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How to expand product of $n$ factors.
up vote
3
down vote
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I have a product say
begin{equation}
F(a,n,x) = prod _{j=0}^{n}(1-{a}^{n-2,j}x)
end{equation}
I want to expand and hope to have general terms of the coefficients. I did for $n= 2,3,4,5,6,7,8...$ I
see it will be different for $n$ even or $n$ odd. We have
begin{equation*}
F(a,2,x)= 1-{x}^{3}+ left( {a}^{2}+1+{a}^{-2} right) {x}^{2}+ left( -{a}^{2}-
1-{a}^{-2} right) x
end{equation*}
begin{equation}
F(a,3,x)= 1+{x}^{4}+ left( -{a}^{3}-a-{a}^{-3}-{a}^{-1} right) {x}^{3}+
left( {a}^{4}+2+{a}^{-2}+{a}^{2}+{a}^{-4} right) {x}^{2}+ left( -{
a}^{3}-a-{a}^{-3}-{a}^{-1} right) x
end{equation}
The coefficients of $a$ gets more interesting as $n$ grows. I am interested in the coefficient of $a.$ Does anyone know how to expand product of $n$ factors.
binomial-coefficients products multinomial-coefficients
asked Nov 11 at 23:31
Learner
409310
add a comment |
up vote
3
down vote
favorite
I have a product say
begin{equation}
F(a,n,x) = prod _{j=0}^{n}(1-{a}^{n-2,j}x)
end{equation}
I want to expand and hope to have general terms of the coefficients. I did for $n= 2,3,4,5,6,7,8...$ I
see it will be different for $n$ even or $n$ odd. We have
begin{equation*}
F(a,2,x)= 1-{x}^{3}+ left( {a}^{2}+1+{a}^{-2} right) {x}^{2}+ left( -{a}^{2}-
1-{a}^{-2} right) x
end{equation*}
begin{equation}
F(a,3,x)= 1+{x}^{4}+ left( -{a}^{3}-a-{a}^{-3}-{a}^{-1} right) {x}^{3}+
left( {a}^{4}+2+{a}^{-2}+{a}^{2}+{a}^{-4} right) {x}^{2}+ left( -{
a}^{3}-a-{a}^{-3}-{a}^{-1} right) x
end{equation}
The coefficients of $a$ gets more interesting as $n$ grows. I am interested in the coefficient of $a.$ Does anyone know how to expand product of $n$ factors.
binomial-coefficients products multinomial-coefficients
asked Nov 11 at 23:31
Learner
409310
By "coefficients of $a$" do you mean "coefficients of $x^k$ in terms of $a$"?
– YiFan
Nov 12 at 1:06
@ YiFan yes you can say that.
– Learner
Nov 12 at 1:23
There should be a development of this product involving the $a^2$-binomial coefficients. See here.
– René Gy
Nov 12 at 18:38
@Thanks René Gy
– Learner
Nov 13 at 2:52
Apply the $q$-binomial formula $prodlimits_{k=0}^{n-1} left(1+q^k tright) = sumlimits_{k=0}^n q^{kleft(k-1right)/2} dbinom{n}{k}_q t^k$ to $n+1$, $a^n x$ and $a^{-2}$ instead of $n$, $t$ and $q$.
– darij grinberg
Nov 14 at 0:44
add a comment |
up vote
3
down vote
favorite
up vote
3
down vote
favorite
I have a product say
begin{equation}
F(a,n,x) = prod _{j=0}^{n}(1-{a}^{n-2,j}x)
end{equation}
I want to expand and hope to have general terms of the coefficients. I did for $n= 2,3,4,5,6,7,8...$ I
see it will be different for $n$ even or $n$ odd. We have
begin{equation*}
F(a,2,x)= 1-{x}^{3}+ left( {a}^{2}+1+{a}^{-2} right) {x}^{2}+ left( -{a}^{2}-
1-{a}^{-2} right) x
end{equation*}
begin{equation}
F(a,3,x)= 1+{x}^{4}+ left( -{a}^{3}-a-{a}^{-3}-{a}^{-1} right) {x}^{3}+
left( {a}^{4}+2+{a}^{-2}+{a}^{2}+{a}^{-4} right) {x}^{2}+ left( -{
a}^{3}-a-{a}^{-3}-{a}^{-1} right) x
end{equation}
The coefficients of $a$ gets more interesting as $n$ grows. I am interested in the coefficient of $a.$ Does anyone know how to expand product of $n$ factors.
