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   Definition of Fourier Series and Typical Examples
Baron Jean Baptiste Joseph Fourier
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.

To consider this idea in more detail, we should introduce some definitions and common terms.
Basic Definitions
A function \(f\left( x \right)\) is said to have period \(P\) if \(f\left( {x + P} \right) = f\left( x \right)\) for all \(x.\) Let the function \(f\left( x \right)\) has period \(2\pi.\) In this case, it is enough to consider behavior of the function on the interval \(\left[ { - \pi ,\pi } \right].\)
  1. Suppose that the function \(f\left( x \right)\) with period \(2\pi\) is absolutely integrable on on \(\left[ { - \pi ,\pi } \right]\) so that the following so-called Dirichlet integral is finite: \[\int\limits_{ - \pi }^\pi {\left| {f\left( x \right)} \right|dx} < \infty ;\]

  2. Suppose also that the function \(f\left( x \right)\) 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 \(f\left( x \right)\) exists and converges to the given function (see also the Convergence of Fourier Series page about convergence conditions.)

At a discontinuity \({x_0}\), the Fouries Series converges to \[\lim\limits_{\varepsilon \to 0} \frac{1}{2}\left[ {f\left( {{x_0} - \varepsilon } \right) - f\left( {{x_0} + \varepsilon } \right)} \right].\] The Fourier series of the function \(f\left( x \right)\) is given by \[f\left( x \right) = \frac{{{a_0}}}{2} + \sum\limits_{n = 1}^\infty {\left\{ {{a_n}\cos nx + {b_n}\sin nx} \right\}} ,\] where the Fourier coefficients \({{a_0}},\) \({{a_n}},\) and \({{b_n}}\) are defined by the integrals \[ {{a_0} = \frac{1}{\pi }\int\limits_{ - \pi }^\pi {f\left( x \right)dx} ,}\;\; {{a_n} = \frac{1}{\pi }\int\limits_{ - \pi }^\pi {f\left( x \right)\cos nx dx} ,}\;\; {{b_n} = \frac{1}{\pi }\int\limits_{ - \pi }^\pi {f\left( x \right)\sin nx dx} .} \] Sometimes alternative forms of the Fourier series are used. Replacing \({{a_n}}\) and \({{b_n}}\) by the new variables \({{d_n}}\) and \({{\varphi_n}}\) or \({{d_n}}\) and \({{\theta_n}},\) where \[ {{d_n} = \sqrt {a_n^2 + b_n^2} ,}\;\; {\tan {\varphi _n} = \frac{{{a_n}}}{{{b_n}}},}\;\; {\tan {\theta _n} = \frac{{{b_n}}}{{{a_n}}},} \] we can write: \[ {f\left( x \right) = \frac{{{a_0}}}{2} + \sum\limits_{n = 1}^\infty {{d_n}\sin \left( {nx + {\varphi _n}} \right)} \;\;\text{or}\;\;} {f\left( x \right) = \frac{{{a_0}}}{2} + \sum\limits_{n = 1}^\infty {{d_n}\cos\left( {nx + {\theta _n}} \right)} .} \]
Fourier Series of Even and Odd Functions
The Fourier series expansion of an even function \(f\left( x \right)\) with the period of \(2\pi\) does not involve the terms with sines and has the form: \[f\left( x \right) = \frac{{{a_0}}}{2} + \sum\limits_{n = 1}^\infty {{a_n}\cos nx} ,\] where the Fourier coefficients are given by the formulas \[ {{a_0} = \frac{2}{\pi }\int\limits_0^\pi {f\left( x \right)dx} ,}\;\; {{a_n} = \frac{2}{\pi }\int\limits_0^\pi {f\left( x \right)\cos nxdx} .} \] Accordingly, the Fourier series expansion of an odd \(2\pi\)-periodic function \(f\left( x \right)\) consists of sine terms only and has the form: \[f\left( x \right) = \sum\limits_{n = 1}^\infty {{b_n}\sin nx} ,\] where the coefficients \({{b_n}}\) are \[{b_n} = \frac{2}{\pi }\int\limits_0^\pi {f\left( x \right)\sin nxdx} .\] Below we consider expansions of \(2\pi\)-periodic functions into their Fourier series, supposing that these expansions exist and are convergent.

