Taylor Series for Rational Functions Near Regular Points

Why Rational Functions Have Taylor Series

A rational function is a quotient of polynomials. At a point where its denominator is nonzero, it is defined and can be expanded locally as a Taylor series. The central task is to find the coefficients in a useful way and determine how far the resulting series converges.

For rational functions, the key information is not just the value of the denominator at the center. The locations of the function's poles—points where the reduced denominator is zero—determine the radius of convergence. Partial fractions make those locations and their contributions to the series visible.

Definition: A point \(a\) is a regular point of a rational function \(f=P/Q\) if, after canceling any common polynomial factors of \(P\) and \(Q\), the remaining denominator satisfies \(Q(a)\ne 0\). A Taylor series about \(a\) is a power series in \(x-a\) that equals \(f(x)\) for \(x\) sufficiently close to \(a\).

Expanding a Partial-Fraction Term

The basic building block is a term with a pole at \(p\). Write \(x=a+(x-a)\) and factor out the value at the center:

\frac{1}{(x-p)^m} = \frac{1}{(a-p)^m} \left(1+\frac{x-a}{a-p}\right)^{-m}.

For a positive integer \(m\), the reciprocal-power series gives

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

When \(m=1\), this is the geometric series. Higher powers follow by differentiating the geometric series or by the reciprocal-power formula. The condition comes from requiring the geometric-series variable \(-(x-a)/(a-p)\) to have absolute value less than one.

A rational function can be divided into a polynomial part and partial fractions. Over the complex numbers, the partial fractions have terms of the form \(A/(x-p)^m\), where \(p\) is a pole and \(m\) is at most its multiplicity. A polynomial has a Taylor series with only finitely many nonzero terms. Each partial-fraction term has the expansion above, so the rational function's series converges at least as far as the nearest pole.

Theorem: For a rational function in reduced form that is not a polynomial, its Taylor series about a regular point \(a\) has radius of convergence equal to the distance from \(a\) to the nearest complex pole. Equivalently, if the poles are \(p_1,\ldots,p_k\), the radius is \(\min_j |a-p_j|\). A polynomial has infinite radius of convergence.

Complex poles matter even when \(x\) and the coefficients of \(f\) are real. A pair of complex-conjugate poles can limit the real Taylor series just as a real pole can. The distance is measured in the complex plane.

Why the Nearest Pole Determines the Radius

Partial fractions show that each pole term has a series converging for \(|x-a|<|a-p|\). Thus, adding the finitely many partial-fraction expansions gives a series that converges inside the distance to the nearest pole. To see why the radius cannot be larger, suppose a Taylor series had radius larger than the distance to a pole.

Write the rational function in reduced form.
Let \(f=P/Q\), with \(P\) and \(Q\) having no common factor. Then a zero \(p\) of \(Q\) is a genuine pole, so \(P(p)\ne 0\).

Use the local series identity.
If \(\sum_{n=0}^{\infty}c_n(x-a)^n\) is the Taylor series, then near \(a\), multiplying by \(Q(x)\) gives \(Q(x)\sum_{n=0}^{\infty}c_n(x-a)^n=P(x)\).

Reach a contradiction at the pole.
If the series radius exceeded \(|p-a|\), the power series identity would hold at \(p\) as well. Substitution would give \(Q(p)\sum c_n(p-a)^n=P(p)\), hence \(0=P(p)\), contrary to \(P(p)\ne0\).

Therefore no power series equal to the rational function near \(a\) can have a radius larger than the distance to a genuine pole. Together with the partial-fraction expansions, this proves the stated radius. The argument also explains why common factors must be canceled before identifying poles: a canceled denominator zero is not a pole of the function.

Worked Examples

Worked Example: Two Real Poles at Unequal Distances

Expand \(f(x)=1/((x-2)(x+1))\) about \(a=0\), and find its radius of convergence.

First decompose into partial fractions. We seek constants \(A\) and \(B\) such that

\frac{1}{(x-2)(x+1)} =\frac{A}{x-2}+\frac{B}{x+1}.

Combining the right-hand side gives numerator \(A(x+1)+B(x-2)\). Matching coefficients with \(1\) gives \(A+B=0\) and \(A-2B=1\). Thus \(B=-A\), \(3A=1\), so \(A=1/3\) and \(B=-1/3\).

For \(|x|<2\), expand the first term geometrically; for \(|x|<1\), expand the second. Both expansions hold together when \(|x|<1\):

\begin{aligned} \frac{1}{x-2} &=-\frac12\frac{1}{1-x/2} =-\sum_{n=0}^{\infty}\frac{x^n}{2^{n+1}},\\ \frac{1}{x+1} &=\frac{1}{1+x} =\sum_{n=0}^{\infty}(-1)^n x^n. \end{aligned}

Therefore

f(x) =-\frac13\sum_{n=0}^{\infty}\frac{x^n}{2^{n+1}} -\frac13\sum_{n=0}^{\infty}(-1)^n x^n, \qquad |x|<1.

