c****r
发帖数: 185 | 1 我想求一个m维数值积分，
F(x_1,...,x_m;n)= \int_{y_i>=x_i, \sum y_i=n} f(y_1,...,y_m;n) dy_1 ...dy_m
x_i>=0, n>0
应该怎么求？
多谢。 | i*******n
发帖数: 166 | 2
蒙特卡罗
【在 c****r 的大作中提到】
: 我想求一个m维数值积分，
: F(x_1,...,x_m;n)= \int_{y_i>=x_i, \sum y_i=n} f(y_1,...,y_m;n) dy_1 ...dy_m
: x_i>=0, n>0
: 应该怎么求？
: 多谢。
| c*******e
发帖数: 8624 | 3 gaussian quadrature
我以前做过4维的,都一样套就可以了
【在 c****r 的大作中提到】
: 我想求一个m维数值积分，
: F(x_1,...,x_m;n)= \int_{y_i>=x_i, \sum y_i=n} f(y_1,...,y_m;n) dy_1 ...dy_m
: x_i>=0, n>0
: 应该怎么求？
: 多谢。
| w**d
发帖数: 2334 | 4
If the dimension is really high, Monte Carlo is the best choice.
【在 i*******n 的大作中提到】
:
: 蒙特卡罗
| n*s
发帖数: 752 | 5 and probably the only choice
【在 w**d 的大作中提到】
:
: If the dimension is really high, Monte Carlo is the best choice.
| w**d
发帖数: 2334 | 6
Is there any good answer to the curse of dimension?
【在 n*s 的大作中提到】
: and probably the only choice
| n*s
发帖数: 752 | 7 the error for MC is
O (N ^ {-1/2})
for ordinary numerical integration, the error is
O (N ^ {-a/d})
d is the dimention of the problem, a is related to the algorithm
【在 w**d 的大作中提到】
:
: Is there any good answer to the curse of dimension?
| c****r
发帖数: 185 | 8 Sounds interesting.
What is MC and N?
Is it an abosolute error or relative error?
【在 n*s 的大作中提到】
: the error for MC is
: O (N ^ {-1/2})
: for ordinary numerical integration, the error is
: O (N ^ {-a/d})
: d is the dimention of the problem, a is related to the algorithm
| n*s
发帖数: 752 | 9 MC is Monte Carlo
N is the # of intervals into which is original intergration interval is divide
【在 c****r 的大作中提到】
: Sounds interesting.
: What is MC and N?
: Is it an abosolute error or relative error?
|
|
