Finding a Power Series Representation
Build power series from 1/(1 − x): substitute, multiply, differentiate and integrate term by term, then state where the new series converges.
Finding a Power Series Representation
A Calc II problem gives you a function such as 1/(1+4x²) and asks for its power series representation. Evaluating a known series at a point is not enough; the task is to build the series from scratch by manipulating a standard expansion. Five moves turn rational, logarithmic, or inverse trigonometric functions into infinite polynomials while preserving the interval of convergence.
Start From the Geometric Series
Every power series representation you construct in Calc II begins with one formula: 1/(1 − x) = Σ_{n=0}^{∞} xⁿ, which converges for |x| < 1. This is the geometric series from OpenStax Calculus Volume 2, Section 6.4. Commit this to memory. From it, you derive representations for 1/(1 + x), 1/(1 − x²), and any rational function that can be rewritten as 1/(1 − r). The radius of convergence for this base series is R = 1. The interval of convergence is (-1, 1), endpoints excluded because the series diverges at x = 1 (harmonic series) and oscillates at x = -1.
Failure case: Students try to force a function into this form without factoring first. Given 1/(2 − x), do not write 1/(2 − x) = Σxⁿ/2. Factor: 1/(2 − x) = 1/2 * 1/(1 − x/2). The series becomes (1/2) Σ (x/2)ⁿ = Σ xⁿ / 2^{n+1}, converging for |x/2| < 1 → |x| < 2.
Substitution
Once you have the geometric form 1/(1, u), substitute any expression for u, provided you adjust the convergence condition accordingly. This is how you find the power series representation of 1/(1 + x²). Write it as 1/(1, (-x²)), so u = -x². The geometric series gives Σ (-x²)ⁿ = Σ (-1)ⁿ x^{2n}. The convergence condition |u| < 1 becomes | -x² | < 1 → |x| < 1.
Worked Example 1: Find the power series for f(x) = 3/(1-2x). Factor: 3 * 1/(1-2x). Let u = 2x. Series: 3 Σ (2x)ⁿ = 3 Σ 2ⁿ xⁿ = Σ 3·2ⁿ xⁿ. Convergence: |2x| < 1 → |x| < 1/2. Radius R = 1/2. Endpoints: at x = 1/2, series is 3 Σ 1ⁿ, diverges; at x = -1/2, series is 3 Σ (-1)ⁿ, diverges by nth term test. Interval: (-1/2, 1/2).
Multiplying by Powers of x
When the function includes an x^k factor, find the series for the rest of the function first, then multiply every term by x^k. This move does not change the radius of convergence, but it may shift the lowest index.
Worked Example 2: Represent f(x) = x²/(1 + x³). Start with 1/(1 + x³) = 1/(1, (-x³)) = Σ (-1)ⁿ x^{3n}, converging for |x| < 1. Multiply by x²: x²/(1 + x³) = Σ (-1)ⁿ x^{3n+2}, n from 0 to ∞. Interval stays (-1, 1). At x = 1, the series is Σ (-1)ⁿ, which diverges by the nth term test. At x = -1, the series is Σ (-1)ⁿ (-1)^{3n+2} = Σ (-1)^{5n+2} = Σ (-1)² = Σ 1, diverges. Interval: (-1, 1).
Differentiating and Integrating Term by Term
Two functions, ln(1+x) and arctan x, do not come from a simple geometric manipulation. Their power series come from integrating or differentiating a known geometric series, which is valid inside the interval of convergence. OpenStax Calculus Volume 2, Section 6.4 confirms the results.
Natural Logarithm
The derivative of ln(1+x) is 1/(1+x). The series for 1/(1+x) = Σ (-1)ⁿ xⁿ, converging for |x| < 1. Integrate term by term: ln(1+x) = C + Σ (-1)ⁿ x^{n+1}/(n+1). Set x=0 to find C = 0. Renaming index: ln(1+x) = Σ_{n=1}^{∞} (-1)^{n+1} xⁿ/n. Convergence: the interval includes x=1 (alternating harmonic series converges conditionally) but excludes x=-1 (negative harmonic diverges). Interval: (-1, 1].
