Mixing
A finite summation of double binomial coefficients
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I find the following identity and have checked on Mathematica, while I have no idea how to prove it:
$$sum_{j=0}^n(-1)^{n-j}binom{p+j}{j}binom{n+beta}{n-j}=binom{p-beta}{n}, quad beta>-1, quad p>beta-1.$$
It seems that proof by induction does not work and because of the gamma function, the transform technique does not work either.
summation binomial-coefficients gamma-function
asked Jan 7 at 22:54
gouwangzhangdonggouwangzhangdong
788
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add a comment |
$begingroup$
I find the following identity and have checked on Mathematica, while I have no idea how to prove it:
$$sum_{j=0}^n(-1)^{n-j}binom{p+j}{j}binom{n+beta}{n-j}=binom{p-beta}{n}, quad beta>-1, quad p>beta-1.$$
It seems that proof by induction does not work and because of the gamma function, the transform technique does not work either.
summation binomial-coefficients gamma-function
asked Jan 7 at 22:54
gouwangzhangdonggouwangzhangdong
788
$endgroup$
add a comment |
$begingroup$
I find the following identity and have checked on Mathematica, while I have no idea how to prove it:
$$sum_{j=0}^n(-1)^{n-j}binom{p+j}{j}binom{n+beta}{n-j}=binom{p-beta}{n}, quad beta>-1, quad p>beta-1.$$
It seems that proof by induction does not work and because of the gamma function, the transform technique does not work either.
summation binomial-coefficients gamma-function
asked Jan 7 at 22:54
gouwangzhangdonggouwangzhangdong
788
$endgroup$
I find the following identity and have checked on Mathematica, while I have no idea how to prove it:
$$sum_{j=0}^n(-1)^{n-j}binom{p+j}{j}binom{n+beta}{n-j}=binom{p-beta}{n}, quad beta>-1, quad p>beta-1.$$
It seems that proof by induction does not work and because of the gamma function, the transform technique does not work either.
summation binomial-coefficients gamma-function
summation binomial-coefficients gamma-function
asked Jan 7 at 22:54
gouwangzhangdonggouwangzhangdong
788
asked Jan 7 at 22:54
gouwangzhangdonggouwangzhangdong
788
edited Jan 8 at 11:07
gouwangzhangdong
asked Jan 7 at 22:54
gouwangzhangdonggouwangzhangdong
788
asked Jan 7 at 22:54
gouwangzhangdonggouwangzhangdong
788
asked Jan 7 at 22:54
gouwangzhangdonggouwangzhangdong
788
788
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
First, if $n,x,y$ are integers, $0le nle xle y$, then
$$binom xn=sum_{k=0}^n(-1)^kbinom{y-x}kbinom{y-k}{n-k}.tag1$$
This is just the inclusion-exclusion principle. If $X,Y$ are sets, $Xsubseteq Y$, $|X|=x$, $|Y|=y$, then $binom xn$ is the number of $n$-element subsets of $X$, which we think of as being sifted out of the collection of all $n$-element subsets of $Y$, and $binom{y-x}k$ is the number of $k$-element subsets of $Ysetminus X$, and $binom{y-k}{n-k}$ is the number of $n$-element subsets of $Y$ containing a given $k$-element subset.
Of course, for a fixed integer $nge0$, since $(1)$ is a polynomial identity in $x$ and
$y$ which holds for all integers $yge xge n$, it also holds for all real or complex values of $x$ and $y$.
Setting $x=p-beta$ and $y=p+n$ in $(1)$, we get
$$binom{p-beta}n=sum_{k=0}^n(-1)^kbinom{n+beta}kbinom{p+n-k}{n-k}.tag2$$
Finally, setting $k=n-j$ in $(2)$, we get your identity:
$$binom{p-beta}n=sum_{j=0}^n(-1)^{n-j}binom{n+beta}{n-j}binom{p+j}j$$
for natural $n$ and arbitrary $beta$ and $p$.
P.S. Here is a detailed explanation of $(1)$. If $S$ is a set, $binom Sn$ is the set of all $n$-element subsets of $S.$ Suppose $n,x,y$ are integers, $0le nle xle y$, and let $X,Y$ be sets, $|X|=x$, $|Y|=y$, $Xsubseteq Y$. Let $m=y-x$, $Ysetminus X={y_1,dots,y_m}$, $[m]={1,dots,m}$. For $kin[m]$ let $mathcal A_k={Ainbinom Yn:y_kin A}$. Then
$$binom xn=left|binom Xnright|=left|binom Yn-bigcup_{k=1}^mmathcal A_kright|=left|binom Ynright|-left|bigcup_{k=1}^mmathcal A_kright|=binom yn-sum_{emptysetne Ksubseteq[m]}(-1)^{k-1}left|bigcap_{kin K}mathcal A_kright|=binom yn-sum_{emptysetne Ksubseteq[m]}(-1)^{k-1}binom{y-|K|}{n-k}=binom yn-sum_{k=1}^m(-1)^{k-1}binom mkbinom{y-k}{n-k}=sum_{k=0}^m(-1)^kbinom mkbinom{y-k}{n-k}=sum_{k=0}^n(-1)^kbinom mkbinom{y-k}{n-k}=sum_{k=0}^n(-1)^kbinom{y-x}kbinom{y-k}{n-k}.$$
answered Jan 8 at 13:56
bofbof
51.3k457120
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add a comment |
$begingroup$
We have
$$
eqalign{
& sumlimits_{j = 0}^n {left( { - 1} right)^{,n - j} left( matrix{
p + j cr
j cr} right)left( matrix{
n + beta cr
n - j cr} right)} = cr
& = sumlimits_{j = 0}^n {left( { - 1} right)^{,n} left( matrix{
- p - 1 cr
j cr} right)left( matrix{
n + beta cr
n - j cr} right)} = cr
& = left( { - 1} right)^{,n} sumlimits_{left( {0, le } right),j,left( { le ,n} right)} {left( matrix{
- p - 1 cr
j cr} right)left( matrix{
n + beta cr
n - j cr} right)} = cr
& = left( { - 1} right)^{,n} left( matrix{
- p - 1 + n + beta cr
n cr} right) = cr
& = left( matrix{
p - beta cr
n cr} right)quad left| matrix{
;n in Z hfill cr
;forall p,beta hfill cr} right. cr}
$$
where the steps are:
- upper negation (always valid for integer $j$);
- we can omit summming limits, because they are implicit in the two binomials;
- convolution;
- upper negation (always valid for integer $n$).
