Polynomial expression for $frac 1{2^n} sum_{i=0}^n binom{n}{i}(2i-n)^{2k}$
Let
$$F (n,k)=frac {1}{2^n}sum_{i=0}^n binom{n}{i}(2i-n)^{2k},$$
where $n,k$ are non-negative integers.
By numerical tests the expression is an integer polynomial in $n $ of order $k $:
$$ F(n,0)=1;
F (n,1)=n;
F (n,2)=n (3n-2),$$
and so on.
Is there a simple general expression for the polynomial?
sequences-and-series combinatorics
|
show 2 more comments
Let
$$F (n,k)=frac {1}{2^n}sum_{i=0}^n binom{n}{i}(2i-n)^{2k},$$
where $n,k$ are non-negative integers.
By numerical tests the expression is an integer polynomial in $n $ of order $k $:
$$ F(n,0)=1;
F (n,1)=n;
F (n,2)=n (3n-2),$$
and so on.
Is there a simple general expression for the polynomial?
sequences-and-series combinatorics
That is a closed form. The sum is finite. I'd try expanding $(2i-n)^{2k}$ or computing some values and plugging them in into OEIS, in any case.
– ajotatxe
Dec 9 at 18:10
@ajotatxe I have edited the question to clarify what I meant by "closed - form expression".
– user
Dec 9 at 19:30
1
This expression comes out in Exercise 7 of UMN Fall 2018 Math 5705 Homework set 4, and also in Corollary 2.5 of Stanley's Algebraic Combinatorics (but these two appearances are actually easily seen to be equivalent: a closed walk of length $n$ on the hypercube is uniquely determined by its starting point and the $n$-tuple of "signless step directions", which $n$-tuple is clearly all-even).
– darij grinberg
Dec 11 at 3:16
1
If there is a closed form, then Stanley could not find it. On the other hand, the polynomial claim is interesting.
– darij grinberg
Dec 11 at 3:18
1
Ah, I see why it is a polynomial. You can count the all-even $k$-tuples in $left[nright]^k$ according to the positions of equal entries (more formally: the set partition of $left[kright]$ that governs which of the entries of the tuple are equal). For any given such choice of positions, the number of tuples is a polynomial in $n$ with degree $k$ (namely, a power of $n$ times a power of $n-1$ times a power of $n-2$ and so on). Feel free to expand on this in an answer -- I am stuck in bed with a flu and not at my most productive.
– darij grinberg
Dec 11 at 3:45
|
show 2 more comments
Let
$$F (n,k)=frac {1}{2^n}sum_{i=0}^n binom{n}{i}(2i-n)^{2k},$$
where $n,k$ are non-negative integers.
By numerical tests the expression is an integer polynomial in $n $ of order $k $:
$$ F(n,0)=1;
F (n,1)=n;
F (n,2)=n (3n-2),$$
and so on.
Is there a simple general expression for the polynomial?
sequences-and-series combinatorics
Let
$$F (n,k)=frac {1}{2^n}sum_{i=0}^n binom{n}{i}(2i-n)^{2k},$$
where $n,k$ are non-negative integers.
By numerical tests the expression is an integer polynomial in $n $ of order $k $:
$$ F(n,0)=1;
F (n,1)=n;
F (n,2)=n (3n-2),$$
and so on.
Is there a simple general expression for the polynomial?
sequences-and-series combinatorics
sequences-and-series combinatorics
edited Dec 13 at 6:06
asked Dec 9 at 18:08
user
3,6361627
3,6361627
That is a closed form. The sum is finite. I'd try expanding $(2i-n)^{2k}$ or computing some values and plugging them in into OEIS, in any case.
– ajotatxe
Dec 9 at 18:10
@ajotatxe I have edited the question to clarify what I meant by "closed - form expression".
– user
Dec 9 at 19:30
1
This expression comes out in Exercise 7 of UMN Fall 2018 Math 5705 Homework set 4, and also in Corollary 2.5 of Stanley's Algebraic Combinatorics (but these two appearances are actually easily seen to be equivalent: a closed walk of length $n$ on the hypercube is uniquely determined by its starting point and the $n$-tuple of "signless step directions", which $n$-tuple is clearly all-even).
– darij grinberg
Dec 11 at 3:16
1
If there is a closed form, then Stanley could not find it. On the other hand, the polynomial claim is interesting.
– darij grinberg
Dec 11 at 3:18
1
Ah, I see why it is a polynomial. You can count the all-even $k$-tuples in $left[nright]^k$ according to the positions of equal entries (more formally: the set partition of $left[kright]$ that governs which of the entries of the tuple are equal). For any given such choice of positions, the number of tuples is a polynomial in $n$ with degree $k$ (namely, a power of $n$ times a power of $n-1$ times a power of $n-2$ and so on). Feel free to expand on this in an answer -- I am stuck in bed with a flu and not at my most productive.
– darij grinberg
Dec 11 at 3:45
|
show 2 more comments
That is a closed form. The sum is finite. I'd try expanding $(2i-n)^{2k}$ or computing some values and plugging them in into OEIS, in any case.
– ajotatxe
Dec 9 at 18:10
@ajotatxe I have edited the question to clarify what I meant by "closed - form expression".
– user
Dec 9 at 19:30
1
This expression comes out in Exercise 7 of UMN Fall 2018 Math 5705 Homework set 4, and also in Corollary 2.5 of Stanley's Algebraic Combinatorics (but these two appearances are actually easily seen to be equivalent: a closed walk of length $n$ on the hypercube is uniquely determined by its starting point and the $n$-tuple of "signless step directions", which $n$-tuple is clearly all-even).
– darij grinberg
Dec 11 at 3:16
1
If there is a closed form, then Stanley could not find it. On the other hand, the polynomial claim is interesting.
