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

Mathematics
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

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Answer

f(x)=∑n=1∞2n(−1)n+1sin⁡(nx)f(x) = \sum_{n=1}^{\infty} \frac{2}{n}(-1)^{n+1} \sin(nx)

Step 1: Tambua kazi na muda. Tunahitaji kupata mfululizo wa Fourier sine wa nusu-masafa kwa kazi f(x)=xf(x) = x kwenye muda 0<x<π0 < x < \pi. Kwa mfululizo wa Fourier sine wa nusu-masafa, urefu wa muda ni L=πL = \pi.

Step 2: Kokotoa vigawo vya Fourier sine bnb_n. Fomula ya vigawo vya Fourier sine ni: bn=2L∫0Lf(x)sin⁡(nπxL)dxb_n = \frac{2}{L} \int_0^L f(x) \sin\left(\frac{n\pi x}{L}\right) dx Badilisha L=πL=\pi na f(x)=xf(x)=x: bn=2π∫0πxsin⁡(nπxπ)dxb_n = \frac{2}{\pi} \int_0^{\pi} x \sin\left(\frac{n\pi x}{\pi}\right) dx bn=2π∫0πxsin⁡(nx)dxb_n = \frac{2}{\pi} \int_0^{\pi} x \sin(nx) dx Tutatumia ujumuishaji kwa sehemu (∫u dv=uv−∫v du\int u\,dv = uv - \int v\,du). Chagua u=x  ⟹  du=dxu = x \implies du = dx. Chagua dv=sin⁡(nx)dx  ⟹  v=∫sin⁡(nx)dx=−1ncos⁡(nx)dv = \sin(nx) dx \implies v = \int \sin(nx) dx = -\frac{1}{n}\cos(nx). bn=2π[−xncos⁡(nx)∣0π−∫0π−1ncos⁡(nx)dx]b_n = \frac{2}{\pi} \left[ \left. -\frac{x}{n}\cos(nx) \right|_0^{\pi} - \int_0^{\pi} -\frac{1}{n}\cos(nx) dx \right] bn=2π[(−πncos⁡(nπ)−(−0ncos⁡(0)))+1n∫0πcos⁡(nx)dx]b_n = \frac{2}{\pi} \left[ \left( -\frac{\pi}{n}\cos(n\pi) - \left(-\frac{0}{n}\cos(0)\right) \right) + \frac{1}{n}\int_0^{\pi} \cos(nx) dx \right] bn=2π[−πncos⁡(nπ)+1n1nsin⁡(nx)∣0π]b_n = \frac{2}{\pi} \left[ -\frac{\pi}{n}\cos(n\pi) + \frac{1}{n} \left. \frac{1}{n}\sin(nx) \right|_0^{\pi} \right] Tunajua kuwa cos⁡(nπ)=(−1)n\cos(n\pi) = (-1)^n na sin⁡(nπ)=0\sin(n\pi) = 0 kwa nambari kamili nn. bn=2π[−πn(−1)n+1n2(sin⁡(nπ)−sin⁡(0))]b_n = \frac{2}{\pi} \left[ -\frac{\pi}{n}(-1)^n + \frac{1}{n^2} (\sin(n\pi) - \sin(0)) \right] bn=2π[−πn(−1)n+1n2(0−0)]b_n = \frac{2}{\pi} \left[ -\frac{\pi}{n}(-1)^n + \frac{1}{n^2} (0 - 0) \right] bn=2π[−πn(−1)n]b_n = \frac{2}{\pi} \left[ -\frac{\pi}{n}(-1)^n \right] bn=−2n(−1)nb_n = -\frac{2}{n}(-1)^n Hii inaweza kuandikwa kama: bn=2n(−1)n+1b_n = \frac{2}{n}(-1)^{n+1}

Step 3: Andika mfululizo wa Fourier sine. Mfululizo wa Fourier sine umetolewa na: f(x)=∑n=1∞bnsin⁡(nπxL)f(x) = \sum_{n=1}^{\infty} b_n \sin\left(\frac{n\pi x}{L}\right) Badilisha L=πL=\pi na bnb_n: f(x)=∑n=1∞2n(−1)n+1sin⁡(nx)f(x) = \sum_{n=1}^{\infty} \frac{2}{n}(-1)^{n+1} \sin(nx) Hivyo, mfululizo wa Fourier sine wa nusu-masafa kwa f(x)=xf(x)=x ni: f(x)=∑n=1∞2n(−1)n+1sin⁡(nx) f(x) = \sum_{n=1^{\infty} \frac{2}{n}(-1)^{n+1} \sin(nx) }

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Tambua kazi na muda. Tunahitaji kupata mfululizo wa Fourier sine wa nusu-masafa kwa kazi f(x) = x kwenye muda 0 < x < .

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
Mathematics

This mathematics problem involves applying core mathematical principles and formulas. Below you will find a complete step-by-step solution with detailed explanations for each step, helping you understand not just the answer but the method behind it.

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Step 1: Tambua kazi na muda. Tunahitaji kupata mfululizo wa Fourier sine wa nusu-masafa kwa kazi f(x) = x kwenye muda 0 < x < . Kwa mfululizo wa Fourier sine wa nusu-masafa, urefu wa muda ni L = . Step 2: Kokotoa vigawo vya Fourier sine b_n. Fomula ya vigawo vya Fourier sine ni: b_n = (2)/(L) _0^L f(x) ((n x)/(L)) dx Badilisha L= na f(x)=x: b_n = (2)/() _0^ x ((n x)/()) dx b_n = (2)/() _0^ x (nx) dx Tutatumia ujumuishaji kwa sehemu ( u\,dv = uv - v\,du). Chagua u = x du = dx. Chagua dv = (nx) dx v = (nx) dx = -(1)/(n)(nx). b_n = (2)/() [ . -(x)/(n)(nx) |_0^ - _0^ -(1)/(n)(nx) dx ] b_n = (2)/() [ ( -()/(n)(n) - (-(0)/(n)(0)) ) + (1)/(n)_0^ (nx) dx ] b_n = (2)/() [ -()/(n)(n) + (1)/(n) . (1)/(n)(nx) |_0^ ] Tunajua kuwa (n) = (-1)^n na (n) = 0 kwa nambari kamili n. b_n = (2)/() [ -()/(n)(-1)^n + (1)/(n^2) ((n) - (0)) ] b_n = (2)/() [ -()/(n)(-1)^n + (1)/(n^2) (0 - 0) ] b_n = (2)/() [ -()/(n)(-1)^n ] b_n = -(2)/(n)(-1)^n Hii inaweza kuandikwa kama: b_n = (2)/(n)(-1)^n+1 Step 3: Andika mfululizo wa Fourier sine. Mfululizo wa Fourier sine umetolewa na: f(x) = _n=1^ b_n ((n x)/(L)) Badilisha L= na b_n: f(x) = _n=1^ (2)/(n)(-1)^n+1 (nx) Hivyo, mfululizo wa Fourier sine wa nusu-masafa kwa f(x)=x ni: f(x) = _n=1^ (2)/(n)(-1)^n+1 (nx) That's 2 down. 3 left today — send the next one.