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Representing Functions by Series · Tutorial 778 of 998

Quadratic Denominators as Geometric Expansions

You will learn how to build and check geometric-series representations of reciprocals of quadratic denominators, and why factoring can give a different interval of convergence.

Advanced 9 min read

What You'll Learn

  • Normalize a quadratic denominator into a constant times one plus a polynomial with zero constant term.
  • Apply the geometric series when the normalized polynomial has absolute value less than one.
  • Find the real intervals where this condition holds, including when the valid set is disconnected.
  • Compare the convergence conditions from direct normalization and linear-factor expansions.
  • Check an expansion by summing it at a chosen point and comparing with the rational function.

Why Normalize a Quadratic Denominator?

A rational function with a quadratic denominator can sometimes be represented by expanding its reciprocal as a geometric series. The essential step is to put the denominator in the form \(a(1+v(x))\), where \(a\) is a nonzero constant and \(v(x)\) has no constant term. The geometric series then applies wherever \(|v(x)|<1\).

This approach does not require the quadratic to factor into real linear factors. It can also converge at points where a representation obtained by expanding individual linear factors does not. The condition depends on the particular expression \(v(x)\), so it must be checked for the quadratic being considered.

Definition: If \(a\ne 0\), a quadratic denominator \(a+bx+cx^2\) is normalized by writing it as \(a(1+v(x))\), where \(v(x)=\frac{b}{a}x+\frac{c}{a}x^2\). The geometric expansion is valid at real \(x\) for which \(|v(x)|<1\).

The Geometric Expansion for a Quadratic

The geometric-series formula is \(\frac{1}{1-r}=\sum_{n=0}^{\infty}r^n\) when \(|r|<1\). Set \(r=-v(x)\), and include the factor \(1/a\) from the normalization. This gives the quadratic-denominator formula.

$$ \frac{1}{a+bx+cx^2} = \frac{1}{a}\frac{1}{1+v(x)} = \frac{1}{a}\sum_{n=0}^{\infty}(-v(x))^n, \qquad v(x)=\frac{b}{a}x+\frac{c}{a}x^2, \qquad |v(x)|<1. $$

Each term is a power of a quadratic polynomial. If expanded and collected by powers of \(x\), the result is a power series in \(x\); the displayed form keeps the geometric structure visible. In particular, the condition is not automatically \(|x|<1\): it is the inequality \(\left|\frac{b}{a}x+\frac{c}{a}x^2\right|<1\).

Why the Formula Works

The geometric expansion is not just a formal substitution. Its finite partial sums leave a remainder that can be calculated exactly. That remainder tends to zero under the stated condition, which proves that the infinite sum equals the reciprocal.

1
Form the finite sum.
For \(N\ge 0\), let \(S_N=\frac{1}{a}\sum_{n=0}^{N}(-v)^n\), with \(v=v(x)\) fixed.
2
Use the finite geometric identity.
Multiplying the sum by \(1+v\) cancels its intermediate terms, so \((1+v)S_N=\frac{1-(-v)^{N+1}}{a}\).
3
Isolate the error.
Since \(a(1+v)=a+bx+cx^2\), subtracting \(S_N\) from the reciprocal gives \(\frac{1}{a+bx+cx^2}-S_N=\frac{(-v)^{N+1}}{a+bx+cx^2}\).
4
Take the limit.
If \(|v|<1\), then \((-v)^{N+1}\to 0\). The denominator is nonzero because \(1+v\ne 0\), so the remainder tends to zero and \(S_N\) tends to the reciprocal.

This also explains the boundary of the method. For a fixed real value of \(v\), the geometric terms \((-v)^n\) fail to tend to zero when \(|v|\ge 1\), so the geometric series does not converge there. Thus the inequality is the exact pointwise convergence condition for this expansion, not just a convenient estimate.

Worked Example: A Quadratic That Does Not Factor Over the Reals

Worked Example: Expand the Reciprocal of \(1+x+x^2\)

Here \(a=1\), and the normalized polynomial is \(v(x)=x+x^2\). Therefore

$$ \frac{1}{1+x+x^2} = \sum_{n=0}^{\infty}(-x-x^2)^n = 1-(x+x^2)+(x+x^2)^2-(x+x^2)^3+\cdots. $$

To find where this expansion converges for real \(x\), solve \(|x+x^2|<1\). The lower inequality is \(-1<x^2+x\). Since \(x^2+x+1=(x+\frac12)^2+\frac34>0\), this inequality holds for every real \(x\). The upper inequality is \(x^2+x<1\), or \(x^2+x-1<0\). Its roots are \(\frac{-1-\sqrt5}{2}\) and \(\frac{-1+\sqrt5}{2}\), so the convergence interval is

$$ \frac{-1-\sqrt5}{2}<x<\frac{-1+\sqrt5}{2}. $$

For a numerical check, take \(x=\frac12\). Then \(v=\frac12+\frac14=\frac34\), and the sum is \(\sum_{n=0}^{\infty}(-\frac34)^n=\frac{1}{1+\frac34}=\frac47\). Direct substitution gives \(1+x+x^2=1+\frac12+\frac14=\frac74\), whose reciprocal is also \(\frac47\).

Now compare with factoring over the complex numbers. If \(\omega=e^{2\pi i/3}\), then \(\omega+\overline{\omega}=-1\) and \(\omega\overline{\omega}=1\), so \(1+x+x^2=(1-\omega x)(1-\overline{\omega}x)\). Expanding each reciprocal separately by a geometric series requires \(|\omega x|<1\) and \(|\overline{\omega}x|<1\), hence \(|x|<1\). The direct quadratic expansion also works at \(x=-\frac65\), since \(v=-\frac65+\frac{36}{25}=\frac6{25}\), which has absolute value less than one. The factor expansions do not work there because \(|x|=\frac65>1\). The same rational function can therefore have different valid regions for different series representations.

