proof error simpsons rule


by georg gill
Tags: error, proof, rule, simpsons
georg gill
georg gill is offline
#1
Dec12-11, 03:57 PM
P: 100
I have tried to figure out a proof for simspons error that I found online

http://rowdy.mscd.edu/~talmanl/PDFs/Misc/Quintics.pdf

it is on page 149

I have sorted out the proof I think to (9) including (9). But I wonder how they could assume that F is continous on [0,h] when F is a different function in 0? It looks like derivative but one has -t to 0 and the other have t to zero would it not give different direction for the derivative?

EDIT: Got it -t in denumerator right?
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georg gill
georg gill is offline
#2
Dec12-11, 05:27 PM
P: 100
I also wonder about a thing in the beginning of the proof. It seems they subdivide an interval of simpson approximation in two from


[tex] \frac{b-a}{n}[/tex] to (a): [tex] \frac{b-a}{2n}[/tex]

How can they just change that? If I were to explain it the best I would think I guess would be to start with derivation of simpson rule and start with parabola centered somewhere else then in origo:

[tex]y=Ax^2+Bx+C[/tex]

Integrate to find real value underneath it

[tex]y=[\frac{A}{3}x^3+\frac{B}{2}x^2+Cx]^h_{-h}[/tex]

(I):

[tex]y=\frac{h}{3}(2ah^2+6C)[/tex]

Use the values on the graph for -h, 0 and h:

[tex]y_0=Ah^2-Bh+C[/tex] [tex]y_0=C[/tex] [tex]y_0=Ah^2+Bh+C[/tex]

and (I) becomes

(II):

[tex]y=\frac{h}{3}(2ah^2+6C)=\frac{h}{3}(y_0+4y_1+y_2)[/tex]

How can we just divide (II) in two like in (a)? When it is derived from something else? It seems that is what they do in the proof.

I am also a bit unsure about if u=h in proof where h is defined in the beginning of the proof. Is that right that u=h?
georg gill
georg gill is offline
#3
Dec15-11, 05:00 PM
P: 100
[tex] \frac{b-a}{n}[/tex] to (a): [tex] \frac{b-a}{2n}[/tex]

I think they only use 2n instead of n and it is just a matter of definition. I only wonder about one thing about this proof now. In the beginning they define

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

in (8) they use limits [0,h] which they talk about in the beginning to approximate error function on [0,h]

So far I get that. But in (11) they go back to using the simpson equation is this kth interval described there [0,h] 0r [-u,u]? I thought it would fit to give it the same interval as in the beginning of the proof, the first formula after proof is written as a semiheader which is [-u,u]. But I dont see how they then would get from [0,h] before (11) to [-u,u]


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