Double Binomial Sum
$begingroup$
I've been tackling the following problem from Concrete Mathematics (Graham et al., chapter 5).
I turned to the answers at some point -- which alas, I have trouble understanding!
Can you provide an answer that elaborates most important steps. For example how is floor-square op replaced by factor $2 k + 1$ and loosened constraint on $j$?
combinatorics summation binomial-coefficients
asked Dec 24 '18 at 17:10
BoLeBoLe
232110
$endgroup$
add a comment |
$begingroup$
I've been tackling the following problem from Concrete Mathematics (Graham et al., chapter 5).
I turned to the answers at some point -- which alas, I have trouble understanding!
Can you provide an answer that elaborates most important steps. For example how is floor-square op replaced by factor $2 k + 1$ and loosened constraint on $j$?
combinatorics summation binomial-coefficients
asked Dec 24 '18 at 17:10
BoLeBoLe
232110
$endgroup$
add a comment |
$begingroup$
I've been tackling the following problem from Concrete Mathematics (Graham et al., chapter 5).
I turned to the answers at some point -- which alas, I have trouble understanding!
Can you provide an answer that elaborates most important steps. For example how is floor-square op replaced by factor $2 k + 1$ and loosened constraint on $j$?
combinatorics summation binomial-coefficients
asked Dec 24 '18 at 17:10
BoLeBoLe
232110
$endgroup$
I've been tackling the following problem from Concrete Mathematics (Graham et al., chapter 5).
I turned to the answers at some point -- which alas, I have trouble understanding!
Can you provide an answer that elaborates most important steps. For example how is floor-square op replaced by factor $2 k + 1$ and loosened constraint on $j$?
combinatorics summation binomial-coefficients
combinatorics summation binomial-coefficients
asked Dec 24 '18 at 17:10
BoLeBoLe
232110
asked Dec 24 '18 at 17:10
BoLeBoLe
232110
asked Dec 24 '18 at 17:10
BoLeBoLe
232110
asked Dec 24 '18 at 17:10
BoLeBoLe
232110
asked Dec 24 '18 at 17:10
BoLeBoLe
232110
232110
add a comment |
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
First, replace $k-j$ by $ell$,
Summing over all integers $0le j le k$ is the same as summing over integers $0le j$ and $0le ell$, where $ell$ is serving the role of $k-j$. The result is
$$sum_{j,ellge 0} binom{-1}{j-lfloorsqrt{ ell} rfloor}binom{j}m frac1{2^j}$$
then replace $lfloor sqrt ell rfloor$ by $k$.
As $ell$ ranges over positive integers, the quantity $lfloor sqrt ell rfloor$ does as well, except that the value of $k$ is attained a total of $2k+1$ times. For example, there are $5$ numbers whose square root rounded down is $2$, and those are $4,5,6,7,8$. In this step, we collect all these repeated terms in to one term; the $k$ represents the square root, and $2k+1$ is the number of $ell$ which make that square root.
Now, sum over $k$,
Ignoring terms which do not depend on $k$, the summation over $k$ looks like
$$
sum_{k} binom{-1}{j-k} (2k+1) = (2j+1)-(2j-1)+(2j-3)-dotspm 1
$$
In words, since $binom{-1}{j-k}=(-1)^{j-k}$ whenever $j-kge 0$, and $0$ otherwise, this summation is the alternating sum of the first $j+1$ odd numbers. A little thought shows that this sum is equal to $j+1$; one way to prove this is to break into cases based on whether $j$ is even or odd, then perform all of the adjacent subtractions, leaving a sum of $lfloor j/2rfloor$ two's.
Absorb the $j+1$ and apply $(5.57)$.
I think you can take it from here? The absorption identity is $binom{n}k(n+1)=binom{n+1}{k+1}(k+1)$.
answered Dec 24 '18 at 17:44
Mike EarnestMike Earnest
28.4k22153
$endgroup$
$begingroup$
This makes everything clear. I wish I hadn't been intimidated by the whole problem. It's not that hard to proceed once you think well about the index domains.
