Difficulty: Intermediate | Prerequisites: Familiarity with geometric series, basic differentiation and integration rules, partial fractions.
Power series let you rewrite functions as infinite polynomials, which turns hard calculus problems (integration, differentiation, limits) into straightforward algebra on polynomials. This material sits at the heart of the second half of Integral Calculus, bridging sequences and series with practical function analysis. You will need a solid grasp of geometric series and basic derivative/integral rules before diving in. Nearly every technique here builds on one core move: start from a known series (usually the geometric series) and adapt it through algebra, substitution, differentiation or integration to match the function you want.
To find a power series for a function, reshape it so it looks like 1/(1 - u), then substitute into the geometric series formula. If the function is a derivative or integral of something with a known series, differentiate or integrate term by term. The radius of convergence tells you where the series actually equals the function.
Power series
An infinite series of the form Σ cₙ(x - a)ⁿ, where a is the centre and cₙ are the coefficients. Think of it as an infinite polynomial built around a chosen point.
Geometric series
The series 1/(1 - u) = Σ uⁿ for |u| < 1. In simple terms, this is the starting recipe you modify to build most other power series.
Radius of convergence
The value R such that the series converges for |x - a| < R and diverges for |x - a| > R. Think of it as the range around the centre where the series actually works as a valid representation of the function.
Term-by-term differentiation
Differentiating each term of a convergent power series individually to obtain the series for the derivative. The radius of convergence stays the same.
Term-by-term integration
Integrating each term of a convergent power series individually. The radius of convergence stays the same, though you may need to check endpoints separately.
Maclaurin series
A Taylor series centred at a = 0. In simple terms, it is the special case of a power series expansion about the origin.
Taylor polynomial of degree n, Tₙ(x)
The partial sum of the first n + 1 terms of a Taylor series. It is the best polynomial approximation of degree n near the centre point.
The foundation of nearly every problem here. Start from:
1/(1 - u) = Σ uⁿ, for |u| < 1 (n = 0 to ∞)
To represent a function as a power series, rewrite it so the denominator matches the form (1 - something).
For f(z) = 1/(3 - z³): factor out 3 to get (1/3) · 1/(1 - z³/3). Now u = z³/3, so f(z) = Σ z³ⁿ / 3ⁿ⁺¹.
For f(z) = z/(9z + 1): rewrite as z · 1/(1 - (-9z)). Now u = -9z, so f(z) = Σ (-1)ⁿ 3²ⁿ zⁿ⁺¹ = Σ (-9z)ⁿ · z.
For f(z) = 1/(3 - z³), notice the cube: you substitute u = z³/3, which means powers jump by 3 (z⁰, z³, z⁶, ...). The exponent pattern in the answer always reflects the substitution.
Sometimes the function needs splitting first.
For (3 + 4x)/(1 + x): do polynomial long division or rewrite as 3/(1 + x) + 4x/(1 + x). Then note that 4x/(1 + x) = 4 - 4/(1 + x). So the whole expression becomes 4 - 1/(1 + x) = 4 - Σ (-1)ⁿ xⁿ, which simplifies to 3 - Σ (-1)ᵏ xᵏ (k = 1 to ∞). The hint in the homework says to separate, then use 1/(1 + x) = Σ (-x)ⁿ.
If f(x) looks like the derivative of a function with a known series, work backwards.
The series for 1/(1 - u)² comes from differentiating 1/(1 - u) = Σ uⁿ term by term: d/du [1/(1 - u)] = 1/(1 - u)² = Σ n uⁿ⁻¹ (n = 1 to ∞), or equivalently Σ (n + 1) uⁿ (n = 0 to ∞).
For f(y) = 1/(6 - y)²: rewrite as (1/36) · 1/(1 - y/6)², then apply the differentiated geometric series with u = y/6. The result is f(y) = Σ (n + 1)/(6ⁿ⁺²) yⁿ, which simplifies depending on how you index.
For the homework's Problem 001, the hint says to differentiate the power series for (x² - 4)⁻¹. Write 1/(x² - 4) as a geometric series in x²/4, differentiate term by term, and evaluate at x = -1.
If f(x) is an integral (or antiderivative) of something with a known series, integrate term by term.
For f(z) = ln(3 - z): note that d/dz [ln(3 - z)] = -1/(3 - z). You already know the series for 1/(3 - z), so integrate it term by term. Don't forget the constant of integration: set z = 0 to find it, giving f(0) = ln(3).
For f(x) = tan⁻¹(3x): since d/dx [tan⁻¹(3x)] = 3/(1 + 9x²), start from 1/(1 + 9x²) = Σ (-9x²)ⁿ = Σ (-1)ⁿ 9ⁿ x²ⁿ. Multiply by 3, then integrate term by term to get f(x) = Σ (-1)ⁿ · 3²ⁿ⁺¹/(2n + 1) · x²ⁿ⁺¹.
For f(t) = ln((1 + 3t)/(1 - 3t)): use log properties to split into ln(1 + 3t) - ln(1 - 3t), then expand each logarithm via the integrated geometric series.
Sometimes you just multiply an existing series by xᵏ.
For f(x) = x³ tan⁻¹(x): first write tan⁻¹(x) = Σ (-1)ⁿ x²ⁿ⁺¹/(2n + 1), then multiply every term by x³. The result is f(x) = Σ (-1)ⁿ/(2n + 1) · x²ⁿ⁺⁴.
A power series and its term-by-term derivative (or integral) share the same radius of convergence. If lim |cₙ₊₁/cₙ| = L, then R = 1/L for both the original series Σ cₙ yⁿ and the derived series Σ n cₙ yⁿ⁻¹.
