https://www.math24.net/fourier-series-definition-typical-examples/
Baron Jean Baptiste Joseph Fourier (1768−1830) introduced the idea that any periodic function can be represented by a series of sines and cosines which are harmonically related.
Fig.1 Baron Jean Baptiste Joseph Fourier (1768−1830)
To consider this idea in more detail, we need to introduce some definitions and common terms.
A function f(x) is said to have period P if f(x+P)=f(x) for all x. Let the function f(x) has period 2π. In this case, it is enough to consider behavior of the function on the interval [−π,π].
Suppose that the function f(x) with period 2π is absolutely integrable on [−π,π] so that the following so-called Dirichlet integral is finite:π∫−π|f(x)|dx<∞;
Suppose also that the function f(x) is a single valued, piecewise continuous (must have a finite number of jump discontinuities), and piecewise monotonic (must have a finite number of maxima and minima).If the conditions 1 and 2 are satisfied, the Fourier series for the function f(x) exists and converges to the given function (see also the Convergence of Fourier Series page about convergence conditions.)
At a discontinuity x0, the Fourier Series converges to
limε→012[f(x0−ε)−f(x0+ε)].
The Fourier series of the function f(x) is given by
f(x)=a02+∞∑n=1{ancosnx+bnsinnx},
where the Fourier coefficients a0, an, and bn are defined by the integrals
a0=1ππ∫−πf(x)dx,an=1ππ∫−πf(x)cosnxdx,bn=1ππ∫−πf(x)sinnxdx.
Sometimes alternative forms of the Fourier series are used. Replacing an and bn by the new variables dn and φn or dn and θn, where
dn=√a2n+b2n,tanφn=anbn,tanθn=bnan,
we can write:
f(x)=a02+∞∑n=1dnsin(nx+φn)orf(x)=a02+∞∑n=1dncos(nx+θn).
The Fourier series expansion of an even function f(x) with the period of 2π does not involve the terms with sines and has the form:
f(x)=a02+∞∑n=1ancosnx,
where the Fourier coefficients are given by the formulas
a0=2ππ∫0f(x)dx,an=2ππ∫0f(x)cosnxdx.
Accordingly, the Fourier series expansion of an odd 2π-periodic function f(x) consists of sine terms only and has the form:
f(x)=∞∑n=1bnsinnx,
where the coefficients bn are
bn=2ππ∫0f(x)sinnxdx.
Below we consider expansions of 2π-periodic functions into their Fourier series, assuming that these expansions exist and are convergent.
Click a problem to see the solution.
Let the function f(x) be 2π-periodic and suppose that it is presented by the Fourier series:
f(x)=a02 + ∞∑n=1{ancosnx+bnsinnx}
Calculate the coefficients a0, an, and bn.
Find the Fourier series for the square 2π-periodic wave defined on the interval [−π,π]:
f(x) = {0,if−π≤x≤01,if0<x≤π.
Find the Fourier series for the sawtooth wave defined on the interval [−π,π] and having period 2π.
Let f(x) be a 2π-periodic function such that f(x)=x2 for x∈[−π,π]. Find the Fourier series for the parabolic wave.
Find the Fourier series for the triangle wave
f(x) = {π2+x,if−π≤x≤0π2−x,if0<x≤π,
defined on the interval [−π,π].
Find the Fourier series for the function
f(x) = ⎧⎪⎨⎪⎩−1,if−π≤x≤−π20,if−π2<x≤π21,ifπ2<x≤π,
defined on the interval [−π,π].
Let the function f(x) be 2π-periodic and suppose that it is presented by the Fourier series:
f(x)=a02 + ∞∑n=1{ancosnx+bnsinnx}
Calculate the coefficients a0, an, and bn.
Solution.
To define a0, we integrate the Fourier series on the interval [−π,π]:
π∫−πf(x)dx=πa0+∞∑n=1⎡⎢⎣anπ∫−πcosnxdx+bnπ∫−πsinnxdx⎤⎥⎦
For all n>0,
π∫−πcosnxdx=(sinnxn)∣∣∣π−π=0andπ∫−πsinnxdx=(−cosnxn)∣∣π−π=0.
Therefore, all the terms on the right of the summation sign are zero, so we obtain
π∫−πf(x)dx=πa0ora0=1ππ∫−πf(x)dx.
In order to find the coefficients an, we multiply both sides of the Fourier series by cosmx and integrate term by term:
π∫−πf(x)cosmxdx=a02π∫−πcosmxdx+∞∑n=1⎡⎢⎣anπ∫−πcosnxcosmxdx+bnπ∫−πsinnxcosmxdx⎤⎥⎦.
The first term on the right side is zero. Then, using the well-known trigonometric identities, we have
π∫−πsinnxcosmxdx=12π∫−π[sin(n+m)x+sin(n−m)x]dx=0,
π∫−πcosnxcosmxdx=12π∫−π[cos(n+m)x+cos(n−m)x]dx=0,
if m≠n.
In case when m=n, we can write:
π∫−πsinnxcosmxdx=12π∫−π[sin2mx+sin0]dx,⇒π∫−πsin2mxdx=12[(−cos2mx2m)∣∣∣π−π]=14m[−cos(2mπ)
+cos(2m(−π))
]=0;
π∫−πcosnxcosmxdx=12π∫−π[cos2mx+cos0]dx,⇒π∫−πcos2mxdx=12[(sin2mx2m)∣∣∣π−π+2π]=14m[sin(2mπ)−sin(2m(−π))]+π=π.
Thus,
π∫−πf(x)cosmxdx=amπ,⇒am=1ππ∫−πf(x)cosmxdx,m=1,2,3,…
Similarly, multiplying the Fourier series by sinmx and integrating term by term, we obtain the expression for bm:
bm=1ππ∫−πf(x)sinmxdx,m=1,2,3,…
Rewriting the formulas for an, bn, we can write the final expressions for the Fourier coefficients:
an=1ππ∫−πf(x)cosnxdx,bn=1ππ∫−πf(x)sinnxdx.
Find the Fourier series for the square 2π-periodic wave defined on the interval [−π,π]:
f(x) = {0,if−π≤x≤01,if0<x≤π.
Solution.
First we calculate the constant a0:
a0=1ππ∫−πf(x)dx=1ππ∫01dx=1π⋅π=1.
Find now the Fourier coefficients for n≠0:
an=1ππ∫−πf(x)cosnxdx=1ππ∫01⋅cosnxdx=1π[(sinnxn)∣∣∣π0]=1πn⋅0=0,
bn=1ππ∫−πf(x)sinnxdx=1ππ∫01⋅sinnxdx=1π[(−cosnxn)∣∣π0]=−1πn⋅(cosnπ−cos0)=1−cosnππn.
As cosnπ=(−1)n, we can write:
bn=1−(−1)nπn.
Thus, the Fourier series for the square wave is
f(x)=12+∞∑n=11−(−1)nπnsinnx.
We can easily find the first few terms of the series. By setting, for example, n=5, we get
f(x)=12+1−(−1)πsinx+1−(−1)22πsin2x+1−(−1)33πsin3x+1−(−1)44πsin4x+1−(−1)55πsin5x+…=12+2πsinx+23πsin3x+25πsin5x+…
The graph of the function and the Fourier series expansion for n=10 is shown below in Figure 2.
Figure 2, n = 10
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Problems 1-2
