Partial Fraction Decomposition

Summary: Euler’s method (§39–§46) for resolving a proper rational function into a sum of simple fractions, one per linear factor of the denominator (with a tower of fractions when factors repeat).

Sources: chapter2

Last updated: 2026-04-23

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Setup

Let $M/N$ be a proper rational function — that is, $\mathrm{deg}M<\mathrm{deg}N$. See improper-rational-function for how to reduce the improper case by polynomial division first.

By factoring-polynomials (and, granted the fundamental-theorem-of-algebra), $N$ factors over $\mathbb{R}$ into linear and quadratic factors. Euler treats the linear case in detail in this chapter; the quadratic case — needed when $N$ has complex roots and a real decomposition is required — is handled in Chapter 12 (see real-partial-fraction-decomposition).

Distinct linear factors (§40–§41)

If $N$ factors into $k$ distinct linear factors, $M/N$ decomposes into $k$ simple fractions, one per factor:

$\frac{M}{N}={\sum }_i\frac{A_i}{p_i-q_{i}z}(\text{source: chapter2, §40}).$

Shortcut formula for $A$ (§41, “cover-up”)

For the factor $p-qz$ of $N$, write $N=(p-qz)S$. Then the numerator of the corresponding simple fraction is

$A={\frac{M}{S}|}_{z=p/q}.$

Euler’s derivation: from $M/N=A/(p-qz)+P/S$ one gets $M-AS=(p-qz)P$, which vanishes at $z=p/q$, forcing $A=M/S$ evaluated there (source: chapter2, §41).

Example (§40–§41)

Decompose $\frac{1+z^2}{z-z^3}$. The denominator factors as $z\cdot (1-z)(1+z)$, so

$\frac{1+z^2}{z-z^3}=\frac{A}{z}+\frac{B}{1-z}+\frac{C}{1+z}.$

Applying the §41 shortcut:

• For factor $z$: $S=1-z^2$, so $A={\frac{1+z^2}{1-z^2}|}_{z=0}=1$.
• For factor $1-z$: $S=z+z^2$, so $B={\frac{1+z^2}{z+z^2}|}_{z=1}=1$.
• For factor $1+z$: $S=z-z^2$, so $C={\frac{1+z^2}{z-z^2}|}_{z=-1}=-1$.

Hence

$\frac{1+z^2}{z-z^3}=\frac{1}{z}+\frac{1}{1-z}-\frac{1}{1+z}.$

Repeated linear factors (§42–§45)

If $(p-qz{)}^n$ divides $N$, that factor contributes the tower

$\frac{A}{(p-qz{)}^n}+\frac{B}{(p-qz{)}^{n-1}}+\frac{C}{(p-qz{)}^{n-2}}+\cdots +\frac{K}{p-qz}.$

The iterative algorithm (§45)

Write $N=(p-qz{)}^{n}Z$. Compute the numerators in sequence, each time substituting $z=p/q$ after dividing by $p-qz$:

1. $A={\frac{M}{Z}|}_{z=p/q}$, then let $P=\frac{M-AZ}{p-qz}$.
2. $B={\frac{P}{Z}|}_{z=p/q}$, then let $Q=\frac{P-BZ}{p-qz}$.
3. $C={\frac{Q}{Z}|}_{z=p/q}$, then let $R=\frac{Q-CZ}{p-qz}$.
4. $D={\frac{R}{Z}|}_{z=p/q}$, then let $S=\frac{R-DZ}{p-qz}$.
5. Continue until all $n$ numerators are obtained (source: chapter2, §45).

Each "$\dots /(p-qz)$" step is an exact polynomial division — the numerator of the previous step is known to be divisible by $p-qz$, so the division clears out before the next substitution is made.

General procedure (§46)

To decompose an arbitrary rational function $M/N$:

1. If $M/N$ is improper, extract the polynomial part by division; see improper-rational-function.
2. Factor $N$ into its linear factors (real or complex).
3. For each distinct linear factor $p-qz$ not repeated elsewhere, use the shortcut (§41) to produce a single simple fraction.
4. For each repeated linear factor $(p-qz{)}^n$, use the iterative algorithm (§45) to produce the tower of $n$ simple fractions.
5. Sum all simple fractions (plus the polynomial part, if any). The result equals $M/N$ in its simplest form (source: chapter2, §46).

Worked example (§46)

Decompose $\frac{1}{z^3(1-z{)}^2(1+z)}$. The denominator has factors $1+z$ (simple), $(1-z{)}^2$, and $z^3$.

• Factor $1+z$ (simple): $Z=z^3-2z^4+z^5$, so at $z=-1$: $A=1/(z^3-2z^4+z^5){|}_{z=-1}=-\frac{1}{4}$. Contribution: $\frac{-1}{4(1+z)}$.

• Factor $(1-z{)}^2$: $Z=z^3+z^4$. At $z=1$: $A=\frac{1}{2}$. Then $P=(M-\frac{1}{2}Z)/(1-z)=1+z+z^2+\frac{1}{2}{z}^3$, and $B=P/Z{|}_{z=1}=\frac{7}{4}$. Contribution: $\frac{1}{2(1-z{)}^2}+\frac{7}{4(1-z)}$.

• Factor $z^3$: $Z=1-z-z^2+z^3$. At $z=0$: $A=1$. Then $P=(M-Z)/z=1+z-z^2$ gives $B=1$. Then $Q=(P-Z)/z=2-z^2$ gives $C=2$. Contribution: $\frac{1}{z^3}+\frac{1}{z^2}+\frac{2}{z}$.

Putting it all together:

$\frac{1}{z^3(1-z{)}^2(1+z)}=\frac{1}{z^3}+\frac{1}{z^2}+\frac{2}{z}+\frac{1}{2(1-z{)}^2}+\frac{7}{4(1-z)}-\frac{1}{4(1+z)}.$

There is no polynomial part because the given function is proper (source: chapter2, §46).

Why this matters

Partial-fraction decomposition turns an arbitrary rational function into a sum of maximally simple pieces. This is what makes rational functions tractable in later chapters of the Introductio — for expansions into series, for summation, and (in the Institutiones calculi integralis) for integration.

Related pages

• real-partial-fraction-decomposition
• improper-rational-function
• factoring-polynomials
• fundamental-theorem-of-algebra
• complex-conjugate-factors
• trinomial-factor
• chapter-2-on-the-transformation-of-functions
• chapter-12-on-the-development-of-real-rational-functions