Mixing
Understanding Binomial coefficient with floored terms
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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
asked Jan 8 at 6:46
Larry FreemanLarry Freeman
3,15821239
$endgroup$
add a comment |
$begingroup$
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
asked Jan 8 at 6:46
Larry FreemanLarry Freeman
3,15821239
$endgroup$
$begingroup$
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
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$begin{Bmatrix}7.4\ 3.5end{Bmatrix} = (4){7choose3}$. 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
add a comment |
$begingroup$
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
asked Jan 8 at 6:46
Larry FreemanLarry Freeman
3,15821239
$endgroup$
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
asked Jan 8 at 6:46
Larry FreemanLarry Freeman
3,15821239
asked Jan 8 at 6:46
Larry FreemanLarry Freeman
3,15821239
edited Jan 8 at 6:56
Larry Freeman
asked Jan 8 at 6:46
Larry FreemanLarry Freeman
3,15821239
asked Jan 8 at 6:46
Larry FreemanLarry Freeman
3,15821239
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
add a comment |
$begingroup$
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
add a comment |
1 Answer
1
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votes
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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:
$$genfrac{{}{}}{0em}{}{x}{y} = frac{prod_{k in (s - r, s] cap mathbb N} k}{prod_{k in (0, r] cap mathbb N} k} = delta(y, x)binom{lfloor xrfloor}{lfloor yrfloor}.$$
answered Jan 8 at 11:56
Anders KaseorgAnders Kaseorg
79349
$endgroup$
$begingroup$
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
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
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:
$$genfrac{{}{}}{0em}{}{x}{y} = frac{prod_{k in (s - r, s] cap mathbb N} k}{prod_{k in (0, r] cap mathbb N} k} = delta(y, x)binom{lfloor xrfloor}{lfloor yrfloor}.$$
answered Jan 8 at 11:56
Anders KaseorgAnders Kaseorg
79349
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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:
$$genfrac{{}{}}{0em}{}{x}{y} = frac{prod_{k in (s - r, s] cap mathbb N} k}{prod_{k in (0, r] cap mathbb N} k} = delta(y, x)binom{lfloor xrfloor}{lfloor yrfloor}.$$
answered Jan 8 at 11:56
Anders KaseorgAnders Kaseorg
79349
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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:
$$genfrac{{}{}}{0em}{}{x}{y} = frac{prod_{k in (s - r, s] cap mathbb N} k}{prod_{k in (0, r] cap mathbb N} k} = delta(y, x)binom{lfloor xrfloor}{lfloor yrfloor}.$$
answered Jan 8 at 11:56
Anders KaseorgAnders Kaseorg
79349
$endgroup$
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:
$$genfrac{{}{}}{0em}{}{x}{y} = frac{prod_{k in (s - r, s] cap mathbb N} k}{prod_{k in (0, r] cap mathbb N} k} = delta(y, x)binom{lfloor xrfloor}{lfloor yrfloor}.$$
answered Jan 8 at 11:56
Anders KaseorgAnders Kaseorg
79349
answered Jan 8 at 11:56
Anders KaseorgAnders Kaseorg
79349
answered Jan 8 at 11:56
Anders KaseorgAnders Kaseorg
79349
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
add a comment |
$begingroup$
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
add a comment |
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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