Sequence from a generating function?
Find the sequence $a_n$ if its generating function is $$A(x) = prod_{n=1}^{infty}(1-q^nx)$$
Well, i need to find the expansion of A(x) in terms of powers of x. For that I take the log of both the sides to get
$$log(A(x)) = sum_{n=1}^{infty}log(1-q^nx)$$
Now how to proceed further? Or is there any other method to expand $A(x)$.
power-series generating-functions infinite-product
asked Nov 18 at 15:32
Mittal G
1,188515
add a comment |
Find the sequence $a_n$ if its generating function is $$A(x) = prod_{n=1}^{infty}(1-q^nx)$$
Well, i need to find the expansion of A(x) in terms of powers of x. For that I take the log of both the sides to get
$$log(A(x)) = sum_{n=1}^{infty}log(1-q^nx)$$
Now how to proceed further? Or is there any other method to expand $A(x)$.
power-series generating-functions infinite-product
asked Nov 18 at 15:32
Mittal G
1,188515
add a comment |
Find the sequence $a_n$ if its generating function is $$A(x) = prod_{n=1}^{infty}(1-q^nx)$$
Well, i need to find the expansion of A(x) in terms of powers of x. For that I take the log of both the sides to get
$$log(A(x)) = sum_{n=1}^{infty}log(1-q^nx)$$
Now how to proceed further? Or is there any other method to expand $A(x)$.
power-series generating-functions infinite-product
asked Nov 18 at 15:32
Mittal G
1,188515
Find the sequence $a_n$ if its generating function is $$A(x) = prod_{n=1}^{infty}(1-q^nx)$$
Well, i need to find the expansion of A(x) in terms of powers of x. For that I take the log of both the sides to get
$$log(A(x)) = sum_{n=1}^{infty}log(1-q^nx)$$
Now how to proceed further? Or is there any other method to expand $A(x)$.
power-series generating-functions infinite-product
power-series generating-functions infinite-product
asked Nov 18 at 15:32
Mittal G
1,188515
asked Nov 18 at 15:32
Mittal G
1,188515
asked Nov 18 at 15:32
Mittal G
1,188515
asked Nov 18 at 15:32
Mittal G
1,188515
asked Nov 18 at 15:32
Mittal G
1,188515
1,188515
add a comment |
add a comment |
1 Answer
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We have $$A(x)=prod_{n geq 1} (1-xq^n)=sum_{n geq 0} frac{(-1)^nx^nq^{frac{n(n+1)}{2}}}{(q;q)_n},$$ where $(q;q)_n := (1-q) cdots (1-q^n).$
Proof with Recursion: Note that $A(x)= (1-xq)A(xq).$ Thus, if $A(x)= sum_{n geq 0} A_n x^n$, then begin{align*} sum_{n geq 0} A_nx^n &= sum_{n geq 0} A_nx^nq^n - sum_{n geq 0} A_nx^{n+1}q^{n+1} \ &= 1 + sum_{n geq 1} x^n(q^nA_n - A_{n-1}q^{n}),end{align*} since $A_0=1$. Thus, $$A_n = q^nA_n - A_{n-1}q^{n} implies A_n = frac{-q^{n}A_{n-1}}{1-q^n}.$$ You can show that this recursion is also satisfied by the coefficients of the RHS, which concludes the proof.
Combinatorial Proof Note that $$A(x) = sum_{lambda in mathcal{D}} (-x)^{ell(lambda)}q^{|lambda|},$$ where $mathcal{D}$ is the set of partitions $lambda$ into distinct parts, $ell(lambda)$ is the number of parts and $|lambda|$ is the size of the partition. Thus, if we can show that the RHS is also this generating function, then the two are equal.
Namely, observe that in each distinct parts partition $lambda$ there is a unique triangle as below $$begin{matrix} {color{red} bullet} & {color{red} bullet} & {color{red} bullet} & bullet & bullet & bullet \ {color{red} bullet} & {color{red} bullet} & bullet & & & \ {color{red} bullet} & & & & & end{matrix}.$$ Here the red dots sum to $1+2+3=frac{3(4)}{2}$ which is counted by $(-x)^{n}q^{frac{n(n+1)}{2}}$ and the black dots form an arbitrary partition counted by $frac{1}{(q;q)_n}$. Summing over $n=ell(lambda)$ gives the claim.
