Calculus 2 Quiz: Taylor And Maclaurin Series
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Taylor And Maclaurin SeriesQuestion 1 of 20

The Maclaurin series for f(x)=(1−x)−2f(x) = (1-x)^{-2} is ∑n=0∞(n+1)xn\sum_{n=0}^{\infty} (n+1)x^n. What is f(10)(0)f^{(10)}(0)?

1010
1111
10!10!
11!11!
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Calculus 2 Quiz

Calculus 2 Quiz: Taylor And Maclaurin Series

Practice Taylor And Maclaurin Series 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 Series, giving you a quick way to practice the rules, question types, and explanations that matter most for Calculus 2.

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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.

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Question 1

The Maclaurin series for f(x)=(1−x)−2f(x) = (1-x)^{-2} is ∑n=0∞(n+1)xn\sum_{n=0}^{\infty} (n+1)x^n. What is f(10)(0)f^{(10)}(0)?

  1. 1010
  2. 1111
  3. 10!10!
  4. 11!11! (correct answer)
Explanation: The coefficient of xnx^n in the Maclaurin series is cn=f(n)(0)n!c_n = \frac{f^{(n)}(0)}{n!}. From the given series, the coefficient of xnx^n is cn=n+1c_n = n+1. We want to find f(10)(0)f^{(10)}(0), so we look at the case where n=10n=10. The coefficient is c10=10+1=11c_{10} = 10+1 = 11. Using the formula, c10=f(10)(0)10!c_{10} = \frac{f^{(10)}(0)}{10!}. So, 11=f(10)(0)10!11 = \frac{f^{(10)}(0)}{10!}. Solving for the derivative gives f(10)(0)=11⋅10!=11!f^{(10)}(0) = 11 \cdot 10! = 11!.

Question 2

The third-degree Taylor polynomial for a function f(x)f(x) centered at a=2a=2 is T3(x)=5−(x−2)+4(x−2)3T_3(x) = 5 - (x-2) + 4(x-2)^3. What is the value of f′′(2)f''(2)?

  1. 44
  2. 11
  3. 00 (correct answer)
  4. 88
Explanation: When you encounter a Taylor polynomial problem, you need to connect the polynomial's coefficients to the function's derivatives at the center point. The general form of a Taylor polynomial centered at aa is: Tn(x)=f(a)+f′(a)(x−a)+f′′(a)2!(x−a)2+f′′′(a)3!(x−a)3+...T_n(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \frac{f'''(a)}{3!}(x-a)^3 + ... Given T3(x)=5−(x−2)+4(x−2)3T_3(x) = 5 - (x-2) + 4(x-2)^3, you can identify each coefficient by comparing terms. The constant term gives f(2)=5f(2) = 5. The coefficient of (x−2)(x-2) gives f′(2)=−1f'(2) = -1. The coefficient of (x−2)2(x-2)^2 is 0, which means f′′(2)2!=0\frac{f''(2)}{2!} = 0, so f′′(2)=0f''(2) = 0. The coefficient of (x−2)3(x-2)^3 gives f′′′(2)3!=4\frac{f'''(2)}{3!} = 4, so f′′′(2)=24f'''(2) = 24. Choice A (4) incorrectly takes the coefficient of the (x−2)3(x-2)^3 term, confusing the third derivative coefficient with the second derivative. Choice B (1) might come from misreading the first derivative coefficient or forgetting about the factorial in the denominator. Choice D (8) could result from incorrectly multiplying the missing second-degree coefficient by 2!, perhaps thinking there's an implied coefficient. The key insight is that when a term is completely missing from a Taylor polynomial, it means that derivative equals zero. Always remember to account for the factorial denominators in Taylor series coefficients—the coefficient of (x−a)n(x-a)^n is f(n)(a)n!\frac{f^{(n)}(a)}{n!}, not just f(n)(a)f^{(n)}(a).

Question 3

The Taylor series for sin⁡(x)\sin(x) centered at x=π4x = \frac{\pi}{4} begins with 22+22(x−π4)+⋯\frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}(x - \frac{\pi}{4}) + \cdots. What is the coefficient of (x−π4)2(x - \frac{\pi}{4})^2?

