2nd vs. 3rd Order Gauss Quadrature

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I have read that using Gauss Quadrature integration,
$$int_{-1}^{1}f(x)dx=sum_i f(x_i)w_i$$
for polynomials of degree $leq2n-1$ (and otherwise it is an approximation). Using weight functions $w_i$ from wikipedia, for 2$^{nd}$ order GQ we have
$$int_{-1}^{1}f(x)dx=fbigg(frac{1}{sqrt{3}}bigg) + fbigg(-frac{1}{sqrt{3}}bigg)$$
and for 3$^{rd}$ order,
$$int_{-1}^{1}f(x)dx=frac 59 fbigg(sqrt{frac{3}{5}}bigg)+ frac 59 fbigg(-sqrt{frac{3}{5}}bigg) $$



My problem is that I don't seem to be getting the same answer for second and third order GQ for polynomials of degree $leq 3$, for which they should both yield exact solutions.



For example, for the function $f(x)=a_1 + a_2x + a_3x^2 + a_4x^3$, from first order GQ I get:
$$int_{-1}^{1}f(x)dx=2a_1 + frac{2}{3}a_3$$
which I believe is the correct solution, but for third order GQ I get



$$int_{-1}^{1}f(x)dx=frac{10}{9}a_1 + frac 23 a_3$$



This error seems to occur for any constant term in any polynomial as a result of the fact that the sum of the weight functions aren't equal for second and third order GQ. Why is this? I can't find my mistake!



Thanks in advance for your help.










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    I have read that using Gauss Quadrature integration,
    $$int_{-1}^{1}f(x)dx=sum_i f(x_i)w_i$$
    for polynomials of degree $leq2n-1$ (and otherwise it is an approximation). Using weight functions $w_i$ from wikipedia, for 2$^{nd}$ order GQ we have
    $$int_{-1}^{1}f(x)dx=fbigg(frac{1}{sqrt{3}}bigg) + fbigg(-frac{1}{sqrt{3}}bigg)$$
    and for 3$^{rd}$ order,
    $$int_{-1}^{1}f(x)dx=frac 59 fbigg(sqrt{frac{3}{5}}bigg)+ frac 59 fbigg(-sqrt{frac{3}{5}}bigg) $$



    My problem is that I don't seem to be getting the same answer for second and third order GQ for polynomials of degree $leq 3$, for which they should both yield exact solutions.



    For example, for the function $f(x)=a_1 + a_2x + a_3x^2 + a_4x^3$, from first order GQ I get:
    $$int_{-1}^{1}f(x)dx=2a_1 + frac{2}{3}a_3$$
    which I believe is the correct solution, but for third order GQ I get



    $$int_{-1}^{1}f(x)dx=frac{10}{9}a_1 + frac 23 a_3$$



    This error seems to occur for any constant term in any polynomial as a result of the fact that the sum of the weight functions aren't equal for second and third order GQ. Why is this? I can't find my mistake!



    Thanks in advance for your help.










    share|cite|improve this question
























      up vote
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      down vote

      favorite









      up vote
      0
      down vote

      favorite











      I have read that using Gauss Quadrature integration,
      $$int_{-1}^{1}f(x)dx=sum_i f(x_i)w_i$$
      for polynomials of degree $leq2n-1$ (and otherwise it is an approximation). Using weight functions $w_i$ from wikipedia, for 2$^{nd}$ order GQ we have
      $$int_{-1}^{1}f(x)dx=fbigg(frac{1}{sqrt{3}}bigg) + fbigg(-frac{1}{sqrt{3}}bigg)$$
      and for 3$^{rd}$ order,
      $$int_{-1}^{1}f(x)dx=frac 59 fbigg(sqrt{frac{3}{5}}bigg)+ frac 59 fbigg(-sqrt{frac{3}{5}}bigg) $$



      My problem is that I don't seem to be getting the same answer for second and third order GQ for polynomials of degree $leq 3$, for which they should both yield exact solutions.



