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Representing Functions by Series · Tutorial 797 of 998

Building Series Representations for Composite Rational Functions

Develop a reliable sequence of transformations for rational functions, then determine where the resulting series is guaranteed to represent the function.

Advanced 10 min read

What You'll Learn

  • Normalize shifted linear factors before substituting into a geometric series.
  • Expand repeated factors using weighted geometric-series coefficients.
  • Multiply series by forming coefficients from finite convolutions.
  • Build a quotient series through coefficient recursion.
  • Combine convergence restrictions from each transformation and distinguish a guaranteed interval from a maximal one.

Why Composite Rational Functions Need a Plan

A rational function may combine several structures: a shifted linear factor, a repeated denominator factor, a numerator, or a denominator that is itself a polynomial. Each structure suggests a different series transformation. Applying the transformations in a deliberate order makes the algebra manageable and helps prevent a common error: using an expansion outside the range where it is valid.

The basic building block is the geometric series. For \(|u|<1\),

$$ \frac{1}{1-u}=\sum_{n=0}^{\infty}u^n. $$

Shifting and scaling a linear factor changes what expression plays the role of \(u\). Repeated factors produce weighted coefficients, while products combine coefficients by convolution. A quotient whose denominator is a power series can be handled by solving for the coefficients of its reciprocal.

Definition: A power-series representation of a function about \(x=0\) is an identity of the form \(f(x)=\sum_{n=0}^{\infty}c_nx^n\) on a stated set of \(x\)-values where the series converges. The convergence conditions belong to the representation and must be checked after the transformations are combined.

A Sequence for Building the Representation

For a composite rational function, first factor or normalize its denominator where useful. Next, identify a known series for each resulting piece. Then combine the series, keeping track of the interval on which every step is justified. In particular, a product of two series is directly justified where both series converge absolutely.

1
Normalize each factor.
Rewrite a shifted factor so its constant term is separated and the remaining expression has the form \(1-u(x)\).
2
Choose an expansion.
Use a geometric series for a simple reciprocal, a weighted geometric series for a repeated factor, or coefficient recursion for a more general reciprocal.
3
Combine the pieces.
For products, calculate each coefficient as a finite sum. For a quotient, multiply by the unknown reciprocal series and match coefficients.
4
Check the conditions.
Intersect the restrictions required by the individual expansions and by any product or quotient operation.

One useful repeated-factor identity follows by differentiating the geometric series inside its interval of convergence. For \(|u|<1\), differentiation gives \(\frac{1}{(1-u)^2}=\sum_{n=0}^{\infty}(n+1)u^n\). The factor \(n+1\) is essential: a repeated denominator does not have the same coefficients as a simple geometric series.

Worked Example: A Shifted Factor Times a Repeated Factor

Worked Example: A Shifted Factor Times a Repeated Factor

Represent \(f(x)=\frac{1}{(2-x)(1+x)^2}\) as a power series about zero, and state where the construction is guaranteed to work.

Normalize both factors:

$$ f(x)=\frac12\frac{1}{1-x/2}\frac{1}{(1+x)^2}. $$

The first factor has geometric expansion \(\sum_{k=0}^{\infty}(x/2)^k\) when \(|x|<2\). For the second factor, substitute \(u=-x\) into the repeated-factor identity. It becomes \(\sum_{j=0}^{\infty}(-1)^j(j+1)x^j\), valid when \(|x|<1\). Thus both expansions, and their product, are justified for \(|x|<1\).

The coefficient of \(x^n\) in the product is the sum of products of terms whose powers add to \(n\):

$$ c_n=\frac12\sum_{k=0}^{n}\left(\frac12\right)^k(-1)^{n-k}(n-k+1). $$

For the first coefficients, compute the finite sums explicitly:

$$ c_0=\frac12,\qquad c_1=\frac12\left(-2+\frac12\right)=-\frac34, $$ $$ c_2=\frac12\left(3-1+\frac14\right)=\frac98,\qquad c_3=\frac12\left(-4+\frac32-\frac12+\frac18\right)=-\frac{23}{16}. $$

Therefore, on the guaranteed interval,

$$ f(x)=\frac12-\frac34x+\frac98x^2-\frac{23}{16}x^3+\cdots,\qquad |x|<1. $$

Check: The denominator simplifies to \((2-x)(1+x)^2=2+3x-x^3\). Multiplying it by the displayed series through degree three gives constant coefficient \(2(1/2)=1\), coefficient of \(x\) equal to \(2(-3/4)+3(1/2)=0\), coefficient of \(x^2\) equal to \(2(9/8)+3(-3/4)=0\), and coefficient of \(x^3\) equal to \(2(-23/16)+3(9/8)-1(1/2)=0\). This agrees with the numerator \(1\) through the calculated order.