  1. −24-\frac{\sqrt{2}}{4} (correct answer)
  2. −22-\frac{\sqrt{2}}{2}
  3. 24\frac{\sqrt{2}}{4}
  4. −28-\frac{\sqrt{2}}{8}
Explanation: For the Taylor series of sin⁡(x)\sin(x) at x=π4x = \frac{\pi}{4}, we need f′′(π4)f''(\frac{\pi}{4}). Since f′(x)=cos⁡(x)f'(x) = \cos(x) and f′′(x)=−sin⁡(x)f''(x) = -\sin(x), we have f′′(π4)=−sin⁡(π4)=−22f''(\frac{\pi}{4}) = -\sin(\frac{\pi}{4}) = -\frac{\sqrt{2}}{2}. The coefficient of (x−π4)2(x - \frac{\pi}{4})^2 is f′′(π4)2!=−222=−24\frac{f''(\frac{\pi}{4})}{2!} = \frac{-\frac{\sqrt{2}}{2}}{2} = -\frac{\sqrt{2}}{4}. Choice B forgets to divide by 2!2!. Choice C has the wrong sign. Choice D incorrectly divides by 4!4! instead of 2!2!.

Question 4

If h(x)=xcos⁡(x3)h(x) = x \cos(x^3), what is the coefficient of x7x^7 in the Maclaurin series for h(x)h(x)?

  1. −16-\frac{1}{6}
  2. −1120-\frac{1}{120}
  3. 16\frac{1}{6}
  4. −12-\frac{1}{2} (correct answer)
Explanation: When finding coefficients in Maclaurin series for products of functions, you need to systematically expand each factor and identify which terms contribute to your target power. Start with the Maclaurin series for cos⁡(u)=1−u22!+u44!−u66!+⋯\cos(u) = 1 - \frac{u^2}{2!} + \frac{u^4}{4!} - \frac{u^6}{6!} + \cdots. Substituting u=x3u = x^3: cos⁡(x3)=1−(x3)22!+(x3)44!−(x3)66!+⋯=1−x62+x1224−x18720+⋯\cos(x^3) = 1 - \frac{(x^3)^2}{2!} + \frac{(x^3)^4}{4!} - \frac{(x^3)^6}{6!} + \cdots = 1 - \frac{x^6}{2} + \frac{x^{12}}{24} - \frac{x^{18}}{720} + \cdots Now multiply by xx: h(x)=xcos⁡(x3)=x(1−x62+x1224−⋯ )=x−x72+x1324−⋯h(x) = x \cos(x^3) = x\left(1 - \frac{x^6}{2} + \frac{x^{12}}{24} - \cdots\right) = x - \frac{x^7}{2} + \frac{x^{13}}{24} - \cdots The coefficient of x7x^7 is −12-\frac{1}{2}, confirming answer D. Let's examine the wrong answers: A) −16-\frac{1}{6} likely comes from incorrectly using 13!\frac{1}{3!} instead of 12!\frac{1}{2!} when expanding cos⁡(x3)\cos(x^3). B) −1120-\frac{1}{120} suggests confusion with 15!\frac{1}{5!}, perhaps from misidentifying which factorial applies. C) 16\frac{1}{6} combines the factorial error from option A with a sign error—forgetting that the x6x^6 term in cos⁡(x3)\cos(x^3) comes from the negative −u22!-\frac{u^2}{2!} term. Study tip: When working with Maclaurin series of composite functions, write out the first several terms explicitly rather than trying to work with general formulas. This prevents factorial and sign errors that are common in these problems.

Question 5

What is the Maclaurin series for the function f(x)=xarctan⁡(x2)f(x) = x \arctan(x^2)?

