$ in radians) simply requires substituting for $l'$, $$ k_{2}=l'\ \theta = \frac{b^{2}}{\sqrt{ b^{2}-a^{2}}} \theta=-i \frac{b^{2}}{\sqrt{ a^{2}-b^{2} }} \theta. $$ Note. This is where the sign discrepancy appears. Square roots of negative real numbers are not well-behaved, so we can't necessarily use normal square root algebra tricks and get the correct result. I'm not 100% on the above expression, yet.

Finally we get the full expression for the surface area of the ellipsoid, $$ S=k_{1}+k_{2} = y\sqrt{ 1+\frac{a^{2}-b^{2}}{b^{4}}y^{2} }-i \frac{b^{2}}{\sqrt{ a^{2}-b^{2} }} \theta. $$

Final Result. Throughout the development above, we obtained two expressions for the surface area of an ellipsoid, which must, of course, agree. They are: $$ \begin{align} S = &\ y\sqrt{ 1+\frac{a^{2}-b^{2}}{b^{4}}y^{2} }-i \frac{b^{2}}{\sqrt{ a^{2}-b^{2} }} \theta \\ S= &\ y\sqrt{ 1+\frac{a^{2}-b^{2}}{b^{4}}y^{2} } + \frac{b^{2}}{\sqrt{ a^{2}-b^{2} }} \ln \left( y \sqrt{ \frac{a^{2}-b^{2}}{b^{4}} }+\sqrt{ 1+\frac{a^{2}-b^{2}}{b^{4}}y^{2} } \right) \end{align} $$ These are the same as Cajori gives. Subtracting these expressions, we must have $$ -i \theta=\ln \left( y \sqrt{ \frac{a^{2}-b^{2}}{b^{4}} }+\sqrt{ 1+\frac{a^{2}-b^{2}}{b^{4}}y^{2} } \right). $$ But we can also show that $\theta$ obeys $$ \sin \theta = y\sqrt{ \frac{b^{2}-a^{2}}{b^4} }=iy\sqrt{ \frac{a^{2}-b^{2}}{b^4} }, $$ and $$ \cos \theta = \sqrt{ 1+\frac{a^{2}-b^{2}}{b^4}y^2 }, $$ and these expressions only depend on quantities shared between the two cases; so that we must have $$ \begin{align} -i\theta = \ln(\cos \theta&+i\sin \theta) \\ &\textit{or} \\ \theta= i\ln(\cos \theta&+i\sin \theta) \end{align} $$ with the latter expression being what Cotes himself wrote. Of course, there is a sign error here, and I would guess it's from the manipulation of square roots and negative real numbers, but hopefully someone can find where this actually happens.


An explanation How logarithm (hence e) took place in this subject.


This derivation is nice as well. It uses the same manipulation as mentioned in the above answer: Flipping the sign to use imaginary number.


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    $\begingroup$ While this link may answer the question, it is better to include the essential parts of the answer here and provide the link for reference. Link-only answers can become invalid if the linked page changes. - From Review $\endgroup$ Commented Nov 30, 2022 at 1:50

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