Representing Functions as Power Series: Study Guide (Calculus BC 10.15)

BC-only material. Representing functions as power series is part of Calculus BC only. Calculus AB students do not need this topic.

Before you start: prerequisites

This topic is the last one in the course. It uses most of Unit 10. If any row is shaky, revisit it from the Calculus BC roadmap first.

Prerequisite	Topic	What you use it for here
Geometric series	10.2	The pattern a + ar + ar² + … = a/(1 − r) for |r| < 1
Radius and interval of convergence	10.13	Finding where a new series is valid, and testing endpoints
Maclaurin series of key functions	10.14	The starting series you change
Antiderivatives of powers	6.8	Integrating a series one term at a time

Notation: Σ from n = 0 to ∞ of aₙ is an infinite series. A series “about x = 0” (a Maclaurin series) uses powers of x. A series “about x = c” uses powers of (x − c).

The idea: change a series you already know

In Topic 10.14 you found a Maclaurin series by working out f(0), f′(0), f″(0) and so on. That works, but the derivatives of a function like ln(4 + x) or 5x/(3 + x²) get messy fast.

There is a quicker route. Start from a series you already know, and change it the same way the function changes. Four series do almost all the work:

Function	Series about x = 0	Valid for
1/(1 − x)	1 + x + x² + x³ + … = Σ xⁿ	−1 < x < 1
eˣ	1 + x + x²/2! + x³/3! + … = Σ xⁿ/n!	all real x
sin x	x − x³/3! + x⁵/5! − … = Σ (−1)ⁿ x^(2n+1)/(2n+1)!	all real x
cos x	1 − x²/2! + x⁴/4! − … = Σ (−1)ⁿ x^(2n)/(2n)!	all real x

You can change them in four ways:

1. Rewrite a rational function in the form a/(1 − r), then use the geometric series.
2. Substitute an expression for x, such as −x², 3x or x/2.
3. Multiply or divide by a power of x (or add and subtract series).
4. Differentiate or integrate term by term.

Why this is allowed. A power series with a positive radius of convergence is the Taylor series of the function it adds up to, on its open interval. So any correct route gives the Maclaurin series. You may then read derivatives from it: the coefficient of xⁿ is f⁽ⁿ⁾(0)/n!.

Rewriting into geometric form

The geometric series needs a 1 in front, then minus something. So divide top and bottom to make the constant term 1.

For 2/(5 − x): 2/(5 − x) = (2/5) · 1/(1 − x/5). Here a = 2/5 and r = x/5. So

2/(5 − x) = Σ (2/5)(x/5)ⁿ = Σ 2xⁿ/5ⁿ⁺¹, valid when |x/5| < 1, that is −5 < x < 5.

For a plus sign, write + as minus a negative: 1/(1 + u) = 1/(1 − (−u)) = Σ (−1)ⁿ uⁿ.

The interval comes from |r| < 1. At an endpoint the terms of a geometric series have size |a|, which does not go to 0, so a geometric series never converges at its endpoints.

Substituting and multiplying by powers of x

Substitution. Replace every x in a known series by the new expression, including inside the powers:

• e^(−x²) = 1 − x² + x⁴/2! − x⁶/3! + … = Σ (−1)ⁿ x^(2n)/n!. Valid for all x.
• sin(3x) = 3x − (3x)³/3! + (3x)⁵/5! − … = 3x − 9x³/2 + 81x⁵/40 − …. Note (3x)³ = 27x³, not 3x³.
• 1/(1 − x³) = 1 + x³ + x⁶ + x⁹ + …, valid when |x³| < 1, that is −1 < x < 1.

A substitution can change the interval. If you replace x by x/2 in 1/(1 − x), you need |x/2| < 1, so the interval becomes −2 < x < 2.

Multiplying or dividing by a power of x. This shifts every power by the same amount and does not change the interval (except that dividing by x leaves out x = 0 from the formula for the function). For example

(1 − cos x)/x² = (x²/2! − x⁴/4! + x⁶/6! − …)/x² = 1/2 − x²/24 + x⁴/720 − …

This also shows that the limit of (1 − cos x)/x² as x → 0 is 1/2, the constant term.