  1. ∑n=0∞(−1)nx4n+22n+1\sum_{n=0}^{\infty} \frac{(-1)^n x^{4n+2}}{2n+1}
  2. ∑n=0∞(−1)nx2n+32n+1\sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+3}}{2n+1}
  3. ∑n=0∞(−1)nx4n+32n+1\sum_{n=0}^{\infty} \frac{(-1)^n x^{4n+3}}{2n+1} (correct answer)
  4. ∑n=0∞x4n+32n+1\sum_{n=0}^{\infty} \frac{x^{4n+3}}{2n+1}
Explanation: When you encounter a Maclaurin series problem involving composite functions like xarctan⁡(x2)x \arctan(x^2), start with known series and use substitution rather than computing derivatives from scratch. The key insight is to recall the Maclaurin series for arctan⁡(u)\arctan(u): arctan⁡(u)=∑n=0∞(−1)nu2n+12n+1\arctan(u) = \sum_{n=0}^{\infty} \frac{(-1)^n u^{2n+1}}{2n+1}. This series is valid for ∣u∣≤1|u| \leq 1. To find the series for arctan⁡(x2)\arctan(x^2), substitute u=x2u = x^2: arctan⁡(x2)=∑n=0∞(−1)n(x2)2n+12n+1=∑n=0∞(−1)nx4n+22n+1\arctan(x^2) = \sum_{n=0}^{\infty} \frac{(-1)^n (x^2)^{2n+1}}{2n+1} = \sum_{n=0}^{\infty} \frac{(-1)^n x^{4n+2}}{2n+1} Now multiply by xx to get f(x)=xarctan⁡(x2)f(x) = x \arctan(x^2): xarctan⁡(x2)=x⋅∑n=0∞(−1)nx4n+22n+1=∑n=0∞(−1)nx4n+32n+1x \arctan(x^2) = x \cdot \sum_{n=0}^{\infty} \frac{(-1)^n x^{4n+2}}{2n+1} = \sum_{n=0}^{\infty} \frac{(-1)^n x^{4n+3}}{2n+1} This confirms answer C. Looking at the wrong answers: A gives you arctan⁡(x2)\arctan(x^2) but forgets the multiplication by xx. B has the wrong exponent pattern—it would come from xarctan⁡(x)x \arctan(x) instead of xarctan⁡(x2)x \arctan(x^2). D is missing the alternating signs (−1)n(-1)^n, which are essential in the arctangent series. Study tip: For composite function series, always start with a known series and substitute carefully. Keep track of every transformation—substitution, then multiplication—and verify your exponent patterns match the operations you performed.

Question 6

What is the coefficient of (x−π/4)2(x - \pi/4)^2 in the Taylor series expansion of f(x)=tan⁡(x)f(x) = \tan(x) centered at a=π/4a = \pi/4?

  1. 11
  2. 22 (correct answer)
  3. 44
  4. 1/21/2
Explanation: The coefficient of (x−a)n(x-a)^n in a Taylor series is cn=f(n)(a)n!c_n = \frac{f^{(n)}(a)}{n!}. We need c2=f′′(π/4)2!c_2 = \frac{f''(\pi/4)}{2!}. First, find the derivatives: f(x)=tan⁡(x)f(x) = \tan(x) f′(x)=sec⁡2(x)f'(x) = \sec^2(x) f′′(x)=2sec⁡(x)⋅(sec⁡(x)tan⁡(x))=2sec⁡2(x)tan⁡(x)f''(x) = 2\sec(x) \cdot (\sec(x)\tan(x)) = 2\sec^2(x)\tan(x) Now, evaluate f′′(π/4)f''(\pi/4): f′′(π/4)=2sec⁡2(π/4)tan⁡(π/4)=2(2)2(1)=2(2)(1)=4f''(\pi/4) = 2\sec^2(\pi/4)\tan(\pi/4) = 2(\sqrt{2})^2(1) = 2(2)(1) = 4. Finally, calculate the coefficient: c2=42!=42=2c_2 = \frac{4}{2!} = \frac{4}{2} = 2.

Question 7

The Maclaurin series for arctan⁡(x)\arctan(x) is ∑n=0∞(−1)nx2n+12n+1\sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{2n+1}.

Use the provided information to find the Maclaurin series for g(x)=11+x2g(x) = \frac{1}{1+x^2}.

  1. ∑n=0∞(−1)nx2n\sum_{n=0}^{\infty} (-1)^n x^{2n} (correct answer)
  2. ∑n=0∞(−1)nx2n2n+1\sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{2n+1}
  3. ∑n=0∞(−1)nx2n+2(2n+1)(2n+2)\sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+2}}{(2n+1)(2n+2)}
  4. ∑n=0∞(−1)nx2n+1\sum_{n=0}^{\infty} (-1)^n x^{2n+1}
Explanation: We know that the derivative of arctan⁡(x)\arctan(x) is 11+x2\frac{1}{1+x^2}. Therefore, we can find the series for g(x)g(x) by differentiating the series for arctan⁡(x)\arctan(x) term-by-term. g(x)=ddx(∑n=0∞(−1)nx2n+12n+1)=∑n=0∞ddx((−1)nx2n+12n+1)g(x) = \frac{d}{dx} \left( \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{2n+1} \right) = \sum_{n=0}^{\infty} \frac{d}{dx} \left( \frac{(-1)^n x^{2n+1}}{2n+1} \right) =∑n=0∞(−1)n(2n+1)x2n2n+1=∑n=0∞(−1)nx2n= \sum_{n=0}^{\infty} \frac{(-1)^n (2n+1)x^{2n}}{2n+1} = \sum_{n=0}^{\infty} (-1)^n x^{2n} This is also recognizable as the geometric series for 11−(−x2)\frac{1}{1-(-x^2)}.

