Editing Classical orthogonal polynomials (section)

=== Orthogonality ===

The differential equation for a particular ''λ'' may be written (omitting explicit dependence on x)

:<math>Q\ddot{f}_n+L\dot{f}_n+\lambda_nf_n=0</math>

multiplying by <math>(R/Q)f_m</math> yields

:<math>Rf_m\ddot{f}_n+\frac{R}{Q}Lf_m\dot{f}_n+\frac{R}{Q}\lambda_nf_mf_n=0</math>

and reversing the subscripts yields

:<math>Rf_n\ddot{f}_m+\frac{R}{Q}Lf_n\dot{f}_m+\frac{R}{Q}\lambda_mf_nf_m=0</math>

subtracting and integrating:

:<math>
\int_a^b \left[R(f_m\ddot{f}_n-f_n\ddot{f}_m)+
\frac{R}{Q}L(f_m\dot{f}_n-f_n\dot{f}_m)\right] \, dx
+(\lambda_n-\lambda_m)\int_a^b \frac{R}{Q}f_mf_n \, dx = 0
</math>

but it can be seen that

:<math>
\frac{d}{dx}\left[R(f_m\dot{f}_n-f_n\dot{f}_m)\right]=
R(f_m\ddot{f}_n-f_n\ddot{f}_m)\,\,+\,\,R\frac{L}{Q}(f_m\dot{f}_n-f_n\dot{f}_m)
</math>

so that:

:<math>\left[R(f_m\dot{f}_n-f_n\dot{f}_m)\right]_a^b\,\,+\,\,(\lambda_n-\lambda_m)\int_a^b \frac{R}{Q}f_mf_n \, dx=0</math>

If the polynomials ''f'' are such that the term on the left is zero, and <math>\lambda_m \ne \lambda_n</math> for <math>m \ne n</math>, then the orthogonality relationship will hold:

:<math>\int_a^b \frac{R}{Q}f_mf_n \, dx=0</math>

for <math>m \ne n</math>.