answered Jan 8 at 15:32
G CabG Cab
18.7k31238
$endgroup$
$begingroup$
$（-1）^n$ should not be in the last equation, right? Could you also pass me a link about this convolution? I understand the convolution in probability, while I cannot see how they are connected.
$endgroup$
– gouwangzhangdong
Jan 8 at 21:40
$begingroup$
@gouwangzhangdong: a) fully right,sorry, typo edited b) is the convolution of binomials $(1+x)^a(1+x)^b=(1+x)^{a+b}$ known as Vandermonde convolution
$endgroup$
– G Cab
Jan 9 at 0:50
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Very nice proof. Many thanks.
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– gouwangzhangdong
Jan 9 at 3:31
$begingroup$
(+1) I posted an almost identical proof before I saw that this answer was essentially the same.
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– robjohn♦
Jan 9 at 5:51
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@robjohn: appreciate your fairness !
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– G Cab
Jan 9 at 17:18
add a comment |
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Evaluating
$$sum_{j=0}^n (-1)^{n-j} {p+jchoose j} {n+betachoose n-j}
= sum_{j=0}^n (-1)^j {p+n-jchoose n-j} {n+betachoose j}$$
we write
$$sum_{j=0}^n (-1)^j {n+betachoose j}
[z^{n-j}] (1+z)^{p+n-j}
\ = [z^n] (1+z)^{p+n} sum_{j=0}^n (-1)^j {n+betachoose j}
z^j (1+z)^{-j}.$$
Now we may extend $j$ beyond $n$ because of the coefficient extractor
in front:
$$[z^n] (1+z)^{p+n} sum_{jge 0} (-1)^j {n+betachoose j}
z^j (1+z)^{-j}
\ = [z^n] (1+z)^{p+n}
left(1-frac{z}{1+z}right)^{n+beta}
= [z^n] (1+z)^{p-beta} = {p-betachoose n}.$$
answered Jan 8 at 15:28
Marko RiedelMarko Riedel
39.7k339108
$endgroup$
$begingroup$
Sorry, I am lost in your answer. First, why we can extend $j$ beyond $n$. For example, when $j=n+1$. the equation is not $0$. Second, why the last equation is true. At least, there is $z$ on the LHS while it is not on the RHS. Sorry for my stupidness.
$endgroup$
– gouwangzhangdong
Jan 8 at 22:37
$begingroup$
Expanding $(1+z)^{p+n} z^j (1+z)^{-j}$ in a formal power series yields $z^j + cdots$ Hence, when $jgt n,$ the contribution to $[z^n]$ is zero. For the second one, we are extracting a coefficient from a formal power series.
$endgroup$
– Marko Riedel
Jan 8 at 23:02
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Ah, I misunderstood your notation. Sorry for that. The last thing is how $[z^{n-j}]$ becomes $[z^n]$ in the second equation.
$endgroup$
– gouwangzhangdong
Jan 8 at 23:12
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If we have for a formal power series $f(z) = cdots + q z^{n-j} + cdots$ then $[z^{n-j}] f(z) = q$ but also $z ^j f(z) = cdots + q z^n +cdots$ so that $[z^n] z^j f(z) = q$ as well.
$endgroup$
– Marko Riedel
Jan 8 at 23:18
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Thanks for you explanation. I got it! Do you know how to add friends on this website so I can just @ someone in the later days?
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– gouwangzhangdong
Jan 8 at 23:28
|
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
First, if $n,x,y$ are integers, $0le nle xle y$, then
$$binom xn=sum_{k=0}^n(-1)^kbinom{y-x}kbinom{y-k}{n-k}.tag1$$
This is just the inclusion-exclusion principle. If $X,Y$ are sets, $Xsubseteq Y$, $|X|=x$, $|Y|=y$, then $binom xn$ is the number of $n$-element subsets of $X$, which we think of as being sifted out of the collection of all $n$-element subsets of $Y$, and $binom{y-x}k$ is the number of $k$-element subsets of $Ysetminus X$, and $binom{y-k}{n-k}$ is the number of $n$-element subsets of $Y$ containing a given $k$-element subset.
Of course, for a fixed integer $nge0$, since $(1)$ is a polynomial identity in $x$ and
$y$ which holds for all integers $yge xge n$, it also holds for all real or complex values of $x$ and $y$.
Setting $x=p-beta$ and $y=p+n$ in $(1)$, we get
$$binom{p-beta}n=sum_{k=0}^n(-1)^kbinom{n+beta}kbinom{p+n-k}{n-k}.tag2$$
Finally, setting $k=n-j$ in $(2)$, we get your identity:
$$binom{p-beta}n=sum_{j=0}^n(-1)^{n-j}binom{n+beta}{n-j}binom{p+j}j$$
for natural $n$ and arbitrary $beta$ and $p$.
