Register to reply

Proof error simpsons rule

by georg gill
Tags: error, proof, rule, simpsons
Share this thread:
georg gill
#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?
Phys.Org News Partner Science news on Phys.org
'Office life' of bacteria may be their weak spot
Lunar explorers will walk at higher speeds than thought
Philips introduces BlueTouch, PulseRelief control for pain relief
georg gill
#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
#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]


Register to reply

Related Discussions
How does simpsons rule work General Math 1
Simpsons Rule Problem Calculus 1
Simpsons/Trapazoidal rule Advanced Physics Homework 2
Question about error theorem for simpsons rule Calculus & Beyond Homework 1