Calculus

Fourier Series

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Orthogonal Polynomials and Generalized Fourier Series

Orthogonal Polynomials

Two polynomials \({p\left( x \right)}\) and \({q\left( x \right)}\) defined on the interval \(\left[ {a,b} \right]\) are orthogonal if

\[\int\limits_a^b {p\left( x \right)q\left( x \right)w\left( x \right)dx} = 0,\]

where \({w\left( x \right)}\) is a nonnegative weight function.

A polynomial sequence \({p_n}\left( x \right),\) \(n = 0,1,2, \ldots ,\) where \(n\) is the degree of \({p_n}\left( x \right),\) is said to be a sequence of orthogonal polynomials if

\[\int\limits_a^b {{p_m}\left( x \right){p_n}\left( x \right)w\left( x \right)dx} = {c_n}{\delta _{mn}},\]

where \({c_n}\) are given constants and \({\delta _{mn}}\) is the Kronecker delta.

Generalized Fourier Series

A generalized Fourier series is a series expansion of a function based on a system of orthogonal polynomials. By using this orthogonality, a piecewise continuous function \({f\left( x \right)}\) can be expressed in the form of generalized Fourier series expansion:

\[ \sum\limits_{n = 0}^\infty {{c_n}{p_n}\left( x \right)} = \begin{cases} f\left( x \right), \;\text{if}\,f\left( x \right)\,\text{is continuous} \\[0.5em] \frac{{f\left( {x - 0} \right) + f\left( {x + 0} \right)}}{2}, \;\text{at a jump discontinuity} \end{cases}.\]

We consider \(4\) types of orthogonal polynomials: Hermite, Laguerre, Legendre and Chebyshev polynomials.

Hermite Polynomials

Hermite Polynomials \({H_n}\left( x \right) = {\left( { - 1} \right)^n}{e^{{x^2}}}{\frac{{{d^n}}}{{d{x^n}}}} {e^{ - {x^2}}}\) are orthogonal on the interval \(\left( { - \infty ,\infty } \right)\) with respect to the weight function \({e^{ - {x^2}}}:\)

\[\int\limits_{ - \infty }^\infty {{e^{ - {x^2}}}{H_m}\left( x \right){H_n}\left( x \right)dx} = \begin{cases} 0, & m \ne n \\ {2^n}n!\sqrt \pi, & m = n \end{cases}.\]

An alternative definition uses the weight function \({e^{ - \frac{{{x^2}}}{2}}}.\) This convention is sometimes preferred in probability theory because \({\frac{1}{{\sqrt {2\pi } }}} {e^{ - \frac{{{x^2}}}{2}}}\) is the probability density function for the normal distribution.

Laguerre Polynomials

Laguerre polynomials \({L_n}\left( x \right) = {\frac{{{e^x}}}{{n!}}} {\frac{{{d^n}\left( {{x^n}{e^{ - x}}} \right)}}{{d{x^n}}}},\) \(n = 0,1,2,3, \ldots \) are orthogonal on the interval \(\left( {0,\infty } \right)\) with the weight function \({{e^{ - x}}}:\)

\[\int\limits_0^\infty {{e^{ - x}}{L_m}\left( x \right){L_n}\left( x \right)dx} = \begin{cases} 0, & m \ne n \\ 1, & m = n \end{cases}.\]

Legendre Polynomials

Legendre Polynomials \({P_n}\left( x \right) = {\frac{1}{{{{2^n}n!}}} {\frac{{{d^n}{{\left( {{x^2} - 1} \right)}^n}}} {d{x^n}}}},\) \(n = 0,1,2,3, \ldots \) are orthogonal on the interval \(\left[ {-1,1} \right]:\)

\[\int\limits_{ - 1}^1 {{P_m}\left( x \right){P_n}\left( x \right)dx} = \begin{cases} 0, & m \ne n \\ \frac{2}{{2n + 1}}, & m = n \end{cases}.\]

Chebyshev Polynomials

Chebyshev Polynomials of the first kind \({T_n}\left( x \right) = \cos \left( {n\arccos x} \right)\) are orthogonal on the interval \(\left[ {-1,1} \right]\) with the weight function \({\frac{1}{{\sqrt {1 - {x^2}} }}} :\)

\[\int\limits_{ - 1}^1 {\frac{{{T_m}\left( x \right){T_n}\left( x \right)}}{{\sqrt {1 - {x^2}} }}dx} = \begin{cases} 0, & m \ne n \\ \pi, & m = n = 0 \\ \frac{\pi }{2}, & m = n \ne 0 \end{cases}.\]

Solved Problems

Click or tap a problem to see the solution.

Example 1

Show that the set of functions

\[1,\cos x,\sin x,\cos 2x,\sin 2x, \ldots ,\cos mx,\sin mx, \ldots\]

is orthogonal on the interval \(\left[ { - \pi ,\pi } \right].\)

Example 2

Find the Fourier-Hermite series expansion of the quadratic function \[f\left( x \right) = A{x^2} + Bx + C.\]

Example 1.

