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Sum of Squares of Binomial Coefficients Using Residue Theorem
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I ran across this interesting question recently that I have an idea for, but am unable to complete. Basically, we use the residue formula to find $$ sumlimits_k=0^n nchoose k^2$$
We define $f$ asbeginalign*
f(z)&= frac1z(1+z)^n(1+frac1z)^n\ &= frac1zleft(nchoose 0 + nchoose 1z + cdots + nchoose n z^n right)left(nchoose 0 + nchoose 1frac1z + cdots + nchoose n frac1z^n right)\
endalign*
We see that the $frac1z$ term will have the coefficient $sumlimits_k=0^n nchoose k^2$. This would imply that $$res_z=0 f(z) = sumlimits_k=0^nnchoose k^2$$
Here's where things start falling apart for me. I choose to integrate $f$ over the unit disc $D$, which contains the pole $z=0$ and according to the residue theorem, I should get that $$ intlimits_D f(z)dz = 2pi i sumlimits_k=0^nnchoose k^2$$
But calculating the integral, I get that beginalign*
intlimits_D f(z)dz &= intlimits_0^2pi fracie^ithetae^itheta(1+e^itheta)^n(1+e^-itheta)^ndtheta\
&= i intlimits_0^2pi (1+e^itheta)^n (1+e^-itheta)^ndtheta
endalign*
However, Wolfram Alpha has that this integral goes to $0$ but we don't have that $sumlimits_k=0^n nchoose k^2 = 0$. Would someone mind pointing out where I messed up in my proof or perhaps point me in a better direction?
integration complex-analysis binomial-coefficients residue-calculus
asked Jul 18 at 4:26
Josh
477
add a comment |Â
up vote
5
down vote
favorite
I ran across this interesting question recently that I have an idea for, but am unable to complete. Basically, we use the residue formula to find $$ sumlimits_k=0^n nchoose k^2$$
We define $f$ asbeginalign*
f(z)&= frac1z(1+z)^n(1+frac1z)^n\ &= frac1zleft(nchoose 0 + nchoose 1z + cdots + nchoose n z^n right)left(nchoose 0 + nchoose 1frac1z + cdots + nchoose n frac1z^n right)\
endalign*
We see that the $frac1z$ term will have the coefficient $sumlimits_k=0^n nchoose k^2$. This would imply that $$res_z=0 f(z) = sumlimits_k=0^nnchoose k^2$$
Here's where things start falling apart for me. I choose to integrate $f$ over the unit disc $D$, which contains the pole $z=0$ and according to the residue theorem, I should get that $$ intlimits_D f(z)dz = 2pi i sumlimits_k=0^nnchoose k^2$$
But calculating the integral, I get that beginalign*
intlimits_D f(z)dz &= intlimits_0^2pi fracie^ithetae^itheta(1+e^itheta)^n(1+e^-itheta)^ndtheta\
&= i intlimits_0^2pi (1+e^itheta)^n (1+e^-itheta)^ndtheta
endalign*
However, Wolfram Alpha has that this integral goes to $0$ but we don't have that $sumlimits_k=0^n nchoose k^2 = 0$. Would someone mind pointing out where I messed up in my proof or perhaps point me in a better direction?
integration complex-analysis binomial-coefficients residue-calculus
asked Jul 18 at 4:26
Josh
477
1
Two easier proofs (but unrelated to the desired technique): 1) Consider $(1+x)^2n = (1+x)^ncdot (x+1)^n$ and apply binomial theorem. 2) Consider an urn with $n$ red balls and $n$ white balls and count how many ways you can select $n$ balls total from the urn in two different ways: directly, or via breaking into cases based on how many red balls were selected.
– JMoravitz
Jul 18 at 4:48
add a comment |Â
up vote
5
down vote
favorite
up vote
5
down vote
favorite
I ran across this interesting question recently that I have an idea for, but am unable to complete. Basically, we use the residue formula to find $$ sumlimits_k=0^n nchoose k^2$$
We define $f$ asbeginalign*
f(z)&= frac1z(1+z)^n(1+frac1z)^n\ &= frac1zleft(nchoose 0 + nchoose 1z + cdots + nchoose n z^n right)left(nchoose 0 + nchoose 1frac1z + cdots + nchoose n frac1z^n right)\
endalign*
We see that the $frac1z$ term will have the coefficient $sumlimits_k=0^n nchoose k^2$. This would imply that $$res_z=0 f(z) = sumlimits_k=0^nnchoose k^2$$
Here's where things start falling apart for me. I choose to integrate $f$ over the unit disc $D$, which contains the pole $z=0$ and according to the residue theorem, I should get that $$ intlimits_D f(z)dz = 2pi i sumlimits_k=0^nnchoose k^2$$
But calculating the integral, I get that beginalign*
intlimits_D f(z)dz &= intlimits_0^2pi fracie^ithetae^itheta(1+e^itheta)^n(1+e^-itheta)^ndtheta\
&= i intlimits_0^2pi (1+e^itheta)^n (1+e^-itheta)^ndtheta
endalign*
However, Wolfram Alpha has that this integral goes to $0$ but we don't have that $sumlimits_k=0^n nchoose k^2 = 0$. Would someone mind pointing out where I messed up in my proof or perhaps point me in a better direction?