$q$-Binomial Theorem Cauchy's $q$-deformation of the binomial theorem states $$sum_{n geq 0} frac{(a;q)_n}{(q;q)_n} z^n = frac{(az;q)_{infty}}{(z;q)_{infty}},$$ which is easy to verify through recursion. For your result, one may substitute $z to -xqa^{-1}$ and then take $a to infty.$
answered Nov 18 at 15:54
Dzoooks
846316
Did $a_n$ of the question become $A_n$ in the answer?
– Mason
Nov 18 at 15:56
@Dzoooks: I think in the starting of the proof we should have $A(x) = (1-xq)A(xq)$
– Mittal G
Nov 18 at 16:02
@MittalG Thanks, fixed.
– Dzoooks
Nov 18 at 16:07
Great thoughts. Thanks for the help.
– Mittal G
Nov 18 at 16:19
add a comment |
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We have $$A(x)=prod_{n geq 1} (1-xq^n)=sum_{n geq 0} frac{(-1)^nx^nq^{frac{n(n+1)}{2}}}{(q;q)_n},$$ where $(q;q)_n := (1-q) cdots (1-q^n).$
Proof with Recursion: Note that $A(x)= (1-xq)A(xq).$ Thus, if $A(x)= sum_{n geq 0} A_n x^n$, then begin{align*} sum_{n geq 0} A_nx^n &= sum_{n geq 0} A_nx^nq^n - sum_{n geq 0} A_nx^{n+1}q^{n+1} \ &= 1 + sum_{n geq 1} x^n(q^nA_n - A_{n-1}q^{n}),end{align*} since $A_0=1$. Thus, $$A_n = q^nA_n - A_{n-1}q^{n} implies A_n = frac{-q^{n}A_{n-1}}{1-q^n}.$$ You can show that this recursion is also satisfied by the coefficients of the RHS, which concludes the proof.
Combinatorial Proof Note that $$A(x) = sum_{lambda in mathcal{D}} (-x)^{ell(lambda)}q^{|lambda|},$$ where $mathcal{D}$ is the set of partitions $lambda$ into distinct parts, $ell(lambda)$ is the number of parts and $|lambda|$ is the size of the partition. Thus, if we can show that the RHS is also this generating function, then the two are equal.
Namely, observe that in each distinct parts partition $lambda$ there is a unique triangle as below $$begin{matrix} {color{red} bullet} & {color{red} bullet} & {color{red} bullet} & bullet & bullet & bullet \ {color{red} bullet} & {color{red} bullet} & bullet & & & \ {color{red} bullet} & & & & & end{matrix}.$$ Here the red dots sum to $1+2+3=frac{3(4)}{2}$ which is counted by $(-x)^{n}q^{frac{n(n+1)}{2}}$ and the black dots form an arbitrary partition counted by $frac{1}{(q;q)_n}$. Summing over $n=ell(lambda)$ gives the claim.
$q$-Binomial Theorem Cauchy's $q$-deformation of the binomial theorem states $$sum_{n geq 0} frac{(a;q)_n}{(q;q)_n} z^n = frac{(az;q)_{infty}}{(z;q)_{infty}},$$ which is easy to verify through recursion. For your result, one may substitute $z to -xqa^{-1}$ and then take $a to infty.$
answered Nov 18 at 15:54
Dzoooks
846316
Did $a_n$ of the question become $A_n$ in the answer?
– Mason
Nov 18 at 15:56
@Dzoooks: I think in the starting of the proof we should have $A(x) = (1-xq)A(xq)$
– Mittal G
Nov 18 at 16:02
@MittalG Thanks, fixed.
– Dzoooks
Nov 18 at 16:07
Great thoughts. Thanks for the help.
– Mittal G
Nov 18 at 16:19
add a comment |
We have $$A(x)=prod_{n geq 1} (1-xq^n)=sum_{n geq 0} frac{(-1)^nx^nq^{frac{n(n+1)}{2}}}{(q;q)_n},$$ where $(q;q)_n := (1-q) cdots (1-q^n).$
Proof with Recursion: Note that $A(x)= (1-xq)A(xq).$ Thus, if $A(x)= sum_{n geq 0} A_n x^n$, then begin{align*} sum_{n geq 0} A_nx^n &= sum_{n geq 0} A_nx^nq^n - sum_{n geq 0} A_nx^{n+1}q^{n+1} \ &= 1 + sum_{n geq 1} x^n(q^nA_n - A_{n-1}q^{n}),end{align*} since $A_0=1$. Thus, $$A_n = q^nA_n - A_{n-1}q^{n} implies A_n = frac{-q^{n}A_{n-1}}{1-q^n}.$$ You can show that this recursion is also satisfied by the coefficients of the RHS, which concludes the proof.
