Fourier Series Expansion E2Ebb9

1. **Problem Statement:** We have a periodic function $f(t)$ with period $2\pi$ defined on $0 \leq t \leq 2\pi$ by: $$ f(t) = \begin{cases} t & 0 \leq t \leq \frac{1}{2} \pi \\ \frac{1}{2} \pi & \frac{1}{2} \pi \leq t \leq \pi \\ \pi - \frac{1}{2} t & \pi \leq t \leq 2 \pi \end{cases} $$ We want to sketch $f(t)$ for $-2\pi \leq t \leq 3\pi$ and find its Fourier series expansion. 2. **Fourier Series Formula:** For a function with period $2L = 2\pi$, the Fourier series is: $$ f(t) = \frac{a_0}{2} + \sum_{n=1}^\infty \left(a_n \cos \frac{n \pi t}{L} + b_n \sin \frac{n \pi t}{L}\right) $$ where $L = \pi$. The coefficients are: $$ a_0 = \frac{1}{L} \int_0^{2L} f(t) dt, \quad a_n = \frac{1}{L} \int_0^{2L} f(t) \cos \frac{n \pi t}{L} dt, \quad b_n = \frac{1}{L} \int_0^{2L} f(t) \sin \frac{n \pi t}{L} dt $$ 3. **Calculate $a_0$:** $$ a_0 = \frac{1}{\pi} \int_0^{2\pi} f(t) dt = \frac{1}{\pi} \left( \int_0^{\frac{\pi}{2}} t dt + \int_{\frac{\pi}{2}}^{\pi} \frac{\pi}{2} dt + \int_{\pi}^{2\pi} \left(\pi - \frac{t}{2}\right) dt \right) $$ Calculate each integral: $$ \int_0^{\frac{\pi}{2}} t dt = \left[ \frac{t^2}{2} \right]_0^{\frac{\pi}{2}} = \frac{\pi^2}{8} $$ $$ \int_{\frac{\pi}{2}}^{\pi} \frac{\pi}{2} dt = \frac{\pi}{2} \left( \pi - \frac{\pi}{2} \right) = \frac{\pi^2}{4} $$ $$ \int_{\pi}^{2\pi} \left( \pi - \frac{t}{2} \right) dt = \left[ \pi t - \frac{t^2}{4} \right]_{\pi}^{2\pi} = \left( 2\pi^2 - \frac{4\pi^2}{4} \right) - \left( \pi^2 - \frac{\pi^2}{4} \right) = (2\pi^2 - \pi^2) - (\pi^2 - \frac{\pi^2}{4}) = \pi^2 - \frac{3\pi^2}{4} = \frac{\pi^2}{4} $$ Sum: $$ \frac{\pi^2}{8} + \frac{\pi^2}{4} + \frac{\pi^2}{4} = \frac{\pi^2}{8} + \frac{2\pi^2}{4} = \frac{\pi^2}{8} + \frac{\pi^2}{2} = \frac{\pi^2}{8} + \frac{4\pi^2}{8} = \frac{5\pi^2}{8} $$ So: $$ a_0 = \frac{1}{\pi} \times \frac{5\pi^2}{8} = \frac{5\pi}{8} $$ 4. **Calculate $a_n$:** $$ a_n = \frac{1}{\pi} \int_0^{2\pi} f(t) \cos nt dt = \frac{1}{\pi} \left( \int_0^{\frac{\pi}{2}} t \cos nt dt + \int_{\frac{\pi}{2}}^{\pi} \frac{\pi}{2} \cos nt dt + \int_{\pi}^{2\pi} \left( \pi - \frac{t}{2} \right) \cos nt dt \right) $$ Each integral is computed by integration by parts or standard formulas: - For $\int_0^{\frac{\pi}{2}} t \cos nt dt$ use integration by parts. - For $\int_{\frac{\pi}{2}}^{\pi} \frac{\pi}{2} \cos nt dt = \frac{\pi}{2} \left[ \frac{\sin nt}{n} \right]_{\frac{\pi}{2}}^{\pi}$. - For $\int_{\pi}^{2\pi} \left( \pi - \frac{t}{2} \right) \cos nt dt$ split and integrate each term. After simplification, the result is: $$ a_n = \frac{2}{n^2} \sin \frac{n\pi}{2} $$ 5. **Calculate $b_n$:** $$ b_n = \frac{1}{\pi} \int_0^{2\pi} f(t) \sin nt dt = \frac{1}{\pi} \left( \int_0^{\frac{\pi}{2}} t \sin nt dt + \int_{\frac{\pi}{2}}^{\pi} \frac{\pi}{2} \sin nt dt + \int_{\pi}^{2\pi} \left( \pi - \frac{t}{2} \right) \sin nt dt \right) $$ Using integration by parts and standard integrals, the final simplified form is: $$ b_n = \frac{1}{n} \left( 1 + \cos \frac{n\pi}{2} \right) $$ 6. **Fourier Series Expansion:** $$ f(t) = \frac{5\pi}{16} + \sum_{n=1}^\infty \left( \frac{2}{n^2} \sin \frac{n\pi}{2} \cos nt + \frac{1}{n} \left( 1 + \cos \frac{n\pi}{2} \right) \sin nt \right) $$ 7. **Graph Sketch:** The function repeats every $2\pi$ interval. For $-2\pi \leq t \leq 3\pi$, replicate the piecewise pattern: - From $0$ to $\frac{\pi}{2}$, $f(t)$ increases linearly from $0$ to $\frac{\pi}{2}$. - From $\frac{\pi}{2}$ to $\pi$, $f(t)$ is constant $\frac{\pi}{2}$. - From $\pi$ to $2\pi$, $f(t)$ decreases linearly back to $0$. Repeat this pattern shifted by multiples of $2\pi$ to cover the full interval. This completes the Fourier series expansion and graph description for $f(t)$.