0

I am trying to determine the node values in the Gaussian type quadrature formula given by:

enter image description here

How do I find node values for the Gaussian quadrature when the weight function is $ w(x) =\frac {1}{\sqrt{1-x^2}} $ (Chebyshev Gauss)?

I know that the quadrature nodes $\phi_i$ are the roots of an orthogonal polynomial.

Would the method of unknown coefficients work in this case?

I am reading here, on Page 12 that the node values are the roots of the Legendre family of polynomials.

Since $L_3 = x^3 - \frac{3x}{5}$, the roots are $x=0$ and $x = \pm \sqrt {\frac{3}{5}}$

This doesn't seem to make sense because the roots are not all within the specified integral range of $\int_{-1}^0$

rrz0
  • 99

1 Answers1

1

Some partial results: we want to work out moments $$\begin{align}M_k&=\int_{-1}^0\frac{x^k}{\sqrt{1-x^2}}dx=(-1)^k\int_0^1\frac{x^k}{\sqrt{1-x^2}}dx\\ &=\frac{(-1)^k}2\cdot2\int_0^{\pi/2}\sin^k\theta\,d\theta=\frac{(-1)^k}2\text{B}\left(\frac{k+1}2,\frac12\right)\end{align}$$ Where $\text{B}(u,v)$ is the Beta function. So for even $k$, $$M_k=\frac{\pi k!}{2^{k+1}\left[\left(\frac k2\right)!\right]^2}$$ While for odd $k$, $$M_k=\frac{-2^k\left(\frac{k-1}2\right)!\left(\frac{k+1}2\right)!}{(k+1)!}$$ Table for the first few $k$: $$\begin{array}{c|c}k&M_k\\ \hline 0&\frac{\pi}2\\ 1&-1\\ 2&\frac{\pi}4\\ 3&-\frac23\\ 4&\frac{3\pi}{16}\\ 5&-\frac8{15}\end{array}$$ We need a cubic polynomial $x^3+ax^2+bx+c$ orthogonal to $x^k$ for $k\in{0,1,2}$. This gives us $3$ equations in $3$ unknowns: $$\begin{array}{rrrr}\frac{\pi}4a&-b&+\frac{\pi}2c&=\frac23\\ -\frac23a&+\frac{\pi}4b&-c&=-\frac{3\pi}{16}\\ \frac{3\pi}{16}a&-\frac23b&+\frac{\pi}4c&=\frac8{15}\end{array}$$ It would be nice to represent $a$, $b$, and $c$ as rational functions of $\pi$, but numerically I get $$\begin{align}a&=1.614477692\\ b&=0.702011721\\ c&=0.064088878\end{align}$$ EDIT: WolframAlpha spits out $$\begin{align}a&=\frac{2\left(256-27\pi^2\right)}{5\pi\left(88-9\pi^2\right)}\\ b&=-\frac{3\left(-448+45\pi^2\right)}{20\left(-88+9\pi^2\right)}\\ c&=-\frac{2048-207\pi^2}{30\pi\left(88-9\pi^2\right)}\end{align}$$ So the cubic equation is $$x^3+1.614477692x^2+0.702011721x+0.064088878=0$$ Roots are $$x\in\{-0.123876429,-0.94052274,-0.550078523\}$$ We want to be exact for $k\in\{0,1,2\}$: $$\begin{array}{rrrr}A_1&+A_2&+A_3&=\frac{\pi}2\\ x_1A_1&+x_2A_2&+x_3A_3&=-1\\ x_1^2A_1&+x_2^2A_2&+x_3^2A_3&=\frac{\pi}4\end{array}$$ I get $$\begin{align}A_1&=0.308760976\\ A_2&=0.685202238\\ A_3&=0.576833113 \end{align}$$ I checked that $$\sum_{j=1}^3A_jx_j^k=M_k$$ For $k\in\{0,1,2,3,4,5\}$ to the accuracy provided by Excel, so the formula is Gaussian.

For Legendre polynomials, the weight function is $w(x)=1$ and the interval is $(-1,1)$, neither of which matches the statement of your problem.

user5713492
  • 15,938
  • Thank you, this helps. Regarding the first half of the answer, why don't you use the normal basis $A_0, A_1, A_2 -> x^0, x^1, x^2$ ? This is what is usually done in the method of undetermined coefficients. – rrz0 Jan 01 '20 at 11:09
  • I'm not sure what you are trying to say here. I think perhaps you mean $$\int_{-1}^0\frac{x^j}{\sqrt{1-x^2}}dx=A_1x_1^j+A_2x_2^j+A_3x_3^j$$ Exact for $j\in{0,1,2,3,4,5}$ and then solve the $6$ nonlinear equations in $6$ unknowns as suggested in the discussion around eq. $(2.11)$ of the reference you provided? That would be really hard... – user5713492 Jan 01 '20 at 18:04
  • ...Instead I am using a $2$-step approach: the first step is to find the orthogonal polynomial $p_3(x)$ via equations $(2.18)$ of your reference and then to find its roots $x_1$, $x_2$, and $x_3$ as in Theorem $1$. Then I applied the equations at the top of my previous comment for $j\in{0,1,2}$, now $3$ linear equations in $3$ unknowns. – user5713492 Jan 01 '20 at 18:05