📂Fourier Analysis

The constant term of the Fourier series is equal to the average of one period of the function.

Theorem

The constant term of the Fourier series of a function with period $2L$, namely $f$, equals the average of one period of the function $f$.

Proof

By definition

The integral over one period of $f(t)$ is

$$\dfrac{1}{2L}\int_{-L}^{L} f(t)dt$$

According to the definition of the Fourier coefficients, this is equal to $\dfrac{1}{2}a_{0}$. Therefore, the integral over one period of $f(t)$ is the same as the constant term of the Fourier series of $f(t)$.

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Direct calculation

We can also prove the above statement by direct calculation. The Fourier series of $f(t)$ is

$$f(t)=\dfrac{1}{2}a_{0} +\sum\limits_{n=1}^{\infty} \left( a_{n} \cos \frac{n\pi}{L}t + b_{n} \sin \frac{n\pi}{L}t \right)$$

Calculating the average over one period of $f(t)$ gives

$$\frac{1}{2L}\int_{-L}^{L} f(t) dt= \frac{1}{2L}\int_{-L}^{L} \dfrac{a_{0}}{2} dt+\sum \limits_{n=1}^{\infty} \left( a_{n} \frac{1}{2L}\int_{-L}^{L} \cos \frac{n\pi}{L}t dt + b_{n}\frac{1}{2L}\int_{-L}^{L} \sin \frac{n\pi}{L}tdt \right)$$

Since the average of one period of a trigonometric function is $0$,

$$\frac{1}{2L}\int_{-L}^{L} f(t) dt= \frac{1}{2L}\int_{-L}^{L} \dfrac{a_{0}}{2} dt=\dfrac{a_{0}}{2}$$

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