Method of Undetermined Coefficients

Summary: Euler’s technique (§60–§61) for expanding a rational function as an infinite series: posit a series $A+Bz+C{z}^2+\cdots$ with unknown coefficients, multiply by the denominator, and match powers of $z$ to determine the coefficients one by one.

Sources: chapter4

Last updated: 2026-04-23

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The method

Given a rational function $\frac{N(z)}{D(z)}$ with $D(0)\ne 0$, write

$\frac{N(z)}{D(z)}=A+Bz+C{z}^2+D{z}^3+\cdots$

with unknown coefficients $A,B,C,\dots$. Multiply through by $D(z)$ to obtain

$N(z)=D(z)\cdot (A+Bz+C{z}^2+\cdots ).$

Expanding the right-hand side and collecting powers of $z$ yields an infinite system of linear equations in the unknowns — the coefficients of matching powers on the two sides must be equal. Each equation determines the next coefficient in terms of its predecessors (source: chapter4, §60–§61).

Euler prefers this method to long division because long division is “tedious and there is no easy way to show the nature of the resulting infinite series” (source: chapter4, §61) — the recurrence produced by matching is more informative than the step-by-step quotient.

Worked examples

Linear denominator (§60)

For $\frac{a}{\alpha +\beta z}=A+Bz+C{z}^2+\cdots$, multiplication gives $a=(\alpha +\beta z)(A+Bz+C{z}^2+\cdots )$, i.e.

$a=\alpha A+(\alpha B+\beta A)z+(\alpha C+\beta B)z^2+(\alpha D+\beta C)z^3+\cdots$

Matching: $A=a/\alpha$ and $\alpha Q+\beta P=0$ for any consecutive $P,Q$. This yields the geometric-series.

Quadratic denominator (§61)

For $\frac{a+bz}{\alpha +\beta z+\gamma z^2}=A+Bz+C{z}^2+\cdots$, the same procedure gives

$a+bz=\alpha A+(\alpha B+\beta A)z+(\alpha C+\beta B+\gamma A)z^2+(\alpha D+\beta C+\gamma B)z^3+\cdots$

Matching: $A=a/\alpha$, $B=b/\alpha -a\beta /{\alpha }^2$, and from the third power onward the three-term recurrence $\alpha R+\beta Q+\gamma P=0$ (source: chapter4, §61). Hence $R=-(\beta Q+\gamma P)/\alpha$.

Lucas-number example (§61)

With $a=1,b=2,\alpha =1,\beta =-1,\gamma =-1$: $A=1,B=3$, then $R=P+Q$, so

$\frac{1+2z}{1-z-z^2}=1+3z+4z^2+7z^3+11z^4+18z^5+\cdots$

— every coefficient is the sum of the two preceding ones (source: chapter4, §61).

Why the method works

Because $D(0)\ne 0$, the power series $A+Bz+C{z}^2+\cdots$ satisfying $N=D\cdot (A+Bz+\cdots )$ is unique: the coefficient of $z^n$ on the left side equals a finite sum involving $A,B,\dots$ through degree at most $n$, and matching these determines each coefficient in turn. Euler treats uniqueness as obvious; a modern statement would invoke the formal power series ring, where $D(0)\ne 0$ makes $D$ a unit.

Relationship to other methods

• Long division produces the same series one term at a time, but does not exhibit the recurrence. See geometric-series §60 for the parallel.
• Once the recurrence is identified, the series is a recurrent-series and the theory of Chapter 4 applies.
• For irrational functions, the analogue is to assume the binomial form $(P+Q{)}^{m/n}$ and derive term-to-term relations; see binomial-series.

Related pages

• geometric-series
• recurrent-series
• binomial-series
• chapter-4-on-the-development-of-functions-in-infinite-series