Absolutely! Here’s your step-by-step solution, with clear explanations at each step and the final answers summarized at the end. --- ### (a) Fourth order Maclaurin polynomial for \(\cos(x^2)\) **Step 1:** Write the Maclaurin series for \(\cos(x)\): \[ \cos(x) = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \cdots \] **Explanation:** The Maclaurin series expands \(\cos(x)\) in powers of \(x\). --- **Step 2:** Substitute \(x^2\) in place of \(x\) and keep terms up to order \(x^4\): \[ \cos(x^2) = 1 - \frac{(x^2)^2}{2!} + \frac{(x^2)^4}{4!} - \cdots = 1 - \frac{x^4}{2} + \cdots \] So, the 4th degree polynomial is: \[ P_4(x) = 1 - \frac{x^4}{2} \] **Explanation:** Only terms up to \(x^4\) are included for the 4th order polynomial. **Final answer (a):** \[ \boxed{P_4(x) = 1 - \frac{x^4}{2}} \] --- ### (b) Error expression for approximating \(\cos(x^2)\) with \(P_4(x)\) **Step 1:** Write the general Lagrange remainder for the Maclaurin series: \[ R_n(x) = \frac{f^{(n+1)}(\xi)}{(n+1)!}x^{n+1} \] **Explanation:** The Lagrange remainder gives the error for a Taylor/Maclaurin polynomial. --- **Step 2:** For \(\cos(x^2)\), after \(x^4\) the next nonzero term is \(x^8\): \[ R_4(x) = \frac{x^8}{4!} \cdot \text{some value involving a derivative at } \xi \] But explicitly, from the series, the next term is \(\frac{x^8}{4!}\): \[ |R(x)| \leq \frac{1}{24} x^8 \] **Explanation:** The error comes from the next term in the series, which is of order \(x^8\). **Final answer (b):** \[ \boxed{|R(x)| \leq \frac{1}{24} x^8} \] --- ### (c) MATLAB-based upper bound on the error **Step 1:** Find the maximum value of \(|\cos^{(4)}(x^2)|\) for \(x \in [0, 1/3]\): For \(x\) in \([0, 1/3]\), \(|\cos^{(4)}(x^2)|\) is maximized at the left endpoint and is at most 12 (since maximal value of cosine and sine is 1, and derivatives of cosine alternate between sine and cosine with appropriate coefficients; for small \(x\), the dominant term is 12). So, \(M = 12\). **Explanation:** The maximal value for the 4th derivative of cosine is 12. --- **Step 2:** Plug into the error formula: \[ |R(x)| \leq \frac{12}{24} x^8 = 0.5 x^8 \] At \(x = 1/3\): \[ |R(1/3)| \leq 0.5 \left(\frac{1}{3}\right)^8 = 0.5 \cdot \frac{1}{6561} \approx 7.62 \times 10^{-5} \] **Explanation:** The error bound uses the maximal value of the derivative and the next power of \(x\). **Final answer (c):** \[ \boxed{7.62 \times 10^{-5}} \] --- ### (d) Rational approximation by integrating \(P_4(x)\) **Step 1:** Integrate \(P_4(x)\) from 0 to \(1/3\): \[ \int_0^{1/3} \left(1 - \frac{x^4}{2}\right) dx = \left[ x - \frac{x^5}{10} \right]_0^{1/3} \] **Explanation:** Integrate term by term. --- **Step 2:** Evaluate at the endpoints: \[ = \frac{1}{3} - \frac{(1/3)^5}{10} = \frac{1}{3} - \frac{1}{2430} = \frac{810 - 1}{2430} = \frac{809}{2430} \] **Explanation:** Substitute \(x = 1/3\) and \(x = 0\) and simplify. **Final answer (d):** \[ \boxed{\frac{809}{2430}} \] --- ### (e) Upper bound on the integration error **Step 1:** Integrate the error bound over the interval: \[ \int_0^{1/3} 0.5 x^8 dx = 0.5 \int_0^{1/3} x^8 dx = 0.5 \left[\frac{x^9}{9}\right]_0^{1/3} \] **Explanation:** Integrate the error bound over the interval. --- **Step 2:** Substitute the limits: \[ 0.5 \cdot \frac{(1/3)^9}{9} = 0.5 \cdot \frac{1}{9 \cdot 19683} = \frac{1}{2} \cdot \frac{1}{177147} = \frac{1}{354294} \approx 2.82 \times 10^{-6} \] **Explanation:** Compute the value explicitly for a tight error bound. **Final answer (e):** \[ \boxed{\frac{1}{354294} \approx 2.82 \times 10^{-6}} \] --- ## **Final summary:** - **(a):** \(P_4(x) = 1 - \frac{x^4}{2}\) - **(b):** \(|R(x)| \leq \frac{1}{24} x^8\) - **(c):** Error upper bound ≈ \(7.62 \times 10^{-5}\) - **(d):** Approximate integral = \(\frac{809}{2430}\) - **(e):** Integration error bound = \(\frac{1}{354294} \approx 2.82 \times 10^{-6}\) Let me know if you’d like to see MATLAB code for part (c)!