Orthogonal Polynomials and Generalized Fourier Series
Solved Problems
Example 3.
Find the Fourier-Laguerre series expansion of the power function \(f\left( x \right) = {x^p},\) \(p \ge 1.\)
Solution.
The expansion is given by the formula
\[f\left( x \right) = \sum\limits_{n = 0}^\infty {{c_n}{L_n}\left( x \right)} .\]
Calculate the coefficients \({{c_n}}.\)
\[{c_0} = \int\limits_0^\infty {f\left( x \right){e^{ - x}}dx} = \int\limits_0^\infty {{x^p}{e^{ - x}}dx} = \Gamma \left( {p + 1} \right) = p!,\]
where \(\Gamma\) is the gamma function.
For \(n \ge 1\) we have
\[{c_n} = \frac{1}{{n!}}\int\limits_0^\infty {{x^p}\frac{{{d^n}\left( {{x^n}{e^{ - x}}} \right)}}{{d{x^n}}}dx}
= \frac{1}{{n!}}\left[ {\left. {\left( {{x^p}\frac{{{d^{n - 1}}\left( {{x^n}{e^{ - x}}} \right)}}{{d{x^{n - 1}}}}} \right)} \right|_0^\infty - \int\limits_0^\infty {p{x^{p - 1}}\frac{{{d^{n - 1}}\left( {{x^n}{e^{ - x}}} \right)}}{{d{x^{n - 1}}}}dx} } \right].\]
Continue integrating by parts to obtain
\[{c_n} = \frac{{{{\left( { - 1} \right)}^n}}}{{n!}}p\left( {p - 1} \right)\left( {p - 2} \right) \cdots \left( {p - n + 1} \right)\int\limits_0^\infty {{x^p}{e^{ - x}}dx}
= \frac{{{{\left( { - 1} \right)}^n}}}{{n!}} \cdot \frac{{p!}}{{\left( {p - n} \right)!}} \cdot \Gamma \left( {p + 1} \right)
= \frac{{{{\left( { - 1} \right)}^n}{{\left( {p!} \right)}^2}}}{{n!\left( {p - n} \right)!}},\;\;
\text{if}\;\;1 \le n \le p.\]
If \(n \gt p,\) then \({c_n} = 0.\)
Hence, the Fourier-Laguerre series expansion of the power function \(f\left( x \right) = {x^p}\) is given by
\[f\left( x \right) = {x^p} = p! + \sum\limits_{n = 1}^p {\frac{{{{\left( { - 1} \right)}^n}{{\left( {p!} \right)}^2}}}{{n!\left( {p - n} \right)!}}{L_n}\left( x \right)} .\]
Since \({L_0}\left( x \right) = 1,\) we can write this result in a more compact form:
\[f\left( x \right) = {x^p}
= \sum\limits_{n = 0}^p {\frac{{{{\left( { - 1} \right)}^n}{{\left( {p!} \right)}^2}}}{{n!\left( {p - n} \right)!}}{L_n}\left( x \right)} .\]
Check the answer, for example, for \(p = 2.\) Then
\[{x^2} = \sum\limits_{n = 0}^2 {\frac{{{{\left( { - 1} \right)}^n}{{\left( {2!} \right)}^2}}}{{n!\left( {2 - n} \right)!}}{L_n}\left( x \right)} = 2{L_0}\left( x \right) - 4{L_1}\left( x \right) + 2{L_2}\left( x \right).\]
Substituting the Laguerre polynomials
\[{L_0}\left( x \right) = 1,\;\; {L_1}\left( x \right) = 1 - x,\;\; {L_2}\left( x \right) = 1 - 2x + \frac{{{x^2}}}{2}\]
into the formula above, we have
\[{x^2} = 2 \cdot 1 - 4\left( {1 - x} \right) + 2\left( {1 - 2x + \frac{{{x^2}}}{2}} \right) \equiv {x^2}.\]
Example 4.
Find the Fourier-Legendre series expansion of the step function
\[
f\left( x \right) =
\begin{cases}
0, & -1 \lt x \lt 0 \\
1, & 0 \lt x \lt 1
\end{cases}.\]
Solution.
