Approximation Error & Bounds - Calculus 2

1. Consider approximating $e$ using the Maclaurin series for $e^x$ evaluated at $x=1$, which is $\sum_{n=0}^\infty \frac{1}{n!}$. We use the partial sum $S_N = \sum_{n=0}^N \frac{1}{n!}$. The Lagrange error bound for this approximation states that $|R_N| \le \frac{M}{(N+1)!}$, where $M$ is the maximum of $e^c$ on $[0,1]$. To use this bound, we need a value for $M$, but $e$ is the number we are approximating. Which is the most effective and valid approach to establish a numerical error bound?
   Answer: Use the fact that $e < 3$ to establish a valid, though not the tightest, upper bound for the error. The problem is that to find the error in approximating $e$, we need to know the value of $e$ to find $M = e^1$. This is circular. However, we do not need the exact maximum $M$, only an upper bound. Since $e^x$ is an increasing function, we know $e < 3$ (e.g., from $S_3 = 1+1+1/2+1/6 = 2.66...$ and knowing the remainder is positive). We can use $M=3$ as a valid upper bound for $e^c$ on $[0,1]$. This gives a concrete, provable error bound $|R_N| \le \frac{3}{(N+1)!}$, allowing us to determine the accuracy without knowing the exact value of $e$.
2. The function $f(x)$ is approximated by its $n$-th degree Taylor polynomial $P_n(x)$ centered at $a$. The Lagrange error bound for $|f(x) - P_n(x)|$ is given by $\frac{M}{(n+1)!}|x-a|^{n+1}$. How is the value of $M$ determined?
   Answer: $M$ is an upper bound for $|f^{(n+1)}(c)|$ for any $c$ in the interval between $a$ and $x$. The value $M$ in the Lagrange error bound formula represents an upper bound on the magnitude of the $(n+1)$-th derivative of $f$. Specifically, $M$ must satisfy $|f^{(n+1)}(c)| \le M$ for all $c$ in the interval containing the center $a$ and the point of approximation $x$. To get the tightest bound, one typically finds the maximum value of $|f^{(n+1)}(c)|$ on this interval.
3. A student uses $n=10$ subintervals to approximate $\int_1^2 \ln(x) dx$. They find the error bound for the Trapezoidal Rule is $B_T$ and the error bound for Simpson's Rule is $B_S$. Which statement correctly compares the bounds?
   Answer: $B_T$ is larger than $B_S$ primarily because the error formula for $B_T$ has a denominator proportional to $n^2$ while $B_S$ has a denominator proportional to $n^4$. The error bound for Trapezoidal Rule is $\propto 1/n^2$ and for Simpson's Rule is $\propto 1/n^4$. For $n=10$, the $n^4$ term (10,000) is much larger than the $n^2$ 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 $\ln(x)$ on $[1,2]$, 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.
4. For the approximation $\sin(x) \approx x - x^3/6$ on the interval $[-0.5, 0.5]$, the Lagrange error bound involves finding the maximum of a derivative of $\sin(x)$. Which derivative should be used and what is its maximum absolute value on the interval?
   Answer: $f^{(5)}(x) = \cos(x)$, with maximum absolute value 1 The polynomial $P_3(x) = x - x^3/6$ is the third-degree Maclaurin polynomial for $\sin(x)$. Because the $x^4$ term is zero, $P_3(x) = P_4(x)$. We can use the error formula for $n=4$, which is $|R_4(x)| \le \frac{M}{5!}|x|^5$, where $M$ is the maximum of $|f^{(5)}(c)|$ on the interval. The fifth derivative is $f^{(5)}(x) = \cos(x)$. On $[-0.5, 0.5]$, $\cos(x)$ is always positive and its maximum occurs at $x=0$, where $\cos(0)=1$. So, the derivative to use is the fifth, and its maximum absolute value is 1.
5. The error bound for approximating $\int_a^b f(x) dx$ with the Trapezoidal Rule is $|E_T| \le \frac{K(b-a)^3}{12n^2}$, where $K$ is an upper bound for $|f''(x)|$. If $f(x)$ is a linear function, what is the value of this error bound?
