11.3: Fourier-Legendre Series
- Page ID
- 8322
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Since Legendre’s equation is self-adjoint, we can show that \(P_n(x)\) forms an orthogonal set of functions. To decompose functions as series in Legendre polynomials we shall need the integrals
\[\int_{-1}^1 P_n^2(x) dx = \frac{2n+1}{2}, \nonumber \]
which can be determined using the relation 5. twice to obtain a recurrence relation
\[\begin{aligned} \int_{-1}^1 P_n^2(x) dx &= \int_{-1}^1 P_n(x) \frac{(2n-1)x P_{n-1}(x)-(n-1)P_{n-2}(x)}{n} dx \nonumber\\ &=\frac{(2n-1)}{n}\int_{-1}^1 x P_n(x) P_{n-1}(x) dx \nonumber\\ &=\frac{(2n-1)}{n}\int_{-1}^1 \frac{(n+1)P_{n+1}(x) + n P_{n-1}(x)}{2n+1} P_{n-1}(x) dx \nonumber\\ &=\frac{(2n-1)}{2n+1}\int_{-1}^1 P_{n-1}^2(x) dx,\end{aligned} \nonumber \]
and the use of a very simple integral to fix this number for \(n=0\),
\[\int_{-1}^1 P_0^2(x) dx = 2. \nonumber \]
So we can now develop any function on \([-1,1]\) in a Fourier-Legendre series
\[\begin{aligned} f(x) & = \sum_n A_n P_n(x) \nonumber\\ A_n & = \frac{2n+1}{2} \int_{-1}^1 f(x) P_n(x) dx\end{aligned} \nonumber \]
Find the Fourier-Legendre series for
\[f(x) = \left\{ \begin{array}{ll} 0, & -1 < x < 0\\ 1, & 0 < x <1 \end{array} \right. . \nonumber \]
- Answer
-
We find \[\begin{aligned} A_0 &= \dfrac{1}{2} \int_0^1 P_0(x) dx = \dfrac{1}{2},\\ A_1 &= \frac{3}{2} \int_0^1 P_1(x) dx= \frac{1}{4}, \\ A_2 &= \frac{5}{2} \int_0^1 P_2(x) dx= 0, \\ A_3 &= \frac{7}{2} \int_0^1 P_3(x) dx= -\frac{7}{16} .\end{aligned} \nonumber \]
All other coefficients for even \(n\) are zero, for odd \(n\) they can be evaluated explicitly.