The poles are \(2\) and \(-1\); their distances from zero are \(2\) and \(1\). The radius is therefore \(1\). As a check, the constant term is \(-1/6-1/3=-1/2\), which equals \(f(0)=1/((-2)(1))=-1/2\). The geometric expansions reconstruct the two partial-fraction terms, whose sum is the original rational function.

Worked Example: A Repeated Pole

Expand \(g(x)=1/(x+2)^2\) about \(a=0\), and identify the radius.

Rewrite the denominator to use the reciprocal-square series:

g(x)=\frac14\left(1+\frac{x}{2}\right)^{-2}.

Using \((1+u)^{-2}=\sum_{n=0}^{\infty}(-1)^n(n+1)u^n\) for \(|u|<1\), we obtain

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

The repeated pole is at \(-2\), whose distance from the center is \(2\), so the radius is \(2\). To check the series, set \(u=x/2\). The reciprocal-square identity sums it to \(\frac14(1+u)^{-2}\), which is exactly \(1/(x+2)^2\). Its constant term is \(1/4=g(0)\), as expected.

Worked Example: Complex Poles at a Nonzero Center

Find the first coefficients and the radius for \(h(x)=1/(x^2+1)\) about \(a=1\). The poles are \(i\) and \(-i\), each at distance \(\sqrt{2}\) from \(1\), so the radius is \(\sqrt{2}\).

Partial fractions over the complex numbers give

\frac{1}{x^2+1} =\frac{1}{2i}\left(\frac{1}{x-i}-\frac{1}{x+i}\right).

For either pole \(p\), the expansion of \(1/(x-p)\) about \(1\) has coefficient \((-1)^n/(1-p)^{n+1}\) on \((x-1)^n\). Consequently the coefficient \(c_n\) of \(h\) is

c_n=\frac{(-1)^n}{2i} \left(\frac{1}{(1-i)^{n+1}}-\frac{1}{(1+i)^{n+1}}\right).

Evaluating for \(n=0,1,2,3\) gives \(c_0=1/2\), \(c_1=-1/2\), \(c_2=1/4\), and \(c_3=0\). Thus the beginning of the expansion is

\frac{1}{x^2+1} =\frac12-\frac12(x-1)+\frac14(x-1)^2 +0(x-1)^3+\cdots,\qquad |x-1|<\sqrt2.

The complex terms combine to give real coefficients, as they must for this real-valued function. The partial-fraction geometric expansions sum back to \(h(x)\) when both are valid, and the constant term \(1/2\) agrees with \(h(1)=1/2\).

A Common Mistake: Checking Only the Center

A nonzero denominator at the center ensures that the rational function is regular there, but it does not determine the radius. For the first example, the denominator at \(0\) is \((0-2)(0+1)=-2\), which is nonzero. That fact alone does not say how far the Taylor series converges.

A tempting but incorrect argument is: “The nearest pole is \(2\), so the radius is \(2\).” This overlooks the pole at \(-1\). The correct distances are \(|2-0|=2\) and \(|-1-0|=1\), so the radius is their minimum, \(1\). Each partial-fraction expansion must be valid, and their shared disk is limited by the nearer pole.

Takeaway: Reduce the rational function first, list all its genuine poles (including complex ones), and measure every pole's distance from the expansion center. The smallest distance is the Taylor radius.

How to Approach a Rational Expansion

For a new rational function, a reliable procedure is:

Cancel common factors and confirm that the chosen center is regular.
Find the poles of the reduced denominator, including complex roots when necessary.
Use polynomial division and partial fractions to separate the polynomial part and pole terms.
Expand each term in powers of \(x-a\), recording its condition of convergence.
Take the smallest pole distance as the exact radius, then check initial coefficients against the function or its defining algebraic identity.

This method is useful because rational functions often look complicated as a single quotient but become simple sums of geometric or reciprocal-power terms. It also makes convergence information part of the calculation rather than an afterthought.

Check Your Understanding

Use the definitions and methods in this tutorial to answer the following.

What must be checked before deciding whether a point is regular for a rational function?
Expand \(1/(x-3)\) about \(a=0\), and state the condition on \(x\) for the geometric expansion.
Why can a complex pole determine the radius of a Taylor series with real coefficients?
For a reduced rational function with poles at \(2+i\) and \(2-i\), what is the radius about \(a=0\)?
Why is the nearest genuine pole, rather than merely a zero in an uncanceled denominator, used to determine the radius?