binomial-coefficients products multinomial-coefficients
asked Nov 11 at 23:31
Learner
409310
I have a product say
begin{equation}
F(a,n,x) = prod _{j=0}^{n}(1-{a}^{n-2,j}x)
end{equation}
I want to expand and hope to have general terms of the coefficients. I did for $n= 2,3,4,5,6,7,8...$ I
see it will be different for $n$ even or $n$ odd. We have
begin{equation*}
F(a,2,x)= 1-{x}^{3}+ left( {a}^{2}+1+{a}^{-2} right) {x}^{2}+ left( -{a}^{2}-
1-{a}^{-2} right) x
end{equation*}
begin{equation}
F(a,3,x)= 1+{x}^{4}+ left( -{a}^{3}-a-{a}^{-3}-{a}^{-1} right) {x}^{3}+
left( {a}^{4}+2+{a}^{-2}+{a}^{2}+{a}^{-4} right) {x}^{2}+ left( -{
a}^{3}-a-{a}^{-3}-{a}^{-1} right) x
end{equation}
The coefficients of $a$ gets more interesting as $n$ grows. I am interested in the coefficient of $a.$ Does anyone know how to expand product of $n$ factors.
binomial-coefficients products multinomial-coefficients
binomial-coefficients products multinomial-coefficients
asked Nov 11 at 23:31
Learner
409310
asked Nov 11 at 23:31
Learner
409310
edited Nov 12 at 16:34
asked Nov 11 at 23:31
Learner
409310
asked Nov 11 at 23:31
Learner
409310
asked Nov 11 at 23:31
Learner
409310
409310
By "coefficients of $a$" do you mean "coefficients of $x^k$ in terms of $a$"?
– YiFan
Nov 12 at 1:06
@ YiFan yes you can say that.
– Learner
Nov 12 at 1:23
There should be a development of this product involving the $a^2$-binomial coefficients. See here.
– René Gy
Nov 12 at 18:38
@Thanks René Gy
– Learner
Nov 13 at 2:52
Apply the $q$-binomial formula $prodlimits_{k=0}^{n-1} left(1+q^k tright) = sumlimits_{k=0}^n q^{kleft(k-1right)/2} dbinom{n}{k}_q t^k$ to $n+1$, $a^n x$ and $a^{-2}$ instead of $n$, $t$ and $q$.
– darij grinberg
Nov 14 at 0:44
add a comment |
By "coefficients of $a$" do you mean "coefficients of $x^k$ in terms of $a$"?
– YiFan
Nov 12 at 1:06
@ YiFan yes you can say that.
– Learner
Nov 12 at 1:23
There should be a development of this product involving the $a^2$-binomial coefficients. See here.
– René Gy
Nov 12 at 18:38
@Thanks René Gy
– Learner
Nov 13 at 2:52
Apply the $q$-binomial formula $prodlimits_{k=0}^{n-1} left(1+q^k tright) = sumlimits_{k=0}^n q^{kleft(k-1right)/2} dbinom{n}{k}_q t^k$ to $n+1$, $a^n x$ and $a^{-2}$ instead of $n$, $t$ and $q$.
– darij grinberg
Nov 14 at 0:44
By "coefficients of $a$" do you mean "coefficients of $x^k$ in terms of $a$"?
– YiFan
Nov 12 at 1:06
By "coefficients of $a$" do you mean "coefficients of $x^k$ in terms of $a$"?
– YiFan
Nov 12 at 1:06
@ YiFan yes you can say that.
– Learner
Nov 12 at 1:23
@ YiFan yes you can say that.
– Learner
Nov 12 at 1:23
There should be a development of this product involving the $a^2$-binomial coefficients. See here.
– René Gy
Nov 12 at 18:38
There should be a development of this product involving the $a^2$-binomial coefficients. See here.
– René Gy
Nov 12 at 18:38
@Thanks René Gy
– Learner
Nov 13 at 2:52
@Thanks René Gy
– Learner
Nov 13 at 2:52
Apply the $q$-binomial formula $prodlimits_{k=0}^{n-1} left(1+q^k tright) = sumlimits_{k=0}^n q^{kleft(k-1right)/2} dbinom{n}{k}_q t^k$ to $n+1$, $a^n x$ and $a^{-2}$ instead of $n$, $t$ and $q$.