Question 8

Find the first three non-zero terms of the Taylor series for g(x)=ln⁡(1+x2)g(x) = \ln(1 + x^2) centered at x=0x = 0.

  1. x2−x42+x63x^2 - \frac{x^4}{2} + \frac{x^6}{3} (correct answer)
  2. x2−x42+x83x^2 - \frac{x^4}{2} + \frac{x^8}{3}
  3. 1+x2−x421 + x^2 - \frac{x^4}{2}
  4. x−x22+x33x - \frac{x^2}{2} + \frac{x^3}{3}
Explanation: Using the known series ln⁡(1+u)=u−u22+u33−u44+⋯\ln(1 + u) = u - \frac{u^2}{2} + \frac{u^3}{3} - \frac{u^4}{4} + \cdots and substituting u=x2u = x^2: ln⁡(1+x2)=x2−(x2)22+(x2)33−⋯=x2−x42+x63−⋯\ln(1 + x^2) = x^2 - \frac{(x^2)^2}{2} + \frac{(x^2)^3}{3} - \cdots = x^2 - \frac{x^4}{2} + \frac{x^6}{3} - \cdots. Choice B incorrectly has x8x^8 instead of x6x^6 in the third term. Choice C includes the constant term 1, which doesn't appear since ln⁡(1)=0\ln(1) = 0. Choice D uses the original ln⁡(1+u)\ln(1+u) series without substituting u=x2u = x^2.

Question 9

Which function is represented by the Maclaurin series ∑n=0∞(−1)n(2x)2n+1(2n+1)!\sum_{n=0}^{\infty} \frac{(-1)^n (2x)^{2n+1}}{(2n+1)!}?

  1. sin⁡(2x)\sin(2x) (correct answer)
  2. 2sin⁡(x)2\sin(x)
  3. cos⁡(2x)\cos(2x)
  4. 12sin⁡(2x)\frac{1}{2}\sin(2x)
Explanation: The standard Maclaurin series for sin⁡(u)\sin(u) is ∑n=0∞(−1)nu2n+1(2n+1)!\sum_{n=0}^{\infty} \frac{(-1)^n u^{2n+1}}{(2n+1)!}. The given series has the same form, ∑n=0∞(−1)n(2x)2n+1(2n+1)!\sum_{n=0}^{\infty} \frac{(-1)^n (2x)^{2n+1}}{(2n+1)!}. By comparing the two series, we can see that u=2xu = 2x. Therefore, the function represented by the series is sin⁡(2x)\sin(2x).

Question 10

Consider the function f(x)=cos⁡(x)sin⁡(x)f(x) = \cos(x) \sin(x). The Maclaurin series for f(x)f(x) has the form ∑n=0∞anxn\sum_{n=0}^{\infty} a_n x^n. Which of the following statements about the coefficients is true?

  1. a2=−13a_2 = -\frac{1}{3} and all even coefficients a2ka_{2k} are zero for k≥2k \geq 2
  2. a1=1a_1 = 1 and all odd coefficients a2k+1a_{2k+1} are zero for k≥1k \geq 1
  3. a1=1a_1 = 1 and a3=−23a_3 = -\frac{2}{3}, with all even coefficients zero (correct answer)
  4. a2=1a_2 = 1 and all coefficients an=0a_n = 0 for n≥3n \geq 3
Explanation: Using the identity cos⁡(x)sin⁡(x)=12sin⁡(2x)\cos(x)\sin(x) = \frac{1}{2}\sin(2x), we can find the Maclaurin series: f(x)=12sin⁡(2x)=12[2x−(2x)33!+(2x)55!−⋯ ]=x−8x312+⋯=x−2x33+⋯f(x) = \frac{1}{2}\sin(2x) = \frac{1}{2}[2x - \frac{(2x)^3}{3!} + \frac{(2x)^5}{5!} - \cdots] = x - \frac{8x^3}{12} + \cdots = x - \frac{2x^3}{3} + \cdots. So a1=1a_1 = 1 and a3=−23a_3 = -\frac{2}{3}. Since sin⁡(2x)\sin(2x) only has odd powers, all even coefficients are zero. Choice A incorrectly gives a non-zero even coefficient. Choice B incorrectly claims odd coefficients are zero for k≥1k \geq 1. Choice D incorrectly has a2=1a_2 = 1.

Question 11

What is the coefficient of x4x^4 in the Maclaurin series expansion for the function f(x)=sin⁡(x)1−xf(x) = \frac{\sin(x)}{1-x}?

