Choosing Expansion Centers for Rational Functions

Why the Expansion Center Matters

A power series is centered at a particular point, and that choice affects where the series converges. For a rational function, its poles limit how far a Taylor series can extend: a series cannot represent the function through a genuine pole. Choosing a center therefore involves both an algebraic task—rewriting the denominator—and a geometric one—measuring distances to the poles.

Here we work with rational functions in a variable $x$, allowing complex values when locating poles. If the function has real coefficients and we are interested only in real $x$, the interval of convergence lies inside the corresponding complex disk. The radius describes the open disk of convergence; behavior at real endpoints must be checked separately.

Key idea: For a non-polynomial rational function, after all common factors have been cancelled, the Taylor series about a point that is not a pole has radius equal to the distance from that center to the nearest pole. A polynomial has no poles and its Taylor series has infinite radius of convergence.

Expanding a Reciprocal Linear Factor

Let $c$ be the expansion center and $p$ a pole, with $c\ne p$. Put $h=x-c$. The factor $x-p$ can be written as $(c-p)+h$, so its reciprocal is a scaled geometric series whenever the displacement from the center is smaller than the center-to-pole distance.

\frac{1}{x-p} = -\frac{1}{p-c}\frac{1}{1-\frac{x-c}{p-c}} = -\sum_{n=0}^{\infty}\frac{(x-c)^n}{(p-c)^{n+1}}, \qquad |x-c|<|p-c|.

The convergence condition is not optional: it is the geometric-series condition $\left|\frac{x-c}{p-c}\right|<1$. In particular, the individual factor’s series has radius $|p-c|$. If a rational function is written as a polynomial plus partial-fraction terms, each term can be expanded around the same center, provided the center is not a pole.

For a pole of order $k$, the corresponding factor has the expansion

\frac{1}{(x-p)^k} = (-1)^k\sum_{n=0}^{\infty} \binom{n+k-1}{k-1} \frac{(x-c)^n}{(p-c)^{n+k}}, \qquad |x-c|<|p-c|.

This follows from the geometric-series formula for $(1-u)^{-k}$. Repeated poles change the coefficients, but not the distance that controls convergence.

Worked Example: One Pole and Two Possible Centers

Worked Example: Expanding a Single Reciprocal Factor

Expand $f(x)=\frac{1}{x-4}$ about $c=1$, and determine the radius of convergence.

Here $p=4$, so $p-c=3$. Substitution into the reciprocal-factor formula gives

\frac{1}{x-4} = -\sum_{n=0}^{\infty}\frac{(x-1)^n}{3^{n+1}}, \qquad |x-1|<3.

The pole is at $x=4$, whose distance from the center $1$ is $|4-1|=3$. The series therefore has radius $3$. To check the expansion at $x=2$, the series becomes

-\sum_{n=0}^{\infty}\frac{1}{3^{n+1}} = -\frac{\frac13}{1-\frac13} = -\frac12.

Direct substitution gives $f(2)=\frac{1}{2-4}=-\frac12$, as required. If the same function were centered at $c=0$, the pole would be $4$ units away and the radius would be $4$. Thus the center changes the radius, even though the function itself has not changed.

Finding the Nearest Pole

A rational function may have several poles, and different poles can impose different distance limits. The relevant distance is the smallest of the distances from the chosen center to the function’s actual poles. For a real center $c$ and a complex pole $p=a+bi$, that distance is

|p-c|=\sqrt{(a-c)^2+b^2}.

Thus a nonreal pole can be the nearest one even when no real singularity is visible on the real line. The radius is a distance in the complex plane, not merely the distance to the nearest real value that makes the denominator zero.

Definition: A pole of a rational function is a point where the function is unbounded because a denominator factor remains after common numerator and denominator factors have been cancelled. A factor that cancels is not an actual pole.

Worked Example: Comparing Centers for Two Simple Poles

Worked Example: Selecting a Center for a Two-Pole Function

Consider $f(x)=\frac{1}{(x-4)(x+2)}$. Choose between centers $c=0$ and $c=1$, and find the expansion about $c=1$.

The poles are $4$ and $-2$. From $c=0$, their distances are $4$ and $2$, so the nearest pole is $2$ units away. From $c=1$, the distances are $|4-1|=3$ and $|-2-1|=3$. The center $1$ therefore gives the larger radius, $3$.

First decompose the rational function. Solving for the constants gives

\frac{1}{(x-4)(x+2)} = \frac{1}{6}\left(\frac{1}{x-4}-\frac{1}{x+2}\right).

Indeed, the numerator on the right is $\frac16((x+2)-(x-4))=\frac16(6)=1$. Set $h=x-1$. Then $x-4=h-3$ and $x+2=h+3$, so the two reciprocal expansions are

\frac{1}{x-4} = -\sum_{n=0}^{\infty}\frac{h^n}{3^{n+1}}, \qquad \frac{1}{x+2} = \sum_{n=0}^{\infty}\frac{(-1)^n h^n}{3^{n+1}}, \qquad |h|<3.

Combining the partial fractions yields

f(x) = -\frac{1}{6}\sum_{n=0}^{\infty} \frac{1+(-1)^n}{3^{n+1}}(x-1)^n, \qquad |x-1|<3.

