“Upper summation” binomial identity: different version from “Concrete Mathematics”
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The book "Concrete Mathematics: A Foundation for Computer Science", 2nd Edition - authored by Ronald L. Graham, Donald E. Knuth, Oren Patashnik - has, in its page 174, a table called: "Table 174 The top ten binomial coefficient identities". The table is also available in Princeton's PlasmaWiki page about "Binomial Coefficient" - http://qed.princeton.edu/main/PlasmaWiki/Binomial_Coefficient
One of the identities is the "Upper Summation" that is stated both in the book and in PlasmaWiki as follows:
$$sum_{0 leq k leq n}binom{k}{m} = binom{n + 1}{m + 1}$$
However, another book - a Portuguese book called "Matemática Finita" (which, literally translated, means "Finite Math") - presents a different formula for the "Upper Summation" - that it calls "Adição do Índice Superior" (literally, "Upper Index Summation").The difference between the two is the lower index of the sum that is $m$ instead of 0 :
$$sum_{k=m}^{n}binom{k}{m} = binom{n + 1}{m + 1}$$
I find it strange that both formulas have the same result, considering that Knuth's apparently has the sum starting from index 0 and the "Matemática Finita" one starts from index $m$. Can anyone help me understand this? Am I missing something?
discrete-mathematics summation binomial-coefficients
add a comment |
up vote
3
down vote
favorite
The book "Concrete Mathematics: A Foundation for Computer Science", 2nd Edition - authored by Ronald L. Graham, Donald E. Knuth, Oren Patashnik - has, in its page 174, a table called: "Table 174 The top ten binomial coefficient identities". The table is also available in Princeton's PlasmaWiki page about "Binomial Coefficient" - http://qed.princeton.edu/main/PlasmaWiki/Binomial_Coefficient
One of the identities is the "Upper Summation" that is stated both in the book and in PlasmaWiki as follows:
$$sum_{0 leq k leq n}binom{k}{m} = binom{n + 1}{m + 1}$$
However, another book - a Portuguese book called "Matemática Finita" (which, literally translated, means "Finite Math") - presents a different formula for the "Upper Summation" - that it calls "Adição do Índice Superior" (literally, "Upper Index Summation").The difference between the two is the lower index of the sum that is $m$ instead of 0 :
$$sum_{k=m}^{n}binom{k}{m} = binom{n + 1}{m + 1}$$
I find it strange that both formulas have the same result, considering that Knuth's apparently has the sum starting from index 0 and the "Matemática Finita" one starts from index $m$. Can anyone help me understand this? Am I missing something?
discrete-mathematics summation binomial-coefficients
I've seen the second before, is it possible using the generalized binomial coefficient formula involving negative terms that the first formula works out? Or maybe they just set them to Zero hmmm
– Evan
Sep 15 '13 at 18:49
1
Oh wait, you get zeros even when you plug them into the generic formula hah, ie 3 choose 5 would be 3*2*1*0*-1/5!=0
– Evan
Sep 15 '13 at 18:51
add a comment |
up vote
3
down vote
favorite
up vote
3
down vote
favorite
The book "Concrete Mathematics: A Foundation for Computer Science", 2nd Edition - authored by Ronald L. Graham, Donald E. Knuth, Oren Patashnik - has, in its page 174, a table called: "Table 174 The top ten binomial coefficient identities". The table is also available in Princeton's PlasmaWiki page about "Binomial Coefficient" - http://qed.princeton.edu/main/PlasmaWiki/Binomial_Coefficient
One of the identities is the "Upper Summation" that is stated both in the book and in PlasmaWiki as follows:
$$sum_{0 leq k leq n}binom{k}{m} = binom{n + 1}{m + 1}$$
However, another book - a Portuguese book called "Matemática Finita" (which, literally translated, means "Finite Math") - presents a different formula for the "Upper Summation" - that it calls "Adição do Índice Superior" (literally, "Upper Index Summation").The difference between the two is the lower index of the sum that is $m$ instead of 0 :
$$sum_{k=m}^{n}binom{k}{m} = binom{n + 1}{m + 1}$$
I find it strange that both formulas have the same result, considering that Knuth's apparently has the sum starting from index 0 and the "Matemática Finita" one starts from index $m$. Can anyone help me understand this? Am I missing something?
