Chapter 5.5, Problem 40E

To determine

To prove : $S_{2n}=\frac{{\text{MRAM}}_n+2T_{2n}}{3}$

Explanation of Solution

Given information:

The statement is:

  $S_{2n}=\frac{{\text{MRAM}}_n+2T_{2n}}{3}$

Calculation:

Hence, $T_n=\frac{h}{2}\left(y_0+2y_1+2y_2+\dots +2y_{n-1}+y_n\right)$

Then,

  $T_{2n}=\frac{h}{2}\left(y_0+2y_1+2y_2+\dots +2y_{2n-1}+y_{2n}\right)$

Hence,

  $\left.S=\frac{h}{3}\left(y_0+4y_1+2y_2+4y_3+\dots +2y_{n-2}+4y_{n-1}\right]+y_n\right)$

Then,

  $S_{2n}=\frac{h}{3}\left(y_0+4y_1+2y_2+4y_3+\dots +2y_{2n-2}+4y_{2n-1}+y_{2n}\right)$

So,

  ${\text{MRAM }}_n=2h\left(y_1+y_3+y_5+\dots .+y_{2n-1}\right)$ because the widths of the rectangles would be twice as wide as those used for the Trap Rule and Simpson's rule since there are only half as many subintervals, and the heights of the rectangles would only be the middle heights of each subinterval.

  $S_{2n}$ can be written as:

  $\begin{array}{c}{S}_{2n}=\frac{h}{3}\times \left(y_0+4y_1+2y_2+4y_3+\dots +2y_{2n-2}+4y_{2n-1}+y_{2n}\right)\\ =\frac{1}{3}\times \left[h\times \left(y_0+2y_1+2y_2+\dots +2y_{2n-1}+y_{2n}\right)+2h\times \left(y_1+y_3+y_5+\dots +y_{2n-1}\right)\right]\\ =\frac{2T_{2n}+{\text{MRAM}}_n}{3}\end{array}$

Therefore, the required given statement $S_{2n}=\frac{{\text{MRAM}}_n+2T_{2n}}{3}$ is $($ proved $)$ .