– darij grinberg
Nov 14 at 0:44
Apply the $q$-binomial formula $prodlimits_{k=0}^{n-1} left(1+q^k tright) = sumlimits_{k=0}^n q^{kleft(k-1right)/2} dbinom{n}{k}_q t^k$ to $n+1$, $a^n x$ and $a^{-2}$ instead of $n$, $t$ and $q$.
– darij grinberg
Nov 14 at 0:44
add a comment |
1 Answer
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When expanding the product $F(a,n,x)$ in terms of $x$ the coefficients of $x^k$ are polynomials $P_{n,k}(a)$ in $a$. Here we expand $F(a,n,x)$ in order to see the coefficients of $a$ in $P_{n,k}(a)$ explicitly.
We obtain
begin{align*}
F(a,n,x)&=prod_{j=0}^nleft(1-a^{n-2j}xright)tag{1}\
&=sum_{Ssubseteq {0,1,ldots,n}}(-x)^{|S|}a^{n|S|}prod_{jin S}a^{-2j}tag{2}\
&=1+sum_{k=1}^{n+1}sum_{{Ssubseteq {0,1,ldots,n}}atop{|S|=k}}(-x)^{|S|}a^{n|S|}prod_{jin S}a^{-2j}tag{3}\
&=1+sum_{k=1}^{n+1}(-x)^ka^{nk}sum_{0leq j_1<cdots<j_kleq n}a^{-2(j_1+cdots+j_k)}tag{4}\
&=1+sum_{k=1}^{n+1}(-x)^ka^{nk}sum_{l=k(k-1)/2}^{k(2n-k+1)/2}sum_{{0leq j_1<cdots<j_kleq n}atop{j_1+cdots +j_k=l}}a^{-2(j_1+cdots +j_k)}tag{5}\
&=1+sum_{k=1}^{n+1}(-1)^ksum_{l=k(k-1)/2}^{k(2n-k+1)/2}left(sum_{{0leq j_1<cdots<j_kleq n}atop{j_1+cdots +j_k=l}}1right)a^{nk-2l}x^ktag{6}\
&=1+sum_{k=1}^{n+1}color{blue}{(-1)^k}sum_{l=0}^{k(n-k+1)}
left(color{blue}{sum_{{0leq j_1<cdots<j_kleq n}atop{j_1+cdots +j_k=l+k(k-1)/2}}1}right)a^{k(n-k+1)-2l}x^ktag{7}\
end{align*}
The coefficients of $a$ in $P_{n,k}(a)$ are given in (7) as the blue marked sum times $(-1)^k$.
Comment:
In (2) we note the product (1) consists of $n+1$ factors and from each factor we choose either $1$ or $-a^{n-2j}x$. We represent each choice as subset $Ssubseteq {0,1,ldots,n}$.
In (3) We reorder the summands according to the size $k$ of $S$. We also extract the term $1$ which represents the case $S=emptyset$. In this case we have chosen always $1$ from each of the $n+1$ factors.
In (4) we can factor out $-x$ and $a^n$ and thanks to $k$ we can explicitly write the elements of $S={j_1,j_2,ldots,j_k}$ for each specific choice.
In (5) we do again a reordering by organizing the summands according to the sum $j_1+j_2+cdots+j_k$ of the $k$-tupels.
We observe the smallest sum comes from the $k$-tupel $(0,1,2,ldots,k-1)$ which gives $$sum_{j=0}^k j=k(k-1)/2$$
while the $k$-tupel with the largest sum is $(n-k+1,n-k+2,ldots,n)$ which gives begin{align*}sum_{j=n-k+1}^n j&=sum_{j=1}^n j-sum_{j=1}^{n-k} j=frac{n(n+1)}{2}-frac{(n-k)(n-k+1)}{2}\
&=frac{k(2n-k+1)}{2}.
end{align*}
In (6) we factor out $a^{-2(j_1+cdots+j_k)}=a^{-2l}$.
In (7) we shift the index of $l$ to start from $0$.
