Convergence of Fourier Series

Some Definitions

A function f (x) defined on an interval [a, b] is said to be piecewise continuous if it is continuous on the interval except for a finite number of jump discontinuities (Figure 1).

A piecewise continuous function — Figure 1.

A function \(f\left( x \right)\) defined on an interval \(\left[ {a,b} \right]\) is said to be piecewise smooth if \(f\left( x \right)\) and its derivative are piecewise continuous.

Partial Sums of Fourier Series

We introduce the Fourier partial sum \({f_N}\left( x \right)\) of the function \(f\left( x \right)\) defined on the interval \(\left[ {-\pi, \pi} \right]\) as

\[{f_N}\left( x \right) = \frac{{{a_0}}}{2} + \sum\limits_{n = 1}^N {\left( {{a_n}\cos nx + {b_n}\sin nx} \right)} .\]

In complex form, the \(n\)th partial sum \({f_N}\left( x \right)\) of a function \(f\left( x \right)\) defined on the interval \(\left[ {-\pi, \pi} \right]\) is given by

\[{f_N}\left( x \right) = \sum\limits_{n = - N}^N {{c_n}{e^{inx}}} = \int\limits_{ - \pi }^\pi {\left( {\frac{1}{{2\pi }}\sum\limits_{n = - N}^N {{e^{in\left( {x - y} \right)}}} } \right)f\left( y \right)dy}.\]

Dirichlet Kernel

The function

\[{D_N}\left( x \right) = \sum\limits_{n = - N}^N {{e^{inx}}} = \frac{{\sin \left( {N + \frac{1}{2}} \right)x}}{{\sin \frac{x}{2}}}\]

is called the Dirichlet kernel. In Figure \(2\) we have graphed Dirichlet kernel for \(n = 10.\)

The Fourier partial sum of \(f\left( x \right)\) can be expressed through the Dirichlet kernel:

\[{f_N}\left( x \right) = \frac{1}{{2\pi }}\int\limits_{ - \pi }^\pi {{D_N}\left( {x - y} \right)f\left( y \right)dy} = \frac{1}{{2\pi }}\int\limits_{ - \pi }^\pi {{D_N}\left( y \right)f\left( {x - y} \right)dy} .\]

In this section, we consider three types of convergence: pointwise, uniform and \({L_2}\)-convergence.

Pointwise Convergence of Fourier Series

Let \(f\left( x \right)\) be a piecewise smooth function on the interval \(\left[ {-\pi, \pi} \right].\) Then for any \({x_0} \in \left[ { - \pi ,\pi } \right]\)

\[\lim\limits_{N \to \infty } {f_N}\left( {{x_0}} \right) = \begin{cases} f\left( {{x_0}} \right), \text{if}\,f\left( x \right)\,\;\text{is continuous on}\, \left[ { - \pi ,\pi } \right] \\[0.6em] \frac{{f\left( {{x_0} - 0} \right) + f\left( {{x_0} + 0} \right)}}{2}, \;\text{if}\,f\left( x \right)\,\text{has a jump discontinuity at}\, {{x_0}} \end{cases}.\]

where \({f\left( {{x_0} - 0} \right)}\) and \({f\left( {{x_0} + 0} \right)}\) represent the left limit and the right limit at the point \({x_0}.\)

Uniform Convergence of Fourier Series

A sequence of the partial sums \(\left\{ {{f_N}\left( x \right)} \right\}\) is said to be uniformly convergent to the function \(f\left( x \right),\) if the speed of convergence of the partial sums \({{f_N}\left( x \right)}\) does not depend on \(x\) (Figure \(3\)).

Uniformly convergent sequence of the partial sums — Figure 3.

We say that the Fourier series of a function \(f\left( x \right)\) converges uniformly to this function if

\[\lim\limits_{N \to \infty } \left[ {\max\limits_{x \in \left[ { - \pi ,\pi } \right]} \left| {f\left( x \right) - {f_N}\left( x \right)} \right|} \right] = 0.\]

Theorem.

