Fourier Series 7Bb5D7

1. **Problem 1: Fourier series expansion of $f(t) = t^2 - t$ with period $2\pi$** 2. The function is periodic with period $2\pi$, defined on $(-\pi, \pi)$. 3. The Fourier series for a function with period $2\pi$ is: $$ f(t) = \frac{a_0}{2} + \sum_{n=1}^\infty \left(a_n \cos(nt) + b_n \sin(nt)\right) $$ where $$ a_0 = \frac{1}{\pi} \int_{-\pi}^\pi f(t) dt, \quad a_n = \frac{1}{\pi} \int_{-\pi}^\pi f(t) \cos(nt) dt, \quad b_n = \frac{1}{\pi} \int_{-\pi}^\pi f(t) \sin(nt) dt. $$ 4. Calculate $a_0$: $$ a_0 = \frac{1}{\pi} \int_{-\pi}^\pi (t^2 - t) dt = \frac{1}{\pi} \left( \int_{-\pi}^\pi t^2 dt - \int_{-\pi}^\pi t dt \right) $$ 5. Evaluate integrals: $$ \int_{-\pi}^\pi t^2 dt = \left[ \frac{t^3}{3} \right]_{-\pi}^\pi = \frac{\pi^3}{3} - \left(-\frac{\pi^3}{3}\right) = \frac{2\pi^3}{3} $$ $$ \int_{-\pi}^\pi t dt = 0 \quad \text{(since $t$ is odd over symmetric interval)} $$ 6. So, $$ a_0 = \frac{1}{\pi} \times \frac{2\pi^3}{3} = \frac{2\pi^2}{3} $$ 7. Calculate $a_n$: $$ a_n = \frac{1}{\pi} \int_{-\pi}^\pi (t^2 - t) \cos(nt) dt = \frac{1}{\pi} \left( \int_{-\pi}^\pi t^2 \cos(nt) dt - \int_{-\pi}^\pi t \cos(nt) dt \right) $$ 8. Note $t^2 \cos(nt)$ is even (since $t^2$ even, $\cos(nt)$ even), $t \cos(nt)$ is odd (since $t$ odd, $\cos(nt)$ even), so $$ \int_{-\pi}^\pi t \cos(nt) dt = 0 $$ 9. Thus, $$ a_n = \frac{1}{\pi} \int_{-\pi}^\pi t^2 \cos(nt) dt = \frac{2}{\pi} \int_0^\pi t^2 \cos(nt) dt $$ 10. Use integration by parts twice to evaluate $\int_0^\pi t^2 \cos(nt) dt$: Let $I = \int_0^\pi t^2 \cos(nt) dt$ First integration by parts: $$ u = t^2, \quad d\omega = \cos(nt) dt \Rightarrow d\nu = 2t dt, \quad \omega = \frac{\sin(nt)}{n} $$ $$ I = \left. t^2 \frac{\sin(nt)}{n} \right|_0^\pi - \int_0^\pi 2t \frac{\sin(nt)}{n} dt = \frac{\pi^2 \sin(n\pi)}{n} - \frac{2}{n} \int_0^\pi t \sin(nt) dt $$ Since $\sin(n\pi) = 0$, first term is zero. 11. Evaluate $J = \int_0^\pi t \sin(nt) dt$ by parts: Let $u = t$, $dv = \sin(nt) dt$, so $du = dt$, $v = -\frac{\cos(nt)}{n}$ $$ J = \left. -t \frac{\cos(nt)}{n} \right|_0^\pi + \int_0^\pi \frac{\cos(nt)}{n} dt = -\frac{\pi \cos(n\pi)}{n} + \frac{1}{n} \int_0^\pi \cos(nt) dt $$ 12. Evaluate $\int_0^\pi \cos(nt) dt$: $$ \int_0^\pi \cos(nt) dt = \left. \frac{\sin(nt)}{n} \right|_0^\pi = \frac{\sin(n\pi) - \sin(0)}{n} = 