Example $F(a,3,x)$:
We evaluate the expression (7) for the case $n=3$. We obtain
begin{align*}
color{blue}{F(a,3,x)}&=
1+sum_{k=1}^4(-1)^ksum_{l=0}^{k(4-k)}left(sum_{{0leq j_1leq cdotsleq j_kleq 3}atop{j_1+cdots j_k=l+k(k-1)/2}} 1right)a^{k(4-k)-2l}x^k\
&=1-sum_{l=0}^3left(sum_{{0leq j_1leq 3}atop{j_1=l}}1right)a^{3-2l}x
+sum_{l=0}^4left(sum_{{0leq j_1<j_2leq 3}atop{j_1+j_2=l+1}} 1right)a^{4-2l}x^2\
&qquad -sum_{l=0}^3left(sum_{{0leq j_1<j_2<j_3leq 3}atop{j_1+j_2+j_3=l+3}} 1right)a^{3-2l}x^3
-sum_{l=0}^0left(sum_{{0leq j_1<j_2<j_3<j_4leq 3}atop{j_1+j_2+j_3j_4=l+6}} 1right)a^{-2l}x^4\
&,,color{blue}{=1-left(a^3+a+a^{-1}+a^{-3}right)x+left(a^4+a^2+2+a^{-2}+a^{-4}right)x^2}\
&qquad,,color{blue}{-left(a^3+a+a^{-1}+a^{-3}right)x^3+x^4}
end{align*}
in accordance with OPs calculation.
answered 19 hours ago
Markus Scheuer
58.9k454140
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1 Answer
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1 Answer
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active
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up vote
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When expanding the product $F(a,n,x)$ in terms of $x$ the coefficients of $x^k$ are polynomials $P_{n,k}(a)$ in $a$. Here we expand $F(a,n,x)$ in order to see the coefficients of $a$ in $P_{n,k}(a)$ explicitly.
We obtain
begin{align*}
F(a,n,x)&=prod_{j=0}^nleft(1-a^{n-2j}xright)tag{1}\
&=sum_{Ssubseteq {0,1,ldots,n}}(-x)^{|S|}a^{n|S|}prod_{jin S}a^{-2j}tag{2}\
&=1+sum_{k=1}^{n+1}sum_{{Ssubseteq {0,1,ldots,n}}atop{|S|=k}}(-x)^{|S|}a^{n|S|}prod_{jin S}a^{-2j}tag{3}\
&=1+sum_{k=1}^{n+1}(-x)^ka^{nk}sum_{0leq j_1<cdots<j_kleq n}a^{-2(j_1+cdots+j_k)}tag{4}\
&=1+sum_{k=1}^{n+1}(-x)^ka^{nk}sum_{l=k(k-1)/2}^{k(2n-k+1)/2}sum_{{0leq j_1<cdots<j_kleq n}atop{j_1+cdots +j_k=l}}a^{-2(j_1+cdots +j_k)}tag{5}\
&=1+sum_{k=1}^{n+1}(-1)^ksum_{l=k(k-1)/2}^{k(2n-k+1)/2}left(sum_{{0leq j_1<cdots<j_kleq n}atop{j_1+cdots +j_k=l}}1right)a^{nk-2l}x^ktag{6}\
&=1+sum_{k=1}^{n+1}color{blue}{(-1)^k}sum_{l=0}^{k(n-k+1)}
left(color{blue}{sum_{{0leq j_1<cdots<j_kleq n}atop{j_1+cdots +j_k=l+k(k-1)/2}}1}right)a^{k(n-k+1)-2l}x^ktag{7}\
end{align*}
The coefficients of $a$ in $P_{n,k}(a)$ are given in (7) as the blue marked sum times $(-1)^k$.
Comment:
In (2) we note the product (1) consists of $n+1$ factors and from each factor we choose either $1$ or $-a^{n-2j}x$. We represent each choice as subset $Ssubseteq {0,1,ldots,n}$.
In (3) We reorder the summands according to the size $k$ of $S$. We also extract the term $1$ which represents the case $S=emptyset$. In this case we have chosen always $1$ from each of the $n+1$ factors.
In (4) we can factor out $-x$ and $a^n$ and thanks to $k$ we can explicitly write the elements of $S={j_1,j_2,ldots,j_k}$ for each specific choice.
In (5) we do again a reordering by organizing the summands according to the sum $j_1+j_2+cdots+j_k$ of the $k$-tupels.
We observe the smallest sum comes from the $k$-tupel $(0,1,2,ldots,k-1)$ which gives $$sum_{j=0}^k j=k(k-1)/2$$
while the $k$-tupel with the largest sum is $(n-k+1,n-k+2,ldots,n)$ which gives begin{align*}sum_{j=n-k+1}^n j&=sum_{j=1}^n j-sum_{j=1}^{n-k} j=frac{n(n+1)}{2}-frac{(n-k)(n-k+1)}{2}\
&=frac{k(2n-k+1)}{2}.
end{align*}
In (6) we factor out $a^{-2(j_1+cdots+j_k)}=a^{-2l}$.
In (7) we shift the index of $l$ to start from $0$.
