Trapezoid rule error












1












$begingroup$


I am trying to compute the error in the trapezoid rule integration for a function $f(x)$
in the interval $[a,b]$.



I believe I have to Taylor-expand $f(x)$ around $x=a$



$f(a) + (x-a)f'(a)+ 1/2 (x-a)^2 f''(a) + ...$



and then do the integral of this, to later use it when comparing with the trapezoid rule:



$(b-a) (f(b) + f(a))/2$



However I am stuck on the integration of the Taylor expansion and on how to deduct from this the trapezoid approximation.



Can someone give me a hint on how to proceed? Thanks.






integration taylor-expansion







share|cite|improve this question











$endgroup$

















    1












    $begingroup$


    I am trying to compute the error in the trapezoid rule integration for a function $f(x)$
    in the interval $[a,b]$.



    I believe I have to Taylor-expand $f(x)$ around $x=a$



    $f(a) + (x-a)f'(a)+ 1/2 (x-a)^2 f''(a) + ...$



    and then do the integral of this, to later use it when comparing with the trapezoid rule:



    $(b-a) (f(b) + f(a))/2$



    However I am stuck on the integration of the Taylor expansion and on how to deduct from this the trapezoid approximation.



    Can someone give me a hint on how to proceed? Thanks.






    integration taylor-expansion







    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      I am trying to compute the error in the trapezoid rule integration for a function $f(x)$
      in the interval $[a,b]$.



      I believe I have to Taylor-expand $f(x)$ around $x=a$



      $f(a) + (x-a)f'(a)+ 1/2 (x-a)^2 f''(a) + ...$



      and then do the integral of this, to later use it when comparing with the trapezoid rule:



      $(b-a) (f(b) + f(a))/2$



      However I am stuck on the integration of the Taylor expansion and on how to deduct from this the trapezoid approximation.



      Can someone give me a hint on how to proceed? Thanks.






      integration taylor-expansion







      share|cite|improve this question











      $endgroup$




      I am trying to compute the error in the trapezoid rule integration for a function $f(x)$
      in the interval $[a,b]$.



      I believe I have to Taylor-expand $f(x)$ around $x=a$



      $f(a) + (x-a)f'(a)+ 1/2 (x-a)^2 f''(a) + ...$



      and then do the integral of this, to later use it when comparing with the trapezoid rule:



      $(b-a) (f(b) + f(a))/2$



      However I am stuck on the integration of the Taylor expansion and on how to deduct from this the trapezoid approximation.



      Can someone give me a hint on how to proceed? Thanks.






      integration taylor-expansion




      integration taylor-expansion






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Feb 18 '16 at 22:05







      Estefy

















      asked Feb 18 '16 at 21:46









      EstefyEstefy

      64




      64






















          1 Answer
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          $begingroup$

          Estimating the error with the Taylor expansion is quite messy.



          For an alternative, consider the local error on a subinterval $[x_n, x_{n+1}]$ with $h = x_{n+1} - x_n$.



          The error in approximating the integral with the trapezoidal formula is



          $$E_n = frac{h}{2}[f(x_n) + f(x_{n+1})] - int_{x_n}^{x_{n+1}} f(x) , dx.$$



          Integration by parts shows this to be



          $$E_n = int_{x_n}^{x_{n+1}}(x-c)f'(x) , dx,$$



          where $c = (x_{n+1}+x_n)/2$ is the midpoint.



          To see this, note that



          $$x_{n+1} - c = c - x_n = frac{x_{n+1} - x_n}{2} = frac{h}{2},$$



          and with $u = (x-c)$ and $dv = f'(x)dx$, integration by parts yields



          $$int_{x_n}^{x_{n+1}}(x-c)f'(x) , dx = (x-c)f(x)|_{x_n}^{x_{n+1}} - int_{x_n}^{x_{n+1}} f(x) , dx = E_n.$$



          If the derivative is bounded as $|f'(x)| leqslant M$, we can bound the error as



          $$|E_n| leqslant int_{x_n}^{x_{n+1}}|x-c||f'(x)| , dx \ leqslant Mint_{x_n}^{x_{n+1}}|x-c| , dx \ = frac{M}{4}(x_{n+1} - x_n)^2 \ = frac{M}{4}h^2.$$



          Approximating the integral over $[a,b]$ with $m$ subintervals of length $h = (b-a)/m$, has a global error bound of



          $$|GE| leqslant sum_{n=1}^{m} |E_n| = mfrac{M}{4}h^2 = frac{M(b-a)^2}{4m}$$



          If the second derivative is bounded as $|f''(x)| leqslant M$, then we can demonstrate $O(h^3)$ local accuracy. An integration by parts of the previous integral for $E_n$ yields



          $$E_n = frac1{2} int_{x_n}^{x_{n+1}} [(h/2)^2 - (x-c)^2]f''(x) , dx.$$



          Using the bound for $f''$ and integrating we obtain the local and global error bounds



          $$|E_n| leqslant frac{M}{12}h^3, \ |GE| leqslant frac{M(b-a)^3}{12m^2}.$$






          share|cite|improve this answer











          $endgroup$













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            $begingroup$

            Estimating the error with the Taylor expansion is quite messy.