Reading a derivative from a built series. Let g(x) = x² sin(x²). Substitute x² into sin x, then multiply by x²:

g(x) = x² (x² − x⁶/3! + x¹⁰/5! − …) = x⁴ − x⁸/6 + x¹²/120 − …

The coefficient of x⁸ is −1/6, and it also equals g⁽⁸⁾(0)/8!. So g⁽⁸⁾(0) = 8! × (−1/6) = −6720. Finding this by differentiating eight times would take a long time.

Worked example 1: a rational function, then a substitution

Question. Let f(x) = 5/(3 + x). (a) Write the first four nonzero terms and the general term of the Maclaurin series for f, and find its interval of convergence. (b) Use (a) to write the Maclaurin series for g(x) = 5x/(3 + x²), with its general term and interval of convergence.

(a)

1. Make the constant 1. 5/(3 + x) = (5/3) · 1/(1 + x/3) = (5/3) · 1/(1 − (−x/3)).
2. Read a and r. a = 5/3 and r = −x/3.
3. Write the series. Σ (5/3)(−x/3)ⁿ = Σ (−1)ⁿ 5xⁿ/3ⁿ⁺¹. First four terms: 5/3 − 5x/9 + 5x²/27 − 5x³/81 + …
4. Interval. The series converges when |−x/3| < 1, so −3 < x < 3. At x = 3 the terms are 5/3, −5/3, 5/3, …; at x = −3 they are all 5/3. Neither tends to 0, so both diverge by the nth term test. The interval of convergence is −3 < x < 3.

(b)

1. Spot the link. 5/(3 + x²) is f with x replaced by x². Then g(x) = x · 5/(3 + x²).
2. Substitute x² into the general term: Σ (−1)ⁿ 5(x²)ⁿ/3ⁿ⁺¹ = Σ (−1)ⁿ 5x^(2n)/3ⁿ⁺¹.
3. Multiply by x: g(x) = Σ (−1)ⁿ 5x^(2n+1)/3ⁿ⁺¹ = 5x/3 − 5x³/9 + 5x⁵/27 − 5x⁷/81 + …
4. Interval. We need |x²| < 3, so |x| < √3. At x = ±√3 the terms have size 5√3/3, which does not tend to 0. The interval is −√3 < x < √3.

Check. At x = 1 (inside the interval), g(1) = 5/4 = 1.25. The partial sums of 5/3 − 5/9 + 5/27 − 5/81 + … are about 1.111 after two terms, 1.235 after four and 1.2498 after eight. They approach 1.25. ✓

Term-by-term differentiation and integration

Inside its interval of convergence, a power series can be differentiated or integrated one term at a time, just like a polynomial.

If f(x) = Σ aₙ xⁿ for |x| < R, then

• f′(x) = Σ from n = 1 to ∞ of n aₙ xⁿ⁻¹ (the constant term differentiates to 0, so the sum starts at n = 1), and
• ∫ from 0 to x of f(t) dt = Σ aₙ xⁿ⁺¹/(n + 1).

The radius of convergence R does not change. In the ratio test, the new factor n (or 1/(n + 1)) changes each ratio only by a factor like (n + 1)/n, which tends to 1, so the limit and R stay the same. The endpoints can change. Integrating divides each term by (n + 1), which can make a divergent endpoint converge. Differentiating multiplies by n, which can make a convergent endpoint diverge. So test both endpoints again every time.

Example: differentiating to sum a series. Differentiate 1/(1 − x) = Σ xⁿ:

1/(1 − x)² = Σ from n = 1 to ∞ of n xⁿ⁻¹ = 1 + 2x + 3x² + …, for |x| < 1.

Multiply by x: x/(1 − x)² = Σ n xⁿ. Put x = 1/2 (inside the interval): Σ from n = 1 to ∞ of n/2ⁿ = (1/2)/(1/2)² = 2. A numerical series has been summed by recognising it as a power series.

Integrating: remember the constant. If you integrate a series as an indefinite integral, a constant C appears. Find it by putting in the centre, where every power of x is 0.

