Simpson method question(numerical analysis)

[nhrock3]

i have been told that simpson with 3 points is a single panel

??

why??

a pannel is the distance between to points
so if we have 3 points then we have 2 panels
??

http://i48.tinypic.com/nlc4t1.jpg

 

[monty37]

can't view the link clearly and what do you mean by pannel?

 

[HallsofIvy]

Simpson's method works by approximating a function by a "piecewise quadratic". If I understand your use of "panel" correctly, it is one of the quadratic pieces.

Since a quadratic, $y= ax^2+ bx+ c$, has three coefficients to be determined, one quadratic, one panel, requires three points, not two, to determine those three coefficients.

The "trapezoid method", where we approximate the function by a piecewise linear function, y= ax+ b, requires only two coefficients to be determined and so each "panel" requires two points.

 

[nhrock3]

so how they use those three points in the formula
i can't see where they use them on the arrow pointed formula

 

[Mark44]

In the last equation, the three points are (-1, f(-1)), (0, f(0)), and (1, f(1)).

The equation could be rewritten as
S₁ = (1/3)[f(-1) + 4f(0) + f(1)].

 

[nhrock3]

so a pannel is not an interval
a panel in simpsons rule is 3 points (which has two intervals between them)

 

[nhrock3]

thanks :)

 

[nhrock3]

In the last equation, the three points are (-1, f(-1)), (0, f(0)), and (1, f(1)).

The equation could be rewritten as
S₁ = (1/3)[f(-1) + 4f(0) + f(1)].

i tried to get to your expression by the formula
if we have three points then we have 2 sub intervals
$$h=\frac{b-a}{m}=2/2=1$$
$$x_h=a+kh=-1+k$$
$$s(f,h)=\frac{h}{3}(f(a)+f(b)+\frac{2h}{3}\sum_{k=1}^{m}f(x_{2k})+\frac{4h}{3}\sum_{k=1}^{m}f(x_{2k-1})=$$
$$\frac{1}{3}(f(-1)+f(1)+\frac{2}{3}\sum_{k=1}^{m}f(x_{2k})+\frac{4}{3}\sum_{k=1}^{m}f(x_{2k-1})=$$

$$\frac{1}{3}(f(-1)+f(1)+\frac{2}{3}[f(x_2)+f(x_4)]+\frac{4}{3}[f(x_1)+f(x_3)]=$$
you see that it requests 4 points
and not
x_0 x_1 x_2

??

 

[Mark44]

In Simpson's rule, you divide an interval [a, b] into n subintervals of equal length. In each subinterval you use function values evaluated at three points: the left endpoint, the middle, and the right endpoint.

Instead of getting lost in the summation symbols, work out for yourself for some simple function and an interval [0, 1] divided into four subintervals. What does Simpson's rule give you for this situation?