discrete-mathematics summation binomial-coefficients
The book "Concrete Mathematics: A Foundation for Computer Science", 2nd Edition - authored by Ronald L. Graham, Donald E. Knuth, Oren Patashnik - has, in its page 174, a table called: "Table 174 The top ten binomial coefficient identities". The table is also available in Princeton's PlasmaWiki page about "Binomial Coefficient" - http://qed.princeton.edu/main/PlasmaWiki/Binomial_Coefficient
One of the identities is the "Upper Summation" that is stated both in the book and in PlasmaWiki as follows:
$$sum_{0 leq k leq n}binom{k}{m} = binom{n + 1}{m + 1}$$
However, another book - a Portuguese book called "Matemática Finita" (which, literally translated, means "Finite Math") - presents a different formula for the "Upper Summation" - that it calls "Adição do Índice Superior" (literally, "Upper Index Summation").The difference between the two is the lower index of the sum that is $m$ instead of 0 :
$$sum_{k=m}^{n}binom{k}{m} = binom{n + 1}{m + 1}$$
I find it strange that both formulas have the same result, considering that Knuth's apparently has the sum starting from index 0 and the "Matemática Finita" one starts from index $m$. Can anyone help me understand this? Am I missing something?
discrete-mathematics summation binomial-coefficients
discrete-mathematics summation binomial-coefficients
asked Sep 15 '13 at 17:58
ricmarques
14310
14310
I've seen the second before, is it possible using the generalized binomial coefficient formula involving negative terms that the first formula works out? Or maybe they just set them to Zero hmmm
– Evan
Sep 15 '13 at 18:49
1
Oh wait, you get zeros even when you plug them into the generic formula hah, ie 3 choose 5 would be 3*2*1*0*-1/5!=0
– Evan
Sep 15 '13 at 18:51
add a comment |
I've seen the second before, is it possible using the generalized binomial coefficient formula involving negative terms that the first formula works out? Or maybe they just set them to Zero hmmm
– Evan
Sep 15 '13 at 18:49
1
Oh wait, you get zeros even when you plug them into the generic formula hah, ie 3 choose 5 would be 3*2*1*0*-1/5!=0
– Evan
Sep 15 '13 at 18:51
I've seen the second before, is it possible using the generalized binomial coefficient formula involving negative terms that the first formula works out? Or maybe they just set them to Zero hmmm
– Evan
Sep 15 '13 at 18:49
I've seen the second before, is it possible using the generalized binomial coefficient formula involving negative terms that the first formula works out? Or maybe they just set them to Zero hmmm
– Evan
Sep 15 '13 at 18:49
1
1
Oh wait, you get zeros even when you plug them into the generic formula hah, ie 3 choose 5 would be 3*2*1*0*-1/5!=0
– Evan
Sep 15 '13 at 18:51
Oh wait, you get zeros even when you plug them into the generic formula hah, ie 3 choose 5 would be 3*2*1*0*-1/5!=0
– Evan
Sep 15 '13 at 18:51
add a comment |
2 Answers
2
active
oldest
votes
up vote
4
down vote
accepted
They’re exactly the same, because $binom{n}m=0$ when $n<m$. This is clear if you define the binomial coefficient $binom{n}m$ for non-negative integers $m$ and $n$ as the number of $m$-element subsets of an $n$-element set. It’s also easily checked if you define $binom{x}k$ for real $x$ and non-negative integers $k$ as
$$binom{x}k=frac{x^{underline k}}{k!}=frac{x(x-1)ldots(x-k+1)}{k!};,$$
where $x^{underline k}$ is a falling factorial: when $x$ is a non-negative integer less than $d$, one of the factors in the numerator is $0$.
