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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) = a

_{0}+ [itex]\sum[/itex]a

_{n}cos(nx)

where a

_{0}= 1/∏[itex]\int[/itex] f(x).dx with bounds ∏ and -∏

and a

_{n}= 1/∏[itex]\int[/itex] f(x)cos(nx).dx with bounds ∏ and -∏

## The Attempt at a Solution

okay, with range L

a

_{0}= 2/L[itex]\int[/itex]f(x).dx with bounds L to 0

and a

_{n}= 2/L[itex]\int[/itex]f(x)cos(n∏x/L).dx with bounds L to 0

starting with a

_{0}:

a

_{0}= 2/L[itex]\int[/itex]f(x).dx with bounds L to 0

a

_{0}= 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 a

_{0}= 0

now a

_{n}:

and a

_{n}= 2/L[itex]\int[/itex]f(x)cos(n∏x/L).dx with bounds L to 0

a

_{n}= 2A/L[itex]\int[/itex]cos(∏x/L)cos(n∏x/L).dx

a

_{n}= 2A/L[itex]\int[/itex]cos

^{2}(n∏x/L).dx

integral of cos

^{2}x = x/2 + (1/4)sin2x

so : a

_{n}= 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,

a

_{n}= 2A/L[n∏/2 + (1/4)sin(2n∏)]

a

_{n}= An∏/L + (2A/L)sin(2n∏)

so then subbing back into expression for fourier series with a

_{0}= 0 gives:

f(x) = [itex]\sum[/itex] [An∏/L + (2A/L)sin(2n∏)](cos(nx))

doing this course by distance education and theres no textbook, and the online notes are fairly vague. so i just wanna check im on the right track or at least the right method of solving etc. sorry about the latex. not the best at that. thanks.