|
楼主 |
发表于 2019-8-18 18:51
|
显示全部楼层
The first is to have the same domain N (≥ 6),
so the functions N/2, r2 (N),
N/2 + r2 (N) are the same domain! No ground for blame!
In the same domain,
1:
Function N/2 has a minimum of 3
2:
At the same time (highlighted here)
Composite function N/2+r2(N) has a minimum of 4
3: Then the equation is valid:
Inf (N/2) +inf (r2 (N)
= inf (N/2 + r2 (N))
3+inf(r2N)=4
Inf(r2N)=4-3=1
Inf (r2 (N) = 1, that is r2 (N) ≥ 1
This is the range of function r2 (N) given at the same time as function N/2 and N/2+r2 (N). |
|