1
$\begingroup$

eThe following is an exercise in the problem section of the Gaussian Quadrature chapter.

The theorem: enter image description here

Derive a formula of the form $$\int_{a}^{b} f(x)dx \approx w_0f(x_0) + w_1f(x_1) + w_2f'(x_2) + w_3f'(x_3)$$

I don't really know how to even start this. I have the Gaussian Quadrature theorem, but...I suppose I don't know how to use it. Can anyone show me how to do this?

We have four weights, $w_0, w_1,w_2,w_3$ and four nodes $x_0, x_1,x_2,x_3$. So our polynomial $q(x)$ will be of degree $4$. (Will the number of weights always determine the degree of the polynomial?). So from $$ \int_{a}^{b} x^kq(x)dx = 0$$ where $0 \leq k \leq n$ and the degree of $q(x)$ is $n+1$, we can write $$\int q(x)dx = \int xq(x) = \int x^2q(x) = 0$$ right?

From here I don't know what to do.

$\endgroup$
2
$\begingroup$

Let's look first at $\int_{-1}^{1} f(t) \ dt$.

We want the formula to evaluate $\int_{-1}^{1} dt, \int_{-1}^{1} t \ dt, \int_{-1}^{1} t^{2} \ dt, \int_{-1}^{1} t^{3} \ dt, \int_{-1}^{1} t^{4} \ dt, \int_{-1}^{1} t^{5} \ dt, \int_{-1}^{1} t^{6} \ dt $ and $\int_{-1}^{1} t^{7} \ dt$ exactly.

That leads to the following system of equations:

$$ 2 = \omega_{0} + \omega_{1}$$

$$ 0 = \omega_{0} x_{0} + \omega_{1} x_{1}+ \omega_{2}+\omega_{3}$$

$$ \frac{2}{3} = \omega_{0} x_{0}^{2} + \omega_{1} x_{1}^{2} + 2 \omega_{2} x_{2} + 2 \omega_{3} x_{3}$$

$$ 0 = \omega_{0} x_{0}^{3} + \omega_{1} x_{1}^{3}+ 3 \omega_{2} x_{2}^{2} + 3 \omega_{3} x_{3}^{2} $$

$$ \frac{2}{5} = \omega_{0} x_{0}^{4} + \omega_{1} x_{1}^{4}+ 4 \omega_{2} x_{2}^{3} + 4 \omega_{3} x_{3}^{3} $$

$$0 = \omega_{0} x_{0}^{5} + \omega_{1} x_{1}^{5} + 5 \omega_{2} x_{2}^{4} + 5 \omega_{3} x_{3}^{4} $$

$$\frac{2}{7} = \omega_{0} x_{0}^{6} + \omega_{1} x_{1}^{6} + 6 \omega_{2} x_{2}^{5} + 6\omega_{3} x_{3}^{5} $$

$$ 0 = \omega_{0} x_{0}^{7} + \omega_{1} x_{1}^{7} + 7 \omega_{2} x_{2}^{6} + 7\omega_{3} x_{3}^{6}$$

Solve the system using a numerical solver.

Then use the the fact that $$\int_{a}^{b} f(x) \ dx = \frac{b-a}{2} \int_{-1}^{1} f \Big(a+(1+t)\frac{b-a}{2} \Big) \ dt$$

$\endgroup$

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.