   Example 1
Let the function \(f\left( x \right)\) be \(2\pi\)-periodic and suppose that it is presented by the Fourier series: \[f\left( x \right) = \frac{{{a_0}}}{2} + \sum\limits_{n = 1}^\infty {\left\{ {{a_n}\cos nx + {b_n}\sin nx} \right\}} .\] Calculate the coefficients \({{a_0}},\) \({{a_n}},\) and \({{b_n}}.\)

Solution.
To define \({{a_0}},\) we integrate the Fourier series on the interval \(\left[ { - \pi ,\pi } \right]:\) \[ {\int\limits_{ - \pi }^\pi {f\left( x \right)dx} } = {\pi {a_0} + \sum\limits_{n = 1}^\infty {\left[ {{a_n}\int\limits_{ - \pi }^\pi {\cos nxdx} + {b_n}\int\limits_{ - \pi }^\pi {\sin nxdx} } \right]} .} \] For all \(n > 0\), \[ {\int\limits_{ - \pi }^\pi {\cos nxdx} = \left. {\left( {\frac{{\sin nx}}{n}} \right)} \right|_{ - \pi }^\pi = 0\;\;\text{and}\;\;} {\int\limits_{ - \pi }^\pi {\sin nxdx} = \left. {\left( { - \frac{{\cos nx}}{n}} \right)} \right|_{ - \pi }^\pi = 0.} \] Therefore, all the terms on the right of the summation sign are zero, so we obtain \[ {\int\limits_{ - \pi }^\pi {f\left( x \right)dx} = \pi {a_0}\;\;\text{or}\;\;} {{a_0} = \frac{1}{\pi }\int\limits_{ - \pi }^\pi {f\left( x \right)dx} .} \] In order to find the coefficients \({{a_n}},\) we multiply both sides of the Fourier series by \(\cos mx\) and integrate term by term: \[ {\int\limits_{ - \pi }^\pi {f\left( x \right)\cos mxdx} } = {\frac{{{a_0}}}{2}\int\limits_{ - \pi }^\pi {\cos mxdx} } + {\sum\limits_{n = 1}^\infty {\left[ {{a_n}\int\limits_{ - \pi }^\pi {\cos nx\cos mxdx} + {b_n}\int\limits_{ - \pi }^\pi {\sin nx\cos mxdx} } \right]} .} \] The first term on the right side is zero. Then, using the well-known trigonometric identities, we have \[ {\int\limits_{ - \pi }^\pi {\sin nx\cos mxdx} } = {\frac{1}{2}\int\limits_{ - \pi }^\pi {\left[ {\sin\left( {n + m} \right)x + \sin \left( {n - m} \right)x} \right]dx} = 0,} \] \[ {\int\limits_{ - \pi }^\pi {\cos nx\cos mxdx} } = {\frac{1}{2}\int\limits_{ - \pi }^\pi {\left[ {\cos \left( {n + m} \right)x + \cos \left( {n - m} \right)x} \right]dx} = 0,} \] if \(m \ne n.\)

In case when \(m = n\), we can write: \[\require{cancel} {\int\limits_{ - \pi }^\pi {\sin nx\cos mxdx} } = {\frac{1}{2}\int\limits_{ - \pi }^\pi {\left[ {\sin 2mx + \sin 0} \right]dx} ,}\;\; {\Rightarrow \int\limits_{ - \pi }^\pi {{\sin^2}mxdx} = \frac{1}{2}\left[ {\left. {\left( { - \frac{{\cos 2mx}}{{2m}}} \right)} \right|_{ - \pi }^\pi } \right] } = {\frac{1}{{4m}}\left[ { - \cancel{\cos \left( {2m\pi } \right)} + \cancel{\cos \left( {2m\left( { - \pi } \right)} \right)}} \right] = 0;} \] \[ {\int\limits_{ - \pi }^\pi {\cos nx\cos mxdx} } = {\frac{1}{2}\int\limits_{ - \pi }^\pi {\left[ {\cos 2mx + \cos 0} \right]dx} ,}\;\; {\Rightarrow \int\limits_{ - \pi }^\pi {{\cos^2}mxdx} } = {\frac{1}{2}\left[ {\left. {\left( {\frac{{\sin 2mx}}{{2m}}} \right)} \right|_{ - \pi }^\pi + 2\pi } \right] } = {\frac{1}{{4m}}\left[ {\sin \left( {2m\pi } \right) - \sin \left( {2m\left( { - \pi } \right)} \right)} \right] + \pi = \pi .