P.S. Here is a detailed explanation of $(1)$. If $S$ is a set, $binom Sn$ is the set of all $n$-element subsets of $S.$ Suppose $n,x,y$ are integers, $0le nle xle y$, and let $X,Y$ be sets, $|X|=x$, $|Y|=y$, $Xsubseteq Y$. Let $m=y-x$, $Ysetminus X={y_1,dots,y_m}$, $[m]={1,dots,m}$. For $kin[m]$ let $mathcal A_k={Ainbinom Yn:y_kin A}$. Then
$$binom xn=left|binom Xnright|=left|binom Yn-bigcup_{k=1}^mmathcal A_kright|=left|binom Ynright|-left|bigcup_{k=1}^mmathcal A_kright|=binom yn-sum_{emptysetne Ksubseteq[m]}(-1)^{k-1}left|bigcap_{kin K}mathcal A_kright|=binom yn-sum_{emptysetne Ksubseteq[m]}(-1)^{k-1}binom{y-|K|}{n-k}=binom yn-sum_{k=1}^m(-1)^{k-1}binom mkbinom{y-k}{n-k}=sum_{k=0}^m(-1)^kbinom mkbinom{y-k}{n-k}=sum_{k=0}^n(-1)^kbinom mkbinom{y-k}{n-k}=sum_{k=0}^n(-1)^kbinom{y-x}kbinom{y-k}{n-k}.$$
answered Jan 8 at 13:56
bofbof
51.3k457120
$endgroup$
add a comment |
$begingroup$
First, if $n,x,y$ are integers, $0le nle xle y$, then
$$binom xn=sum_{k=0}^n(-1)^kbinom{y-x}kbinom{y-k}{n-k}.tag1$$
This is just the inclusion-exclusion principle. If $X,Y$ are sets, $Xsubseteq Y$, $|X|=x$, $|Y|=y$, then $binom xn$ is the number of $n$-element subsets of $X$, which we think of as being sifted out of the collection of all $n$-element subsets of $Y$, and $binom{y-x}k$ is the number of $k$-element subsets of $Ysetminus X$, and $binom{y-k}{n-k}$ is the number of $n$-element subsets of $Y$ containing a given $k$-element subset.
Of course, for a fixed integer $nge0$, since $(1)$ is a polynomial identity in $x$ and
$y$ which holds for all integers $yge xge n$, it also holds for all real or complex values of $x$ and $y$.
Setting $x=p-beta$ and $y=p+n$ in $(1)$, we get
$$binom{p-beta}n=sum_{k=0}^n(-1)^kbinom{n+beta}kbinom{p+n-k}{n-k}.tag2$$
Finally, setting $k=n-j$ in $(2)$, we get your identity:
$$binom{p-beta}n=sum_{j=0}^n(-1)^{n-j}binom{n+beta}{n-j}binom{p+j}j$$
for natural $n$ and arbitrary $beta$ and $p$.
P.S. Here is a detailed explanation of $(1)$. If $S$ is a set, $binom Sn$ is the set of all $n$-element subsets of $S.$ Suppose $n,x,y$ are integers, $0le nle xle y$, and let $X,Y$ be sets, $|X|=x$, $|Y|=y$, $Xsubseteq Y$. Let $m=y-x$, $Ysetminus X={y_1,dots,y_m}$, $[m]={1,dots,m}$. For $kin[m]$ let $mathcal A_k={Ainbinom Yn:y_kin A}$. Then
$$binom xn=left|binom Xnright|=left|binom Yn-bigcup_{k=1}^mmathcal A_kright|=left|binom Ynright|-left|bigcup_{k=1}^mmathcal A_kright|=binom yn-sum_{emptysetne Ksubseteq[m]}(-1)^{k-1}left|bigcap_{kin K}mathcal A_kright|=binom yn-sum_{emptysetne Ksubseteq[m]}(-1)^{k-1}binom{y-|K|}{n-k}=binom yn-sum_{k=1}^m(-1)^{k-1}binom mkbinom{y-k}{n-k}=sum_{k=0}^m(-1)^kbinom mkbinom{y-k}{n-k}=sum_{k=0}^n(-1)^kbinom mkbinom{y-k}{n-k}=sum_{k=0}^n(-1)^kbinom{y-x}kbinom{y-k}{n-k}.$$
answered Jan 8 at 13:56
bofbof
51.3k457120
$endgroup$
add a comment |
$begingroup$
First, if $n,x,y$ are integers, $0le nle xle y$, then
$$binom xn=sum_{k=0}^n(-1)^kbinom{y-x}kbinom{y-k}{n-k}.tag1$$
This is just the inclusion-exclusion principle. If $X,Y$ are sets, $Xsubseteq Y$, $|X|=x$, $|Y|=y$, then $binom xn$ is the number of $n$-element subsets of $X$, which we think of as being sifted out of the collection of all $n$-element subsets of $Y$, and $binom{y-x}k$ is the number of $k$-element subsets of $Ysetminus X$, and $binom{y-k}{n-k}$ is the number of $n$-element subsets of $Y$ containing a given $k$-element subset.
Of course, for a fixed integer $nge0$, since $(1)$ is a polynomial identity in $x$ and
$y$ which holds for all integers $yge xge n$, it also holds for all real or complex values of $x$ and $y$.
Setting $x=p-beta$ and $y=p+n$ in $(1)$, we get
$$binom{p-beta}n=sum_{k=0}^n(-1)^kbinom{n+beta}kbinom{p+n-k}{n-k}.tag2$$
Finally, setting $k=n-j$ in $(2)$, we get your identity:
$$binom{p-beta}n=sum_{j=0}^n(-1)^{n-j}binom{n+beta}{n-j}binom{p+j}j$$
for natural $n$ and arbitrary $beta$ and $p$.
