10楼: Originally posted by hulu1528 at 2015-04-09 19:26:03
感谢指点，请问这个m和n表示的是细分的吗？为何不直接使用x或者y了，而要使用m*x和n*y了？...

11楼: Originally posted by feixiaolin at 2015-04-09 23:01:30
只是调节而已
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12楼: Originally posted by hulu1528 at 2015-04-11 08:44:40
请问一下，如果平面方程是 ax + by + cz + d = 0的话，请问这个曲面可以如何构建了？...

13楼: Originally posted by feixiaolin at 2015-04-11 12:47:16
z*(1+sin(k*z)) 代替 z 即可...

13楼: Originally posted by feixiaolin at 2015-04-11 12:47:16
z*(1+sin(k*z)) 代替 z 即可...

13楼: Originally posted by feixiaolin at 2015-04-11 12:47:16
z*(1+sin(k*z)) 代替 z 即可...

14楼: Originally posted by hulu1528 at 2015-04-11 20:41:13
也就是说对于平面方程ax+by+cz+d = 0
其振幅不超过A的曲面方程是否就是由下式得出了？
a*k1*sin(m*x) + b*sin(n*y) + c*z*(1+sin(k*z)) + d= 0
a*k1 + b + c = A
-a*k1 - b + c = -A

这个 x 》 k1*sin(m*x) ...

16楼: Originally posted by lyzmeteor at 2015-04-11 21:25:15
改为二维平面来看，如果直线y=0上下波动1，那么其实在y=-1与y=+1之间是可以无限复杂的，你可以说用三角级数即傅里叶级数去逼近它但，你只用有限三角函数去代替它是不对的   我是这么认为...

17楼: Originally posted by feixiaolin at 2015-04-11 21:26:52
x+kx*sin(k*x)代替 x
y+ky*sin(l*x)代替 y
z+kz*sin(m*x)代替 z
a*kx+b*ky+c*kz=A
...

19楼: Originally posted by hulu1528 at 2015-04-12 10:00:28
平面方程ax+by+cz+d = 0，振幅不超过A的曲面的曲面用下列来表示：
x+kx*sin(k*x)代替 x，  y+ky*sin(l*x)代替 y， z+kz*sin(m*x)代替 z
a*kx+b*ky+c*kz = A
这里kx，ky，kz是三个系数，会有很多中情况，是否根据 ...