The series expansion is written in the form
\[f\left( x \right) = \sum\limits_{n = 0}^\infty {{c_n}{P_n}\left( x \right)} .\]
Substituting the explicit expressions for the Legendre polynomials, we have
\[f\left( x \right) = \frac{{2n + 1}}{2}\int\limits_{ - 1}^1 {f\left( x \right){P_n}\left( x \right)dx}
= \frac{{2n + 1}}{2}\int\limits_0^1 {{P_n}\left( x \right)dx}
= \frac{{2n + 1}}{2} \int\limits_0^1 {\frac{1}{{{2^n}n!}}\frac{{{d^n}{{\left( {{x^2} - 1} \right)}^n}}}{{d{x^n}}}dx}
= \frac{{2n + 1}}{{{2^{n + 1}}n!}} \left[ {\left. {\left( {\frac{{{d^{n - 1}}{{\left( {{x^2} - 1} \right)}^n}}}{{d{x^{n - 1}}}}} \right)} \right|_0^1} \right],\;\;n = 1,2,3, \ldots \]
Calculate the coefficients \({c_n}.\) For \(n = 0\), we find that \({P_0}\left( x \right) = 0.\) Then
\[{c_0} = \frac{1}{2}\int\limits_{ - 1}^1 {f\left( x \right){P_0}\left( x \right)dx} = \frac{1}{2}\int\limits_0^1 {dx} = \frac{1}{2}.\]
Compute now the value of the derivative \({\left. {\left( {\frac{{{d^{n - 1}}{{\left( {{x^2} - 1} \right)}^n}}}{{d{x^{n - 1}}}}} \right)} \right|_0^1}\) to find the coefficients \({c_n}\) for \(n \ge 1.\) It is evident that this expression is zero at \(x = 1\) for any numbers \(n \ge 1.\) To get its value at the point \(x = 0,\) we use Newton's binomial formula:
\[\frac{{{d^{n - 1}}{{\left( {{x^2} - 1} \right)}^n}}}{{d{x^{n - 1}}}}
= \frac{{d{{\left( {\sum\limits_{m = 0}^\infty {C_n^m{{\left( { - 1} \right)}^m}{x^{2n - 2m}}} } \right)}^{n - 1}}}}{{d{x^{n - 1}}}}
= \sum\limits_{m = 0}^{m \le \frac{{n + 1}}{2}} {\left[ {C_n^m{{\left( { - 1} \right)}^m}\left( {2n - 2m} \right) \left( {2n - 2m - 1} \right)} \right.} \cdots \left. {\left( { - 2m + n + 2} \right){x^{n - 2m + 1}}} \right].\]
As you can see from this expression, the sum is zero at \(x = 0\) for even numbers \(n = 2k,\) \(k = 0,1,2,3, \ldots \) For odd numbers, the sum at \(x = 0\) is
\[C_{2k + 1}^{k + 1}{\left( { - 1} \right)^{k + 1}}2k\left( {2k - 1} \right) \left( {2k - 2} \right) \cdots 3 \cdot 2 = C_{2k + 1}^{k + 1}{\left( { - 1} \right)^{k + 1}}\left( {2k} \right)!\]
We used here the fact that for \(n = 2k + 1\) and \(m = k + 1,\)
\[x^{n - 2m + 1} = x^{2k + 1 - 2\left( {k + 1} \right) + 1} = {x^0} = 1\]
as \(x \to 0.\) For other values of \(m\) and \(n\), the terms are zero. Hence,
\[{c_{2k}} = 0,\]
\[{c_{2k + 1}} = \frac{{4k + 3}}{{{2^{2k + 2}}\left( {2k + 1} \right)!}} \cdot \frac{{\left( {2k + 1} \right)!}}{{\left( {k + 1} \right)!k!}}{\left( { - 1} \right)^k}\left( {2k} \right)! = \frac{{{{\left( { - 1} \right)}^k}\left( {4k + 3} \right)\left( {2k} \right)!}}{{{2^{2k + 2}}\left( {k + 1} \right)!k!}}.\]
Thus, the Fourier-Legendre series expansion of the step function is given by
\[f\left( x \right) = \frac{1}{2} + \sum\limits_{k = 0}^\infty {\frac{{{{\left( { - 1} \right)}^k}\left( {4k + 3} \right)\left( {2k} \right)!}}{{{2^{2k + 2}}\left( {k + 1} \right)!k!}} {P_{2k + 1}}\left( x \right)} .\]
Approximation of the step function by the Fourier-Legendre series for \(n = 5, 10\) and \(15\) is shown in Figure \(1.\)
Figure 1.
