User:Egm6322.s12.team2.Xia/RP11.4

= R 11.4 - Verify the weights of GL quadrature =

Given
Give the following table

with

$$\displaystyle w_j = \frac{2}{\left( 1-x_j^2 \right) [P'_n(x_j)]^2} \,\!$$

and

$$\displaystyle P_n(x) = {1 \over 2^n n!} {d^n \over dx^n } \left[ (x^2 -1)^n \right].$$

Find
Verify

1, $$\displaystyle w_{j}>0,\; j=1,\ldots,n$$ (11.4.1) 2, $$\displaystyle \sum_{j=1}^{n}w_{j}=2$$ (11.4.2)

Solution
Part I

From the given conditions, Using Maxima with the following commands:

p(n,x):=(1/(2^n*n!))*diff((x^2-1)^n,x,n);  %%$$\displaystyle \mathrm{p}\left( n,x\right) :=\frac{1}{{2}^{n}\,n!}\,\mathrm{diff}\left( {\left( {x}^{2}-1\right) }^{n},x,n\right) $$

w(n,x):=2/((1-x^2)*(diff(p(n,x),x))^2);    %% $$\displaystyle \mathrm{w}\left( n,x\right) :=\frac{2}{\left( 1-{x}^{2}\right) \,{\mathrm{diff}\left( \mathrm{p}\left( n,x\right) ,x\right) }^{2}}$$

we can obtain

$$\displaystyle w(1,x)= \frac{2}{1-{x}^{2}}$$ $$\displaystyle w(2,x)= \frac{2}{9\,{x}^{2}\,\left( 1-{x}^{2}\right) }$$ $$\displaystyle w(3,x)= \frac{4608}{\left( 1-{x}^{2}\right) \,{\left( 288\,{x}^{2}+72\,\left( {x}^{2}-1\right) \right) }^{2}}$$ $$\displaystyle w(4,x)= \frac{294912}{\left( 1-{x}^{2}\right) \,{\left( 3840\,{x}^{3}+2880\,x\,\left( {x}^{2}-1\right) \right) }^{2}}$$ $$\displaystyle w(5,x)= \frac{29491200}{\left( 1-{x}^{2}\right) \,{\left( 7200\,{\left( {x}^{2}-1\right) }^{2}+57600\,{x}^{4}+86400\,{x}^{2}\,\left( {x}^{2}-1\right) \right) }^{2}}$$ Then, use the command subst(x_j,x,w(n,x))

we have

$$\displaystyle w(1,0)=2$$

$$\displaystyle w(2,\pm1/\sqrt{3})=1$$

$$\displaystyle w(3,0)=\frac{8}{9}$$

$$\displaystyle w(3,\pm\sqrt{3/5})=\frac{5}{9}$$

$$\displaystyle w(4,\pm\sqrt{(3-2\sqrt{6/5})/7})=-\frac{49}{6\,\left( \sqrt{6}+2\,\sqrt{5}\right) \,\left( 2\,\sqrt{6}-3\,\sqrt{5}\right) }=\frac{18+\sqrt{30}}{36}$$

$$\displaystyle w(4,\pm\sqrt{(3+2\sqrt{6/5})/7})=-\frac{49}{6\,\left( \sqrt{6}-2\,\sqrt{5}\right) \,\left( 2\,\sqrt{6}+3\,\sqrt{5}\right) }=\frac{18-\sqrt{30}}{36}$$

$$\displaystyle w(5,0)=\frac{128}{225}$$

$$\displaystyle w(5,\pm\frac{1}{3}\sqrt{(5-2\sqrt{10/7})})=\frac{729\,\sqrt{7}}{25\,\left( \sqrt{10}+2\,\sqrt{7}\right) \,{\left( \sqrt{7}\,\sqrt{10}-4\right) }^{2}}=\frac{322+13\sqrt{70}}{900}$$

$$\displaystyle w(5,\pm\frac{1}{3}\sqrt{(5+2\sqrt{10/7})})=-\frac{729\,\sqrt{7}}{25\,\left( \sqrt{10}-2\,\sqrt{7}\right) \,{\left( \sqrt{7}\,\sqrt{10}+4\right) }^{2}}=\frac{322-13\sqrt{70}}{900}$$

Part II

Thus from the above solution we can easily obtain

$$\displaystyle n=1$$

$$\displaystyle \sum w_j =w_1=2$$

$$\displaystyle n=2$$

$$\displaystyle \sum w_j =w_1+w_2=1+1=2$$

$$\displaystyle n=3$$

$$\displaystyle \sum w_j =w_1+w_2+w_3=\frac{8}{9}+\frac{5}{9}+\frac{5}{9}=2$$

$$\displaystyle n=4$$

$$\displaystyle \sum w_j =w_1+w_2+w_3+w_4=2 \frac{18+\sqrt{30}}{36}+2 \frac{18-\sqrt{30}}{36}=2$$

$$\displaystyle n=5$$

$$\displaystyle \sum w_j =w_1+w_2+w_3+w_4+w_5=\frac{128}{225}+2 \frac{322+13\sqrt{70}}{900}+2 \frac{322-13\sqrt{70}}{900}=2$$