Chapter 12: On the Development of Real Rational Functions

Summary: Euler extends Chapter 2’s partial-fraction algorithm from real linear factors to real quadratic factors of the form $p^2-2pqz\cos \varphi +q^2{z}^2$ (the trinomial-factor of Chapter 9). The result keeps the decomposition entirely real even when the denominator has complex roots, and an iterative variant handles repeated quadratic factors.

Sources: chapter12

Last updated: 2026-05-04

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Overview

Chapter 2 decomposed a proper rational function $M/N$ into one simple fraction per linear factor of $N$. When $N$ has complex linear factors the resulting partial fractions are also complex — useless if the goal is a real decomposition. Chapter 9 supplied the cure: every real polynomial factors into real linear factors and real quadratic trinomial factors $p^2-2pqz\cos \varphi +q^2{z}^2$. Chapter 12 makes this cure operational: it gives an algorithm for the partial fraction with denominator equal to a quadratic trinomial, and then extends the algorithm to the case where that trinomial is repeated (source: chapter12, §199).

The chapter has two movements:

1. §199–§205 — distinct quadratic factors. For a single trinomial factor $p^2-2pqz\cos \varphi +q^2{z}^2$ in $N$, the corresponding partial fraction is $(P+Qz)/(p^2-2pqz\cos \varphi +q^2{z}^2)$ with two unknown coefficients $P$ and $Q$. Substituting the complex roots $z=(p/q)(\cos \varphi \pm i\sin \varphi )$ via De Moivre turns the divisibility condition into two real equations, solvable for $P$ and $Q$ in closed form. §204–§205 streamline the formula so only one substitution is needed instead of two. See real-partial-fraction-decomposition.
2. §206–§210 — repeated quadratic factors. When $(p^2-2pqz\cos \varphi +q^2{z}^2{)}^k$ divides $N$, the contribution is a tower of $k$ partial fractions with quadratic denominators of decreasing power. Each numerator is determined by an iterative procedure that mirrors §42–§45: extract the top numerator, divide out one factor of the trinomial, repeat. The last “remainder” polynomial is the numerator of the complementary fraction with denominator $Z$ (the cofactor of the trinomial tower in $N$). See real-partial-fraction-decomposition.

Structure of the chapter

§199 — Why real quadratic denominators

If the linear factors of the denominator are complex, the partial fractions of Chapter 2 are also complex — “of little use to express a real rational function in terms of complex fractions” (source: chapter12, §199). Since every real polynomial admits real factorization into linear and quadratic pieces (fundamental-theorem-of-algebra, constructively given by Chapter 9), a real rational function decomposes into partial fractions whose denominators are real linear or real quadratic factors.

§200 — The form of the partial fraction

For one trinomial factor $p^2-2pqz\cos \varphi +q^2{z}^2$ of $N$, the corresponding partial fraction has the form

$\frac{P+Qz}{p^2-2pqz\cos \varphi +q^2{z}^2},$

with a first-degree numerator. The numerator must be exactly first degree: a higher degree would leave behind a polynomial part (which should already have been removed; see improper-rational-function) (source: chapter12, §200).

§201 — The divisibility condition

Write $N=(p^2-2pqz\cos \varphi +q^2{z}^2)Z$ and let $M$ be the numerator. Subtracting the $(P+Qz)/(\cdots )$ piece leaves $Y/Z$ where

$Y=\frac{M-PZ-QZz}{p^2-2pqz\cos \varphi +q^2{z}^2}.$

For $Y$ to be a polynomial, $M-PZ-QZz$ must vanish at the two roots of the trinomial — namely $z=(p/q)(\cos \varphi \pm i\sin \varphi )$. Setting $f=p/q$, De Moivre gives $z^n=f^n(\cos n\varphi \pm i\sin n\varphi )$ at each root (source: chapter12, §201).

