- #1

- 415

- 0

??

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

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- Thread starter nhrock3
- Start date

- #1

- 415

- 0

??

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

- #2

- 225

- 1

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

- #3

HallsofIvy

Science Advisor

Homework Helper

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Since a quadratic, [itex]y= ax^2+ bx+ c[/itex], 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.

- #4

- 415

- 0

i cant see where they use them on the arrow pointed formula

- #5

Mark44

Mentor

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The equation could be rewritten as

S

- #6

- 415

- 0

a panel in simpsons rule is 3 points (which has two intervals between them)

- #7

- 415

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thanks :)

- #8

- 415

- 0

The equation could be rewritten as

S_{1}= (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

[tex]h=\frac{b-a}{m}=2/2=1[/tex]

[tex]x_h=a+kh=-1+k[/tex]

[tex]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})=[/tex]

[tex]\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})=[/tex]

[tex]\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)]=

[/tex]

you see that it requests 4 points

and not

x_0 x_1 x_2

???????

Last edited:

- #9

Mark44

Mentor

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- 6,758

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?

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