Redian新闻
>
请教一个Gaussian quadrature的问题
avatar
请教一个Gaussian quadrature的问题# Computation - 科学计算
r*t
1
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!
avatar
w*d
2
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!

avatar
r*t
3
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

avatar
w*d
4

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.
: ,
: ,

相关阅读
logo
联系我们隐私协议©2024 redian.news
Redian新闻
Redian.news刊载任何文章,不代表同意其说法或描述,仅为提供更多信息,也不构成任何建议。文章信息的合法性及真实性由其作者负责,与Redian.news及其运营公司无关。欢迎投稿,如发现稿件侵权,或作者不愿在本网发表文章,请版权拥有者通知本网处理。