Calculus 2 Quiz: Approximation Error And Bounds
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Approximation Error And BoundsQuestion 1 of 20

The trapezoidal rule with n=4n = 4 subintervals is used to approximate 02ex2dx\int_0^2 e^{x^2} dx. Given that f(x)400|f''(x)| \leq 400 on [0,2][0,2], which statement about the approximation error is correct?

The error is bounded by (20)34001242=8400192=503\frac{(2-0)^3 \cdot 400}{12 \cdot 4^2} = \frac{8 \cdot 400}{192} = \frac{50}{3}
The error is bounded by (20)2400124=440048=1003\frac{(2-0)^2 \cdot 400}{12 \cdot 4} = \frac{4 \cdot 400}{48} = \frac{100}{3}
The error is bounded by (20)3400124=840048=2003\frac{(2-0)^3 \cdot 400}{12 \cdot 4} = \frac{8 \cdot 400}{48} = \frac{200}{3}
The error is bounded by (20)44001243=16400768=253\frac{(2-0)^4 \cdot 400}{12 \cdot 4^3} = \frac{16 \cdot 400}{768} = \frac{25}{3}
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Calculus 2 Quiz

Calculus 2 Quiz: Approximation Error And Bounds

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

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

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Question 1

The trapezoidal rule with n=4n = 4 subintervals is used to approximate 02ex2dx\int_0^2 e^{x^2} dx. Given that f(x)400|f''(x)| \leq 400 on [0,2][0,2], which statement about the approximation error is correct?

  1. The error is bounded by (20)34001242=8400192=503\frac{(2-0)^3 \cdot 400}{12 \cdot 4^2} = \frac{8 \cdot 400}{192} = \frac{50}{3} (correct answer)
  2. The error is bounded by (20)2400124=440048=1003\frac{(2-0)^2 \cdot 400}{12 \cdot 4} = \frac{4 \cdot 400}{48} = \frac{100}{3}
  3. The error is bounded by (20)3400124=840048=2003\frac{(2-0)^3 \cdot 400}{12 \cdot 4} = \frac{8 \cdot 400}{48} = \frac{200}{3}
  4. The error is bounded by (20)44001243=16400768=253\frac{(2-0)^4 \cdot 400}{12 \cdot 4^3} = \frac{16 \cdot 400}{768} = \frac{25}{3}
Explanation: The error bound for the trapezoidal rule is ET(ba)3M12n2|E_T| \leq \frac{(b-a)^3 M}{12n^2} where M is the maximum of f(x)|f''(x)| on the interval. Here, ba=2b-a = 2, M=400M = 400, and n=4n = 4, so the error bound is 234001242=8400192=503\frac{2^3 \cdot 400}{12 \cdot 4^2} = \frac{8 \cdot 400}{192} = \frac{50}{3}. Choice B incorrectly uses (ba)2(b-a)^2 and nn instead of n2n^2. Choice C uses nn instead of n2n^2. Choice D incorrectly uses (ba)4(b-a)^4 and n3n^3.

Question 2

Simpson's rule is applied to approximate 131xdx\int_1^3 \frac{1}{x} dx with n=6n = 6 subintervals. If the fourth derivative satisfies f(4)(x)24x5|f^{(4)}(x)| \leq \frac{24}{x^5} for x[1,3]x \in [1,3], what is the error bound for this approximation?

