AP® Calculus BC review sheet from Aim for Five (aimforfive.com/calc-bc/units/10/10-15)
Unit 10 · Topic 10.15
BC only10.15 Representing Functions as Power Series
BC only. You can build new power series from ones you know by substituting, multiplying by powers of x, or differentiating or integrating term by term. Differentiating or integrating keeps the same radius of convergence, but the series may behave differently at the endpoints.
Key terms
- power series
- substitution
- term-by-term differentiation
- term-by-term integration
- geometric series
Why build series instead of differentiating
This whole unit is BC only. Finding a Taylor series from derivatives can get messy fast. Try finding the tenth derivative of x e^(−x²). Instead, start from a series you know (eˣ, sin x, cos x, 1/(1 − x)) and change it with algebra or calculus. Because a function's power series is unique, any correct method gives the Taylor series.
The four moves
- Substitute: replace x with another expression. From eˣ = Σ xⁿ/n!, get e^(−x²) = Σ (−x²)ⁿ/n! = Σ (−1)ⁿ x²ⁿ/n!.
- Multiply or divide by a power of x: x e^(−x²) = Σ (−1)ⁿ x²ⁿ⁺¹/n!.
- Differentiate term by term: d/dx [1/(1 − x)] = 1/(1 − x)², so 1/(1 − x)² = Σ (n = 1 to ∞) n xⁿ⁻¹.
- Integrate term by term: ∫₀ˣ 1/(1 + t²) dt = arctan x, so arctan x = Σ (−1)ⁿ x²ⁿ⁺¹/(2n + 1).
Intervals of convergence after each move
Substituting changes where the series converges. Since 1/(1 − x) needs |x| < 1, the series for 1/(1 + x²) (substituting −x²) needs |−x²| < 1, which is |x| < 1. The series for 1/(1 − 3x) needs |3x| < 1, which is |x| < 1/3.
Differentiating or integrating keeps the radius the same. Endpoints can change, though, so check them again. For example, Σ xⁿ diverges at both x = 1 and x = −1. Integrating gives −ln(1 − x) = Σ (n = 1 to ∞) xⁿ/n, which converges at x = −1 (alternating harmonic) but still diverges at x = 1. Integrating 1/(1 + x²) to get arctan x adds both endpoints: the arctan series converges on [−1, 1].
Results worth recognizing
- ln(1 + x) = x − x²/2 + x³/3 − … = Σ (n = 1 to ∞) (−1)ⁿ⁺¹ xⁿ/n, for −1 < x ≤ 1.
- arctan x = x − x³/3 + x⁵/5 − … = Σ (n = 0 to ∞) (−1)ⁿ x²ⁿ⁺¹/(2n + 1), for −1 ≤ x ≤ 1.
- Using a series, you can also find a derivative at the center: f⁽ᵏ⁾(a) = k! × (coefficient of (x − a)ᵏ).
Worked examples
Try each one yourself first, then open the solution.
- Example 1
Integrating a geometric series
Find a power series for arctan x, and give its interval of convergence.
Show the solutionHide the solution
- Step 1: Start with 1/(1 − u) = Σ uⁿ for |u| < 1. Substitute u = −t²: 1/(1 + t²) = Σ (−1)ⁿ t²ⁿ, for |t| < 1.
- Step 2: Integrate from 0 to x term by term: arctan x = Σ (−1)ⁿ x²ⁿ⁺¹/(2n + 1) = x − x³/3 + x⁵/5 − … .
- Step 3: The radius stays 1. Check endpoints: at x = 1, the series is 1 − 1/3 + 1/5 − …, which converges by the alternating series test. At x = −1, it's the negative of that, which also converges.
- Step 4: Interval: [−1, 1].
Answer: arctan x = Σ (n = 0 to ∞) (−1)ⁿ x²ⁿ⁺¹/(2n + 1), for −1 ≤ x ≤ 1
- Example 2
Substitution, then a derivative value
Let f(x) = x e^(−x²). Find the first four nonzero terms of its Maclaurin series, and find f⁽⁵⁾(0).
Show the solutionHide the solution
- Step 1: eᵘ = 1 + u + u²/2! + u³/3! + … . With u = −x²: e^(−x²) = 1 − x² + x⁴/2 − x⁶/6 + … .
- Step 2: Multiply by x: f(x) = x − x³ + x⁵/2 − x⁷/6 + … .
- Step 3: The coefficient of x⁵ is 1/2, so f⁽⁵⁾(0) = 5! · (1/2) = 120/2 = 60.
- Step 4: Doing this by differentiating five times would be slow and error-prone.
Answer: x − x³ + x⁵/2 − x⁷/6 + …; f⁽⁵⁾(0) = 60
- Example 3
Trap: forgetting the interval when you plug in
Use 1/(1 − x)² = Σ (n = 1 to ∞) n xⁿ⁻¹ to find Σ (n = 1 to ∞) n/2ⁿ. Can you use the same series to find Σ n·2ⁿ⁻¹?
Show the solutionHide the solution
- Step 1: The series is the derivative of Σ xⁿ, so it has radius 1 and is valid for |x| < 1.
- Step 2: At x = 1/2: Σ n(1/2)ⁿ⁻¹ = 1/(1 − 1/2)² = 4.
- Step 3: Σ n/2ⁿ = (1/2)Σ n(1/2)ⁿ⁻¹ = (1/2)(4) = 2.
- Step 4: For Σ n·2ⁿ⁻¹, you'd need x = 2, which is outside |x| < 1. Plugging in would give 1/(1 − 2)² = 1, which is nonsense: that series diverges (its terms grow).
Answer: Σ n/2ⁿ = 2. The same method fails for Σ n·2ⁿ⁻¹, which diverges because x = 2 is outside the interval of convergence.
Common mistakes
- Losing the sign when substituting −x²: (−x²)ⁿ = (−1)ⁿ x²ⁿ, so the signs alternate.
- Forgetting the constant of integration when integrating a series. Find it by plugging in the center (for arctan x, it's 0).
- Assuming the endpoints stay the same after differentiating or integrating.
- Using a series outside its interval of convergence.
On the exam
- BC free-response Taylor questions often say “use the Maclaurin series for eˣ (or sin x, cos x) to write the first four nonzero terms of the series for …”. Use the known series; don't differentiate from scratch.
- A frequent follow-up asks for the series of f′ or of ∫₀ˣ f(t) dt. Differentiate or integrate term by term, and watch for a constant term that disappears when you differentiate, so you may need one more term to get the number of nonzero terms asked for.
Connected topics
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Check yourself
4 questions on 10.15 Representing Functions as Power Series. Pick an answer to see if you got it, and why.
For |x| < 1, 1/(1 − x) = Σ (n = 0 to ∞) xⁿ. Using this, what is the value of Σ (n = 1 to ∞) n/3ⁿ?
The first three nonzero terms of the Maclaurin series for e^(−t²) are used to approximate ∫ from 0 to 0.8 of e^(−t²) dt. What is the approximation?
The function f is defined by the power series f(x) = Σ (n = 1 to ∞) (x − 3)ⁿ/(n · 2ⁿ) = (x − 3)/2 + (x − 3)²/8 + (x − 3)³/24 + ⋯ for all x for which the series converges.
Described function
What is the interval of convergence of the series obtained by differentiating the series for f term by term?
The function g is defined for all real x by the Maclaurin series g(x) = 1 − x²/2! + x⁴/3! − x⁶/4! + ⋯ + (−1)ⁿ x²ⁿ/(n + 1)! + ⋯.
Described function
For x ≠ 0, which of the following is an expression for g(x)?
0 of 4 answered