find coefficients of a numerical integration formula












0












$begingroup$


we have this formula for calculating integral
$int_{0}^{h} xf(x) dx = (h)^2 [A_0 f(0) + A_1f(h) + A_2f(2h)] + R \$

the goal is to find $A_0,A_1, A_2$ and $R$, when $f^{primeprimeprime}$ is constant.



I tried to get to this formula by taking integral of Taylor Expansion of $xf(x)$ but it didn't work, or I messed up in it.






integration numerical-methods







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$endgroup$

















    0












    $begingroup$


    we have this formula for calculating integral
    $int_{0}^{h} xf(x) dx = (h)^2 [A_0 f(0) + A_1f(h) + A_2f(2h)] + R \$

    the goal is to find $A_0,A_1, A_2$ and $R$, when $f^{primeprimeprime}$ is constant.



    I tried to get to this formula by taking integral of Taylor Expansion of $xf(x)$ but it didn't work, or I messed up in it.






    integration numerical-methods







    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      we have this formula for calculating integral
      $int_{0}^{h} xf(x) dx = (h)^2 [A_0 f(0) + A_1f(h) + A_2f(2h)] + R \$

      the goal is to find $A_0,A_1, A_2$ and $R$, when $f^{primeprimeprime}$ is constant.



      I tried to get to this formula by taking integral of Taylor Expansion of $xf(x)$ but it didn't work, or I messed up in it.






      integration numerical-methods







      share|cite|improve this question









      $endgroup$




      we have this formula for calculating integral
      $int_{0}^{h} xf(x) dx = (h)^2 [A_0 f(0) + A_1f(h) + A_2f(2h)] + R \$

      the goal is to find $A_0,A_1, A_2$ and $R$, when $f^{primeprimeprime}$ is constant.



      I tried to get to this formula by taking integral of Taylor Expansion of $xf(x)$ but it didn't work, or I messed up in it.






      integration numerical-methods




      integration numerical-methods






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Jan 8 at 16:47









      SajadSajad

      103




      103






















          1 Answer
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          $begingroup$

          The third derivative $f$ being constant means that $f$ is a cubic, so let $f(x)=ax^3+b^2x+cx+d$. Plug that into your equation. The result needs to be true for all $a,b,c,d$. You will get simultaneous equations for $A_0,A_1,A_2$. Note that $f(x)=0$ tells us that $R=0$. It will fail because you have four equations in three unknowns.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            so it's ok that for R to be 0? cause I did this too but I thought it's wrong :))
            $endgroup$
            – Sajad
            Jan 8 at 16:58










          • $begingroup$
            Yes, $R$ has to be $0$. If it weren't, you could let $h$ go to zero and have the left be very small, smaller than $R$. Some of the $A$s are $0$ in formulas like this as well. There is no solution to the problem as stated. You need one more point and one more $A$ on the right to make it work, or to demand that $f''(x)$ is constant so you only have a quadratic.
            $endgroup$
            – Ross Millikan
            Jan 8 at 17:01













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          1 Answer
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          1 Answer
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          active

          oldest

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          1












          $begingroup$

          The third derivative $f$ being constant means that $f$ is a cubic, so let $f(x)=ax^3+b^2x+cx+d$. Plug that into your equation. The result needs to be true for all $a,b,c,d$. You will get simultaneous equations for $A_0,A_1,A_2$. Note that $f(x)=0$ tells us that $R=0$. It will fail because you have four equations in three unknowns.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            so it's ok that for R to be 0? cause I did this too but I thought it's wrong :))
            $endgroup$
            – Sajad
            Jan 8 at 16:58










          • $begingroup$
            Yes, $R$ has to be $0$. If it weren't, you could let $h$ go to zero and have the left be very small, smaller than $R$. Some of the $A$s are $0$ in formulas like this as well. There is no solution to the problem as stated. You need one more point and one more $A$ on the right to make it work, or to demand that $f''(x)$ is constant so you only have a quadratic.
            $endgroup$
            – Ross Millikan
            Jan 8 at 17:01


















