# Proof error simpsons rule

by georg gill
Tags: error, proof, rule, simpsons
 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 $$\frac{b-a}{n}$$ to (a): $$\frac{b-a}{2n}$$ 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: $$y=Ax^2+Bx+C$$ Integrate to find real value underneath it $$y=[\frac{A}{3}x^3+\frac{B}{2}x^2+Cx]^h_{-h}$$ (I): $$y=\frac{h}{3}(2ah^2+6C)$$ Use the values on the graph for -h, 0 and h: $$y_0=Ah^2-Bh+C$$ $$y_0=C$$ $$y_0=Ah^2+Bh+C$$ and (I) becomes (II): $$y=\frac{h}{3}(2ah^2+6C)=\frac{h}{3}(y_0+4y_1+y_2)$$ 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?
 P: 100 $$\frac{b-a}{n}$$ to (a): $$\frac{b-a}{2n}$$ 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 $$h=\frac{b-a}{2n}$$ 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]