P.S. Here is a detailed explanation of $(1)$. If $S$ is a set, $binom Sn$ is the set of all $n$-element subsets of $S.$ Suppose $n,x,y$ are integers, $0le nle xle y$, and let $X,Y$ be sets, $|X|=x$, $|Y|=y$, $Xsubseteq Y$. Let $m=y-x$, $Ysetminus X={y_1,dots,y_m}$, $[m]={1,dots,m}$. For $kin[m]$ let $mathcal A_k={Ainbinom Yn:y_kin A}$. Then
$$binom xn=left|binom Xnright|=left|binom Yn-bigcup_{k=1}^mmathcal A_kright|=left|binom Ynright|-left|bigcup_{k=1}^mmathcal A_kright|=binom yn-sum_{emptysetne Ksubseteq[m]}(-1)^{k-1}left|bigcap_{kin K}mathcal A_kright|=binom yn-sum_{emptysetne Ksubseteq[m]}(-1)^{k-1}binom{y-|K|}{n-k}=binom yn-sum_{k=1}^m(-1)^{k-1}binom mkbinom{y-k}{n-k}=sum_{k=0}^m(-1)^kbinom mkbinom{y-k}{n-k}=sum_{k=0}^n(-1)^kbinom mkbinom{y-k}{n-k}=sum_{k=0}^n(-1)^kbinom{y-x}kbinom{y-k}{n-k}.$$
answered Jan 8 at 13:56
bofbof
51.3k457120
$endgroup$
First, if $n,x,y$ are integers, $0le nle xle y$, then
$$binom xn=sum_{k=0}^n(-1)^kbinom{y-x}kbinom{y-k}{n-k}.tag1$$
This is just the inclusion-exclusion principle. If $X,Y$ are sets, $Xsubseteq Y$, $|X|=x$, $|Y|=y$, then $binom xn$ is the number of $n$-element subsets of $X$, which we think of as being sifted out of the collection of all $n$-element subsets of $Y$, and $binom{y-x}k$ is the number of $k$-element subsets of $Ysetminus X$, and $binom{y-k}{n-k}$ is the number of $n$-element subsets of $Y$ containing a given $k$-element subset.
Of course, for a fixed integer $nge0$, since $(1)$ is a polynomial identity in $x$ and
$y$ which holds for all integers $yge xge n$, it also holds for all real or complex values of $x$ and $y$.
Setting $x=p-beta$ and $y=p+n$ in $(1)$, we get
$$binom{p-beta}n=sum_{k=0}^n(-1)^kbinom{n+beta}kbinom{p+n-k}{n-k}.tag2$$
Finally, setting $k=n-j$ in $(2)$, we get your identity:
$$binom{p-beta}n=sum_{j=0}^n(-1)^{n-j}binom{n+beta}{n-j}binom{p+j}j$$
for natural $n$ and arbitrary $beta$ and $p$.
P.S. Here is a detailed explanation of $(1)$. If $S$ is a set, $binom Sn$ is the set of all $n$-element subsets of $S.$ Suppose $n,x,y$ are integers, $0le nle xle y$, and let $X,Y$ be sets, $|X|=x$, $|Y|=y$, $Xsubseteq Y$. Let $m=y-x$, $Ysetminus X={y_1,dots,y_m}$, $[m]={1,dots,m}$. For $kin[m]$ let $mathcal A_k={Ainbinom Yn:y_kin A}$. Then
$$binom xn=left|binom Xnright|=left|binom Yn-bigcup_{k=1}^mmathcal A_kright|=left|binom Ynright|-left|bigcup_{k=1}^mmathcal A_kright|=binom yn-sum_{emptysetne Ksubseteq[m]}(-1)^{k-1}left|bigcap_{kin K}mathcal A_kright|=binom yn-sum_{emptysetne Ksubseteq[m]}(-1)^{k-1}binom{y-|K|}{n-k}=binom yn-sum_{k=1}^m(-1)^{k-1}binom mkbinom{y-k}{n-k}=sum_{k=0}^m(-1)^kbinom mkbinom{y-k}{n-k}=sum_{k=0}^n(-1)^kbinom mkbinom{y-k}{n-k}=sum_{k=0}^n(-1)^kbinom{y-x}kbinom{y-k}{n-k}.$$
answered Jan 8 at 13:56
bofbof
51.3k457120
edited Jan 9 at 4:11
answered Jan 8 at 13:56
bofbof
51.3k457120
answered Jan 8 at 13:56
bofbof
51.3k457120
answered Jan 8 at 13:56
bofbof
51.3k457120
51.3k457120
add a comment |
add a comment |
$begingroup$
We have
$$
eqalign{
& sumlimits_{j = 0}^n {left( { - 1} right)^{,n - j} left( matrix{
p + j cr
j cr} right)left( matrix{
n + beta cr
n - j cr} right)} = cr
& = sumlimits_{j = 0}^n {left( { - 1} right)^{,n} left( matrix{
- p - 1 cr
j cr} right)left( matrix{
n + beta cr
n - j cr} right)} = cr
& = left( { - 1} right)^{,n} sumlimits_{left( {0, le } right),j,left( { le ,n} right)} {left( matrix{
- p - 1 cr
j cr} right)left( matrix{
n + beta cr
n - j cr} right)} = cr
& = left( { - 1} right)^{,n} left( matrix{
- p - 1 + n + beta cr
n cr} right) = cr
& = left( matrix{
p - beta cr
n cr} right)quad left| matrix{
;n in Z hfill cr
;forall p,beta hfill cr} right. cr}
$$
where the steps are:
- upper negation (always valid for integer $j$);
- we can omit summming limits, because they are implicit in the two binomials;
- convolution;
- upper negation (always valid for integer $n$).
answered Jan 8 at 15:32
G CabG Cab
18.7k31238
$endgroup$
$begingroup$
$（-1）^n$ should not be in the last equation, right? Could you also pass me a link about this convolution? I understand the convolution in probability, while I cannot see how they are connected.