§202–§203 — Two real equations, closed-form solution

Substituting both roots into $M=PZ+QZz$ and separating real and imaginary parts yields two real equations $R=PS+QT$ and $r=Ps+Qt$, where (for $M=A+Bz+C{z}^2+\cdots$ and $Z=\alpha +\beta z+\gamma z^2+\cdots$)

$R=A+Bf\cos \varphi +C{f}^2\cos 2\varphi +\cdots ,r=Bf\sin \varphi +C{f}^2\sin 2\varphi +\cdots ,$ $S=\alpha +\beta f\cos \varphi +\gamma f^2\cos 2\varphi +\cdots ,s=\beta f\sin \varphi +\gamma f^2\sin 2\varphi +\cdots ,$ $T=\alpha f\cos \varphi +\beta f^2\cos 2\varphi +\cdots ,t=\alpha f\sin \varphi +\beta f^2\sin 2\varphi +\cdots .$

(That is: $R,r$ come from $M$; $S,s$ come from $Z$; $T,t$ come from $Zz$.) Solving the linear system,

$P=\frac{Rt-rT}{St-sT},Q=\frac{Rs-rS}{sT-St}.$

(source: chapter12, §203). See real-partial-fraction-decomposition for derivation and worked examples.

§204–§205 — Streamlined formula

A direct calculation shows $T=f(S\cos \varphi -s\sin \varphi )$ and $t=f(S\sin \varphi +s\cos \varphi )$, so $T$ and $t$ are determined by $S,s,\cos \varphi ,\sin \varphi$ alone. Substituting into the §203 expressions collapses to

$\frac{P+Qz}{p^2-2pqz\cos \varphi +q^2{z}^2}=\frac{(RS+rs)p\sin \varphi +(Rs-rS)(p\cos \varphi -qz)}{(p^2-2pqz\cos \varphi +q^2{z}^2)(S^2+s^2)p\sin \varphi }.$

Only the four scalars $R,r,S,s$ are needed — half the computation of §203 (source: chapter12, §205).

§206 — Why repeated factors need a different rule

If the denominator contains $(p^2-2pqz\cos \varphi +q^2{z}^2{)}^2$ or higher, then after substituting $z=f(\cos \varphi \pm i\sin \varphi )$ both $M-PZ-QZz$ and $Z$ vanish, so the §203 system degenerates and $P,Q$ cannot be solved (source: chapter12, §206). A new procedure is needed.

§207–§209 — Iterative algorithm for $(p^2-2pqz\cos \varphi +q^2{z}^2{)}^k$

The repeated-quadratic factor contributes the tower

$\frac{U+uz}{(p^2-2pqz\cos \varphi +q^2{z}^2{)}^k}+\frac{V+vz}{(p^2-2pqz\cos \varphi +q^2{z}^2{)}^{k-1}}+\cdots +\frac{X+xz}{p^2-2pqz\cos \varphi +q^2{z}^2}.$

Each numerator is determined by a one-step substitution analogous to §45: substitute $z=f(\cos \varphi \pm i\sin \varphi )$ into a numerator expression and divide by the trinomial to expose the next numerator. Concretely (writing $N$ for the value of $Z$ at the root and $n$ for its imaginary part):

$U=\frac{YN+yn}{N^2+n^2}+\frac{Yn-yN}{N^2+n^2}\cdot \frac{\cos \varphi }{\sin \varphi },u=-\frac{Yn-yN}{N^2+n^2}\cdot \frac{q}{p\sin \varphi },$

with $Y,y$ the values of $M$ at the roots. Then $F=(M-(U+uz)Z)/(p^2-2pqz\cos \varphi +q^2{z}^2)$ is the next polynomial; the same formula on $F$ produces $V,v$; and so on through $G,H,I,K,\dots$ (source: chapter12, §207–§209). See real-partial-fraction-decomposition for the full derivation and worked example.

§210 — The complementary fraction falls out for free

The sequence of polynomials $F,G,H,I,K,\dots$ used to extract numerators is automatically the right object for the complementary fraction with denominator $Z$ (the cofactor of the trinomial tower in $N$). For $k=1$, $F/Z$ is the complement; for $k=2$, $G/Z$; and so on. The complement, having denominator $Z$ which contains no further power of this trinomial, can itself be expressed in partial fractions by the rules above (source: chapter12, §210).

Worked examples in the chapter

Three examples in §203 plus one in §209 illustrate the procedure.

