Calculus 2 Quiz: Taylor And Maclaurin Expansions
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Taylor And Maclaurin ExpansionsQuestion 1 of 7

Let p(x)=1+2x−x2+x32p(x) = 1 + 2x - x^2 + \frac{x^3}{2} be the third-degree Taylor polynomial for f(x)f(x) centered at x=0x = 0. If g(x)=f(2x)g(x) = f(2x), what is the coefficient of x2x^2 in the third-degree Taylor polynomial for g(x)g(x) centered at x=0x = 0?

−1-1
−2-2
−4-4
44
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Calculus 2 Quiz

Calculus 2 Quiz: Taylor And Maclaurin Expansions

Practice Taylor And Maclaurin Expansions in Calculus 2 with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Taylor And Maclaurin Expansions, giving you a quick way to practice the rules, question types, and explanations that matter most for Calculus 2.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

Let p(x)=1+2x−x2+x32p(x) = 1 + 2x - x^2 + \frac{x^3}{2} be the third-degree Taylor polynomial for f(x)f(x) centered at x=0x = 0. If g(x)=f(2x)g(x) = f(2x), what is the coefficient of x2x^2 in the third-degree Taylor polynomial for g(x)g(x) centered at x=0x = 0?

  1. −1-1
  2. −2-2
  3. −4-4 (correct answer)
  4. 44
Explanation: This question tests your understanding of Taylor polynomials and function composition. When you see a function transformed by composition (like g(x)=f(2x)g(x) = f(2x)), you need to find how the Taylor polynomial transforms accordingly. Given that p(x)=1+2x−x2+x32p(x) = 1 + 2x - x^2 + \frac{x^3}{2} is the third-degree Taylor polynomial for f(x)f(x) at x=0x = 0, we can find the Taylor polynomial for g(x)=f(2x)g(x) = f(2x) by substituting 2x2x for xx in p(x)p(x). The Taylor polynomial for g(x)g(x) becomes: p(2x)=1+2(2x)−(2x)2+(2x)32p(2x) = 1 + 2(2x) - (2x)^2 + \frac{(2x)^3}{2} =1+4x−4x2+8x32= 1 + 4x - 4x^2 + \frac{8x^3}{2} =1+4x−4x2+4x3= 1 + 4x - 4x^2 + 4x^3 The coefficient of x2x^2 is −4-4, making C the correct answer. Answer A (−1-1) comes from mistakenly using the original coefficient without applying the transformation. Answer B (−2-2) results from incorrectly multiplying the original coefficient by 2 instead of 222^2. Answer D (44) makes the sign error of forgetting the negative sign in the original polynomial. Strategy tip: When dealing with composition f(ax)f(ax) in Taylor polynomials, substitute axax directly into the given polynomial. Remember that the coefficient of xnx^n gets multiplied by ana^n due to the power rule. Practice recognizing this pattern to avoid coefficient errors on exams.

Question 2

Suppose f(x)f(x) has the Taylor series f(x)=∑n=0∞f(n)(2)n!(x−2)nf(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(2)}{n!}(x-2)^n that converges for ∣x−2∣<3|x-2| < 3. What is the radius of convergence for the Taylor series of h(x)=f(x2+2)h(x) = f(x^2 + 2) centered at x=0x = 0?

  1. R=3R = 3
  2. R=5R = \sqrt{5}
  3. R=9R = 9
  4. R=3R = \sqrt{3} (correct answer)
Explanation: When you encounter a problem involving the Taylor series of a composite function, you need to think about how the radius of convergence changes when you substitute one function into another. Given that f(x)f(x) has a Taylor series centered at x=2x = 2 with radius of convergence 3, this means the series converges when ∣x−2∣<3|x - 2| < 3. Now you need to find the radius of convergence for h(x)=f(x2+2)h(x) = f(x^2 + 2) centered at x=0x = 0. The key insight is that h(x)=f(x2+2)h(x) = f(x^2 + 2) will converge when the input to ff satisfies the original convergence condition. Since ff converges when its input is within distance 3 of the center 2, you need ∣(x2+2)−2∣<3|(x^2 + 2) - 2| < 3, which simplifies to ∣x2∣<3|x^2| < 3 or ∣x∣<3|x| < \sqrt{3}. Therefore, the radius of convergence is 3\sqrt{3}, making D correct. Let's examine why the other options are wrong: A) R=3R = 3 incorrectly assumes the radius stays the same under composition. B) R=5R = \sqrt{5} might come from incorrectly solving x2+2<5x^2 + 2 < 5 instead of using the proper convergence condition. C) R=9R = 9 likely results from squaring the original radius instead of taking the square root. Remember this pattern: when you have f(g(x))f(g(x)) and need the radius of convergence, set up the inequality for when g(x)g(x) falls within the convergence region of ff, then solve for the values of xx that satisfy this condition.