$endgroup$
– gouwangzhangdong
Jan 8 at 21:40
$begingroup$
@gouwangzhangdong: a) fully right,sorry, typo edited b) is the convolution of binomials $(1+x)^a(1+x)^b=(1+x)^{a+b}$ known as Vandermonde convolution
$endgroup$
– G Cab
Jan 9 at 0:50
$begingroup$
Very nice proof. Many thanks.
$endgroup$
– gouwangzhangdong
Jan 9 at 3:31
$begingroup$
(+1) I posted an almost identical proof before I saw that this answer was essentially the same.
$endgroup$
– robjohn♦
Jan 9 at 5:51
$begingroup$
@robjohn: appreciate your fairness !
$endgroup$
– G Cab
Jan 9 at 17:18
add a comment |
$begingroup$
We have
$$
eqalign{
& sumlimits_{j = 0}^n {left( { - 1} right)^{,n - j} left( matrix{
p + j cr
j cr} right)left( matrix{
n + beta cr
n - j cr} right)} = cr
& = sumlimits_{j = 0}^n {left( { - 1} right)^{,n} left( matrix{
- p - 1 cr
j cr} right)left( matrix{
n + beta cr
n - j cr} right)} = cr
& = left( { - 1} right)^{,n} sumlimits_{left( {0, le } right),j,left( { le ,n} right)} {left( matrix{
- p - 1 cr
j cr} right)left( matrix{
n + beta cr
n - j cr} right)} = cr
& = left( { - 1} right)^{,n} left( matrix{
- p - 1 + n + beta cr
n cr} right) = cr
& = left( matrix{
p - beta cr
n cr} right)quad left| matrix{
;n in Z hfill cr
;forall p,beta hfill cr} right. cr}
$$
where the steps are:
- upper negation (always valid for integer $j$);
- we can omit summming limits, because they are implicit in the two binomials;
- convolution;
- upper negation (always valid for integer $n$).
answered Jan 8 at 15:32
G CabG Cab
18.7k31238
$endgroup$
$begingroup$
$（-1）^n$ should not be in the last equation, right? Could you also pass me a link about this convolution? I understand the convolution in probability, while I cannot see how they are connected.
$endgroup$
– gouwangzhangdong
Jan 8 at 21:40
$begingroup$
@gouwangzhangdong: a) fully right,sorry, typo edited b) is the convolution of binomials $(1+x)^a(1+x)^b=(1+x)^{a+b}$ known as Vandermonde convolution
$endgroup$
– G Cab
Jan 9 at 0:50
$begingroup$
Very nice proof. Many thanks.
$endgroup$
– gouwangzhangdong
Jan 9 at 3:31
$begingroup$
(+1) I posted an almost identical proof before I saw that this answer was essentially the same.
$endgroup$
– robjohn♦
Jan 9 at 5:51
$begingroup$
@robjohn: appreciate your fairness !
$endgroup$
– G Cab
Jan 9 at 17:18
add a comment |
$begingroup$
We have
$$
eqalign{
& sumlimits_{j = 0}^n {left( { - 1} right)^{,n - j} left( matrix{
p + j cr
j cr} right)left( matrix{
n + beta cr
n - j cr} right)} = cr
& = sumlimits_{j = 0}^n {left( { - 1} right)^{,n} left( matrix{
- p - 1 cr
j cr} right)left( matrix{
n + beta cr
n - j cr} right)} = cr
& = left( { - 1} right)^{,n} sumlimits_{left( {0, le } right),j,left( { le ,n} right)} {left( matrix{
- p - 1 cr
j cr} right)left( matrix{
n + beta cr
n - j cr} right)} = cr
& = left( { - 1} right)^{,n} left( matrix{
- p - 1 + n + beta cr
n cr} right) = cr
& = left( matrix{
p - beta cr
n cr} right)quad left| matrix{
;n in Z hfill cr
;forall p,beta hfill cr} right. cr}
$$
where the steps are:
- upper negation (always valid for integer $j$);
- we can omit summming limits, because they are implicit in the two binomials;
- convolution;
- upper negation (always valid for integer $n$).
answered Jan 8 at 15:32
G CabG Cab
18.7k31238
$endgroup$
We have
$$
eqalign{
& sumlimits_{j = 0}^n {left( { - 1} right)^{,n - j} left( matrix{
p + j cr
j cr} right)left( matrix{
n + beta cr
n - j cr} right)} = cr
& = sumlimits_{j = 0}^n {left( { - 1} right)^{,n} left( matrix{
- p - 1 cr
j cr} right)left( matrix{
n + beta cr
n - j cr} right)} = cr
& = left( { - 1} right)^{,n} sumlimits_{left( {0, le } right),j,left( { le ,n} right)} {left( matrix{
- p - 1 cr
j cr} right)left( matrix{
n + beta cr
n - j cr} right)} = cr
& = left( { - 1} right)^{,n} left( matrix{
- p - 1 + n + beta cr
n cr} right) = cr
& = left( matrix{
p - beta cr
n cr} right)quad left| matrix{
;n in Z hfill cr
;forall p,beta hfill cr} right. cr}
$$
where the steps are:
- upper negation (always valid for integer $j$);
- we can omit summming limits, because they are implicit in the two binomials;
- convolution;
- upper negation (always valid for integer $n$).
answered Jan 8 at 15:32
G CabG Cab
18.7k31238
edited Jan 9 at 0:46
answered Jan 8 at 15:32
G CabG Cab
18.7k31238
answered Jan 8 at 15:32
G CabG Cab
18.7k31238
answered Jan 8 at 15:32
G CabG Cab
18.7k31238
18.7k31238
$begingroup$
$（-1）^n$ should not be in the last equation, right? Could you also pass me a link about this convolution? I understand the convolution in probability, while I cannot see how they are connected.
