Determine the half-range Fourier sine series of f(x) = x, 0<x<1. Sketch the function f(t) for 2 periods. Determine the Fourier series of f(t). By using a suitable value of t, show that f(t+2) = t−1 for 0<t<1, -t+1 for −1<t<0 and Sum_n=1^ (1 / (2n−1)2) = pi2 / 8

Tunahitaji kupata mfululizo wa Fourier sine wa nusu-masafa kwa kazi $f(x) = x$ kwenye muda $0 < x < \pi$.
Kwa mfululizo wa Fourier sine wa nusu-masafa, urefu wa muda ni $L = \pi$.

Fomula ya vigawo vya Fourier sine ni:

Badilisha $L=\pi$ na $f(x)=x$:

Tutatumia ujumuishaji kwa sehemu ($\int u\,dv = uv - \int v\,du$).
Chagua $u = x \implies du = dx$.
Chagua $dv = \sin(nx) dx \implies v = \int \sin(nx) dx = -\frac{1}{n}\cos(nx)$.

Tunajua kuwa $\cos(n\pi) = (-1)^n$ na $\sin(n\pi) = 0$ kwa nambari kamili $n$.

Hii inaweza kuandikwa kama:

Mfululizo wa Fourier sine umetolewa na:

Badilisha $L=\pi$ na $b_n$:

Hivyo, mfululizo wa Fourier sine wa nusu-masafa kwa $f(x)=x$ ni: