Can Simpson's Rule Error Proof Assume Continuity on Varied Interval Bounds?

  • Context: Graduate 
  • Thread starter Thread starter georg gill
  • Start date Start date
  • Tags Tags
    Error Proof
Click For Summary
SUMMARY

The forum discussion centers on the proof of Simpson's Rule error, specifically addressing the continuity assumption of the function F on the interval [0,h]. The user questions the validity of this assumption given that F behaves differently at the endpoints. They also express confusion regarding the subdivision of the interval for Simpson's approximation and the transition between intervals [0,h] and [-u,u]. The discussion highlights the need for clarity in defining the intervals and the implications of changing the bounds in the proof.

PREREQUISITES
  • Understanding of Simpson's Rule and its application in numerical integration
  • Familiarity with continuity and differentiability of functions
  • Basic knowledge of calculus, particularly integration techniques
  • Ability to interpret mathematical proofs and notation
NEXT STEPS
  • Study the derivation of Simpson's Rule in detail
  • Research the implications of continuity on numerical integration methods
  • Learn about error analysis in numerical methods, focusing on Simpson's Rule
  • Examine the transition between different interval notations in mathematical proofs
USEFUL FOR

Mathematicians, students of calculus, and anyone involved in numerical analysis or mathematical proofs will benefit from this discussion.

georg gill
Messages
151
Reaction score
6
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 continuous 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?
 
Last edited by a moderator:
Physics news on Phys.org
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?
 
[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 don't see how they then would get from [0,h] before (11) to [-u,u]
 

Similar threads

Replies
6
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 3 ·
Replies
3
Views
4K
  • · Replies 13 ·
Replies
13
Views
3K
  • · Replies 11 ·
Replies
11
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 3 ·
Replies
3
Views
4K
  • · Replies 11 ·
Replies
11
Views
3K