取整函数凑成恒等

hbghlyj

对任意实数x,y,证明下面两个函数恒等:
f(x,y)=1-[{x+0.5}+{y+0.5}]
g(x,y)=[2x]+[2y]-[x]-[y]-[x+y]

hbghlyj

记$x=n+r$和$y=m+s$,其中$n,m\in\Bbb Z$和$r,s\in\left[0,1\right),r\ge s$.
则$f(x,y)=f(r,s)$,$g(x,y)=g(r,s)=[2x]+[2y]-[x+y]$.
Case 1. $r,s\in\left[0,0.5\right)$.
$\{r+0.5\}+\{s+0.5\}=r+s+1\in[1,2)⇒f(r,s)=0$.
$[2r]=[2s]=[r+s]=0⇒g(r,s)=0$.
Case 2. $r\in[0.5,1),s\in[0,0.5)$.
$\{r+0.5\}+\{s+0.5\}=r+s\in[0.5,1.5)⇒f(r,s)=1-[r+s]$.
$[2r]=1,[2s]=0⇒g(r,s)=1-[r+s]$.
Case 3. $r,s\in[0.5,1)$.
$\{r+0.5\}+\{s+0.5\}=r+s-1⇒f(r,s)=2-[r+s]$.
$[2r]=[2s]=1⇒g(r,s)=2-[r+s]$.

hbghlyj

从$g(x,y)=f(x,y)≥0$得$[2x]+[2y]\ge[x]+[y]+[x+y]$ (华罗庚“数论导引”第1章习题3)
由以上不等式和Legendre's formula就可以推得
$m!n!(m + n)!{\Large|}(2m)!(2n)!$
对任意正整数$m,n$成立.这是这帖中的结论.(math.stackexchange.com)
满足这帖中的递推关系:$$f(n,m)=\frac{(2n)!(2m)!}{n!(n+m)!m!}.$$  $f(0,t)=\binom{2t}{t}$ and  $f(i+1,j)=4f(i,j)-f(i,j+1).$