Gauss-Jacobi Numerical Integration
Calculate the integral over the finite interval (a, b) using the Gauss-Jacobi integration method.
| \(\normalsize Gauss-Jacobi\ quadrature\\ {\large\int_{\small -1}^{\small 1}}(1-x)^{\alpha}(1+x)^{\beta}f(x)dx\simeq{\large\displaystyle \sum_{\small i=1}^{n}}w_{i}f(x_i)\\ {\large\int_{\small -1}^{\small 1}}g(x)dx\simeq{\large\displaystyle \sum_{\small i=1}^{n}}{\large\frac{w_i}{(1-x_i)^{\alpha}(1+x_i)^{\beta}}}g(x_i)\\ \hspace{130px}g(x)=(1-x)^{\alpha}(1+x)^{\beta}f(x)\\\) |
The integrand f(x) is assumed to be analytic and not a periodic function.
| \(\normalsize Gaussian\ quadrature\\ \hspace{20px} {\large\int_{\small a}^{\small b}}w(x)f(x)dx\simeq{\large\displaystyle \sum_{\small i=1}^{n}}w_{i}f(x_i), \hspace{20px} {\large\int_{\small a}^{\small b}}g(x)dx\simeq{\large\displaystyle \sum_{\small i=1}^{n}}{\large\frac{w_{i}}{w(x_i)}}g(x_i)\\ Gauss-Jacobi\ quadrature\\ \hspace{30px} interval(a,b):\hspace{20px} (-1,\ 1)\\ \hspace{30px} w(x):\hspace{80px} (1-x)^{\alpha}(1+x)^{\beta}\\ \hspace{30px} polynomialsl:\hspace{10px} J_n^{\alpha,\beta} (x)\\ \) |
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