How Does Simpson's Rule Approximate Areas Under Curves?

[brandy]

can anyone tell me how it works, like visually, what does it do. I am curious.
iv heard its a three point approximation whereas the trapazoidal rule is a two point approximation. But I am not entirely sure i understand.

 

[HallsofIvy]

The trapezoid method approximates the curve, between successive points, by a straight line while Simpson's rule uses three points at a time and approximates the curve by the parabola passing through the three points.

Suppose your function is f(x) and you take three points, [itex](x_0, f(x_0), (x_1, f(x_1), (x_2, f(x_2)), equally spaced on the x-axis. For simplicity, take the x-values to be $x_0= x_1- h$, $x_1$, and $x_2= x_1+ h$ where h is the distance between succesive x values so the points themselves are $(x_1- h, f(x_0)), (x_1, f(x_1)), (x_1+ h,f(x_2))$.<br /> <br /> The equation of a parabola through $(x_1, f(x_1))$ can be written $y= a(x- x_1)^2+ b(x- x_1)+ f(x_1)$. Taking $x= x_0= x_1- h$ we get $y= f(x_0)= ah^2+ bh+ f(x_1)$. Taking $x= x_2= x_1+ h$ we get $y= f(x_2)= ah^2- bh+ f(x_1)$.<br /> <br /> Adding those two equations, $f(x_2)+ f(x_0)= 2ah^2+ 2f(x_1)$ so that <br /> $$ah^2= \frac{f(x_2)+ f(x_0)}{2}- f(x_1}= \frac{f(x_2)- 2f(x_1)+ f(x_0)}{2}$$]<br /> and<br /> $$a= \frac{f(x_0)+ 2f(x_1)+ f(x_2)}{2h^2}$$[/itex][tex] <br /> Subtracting those two equations, $f(x_2)- f(x_0)= 2bh$ so that <br /> $$b= \frac{f(x_2)- f(x_0)}{2h}$$<br /> <br /> Now, $\int_{x_0}^{x_2}a(x-x_1)^2+ b(x-x_1)+ f(x_1) dx= \int_{x_1-h}^{x_1+h} a(x-x_1)^2+ b(x-x_1)+ f(x_1) dx$[itex]= (1/3)ah^3+ (1/2)bh^2+ f(x_1)h- (1/3)a(-h)^3- b(-h)^2- f(x_1)(-h)= (2/3)ah^3+ 2f(x_1)h[itex]so b really isn't important. The integral of the parabola through $(x_1-h, f(x_0)), (x_1, f(x_1)), (x_1+h,f(x_2))$ is <br /> $$(2/3)ah^3+ 2f(x_1)h= (2/3)\frac{f(x_0)- 2f(x_1)+ f(x_2)}{2h^2}h^3+ 2f(x_1)h$$<br /> $$= \frac{2(f(x_0)- 2f(x_1)+ f(x_2)}{3}h+ 2f(x_1)h= h\frac{2f(x_0)- 4f(x_1)+ 2f(x_2)}{3}$$[/itex][tex] <br /> That is the formula for exactly 3 data points. If we have more, we can separte into groups of three and do that for each such interval. Bacause the overlap at the ends, we must have an <b>odd</b> number of points to do that.<br /> <br /> For example, if we had only an even number of points, $x_0, x_1, x_2, x_3$, our first parabola would be over $x_0, x_1, x_2$ but, even including $x_3$ as our first endpoint in the next interval we would have only $x_2, x_3$, not enough for a parabola. But with 5 points, we could have $x_0, x_1, x_2$, $x_2, x_3, x_4$. Similarly, with 6 points, an even number, we would have $x_0, x_1, x_2$, $x_2, x_3, x_4$, $x_4, x_4$ and not be able to complete the last parabola but with 7, an odd number, we would have $x_0, x_1, x_2$, $x_2, x_3, x_4$, $x_4, x_5, x_6$.<br /> <br /> Now see what happens when we "attach" parabolas: on $x_0, x_1, x_2$ we have <br /> $$h\frac{2f(x_0)- 4f(x_1)+ 2f(x_2)}{3}$$[/tex][/itex][tex][tex] while on $x_2, x_3, x_4$<br /> [tex]h\frac{2f(x_2)- 4f(x_3)+ 2f(x_4)}{3}[/itex]<br /> <br /> Those add to give<br /> [tex]h\frac{2f(x_0)- 4f(x_1)+ 4f(x_2)- 4f(x_3)+ 2f(x_4)}{3}[/itex]<br /> so you see why we have "4" multiplying everything except the first and last terms. Taking the integral to be from a to b and using n intervals, h= (b-a)/n so the formula becomes <br /> [tex](b-a)\frac{2f(x_0)- 4f(x_1)+ 4f(x_2)- 4f(x_3)+ 2f(x_4)}{3n}[/itex]<br /> Simpson's rule.[/tex][/tex][/tex][/tex][/tex][/tex]