   Answer: Zero If $f(x)$ is a linear function, say $f(x) = mx+c$, its first derivative is $f'(x) = m$ and its second derivative is $f''(x) = 0$. Therefore, the maximum value of $|f''(x)|$ on any interval is $K=0$. Substituting $K=0$ into the error bound formula gives $|E_T| \le \frac{0 \cdot (b-a)^3}{12n^2} = 0$. This means the Trapezoidal Rule is exact for linear functions.
6. To approximate $\ln(1.5)$ using a Taylor polynomial for $f(x) = \ln(x)$ centered at $a=1$, what is the minimum degree $n$ required to guarantee an absolute error of no more than $\frac{1}{3000}$?
   Answer: 8 The Lagrange error bound is $|R_n(1.5)| \le \frac{M}{(n+1)!} |1.5-1|^{n+1}$, where $M$ is an upper bound for $|f^{(n+1)}(c)|$ on $[1, 1.5]$. The $(n+1)$-th derivative of $\ln(x)$ is $f^{(n+1)}(x) = (-1)^n n! x^{-(n+1)}$. On $[1, 1.5]$, $|f^{(n+1)}(c)| = \frac{n!}{c^{n+1}}$ is maximized at $c=1$, so $M=n!$. The inequality becomes $\frac{n!}{(n+1)!} (0.5)^{n+1} \le \frac{1}{3000}$, which simplifies to $\frac{1}{n+1} \frac{1}{2^{n+1}} \le \frac{1}{3000}$, or $(n+1)2^{n+1} \ge 3000$. Testing values: for $n=7$, $8 \cdot 2^8 = 2048$. For $n=8$, $9 \cdot 2^9 = 4608$. Thus, the minimum degree is $n=8$.
7. The sum of the series $\sum_{n=1}^{\infty} \frac{1}{n^3}$ is approximated by its 10th partial sum, $S_{10}$. Let $R_{10}$ be the remainder $S - S_{10}$. According to the Integral Test remainder estimate, which of the following is a correct statement about $R_{10}$?
   Answer: $R_{10} < \frac{1}{200}$ The Integral Test remainder estimate states that $\int_{n+1}^{\infty} f(x) \, dx < R_n < \int_{n}^{\infty} f(x) \, dx$. For this problem, $f(x) = 1/x^3$ and $n=10$. We can find an upper bound for $R_{10}$ by evaluating $\int_{10}^{\infty} \frac{1}{x^3} \, dx$. The integral is $\left[ -\frac{1}{2x^2} \right]_{10}^{\infty} = 0 - \left(-\frac{1}{2(10)^2}\right) = \frac{1}{200}$. Thus, we are guaranteed that $R_{10} < \frac{1}{200}$.
8. The integral $\int_0^4 (x^3 - 6x^2 + 1) dx$ is approximated with Simpson's Rule using $n=100$ subintervals. What is the value of the approximation error $E_S$?
   Answer: Exactly 0 The error bound for Simpson's Rule depends on the fourth derivative of the integrand. Let $f(x) = x^3 - 6x^2 + 1$. Then $f'(x) = 3x^2 - 12x$, $f''(x) = 6x - 12$, $f'''(x) = 6$, and $f^{(4)}(x) = 0$. Since the fourth derivative is identically zero, the error bound $|E_S| \le \frac{M(b-a)^5}{180n^4}$ becomes $|E_S| \le \frac{0 \cdot (4-0)^5}{180(100)^4} = 0$. Therefore, the error is exactly 0. Simpson's rule is exact for all polynomials of degree 3 or less.
9. Simpson's rule is applied to approximate $\int_1^3 \frac{1}{x} dx$ with $n = 6$ subintervals. If the fourth derivative satisfies $|f^{(4)}(x)| \leq \frac{24}{x^5}$ for $x \in [1,3]$, what is the error bound for this approximation?
