chapter-7-on-exponentials-and-logarithms-expressed-through-series

Chapter 7: Exponentials and Logarithms Expressed through Series

Summary: Euler returns to $a^z$ and $\log y$ from Chapter 6 and now develops them analytically. Using a single heuristic — let $\omega$ be infinitely small and $j=z/\omega$ infinitely large, so $a^z=(1+k\omega {)}^j=(1+kz/j{)}^j$ — he produces the exponential-series $a^z=\sum (kz{)}^n/n!$ and the logarithmic-series $\log (1+x)=(1/k)(x-x^2/2+\cdots )$. Choosing the base so that $k=1$ defines [[eulers-number|$e$]]; the resulting natural-logarithm system is the simplest, and $k={\log }_{e}a$ is the modern conversion factor of change-of-base.

Sources: chapter7

Last updated: 2026-04-26

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Overview

Chapter 6 set the stage: $a^z$ exists, its inverse $\log y$ exists, and most of their values are transcendental. The two computational tools available there were ad hoc — iterated geometric means (§106) for individual logarithms and table lookup (§109–§111) for everything else. Chapter 7 replaces those with infinite series that compute exponentials and logarithms to any precision.

The chapter has three movements:

1. §114–§117 — the exponential series. Set $a^{\omega }=1+k\omega$ with $\omega$ infinitely small; the proportionality constant $k$ depends on the base. Raise to the $j$-th power, let $j=z/\omega$ be infinitely large, and the binomial expansion of $(1+kz/j{)}^j$ collapses (since $(j-m)/j=1$ for finite $m$) into $a^z=\sum (kz{)}^n/n!$. The general $b^z$ follows from $b=a^{\log b}$.
2. §118–§121 — the logarithmic series. Run the same machinery in reverse. From $a^{\omega }=1+k\omega$ get $\omega =\log (1+k\omega )$; setting $(1+k\omega {)}^j=1+x$ and expanding $(1+x{)}^{1/j}$ binomially yields $\log (1+x)=(1/k)(x-x^2/2+x^3/3-\cdots )$. The natural derivative — set $1+x=a$ to extract $k$ — diverges for $a=10$, but the trick of subtracting $\log (1-x)$ from $\log (1+x)$ gives the rapidly convergent $\log \frac{1+x}{1-x}=(2/k)(x+x^3/3+x^5/5+\cdots )$, from which $k$ can be computed for any base.
3. §122–§125 — the natural logarithm and $e$. Choose the base so $k=1$. Then $a=1+1/1!+1/2!+1/3!+\cdots$, which Euler computes as $2.71828182845904523536028\dots$, denotes $e$, and calls “the base of natural or hyperbolic logarithms” (§122). The series collapse to the canonical $e^z=\sum z^n/n!$, $\log (1+x)=x-x^2/2+x^3/3-\cdots$. Euler tabulates $\log 1,\dots ,\log 10$ to twenty digits and shows that for any other base $a$, $k={\log }_{e}a$ is the conversion factor.

See also: infinitesimal-and-infinite-numbers, exponential-series, logarithmic-series, eulers-number, natural-logarithm, change-of-base.

Structure of the chapter

§114 — Setup: $a^{\omega }=1+k\omega$ and the constant $k$

For $a>1$ and any infinitely small positive $\omega$, $a^{\omega }$ exceeds 1 by an infinitely small amount: $a^{\omega }=1+\psi$. Since $\omega$ being infinitely small forces $\psi$ to be infinitely small (and vice versa), they are commensurable and Euler writes $\psi =k\omega$, so

$a^{\omega }=1+k\omega ,\omega =\log (1+k\omega ).$

The constant $k$ is finite and depends on $a$. Worked example: with $a=10$ and $1+k\omega =1+1/1,000,000$, the common log table gives $\omega =\log (1+10^{-6})=0.00000043429$, so $1/k=0.43429$ and $k\approx 2.30258$ (source: chapter7, §114). This is the same $k$ that turns out to equal ${\log }_{e}10$ — but at this stage Euler only knows it as a base-dependent finite quantity.

§115 — Binomial expansion of $a^{j\omega }$

Raising $a^{\omega }=1+k\omega$ to the $j$-th power gives $a^{j\omega }=(1+k\omega {)}^j$. Expanding the right side by Newton’s binomial (Chapter 4, binomial-series):

$a^{j\omega }=1+\frac{j}{1}k\omega +\frac{j(j-1)}{1\cdot 2}{k}^2{\omega }^2+\frac{j(j-1)(j-2)}{1\cdot 2\cdot 3}{k}^3{\omega }^3+\cdots$

Now substitute $j=z/\omega$, so $j$ is infinitely large and $\omega =z/j$ is infinitely small:

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

(source: chapter7, §115). This is true because $j$ is infinitely large.

