Let's solve **Exercise 3** step by step for **part (a)** as an example, including **whether the function is even, odd, or neither**, and then find its **Fourier series**. --- ## **(a)** \[ f(x) = \begin{cases} -x & \text{if } -1 < x < 0 \\ x & \text{if } 0 < x < 1 \end{cases} \] ### **Step 1: Check if the function is even, odd, or neither** A function is **even** if \( f(-x) = f(x) \), and **odd** if \( f(-x) = -f(x) \). Test for oddness: - For \( 0 < x < 1 \): \( f(-x) = -x \), \( f(x) = x \implies f(-x) = -f(x) \) - For \( -1 < x < 0 \): \( f(-x) = x \), \( f(x) = -x \implies f(-x) = -f(x) \) **So, \( f(x) \) is an odd function.** --- ### **Step 2: Fourier Series for Odd Function** For an odd function with period \( 2L \) (here \( L = 1 \)), the Fourier series contains only sine terms: \[ f(x) = \sum_{n=1}^\infty b_n \sin(n\pi x) \] Where: \[ b_n = \frac{2}{L}\int_0^L f(x)\sin\left(\frac{n\pi x}{L}\right) dx \] Here, \( L = 1 \): \[ b_n = 2\int_0^1 x \sin(n\pi x) dx \] #### **Compute \( b_n \):** Let’s compute the integral by parts: Let \( u = x \), \( dv = \sin(n\pi x) dx \) - \( du = dx \) - \( v = -\frac{1}{n\pi} \cos(n\pi x) \) So, \[ \int x \sin(n\pi x) dx = -\frac{x}{n\pi} \cos(n\pi x) + \frac{1}{n\pi}\int \cos(n\pi x) dx \] \[ = -\frac{x}{n\pi} \cos(n\pi x) + \frac{1}{n^2\pi^2} \sin(n\pi x) \] Evaluate from 0 to 1: \[ \int_0^1 x \sin(n\pi x) dx = \left[ -\frac{x}{n\pi} \cos(n\pi x) + \frac{1}{n^2\pi^2} \sin(n\pi x) \right]_0^1 \] At \( x = 1 \): - \( \cos(n\pi) = (-1)^n \) - \( \sin(n\pi) = 0 \) At \( x = 0 \): - \( \cos(0) = 1 \) - \( \sin(0) = 0 \) So, \[ = \left( -\frac{1}{n\pi} (-1)^n + 0 \right) - \left( 0 + 0 \right) = \frac{(-1)^{n+1}}{n\pi} \] So, \[ b_n = 2 \cdot \frac{(-1)^{n+1}}{n\pi} \] --- ### **Step 3: Write the Fourier series** \[ f(x) = \sum_{n=1}^\infty \frac{2(-1)^{n+1}}{n\pi} \sin(n\pi x) \] --- ## **Final Answer for (a):** - **The function is odd.** - **Fourier Series:** \[ f(x) = \sum_{n=1}^\infty \frac{2(-1)^{n+1}}{n\pi} \sin(n\pi x) \] --- Would you like solutions for parts (b), (c), or (d) as well?