– darij grinberg
Dec 11 at 3:18
1
Ah, I see why it is a polynomial. You can count the all-even $k$-tuples in $left[nright]^k$ according to the positions of equal entries (more formally: the set partition of $left[kright]$ that governs which of the entries of the tuple are equal). For any given such choice of positions, the number of tuples is a polynomial in $n$ with degree $k$ (namely, a power of $n$ times a power of $n-1$ times a power of $n-2$ and so on). Feel free to expand on this in an answer -- I am stuck in bed with a flu and not at my most productive.
– darij grinberg
Dec 11 at 3:45
That is a closed form. The sum is finite. I'd try expanding $(2i-n)^{2k}$ or computing some values and plugging them in into OEIS, in any case.
– ajotatxe
Dec 9 at 18:10
That is a closed form. The sum is finite. I'd try expanding $(2i-n)^{2k}$ or computing some values and plugging them in into OEIS, in any case.
– ajotatxe
Dec 9 at 18:10
@ajotatxe I have edited the question to clarify what I meant by "closed - form expression".
– user
Dec 9 at 19:30
@ajotatxe I have edited the question to clarify what I meant by "closed - form expression".
– user
Dec 9 at 19:30
1
1
This expression comes out in Exercise 7 of UMN Fall 2018 Math 5705 Homework set 4, and also in Corollary 2.5 of Stanley's Algebraic Combinatorics (but these two appearances are actually easily seen to be equivalent: a closed walk of length $n$ on the hypercube is uniquely determined by its starting point and the $n$-tuple of "signless step directions", which $n$-tuple is clearly all-even).
– darij grinberg
Dec 11 at 3:16
This expression comes out in Exercise 7 of UMN Fall 2018 Math 5705 Homework set 4, and also in Corollary 2.5 of Stanley's Algebraic Combinatorics (but these two appearances are actually easily seen to be equivalent: a closed walk of length $n$ on the hypercube is uniquely determined by its starting point and the $n$-tuple of "signless step directions", which $n$-tuple is clearly all-even).
– darij grinberg
Dec 11 at 3:16
1
1
If there is a closed form, then Stanley could not find it. On the other hand, the polynomial claim is interesting.
– darij grinberg
Dec 11 at 3:18
If there is a closed form, then Stanley could not find it. On the other hand, the polynomial claim is interesting.
– darij grinberg
Dec 11 at 3:18
1
1
Ah, I see why it is a polynomial. You can count the all-even $k$-tuples in $left[nright]^k$ according to the positions of equal entries (more formally: the set partition of $left[kright]$ that governs which of the entries of the tuple are equal). For any given such choice of positions, the number of tuples is a polynomial in $n$ with degree $k$ (namely, a power of $n$ times a power of $n-1$ times a power of $n-2$ and so on). Feel free to expand on this in an answer -- I am stuck in bed with a flu and not at my most productive.
– darij grinberg
Dec 11 at 3:45
Ah, I see why it is a polynomial. You can count the all-even $k$-tuples in $left[nright]^k$ according to the positions of equal entries (more formally: the set partition of $left[kright]$ that governs which of the entries of the tuple are equal). For any given such choice of positions, the number of tuples is a polynomial in $n$ with degree $k$ (namely, a power of $n$ times a power of $n-1$ times a power of $n-2$ and so on). Feel free to expand on this in an answer -- I am stuck in bed with a flu and not at my most productive.
– darij grinberg
Dec 11 at 3:45
|
show 2 more comments
3 Answers
3
active
oldest
votes
It is convenient to use the coefficient of operator $[z^n]$ to denote the coefficient of $z^n$ of a series. This way we can write for instance
begin{align*}
n![z^n]e^{jz}=j^ntag{1}
end{align*}
We obtain
begin{align*}
color{blue}{frac{1}{2^n}}&color{blue}{sum_{j=0}^nbinom{n}{j}(2j-n)^{2k}}\
&=frac{1}{2^n}sum_{j=0}^nbinom{n}{j}(2k)![z^{2k}]e^{(2j-n)z}tag{2}\
&=frac{(2k)!}{2^n}[z^{2k}]e^{-nz}sum_{j=0}^nbinom{n}{j}left(e^{2z}right)^jtag{3}\
&=frac{(2k)!}{2^n}[z^{2k}]e^{-nz}left(1+e^{2z}right)^ntag{4}\
&=(2k)![z^{2k}]left(frac{e^{z}+e^{-z}}{2}right)^ntag{5}\
&,,color{blue}{=(2k)![z^{2k}]left(cosh zright)^n}
end{align*}
We see OPs formula is essentially the coefficient of $z^{2k}$ of $left(cosh zright)^n$ which does not have a closed formula as far as I know.
Comment:
In (2) we apply the coefficient of operator according to (1).
In (3) use the linearity of the coefficient of operator.
In (4) we apply the binomial theorem.
In (5) we write the expression somewhat more conveniently.
I think I have found such a closed polynomial formula. See my answer below.
– user
Dec 27 at 12:06
@user: A closed formula means there is no $Sigma$ involved. Somewhat more concrete a closed formula in this context is a function in $n$ with a constant number of terms. When using something like $Sigma_{j=0}^n$ the number of terms is not constant but increases with increasing $n$.
– Markus Scheuer
Dec 27 at 14:05
What was meant was a closed formula for the coefficients of the polynomial (which do not depend on $n $).
– user
Dec 27 at 16:14
@user: I agree, you're right. (+1) for your nice approach.
– Markus Scheuer
Dec 27 at 16:27
add a comment |
I'll play with the cosh and
see if I get
anything other than
the original problem.