  1. (31)42418064=16241801296=384233280\frac{(3-1)^4 \cdot 24}{180 \cdot 6^4} = \frac{16 \cdot 24}{180 \cdot 1296} = \frac{384}{233280}
  2. (31)5249064=3224901296=768116640\frac{(3-1)^5 \cdot 24}{90 \cdot 6^4} = \frac{32 \cdot 24}{90 \cdot 1296} = \frac{768}{116640}
  3. (31)52418064=32241801296=768233280\frac{(3-1)^5 \cdot 24}{180 \cdot 6^4} = \frac{32 \cdot 24}{180 \cdot 1296} = \frac{768}{233280} (correct answer)
  4. (31)52418065=32241807776=7681399680\frac{(3-1)^5 \cdot 24}{180 \cdot 6^5} = \frac{32 \cdot 24}{180 \cdot 7776} = \frac{768}{1399680}
Explanation: When you encounter Simpson's rule error bounds, you're working with numerical integration accuracy. The error bound formula for Simpson's rule is: ES(ba)5180n4maxx[a,b]f(4)(x)E_S \leq \frac{(b-a)^5}{180n^4} \max_{x \in [a,b]} |f^{(4)}(x)| Let's apply this formula step by step. You have f(x)=1xf(x) = \frac{1}{x}, integration bounds [1,3][1,3], so ba=31=2b-a = 3-1 = 2, and n=6n = 6 subintervals. The maximum value of f(4)(x)=24x5|f^{(4)}(x)| = \frac{24}{x^5} occurs at x=1x = 1 (since the function decreases as xx increases), giving us maxf(4)(x)=2415=24\max |f^{(4)}(x)| = \frac{24}{1^5} = 24. Substituting into the formula: ES(2)52418064=32241801296=768233280E_S \leq \frac{(2)^5 \cdot 24}{180 \cdot 6^4} = \frac{32 \cdot 24}{180 \cdot 1296} = \frac{768}{233280} Choice A incorrectly uses (ba)4(b-a)^4 instead of (ba)5(b-a)^5 in the numerator – this confuses Simpson's rule with the trapezoidal rule error formula. Choice B uses the wrong denominator coefficient (90 instead of 180), which would apply to a different numerical method. Choice D correctly uses (ba)5(b-a)^5 and 180 but mistakenly puts n5n^5 instead of n4n^4 in the denominator. The correct answer is C. Study tip: Memorize that Simpson's rule error has the pattern (ba)5180n4\frac{(b-a)^5}{180n^4} – the "5-4-180" combination is unique to Simpson's rule and frequently tested.

Question 3

A function f(x)f(x) is approximated using its third-degree Taylor polynomial P3(x)P_3(x) centered at a=2a = 2. If f(4)(x)120|f^{(4)}(x)| \leq 120 for all xx in [1.5,2.5][1.5, 2.5], what is the maximum possible error when using P3(x)P_3(x) to approximate f(1.8)f(1.8)?

  1. 120(0.2)44!=1200.001624=0.008\frac{120 \cdot (0.2)^4}{4!} = \frac{120 \cdot 0.0016}{24} = 0.008 (correct answer)
  2. 120(0.2)33!=1200.0086=0.16\frac{120 \cdot (0.2)^3}{3!} = \frac{120 \cdot 0.008}{6} = 0.16
  3. 120(0.2)43!=1200.00166=0.032\frac{120 \cdot (0.2)^4}{3!} = \frac{120 \cdot 0.0016}{6} = 0.032
  4. 120(0.2)54!=1200.0003224=0.0016\frac{120 \cdot (0.2)^5}{4!} = \frac{120 \cdot 0.00032}{24} = 0.0016
Explanation: Taylor's theorem with remainder states that the error in the nth-degree Taylor approximation is Rn(x)=f(n+1)(c)(n+1)!xan+1|R_n(x)| = \frac{|f^{(n+1)}(c)|}{(n+1)!}|x-a|^{n+1} for some c between x and a. Here, n=3, so the error is bounded by f(4)(c)4!xa41204!1.824=1200.001624=0.008\frac{|f^{(4)}(c)|}{4!}|x-a|^4 \leq \frac{120}{4!}|1.8-2|^4 = \frac{120 \cdot 0.0016}{24} = 0.008. Choice B incorrectly uses the third derivative term. Choice C uses the wrong factorial. Choice D incorrectly uses the fifth power.

Question 4

The power series n=0(1)nx2n+1(2n+1)!\sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{(2n+1)!} converges to sin(x)\sin(x) for all real xx. When this series is truncated after the first 4 terms to approximate sin(1.2)\sin(1.2), what type of error bound applies, and what is the bound?