Certainly! Let's proceed step-by-step for parts (b), (c), and (d), analyzing each function's symmetry and then deriving their Fourier series. --- ## **Part (b)** \[ w(z) = \begin{cases} 1 & \text{if } -2 < z < 0 \\ 0 & \text{if } 0 < z < 2 \end{cases} \] --- ### **Step 1: Symmetry check** - Find \( w(-z) \): \[ w(-z) = \begin{cases} 1 & \text{if } -2 < -z < 0 \Rightarrow 0 < z < 2 \\ 0 & \text{if } 0 < -z < 2 \Rightarrow -2 < z < 0 \end{cases} \] Compare to \( w(z) \): - \( w(z) = 1 \) for \( -2 < z < 0 \) - \( w(z) = 0 \) for \( 0 < z < 2 \) - \( w(-z) = 1 \) for \( 0 < z < 2 \) - \( w(-z) = 0 \) for \( -2 < z < 0 \) Thus, \[ w(-z) \neq w(z), \quad w(-z) \neq -w(z) \] **The function is neither even nor odd.** --- ### **Step 2: Fourier series** Since the function is neither even nor odd, the Fourier series will contain sine and cosine terms: \[ w(z) = \frac{a_0}{2} + \sum_{n=1}^\infty \left( a_n \cos \frac{n\pi z}{L} + b_n \sin \frac{n\pi z}{L} \right) \] Assuming periodicity over \( -2 < z < 2 \), the period \( T = 4 \), so \( L = T/2= 2 \). --- ### **Step 3: Compute coefficients** \[ a_0 = \frac{1}{L} \int_{-L}^L w(z) dz = \frac{1}{2} \int_{-2}^2 w(z) dz \] - \( w(z) = 1 \) on \( -2 < z < 0 \) - \( w(z) = 0 \) on \( 0 < z < 2 \) \[ a_0 = \frac{1}{2} \left( \int_{-2}^0 1\, dz + \int_0^2 0\, dz \right) = \frac{1}{2} (0 - (-2)) = \frac{1}{2} \times 2 = 1 \] --- ### **Coefficients \( a_n \):** \[ a_n = \frac{1}{L} \int_{-L}^L w(z) \cos \left( \frac{n\pi z}{L} \right) dz = \frac{1}{2} \int_{-2}^2 w(z) \cos \left( \frac{n\pi z}{2} \right) dz \] Since \( w(z) = 1 \) on \( -2 < z < 0 \), zero elsewhere: \[ a_n = \frac{1}{2} \int_{-2}^0 \cos \left( \frac{n\pi z}{2} \right) dz \] Calculate: \[ a_n = \frac{1}{2} \left[ \frac{2}{n\pi} \sin \left( \frac{n\pi z}{2} \right) \right]_{z=-2}^{z=0} \] At \( z=0 \): \[ \sin(0) = 0 \] At \( z=-2 \): \[ \sin \left( \frac{n\pi (-2)}{2} \right) = \sin(- n\pi) = - \sin(n\pi) = 0 \] Thus, \[ a_n = \frac{1}{2} \times \left( 0 - 0 \right) = 0 \] Similarly, for \( b_n \): \[ b_n = \frac{1}{L} \int_{-L}^L w(z) \sin \left( \frac{n\pi z}{L} \right) dz = \frac{1}{2} \int_{-2}^0 \sin \left( \frac{n\pi z}{2} \right) dz \] Calculate: \[ b_n = \frac{1}{2} \left[ -\frac{2}{n\pi} \cos \left( \frac{n\pi z}{2} \right) \right]_{z=-2}^{z=0} \] At \( z=0 \): \[ \cos 0 = 1 \] At \( z=-2 \): \[ \cos \left( - n\pi \right) = \cos(n\pi) = (-1)^n \] So, \[ b_n = \frac{1}{2} \left[ -\frac{2}{n\pi} (1 - (-1)^n) \right] = -\frac{1}{n\pi} (1 - (-1)^n) \] Note that: - If \( n \) is even, \( (-1)^n = 1 \), then: \[ b_n = -\frac{1}{n\pi} (1 - 1) = 0 \] - If \( n \) is odd, \( (-1)^n = -1 \): \[ b_n = -\frac{1}{n\pi} (1 - (-1)) = -\frac{1}{n\pi} \times 2 = -\frac{2}{n\pi} \] --- ### **Final Fourier series for (b):** \[ w(z) = \frac{a_0}{2} + \sum_{n=1}^\infty \left( a_n \cos \frac{n\pi z}{2} + b_n \sin \frac{n\pi z}{2} \right) \] with: \[ a_0 = 1, \quad a_n=0, \quad b_n = \begin{cases} 0, & \text{if } n \text{ even} \\ -\frac{2}{n\pi}, & \text{if } n \text{ odd} \end{cases} \] **Thus:** \[ \boxed{ w(z) = \frac{1}{2} - \sum_{\substack{n=1 \\ n \text{ odd}}}^\infty \frac{2}{n\pi} \sin \left( \frac{n\pi z}{2} \right) } \] --- ## **Part (c)** \[ f(z) = \begin{cases} 0 & -1 < z < 0 \\ w(z) & 0 < z < 1/2 \end{cases} \] *(Assuming \( w(z) \) is the same as in part (b), or similar?)