integration complex-analysis binomial-coefficients residue-calculus
asked Jul 18 at 4:26
Josh
477
I ran across this interesting question recently that I have an idea for, but am unable to complete. Basically, we use the residue formula to find $$ sumlimits_k=0^n nchoose k^2$$
We define $f$ asbeginalign*
f(z)&= frac1z(1+z)^n(1+frac1z)^n\ &= frac1zleft(nchoose 0 + nchoose 1z + cdots + nchoose n z^n right)left(nchoose 0 + nchoose 1frac1z + cdots + nchoose n frac1z^n right)\
endalign*
We see that the $frac1z$ term will have the coefficient $sumlimits_k=0^n nchoose k^2$. This would imply that $$res_z=0 f(z) = sumlimits_k=0^nnchoose k^2$$
Here's where things start falling apart for me. I choose to integrate $f$ over the unit disc $D$, which contains the pole $z=0$ and according to the residue theorem, I should get that $$ intlimits_D f(z)dz = 2pi i sumlimits_k=0^nnchoose k^2$$
But calculating the integral, I get that beginalign*
intlimits_D f(z)dz &= intlimits_0^2pi fracie^ithetae^itheta(1+e^itheta)^n(1+e^-itheta)^ndtheta\
&= i intlimits_0^2pi (1+e^itheta)^n (1+e^-itheta)^ndtheta
endalign*
However, Wolfram Alpha has that this integral goes to $0$ but we don't have that $sumlimits_k=0^n nchoose k^2 = 0$. Would someone mind pointing out where I messed up in my proof or perhaps point me in a better direction?
integration complex-analysis binomial-coefficients residue-calculus
asked Jul 18 at 4:26
Josh
477
asked Jul 18 at 4:26
Josh
477
asked Jul 18 at 4:26
Josh
477
asked Jul 18 at 4:26
Josh
477
477
1
Two easier proofs (but unrelated to the desired technique): 1) Consider $(1+x)^2n = (1+x)^ncdot (x+1)^n$ and apply binomial theorem. 2) Consider an urn with $n$ red balls and $n$ white balls and count how many ways you can select $n$ balls total from the urn in two different ways: directly, or via breaking into cases based on how many red balls were selected.
– JMoravitz
Jul 18 at 4:48
add a comment |Â
1
Two easier proofs (but unrelated to the desired technique): 1) Consider $(1+x)^2n = (1+x)^ncdot (x+1)^n$ and apply binomial theorem. 2) Consider an urn with $n$ red balls and $n$ white balls and count how many ways you can select $n$ balls total from the urn in two different ways: directly, or via breaking into cases based on how many red balls were selected.
– JMoravitz
Jul 18 at 4:48
1
1
Two easier proofs (but unrelated to the desired technique): 1) Consider $(1+x)^2n = (1+x)^ncdot (x+1)^n$ and apply binomial theorem. 2) Consider an urn with $n$ red balls and $n$ white balls and count how many ways you can select $n$ balls total from the urn in two different ways: directly, or via breaking into cases based on how many red balls were selected.
– JMoravitz
Jul 18 at 4:48
Two easier proofs (but unrelated to the desired technique): 1) Consider $(1+x)^2n = (1+x)^ncdot (x+1)^n$ and apply binomial theorem. 2) Consider an urn with $n$ red balls and $n$ white balls and count how many ways you can select $n$ balls total from the urn in two different ways: directly, or via breaking into cases based on how many red balls were selected.
– JMoravitz
Jul 18 at 4:48
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
6
down vote
accepted
Using the reduction formula
$$int_0^picos^2nphi,dphi
=frac2n-12nint_0^picos^2n-2phi,dphi ,$$
your sum is
$$eqalignS
&=frac12piint_0^2pi (1+e^itheta)^n(1+e^-itheta)^n,dthetacr
&=frac12piint_0^2pi(2+2costheta)^n,dthetacr
&=frac12piint_0^2pi 2^2ncos^2nfractheta2,dthetacr
&=frac1piint_0^pi2^2ncos^2nphi,dphicr
&=frac1pi 2^2nfrac2n-12nfrac2n-32n-2cdotsfrac12picr
&=frac(2n)!(n!)^2cr
&=binom2nn .cr$$
answered Jul 18 at 4:57
David
66k662125
add a comment |Â
up vote
6
down vote
Note, that applying the residue theorem is extracting the coefficient of $z^-1$ of the Laurent-series expansion.