} \] Thus, \[ {\int\limits_{ - \pi }^\pi {f\left( x \right)\cos mxdx} = {a_m}\pi ,}\;\; {\Rightarrow {a_m} = \frac{1}{\pi }\int\limits_{ - \pi }^\pi {f\left( x \right)\cos mxdx} ,}\;\; {m = 1,2,3, \ldots } \] Similarly, multiplying the Fourier series by \(\sin mx\) and integrating term by term, we obtain the expression for \({{b_m}}:\) \[ {{b_m} = \frac{1}{\pi }\int\limits_{ - \pi }^\pi {f\left( x \right)\sin mxdx} ,}\;\; {m = 1,2,3, \ldots } \] Rewriting the formulas for \({{a_n}},\) \({{b_n}},\) we can write the final expressions for the Fourier coefficients: \[ {{a_n} = \frac{1}{\pi }\int\limits_{ - \pi }^\pi {f\left( x \right)\cos nxdx} ,}\;\; {{b_n} = \frac{1}{\pi }\int\limits_{ - \pi }^\pi {f\left( x \right)\sin nxdx} .} \]
   Example 2
Find the Fourier series for the square \(2\pi\)-periodic wave defined on the interval \(\left[ { - \pi ,\pi } \right]:\) \[ f\left( x \right) = \begin{cases} 0, & \text{if} & - \pi \le x \le 0 \\ 1, & \text{if} & 0 < x \le \pi \end{cases}. \]

Solution.
First we calculate the constant \({{a_0}}:\) \[ {{a_0} = \frac{1}{\pi }\int\limits_{ - \pi }^\pi {f\left( x \right)dx} } = {\frac{1}{\pi }\int\limits_0^\pi {1dx} } = {\frac{1}{\pi } \cdot \pi = 1.} \] Find now the Fourier coefficients for \(n \ne 0:\) \[ {{a_n} = \frac{1}{\pi }\int\limits_{ - \pi }^\pi {f\left( x \right)\cos nxdx} } = {\frac{1}{\pi }\int\limits_0^\pi {1 \cdot \cos nxdx} } = {\frac{1}{\pi }\left[ {\left. {\left( {\frac{{\sin nx}}{n}} \right)} \right|_0^\pi } \right] } = {\frac{1}{{\pi n}} \cdot 0 = 0,} \] \[ {{b_n} = \frac{1}{\pi }\int\limits_{ - \pi }^\pi {f\left( x \right)\sin nxdx} } = {\frac{1}{\pi }\int\limits_0^\pi {1 \cdot \sin nxdx} } = {\frac{1}{\pi }\left[ {\left. {\left( { - \frac{{\cos nx}}{n}} \right)} \right|_0^\pi } \right] } = { - \frac{1}{{\pi n}} \cdot \left( {\cos n\pi - \cos 0} \right) } = {\frac{{1 - \cos n\pi }}{{\pi n}}.} \] As \(\cos n\pi = {\left( { - 1} \right)^n},\) we can write: \[{b_n} = \frac{{1 - {{\left( { - 1} \right)}^n}}}{{\pi n}}.\] Thus, the Fourier series for the square wave is \[f\left( x \right) = \frac{1}{2} + \sum\limits_{n = 1}^\infty {\frac{{1 - {{\left( { - 1} \right)}^n}}}{{\pi n}}\sin nx} .\] We can easily find the first few terms of the series. By setting, for example, \(n = 5,\) we get \[ {f\left( x \right) = \frac{1}{2} + \frac{{1 - \left( { - 1} \right)}}{\pi }\sin x } + {\frac{{1 - {{\left( { - 1} \right)}^2}}}{{2\pi }}\sin 2x } + {\frac{{1 - {{\left( { - 1} \right)}^3}}}{{3\pi }}\sin 3x } + {\frac{{1 - {{\left( { - 1} \right)}^4}}}{{4\pi }}\sin 4x } + {\frac{{1 - {{\left( { - 1} \right)}^5}}}{{5\pi }}\sin 5x + \ldots } = {\frac{1}{2} + \frac{2}{\pi }\sin x } + {\frac{2}{{3\pi }}\sin 3x } + {\frac{2}{{5\pi }}\sin 5x + \ldots } \] The graph of the function and the Fourier series expansion for \(n = 10\) is shown in Figure \(1.\)
Fourier series of the square wave
Fourier series of the sawtooth wave
Fig.1, n = 10
Fig.2, n = 5, n = 10
   Example 3
Find the Fourier series for the sawtooth wave defined on the interval \(\left[ { - \pi ,\pi } \right]\) and having period \(2\pi.\)

Solution.
