What do you mean by 2^n? Is 'n' the order of quadrature ? Could you be more specific? For 1D, n classical (legendre) gauss points (& weights) will give a quadrature with 2n order accuracy. If you don't care about the sign of the weights, you can always find n-th order quadrature with any n distinct points. To get the weights, you only need to solve a linear algebra equation. So it is easy to make that 低阶时的积分点包含于高阶的积分点中. But if you want all weights are non-negative, you might need to choose the
是的,n是the order of quadrature. 2^n是quadrature point的个数,当然不一定 非要是2^n,只要个数随n增长的不快就行。 我最关心的是低阶的积分点包含于高阶的积分点中。legendre积分点是自己随便取的吗? 如果我希望有比较多的积分点位于积分区间的两个端点处,能不能实现呢? 能不能推荐一本讲地比较好的书?谢谢
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【在 w**d 的大作中提到】 : What do you mean by 2^n? Is 'n' the order of quadrature ? : Could you be more specific? : For 1D, n classical (legendre) gauss points (& weights) will give : a quadrature with 2n order accuracy. If you don't care about the sign of : the weights, you can always find n-th order quadrature with any n distinct : points. To get the weights, you only need to solve a linear algebra equation. : So it is easy to make that 低阶时的积分点包含于高阶的积分点中. : But if you want all weights are non-negative, you might need to choose the
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If you don't care the sign of the weights. It should be easy. Just follow the definition of the quadrature. E.g., given n points, x1, ..., xn, assume the weights are w1, ..., wn, If you want M-th order quadrature, i.e. for any fi = x^i, i=0, ..., M, the following fact holds:(def of quadrature) integral( fi ) = fi(x1)*w1 + .... + fi(xn) *wn when x1, ..., xn are distinct and M = n-1. The weights always exist. So if you don't care the sign of the weights, it is always possible to choose 低阶的积分