Using the coefficient of operator $[z^n]$ to denote the coefficient of $z^n$ in a series we obtain
beginalign*
colorbluemathrmres_z=0f(z)&=[z^-1]frac1z(1+z)^nleft(1+frac1zright)^n\
&=[z^-1]frac1z^n+1(1+z)^2n\
&=[z^n](1+z)^2n\
&,,colorblue=binom2nn
endalign*
answered Jul 18 at 10:14
Markus Scheuer
56k450135
add a comment |Â
up vote
0
down vote
What a cute problem! Hint: using Cauchy formula, $int_c g(z)/ (z-a)^n+1= frac2pi in! g^(n)(a)$ where $g(z)=(1+z)^2n$ it reduces to finding the $n$th derivative of $g$ at 0.
answered Jul 18 at 4:42
baharampuri
878717
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
6
down vote
accepted
Using the reduction formula
$$int_0^picos^2nphi,dphi
=frac2n-12nint_0^picos^2n-2phi,dphi ,$$
your sum is
$$eqalignS
&=frac12piint_0^2pi (1+e^itheta)^n(1+e^-itheta)^n,dthetacr
&=frac12piint_0^2pi(2+2costheta)^n,dthetacr
&=frac12piint_0^2pi 2^2ncos^2nfractheta2,dthetacr
&=frac1piint_0^pi2^2ncos^2nphi,dphicr
&=frac1pi 2^2nfrac2n-12nfrac2n-32n-2cdotsfrac12picr
&=frac(2n)!(n!)^2cr
&=binom2nn .cr$$
answered Jul 18 at 4:57
David
66k662125
add a comment |Â
up vote
6
down vote
accepted
Using the reduction formula
$$int_0^picos^2nphi,dphi
=frac2n-12nint_0^picos^2n-2phi,dphi ,$$
your sum is
$$eqalignS
&=frac12piint_0^2pi (1+e^itheta)^n(1+e^-itheta)^n,dthetacr
&=frac12piint_0^2pi(2+2costheta)^n,dthetacr
&=frac12piint_0^2pi 2^2ncos^2nfractheta2,dthetacr
&=frac1piint_0^pi2^2ncos^2nphi,dphicr
&=frac1pi 2^2nfrac2n-12nfrac2n-32n-2cdotsfrac12picr
&=frac(2n)!(n!)^2cr
&=binom2nn .cr$$
answered Jul 18 at 4:57
David
66k662125
add a comment |Â
up vote
6
down vote
accepted
up vote
6
down vote
accepted
Using the reduction formula
$$int_0^picos^2nphi,dphi
=frac2n-12nint_0^picos^2n-2phi,dphi ,$$
your sum is
$$eqalignS
&=frac12piint_0^2pi (1+e^itheta)^n(1+e^-itheta)^n,dthetacr
&=frac12piint_0^2pi(2+2costheta)^n,dthetacr
&=frac12piint_0^2pi 2^2ncos^2nfractheta2,dthetacr
&=frac1piint_0^pi2^2ncos^2nphi,dphicr
&=frac1pi 2^2nfrac2n-12nfrac2n-32n-2cdotsfrac12picr
&=frac(2n)!(n!)^2cr
&=binom2nn .cr$$
answered Jul 18 at 4:57
David
66k662125
Using the reduction formula
$$int_0^picos^2nphi,dphi
=frac2n-12nint_0^picos^2n-2phi,dphi ,$$
your sum is
$$eqalignS
&=frac12piint_0^2pi (1+e^itheta)^n(1+e^-itheta)^n,dthetacr
&=frac12piint_0^2pi(2+2costheta)^n,dthetacr
&=frac12piint_0^2pi 2^2ncos^2nfractheta2,dthetacr
&=frac1piint_0^pi2^2ncos^2nphi,dphicr
&=frac1pi 2^2nfrac2n-12nfrac2n-32n-2cdotsfrac12picr
&=frac(2n)!(n!)^2cr
&=binom2nn .cr$$
answered Jul 18 at 4:57
David
66k662125
answered Jul 18 at 4:57
David
66k662125
answered Jul 18 at 4:57
David
66k662125
answered Jul 18 at 4:57
David
66k662125
66k662125
add a comment |Â
add a comment |Â
up vote
6
down vote
Note, that applying the residue theorem is extracting the coefficient of $z^-1$ of the Laurent-series expansion.