Example 5.
Find the Fourier-Chebyshev series expansion of the function \(f\left( x \right) = {x^3}\) on the interval \(\left[ { - 1,1} \right].\)
Solution.
According to the general representation, we can write:
\[{x^3} = \sum\limits_{n = 0}^\infty {{c_n}{T_n}\left( x \right)} = {c_0} + \sum\limits_{n = 1}^\infty {{c_n}{T_n}\left( x \right)} .\]
To calculate the coefficients \({{c_n}},\) we use the orthogonality property of Chebyshev polynomials on the interval \(\left[ { - 1,1} \right]\) with the weight function \({\frac{1}{{\sqrt {1 - {x^2}} }}}.\)
Multiplying both sides of the above named expansion by \(\frac{1}{{\sqrt {1 - {x^2}} }}\) and integrating on \(\left[ { - 1,1} \right],\) we obtain
\[\int\limits_{ - 1}^1 {\frac{{{x^3}dx}}{{\sqrt {1 - {x^2}} }}} = \int\limits_{ - 1}^1 {\left( {\sum\limits_{n = 0}^\infty {{c_n}{T_n}\left( x \right)} } \right)\frac{{dx}}{{\sqrt {1 - {x^2}} }}} .\]
Since the function \(f\left( x \right) = {x^3}\) is odd and we integrate on the symmetric interval \(\left[ { - 1,1} \right],\) the left-side integral is zero:
\[\int\limits_{ - 1}^1 {\frac{{{x^3}dx}}{{\sqrt {1 - {x^2}} }}} = 0.\]
Convert the right side:
\[\int\limits_{ - 1}^1 {\left( {\sum\limits_{n = 0}^\infty {{c_n}{T_n}\left( x \right)} } \right)\frac{{dx}}{{\sqrt {1 - {x^2}} }}}
= \int\limits_{ - 1}^1 {\left( {{c_0} + \sum\limits_{n = 1}^\infty {{c_n}{T_n}\left( x \right)} } \right)\frac{{dx}}{{\sqrt {1 - {x^2}} }}}
= {c_0}\int\limits_{ - 1}^1 {\frac{{dx}}{{\sqrt {1 - {x^2}} }}} + \sum\limits_{n = 1}^\infty {\left[ {{c_n}\int\limits_{ - 1}^1 {\frac{{{T_n}\left( x \right)}}{{\sqrt {1 - {x^2}} }}dx} } \right]} .\]
Multiplying the numerator of the integrand of the latter integral by \({T_0}\left( x \right) = 1,\) we see that
\[\int\limits_{ - 1}^1 {\frac{{{T_n}\left( x \right){T_0}\left( x \right)}}{{\sqrt {1 - {x^2}} }}dx} = 0\]
due to orthogonality of Chebyshev polynomials.
Thus,
\[{c_0}\int\limits_{ - 1}^1 {\frac{{dx}}{{\sqrt {1 - {x^2}} }}} = 0.\]
Calculating, we find that
\[\int\limits_{ - 1}^1 {\frac{{dx}}{{\sqrt {1 - {x^2}} }}} = \left. {\left( {\arcsin x} \right)} \right|_{ - 1}^1 = \arcsin 1 - \arcsin \left( { - 1} \right) = \frac{\pi }{2} - \left( { - \frac{\pi }{2}} \right) = \pi .\]
Hence, \({c_0} = 0.\)
Similarly, we calculate the coefficients \({c_n}.\)
Multiply the equation \({x^3} = \sum\limits_{n = 0}^\infty {{c_n}{T_n}\left( x \right)} \) by \({\frac{{{T_m}\left( x \right)}}{{\sqrt {1 - {x^2}} }}},\) \(m = 1,2,3, \ldots \) and integrate it from \(-1\) to \(1\) to get
\[\int\limits_{ - 1}^1 {\frac{{{x^3}{T_m}\left( x \right)}}{{\sqrt {1 - {x^2}} }}dx
= \int\limits_{ - 1}^1 {\left( {\sum\limits_{n - 0}^\infty {{c_n}{T_n}\left( x \right)} } \right)\frac{{{T_m}\left( x \right)}}{{\sqrt {1 - {x^2}} }}dx} ,\;\;} \Rightarrow
\int\limits_{ - 1}^1 {\frac{{{x^3}{T_m}\left( x \right)}}{{\sqrt {1 - {x^2}} }}dx}
= \sum\limits_{n = 0}^\infty {\left[ {{c_n}\int\limits_{ - 1}^1 {\frac{{{T_n}\left( x \right){T_m}\left( x \right)}}{{\sqrt {1 - {x^2}} }}dx} } \right]} ,\;\; \Rightarrow
\int\limits_{ - 1}^1 {\frac{{{x^3}{T_m}\left( x \right)}}{{\sqrt {1 - {x^2}} }}dx} = \frac{\pi }{2}{c_m}\]
by the orthogonality property.