  1. Taylor remainder applies; error (1.2)818!=1.28403201.29×104\leq \frac{(1.2)^8 \cdot 1}{8!} = \frac{1.2^8}{40320} ≈ 1.29 × 10^{-4}
  2. Lagrange remainder applies; error (1.2)99!=1.293628801.38×105\leq \frac{(1.2)^9}{9!} = \frac{1.2^9}{362880} ≈ 1.38 × 10^{-5}
  3. Alternating series test applies; error (1.2)88!=1.28403201.29×104\leq \frac{(1.2)^8}{8!} = \frac{1.2^8}{40320} ≈ 1.29 × 10^{-4}
  4. Alternating series test applies; error (1.2)99!=1.293628801.38×105\leq \frac{(1.2)^9}{9!} = \frac{1.2^9}{362880} ≈ 1.38 × 10^{-5} (correct answer)
Explanation: When you encounter a power series approximation problem, you need to identify which error bound theorem applies based on the series structure. This series has alternating signs with terms (1)nx2n+1(2n+1)!\frac{(-1)^n x^{2n+1}}{(2n+1)!}, making it an alternating series. For alternating series where terms decrease in absolute value, the Alternating Series Test provides the tightest error bound: the error is at most the absolute value of the first omitted term. The first four terms correspond to n=0,1,2,3n = 0, 1, 2, 3, so the first omitted term has n=4n = 4. When n=4n = 4: (1)4x99!=x99!\frac{(-1)^4 x^{9}}{9!} = \frac{x^9}{9!} At x=1.2x = 1.2, the error bound is (1.2)99!=1.293628801.38×105\frac{(1.2)^9}{9!} = \frac{1.2^9}{362880} ≈ 1.38 × 10^{-5} Option A incorrectly applies "Taylor remainder" (not the specific theorem name) and uses the wrong power (8 instead of 9). Option B correctly calculates the bound but misnames the theorem as "Lagrange remainder" - while technically applicable, it's unnecessarily complex since the Alternating Series Test gives the same result more directly. Option C correctly identifies the alternating series test but uses x8x^8 instead of x9x^9, missing that the first omitted term has n=4n = 4, giving 2n+1=92n+1 = 9. Remember: for alternating series approximations, always use the first omitted term for your error bound, and count carefully to identify which term that is.

Question 5

The alternating series n=1(1)n+1n4\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n^4} converges to a sum SS. What is the smallest integer NN such that the partial sum SNS_N is guaranteed by the Alternating Series Estimation Theorem to approximate SS with an error less than 10310^{-3}?

  1. 4
  2. 5 (correct answer)
  3. 6
  4. 7
Explanation: According to the Alternating Series Estimation Theorem, the error in approximating the sum SS with the partial sum SNS_N is bounded by the magnitude of the first neglected term, RNaN+1|R_N| \le |a_{N+1}|. We need aN+1<103|a_{N+1}| < 10^{-3}. Here, an=1n4a_{n} = \frac{1}{n^4}, so we need 1(N+1)4<11000\frac{1}{(N+1)^4} < \frac{1}{1000}. This implies (N+1)4>1000(N+1)^4 > 1000. We can test integer values: 54=6255^4 = 625 and 64=12966^4 = 1296. Therefore, we must have N+16N+1 \ge 6, which means N5N \ge 5. The smallest integer value for NN is 5.

Question 6

The sum of the series n=11n3\sum_{n=1}^{\infty} \frac{1}{n^3} is approximated by its 10th partial sum, S10S_{10}. Let R10R_{10} be the remainder SS10S - S_{10}. According to the Integral Test remainder estimate, which of the following is a correct statement about R10R_{10}?

  1. R10<1100R_{10} < \frac{1}{100}
  2. R10<1200R_{10} < \frac{1}{200} (correct answer)
  3. R10>1200R_{10} > \frac{1}{200}
  4. R10<1242R_{10} < \frac{1}{242}
Explanation: The Integral Test remainder estimate states that n+1f(x)dx<Rn<nf(x)dx\int_{n+1}^{\infty} f(x) \, dx < R_n < \int_{n}^{\infty} f(x) \, dx. For this problem, f(x)=1/x3f(x) = 1/x^3 and n=10n=10. We can find an upper bound for R10R_{10} by evaluating 101x3dx\int_{10}^{\infty} \frac{1}{x^3} \, dx. The integral is [12x2]10=0(12(10)2)=1200\left[ -\frac{1}{2x^2} \right]_{10}^{\infty} = 0 - \left(-\frac{1}{2(10)^2}\right) = \frac{1}{200}. Thus, we are guaranteed that R10<1200R_{10} < \frac{1}{200}.

