Let's solve the problem step by step. The question asks for the **Taylor series expansion around \( z_0 = 1 \)** for the following functions: a) \( \ln(z) \) b) \( z^6 \) c) \( e^{-jz^2} \) The Taylor series of a function \( f(z) \) centered at \( z_0 \) is: \[ f(z) = \sum_{n=0}^{\infty} \frac{f^{(n)}(z_0)}{n!} (z-z_0)^n \] Let’s do each case: --- ### a) \( \ln(z) \) around \( z_0 = 1 \) #### Step 1: Find derivatives - \( f(z) = \ln(z) \) - \( f'(z) = \frac{1}{z} \) - \( f''(z) = -\frac{1}{z^2} \) - \( f^{(3)}(z) = \frac{2}{z^3} \) - \( f^{(4)}(z) = -\frac{6}{z^4} \) - In general, \( f^{(n)}(z) = (-1)^{n+1} (n-1)! / z^n \) #### Step 2: Evaluate at \( z_0 = 1 \) - \( f(1) = 0 \) - \( f'(1) = 1 \) - \( f''(1) = -1 \) - \( f^{(3)}(1) = 2 \) - \( f^{(4)}(1) = -6 \) #### Step 3: Write the series \[ \ln(z) = 0 + 1(z-1) + \frac{-1}{2!}(z-1)^2 + \frac{2}{3!}(z-1)^3 + \frac{-6}{4!}(z-1)^4 + \cdots \] \[ \ln(z) = (z-1) - \frac{1}{2}(z-1)^2 + \frac{1}{3}(z-1)^3 - \frac{1}{4}(z-1)^4 + \cdots \] --- ### b) \( z^6 \) around \( z_0 = 1 \) #### Step 1: Expand binomially \[ z^6 = (1 + (z-1))^6 \] The binomial expansion: \[ z^6 = \sum_{n=0}^{6} \binom{6}{n} (z-1)^n \] #### Step 2: Write out the first terms \[ z^6 = 1 + 6(z-1) + 15(z-1)^2 + 20(z-1)^3 + 15(z-1)^4 + 6(z-1)^5 + (z-1)^6 \] --- ### c) \( e^{-jz^2} \) around \( z_0 = 1 \) #### Step 1: Substitute \( z = 1 + (z-1) \), expand \( z^2 \): \[ z^2 = (1 + (z-1))^2 = 1 + 2(z-1) + (z-1)^2 \] So, \[ e^{-jz^2} = e^{-j[1 + 2(z-1) + (z-1)^2]} \] \[ = e^{-j} \cdot e^{-j2(z-1)} \cdot e^{-j(z-1)^2} \] #### Step 2: Expand each exponential The Taylor expansion of \( e^{a x} \) is \( \sum_{n=0}^{\infty} \frac{(a x)^n}{n!} \). - For \( e^{-j2(z-1)} \): \( \sum_{n=0}^{\infty} \frac{(-j2(z-1))^n}{n!} \) - For \( e^{-j(z-1)^2} \): \( \sum_{m=0}^{\infty} \frac{[-j(z-1)^2]^m}{m!} \) #### Step 3: Series up to second order (optional): Expand up to \( (z-1)^2 \): - \( e^{-j2(z-1)} \approx 1 - j2(z-1) + \frac{(-j2(z-1))^2}{2!} = 1 - j2(z-1) - 2(z-1)^2 \) - \( e^{-j(z-1)^2} \approx 1 - j(z-1)^2 \) Multiply: \[ e^{-jz^2} \approx e^{-j}[1 - j2(z-1) - 2(z-1)^2][1 - j(z-1)^2] \] Multiply out and keep terms up to \( (z-1)^2 \): \[ = e^{-j}\left[1 - j2(z-1) - 2(z-1)^2 - j(z-1)^2 \right] \] \[ = e^{-j}\left[1 - j2(z-1) - (2 + j)(z-1)^2 \right] \] --- ## **Final Answers (Taylor Series around \( z_0 = 1 \)):** **a)** \[ \ln(z) = (z-1) - \frac{1}{2}(z-1)^2 + \frac{1}{3}(z-1)^3 - \frac{1}{4}(z-1)^4 + \cdots \] **b)** \[ z^6 = 1 + 6(z-1) + 15(z-1)^2 + 20(z-1)^3 + 15(z-1)^4 + 6(z-1)^5 + (z-1)^6 \] **c)** \[ e^{-jz^2} = e^{-j}\left[1 - j2(z-1) - (2 + j)(z-1)^2 + \cdots \right] \] Let me know if you need more terms or further clarification!