Worked example 2: integrating to get a logarithm

Question. (a) Find the Maclaurin series for ln(4 + x): the first four nonzero terms and the general term. (b) Find its interval of convergence. (c) Use three terms after the constant to estimate ln(4.4) − ln 4, and say how accurate the estimate is.

(a)

1. Link to a known series. d/dx [ln(4 + x)] = 1/(4 + x), and 1/(4 + x) is geometric.
2. Series for 1/(4 + x). 1/(4 + x) = (1/4) · 1/(1 − (−x/4)) = Σ (−1)ⁿ xⁿ/4ⁿ⁺¹ = 1/4 − x/16 + x²/64 − x³/256 + …, for |x| < 4.
3. Integrate term by term from 0 to x. Since ln(4 + x) = ln 4 + ∫ from 0 to x of 1/(4 + t) dt, ln(4 + x) = ln 4 + Σ (−1)ⁿ xⁿ⁺¹/((n + 1) 4ⁿ⁺¹) = ln 4 + x/4 − x²/32 + x³/192 − x⁴/1024 + …
4. The constant. At x = 0 every power term is 0, so the constant must be ln(4 + 0) = ln 4. Leaving it out is the most common error here.

(b)

1. Radius. Integration keeps the radius, so R = 4. The series converges for −4 < x < 4.
2. x = 4. The terms become (−1)ⁿ 4ⁿ⁺¹/((n + 1) 4ⁿ⁺¹) = (−1)ⁿ/(n + 1): 1 − 1/2 + 1/3 − 1/4 + …. This is the alternating harmonic series, which converges.
3. x = −4. The terms become (−1)ⁿ (−4)ⁿ⁺¹/((n + 1) 4ⁿ⁺¹) = −1/(n + 1): −(1 + 1/2 + 1/3 + …). This is a multiple of the harmonic series, which diverges. (Also, ln(4 + x) is not defined at x = −4.)
4. Interval of convergence: −4 < x ≤ 4. The series for 1/(4 + x) converged on −4 < x < 4 only, so integrating has gained the endpoint x = 4.

(c) ln(4.4) − ln 4 is the series without its constant, at x = 0.4:

0.4/4 − 0.4²/32 + 0.4³/192 = 0.1 − 0.005 + 0.000333… ≈ 0.095333.

The series is alternating with terms decreasing to 0, so the error is less than the next term, 0.4⁴/1024 = 0.000025. (The true value, ln 1.1 ≈ 0.095310, is indeed within 0.000025.)

Common misconceptions

• “Any fraction is geometric as it stands.” 5/(3 + x) is not Σ 5(−x)ⁿ. You must first make the constant 1: (5/3) · 1/(1 + x/3).
• Substituting into x but not into the powers. sin(2x) starts 2x − (2x)³/3! = 2x − 8x³/6, not 2x − 2x³/6.
• Keeping the old interval after a substitution. Replacing x by x² or x/2 changes the condition |r| < 1, so it changes the interval.
• Forgetting the constant after integrating. ln(4 + x) needs ln 4 in front. Always check the value at the centre.
• “The interval is the same after integrating or differentiating.” Only the radius is guaranteed to stay the same. Test the endpoints again.
• Starting a differentiated series at n = 0. The constant term becomes 0, so Σ n aₙ xⁿ⁻¹ starts at n = 1 (writing n = 0 adds a zero term but can confuse the general term).
• Using the series outside its interval. 1/(1 − x) = −1 at x = 2, but 1 + 2 + 4 + 8 + … diverges. The function exists there; the series does not represent it.
• Confusing the coefficient with the derivative. The coefficient of xⁿ is f⁽ⁿ⁾(0)/n!. Multiply by n! to get the derivative.

Where this leads

This is the final topic of Calculus BC. Power series tie the course together: they turn hard functions into polynomials, so you can approximate values, evaluate limits and estimate integrals that have no elementary antiderivative, and then bound the error with the alternating series error bound or the Lagrange error bound from earlier in Unit 10. For a full review of the unit and the course, return to the Calculus BC roadmap, or look back at Topic 10.14, Finding Taylor or Maclaurin Series for a Function.

Try the practice questions now, then use the revision notes and the checklist to consolidate.