Question 7

The function f(x)=e3xf(x) = e^{-3x} is approximated by its second-degree Taylor polynomial, P2(x)P_2(x), centered at a=0a=0. Using Taylor's inequality, what is the tightest upper bound for the error f(x)P2(x)|f(x) - P_2(x)| on the interval [0,0.2][0, 0.2]?

  1. 0.036 (correct answer)
  2. 0.180
  3. 0.108
  4. 0.216
Explanation: Taylor's inequality states that the error R2(x)|R_2(x)| is bounded by M(2+1)!x03\frac{M}{(2+1)!}|x-0|^3, where MM is an upper bound for f(3)(c)|f^{(3)}(c)| on the interval [0,0.2][0, 0.2]. The derivatives are f(x)=3e3xf'(x) = -3e^{-3x}, f(x)=9e3xf''(x) = 9e^{-3x}, and f(x)=27e3xf'''(x) = -27e^{-3x}. The function f(x)=27e3x|f'''(x)| = 27e^{-3x} is a decreasing function, so its maximum on [0,0.2][0, 0.2] occurs at x=0x=0, giving M=27e0=27M=27e^0=27. The maximum value of x3|x|^3 on [0,0.2][0, 0.2] is (0.2)3=0.008(0.2)^3 = 0.008. Therefore, the error bound is R2(x)273!(0.2)3=276(0.008)=4.5(0.008)=0.036|R_2(x)| \le \frac{27}{3!} (0.2)^3 = \frac{27}{6}(0.008) = 4.5(0.008) = 0.036.

Question 8

To approximate ln(1.5)\ln(1.5) using a Taylor polynomial for f(x)=ln(x)f(x) = \ln(x) centered at a=1a=1, what is the minimum degree nn required to guarantee an absolute error of no more than 13000\frac{1}{3000}?

  1. 5
  2. 7
  3. 8 (correct answer)
  4. 9
Explanation: The Lagrange error bound is Rn(1.5)M(n+1)!1.51n+1|R_n(1.5)| \le \frac{M}{(n+1)!} |1.5-1|^{n+1}, where MM is an upper bound for f(n+1)(c)|f^{(n+1)}(c)| on [1,1.5][1, 1.5]. The (n+1)(n+1)-th derivative of ln(x)\ln(x) is f(n+1)(x)=(1)nn!x(n+1)f^{(n+1)}(x) = (-1)^n n! x^{-(n+1)}. On [1,1.5][1, 1.5], f(n+1)(c)=n!cn+1|f^{(n+1)}(c)| = \frac{n!}{c^{n+1}} is maximized at c=1c=1, so M=n!M=n!. The inequality becomes n!(n+1)!(0.5)n+113000\frac{n!}{(n+1)!} (0.5)^{n+1} \le \frac{1}{3000}, which simplifies to 1n+112n+113000\frac{1}{n+1} \frac{1}{2^{n+1}} \le \frac{1}{3000}, or (n+1)2n+13000(n+1)2^{n+1} \ge 3000. Testing values: for n=7n=7, 828=20488 \cdot 2^8 = 2048. For n=8n=8, 929=46089 \cdot 2^9 = 4608. Thus, the minimum degree is n=8n=8.

Question 9

Let BTB_T be the error bound for the Trapezoidal Rule and BSB_S be the error bound for Simpson's Rule when approximating 01(x4+x)dx\int_0^1 (x^4+x) dx with n=4n=4. What is the value of the ratio BT/BSB_T / B_S?

