自适应Simpson积分
摘要:用$g(x)=A*x^2+B*x+C$来代替原函数$f(x)$,设$g(x)$原函数为$G(x)$,显然$G(x)=\frac{1}{3}*A*x^3+\frac{1}{2}*B*x^2+C*x$ $\int_a^b f(x) dx \approx \int_a^b g(x) dx=G(b)-G(a
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posted @ 2020-08-14 16:01
posted @ 2020-08-14 16:01
posted @ 2020-08-08 17:10
posted @ 2020-08-07 08:40