   Answer: $\frac{(3-1)^5 \cdot 24}{180 \cdot 6^4} = \frac{32 \cdot 24}{180 \cdot 1296} = \frac{768}{233280}$ When you encounter Simpson's rule error bounds, you're working with numerical integration accuracy. The error bound formula for Simpson's rule is: $E_S \leq \frac{(b-a)^5}{180n^4} \max_{x \in [a,b]} |f^{(4)}(x)|$ Let's apply this formula step by step. You have $f(x) = \frac{1}{x}$, integration bounds $[1,3]$, so $b-a = 3-1 = 2$, and $n = 6$ subintervals. The maximum value of $|f^{(4)}(x)| = \frac{24}{x^5}$ occurs at $x = 1$ (since the function decreases as $x$ increases), giving us $\max |f^{(4)}(x)| = \frac{24}{1^5} = 24$. Substituting into the formula: $E_S \leq \frac{(2)^5 \cdot 24}{180 \cdot 6^4} = \frac{32 \cdot 24}{180 \cdot 1296} = \frac{768}{233280}$ Choice A incorrectly uses $(b-a)^4$ instead of $(b-a)^5$ 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 $(b-a)^5$ and 180 but mistakenly puts $n^5$ instead of $n^4$ in the denominator. The correct answer is C. Study tip: Memorize that Simpson's rule error has the pattern $\frac{(b-a)^5}{180n^4}$ – the "5-4-180" combination is unique to Simpson's rule and frequently tested.
10. The function $f(x) = e^{-3x}$ is approximated by its second-degree Taylor polynomial, $P_2(x)$, centered at $a=0$. Using Taylor's inequality, what is the tightest upper bound for the error $|f(x) - P_2(x)|$ on the interval $[0, 0.2]$?
    Answer: 0.036 Taylor's inequality states that the error $|R_2(x)|$ is bounded by $\frac{M}{(2+1)!}|x-0|^3$, where $M$ is an upper bound for $|f^{(3)}(c)|$ on the interval $[0, 0.2]$. The derivatives are $f'(x) = -3e^{-3x}$, $f''(x) = 9e^{-3x}$, and $f'''(x) = -27e^{-3x}$. The function $|f'''(x)| = 27e^{-3x}$ is a decreasing function, so its maximum on $[0, 0.2]$ occurs at $x=0$, giving $M=27e^0=27$. The maximum value of $|x|^3$ on $[0, 0.2]$ is $(0.2)^3 = 0.008$. Therefore, the error bound is $|R_2(x)| \le \frac{27}{3!} (0.2)^3 = \frac{27}{6}(0.008) = 4.5(0.008) = 0.036$.
11. The trapezoidal rule with $n = 4$ subintervals is used to approximate $\int_0^2 e^{x^2} dx$. Given that $|f''(x)| \leq 400$ on $[0,2]$, which statement about the approximation error is correct?
    Answer: The error is bounded by $\frac{(2-0)^3 \cdot 400}{12 \cdot 4^2} = \frac{8 \cdot 400}{192} = \frac{50}{3}$ The error bound for the trapezoidal rule is $|E_T| \leq \frac{(b-a)^3 M}{12n^2}$ where M is the maximum of $|f''(x)|$ on the interval. Here, $b-a = 2$, $M = 400$, and $n = 4$, so the error bound is $\frac{2^3 \cdot 400}{12 \cdot 4^2} = \frac{8 \cdot 400}{192} = \frac{50}{3}$. Choice B incorrectly uses $(b-a)^2$ and $n$ instead of $n^2$. Choice C uses $n$ instead of $n^2$. Choice D incorrectly uses $(b-a)^4$ and $n^3$.
12. The integral $\int_0^{\pi} \sin(x) dx$ is approximated using the Trapezoidal Rule with $n=2$ subintervals. What is the absolute value of the exact error of this approximation?
    Answer: $2 - \pi/2$ This question requires calculating the exact error, not just the bound. First, find the exact value of the integral: $\int_0^{\pi} \sin(x) dx = [-\cos(x)]_0^{\pi} = -\cos(\pi) - (-\cos(0)) = -(-1) - (-1) = 2$. Next, apply the Trapezoidal Rule with $n=2$. The step size is $\Delta x = (\pi-0)/2 = \pi/2$. The grid points are $0, \pi/2, \pi$. The approximation is $T_2 = \frac{\Delta x}{2}[f(0) + 2f(\pi/2) + f(\pi)] = \frac{\pi/2}{2}[\sin(0) + 2\sin(\pi/2) + \sin(\pi)] = \frac{\pi}{4}[0 + 2(1) + 0] = \frac{\pi}{2}$. The exact error is $|\text{Exact} - \text{Approx}| = |2 - \frac{\pi}{2}|$. Since $\pi \approx 3.14$, $\pi/2 \approx 1.57$, so $2 - \pi/2 > 0$. The error is $2 - \pi/2$.