$begin{array}\
cosh^n(x)
&=frac1{2^n}(e^x+e^{-x})^n\
&=frac1{2^n}sum_{k=0}^n binom{n}{k}e^ke^{(n-k)x}\
&=frac1{2^n}e^{nx}sum_{k=0}^n binom{n}{k}e^{-2kx}\
&=frac1{2^n}sum_{i=0}^{infty} dfrac{(nx)^i}{i!}sum_{k=0}^n binom{n}{k}sum_{j=0}^{infty} dfrac{(-2kx)^j}{j!}\
&=frac1{2^n}sum_{k=0}^n binom{n}{k}sum_{i=0}^{infty} dfrac{(nx)^i}{i!}sum_{j=0}^{infty} dfrac{(-2kx)^j}{j!}\
&=frac1{2^n}sum_{k=0}^n binom{n}{k}sum_{m=0}^{infty}sum_{i=0}^{m} dfrac{(nx)^i}{i!} dfrac{(-2kx)^{m-i}}{(m-i)!}\
&=frac1{2^n}sum_{k=0}^n binom{n}{k}sum_{m=0}^{infty}x^msum_{i=0}^{m} dfrac{(n)^i}{i!} dfrac{(-2k)^{m-i}}{(m-i)!}\
&=frac1{2^n}sum_{m=0}^{infty}x^msum_{k=0}^n binom{n}{k}sum_{i=0}^{m} dfrac{(n)^i}{i!} dfrac{(-2k)^{m-i}}{(m-i)!}\
&=frac1{2^n}sum_{m=0}^{infty}x^msum_{i=0}^{m}dfrac{(-2)^{m-i}(n)^i}{(m-i)!i!}sum_{k=0}^n binom{n}{k} k^{m-i}\
&=frac1{2^n}sum_{m=0}^{infty}dfrac{x^m}{m!}sum_{i=0}^{m}binom{m}{i}(-2)^{m-i}n^isum_{k=0}^n binom{n}{k} k^{m-i}\
&=frac1{2^n}sum_{m=0}^{infty}dfrac{(-2)^mx^m}{m!}sum_{i=0}^{m}binom{m}{i}(-n/2)^isum_{k=0}^n binom{n}{k} k^{m-i}\
&=frac1{2^n}sum_{m=0}^{infty}dfrac{(-2)^mx^m}{m!}sum_{k=0}^n binom{n}{k}sum_{i=0}^{m}binom{m}{i}(-n/2)^i k^{m-i}\
&=frac1{2^n}sum_{m=0}^{infty}dfrac{(-2)^mx^m}{m!}sum_{k=0}^n binom{n}{k}(-n/2+k)^m\
&=frac1{2^n}sum_{m=0}^{infty}dfrac{x^m}{m!}sum_{k=0}^n binom{n}{k}(2k-n)^m\
end{array}
$
And this is the OP.
Oh well.
add a comment |
Let $omega $ be a $n$-dimensional vector with binary components $omega_i=pm1$ and $Omega_n $ be a set of all such vectors, the size of the set obviously being $2^n $. The sum of elements of a vector with $i$
positive and $n-i $ negative components is $2i-n $ and the number of such vectors is $binom {n}{i}$. Thus
$$
F (n,k)=frac {1}{2^n}sum_{omegainOmega_n } left (sum_{i=1}^nomega_i right)^{2k}
=frac {1}{2^n}sum_{omegainOmega_n } sum_{p_ige0}^{sum_i p_i=2k}binom {2k} {p_1,p_2,dots,p_n}prod_i omega_i^{p_i}
=sum_{p_ige0}^{sum_i p_i=2k}binom {2k} {p_1,p_2,dots,p_n}left (frac {1}{2^n}sum_{omegainOmega_n }prod_i omega_i^{p_i}right)
=sum_{p_ige0,;p_i,text {mod},2=0}^{sum_i p_i=2k}binom {2k} {p_1,p_2,dots,p_n}.
$$
To proceed further one splits the last sum into partial ones over terms with particular count $l$ of non-zero $p_i$ and ends up with:
$$
F (n,k)=sum_{l=1}^n T (k,l)frac{n!}{(n-l)!},tag {1}
$$
where $T (k,l)$ is the number of partitions of a set of size $2k$ into $l$ blocks of even size. It is the OEIS sequence A156289 with known close-form and recurrence expressions.
By numerical evidence the polynomial (1) can be expressed in the terms of usual powers as:
$$
F (n,k)=sum_{l=1}^n A (k,l)n^l,tag {2}
$$
with $A (k,l) $ being the OEIS sequence A318146. In other words $F(n,k) $ is in fact the so called Omega polynomial.
add a comment |
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3 Answers
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3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
It is convenient to use the coefficient of operator $[z^n]$ to denote the coefficient of $z^n$ of a series. This way we can write for instance
begin{align*}
n![z^n]e^{jz}=j^ntag{1}
end{align*}
We obtain
begin{align*}
color{blue}{frac{1}{2^n}}&color{blue}{sum_{j=0}^nbinom{n}{j}(2j-n)^{2k}}\
&=frac{1}{2^n}sum_{j=0}^nbinom{n}{j}(2k)![z^{2k}]e^{(2j-n)z}tag{2}\
&=frac{(2k)!}{2^n}[z^{2k}]e^{-nz}sum_{j=0}^nbinom{n}{j}left(e^{2z}right)^jtag{3}\
&=frac{(2k)!}{2^n}[z^{2k}]e^{-nz}left(1+e^{2z}right)^ntag{4}\
&=(2k)![z^{2k}]left(frac{e^{z}+e^{-z}}{2}right)^ntag{5}\
&,,color{blue}{=(2k)![z^{2k}]left(cosh zright)^n}
end{align*}
We see OPs formula is essentially the coefficient of $z^{2k}$ of $left(cosh zright)^n$ which does not have a closed formula as far as I know.
Comment:
In (2) we apply the coefficient of operator according to (1).