Arctangent
Start with 1/(1+x²) = Σ (-1)ⁿ x^{2n} for |x| < 1. Integrate: arctan x = Σ (-1)ⁿ x^{2n+1}/(2n+1). The interval includes both endpoints. At x=1, the alternating series Σ (-1)ⁿ/(2n+1) converges by the alternating series test. At x=-1, the series is Σ (-1)ⁿ (-1)^{2n+1}/(2n+1) = Σ -(-1)ⁿ/(2n+1), which also converges. Interval: [-1, 1].
Using Taylor's Formula When Manipulation Won't Work
When the function does not fit the geometric pattern, things like e^x, sin x, cos x, or a function with no algebraic simplification, you must fall back on Taylor's formula. The Maclaurin series (Taylor series centered at 0) is f(x) = Σ f^{ (n) }(0) xⁿ / n!.For sin x, f^{ (2n+1) }(0)=(-1)ⁿ, f^{ (2n) }(0)=0, so sin x = Σ (-1)ⁿ x^{2n+1}/(2n+1)!, also R = ∞. This method is general but tedious; use it only when the algebraic moves fail. The Lagrange error bound from OpenStax Section 6.3 lets you bound the remainder: |R_N(x)| ≤ M|x|^{N+1}/(N+1)!, where M is the maximum of |f^{(N+1)}(t)| on [0, x].
Failure case: Students apply Taylor's formula to a rational function, e.g., 1/(1-x), and compute 40 derivatives. That is wasted work. Manipulate the geometric series instead, it is faster and you already know the coefficients.
Stating the Interval of Convergence
Every power series representation must include its interval of convergence. […] You must test the endpoints x = a ± R separately using the alternating series test, p-series test, or comparison test from OpenStax Chapter 5. A series may converge at both endpoints, one, or neither. The College Board AP Calculus BC Course and Exam Description (Unit 10) requires this endpoint check on free-response questions.
Failure case: Students find R = 1 and report (-1, 1) without testing endpoints. […] Interval: (-1/2, 1/2].
The One Thing That Most Often Goes Wrong
You find the series, you state the radius, you skip the endpoints. […] The extra two minutes at x = a ± R, applying the alternating series test or the p-series test, is what separates a complete representation from a partial one. Do not leave the interval open unless the series diverges at both endpoints.
| Function | Power Series | Interval of Convergence |
|---|---|---|
| 1/(1-x) | Σ xⁿ, n=0 to ∞ | (-1, 1) |
| 1/(1+x) | Σ (-1)ⁿ xⁿ, n=0 to ∞ | (-1, 1) |
| 1/(1+x²) | Σ (-1)ⁿ x^{2n}, n=0 to ∞ | (-1, 1) |
| ln(1+x) | Σ (-1)^{n+1} xⁿ/n, n=1 to ∞ | (-1, 1] |
| arctan x | Σ (-1)ⁿ x^{2n+1}/(2n+1), n=0 to ∞ | [-1, 1] |
| eˣ | Σ xⁿ/n!, n=0 to ∞ | (-∞, ∞) |
| sin x | Σ (-1)ⁿ x^{2n+1}/(2n+1)!, n=0 to ∞ | (-∞, ∞) |
Common Questions
How do I find the power series representation if my function is not in the standard list?
Rewrite the function algebraically to match a known form, usually 1/(1-u). […] If none work, apply Taylor's formula directly.
What is the actual error bound for my series, not just the calculator's displayed error?
Use the Lagrange error bound: |R_N(x)| ≤ M|x-a|^{N+1}/(N+1)!, where M is the maximum […] This requires computing a derivative bound, not just a numeric difference.
Why should I test endpoints separately if the ratio test already gives R?
The ratio test only tells you where absolute convergence fails. […] The alternating series test or p-series test at x = a ± R determines inclusion.
Can I integrate a power series term by term, and does the interval change?
Yes, inside the open interval of convergence. Integration preserves the radius R but may include endpoints that were excluded before. For example, integrating 1/(1+x) gives ln(1+x) which converges at x=1, whereas the original geometric series did not.