Infinitesimal and Infinite Numbers

Summary: Euler’s working device throughout Chapter 7 (and most of Book I from this point on): introduce an infinitely small positive number $\omega$ and an infinitely large number $j$, linked by $j\omega =z$ for some finite $z$. Coefficients of the form $(j-m)/(nj)$ with $m,n$ finite are then treated as the algebraic identity $1/n$, since $j-m=j$ when $j$ is infinite. This collapses binomial expansions of $(1+k\omega {)}^j=(1+kz/j{)}^j$ into power series for $a^z$ and $\log (1+x)$.

Sources: chapter7 (§114–§125)

Last updated: 2026-04-26

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The setup

In §114 Euler writes:

Let $\omega$ be an infinitely small number, or a fraction so small that, although not equal to zero, still $a^{\omega }=1+\psi$, where $\psi$ is also an infinitely small number.

and proceeds to set $\psi =k\omega$ — i.e. he treats two infinitely small quantities as commensurable, with a finite ratio $k$ depending on $a$.

The companion device is the infinitely large number. In §115 he raises $a^{\omega }=1+k\omega$ to a power $j$:

$a^{j\omega }=(1+k\omega {)}^j$

and then sets $j=z/\omega$. Since $\omega$ is infinitely small and $z$ is finite, $j=z/\omega$ is infinitely large. The product $j\omega =z$ is a finite, ordinary real number.

So the recipe has three actors:

Symbol	Status	Role
$\omega$	infinitely small ($>0$)	step size
$j$	infinitely large	repetition count
$z=j\omega$	finite	the variable of interest

The whole of Chapter 7 manipulates this triple.

The key algebraic move

When $j$ is infinitely large and $m,n$ are finite,

$\frac{j-m}{nj}=\frac{1}{n}.$

Euler defends this in §116:

Since $j$ is infinitely large, $(j-1)/j=1$, and the larger the number we substitute for $j$, the closer the value of the fraction $(j-1)/j$ comes to 1. Therefore, if $j$ is a number larger than any assignable number, then $(j-1)/j$ is equal to 1. For the same reason $(j-2)/j=1$, $(j-3)/j=1$, and so forth.

This is treated as an equality, not a limit relation. Inside any binomial coefficient, every $(j-m)/j$ collapses to 1 and every $(j-m)/(nj)$ collapses to $1/n$.

Why this collapses series

Apply the binomial theorem (binomial-series) to $(1+kz/j{)}^j$:

$(1+kz/j{)}^j=1+\frac{j}{1}\cdot \frac{kz}{j}+\frac{j(j-1)}{1\cdot 2}\cdot \frac{k^2{z}^2}{j^2}+\frac{j(j-1)(j-2)}{1\cdot 2\cdot 3}\cdot \frac{k^3{z}^3}{j^3}+\cdots$

Group the $j$‘s. The $n$-th term is

$\frac{j(j-1)(j-2)\cdots (j-n+1)}{n!}\cdot \frac{k^n{z}^n}{j^n}=\frac{k^n{z}^n}{n!}\cdot \frac{j}{j}\cdot \frac{j-1}{j}\cdot \frac{j-2}{j}\cdots \frac{j-n+1}{j}.$

By the §116 collapse, every factor $(j-m)/j$ equals 1 for finite $m$. So the $n$-th term reduces to $k^n{z}^n/n!$, and

$a^z={\sum }_{n=0}^{\infty }\frac{(kz{)}^n}{n!}.$

This is the exponential-series. The same collapse, applied to the binomial expansion of $(1+x{)}^{1/j}$, produces the logarithmic-series in §119.

What status does this argument have?

Euler is comfortable treating $\omega$, $j$, and the collapse $(j-m)/j=1$ as legitimate algebraic operations, not as approximations. He distinguishes:

• Infinitely small — smaller than any assignable positive quantity, but not zero. So $\omega \ne 0$ and division by $\omega$ is permitted.
• Infinitely large — larger than any assignable number; the reciprocal of an infinitely small. The product $j\omega$ can be finite, infinite, or infinitely small depending on the relationship between them.
• Finite — an ordinary real number, possibly the product $j\omega$ when $j$ and $\omega$ are inversely commensurable.

These are not the modern $\epsilon$–$\delta$ concepts. They behave as a separate algebraic system, in which the rules "$j-m=j$" and ”$j\omega =z$ is finite when $z$ is finite” are postulates one accepts as part of how the calculus operates. (A rigorous reconstruction is the modern non-standard analysis of Robinson, but Euler does not need it: his series have all the right coefficients, which is what counts.)

A second use: defining $e$ as a limit (§125)

The same machinery represents the exponential as a “power”:

$e^z={\left(1+\frac{z}{j}\right)}^j,a^y={\left(1+\frac{y\log a}{j}\right)}^j$

with $j$ infinitely large (source: chapter7, §125). In modern notation this is the limit definition $e^z={\lim }_{n\to \infty }(1+z/n{)}^n$, but for Euler it is an equality between an infinite-$j$ power and the corresponding series.

Where else this appears

The technique is reused throughout Book I:

• Chapter 7 itself uses it for both $a^z$ (§115) and $\log (1+x)$ (§119), and for the closed-form $e^z=(1+z/j{)}^j$ (§125).
• Chapter 8 will reuse it with imaginary increments to derive the trigonometric series and the formula $e^{ix}=\cos x+i\sin x$.
• Later chapters apply it to angle multiplication, partial fractions of $\sin$ and $\cos$, and the product expansions for $\sin x$ and $\cos x$.

Caveats

• The argument that $\omega$ being infinitely small forces $\psi =a^{\omega }-1$ to be infinitely small (§114) is not proved; Euler appeals to continuity from Chapter 6. In modern terms it is the continuity of $a^z$ at $z=0$.
• The collapse $(j-m)/j=1$ is uniform in the position $m$ within a fixed term, but Euler implicitly applies it across all terms simultaneously — equivalent to swapping a limit with an infinite sum. This is where modern analysis would demand a uniform-convergence justification. The series produced are nonetheless correct.
• Different choices of how $\omega$ and $j$ go to their limits — i.e. different orders or rates — would in general produce different answers in modern analysis. Euler tacitly chooses the rate $j\omega =z$ (constant), which is the choice that produces the analytic series.

Related pages

• exponential-series
• logarithmic-series
• eulers-number
• binomial-series
• chapter-7-on-exponentials-and-logarithms-expressed-through-series