Fourier Series | Basic Physics for IIT JAM PDF Download

Consider a periodic function f(x) with a period L and let us calculate the Fourier transform of it. We define a new function f₀(x)=f(x) in the [0, L] interval and zero otherwise. Then:

Apply Fourier transform:

where f_n are called Fourier coefficients:

We can see that the Fourier transform is zero for . For  it is equal to a delta function times a 2π multiple of a Fourier series coefficient. The delta functions structure is given by the period L of the function f(x). All the information that is stored in the answer is inside the f_n coefficients, so those are the only ones that we need to calculate and store.
The function f(x) is calculated from the f_n coefficients by applying the inverse Fourier transform to the final result of as follows:

The expansion is called a Fourier series. It is given by the Fourier coefficients f_n. The equation provides the relation between a Fourier transform and a Fourier series.
For example for f(x) = sin(x), the only nonzero Fourier coefficients for L=2π are f_-1 = i/2 and f₁ =-i/2  . The Fourier transform then is:

For f(x) = 1 the only nonzero Fourier coefficient is f₀=1, the Fourier transform then is:

For f(x) = e^3ix the only nonzero Fourier coefficient for L=2π is f₃= 1, the Fourier transform then is:

For the Fourier coefficients for L=2\pi are all equal to f_n = 1/2π and the Fourier transform is:

Note: if we start from, for simplicity on an interval [-π,π]:

To calculate the Fourier coefficients f_n, we can just multiply both sides of by e^-imx and integrate:

so