13. The sum of a convergent alternating series $\sum_{n=1}^{\infty} (-1)^{n+1} a_n$ (with $a_n > 0$) is approximated by the partial sum $S_N$. The error bound is given by $|R_N| \le a_{N+1}$. Which of the following conditions is necessary for this error bound to be valid?
    Answer: The sequence $\{a_n\}$ must be non-increasing for $n \ge 1$. The Alternating Series Estimation Theorem requires two conditions on the sequence of magnitudes $\{a_n\}$: 1) $\lim_{n\to\infty} a_n = 0$ and 2) the sequence $\{a_n\}$ is non-increasing, i.e., $a_{n+1} \le a_n$ for all $n$ (or for all $n$ 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.
14. An integral is approximated using Simpson's Rule with $n$ subintervals, resulting in a theoretical error bound $B$. If the number of subintervals is tripled (to $3n$), what will the new theoretical error bound be in terms of $B$?
    Answer: $B/81$ The error bound for Simpson's Rule is given by $|E_S| \le \frac{M(b-a)^5}{180n^4}$. Let this bound be $B$. If the number of subintervals is changed to $n' = 3n$, the new error bound $B'$ will be $B' = \frac{M(b-a)^5}{180(3n)^4} = \frac{M(b-a)^5}{180 \cdot 3^4 n^4} = \frac{1}{81} \left( \frac{M(b-a)^5}{180n^4} \right) = \frac{B}{81}$.
15. The Taylor series for $\sin(x)$ centered at $x = 0$ is used to approximate $\sin(0.3)$. If the alternating series estimation theorem is applied and we want the approximation error to be less than $10^{-4}$, what is the minimum number of non-zero terms needed?
    Answer: 2 terms (up to the $x^3$ term) The Taylor series for sin(x) is $x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots$. For x = 0.3, by the alternating series estimation theorem, the error after n terms is bounded by the absolute value of the (n+1)th term. After 2 terms, the error is bounded by the third term: $\frac{(0.3)^5}{5!} = \frac{0.00243}{120} ≈ 2.03 × 10^{-5} < 10^{-4}$. So 2 terms suffice. Choice A uses more terms than necessary. Choice B uses 4 terms, which is excessive. Choice D is also excessive.
16. The power series $\sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{(2n+1)!}$ converges to $\sin(x)$ for all real $x$. When this series is truncated after the first 4 terms to approximate $\sin(1.2)$, what type of error bound applies, and what is the bound?
    Answer: Alternating series test applies; error $\leq \frac{(1.2)^9}{9!} = \frac{1.2^9}{362880} ≈ 1.38 × 10^{-5}$ 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 $\frac{(-1)^n x^{2n+1}}{(2n+1)!}$, 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 $n = 0, 1, 2, 3$, so the first omitted term has $n = 4$. When $n = 4$: $\frac{(-1)^4 x^{9}}{9!} = \frac{x^9}{9!}$ At $x = 1.2$, the error bound is $\frac{(1.2)^9}{9!} = \frac{1.2^9}{362880} ≈ 1.38 × 10^{-5}$ 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 $x^8$ instead of $x^9$, missing that the first omitted term has $n = 4$, giving $2n+1 = 9$. Remember: for alternating series approximations, always use the first omitted term for your error bound, and count carefully to identify which term that is.
17. The function $g(x) = \cos(x^2)$ is approximated by its fourth-degree Taylor polynomial $P_4(x)$ centered at $x = 0$. For $x \in [0, 0.5]$, if $|g^{(5)}(x)| \leq M$ and $|g^{(6)}(x)| \leq N$, which bound gives the tightest error estimate for $g(0.3)$?