  1. 56\frac{5}{6} (correct answer)
  2. 76\frac{7}{6}
  3. 11
  4. −16-\frac{1}{6}
Explanation: To find the coefficient, multiply the Maclaurin series for sin⁡(x)\sin(x) and 11−x\frac{1}{1-x}. sin⁡(x)=x−x33!+x55!−…\sin(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \dots 11−x=1+x+x2+x3+x4+…\frac{1}{1-x} = 1 + x + x^2 + x^3 + x^4 + \dots Multiplying these two series, we look for the terms that produce x4x^4: (x)(x3)+(−x36)(x)=x4−16x4=(1−16)x4=56x4(x)(x^3) + \left(-\frac{x^3}{6}\right)(x) = x^4 - \frac{1}{6}x^4 = \left(1 - \frac{1}{6}\right)x^4 = \frac{5}{6}x^4 The coefficient is 56\frac{5}{6}.

Question 12

Which of the following is the Taylor series for f(x)=ln⁡(x)f(x) = \ln(x) centered at a=2a = 2?

  1. ln⁡(2)+∑n=1∞(−1)n−1(x−2)nn⋅2n\ln(2) + \sum_{n=1}^{\infty} \frac{(-1)^{n-1}(x-2)^n}{n \cdot 2^n} (correct answer)
  2. ∑n=1∞(−1)n−1(x−2)nn⋅2n\sum_{n=1}^{\infty} \frac{(-1)^{n-1}(x-2)^n}{n \cdot 2^n}
  3. ln⁡(2)+∑n=1∞(−1)n(x−2)nn⋅2n\ln(2) + \sum_{n=1}^{\infty} \frac{(-1)^{n}(x-2)^n}{n \cdot 2^n}
  4. ln⁡(2)+∑n=1∞(−1)n−1(x−2)n2n\ln(2) + \sum_{n=1}^{\infty} \frac{(-1)^{n-1}(x-2)^n}{2^n}
Explanation: To find the Taylor series, we compute the derivatives of f(x)=ln⁡(x)f(x) = \ln(x) at a=2a=2. f(2)=ln⁡(2)f(2) = \ln(2). f′(x)=x−1f'(x) = x^{-1}, f′′(x)=−x−2f''(x) = -x^{-2}, f′′′(x)=2x−3f'''(x) = 2x^{-3}, and in general, f(n)(x)=(−1)n−1(n−1)!x−nf^{(n)}(x) = (-1)^{n-1}(n-1)!x^{-n} for n≥1n \ge 1. At a=2a=2, f(n)(2)=(−1)n−1(n−1)!2nf^{(n)}(2) = \frac{(-1)^{n-1}(n-1)!}{2^n}. The Taylor series coefficient is cn=f(n)(2)n!=(−1)n−1(n−1)!n!⋅2n=(−1)n−1n⋅2nc_n = \frac{f^{(n)}(2)}{n!} = \frac{(-1)^{n-1}(n-1)!}{n! \cdot 2^n} = \frac{(-1)^{n-1}}{n \cdot 2^n} for n≥1n \ge 1. The series is f(2)+∑n=1∞cn(x−2)nf(2) + \sum_{n=1}^{\infty} c_n (x-2)^n, which is ln⁡(2)+∑n=1∞(−1)n−1(x−2)nn⋅2n\ln(2) + \sum_{n=1}^{\infty} \frac{(-1)^{n-1}(x-2)^n}{n \cdot 2^n}.

Question 13

Let P4(x)P_4(x) be the fourth-degree Maclaurin polynomial for f(x)=cos⁡(3x)f(x) = \cos(3x). What is the value of P4(π/6)P_4(\pi/6)?