The Fourier series of a \(2\pi\)-periodic continuous and piecewise smooth function converges uniformly.

Convergence of Fourier Series in \({L_2}\)-Norm

The space \({L_2}\left( { - \pi ,\pi } \right)\) is formed by those functions for which

\[\int\limits_{ - \pi }^\pi {{{\left| {f\left( x \right)} \right|}^2}dx} < \infty .\]

We will say that a function \(f\left( x \right)\) is square-integrable if it belongs to the space \({L_2}.\) If a function \(f\left( x \right)\) is square-integrable, then

\[\lim\limits_{N \to \infty } \frac{1}{{2\pi }}\int\limits_{ - \pi }^\pi {{{\left| {f\left( x \right) - {f_N}\left( x \right)} \right|}^2}dx} = 0,\]

that is the partial sums \({f_N}\left( x \right)\) converge to \(f\left( x \right)\) in the norm \({L_2}.\)

The uniform convergence implies both pointwise and \({L_2}\)-convergence. But the opposite is not true: the \({L_2}\)-convergence implies neither pointwise nor uniform convergence, and the pointwise convergence implies neither uniform nor \({L_2}\)-convergence.

Gibbs Phenomenon

If there is a jump discontinuity, the partial sum of the Fourier series has oscillations near the jump, which might increase the maximum of the partial sum above the function itself. This phenomenon is called Gibbs phenomenon. The amplitude of the "overshoot" at any jump point of a piecewise smooth function is about \(18\%\) larger (as \(n \to \infty\)) than the jump in the original function (Figure \(4\)).

Solved Problems

Example 1.

Calculate the integral \[\int\limits_{ - \pi }^\pi {{D_N}\left( z \right)dz}.\]

Solution.

It is known that

\[{f_N}\left( x \right) = \frac{1}{{2\pi }}\int\limits_{ - \pi }^\pi {{D_N}\left( {x - y} \right)f\left( y \right)dy} .\]

The Dirichlet kernel \({D_N}\left( x \right)\) is an even and \(2\pi\)-periodic function, so that we may write:

\[{f_N}\left( x \right) = \frac{1}{\pi }\int\limits_0^\pi {{D_N}\left( {x - y} \right)f\left( y \right)dy} .\]

Suppose that \({f_N}\left( x \right) = f\left( x \right) = 1\) and plug in this function into the formula above. We obtain

\[1 = \frac{1}{\pi }\int\limits_0^\pi {{D_N}\left( {x - y} \right)dy} .\]

Make the substitution: \(z = x - y.\) Then \(y = x - z,\) \(dy = dz.\) Find the new limits of integration: when \(y = 0,\) we get \(z = x,\) and when \(y = \pi,\) we have \(z = x - \pi.\) As a result, we have

\[1 = \frac{1}{\pi }\int\limits_x^{x - \pi } {{D_N}\left( z \right)\left( { - dz} \right)} \;\;\text{or}\;\; 1 = \frac{1}{\pi }\int\limits_{x - \pi}^x {{D_N}\left( z \right)dz} .\]

Due to periodicity of \({{D_N}\left( x \right)}\) we may write:

\[1 = \frac{1}{\pi }\int\limits_{ - \pi }^0 {{D_N}\left( z \right)dz} .\]

Hence,

\[\int\limits_{ - \pi }^\pi {{D_N}\left( z \right)dz} = 2\int\limits_{ - \pi }^0 {{D_N}\left( z \right)dz} = 2\pi .\]

There is another way to calculate this integral. Rewrite it in the form

\[I = \int\limits_{ - \pi }^\pi {{D_N}\left( z \right)dz} = 2\int\limits_0^\pi {{D_N}\left( z \right)dz} .\]

Since

\[{D_N}\left( z \right) = \frac{{\sin \left( {N + \frac{1}{2}} \right)z}}{{\sin \frac{z}{2}}} = 2\left( {\frac{1}{2} + \sum\limits_{n = 1}^N {\cos nz} } \right),\]

we can integrate this series term by term. Then

\[I = 2\int\limits_0^\pi {{D_N}\left( z \right)dz} = 4\int\limits_0^\pi {\left( {\frac{1}{2} + \sum\limits_{n = 1}^N {\cos nz} } \right)dz} = 4\left[ {\left. {\left( {\frac{z}{2} + \sum\limits_{n = 1}^N {\frac{{\sin nz}}{n}} } \right)} \right|_0^\pi } \right].\]

Here \(\sin {nz} = 0\) at \(z = 0, \pi.\) Consequently,

\[I = 4 \cdot \frac{\pi }{2} = 2\pi .\]

Example 2.

