11.3: Fourier-Legendre Series

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

\[\begin{aligned} f(x) & = \sum_n A_n P_n(x) \nonumber\\ A_n & = \frac{2n+1}{2} \int_{-1}^1 f(x) P_n(x) dx\end{aligned} \nonumber \]

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