91Ó°ÊÓ

The given data is Simpson’s Rule is equivalent to the weighted average of the trapezoid and midpoint sums:

Consider an integrable function f on [a,b] and a positive integer n.

Define $∆x=\frac{b-a}{n}and{x}_k=a+k∆x$

The simpson's rule is stated as the integral is approximated using $\frac{n}{2}$parabola topped rectangle as follows,

role="math" localid="1649524533386" $SIMP\left(n\right)=\left(f\left(x_0\right)+4f\left(x_1\right)+2f\left(x_2\right)+....4f\left(x_{n-1}\right)+f\left(x_n\right)\right)\frac{∆x}{3}$

The n-rectangle midpoint sum and n-rectangle trapezoid sum is determined as$MID\left(n\right)=\underset{K=1}{\overset{n}{∑}}f\left(\frac{x_{k-1}+x_k}{2}\right)∆xandTRAP\left(n\right)=\underset{K=1}{\overset{n}{∑}}\left(\frac{f\left(x_{k-1}\right)+f\left(x_k\right)}{2}\right)∆x$

$$\begin{gathered} \mathrm{For}\mathrm{any}\mathrm{integer}i,\mathrm{where}1≤i≤n-1,\mathrm{observe}, \\ {x}_{i-1}=a+(i-1)∆x\mathrm{and}{x}_{i+1}=a+(i+1)∆x \\ \mathrm{Compute}\frac{x_{i-1}+x_{i+1}}{2}, \\ \frac{x_{i-1}+x_{i+1}}{2}=\frac{a+(i-1)∆x+a+(i+1)∆x}{2} \\ =\frac{2a+2i∆x}{2} \\ =a+i∆x \\ =x_i \end{gathered}$$

role="math" localid="1649525299160" $$\begin{gathered} \mathrm{Compute}SIMP\left(2n\right), \\ SIMP\left(2n\right)=\left(f\left(x_0\right)+4f\left(x_1\right)+2f\left(x_2\right)+....4f\left(x_{2n-2}\right)+f\left(x_{2n}\right)\right)\frac{∆x}{3} \\ SIMP\left(2n\right)=\left(f\left(x_0\right)+2f\left(x_2\right)+....2f\left(x_{2n-2}\right)+f\left(x_{2n}\right)\right)\frac{∆x}{3}+\left(4f\left(x_1\right)+4f\left(x_3\right)+.......4f\left(x_{2n-1}\right)\right)\frac{∆x}{3} \\ \mathrm{For}\mathrm{the}\mathrm{terms}\mathrm{in}\mathrm{second}\mathrm{sum}\mathrm{use}\frac{x_{i-1}+x_{i+1}}{2}=x_i, \\ SIMP\left(2n\right)=\left(\frac{2∆x}{3}\left(\frac{f\left(x_0\right)}{2}+\frac{f\left(x_2\right)}{2}+....\frac{f\left(x_{2n-2}\right)}{2}+\frac{f\left(x_{2n}\right)}{2}\right)\right)+\left(\frac{4∆x}{3}\left(f\left(\frac{x_0+X_2}{2}\right)+f\left(\frac{x_2+X_4}{2}\right)+.......f\left(\frac{x_{2n-2}+x_{2n}}{2}\right)\right)\right) \\ SIMP\left(2n\right)=\left(\frac{∆x}{3}2\left(\frac{f\left(x_0\right)+f\left(x_2\right)}{2}+....\frac{f\left(x_{2n-2}\right)+f\left(x_{2n}\right)}{2}\right)\right)+\left(\frac{2∆x}{3}2\left(f\left(\frac{x_0+X_2}{2}\right)+f\left(\frac{x_2+X_4}{2}\right)+.......f\left(\frac{x_{2n-2}+x_{2n}}{2}\right)\right)\right) \end{gathered}$$

$$\begin{gathered} \mathrm{Calculate}TRAP\left(n\right), \\ TRAP\left(n\right)=\underset{k=1}{\overset{n}{∑}}\left(\frac{f\left(x_{2k-2}\right)+f\left(x_{2k}\right)}{2}\right)2∆x \\ TRAP\left(n\right)=2∆x\left(\frac{f\left(x_0\right)+f\left(x_2\right)}{2}+\frac{f\left(x_2\right)+f\left(x_4\right)}{2}+.....\frac{f\left(x_{2n-2}\right)+f\left(x_{2n}\right)}{2}\right) \\ \mathrm{Calculate}MID\left(n\right), \\ MID\left(n\right)=\underset{k=1}{\overset{n}{∑f}}\left(\frac{\left(x_{2k-2}\right)+\left(x_{2k}\right)}{2}\right)2∆x \\ MID\left(n\right)=2∆x\left(f\left(\frac{\left(x_0\right)+\left(x_2\right)}{2}\right)+f\left(\frac{\left(x_2\right)+\left(x_4\right)}{2}\right)+.....f\left(\frac{\left(x_{2n-2}\right)+\left(x_{2n}\right)}{2}\right)\right) \end{gathered}$$

Substitute the terms in the right side of the equation, we get,

$$\begin{gathered} SIMP\left(2n\right)=\left(\frac{1}{3}2∆x\left(\frac{f\left(x_0\right)+f\left(x_2\right)}{2}+....\frac{f\left(x_{2n-2}\right)+f\left(x_{2n}\right)}{2}\right)\right)+\left(\frac{2}{3}2∆x\left(f\left(\frac{x_0+X_2}{2}\right)+f\left(\frac{x_2+X_4}{2}\right)+.......f\left(\frac{x_{2n-2}+x_{2n}}{2}\right)\right)\right) \\ SIMP\left(2n\right)=\frac{1}{3}TRAP\left(n\right)+\frac{2}{3}MID\left(n\right) \end{gathered}$$

Hence, Proved.