In (3) use the linearity of the coefficient of operator.
In (4) we apply the binomial theorem.
In (5) we write the expression somewhat more conveniently.
I think I have found such a closed polynomial formula. See my answer below.
– user
Dec 27 at 12:06
@user: A closed formula means there is no $Sigma$ involved. Somewhat more concrete a closed formula in this context is a function in $n$ with a constant number of terms. When using something like $Sigma_{j=0}^n$ the number of terms is not constant but increases with increasing $n$.
– Markus Scheuer
Dec 27 at 14:05
What was meant was a closed formula for the coefficients of the polynomial (which do not depend on $n $).
– user
Dec 27 at 16:14
@user: I agree, you're right. (+1) for your nice approach.
– Markus Scheuer
Dec 27 at 16:27
add a comment |
It is convenient to use the coefficient of operator $[z^n]$ to denote the coefficient of $z^n$ of a series. This way we can write for instance
begin{align*}
n![z^n]e^{jz}=j^ntag{1}
end{align*}
We obtain
begin{align*}
color{blue}{frac{1}{2^n}}&color{blue}{sum_{j=0}^nbinom{n}{j}(2j-n)^{2k}}\
&=frac{1}{2^n}sum_{j=0}^nbinom{n}{j}(2k)![z^{2k}]e^{(2j-n)z}tag{2}\
&=frac{(2k)!}{2^n}[z^{2k}]e^{-nz}sum_{j=0}^nbinom{n}{j}left(e^{2z}right)^jtag{3}\
&=frac{(2k)!}{2^n}[z^{2k}]e^{-nz}left(1+e^{2z}right)^ntag{4}\
&=(2k)![z^{2k}]left(frac{e^{z}+e^{-z}}{2}right)^ntag{5}\
&,,color{blue}{=(2k)![z^{2k}]left(cosh zright)^n}
end{align*}
We see OPs formula is essentially the coefficient of $z^{2k}$ of $left(cosh zright)^n$ which does not have a closed formula as far as I know.
Comment:
In (2) we apply the coefficient of operator according to (1).
In (3) use the linearity of the coefficient of operator.
In (4) we apply the binomial theorem.
In (5) we write the expression somewhat more conveniently.
I think I have found such a closed polynomial formula. See my answer below.
– user
Dec 27 at 12:06
@user: A closed formula means there is no $Sigma$ involved. Somewhat more concrete a closed formula in this context is a function in $n$ with a constant number of terms. When using something like $Sigma_{j=0}^n$ the number of terms is not constant but increases with increasing $n$.
– Markus Scheuer
Dec 27 at 14:05
What was meant was a closed formula for the coefficients of the polynomial (which do not depend on $n $).
– user
Dec 27 at 16:14
@user: I agree, you're right. (+1) for your nice approach.
– Markus Scheuer
Dec 27 at 16:27
add a comment |
It is convenient to use the coefficient of operator $[z^n]$ to denote the coefficient of $z^n$ of a series. This way we can write for instance
begin{align*}
n![z^n]e^{jz}=j^ntag{1}
end{align*}
We obtain
begin{align*}
color{blue}{frac{1}{2^n}}&color{blue}{sum_{j=0}^nbinom{n}{j}(2j-n)^{2k}}\
&=frac{1}{2^n}sum_{j=0}^nbinom{n}{j}(2k)![z^{2k}]e^{(2j-n)z}tag{2}\
&=frac{(2k)!}{2^n}[z^{2k}]e^{-nz}sum_{j=0}^nbinom{n}{j}left(e^{2z}right)^jtag{3}\
&=frac{(2k)!}{2^n}[z^{2k}]e^{-nz}left(1+e^{2z}right)^ntag{4}\
&=(2k)![z^{2k}]left(frac{e^{z}+e^{-z}}{2}right)^ntag{5}\
&,,color{blue}{=(2k)![z^{2k}]left(cosh zright)^n}
end{align*}
We see OPs formula is essentially the coefficient of $z^{2k}$ of $left(cosh zright)^n$ which does not have a closed formula as far as I know.
Comment:
In (2) we apply the coefficient of operator according to (1).
In (3) use the linearity of the coefficient of operator.
In (4) we apply the binomial theorem.
In (5) we write the expression somewhat more conveniently.
It is convenient to use the coefficient of operator $[z^n]$ to denote the coefficient of $z^n$ of a series. This way we can write for instance
begin{align*}
n![z^n]e^{jz}=j^ntag{1}
end{align*}
We obtain
begin{align*}
color{blue}{frac{1}{2^n}}&color{blue}{sum_{j=0}^nbinom{n}{j}(2j-n)^{2k}}\
&=frac{1}{2^n}sum_{j=0}^nbinom{n}{j}(2k)![z^{2k}]e^{(2j-n)z}tag{2}\
&=frac{(2k)!}{2^n}[z^{2k}]e^{-nz}sum_{j=0}^nbinom{n}{j}left(e^{2z}right)^jtag{3}\
&=frac{(2k)!}{2^n}[z^{2k}]e^{-nz}left(1+e^{2z}right)^ntag{4}\
&=(2k)![z^{2k}]left(frac{e^{z}+e^{-z}}{2}right)^ntag{5}\
&,,color{blue}{=(2k)![z^{2k}]left(cosh zright)^n}
end{align*}
We see OPs formula is essentially the coefficient of $z^{2k}$ of $left(cosh zright)^n$ which does not have a closed formula as far as I know.
Comment:
In (2) we apply the coefficient of operator according to (1).
In (3) use the linearity of the coefficient of operator.
In (4) we apply the binomial theorem.
In (5) we write the expression somewhat more conveniently.
edited Dec 9 at 19:53
answered Dec 9 at 19:40
Markus Scheuer
60k455143
60k455143
I think I have found such a closed polynomial formula. See my answer below.
