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trap101
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Find the Fourier SIne Series for f(x) = x on -L < x < L (Full Fourier)
Ok, so my issue is in calculating the coefficients for the sine and cosine parts, more so an interpretation. So I have calulated the sine and cosine series to this point:
let An: Cosine series Bn: sine series:
An = l/(n[itex]\pi[/itex])) [ sin(n[itex]\pi[/itex] + sin(-n[itex]\pi[/itex])] - l/(n2[itex]\pi[/itex]2) [sin(n[itex]\pi[/itex]) - sin(-n[itex]\pi[/itex])]
Bn = l/(n[itex]\pi[/itex])) [ cos(n[itex]\pi[/itex] + cos(-n[itex]\pi[/itex])] - l/(n2[itex]\pi[/itex]2) [cos(n[itex]\pi[/itex]) - cos(-n[itex]\pi[/itex])]
My questions are these: How do I interpret the relationship between cos(n[itex]\pi[/itex]) and cos(-n[itex]\pi[/itex])? and the same with the sine values?
My take is this: for the sine function, since all of the sine coeffcients have a [itex]\pi[/itex] in them then it doesn't matter what value I select for "n", they will all go to 0.
As for the cosine: Since cos(n[itex]\pi[/itex]) and cos(-n[itex]\pi[/itex]) end up going to the same value, just in opposite directions, I will have 2cos(ncos(n[itex]\pi[/itex]) since cos(n[itex]\pi[/itex]) = cos(-n[itex]\pi[/itex]) in values.
Ok, so my issue is in calculating the coefficients for the sine and cosine parts, more so an interpretation. So I have calulated the sine and cosine series to this point:
let An: Cosine series Bn: sine series:
An = l/(n[itex]\pi[/itex])) [ sin(n[itex]\pi[/itex] + sin(-n[itex]\pi[/itex])] - l/(n2[itex]\pi[/itex]2) [sin(n[itex]\pi[/itex]) - sin(-n[itex]\pi[/itex])]
Bn = l/(n[itex]\pi[/itex])) [ cos(n[itex]\pi[/itex] + cos(-n[itex]\pi[/itex])] - l/(n2[itex]\pi[/itex]2) [cos(n[itex]\pi[/itex]) - cos(-n[itex]\pi[/itex])]
My questions are these: How do I interpret the relationship between cos(n[itex]\pi[/itex]) and cos(-n[itex]\pi[/itex])? and the same with the sine values?
My take is this: for the sine function, since all of the sine coeffcients have a [itex]\pi[/itex] in them then it doesn't matter what value I select for "n", they will all go to 0.
As for the cosine: Since cos(n[itex]\pi[/itex]) and cos(-n[itex]\pi[/itex]) end up going to the same value, just in opposite directions, I will have 2cos(ncos(n[itex]\pi[/itex]) since cos(n[itex]\pi[/itex]) = cos(-n[itex]\pi[/itex]) in values.