How to find the weight coefficients in the Gaussian type quadrature formula?

Question

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

I need to find the weight coefficients for the Gaussian quadrature when the weight function is $ w(x) =\frac {1}{\sqrt{1-x^2}} $

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In order to do so, I use the Gram-Schmidt process on the standard polynomial basis to get a second degree orthogoanl polynomial. The roots of this polynomial are the nodes.

I get -0.877 and -0.242 as the nodes which are within the integral range of -1 to 0.

How can I proceed from here to find the wieght coefficients?

Searching online I found this formula from Atkinson's 'An Introduction to Numerical Analysis'

During the calculation here, obne of the weight coefficients returns negative, which is impossible. I am therefore confused as to what I am doing wrong and if this is the right way to proceed. Any suggestoins are appreciated.

Answer 1

It's kind of amusing that this problem is copied from this question only the upper bound of the summation has been changed by hand, forgetting to change the subscript on $R_3(f)$. Did you complain to your instructor for not having properly composed his problem in MathJax? In my answer I worked out a table of moments $$M_k=\int_{-1}^0\frac{x^k}{\sqrt{1-x^2}}dx$$ $$\begin{array}{c|c}k&M_k\\ \hline 0&\frac{\pi}2\\ 1&-1\\ 2&\frac{\pi}4\\ 3&-\frac23\\ 4&\frac{3\pi}{16}\\ 5&-\frac8{15}\end{array}$$ We need a polynomial $\varphi_2(x)=x^2+ux+v$ orthogonal to $x^k$ for $k\in\{0,1\}$. These leads to the $2$ equations in $2$ unknowns $$\begin{array}{rrr}-u&\frac{\pi}2v&=-\frac{\pi}4\\ \frac{\pi}4u&-v&=\frac23\end{array}$$ With solutions $$\begin{align}u&=\frac{2\pi}{3\pi^2-24}\\ v&=\frac{32-3\pi^2}{6\pi^2-48}\end{align}$$ Numerically, $$\begin{align}u&=1.12023436678521858941728667302911065\\ v&=0.213163347581147484467276859105524958\end{align}$$ We can solve via the quadratic formula to get $$\begin{align}x_1&=-0.877241625135156835342909006193109829\\ x_2&=-0.242992741650061754074377666836000821\end{align}$$ Now, in my previous answer I applied the formula to $x^k$ for $k\in\{0,1\}$ to get $$\begin{array}{rrr}A_1&+A_2&=\frac{\pi}2\\ x_1A_1&+x_2A_2&=-1\end{array}$$ With solutions $$\begin{align}A_1&=0.974866349911031438041049989239156851\\ A_2&=0.595929976883865181190271702400594548\end{align}$$ And we could check that the integration formula was exact for $f(x)=x^k$ for $k\in\{0,1,2,3\}$.

So now we want to apply formula $5.3.11$ from Atkinson p. 272. We will need $a_2=\frac{A_3}{A_2}$ which is $1$ because we are using monic polynomials. Then we need $$\begin{align}\gamma_2&=\int_{-1}^0\frac{\left(\varphi_2(x)\right)^2}{\sqrt{1-x^2}}dx\\ &=\int_{-1}^0\frac{x^4+2ux^3+(u^2+2v)x^2+2uvx+v^2}{\sqrt{1-x^2}}dx\\ &=M_4+2uM_3+(u^2+2v)M_2+2uvM_1+v^2M_0\\ &=0.00964381305182564583436743261678388874\end{align}$$ Fortunately in my previous answer I worked out $$\varphi_3(x)=x^3+ax^2+bx+c$$ Where $$\begin{align}a&=\frac{2\left(256-27\pi^2\right)}{5\pi\left(88-9\pi^2\right)}\\ b&=-\frac{3\left(-448+45\pi^2\right)}{20\left(-88+9\pi^2\right)}\\ c&=-\frac{2048-207\pi^2}{30\pi\left(88-9\pi^2\right)}\end{align}$$ Numerically, $$\begin{align}a&=1.61447769167994311356947744905749821\\ b&=0.702011720837368702218533133421594847\\ c&=0.0640888776572757287520840004030105744\end{align}$$ So now we can apply the formula from Atkinson $$A_k=\frac{-a_2\gamma_2}{\varphi_2^{\prime}(x_k)\varphi_3(x+k)}$$ To get $$\begin{align}A_1&=0.974866349911031438041049989239165999\\ A_2&=0.595929976883865181190271702400598399\end{align}$$ Wikipedia seems to have a nicer formula $$A_k=\frac{a_{n-1}\gamma_{n-1}}{\varphi_n^{\prime}(x_k)\varphi_{n-1}(x_k)}=\frac{a_1\gamma_1}{\varphi_2^{\prime}(x_k)\varphi_1(x_k)}$$ Again because we are working in monic polynomials, $a_1=A_2/A_1=1$ and if we let $\varphi_1(x)=x+w$ then the condition of orthogonality to $f(x)=1$ is $-1+\frac{\pi}2w=0$ so $$w=\frac2{\pi}=0.636619772367581343075535053490057474$$ and then $$\begin{align}\gamma_1&=\int_{-1}^0\frac{\left(\varphi_1(x)\right)^2}{\sqrt{1-x^2}}dx\\ &=\int_{-1}^0\frac{x^2+2wx+w^2}{\sqrt{1-x^2}}dx\\ &=M_2+2wM_1+w^2M_0\\ &=0.148778391029866966540125792329818225\end{align}$$ And now Wikipedia producees $$\begin{align}A_1&=0.974866349911031438041049989239157236\\ A_2&=0.595929976883865181190271702400594740\end{align}$$ All of these results are consistent to IEEE-$754$ quadruple precision. Wikipedia's formula was a little nicer than Atkinson's in that it used the lower degree polynomial $\varphi_{n-1}(x)$ rather than $\varphi_{n+1}(x)$.