Let's break down and solve **Question 1** in three steps, with explanations at each step and a final answer at the end: --- ### **Step 1: Write the Power Series and Truncate to Second Degree** The series expansion given is: \[ \left(1 + \frac{x}{5}\right)^{1/2} = 1 + \frac{x}{10} - \frac{x^2}{200} + \frac{x^3}{2000} - \frac{x^4}{16000} + \frac{7x^5}{800000} + \cdots \] **Truncate to second degree:** \[ f(x) \approx 1 + \frac{x}{10} - \frac{x^2}{200} \] **Explanation:** We take terms up to \(x^2\) because we're asked to compare truncation errors to the second degree. --- ### **Step 2: Convert Truncated Power Series to Chebyshev Series (up to \(T_2\))** Recall: - Chebyshev polynomials: \(T_0(x) = 1\), \(T_1(x) = x\), \(T_2(x) = 2x^2 - 1\) - Express \(x^2\) in terms of Chebyshev polynomials: \(x^2 = \frac{T_2(x)+T_0(x)}{2}\) Rewrite the truncated series in terms of Chebyshev polynomials: \[ f(x) \approx 1 + \frac{x}{10} - \frac{1}{200}x^2 \] \[ x^2 = \frac{T_2(x) + 1}{2} \Rightarrow -\frac{1}{200}x^2 = -\frac{1}{200} \cdot \frac{T_2(x) + 1}{2} = -\frac{1}{400} T_2(x) - \frac{1}{400} \] So, \[ f(x) \approx 1 - \frac{1}{400} + \frac{x}{10} - \frac{1}{400}T_2(x) \] Combine constants: \[ f(x) \approx \left(1 - \frac{1}{400}\right) + \frac{x}{10} - \frac{1}{400}T_2(x) \] \[ f(x) \approx \frac{399}{400} + \frac{x}{10} - \frac{1}{400}T_2(x) \] So, **Chebyshev series up to \(T_2\):** \[ f(x) \approx \frac{399}{400} T_0(x) + \frac{1}{10} T_1(x) - \frac{1}{400} T_2(x) \] **Explanation:** We rewrote the power series in the Chebyshev polynomial basis up to \(T_2\). --- ### **Step 3: Calculate Maximum Error for Truncated Series** #### **A. Power Series Error (truncated at \(x^2\)):** The next term is \(\frac{x^3}{2000}\). On \([-1,1]\), the maximum error is at \(x=1\): \[ \text{Error}_{\text{Power}} \approx \left| \frac{1^3}{2000} \right| = 0.0005 \] #### **B. Chebyshev Series Error (truncated at \(T_2\)):** The next term in the Chebyshev expansion would correspond to \(x^3\) or higher. The coefficient for \(x^3\) in the power expansion is \(\frac{1}{2000}\). Its maximum absolute contribution for \(x \in [-1,1]\) is at \(x=1\): \[ \text{Error}_{\text{Cheb}} \approx \left| \frac{1}{2000} \right| = 0.0005 \] **Explanation:** Both series, when truncated at second degree, have their leading neglected term as \(x^3/2000\), so their maximum errors on \([-1,1]\) are both \(0.0005\). --- ### **Final Answer** **Chebyshev series up to \(T_2\):** \[ \boxed{f(x) \approx \frac{399}{400} T_0(x) + \frac{1}{10} T_1(x) - \frac{1}{400} T_2(x)} \] **Maximum error of the truncated Chebyshev series (up to \(T_2\)) on \([-1,1]\):** \[ \boxed{0.0005} \] **Comparison:** The maximum error of the truncated Chebyshev series is the same as that of the truncated power series to second degree: **0.0005** on \([-1,1]\). --- **Summary Table:** | Series | Approximation (deg 2) | Max Error (\([-1,1]\)) | |---------------------|-------------------------------------------------------------|------------------------| | Power Series | \(1 + \frac{x}{10} - \frac{x^2}{200}\) | 0.0005 | | Chebyshev Series | \(\frac{399}{400}T_0 + \frac{1}{10}T_1 - \frac{1}{400}T_2\)| 0.0005 |