Convergence of Fourier Series

Solved Problems

Example 3.

Prove that the Fourier series of the function \(f\left( x \right) = {x^2}\) converges uniformly to \(f\left( x \right)\) on the interval \(\left[ {-\pi, \pi} \right].\)

Solution.

The Fourier series expansion of \(f\left( x \right) = {x^2}\) on the interval \(\left[ {-\pi, \pi} \right]\) is given by

\[f\left( x \right) = {x^2} = \frac{{{\pi ^2}}}{3} + 4\sum\limits_{n = 1}^\infty {\frac{{{{\left( { - 1} \right)}^n}}}{{{n^2}}}\cos nx} .\]

(See Example \(4\) in the section Definition of Fourier Series and Typical Examples.)

The partial sums are defined as

\[{f_N}\left( x \right) = \frac{{{\pi ^2}}}{3} + 4\sum\limits_{n = 1}^N {\frac{{{{\left( { - 1} \right)}^n}}}{{{n^2}}}\cos nx} .\]

Then

\[\left| {f\left( x \right) - {f_N}\left( x \right)} \right| = \left| {4\sum\limits_{n = 1}^\infty {\frac{{{{\left( { - 1} \right)}^n}}}{{{n^2}}}\cos nx} - 4\sum\limits_{n = 1}^N {\frac{{{{\left( { - 1} \right)}^n}}}{{{n^2}}}\cos nx} } \right| = \left| {4\sum\limits_{n = N + 1}^\infty {\frac{{{{\left( { - 1} \right)}^n}}}{{{n^2}}}\cos nx} } \right| \le 4\sum\limits_{n = N + 1}^\infty {\left| {\frac{{{{\left( { - 1} \right)}^n}}}{{{n^2}}}\cos nx} \right|} \le 4\sum\limits_{n = N + 1}^\infty {\frac{1}{{{n^2}}}} .\]

The last sum converges to zero as \(N \to \infty.\) Indeed, applying the integral test, we find that

\[\lim\limits_{N \to \infty } \sum\limits_{n = N + 1}^\infty {\frac{1}{{{n^2}}}} = \lim\limits_{N \to \infty } \int\limits_{N + 1}^\infty {\frac{{dx}}{{{x^2}}}} = \lim\limits_{N \to \infty } \left[ {\left. {\left( { - \frac{1}{x}} \right)} \right|_{N + 1}^\infty } \right] = \lim\limits_{N \to \infty } \frac{1}{{N + 1}} = 0.\]

Thus,

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

which implies that the Fourier series of \(f\left( x \right) = {x^2}\) converges uniformly.

Example 4.

Prove that the Fourier series of the function \(f\left( x \right) = x\) converges to \(f\left( x \right)\) in the norm \({L_2}\) on the interval \(\left[ {-\pi, \pi} \right].\)

Solution.

The Fourier series of the function \(f\left( x \right) = x\) on the interval \(\left[ {-\pi, \pi} \right]\) is given by

\[f\left( x \right) = x = 2\sum\limits_{n = 1}^\infty {\frac{{{{\left( { - 1} \right)}^{n + 1}}}}{n}\sin nx} .\]

(See Example \(3\) on the page Definition of Fourier Series and Typical Examples.)

Then the partial sums are defined by the expression

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

Calculate the limit

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

where \(\left\| {f\left( x \right)} \right\|\) denotes \({L_2}\)-norm of the function \({f\left( x \right)}\).

We find the norm \(\left\| {f\left( x \right) - {f_N}\left( x \right)} \right\|:\)

\[\left| {f\left( x \right) - {f_N}\left( x \right)} \right| = {\left[ {\frac{1}{{2\pi }}{{\int\limits_{ - \pi }^\pi {\left| {\sum\limits_{n = 1}^\infty {\frac{{2{{\left( { - 1} \right)}^{n + 1}}}}{n}\sin nx} - \sum\limits_{n = 1}^N {\frac{{2{{\left( { - 1} \right)}^{n + 1}}}}{n}\sin nx} } \right|} }^2}dx} \right]}^{\frac{1}{2}} = \left| {\sum\limits_{n = N + 1}^\infty {\frac{{2{{\left( { - 1} \right)}^{n + 1}}}}{n}\sin nx} } \right|.\]

Using the triangle inequality \(\left\| {f + g} \right\| \) \(\le \left\| f \right\| + \left\| g \right\|\) for functions in \({L_2}\)-space, we can write:

\[\left| {f\left( x \right) - {f_N}\left( x \right)} \right| = \left| {\sum\limits_{n = N + 1}^\infty {\frac{{2{{\left( { - 1} \right)}^{n + 1}}}}{n}\sin nx} } \right| \le \sum\limits_{n = N + 1}^\infty {\left| {\frac{{2{{\left( { - 1} \right)}^{n + 1}}}}{n}\sin nx} \right|} \le \sum\limits_{n = N + 1}^\infty {\left| {\frac{2}{n}} \right|} = \lim\limits_{N \to \infty } \sum\limits_{n = N + 1}^\infty {{{\left( {\frac{2}{n}} \right)}^2}} = \lim\limits_{N \to \infty } \sum\limits_{n = N + 1}^\infty {\frac{4}{{{n^2}}}}. \]

The last limit is equal to zero:

\[\lim\limits_{N \to \infty } \left| {f\left( x \right) - {f_N}\left( x \right)} \right| = \lim\limits_{N \to \infty } \sum\limits_{n = N + 1}^\infty {\frac{4}{{{n^2}}}} = 0.\]

Thus, we have proved that the Fourier series of the function \(f\left( x \right) = x\) converges to \(f\left( x \right)\) in \({L_2}\)-norm.