Why Multiplication Produces Convolution Coefficients

Suppose \(\sum_{n=0}^{\infty}a_nx^n\) and \(\sum_{n=0}^{\infty}b_nx^n\) both converge absolutely at a particular \(x\). Their product can be written as a double sum of terms \(a_ib_jx^{i+j}\). Absolute convergence allows these terms to be regrouped by their total power \(i+j\). For total power \(n\), the possible pairs are \((0,n),(1,n-1),\ldots,(n,0)\). Consequently, the product coefficient is the finite sum \(c_n=\sum_{k=0}^{n}a_kb_{n-k}\).

Why this works: Absolute convergence ensures that regrouping the terms of the product does not change its sum. Grouping by total power gives the convolution formula, and each coefficient involves only finitely many terms.

This also explains why convergence restrictions should be tracked during the calculation. The product formula is justified wherever both original series converge absolutely; a formal multiplication of coefficients alone does not establish that the resulting series represents the product at other values.

Worked Example: Scaling a Repeated Factor and Multiplying by a Numerator

Worked Example: Scaling a Repeated Factor and Multiplying by a Numerator

Find a power-series representation of \(g(x)=\frac{1+x}{(2+x)^2}\) and its guaranteed interval of convergence.

Factor the constant from the denominator and express the repeated factor using the weighted geometric series:

$$ g(x)=\frac14(1+x)\frac{1}{(1+x/2)^2} =\frac14(1+x)\sum_{n=0}^{\infty}(-1)^n(n+1)\left(\frac{x}{2}\right)^n. $$

This expansion requires \(|x/2|<1\), or \(|x|<2\). Before multiplying by \(1+x\), let \(d_n=\frac14(-1)^n(n+1)/2^n\) denote the coefficient of \(x^n\) in the repeated-factor series including the factor \(1/4\). The coefficient after multiplication is \(c_0=d_0\), and \(c_n=d_n+d_{n-1}\) for \(n\geq1\). The first values are

$$ d_0=\frac14,\quad d_1=-\frac14,\quad d_2=\frac{3}{16},\quad d_3=-\frac18, $$ $$ c_0=\frac14,\quad c_1=-\frac14+\frac14=0,\quad c_2=\frac{3}{16}-\frac14=-\frac{1}{16},\quad c_3=-\frac18+\frac{3}{16}=\frac{1}{16}. $$

Hence

$$ g(x)=\frac14-\frac{1}{16}x^2+\frac{1}{16}x^3+\cdots,\qquad |x|<2. $$

Check: The denominator is \((2+x)^2=4+4x+x^2\). Multiplying the displayed series by it gives constant term \(4(1/4)=1\), coefficient of \(x\) equal to \(4(0)+4(1/4)=1\), coefficient of \(x^2\) equal to \(4(-1/16)+4(0)+1(1/4)=0\), and coefficient of \(x^3\) equal to \(4(1/16)+4(-1/16)+0=0\). These are the coefficients of \(1+x\) through degree three.

Quotients Through Coefficient Recursion

When a denominator is not already a single linear factor or repeated factor, its reciprocal can be found by coefficient recursion. Suppose \(D(x)=1+d_1x+d_2x^2+\cdots\), and seek \(1/D(x)=\sum_{n=0}^{\infty}a_nx^n\). The identity \(D(x)\sum a_nx^n=1\) requires \(a_0=1\). For each \(n\geq1\), matching the coefficient of \(x^n\) gives \(a_n+d_1a_{n-1}+\cdots+d_na_0=0\). This determines each new coefficient from earlier ones. The coefficient calculation is necessary, but convergence must still be established for the resulting series.