Question 3

If g(x)=∫0xsin⁡(t2)dtg(x) = \int_0^x \sin(t^2) dt, what is the coefficient of x7x^7 in the Maclaurin series for g(x)g(x)?

  1. 142\frac{1}{42}
  2. −142-\frac{1}{42} (correct answer)
  3. 15040\frac{1}{5040}
  4. −15040-\frac{1}{5040}
Explanation: First, find the Maclaurin series for sin⁡(t2)\sin(t^2). Since sin⁡(u)=u−u36+u5120−...\sin(u) = u - \frac{u^3}{6} + \frac{u^5}{120} - ... and u=t2u = t^2, we get sin⁡(t2)=t2−t66+t10120−...\sin(t^2) = t^2 - \frac{t^6}{6} + \frac{t^{10}}{120} - ... Integrating term by term: g(x)=∫0x(t2−t66+...)dt=x33−x76⋅7+...=x33−x742+...g(x) = \int_0^x (t^2 - \frac{t^6}{6} + ...) dt = \frac{x^3}{3} - \frac{x^7}{6 \cdot 7} + ... = \frac{x^3}{3} - \frac{x^7}{42} + ... The coefficient of x7x^7 is −142-\frac{1}{42}. Choice A has the wrong sign. Choices C and D incorrectly use 7!7! instead of 6⋅76 \cdot 7.

Question 4

The function h(x)=11+x2h(x) = \frac{1}{1+x^2} has the Maclaurin series ∑n=0∞(−1)nx2n\sum_{n=0}^{\infty} (-1)^n x^{2n}. Using this series, find an approximation for ∫00.511+x2dx\int_0^{0.5} \frac{1}{1+x^2} dx using the first three non-zero terms.

  1. 0.5−124+1160≈0.4640.5 - \frac{1}{24} + \frac{1}{160} \approx 0.464 (correct answer)
  2. 0.5−112+180≈0.4290.5 - \frac{1}{12} + \frac{1}{80} \approx 0.429
  3. 0.5−124+180≈0.4710.5 - \frac{1}{24} + \frac{1}{80} \approx 0.471
  4. 0.5−112+1160≈0.4230.5 - \frac{1}{12} + \frac{1}{160} \approx 0.423
Explanation: The Maclaurin series is 11+x2=1−x2+x4−x6+...\frac{1}{1+x^2} = 1 - x^2 + x^4 - x^6 + ... Integrating from 0 to 0.5: ∫00.5(1−x2+x4)dx=[x−x33+x55]00.5=0.5−(0.5)33+(0.5)55=0.5−0.1253+0.031255=0.5−124+1160\int_0^{0.5} (1 - x^2 + x^4) dx = [x - \frac{x^3}{3} + \frac{x^5}{5}]_0^{0.5} = 0.5 - \frac{(0.5)^3}{3} + \frac{(0.5)^5}{5} = 0.5 - \frac{0.125}{3} + \frac{0.03125}{5} = 0.5 - \frac{1}{24} + \frac{1}{160}. Choice B incorrectly uses x22\frac{x^2}{2} instead of x33\frac{x^3}{3}. Choice C uses x44\frac{x^4}{4} instead of x55\frac{x^5}{5}. Choice D has both errors.

Question 5

The Taylor series for f(x)f(x) centered at a=2a = 2 is ∑n=0∞(−1)n3n(x−2)n\sum_{n=0}^{\infty} \frac{(-1)^n}{3^n}(x-2)^n. What is f′′′(2)f'''(2)?

