Exponential Series

Summary: §115–§117 of Chapter 7. From the Chapter-6 exponential-function $a^z$ Euler derives the power series

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

where the constant $k$ depends on the base $a$ via $a^{\omega }=1+k\omega$ for infinitely small $\omega$. Setting $z=1$ gives the relation $a=\sum k^n/n!$ between $a$ and $k$. Choosing the base so $k=1$ produces $e^z=\sum z^n/n!$ — see eulers-number. The general $b^z$ follows by substituting $b=a^{\log b}$ and reads $b^z=\sum (kz\log b{)}^n/n!$ (§117).

Sources: chapter7 (§115–§117)

Last updated: 2026-04-26

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Setup: $a^{\omega }=1+k\omega$

Fix a base $a>1$. For an infinitely small positive $\omega$, the value $a^{\omega }$ exceeds 1 by an infinitely small amount, written $a^{\omega }=1+k\omega$ with $k$ a finite, base-dependent constant (source: chapter7, §114). For $a=10$ Euler computes $k\approx 2.30258$; in modern notation $k={\log }_{e}a$ — see natural-logarithm. See infinitesimal-and-infinite-numbers for the status of $\omega$ and the next section’s $j$.

Derivation (§115–§116)

Raise $a^{\omega }=1+k\omega$ to a power $j$:

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

Expand the right side by the binomial theorem (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 let $j=z/\omega$. Then $\omega =z/j$, and substituting:

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

Since $j$ is infinitely large, $(j-m)/(nj)=1/n$ for every finite $m,n$ — see infinitesimal-and-infinite-numbers — and each coefficient collapses:

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

This is the exponential series (source: chapter7, §116).

The defining relation between $a$ and $k$

Setting $z=1$ in the boxed series:

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

This is the implicit equation that links $a$ and $k$. For $a=10$, the series in $k$ must equal 10, recovering $k\approx 2.30258$ from §114 (source: chapter7, §116). For $k=1$, the sum is $1+1+1/2+1/6+1/24+\cdots =2.71828\dots =e$ — see eulers-number.

The general exponential $b^z$ (§117)

Suppose $b=a^n$, so ${\log }_{a}b=n$. Then $b^z=a^{nz}$, and substituting $nz$ for $z$ in the boxed series:

$b^z=1+\frac{knz}{1}+\frac{k^2{n}^2{z}^2}{1\cdot 2}+\frac{k^3{n}^3{z}^3}{1\cdot 2\cdot 3}+\cdots$

Now $n=\log b$ (in base $a$), giving Euler’s general form:

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

(source: chapter7, §117). One $k$ — the constant of the chosen base $a$ — and one $\log b$ (in that base) suffice to compute $b^z$ for every $b$.

When the chosen base is $e$ (so $k=1$), $\log ={\log }_e$ and the formula reduces to $b^z=\sum (z\log b{)}^n/n!$, which is just $e^{z\log b}$. This is the modern $b^z=e^{z\ln b}$.

Worked specializations

Substitute	Series for	Result
$k=1$ (so $a=e$)	$e^z$	${\sum }_{n\ge 0}{z}^n/n!$
$k=1$, $z=1$	$e$	${\sum }_{n\ge 0}1/n!$
$k=\log 10\approx 2.30258$, $a=10$	$10^z$	${\sum }_n(kz{)}^n/n!$
$k=1$, base $e$, $b$ arbitrary	$b^z=e^{z\log b}$	${\sum }_n(z\log b{)}^n/n!$

Why the series converges

Euler does not state convergence as a theorem in this chapter, but one can read off two structural reasons it works:

• The coefficient of $z^n$ is $k^n/n!$, and $n!$ grows faster than any geometric, so the series converges for every $z$ — uniformly on compacts in modern terms.
• The series agrees with $a^z$ at $z=0$ (both equal 1) and is the formal expansion forced by the algebraic identity $a^z=(1+kz/j{)}^j$ as $j$ becomes infinite. Whatever subtleties are lurking, the result must equal $a^z$ on whatever domain the manipulation is valid.

In Chapter 8 Euler will use this series with imaginary $z$ to derive the trigonometric series, and the unbounded convergence radius is exactly what licenses that step.

Status of the derivation

The derivation rests on the infinite identification: $j=z/\omega$ is treated as both infinite and as a definite quantity in the binomial coefficients, and $(j-m)/(nj)=1/n$ is treated as an algebraic identity rather than a limit. Modern analysis recovers the same series via ${\lim }_{n\to \infty }(1+z/n{)}^n=e^z$ and termwise comparison; Euler’s argument is the formal antecedent of that limit.

Related pages

• exponential-function
• logarithmic-series
• eulers-number
• natural-logarithm
• infinitesimal-and-infinite-numbers
• binomial-series
• chapter-7-on-exponentials-and-logarithms-expressed-through-series
• chapter-6-on-exponentials-and-logarithms