need help
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BigGlenntheHeavy
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Student's poser doesn't look that hard, however I don't understand exactly what he is aiming at, any takers?
\(\displaystyle f(\theta) \ = \ sin(\theta)\)
\(\displaystyle f(\theta+2\pi) \ = \ sin(\theta+2\pi)\)
\(\displaystyle Hence, \ f(\theta) \ = \ f(\theta+2\pi), \ something \ on \ this \ order?\)
\(\displaystyle f(\theta) \ = \ sin(\theta)\)
\(\displaystyle f(\theta+2\pi) \ = \ sin(\theta+2\pi)\)
\(\displaystyle Hence, \ f(\theta) \ = \ f(\theta+2\pi), \ something \ on \ this \ order?\)
mmm4444bot
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Ted S is a taker, at another site where this exercise was posted, 1 hour ago. Ted writes:
you 1st extend F to be an odd function on [-? , ?] so that b_n = 2/? int over x in [0,?] of { F(x) sin nx dx} = 2 /? { 4 [(1/n²) sin(nx) - (x/n)cos(nx) ] - (1/n) cos (nx) } ..evaluate at ? and 0....you certainly can do that...? { n = 1,2,3,..} b_n sin (nx) = F(x)
So, perhaps the half-range sine series is just the Fourier series for the odd periodic extension of f(x).
I'm not sure.
you 1st extend F to be an odd function on [-? , ?] so that b_n = 2/? int over x in [0,?] of { F(x) sin nx dx} = 2 /? { 4 [(1/n²) sin(nx) - (x/n)cos(nx) ] - (1/n) cos (nx) } ..evaluate at ? and 0....you certainly can do that...? { n = 1,2,3,..} b_n sin (nx) = F(x)
So, perhaps the half-range sine series is just the Fourier series for the odd periodic extension of f(x).
I'm not sure.