Calculate the Fourier coefficients for the sawtooth wave. Since this function is odd (Figure \(2\)), then \({a_0} = {a_n} = 0.\) Find the coefficients \({b_n}:\) \[ {{b_n} = \frac{1}{\pi }\int\limits_{ - \pi }^\pi {f\left( x \right)\sin nxdx} } = {\frac{1}{\pi }\int\limits_{ - \pi }^\pi {x\sin nxdx} .} \] To calculate the latter integral we use integration by parts formula: \[\int\limits_{ - \pi }^\pi {udv} = \left. {\left( {uv} \right)} \right|_{ - \pi }^\pi - \int\limits_{ - \pi }^\pi {vdu} .\] Let \(u = x,\;dv = \sin nxdx.\) Then \(du = dx,\;v = \int {\sin nxdx} = - {\large\frac{{\cos nx}}{n}\normalsize},\) so the integral becomes \[ {{b_n} = \frac{1}{\pi }\int\limits_{ - \pi }^\pi {x\sin nxdx} } = {\frac{1}{\pi }\left[ {\left. {\left( { - \frac{{\cos nx}}{n}} \right)} \right|_{ - \pi }^\pi - \int\limits_{ - \pi }^\pi {\left( { - \frac{{\cos nx}}{n}} \right)dx} } \right] } = {\frac{1}{\pi }\left[ {\left( { - \frac{{\pi \cos n\pi }}{n} + \frac{{\left( { - \pi } \right)\cos \left( { - n\pi } \right)}}{n}} \right) + \frac{1}{n}\int\limits_{ - \pi }^\pi {\cos nxdx} } \right] } = {\frac{1}{{n\pi }}\left[ { - 2\pi \cos n\pi + \left. {\left( {\frac{{\sin nx}}{n}} \right)} \right|_{ - \pi }^\pi } \right] } = {\frac{1}{{n\pi }}\left[ { - 2\pi \cos n\pi + \frac{1}{n}\left( {\sin n\pi - \sin \left( { - n\pi } \right)} \right)} \right] } = {\frac{1}{{n\pi }}\left[ { - 2\pi \cos n\pi + \frac{{2\sin n\pi }}{n}} \right] } = {\frac{2}{{n\pi }}\left[ {\frac{{\sin n\pi }}{n} - \pi \cos n\pi } \right].} \] Substituting \(\sin n\pi = 0\) and \(\cos n\pi = {\left( { - 1} \right)^n}\) for all integer values of \(n,\) we obtain \[ {{b_n} = \frac{2}{{n\pi }}\left( { - \pi {{\left( { - 1} \right)}^n}} \right) } = { - \frac{2}{n}{\left( { - 1} \right)^n} } = {\frac{2}{n}{\left( { - 1} \right)^{n + 1}}.} \] Thus, the Fourier series expansion of the sawtooth wave is (Figure \(2\) above) \[x = \sum\limits_{n = 1}^\infty {\frac{2}{n}{{\left( { - 1} \right)}^{n + 1}}\sin nx} .\]
   Example 4
Let \(f\left( x \right)\) be a \(2\pi\)-periodic function such that \(f\left( x \right) = {x^2}\) for \(x \in \left[ { - \pi ,\pi } \right].\) Find the Fourier series for the parabolic wave.

Solution.
