Let's solve the problem step-by-step: ### Step 1: **Understand the function and its period** Given: \[ f(t) = \begin{cases} -t & \text{if } -1 < t < 0 \\ t & \text{if } 0 \leq t < 1 \end{cases} \] This function is defined on the interval \((-1, 1)\), and is **odd** (since \( f(-t) = -f(t) \)). We can assume the period is \(2\) (from \(-1\) to \(1\)), so \(L = 1\). **Explanation:** Recognizing that the function is odd and defined over a symmetric interval simplifies the Fourier series: only sine terms (\(b_n\)) will be nonzero. --- ### Step 2: **Find the Fourier sine coefficients \(b_n\)** The general Fourier series for period \(2L = 2\) is: \[ f(t) = \sum_{n=1}^{\infty} b_n \sin\left(\frac{n\pi t}{L}\right) \] with \[ b_n = \frac{1}{L} \int_{-L}^{L} f(t) \sin\left(\frac{n\pi t}{L}\right) \, dt \] Here, \(L = 1\): Calculate \(b_n\): \[ b_n = \int_{-1}^{1} f(t) \sin(n\pi t) \, dt \] Split the integral: \[ b_n = \int_{-1}^0 (-t) \sin(n\pi t) dt + \int_0^1 t \sin(n\pi t) dt \] Let’s substitute \(u = -t\) in the first integral (\(t\) from \(-1\) to \(0\), so \(u\) goes from \(1\) to \(0\)): \[ \int_{-1}^0 (-t) \sin(n\pi t) dt = \int_{1}^{0} u \sin(-n\pi u) (-du) = \int_0^1 u \sin(-n\pi u) du \] But \(\sin(-x) = -\sin(x)\), so: \[ \int_0^1 u \sin(-n\pi u) du = -\int_0^1 u \sin(n\pi u) du \] So, \[ b_n = -\int_0^1 u \sin(n\pi u) du + \int_0^1 u \sin(n\pi u) du = 2\int_0^1 t \sin(n\pi t) dt \] Now, integrate by parts: Let \(u = t\), \(dv = \sin(n\pi t) dt\). Then, \(du = dt\), \(v = -\frac{1}{n\pi} \cos(n\pi t)\): \[ \int t \sin(n\pi t) dt = -\frac{t}{n\pi} \cos(n\pi t) + \frac{1}{n\pi} \int \cos(n\pi t) dt \] \[ = -\frac{t}{n\pi} \cos(n\pi t) + \frac{1}{n^2\pi^2} \sin(n\pi t) + C \] Evaluate from \(0\) to \(1\): \[ \int_0^1 t \sin(n\pi t) dt = \left[ -\frac{t}{n\pi} \cos(n\pi t) + \frac{1}{n^2\pi^2} \sin(n\pi t) \right]_0^1 \] At \(t=1\): \(\cos(n\pi) = (-1)^n\), \(\sin(n\pi) = 0\). At \(t=0\): \(\cos(0) = 1\), \(\sin(0) = 0\). So, \[ = \left(-\frac{1}{n\pi}(-1)^n + 0 \right) - \left(0 + 0\right) = \frac{(-1)^{n+1}}{n\pi} \] Thus, \[ b_n = 2 \cdot \frac{(-1)^{n+1}}{n\pi} \] --- ### **Final Answer: Fourier Series** \[ f(t) = \sum_{n=1}^{\infty} \frac{2(-1)^{n+1}}{n\pi} \sin(n\pi t) \] --- #### **Summary** - **Step 1:** Recognized the function is odd and period 2, so only sine terms are needed. - **Step 2:** Calculated the sine coefficients, giving the final Fourier series as above.