  1. −29-\frac{2}{9} (correct answer)
  2. −127-\frac{1}{27}
  3. 127\frac{1}{27}
  4. −227-\frac{2}{27}
Explanation: From the Taylor series f(x)=∑n=0∞(−1)n3n(x−2)nf(x) = \sum_{n=0}^{\infty} \frac{(-1)^n}{3^n}(x-2)^n, the coefficient of (x−2)3(x-2)^3 is (−1)333=−127\frac{(-1)^3}{3^3} = -\frac{1}{27}. Since the coefficient of (x−2)3(x-2)^3 in a Taylor series equals f′′′(2)3!\frac{f'''(2)}{3!}, we have f′′′(2)6=−127\frac{f'''(2)}{6} = -\frac{1}{27}, so f′′′(2)=−627=−29f'''(2) = -\frac{6}{27} = -\frac{2}{9}. Choice B forgets to multiply by 3!3!. Choice C has a sign error. Choice D incorrectly uses 2⋅3!2 \cdot 3! instead of 3!3!.

Question 6

The Maclaurin series for 11−x\frac{1}{1-x} is ∑n=0∞xn=1+x+x2+x3+...\sum_{n=0}^{\infty} x^n = 1 + x + x^2 + x^3 + ... for ∣x∣<1|x| < 1. Find the coefficient of x5x^5 in the Maclaurin series for x2(1−x)2\frac{x^2}{(1-x)^2}.

  1. 33
  2. 44 (correct answer)
  3. 55
  4. 66
Explanation: First, find the series for 1(1−x)2\frac{1}{(1-x)^2}. Since ddx[11−x]=1(1−x)2\frac{d}{dx}[\frac{1}{1-x}] = \frac{1}{(1-x)^2} and 11−x=1+x+x2+x3+...\frac{1}{1-x} = 1 + x + x^2 + x^3 + ..., we get 1(1−x)2=1+2x+3x2+4x3+5x4+6x5+...\frac{1}{(1-x)^2} = 1 + 2x + 3x^2 + 4x^3 + 5x^4 + 6x^5 + ... Therefore, x2(1−x)2=x2(1+2x+3x2+4x3+5x4+6x5+...)=x2+2x3+3x4+4x5+5x6+...\frac{x^2}{(1-x)^2} = x^2(1 + 2x + 3x^2 + 4x^3 + 5x^4 + 6x^5 + ...) = x^2 + 2x^3 + 3x^4 + 4x^5 + 5x^6 + ... The coefficient of x5x^5 is 44. Choice A uses the coefficient of x4x^4. Choice C uses the coefficient of x6x^6. Choice D uses the coefficient from the original series without the x2x^2 shift.

Question 7

The Maclaurin series for ln⁡(1+x)\ln(1+x) is x−x22+x33−x44+...x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + ... for ∣x∣<1|x| < 1. What is the radius of convergence for the Maclaurin series of ln⁡(1+x3)\ln(1+x^3)?

  1. R=13R = \frac{1}{3}
  2. R=3R = 3
  3. R=1R = 1 (correct answer)
  4. R=13=1R = \sqrt[3]{1} = 1
Explanation: When working with power series transformations, the key principle is understanding how substitutions affect the radius of convergence. If you have a series for f(x)f(x) with radius RR, then the series for f(g(x))f(g(x)) converges when ∣g(x)∣<R|g(x)| < R. Starting with the given Maclaurin series for ln⁡(1+x)=x−x22+x33−x44+...\ln(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + ... which converges for ∣x∣<1|x| < 1, we can find the series for ln⁡(1+x3)\ln(1+x^3) by substituting x3x^3 for every xx: ln⁡(1+x3)=x3−(x3)22+(x3)33−(x3)44+...=x3−x62+x93−x124+...\ln(1+x^3) = x^3 - \frac{(x^3)^2}{2} + \frac{(x^3)^3}{3} - \frac{(x^3)^4}{4} + ... = x^3 - \frac{x^6}{2} + \frac{x^9}{3} - \frac{x^{12}}{4} + ... For this new series to converge, we need ∣x3∣<1|x^3| < 1, which means ∣x∣<1|x| < 1. Therefore, the radius of convergence is R=1R = 1. Answer A (R=13R = \frac{1}{3}) incorrectly assumes you divide the original radius by the exponent. Answer B (R=3R = 3) mistakenly multiplies the original radius by the exponent. Answer D, while mathematically equivalent to C, suggests unnecessary complexity with the cube root notation when the answer is simply 1. Remember this pattern: when substituting xnx^n into a power series, the new convergence condition becomes ∣xn∣<Roriginal|x^n| < R_{original}, so you solve ∣x∣<Roriginal1/n|x| < R_{original}^{1/n} to find the new radius. Here, ∣x∣<11/3=1|x| < 1^{1/3} = 1.