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11.3 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
{\mathop{\mathop{\mathop{∫
}\nolimits }}\nolimits }_{−1}^{1}{P}_{
n}^{2}(x)dx = {2n + 1\over
2} ,
(11.27)
which can be determined using the relation 5. twice to obtain a recurrence relation
\begin{eqnarray}
{\mathop{\mathop{\mathop{∫
}\nolimits }}\nolimits }_{−1}^{1}{P}_{
n}^{2}(x)dx& =& {\mathop{\mathop{\mathop{∫
}\nolimits }}\nolimits }_{−1}^{1}{P}_{
n}(x){(2n − 1)x{P}_{n−1}(x) − (n − 1){P}_{n−2}(x)\over
n} dx %&
\\
& =& {(2n − 1)\over
n} {\mathop{\mathop{\mathop{∫
}\nolimits }}\nolimits }_{−1}^{1}x{P}_{
n}(x){P}_{n−1}(x)dx %&
\\
& =& {(2n − 1)\over
n} {\mathop{\mathop{\mathop{∫
}\nolimits }}\nolimits }_{−1}^{1}{(n + 1){P}_{n+1}(x) + n{P}_{n−1}(x)\over
2n + 1} {P}_{n−1}(x)dx%&
\\
& =& {(2n − 1)\over
2n + 1} {\mathop{\mathop{\mathop{∫
}\nolimits }}\nolimits }_{−1}^{1}{P}_{
n−1}^{2}(x)dx, %&(11.28) \\
\end{eqnarray}
and the use of a very simple integral to fix this number for
n = 0 ,
{\mathop{\mathop{\mathop{∫
}\nolimits }}\nolimits }_{−1}^{1}{P}_{
0}^{2}(x)dx = 2.
(11.29)
So we can now develop any function on [−1,1]
in a Fourier-Legendre series
\begin{eqnarray}
f(x)& =& {\mathop{∑
}}_{n}{A}_{n}{P}_{n}(x) %&
\\
{A}_{n}& =& {2n + 1\over
2} {\mathop{\mathop{\mathop{∫
}\nolimits }}\nolimits }_{−1}^{1}f(x){P}_{
n}(x)dx%&(11.30) \\
\end{eqnarray}
Example 11.4:
Find the Fourier-Legendre series for
f(x) = \left \{\array{
0,& − 1 < x < 0
\cr
1,&0 < x < 1} \right ..
(11.31)
Solution:
We find
\begin{eqnarray}{
A}_{0}& =& {1\over
2}{\mathop{ \mathop{\mathop{∫
}\nolimits }}\nolimits }_{0}^{1}{P}_{
0}(x)dx = {1\over
2}, %&(11.32)
\\
{A}_{1}& =& {3\over
2}{\mathop{ \mathop{\mathop{∫
}\nolimits }}\nolimits }_{0}^{1}{P}_{
1}(x)dx = {1\over
4}, %&(11.33)
\\
{A}_{2}& =& {5\over
2}{\mathop{ \mathop{\mathop{∫
}\nolimits }}\nolimits }_{0}^{1}{P}_{
2}(x)dx = 0, %&(11.34)
\\
{A}_{3}& =& {7\over
2}{\mathop{ \mathop{\mathop{∫
}\nolimits }}\nolimits }_{0}^{1}{P}_{
3}(x)dx = −{7\over
16}.%&(11.35) \\
\end{eqnarray}
All other coefficients for even n
are zero, for odd n
they can be evaluated explicitly.