  1. 00
  2. 1−π281 - \frac{\pi^2}{8}
  3. 1−π28+π43841 - \frac{\pi^2}{8} + \frac{\pi^4}{384} (correct answer)
  4. 1−π22+π4241 - \frac{\pi^2}{2} + \frac{\pi^4}{24}
Explanation: The Maclaurin series for cos⁡(u)\cos(u) is 1−u22!+u44!−…1 - \frac{u^2}{2!} + \frac{u^4}{4!} - \dots. Let u=3xu=3x. The series for cos⁡(3x)\cos(3x) is 1−(3x)22!+(3x)44!−⋯=1−9x22+81x424−…1 - \frac{(3x)^2}{2!} + \frac{(3x)^4}{4!} - \dots = 1 - \frac{9x^2}{2} + \frac{81x^4}{24} - \dots. The fourth-degree polynomial is P4(x)=1−92x2+278x4P_4(x) = 1 - \frac{9}{2}x^2 + \frac{27}{8}x^4. Evaluating at x=π/6x = \pi/6: P4(π/6)=1−92(π6)2+278(π6)4=1−9π22⋅36+27π48⋅1296=1−π28+π4384P_4(\pi/6) = 1 - \frac{9}{2}\left(\frac{\pi}{6}\right)^2 + \frac{27}{8}\left(\frac{\pi}{6}\right)^4 = 1 - \frac{9\pi^2}{2 \cdot 36} + \frac{27\pi^4}{8 \cdot 1296} = 1 - \frac{\pi^2}{8} + \frac{\pi^4}{384} Distractor (A) is f(π/6)=cos⁡(π/2)=0f(\pi/6) = \cos(\pi/2) = 0, a common trap. Distractor (B) is the value of the second-degree polynomial. Distractor (D) incorrectly uses xx instead of 3x3x in the series.

Question 14

The Maclaurin series for a function f(x)f(x) is given by ∑n=0∞(n+1)2nxn\sum_{n=0}^{\infty} (n+1)2^n x^n. What is the value of f′′′(0)f'''(0)?

  1. 3232
  2. 4848
  3. 144144
  4. 192192 (correct answer)
Explanation: The coefficient of xnx^n in a Maclaurin series is given by cn=f(n)(0)n!c_n = \frac{f^{(n)}(0)}{n!}. We want f′′′(0)f'''(0), so we need the coefficient of the n=3n=3 term, c3c_3. From the given series, cn=(n+1)2nc_n = (n+1)2^n. For n=3n=3, c3=(3+1)23=4⋅8=32c_3 = (3+1)2^3 = 4 \cdot 8 = 32. Now, we solve for the derivative: f′′′(0)=c3⋅3!=32⋅6=192f'''(0) = c_3 \cdot 3! = 32 \cdot 6 = 192.

Question 15

Let f(x)f(x) be a function such that f(1)=2f(1) = 2, f′(1)=−1f'(1) = -1, f′′(1)=6f''(1) = 6, and f′′′(1)=−12f'''(1) = -12. What is the third-degree Taylor polynomial, T3(x)T_3(x), for f(x)f(x) centered at a=1a = 1?

  1. 2−(x−1)+6(x−1)2−12(x−1)32 - (x-1) + 6(x-1)^2 - 12(x-1)^3
  2. 2−(x−1)+3(x−1)2−2(x−1)32 - (x-1) + 3(x-1)^2 - 2(x-1)^3 (correct answer)
  3. 2−x+3x2−2x32 - x + 3x^2 - 2x^3
  4. 2−(x−1)+3(x−1)2−4(x−1)32 - (x-1) + 3(x-1)^2 - 4(x-1)^3
Explanation: The formula for the third-degree Taylor polynomial centered at a=1a=1 is: T3(x)=f(1)+f′(1)1!(x−1)+f′′(1)2!(x−1)2+f′′′(1)3!(x−1)3T_3(x) = f(1) + \frac{f'(1)}{1!}(x-1) + \frac{f''(1)}{2!}(x-1)^2 + \frac{f'''(1)}{3!}(x-1)^3 Substituting the given values: T3(x)=2+−11(x−1)+62(x−1)2+−126(x−1)3T_3(x) = 2 + \frac{-1}{1}(x-1) + \frac{6}{2}(x-1)^2 + \frac{-12}{6}(x-1)^3 T3(x)=2−(x−1)+3(x−1)2−2(x−1)3T_3(x) = 2 - (x-1) + 3(x-1)^2 - 2(x-1)^3 Distractor (A) forgets to divide by the factorials. Distractor (C) incorrectly centers the polynomial at 0. Distractor (D) makes an error in calculating 3!3!.

Question 16

The Taylor series for a function f(x)f(x) around x=0x=0 is 1−2x2+3x4−4x6+…1 - 2x^2 + 3x^4 - 4x^6 + \dots. What is the value of f(4)(0)f^{(4)}(0)?

  1. 33
  2. 1212
  3. 7272 (correct answer)
  4. −48-48
Explanation: The general formula for the coefficient of xnx^n in a Maclaurin series is cn=f(n)(0)n!c_n = \frac{f^{(n)}(0)}{n!}. We are interested in f(4)(0)f^{(4)}(0), so we look at the x4x^4 term in the series. The term is 3x43x^4, so its coefficient is c4=3c_4 = 3. Using the formula, we have c4=f(4)(0)4!c_4 = \frac{f^{(4)}(0)}{4!}. Substituting c4=3c_4 = 3, we get 3=f(4)(0)4!3 = \frac{f^{(4)}(0)}{4!}. Solving for the derivative: f(4)(0)=3⋅4!=3⋅24=72f^{(4)}(0) = 3 \cdot 4! = 3 \cdot 24 = 72.