@Marc: I didn’t offer any generalization. I gave a combinatorial definition and an algebraic one and pointed out that both show that the two versions of the identity are essentially the same.
– Brian M. Scott
Sep 16 '13 at 16:21
add a comment |
up vote
3
down vote
The formula is qualified in Concrete Mathematics by "integers $m,ngeq0$", so the difference between the two quoted right hand sides is $sum_{0leq k<m}binom km=0$ (all terms in this sum are$~0$).
I think the point of starting the summation at $0$ is that this is more natural in practice: you typically consider all partial sums down a "column" of Pascal's triangle, which starts to the right of the triangle with $m$ entries equal to$~0$ before reaching the right edge. Then for the formula in Concrete Mathematics, there is nothing special for $n<m$, apart from the fact that one hasn't got to the nonzero entries yet; on the other hand the formula $sum_{mleq kleq n}binom km=binom{n+1}{m+1}$ might suggest something fishy is going on for $n<m-1$, as the sum is over a "negative length interval"; by analogy with integrals, one might expect that subtracting off terms encountered during backing up would be the right thing to do. In fact the stated sum is empty, and correctly equated to$~0$ by the formula, because the terms that would have been "backed up" over are all$~0$; it is better however not to have such worries in the first place.
I note that in the "finite calculus" form (2.48) of the summation, we can allow negative integer $n$; then it is convenient to replace $n+1$ by $n$, and one gets the nice formula
$$
sumnolimits_0^nbinom kmdelta k = binom n{m+1}, quadtext{for all integers $n$, and integer $mneq-1$.}
$$
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
They’re exactly the same, because $binom{n}m=0$ when $n<m$. This is clear if you define the binomial coefficient $binom{n}m$ for non-negative integers $m$ and $n$ as the number of $m$-element subsets of an $n$-element set. It’s also easily checked if you define $binom{x}k$ for real $x$ and non-negative integers $k$ as
$$binom{x}k=frac{x^{underline k}}{k!}=frac{x(x-1)ldots(x-k+1)}{k!};,$$
where $x^{underline k}$ is a falling factorial: when $x$ is a non-negative integer less than $d$, one of the factors in the numerator is $0$.
@Marc: I didn’t offer any generalization. I gave a combinatorial definition and an algebraic one and pointed out that both show that the two versions of the identity are essentially the same.
– Brian M. Scott
Sep 16 '13 at 16:21
add a comment |
up vote
4
down vote
accepted
They’re exactly the same, because $binom{n}m=0$ when $n<m$. This is clear if you define the binomial coefficient $binom{n}m$ for non-negative integers $m$ and $n$ as the number of $m$-element subsets of an $n$-element set. It’s also easily checked if you define $binom{x}k$ for real $x$ and non-negative integers $k$ as
$$binom{x}k=frac{x^{underline k}}{k!}=frac{x(x-1)ldots(x-k+1)}{k!};,$$
where $x^{underline k}$ is a falling factorial: when $x$ is a non-negative integer less than $d$, one of the factors in the numerator is $0$.
@Marc: I didn’t offer any generalization. I gave a combinatorial definition and an algebraic one and pointed out that both show that the two versions of the identity are essentially the same.
– Brian M. Scott
Sep 16 '13 at 16:21
add a comment |
up vote
4
down vote
accepted
up vote
4
down vote
accepted
They’re exactly the same, because $binom{n}m=0$ when $n<m$. This is clear if you define the binomial coefficient $binom{n}m$ for non-negative integers $m$ and $n$ as the number of $m$-element subsets of an $n$-element set. It’s also easily checked if you define $binom{x}k$ for real $x$ and non-negative integers $k$ as
$$binom{x}k=frac{x^{underline k}}{k!}=frac{x(x-1)ldots(x-k+1)}{k!};,$$
where $x^{underline k}$ is a falling factorial: when $x$ is a non-negative integer less than $d$, one of the factors in the numerator is $0$.
