0
$\begingroup$

What are the steps one needs to follow to find a quadrature formula of a certain shape with maximal degree of precision.

For example: Find a quadrature formula of the following shape

$\int_1^2 (x^4 - 1)f(x)dx = A_1f(x_1) + A_2f(x_2) + A_3f(x_3) + R_3(f)$

Should I start by changing the interval to [-1, 1] and then somehow change the formula to a known orthogonal polynomial, Legendre for example ?

$\endgroup$
2
$\begingroup$

You need to construct the Gaussian quadrature for a specific weight function, $w(x) = x^4 - 1$ in your case.

Start with constructing polynomials orthogonal with that weight, i.e. $$ \int_1^2 P_m(x) P_n(x) w(x) = C_n \delta_{nm},\qquad \operatorname{deg} P_n(x) = n. $$ Since you need a quadrature with three points, the nodes of the quadrature are roots of the $P_3(x)$ polynomial. The easiest way to build the orthogonal set by hand is to apply Gramm-Schmidt process to the $1, x, x^2, x^3$ set.

$P_0(x)$ is constant and could be taken as $1$ without loosing any generality. $P_1(x) = x - \alpha$. Since $P_1(x)$ should be orthogonal to $P_0(x)$ we have $$ \int_1^2 P_1(x) P_0(x) (x^4-1)dx = \int_1^2 (x-\alpha)(x^4-1) dx = 9 - \frac{26}{5}\alpha = 0, $$ so $\alpha = \frac{45}{26}$.

Next, $P_2(x) = x^2 - \beta x - \gamma$. Integrating with $P_0(x)$ and $P_1(x)$ we have $$ \int_1^2 P_2(x) P_0(x) (x^4-1)dx = \frac{332}{21} - 9\beta -\frac{26}{5}\gamma = 0\\ \int_1^2 P_2(x) P_1(x) (x^4-1)dx = \frac{555}{728} - \frac{127\beta}{546} = 0, $$ so $\beta = \frac{1665}{508}, \gamma = -\frac{56165}{21336}$. (Ewww)

Finally, $P_3(x) = x^3 - \delta x^2 - \epsilon x - \zeta$. $$ \int_1^2 P_3(x) P_0(x)(x^4-1)dx = -\frac{332 \delta }{21}-9 \varepsilon -\frac{26 \zeta }{5}+\frac{225}{8}\\ \int_1^2 P_3(x) P_1(x)(x^4-1)dx = -\frac{555 \delta }{728}-\frac{127 \varepsilon }{546}+\frac{17783}{9360}\\ \int_1^2 P_3(x) P_2(x)(x^4-1)dx = \frac{18419}{284480}-\frac{40337 \delta }{2987040} $$ And the $P_3(x)$ is $$P_3(x) = x^3 - \frac{386799}{80674}x^2 +\frac{4566058}{605055}x - \frac{2512489}{645392}$$ with roots $$ x_1 \approx 1.24311506656362394748261056490...\\ x_2 \approx 1.62867605333517443866228087173...\\ x_3 \approx 1.92280193362526141007198140964... $$

Now we are able to compute the weights by integrating Lagrange basis polynomials $$ w_i = \int_1^2 \ell_i(x) (x^4 - 1) dx, \qquad \ell_i = \prod_{j \neq i} \frac{x-x_j}{x_i - x_j}\\ w_1 \approx 0.498562063242571816059749610...\\ w_2 \approx 2.242930705042292632885672141...\\ w_3 \approx 2.458507231715135551054578248... $$

Checking that it actually works for integrating every polynomial of degree up to five including: $$ \int_1^2 1 (x^4 - 1)dx = 5.2 = w_1 + w_2 + w_3\\ \int_1^2 x (x^4 - 1)dx = 9 = w_1 x_1 + w_2 x_2 + w_3 x_3\\ \int_1^2 x^2 (x^4 - 1)dx = 15.809523 = w_1 x_1^2 + w_2 x_2^2 + w_3 x_3^2\\ \int_1^2 x^3 (x^4 - 1)dx = 28.125 = w_1 x_1^3 + w_2 x_2^3 + w_3 x_3^3\\ \int_1^2 x^4 (x^4 - 1)dx = 50.577777 = w_1 x_1^4 + w_2 x_2^4 + w_3 x_3^4\\ \int_1^2 x^5 (x^4 - 1)dx = 91.8 = w_1 x_1^5 + w_2 x_2^5 + w_3 x_3^5\\ \int_1^2 x^6 (x^4 - 1)dx = 167.9480519 \neq 167.947233 = w_1 x_1^6 + w_2 x_2^6 + w_3 x_3^6. $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.