
The trapezoid rule
Hi. Need some guidance to make the trapezoid rule more effective.
Let $\displaystyle f:[a,b]\to\mathbb{R}$ be a function. I want to find an approximation to the integral
$\displaystyle \int_{a}^b f(x)dx$.
I know that the trapezoid rule is given by
$\displaystyle I_n =h\sum_{i=1}^{2^n} \frac{f(x_{i1})+f(x_i)}{2}$ where $\displaystyle h=\frac{ba}{2^n}$.
(Notice that I want $\displaystyle 2^n$ trapezoids/parts)
$\displaystyle I_n$ can be rewritten as
$\displaystyle I_n =h[\frac{f(a)+f(b)}{2}+\sum_{i=1}^{2^n 1}f(a+ih)]$.
Ok, now over to the problem. To make the method more effective I use the approximation to integral from $\displaystyle I_{n1}$ in $\displaystyle I_n$ in this way
$\displaystyle I_n =\frac{I_{n1}}{2}+h\sum_{i=1}^{2^{n1}}f(a+(2i1)h)$.
The part that I dont understand in this new formula is why you just can divide $\displaystyle \frac{I_{n1}}{2}$? The other part I do understand, because in $\displaystyle I_{n1}$ you have already calculated $\displaystyle f(x_i)$ where $\displaystyle i$ is even and $\displaystyle i\in[0,2^n]$, so in $\displaystyle I_n$ you only need to calculate $\displaystyle f(x_i)$ with odd indexes $\displaystyle i$.
I hope the problem is clear!? And I hope someone can help me explain and understand this formula. Maby it's a good idea to prove that it works? I dont know how, but maby a hint could help?