$endgroup$
– BoLe
Dec 25 '18 at 13:50
add a comment |
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1 Answer
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1 Answer
1
active
oldest
votes
active
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active
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votes
$begingroup$
First, replace $k-j$ by $ell$,
Summing over all integers $0le j le k$ is the same as summing over integers $0le j$ and $0le ell$, where $ell$ is serving the role of $k-j$. The result is
$$sum_{j,ellge 0} binom{-1}{j-lfloorsqrt{ ell} rfloor}binom{j}m frac1{2^j}$$
then replace $lfloor sqrt ell rfloor$ by $k$.
As $ell$ ranges over positive integers, the quantity $lfloor sqrt ell rfloor$ does as well, except that the value of $k$ is attained a total of $2k+1$ times. For example, there are $5$ numbers whose square root rounded down is $2$, and those are $4,5,6,7,8$. In this step, we collect all these repeated terms in to one term; the $k$ represents the square root, and $2k+1$ is the number of $ell$ which make that square root.
Now, sum over $k$,
Ignoring terms which do not depend on $k$, the summation over $k$ looks like
$$
sum_{k} binom{-1}{j-k} (2k+1) = (2j+1)-(2j-1)+(2j-3)-dotspm 1
$$
In words, since $binom{-1}{j-k}=(-1)^{j-k}$ whenever $j-kge 0$, and $0$ otherwise, this summation is the alternating sum of the first $j+1$ odd numbers. A little thought shows that this sum is equal to $j+1$; one way to prove this is to break into cases based on whether $j$ is even or odd, then perform all of the adjacent subtractions, leaving a sum of $lfloor j/2rfloor$ two's.
Absorb the $j+1$ and apply $(5.57)$.
I think you can take it from here? The absorption identity is $binom{n}k(n+1)=binom{n+1}{k+1}(k+1)$.
answered Dec 24 '18 at 17:44
Mike EarnestMike Earnest
28.4k22153
$endgroup$
$begingroup$
This makes everything clear. I wish I hadn't been intimidated by the whole problem. It's not that hard to proceed once you think well about the index domains.
$endgroup$
– BoLe
Dec 25 '18 at 13:50
add a comment |
$begingroup$
First, replace $k-j$ by $ell$,
Summing over all integers $0le j le k$ is the same as summing over integers $0le j$ and $0le ell$, where $ell$ is serving the role of $k-j$. The result is
$$sum_{j,ellge 0} binom{-1}{j-lfloorsqrt{ ell} rfloor}binom{j}m frac1{2^j}$$
then replace $lfloor sqrt ell rfloor$ by $k$.
As $ell$ ranges over positive integers, the quantity $lfloor sqrt ell rfloor$ does as well, except that the value of $k$ is attained a total of $2k+1$ times. For example, there are $5$ numbers whose square root rounded down is $2$, and those are $4,5,6,7,8$. In this step, we collect all these repeated terms in to one term; the $k$ represents the square root, and $2k+1$ is the number of $ell$ which make that square root.
Now, sum over $k$,
Ignoring terms which do not depend on $k$, the summation over $k$ looks like
$$
sum_{k} binom{-1}{j-k} (2k+1) = (2j+1)-(2j-1)+(2j-3)-dotspm 1
$$
In words, since $binom{-1}{j-k}=(-1)^{j-k}$ whenever $j-kge 0$, and $0$ otherwise, this summation is the alternating sum of the first $j+1$ odd numbers. A little thought shows that this sum is equal to $j+1$; one way to prove this is to break into cases based on whether $j$ is even or odd, then perform all of the adjacent subtractions, leaving a sum of $lfloor j/2rfloor$ two's.
Absorb the $j+1$ and apply $(5.57)$.
I think you can take it from here? The absorption identity is $binom{n}k(n+1)=binom{n+1}{k+1}(k+1)$.
answered Dec 24 '18 at 17:44
Mike EarnestMike Earnest
28.4k22153
$endgroup$
$begingroup$
This makes everything clear. I wish I hadn't been intimidated by the whole problem. It's not that hard to proceed once you think well about the index domains.
$endgroup$
– BoLe
Dec 25 '18 at 13:50
add a comment |
$begingroup$
First, replace $k-j$ by $ell$,
Summing over all integers $0le j le k$ is the same as summing over integers $0le j$ and $0le ell$, where $ell$ is serving the role of $k-j$. The result is
$$sum_{j,ellge 0} binom{-1}{j-lfloorsqrt{ ell} rfloor}binom{j}m frac1{2^j}$$
then replace $lfloor sqrt ell rfloor$ by $k$.