0 $$ 13. So, $$ J = -\frac{\pi \cos(n\pi)}{n} $$ 14. Substitute back into $I$: $$ I = -\frac{2}{n} J = -\frac{2}{n} \left(-\frac{\pi \cos(n\pi)}{n}\right) = \frac{2\pi \cos(n\pi)}{n^2} $$ 15. Therefore, $$ a_n = \frac{2}{\pi} I = \frac{2}{\pi} \times \frac{2\pi \cos(n\pi)}{n^2} = \frac{4 \cos(n\pi)}{n^2} $$ 16. Since $\cos(n\pi) = (-1)^n$, we have $$ a_n = \frac{4 (-1)^n}{n^2} $$ 17. Calculate $b_n$: $$ b_n = \frac{1}{\pi} \int_{-\pi}^\pi (t^2 - t) \sin(nt) dt = \frac{1}{\pi} \left( \int_{-\pi}^\pi t^2 \sin(nt) dt - \int_{-\pi}^\pi t \sin(nt) dt \right) $$ 18. $t^2 \sin(nt)$ is odd (since $t^2$ even, $\sin(nt)$ odd), integral over symmetric interval is zero. 19. $t \sin(nt)$ is even (since $t$ odd, $\sin(nt)$ odd, product even), so $$ \int_{-\pi}^\pi t \sin(nt) dt = 2 \int_0^\pi t \sin(nt) dt = 2J $$ 20. From step 13, $J = -\frac{\pi \cos(n\pi)}{n}$, so $$ b_n = \frac{1}{\pi} (0 - 2J) = -\frac{2}{\pi} J = -\frac{2}{\pi} \left(-\frac{\pi \cos(n\pi)}{n}\right) = \frac{2 \cos(n\pi)}{n} = \frac{2 (-1)^n}{n} $$ 21. **Final Fourier series:** $$ f(t) = \frac{a_0}{2} + \sum_{n=1}^\infty \left(a_n \cos(nt) + b_n \sin(nt)\right) = \frac{\pi^2}{3} + \sum_{n=1}^\infty \left( \frac{4 (-1)^n}{n^2} \cos(nt) + \frac{2 (-1)^n}{n} \sin(nt) \right) $$ --- 22. **Problem 2: Function $f(t)$ defined on $0 < t < 6$ by** $$ f(t) = \begin{cases} t, & 0 < t < 3 \\ 6 - t, & 3 < t < 6 \end{cases} $$ 23. **2.1 Half-range cosine Fourier expansion on $0 < t < 6$** 24. The half-range cosine expansion assumes even extension on $(-6,6)$ with period $12$: $$ f(t) \sim \frac{a_0}{2} + \sum_{n=1}^\infty a_n \cos\left( \frac{n \pi t}{6} \right) $$ where $$ a_0 = \frac{2}{6} \int_0^6 f(t) dt = \frac{1}{3} \int_0^6 f(t) dt, \quad a_n = \frac{2}{6} \int_0^6 f(t) \cos\left( \frac{n \pi t}{6} \right) dt = \frac{1}{3} \int_0^6 f(t) \cos\left( \frac{n \pi t}{6} \right) dt $$ 25. Calculate $a_0$: $$ \int_0^6 f(t) dt = \int_0^3 t dt + \int_3^6 (6 - t) dt = \left[ \frac{t^2}{2} \right]_0^3 + \left[ 6t - \frac{t^2}{2} \right]_3^6 = \frac{9}{2} + \left(36 - 18 - (18 - \frac{9}{2})\right) $$ Calculate second integral: $$ 36 - 18 = 18, \quad 18 - \frac{9}{2} = \frac{27}{2}, \quad 18 - \frac{27}{2} = -\frac{9}{2} $$ So second integral is $-\frac{9}{2}$? Recheck carefully: $$ \int_3^6 (6 - t) dt = \left[6t - \frac{t^2}{2}\right]_3^6 = (36 - 18) - (18 - \frac{9}{2}) = 18 - (18 - 4.5) = 18 - 13.5 = 4.5 = \frac{9}{2} $$ Corrected second integral is $\frac{9}{2}$. 