Since this function is even, the coefficients \({b_n} = 0.\) Then \[ {{a_0} = \frac{1}{\pi }\int\limits_{ - \pi }^\pi {f\left( x \right)dx} } = {\frac{1}{\pi }\int\limits_{ - \pi }^\pi {{x^2}dx} = \frac{2}{\pi }\int\limits_0^\pi {{x^2}dx} } = {\frac{2}{\pi } \cdot \left[ {\left. {\left( {\frac{{{x^3}}}{3}} \right)} \right|_0^\pi } \right] } = {\frac{2}{\pi } \cdot \frac{{{\pi ^3}}}{3} } = {\frac{{2{\pi ^2}}}{3},} \] \[ {{a_n} = \frac{1}{\pi }\int\limits_{ - \pi }^\pi {f\left( x \right)\cos nxdx} } = {\frac{2}{\pi }\int\limits_0^\pi {f\left( x \right)\cos nxdx} } = {\frac{2}{\pi }\int\limits_0^\pi {{x^2}\cos nxdx} .} \] Apply integration by parts twice to find: \[ {{a_n} = \frac{2}{\pi }\int\limits_0^\pi {{x^2}\cos nxdx} } = {\left[ {\begin{array}{*{20}{l}} {u = {x^2}}\\ {dv = \cos nxdx}\\ {du = 2xdx}\\ {v = \int {\cos nxdx} = \frac{{\sin nx}}{n}} \end{array}} \right] } = {\frac{2}{\pi }\left[ {\left. {\left( {\frac{{{x^2}\sin nx}}{n}} \right)} \right|_0^\pi - \int\limits_0^\pi {2x\frac{{\sin nx}}{n}dx} } \right] } = {\frac{2}{{\pi n}}\left[ {{\pi ^2}\sin n\pi - {{\left( { - \pi } \right)}^2}\sin \left( { - n\pi } \right) - 2\int\limits_0^\pi {x\sin nxdx} } \right] } = {\frac{2}{{\pi n}}\left[ {2{\pi ^2}\sin n\pi - 2\int\limits_0^\pi {x\sin nxdx} } \right] } = { - \frac{4}{{\pi n}}\int\limits_0^\pi {x\sin nxdx} } = {\left[ {\begin{array}{*{20}{l}} {u = x}\\ {dv = \sin nxdx}\\ {du = dx}\\ {v = \int {\sin nxdx} = - \frac{{\cos nx}}{n}} \end{array}} \right] } = { - \frac{4}{{\pi n}}\left[ {\left. {\left( { - \frac{{x\cos nx}}{n}} \right)} \right|_0^\pi - \int\limits_0^\pi {\left( { - \frac{{\cos nx}}{n}} \right)dx} } \right] } = {\frac{4}{{\pi {n^2}}}\left[ {\pi \cos n\pi - \int\limits_0^\pi {\cos nxdx} } \right] } = {\frac{4}{{\pi {n^2}}}\left[ {\pi \cos n\pi - \left. {\left( {\frac{{\sin nx}}{n}} \right)} \right|_0^\pi } \right] } = {\frac{4}{{\pi {n^2}}}\left[ {\pi \cos n\pi - \frac{{\sin n\pi }}{n}} \right].} \] As \(\sin n\pi = 0\) and \(\cos n\pi = {\left( { - 1} \right)^n}\) for integer \(n,\) we have \[ {{a_n} = \frac{4}{{\pi {n^2}}} \cdot \pi {\left( { - 1} \right)^n} } = {\frac{4}{{{n^2}}}{\left( { - 1} \right)^n}.} \] Then the Fourier series expansion for the parabolic wave is (Figure \(3\)) \[{x^2} = \frac{{{\pi ^2}}}{3} + \sum\limits_{n = 1}^\infty {\frac{4}{{{n^2}}}{{\left( { - 1} \right)}^n}\cos nx} .\]
Fourier series of the parabolic wave
Fourier series of the triangle wave
Fig.3, n = 2, n = 5
Fig.4, n = 1, n = 2
   Example 5
Find the Fourier series for the triangle wave \[ f\left( x \right) = \begin{cases} \frac{\pi }{2} + x, & \text{if} & - \pi \le x \le 0 \\ \frac{\pi }{2} - x, & \text{if} & 0 < x \le \pi \end{cases}, \] defined on the interval \(\left[ { - \pi ,\pi } \right].\)

Solution.
