Simpson's rule



         


computer science, in the field of numerical analysis, Simpson's Rule is a way to get an approximation of an integral:

<math> \int_{a}^{b} f(x) dx<math>

using an interpolating polynomial of higher degree. Simpson's rule belong to the family of rules derived from Newton-Cotes formulas. The most common is a quadratic polynomial interpolating at a, (a+b)/2, and b which gives us the polynomial:

<math> P(x) = f(a) + (x-a)f\left[a,\left( \frac{a+b}{2} \right)\right] + \left(x-\left( \frac{a+b}{2} \right)\right)(x-a)f\left[a,\left( \frac{a+b}{2} \right), b\right] <math>

From this Simpson's Rule is:

<math> \int_{a}^{b} f(x) dx \approx S(f) = \frac{b-a}{6}\left[f(a) + 4f\left(\frac{a+b}{2}\right)+f(b)\right]

<math>

[Top]

Proof

We want to have our polynomial on the form:

<math>P(x) = \alpha x^2 + \beta x + \gamma<math>

Assume we have the function values <math>a=x_0<math>, <math>\frac{a+b}{2}=x_1<math> and <math>b=x_2<math>. The situation will look like this, with our sampled function values at <math>f(a)<math>, <math>f\left(\frac{a+b}{2}\right)<math> and <math>f(b)<math>:

As this Simpson's rule apply to equidistant points, we know that <math>x_0 < x_1 < x_2<math> and that <math>x_1-x_0 = x_2-x_1<math>. This means we may transport our solution to the intervals formed by <math>-h, 0, h<math> such that

<math>h \equiv \frac{a+b}{2}<math>

We need to interpolate these values and function values with a polynomial and form our equations:

<math>f(-h) = \alpha h^2 - \beta h + \gamma<math>
<math>f(0) = \gamma<math>
<math>f(h) = \alpha h^2 + \beta h + \gamma<math>

Which yields:

<math>\alpha = \frac{f(-h) - 2f(0) + f(h)}{2h^2}<math>
<math>\beta = \frac{f(h) - f(-h)}{2h}<math>
<math>\gamma = f(0)<math>

We then integrate our polynomial:

<math>\int_{-h}^h P(x) dx =<math>
<math>\int_{-h}^h \frac{f(-h) - 2f(0) + f(h)}{2h^2}x^2 +

\frac{f(h) - f(-h)}{2h}x + f(0) dx =<math>

<math>\left[\frac{f(-h) - 2f(0) + f(h)}{6h^2}x^3 +

\frac{f(h) - f(-h)}{4h}x^2 + f(0)x \right]_{-h}^h =<math>

<math>\frac{f(-h) - 2f(0) + f(h)}{6h^2}h^3 +

\frac{f(h) - f(-h)}{4h}h^2 + f(0)h +<math>

<math>\frac{f(-h) - 2f(0) + f(h)}{6h^2}h^3 -

\frac{f(h) - f(-h)}{4h}h^2 + f(0)h =<math>

<math>\frac{f(-h) - 2f(0) + f(h)}{3h^2}h^3 +

2f(0)h<math>

Substitute back our original values:

<math>\frac{f(a) - 2f\left(\frac{a+b}{2}\right) + f(b)}{3\left(\frac{b-a}{2}\right)^2}\left(\frac{b-a}{2}\right)^3 +

2f\left(\frac{a+b}{2}\right) \left(\frac{b-a}{2}\right) =<math>

<math>\frac{f(a) - 2f\left(\frac{a+b}{2}\right) + f(b)}{3}\left(\frac{b-a}{2}\right) +

2f\left(\frac{a+b}{2}\right) \left(\frac{b-a}{2}\right) = <math>

<math>\frac{b-a}{6} \left(f(a) - 2f\left(\frac{a+b}{2}\right) + f(b) +

6f\left(\frac{a+b}{2}\right) \right) = <math>

<math>\frac{b-a}{6} \left(f(a) + 4f\left(\frac{a+b}{2}\right) + f(b) \right)<math>

Q.E.D.


[Top]

Error of Simpson's Rule

To examine the accuracy of the rule, take <math>c = \frac{a+b}{2}<math>, so

<math> \int_{a}^{b} f(x) dx = \int_{a}^{c} f(x) dx + \int_{c}^{b} f(x) dx<math>

Using integration by parts we get:

<math> \int_{a}^{c} f(x) dx = f(x)(x-\alpha)|^c_a - \int_{a}^{c} f'(x)(x-\alpha) dx <math>

and

<math>\int_{c}^{b} f(x) dx = f(x)(x-\beta)|^b_c - \int_{c}^{b} f'(x)(x-\beta) dx<math>

where α and β are constants that we can choose. Adding these expressions, we get:

<math> \int_{a}^{b} f(x) dx = f(b)(b-\beta)+f(c)(\beta-\alpha)+f(a)(\alpha-a) - \int_{a}^{c} f'(x)(x-\alpha) dx + \int_{c}^{b} f'(x)(x-\beta) dx<math>


Let's take α and β, so as to get Simpson's Rule:

<math>\alpha = \frac{b+5a}{6}, \beta = \frac{5b+a}{6}<math>

and defining the function Py(x) by:


<math>P_y(x)=\left\{\begin{matrix} x-\alpha, & \mbox{if }a\le x \le c \\ x-\beta, & \mbox{if }c < x \le b \end{matrix}\right.<math>

we have

<math> \int_{a}^{b} f(x) dx = S(f) - \int_{a}^{b}f'(x)P_y(x)dx<math>

where

<math>\int_{a}^{b}f'(x)P_y(x)dx<math>

is the error value.





  View Live Article   This article is from Wikipedia. All text is available under the terms of the GNU Free Documentation License