Let the function \[f\left( x \right) = {\frac{{\pi - x}}{2}}\] be defined on the interval \(\left[ {0,2\pi } \right].\) Find the Fourier series expansion of the function on the given interval and calculate the approximate value of \(\pi.\)

Solution.

We compute the Fourier coefficients.

\[{a_0} = \frac{1}{\pi }\int\limits_0^{2\pi } {f\left( x \right)dx} = \frac{1}{\pi }\int\limits_0^{2\pi } {\frac{{\pi - x}}{2}dx} = \frac{1}{{2\pi }}\left[ {\left. {\left( {\pi x - \frac{{{x^2}}}{2}} \right)} \right|_0^{2\pi }} \right] = 0.\]

For \(n \ge 1:\)

\[{a_n} = \frac{1}{\pi }\int\limits_0^{2\pi } {f\left( x \right)\cos nxdx} = \frac{1}{\pi }\int\limits_0^{2\pi } {\frac{{\pi - x}}{2}\cos nxdx} = \left. {\left( {\frac{{\pi - x}}{2}\frac{{\sin nx}}{{n\pi }}} \right)} \right|_0^{2\pi } + \frac{1}{{2\pi n}}\int\limits_0^{2\pi } {\sin nxdx} = 0 - \frac{1}{{2\pi n}}\left[ {\left. {\left( {\frac{{\cos nx}}{n}} \right)} \right|_0^{2\pi }} \right] = 0,\]

\[{b_n} = \frac{1}{\pi }\int\limits_0^{2\pi } {f\left( x \right)\sin nxdx} = \frac{1}{\pi }\int\limits_0^{2\pi } {\frac{{\pi - x}}{2}\sin nxdx} = \left. {\left( { - \frac{{\pi - x}}{2}\frac{{\cos nx}}{{n\pi }}} \right)} \right|_0^{2\pi } - \frac{1}{{2\pi n}}\int\limits_0^{2\pi } {\cos nxdx} = \left( { - \frac{{\frac{{\pi - 2\pi }}{2}\cos 2\pi n}}{{n\pi }} + \frac{{\frac{\pi }{2}\cos 0}}{{n\pi }}} \right) - \frac{1}{{2\pi n}}\left[ {\left. {\left( {\frac{{\sin nx}}{n}} \right)} \right|_0^{2\pi }} \right] = \frac{1}{{2n}} + \frac{1}{{2n}} = \frac{1}{n}.\]

Thus, the Fourier series expansion is

\[\frac{{\pi - x}}{2} = \sum\limits_{n = 1}^\infty {\frac{{\sin nx}}{n}} \;\; \text{for}\;\;x \in \left[ {0,2\pi } \right].\]

By setting \(x = {\frac{\pi }{2}},\) we obtain an alternating series for \({\frac{\pi }{4}}:\)

\[\frac{\pi }{4} = \sum\limits_{n = 1}^\infty {\frac{{\sin \frac{{n\pi }}{2}}}{n}} = 1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \ldots = \sum\limits_{n = 1}^\infty {\frac{{{{\left( { - 1} \right)}^{n + 1}}}}{{2n - 1}}} .\]

From here we find the following infinite series representation for \(\pi:\)

\[\pi = 4\sum\limits_{n = 1}^\infty {\frac{{{{\left( { - 1} \right)}^{n + 1}}}}{{2n - 1}}} = 4\left( {1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \ldots } \right).\]

See more problems on Page 2.