Certainly! Here are three key theoretical points about Taylor series expansions: 1. **Definition of Taylor Series**: The Taylor series of a function \(f(z)\) centered at a point \(z_0\) is an infinite sum of terms calculated from the derivatives of \(f(z)\) at \(z_0\). It provides a polynomial approximation of the function near \(z_0\): \[ f(z) = \sum_{n=0}^{\infty} \frac{f^{(n)}(z_0)}{n!} (z - z_0)^n \] This expansion is valid within the radius of convergence determined by the function's behavior. 2. **Purpose and Usage**: Taylor series are used to approximate complex functions with polynomials that are easier to analyze and compute, especially near the point of expansion. They are fundamental in numerical analysis, differential equations, and engineering for simplifying functions and analyzing their local behavior. 3. **Convergence and Error**: The Taylor series converges to the actual function within its radius of convergence. The difference between the function and its Taylor polynomial approximation is called the remainder or error term, which diminishes as more terms are included, provided the series converges at the point of interest. Let me know if you'd like more detailed explanations or examples!

Certainly! Let's provide detailed step-by-step calculations for each function's Taylor series expansion around \( z_0 = 1 \), along with the final expressions. --- ### **Part a) Expand \( \ln(z) \) around \( z_0 = 1 \)** **Step 1: Derivatives of \( \ln(z) \)** \[ f(z) = \ln(z) \] \[ f'(z) = \frac{1}{z} \] \[ f''(z) = -\frac{1}{z^2} \] \[ f^{(3)}(z) = \frac{2!}{z^3} = \frac{2}{z^3} \] \[ f^{(4)}(z) = -\frac{3!}{z^4} = -\frac{6}{z^4} \] **General formula for the \( n^{th} \) derivative:** \[ f^{(n)}(z) = (-1)^{n-1} (n-1)! \frac{1}{z^n} \] **Step 2: Evaluate derivatives at \( z_0 = 1 \)** \[ f(1) = \ln(1) = 0 \] \[ f'(1) = 1 \] \[ f''(1) = -1 \] \[ f^{(3)}(1) = 2 \] \[ f^{(4)}(1) = -6 \] **Step 3: Taylor series formula:** \[ \ln(z) = \sum_{n=0}^{\infty} \frac{f^{(n)}(1)}{n!} (z-1)^n \] **Calculating terms:** - \( n=0 \): \[ f(1) = 0 \] - \( n=1 \): \[ \frac{f'(1)}{1!} (z-1) = 1 \cdot (z-1) = (z-1) \] - \( n=2 \): \[ \frac{f''(1)}{2!} (z-1)^2 = \frac{-1}{2} (z-1)^2 \] - \( n=3 \): \[ \frac{f^{(3)}(1)}{3!} (z-1)^3 = \frac{2}{6} (z-1)^3 = \frac{1}{3}(z-1)^3 \] - \( n=4 \): \[ \frac{f^{(4)}(1)}{4!