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...

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...

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.