Example $F(a,3,x)$:
We evaluate the expression (7) for the case $n=3$. We obtain
begin{align*}
color{blue}{F(a,3,x)}&=
1+sum_{k=1}^4(-1)^ksum_{l=0}^{k(4-k)}left(sum_{{0leq j_1leq cdotsleq j_kleq 3}atop{j_1+cdots j_k=l+k(k-1)/2}} 1right)a^{k(4-k)-2l}x^k\
&=1-sum_{l=0}^3left(sum_{{0leq j_1leq 3}atop{j_1=l}}1right)a^{3-2l}x
+sum_{l=0}^4left(sum_{{0leq j_1<j_2leq 3}atop{j_1+j_2=l+1}} 1right)a^{4-2l}x^2\
&qquad -sum_{l=0}^3left(sum_{{0leq j_1<j_2<j_3leq 3}atop{j_1+j_2+j_3=l+3}} 1right)a^{3-2l}x^3
-sum_{l=0}^0left(sum_{{0leq j_1<j_2<j_3<j_4leq 3}atop{j_1+j_2+j_3j_4=l+6}} 1right)a^{-2l}x^4\
&,,color{blue}{=1-left(a^3+a+a^{-1}+a^{-3}right)x+left(a^4+a^2+2+a^{-2}+a^{-4}right)x^2}\
&qquad,,color{blue}{-left(a^3+a+a^{-1}+a^{-3}right)x^3+x^4}
end{align*}
in accordance with OPs calculation.
answered 19 hours ago
Markus Scheuer
58.9k454140
add a comment |
up vote
0
down vote
When expanding the product $F(a,n,x)$ in terms of $x$ the coefficients of $x^k$ are polynomials $P_{n,k}(a)$ in $a$. Here we expand $F(a,n,x)$ in order to see the coefficients of $a$ in $P_{n,k}(a)$ explicitly.
We obtain
begin{align*}
F(a,n,x)&=prod_{j=0}^nleft(1-a^{n-2j}xright)tag{1}\
&=sum_{Ssubseteq {0,1,ldots,n}}(-x)^{|S|}a^{n|S|}prod_{jin S}a^{-2j}tag{2}\
&=1+sum_{k=1}^{n+1}sum_{{Ssubseteq {0,1,ldots,n}}atop{|S|=k}}(-x)^{|S|}a^{n|S|}prod_{jin S}a^{-2j}tag{3}\
&=1+sum_{k=1}^{n+1}(-x)^ka^{nk}sum_{0leq j_1<cdots<j_kleq n}a^{-2(j_1+cdots+j_k)}tag{4}\
&=1+sum_{k=1}^{n+1}(-x)^ka^{nk}sum_{l=k(k-1)/2}^{k(2n-k+1)/2}sum_{{0leq j_1<cdots<j_kleq n}atop{j_1+cdots +j_k=l}}a^{-2(j_1+cdots +j_k)}tag{5}\
&=1+sum_{k=1}^{n+1}(-1)^ksum_{l=k(k-1)/2}^{k(2n-k+1)/2}left(sum_{{0leq j_1<cdots<j_kleq n}atop{j_1+cdots +j_k=l}}1right)a^{nk-2l}x^ktag{6}\
&=1+sum_{k=1}^{n+1}color{blue}{(-1)^k}sum_{l=0}^{k(n-k+1)}
left(color{blue}{sum_{{0leq j_1<cdots<j_kleq n}atop{j_1+cdots +j_k=l+k(k-1)/2}}1}right)a^{k(n-k+1)-2l}x^ktag{7}\
end{align*}
The coefficients of $a$ in $P_{n,k}(a)$ are given in (7) as the blue marked sum times $(-1)^k$.
Comment:
In (2) we note the product (1) consists of $n+1$ factors and from each factor we choose either $1$ or $-a^{n-2j}x$. We represent each choice as subset $Ssubseteq {0,1,ldots,n}$.
In (3) We reorder the summands according to the size $k$ of $S$. We also extract the term $1$ which represents the case $S=emptyset$. In this case we have chosen always $1$ from each of the $n+1$ factors.
In (4) we can factor out $-x$ and $a^n$ and thanks to $k$ we can explicitly write the elements of $S={j_1,j_2,ldots,j_k}$ for each specific choice.
In (5) we do again a reordering by organizing the summands according to the sum $j_1+j_2+cdots+j_k$ of the $k$-tupels.