\frac{1}{1-u} = \sum_{n=0}^{\infty} u^n, \quad |u| < 1\frac{1}{(1-u)^2} = \sum_{n=0}^{\infty} (n+1)\, u^n, \quad |u| < 1\ln(1-u) = -\sum_{n=1}^{\infty} \frac{u^n}{n}, \quad |u| < 1\tan^{-1}(u) = \sum_{n=0}^{\infty} \frac{(-1)^n}{2n+1}\, u^{2n+1}, \quad |u| \le 1e^u = \sum_{n=0}^{\infty} \frac{u^n}{n!}, \quad \text{all } u\cos(u) = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n)!}\, u^{2n}, \quad \text{all } u\sin^2(u) = \frac{1 - \cos(2u)}{2}The last identity is the standard approach for finding the Maclaurin series of sin²(u): convert to a cosine series, then substitute.
Power series representations are how calculators and computers evaluate functions like ln, sin, cos, and arctan internally: they compute a truncated series to the desired number of decimal places. Engineers use Taylor polynomial approximations to linearise nonlinear systems (for example, small-angle approximations in pendulum mechanics come from truncating the sine series).
Students often forget to factor out the constant before applying the geometric series. If the denominator is (3 - z³), you must write it as 3(1 - z³/3), not try to use 1/(1 - z³) directly.
Confusing the radius of convergence of the original series with that of the differentiated or integrated series. They are the same. The endpoints may differ, but R does not change.
Forgetting the constant of integration when building a series for a logarithm or arctangent. Always evaluate at a convenient point (usually x = 0) to pin down the constant.
Mixing up 1/(1 + x) and 1/(1 - x). The series for 1/(1 + x) is Σ (-1)ⁿ xⁿ (alternating signs). Dropping the (-1)ⁿ gives the wrong series.
⚠️ Expect at least one problem that requires you to recognise a function as the derivative or integral of a simpler one, then use term-by-term operations. This is the most commonly tested technique.
⚠️ The geometric series formula must be memorised cold. Every other series derivation in this unit flows from it.
⚠️ Problems often test whether you can correctly identify u in the substitution 1/(1 - u). Watch the sign and any powers (u = -x, u = x², u = z³/3, etc.).
⚠️ Evaluating a series at a specific point (like f(-1)) is a favourite exam question. You need both the series representation and the ability to substitute and simplify.
True or False: The radius of convergence changes when you differentiate a power series term by term. (False, it stays the same.)
Fill in the blank: 1/(1 + x) = Σ ______ for |x| < 1. ((-1)ⁿ xⁿ, n = 0 to ∞.)
True or False: To find the power series for ln(3 - z), you integrate the series for 1/(3 - z). (False; you integrate -1/(3 - z), because d/dz[ln(3 - z)] = -1/(3 - z).)
Fill in the blank: The series for 1/(1 - u)² is obtained by ______ the geometric series. (Differentiating.)
True or False: When building a power series for tan⁻¹(3x), the resulting series has only odd powers of x. (True.)
Q: Find the power series for f(z) = 1/(3 - z³) centred at the origin.
A: Factor out 3: f(z) = (1/3) · 1/(1 - z³/3). Apply the geometric series with u = z³/3: f(z) = Σ z³ⁿ / 3ⁿ⁺¹ (n = 0 to ∞).
Q: Find the power series for f(y) = 1/(6 - y)². State the approach.
A: Recognise that 1/(6 - y)² is the derivative of 1/(6 - y) (up to a constant). Start from 1/(6 - y) = (1/6) · 1/(1 - y/6) = (1/6) Σ (y/6)ⁿ. Differentiate with respect to y: 1/(6 - y)² = Σ (n + 1) yⁿ / 6ⁿ⁺² (n = 0 to ∞).
Q: Find the power series for f(x) = tan⁻¹(3x).
A: Start from 1/(1 + 9x²) = Σ (-1)ⁿ (9x²)ⁿ = Σ (-1)ⁿ 9ⁿ x²ⁿ. Multiply by 3 and integrate term by term: f(x) = Σ (-1)ⁿ · 3²ⁿ⁺¹/(2n + 1) · x²ⁿ⁺¹ (n = 0 to ∞).
Q: If lim |cₙ₊₁/cₙ| = 6, what are the radii of convergence R₁ (for Σ cₙ yⁿ) and R₂ (for Σ n cₙ yⁿ⁻¹)?
A: Both equal 1/6. Differentiation does not change the radius of convergence. R₁ = R₂ = 1/6.
Q: Find the power series for f(z) = ln(3 - z).
A: Note d/dz[ln(3 - z)] = -1/(3 - z) = -(1/3) · 1/(1 - z/3) = -(1/3) Σ (z/3)ⁿ. Integrate: ln(3 - z) = C - Σ zⁿ/(n · 3ⁿ) (n = 1 to ∞). Set z = 0: C = ln(3). So f(z) = ln(3) - Σ zⁿ/(n · 3ⁿ) (n = 1 to ∞).
This material connects directly to Taylor and Maclaurin series (the next set of study notes), where you build series from scratch using derivatives at a point rather than manipulating known series. The integration techniques here also tie back to the Fundamental Theorem of Calculus: term-by-term integration is valid precisely because power series converge uniformly inside their radius of convergence. These representations will return in differential equations courses, where power series methods solve ODEs that resist closed-form answers.
Power series representation, geometric series substitution, term-by-term differentiation, term-by-term integration, radius of convergence, interval of convergence, Maclaurin series, Taylor series, arctangent series, logarithm series, 1/(1-x) expansion, power series for rational functions, integral calculus UT Austin, HW 13 integral calculus