Let's analyze each function for **evenness/oddness** and find its **Fourier series**. --- ## **(b)** \[ f(x) = \begin{cases} -1 & \text{if } -2 < x < 0 \\ 1 & \text{if } 0 < x < 2 \end{cases} \] ### **Even, Odd, or Neither?** Check \( f(-x) \): - For \( 0 < x < 2 \): \( f(-x) = f(-x) = -1 \) (since \(-x \in (-2,0)\)) - For \( -2 < x < 0 \): \( f(-x) = f(-x) = 1 \) (since \(-x \in (0,2)\)) So, \[ f(-x) = -f(x) \] Thus, **the function is odd**. ### **Fourier Series** Since it's odd and period \( p = 2 \), \( L = 2 \). The Fourier series (since \( f(x) \) is odd, only sine terms): \[ f(x) = \sum_{n=1}^\infty b_n \sin\left( \frac{n\pi x}{2} \right) \] where \[ b_n = \frac{1}{L} \int_{-L}^{L} f(x) \sin \left(\frac{n\pi x}{L}\right) dx \] Here, \( L = 2 \): \[ b_n = \frac{1}{2} \int_{-2}^{2} f(x) \sin\left(\frac{n\pi x}{2}\right) dx \] Split at 0: \[ b_n = \frac{1}{2} \left( \int_{-2}^{0} (-1) \sin\left(\frac{n\pi x}{2}\right) dx + \int_{0}^{2} (1) \sin\left(\frac{n\pi x}{2}\right) dx \right) \] \[ b_n = \frac{1}{2} \left( -\int_{-2}^{0} \sin\left(\frac{n\pi x}{2}\right) dx + \int_{0}^{2} \sin\left(\frac{n\pi x}{2}\right) dx \right) \] Let \( I = \int_{a}^{b} \sin(ax) dx = -\frac{1}{a} \cos(ax) \bigg|_{a}^{b} \). Calculate both integrals: \[ \int \sin \left(\frac{n\pi x}{2}\right) dx = -\frac{2}{n\pi} \cos \left(\frac{n\pi x}{2}\right) \] So, \[ \int_{-2}^{0} \sin \left(\frac{n\pi x}{2}\right) dx = -\frac{2}{n\pi} \left[\cos(0) - \cos\left(-n\pi\right)\right] = -\frac{2}{n\pi} \left[1 - \cos(n\pi)\right] \] But \(\cos(-n\pi) = \cos(n\pi) = (-1)^n\). Similarly, \[ \int_{0}^{2} \sin \left(\frac{n\pi x}{2}\right) dx = -\frac{2}{n\pi} \left[\cos(n\pi) - \cos(0)\right] = -\frac{2}{n\pi} \left[(-1)^n - 1\right] \] So, \[ b_n = \frac{1}{2} \left( -\left(-\frac{2}{n\pi}[1 - (-1)^n]\right) + -\frac{2}{n\pi}[(-1)^n - 1] \right) \] \[ = \frac{1}{2} \left(\frac{2}{n\pi}[1 - (-1)^n] - \frac{2}{n\pi}[(-1)^n - 1]\right) \] \[ = \frac{1}{2} \cdot \frac{2}{n\pi}[1 - (-1)^n - (-1)^n + 1] \] \[ = \frac{1}{n\pi}[2 - 2(-1)^n] \] \[ = \frac{2}{n\pi}[1 - (-1)^n] \] So, - When \( n \) is even: \( (-1)^n = 1 \implies b_n = 0 \) - When \( n \) is odd: \( (-1)^n = -1 \implies b_n = \frac{4}{n\pi} \) **Final Fourier Series:** \[ \boxed{ f(x) = \sum_{k=1}^\infty \frac{4}{(2k-1)\pi} \sin\left(\frac{(2k-1)\pi x}{2}\right) } \] --- ## **(c)** \[ f(x) = \begin{cases} 0 & \text{if } -1/2 < x < 0 \\ x & \text{if } 0 < x < 1/2 \end{cases} \] Period = 1. ### **Even, Odd, or Neither?** For \( x \in (0, 1/2) \), \( f(-x) = 0 \) but \( f(x) = x \), so not odd or even. ### **Fourier Series:** \[ f(x) = a_0 + \sum_{n=1}^{\infty} a_n \cos(2n\pi x) + b_n \sin(2n\pi x) \] where \[ a_0 = \int_{-1/2}^{1/2} f(x) dx \] \[ a_n = 2 \int_{-1/2}^{1/2} f(x) \cos(2n\pi x) dx \] \[ b_n = 2 \int_{-1/2}^{1/2} f(x) \sin(2n\pi x) dx \] But \( f(x) = 0 \) for \( -1/2 < x < 0 \), so all integrals reduce to \( 0 < x < 1/2 \): \[ a_0 = \int_0^{1/2} x dx = \left. \frac{x^2}{2} \right|_0^{1/2} = \frac{1}{8} \] \[ a_n = 2 \int_0^{1/2} x \cos(2n\pi x) dx \] Use integration by parts: Let \( u = x, dv = \cos(2n\pi x) dx \), \( du = dx \), \( v = \frac{1}{2 n \pi} \sin(2n\pi x) \): \[ a_n = 2 \left[ x \cdot \frac{1}{2 n \pi} \sin(2n\pi x) \Big|_0^{1/2} - \int_0^{1/2} \frac{1}{2 n \pi} \sin(2n\pi x) dx \right] \] \[ = 2 \left[ \frac{1}{2 n \pi} \left( \frac{1}{2} \sin(n\pi) - 0 \right) - \frac{1}{2 n \pi} \int_0^{1/2} \sin(2n\pi x) dx \right] \] But \( \sin(n\pi) = 0 \); \[ \int_0^{1/2} \sin(2n\pi x) dx = -\frac{1}{2n\pi} [\cos(2n\pi x)]_0^{1/2} = -\frac{1}{2n\pi} [\cos(n\pi) - 1] \] \[ = -\frac{1}{2n\pi}[(-1)^n - 1] \] So, \[ a_n = 2 \left[ 0 - \frac{1}{2 n \pi} \cdot -\frac{1}{2n\pi}[(-1)^n - 1] \right] = 2 \cdot \frac{1}{4 n^2 \pi^2} [(-1)^n - 1] = \frac{1}{2 n^2 \pi^2} [(-1)^n - 1] \] \[ b_n = 2 \int_0^{1/2} x \sin(2n\pi x) dx \] Integration by parts: Let \( u = x, dv = \sin(2n\pi x) dx \), \( du = dx \), \( v = -\frac{1}{2n\pi} \cos(2n\pi x) \): \[ b_n = 2 \left[ -x \cdot \frac{1}{2n\pi} \cos(2n\pi x) \Big|_0^{1/2} + \int_0^{1/2} \frac{1}{2 n \pi} \cos(2n\pi x) dx \right] \] \[ = 2 \left[ -\frac{1}{2n\pi} \cdot \frac{1}{2} \cos(n\pi) + 0 + \frac{1}{2 n \pi} \int_0^{1/2} \cos(2n\pi x) dx \right] \] \[ \int_0^{1/2} \cos(2 n \pi x) dx = \frac{1}{2 n \pi} [\sin(2 n \pi x)]_0^{1/2} = \frac{1}{2 n \pi} [\sin(n \pi) - 0] = 0 \] \[ b_n = 2 \left[ -\frac{1}{4 n \pi} (-1)^n \right] = \frac{(-1)^{n+1}}{2 n \pi} \] --- **Final Fourier Series:** \[ \boxed{ f(x) = \frac{1}{8} + \sum_{n=1}^\infty \left( \frac{1}{2 n^2 \pi^2} [(-1)^n - 1] \cos(2 n \pi x) + \frac{(-1)^{n+1}}{2 n \pi} \sin(2 n \pi x) \right) } \] --- ## **(d)** \[ f(x) = x|x|, \quad -1 < x < 1, \quad p = 2 \] ### **Even, Odd, or Neither?** \( f(-x) = (-x) | -x | = -x | x | = -f(x) \) So, **odd function**. ### **Fourier Series:** Period \( p = 2 \), \( L = 1 \). Only sine terms: \[ f(x) = \sum_{n=1}^\infty b_n \sin(n\pi x) \] \[ b_n = \int_{-1}^{1} f(x) \sin(n\pi x) dx \] But since \( f(x) \) is odd and \( \sin \) is odd, \( f(x)\sin(n\pi x) \) is even, so: \[ b_n = 2 \int_{0}^{1} f(x) \sin(n\pi x) dx \] But for \( 0 < x < 1 \), \( f(x) = x^2 \): \[ b_n = 2 \int_0^1 x^2 \sin(n\pi x) dx \] Integrate by parts twice: Let \( u = x^2, dv = \sin(n\pi x) dx \), \( du = 2x dx, v = -\frac{1}{n\pi} \cos(n\pi x) \): \[ \int x^2 \sin(n\pi x) dx = -x^2 \frac{\cos(n\pi x)}{n\pi} + \int 2x \frac{\cos(n\pi x)}{n\pi} dx \] Now, \( \int x \cos(n\pi x) dx \): Let \( u = x, dv = \cos(n\pi x) dx \), \( du = dx, v = \frac{1}{n\pi} \sin(n\pi x) \): \[ \int x \cos(n\pi x) dx = x \frac{\sin(n\pi x)}{n\pi} - \int \frac{\sin(n\pi x)}{n\pi} dx = x \frac{\sin(n\pi x)}{n\pi} + \frac{1}{(n\pi)^2} \cos(n\pi x) \] So, \[ \int x^2 \sin(n\pi x) dx = -x^2 \frac{\cos(n\pi x)}{n\pi} + 2 \left[ x \frac{\sin(n\pi x)}{(n\pi)^2