We observe the smallest sum comes from the $k$-tupel $(0,1,2,ldots,k-1)$ which gives $$sum_{j=0}^k j=k(k-1)/2$$
while the $k$-tupel with the largest sum is $(n-k+1,n-k+2,ldots,n)$ which gives begin{align*}sum_{j=n-k+1}^n j&=sum_{j=1}^n j-sum_{j=1}^{n-k} j=frac{n(n+1)}{2}-frac{(n-k)(n-k+1)}{2}\
&=frac{k(2n-k+1)}{2}.
end{align*}
In (6) we factor out $a^{-2(j_1+cdots+j_k)}=a^{-2l}$.
In (7) we shift the index of $l$ to start from $0$.
Example $F(a,3,x)$:
We evaluate the expression (7) for the case $n=3$. We obtain
begin{align*}
color{blue}{F(a,3,x)}&=
1+sum_{k=1}^4(-1)^ksum_{l=0}^{k(4-k)}left(sum_{{0leq j_1leq cdotsleq j_kleq 3}atop{j_1+cdots j_k=l+k(k-1)/2}} 1right)a^{k(4-k)-2l}x^k\
&=1-sum_{l=0}^3left(sum_{{0leq j_1leq 3}atop{j_1=l}}1right)a^{3-2l}x
+sum_{l=0}^4left(sum_{{0leq j_1<j_2leq 3}atop{j_1+j_2=l+1}} 1right)a^{4-2l}x^2\
&qquad -sum_{l=0}^3left(sum_{{0leq j_1<j_2<j_3leq 3}atop{j_1+j_2+j_3=l+3}} 1right)a^{3-2l}x^3
-sum_{l=0}^0left(sum_{{0leq j_1<j_2<j_3<j_4leq 3}atop{j_1+j_2+j_3j_4=l+6}} 1right)a^{-2l}x^4\
&,,color{blue}{=1-left(a^3+a+a^{-1}+a^{-3}right)x+left(a^4+a^2+2+a^{-2}+a^{-4}right)x^2}\
&qquad,,color{blue}{-left(a^3+a+a^{-1}+a^{-3}right)x^3+x^4}
end{align*}
in accordance with OPs calculation.
answered 19 hours ago
Markus Scheuer
58.9k454140
add a comment |
up vote
0
down vote
up vote
0
down vote
When expanding the product $F(a,n,x)$ in terms of $x$ the coefficients of $x^k$ are polynomials $P_{n,k}(a)$ in $a$. Here we expand $F(a,n,x)$ in order to see the coefficients of $a$ in $P_{n,k}(a)$ explicitly.
We obtain
begin{align*}
F(a,n,x)&=prod_{j=0}^nleft(1-a^{n-2j}xright)tag{1}\
&=sum_{Ssubseteq {0,1,ldots,n}}(-x)^{|S|}a^{n|S|}prod_{jin S}a^{-2j}tag{2}\
&=1+sum_{k=1}^{n+1}sum_{{Ssubseteq {0,1,ldots,n}}atop{|S|=k}}(-x)^{|S|}a^{n|S|}prod_{jin S}a^{-2j}tag{3}\
&=1+sum_{k=1}^{n+1}(-x)^ka^{nk}sum_{0leq j_1<cdots<j_kleq n}a^{-2(j_1+cdots+j_k)}tag{4}\
&=1+sum_{k=1}^{n+1}(-x)^ka^{nk}sum_{l=k(k-1)/2}^{k(2n-k+1)/2}sum_{{0leq j_1<cdots<j_kleq n}atop{j_1+cdots +j_k=l}}a^{-2(j_1+cdots +j_k)}tag{5}\
&=1+sum_{k=1}^{n+1}(-1)^ksum_{l=k(k-1)/2}^{k(2n-k+1)/2}left(sum_{{0leq j_1<cdots<j_kleq n}atop{j_1+cdots +j_k=l}}1right)a^{nk-2l}x^ktag{6}\
&=1+sum_{k=1}^{n+1}color{blue}{(-1)^k}sum_{l=0}^{k(n-k+1)}
left(color{blue}{sum_{{0leq j_1<cdots<j_kleq n}atop{j_1+cdots +j_k=l+k(k-1)/2}}1}right)a^{k(n-k+1)-2l}x^ktag{7}\
end{align*}
The coefficients of $a$ in $P_{n,k}(a)$ are given in (7) as the blue marked sum times $(-1)^k$.
Comment:
In (2) we note the product (1) consists of $n+1$ factors and from each factor we choose either $1$ or $-a^{n-2j}x$. We represent each choice as subset $Ssubseteq {0,1,ldots,n}$.
