请教一个Gaussian quadrature的问题# Computation - 科学计算
r*t
1 楼
如果需要一种Gaussian quadrature scheme：低阶时的积分点包含于高阶的积分点中，
同时积分点的个数虽阶数的升高而增加得不快，比如以2^n的速度增加是可以接受的，
请问有没有这样的scheme? Thank you!
同时积分点的个数虽阶数的升高而增加得不快，比如以2^n的速度增加是可以接受的，
请问有没有这样的scheme? Thank you!
w*d
2 楼
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
【在 r****t 的大作中提到】
: 如果需要一种Gaussian quadrature scheme：低阶时的积分点包含于高阶的积分点中，
: 同时积分点的个数虽阶数的升高而增加得不快，比如以2^n的速度增加是可以接受的，
: 请问有没有这样的scheme? Thank you!
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
【在 r****t 的大作中提到】
: 如果需要一种Gaussian quadrature scheme：低阶时的积分点包含于高阶的积分点中，
: 同时积分点的个数虽阶数的升高而增加得不快，比如以2^n的速度增加是可以接受的，
: 请问有没有这样的scheme? Thank you!
r*t
3 楼
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
equation.
，
，
【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
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
equation.
，
，
【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
w*d
4 楼
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
【在 r****t 的大作中提到】
: 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
:
: equation.
: ，
: ，
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