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Proof error simpsons rule

  1. Dec 12, 2011 #1
    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?
     
    Last edited: Dec 12, 2011
  2. jcsd
  3. Dec 12, 2011 #2
    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?
     
  4. Dec 15, 2011 #3
    [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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