Worked Example: A Quadratic That Factors Into Real Linear Terms

Worked Example: Expand the Reciprocal of \(1-x^2\)

Write the denominator as \(1+v(x)\), where \(v(x)=-x^2\). The direct expansion is

$$ \frac{1}{1-x^2} = \sum_{n=0}^{\infty}x^{2n} = 1+x^2+x^4+\cdots, \qquad |x^2|<1. $$

For real \(x\), the condition is \(|x|<1\). At \(x=\frac12\), the geometric ratio is \(\frac14\), so the sum is \(\frac{1}{1-\frac14}=\frac43\). Direct substitution gives \(1-(\frac12)^2=\frac34\), and its reciprocal is \(\frac43\), as required.

The same denominator factors as \((1-x)(1+x)\). Each reciprocal factor has a geometric expansion about zero when \(|x|<1\), so the factor-based representation has the same convergence condition here. This example shows that the direct method need not give a larger region; the comparison depends on the quadratic.

Worked Example: A Disconnected Convergence Set

Worked Example: Expand the Reciprocal of \(2+3x+x^2\)

Factor out the constant term \(2\):

$$ 2+3x+x^2 = 2\left(1+\frac32x+\frac12x^2\right). $$

Thus \(v(x)=\frac32x+\frac12x^2\), and the expansion is

$$ \frac{1}{2+3x+x^2} = \frac12\sum_{n=0}^{\infty} \left(-\frac32x-\frac12x^2\right)^n, \qquad \left|\frac32x+\frac12x^2\right|<1. $$

To solve the condition, first require \(v<1\). This gives \(x^2+3x-2<0\), so \(\frac{-3-\sqrt{17}}2<x<\frac{-3+\sqrt{17}}2\). Next require \(v>-1\). This gives \(x^2+3x+2>0\), or \((x+1)(x+2)>0\), so \(x<-2\) or \(x>-1\). Intersecting these sets gives

$$ \frac{-3-\sqrt{17}}2<x<-2 \qquad\text{or}\qquad -1<x<\frac{-3+\sqrt{17}}2. $$

The convergence set has two intervals. For a check, take \(x=-\frac12\). Then \(v=-\frac34+\frac18=-\frac58\), so the series sums to \(\frac12\sum_{n=0}^{\infty}(\frac58)^n=\frac12\cdot\frac{1}{1-\frac58}=\frac43\). The denominator itself is \(2-\frac32+\frac14=\frac34\), giving reciprocal \(\frac43\).

The denominator also factors as \((x+1)(x+2)\), and

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

Expanding these two factors about zero requires both \(|x|<1\) and \(|x/2|<1\), so their common interval is \(|x|<1\). The direct quadratic expansion also works at \(x=-3\): there \(v=\frac12(9-9)=0\), and the series gives \(\frac12\). Direct substitution gives \(2-9+9=2\), whose reciprocal is \(\frac12\). The factor expansions cannot be used at \(x=-3\), because their individual geometric ratios do not have absolute value less than one.

A Common Sign Error

The geometric formula is \(\frac{1}{1-r}=\sum r^n\), not \(\frac{1}{1+r}=\sum r^n\). When the denominator is \(1+v\), the ratio must be \(r=-v\). Losing this minus sign changes the function being represented.

Common mistake: For \(\frac{1}{1+x+x^2}\), an incorrect substitution gives \(\sum_{n=0}^{\infty}(x+x^2)^n=\frac{1}{1-(x+x^2)}\). At \(x=\frac{1}{10}\), this incorrect sum is \(\frac{1}{1-\frac{11}{100}}=\frac{100}{89}\), while the target function is \(\frac{1}{1+\frac{11}{100}}=\frac{100}{111}\). The correction is to use \(\sum_{n=0}^{\infty}(-x-x^2)^n\), whose sum at that point is \(\frac{100}{111}\).

How to Choose Between the Representations

Direct normalization is especially useful when the quadratic is irreducible over the reals or when its normalized polynomial is small at points of interest. Factoring can be useful when the linear-factor expansions are simple. Neither method should be assumed to give the widest possible interval before its convergence conditions are checked.

For each proposed representation, apply the condition belonging to that representation. For direct normalization, test \(|v(x)|<1\). For separate geometric expansions of linear factors, test the absolute-value condition for each factor and take their common region. The rational function’s domain must also be respected: a series cannot represent the function at a point where its denominator is zero.

Key takeaway: Normalize a quadratic denominator as \(a(1+v(x))\), then use \(\frac{1}{a}\sum_{n=0}^{\infty}(-v(x))^n\) exactly where \(|v(x)|<1\). Factoring gives another possible representation, with its own convergence conditions.

Check Your Understanding

For each question, state the normalization and check the convergence condition before evaluating or comparing representations.

  1. Write the geometric expansion of \(\frac{1}{3-2x+x^2}\) in terms of a normalized polynomial \(v(x)\).
  2. For that expansion, what inequality must real \(x\) satisfy for convergence?
  3. For \(\frac{1}{1+2x+x^2}\), identify the ratio used in the geometric series and explain why its sign matters.
  4. Explain why separate geometric expansions of linear factors can have a narrower region of convergence than direct normalization.
  5. At a point where \(v(x)=0\), what value does the direct geometric expansion give, and why must the denominator be nonzero there?

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