Relation between Simpson's Rule, Trapezoid Rule and Midpoint Rule. How 2 n numbers come out at the left side while only n numbers at the right side.

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Relation between Simpson&#039;s Rule, Trapezoid Rule and Midpoint Rule. How   2  n numbers come out at the left side while only   n numbers at the right side.

aletantas1x · Accepted Answer

For the Trapezoidal Rule, you actually use   n  +  1 points. For example, in the simple case where you are integrating   f  (  x  ) from   0 to   1, and you want       T    4  , you evaluate   f at the points   0      /    4,   1      /    4,   2      /    4,   3      /    4, and   4      /    4. It is   n  +  1 points because we use the endpoints.For the Midpoint Rule, you use   n points, but these are not the same points as for the Trapezoidal Rule. They are the midpoints of our intervals. So in the example discussed above, for       M    4   you would be evaluating   f at   1      /    8,   3      /    8,   5      /    8, and   7      /    8.The Simpson Rule       S          2      n       uses evaluation of   f at   2  n  +  1 points. If for example   n  =  4, then you are dividing the interval into 8 subintervals. With   n  =  4 and the interval   8, you would be using the points   0      /    8,   1      /    8,   2      /    8,   3      /    8,   4      /    8,   5      /    8,   6      /    8,   7      /    8, and   8      /    8.Note that   1      /    8,   3      /    8,   5      /    8 and   7      /    8 are the points that were used for       M    4  . The points   0      /    8,   2      /    8,   4      /    8,   6      /    8, and   8      /    8 are just   0      /    4,   1      /    4,   2      /    4,   3      /    4, and   4      /    4, exactly the points that were used for       T    4  .A more abstract summary:       T    n   uses   n  +  1 points, and       M    n   uses   n points. But the   n points used by Mn are completely different from the points used for       M    n  . So altogether,       T    n   and       M    n   carry information about function evaluation at   2  n  +  1 points, which is exactly what       S          2      n       does.I have not written out a proof of the formula, only tried to deal with your discomfort with the   2  n on one side and   n&#039;s on the other. The formula is not hard to verify. Let&#039;s do it explicitly for   n  =  4. Write down, say for the interval   [  0  ,  1  ], what       T    4   is. We have      T    4    =      1    8    (  f  (  0  )  +  2  f  (  1      /    4  )  +  2  f  (  1      /    2  )  +  2  f  (  3      /    4  )  +  f  (  1  )  )  .Now write down       M    4  :      M    4    =      1    4    (  f  (  1      /    8  )  +  f  (  3      /    8  )  +  f  (  5      /    8  )  +  f  (  7      /    8  )  )  .Now calculate       T    4    +  2      M    4  . It is convenient for the addition to make sure that       M    4   has denominator   8, so write   2      M    4   as       1    8    (  4  f  (  1      /    8  )  +  4  f  (  3      /    8  )  +  4  f  (  5      /    8  )  +  4  f  (  7      /    8  )  ), and add. Divide by   3 and you will get the expression you would get in       S          8      . The same method works in general.

Relation between Simpson's Rule, Trapezoid Rule and Midpoint Rule. How

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