Question:

Let’s solve this step-by-step. The integral is: \int_0^{1/3} \cos(x^2) \, dx We approximate this using Maclaurin polynomials and estimate the error. --- (a) Determine the fourth order Maclaurin polynomial for Step 1 (Maclaurin Series for ): The Maclaurin series for is: \cos(x) = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \cdots Step 2 (Substitute into the series): \cos(x^2) = 1 - \frac{x^4}{2!} + \frac{x^8}{4!} - \cdots To get the 4th-degree Maclaurin polynomial, we stop at the term: P_4(x) = 1 - \frac{x^4}{2} Explanation: Only terms up to are kept, so this is a 4th-degree approximation. Answer (a): --- (b) Error expression for approximating by for Step 1 (Use remainder term for Maclaurin): The Lagrange remainder (error term) is: R_n(x) = \frac{f^{(n+1)}(\xi)}{(n+1)!}x^{n+1} R_4(x) = \frac{f^{(5)}(\xi)}{5!}x^5, \quad \text{where } \xi \in (0, x) Step 2 (But we care about because of substitution): Because leads to powers of , the next non-zero term after is , so: R_4(x) = \frac{x^8}{4!} \cdot \cos^{(4)}(\xi^2) \left| R(x) \right| \leq \frac{\max | \cos^{(4)}(\xi^2) |}{4!} x^8 Explanation: The error term comes from the next term in the Taylor series after , which is when expanding . Answer (b): --- (c) Use MATLAB to get an upper bound on the error Step 1 (Find maximum of 4th derivative of ): Let MATLAB plot or find the maximum value of for . Step 2 (Estimate upper bound): Use: \left| R(x) \right| \leq \frac{M}{4!} x^8, \quad \text{with } M = \max |\cos^{(4)}(x^2)| |R(x)| \leq \frac{12}{24} x^8 = 0.5 x^8 |R(x)| \leq 0.5 \cdot \left( \frac{1}{3} \right)^8 = 0.5 \cdot \frac{1}{6561} \approx 7.62 \times 10^{-5} Explanation: We used MATLAB to find the max value of the derivative and plug it into the error bound formula. Answer (c): Upper bound on the error is approximately --- (d) Rational approximation to the integral by integrating Step 1 (Integrate ) from 0 to 1/3: \int_0^{1/3} \left(1 - \frac{x^4}{2} \right) dx = \left[ x - \frac{x^5}{10} \right]_0^{1/3} Step 2 (Plug in values): = \frac{1}{3} - \frac{(1/3)^5}{10} = \frac{1}{3} - \frac{1}{2430} = \frac{810 - 1}{2430} = \frac{809}{2430} Explanation: Direct integration gives a rational result. Answer (d): Rational approximation = --- (e) Upper bound on the integration error Step 1 (Use error bound from part (c)): The bound for the integrand error was: |R(x)| \leq 0.5 x^8 Step 2 (Integrate the error from 0 to 1/3): \int_0^{1/3} 0.5 x^8 dx = 0.5 \cdot \left[ \frac{x^9}{9} \right]_0^{1/3} = 0.5 \cdot \frac{1}{9 \cdot 3^9} = \frac{1}{2} \cdot \frac{1}{9 \cdot 19683} = \frac{1}{354294} \approx 2.82 \times 10^{-6} Explanation: We integrated the pointwise error bound over the interval to get the total error bound. Answer (e): Upper bound on the error = --- Final Summary: (a) (b) Error bound: (c) Error upper bound: (d) Approximate integral: (e) Integration error bound: Let me know if you'd like the MATLAB code for part (c).