$endgroup$
– gouwangzhangdong
Jan 8 at 21:40
$begingroup$
@gouwangzhangdong: a) fully right,sorry, typo edited b) is the convolution of binomials $(1+x)^a(1+x)^b=(1+x)^{a+b}$ known as Vandermonde convolution
$endgroup$
– G Cab
Jan 9 at 0:50
$begingroup$
Very nice proof. Many thanks.
$endgroup$
– gouwangzhangdong
Jan 9 at 3:31
$begingroup$
(+1) I posted an almost identical proof before I saw that this answer was essentially the same.
$endgroup$
– robjohn♦
Jan 9 at 5:51
$begingroup$
@robjohn: appreciate your fairness !
$endgroup$
– G Cab
Jan 9 at 17:18
add a comment |
$begingroup$
$（-1）^n$ should not be in the last equation, right? Could you also pass me a link about this convolution? I understand the convolution in probability, while I cannot see how they are connected.
$endgroup$
– gouwangzhangdong
Jan 8 at 21:40
$begingroup$
@gouwangzhangdong: a) fully right,sorry, typo edited b) is the convolution of binomials $(1+x)^a(1+x)^b=(1+x)^{a+b}$ known as Vandermonde convolution
$endgroup$
– G Cab
Jan 9 at 0:50
$begingroup$
Very nice proof. Many thanks.
$endgroup$
– gouwangzhangdong
Jan 9 at 3:31
$begingroup$
(+1) I posted an almost identical proof before I saw that this answer was essentially the same.
$endgroup$
– robjohn♦
Jan 9 at 5:51
$begingroup$
@robjohn: appreciate your fairness !
$endgroup$
– G Cab
Jan 9 at 17:18
$begingroup$
$（-1）^n$ should not be in the last equation, right? Could you also pass me a link about this convolution? I understand the convolution in probability, while I cannot see how they are connected.
$endgroup$
– gouwangzhangdong
Jan 8 at 21:40
$begingroup$
$（-1）^n$ should not be in the last equation, right? Could you also pass me a link about this convolution? I understand the convolution in probability, while I cannot see how they are connected.
$endgroup$
– gouwangzhangdong
Jan 8 at 21:40
$begingroup$
@gouwangzhangdong: a) fully right,sorry, typo edited b) is the convolution of binomials $(1+x)^a(1+x)^b=(1+x)^{a+b}$ known as Vandermonde convolution
$endgroup$
– G Cab
Jan 9 at 0:50
$begingroup$
@gouwangzhangdong: a) fully right,sorry, typo edited b) is the convolution of binomials $(1+x)^a(1+x)^b=(1+x)^{a+b}$ known as Vandermonde convolution
$endgroup$
– G Cab
Jan 9 at 0:50
$begingroup$
Very nice proof. Many thanks.
$endgroup$
– gouwangzhangdong
Jan 9 at 3:31
$begingroup$
Very nice proof. Many thanks.
$endgroup$
– gouwangzhangdong
Jan 9 at 3:31
$begingroup$
(+1) I posted an almost identical proof before I saw that this answer was essentially the same.
$endgroup$
– robjohn♦
Jan 9 at 5:51
$begingroup$
(+1) I posted an almost identical proof before I saw that this answer was essentially the same.
$endgroup$
– robjohn♦
Jan 9 at 5:51
$begingroup$
@robjohn: appreciate your fairness !
$endgroup$
– G Cab
Jan 9 at 17:18
$begingroup$
@robjohn: appreciate your fairness !
$endgroup$
– G Cab
Jan 9 at 17:18
add a comment |
$begingroup$
Evaluating
$$sum_{j=0}^n (-1)^{n-j} {p+jchoose j} {n+betachoose n-j}
= sum_{j=0}^n (-1)^j {p+n-jchoose n-j} {n+betachoose j}$$
we write
$$sum_{j=0}^n (-1)^j {n+betachoose j}
[z^{n-j}] (1+z)^{p+n-j}
\ = [z^n] (1+z)^{p+n} sum_{j=0}^n (-1)^j {n+betachoose j}
z^j (1+z)^{-j}.$$
Now we may extend $j$ beyond $n$ because of the coefficient extractor
in front:
$$[z^n] (1+z)^{p+n} sum_{jge 0} (-1)^j {n+betachoose j}
z^j (1+z)^{-j}
\ = [z^n] (1+z)^{p+n}
left(1-frac{z}{1+z}right)^{n+beta}
= [z^n] (1+z)^{p-beta} = {p-betachoose n}.$$
answered Jan 8 at 15:28
Marko RiedelMarko Riedel
39.7k339108
$endgroup$
$begingroup$
Sorry, I am lost in your answer. First, why we can extend $j$ beyond $n$. For example, when $j=n+1$. the equation is not $0$. Second, why the last equation is true. At least, there is $z$ on the LHS while it is not on the RHS. Sorry for my stupidness.
$endgroup$
– gouwangzhangdong
Jan 8 at 22:37
$begingroup$
Expanding $(1+z)^{p+n} z^j (1+z)^{-j}$ in a formal power series yields $z^j + cdots$ Hence, when $jgt n,$ the contribution to $[z^n]$ is zero. For the second one, we are extracting a coefficient from a formal power series.
$endgroup$
– Marko Riedel
Jan 8 at 23:02
$begingroup$
Ah, I misunderstood your notation. Sorry for that. The last thing is how $[z^{n-j}]$ becomes $[z^n]$ in the second equation.
