b****d
发帖数: 30 | |
E*****y
发帖数: 33 | 2
就老老实实做就可以了，然后用用gamma函数的定义
如:
v(n)=2^n \int_0^1 dx_1 \int_0^\sqrt{1-x_1^2} dx_2...
\int_0^\sqrt{1-x_1^2-...-x_{n-1}^2} dx_n
从后往前一步步做==>
v(n)=2^n \int_0^1 (1-x^2)^{(n-1)/2} dx \int_0^1 (1-x^2)^{(n-2)/2}...
\int_0^{1/2} (1-x^2)^{1/2}
\int_0^1 (1-x^2)^{(n-1)/2}=\sqrt{\pi} \Gamma((1+n)/2)/(n \Gamma(n/2))
.....
v(n)=... |
C******a
发帖数: 115 | 3
You can use software to compute 2*\int_0.5^1 (1-x^2)^{(n-1)/2}dx
and then times the volume of n-1-dimensinal unit-radius ball. |
C******a
发帖数: 115 | 4
不用伽玛函数也可做。从N维推导N+2维，即在平面圆盘上做一个积分。
采用极坐标，消去角度参量，再做变量替换t=r*r即可。
【在 E*****y 的大作中提到】
:
: 就老老实实做就可以了，然后用用gamma函数的定义
: 如:
: v(n)=2^n \int_0^1 dx_1 \int_0^\sqrt{1-x_1^2} dx_2...
: \int_0^\sqrt{1-x_1^2-...-x_{n-1}^2} dx_n
: 从后往前一步步做==>
: v(n)=2^n \int_0^1 (1-x^2)^{(n-1)/2} dx \int_0^1 (1-x^2)^{(n-2)/2}...
: \int_0^{1/2} (1-x^2)^{1/2}
: \int_0^1 (1-x^2)^{(n-1)/2}=\sqrt{\pi} \Gamma((1+n)/2)/(n \Gamma(n/2))
: .....
|
s****g
发帖数: 3 | 5 That is an interesting problem.
A point x (2d vector) in the large photo is mapped into
a point y (another 2d vector) in the small photo.
The relation between y and x is given by combination of
rescaling r<1 , rotation O (2x2 orthogonal matrix) and
translation t(2d vector). All these three quantities
can be measured. That is:
y = r O x + t
The point we are looking for satifies the above equation
as well as
y=x
Therefore:
x = r O x + t
x = (I - r O)^(-1) t
where I is the (2x2) identity matrix. ^( |