Question 17

If the Maclaurin series for f(x)f(x) is ∑n=1∞(−1)n+1xnn2\sum_{n=1}^{\infty} \frac{(-1)^{n+1} x^n}{n^2}, what is the Maclaurin series for f′(x)f'(x)?

  1. ∑k=0∞(−1)kxkk+1\sum_{k=0}^{\infty} \frac{(-1)^k x^k}{k+1} (correct answer)
  2. ∑n=1∞(−1)n+1xn−1n2\sum_{n=1}^{\infty} \frac{(-1)^{n+1} x^{n-1}}{n^2}
  3. ∑n=1∞(−1)n+1xn+1n2(n+1)\sum_{n=1}^{\infty} \frac{(-1)^{n+1} x^{n+1}}{n^2(n+1)}
  4. ∑k=1∞(−1)kxkk\sum_{k=1}^{\infty} \frac{(-1)^k x^k}{k}
Explanation: We differentiate the series for f(x)f(x) term-by-term to find the series for f′(x)f'(x). f′(x)=ddx(∑n=1∞(−1)n+1xnn2)=∑n=1∞(−1)n+1(nxn−1)n2=∑n=1∞(−1)n+1xn−1nf'(x) = \frac{d}{dx} \left( \sum_{n=1}^{\infty} \frac{(-1)^{n+1} x^n}{n^2} \right) = \sum_{n=1}^{\infty} \frac{(-1)^{n+1} (n x^{n-1})}{n^2} = \sum_{n=1}^{\infty} \frac{(-1)^{n+1} x^{n-1}}{n} To match the format of the answer choices, we re-index the series. Let k=n−1k = n-1, so n=k+1n = k+1. When n=1n=1, k=0k=0. The sum becomes: ∑k=0∞(−1)(k+1)+1xkk+1=∑k=0∞(−1)k+2xkk+1=∑k=0∞(−1)kxkk+1\sum_{k=0}^{\infty} \frac{(-1)^{(k+1)+1} x^k}{k+1} = \sum_{k=0}^{\infty} \frac{(-1)^{k+2} x^k}{k+1} = \sum_{k=0}^{\infty} \frac{(-1)^k x^k}{k+1} This is the Maclaurin series for ln⁡(1+x)x\frac{\ln(1+x)}{x}.

Question 18

Find the Taylor series representation for f(x)=1x2f(x) = \frac{1}{x^2} centered at a=−1a = -1.

  1. ∑n=0∞(n+1)(x+1)n\sum_{n=0}^{\infty} (n+1) (x+1)^n (correct answer)
  2. ∑n=0∞(−1)n(n+1)(x+1)n\sum_{n=0}^{\infty} (-1)^n (n+1) (x+1)^n
  3. ∑n=0∞(x+1)n\sum_{n=0}^{\infty} (x+1)^n
  4. ∑n=0∞(−1)n(x+1)n\sum_{n=0}^{\infty} (-1)^n (x+1)^n
Explanation: We need to find the Taylor series for f(x)=x−2f(x) = x^{-2} centered at a=−1a = -1. We'll compute the derivatives and evaluate them at x=−1x = -1. f(x)=x−2  ⟹  f(−1)=1f(x) = x^{-2} \implies f(-1) = 1 f′(x)=−2x−3  ⟹  f′(−1)=2f'(x) = -2x^{-3} \implies f'(-1) = 2 f′′(x)=6x−4  ⟹  f′′(−1)=6f''(x) = 6x^{-4} \implies f''(-1) = 6 f′′′(x)=−24x−5  ⟹  f′′′(−1)=24f'''(x) = -24x^{-5} \implies f'''(-1) = 24 In general, f(n)(x)=(−1)n(n+1)!x−(n+2)f^{(n)}(x) = (-1)^n (n+1)! x^{-(n+2)}. Evaluating at x=−1x = -1: f(n)(−1)=(−1)n(n+1)!(−1)−(n+2)=(n+1)!f^{(n)}(-1) = (-1)^n (n+1)! (-1)^{-(n+2)} = (n+1)!. The Taylor series coefficient is cn=f(n)(−1)n!=(n+1)!n!=n+1c_n = \frac{f^{(n)}(-1)}{n!} = \frac{(n+1)!}{n!} = n+1. Therefore, the Taylor series is ∑n=0∞(n+1)(x+1)n\sum_{n=0}^{\infty} (n+1)(x+1)^n.

