Can anyone help me with what seems like an easy problem?
Show that:
$\displaystyle (1/3)T_n + (2/3)M_n = S_2_n$
I think he's using Simpson's rule. If I'm correct, it looks like he's trying to show the relation between the midpoint rule and the trapezoidal rule with Simpson's rule.
Midpoint rule: $\displaystyle M = (b-a)f(\frac{a+b}{s})$
Trapezoidal rule: $\displaystyle T = \frac{1}{2}(b-a)(f(a)+f(b))$
Weighted average:
$\displaystyle \frac{2M+T}{3}$
This weighted average is Simpson's rule.
Simpson's rule:
$\displaystyle \int_{a}^{b} f(x) dx \approx \frac{b-a}{6} \left[ f(a) + 4f\left(\frac{a+b}{2}\right)+f(b)\right]$
Looking at his equation,
$\displaystyle \frac{1}{3}T_{n} + \frac{2}{3}M_{n} = \frac{2M+T}{3}$ = weighted average = Simpson's Rule
Not totally sure what these subscripts are, but there's a start.