In (3) We reorder the summands according to the size $k$ of $S$. We also extract the term $1$ which represents the case $S=emptyset$. In this case we have chosen always $1$ from each of the $n+1$ factors.
In (4) we can factor out $-x$ and $a^n$ and thanks to $k$ we can explicitly write the elements of $S={j_1,j_2,ldots,j_k}$ for each specific choice.
In (5) we do again a reordering by organizing the summands according to the sum $j_1+j_2+cdots+j_k$ of the $k$-tupels.
We observe the smallest sum comes from the $k$-tupel $(0,1,2,ldots,k-1)$ which gives $$sum_{j=0}^k j=k(k-1)/2$$
while the $k$-tupel with the largest sum is $(n-k+1,n-k+2,ldots,n)$ which gives begin{align*}sum_{j=n-k+1}^n j&=sum_{j=1}^n j-sum_{j=1}^{n-k} j=frac{n(n+1)}{2}-frac{(n-k)(n-k+1)}{2}\
&=frac{k(2n-k+1)}{2}.
end{align*}
In (6) we factor out $a^{-2(j_1+cdots+j_k)}=a^{-2l}$.
In (7) we shift the index of $l$ to start from $0$.
Example $F(a,3,x)$:
We evaluate the expression (7) for the case $n=3$. We obtain
begin{align*}
color{blue}{F(a,3,x)}&=
1+sum_{k=1}^4(-1)^ksum_{l=0}^{k(4-k)}left(sum_{{0leq j_1leq cdotsleq j_kleq 3}atop{j_1+cdots j_k=l+k(k-1)/2}} 1right)a^{k(4-k)-2l}x^k\
&=1-sum_{l=0}^3left(sum_{{0leq j_1leq 3}atop{j_1=l}}1right)a^{3-2l}x
+sum_{l=0}^4left(sum_{{0leq j_1<j_2leq 3}atop{j_1+j_2=l+1}} 1right)a^{4-2l}x^2\
&qquad -sum_{l=0}^3left(sum_{{0leq j_1<j_2<j_3leq 3}atop{j_1+j_2+j_3=l+3}} 1right)a^{3-2l}x^3
-sum_{l=0}^0left(sum_{{0leq j_1<j_2<j_3<j_4leq 3}atop{j_1+j_2+j_3j_4=l+6}} 1right)a^{-2l}x^4\
&,,color{blue}{=1-left(a^3+a+a^{-1}+a^{-3}right)x+left(a^4+a^2+2+a^{-2}+a^{-4}right)x^2}\
&qquad,,color{blue}{-left(a^3+a+a^{-1}+a^{-3}right)x^3+x^4}
end{align*}
in accordance with OPs calculation.
answered 19 hours ago
Markus Scheuer
58.9k454140
When expanding the product $F(a,n,x)$ in terms of $x$ the coefficients of $x^k$ are polynomials $P_{n,k}(a)$ in $a$. Here we expand $F(a,n,x)$ in order to see the coefficients of $a$ in $P_{n,k}(a)$ explicitly.
We obtain
begin{align*}
F(a,n,x)&=prod_{j=0}^nleft(1-a^{n-2j}xright)tag{1}\
&=sum_{Ssubseteq {0,1,ldots,n}}(-x)^{|S|}a^{n|S|}prod_{jin S}a^{-2j}tag{2}\
&=1+sum_{k=1}^{n+1}sum_{{Ssubseteq {0,1,ldots,n}}atop{|S|=k}}(-x)^{|S|}a^{n|S|}prod_{jin S}a^{-2j}tag{3}\
&=1+sum_{k=1}^{n+1}(-x)^ka^{nk}sum_{0leq j_1<cdots<j_kleq n}a^{-2(j_1+cdots+j_k)}tag{4}\
&=1+sum_{k=1}^{n+1}(-x)^ka^{nk}sum_{l=k(k-1)/2}^{k(2n-k+1)/2}sum_{{0leq j_1<cdots<j_kleq n}atop{j_1+cdots +j_k=l}}a^{-2(j_1+cdots +j_k)}tag{5}\
&=1+sum_{k=1}^{n+1}(-1)^ksum_{l=k(k-1)/2}^{k(2n-k+1)/2}left(sum_{{0leq j_1<cdots<j_kleq n}atop{j_1+cdots +j_k=l}}1right)a^{nk-2l}x^ktag{6}\
&=1+sum_{k=1}^{n+1}color{blue}{(-1)^k}sum_{l=0}^{k(n-k+1)}
left(color{blue}{sum_{{0leq j_1<cdots<j_kleq n}atop{j_1+cdots +j_k=l+k(k-1)/2}}1}right)a^{k(n-k+1)-2l}x^ktag{7}\
end{align*}
The coefficients of $a$ in $P_{n,k}(a)$ are given in (7) as the blue marked sum times $(-1)^k$.
