Understanding Binomial coefficient with floored terms












0












$begingroup$


I was reading through the notation used in a paper on arxiv.org when I came across this on page 6:



$[x]$ the floor of $x$



${x}$ the sawtooth function of $x$. That is ${x} = x - [x]$



$begin{Bmatrix}x\y end{Bmatrix}$ Binomial coefficient with floored terms.



Here is the explanation:



That is, $begin{Bmatrix}x\yend{Bmatrix} = delta(y,x){{[x]}choose{[y]}}$



where:




  • $delta(y,x)=1$ if ${x} ge {y}$


  • $delta(y,x)=[x-y]+1$ if ${x} < {y}$



Does this definition make sense? If so, could someone help me to understand what it means when $delta(y,x) neq 1$?






binomial-coefficients floor-function







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  • $begingroup$
    What part doesn’t make sense to you? For example, try computing $genfrac{{}{}}{0pt}{}{7.4}{3.6}$ and tell us where you get stuck.
    $endgroup$
    – Anders Kaseorg
    Jan 8 at 9:19










  • $begingroup$
    $begin{Bmatrix}7.4\ 3.5end{Bmatrix} = (4){7choose3}$. I am clear on the computation. I'm not clear why the $4$ is needed. Why not just $begin{Bmatrix}7.4\ 3.5end{Bmatrix} = {7choose3}$
    $endgroup$
    – Larry Freeman
    Jan 8 at 11:40


















0












$begingroup$


I was reading through the notation used in a paper on arxiv.org when I came across this on page 6:



$[x]$ the floor of $x$



${x}$ the sawtooth function of $x$. That is ${x} = x - [x]$



$begin{Bmatrix}x\y end{Bmatrix}$ Binomial coefficient with floored terms.



Here is the explanation:



That is, $begin{Bmatrix}x\yend{Bmatrix} = delta(y,x){{[x]}choose{[y]}}$



where:




  • $delta(y,x)=1$ if ${x} ge {y}$


  • $delta(y,x)=[x-y]+1$ if ${x} < {y}$



Does this definition make sense? If so, could someone help me to understand what it means when $delta(y,x) neq 1$?






binomial-coefficients floor-function







share|cite|improve this question











$endgroup$












  • $begingroup$
    What part doesn’t make sense to you? For example, try computing $genfrac{{}{}}{0pt}{}{7.4}{3.6}$ and tell us where you get stuck.
    $endgroup$
    – Anders Kaseorg
    Jan 8 at 9:19










  • $begingroup$
    $begin{Bmatrix}7.4\ 3.5end{Bmatrix} = (4){7choose3}$. I am clear on the computation. I'm not clear why the $4$ is needed. Why not just $begin{Bmatrix}7.4\ 3.5end{Bmatrix} = {7choose3}$
    $endgroup$
    – Larry Freeman
    Jan 8 at 11:40
















0












0








0


2



$begingroup$


I was reading through the notation used in a paper on arxiv.org when I came across this on page 6:



$[x]$ the floor of $x$



${x}$ the sawtooth function of $x$. That is ${x} = x - [x]$



$begin{Bmatrix}x\y end{Bmatrix}$ Binomial coefficient with floored terms.



Here is the explanation:



That is, $begin{Bmatrix}x\yend{Bmatrix} = delta(y,x){{[x]}choose{[y]}}$



where:




  • $delta(y,x)=1$ if ${x} ge {y}$


  • $delta(y,x)=[x-y]+1$ if ${x} < {y}$



Does this definition make sense? If so, could someone help me to understand what it means when $delta(y,x) neq 1$?






binomial-coefficients floor-function







share|cite|improve this question











$endgroup$




I was reading through the notation used in a paper on arxiv.org when I came across this on page 6:



$[x]$ the floor of $x$



${x}$ the sawtooth function of $x$. That is ${x} = x - [x]$



$begin{Bmatrix}x\y end{Bmatrix}$ Binomial coefficient with floored terms.



Here is the explanation:



That is, $begin{Bmatrix}x\yend{Bmatrix} = delta(y,x){{[x]}choose{[y]}}$



where:




  • $delta(y,x)=1$ if ${x} ge {y}$


  • $delta(y,x)=[x-y]+1$ if ${x} < {y}$



Does this definition make sense? If so, could someone help me to understand what it means when $delta(y,x) neq 1$?






binomial-coefficients floor-function




binomial-coefficients floor-function






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share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 8 at 6:56







Larry Freeman

















asked Jan 8 at 6:46









Larry FreemanLarry Freeman

3,15821239




3,15821239












  • $begingroup$
    What part doesn’t make sense to you? For example, try computing $genfrac{{}{}}{0pt}{}{7.4}{3.6}$ and tell us where you get stuck.
    $endgroup$
    – Anders Kaseorg
    Jan 8 at 9:19










  • $begingroup$
    $begin{Bmatrix}7.4\ 3.5end{Bmatrix} = (4){7choose3}$. I am clear on the computation. I'm not clear why the $4$ is needed. Why not just $begin{Bmatrix}7.4\ 3.5end{Bmatrix} = {7choose3}$
    $endgroup$
    – Larry Freeman
    Jan 8 at 11:40




















  • $begingroup$
    What part doesn’t make sense to you? For example, try computing $genfrac{{}{}}{0pt}{}{7.4}{3.6}$ and tell us where you get stuck.
    $endgroup$
    – Anders Kaseorg
    Jan 8 at 9:19










  • $begingroup$
    $begin{Bmatrix}7.4\ 3.5end{Bmatrix} = (4){7choose3}$. I am clear on the computation. I'm not clear why the $4$ is needed. Why not just $begin{Bmatrix}7.4\ 3.5end{Bmatrix} = {7choose3}$
    $endgroup$
    – Larry Freeman
    Jan 8 at 11:40


















