Question

Show that \(\frac{1}{3}{T_n} + \frac{2}{3}{M_n} = {S_{2n}}\).

Short answer

Hence, the prove for \(\frac{1}{3}{T_n} + \frac{2}{3}{M_n} = {S_{2n}}\) is obtained.

Step-by-step solution

Definitions:

Trapezoid Rule

If\((a,b)\)is divided into\(n\)subintervals, each of length\(\Delta x\), with endpoints at\(\left\{ {{x_0},{x_1},{x_2}, \ldots ,{x_n}} \right\}\), then\({T_n} = \frac{{\Delta x}}{2}\left( {f\left( {{x_0}} \right) + 2f\left( {{x_1}} \right) + 2f\left( {{x_2}} \right) + \cdots + 2f\left( {{x_{n - 2}}} \right) + 2f\left( {{x_{n - 1}}} \right) + f\left( {{x_n}} \right)} \right){\rm{ }}\)

Midpoint Rule

If\((a,b)\)is divided into\(n\)subintervals, each of length\(\Delta x\)and\({m_i}\)is the midpoint of the\(i\)th subinterval, then\({M_n} = \sum\limits_{i = 1}^n f \left( {{m_i}} \right)\Delta x\).

Determine the Simpson’s approximation.

Simpson's approximation for the interval \(\left( {{x_0},{x_n}} \right)\) is given by:

\({S_n} = \frac{{\Delta x}}{3}\left( {f\left( {{x_0}} \right) + 4f\left( {{x_1}} \right) + 2f\left( {{x_2}} \right)4f\left( {{x_3}} \right) + \ldots .. + 4f\left( {{x_{n - 1}}} \right) + f\left( {{x_n}} \right)} \right)\)

Doubling the number \({\rm{n}}\) results subintervals\(\left( {{x_0},{{\bar x}_1}} \right),\left( {{{\bar x}_1},{x_1}} \right),\left( {{x_1},{{\bar x}_2}} \right),\left( {{{\bar x}_2},{x_2}} \right), \ldots \left( {{{\bar x}_n},{x_n}} \right)\).

Where \({\bar x_1},{\bar x_2},{\bar x_3}, \ldots \ldots \)are the mid points of the intervals\(\left( {{x_0},{x_1}} \right),\left( {{x_1},{x_2}} \right), \ldots \ldots .,\left( {{x_{n - 1}},{x_n}} \right)\).

Evaluate Right hand side.

Now expand right hand side i.e; \({S_{2n}}\)using Simpson’s definition for \(n = 2n\).

Using Simpson’s definition:

\(\begin{aligned}{c}{S_{2n}} = \frac{1}{2} \cdot \frac{{\Delta x}}{3}\left( {f\left( {{x_0}} \right) + 4f\left( {{{\bar x}_1}} \right) + 2f\left( {{x_1}} \right) + 4f\left( {{{\bar x}_2}} \right) + 2f\left( {{x_2}} \right) + \ldots \ldots + 4f\left( {{{\bar x}_n}} \right) + f\left( {{x_n}} \right)} \right)\\ = \frac{1}{6}\Delta x\left( {f\left( {{x_0}} \right) + 2f\left( {{x_1}} \right) + 2f\left( {{x_2}} \right) + \ldots .. + f\left( {{x_n}} \right) + 4f\left( {{{\bar x}_1}} \right) + 4f\left( {{{\bar x}_2}} \right) + - - + 4f\left( {{{\bar x}_n}} \right)} \right)\\ = \frac{1}{2}\left( {\frac{{\Delta x}}{3}\left( {f\left( {{x_0}} \right) + 2f\left( {{x_1}} \right) + 2f\left( {{x_2}} \right) + \ldots . + f\left( {{x_n}} \right)} \right)} \right) + \frac{{4\Delta x}}{6}\left( {4f\left( {{{\bar x}_1}} \right) + f\left( {{{\bar x}_2}} \right) + \ldots \ldots . + f\left( {{{\bar x}_n}} \right)} \right)\end{aligned}\)

Substitute the values and solve as:

\({S_{2n}} = \frac{1}{2}{T_n} + \frac{2}{3}{M_n}\)

Hence, proved.