  1. 16
  2. 60
  3. 120 (correct answer)
  4. 240
Explanation: For f(x)=x4+xf(x)=x^4+x, we find the derivatives: f(x)=12x2f''(x) = 12x^2 and f(4)(x)=24f^{(4)}(x) = 24. On [0,1][0,1], the max of f(x)|f''(x)| is K=12K=12 and the max of f(4)(x)|f^{(4)}(x)| is M=24M=24. For a=0,b=1,n=4a=0, b=1, n=4:BT=K(ba)312n2=12(1)312(42)=116B_T = \frac{K(b-a)^3}{12n^2} = \frac{12(1)^3}{12(4^2)} = \frac{1}{16}. BS=M(ba)5180n4=24(1)5180(44)=24180256=11920B_S = \frac{M(b-a)^5}{180n^4} = \frac{24(1)^5}{180(4^4)} = \frac{24}{180 \cdot 256} = \frac{1}{1920}. The ratio is BT/BS=(1/16)/(1/1920)=1920/16=120B_T / B_S = (1/16) / (1/1920) = 1920/16 = 120.

Question 10

The sum of a convergent alternating series n=1(1)n+1an\sum_{n=1}^{\infty} (-1)^{n+1} a_n (with an>0a_n > 0) is approximated by the partial sum SNS_N. The error bound is given by RNaN+1|R_N| \le a_{N+1}. Which of the following conditions is necessary for this error bound to be valid?

  1. The series n=1an\sum_{n=1}^{\infty} a_n must converge.
  2. The sequence {an}\{a_n\} must be strictly decreasing.
  3. The sequence {an}\{a_n\} must be non-increasing for n1n \ge 1. (correct answer)
  4. The terms ana_n must be rational numbers for all nn.
Explanation: The Alternating Series Estimation Theorem requires two conditions on the sequence of magnitudes {an}\{a_n\}: 1) limnan=0\lim_{n\to\infty} a_n = 0 and 2) the sequence {an}\{a_n\} is non-increasing, i.e., an+1ana_{n+1} \le a_n for all nn (or for all nn beyond some integer). This non-increasing condition is what ensures the remainder is trapped between 0 and the next term. Strict decrease is sufficient but not necessary. Absolute convergence (A) is also sufficient for convergence of the alternating series but not necessary for the error bound. Rationality of terms (D) is irrelevant.

Question 11

The sum of the convergent alternating series n=1(1)n+1n2+1\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n^2 + 1} is SS. This sum is approximated by its fourth partial sum, S4S_4. Which of the following statements is true about the error, R4=SS4R_4 = S - S_4?

  1. The approximation S4S_4 is an overestimate of SS, and R4<117|R_4| < \frac{1}{17}.
  2. The approximation S4S_4 is an underestimate of SS, and R4<117|R_4| < \frac{1}{17}.
  3. The approximation S4S_4 is an overestimate of SS, and R4<126|R_4| < \frac{1}{26}.
  4. The approximation S4S_4 is an underestimate of SS, and R4<126|R_4| < \frac{1}{26}. (correct answer)
Explanation: The error is R4=SS4=a5a6+a7R_4 = S - S_4 = a_5 - a_6 + a_7 - \dots. The first term of the series for the remainder is a5=(1)5+152+1=126a_5 = \frac{(-1)^{5+1}}{5^2 + 1} = \frac{1}{26}, which is positive. For a convergent alternating series, the sign of the remainder is the same as the sign of its first term. Since a5a_5 is positive, R4>0R_4 > 0, which means SS4>0S - S_4 > 0, so S>S4S > S_4. Thus, S4S_4 is an underestimate. The Alternating Series Estimation Theorem states that R4<a5|R_4| < |a_5|, so R4<126|R_4| < \frac{1}{26}.

Question 12

The error bound for approximating abf(x)dx\int_a^b f(x) dx with the Trapezoidal Rule is ETK(ba)312n2|E_T| \le \frac{K(b-a)^3}{12n^2}, where KK is an upper bound for f(x)|f''(x)|. If f(x)f(x) is a linear function, what is the value of this error bound?

  1. Dependent on nn
  2. Infinite
  3. Dependent on bab-a
  4. Zero (correct answer)
Explanation: If f(x)f(x) is a linear function, say f(x)=mx+cf(x) = mx+c, its first derivative is f(x)=mf'(x) = m and its second derivative is f(x)=0f''(x) = 0. Therefore, the maximum value of f(x)|f''(x)| on any interval is K=0K=0. Substituting K=0K=0 into the error bound formula gives ET0(ba)312n2=0|E_T| \le \frac{0 \cdot (b-a)^3}{12n^2} = 0. This means the Trapezoidal Rule is exact for linear functions.

