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Representing Functions by Series · Tutorial 784 of 998

Composing Functions Inside a Geometric Series

Substitute a function into the geometric series, determine where its value has absolute value less than one, and verify the resulting representation.

Advanced 10 min read

What You'll Learn

  • State the geometric-series composition rule and its convergence condition
  • Derive the condition on x by solving |g(x)| < 1
  • Represent a rational function as a geometric series in a composed ratio
  • Distinguish convergence of the composed series from a power series in x
  • Check a composed representation using its finite partial sums

Why Compose a Geometric Series?

A familiar way to represent a function is to begin with the geometric series and replace its variable by an expression involving \(x\). This can turn a simple series into a representation for a polynomial or rational function. The key is to track the value of the substituted expression: the geometric series converges only when that value has absolute value less than one.

For a function \(g\), the substitution is not automatically valid at every \(x\) where the resulting algebraic expression makes sense. We must first require \(|g(x)|<1\), then simplify the sum. This tutorial develops that rule and applies it to polynomial and rational compositions.

Central rule: If \(g(x)\) is defined and \(|g(x)|<1\), then the geometric series with ratio \(g(x)\) converges, and $$ \sum_{n=0}^{\infty}g(x)^n=\frac{1}{1-g(x)}. $$ The condition is on the value of the entire substituted expression, not on its separate pieces.

The Geometric-Series Composition Rule

The geometric series with real ratio \(r\) is \(\sum_{n=0}^{\infty}r^n\). It converges exactly when \(|r|<1\), and then its sum is \(1/(1-r)\). Substituting \(r=g(x)\) gives a pointwise statement: fix a real \(x\) at which \(g(x)\) is defined, and apply the geometric-series result to the resulting real number \(g(x)\).

Definition and condition: The composed geometric series associated with \(g\) is $$ \sum_{n=0}^{\infty} [g(x)]^n. $$ At a particular real \(x\), this series converges if and only if \(|g(x)|<1\). Wherever it converges, its sum is \(1/(1-g(x))\).

The condition is both necessary and sufficient. If \(|g(x)|<1\), its powers tend to zero and the geometric-series sum formula applies. If \(|g(x)|\geq 1\), the terms \([g(x)]^n\) do not tend to zero, so the series cannot converge. In particular, at \(g(x)=1\) the partial sums grow without bound, and at \(g(x)=-1\) they oscillate.

Why the Sum Formula Still Works After Substitution

The justification is the finite geometric-sum identity. For a real number \(r\neq 1\), multiplying the finite sum by \(1-r\) makes its middle terms cancel:

$$ (1-r)(1+r+r^2+\cdots+r^N)=1-r^{N+1}. $$

Dividing by \(1-r\) gives the finite sum, and when \(|r|<1\), the remainder term \(r^{N+1}\) tends to zero. The same argument applies after setting \(r=g(x)\), provided \(x\) is fixed and \(|g(x)|<1\).

1
Fix an input.
Choose a real \(x\) for which \(g(x)\) is defined.
2
Check convergence.
Require \(|g(x)|<1\), so the powers of this fixed ratio tend to zero.
3
Apply the finite identity and take a limit.
The finite sums tend to \(1/(1-g(x))\), giving the infinite-series representation.
$$ \sum_{n=0}^{N}[g(x)]^n = \frac{1-[g(x)]^{N+1}}{1-g(x)} \quad\longrightarrow\quad \frac{1}{1-g(x)} \qquad\text{when } |g(x)|<1. $$

This is a pointwise argument: it establishes convergence for each fixed input satisfying the condition. It does not assert that the series converges at inputs where the condition fails.

Worked Example: A Polynomial Inside the Geometric Series

Worked Example: Compose with \(g(x)=x+x^2\)

Represent \(1/(1-x-x^2)\) using the geometric series, and determine where the representation converges.

Here \(g(x)=x+x^2\), so the required inequality is

$$ |x+x^2|<1, \qquad\text{equivalently}\qquad -1<x+x^2<1. $$

For the lower inequality, complete the square:

$$ x+x^2+\frac14=\left(x+\frac12\right)^2\geq 0, \qquad x+x^2\geq-\frac14>-1. $$

Thus the lower inequality holds for every real \(x\). The upper inequality is

$$ x^2+x-1<0. $$

The roots of \(x^2+x-1=0\) are \((-1-\sqrt5)/2\) and \((-1+\sqrt5)/2\). Since the quadratic opens upward, it is negative between its roots. Combining the two inequalities gives

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

Therefore the composed geometric series is

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

on that open interval. Its first terms can be found by expanding the first few powers:

$$ \begin{aligned} 1+(x+x^2)+(x+x^2)^2+(x+x^2)^3+(x+x^2)^4+\cdots &=1+(x+x^2)+(x^2+2x^3+x^4)\\ &\quad +(x^3+3x^4+3x^5+x^6)\\ &\quad +(x^4+4x^5+6x^6+4x^7+x^8)+\cdots\\ &=1+x+2x^2+3x^3+5x^4+\cdots. \end{aligned} $$

For instance, the coefficient of \(x^4\) is \(1+3+1=5\): it comes from the second, third, and fourth powers of \(x+x^2\), respectively. The composition rule verifies the complete representation: the ratio is \(x+x^2\), it has absolute value less than one on the interval just found, and the sum is \(1/[1-(x+x^2)]=1/(1-x-x^2)\). At either endpoint, \(|x+x^2|=1\), so the geometric series does not converge there.

