# a) showing details of the work, write an expansion in Fourier series of the signal f(x) which is assumed to have the period 2`pi` f(x) = "Please refer to the attached image" Use Integration by...

a) showing details of the work, write an expansion in Fourier series of the signal f(x) which is assumed to have the period 2`pi`

f(x) = "Please refer to the attached image"

Use Integration by parts show detailed work for each calculation of a0,an,bn. also show the values of all coefficients a1, b1, a2, b2, a3, b3, a4, b4, a5, b5

b) Graph both the f(x) and the partial sums up to that including cos 5x and sin 5x, for the interval −4`pi` `<=` x `<=` 4`pi` . Compare the graphs. (you can use Matlab for plots)

*print*Print*list*Cite

This function is odd, thus its Fourier expansions contains only `sin(nx)` terms, i.e. `a_n=0, ngt=0.` This expansion has the form `sum_(n=1)^(oo) b_n sin(nx)` where `b_n = 1/pi int_(-pi)^(pi) f(x) sin(nx) dx.`

Find these coefficients:

`b_n = 1/pi int_(-pi)^(pi) f(x) sin(nx) dx = 2/pi int_0^(pi) f(x) sin(nx) dx =`

`= 2/pi int_0^(pi/2) f(x) sin(nx) dx + 2/piint_(pi/2)^(pi) f(x) sin(nx) dx) =`

`= 2/pi int_0^(pi/2) x sin(nx) dx + 2/piint_(pi/2)^(pi) pi/2 sin(nx) dx.`

The second integral is obviously

`-(cos(nx))|_(x=pi/2)^pi = -1/n (cos(n pi)-cos((n pi)/2)).`

To find the second, use integration by parts:

`2/pi int_0^(pi/2) x sin(nx) dx =`

`= |u=x, du=dx, dv=sin(nx)dx, v=-1/n cos(nx)| =`

`= -2/(n pi) (x cos(nx))|_(x=0)^(pi/2) + 2/(n pi) int_0^(pi/2) cos(nx) dx =`

`= -1/n cos((n pi)/2) + 2/(n^2 pi) (sin(nx))|_(x=0)^(pi/2) =`

`= 2/(n^2 pi)sin((n pi)/2)-1/n cos((n pi)/2).`

This way `b_n =2/(n^2 pi)sin((n pi)/2)-1/n cos(n pi).` Therefore `b_1=1+2/pi, b_3=1/3 -2/(9 pi), b_5=1/5+2/(25 pi).`

The graphs are attached (the function is in blue, the approximation is in green). They are not so close but are somewhat similar. Note that on `(-4pi,4pi)` the function is `2pi` -periodic.