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
|Mathematics
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
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Answer
f(t)=∑n=1∞nπ23sin(nt)
Step 1: Pata mfululizo wa Fourier sine wa nusu-masafa kwa f(t)=3(1−πt) kwa 0<t<π.
Urefu wa muda ni L=π.
Vigawo vya Fourier sine bn hutolewa na fomula:
bn=L2∫0Lf(t)sin(Lnπt)dt
Badilisha L=π na f(t)=3(1−πt):
bn=π2∫0π3(1−πt)sin(πnπt)dtbn=π23∫0π(1−πt)sin(nt)dt
Tumia ujumuishaji kwa sehemu (∫udv=uv−∫vdu).
Chagua u=1−πt⟹du=−π1dt.
Chagua dv=sin(nt)dt⟹v=−n1cos(nt).
bn=π23[(1−πt)(−n1cos(nt))0π−∫0π(−n1cos(nt))(−π1)dt]
Tathmini sehemu ya kwanza:
Kwa t=π: (1−ππ)(−n1cos(nπ))=(0)(−n1(−1)n)=0.
Kwa t=0: (1−π0)(−n1cos(0))=(1)(−n1(1))=−n1.
Hivyo, sehemu ya kwanza ni 0−(−n1)=n1.
Tathmini sehemu ya pili:
−∫0π(−n1cos(nt))(−π1)dt=−nπ1∫0πcos(nt)dt=−nπ1[n1sin(nt)]0π=−n2π1(sin(nπ)−sin(0))
Kwa kuwa sin(nπ)=0 na sin(0)=0, sehemu hii ni 0.
Kwa hiyo, bn=π23[n1−0]=nπ23.
Mfululizo wa Fourier sine ni:
f(t)=∑n=1∞nπ23sin(nt)
Step 2: Onyesha utambulisho 4π=1−31+51−71+91−….
Utambulisho huu unatokana na mfululizo wa Fourier wa kazi ya wimbi la mraba.
Hebu tuchunguze kazi g(x) iliyofafanuliwa kama:
g(x)={1−1kwa0<x<πkwa−π<x<0
Na ina kipindi cha 2π. Hii ni kazi isiyo ya kawaida (odd function), hivyo vigawo vya cosine an=0.
Vigawo vya sine bn ni:
bn=π1∫−ππg(x)sin(nx)dx=π2∫0π1⋅sin(nx)dxbn=π2[−n1cos(nx)]0π=π2(−n1cos(nπ)−(−n1cos(0)))bn=π2(−n1(−1)n+n1)=nπ2(1−(−1)n)
Ikiwa n ni nambari shufwa, 1−(−1)n=1−1=0, hivyo bn=0.
Ikiwa n ni nambari witiri, 1−(−1)n=1−(−1)=2, hivyo bn=nπ2⋅2=nπ4.
Mfululizo wa Fourier kwa g(x) ni:
g(x)=∑nwitiri∞nπ4sin(nx)
Tunaweza kuandika n witiri kama 2k−1 kwa k=1,2,3,…:
g(x)=∑k=1∞(2k−1)π4sin((2k−1)x)
Sasa, badilisha x=2π (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=2π, na g(2π)=1.
1=∑k=1∞(2k−1)π4sin((2k−1)2π)
Tunajua kuwa sin((2k−1)2π)=(−1)k−1.
1=∑k=1∞(2k−1)π4(−1)k−11=π4(11(−1)0+31(−1)1+51(−1)2+71(−1)3+…)1=π4(1−31+51−71+…)
Zidisha pande zote mbili kwa 4π:
\frac{\pi{4} = 1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \dots }
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Pata mfululizo wa Fourier sine wa nusu-masafa kwa f(t) = sqrt(3)(1 - (t)/()) kwa 0 < t < .