Question 13

The function f(x)=1+xf(x) = \sqrt{1+x} is approximated by its first-degree Taylor polynomial P1(x)P_1(x) centered at a=0a=0. What is the Lagrange error bound for this approximation on the interval [0,0.2][0, 0.2]?

  1. 0.005 (correct answer)
  2. 0.010
  3. 0.020
  4. 0.025
Explanation: The Lagrange error bound for P1(x)P_1(x) is R1(x)M2!x2|R_1(x)| \le \frac{M}{2!}|x|^2, where Mf(c)M \ge |f''(c)| for cc in [0,0.2][0, 0.2]. We have f(x)=(1+x)1/2f(x) = (1+x)^{1/2}, f(x)=12(1+x)1/2f'(x) = \frac{1}{2}(1+x)^{-1/2}, and f(x)=14(1+x)3/2f''(x) = -\frac{1}{4}(1+x)^{-3/2}. The magnitude is f(x)=14(1+x)3/2|f''(x)| = \frac{1}{4(1+x)^{3/2}}. This is a decreasing function on [0,0.2][0, 0.2], so its maximum occurs at x=0x=0, which is M=14(1)3/2=14M = \frac{1}{4(1)^ {3/2}} = \frac{1}{4}. The maximum value of x2|x|^2 on the interval is (0.2)2=0.04(0.2)^2 = 0.04. The error bound is 1/42(0.04)=18(0.04)=0.005\frac{1/4}{2} (0.04) = \frac{1}{8}(0.04) = 0.005.

Question 14

The series n=11n(n+1)\sum_{n=1}^{\infty} \frac{1}{n(n+1)} is a telescoping series that sums to 1. If we approximate this sum with SN=n=1N1n(n+1)S_N = \sum_{n=1}^{N} \frac{1}{n(n+1)}, the exact error is RN=1SNR_N = 1 - S_N. By using partial fractions, SN=11N+1S_N = 1 - \frac{1}{N+1}. Thus, the exact error is RN=1N+1R_N = \frac{1}{N+1}. What upper bound does the Integral Test provide for this remainder?

  1. 1N\frac{1}{N} (correct answer)
  2. 1N+1\frac{1}{N+1}
  3. ln(N+1)\ln(N+1)
  4. 12N2\frac{1}{2N^2}
Explanation: The integral test provides an upper bound for the remainder RNR_N given by Nf(x)dx\int_N^\infty f(x) dx. Here, f(x)=1x(x+1)=1x1x+1f(x) = \frac{1}{x(x+1)} = \frac{1}{x} - \frac{1}{x+1}. The integral is N(1x1x+1)dx=[lnxlnx+1]N=[lnxx+1]N\int_N^\infty (\frac{1}{x} - \frac{1}{x+1}) dx = [\ln|x| - \ln|x+1|]_N^\infty = [\ln|\frac{x}{x+1}|]_N^\infty. As xx \to \infty, lnxx+1ln(1)=0\ln|\frac{x}{x+1}| \to \ln(1) = 0. So the integral is 0ln(NN+1)=ln(NN+1)=ln(N+1N)0 - \ln(\frac{N}{N+1}) = -\ln(\frac{N}{N+1}) = \ln(\frac{N+1}{N}). This is not among the answers. Let's use a simpler bounding function. Since 1x(x+1)<1x2\frac{1}{x(x+1)} < \frac{1}{x^2}, we can say RN<N1x2dx=1NR_N < \int_N^\infty \frac{1}{x^2} dx = \frac{1}{N}. This shows the integral test bound is not always tight but provides a correct upper bound. The exact error is 1/(N+1)1/(N+1), which is indeed less than 1/N1/N.

Question 15

For the approximation sin(x)xx3/6\sin(x) \approx x - x^3/6 on the interval [0.5,0.5][-0.5, 0.5], the Lagrange error bound involves finding the maximum of a derivative of sin(x)\sin(x). Which derivative should be used and what is its maximum absolute value on the interval?

