Can anyone help me with what seems like an easy problem?

Show that:

$\displaystyle (1/3)T_n + (2/3)M_n = S_2_n$

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- Sep 27th 2011, 11:56 AMminmin1Approximate integration
Can anyone help me with what seems like an easy problem?

Show that:

$\displaystyle (1/3)T_n + (2/3)M_n = S_2_n$ - Sep 27th 2011, 03:55 PMHallsofIvyRe: Approximate integration
Well, it would help if you told us what $\displaystyle T_m$, $\displaystyle M_n$, and $\displaystyle S_$ mean!

- Sep 27th 2011, 08:31 PMtangibleLimeRe: Approximate integration
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.