- #1
stunner5000pt
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Find the Fourier Series for f(x) =1 on the interval 0<= x <= pi in terms of [itex] \phi_{n} = \sin{nx} [/itex]. Bu integrating this series find a convergent series for the function g(x) =x on this interval assuming the set {sin nx} is complete
Ok for the FOurier Coefficients
[tex] c_{n} = \frac{\int_{0}^{\pi} f \phi \rho dx}{\int_{0}^{\pi} \phi^2} dx [/tex]
this is how it is in my test
rho is suppsod to be the weight
ok for the numerator
[tex] \int_{0}^{\pi} \sin{nx} dx = \frac{1}{n} [- \cos{nx}]_{0}^{\pi} = (-1)^n + 1 [/tex]
for hte denominator
[tex] \int_{0}^{\pi} (\sin{nx})^2 dx = \frac{\pi}{2} - \frac{\sin{2n \pi}}{4} =\frac{\pi}{2} [/tex]
so the Fourier series is
[tex] \frac{2}{\pi} \sum_{n=1}^{\infty} \frac{\sin{nx}}{n} (-1^n + 1) [/tex]
what do they mean by integrate the series? Does it mena i should integrate the argument of this sum? ANd how would one find a convergent series for the function g(x) =x??
Please help!
Ok for the FOurier Coefficients
[tex] c_{n} = \frac{\int_{0}^{\pi} f \phi \rho dx}{\int_{0}^{\pi} \phi^2} dx [/tex]
this is how it is in my test
rho is suppsod to be the weight
ok for the numerator
[tex] \int_{0}^{\pi} \sin{nx} dx = \frac{1}{n} [- \cos{nx}]_{0}^{\pi} = (-1)^n + 1 [/tex]
for hte denominator
[tex] \int_{0}^{\pi} (\sin{nx})^2 dx = \frac{\pi}{2} - \frac{\sin{2n \pi}}{4} =\frac{\pi}{2} [/tex]
so the Fourier series is
[tex] \frac{2}{\pi} \sum_{n=1}^{\infty} \frac{\sin{nx}}{n} (-1^n + 1) [/tex]
what do they mean by integrate the series? Does it mena i should integrate the argument of this sum? ANd how would one find a convergent series for the function g(x) =x??
Please help!
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