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Consider a periodic function f(x) with a period L and let us calculate the Fourier transform of it. We define a new function f_{0}(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 2Ï€ 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

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=2Ï€ are f_{-1} = *i*/2 and f_{1} =-*i*/2 . The Fourier transform then is:

For f(x) = 1 the only nonzero Fourier coefficient is f_{0}=1, the Fourier transform then is:

For f(x) = e^{3ix} the only nonzero Fourier coefficient for L=2Ï€ is f_{3}= 1, the Fourier transform then is:

For the Fourier coefficients for L=2\pi are all equal to f_{n} = 1/2Ï€ and the Fourier transform is:

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

To calculate the Fourier coefficients f_{n}, we can just multiply both sides of by e^{-}* ^{imx}* and integrate:

so

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