$begingroup$
What part doesn’t make sense to you? For example, try computing $genfrac{{}{}}{0pt}{}{7.4}{3.6}$ and tell us where you get stuck.
$endgroup$
– Anders Kaseorg
Jan 8 at 9:19




$begingroup$
What part doesn’t make sense to you? For example, try computing $genfrac{{}{}}{0pt}{}{7.4}{3.6}$ and tell us where you get stuck.
$endgroup$
– Anders Kaseorg
Jan 8 at 9:19












$begingroup$
$begin{Bmatrix}7.4\ 3.5end{Bmatrix} = (4){7choose3}$. I am clear on the computation. I'm not clear why the $4$ is needed. Why not just $begin{Bmatrix}7.4\ 3.5end{Bmatrix} = {7choose3}$
$endgroup$
– Larry Freeman
Jan 8 at 11:40






$begingroup$
$begin{Bmatrix}7.4\ 3.5end{Bmatrix} = (4){7choose3}$. I am clear on the computation. I'm not clear why the $4$ is needed. Why not just $begin{Bmatrix}7.4\ 3.5end{Bmatrix} = {7choose3}$
$endgroup$
– Larry Freeman
Jan 8 at 11:40












1 Answer
1






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oldest

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$begingroup$

The author is of course free to make any definition they want, and apparently they found $delta(y, x)binom{lfloor xrfloor}{lfloor yrfloor}$ to be useful for their purposes in a way that $binom{lfloor xrfloor}{lfloor yrfloor}$ alone was not. See Lemma 2.0.2 for a hint of why that might be:



$$genfrac{{}{}}{0em}{}{x}{y} = frac{prod_{k in (s - r, s] cap mathbb N} k}{prod_{k in (0, r] cap mathbb N} k} = delta(y, x)binom{lfloor xrfloor}{lfloor yrfloor}.$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks. I was trying to understand if this was a standard definition or a definition specific to this paper.
    $endgroup$
    – Larry Freeman
    Jan 8 at 12:51











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1 Answer
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$begingroup$

The author is of course free to make any definition they want, and apparently they found $delta(y, x)binom{lfloor xrfloor}{lfloor yrfloor}$ to be useful for their purposes in a way that $binom{lfloor xrfloor}{lfloor yrfloor}$ alone was not. See Lemma 2.0.2 for a hint of why that might be:



$$genfrac{{}{}}{0em}{}{x}{y} = frac{prod_{k in (s - r, s] cap mathbb N} k}{prod_{k in (0, r] cap mathbb N} k} = delta(y, x)binom{lfloor xrfloor}{lfloor yrfloor}.$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks. I was trying to understand if this was a standard definition or a definition specific to this paper.
    $endgroup$
    – Larry Freeman
    Jan 8 at 12:51
















1












$begingroup$

The author is of course free to make any definition they want, and apparently they found $delta(y, x)binom{lfloor xrfloor}{lfloor yrfloor}$ to be useful for their purposes in a way that $binom{lfloor xrfloor}{lfloor yrfloor}$ alone was not. See Lemma 2.0.2 for a hint of why that might be:



$$genfrac{{}{}}{0em}{}{x}{y} = frac{prod_{k in (s - r, s] cap mathbb N} k}{prod_{k in (0, r] cap mathbb N} k} = delta(y, x)binom{lfloor xrfloor}{lfloor yrfloor}.$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks. I was trying to understand if this was a standard definition or a definition specific to this paper.
    $endgroup$
    – Larry Freeman
    Jan 8 at 12:51














1












1








1





$begingroup$

The author is of course free to make any definition they want, and apparently they found $delta(y, x)binom{lfloor xrfloor}{lfloor yrfloor}$ to be useful for their purposes in a way that $binom{lfloor xrfloor}{lfloor yrfloor}$ alone was not. See Lemma 2.0.2 for a hint of why that might be:



$$genfrac{{}{}}{0em}{}{x}{y} = frac{prod_{k in (s - r, s] cap mathbb N} k}{prod_{k in (0, r] cap mathbb N} k} = delta(y, x)binom{lfloor xrfloor}{lfloor yrfloor}.$$






share|cite|improve this answer









$endgroup$



The author is of course free to make any definition they want, and apparently they found $delta(y, x)binom{lfloor xrfloor}{lfloor yrfloor}$ to be useful for their purposes in a way that $binom{lfloor xrfloor}{lfloor yrfloor}$ alone was not. See Lemma 2.0.2 for a hint of why that might be:



$$genfrac{{}{}}{0em}{}{x}{y} = frac{prod_{k in (s - r, s] cap mathbb N} k}{prod_{k in (0, r] cap mathbb N} k} = delta(y, x)binom{lfloor xrfloor}{lfloor yrfloor}.$$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 8 at 11:56









Anders KaseorgAnders Kaseorg

79349




79349












  • $begingroup$
    Thanks. I was trying to understand if this was a standard definition or a definition specific to this paper.
    $endgroup$
    – Larry Freeman
    Jan 8 at 12:51


















  • $begingroup$
    Thanks. I was trying to understand if this was a standard definition or a definition specific to this paper.
    $endgroup$
    – Larry Freeman
    Jan 8 at 12:51
















$begingroup$
Thanks. I was trying to understand if this was a standard definition or a definition specific to this paper.
$endgroup$
– Larry Freeman
Jan 8 at 12:51




$begingroup$
Thanks. I was trying to understand if this was a standard definition or a definition specific to this paper.
$endgroup$
– Larry Freeman
Jan 8 at 12:51


















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