            For an alternative, consider the local error on a subinterval $[x_n, x_{n+1}]$ with $h = x_{n+1} - x_n$.



            The error in approximating the integral with the trapezoidal formula is



            $$E_n = frac{h}{2}[f(x_n) + f(x_{n+1})] - int_{x_n}^{x_{n+1}} f(x) , dx.$$



            Integration by parts shows this to be



            $$E_n = int_{x_n}^{x_{n+1}}(x-c)f'(x) , dx,$$



            where $c = (x_{n+1}+x_n)/2$ is the midpoint.



            To see this, note that



            $$x_{n+1} - c = c - x_n = frac{x_{n+1} - x_n}{2} = frac{h}{2},$$



            and with $u = (x-c)$ and $dv = f'(x)dx$, integration by parts yields



            $$int_{x_n}^{x_{n+1}}(x-c)f'(x) , dx = (x-c)f(x)|_{x_n}^{x_{n+1}} - int_{x_n}^{x_{n+1}} f(x) , dx = E_n.$$



            If the derivative is bounded as $|f'(x)| leqslant M$, we can bound the error as



            $$|E_n| leqslant int_{x_n}^{x_{n+1}}|x-c||f'(x)| , dx \ leqslant Mint_{x_n}^{x_{n+1}}|x-c| , dx \ = frac{M}{4}(x_{n+1} - x_n)^2 \ = frac{M}{4}h^2.$$



            Approximating the integral over $[a,b]$ with $m$ subintervals of length $h = (b-a)/m$, has a global error bound of



            $$|GE| leqslant sum_{n=1}^{m} |E_n| = mfrac{M}{4}h^2 = frac{M(b-a)^2}{4m}$$



            If the second derivative is bounded as $|f''(x)| leqslant M$, then we can demonstrate $O(h^3)$ local accuracy. An integration by parts of the previous integral for $E_n$ yields



            $$E_n = frac1{2} int_{x_n}^{x_{n+1}} [(h/2)^2 - (x-c)^2]f''(x) , dx.$$



            Using the bound for $f''$ and integrating we obtain the local and global error bounds



            $$|E_n| leqslant frac{M}{12}h^3, \ |GE| leqslant frac{M(b-a)^3}{12m^2}.$$






            share|cite|improve this answer











            $endgroup$


















              0












              $begingroup$

              Estimating the error with the Taylor expansion is quite messy.



              For an alternative, consider the local error on a subinterval $[x_n, x_{n+1}]$ with $h = x_{n+1} - x_n$.



              The error in approximating the integral with the trapezoidal formula is



              $$E_n = frac{h}{2}[f(x_n) + f(x_{n+1})] - int_{x_n}^{x_{n+1}} f(x) , dx.$$



              Integration by parts shows this to be



              $$E_n = int_{x_n}^{x_{n+1}}(x-c)f'(x) , dx,$$



              where $c = (x_{n+1}+x_n)/2$ is the midpoint.



              To see this, note that



              $$x_{n+1} - c = c - x_n = frac{x_{n+1} - x_n}{2} = frac{h}{2},$$



              and with $u = (x-c)$ and $dv = f'(x)dx$, integration by parts yields



              $$int_{x_n}^{x_{n+1}}(x-c)f'(x) , dx = (x-c)f(x)|_{x_n}^{x_{n+1}} - int_{x_n}^{x_{n+1}} f(x) , dx = E_n.$$



              If the derivative is bounded as $|f'(x)| leqslant M$, we can bound the error as



              $$|E_n| leqslant int_{x_n}^{x_{n+1}}|x-c||f'(x)| , dx \ leqslant Mint_{x_n}^{x_{n+1}}|x-c| , dx \ = frac{M}{4}(x_{n+1} - x_n)^2 \ = frac{M}{4}h^2.$$



              Approximating the integral over $[a,b]$ with $m$ subintervals of length $h = (b-a)/m$, has a global error bound of



              $$|GE| leqslant sum_{n=1}^{m} |E_n| = mfrac{M}{4}h^2 = frac{M(b-a)^2}{4m}$$



              If the second derivative is bounded as $|f''(x)| leqslant M$, then we can demonstrate $O(h^3)$ local accuracy. An integration by parts of the previous integral for $E_n$ yields



              $$E_n = frac1{2} int_{x_n}^{x_{n+1}} [(h/2)^2 - (x-c)^2]f''(x) , dx.$$



              Using the bound for $f''$ and integrating we obtain the local and global error bounds



              $$|E_n| leqslant frac{M}{12}h^3, \ |GE| leqslant frac{M(b-a)^3}{12m^2}.$$






              share|cite|improve this answer











              $endgroup$
















                0












                0








                0





                $begingroup$

                Estimating the error with the Taylor expansion is quite messy.



