What is the Fourierseries of the following equation?

#f(x)=|sin(x)|#

1 Answer
Jul 20, 2018

Answer:

#abs(sin x)= 2/pi - 4/pisum_(n=1)^infty (cos ( 2 n x))/( 4n^2 -1) #

Explanation:

graph{y = abs(sin x) [-10, 10, -1, 3]}

The function has period #pi# and is even, ie #f(x) = f("-"x)#, which means a cosine series. Using this definition:

  • #{(a_0 = 1/L int_0^(2L)f(x )dx),(a_n = 1/L int_0^(2L)f(x )cos((npix )/L)dx), (f(x)=1/2a_0+sum_(n=1)^infty a_n cos((npi x)/L)),(2L = pi):} #

#a_0 = 2/pi int_0^(pi )sin x dx = 4/pi#

#a_n = 2/pi int_0^(pi) sin x cos2nx \ dx #

Using this ID:

  • #2\sin \theta \cos \varphi =\sin(\theta +\varphi )+\sin(\theta -\varphi )#

#:. a_n = 1/pi int_0^(pi) sin ( 2n+1)x + sin(1 - 2n)x \dx #

#= 1/pi [ -(cos ( 2n+1)x)/(2n+1) + (cos(2n - 1)x)/(2n - 1) ]_0^(pi)#

#= 1/pi [ ( (2n+1)cos(2n - 1)x -(2n-1)cos ( 2n+1)x ) /(4n^2 -1) \]_0^(pi) #

#2n-1# and #2n+1 # are both odd, #implies cos(2n - 1)pi = cos(2n + 1)pi = -1#

#= 1/pi\{ [ ( (2n+1)(-1) -(2n-1)(-1) ) /(4n^2 -1) \] - [ ( (2n+1) -(2n-1)) /(4n^2 -1) \]\}#

#= 1/pi\{ [ ( -2n-1 + 2n-1 ) /(4n^2 -1) \] - [ ( 2n+1 - 2n+1 ) /(4n^2 -1) \]\}#

#= - (4)/(pi(4n^2 -1)) #

From the definition:

#abs(sin x)= 2/pi - 4/pi sum_(n=1)^infty (cos ( 2 n x))/( 4n^2 -1) #

Looks like this for first few terms:

graph{(y - 2/pi + 4/pi ( ( cos (2x) )/3+ ( cos (4 x) ) / 15 + ( cos(6x) ) / 35 ))(y - |sin x|) = 0 [0, 5, -0.1 , 1.25]}