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】![](/moin_static193/solenoid/img/up.png) : : 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)) : .....
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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. ^( |