§116 — Collapsing the coefficients

Since $j$ is infinitely large, $(j-m)/j=1$ and $(j-m)/(nj)=1/n$ for every finite $m,n$ (source: chapter7, §116). Each binomial coefficient of $(1+kz/j{)}^j$ collapses to a reciprocal factorial:

$a^z=1+\frac{kz}{1}+\frac{k^2{z}^2}{1\cdot 2}+\frac{k^3{z}^3}{1\cdot 2\cdot 3}+\frac{k^4{z}^4}{1\cdot 2\cdot 3\cdot 4}+\cdots$

Setting $z=1$ gives the defining relation between $a$ and $k$:

$a=1+\frac{k}{1}+\frac{k^2}{1\cdot 2}+\frac{k^3}{1\cdot 2\cdot 3}+\frac{k^4}{1\cdot 2\cdot 3\cdot 4}+\cdots$

For $a=10$, this series in $k$ must equal 10, recovering $k\approx 2.30258$. See exponential-series.

§117 — The general exponential $b^z$

If $b=a^n$ then ${\log }_{a}b=n$, and $b^z=a^{nz}$. Substituting $nz$ for $z$ in §116 and then writing $n=\log b$:

$b^z=1+\frac{kz\log b}{1}+\frac{k^2{z}^2(\log b{)}^2}{1\cdot 2}+\frac{k^3{z}^3(\log b{)}^3}{1\cdot 2\cdot 3}+\cdots$

So once $k$ is known for a chosen base $a$, every other exponential $b^z$ has a power series in $z$ whose coefficients are powers of $\log b$ (source: chapter7, §117).

§118–§119 — Inverting: the logarithmic series

§118 starts the inversion. From $a^{\omega }=1+k\omega$, $\omega =\log (1+k\omega )$ and $j\omega =\log (1+k\omega {)}^j$. Setting

$(1+k\omega {)}^j=1+x⟹\log (1+x)=j\omega .$

For $j\omega$ to remain a finite number (the logarithm), $j$ must be infinitely large (and so $\omega$ infinitely small).

§119 inverts the relation: $1+k\omega =(1+x{)}^{1/j}$, so $k\omega =(1+x{)}^{1/j}-1$ and $j\omega =(j/k)((1+x{)}^{1/j}-1)$. Hence

$\log (1+x)=\frac{j}{k}(1+x{)}^{1/j}-\frac{j}{k}.$

Expand $(1+x{)}^{1/j}$ by the binomial series:

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

For $j$ infinite, $(j-1)/(2j)=1/2$, $(2j-1)/(3j)=2/3$, etc. — the same collapse as §116 — and the leading $j/k$ cancels the constant term, leaving

$\log (1+x)=\frac{1}{k}\left(\frac{x}{1}-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\cdots \right).$

See logarithmic-series.

§120 — The divergence paradox

Set $1+x=a$ in the §119 series. Since $\log a=1$,

$k=\frac{a-1}{1}-\frac{(a-1{)}^2}{2}+\frac{(a-1{)}^3}{3}-\frac{(a-1{)}^4}{4}+\cdots$

For $a=10$ this reads $2.30258=9/1-81/2+729/3-6561/4+\cdots$ — manifestly divergent in any pre-modern reading (source: chapter7, §120). Euler flags this as a paradox to be resolved in §121.

§121 — The fast-convergent $\log \frac{1+x}{1-x}$

Substituting $-x$ for $x$ in the §119 series:

$\log (1-x)=-\frac{1}{k}\left(\frac{x}{1}+\frac{x^2}{2}+\frac{x^3}{3}+\frac{x^4}{4}+\cdots \right).$

Subtract from $\log (1+x)$: even-power terms cancel, odd-power terms double:

$\log \frac{1+x}{1-x}=\frac{2}{k}\left(\frac{x}{1}+\frac{x^3}{3}+\frac{x^5}{5}+\frac{x^7}{7}+\cdots \right).$

Now solve $(1+x)/(1-x)=a$ for $x$: $x=(a-1)/(a+1)$. For $a=10$, $x=9/11<1$, so

$k=2\left(\frac{9}{11}+\frac{9^3}{3\cdot 11^3}+\frac{9^5}{5\cdot 11^5}+\cdots \right)$

— a series whose terms decrease geometrically, giving fast convergence to $k\approx 2.30258$ (source: chapter7, §121). The paradox of §120 is resolved: the divergent series of §120 is the wrong tool for $a>2$; the §121 series is the right one.

§122 — Defining $e$ as the base where $k=1$

The base $a$ is at the analyst’s disposal — choose it so $k=1$. Then the §116 defining series collapses to

$a=1+\frac{1}{1}+\frac{1}{1\cdot 2}+\frac{1}{1\cdot 2\cdot 3}+\frac{1}{1\cdot 2\cdot 3\cdot 4}+\cdots =2.71828182845904523536028\dots$

(source: chapter7, §122). Euler denotes this number $e$ — the first appearance of the symbol — and calls the resulting logarithms natural or hyperbolic (the latter “since the quadrature of a hyperbola can be expressed through these logarithms”). See eulers-number.

