Tag Archives: Simpsons Rule

Equation for Simpsons Rule: Proof and Derivation

We get in the book that an equation for a parabola for any three points is

$ q(x)=A\frac {(x-b)(x-c)}{(a-b)(a-c)}+ B \frac{(x-a)(x-c)}{(b-a)(b-c)} +C\frac{(x-a)(x-b)}{(c-a)(c-b)}$

This is the formula for some parabola $ q= mx^2+nx+p$ through the points, (a, A), (b,B), (c,C), we will take this as a given and go from there.

First we need to say we are working on a interval [-h, h]. Then our three points will be (-h, f(-h)), (0,f(0)), (h,f(h)).

Photo on 2013-11-18 at 09.09

First we write a generic equation for this quadratic. We can do this using the formula given and replacing the values:
$ q(x)=f(-h)\frac {(x)(x-h)}{(-h)(-2h)}+ f(0) \frac{(x+h)(x-h)}{(h)(-h)} +f(h)\frac{(x+h)(x)}{(2h)(h)}$

Reducing this we get:

$ q(x)=f(-h)\frac {x^2-xh}{2h^2}+ f(0) \frac{(x^2-h^2)}{-h^2} +f(h)\frac{x^2+xh}{2h^2}$

Now we integrate from [-h, h].

Photo on 2013-11-18 at 09.20

$\int^h_{-h} f(-h)\frac {x^2-xh}{2h^2}+ \int^h_{-h} f(0) \frac{(x^2-h^2)}{-h^2} +\int^h_{-h} f(h)\frac{x^2+xh}{2h^2}$

$=f(-h) (\frac{x^3}{6h^2} – \frac{x^2}{4h}) |_{-h}^h + f(0) (\frac{x^3}{-h^2} + x) |_{-h}^h + f(h) (\frac{x^3}{6h^2} + \frac{x^2}{4h}) |_{-h}^h$

Just taking the first integral:

$f(-h)((\frac{h^3}{6h^2} – \frac{h^2}{4h})-(\frac{-h^3}{6h^2} – \frac{h^2}{4h}))$


Evaluating the other integrals and reducing we get

$ f(0)\frac{4}{3}h$ and $f(h)\frac{h}{3}$

So we have: $f(-h)\frac{h}{3} + f(0)\frac{4}{3}h +f(h)\frac{h}{3}$

Now the points if you change the points we were working with to

$(x_{i-1},f(x_{i-1})),(x_i,f(x_{i})), (x_{i+1},f(x_{i+1}))$

we get

This is the formula for Simpson’s Rule. This has major implications. It shoes how accurate it can be. This is accurate because when you have three points and a function that goes through them Simpsons rule finds a best fit quadratic for that region. Then we see that the formula does not depend on the actual equations themselves, but the points that one is evaluating them at and the interval they are going across.