From Geometric Series to Rational Functions
A rational function can sometimes be represented by a power series without starting from a new formula. The key is to break the function into simpler fractions, then rewrite each fraction as a geometric series centered at the desired point.
This method is especially useful when the denominator factors into distinct linear factors. Each resulting fraction has one simple denominator, and its geometric expansion is determined by the location of that denominator’s zero relative to the center.
The Geometric Expansion of a Simple Fraction
The geometric-series identity is
For a simple fraction \(A/(x-r)\), choose a center \(c\) and write \(x-r=(c-r)+(x-c)\). If \(r\ne c\), factoring out \(c-r\) gives a constant multiple of the geometric form:
Substitute \(u=-(x-c)/(c-r)\). The geometric-series condition \(|u|<1\) becomes \(|x-c|<|c-r|\).
Why the Geometric Expansion Works
The identity used above follows from the finite geometric-sum formula. For \(u\ne1\),
When \(|u|<1\), \(u^{N+1}\) approaches \(0\) as \(N\) increases. Therefore the finite sums approach \(1/(1-u)\). With \(u=-(x-c)/(c-r)\), this proves the stated expansion under its required condition.
Partial fractions let us apply this argument one component at a time. Suppose a rational function has a decomposition
where \(P(x)\) is a polynomial, the \(r_j\) are distinct real numbers, and no \(r_j\) equals the center \(c\). Each fraction has a geometric expansion when \(|x-c|<|c-r_j|\). Thus all component expansions converge together when
The polynomial part remains unchanged. The displayed interval is the region guaranteed by expanding each fraction separately. At an endpoint, one or more component series may no longer converge; check the resulting series there instead of including the endpoint automatically.
Worked Example: One Fraction About a Nonzero Center
Worked Example: Expand a Simple Fraction About \(c=1\)
Find a power-series representation of \(f(x)=3/(x+1)\) centered at \(c=1\), and state its interval of convergence.
Step 1: Express the denominator using \(x-1\). Since \(x+1=2+(x-1)\),
Step 2: Apply the geometric identity. Use \(u=-(x-1)/2\):
The condition is \(\left|-(x-1)/2\right|<1\), or \(|x-1|<2\). Hence the open interval is \((-1,3)\).
Step 3: Check the representation. Summing the geometric series gives
At either endpoint, the geometric ratio has absolute value \(1\), so the terms do not approach \(0\). The series diverges at \(x=-1\) and \(x=3\); neither endpoint is included.
Worked Example: Two Partial Fractions and Coefficient Cancellation
Worked Example: Expand a Rational Function About \(c=1\)
Represent \(f(x)=1/((x+1)(x-3))\) as a power series centered at \(c=1\).
Step 1: Decompose into partial fractions. Seek constants \(A\) and \(B\) such that
Multiplying by \((x+1)(x-3)\) gives \(1=A(x-3)+B(x+1)\). Set \(x=-1\): \(1=-4A\), so \(A=-1/4\). Set \(x=3\): \(1=4B\), so \(B=1/4\). Therefore
Step 2: Expand each fraction about \(1\). Let \(h=x-1\), so \(x+1=h+2\) and \(x-3=h-2\). Then
Step 3: Combine corresponding terms. Both series converge when \(|h|<2\). Adding them yields
For odd \(n\), \((-1)^n+1=0\); for even \(n\), it equals \(2\). The remaining series is geometric, so
Step 4: Check against the original function. Since \(h=x-1\), we have \((x+1)(x-3)=(h+2)(h-2)=h^2-4\). Therefore
This agrees with the summed series. The endpoint values \(h=\pm2\) make the terms of the final series fail to approach \(0\), so the interval of convergence is \((-1,3)\).
Worked Example: Unequal Distances from the Center
Worked Example: Find the Shared Convergence Interval
Expand \(g(x)=1/((x+1)(x-4))\) about \(c=0\), and determine its interval of convergence.
Step 1: Find the partial fractions. Write \(1/((x+1)(x-4))=A/(x+1)+B/(x-4)\). Then \(1=A(x-4)+B(x+1)\). Setting \(x=-1\) gives \(1=-5A\), so \(A=-1/5\). Setting \(x=4\) gives \(1=5B\), so \(B=1/5\).
Step 2: Expand each part about zero.
Both expansions converge when their conditions hold, so their shared interval is \(|x|<1\). Adding them gives
Step 3: Check the function and endpoints. Summing the two geometric series returns \(-1/[5(1+x)]-1/[20(1-x/4)]\). Combining these fractions gives
At \(x=1\), the first series has terms of constant magnitude \(1/5\); at \(x=-1\), it also has terms of constant magnitude \(1/5\). Thus its terms do not approach \(0\) at either endpoint, and the interval is exactly \((-1,1)\).
A Reliable Process and a Common Sign Error
For a rational function with distinct real linear factors, the following process keeps the algebra and convergence conditions visible:
Write each denominator in terms of \(x-c\), rather than expanding about zero by habit.
Find the partial fractions and verify their sum reproduces the original rational function.
Identify the ratio \(u\) and impose \(|u|<1\) separately for every fraction.
Add the series only on their shared interval, simplify coefficients if useful, then sum or otherwise verify the result.
A frequent error is to lose a negative sign when the denominator is written as a constant plus \(x-c\). For example, about \(c=0\),
The incorrect calculation \(1/(1+x)=\sum_{n=0}^{\infty}x^n\) uses the wrong ratio: the geometric ratio is \(-x\), not \(x\). A quick check is to compare the first two terms. The correct series begins \(1-x+x^2-\cdots\); the incorrect one begins \(1+x+x^2+\cdots\). Only the first agrees with \(1/(1+x)\) for small \(x\).
Using the Representation Carefully
The method provides more than a way to rewrite a formula. It makes the center and convergence restriction explicit, and it can reveal cancellations among coefficients, as in the example with only even powers. The partial fractions also show why the location of each denominator’s zero matters: a zero farther from the center permits a wider geometric expansion, but the shared interval must satisfy every component’s condition.
Do not infer endpoint convergence from the open interval. At an endpoint the geometric ratio may have absolute value \(1\), so the geometric identity used in the derivation no longer applies. Substitute the endpoint into the resulting series and test convergence directly. Also distinguish the original rational function’s domain from the series interval: the function may be defined at points where a particular series representation does not converge.
Check Your Understanding
For each question, keep the center and each geometric-series condition explicit.
- Write \(2/(x-5)\) as a power series centered at \(c=1\), and state its interval of convergence.
- Find the partial fractions of \(1/((x-2)(x+4))\), then identify the shared convergence interval about \(c=0\).
- Why must the geometric-series condition be checked for every partial-fraction term?
- For \(1/(1-x)\), explain why the geometric expansion about zero does not include \(x=1\).
- In a sum of two geometric expansions, what does a zero coefficient for every odd power tell you about the resulting series?