Using the coefficient of operator $[z^n]$ to denote the coefficient of $z^n$ in a series we obtain
beginalign*
colorbluemathrmres_z=0f(z)&=[z^-1]frac1z(1+z)^nleft(1+frac1zright)^n\
&=[z^-1]frac1z^n+1(1+z)^2n\
&=[z^n](1+z)^2n\
&,,colorblue=binom2nn
endalign*
answered Jul 18 at 10:14
Markus Scheuer
56k450135
add a comment |Â
up vote
6
down vote
Note, that applying the residue theorem is extracting the coefficient of $z^-1$ of the Laurent-series expansion.
Using the coefficient of operator $[z^n]$ to denote the coefficient of $z^n$ in a series we obtain
beginalign*
colorbluemathrmres_z=0f(z)&=[z^-1]frac1z(1+z)^nleft(1+frac1zright)^n\
&=[z^-1]frac1z^n+1(1+z)^2n\
&=[z^n](1+z)^2n\
&,,colorblue=binom2nn
endalign*
answered Jul 18 at 10:14
Markus Scheuer
56k450135
add a comment |Â
up vote
6
down vote
up vote
6
down vote
Note, that applying the residue theorem is extracting the coefficient of $z^-1$ of the Laurent-series expansion.
Using the coefficient of operator $[z^n]$ to denote the coefficient of $z^n$ in a series we obtain
beginalign*
colorbluemathrmres_z=0f(z)&=[z^-1]frac1z(1+z)^nleft(1+frac1zright)^n\
&=[z^-1]frac1z^n+1(1+z)^2n\
&=[z^n](1+z)^2n\
&,,colorblue=binom2nn
endalign*
answered Jul 18 at 10:14
Markus Scheuer
56k450135
Note, that applying the residue theorem is extracting the coefficient of $z^-1$ of the Laurent-series expansion.
Using the coefficient of operator $[z^n]$ to denote the coefficient of $z^n$ in a series we obtain
beginalign*
colorbluemathrmres_z=0f(z)&=[z^-1]frac1z(1+z)^nleft(1+frac1zright)^n\
&=[z^-1]frac1z^n+1(1+z)^2n\
&=[z^n](1+z)^2n\
&,,colorblue=binom2nn
endalign*
answered Jul 18 at 10:14
Markus Scheuer
56k450135
answered Jul 18 at 10:14
Markus Scheuer
56k450135
answered Jul 18 at 10:14
Markus Scheuer
56k450135
answered Jul 18 at 10:14
Markus Scheuer
56k450135
56k450135
add a comment |Â
add a comment |Â
up vote
0
down vote
What a cute problem! Hint: using Cauchy formula, $int_c g(z)/ (z-a)^n+1= frac2pi in! g^(n)(a)$ where $g(z)=(1+z)^2n$ it reduces to finding the $n$th derivative of $g$ at 0.
answered Jul 18 at 4:42
baharampuri
878717
add a comment |Â
up vote
0
down vote
What a cute problem! Hint: using Cauchy formula, $int_c g(z)/ (z-a)^n+1= frac2pi in! g^(n)(a)$ where $g(z)=(1+z)^2n$ it reduces to finding the $n$th derivative of $g$ at 0.
answered Jul 18 at 4:42
baharampuri
878717
add a comment |Â
up vote
0
down vote
up vote
0
down vote
What a cute problem! Hint: using Cauchy formula, $int_c g(z)/ (z-a)^n+1= frac2pi in! g^(n)(a)$ where $g(z)=(1+z)^2n$ it reduces to finding the $n$th derivative of $g$ at 0.
answered Jul 18 at 4:42
baharampuri
878717
What a cute problem! Hint: using Cauchy formula, $int_c g(z)/ (z-a)^n+1= frac2pi in! g^(n)(a)$ where $g(z)=(1+z)^2n$ it reduces to finding the $n$th derivative of $g$ at 0.
answered Jul 18 at 4:42
baharampuri
878717
answered Jul 18 at 4:42
baharampuri
878717
answered Jul 18 at 4:42
baharampuri
878717
answered Jul 18 at 4:42
baharampuri
878717
878717
add a comment |Â
add a comment |Â
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1
Two easier proofs (but unrelated to the desired technique): 1) Consider $(1+x)^2n = (1+x)^ncdot (x+1)^n$ and apply binomial theorem. 2) Consider an urn with $n$ red balls and $n$ white balls and count how many ways you can select $n$ balls total from the urn in two different ways: directly, or via breaking into cases based on how many red balls were selected.
– JMoravitz
Jul 18 at 4:48