By substituting x=0 show that 4 = 1 − 1/3 + 1/5 - 1/7 + 1/9. Find the fourier sine series of f(t) = (3(1−t/π)) for 0<t<π. Sketch the corresponding graph of the function f(t) on interval -π to π

Urefu wa muda ni $L = \pi$.
Vigawo vya Fourier sine $b_n$ hutolewa na fomula:

$$b_n = \frac{2}{L} \int_0^L f(t) \sin\left(\frac{n\pi t}{L}\right) dt$$

Badilisha $L=\pi$ na $f(t) = \sqrt{3}\left(1 - \frac{t}{\pi}\right)$:

$$b_n = \frac{2}{\pi} \int_0^{\pi} \sqrt{3}\left(1 - \frac{t}{\pi}\right) \sin\left(\frac{n\pi t}{\pi}\right) dt$$ $$b_n = \frac{2\sqrt{3}}{\pi} \int_0^{\pi} \left(1 - \frac{t}{\pi}\right) \sin(nt) dt$$

Tumia ujumuishaji kwa sehemu ($\int u\,dv = uv - \int v\,du$).
Chagua $u = 1 - \frac{t}{\pi} \implies du = -\frac{1}{\pi} dt$.
Chagua $dv = \sin(nt) dt \implies v = -\frac{1}{n}\cos(nt)$.

$$b_n = \frac{2\sqrt{3}}{\pi} \left[ \left. \left(1 - \frac{t}{\pi}\right) \left(-\frac{1}{n}\cos(nt)\right) \right|_0^{\pi} - \int_0^{\pi} \left(-\frac{1}{n}\cos(nt)\right) \left(-\frac{1}{\pi}\right) dt \right]$$

Tathmini sehemu ya kwanza:
Kwa $t=\pi$: $\left(1 - \frac{\pi}{\pi}\right) \left(-\frac{1}{n}\cos(n\pi)\right) = (0) \left(-\frac{1}{n}(-1)^n\right) = 0$.
Kwa $t=0$: $\left(1 - \frac{0}{\pi}\right) \left(-\frac{1}{n}\cos(0)\right) = (1) \left(-\frac{1}{n}(1)\right) = -\frac{1}{n}$.
Hivyo, sehemu ya kwanza ni $0 - \left(-\frac{1}{n}\right) = \frac{1}{n}$.

Tathmini sehemu ya pili:

$$- \int_0^{\pi} \left(-\frac{1}{n}\cos(nt)\right) \left(-\frac{1}{\pi}\right) dt = - \frac{1}{n\pi} \int_0^{\pi} \cos(nt) dt$$ $$= - \frac{1}{n\pi} \left[ \frac{1}{n}\sin(nt) \right]_0^{\pi} = - \frac{1}{n^2\pi} (\sin(n\pi) - \sin(0))$$

Kwa kuwa $\sin(n\pi) = 0$ na $\sin(0) = 0$, sehemu hii ni 0.

Kwa hiyo, $b_n = \frac{2\sqrt{3}}{\pi} \left[ \frac{1}{n} - 0 \right] = \frac{2\sqrt{3}}{n\pi}$.
Mfululizo wa Fourier sine ni:

$$\boxed{ f(t) = \sum_{n=1}^{\infty} \frac{2\sqrt{3}}{n\pi} \sin(nt) }$$

Utambulisho huu unatokana na mfululizo wa Fourier wa kazi ya wimbi la mraba.
Hebu tuchunguze kazi $g(x)$ iliyofafanuliwa kama:

$$g(x) = \begin{cases} 1 & \text{kwa } 0 < x < \pi \\ -1 & \text{kwa } -\pi < x < 0 \end{cases}$$

Na ina kipindi cha $2\pi$. Hii ni kazi isiyo ya kawaida (odd function), hivyo vigawo vya cosine $a_n=0$.
Vigawo vya sine $b_n$ ni:

$$b_n = \frac{1}{\pi} \int_{-\pi}^{\pi} g(x) \sin(nx) dx = \frac{2}{\pi} \int_0^{\pi} 1 \cdot \sin(nx) dx$$ $$b_n = \frac{2}{\pi} \left[ -\frac{1}{n}\cos(nx) \right]_0^{\pi} = \frac{2}{\pi} \left( -\frac{1}{n}\cos(n\pi) - \left(-\frac{1}{n}\cos(0)\right) \right)$$ $$b_n = \frac{2}{\pi} \left( -\frac{1}{n}(-1)^n + \frac{1}{n} \right) = \frac{2}{n\pi} (1 - (-1)^n)$$

• Ikiwa $n$ ni nambari shufwa, $1 - (-1)^n = 1 - 1 = 0$, hivyo $b_n = 0$.
• Ikiwa $n$ ni nambari witiri, $1 - (-1)^n = 1 - (-1) = 2$, hivyo $b_n = \frac{2}{n\pi} \cdot 2 = \frac{4}{n\pi}$. Mfululizo wa Fourier kwa $g(x)$ ni:

$$g(x) = \sum_{n \text{ witiri}}^{\infty} \frac{4}{n\pi} \sin(nx)$$

Tunaweza kuandika $n$ witiri kama $2k-1$ kwa $k=1, 2, 3, \dots$:

$$g(x) = \sum_{k=1}^{\infty} \frac{4}{(2k-1)\pi} \sin((2k-1)x)$$

Sasa, badilisha $x = \frac{\pi}{2}$ (kumbuka, swali linaweza kuwa na makosa kwa kutaja $x=0$, kwani $x=0$ hutoa $0=0$ kwa mfululizo huu). Kazi $g(x)$ ni endelevu kwa $x=\frac{\pi}{2}$, na $g(\frac{\pi}{2}) = 1$.

$$1 = \sum_{k=1}^{\infty} \frac{4}{(2k-1)\pi} \sin\left((2k-1)\frac{\pi}{2}\right)$$

Tunajua kuwa $\sin\left((2k-1)\frac{\pi}{2}\right) = (-1)^{k-1}$.

$$1 = \sum_{k=1}^{\infty} \frac{4}{(2k-1)\pi} (-1)^{k-1}$$ $$1 = \frac{4}{\pi} \left( \frac{1}{1}(-1)^0 + \frac{1}{3}(-1)^1 + \frac{1}{5}(-1)^2 + \frac{1}{7}(-1)^3 + \dots \right)$$ $$1 = \frac{4}{\pi} \left( 1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \dots \right)$$

Zidisha pande zote mbili kwa $\frac{\pi}{4}$:

$$\boxed{ \frac{\pi}{4} = 1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \dots }$$