The constant \({a_0}\) is \[ {{a_0} = \frac{1}{\pi }\int\limits_{ - \pi }^\pi {f\left( x \right)dx} } = {\frac{1}{\pi }\left[ {\int\limits_{ - \pi }^0 {\left( {\frac{\pi }{2} + x} \right)dx} + \int\limits_0^\pi {\left( {\frac{\pi }{2} - x} \right)dx} } \right] } = {\frac{1}{\pi }\left[ {\left. {\left( {\frac{\pi }{2}x + \frac{{{x^2}}}{2}} \right)} \right|_{ - \pi }^0 + \left. {\left( {\frac{\pi }{2}x - \frac{{{x^2}}}{2}} \right)} \right|_0^\pi } \right] } = {\frac{1}{\pi }\left[ {0 - \left( { - \cancel{\frac{{{\pi ^2}}}{2}} + \cancel{\frac{{{{\left( { - \pi } \right)}^2}}}{2}}} \right) + \left( {\cancel{\frac{{{\pi ^2}}}{2}} - \cancel{\frac{{{\pi ^2}}}{2}}} \right) - 0} \right] = 0.} \] Determine the coefficients \({a_n}:\) \[ {{a_n} = \frac{1}{\pi }\int\limits_{ - \pi }^\pi {f\left( x \right)\cos nxdx} } = {\frac{1}{\pi }\left[ {\int\limits_{ - \pi }^0 {\left( {\frac{\pi }{2} + x} \right)\cos nxdx} + \int\limits_0^\pi {\left( {\frac{\pi }{2} - x} \right)\cos nxdx} } \right] } = {\frac{1}{\pi }\left[ {\int\limits_{ - \pi }^0 {\frac{\pi }{2}\cos nxdx} + \int\limits_{ - \pi }^0 {x\cos nxdx} } \right. } + {\left. {\int\limits_0^\pi {\frac{\pi }{2}\cos nxdx} - \int\limits_0^\pi {x\cos nxdx} } \right].} \] Integrating by parts, we can write \[ {\int {x\cos nxdx} = \frac{{x\sin nx}}{n} - \int {\frac{{x\sin nx}}{n}dx} } = {\frac{{x\sin nx}}{n} + \frac{{\cos nx}}{{{n^2}}}.} \] Then \[ {{a_n} = \frac{1}{\pi }\left[ {\frac{\pi }{2}\left. {\left( {\frac{{\sin nx}}{n}} \right)} \right|_{ - \pi }^0 + \left. {\left( {\frac{{x\sin nx}}{n} + \frac{{\cos nx}}{{{n^2}}}} \right)} \right|_{ - \pi }^0} \right. } + {\;\left. {\frac{\pi }{2}\left. {\left( {\frac{{\sin nx}}{n}} \right)} \right|_0^\pi - \left. {\left( {\frac{{x\sin nx}}{n} + \frac{{\cos nx}}{{{n^2}}}} \right)} \right|_0^\pi } \right].} \] The values of \(\sin nx\) at \(x = 0\) or \(x = \pm \pi\) are zero. Therefore, \[ {{a_n} = \frac{1}{{\pi {n^2}}}\left[ {\left. {\left( {\cos nx} \right)} \right|_{ - \pi }^0 - \left. {\left( {\cos nx} \right)} \right|_0^\pi } \right] } = {\frac{1}{{\pi {n^2}}}\left[ {\cos 0 - \cos \left( { - \pi n} \right) - \cos \pi n + \cos 0} \right] } = {\frac{2}{{\pi {n^2}}}\left[ {1 - \cos \pi n} \right] } = {\frac{2}{{\pi {n^2}}}\left[ {1 - {{\left( { - 1} \right)}^n}} \right].} \] When \(n = 2k,\) then \({a_{2k}} = 0.\) When \(n = 2k + 1,\) then \({a_{2k + 1}} = {\large\frac{4}{{\pi {n^2}}}\normalsize},\;k = 0,1,2,3, \ldots \)
Since the function \(f\left( x \right)\) is even, the Fourier coefficients \({b_n}\) are zero. Therefore, the complete Fourier expansion for the triangle wave (see Figure \(4\) above) is \[f\left( x \right) = \frac{4}{\pi }\sum\limits_{k = 0}^\infty {\frac{{\cos \left( {2k + 1} \right)x}}{{{{\left( {2k + 1} \right)}^2}}}} .\]
   Example 6
Find the Fourier series for the function \[ f\left( x \right) = \begin{cases} 0, & \text{if} & - \pi \le x \le 0 \\ \sin x, & \text{if} & 0 < x \le \pi \end{cases}, \] defined on the interval \(\left[ { - \pi ,\pi } \right].\)

Solution.