They’re exactly the same, because $binom{n}m=0$ when $n<m$. This is clear if you define the binomial coefficient $binom{n}m$ for non-negative integers $m$ and $n$ as the number of $m$-element subsets of an $n$-element set. It’s also easily checked if you define $binom{x}k$ for real $x$ and non-negative integers $k$ as
$$binom{x}k=frac{x^{underline k}}{k!}=frac{x(x-1)ldots(x-k+1)}{k!};,$$
where $x^{underline k}$ is a falling factorial: when $x$ is a non-negative integer less than $d$, one of the factors in the numerator is $0$.
edited Sep 17 '13 at 5:11
answered Sep 15 '13 at 19:36
Brian M. Scott
454k38505906
454k38505906
@Marc: I didn’t offer any generalization. I gave a combinatorial definition and an algebraic one and pointed out that both show that the two versions of the identity are essentially the same.
– Brian M. Scott
Sep 16 '13 at 16:21
add a comment |
@Marc: I didn’t offer any generalization. I gave a combinatorial definition and an algebraic one and pointed out that both show that the two versions of the identity are essentially the same.
– Brian M. Scott
Sep 16 '13 at 16:21
@Marc: I didn’t offer any generalization. I gave a combinatorial definition and an algebraic one and pointed out that both show that the two versions of the identity are essentially the same.
– Brian M. Scott
Sep 16 '13 at 16:21
@Marc: I didn’t offer any generalization. I gave a combinatorial definition and an algebraic one and pointed out that both show that the two versions of the identity are essentially the same.
– Brian M. Scott
Sep 16 '13 at 16:21
add a comment |
up vote
3
down vote
The formula is qualified in Concrete Mathematics by "integers $m,ngeq0$", so the difference between the two quoted right hand sides is $sum_{0leq k<m}binom km=0$ (all terms in this sum are$~0$).
I think the point of starting the summation at $0$ is that this is more natural in practice: you typically consider all partial sums down a "column" of Pascal's triangle, which starts to the right of the triangle with $m$ entries equal to$~0$ before reaching the right edge. Then for the formula in Concrete Mathematics, there is nothing special for $n<m$, apart from the fact that one hasn't got to the nonzero entries yet; on the other hand the formula $sum_{mleq kleq n}binom km=binom{n+1}{m+1}$ might suggest something fishy is going on for $n<m-1$, as the sum is over a "negative length interval"; by analogy with integrals, one might expect that subtracting off terms encountered during backing up would be the right thing to do. In fact the stated sum is empty, and correctly equated to$~0$ by the formula, because the terms that would have been "backed up" over are all$~0$; it is better however not to have such worries in the first place.
I note that in the "finite calculus" form (2.48) of the summation, we can allow negative integer $n$; then it is convenient to replace $n+1$ by $n$, and one gets the nice formula
$$
sumnolimits_0^nbinom kmdelta k = binom n{m+1}, quadtext{for all integers $n$, and integer $mneq-1$.}
$$
add a comment |
up vote
3
down vote
The formula is qualified in Concrete Mathematics by "integers $m,ngeq0$", so the difference between the two quoted right hand sides is $sum_{0leq k<m}binom km=0$ (all terms in this sum are$~0$).
I think the point of starting the summation at $0$ is that this is more natural in practice: you typically consider all partial sums down a "column" of Pascal's triangle, which starts to the right of the triangle with $m$ entries equal to$~0$ before reaching the right edge. Then for the formula in Concrete Mathematics, there is nothing special for $n<m$, apart from the fact that one hasn't got to the nonzero entries yet; on the other hand the formula $sum_{mleq kleq n}binom km=binom{n+1}{m+1}$ might suggest something fishy is going on for $n<m-1$, as the sum is over a "negative length interval"; by analogy with integrals, one might expect that subtracting off terms encountered during backing up would be the right thing to do. In fact the stated sum is empty, and correctly equated to$~0$ by the formula, because the terms that would have been "backed up" over are all$~0$; it is better however not to have such worries in the first place.