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
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: Pata mfululizo wa Fourier sine wa nusu-masafa kwa f(t) = sqrt(3)(1 - (t)/()) kwa 0 < t < . Urefu wa muda ni L = . Vigawo vya Fourier sine b_n hutolewa na fomula: b_n = (2)/(L) _0^L f(t) ((n t)/(L)) dt Badilisha L= na f(t) = sqrt(3)(1 - (t)/()): b_n = (2)/() _0^ sqrt(3)(1 - (t)/()) ((n t)/()) dt b_n = 2sqrt(3) _0^ (1 - (t)/()) (nt) dt Tumia ujumuishaji kwa sehemu ( u\,dv = uv - v\,du). Chagua u = 1 - (t)/() du = -(1)/() dt. Chagua dv = (nt) dt v = -(1)/(n)(nt). b_n = 2sqrt(3) [ . (1 - (t)/()) (-(1)/(n)(nt)) |_0^ - _0^ (-(1)/(n)(nt)) (-(1)/()) dt ] Tathmini sehemu ya kwanza: Kwa t=: (1 - ()/()) (-(1)/(n)(n)) = (0) (-(1)/(n)(-1)^n) = 0. Kwa t=0: (1 - (0)/()) (-(1)/(n)(0)) = (1) (-(1)/(n)(1)) = -(1)/(n). Hivyo, sehemu ya kwanza ni 0 - (-(1)/(n)) = (1)/(n). Tathmini sehemu ya pili: _0^ (-(1)/(n)(nt)) (-(1)/()) dt = - (1)/(n) _0^ (nt) dt = - (1)/(n) [ (1)/(n)(nt) ]_0^ = - (1)/(n^2) ((n) - (0)) Kwa kuwa (n) = 0 na (0) = 0, sehemu hii ni 0. Kwa hiyo, b_n = 2sqrt(3) [ (1)/(n) - 0 ] = 2sqrt(3)n. Mfululizo wa Fourier sine ni: f(t) = _n=1^ 2sqrt(3)n (nt) Step 2: Onyesha utambulisho ()/(4) = 1 - (1)/(3) + (1)/(5) - (1)/(7) + (1)/(9) - . Utambulisho huu unatokana na mfululizo wa Fourier wa kazi ya wimbi la mraba. Hebu tuchunguze kazi g(x) iliyofafanuliwa kama: g(x) = 1 & kwa 0 < x < \\ -1 & kwa - < x < 0 Na ina kipindi cha 2. Hii ni kazi isiyo ya kawaida (odd function), hivyo vigawo vya cosine a_n=0. Vigawo vya sine b_n ni: b_n = (1)/() _-^ g(x) (nx) dx = (2)/() _0^ 1 · (nx) dx b_n = (2)/() [ -(1)/(n)(nx) ]_0^ = (2)/() ( -(1)/(n)(n) - (-(1)/(n)(0)) ) b_n = (2)/() ( -(1)/(n)(-1)^n + (1)/(n) ) = (2)/(n) (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 = (2)/(n) · 2 = (4)/(n). Mfululizo wa Fourier kwa g(x) ni: g(x) = _n witiri^ (4)/(n) (nx) Tunaweza kuandika n witiri kama 2k-1 kwa k=1, 2, 3, : g(x) = _k=1^ (4)/((2k-1)) ((2k-1)x) Sasa, badilisha x = ()/(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=()/(2), na g(()/(2)) = 1. 1 = _k=1^ (4)/((2k-1)) ((2k-1)()/(2)) Tunajua kuwa ((2k-1)()/(2)) = (-1)^k-1. 1 = _k=1^ (4)/((2k-1)) (-1)^k-1 1 = (4)/() ( (1)/(1)(-1)^0 + (1)/(3)(-1)^1 + (1)/(5)(-1)^2 + (1)/(7)(-1)^3 + ) 1 = (4)/() ( 1 - (1)/(3) + (1)/(5) - (1)/(7) + ) Zidisha pande zote mbili kwa ()/(4): ()/(4) = 1 - (1)/(3) + (1)/(5) - (1)/(7) + That's 4 down, 1 left today. What's next?
