1. **Problem Statement:** Obtain the trigonometric Fourier series for the periodic triangular wave shown in Figure 2.5 with period $T=4$, peaks of height 1 at $t=-2$ and $t=2$, and zeros at $t=-4,0,4$. 2. **Fourier Series Formula:** The trigonometric Fourier series for a periodic function $x(t)$ with period $T$ is: $$x(t) = a_0 + \sum_{n=1}^\infty \left(a_n \cos\left(\frac{2\pi n}{T}t\right) + b_n \sin\left(\frac{2\pi n}{T}t\right)\right)$$ where $$a_0 = \frac{1}{T} \int_{t_0}^{t_0+T} x(t) dt$$ $$a_n = \frac{2}{T} \int_{t_0}^{t_0+T} x(t) \cos\left(\frac{2\pi n}{T}t\right) dt$$ $$b_n = \frac{2}{T} \int_{t_0}^{t_0+T} x(t) \sin\left(\frac{2\pi n}{T}t\right) dt$$ 3. **Important Notes:** - The function is even (triangular wave symmetric about vertical axis), so all $b_n=0$. - We only need to compute $a_0$ and $a_n$. 4. **Define the function over one period:** Choose $t \in [-2,2]$ for one period. The triangular wave rises linearly from 0 at $t=-2$ to 1 at $t=0$, then falls linearly back to 0 at $t=2$. So, $$x(t) = \begin{cases} 1 + \frac{t}{2}, & -2 \leq t < 0 \\ 1 - \frac{t}{2}, & 0 \leq t \leq 2 \end{cases}$$ 5. **Calculate $a_0$:** $$a_0 = \frac{1}{4} \int_{-2}^2 x(t) dt = \frac{1}{4} \left( \int_{-2}^0 \left(1 + \frac{t}{2}\right) dt + \int_0^2 \left(1 - \frac{t}{2}\right) dt \right)$$ Calculate each integral: $$\int_{-2}^0 \left(1 + \frac{t}{2}\right) dt = \left[t + \frac{t^2}{4}\right]_{-2}^0 = (0 + 0) - (-2 + 1) = 1$$ $$\int_0^2 \left(1 - \frac{t}{2}\right) dt = \left[t - \frac{t^2}{4}\right]_0^2 = (2 - 1) - 0 = 1$$ So, $$a_0 = \frac{1}{4} (1 + 1) = \frac{1}{2}$$ 6. **Calculate $a_n$:** $$a_n = \frac{2}{4} \int_{-2}^2 x(t) \cos\left(\frac{\pi n}{2} t\right) dt = \frac{1}{2} \int_{-2}^2 x(t) \cos\left(\frac{\pi n}{2} t\right) dt$$ Split integral: $$a_n = \frac{1}{2} \left( \int_{-2}^0 \left(1 + \frac{t}{2}\right) \cos\left(\frac{\pi n}{2} t\right) dt + \int_0^2 \left(1 - \frac{t}{2}\right) \cos\left(\frac{\pi n}{2} t\right) dt \right)$$ 7. **Use evenness of $x(t)$ and cosine:** Since $x(t)$ is even and cosine is even, the integral over $[-2,2]$ is twice the integral over $[0,2]$: $$a_n = \int_0^2 x(t) \cos\left(\frac{\pi n}{2} t\right) dt$$ On $[0,2]$, $x(t) = 1 - \frac{t}{2}$, so $$a_n = \int_0^2 \left(1 - \frac{t}{2}\right) \cos\left(\frac{\pi n}{2} t\right) dt$$ 8. **Calculate $a_n$ integral:** Split integral: $$a_n = \int_0^2 \cos\left(\frac{\pi n}{2} t\right) dt - \frac{1}{2} \int_0^2 t \cos\left(\frac{\pi n}{2} t\right) dt$$ First integral: $$I_1 = \int_0^2 \cos\left(\frac{\pi n}{2} t\right) dt = \left[ \frac{2}{\pi n} \sin\left(\frac{\pi n}{2} t\right) \right]_0^2 = \frac{2}{\pi n} \sin(\pi n) = 0$$ Second integral by parts: Let $u = t$, $dv = \cos\left(\frac{\pi n}{2} t\right) dt$. Then $du = dt$, and $$v = \frac{2}{\pi n} \sin\left(\frac{\pi n}{2} t\right)$$ So, $$I_2 = \int_0^2 t \cos\left(\frac{\pi n}{2} t\right) dt = \left. t \cdot \frac{2}{\pi n} \sin\left(\frac{\pi n}{2} t\right) \right|_0^2 - \int_0^2 \frac{2}{\pi n} \sin\left(\frac{\pi n}{2} t\right) dt$$ Evaluate boundary term: $$2 \cdot \frac{2}{\pi n} \sin(\pi n) = 0$$ Integral of sine: $$\int_0^2 \sin\left(\frac{\pi n}{2} t\right) dt = \left[ -\frac{2}{\pi n} \cos\left(\frac{\pi n}{2} t\right) \right]_0^2 = -\frac{2}{\pi n} (\cos(\pi n) - 1) = -\frac{2}{\pi n}((-1)^n - 1)$$ So, $$I_2 = 0 - \frac{2}{\pi n} \cdot \left(-\frac{2}{\pi n}((-1)^n - 1)\right) = \frac{4}{\pi^2 n^2} ((-1)^n - 1)$$ 9. **Combine results:** $$a_n = 0 - \frac{1}{2} I_2 = -\frac{1}{2} \cdot \frac{4}{\pi^2 n^2} ((-1)^n - 1) = -\frac{2}{\pi^2 n^2} ((-1)^n - 1)$$ 10. **Simplify $a_n$:** For even $n$, $(-1)^n = 1$, so $a_n=0$. For odd $n$, $(-1)^n = -1$, so $$a_n = -\frac{2}{\pi^2 n^2} (-1 - 1) = -\frac{2}{\pi^2 n^2} (-2) = \frac{4}{\pi^2 n^2}$$ 11. **Final Fourier series:** Only odd harmonics contribute: $$x(t) = \frac{1}{2} + \sum_{n=1,3,5,...}^\infty \frac{4}{\pi^2 n^2} \cos\left(\frac{\pi n}{2} t\right)$$ **Answer:** $$\boxed{x(t) = \frac{1}{2} + \frac{4}{\pi^2} \sum_{n=1,3,5,...}^\infty \frac{1}{n^2} \cos\left(\frac{\pi n}{2} t\right)}$$ This is the trigonometric Fourier series for the given triangular wave.