Show that the set of functions

\[1,\cos x,\sin x,\cos 2x,\sin 2x, \ldots ,\cos mx,\sin mx, \ldots\]

is orthogonal on the interval \(\left[ { - \pi ,\pi } \right].\)

Solution.

We evaluate the integrals

\[{I_1} = \int\limits_{ - \pi }^\pi {\sin mx\sin nxdx} ,\;\; {I_2} = \int\limits_{ - \pi }^\pi {\cos mx\cos nxdx} ,\;\; {I_3} = \int\limits_{ - \pi }^\pi {\sin mx\cos nxdx} .\]

The first integral is

\[{I_1} = \int\limits_{ - \pi }^\pi {\sin mx\sin nxdx} = \frac{1}{2}\int\limits_{ - \pi }^\pi {\left[ {\cos \left( {mx - nx} \right) - \cos \left( {mx + nx} \right)} \right]dx} = \frac{1}{2}\int\limits_{ - \pi }^\pi {\left[ {\cos \left( {m - n} \right)x - \cos \left( {m + n} \right)x} \right]dx} = \frac{1}{2}\left[ {\left. {\Big( {\frac{{\sin \left( {m - n} \right)x}}{{m - n}} - \frac{{\sin \left( {m + n} \right)x}}{{m + n}}} \Big)} \right|_{ - \pi }^\pi } \right].\]

For \(m \ne n,\)

\[{I_1} = \frac{{\sin \left( {m - n} \right)\pi }}{{m - n}} - \frac{{\sin \left( {m + n} \right)\pi }}{{m + n}} = 0.\]

For \(m = n\), we obtain

\[{I_1} = \int\limits_{ - \pi }^\pi {{{\sin }^2}xdx} = \frac{1}{2}\int\limits_{ - \pi }^\pi {\left( {1 - \cos 2nx} \right)dx} = \frac{1}{2}\left[ {\left. {\left( {x - \frac{{\sin 2nx}}{{2n}}} \right)} \right|_{ - \pi }^\pi } \right] = \frac{1}{2}\left[ {\pi - \frac{{\sin 2n\pi }}{{2n}} - \left( { - \pi } \right) - \frac{{\sin \left( { - 2n\pi } \right)}}{{2n}}} \right] = \pi .\]

Thus,

\[{I_1} = \int\limits_{ - \pi }^\pi {\sin mx\sin nxdx} = \begin{cases} 0, & m \ne n\\ \pi, & m = n\end{cases}.\]

Similarly, we can find that

\[{I_2} = \int\limits_{ - \pi }^\pi {\cos mx\cos nxdx} = \begin{cases} 0, & m \ne n \\ \pi, & m = n \end{cases},\]
\[{I_3} = \int\limits_{ - \pi }^\pi {\sin mx\cos nxdx} = \begin{cases} 0, & m \ne n \\ \pi, & m = n \end{cases}.\]

This means that the set of functions

\[1,\cos x,\sin x,\cos 2x,\sin 2x, \ldots ,\cos mx,\sin mx, \ldots\]

form the orthogonal system on the interval \(\left[ { - \pi ,\pi } \right].\)

Example 2.

Find the Fourier-Hermite series expansion of the quadratic function \[f\left( x \right) = A{x^2} + Bx + C.\]

Solution.

We use the explicit expressions for Hermite polynomials:

\[{H_0}\left( x \right) = 1,\;\; {H_1}\left( x \right) = 2x,\;\; {H_2}\left( x \right) = 4{x^2} - 2.\]

Apply the method of undetermined coefficients.

\[A{x^2} + Bx + C = {c_0}{H_0}\left( x \right) + {c_1}{H_1}\left( x \right) + {c_2}{H_2}\left( x \right).\]

Substituting the Hermite polynomials and equating the coefficients, we obtain

\[A{x^2} + Bx + C = {c_0} \cdot 1 + {c_1} \cdot 2x + {c_2} \cdot \left( {4{x^2} - 2} \right),\;\; \Rightarrow A{x^2} + Bx + C = {c_0} + 2{c_1}x + 4{c_2}{x^2} - 2{c_2},\;\; \Rightarrow \left\{ {\begin{array}{*{20}{l}}{4{c^2} = A}\\{2{c_1} = B}\\{{c_0} - 2{c_2} = C}\end{array}} \right.,\;\; \Rightarrow {c_0} = C + \frac{A}{2},\;\; {c_1} = \frac{B}{2},\;\; {c_2} = \frac{A}{4}.\]

Hence, the Fourier-Hermite series expansion of the given function is defined by the expression

\[f\left( x \right) = A{x^2} + Bx + C = \left( {C + \frac{A}{2}} \right){H_0}\left( x \right) + \frac{B}{2}{H_1}\left( x \right) + \frac{A}{4}{H_2}\left( x \right).\]

See more problems on Page 2.

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