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lampCable
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Homework Statement
Consider the ansatz
[tex] \frac{1}{b-a}\int_a^bf(x)dx \approx \alpha_1f(a)+\alpha_2f(\frac{b-a}{2})+\alpha_3f(b)[/tex]
We can determine the values of [itex]\alpha_1,\alpha_2,\alpha_3[/itex] by requiring the approximation to be exact for quadratic polynomials, which yeilds Simpsons rule.
Why there is no loss of generality in assuming that [itex]a = -1[/itex] and [itex]b = 1[/itex]?