  1. f(4)(x)=sin(x)f^{(4)}(x) = \sin(x), with maximum absolute value sin(0.5)\sin(0.5)
  2. f(4)(x)=sin(x)f^{(4)}(x) = \sin(x), with maximum absolute value 1
  3. f(5)(x)=cos(x)f^{(5)}(x) = \cos(x), with maximum absolute value cos(0.5)\cos(0.5)
  4. f(5)(x)=cos(x)f^{(5)}(x) = \cos(x), with maximum absolute value 1 (correct answer)
Explanation: The polynomial P3(x)=xx3/6P_3(x) = x - x^3/6 is the third-degree Maclaurin polynomial for sin(x)\sin(x). Because the x4x^4 term is zero, P3(x)=P4(x)P_3(x) = P_4(x). We can use the error formula for n=4n=4, which is R4(x)M5!x5|R_4(x)| \le \frac{M}{5!}|x|^5, where MM is the maximum of f(5)(c)|f^{(5)}(c)| on the interval. The fifth derivative is f(5)(x)=cos(x)f^{(5)}(x) = \cos(x). On [0.5,0.5][-0.5, 0.5], cos(x)\cos(x) is always positive and its maximum occurs at x=0x=0, where cos(0)=1\cos(0)=1. So, the derivative to use is the fifth, and its maximum absolute value is 1.

Question 16

A function f(x)f(x) has a fourth derivative f(4)(x)=cos(x2)10f^{(4)}(x) = \frac{\cos(x^2)}{10}. We approximate 11f(x)dx\int_{-1}^{1} f(x) dx using Simpson's rule with nn subintervals. What is the smallest even integer nn that guarantees an error no more than 19000\frac{1}{9000}?

  1. 2
  2. 4 (correct answer)
  3. 8
  4. 10
Explanation: The error bound for Simpson's Rule is ESM(ba)5180n4|E_S| \le \frac{M(b-a)^5}{180n^4}, where MM is an upper bound for f(4)(x)|f^{(4)}(x)| on [1,1][-1, 1]. We are given f(4)(x)=cos(x2)10f^{(4)}(x) = \frac{\cos(x^2)}{10}. The maximum value of cos(u)|\cos(u)| is 1. Thus, the maximum value of f(4)(x)|f^{(4)}(x)| is M=110M = \frac{1}{10}. We have a=1,b=1a=-1, b=1, so ba=2b-a=2. We need (1/10)(2)5180n419000\frac{(1/10)(2)^5}{180n^4} \le \frac{1}{9000}. This simplifies to 3.2180n419000\frac{3.2}{180n^4} \le \frac{1}{9000}, which is 321800n419000\frac{32}{1800n^4} \le \frac{1}{9000} or 4225n419000\frac{4}{225n^4} \le \frac{1}{9000}. Cross-multiplying gives 49000225n44 \cdot 9000 \le 225n^4, so 36000225n436000 \le 225n^4. Dividing by 225 gives 160n4160 \le n^4. We need to find the smallest even integer nn satisfying this. Test values: 24=162^4=16 (too small). 44=2564^4=256 (works). So the smallest even integer is n=4n=4.

Question 17

The function f(x)f(x) is approximated by its nn-th degree Taylor polynomial Pn(x)P_n(x) centered at aa. The Lagrange error bound for f(x)Pn(x)|f(x) - P_n(x)| is given by M(n+1)!xan+1\frac{M}{(n+1)!}|x-a|^{n+1}. How is the value of MM determined?

  1. MM is the exact value of f(n+1)(x)|f^{(n+1)}(x)| at the point of approximation xx.
  2. MM is an upper bound for f(n+1)(c)|f^{(n+1)}(c)| for any cc in the interval between aa and xx. (correct answer)
  3. MM is the average value of f(n+1)(c)|f^{(n+1)}(c)| on the interval between aa and xx.
  4. MM is the exact value of f(n+1)(a)|f^{(n+1)}(a)| at the center of the approximation aa.
Explanation: The value MM in the Lagrange error bound formula represents an upper bound on the magnitude of the (n+1)(n+1)-th derivative of ff. Specifically, MM must satisfy f(n+1)(c)M|f^{(n+1)}(c)| \le M for all cc in the interval containing the center aa and the point of approximation xx. To get the tightest bound, one typically finds the maximum value of f(n+1)(c)|f^{(n+1)}(c)| on this interval.

