Let's break down each part, step by step, with explanations: --- ### **(a) Fourth Order Maclaurin Polynomial \(P_4(x)\) for \(\cos(x^2)\)** #### **Step 1:** Write the Maclaurin series for \(\cos(x)\): \[ \cos(x) = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \cdots \] #### **Step 2:** Substitute \(x^2\) for \(x\) to get the Maclaurin series for \(\cos(x^2)\) up to \(x^4\): \[ \cos(x^2) = 1 - \frac{(x^2)^2}{2!} + \frac{(x^2)^4}{4!} - \cdots = 1 - \frac{x^4}{2} + \cdots \] **Explanation:** The fourth order polynomial \(P_4(x)\) includes all terms up to \(x^4\). Since the next term, \(\frac{(x^2)^6}{6!}\), is of order \(x^{12}\), it is not included in \(P_4(x)\). --- **Final Answer (a):** \[ \boxed{P_4(x) = 1 - \frac{x^4}{2}} \] --- ### **(b) Expression for the Error in Approximating \(\cos(x^2)\) by \(P_4(x)\) for \(0 \leq x \leq \frac{1}{3}\)** #### **Step 1:** Recall the general Maclaurin polynomial error (remainder) formula: For a function \(f(x)\) approximated by its \(n\)th order Maclaurin polynomial, the error is: \[ R_n(x) = \frac{f^{(n+1)}(\xi)}{(n+1)!} x^{n+1} \] for some \(\xi\) between 0 and \(x\). #### **Step 2:** For \(P_4(x)\), \(n = 4\). So, the error is: \[ R_4(x) = \frac{f^{(5)}(\xi)}{5!} x^5 \] where \(f(x) = \cos(x^2)\). **Explanation:** The error term uses the next derivative (the 5th derivative) evaluated at some point \(\xi\) between 0 and \(x\). --- **Final Answer (b):** \[ \boxed{ R_4(x) = \frac{d^5}{dx^5}\left[\cos(x^2)\right]\Bigg|_{x=\xi} \frac{x^5}{5!} } \] for some \(\xi \in [0, x]\). --- **Would you like to proceed to (c), (d), and (e) step-by-step as well, or do you want the answers for all parts at once?**

Certainly! Let's proceed step-by-step for parts (c), (d), and (e): --- ### **(c) Use MATLAB to find an upper bound on the error in part (b).** #### **Step 1:** Find the 5th derivative \(f^{(5)}(x) = \frac{d^5}{dx^5} \cos(x^2)\). **Explanation:** The error depends on the maximum of \(|f^{(5)}(\xi)|\) for \(\xi \in [0, x]\). To find an upper bound, we analyze the magnitude of the 5th derivative. --- #### **Step 2:** Derive or numerically compute the 5th derivative using MATLAB. **MATLAB code snippet:** ```matlab syms x f = cos(x^2); f5 = diff(f, x, 5); % 5th derivative % To find maximum absolute value on [0, 1/3] x_vals = linspace(0, 1/3, 1000); f5_vals = double(subs(f5, x, x_vals)); max_f5 = max(abs(f5_vals)); disp(['Maximum |f^{(5)}(x)| on [0, 1/3]: ', num2str(max_f5)]); ``` **Explanation:** Running this code will give the maximum absolute value of the 5th derivative over the interval, which serves as an upper bound. --- #### **Step 3:** Calculate the error bound: \[ |R_4(x)| \leq \frac{\max_{\xi \in [0, x]} |f^{(5)}(\xi)|}{5!} |x|^5 \] Given \(x \leq 1/3\), plug in the maximum derivative value and \(x=1/3\): \[ \boxed{ \text{Error bound} \leq \frac{\text{max\_f5}}{120} \times \left(\frac{1}{3}\right)^5 } \] --- ### **(d) Find a rational approximation to the integral by integrating \(P_4(z)\).** #### **Step 1:** Recall \(P_4(z) = 1 - \frac{z^4}{2}\). #### **Step 2:** Integrate \(P_4(z)\) from 0 to 1/3: \[ I \approx \int_0^{1/3} P_4(z) dz = \int_0^{1/3} \left(1 - \frac{z^4}{2}\right) dz \] **MATLAB code snippet:** ```matlab syms z P4 = 1 - z^4/2; approx_integral = int(P4, z, 0, 1/3); disp(['Approximate integral: ', char(vpa(approx_integral, 6))]); ``` **Analytically:** \[ I \approx \left[ z - \frac{z^5}{10} \right]_0^{1/3} = \frac{1}{3} - \frac{(1/3)^5}{10} \] --- ### **(e) Obtain an upper bound on the error in the integration (from part d).** #### **Step 1:** The error in approximating the integral is bounded by the integral of the remainder term: \[ |E| \leq \int_0^{1/3} |R_4(z)| dz \] From part (b), the error in the integrand is \(R_4(z)\), which is bounded by: \[ |R_4(z)| \leq \frac{\max_{\xi \in [0,z]} |f^{(5)}(\xi)|}{120} z^5 \] Assuming \(\max |f^{(5)}(\xi)|\) is constant (from part (c)), the worst-case error bound is: \[ |E| \leq \frac{\max |f^{(5)}(\xi)|}{120} \int_0^{1/3} z^5 dz = \frac{\max |f^{(5)}(\xi)|}{120} \times \frac{(1/3)^6}{6} \] **Simplified:** \[ |E| \leq \frac{\max |f^{(5)}(\xi)|}{720} \times \left(\frac{1}{3}\right)^6 \] **Answer:** Calculate numerically using the maximum derivative value from part (c). --- ## **Summary of final answers:** - **(a)** \[ \boxed{P_4(x) = 1 - \frac{x^4}{2}} \] - **(b)** \[ \boxed{ R_4(x) = \frac{f^{(5)}(\xi)}{120} x^5,\quad \text{for some } \xi \in [0,x] } \] - **(c)** Use MATLAB to compute \(\max_{\xi \in [0, 1/3]} |f^{(5)}(\xi)|\). Error bound: \[ |R_4(x)| \leq \frac{\text{max\_f5}}{120} \times \left(\frac{1}{3}\right)^5 \] - **(d)** Approximate integral: \[ I \approx \frac{1}{3} - \frac{(1/3)^5}{10} \] - **(e)** Error in the integral: \[ |E| \leq \frac{\text{max\_f5}}{720} \times \left(\frac{1}{3}\right)^6 \] --- Would you like me to perform the numerical calculations or provide MATLAB code outputs?