Worked Example: A Rational Function as the Ratio

Worked Example: Compose with \(g(x)=x/(2-x)\)

Find where the composed series converges and identify its sum.

The ratio is defined for \(x\neq 2\). Its convergence condition is

$$ \left|\frac{x}{2-x}\right|<1. $$

Both sides of the equivalent inequality \(|x|<|2-x|\) are nonnegative, so squaring preserves the inequality:

$$ x^2<(2-x)^2 =x^2-4x+4 \quad\Longleftrightarrow\quad 0<4-4x \quad\Longleftrightarrow\quad x<1. $$

The condition \(x\neq2\) is already satisfied whenever \(x<1\). Thus

$$ \sum_{n=0}^{\infty}\left(\frac{x}{2-x}\right)^n = \frac{1}{1-\frac{x}{2-x}} = \frac{2-x}{2-2x}, \qquad x<1. $$

To check the simplification, combine the terms in the denominator:

$$ 1-\frac{x}{2-x} = \frac{2-x-x}{2-x} = \frac{2-2x}{2-x}, \qquad \frac{1}{1-\frac{x}{2-x}} = \frac{2-x}{2-2x}. $$

For a direct check of convergence and the sum, the finite partial sum is

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

When \(x<1\), the ratio has absolute value less than one, so the power in the numerator tends to zero. The limit is exactly \((2-x)/(2-2x)\), as claimed.

This composition also illustrates an important distinction. The displayed series in powers of \(x/(2-x)\) converges for every \(x<1\). If instead we rewrite its sum as a power series centered at zero, we obtain

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

The latter power series has the narrower condition \(|x|<1\). It is a different series representation, so its convergence condition should not be confused with that of the original composed geometric series.

Worked Example: A Shifted Rational Ratio

Worked Example: Compose with \(g(x)=(x-1)/(x+2)\)

Determine the convergence interval and sum of the composed series.

The ratio is defined when \(x\neq-2\). To solve its absolute-value condition, use \(|x-1|<|x+2|\) and square both nonnegative sides:

$$ (x-1)^2<(x+2)^2 \quad\Longleftrightarrow\quad x^2-2x+1<x^2+4x+4 \quad\Longleftrightarrow\quad -6x<3 \quad\Longleftrightarrow\quad x>-\frac12. $$

Every \(x>-\tfrac12\) differs from \(-2\), so the ratio is defined throughout this interval. The series representation is

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

Indeed, the denominator in the sum simplifies as follows:

$$ 1-\frac{x-1}{x+2} = \frac{x+2-(x-1)}{x+2} = \frac{3}{x+2}. $$

Its reciprocal is \((x+2)/3\), which verifies the sum formula. Also, the finite partial sum has remainder equal to \([ (x-1)/(x+2)]^{N+1}\) divided by \(1-(x-1)/(x+2)\). Since the ratio has absolute value less than one for \(x>-\tfrac12\), that remainder tends to zero, confirming the claimed limit. At \(x=-\tfrac12\), the ratio is \(-1\), so the series fails to converge at the endpoint.

A Common Mistake: Checking the Wrong Quantity

A frequent error is to use the geometric-series formula without testing the substituted ratio. For example, someone might write

$$ \sum_{n=0}^{\infty}(2x)^n=\frac{1}{1-2x} \quad\text{for every }x\neq\frac12. $$

The algebraic expression \(1/(1-2x)\) is defined for many values where the series does not converge. The geometric condition is \(|2x|<1\), not merely \(2x\neq1\). Therefore the corrected representation is

$$ \sum_{n=0}^{\infty}(2x)^n=\frac{1}{1-2x}, \qquad -\frac12<x<\frac12. $$

For example, at \(x=1\), the terms are \(1,2,4,8,\ldots\), which do not tend to zero. The expression \(1/(1-2x)=-1\) exists there, but it is not the sum of that divergent series.

Takeaway: Simplifying \(1/(1-g(x))\) does not establish convergence. First verify that \(g(x)\) is defined and satisfies \(|g(x)|<1\).

How to Use the Composition

For a proposed composition, the reliable procedure is to identify the full ratio \(g(x)\), determine where it is defined, and solve \(-1<g(x)<1\) or the equivalent absolute-value inequality. Only then should you apply the geometric-series sum formula. This sequence keeps the convergence condition attached to the actual series rather than to a simplified expression that might have a larger domain.

The result can be useful even when the ratio is rational rather than polynomial. In that case, the terms of the composed series need not be powers of \(x\), but the geometric-series identity still gives a valid representation wherever the ratio has absolute value less than one. If the goal is a power series centered at a particular point, that is a separate representation whose convergence condition must be established for that series.

Check Your Understanding

For each question, begin by identifying the ratio being substituted and applying its absolute-value condition.

  1. For \(g(x)=3x^2\), on what set of real \(x\) does \(\sum_{n=0}^{\infty}(3x^2)^n\) converge, and what is its sum there?
  2. For \(g(x)=x-2\), solve \(|g(x)|<1\) and write the corresponding geometric-series representation.
  3. Why does the expression \(1/(1-g(x))\) existing at an input not by itself prove that the composed series converges there?
  4. For \(g(x)=x/(1+x)\), find the real inputs where \(g(x)\) is defined and satisfies \(|g(x)|<1\).
  5. State the finite partial-sum identity for a composed geometric series and explain why its remainder tends to zero under the convergence condition.

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