    Answer: $\frac{M \cdot (0.3)^5}{5!}$ using the 5th derivative bound with Lagrange remainder When you encounter Taylor polynomial error estimation problems, you need to identify which remainder formula applies and match it with the correct derivative bound. For a fourth-degree Taylor polynomial $P_4(x)$, the error is $|g(x) - P_4(x)|$. The Lagrange remainder formula gives us $R_n(x) = \frac{g^{(n+1)}(c)}{(n+1)!}x^{n+1}$ for some $c$ between 0 and $x$. Since we're using $P_4(x)$, we have $n = 4$, so the remainder involves the 5th derivative: $R_4(x) = \frac{g^{(5)}(c)}{5!}x^5$. The error bound becomes $|R_4(0.3)| \leq \frac{M \cdot (0.3)^5}{5!}$ since $|g^{(5)}(x)| \leq M$. This matches option B exactly. Option A incorrectly uses the 6th derivative with "integral remainder" - there's no standard integral remainder formula that would use the 6th derivative for a 4th-degree polynomial error. Option C uses the 6th derivative with Lagrange remainder, but Lagrange remainder for $P_4(x)$ specifically requires the 5th derivative, not the 6th. Option D suggests choosing the smaller bound, but this misunderstands that only the 5th derivative bound is mathematically valid for this situation. Study tip: For Taylor polynomial error estimation, always remember that an $n$th degree polynomial's Lagrange remainder uses the $(n+1)$th derivative. Don't be tempted by higher derivative bounds - stick to the formula that matches your polynomial degree.
18. The value of $\cos(0.4)$ is approximated using the Maclaurin series for $\cos(x)$. What is the smallest degree $N$ of the Maclaurin polynomial $P_N(x)$ that guarantees an error of less than $10^{-4}$ when used to estimate $\cos(0.4)$?
    Answer: 4 The Maclaurin series for $\cos(x)$ is $1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \dots$. At $x=0.4$, this is a convergent alternating series. We can use the Alternating Series Estimation Theorem, where the error is bounded by the magnitude of the first neglected term. If we use $P_2(x) = 1 - x^2/2$, the error is bounded by $|a_3| = \frac{(0.4)^4}{4!} = \frac{0.0256}{24} \approx 0.00107$, which is not less than $10^{-4}$. If we use $P_4(x) = 1 - x^2/2 + x^4/24$, the error is bounded by $|a_4| = \frac{(0.4)^6}{6!} = \frac{0.004096}{720} \approx 5.7 \times 10^{-6}$, which is less than $10^{-4}$. Thus, the smallest degree required is $N=4$.
19. The sum of the convergent alternating series $\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n^2 + 1}$ is $S$. This sum is approximated by its fourth partial sum, $S_4$. Which of the following statements is true about the error, $R_4 = S - S_4$?
    Answer: The approximation $S_4$ is an underestimate of $S$, and $|R_4| < \frac{1}{26}$. The error is $R_4 = S - S_4 = a_5 - a_6 + a_7 - \dots$. The first term of the series for the remainder is $a_5 = \frac{(-1)^{5+1}}{5^2 + 1} = \frac{1}{26}$, which is positive. For a convergent alternating series, the sign of the remainder is the same as the sign of its first term. Since $a_5$ is positive, $R_4 > 0$, which means $S - S_4 > 0$, so $S > S_4$. Thus, $S_4$ is an underestimate. The Alternating Series Estimation Theorem states that $|R_4| < |a_5|$, so $|R_4| < \frac{1}{26}$.
20. The function $f(x) = \sqrt{1+x}$ is approximated by its first-degree Taylor polynomial $P_1(x)$ centered at $a=0$. What is the Lagrange error bound for this approximation on the interval $[0, 0.2]$?
    Answer: 0.005 The Lagrange error bound for $P_1(x)$ is $|R_1(x)| \le \frac{M}{2!}|x|^2$, where $M \ge |f''(c)|$ for $c$ in $[0, 0.2]$. We have $f(x) = (1+x)^{1/2}$, $f'(x) = \frac{1}{2}(1+x)^{-1/2}$, and $f''(x) = -\frac{1}{4}(1+x)^{-3/2}$. The magnitude is $|f''(x)| = \frac{1}{4(1+x)^{3/2}}$. This is a decreasing function on $[0, 0.2]$, so its maximum occurs at $x=0$, which is $M = \frac{1}{4(1)^ {3/2}} = \frac{1}{4}$. The maximum value of $|x|^2$ on the interval is $(0.2)^2 = 0.04$. The error bound is $\frac{1/4}{2} (0.04) = \frac{1}{8}(0.04) = 0.005$.

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