– user
Dec 27 at 12:06
@user: A closed formula means there is no $Sigma$ involved. Somewhat more concrete a closed formula in this context is a function in $n$ with a constant number of terms. When using something like $Sigma_{j=0}^n$ the number of terms is not constant but increases with increasing $n$.
– Markus Scheuer
Dec 27 at 14:05
What was meant was a closed formula for the coefficients of the polynomial (which do not depend on $n $).
– user
Dec 27 at 16:14
@user: I agree, you're right. (+1) for your nice approach.
– Markus Scheuer
Dec 27 at 16:27
add a comment |
I think I have found such a closed polynomial formula. See my answer below.
– user
Dec 27 at 12:06
@user: A closed formula means there is no $Sigma$ involved. Somewhat more concrete a closed formula in this context is a function in $n$ with a constant number of terms. When using something like $Sigma_{j=0}^n$ the number of terms is not constant but increases with increasing $n$.
– Markus Scheuer
Dec 27 at 14:05
What was meant was a closed formula for the coefficients of the polynomial (which do not depend on $n $).
– user
Dec 27 at 16:14
@user: I agree, you're right. (+1) for your nice approach.
– Markus Scheuer
Dec 27 at 16:27
I think I have found such a closed polynomial formula. See my answer below.
– user
Dec 27 at 12:06
I think I have found such a closed polynomial formula. See my answer below.
– user
Dec 27 at 12:06
@user: A closed formula means there is no $Sigma$ involved. Somewhat more concrete a closed formula in this context is a function in $n$ with a constant number of terms. When using something like $Sigma_{j=0}^n$ the number of terms is not constant but increases with increasing $n$.
– Markus Scheuer
Dec 27 at 14:05
@user: A closed formula means there is no $Sigma$ involved. Somewhat more concrete a closed formula in this context is a function in $n$ with a constant number of terms. When using something like $Sigma_{j=0}^n$ the number of terms is not constant but increases with increasing $n$.
– Markus Scheuer
Dec 27 at 14:05
What was meant was a closed formula for the coefficients of the polynomial (which do not depend on $n $).
– user
Dec 27 at 16:14
What was meant was a closed formula for the coefficients of the polynomial (which do not depend on $n $).
– user
Dec 27 at 16:14
@user: I agree, you're right. (+1) for your nice approach.
– Markus Scheuer
Dec 27 at 16:27
@user: I agree, you're right. (+1) for your nice approach.
– Markus Scheuer
Dec 27 at 16:27
add a comment |
I'll play with the cosh and
see if I get
anything other than
the original problem.
$begin{array}\
cosh^n(x)
&=frac1{2^n}(e^x+e^{-x})^n\
&=frac1{2^n}sum_{k=0}^n binom{n}{k}e^ke^{(n-k)x}\
&=frac1{2^n}e^{nx}sum_{k=0}^n binom{n}{k}e^{-2kx}\
&=frac1{2^n}sum_{i=0}^{infty} dfrac{(nx)^i}{i!}sum_{k=0}^n binom{n}{k}sum_{j=0}^{infty} dfrac{(-2kx)^j}{j!}\
&=frac1{2^n}sum_{k=0}^n binom{n}{k}sum_{i=0}^{infty} dfrac{(nx)^i}{i!}sum_{j=0}^{infty} dfrac{(-2kx)^j}{j!}\
&=frac1{2^n}sum_{k=0}^n binom{n}{k}sum_{m=0}^{infty}sum_{i=0}^{m} dfrac{(nx)^i}{i!} dfrac{(-2kx)^{m-i}}{(m-i)!}\
&=frac1{2^n}sum_{k=0}^n binom{n}{k}sum_{m=0}^{infty}x^msum_{i=0}^{m} dfrac{(n)^i}{i!} dfrac{(-2k)^{m-i}}{(m-i)!}\
&=frac1{2^n}sum_{m=0}^{infty}x^msum_{k=0}^n binom{n}{k}sum_{i=0}^{m} dfrac{(n)^i}{i!} dfrac{(-2k)^{m-i}}{(m-i)!}\
&=frac1{2^n}sum_{m=0}^{infty}x^msum_{i=0}^{m}dfrac{(-2)^{m-i}(n)^i}{(m-i)!i!}sum_{k=0}^n binom{n}{k} k^{m-i}\
&=frac1{2^n}sum_{m=0}^{infty}dfrac{x^m}{m!}sum_{i=0}^{m}binom{m}{i}(-2)^{m-i}n^isum_{k=0}^n binom{n}{k} k^{m-i}\
&=frac1{2^n}sum_{m=0}^{infty}dfrac{(-2)^mx^m}{m!}sum_{i=0}^{m}binom{m}{i}(-n/2)^isum_{k=0}^n binom{n}{k} k^{m-i}\
&=frac1{2^n}sum_{m=0}^{infty}dfrac{(-2)^mx^m}{m!}sum_{k=0}^n binom{n}{k}sum_{i=0}^{m}binom{m}{i}(-n/2)^i k^{m-i}\
&=frac1{2^n}sum_{m=0}^{infty}dfrac{(-2)^mx^m}{m!}sum_{k=0}^n binom{n}{k}(-n/2+k)^m\
&=frac1{2^n}sum_{m=0}^{infty}dfrac{x^m}{m!}sum_{k=0}^n binom{n}{k}(2k-n)^m\
end{array}
$
And this is the OP.
Oh well.
add a comment |
I'll play with the cosh and
see if I get
anything other than
the original problem.