§123 — The canonical natural-log series and a logarithm table

With $k=1$ the three master identities take their cleanest form (source: chapter7, §123):

$e^z=1+z+\frac{z^2}{2!}+\frac{z^3}{3!}+\cdots ,$

$\log (1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\frac{x^5}{5}-\cdots ,$

$\log \frac{1+x}{1-x}=\frac{2x}{1}+\frac{2x^3}{3}+\frac{2x^5}{5}+\frac{2x^7}{7}+\cdots .$

Sample applications: $x=1/5$ gives $\log (3/2)=2/(1\cdot 5)+2/(3\cdot 5^3)+\cdots$; $x=1/7$ gives $\log (4/3)$; $x=1/9$ gives $\log (5/4)$. Combining these by the Chapter 6 algebraic rules ($\log (3/2)+\log (4/3)=\log 2$, etc.) yields all integer logs $\log 2,\dots ,\log 6,\log 8,\log 9,\log 10$. For $\log 7$, set $x=1/99$ to get $\log (50/49)=\log 50-2\log 7$, then $\log 7=(\log 50-\log (50/49))/2$. Euler writes out the resulting table to twenty decimal places (e.g. $\log 2=0.6931471805599453094172321$, $\log 10=2.3025850929940456840179914$).

§124 — $k$ is the natural log of $a$

For an arbitrary base $a$, let $y={\log }_e(1+x)$ and $v={\log }_a(1+x)$. The §119 series for $v$ differs from the §123 series for $y$ only by the factor $1/k$, so $v=y/k$ and

$k=\frac{y}{v}=\frac{{\log }_e(1+x)}{{\log }_a(1+x)}.$

Setting $1+x=a$ gives $v=1$ and $k={\log }_{e}a$ (source: chapter7, §124). For $a=10$: $k={\log }_{e}10=2.3025850929940456840179914$, exactly the value computed in §114, §121, §123. Conversely, dividing every natural log by $k$ (or multiplying by $1/k=0.4342944819032518276511289$) gives common logs — recovering change-of-base from Chapter 6.

§125 — Two more identities

From $e^z$ and $a^y=e^{y\log a}$ (with $\log ={\log }_e$, since $\log e=1$):

$a^y=1+\frac{y\log a}{1}+\frac{y^2(\log a{)}^2}{1\cdot 2}+\frac{y^3(\log a{)}^3}{1\cdot 2\cdot 3}+\cdots .$

And from the underlying $j$-form:

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

with $j$ infinitely large (source: chapter7, §125). Euler closes the chapter by noting that further uses of natural logs are deferred to integral calculus.

Notable points

• The chapter’s logical machine is Euler’s infinitesimal–infinite identification. Every derivation rests on simultaneously using $\omega \to 0$ and $j=z/\omega \to \infty$, then setting $(j-m)/j=1$ for finite $m$. This is not modern limit-taking; it is treated as a transparent algebraic identity. The whole power-series theory of $e^z$ and $\log$ falls out by one move.
• $e$ is defined analytically, not geometrically. Unlike $\pi$, which Euler will reach via the circle, $e$ enters as the unique base for which the multiplicative constant $k$ in $a^{\omega }=1+k\omega$ equals 1. The geometric (“hyperbolic”) interpretation is mentioned only as etymology; the working definition is the series $\sum 1/n!$.
• The §120 divergence is honest. Euler does not hide that his most natural inversion of the exponential series fails for $a=10$. He resolves it not by qualifying the original series but by deriving a different series (the $\log \frac{1+x}{1-x}$ one) whose convergence properties are favorable. The interplay of two series, one slow/divergent and one fast, computing the same quantity, will recur throughout the Introductio.
• Every transcendental computation in Chapter 6 now has a series counterpart. Geometric means still work but are no longer needed: §123 produces $\log 2,\log 3,\log 5,\log 7$ from $x=1/5,1/7,1/9,1/99$ in a few quickly-converging terms, and the rest follow by the algebraic rules of §104.
• Chapter 7 is the analytic counterpart of Chapter 6 in the same sense Chapter 4 was for Chapters 2–3. Where Chapter 4 expanded algebraic functions (rational, then irrational via Newton’s binomial) in series, Chapter 7 expands the transcendental exponential and logarithm. The (1 + z/j)^j representation of $e^z$ is the bridge: a transcendental function exhibited as the limit of an algebraic family.

Why this chapter matters

Chapters 6 and 7 together are the foundation of every later chapter that mentions $e$, $\log$, sines, cosines, or $\pi$. Chapter 6 secured existence; Chapter 7 secures computation — every value of $a^z$ and $\log y$ is now reachable by summing a power series. The exponential series $e^z=\sum z^n/n!$ in particular will be the input to Chapter 8’s derivation of the trigonometric series and the Euler formula $e^{ix}=\cos x+i\sin x$.

The chapter also installs Euler’s main analytic technique — the simultaneous use of an infinitely small $\omega$ and an infinitely large $j$ with $j\omega$ finite — as the standard tool of the Introductio. The next several chapters will reuse this device almost verbatim, with $z$ replaced by $iz$, by trigonometric arguments, and by other parameters.

Related pages

• infinitesimal-and-infinite-numbers
• exponential-series
• logarithmic-series
• eulers-number
• natural-logarithm
• exponential-function
• logarithm
• change-of-base
• binomial-series
• geometric-mean-method-for-logarithms
• chapter-6-on-exponentials-and-logarithms
• chapter-4-on-the-development-of-functions-in-infinite-series