I note that in the "finite calculus" form (2.48) of the summation, we can allow negative integer $n$; then it is convenient to replace $n+1$ by $n$, and one gets the nice formula
$$
sumnolimits_0^nbinom kmdelta k = binom n{m+1}, quadtext{for all integers $n$, and integer $mneq-1$.}
$$
add a comment |
up vote
3
down vote
up vote
3
down vote
The formula is qualified in Concrete Mathematics by "integers $m,ngeq0$", so the difference between the two quoted right hand sides is $sum_{0leq k<m}binom km=0$ (all terms in this sum are$~0$).
I think the point of starting the summation at $0$ is that this is more natural in practice: you typically consider all partial sums down a "column" of Pascal's triangle, which starts to the right of the triangle with $m$ entries equal to$~0$ before reaching the right edge. Then for the formula in Concrete Mathematics, there is nothing special for $n<m$, apart from the fact that one hasn't got to the nonzero entries yet; on the other hand the formula $sum_{mleq kleq n}binom km=binom{n+1}{m+1}$ might suggest something fishy is going on for $n<m-1$, as the sum is over a "negative length interval"; by analogy with integrals, one might expect that subtracting off terms encountered during backing up would be the right thing to do. In fact the stated sum is empty, and correctly equated to$~0$ by the formula, because the terms that would have been "backed up" over are all$~0$; it is better however not to have such worries in the first place.
I note that in the "finite calculus" form (2.48) of the summation, we can allow negative integer $n$; then it is convenient to replace $n+1$ by $n$, and one gets the nice formula
$$
sumnolimits_0^nbinom kmdelta k = binom n{m+1}, quadtext{for all integers $n$, and integer $mneq-1$.}
$$
The formula is qualified in Concrete Mathematics by "integers $m,ngeq0$", so the difference between the two quoted right hand sides is $sum_{0leq k<m}binom km=0$ (all terms in this sum are$~0$).
I think the point of starting the summation at $0$ is that this is more natural in practice: you typically consider all partial sums down a "column" of Pascal's triangle, which starts to the right of the triangle with $m$ entries equal to$~0$ before reaching the right edge. Then for the formula in Concrete Mathematics, there is nothing special for $n<m$, apart from the fact that one hasn't got to the nonzero entries yet; on the other hand the formula $sum_{mleq kleq n}binom km=binom{n+1}{m+1}$ might suggest something fishy is going on for $n<m-1$, as the sum is over a "negative length interval"; by analogy with integrals, one might expect that subtracting off terms encountered during backing up would be the right thing to do. In fact the stated sum is empty, and correctly equated to$~0$ by the formula, because the terms that would have been "backed up" over are all$~0$; it is better however not to have such worries in the first place.
I note that in the "finite calculus" form (2.48) of the summation, we can allow negative integer $n$; then it is convenient to replace $n+1$ by $n$, and one gets the nice formula
$$
sumnolimits_0^nbinom kmdelta k = binom n{m+1}, quadtext{for all integers $n$, and integer $mneq-1$.}
$$
edited Dec 3 at 13:01
answered Sep 16 '13 at 14:47
Marc van Leeuwen
86.1k5105218
86.1k5105218
add a comment |
add a comment |
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I've seen the second before, is it possible using the generalized binomial coefficient formula involving negative terms that the first formula works out? Or maybe they just set them to Zero hmmm
– Evan
Sep 15 '13 at 18:49
1
Oh wait, you get zeros even when you plug them into the generic formula hah, ie 3 choose 5 would be 3*2*1*0*-1/5!=0
– Evan
Sep 15 '13 at 18:51