$begin{array}\
cosh^n(x)
&=frac1{2^n}(e^x+e^{-x})^n\
&=frac1{2^n}sum_{k=0}^n binom{n}{k}e^ke^{(n-k)x}\
&=frac1{2^n}e^{nx}sum_{k=0}^n binom{n}{k}e^{-2kx}\
&=frac1{2^n}sum_{i=0}^{infty} dfrac{(nx)^i}{i!}sum_{k=0}^n binom{n}{k}sum_{j=0}^{infty} dfrac{(-2kx)^j}{j!}\
&=frac1{2^n}sum_{k=0}^n binom{n}{k}sum_{i=0}^{infty} dfrac{(nx)^i}{i!}sum_{j=0}^{infty} dfrac{(-2kx)^j}{j!}\
&=frac1{2^n}sum_{k=0}^n binom{n}{k}sum_{m=0}^{infty}sum_{i=0}^{m} dfrac{(nx)^i}{i!} dfrac{(-2kx)^{m-i}}{(m-i)!}\
&=frac1{2^n}sum_{k=0}^n binom{n}{k}sum_{m=0}^{infty}x^msum_{i=0}^{m} dfrac{(n)^i}{i!} dfrac{(-2k)^{m-i}}{(m-i)!}\
&=frac1{2^n}sum_{m=0}^{infty}x^msum_{k=0}^n binom{n}{k}sum_{i=0}^{m} dfrac{(n)^i}{i!} dfrac{(-2k)^{m-i}}{(m-i)!}\
&=frac1{2^n}sum_{m=0}^{infty}x^msum_{i=0}^{m}dfrac{(-2)^{m-i}(n)^i}{(m-i)!i!}sum_{k=0}^n binom{n}{k} k^{m-i}\
&=frac1{2^n}sum_{m=0}^{infty}dfrac{x^m}{m!}sum_{i=0}^{m}binom{m}{i}(-2)^{m-i}n^isum_{k=0}^n binom{n}{k} k^{m-i}\
&=frac1{2^n}sum_{m=0}^{infty}dfrac{(-2)^mx^m}{m!}sum_{i=0}^{m}binom{m}{i}(-n/2)^isum_{k=0}^n binom{n}{k} k^{m-i}\
&=frac1{2^n}sum_{m=0}^{infty}dfrac{(-2)^mx^m}{m!}sum_{k=0}^n binom{n}{k}sum_{i=0}^{m}binom{m}{i}(-n/2)^i k^{m-i}\
&=frac1{2^n}sum_{m=0}^{infty}dfrac{(-2)^mx^m}{m!}sum_{k=0}^n binom{n}{k}(-n/2+k)^m\
&=frac1{2^n}sum_{m=0}^{infty}dfrac{x^m}{m!}sum_{k=0}^n binom{n}{k}(2k-n)^m\
end{array}
$
And this is the OP.
Oh well.
add a comment |
I'll play with the cosh and
see if I get
anything other than
the original problem.
$begin{array}\
cosh^n(x)
&=frac1{2^n}(e^x+e^{-x})^n\
&=frac1{2^n}sum_{k=0}^n binom{n}{k}e^ke^{(n-k)x}\
&=frac1{2^n}e^{nx}sum_{k=0}^n binom{n}{k}e^{-2kx}\
&=frac1{2^n}sum_{i=0}^{infty} dfrac{(nx)^i}{i!}sum_{k=0}^n binom{n}{k}sum_{j=0}^{infty} dfrac{(-2kx)^j}{j!}\
&=frac1{2^n}sum_{k=0}^n binom{n}{k}sum_{i=0}^{infty} dfrac{(nx)^i}{i!}sum_{j=0}^{infty} dfrac{(-2kx)^j}{j!}\
&=frac1{2^n}sum_{k=0}^n binom{n}{k}sum_{m=0}^{infty}sum_{i=0}^{m} dfrac{(nx)^i}{i!} dfrac{(-2kx)^{m-i}}{(m-i)!}\
&=frac1{2^n}sum_{k=0}^n binom{n}{k}sum_{m=0}^{infty}x^msum_{i=0}^{m} dfrac{(n)^i}{i!} dfrac{(-2k)^{m-i}}{(m-i)!}\
&=frac1{2^n}sum_{m=0}^{infty}x^msum_{k=0}^n binom{n}{k}sum_{i=0}^{m} dfrac{(n)^i}{i!} dfrac{(-2k)^{m-i}}{(m-i)!}\
&=frac1{2^n}sum_{m=0}^{infty}x^msum_{i=0}^{m}dfrac{(-2)^{m-i}(n)^i}{(m-i)!i!}sum_{k=0}^n binom{n}{k} k^{m-i}\
&=frac1{2^n}sum_{m=0}^{infty}dfrac{x^m}{m!}sum_{i=0}^{m}binom{m}{i}(-2)^{m-i}n^isum_{k=0}^n binom{n}{k} k^{m-i}\
&=frac1{2^n}sum_{m=0}^{infty}dfrac{(-2)^mx^m}{m!}sum_{i=0}^{m}binom{m}{i}(-n/2)^isum_{k=0}^n binom{n}{k} k^{m-i}\
&=frac1{2^n}sum_{m=0}^{infty}dfrac{(-2)^mx^m}{m!}sum_{k=0}^n binom{n}{k}sum_{i=0}^{m}binom{m}{i}(-n/2)^i k^{m-i}\
&=frac1{2^n}sum_{m=0}^{infty}dfrac{(-2)^mx^m}{m!}sum_{k=0}^n binom{n}{k}(-n/2+k)^m\
&=frac1{2^n}sum_{m=0}^{infty}dfrac{x^m}{m!}sum_{k=0}^n binom{n}{k}(2k-n)^m\
end{array}
$
And this is the OP.
Oh well.
I'll play with the cosh and
see if I get
anything other than
the original problem.