Substitute the explicit expressions for \({{T_m}\left( x \right)}\) and change the variable:
\[x = \cos t,\;\; \Rightarrow \arccos x = t,\;\; \Rightarrow - \frac{{dx}}{{\sqrt {1 - {x^2}} }} = dt.\]
The limits of integration will be as follows:
Then
\[{c_m} = \frac{2}{\pi }\int\limits_{ - 1}^1 {\frac{{{x^3}{T_m}\left( x \right)}}{{\sqrt {1 - {x^2}} }}dx}
= \frac{2}{\pi }\int\limits_0^1 {{{\cos }^3}t\cos mtdt}
= \frac{2}{\pi }\int\limits_0^\pi {\frac{1}{4}\left( {3\cos t + \cos 3t} \right)\cos mtdt}
= \frac{3}{{2\pi }}\int\limits_0^\pi {\cos t\cos mtdt} + \frac{1}{{2\pi }}\int\limits_0^\pi {\cos 3t\cos mtdt} .\]
We calculate these integrals separately.
\[\int\limits_0^\pi {\cos t\cos mtdt} =
\frac{1}{2}\int\limits_0^\pi {\left[ {\cos \left( {t - mt} \right) + \cos \left( {t + mt} \right)} \right]dt}
= \frac{1}{2}\left[ {\left. {\Big( {\frac{{\sin \left( {m - 1} \right)t}}{{m - 1}} + \frac{{\sin \left( {m + 1} \right)t}}{{m + 1}}} \Big)} \right|_0^\pi } \right] = 0,\;\;\text{if}\;\;m \ne 1.\]
For \(m = 1\),
\[\int\limits_0^\pi {{{\cos }^2}tdt}
= \frac{1}{2}\int\limits_0^\pi {\left( {1 + \cos 2t} \right)dt}
= \frac{1}{2}\left[ {\left. {\left( {t + \frac{{\sin 2t}}{2}} \right)} \right|_0^\pi } \right] = \frac{\pi }{2}.\]
Similarly, we compute the second integral:
\[\int\limits_0^\pi {\cos 3t\cos mtdt} = 0,\;\; \text{if}\;\;m \ne 3.\]
For \(m = 3,\) we have
\[\int\limits_0^\pi {{{\cos }^2}3tdt} = \frac{1}{2}\int\limits_0^\pi {\left( {1 + \cos 6t} \right)dt} = \frac{1}{2}\left[ {\left. {\left( {t + \frac{{\sin 6t}}{6}} \right)} \right|_0^\pi } \right] = \frac{\pi }{2}.\]
As one can see, the set of functions \(1,\cos t,\cos 2t,\cos 3t, \ldots ,\) \(\cos mt, \ldots \) is orthogonal on the interval \(\left[ {0,\pi } \right].\) Thus, the coefficients \({c_m}\) are
\[
{c_m} =
\begin{cases}
0, & m \ne 1,3 \\
\frac{3}{4}, & m = 1 \\
\frac{1}{4}, & m = 3
\end{cases}.
\]
Hence, the Fourier-Chebyshev series expansion of the function \(f\left( x \right) = {x^3}\) on the interval \(\left[ { - 1,1} \right]\) is
\[f\left( x \right) = {x^3} = \frac{3}{4}{T_1}\left( x \right) + \frac{1}{4}{T_3}\left( x \right).\]