Example 5.

The Fourier series of the function \[f\left( x \right) = \frac{{\pi - x}}{2}\] defined on the interval \(\left[ {0,2\pi } \right]\) is given by the formula \[f\left( x \right) = \frac{{\pi - x}}{2} = \sum\limits_{n = 1}^\infty {\frac{{\sin nx}}{n}} \] (see Example \(2\)). Investigate behavior of the partial sums \({f_N}\left( x \right)\) of the Fourier series.

Solution.

The partial sums of the Fourier series are given by the formula

\[{f_N}\left( x \right) = \sum\limits_{n = 1}^N {\frac{{\sin nx}}{n}} .\]

Figures \(5-8\) show how the partial sums approximate the function at different values of \(N.\) As one can see, the overshoot caused by Gibbs phenomenon occurs over smaller and smaller intervals with increasing \(N.\)

Gibbs Phenomenon - partial sums approximate at n=2 — Figure 5, n = 2

Gibbs Phenomenon - partial sums approximate at n=5 — Figure 6, n = 5

Gibbs Phenomenon - partial sums approximate at n=10 — Figure 7, n = 10

Gibbs Phenomenon - partial sums approximate at n=50 — Figure 8, n = 50

We examine the amplitude of the overshoot as \(N \to \infty .\) Integrating term by term, we obtain

\[{f_N}\left( x \right) = \int\limits_0^x {\left( {\sum\limits_{n = 1}^N {\cos nt} } \right)dt} .\]

Using the relationship

\[\frac{1}{2} + \sum\limits_{n = 1}^N {\cos nt} = \frac{1}{2} + \cos t + \cos 2t + \ldots + \cos nt = \frac{{\sin \frac{{2n + 1}}{2}t}}{{2\sin \frac{t}{2}}},\]

we have

\[{f_N}\left( x \right) = \int\limits_0^x {\left( {\frac{1}{2} - \frac{{\sin \frac{{2n + 1}}{2}t}}{{2\sin \frac{t}{2}}}} \right)dt} = - \frac{x}{2} + \int\limits_0^x {\frac{{\sin \frac{{2n + 1}}{2}t}}{{2\sin \frac{t}{2}}}dt} .\]

We put \({x_N} = \frac{{2\pi }}{{2N + 1}}.\) Then

\[{f_N}\left( {{x_N}} \right) + \frac{{{x_N}}}{2} = \int\limits_0^{{x_N}} {\frac{{\sin \frac{{2N + 1}}{2}t}}{{2\sin \frac{t}{2}}}dt} .\]

Make the substitution: \({\frac{{2N + 1}}{2}} t = z,\) \(dt = {\frac{2}{{2N + 1}}} dz.\) Here \(z = 0\) when \(t = 0,\) and \(z = {\frac{{2N + 1}}{2}} \cdot {\frac{{2\pi }}{{2N + 1}}} = \pi \) when \(t = {x_N} = {\frac{{2\pi }}{{2N + 1}}}.\)

Hence, we get

\[{f_N}\left( {{x_N}} \right) + \frac{{{x_N}}}{2} = \int\limits_0^\pi {\frac{{\sin z}}{{2\sin \frac{z}{{2N + 1}}}} \cdot \frac{{2dz}}{{2N + 1}}} = \int\limits_0^\pi {\frac{{\sin z}}{{\sin \frac{z}{{2N + 1}}\left( {2N + 1} \right)}}dz} = \int\limits_0^\pi {\frac{{\sin z}}{{\frac{{z \cdot \sin \frac{z}{{2N + 1}}}}{{\frac{z}{{2N + 1}}}}}}dz.} \]

We find from here that \({f_N}\left( x \right) = \int\limits_0^\pi {{\frac{{\sin z}}{z}} dz} \) as \(N \to \infty\) since

\[\lim\limits_{N \to \infty } {x_N} = \lim\limits_{N \to \infty } \frac{{2\pi }}{{2N + 1}} = 0\;\; \text{and}\;\; \lim\limits_{N \to \infty } \frac{{\sin \frac{z}{{2N + 1}}}}{{\frac{z}{{2N + 1}}}} = 1.\]

The integral \(\int\limits_0^x {{\frac{{\sin z}}{z}} dz} \) is called sine integral and denoted by

\[\text{Si}\left( x \right) = \int\limits_0^x {\frac{{\sin z}}{z}dz} .\]

Returning to the beginning of the solution, we see that

\[\lim\limits_{N \to \infty } \sum\limits_{n = 1}^N {\frac{{\sin n{x_N}}}{n}} = \int\limits_0^\pi {\frac{{\sin z}}{z}dz} = \text{Si}\left( \pi \right),\]

where \(\text{Si}\left( \pi \right) \approx {\frac{\pi }{2}} \cdot 1,17898.\)

Thus, the amplitude of the overshoot is approximately \(18\%.\)