Fourier Series: Definition, Formula, Solved Examples

A Fourier series of a function f(x) with period 2π is an infinite trigonometric series given by f(x) = a₀ + ∑_n=1[ a_ncos(nx) + b_n sin(nx) ] if it exists. The constants a₀, a_n, b_n are called Fourier coefficients of f(x). In this article, let us learn Fourier series along with its formula and examples.

Before we study Fourier series, let us learn the following.

The set {1, cosx, sinx, cos2x, sin2x, …} containing functions of period 2π, forms the trigonometric system. Using this system, let us now define the trigonometric series below.

Trigonometric Series

A series obtained from the trigonometric system and of the form

a₀ + a₁cosx + b₁sinx + a₂cos2x + b₂sin2x + a₃cos3x + b₃sin3x + …

= $a_0 + \sum_{n=1}^\infty \left[ a_n \cos(nx) + b_n \sin(nx) \right]$

where a₀, a₁, b₁, a₂, b₂, … are constants, is called a trigonometric series. The constants a₀, a_n, b_n for n = 1, 2, 3, … are called the coefficients of the series.

Note: All the terms in a trigonometric series have period 2π. Thus, if the series converges, then the sum will be a function of period 2π. This gives the idea for the formation of a Fourier series.

Definition of Fourier Series

Let f(x) be a periodic function with period 2π. The Fourier series representation of f(x) is an infinite trigonometric series given by

$\boxed{f(x) = a_0 + \sum_{n=1}^\infty \left[ a_n \cos(nx) + b_n \sin(nx) \right]}$ …(I)

provided the series converges.

Coefficient of Fourier Series

The constants a₀, a_n, b_n for n = 1, 2, 3, … are called Fourier coefficients of f(x) and given by the Euler’s formula:

$a_0=\dfrac{1}{2\pi}\int_{-\pi}^\pi f(x) ~ dx$ $a_n=\dfrac{1}{\pi} \int_{-\pi}^\pi f(x) \cos(nx) ~ dx$, n = 1, 2, … $b_n=\dfrac{1}{\pi} \int_{-\pi}^\pi f(x) \sin(nx) ~ dx$, n = 1, 2, …

Proof:

Step 1:

Assuming the termwise integration is possible in the Equation (I), and taking the integration over the interval [-π, π] in both sides of Equ (I), we get that

$\int_{-\pi}^\pi f(x) ~ dx$

= $\int_{-\pi}^\pi a_0 ~ dx$ + $\sum_{n=1}^\infty [ a_n \int_{-\pi}^\pi \cos(nx) ~ dx$ $+ b_n \int_{-\pi}^\pi \sin(nx) ~ dx ]$ as termwise integration is allowed.

= $2a_0 \int_{0}^\pi dx$ + $\sum_{n=1}^\infty [ 2a_n \int_{0}^\pi \cos(nx) ~ dx$ $+ 0]$

= 2πa₀ + 0

= 2πa₀.

This implies that

$\boxed{a_0=\dfrac{1}{2\pi}\int_{-\pi}^\pi f(x) ~ dx}$

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Step 2:

Multiplying Equation (I) by cos(mx) and integrating over [-π, π] it follows that

$\int_{-\pi}^\pi f(x) \cos(mx) ~ dx$

= $\int_{-\pi}^\pi a_0 \cos(mx) ~ dx$ + $\sum_{n=1}^\infty [ a_n \int_{-\pi}^\pi \cos(nx) \cos(mx)~ dx$ $+ b_n \int_{-\pi}^\pi \sin(nx) \cos(mx) ~ dx ]$

= $2a_0 \int_{0}^\pi \cos(mx)~dx$ + $\sum_{n=1}^\infty [ 2a_n \int_{0}^\pi \cos(nx) \cos(mx) ~ dx$ $+ 0]$, because sin(nx) cos(mx) is an odd function.

Now, since ∫₀^π cos(nx) dx = 0, from above we obtain that

$\int_{-\pi}^\pi f(x) \cos(nx) ~ dx$ = 2a0 × 0 + an ∫0π 2cos2(nx) dx ⇒ ∫-ππ f(x) cos(nx) dx = an ∫0π {1+cos(2nx)} dx as we know 2cos2θ = 1+cos2θ. ⇒ $\int_{-\pi}^\pi f(x) \cos(nx) dx$ $= a_n \left[ x+\dfrac{\sin(2nx)}{2n} \right]_0^\pi$ $= a_n \pi$

Therefore,

$\boxed{a_n=\dfrac{1}{\pi} \int_{-\pi}^\pi f(x) \cos(nx) ~ dx}$

Step 3:

Multiplying Equation (I) by sin(mx) and integrating over the integral [-π, π], we get that

$\int_{-\pi}^\pi f(x) \sin(mx) ~ dx$

= $\int_{-\pi}^\pi a_0 \sin(mx) ~ dx$ + $\sum_{n=1}^\infty [ a_n \int_{-\pi}^\pi \cos(nx) \sin(mx)~ dx$ $+ b_n \int_{-\pi}^\pi \sin(nx) \sin(mx) ~ dx ]$

Now, note that

1. $\int_{-\pi}^\pi \sin(mx)~dx=0$ as sin(mx) is an odd function.
2. $\int_{-\pi}^\pi \cos(nx) \sin(mx)~dx=0$ as cos(nx) and sin(mx) are orthogonal.
3. $\int_{-\pi}^\pi \sin(nx) \sin(mx)~dx=0$ for all m≠n, because sin(mx) and sin(nx) are orthogonal for m≠n.

Therefore, from above we obtain that

$\int_{-\pi}^\pi f(x) \sin(nx) ~ dx$ = bn ∫-ππ sin2(nx) dx = bn ∫0π 2sin2(nx) dx = bn ∫0π [1-cos(2nx)] dx = bnπ.

Therefore,

$\boxed{b_n=\dfrac{1}{\pi} \int_{-\pi}^\pi f(x) \sin(nx) ~ dx}$

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