Combinatorial Proof Note that $$A(x) = sum_{lambda in mathcal{D}} (-x)^{ell(lambda)}q^{|lambda|},$$ where $mathcal{D}$ is the set of partitions $lambda$ into distinct parts, $ell(lambda)$ is the number of parts and $|lambda|$ is the size of the partition. Thus, if we can show that the RHS is also this generating function, then the two are equal.
Namely, observe that in each distinct parts partition $lambda$ there is a unique triangle as below $$begin{matrix} {color{red} bullet} & {color{red} bullet} & {color{red} bullet} & bullet & bullet & bullet \ {color{red} bullet} & {color{red} bullet} & bullet & & & \ {color{red} bullet} & & & & & end{matrix}.$$ Here the red dots sum to $1+2+3=frac{3(4)}{2}$ which is counted by $(-x)^{n}q^{frac{n(n+1)}{2}}$ and the black dots form an arbitrary partition counted by $frac{1}{(q;q)_n}$. Summing over $n=ell(lambda)$ gives the claim.
$q$-Binomial Theorem Cauchy's $q$-deformation of the binomial theorem states $$sum_{n geq 0} frac{(a;q)_n}{(q;q)_n} z^n = frac{(az;q)_{infty}}{(z;q)_{infty}},$$ which is easy to verify through recursion. For your result, one may substitute $z to -xqa^{-1}$ and then take $a to infty.$
answered Nov 18 at 15:54
Dzoooks
846316
Did $a_n$ of the question become $A_n$ in the answer?
– Mason
Nov 18 at 15:56
@Dzoooks: I think in the starting of the proof we should have $A(x) = (1-xq)A(xq)$
– Mittal G
Nov 18 at 16:02
@MittalG Thanks, fixed.
– Dzoooks
Nov 18 at 16:07
Great thoughts. Thanks for the help.
– Mittal G
Nov 18 at 16:19
add a comment |
We have $$A(x)=prod_{n geq 1} (1-xq^n)=sum_{n geq 0} frac{(-1)^nx^nq^{frac{n(n+1)}{2}}}{(q;q)_n},$$ where $(q;q)_n := (1-q) cdots (1-q^n).$
Proof with Recursion: Note that $A(x)= (1-xq)A(xq).$ Thus, if $A(x)= sum_{n geq 0} A_n x^n$, then begin{align*} sum_{n geq 0} A_nx^n &= sum_{n geq 0} A_nx^nq^n - sum_{n geq 0} A_nx^{n+1}q^{n+1} \ &= 1 + sum_{n geq 1} x^n(q^nA_n - A_{n-1}q^{n}),end{align*} since $A_0=1$. Thus, $$A_n = q^nA_n - A_{n-1}q^{n} implies A_n = frac{-q^{n}A_{n-1}}{1-q^n}.$$ You can show that this recursion is also satisfied by the coefficients of the RHS, which concludes the proof.
Combinatorial Proof Note that $$A(x) = sum_{lambda in mathcal{D}} (-x)^{ell(lambda)}q^{|lambda|},$$ where $mathcal{D}$ is the set of partitions $lambda$ into distinct parts, $ell(lambda)$ is the number of parts and $|lambda|$ is the size of the partition. Thus, if we can show that the RHS is also this generating function, then the two are equal.
Namely, observe that in each distinct parts partition $lambda$ there is a unique triangle as below $$begin{matrix} {color{red} bullet} & {color{red} bullet} & {color{red} bullet} & bullet & bullet & bullet \ {color{red} bullet} & {color{red} bullet} & bullet & & & \ {color{red} bullet} & & & & & end{matrix}.$$ Here the red dots sum to $1+2+3=frac{3(4)}{2}$ which is counted by $(-x)^{n}q^{frac{n(n+1)}{2}}$ and the black dots form an arbitrary partition counted by $frac{1}{(q;q)_n}$. Summing over $n=ell(lambda)$ gives the claim.