$endgroup$
– gouwangzhangdong
Jan 8 at 23:12
$begingroup$
If we have for a formal power series $f(z) = cdots + q z^{n-j} + cdots$ then $[z^{n-j}] f(z) = q$ but also $z ^j f(z) = cdots + q z^n +cdots$ so that $[z^n] z^j f(z) = q$ as well.
$endgroup$
– Marko Riedel
Jan 8 at 23:18
$begingroup$
Thanks for you explanation. I got it! Do you know how to add friends on this website so I can just @ someone in the later days?
$endgroup$
– gouwangzhangdong
Jan 8 at 23:28
|
show 1 more comment
$begingroup$
Evaluating
$$sum_{j=0}^n (-1)^{n-j} {p+jchoose j} {n+betachoose n-j}
= sum_{j=0}^n (-1)^j {p+n-jchoose n-j} {n+betachoose j}$$
we write
$$sum_{j=0}^n (-1)^j {n+betachoose j}
[z^{n-j}] (1+z)^{p+n-j}
\ = [z^n] (1+z)^{p+n} sum_{j=0}^n (-1)^j {n+betachoose j}
z^j (1+z)^{-j}.$$
Now we may extend $j$ beyond $n$ because of the coefficient extractor
in front:
$$[z^n] (1+z)^{p+n} sum_{jge 0} (-1)^j {n+betachoose j}
z^j (1+z)^{-j}
\ = [z^n] (1+z)^{p+n}
left(1-frac{z}{1+z}right)^{n+beta}
= [z^n] (1+z)^{p-beta} = {p-betachoose n}.$$
answered Jan 8 at 15:28
Marko RiedelMarko Riedel
39.7k339108
$endgroup$
$begingroup$
Sorry, I am lost in your answer. First, why we can extend $j$ beyond $n$. For example, when $j=n+1$. the equation is not $0$. Second, why the last equation is true. At least, there is $z$ on the LHS while it is not on the RHS. Sorry for my stupidness.
$endgroup$
– gouwangzhangdong
Jan 8 at 22:37
$begingroup$
Expanding $(1+z)^{p+n} z^j (1+z)^{-j}$ in a formal power series yields $z^j + cdots$ Hence, when $jgt n,$ the contribution to $[z^n]$ is zero. For the second one, we are extracting a coefficient from a formal power series.
$endgroup$
– Marko Riedel
Jan 8 at 23:02
$begingroup$
Ah, I misunderstood your notation. Sorry for that. The last thing is how $[z^{n-j}]$ becomes $[z^n]$ in the second equation.
$endgroup$
– gouwangzhangdong
Jan 8 at 23:12
$begingroup$
If we have for a formal power series $f(z) = cdots + q z^{n-j} + cdots$ then $[z^{n-j}] f(z) = q$ but also $z ^j f(z) = cdots + q z^n +cdots$ so that $[z^n] z^j f(z) = q$ as well.
$endgroup$
– Marko Riedel
Jan 8 at 23:18
$begingroup$
Thanks for you explanation. I got it! Do you know how to add friends on this website so I can just @ someone in the later days?
$endgroup$
– gouwangzhangdong
Jan 8 at 23:28
|
show 1 more comment
$begingroup$
Evaluating
$$sum_{j=0}^n (-1)^{n-j} {p+jchoose j} {n+betachoose n-j}
= sum_{j=0}^n (-1)^j {p+n-jchoose n-j} {n+betachoose j}$$
we write
$$sum_{j=0}^n (-1)^j {n+betachoose j}
[z^{n-j}] (1+z)^{p+n-j}
\ = [z^n] (1+z)^{p+n} sum_{j=0}^n (-1)^j {n+betachoose j}
z^j (1+z)^{-j}.$$
Now we may extend $j$ beyond $n$ because of the coefficient extractor
in front:
$$[z^n] (1+z)^{p+n} sum_{jge 0} (-1)^j {n+betachoose j}
z^j (1+z)^{-j}
\ = [z^n] (1+z)^{p+n}
left(1-frac{z}{1+z}right)^{n+beta}
= [z^n] (1+z)^{p-beta} = {p-betachoose n}.$$
answered Jan 8 at 15:28
Marko RiedelMarko Riedel
39.7k339108
$endgroup$
Evaluating
$$sum_{j=0}^n (-1)^{n-j} {p+jchoose j} {n+betachoose n-j}
= sum_{j=0}^n (-1)^j {p+n-jchoose n-j} {n+betachoose j}$$
we write
$$sum_{j=0}^n (-1)^j {n+betachoose j}
[z^{n-j}] (1+z)^{p+n-j}
\ = [z^n] (1+z)^{p+n} sum_{j=0}^n (-1)^j {n+betachoose j}
z^j (1+z)^{-j}.$$
Now we may extend $j$ beyond $n$ because of the coefficient extractor
in front:
$$[z^n] (1+z)^{p+n} sum_{jge 0} (-1)^j {n+betachoose j}
z^j (1+z)^{-j}
\ = [z^n] (1+z)^{p+n}
left(1-frac{z}{1+z}right)^{n+beta}
= [z^n] (1+z)^{p-beta} = {p-betachoose n}.$$
answered Jan 8 at 15:28
Marko RiedelMarko Riedel
39.7k339108
answered Jan 8 at 15:28
Marko RiedelMarko Riedel
39.7k339108
answered Jan 8 at 15:28
Marko RiedelMarko Riedel
39.7k339108
answered Jan 8 at 15:28
Marko RiedelMarko Riedel
39.7k339108
39.7k339108
$begingroup$
Sorry, I am lost in your answer. First, why we can extend $j$ beyond $n$. For example, when $j=n+1$. the equation is not $0$. Second, why the last equation is true. At least, there is $z$ on the LHS while it is not on the RHS. Sorry for my stupidness.