Worked Example: A Quotient with a Quadratic Denominator

Worked Example: A Quotient with a Quadratic Denominator

Find a series representation for \(h(x)=\frac{1+x}{1-x-x^2}\) about zero and give an interval where it converges.

Let \(1/(1-x-x^2)=\sum_{n=0}^{\infty}a_nx^n\). Multiplying by \(1-x-x^2\) and matching coefficients gives \(a_0=1\), \(a_1-a_0=0\), and \(a_n-a_{n-1}-a_{n-2}=0\) for \(n\geq2\). Thus \(a_1=1\), and each later coefficient is the sum of the preceding two. With \(F_0=0,F_1=1\), and \(F_{n+2}=F_{n+1}+F_n\), the Fibonacci sequence, this says \(a_n=F_{n+1}\).

Multiplication by the numerator \(1+x\) changes the coefficient of degree \(n\geq1\) to \(a_n+a_{n-1}=F_{n+1}+F_n=F_{n+2}\). The constant coefficient is \(1=F_2\). Therefore the candidate representation is

$$ h(x)=\sum_{n=0}^{\infty}F_{n+2}x^n =1+2x+3x^2+5x^3+8x^4+\cdots. $$

To give a convergence interval, set \(\phi=(1+\sqrt5)/2\), so that \(\phi^2=\phi+1\). The recurrence and induction give \(F_{n+2}\leq\phi^{n+1}\) for \(n\geq0\). Therefore, if \(|x|<1/\phi=(\sqrt5-1)/2\), the terms are bounded by \(\phi(\phi|x|)^n\), a convergent geometric series. This proves convergence on the stated interval.

Check: Multiplying the first terms by the denominator gives \((1-x-x^2)(1+2x+3x^2+5x^3+8x^4+\cdots)\). The constant coefficient is \(1\); the coefficient of \(x\) is \(2-1=1\); the coefficient of \(x^2\) is \(3-2-1=0\); and those of \(x^3\) and \(x^4\) are \(5-3-2=0\) and \(8-5-3=0\). This matches \(1+x\), as required.

Common Pitfall: Forgetting the Substitution Condition

Consider the incorrect calculation \(\frac{1}{2-x}=\frac12\sum_{n=0}^{\infty}(x/2)^n\), followed by the claim that it is valid for every real \(x\). The algebraic normalization is correct, but the geometric series requires \(|x/2|<1\). For example, at \(x=3\), its terms are \((-3/2)^n\), which do not tend to zero, so the series diverges. The corrected statement is

$$ \frac{1}{2-x}=\frac12\sum_{n=0}^{\infty}\left(\frac{x}{2}\right)^n \quad\text{only for } |x|<2. $$

For a composite expression, collect all such restrictions. If one factor requires \(|x|<2\) and another requires \(|x|<1\), the product construction is guaranteed only where both conditions hold, namely \(|x|<1\). Endpoint behavior requires a separate convergence check; the open interval guaranteed by the transformations should not automatically be presented as the exact full interval of convergence.

Putting the Method to Work

Series representations of rational functions are useful when a function must be manipulated algebraically as a sequence of coefficients, or when the terms of a series are later used in a sum, limit, or integral. The important skill here is not merely recognizing a familiar expansion. It is choosing transformations in a workable order, making coefficient calculations explicit, and preserving the conditions that justify each identity.

Takeaway: Normalize first, expand each factor using a valid known series, combine coefficients by convolution or recursion, and then state the common convergence conditions. A correct formula without its conditions is incomplete.

Check Your Understanding

Answer each question by identifying both the coefficient operation and the relevant convergence condition.

  1. What substitution converts \(1/(1+x)^2\) into a weighted geometric series, and what restriction does it impose on \(x\)?
  2. Write the convolution formula for the coefficient of \(x^n\) in the product of two power series.
  3. For a denominator \(1+d_1x+d_2x^2+\cdots\), what equation determines the coefficient \(a_n\) of its reciprocal series?
  4. Why is absolute convergence important when multiplying two power series and regrouping terms by total degree?
  5. If two component expansions have different convergence restrictions, how do you find the interval guaranteed for their product?

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