                For an alternative, consider the local error on a subinterval $[x_n, x_{n+1}]$ with $h = x_{n+1} - x_n$.



                The error in approximating the integral with the trapezoidal formula is



                $$E_n = frac{h}{2}[f(x_n) + f(x_{n+1})] - int_{x_n}^{x_{n+1}} f(x) , dx.$$



                Integration by parts shows this to be



                $$E_n = int_{x_n}^{x_{n+1}}(x-c)f'(x) , dx,$$



                where $c = (x_{n+1}+x_n)/2$ is the midpoint.



                To see this, note that



                $$x_{n+1} - c = c - x_n = frac{x_{n+1} - x_n}{2} = frac{h}{2},$$



                and with $u = (x-c)$ and $dv = f'(x)dx$, integration by parts yields



                $$int_{x_n}^{x_{n+1}}(x-c)f'(x) , dx = (x-c)f(x)|_{x_n}^{x_{n+1}} - int_{x_n}^{x_{n+1}} f(x) , dx = E_n.$$



                If the derivative is bounded as $|f'(x)| leqslant M$, we can bound the error as



                $$|E_n| leqslant int_{x_n}^{x_{n+1}}|x-c||f'(x)| , dx \ leqslant Mint_{x_n}^{x_{n+1}}|x-c| , dx \ = frac{M}{4}(x_{n+1} - x_n)^2 \ = frac{M}{4}h^2.$$



                Approximating the integral over $[a,b]$ with $m$ subintervals of length $h = (b-a)/m$, has a global error bound of



                $$|GE| leqslant sum_{n=1}^{m} |E_n| = mfrac{M}{4}h^2 = frac{M(b-a)^2}{4m}$$



                If the second derivative is bounded as $|f''(x)| leqslant M$, then we can demonstrate $O(h^3)$ local accuracy. An integration by parts of the previous integral for $E_n$ yields



                $$E_n = frac1{2} int_{x_n}^{x_{n+1}} [(h/2)^2 - (x-c)^2]f''(x) , dx.$$



                Using the bound for $f''$ and integrating we obtain the local and global error bounds



                $$|E_n| leqslant frac{M}{12}h^3, \ |GE| leqslant frac{M(b-a)^3}{12m^2}.$$






                share|cite|improve this answer











                $endgroup$



                Estimating the error with the Taylor expansion is quite messy.



                For an alternative, consider the local error on a subinterval $[x_n, x_{n+1}]$ with $h = x_{n+1} - x_n$.



                The error in approximating the integral with the trapezoidal formula is



                $$E_n = frac{h}{2}[f(x_n) + f(x_{n+1})] - int_{x_n}^{x_{n+1}} f(x) , dx.$$



                Integration by parts shows this to be



                $$E_n = int_{x_n}^{x_{n+1}}(x-c)f'(x) , dx,$$



                where $c = (x_{n+1}+x_n)/2$ is the midpoint.



                To see this, note that



                $$x_{n+1} - c = c - x_n = frac{x_{n+1} - x_n}{2} = frac{h}{2},$$



                and with $u = (x-c)$ and $dv = f'(x)dx$, integration by parts yields



                $$int_{x_n}^{x_{n+1}}(x-c)f'(x) , dx = (x-c)f(x)|_{x_n}^{x_{n+1}} - int_{x_n}^{x_{n+1}} f(x) , dx = E_n.$$



                If the derivative is bounded as $|f'(x)| leqslant M$, we can bound the error as



                $$|E_n| leqslant int_{x_n}^{x_{n+1}}|x-c||f'(x)| , dx \ leqslant Mint_{x_n}^{x_{n+1}}|x-c| , dx \ = frac{M}{4}(x_{n+1} - x_n)^2 \ = frac{M}{4}h^2.$$



                Approximating the integral over $[a,b]$ with $m$ subintervals of length $h = (b-a)/m$, has a global error bound of



                $$|GE| leqslant sum_{n=1}^{m} |E_n| = mfrac{M}{4}h^2 = frac{M(b-a)^2}{4m}$$



                If the second derivative is bounded as $|f''(x)| leqslant M$, then we can demonstrate $O(h^3)$ local accuracy. An integration by parts of the previous integral for $E_n$ yields



                $$E_n = frac1{2} int_{x_n}^{x_{n+1}} [(h/2)^2 - (x-c)^2]f''(x) , dx.$$



                Using the bound for $f''$ and integrating we obtain the local and global error bounds



                $$|E_n| leqslant frac{M}{12}h^3, \ |GE| leqslant frac{M(b-a)^3}{12m^2}.$$







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Feb 19 '16 at 4:04

























                answered Feb 18 '16 at 22:18









                RRLRRL

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