26. So total integral: $$ \frac{9}{2} + \frac{9}{2} = 9 $$ 27. Thus, $$ a_0 = \frac{1}{3} \times 9 = 3 $$ 28. Calculate $a_n$: $$ a_n = \frac{1}{3} \left( \int_0^3 t \cos\left( \frac{n \pi t}{6} \right) dt + \int_3^6 (6 - t) \cos\left( \frac{n \pi t}{6} \right) dt \right) $$ 29. Use integration by parts for $\int t \cos(kt) dt$ with $k = \frac{n \pi}{6}$: Let $I_1 = \int_0^3 t \cos(kt) dt$ Set $u = t$, $dv = \cos(kt) dt$, so $du = dt$, $v = \frac{\sin(kt)}{k}$ $$ I_1 = \left. t \frac{\sin(kt)}{k} \right|_0^3 - \int_0^3 \frac{\sin(kt)}{k} dt = \frac{3 \sin(3k)}{k} - \frac{1}{k} \int_0^3 \sin(kt) dt $$ 30. Evaluate $\int_0^3 \sin(kt) dt$: $$ \int_0^3 \sin(kt) dt = \left. -\frac{\cos(kt)}{k} \right|_0^3 = -\frac{\cos(3k) - 1}{k} = \frac{1 - \cos(3k)}{k} $$ 31. Substitute back: $$ I_1 = \frac{3 \sin(3k)}{k} - \frac{1}{k} \times \frac{1 - \cos(3k)}{k} = \frac{3 \sin(3k)}{k} - \frac{1 - \cos(3k)}{k^2} $$ 32. Similarly, for $I_2 = \int_3^6 (6 - t) \cos(kt) dt$, rewrite as: $$ I_2 = 6 \int_3^6 \cos(kt) dt - \int_3^6 t \cos(kt) dt $$ 33. Evaluate $\int_3^6 \cos(kt) dt$: $$ \int_3^6 \cos(kt) dt = \left. \frac{\sin(kt)}{k} \right|_3^6 = \frac{\sin(6k) - \sin(3k)}{k} $$ 34. Evaluate $\int_3^6 t \cos(kt) dt$ by parts: Let $u = t$, $dv = \cos(kt) dt$, so $du = dt$, $v = \frac{\sin(kt)}{k}$ $$ \int_3^6 t \cos(kt) dt = \left. t \frac{\sin(kt)}{k} \right|_3^6 - \int_3^6 \frac{\sin(kt)}{k} dt = \frac{6 \sin(6k)}{k} - \frac{3 \sin(3k)}{k} - \frac{1}{k} \int_3^6 \sin(kt) dt $$ 35. Evaluate $\int_3^6 \sin(kt) dt$: $$ \int_3^6 \sin(kt) dt = \left. -\frac{\cos(kt)}{k} \right|_3^6 = -\frac{\cos(6k) - \cos(3k)}{k} = \frac{\cos(3k) - \cos(6k)}{k} $$ 36. Substitute back: $$ \int_3^6 t \cos(kt) dt = \frac{6 \sin(6k)}{k} - \frac{3 \sin(3k)}{k} - \frac{1}{k} \times \frac{\cos(3k) - \cos(6k)}{k} = \frac{6 \sin(6k) - 3 \sin(3k)}{k} - \frac{\cos(3k) - \cos(6k)}{k^2} $$ 37. Therefore, $$ I_2 = 6 \times \frac{\sin(6k) - \sin(3k)}{k} - \left( \frac{6 \sin(6k) - 3 \sin(3k)}{k} - \frac{\cos(3k) - \cos(6k)}{k^2} \right) = \frac{3 \sin(3k)}{k} + \frac{\cos(3k) - \cos(6k)}{k^2} $$ 38. Sum $I_1 + I_2$: $$ I_1 + I_2 = \left( \frac{3 \sin(3k)}{k} - \frac{1 - \cos(3k)}{k^2} \right) + \left( \frac{3 \sin(3k)}{k} + \frac{\cos(3k) - \cos(6k)}{k^2} \right) = \frac{6 \sin(3k)}{k} + \frac{2 \cos(3k) - 1 - \cos(6k)}{k^2} $$ 39. Recall $k = \frac{n \pi}{6}$, so $$ a_n = \frac{1}{3} \left( I_1 + I_2 \right) = \frac{1}{3} \left( \frac{6 \sin\left( \frac{n \pi}{2} \right)}{\frac{n \pi}{6}} + \frac{2 \cos\left( \frac{n \pi}{2} \right) - 1 - \cos\left( \frac{n \pi}{1} \right)}{\left( \frac{n \pi}{6} \right)^2} \right) $$ Simplify denominators: $$ \frac{6}{\frac{n \pi}{6}} = \frac{6 \times 6}{n \pi} = \frac{36}{n \pi}, \quad \left( \frac{n \pi}{6} \right)^2 = \frac{n^2 \pi^2}{36} $$ So, $$ a_n = \frac{1}{3} \left( \frac{36}{n \pi} \sin\left( \frac{n \pi}{2} \right) + \frac{36}{n^2 \pi^2} \left( 2 \cos\left( \frac{n \pi}{2} \right) - 1 - \cos(n \pi) \right) \right) = \frac{12}{n \pi} \sin\left( \frac{n \pi}{2} \right) + \frac{12}{n^2 \pi^2} \left( 2 \cos\left( \frac{n \pi}{2} \right) - 1 - (-1)^n \right) $$ 40. **2.2 Half-range sine Fourier expansion on $0 < t < 6$** 41. The half-range sine expansion assumes odd extension on $(-6,6)$ with period $12$: $$ f(t) \sim \sum_{n=1}^\infty b_n \sin\left( \frac{n \pi t}{6} \right) $$ where $$ b_n = \frac{2}{6} \int_0^6 f(t) \sin\left( \frac{n \pi t}{6} \right) dt = \frac{1}{3} \int_0^6 f(t) \sin\left( \frac{n \pi t}{6} \right) dt $$ 42. Calculate $b_n$: $$ b_n = \frac{1}{3} \left( \int_0^3 t \sin\left( \frac{n \pi t}{6} \right) dt + \int_3^6 (6 - t) \sin\left( \frac{n \pi t}{6} \right) dt \right) $$ 43. Use integration by parts for $\int t \sin(kt) dt$ with $k = \frac{n \pi}{6}$: Let $I_3 = \int_0^3 t \sin(kt) dt$ Set $u = t$, $dv = \sin(kt) dt$, so $du = dt$, $v = -\frac{\cos(kt)}{k}$ $$ I_3 = \left. -t \frac{\cos(kt)}{k} \right|_0^3 + \int_0^3 \frac{\cos(kt)}{k} dt = -\frac{3 \cos(3k)}{k} + \frac{1}{k} \int_0^3 \cos(kt) dt $$ 44. Evaluate $\int_0^3 \cos(kt) dt$: $$ \int_0^3 \cos(kt) dt = \left. \frac{\sin(kt)}{k} \right|_0^3 = \frac{\sin(3k)}{k} $$ 45. Substitute back: $$ I_3 = -\frac{3 \cos(3k)}{k} + \frac{\sin(3k)}{k^2} $$ 46. For $I_4 = \int_3^6 (6 - t) \sin(kt) dt$, rewrite as: $$ I_4 = 6 \int_3^6 \sin(kt) dt - \int_3^6 t \sin(kt) dt $$ 47. Evaluate $\int_3^6 \sin(kt) dt$: $$ \int_3^6 \sin(kt) dt = \left. -\frac{\cos(kt)}{k} \right|_3^6 = -\frac{\cos(6k) - \cos(3k)}{k} = \frac{\cos(3k) - \cos(6k)}{k} $$ 