I know that Simpson's rule is exact on all intervals for polynomials with $\deg(f) \leq 3$, but are these the only functions with the rule is exact for on all intervals? If so how would I prove this?
Simpson's rule with error $0$
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calculus
integration
simpsons-rule
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0@user49640 Thats true, but I'm looking for classes of functions for which this is true on every interval, i.e for ever $a,b$ in $f$'s domain with $a < b$ we have $\int_a^b f(x) dx = \frac{b - a}{6}(f(a) + f(b) + 4f(\frac{a + b}{2}))$. So even if I bend it a certain way to make the integrals agree on $[0,1]$, on $[0,\frac{1}{2}]$ the result may not hold. – 2017-02-23
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0Okay, thanks for the clarification. – 2017-02-23
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1Simpson's rule requires it to be 4th derivative. Do each case: constant, linear, quadratic and cubic. – 2017-02-23
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0@user29418 Oh I see, so I can use the error bound of Simpson's approximation. Thanks! – 2017-02-23
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0If $f$ has a continuous fourth derivative, then the result will be true, since the error term is $\frac{1}{2880}(b-a)^5 f^{(4)}(\xi)$ with $\xi \in (a,b)$. Thus $f^{(4)}$ must be zero on a dense set, hence everywhere. – 2017-02-23
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0@user49640 I don't quite understand the density argument. For example suppose $f(x) = 0$ for $x \in \mathbb{Q}$ and $f(x) = x$ otherwise, then $f(x)$ would be zero on a dense set, but is not zero everywhere. If $f^{(4)}$ is continuous then the argument could work, but I don't see it working otherwise. – 2017-02-23
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0I qualified my statement by the assumption that $f^{(4)}$ was continuous. – 2017-02-23