$begin{array}\
cosh^n(x)
&=frac1{2^n}(e^x+e^{-x})^n\
&=frac1{2^n}sum_{k=0}^n binom{n}{k}e^ke^{(n-k)x}\
&=frac1{2^n}e^{nx}sum_{k=0}^n binom{n}{k}e^{-2kx}\
&=frac1{2^n}sum_{i=0}^{infty} dfrac{(nx)^i}{i!}sum_{k=0}^n binom{n}{k}sum_{j=0}^{infty} dfrac{(-2kx)^j}{j!}\
&=frac1{2^n}sum_{k=0}^n binom{n}{k}sum_{i=0}^{infty} dfrac{(nx)^i}{i!}sum_{j=0}^{infty} dfrac{(-2kx)^j}{j!}\
&=frac1{2^n}sum_{k=0}^n binom{n}{k}sum_{m=0}^{infty}sum_{i=0}^{m} dfrac{(nx)^i}{i!} dfrac{(-2kx)^{m-i}}{(m-i)!}\
&=frac1{2^n}sum_{k=0}^n binom{n}{k}sum_{m=0}^{infty}x^msum_{i=0}^{m} dfrac{(n)^i}{i!} dfrac{(-2k)^{m-i}}{(m-i)!}\
&=frac1{2^n}sum_{m=0}^{infty}x^msum_{k=0}^n binom{n}{k}sum_{i=0}^{m} dfrac{(n)^i}{i!} dfrac{(-2k)^{m-i}}{(m-i)!}\
&=frac1{2^n}sum_{m=0}^{infty}x^msum_{i=0}^{m}dfrac{(-2)^{m-i}(n)^i}{(m-i)!i!}sum_{k=0}^n binom{n}{k} k^{m-i}\
&=frac1{2^n}sum_{m=0}^{infty}dfrac{x^m}{m!}sum_{i=0}^{m}binom{m}{i}(-2)^{m-i}n^isum_{k=0}^n binom{n}{k} k^{m-i}\
&=frac1{2^n}sum_{m=0}^{infty}dfrac{(-2)^mx^m}{m!}sum_{i=0}^{m}binom{m}{i}(-n/2)^isum_{k=0}^n binom{n}{k} k^{m-i}\
&=frac1{2^n}sum_{m=0}^{infty}dfrac{(-2)^mx^m}{m!}sum_{k=0}^n binom{n}{k}sum_{i=0}^{m}binom{m}{i}(-n/2)^i k^{m-i}\
&=frac1{2^n}sum_{m=0}^{infty}dfrac{(-2)^mx^m}{m!}sum_{k=0}^n binom{n}{k}(-n/2+k)^m\
&=frac1{2^n}sum_{m=0}^{infty}dfrac{x^m}{m!}sum_{k=0}^n binom{n}{k}(2k-n)^m\
end{array}
$
And this is the OP.
Oh well.
answered Dec 9 at 21:48
marty cohen
72.5k549127
72.5k549127
add a comment |
add a comment |
Let $omega $ be a $n$-dimensional vector with binary components $omega_i=pm1$ and $Omega_n $ be a set of all such vectors, the size of the set obviously being $2^n $. The sum of elements of a vector with $i$
positive and $n-i $ negative components is $2i-n $ and the number of such vectors is $binom {n}{i}$. Thus
$$
F (n,k)=frac {1}{2^n}sum_{omegainOmega_n } left (sum_{i=1}^nomega_i right)^{2k}
=frac {1}{2^n}sum_{omegainOmega_n } sum_{p_ige0}^{sum_i p_i=2k}binom {2k} {p_1,p_2,dots,p_n}prod_i omega_i^{p_i}
=sum_{p_ige0}^{sum_i p_i=2k}binom {2k} {p_1,p_2,dots,p_n}left (frac {1}{2^n}sum_{omegainOmega_n }prod_i omega_i^{p_i}right)
=sum_{p_ige0,;p_i,text {mod},2=0}^{sum_i p_i=2k}binom {2k} {p_1,p_2,dots,p_n}.
$$
To proceed further one splits the last sum into partial ones over terms with particular count $l$ of non-zero $p_i$ and ends up with:
$$
F (n,k)=sum_{l=1}^n T (k,l)frac{n!}{(n-l)!},tag {1}
$$
where $T (k,l)$ is the number of partitions of a set of size $2k$ into $l$ blocks of even size. It is the OEIS sequence A156289 with known close-form and recurrence expressions.
By numerical evidence the polynomial (1) can be expressed in the terms of usual powers as:
$$
F (n,k)=sum_{l=1}^n A (k,l)n^l,tag {2}
$$
with $A (k,l) $ being the OEIS sequence A318146. In other words $F(n,k) $ is in fact the so called Omega polynomial.
add a comment |
Let $omega $ be a $n$-dimensional vector with binary components $omega_i=pm1$ and $Omega_n $ be a set of all such vectors, the size of the set obviously being $2^n $. The sum of elements of a vector with $i$
positive and $n-i $ negative components is $2i-n $ and the number of such vectors is $binom {n}{i}$. Thus
$$
F (n,k)=frac {1}{2^n}sum_{omegainOmega_n } left (sum_{i=1}^nomega_i right)^{2k}
=frac {1}{2^n}sum_{omegainOmega_n } sum_{p_ige0}^{sum_i p_i=2k}binom {2k} {p_1,p_2,dots,p_n}prod_i omega_i^{p_i}
=sum_{p_ige0}^{sum_i p_i=2k}binom {2k} {p_1,p_2,dots,p_n}left (frac {1}{2^n}sum_{omegainOmega_n }prod_i omega_i^{p_i}right)
=sum_{p_ige0,;p_i,text {mod},2=0}^{sum_i p_i=2k}binom {2k} {p_1,p_2,dots,p_n}.
$$
To proceed further one splits the last sum into partial ones over terms with particular count $l$ of non-zero $p_i$ and ends up with:
$$
F (n,k)=sum_{l=1}^n T (k,l)frac{n!}{(n-l)!},tag {1}
$$
where $T (k,l)$ is the number of partitions of a set of size $2k$ into $l$ blocks of even size. It is the OEIS sequence A156289 with known close-form and recurrence expressions.