Both poles are actual poles, and each is at distance $3$, so the radius is exactly $3$. As a check, at the center $x=1$, only the constant term contributes: $-\frac16\frac{2}{3}=-\frac19$. Direct substitution gives $f(1)=\frac{1}{(-3)(3)}=-\frac19$. A second check uses $x=0$, which lies inside the disk. The two partial-fraction series evaluate to $\frac{1}{x-4}=-\frac14$ and $\frac{1}{x+2}=\frac12$; hence their combination is $\frac16(-\frac14-\frac12)=-\frac18$, agreeing with $f(0)=\frac{1}{(-4)(2)}=-\frac18$.

Why the Nearest Actual Pole Determines the Radius

The next result explains why distances, rather than just a convenient algebraic form, determine the convergence radius. A rational function can be decomposed over the complex numbers into a polynomial and partial-fraction terms. In the general case, a pole $p_j$ of order $m_j$ contributes terms with denominators $(x-p_j)^k$ for $1\le k\le m_j$. A pole need not be simple merely because it is distinct from the other poles.

Theorem: Let $f$ be a non-polynomial rational function in reduced form, and let $c$ be a point that is not a pole. Its Taylor series about $c$ has radius of convergence $$ R=\min_j |p_j-c|, $$ where the $p_j$ are all the actual poles of $f$. If $f$ is a polynomial, its Taylor series has infinite radius of convergence.

Justification. For each partial-fraction term with pole $p_j$, the reciprocal-factor or repeated-pole expansion converges whenever $|x-c|<|p_j-c|$. A finite sum of such expansions and a polynomial therefore represent $f$ throughout the disk $|x-c|<R$, where $R=\min_j|p_j-c|$. This gives convergence throughout that disk.

To see why the radius cannot be larger, suppose the Taylor series converged in a disk reaching a pole $p_j$ at distance $R$. A convergent power series defines an analytic, and therefore bounded on a sufficiently small closed disk, function near its center and throughout its disk of convergence. Within the disk, where the rational function has no poles, the series agrees with $f$. Since $p_j$ is an actual pole, $f$ becomes unbounded as $x$ approaches $p_j$; the series cannot extend the same function analytically through that point. This contradiction shows that the radius is at most $R$. Together with convergence for $|x-c|<R$, it proves the formula. A polynomial has a finite Taylor expansion, so its radius is infinite.

Worked Example: A Repeated Pole

Worked Example: Repeated Pole and Its Convergence Radius

Expand $f(x)=\frac{1}{(x-2)^2}$ about $c=1$, and check the series at $x=\frac12$.

The pole is $p=2$, with order $2$, and its distance from the center is $|2-1|=1$. With $h=x-1$, the repeated-pole formula gives

\frac{1}{(x-2)^2} = \frac{1}{(1-h)^2} = \sum_{n=0}^{\infty}(n+1)h^n, \qquad |h|<1.

For the check, take $x=\frac12$, so $h=-\frac12$. The geometric-series derivative identity $\sum_{n=0}^{\infty}(n+1)u^n=\frac{1}{(1-u)^2}$, valid for $|u|<1$, gives

\sum_{n=0}^{\infty}(n+1)\left(-\frac12\right)^n = \frac{1}{(1+\frac12)^2} = \frac49.

Direct substitution confirms $f(\frac12)=\frac{1}{(\frac12-2)^2}=\frac{1}{(-\frac32)^2}=\frac49$. The radius is $1$, exactly the distance to the repeated pole; the pole’s order affects the coefficients, not the radius.

Common Pitfalls

Using a cancelled factor as a pole. The expression $\frac{x^2-1}{x-1}$ simplifies to $x+1$ wherever the original quotient is defined. After cancellation there is no pole at $x=1$; the simplified function has a polynomial Taylor expansion with infinite radius. Treating the uncancelled denominator factor as a genuine pole would give the wrong radius.

Using the wrong distance. The distance is measured from the chosen center, not from zero unless zero is the center. For example, the pole $p=4$ is $3$ units from the center $c=1$, not $4$ units away. The corresponding convergence condition is $|x-1|<3$.

Confusing the open disk with endpoint behavior. A radius $R$ guarantees convergence when $|x-c|<R$. It does not, by itself, say whether the series converges at points where $|x-c|=R$. Those boundary points require separate analysis.

Choosing a Center in Practice

When a problem specifies a center, use it consistently: check that it is not a pole, calculate every pole’s distance from it, and expand each partial-fraction term with that same center. When you can choose the center, consider the region where the series will be used. The best center is not necessarily the one with the largest radius overall; it is one that makes the desired points lie inside its convergence disk and leaves the algebra manageable.

Reduce and locate poles.
Cancel common factors, then identify the actual poles and their orders.

Measure from the center.
For each pole $p_j$, calculate $|p_j-c|$; the smallest distance sets the radius.

Expand each term.
Rewrite the partial-fraction terms using geometric or repeated-geometric expansions, valid inside their stated disks.

State the convergence region.
Give the open disk $|x-c|<R$, and examine boundary points separately if needed.

Check Your Understanding

Use the pole distances and expansion conditions to answer these questions.

For $\frac{1}{x+5}$ centered at $c=1$, what is the radius of convergence, and what is the geometric-series condition?
A rational function has actual poles at $2$ and $-3$. What radius does its Taylor series have when centered at $c=0$?
Why does a repeated pole change the coefficients of an expansion without changing the distance-based radius?
What must be checked before treating a denominator zero as a pole?
If a series has radius $R$, what additional work is needed to decide convergence at a point on the boundary?