Mixing
Identity involving floor, ceiling and nearest integer functions
$begingroup$
For $ngeqslant0$, $m>0$, $s>tgeqslant0$, $n,m,s,t$ - integers we have
$$sumlimits_{k=0}^{m-1}leftlfloor{n+ks+tover ms}rightrfloor=leftlfloor{n+tover s}rightrfloor$$
$$sumlimits_{k=0}^{m-1}leftlceil{n+ks+tover ms}rightrceil=leftlceil{n+tover s}rightrceil+m-1$$
$$sumlimits_{k=0}^{m-1}left[{n+ks+tover ms}right]=left[{n+t+lceil{sover 2}rceil lfloor{mover 2}rfloor+lceil{s-1over 2}rceil lfloor{m-1over 2}rfloor over s}right]$$
All of them are self-discovered. I'm looking for nice and simple closed form (and also maybe more simple proof) for the last.
summation integers floor-function ceiling-function
asked Jan 2 at 19:35
user514787user514787
700210
$endgroup$
add a comment |
$begingroup$
For $ngeqslant0$, $m>0$, $s>tgeqslant0$, $n,m,s,t$ - integers we have
$$sumlimits_{k=0}^{m-1}leftlfloor{n+ks+tover ms}rightrfloor=leftlfloor{n+tover s}rightrfloor$$
$$sumlimits_{k=0}^{m-1}leftlceil{n+ks+tover ms}rightrceil=leftlceil{n+tover s}rightrceil+m-1$$
$$sumlimits_{k=0}^{m-1}left[{n+ks+tover ms}right]=left[{n+t+lceil{sover 2}rceil lfloor{mover 2}rfloor+lceil{s-1over 2}rceil lfloor{m-1over 2}rfloor over s}right]$$
All of them are self-discovered. I'm looking for nice and simple closed form (and also maybe more simple proof) for the last.
summation integers floor-function ceiling-function
asked Jan 2 at 19:35
user514787user514787
700210
$endgroup$
1
$begingroup$
Thanks for the upvote. I do realize that $n, s, t$ come from some context you work in, but as $n$ is unbounded, in your expression you can redefine $n := n+t $ and redefine $t:=0$. That should not affect the formulas.
$endgroup$
– JAskgaard
Jan 2 at 21:37
$begingroup$
@JAskgaard, you are absolutely right, but I sure it is not so significant.
$endgroup$
– user514787
Jan 2 at 21:41
add a comment |
$begingroup$
For $ngeqslant0$, $m>0$, $s>tgeqslant0$, $n,m,s,t$ - integers we have
$$sumlimits_{k=0}^{m-1}leftlfloor{n+ks+tover ms}rightrfloor=leftlfloor{n+tover s}rightrfloor$$
$$sumlimits_{k=0}^{m-1}leftlceil{n+ks+tover ms}rightrceil=leftlceil{n+tover s}rightrceil+m-1$$
$$sumlimits_{k=0}^{m-1}left[{n+ks+tover ms}right]=left[{n+t+lceil{sover 2}rceil lfloor{mover 2}rfloor+lceil{s-1over 2}rceil lfloor{m-1over 2}rfloor over s}right]$$
All of them are self-discovered. I'm looking for nice and simple closed form (and also maybe more simple proof) for the last.
summation integers floor-function ceiling-function
asked Jan 2 at 19:35
user514787user514787
700210
$endgroup$
For $ngeqslant0$, $m>0$, $s>tgeqslant0$, $n,m,s,t$ - integers we have
$$sumlimits_{k=0}^{m-1}leftlfloor{n+ks+tover ms}rightrfloor=leftlfloor{n+tover s}rightrfloor$$
$$sumlimits_{k=0}^{m-1}leftlceil{n+ks+tover ms}rightrceil=leftlceil{n+tover s}rightrceil+m-1$$
$$sumlimits_{k=0}^{m-1}left[{n+ks+tover ms}right]=left[{n+t+lceil{sover 2}rceil lfloor{mover 2}rfloor+lceil{s-1over 2}rceil lfloor{m-1over 2}rfloor over s}right]$$
All of them are self-discovered. I'm looking for nice and simple closed form (and also maybe more simple proof) for the last.
summation integers floor-function ceiling-function
summation integers floor-function ceiling-function
asked Jan 2 at 19:35
user514787user514787
700210
asked Jan 2 at 19:35
user514787user514787
700210
edited Jan 2 at 20:28
user514787
asked Jan 2 at 19:35
user514787user514787
700210
asked Jan 2 at 19:35
user514787user514787
700210
asked Jan 2 at 19:35
user514787user514787
700210
700210
1
$begingroup$
Thanks for the upvote. I do realize that $n, s, t$ come from some context you work in, but as $n$ is unbounded, in your expression you can redefine $n := n+t $ and redefine $t:=0$. That should not affect the formulas.
$endgroup$
– JAskgaard
Jan 2 at 21:37
$begingroup$
@JAskgaard, you are absolutely right, but I sure it is not so significant.
$endgroup$
– user514787
Jan 2 at 21:41
add a comment |
1
$begingroup$
Thanks for the upvote. I do realize that $n, s, t$ come from some context you work in, but as $n$ is unbounded, in your expression you can redefine $n := n+t $ and redefine $t:=0$. That should not affect the formulas.
$endgroup$
– JAskgaard
Jan 2 at 21:37
$begingroup$
@JAskgaard, you are absolutely right, but I sure it is not so significant.
$endgroup$
– user514787
Jan 2 at 21:41
1
1
$begingroup$
Thanks for the upvote. I do realize that $n, s, t$ come from some context you work in, but as $n$ is unbounded, in your expression you can redefine $n := n+t $ and redefine $t:=0$. That should not affect the formulas.
$endgroup$
– JAskgaard
Jan 2 at 21:37
$begingroup$
Thanks for the upvote. I do realize that $n, s, t$ come from some context you work in, but as $n$ is unbounded, in your expression you can redefine $n := n+t $ and redefine $t:=0$. That should not affect the formulas.
$endgroup$
– JAskgaard
Jan 2 at 21:37
$begingroup$
@JAskgaard, you are absolutely right, but I sure it is not so significant.
$endgroup$
– user514787
Jan 2 at 21:41
$begingroup$
@JAskgaard, you are absolutely right, but I sure it is not so significant.
$endgroup$
– user514787
Jan 2 at 21:41
add a comment |
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$begingroup$
Thanks for the upvote. I do realize that $n, s, t$ come from some context you work in, but as $n$ is unbounded, in your expression you can redefine $n := n+t $ and redefine $t:=0$. That should not affect the formulas.
$endgroup$
– JAskgaard
Jan 2 at 21:37
$begingroup$
@JAskgaard, you are absolutely right, but I sure it is not so significant.
$endgroup$
– user514787
Jan 2 at 21:41