          1












          $begingroup$

          The third derivative $f$ being constant means that $f$ is a cubic, so let $f(x)=ax^3+b^2x+cx+d$. Plug that into your equation. The result needs to be true for all $a,b,c,d$. You will get simultaneous equations for $A_0,A_1,A_2$. Note that $f(x)=0$ tells us that $R=0$. It will fail because you have four equations in three unknowns.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            so it's ok that for R to be 0? cause I did this too but I thought it's wrong :))
            $endgroup$
            – Sajad
            Jan 8 at 16:58










          • $begingroup$
            Yes, $R$ has to be $0$. If it weren't, you could let $h$ go to zero and have the left be very small, smaller than $R$. Some of the $A$s are $0$ in formulas like this as well. There is no solution to the problem as stated. You need one more point and one more $A$ on the right to make it work, or to demand that $f''(x)$ is constant so you only have a quadratic.
            $endgroup$
            – Ross Millikan
            Jan 8 at 17:01
















          1












          1








          1





          $begingroup$

          The third derivative $f$ being constant means that $f$ is a cubic, so let $f(x)=ax^3+b^2x+cx+d$. Plug that into your equation. The result needs to be true for all $a,b,c,d$. You will get simultaneous equations for $A_0,A_1,A_2$. Note that $f(x)=0$ tells us that $R=0$. It will fail because you have four equations in three unknowns.






          share|cite|improve this answer











          $endgroup$



          The third derivative $f$ being constant means that $f$ is a cubic, so let $f(x)=ax^3+b^2x+cx+d$. Plug that into your equation. The result needs to be true for all $a,b,c,d$. You will get simultaneous equations for $A_0,A_1,A_2$. Note that $f(x)=0$ tells us that $R=0$. It will fail because you have four equations in three unknowns.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 8 at 17:01

























          answered Jan 8 at 16:53









          Ross MillikanRoss Millikan

          294k23198371




          294k23198371












          • $begingroup$
            so it's ok that for R to be 0? cause I did this too but I thought it's wrong :))
            $endgroup$
            – Sajad
            Jan 8 at 16:58










          • $begingroup$
            Yes, $R$ has to be $0$. If it weren't, you could let $h$ go to zero and have the left be very small, smaller than $R$. Some of the $A$s are $0$ in formulas like this as well. There is no solution to the problem as stated. You need one more point and one more $A$ on the right to make it work, or to demand that $f''(x)$ is constant so you only have a quadratic.
            $endgroup$
            – Ross Millikan
            Jan 8 at 17:01




















          • $begingroup$
            so it's ok that for R to be 0? cause I did this too but I thought it's wrong :))
            $endgroup$
            – Sajad
            Jan 8 at 16:58










          • $begingroup$
            Yes, $R$ has to be $0$. If it weren't, you could let $h$ go to zero and have the left be very small, smaller than $R$. Some of the $A$s are $0$ in formulas like this as well. There is no solution to the problem as stated. You need one more point and one more $A$ on the right to make it work, or to demand that $f''(x)$ is constant so you only have a quadratic.
            $endgroup$
            – Ross Millikan
            Jan 8 at 17:01


















          $begingroup$
          so it's ok that for R to be 0? cause I did this too but I thought it's wrong :))
          $endgroup$
          – Sajad
          Jan 8 at 16:58




          $begingroup$
          so it's ok that for R to be 0? cause I did this too but I thought it's wrong :))
          $endgroup$
          – Sajad
          Jan 8 at 16:58












          $begingroup$
          Yes, $R$ has to be $0$. If it weren't, you could let $h$ go to zero and have the left be very small, smaller than $R$. Some of the $A$s are $0$ in formulas like this as well. There is no solution to the problem as stated. You need one more point and one more $A$ on the right to make it work, or to demand that $f''(x)$ is constant so you only have a quadratic.
          $endgroup$
          – Ross Millikan
          Jan 8 at 17:01






          $begingroup$
          Yes, $R$ has to be $0$. If it weren't, you could let $h$ go to zero and have the left be very small, smaller than $R$. Some of the $A$s are $0$ in formulas like this as well. There is no solution to the problem as stated. You need one more point and one more $A$ on the right to make it work, or to demand that $f''(x)$ is constant so you only have a quadratic.
          $endgroup$
          – Ross Millikan
          Jan 8 at 17:01




















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