I want to understand Fourier series

Question

{ x, 0<x<π}

f(x)={ }

{π, π<x<2π}

Find the Fourier series for 2π period periodic function

Answer 1

Fourier series are used to change a function into a sum of a constant + cos functions + sin functions.

The cos, sin, and constant terms are orthogonal to one another and are normalized so that each integrates to 1 if the true function were only the constant, sin, or cos term.

This is why x = a₀ + the sum for n = 1 to infinity of a_n cos nx + b_n cos nx

Use I[a,b] for integration from a to b and E[a,b] for the evaluation from a to b

Then,

a₀ = 1/2π I[o, 2π ] f(x) dx

a_n = 1/π I[o, 2π ] f(x) cos(nx) dx , n > 0

b_n = 1/π I[o, 2π ] f(x) sin(nx) dx , n > 0

f(x) = x , 0 <= x <= π

f(x) = π , π < x <= 2π

Then,

a₀ = 1/2π (I[o, π ] x dx + I[π, 2π] π dx) =

1/2π(x²/2 E[0, π] + πx E[π, 2π]) =

1/2π(π²/2 + π(2π - π)) =

1/2π(3π²/2) =

3π/4

a_n = 1/π I[o, π ] x cos(nx) dx + 1/π I[o, π ] π cos(nx) dx =

1/π (x/n sin (nx) + 1/n² cos (nx)) E[o, π ] + 1/πn sin(nx) E[π, 2π]) =

1/(πn²) (cos n π - 1) =

-2/(πn²) , n odd

b_n = 1/π I[o, π ] x sin(nx) dx + 1/π I[o, π ] π sin(nx) dx =

1/π ((-x/n cos nx + 1/n² sin(nx)) E[0,π ] - π/n cos(nx) E[π, 2π]) =

1/n(-cos n π -cos 2n π + cos n π) =

-1/n, all n

f(x) = 3π/4 + ¨∑-2/(π(2n-1)^2) cos(2n-1)x + ¨∑-1/n sin nx , where n is summed from 1 to infinity