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ProPatto16
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Homework Statement
Given the function f(x) = Acos(∏x/L), find its Fourier series
Homework Equations
Okay so, f(x) is even, so the Fourier series is given by:
f(x) = a0 + [itex]\sum[/itex]ancos(nx)
where a0 = 1/∏[itex]\int[/itex] f(x).dx with bounds ∏ and -∏
and an = 1/∏[itex]\int[/itex] f(x)cos(nx).dx with bounds ∏ and -∏
The Attempt at a Solution
okay, with range L
a0 = 2/L[itex]\int[/itex]f(x).dx with bounds L to 0
and an = 2/L[itex]\int[/itex]f(x)cos(n∏x/L).dx with bounds L to 0
starting with a0:
a0 = 2/L[itex]\int[/itex]f(x).dx with bounds L to 0
a0 = 2A/L[itex]\int[/itex]cos(∏x/L).dx
integrates to a sin, first sub in L gives sin∏ which = 0 and then sub the 0 gives sin0 which = 0 so therefore a0 = 0
now an:
and an = 2/L[itex]\int[/itex]f(x)cos(n∏x/L).dx with bounds L to 0
an = 2A/L[itex]\int[/itex]cos(∏x/L)cos(n∏x/L).dx
an = 2A/L[itex]\int[/itex]cos2(n∏x/L).dx
integral of cos2x = x/2 + (1/4)sin2x
so : an = 2A/L[n∏x/2L + (1/4)sin(2n∏x/L)] between L and 0
sub in 0 gives 0 + sin0, so only need to sub L,
an = 2A/L[n∏/2 + (1/4)sin(2n∏)]
an = An∏/L + (2A/L)sin(2n∏)
so then subbing back into expression for Fourier series with a0 = 0 gives:
f(x) = [itex]\sum[/itex] [An∏/L + (2A/L)sin(2n∏)](cos(nx))
doing this course by distance education and there's no textbook, and the online notes are fairly vague. so i just want to check I am on the right track or at least the right method of solving etc. sorry about the latex. not the best at that. thanks.