* **Note:** The question seems incomplete or may have a typo in the statement, but based on typical patterns, let's interpret: Suppose the function is: \[ f(z) = \begin{cases} 0 & -1 < z < 0 \\ w(z) & 0 < z < 1/2 \end{cases} \] with \( w(z) \) as in part (b), or similar. --- ### **Step 1: Symmetry** - \( f(z) \) is zero on \( -1 < z < 0 \), and non-zero on \( 0 < z < 1/2 \). It is neither even nor odd because it is not symmetric about the y-axis or origin. ### **Step 2: Fourier series** Given the limited interval and piecewise definition, the Fourier series can be computed over \( -1 < z < 1/2 \). For a precise Fourier series, we need to specify the period or the interval. *Given the ambiguity, and to keep consistency, I will omit the detailed derivation for (c) since the question appears incomplete or inconsistent.* --- ## **Part (d)** \[ f(z) = z, \quad -1 < z < 1 \] with period \( p = 2 \). --- ### **Step 1: Symmetry** - \( f(z) = z \): odd function. \[ f(-z) = -z = -f(z) \] **Thus, it is an odd function.** ### **Step 2: Fourier series** Since \( f(z) \) is odd over \([-1, 1]\), the Fourier series contains only sine terms: \[ f(z) = \sum_{n=1}^\infty b_n \sin \left( n \pi z \right) \] where: \[ b_n = \frac{2}{T} \int_0^{T/2} f(z) \sin(n \pi z) dz \] with \( T=2 \), so \( T/2=1 \). Compute: \[ b_n = 2 \int_0^1 z \sin(n \pi z) dz \] This is similar to the earlier integral: Using integration by parts: - \( u = z \), \( dv = \sin(n\pi z) dz \) - \( du = dz \), \( v= - \frac{1}{n\pi} \cos(n\pi z) \) \[ b_n = 2 \left[ -\frac{z}{n\pi} \cos(n\pi z) + \frac{1}{n\pi} \int \cos(n\pi z) dz \right]_0^1 \] Calculate: \[ b_n = 2 \left[ -\frac{z}{n\pi} \cos(n\pi z) + \frac{1}{n^2 \pi^2} \sin(n\pi z) \right]_0^1 \] At \( z=1 \): \[ \cos(n\pi) = (-1)^n, \quad \sin(n\pi) = 0 \] At \( z=0 \): \[ \cos 0 = 1, \quad \sin 0=0 \] So, \[ b_n = 2 \left[ -\frac{1}{n\pi} (-1)^n + 0 - (0 + 0) \right] = 2 \times \left( - \frac{1}{n\pi} (-1)^n \right) = - \frac{2}{n\pi} (-1)^n \] --- ### **Final Fourier series for (d):** \[ f(z) = \sum_{n=1}^\infty - \frac{2}{n\pi} (-1)^n \sin(n \pi z) \] or equivalently, \[ \boxed{ f(z) = - \frac{2}{\pi} \sum_{n=1}^\infty \frac{(-1)^n}{n} \sin(n \pi z) } \] --- ## **Summary of Final Results:** | Part | Function | Symmetry | Fourier Series | |---|---|---|---| | (a) | piecewise \( -x \) and \( x \) | **Odd** | \(\displaystyle \sum_{n=1}^\infty \frac{2(-1)^{n+1}}{n\pi} \sin(n\pi x)\) | | (b) | step function over \(-2, 2\) | **Neither** | \(\displaystyle \frac{1}{2} - \sum_{\substack{n=1 \\ n \text{ odd}}}^\infty \frac{2}{n\pi} \sin \left( \frac{n\pi z}{2} \right)\) | | (c) | incomplete/uncertain, skipped detailed derivation | — | — | | (d) | \(f(z) = z\) over \([-1,1]\) | **Odd** | \(\displaystyle - \frac{2}{\pi} \sum_{n=1}^\infty \frac{(-1)^n}{n} \sin(n \pi z)\) | If you need further details or clarification on any part, feel free to ask!