By numerical evidence the polynomial (1) can be expressed in the terms of usual powers as:
$$
F (n,k)=sum_{l=1}^n A (k,l)n^l,tag {2}
$$
with $A (k,l) $ being the OEIS sequence A318146. In other words $F(n,k) $ is in fact the so called Omega polynomial.
add a comment |
Let $omega $ be a $n$-dimensional vector with binary components $omega_i=pm1$ and $Omega_n $ be a set of all such vectors, the size of the set obviously being $2^n $. The sum of elements of a vector with $i$
positive and $n-i $ negative components is $2i-n $ and the number of such vectors is $binom {n}{i}$. Thus
$$
F (n,k)=frac {1}{2^n}sum_{omegainOmega_n } left (sum_{i=1}^nomega_i right)^{2k}
=frac {1}{2^n}sum_{omegainOmega_n } sum_{p_ige0}^{sum_i p_i=2k}binom {2k} {p_1,p_2,dots,p_n}prod_i omega_i^{p_i}
=sum_{p_ige0}^{sum_i p_i=2k}binom {2k} {p_1,p_2,dots,p_n}left (frac {1}{2^n}sum_{omegainOmega_n }prod_i omega_i^{p_i}right)
=sum_{p_ige0,;p_i,text {mod},2=0}^{sum_i p_i=2k}binom {2k} {p_1,p_2,dots,p_n}.
$$
To proceed further one splits the last sum into partial ones over terms with particular count $l$ of non-zero $p_i$ and ends up with:
$$
F (n,k)=sum_{l=1}^n T (k,l)frac{n!}{(n-l)!},tag {1}
$$
where $T (k,l)$ is the number of partitions of a set of size $2k$ into $l$ blocks of even size. It is the OEIS sequence A156289 with known close-form and recurrence expressions.
By numerical evidence the polynomial (1) can be expressed in the terms of usual powers as:
$$
F (n,k)=sum_{l=1}^n A (k,l)n^l,tag {2}
$$
with $A (k,l) $ being the OEIS sequence A318146. In other words $F(n,k) $ is in fact the so called Omega polynomial.
Let $omega $ be a $n$-dimensional vector with binary components $omega_i=pm1$ and $Omega_n $ be a set of all such vectors, the size of the set obviously being $2^n $. The sum of elements of a vector with $i$
positive and $n-i $ negative components is $2i-n $ and the number of such vectors is $binom {n}{i}$. Thus
$$
F (n,k)=frac {1}{2^n}sum_{omegainOmega_n } left (sum_{i=1}^nomega_i right)^{2k}
=frac {1}{2^n}sum_{omegainOmega_n } sum_{p_ige0}^{sum_i p_i=2k}binom {2k} {p_1,p_2,dots,p_n}prod_i omega_i^{p_i}
=sum_{p_ige0}^{sum_i p_i=2k}binom {2k} {p_1,p_2,dots,p_n}left (frac {1}{2^n}sum_{omegainOmega_n }prod_i omega_i^{p_i}right)
=sum_{p_ige0,;p_i,text {mod},2=0}^{sum_i p_i=2k}binom {2k} {p_1,p_2,dots,p_n}.
$$
To proceed further one splits the last sum into partial ones over terms with particular count $l$ of non-zero $p_i$ and ends up with:
$$
F (n,k)=sum_{l=1}^n T (k,l)frac{n!}{(n-l)!},tag {1}
$$
where $T (k,l)$ is the number of partitions of a set of size $2k$ into $l$ blocks of even size. It is the OEIS sequence A156289 with known close-form and recurrence expressions.
By numerical evidence the polynomial (1) can be expressed in the terms of usual powers as:
$$
F (n,k)=sum_{l=1}^n A (k,l)n^l,tag {2}
$$
with $A (k,l) $ being the OEIS sequence A318146. In other words $F(n,k) $ is in fact the so called Omega polynomial.
edited Dec 27 at 11:41
answered Dec 26 at 19:53
user
3,6361627
3,6361627
add a comment |
add a comment |
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That is a closed form. The sum is finite. I'd try expanding $(2i-n)^{2k}$ or computing some values and plugging them in into OEIS, in any case.
– ajotatxe
Dec 9 at 18:10
@ajotatxe I have edited the question to clarify what I meant by "closed - form expression".
– user
Dec 9 at 19:30
1
This expression comes out in Exercise 7 of UMN Fall 2018 Math 5705 Homework set 4, and also in Corollary 2.5 of Stanley's Algebraic Combinatorics (but these two appearances are actually easily seen to be equivalent: a closed walk of length $n$ on the hypercube is uniquely determined by its starting point and the $n$-tuple of "signless step directions", which $n$-tuple is clearly all-even).
– darij grinberg
Dec 11 at 3:16
1
If there is a closed form, then Stanley could not find it. On the other hand, the polynomial claim is interesting.
– darij grinberg
Dec 11 at 3:18
1
Ah, I see why it is a polynomial. You can count the all-even $k$-tuples in $left[nright]^k$ according to the positions of equal entries (more formally: the set partition of $left[kright]$ that governs which of the entries of the tuple are equal). For any given such choice of positions, the number of tuples is a polynomial in $n$ with degree $k$ (namely, a power of $n$ times a power of $n-1$ times a power of $n-2$ and so on). Feel free to expand on this in an answer -- I am stuck in bed with a flu and not at my most productive.
– darij grinberg
Dec 11 at 3:45