Question 18

A student uses n=10n=10 subintervals to approximate 12ln(x)dx\int_1^2 \ln(x) dx. They find the error bound for the Trapezoidal Rule is BTB_T and the error bound for Simpson's Rule is BSB_S. Which statement correctly compares the bounds?

  1. BSB_S is larger than BTB_T because the fourth derivative of ln(x)\ln(x) is larger in magnitude than the second derivative.
  2. BTB_T is larger than BSB_S primarily because the error formula for BTB_T has a denominator proportional to n2n^2 while BSB_S has a denominator proportional to n4n^4. (correct answer)
  3. BTB_T and BSB_S cannot be computed because ln(x)\ln(x) is undefined at x=0x=0.
  4. BTB_T is larger than BSB_S because the Trapezoidal Rule always overestimates for concave down functions, while Simpson's Rule does not.
Explanation: The error bound for Trapezoidal Rule is 1/n2\propto 1/n^2 and for Simpson's Rule is 1/n4\propto 1/n^4. For n=10n=10, the n4n^4 term (10,000) is much larger than the n2n^2 term (100), leading to a much smaller error bound for Simpson's Rule, provided the derivative factors (K and M) are of a similar order of magnitude. For ln(x)\ln(x) on [1,2][1,2], this is the case. The interval of integration does not include 0, so C is incorrect. While D contains a correct statement about the Trapezoidal Rule, it's not the primary reason for the difference in the error bounds.

Question 19

An integral is approximated using Simpson's Rule with nn subintervals, resulting in a theoretical error bound BB. If the number of subintervals is tripled (to 3n3n), what will the new theoretical error bound be in terms of BB?

  1. B/3B/3
  2. B/9B/9
  3. B/27B/27
  4. B/81B/81 (correct answer)
Explanation: The error bound for Simpson's Rule is given by ESM(ba)5180n4|E_S| \le \frac{M(b-a)^5}{180n^4}. Let this bound be BB. If the number of subintervals is changed to n=3nn' = 3n, the new error bound BB' will be B=M(ba)5180(3n)4=M(ba)518034n4=181(M(ba)5180n4)=B81B' = \frac{M(b-a)^5}{180(3n)^4} = \frac{M(b-a)^5}{180 \cdot 3^4 n^4} = \frac{1}{81} \left( \frac{M(b-a)^5}{180n^4} \right) = \frac{B}{81}.

Question 20

Consider approximating ee using the Maclaurin series for exe^x evaluated at x=1x=1, which is n=01n!\sum_{n=0}^\infty \frac{1}{n!}. We use the partial sum SN=n=0N1n!S_N = \sum_{n=0}^N \frac{1}{n!}. The Lagrange error bound for this approximation states that RNM(N+1)!|R_N| \le \frac{M}{(N+1)!}, where MM is the maximum of ece^c on [0,1][0,1]. To use this bound, we need a value for MM, but ee is the number we are approximating. Which is the most effective and valid approach to establish a numerical error bound?

  1. Use the approximation e2.718e \approx 2.718 to calculate MM, since high precision is not needed for the bound.
  2. Use the fact that e<3e < 3 to establish a valid, though not the tightest, upper bound for the error. (correct answer)
  3. Use a lower bound for ee, such as e>2e > 2, since the error bound is insensitive to small changes in MM.
  4. The Lagrange error bound cannot be used in this case because the value of MM depends on the quantity being calculated.
Explanation: The problem is that to find the error in approximating ee, we need to know the value of ee to find M=e1M = e^1. This is circular. However, we do not need the exact maximum MM, only an upper bound. Since exe^x is an increasing function, we know e<3e < 3 (e.g., from S3=1+1+1/2+1/6=2.66...S_3 = 1+1+1/2+1/6 = 2.66... and knowing the remainder is positive). We can use M=3M=3 as a valid upper bound for ece^c on [0,1][0,1]. This gives a concrete, provable error bound RN3(N+1)!|R_N| \le \frac{3}{(N+1)!}, allowing us to determine the accuracy without knowing the exact value of ee.