$endgroup$
– gouwangzhangdong
Jan 8 at 22:37
$begingroup$
Expanding $(1+z)^{p+n} z^j (1+z)^{-j}$ in a formal power series yields $z^j + cdots$ Hence, when $jgt n,$ the contribution to $[z^n]$ is zero. For the second one, we are extracting a coefficient from a formal power series.
$endgroup$
– Marko Riedel
Jan 8 at 23:02
$begingroup$
Ah, I misunderstood your notation. Sorry for that. The last thing is how $[z^{n-j}]$ becomes $[z^n]$ in the second equation.
$endgroup$
– gouwangzhangdong
Jan 8 at 23:12
$begingroup$
If we have for a formal power series $f(z) = cdots + q z^{n-j} + cdots$ then $[z^{n-j}] f(z) = q$ but also $z ^j f(z) = cdots + q z^n +cdots$ so that $[z^n] z^j f(z) = q$ as well.
$endgroup$
– Marko Riedel
Jan 8 at 23:18
$begingroup$
Thanks for you explanation. I got it! Do you know how to add friends on this website so I can just @ someone in the later days?
$endgroup$
– gouwangzhangdong
Jan 8 at 23:28
|
show 1 more comment
$begingroup$
Sorry, I am lost in your answer. First, why we can extend $j$ beyond $n$. For example, when $j=n+1$. the equation is not $0$. Second, why the last equation is true. At least, there is $z$ on the LHS while it is not on the RHS. Sorry for my stupidness.
$endgroup$
– gouwangzhangdong
Jan 8 at 22:37
$begingroup$
Expanding $(1+z)^{p+n} z^j (1+z)^{-j}$ in a formal power series yields $z^j + cdots$ Hence, when $jgt n,$ the contribution to $[z^n]$ is zero. For the second one, we are extracting a coefficient from a formal power series.
$endgroup$
– Marko Riedel
Jan 8 at 23:02
$begingroup$
Ah, I misunderstood your notation. Sorry for that. The last thing is how $[z^{n-j}]$ becomes $[z^n]$ in the second equation.
$endgroup$
– gouwangzhangdong
Jan 8 at 23:12
$begingroup$
If we have for a formal power series $f(z) = cdots + q z^{n-j} + cdots$ then $[z^{n-j}] f(z) = q$ but also $z ^j f(z) = cdots + q z^n +cdots$ so that $[z^n] z^j f(z) = q$ as well.
$endgroup$
– Marko Riedel
Jan 8 at 23:18
$begingroup$
Thanks for you explanation. I got it! Do you know how to add friends on this website so I can just @ someone in the later days?
$endgroup$
– gouwangzhangdong
Jan 8 at 23:28
$begingroup$
Sorry, I am lost in your answer. First, why we can extend $j$ beyond $n$. For example, when $j=n+1$. the equation is not $0$. Second, why the last equation is true. At least, there is $z$ on the LHS while it is not on the RHS. Sorry for my stupidness.
$endgroup$
– gouwangzhangdong
Jan 8 at 22:37
$begingroup$
Sorry, I am lost in your answer. First, why we can extend $j$ beyond $n$. For example, when $j=n+1$. the equation is not $0$. Second, why the last equation is true. At least, there is $z$ on the LHS while it is not on the RHS. Sorry for my stupidness.
$endgroup$
– gouwangzhangdong
Jan 8 at 22:37
$begingroup$
Expanding $(1+z)^{p+n} z^j (1+z)^{-j}$ in a formal power series yields $z^j + cdots$ Hence, when $jgt n,$ the contribution to $[z^n]$ is zero. For the second one, we are extracting a coefficient from a formal power series.
$endgroup$
– Marko Riedel
Jan 8 at 23:02
$begingroup$
Expanding $(1+z)^{p+n} z^j (1+z)^{-j}$ in a formal power series yields $z^j + cdots$ Hence, when $jgt n,$ the contribution to $[z^n]$ is zero. For the second one, we are extracting a coefficient from a formal power series.
$endgroup$
– Marko Riedel
Jan 8 at 23:02
$begingroup$
Ah, I misunderstood your notation. Sorry for that. The last thing is how $[z^{n-j}]$ becomes $[z^n]$ in the second equation.
$endgroup$
– gouwangzhangdong
Jan 8 at 23:12
$begingroup$
Ah, I misunderstood your notation. Sorry for that. The last thing is how $[z^{n-j}]$ becomes $[z^n]$ in the second equation.
$endgroup$
– gouwangzhangdong
Jan 8 at 23:12
$begingroup$
If we have for a formal power series $f(z) = cdots + q z^{n-j} + cdots$ then $[z^{n-j}] f(z) = q$ but also $z ^j f(z) = cdots + q z^n +cdots$ so that $[z^n] z^j f(z) = q$ as well.
$endgroup$
– Marko Riedel
Jan 8 at 23:18
$begingroup$
If we have for a formal power series $f(z) = cdots + q z^{n-j} + cdots$ then $[z^{n-j}] f(z) = q$ but also $z ^j f(z) = cdots + q z^n +cdots$ so that $[z^n] z^j f(z) = q$ as well.
$endgroup$
– Marko Riedel
Jan 8 at 23:18
$begingroup$
Thanks for you explanation. I got it! Do you know how to add friends on this website so I can just @ someone in the later days?
$endgroup$
– gouwangzhangdong
Jan 8 at 23:28
$begingroup$
Thanks for you explanation. I got it! Do you know how to add friends on this website so I can just @ someone in the later days?
$endgroup$
– gouwangzhangdong
Jan 8 at 23:28
|
show 1 more comment
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