Comment:
In (2) we note the product (1) consists of $n+1$ factors and from each factor we choose either $1$ or $-a^{n-2j}x$. We represent each choice as subset $Ssubseteq {0,1,ldots,n}$.
In (3) We reorder the summands according to the size $k$ of $S$. We also extract the term $1$ which represents the case $S=emptyset$. In this case we have chosen always $1$ from each of the $n+1$ factors.
In (4) we can factor out $-x$ and $a^n$ and thanks to $k$ we can explicitly write the elements of $S={j_1,j_2,ldots,j_k}$ for each specific choice.
In (5) we do again a reordering by organizing the summands according to the sum $j_1+j_2+cdots+j_k$ of the $k$-tupels.
We observe the smallest sum comes from the $k$-tupel $(0,1,2,ldots,k-1)$ which gives $$sum_{j=0}^k j=k(k-1)/2$$
while the $k$-tupel with the largest sum is $(n-k+1,n-k+2,ldots,n)$ which gives begin{align*}sum_{j=n-k+1}^n j&=sum_{j=1}^n j-sum_{j=1}^{n-k} j=frac{n(n+1)}{2}-frac{(n-k)(n-k+1)}{2}\
&=frac{k(2n-k+1)}{2}.
end{align*}
In (6) we factor out $a^{-2(j_1+cdots+j_k)}=a^{-2l}$.
In (7) we shift the index of $l$ to start from $0$.
Example $F(a,3,x)$:
We evaluate the expression (7) for the case $n=3$. We obtain
begin{align*}
color{blue}{F(a,3,x)}&=
1+sum_{k=1}^4(-1)^ksum_{l=0}^{k(4-k)}left(sum_{{0leq j_1leq cdotsleq j_kleq 3}atop{j_1+cdots j_k=l+k(k-1)/2}} 1right)a^{k(4-k)-2l}x^k\
&=1-sum_{l=0}^3left(sum_{{0leq j_1leq 3}atop{j_1=l}}1right)a^{3-2l}x
+sum_{l=0}^4left(sum_{{0leq j_1<j_2leq 3}atop{j_1+j_2=l+1}} 1right)a^{4-2l}x^2\
&qquad -sum_{l=0}^3left(sum_{{0leq j_1<j_2<j_3leq 3}atop{j_1+j_2+j_3=l+3}} 1right)a^{3-2l}x^3
-sum_{l=0}^0left(sum_{{0leq j_1<j_2<j_3<j_4leq 3}atop{j_1+j_2+j_3j_4=l+6}} 1right)a^{-2l}x^4\
&,,color{blue}{=1-left(a^3+a+a^{-1}+a^{-3}right)x+left(a^4+a^2+2+a^{-2}+a^{-4}right)x^2}\
&qquad,,color{blue}{-left(a^3+a+a^{-1}+a^{-3}right)x^3+x^4}
end{align*}
in accordance with OPs calculation.
answered 19 hours ago
Markus Scheuer
58.9k454140
edited 7 hours ago
answered 19 hours ago
Markus Scheuer
58.9k454140
answered 19 hours ago
Markus Scheuer
58.9k454140
answered 19 hours ago
Markus Scheuer
58.9k454140
58.9k454140
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By "coefficients of $a$" do you mean "coefficients of $x^k$ in terms of $a$"?
– YiFan
Nov 12 at 1:06
@ YiFan yes you can say that.
– Learner
Nov 12 at 1:23
There should be a development of this product involving the $a^2$-binomial coefficients. See here.
– René Gy
Nov 12 at 18:38
@Thanks René Gy
– Learner
Nov 13 at 2:52
Apply the $q$-binomial formula $prodlimits_{k=0}^{n-1} left(1+q^k tright) = sumlimits_{k=0}^n q^{kleft(k-1right)/2} dbinom{n}{k}_q t^k$ to $n+1$, $a^n x$ and $a^{-2}$ instead of $n$, $t$ and $q$.
– darij grinberg
Nov 14 at 0:44