• Example I (§203): $\frac{z^2}{(1-z+z^2)(1+z^4)}$. The factor $1-z+z^2$ has $p=q=1$, $\cos \varphi =1/2$, $\varphi =\pi /3$. Computation gives $P=-1$, $Q=0$, so the partial fraction is $-1/(1-z+z^2)$ and the complement is $(1+z+z^2)/(1+z^4)$.
• Example II (§203): Continues by decomposing the complement $(1+z+z^2)/((1+\sqrt{2}z+z^2)(1-\sqrt{2}z+z^2))$. Both factors have $\varphi =\pi /4$, with $f=-1$ and $f=+1$ respectively. The two partial fractions are $(\sqrt{2}-1)\sqrt{2}/(2(1+\sqrt{2}z+z^2))$ and $(\sqrt{2}+1)\sqrt{2}/(2(1-\sqrt{2}z+z^2))$.
• Example III (§203): $\frac{1+2z+z^2}{(1-\frac{8}{5}z+z^2)(1+2z+3z^2)}$. The first factor has $\cos \varphi =4/5$ — not a fractional part of a right angle, so the multiples $\cos n\varphi ,\sin n\varphi$ must be computed by the addition formulas. The result is $\frac{9(17-5z)/178}{1-\frac{8}{5}z+z^2}+\frac{5(5+27z)/178}{1+2z+3z^2}$.
• Example (§209): $\frac{z-z^3}{(1+z^2{)}^4(1+z^4)}$. The factor $(1+z^2{)}^4$ has $\cos \varphi =0$, $\varphi =\pi /2$. Iterating §207 gives the four partial fractions $\frac{z}{(1+z^2{)}^4}+\frac{z}{2(1+z^2{)}^3}+0\cdot \frac{1}{(1+z^2{)}^2}-\frac{z}{4(1+z^2)}$. The complement, with denominator $1+z^4$, is $(-z+z^3)/(4(1+z^4))$.

Notable points

• The chapter is the missing complement to Chapter 2. Chapter 2 promised a partial-fraction decomposition algorithm and gave one for linear factors. Chapter 12 finishes the job by handling the quadratic-factor case that arises whenever the polynomial has complex roots. After this chapter, every real rational function admits a fully real partial-fraction decomposition.
• The §201 device is the same one as Chapter 9. Substituting $z=f(\cos \varphi \pm i\sin \varphi )$ and separating real/imaginary parts is exactly the §148 trick for finding trinomial factors. Chapter 9 used the trick to find the factor; Chapter 12 uses it to find the coefficient of the partial fraction whose denominator is that factor. The unified mechanism explains why the de-moivre-formula is the workhorse of both chapters.
• §204 is a small but real economy. Without §204, one needs three substitutions (for $R/r$, $S/s$, $T/t$) — each a separate calculation of trigonometric multiples. §204 cuts the work by a third. Euler does not state a corresponding economy for the §207 repeated-factor formula, but the same observation applies there.
• The “complementary fraction is free” remark of §210 is structural. In the iterative procedure, the polynomial being divided at the last step is already the numerator of the leftover fraction. Modern implementations of partial-fraction decomposition exploit the same fact.

Why this chapter matters

For integration, the goal is a sum of pieces each of which has a known antiderivative. Real linear pieces $A/(p-qz)$ integrate to logarithms; real quadratic pieces $(P+Qz)/(p^2-2pqz\cos \varphi +q^2{z}^2)$ integrate to a logarithm plus an arctangent. By insisting on real denominators, Chapter 12 is the prerequisite for the partial-fractions integration technique developed in Euler’s later Institutiones calculi integralis — and for everything in modern calculus textbooks under the same heading.

For series expansion, the recurrent-series machinery of Chapter 4 reads coefficients off the denominator. A real partial-fraction decomposition lets the same machinery operate even when the denominator has complex roots — each real quadratic piece contributes a recurrent series with denominator $1-2(\cos \varphi )Z+Z^2$, exactly the §129 trigonometric recurrent progression.

Related pages

• real-partial-fraction-decomposition
• partial-fraction-decomposition
• improper-rational-function
• trinomial-factor
• factoring-polynomials
• fundamental-theorem-of-algebra
• de-moivre-formula
• trigonometric-addition-formulas
• trigonometric-recurrent-progression
• chapter-2-on-the-transformation-of-functions
• chapter-9-on-trinomial-factors