Question 19

The function f(x)=ln⁡(cos⁡(x))f(x) = \ln(\cos(x)) has a Maclaurin series. What are the first two non-zero terms of this series?

  1. x−x22x - \frac{x^2}{2}
  2. −x22−x412-\frac{x^2}{2} - \frac{x^4}{12} (correct answer)
  3. 1−x221 - \frac{x^2}{2}
  4. −x22+x424-\frac{x^2}{2} + \frac{x^4}{24}
Explanation: We use composition of series. Let u=cos⁡(x)−1u = \cos(x) - 1. Then ln⁡(cos⁡x)=ln⁡(1+u)\ln(\cos x) = \ln(1 + u). The series for cos⁡(x)\cos(x) is 1−x22!+x44!−…1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \dots. So, u=−x22+x424−…u = -\frac{x^2}{2} + \frac{x^4}{24} - \dots. The series for ln⁡(1+u)\ln(1+u) is u−u22+…u - \frac{u^2}{2} + \dots. Substitute the series for uu into the series for ln⁡(1+u)\ln(1+u): (−x22+x424−… )−12(−x22+… )2+…\left(-\frac{x^2}{2} + \frac{x^4}{24} - \dots\right) - \frac{1}{2}\left(-\frac{x^2}{2} + \dots\right)^2 + \dots =−x22+x424−12(x44)+⋯=−x22+x424−x48+⋯=−x22−2x424=−x22−x412= -\frac{x^2}{2} + \frac{x^4}{24} - \frac{1}{2}\left(\frac{x^4}{4}\right) + \dots = -\frac{x^2}{2} + \frac{x^4}{24} - \frac{x^4}{8} + \dots = -\frac{x^2}{2} - \frac{2x^4}{24} = -\frac{x^2}{2} - \frac{x^4}{12} The first two non-zero terms are −x22−x412-\frac{x^2}{2} - \frac{x^4}{12}.

Question 20

Find the Taylor series for f(x)=e3xf(x) = e^{3x} centered at a=−1a=-1.

  1. ∑n=0∞e−3(x+1)nn!\sum_{n=0}^{\infty} \frac{e^{-3} (x+1)^n}{n!}
  2. ∑n=0∞3ne3(x+1)nn!\sum_{n=0}^{\infty} \frac{3^n e^{3} (x+1)^n}{n!}
  3. ∑n=0∞3ne−3(x+1)nn!\sum_{n=0}^{\infty} \frac{3^n e^{-3} (x+1)^n}{n!} (correct answer)
  4. ∑n=0∞3n(x−1)nn!\sum_{n=0}^{\infty} \frac{3^n (x-1)^n}{n!}
Explanation: When finding a Taylor series centered at a point other than zero, you need to use the general Taylor series formula and carefully evaluate all derivatives at the given center point. For f(x)=e3xf(x) = e^{3x} centered at a=−1a = -1, start by finding the pattern of derivatives. The first few derivatives are: f′(x)=3e3xf'(x) = 3e^{3x}, f′′(x)=9e3xf''(x) = 9e^{3x}, f′′′(x)=27e3xf'''(x) = 27e^{3x}. In general, f(n)(x)=3ne3xf^{(n)}(x) = 3^n e^{3x}. Next, evaluate each derivative at the center point x=−1x = -1: f(n)(−1)=3ne3(−1)=3ne−3f^{(n)}(-1) = 3^n e^{3(-1)} = 3^n e^{-3}. The Taylor series formula gives us: f(x)=∑n=0∞f(n)(−1)n!(x−(−1))n=∑n=0∞3ne−3n!(x+1)nf(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(-1)}{n!}(x-(-1))^n = \sum_{n=0}^{\infty} \frac{3^n e^{-3}}{n!}(x+1)^n This matches choice C exactly. Looking at the wrong answers: Choice A is missing the 3n3^n factor that comes from the chain rule when differentiating e3xe^{3x}. Choice B incorrectly uses e3e^3 instead of e−3e^{-3} - this happens when you forget to substitute x=−1x = -1 and mistakenly use x=1x = 1. Choice D has the wrong center point, using (x−1)(x-1) instead of (x+1)(x+1), and is missing the exponential factor entirely. Study tip: Always substitute the center point into your derivatives carefully, and remember that centering at a=−1a = -1 means your series involves (x+1)(x+1) terms, not (x−1)(x-1).