      For example, for the function $f(x)=a_1 + a_2x + a_3x^2 + a_4x^3$, from first order GQ I get:
      $$int_{-1}^{1}f(x)dx=2a_1 + frac{2}{3}a_3$$
      which I believe is the correct solution, but for third order GQ I get



      $$int_{-1}^{1}f(x)dx=frac{10}{9}a_1 + frac 23 a_3$$



      This error seems to occur for any constant term in any polynomial as a result of the fact that the sum of the weight functions aren't equal for second and third order GQ. Why is this? I can't find my mistake!



      Thanks in advance for your help.










      share|cite|improve this question













      I have read that using Gauss Quadrature integration,
      $$int_{-1}^{1}f(x)dx=sum_i f(x_i)w_i$$
      for polynomials of degree $leq2n-1$ (and otherwise it is an approximation). Using weight functions $w_i$ from wikipedia, for 2$^{nd}$ order GQ we have
      $$int_{-1}^{1}f(x)dx=fbigg(frac{1}{sqrt{3}}bigg) + fbigg(-frac{1}{sqrt{3}}bigg)$$
      and for 3$^{rd}$ order,
      $$int_{-1}^{1}f(x)dx=frac 59 fbigg(sqrt{frac{3}{5}}bigg)+ frac 59 fbigg(-sqrt{frac{3}{5}}bigg) $$



      My problem is that I don't seem to be getting the same answer for second and third order GQ for polynomials of degree $leq 3$, for which they should both yield exact solutions.



      For example, for the function $f(x)=a_1 + a_2x + a_3x^2 + a_4x^3$, from first order GQ I get:
      $$int_{-1}^{1}f(x)dx=2a_1 + frac{2}{3}a_3$$
      which I believe is the correct solution, but for third order GQ I get



      $$int_{-1}^{1}f(x)dx=frac{10}{9}a_1 + frac 23 a_3$$



      This error seems to occur for any constant term in any polynomial as a result of the fact that the sum of the weight functions aren't equal for second and third order GQ. Why is this? I can't find my mistake!



      Thanks in advance for your help.







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      asked Nov 13 at 1:46









      lamdoug

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          For 3 point quadrature I believe you are missing $frac{8}{9} f(0)$ in your sum.



          It would also introduce a $frac{8}{9}a_{1}$ which would correct your toy example.






          share|cite|improve this answer





















          • Of course. I saw the zero and neglected it but obviously f(0) is not zero! Thanks.
            – lamdoug
            Nov 13 at 1:52











          Your Answer





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          up vote
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          down vote



          accepted










          For 3 point quadrature I believe you are missing $frac{8}{9} f(0)$ in your sum.



          It would also introduce a $frac{8}{9}a_{1}$ which would correct your toy example.






          share|cite|improve this answer





















          • Of course. I saw the zero and neglected it but obviously f(0) is not zero! Thanks.
            – lamdoug
            Nov 13 at 1:52















          up vote
          0
          down vote



          accepted










          For 3 point quadrature I believe you are missing $frac{8}{9} f(0)$ in your sum.



          It would also introduce a $frac{8}{9}a_{1}$ which would correct your toy example.






          share|cite|improve this answer





















          • Of course. I saw the zero and neglected it but obviously f(0) is not zero! Thanks.
            – lamdoug
            Nov 13 at 1:52













          up vote
          0
          down vote



          accepted







          up vote
          0
          down vote



          accepted






          For 3 point quadrature I believe you are missing $frac{8}{9} f(0)$ in your sum.



          It would also introduce a $frac{8}{9}a_{1}$ which would correct your toy example.






          share|cite|improve this answer












          For 3 point quadrature I believe you are missing $frac{8}{9} f(0)$ in your sum.



          It would also introduce a $frac{8}{9}a_{1}$ which would correct your toy example.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 13 at 1:50









          Nebu

          585




          585












          • Of course. I saw the zero and neglected it but obviously f(0) is not zero! Thanks.
            – lamdoug
            Nov 13 at 1:52


















          • Of course. I saw the zero and neglected it but obviously f(0) is not zero! Thanks.
            – lamdoug
            Nov 13 at 1:52
















          Of course. I saw the zero and neglected it but obviously f(0) is not zero! Thanks.
          – lamdoug
          Nov 13 at 1:52




          Of course. I saw the zero and neglected it but obviously f(0) is not zero! Thanks.
          – lamdoug
          Nov 13 at 1:52


















           

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