Apologies for the interruption. Let's continue the calculation for the Fourier coefficient \(b_n\). We have: \[ b_n = 2 \int_0^1 x^2 \sin(n\pi x) dx \] From previous steps, the integral: \[ I = \int x^2 \sin(n\pi x) dx = -x^2 \frac{\cos(n\pi x)}{n\pi} + 2 \left[ x \frac{\sin(n\pi x)}{(n\pi)^2} + \frac{\cos(n\pi x)}{(n\pi)^3} \right] \] Now, evaluate from 0 to 1: At \(x=1\): \[ I(1) = -1^2 \frac{\cos(n\pi)}{n\pi} + 2 \left[ 1 \frac{\sin(n\pi)}{(n\pi)^2} + \frac{\cos(n\pi)}{(n\pi)^3} \right] \] \[ = - \frac{\cos(n\pi)}{n\pi} + 2 \left[ 0 + \frac{\cos(n\pi)}{(n\pi)^3} \right] = - \frac{\cos(n\pi)}{n\pi} + \frac{2 \cos(n\pi)}{(n\pi)^3} \] At \(x=0\): \[ I(0) = 0 - 0 + 2 \left[ 0 + \frac{\cos 0}{(n\pi)^3} \right] = \frac{2}{(n\pi)^3} \] Therefore, \[ \int_0^1 x^2 \sin(n\pi x) dx = I(1) - I(0) = \left(- \frac{\cos(n\pi)}{n\pi} + \frac{2 \cos(n\pi)}{(n\pi)^3}\right) - \frac{2}{(n\pi)^3} \] Recall \(\cos(n\pi) = (-1)^n\): \[ = - \frac{(-1)^n}{n\pi} + \frac{2 (-1)^n}{(n\pi)^3} - \frac{2}{(n\pi)^3} \] Multiply by 2 to find \(b_n\): \[ b_n = 2 \times \left[ - \frac{(-1)^n}{n\pi} + \frac{2 (-1)^n}{(n\pi)^3} - \frac{2}{(n\pi)^3} \right] \] \[ b_n = - \frac{2 (-1)^n}{n\pi} + \frac{4 (-1)^n}{(n\pi)^3} - \frac{4}{(n\pi)^3} \] **Final Fourier sine coefficient:** \[ \boxed{ b_n = - \frac{2 (-1)^n}{n\pi} + \frac{4 (-1)^n}{(n\pi)^3} - \frac{4}{(n\pi)^3} } \] The Fourier series for \(f(x)\) in \(-1 < x < 1\): \[ f(x) = \sum_{n=1}^\infty b_n \sin(n \pi x) \] with \(b_n\) as above. --- ## **Summary** | Part | Function | Symmetry | Fourier Series Type | Key Coefficients | |--------|--------------|--------------|------------------------------|---------------------| | (b) | Piecewise \(-1\) and \(1\) | Odd | Sine series | \(b_n = \frac{4}{(2k-1)\pi}\) for odd \(n\); zero for even \(n\) | | (c) | Piecewise \(0\) and \(x\) | Neither | Full Fourier series | \(a_0 = \frac{1}{8}\), \(a_n\) and \(b_n\) as derived | | (d) | \(x|x|\) | Odd | Sine series | \(b_n = - \frac{2 (-1)^n}{n\pi} + \frac{4 (-1)^n}{(n\pi)^3} - \frac{4}{(n\pi)^3}\) | --- **Note:** The functions' symmetry determines whether their Fourier series contain only sine or cosine terms, simplifying the calculations accordingly. --- If you need further clarification or detailed steps on any part, feel free to ask!

**(b)** Function: \[ f(x) = \begin{cases} -1, & -2 < x < 0 \\ 1, & 0 < x < 2 \end{cases} \] Period: \( p=2 \) *Symmetry:* Odd *Fourier Series:* \[ \boxed{ f(x) = \sum_{k=1}^\infty \frac{4}{(2k-1)\pi} \sin\left( \frac{(2k-1)\pi x}{2} \right) } \] --- **(c)** Function: \[ f(x) = \begin{cases} 0, & -\frac{1}{2} < x < 0 \\ x, & 0 < x < \frac{1}{2} \end{cases} \] Period: \( p=1 \) *Symmetry:* Neither even nor odd *Fourier Series:* \[ \boxed{ f(x) = \frac{1}{8} + \sum_{n=1}^\infty \left[ \frac{1}{2 n^2 \pi^2} \left( (-1)^n - 1 \right) \cos(2 n \pi x) + \frac{(-1)^{n+1}}{2 n \pi} \sin(2 n \pi x) \right] } \] --- **(d)** Function: \[ f(x) = x|x|,\quad -1 < x < 1,\quad p=2 \] *Symmetry:* Odd *Fourier Series:* \[ \boxed{ f(x) = \sum_{n=1}^\infty \left[ - \frac{2 (-1)^n}{n\pi} + \frac{4 (-1)^n}{(n\pi)^3} - \frac{4}{(n\pi)^3} \right] \sin(n \pi x) } \]