} (z-1)^4 = \frac{-6}{24} (z-1)^4 = -\frac{1}{4}(z-1)^4 \] **Final Series:** \[ \boxed{ \ln(z) \approx (z-1) - \frac{1}{2}(z-1)^2 + \frac{1}{3}(z-1)^3 - \frac{1}{4}(z-1)^4 + \cdots } \] --- ### **Part b) Expand \( z^6 \) around \( z_0 = 1 \)** **Step 1: Use binomial expansion** \[ z^6 = (1 + (z-1))^6 \] \[ z^6 = \sum_{n=0}^{6} \binom{6}{n} (z-1)^n \] **Step 2: Compute terms** - \( n=0 \): \[ \binom{6}{0} (z-1)^0 = 1 \] - \( n=1 \): \[ \binom{6}{1} (z-1) = 6(z-1) \] - \( n=2 \): \[ \binom{6}{2} (z-1)^2 = 15 (z-1)^2 \] - \( n=3 \): \[ \binom{6}{3} (z-1)^3 = 20 (z-1)^3 \] - \( n=4 \): \[ \binom{6}{4} (z-1)^4 = 15 (z-1)^4 \] - \( n=5 \): \[ \binom{6}{5} (z-1)^5 = 6 (z-1)^5 \] - \( n=6 \): \[ \binom{6}{6} (z-1)^6 = (z-1)^6 \] **Final series:** \[ \boxed{ z^6 \approx 1 + 6(z-1) + 15 (z-1)^2 + 20 (z-1)^3 + 15 (z-1)^4 + 6 (z-1)^5 + (z-1)^6 } \] --- ### **Part c) Expand \( e^{-jz^2} \) around \( z_0=1 \)** **Step 1: Rewrite \( z^2 \) near \( z=1 \):** \[ z = 1 + (z-1) \] \[ z^2 = (1 + (z-1))^2 = 1 + 2(z-1) + (z-1)^2 \] Thus, \[ e^{-jz^2} = e^{-j [1 + 2(z-1) + (z-1)^2]} = e^{-j} \cdot e^{-j 2(z-1)} \cdot e^{-j (z-1)^2} \] --- **Step 2: Expand each exponential to second order** - \( e^{-j 2(z-1)} \): \[ = 1 - j 2 (z-1) + \frac{(-j 2 (z-1))^2}{2} = 1 - j 2 (z-1) - 2 (z-1)^2 \] (since \((-j)^2 = -1\) and \( (-j 2)^2 = -4 \), divided by 2 gives \(-2\).) - \( e^{-j (z-1)^2} \): \[ = 1 - j (z-1)^2 \] **Step 3: Multiply the expansions** \[ e^{-jz^2} \approx e^{-j} \times [1 - j 2 (z-1) - 2 (z-1)^2] \times [1 - j (z-1)^2] \] First, multiply the last two: \[ (1 - j 2 (z-1) - 2 (z-1)^2) (1 - j (z-1)^2) \] Distribute: \[ = 1 \times (1 - j (z-1)^2) - j 2 (z-1) \times (1 - j (z-1)^2) - 2 (z-1)^2 \times (1 - j (z-1)^2) \] Calculate each term: 1. \( 1 - j (z-1)^2 \) 2. \( - j 2 (z-1) + j^2 2 (z-1)^3 = - j 2 (z-1) - 2 (z-1)^3 \) (since \( j^2 = -1 \)) 3. \( - 2 (z-1)^2 + 2 j (z-1)^4 \) Now, sum all: \[ 1 - j (z-1)^2 - j 2 (z-1) - 2 (z-1)^3 - 2 (z-1)^2 + 2 j (z-1)^4 \] Combine like terms: \[ = 1 - j 2 (z-1) - j (z-1)^2 - 2 (z-1)^2 - 2 (z-1)^3 + 2 j (z-1)^4 \] Group similar terms: \[ = 1 - 2 j (z-1) - (j + 2) (z-1)^2 - 2 (z-1)^3 + 2 j (z-1)^4 \] Finally, multiply by \( e^{-j} \): \[ e^{-jz^2} \approx e^{-j} \left[ 1 - 2 j (z-1) - (j + 2) (z-1)^2 + \text{higher order terms} \right] \] --- ## **Final approximate series:** \[ \boxed{ e^{-jz^2} \approx e^{-j} \left[ 1 - 2 j (z-1) - (j + 2)(z-1)^2 + \cdots \right] } \] --- **Summary of Final Results:** | Function | Taylor Series around \( z=1 \) (up to 4th order) | |---|---| | \( \ln(z) \) | \(\displaystyle (z-1) - \frac{1}{2}(z-1)^2 + \frac{1}{3}(z-1)^3 - \frac{1}{4}(z-1)^4 + \cdots \) | | \( z^6 \) | \(\displaystyle 1 + 6(z-1) + 15 (z-1)^2 + 20 (z-1)^3 + 15 (z-1)^4 + 6 (z-1)^5 + (z-1)^6 \) | | \( e^{-jz^2} \) | \(\displaystyle e^{-j} \left[ 1 - 2 j (z-1) - (j + 2)(z-1)^2 + \cdots \right] \) | Let me know if you'd like additional terms or further clarifications!