First we find the constant \({a_0}:\) \[ {{a_0} = \frac{1}{\pi }\int\limits_{ - \pi }^\pi {f\left( x \right)dx} } = {\frac{1}{\pi }\int\limits_0^\pi {\sin xdx} } = {\frac{1}{\pi }\left[ {\left. {\left( { - \cos x} \right)} \right|_0^\pi } \right] } = {\frac{1}{\pi }\left( { - \cos \pi + \cos 0} \right) } = {\frac{2}{\pi }.} \] Now we calculate the coefficients \({a_n}:\) \[ {{a_n} = \frac{1}{\pi }\int\limits_{ - \pi }^\pi {f\left( x \right)\cos nxdx} } = {\frac{1}{\pi }\int\limits_0^\pi {\sin x\cos nxdx} } = {\frac{1}{{2\pi }}\int\limits_0^\pi {\left[ {\sin \left( {x + nx} \right) + \sin\left( {x - nx} \right)} \right]dx} } = {\frac{1}{{2\pi }}\int\limits_0^\pi {\left[ {\sin \left( {n + 1} \right)x + \sin\left( {n - 1} \right)x} \right]dx} } = {\frac{1}{{2\pi }}\left[ {\left. {\left( { - \frac{{\cos \left( {n + 1} \right)x}}{{n + 1}} + \frac{{\cos \left( {n - 1} \right)x}}{{n - 1}}} \right)} \right|_0^\pi } \right] } = {\frac{1}{{2\pi }}\left[ {\frac{{\cos \left( {n - 1} \right)\pi }}{{n - 1}} - \frac{{\cos \left( {n + 1} \right)\pi }}{{n + 1}} - \frac{1}{{n - 1}} + \frac{1}{{n + 1}}} \right] } = {\frac{1}{{2\pi }}\left[ {\frac{{\cos \left( {n - 1} \right)\pi }}{{n - 1}} - \frac{{\cos \left( {n + 1} \right)\pi }}{{n + 1}} - \frac{2}{{{n^2} - 1}}} \right].} \] Notice that \[ {\cos \left( {n + 1} \right)\pi } = {\cos \left( {\pi n + \pi } \right) } = {\cos \left( {\pi n - \pi + 2\pi } \right) } = {\cos \left( {\left( {n - 1} \right)\pi + 2\pi } \right) } = {\cos \left( {n - 1} \right)\pi .} \] Since \(\cos \left( {n - 1} \right)\pi = {\left( { - 1} \right)^{n - 1}},\) we get the following expression for the coefficients \({a_n}:\) \[{a_n} = \frac{1}{{2\pi }} \cdot \frac{2}{{{n^2} - 1}} \cdot \left[ {{{\left( { - 1} \right)}^{n - 1}} - 1} \right].\] It can be seen that \({a_n} = 0\) for odd \(n.\) For even \(n,\) when \(n = 2k\;\left( {k = 1,2,3, \ldots } \right),\) we have \[{a_{2k}} = \frac{1}{{2\pi }} \cdot \left( { - \frac{4}{{{k^2} - 1}}} \right) = - \frac{1}{\pi } \cdot \frac{2}{{{k^2} - 1}}.\] Calculate the coefficients \({b_n}.\) Start with \({b_1}:\) \[ {{b_1} = \frac{1}{\pi }\int\limits_{ - \pi }^\pi {f\left( x \right)\sin xdx} } = {\frac{1}{\pi }\int\limits_0^\pi {\sin x\sin xdx} } = {\frac{1}{\pi }\int\limits_0^\pi {{{\sin }^2}xdx} } = {\frac{1}{{2\pi }}\int\limits_0^\pi {\left( {1 - \cos 2x} \right)dx} } = {\frac{1}{{2\pi }}\left[ {\left. {\left( {x - \frac{{\sin 2x}}{2}} \right)} \right|_0^\pi } \right] } = {\frac{1}{{2\pi }} \cdot \pi = \frac{1}{2}.} \] The other coefficients \({b_n}\) for \(n > 1\) are zero. Indeed, \[ {{b_n} = \frac{1}{\pi }\int\limits_{ - \pi }^\pi {f\left( x \right)\sin nxdx} } = {\frac{1}{\pi }\int\limits_0^\pi {\sin x\sin nxdx} } = {\frac{1}{{2\pi }}\int\limits_0^\pi {\left[ {\cos \left( {x - nx} \right) - \cos \left( {x + nx} \right)} \right]dx} } = {\frac{1}{{2\pi }}\int\limits_0^\pi {\left[ {\cos \left( {n - 1} \right)x - \cos \left( {n + 1} \right)x} \right]dx} } = {\frac{1}{{2\pi }}\left[ {\left. {\left( {\frac{{\sin \left( {n - 1} \right)x}}{{n - 1}} - \frac{{\sin \left( {n + 1} \right)x}}{{n + 1}}} \right)} \right|_0^\pi } \right] = 0} \] Thus, the Fourier series of the given function is given by \[ {f\left( x \right) = \frac{1}{\pi } + \frac{1}{2}\sin x } - {\frac{2}{\pi }\sum\limits_{k = 1}^\infty {\frac{1}{{4{k^2} - 1}}\cos \left( {2kx} \right)} .} \] Graphs of the function and its Fourier expansions for \(n = 2\) and \(n = 8\) are shown in Figure \(5.\)
Fig.5, n = 2, n = 8
Fig.6, n = 10
   Example 7
Find the Fourier series for the function \[ f\left( x \right) = \begin{cases} -1, & \text{if} & - \pi \le x \le - \frac{\pi }{2} \\ 0, & \text{if} & - \frac{\pi }{2} \lt x \le \frac{\pi }{2} \\ 1, & \text{if} & \frac{\pi }{2} < x \le \pi \end{cases}, \] defined on the interval \(\left[ { - \pi ,\pi } \right].\)

Solution.