As $ell$ ranges over positive integers, the quantity $lfloor sqrt ell rfloor$ does as well, except that the value of $k$ is attained a total of $2k+1$ times. For example, there are $5$ numbers whose square root rounded down is $2$, and those are $4,5,6,7,8$. In this step, we collect all these repeated terms in to one term; the $k$ represents the square root, and $2k+1$ is the number of $ell$ which make that square root.
Now, sum over $k$,
Ignoring terms which do not depend on $k$, the summation over $k$ looks like
$$
sum_{k} binom{-1}{j-k} (2k+1) = (2j+1)-(2j-1)+(2j-3)-dotspm 1
$$
In words, since $binom{-1}{j-k}=(-1)^{j-k}$ whenever $j-kge 0$, and $0$ otherwise, this summation is the alternating sum of the first $j+1$ odd numbers. A little thought shows that this sum is equal to $j+1$; one way to prove this is to break into cases based on whether $j$ is even or odd, then perform all of the adjacent subtractions, leaving a sum of $lfloor j/2rfloor$ two's.
Absorb the $j+1$ and apply $(5.57)$.
I think you can take it from here? The absorption identity is $binom{n}k(n+1)=binom{n+1}{k+1}(k+1)$.
answered Dec 24 '18 at 17:44
Mike EarnestMike Earnest
28.4k22153
$endgroup$
First, replace $k-j$ by $ell$,
Summing over all integers $0le j le k$ is the same as summing over integers $0le j$ and $0le ell$, where $ell$ is serving the role of $k-j$. The result is
$$sum_{j,ellge 0} binom{-1}{j-lfloorsqrt{ ell} rfloor}binom{j}m frac1{2^j}$$
then replace $lfloor sqrt ell rfloor$ by $k$.
As $ell$ ranges over positive integers, the quantity $lfloor sqrt ell rfloor$ does as well, except that the value of $k$ is attained a total of $2k+1$ times. For example, there are $5$ numbers whose square root rounded down is $2$, and those are $4,5,6,7,8$. In this step, we collect all these repeated terms in to one term; the $k$ represents the square root, and $2k+1$ is the number of $ell$ which make that square root.
Now, sum over $k$,
Ignoring terms which do not depend on $k$, the summation over $k$ looks like
$$
sum_{k} binom{-1}{j-k} (2k+1) = (2j+1)-(2j-1)+(2j-3)-dotspm 1
$$
In words, since $binom{-1}{j-k}=(-1)^{j-k}$ whenever $j-kge 0$, and $0$ otherwise, this summation is the alternating sum of the first $j+1$ odd numbers. A little thought shows that this sum is equal to $j+1$; one way to prove this is to break into cases based on whether $j$ is even or odd, then perform all of the adjacent subtractions, leaving a sum of $lfloor j/2rfloor$ two's.
Absorb the $j+1$ and apply $(5.57)$.
I think you can take it from here? The absorption identity is $binom{n}k(n+1)=binom{n+1}{k+1}(k+1)$.
answered Dec 24 '18 at 17:44
Mike EarnestMike Earnest
28.4k22153
answered Dec 24 '18 at 17:44
Mike EarnestMike Earnest
28.4k22153
answered Dec 24 '18 at 17:44
Mike EarnestMike Earnest
28.4k22153
answered Dec 24 '18 at 17:44
Mike EarnestMike Earnest
28.4k22153
28.4k22153
$begingroup$
This makes everything clear. I wish I hadn't been intimidated by the whole problem. It's not that hard to proceed once you think well about the index domains.
$endgroup$
– BoLe
Dec 25 '18 at 13:50
add a comment |
$begingroup$
This makes everything clear. I wish I hadn't been intimidated by the whole problem. It's not that hard to proceed once you think well about the index domains.
$endgroup$
– BoLe
Dec 25 '18 at 13:50
$begingroup$
This makes everything clear. I wish I hadn't been intimidated by the whole problem. It's not that hard to proceed once you think well about the index domains.
$endgroup$
– BoLe
Dec 25 '18 at 13:50
$begingroup$
This makes everything clear. I wish I hadn't been intimidated by the whole problem. It's not that hard to proceed once you think well about the index domains.
$endgroup$
– BoLe
Dec 25 '18 at 13:50
add a comment |
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