Two Point Gauss-Laguerre Integration











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I have the following question and in the notes we are taught about general gaussian integration and gauss-legendre, but only briefly about gauss-laguerre so I am a bit stuck.



The question is as follows:




$$ text{Using the two point Gauss-laguerre integration method estimate:}$$
$$frac{1}{e^pi}int_{-pi}^{infty} (ye^{-y}sin(y) + pi e^{-y} sin(y))text{ }dy$$




From the notes I know that: $int_{0}^{infty}e^{-x}f(x) dx approx sum_{i=1}^{n}w_if(x_i)$ where $x_i$ is the i-th root of Laguerre polynomial $L_n(x).$



So what I have done so far is re arrange the original equation to get it into the format $int_{-pi}^{infty} e^{-y}f(y)dy$ and work out the roots of $L_2(x)$:



$$frac{1}{e^pi}int_{-pi}^{infty} e^{-y}(ysin(y) + pi sin(y))text{ }dy$$
$$L_{2}(x)=frac{x^2}{2}-2x+1$$ where the roots are: $$x_{0}=2-sqrt{2} text{ , } x_{1}=2+sqrt{2}$$



Not really sure where to go from here or even if I have started correctly. From putting the question into wolframalpha, I know that the value is -0.5.



Any help would be appreciated.










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  • 1




    What seems to be the issue? Do you not know the expression for the weights $w_i$? Wikipedia page has it
    – Yuriy S
    Nov 12 at 16:57















up vote
0
down vote

favorite












I have the following question and in the notes we are taught about general gaussian integration and gauss-legendre, but only briefly about gauss-laguerre so I am a bit stuck.



The question is as follows:




$$ text{Using the two point Gauss-laguerre integration method estimate:}$$
$$frac{1}{e^pi}int_{-pi}^{infty} (ye^{-y}sin(y) + pi e^{-y} sin(y))text{ }dy$$




From the notes I know that: $int_{0}^{infty}e^{-x}f(x) dx approx sum_{i=1}^{n}w_if(x_i)$ where $x_i$ is the i-th root of Laguerre polynomial $L_n(x).$



So what I have done so far is re arrange the original equation to get it into the format $int_{-pi}^{infty} e^{-y}f(y)dy$ and work out the roots of $L_2(x)$:



$$frac{1}{e^pi}int_{-pi}^{infty} e^{-y}(ysin(y) + pi sin(y))text{ }dy$$
$$L_{2}(x)=frac{x^2}{2}-2x+1$$ where the roots are: $$x_{0}=2-sqrt{2} text{ , } x_{1}=2+sqrt{2}$$



Not really sure where to go from here or even if I have started correctly. From putting the question into wolframalpha, I know that the value is -0.5.



Any help would be appreciated.










share|cite|improve this question


















  • 1




    What seems to be the issue? Do you not know the expression for the weights $w_i$? Wikipedia page has it
    – Yuriy S
    Nov 12 at 16:57













up vote
0
down vote

favorite









up vote
0
down vote

favorite











I have the following question and in the notes we are taught about general gaussian integration and gauss-legendre, but only briefly about gauss-laguerre so I am a bit stuck.



The question is as follows:




$$ text{Using the two point Gauss-laguerre integration method estimate:}$$
$$frac{1}{e^pi}int_{-pi}^{infty} (ye^{-y}sin(y) + pi e^{-y} sin(y))text{ }dy$$




From the notes I know that: $int_{0}^{infty}e^{-x}f(x) dx approx sum_{i=1}^{n}w_if(x_i)$ where $x_i$ is the i-th root of Laguerre polynomial $L_n(x).$



So what I have done so far is re arrange the original equation to get it into the format $int_{-pi}^{infty} e^{-y}f(y)dy$ and work out the roots of $L_2(x)$:



$$frac{1}{e^pi}int_{-pi}^{infty} e^{-y}(ysin(y) + pi sin(y))text{ }dy$$
$$L_{2}(x)=frac{x^2}{2}-2x+1$$ where the roots are: $$x_{0}=2-sqrt{2} text{ , } x_{1}=2+sqrt{2}$$



Not really sure where to go from here or even if I have started correctly. From putting the question into wolframalpha, I know that the value is -0.5.



Any help would be appreciated.










share|cite|improve this question













I have the following question and in the notes we are taught about general gaussian integration and gauss-legendre, but only briefly about gauss-laguerre so I am a bit stuck.



The question is as follows:




$$ text{Using the two point Gauss-laguerre integration method estimate:}$$
$$frac{1}{e^pi}int_{-pi}^{infty} (ye^{-y}sin(y) + pi e^{-y} sin(y))text{ }dy$$




From the notes I know that: $int_{0}^{infty}e^{-x}f(x) dx approx sum_{i=1}^{n}w_if(x_i)$ where $x_i$ is the i-th root of Laguerre polynomial $L_n(x).$



So what I have done so far is re arrange the original equation to get it into the format $int_{-pi}^{infty} e^{-y}f(y)dy$ and work out the roots of $L_2(x)$:



$$frac{1}{e^pi}int_{-pi}^{infty} e^{-y}(ysin(y) + pi sin(y))text{ }dy$$
$$L_{2}(x)=frac{x^2}{2}-2x+1$$ where the roots are: $$x_{0}=2-sqrt{2} text{ , } x_{1}=2+sqrt{2}$$



Not really sure where to go from here or even if I have started correctly. From putting the question into wolframalpha, I know that the value is -0.5.



Any help would be appreciated.







numerical-methods gaussian-integral






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asked Nov 12 at 16:28









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  • 1




    What seems to be the issue? Do you not know the expression for the weights $w_i$? Wikipedia page has it
    – Yuriy S
    Nov 12 at 16:57














  • 1




    What seems to be the issue? Do you not know the expression for the weights $w_i$? Wikipedia page has it
    – Yuriy S
    Nov 12 at 16:57








1




1




What seems to be the issue? Do you not know the expression for the weights $w_i$? Wikipedia page has it
– Yuriy S
Nov 12 at 16:57




What seems to be the issue? Do you not know the expression for the weights $w_i$? Wikipedia page has it
– Yuriy S
Nov 12 at 16:57















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