\[ {{a_0} = \frac{1}{\pi }\int\limits_{ - \pi }^\pi {f\left( x \right)dx} } = {\frac{1}{\pi }\left[ {\int\limits_{ - \pi }^{ - \large\frac{\pi }{2}\normalsize} {\left( { - 1} \right)dx} + \int\limits_{ - \large\frac{\pi }{2}\normalsize}^{\large\frac{\pi }{2}\normalsize} {0dx} + \int\limits_{\large\frac{\pi }{2}\normalsize}^\pi {1dx} } \right] } = {\frac{1}{\pi }\left( { - \cancel{\frac{\pi }{2}} + 0 + \cancel{\frac{\pi }{2}}} \right) = 0.} \] Compute the coefficients \({a_n}:\) \[ {{a_n} = \frac{1}{\pi }\int\limits_{ - \pi }^\pi {f\left( x \right)\cos nxdx} } = {\frac{1}{\pi }\left[ {\int\limits_{ - \pi }^{ - \large\frac{\pi }{2}\normalsize} {\left( { - \cos nx} \right)dx} + \int\limits_{\large\frac{\pi }{2}\normalsize}^\pi {\cos nxdx} } \right] } = {\frac{1}{\pi }\left[ {\left. {\left( { - \frac{{\sin nx}}{n}} \right)} \right|_{ - \pi }^{ - \large\frac{\pi }{2}\normalsize} + \left. {\left( {\frac{{\sin nx}}{n}} \right)} \right|_{\large\frac{\pi }{2}\normalsize}^\pi } \right] } = {\frac{1}{{\pi n}}\left[ { - \sin \left( { - \frac{{n\pi }}{2}} \right) + \sin \left( { - n\pi } \right) + \sin n\pi - \sin \frac{{n\pi }}{2}} \right] } = {\frac{1}{{\pi n}}\left[ {\cancel{\sin \frac{{n\pi }}{2}} - \cancel{\sin n\pi} + \cancel{\sin n\pi} - \cancel{\sin \frac{{n\pi }}{2}}} \right] = 0.} \] (These results are obvious since this function is odd.)

Calculate the coefficients \({b_n}:\) \[ {{b_n} = \frac{1}{\pi }\int\limits_{ - \pi }^\pi {f\left( x \right)\sin nxdx} } = {\frac{1}{\pi }\left[ {\int\limits_{ - \pi }^{ - \large\frac{\pi }{2}\normalsize} {\left( { - \sin nx} \right)dx} + \int\limits_{\large\frac{\pi }{2}\normalsize}^\pi {\sin nxdx} } \right] } = {\frac{1}{\pi }\left[ {\left. {\left( {\frac{{\cos nx}}{n}} \right)} \right|_{ - \pi }^{ - \large\frac{\pi }{2}\normalsize} - \left. {\left( {\frac{{\cos nx}}{n}} \right)} \right|_{\large\frac{\pi }{2}\normalsize}^\pi } \right] } = {\frac{1}{{\pi n}}\left[ {\cos\left( { - \frac{{n\pi }}{2}} \right) - \cos \left( { - n\pi } \right) - \cos n\pi + \cos \frac{{n\pi }}{2}} \right] } = {\frac{1}{{\pi n}}\left[ {\cos\frac{{n\pi }}{2} - \cos n\pi - \cos n\pi + \cos \frac{{n\pi }}{2}} \right] } = {\frac{2}{{\pi n}}\left( {\cos\frac{{n\pi }}{2} - \cos n\pi } \right).} \] Thus, the Fourier series expansion of the function is given by \[f\left( x \right) = \frac{2}{\pi }\sum\limits_{n = 1}^\infty {\frac{1}{n}\left( {\cos\frac{{n\pi }}{2} - \cos n\pi } \right)\sin nx} .\] The graph of the function and the Fourier series expansion for \(n = 10\) are shown in Figure \(6\) above.

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