48. Evaluate $\int_3^6 t \sin(kt) dt$ by parts: Let $u = t$, $dv = \sin(kt) dt$, so $du = dt$, $v = -\frac{\cos(kt)}{k}$ $$ \int_3^6 t \sin(kt) dt = \left. -t \frac{\cos(kt)}{k} \right|_3^6 + \int_3^6 \frac{\cos(kt)}{k} dt = -\frac{6 \cos(6k)}{k} + \frac{3 \cos(3k)}{k} + \frac{1}{k} \int_3^6 \cos(kt) dt $$ 49. Evaluate $\int_3^6 \cos(kt) dt$: $$ \int_3^6 \cos(kt) dt = \left. \frac{\sin(kt)}{k} \right|_3^6 = \frac{\sin(6k) - \sin(3k)}{k} $$ 50. Substitute back: $$ \int_3^6 t \sin(kt) dt = -\frac{6 \cos(6k)}{k} + \frac{3 \cos(3k)}{k} + \frac{\sin(6k) - \sin(3k)}{k^2} $$ 51. Therefore, $$ I_4 = 6 \times \frac{\cos(3k) - \cos(6k)}{k} - \left( -\frac{6 \cos(6k)}{k} + \frac{3 \cos(3k)}{k} + \frac{\sin(6k) - \sin(3k)}{k^2} \right) = \frac{6 \cos(3k) - 6 \cos(6k) + 6 \cos(6k) - 3 \cos(3k)}{k} - \frac{\sin(6k) - \sin(3k)}{k^2} $$ Simplify numerator: $$ (6 \cos(3k) - 3 \cos(3k)) = 3 \cos(3k) $$ So, $$ I_4 = \frac{3 \cos(3k)}{k} - \frac{\sin(6k) - \sin(3k)}{k^2} $$ 52. Sum $I_3 + I_4$: $$ I_3 + I_4 = \left(-\frac{3 \cos(3k)}{k} + \frac{\sin(3k)}{k^2}\right) + \left( \frac{3 \cos(3k)}{k} - \frac{\sin(6k) - \sin(3k)}{k^2} \right) = \frac{\sin(3k) - \sin(6k) + \sin(3k)}{k^2} = \frac{2 \sin(3k) - \sin(6k)}{k^2} $$ 53. Substitute $k = \frac{n \pi}{6}$: $$ b_n = \frac{1}{3} \times \frac{2 \sin\left( \frac{n \pi}{2} \right) - \sin\left( \frac{n \pi}{1} \right)}{\left( \frac{n \pi}{6} \right)^2} = \frac{1}{3} \times \frac{2 \sin\left( \frac{n \pi}{2} \right) - \sin(n \pi)}{\frac{n^2 \pi^2}{36}} = \frac{12}{n^2 \pi^2} \left( 2 \sin\left( \frac{n \pi}{2} \right) - 0 \right) = \frac{24}{n^2 \pi^2} \sin\left( \frac{n \pi}{2} \right) $$ 54. **Summary:** - Fourier series for Problem 1: $$ f(t) = \frac{\pi^2}{3} + \sum_{n=1}^\infty \left( \frac{4 (-1)^n}{n^2} \cos(nt) + \frac{2 (-1)^n}{n} \sin(nt) \right) $$ - Half-range cosine expansion for Problem 2: $$ f(t) \sim \frac{3}{2} + \sum_{n=1}^\infty \left[ \frac{12}{n \pi} \sin\left( \frac{n \pi}{2} \right) + \frac{12}{n^2 \pi^2} \left( 2 \cos\left( \frac{n \pi}{2} \right) - 1 - (-1)^n \right) \right] \cos\left( \frac{n \pi t}{6} \right) $$ - Half-range sine expansion for Problem 2: $$ f(t) \sim \sum_{n=1}^\infty \frac{24}{n^2 \pi^2} \sin\left( \frac{n \pi}{2} \right) \sin\left( \frac{n \pi t}{6} \right) $$