Romberg Integration (Recurrence Table)
Compute the integral over the finite interval (a, b) using the Romberg method and derive the recurrence table.
Create a Romberg recurrence table using Richardson extrapolation, which forms the calculation process for the Romberg integration.
The last diagonal element represents the optimal solution for the desired integral value.
The last diagonal element represents the optimal solution for the desired integral value.
| \(I={\large\int_a^{b}}f(x)dx=R_{\small n}^{\small\ k}+O((\frac{b-a}{2^n})^{\small 2k+2})\\ \hspace{10px} R_{\small 0}^{\small\ 0}\\ \hspace{10px} R_{\small 1}^{\small\ 0}\hspace{30px} R_{\small 1}^{\small\ 1}\\ \hspace{10px} \cdots\hspace{40px} \cdots\hspace{30px} \ddots \\ \hspace{10px} R_{\small n-1}^{\small\ 0}\hspace{15px} R_{\small n-1}^{\small\ 1}\hspace{15px} \cdots \hspace{10px} R_{\small n-1}^{\small\ n-1}\\ \hspace{10px} R_{\small n}^{\small\ 0}\hspace{30px} R_{\small n}^{\small\ 1}\hspace{30px} \cdots \hspace{10px} R_{\small n}^{\small\ n-1}\hspace{20px} R_{\small n}^{\small\ n}\\ \) |
With the Romberg integration method, the integral value converges as n increases.
The number of subintervals is calculated by doubling from 2, 4, 8, 16, ..., N=2n, doubling the number of subintervals at each step.
If the endpoints are not analytic, the value is determined using the midpoint rule. Example: sin(x)/x, [0.π/2]
For periodic functions, the values are determined using a nonlinear transformation. Example: (cos(64x))^(2), [0, π]
The number of subintervals is calculated by doubling from 2, 4, 8, 16, ..., N=2n, doubling the number of subintervals at each step.
If the endpoints are not analytic, the value is determined using the midpoint rule. Example: sin(x)/x, [0.π/2]
For periodic functions, the values are determined using a nonlinear transformation. Example: (cos(64x))^(2), [0, π]
| \(\normalsize Romberg\ integration\\ (1)\ Trapezoidal\ rule\\ \hspace{10px} R_{\small 0}^{\small\ 0}= {\large\frac{h_0}{2}}\{f(a)+f(b)\}, \hspace{20px}h_0=b-a\\ \hspace{8px} R_{\small n}^{\small\ 0}= {\large\frac{R_{\small n-1}^{\small\ 0}}{2}}+h_n{\large \displaystyle \sum_{\tiny j=1}^{\small 2^{n-1}}}f(a+(2j-1)h_n)\\ \hspace{190px}h_n={\large\frac{b-a}{2^n}}\\ (2)\ Midpoint\ rule\\ \hspace{10px} R_{\small 0}^{\small\ 0}= h_0f(a+{\large\frac{h_0}{2}}),\hspace{20px}h_0=b-a\\ \hspace{10px} R_{\small n}^{\small\ 0}= h_n{\large \displaystyle \sum_{\tiny j=1}^{\small 2^n}}f(a+(j-\frac{1}{2})h_n), \hspace{15px}h_n={\large\frac{b-a}{2^n}}\\ (3)\ R_{\small n}^{\small\ k}= {\large\frac{4^{\small k} R_{\small n}^{\small\ k-1}-R_{\small n-1}^{\small\ k-1}}{4^{\small k}-1}}\\ (4)\ Relative\ Error\hspace{30px} \epsilon_{\small n}=\left|\frac{R_{\small n}^{\small\ n}-R_{\small n-1}^{\small\ n-1}}{R_{\small n}^{\small\ n}}\right|\\ Non-linear\ substitution\ in\ x\\ \hspace{25px} I={\large\int_a^{b}}f(x)dx\ ={\large\int_{\small-1}^{\small1}}f(x)\ \frac{b-a}{2}\ \frac{3(1-u^2)}{2}du\\ \hspace{100px}x=\frac{b-a}{2}t+\frac{b+a}{2},\hspace{20px}t=\frac{u}{2}(3-u^2)\) |
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