I am stucked at this problem:
Gauss-Lobatto quadrature is defined as:
$\int_{-1}^1 f(x)dx\approx w_1 f(-1)+w_n f(1) + \Sigma_{k=2}^{n-1}w_k f(x_k)$
($2\leq n\in\Bbb{N}$)
Where the nodes $x_k$ and the coefficients $w_k$ are choosen such that $\int_{-1}^1 f(x)dx$ is exact if $f$ is a polynomial of degree at most $2n-3$.
Find the $x_k$'s and $w_k$'s for $n=4$.
I've tried to solve it by substituting the polynomials $1,x,x^2,x^3,x^4,x^5$ for $f$, but I got stucked when I've tried to solve the resulting non-linear system of 6 equations in 6 variables.
If I haven't made any mistake, the resulting non-linear system is:
$$\begin{cases}
w_1+w_2+w_3+w_4=2 \\
-w_1+x_2w_2+x_3w_3+w_4=0 \\
w_1+x_2^2w_2+x_3^2w_3+w_4=\frac{2}{3} \\
-w_1+x_2^3w_2+x_3^3w_3+w_4=0 \\
w_1+x_2^4w_2+x_3^4w_3+w_4=\frac{2}{5} \\
-w_1+x_2^5w_2+x_3^5w_3+w_4=0
\end{cases}
$$
Thanks for any hint or help.
Best Answer
Note that $$ J[f(x)] \equiv \int_{-1}^{1} f(x) dx. $$ in invariant under change $f(x) \to f(-x)$: $$J[f(x)] = J[f(-x)]. \tag1 $$ Let's restrict the quardature to share this property $$ Q[f(x)] \equiv w_1 f(-1) + w_n f(1) + \sum_{k=2}^{n-1} w_k f(x_k)\\ Q[f(x)] = Q[f(-x)] $$ Expanding the $Q[f(-x)]$ we obtain $$ w_n f(-1) + w_1 f(-1) + \sum_{k=2}^{n-1} w_k f(-x_k) = Q[f(-x)] = Q[f(x)] = w_1 f(-1) + w_n f(1) + \sum_{k=2}^{n-1} w_k f(x_k) $$ and the condition is guaranteed if $$ w_1 = w_n\\ w_k = w_{n+1-k}\\ x_k = -x_{n+1-k}. $$ Next, every function $f(x)$ can be decomposed into symmetric and asymmetric parts $$ f(x) = f_s(x) + f_a(x)\\ f_s(x) \equiv \frac{f(x) + f(-x)}{2} = f_s(-x)\\ f_a(x) \equiv \frac{f(x) - f(-x)}{2} = -f_a(-x). $$ Note that $$ J[f(x)] = J[f_s(x) + f_a(x)] = J[f_s(x)] + J[f_a(x)] \overset{(1)}{=} J[f_s(-x)] + J[f_a(-x)] = J[f_s(x)] - J[f_a(x)]\\ J[f_a(x)] = 0\\ J[f(x)] = J[f_s(x)]. $$ Just the same is true for $Q[f]$: $$ Q[f_a(x)] = 0\\ Q[f(x)] = Q[f_s(x)]. $$ So it really sufficies to require $Q[f(x)] = J[f(x)]$ only for symmetric functions $f(x) = f(-x)$, so you can exclude odd powers of $x$ (you can still left them, though, but that would be equations $0 = 0$): $$ w_1 + w_2 = 1\\ w_1 + w_2 x_2^2 = \frac{1}{3}\\ w_1 + w_2 x_2^4 = \frac{1}{5} $$ Subtracting adjacent equations one gets $$ w_2 (1 - x_2^2) = \frac{2}{3}\\ w_2 x_2^2(1 -x_2^2) = \frac{2}{15} $$ and dividing gives $$ x_2 = \pm \sqrt\frac{1}{5}. $$ You can pick up any sign, since $x_3 = -x_2$, thus chaging sign only rearranges the points in quadrature. The weights are $$ w_2 = \frac{5}{6}, \quad w_1 = \frac{1}{6} $$ and the whole rule is $$ Q[f(x)] = \frac{f(1) + f(-1)}{6} + \frac{5}{6}\left(f\left(\sqrt\frac{1}{5}\right)+f\left(-\sqrt\frac{1}{5}\right)\right) $$