$q$-Binomial Theorem Cauchy's $q$-deformation of the binomial theorem states $$sum_{n geq 0} frac{(a;q)_n}{(q;q)_n} z^n = frac{(az;q)_{infty}}{(z;q)_{infty}},$$ which is easy to verify through recursion. For your result, one may substitute $z to -xqa^{-1}$ and then take $a to infty.$
answered Nov 18 at 15:54
Dzoooks
846316
We have $$A(x)=prod_{n geq 1} (1-xq^n)=sum_{n geq 0} frac{(-1)^nx^nq^{frac{n(n+1)}{2}}}{(q;q)_n},$$ where $(q;q)_n := (1-q) cdots (1-q^n).$
Proof with Recursion: Note that $A(x)= (1-xq)A(xq).$ Thus, if $A(x)= sum_{n geq 0} A_n x^n$, then begin{align*} sum_{n geq 0} A_nx^n &= sum_{n geq 0} A_nx^nq^n - sum_{n geq 0} A_nx^{n+1}q^{n+1} \ &= 1 + sum_{n geq 1} x^n(q^nA_n - A_{n-1}q^{n}),end{align*} since $A_0=1$. Thus, $$A_n = q^nA_n - A_{n-1}q^{n} implies A_n = frac{-q^{n}A_{n-1}}{1-q^n}.$$ You can show that this recursion is also satisfied by the coefficients of the RHS, which concludes the proof.
Combinatorial Proof Note that $$A(x) = sum_{lambda in mathcal{D}} (-x)^{ell(lambda)}q^{|lambda|},$$ where $mathcal{D}$ is the set of partitions $lambda$ into distinct parts, $ell(lambda)$ is the number of parts and $|lambda|$ is the size of the partition. Thus, if we can show that the RHS is also this generating function, then the two are equal.
Namely, observe that in each distinct parts partition $lambda$ there is a unique triangle as below $$begin{matrix} {color{red} bullet} & {color{red} bullet} & {color{red} bullet} & bullet & bullet & bullet \ {color{red} bullet} & {color{red} bullet} & bullet & & & \ {color{red} bullet} & & & & & end{matrix}.$$ Here the red dots sum to $1+2+3=frac{3(4)}{2}$ which is counted by $(-x)^{n}q^{frac{n(n+1)}{2}}$ and the black dots form an arbitrary partition counted by $frac{1}{(q;q)_n}$. Summing over $n=ell(lambda)$ gives the claim.
$q$-Binomial Theorem Cauchy's $q$-deformation of the binomial theorem states $$sum_{n geq 0} frac{(a;q)_n}{(q;q)_n} z^n = frac{(az;q)_{infty}}{(z;q)_{infty}},$$ which is easy to verify through recursion. For your result, one may substitute $z to -xqa^{-1}$ and then take $a to infty.$
answered Nov 18 at 15:54
Dzoooks
846316
edited Nov 18 at 16:20
answered Nov 18 at 15:54
Dzoooks
846316
answered Nov 18 at 15:54
Dzoooks
846316
answered Nov 18 at 15:54
Dzoooks
846316
846316
Did $a_n$ of the question become $A_n$ in the answer?
– Mason
Nov 18 at 15:56
@Dzoooks: I think in the starting of the proof we should have $A(x) = (1-xq)A(xq)$
– Mittal G
Nov 18 at 16:02
@MittalG Thanks, fixed.
– Dzoooks
Nov 18 at 16:07
Great thoughts. Thanks for the help.
– Mittal G
Nov 18 at 16:19
add a comment |
Did $a_n$ of the question become $A_n$ in the answer?
– Mason
Nov 18 at 15:56
@Dzoooks: I think in the starting of the proof we should have $A(x) = (1-xq)A(xq)$
– Mittal G
Nov 18 at 16:02
@MittalG Thanks, fixed.
– Dzoooks
Nov 18 at 16:07
Great thoughts. Thanks for the help.
– Mittal G
Nov 18 at 16:19
Did $a_n$ of the question become $A_n$ in the answer?
– Mason
Nov 18 at 15:56
Did $a_n$ of the question become $A_n$ in the answer?
– Mason
Nov 18 at 15:56
@Dzoooks: I think in the starting of the proof we should have $A(x) = (1-xq)A(xq)$
– Mittal G
Nov 18 at 16:02
@Dzoooks: I think in the starting of the proof we should have $A(x) = (1-xq)A(xq)$
– Mittal G
Nov 18 at 16:02
@MittalG Thanks, fixed.
– Dzoooks
Nov 18 at 16:07
@MittalG Thanks, fixed.
– Dzoooks
Nov 18 at 16:07
Great thoughts. Thanks for the help.
– Mittal G
Nov 18 at 16:19
Great thoughts. Thanks for the help.
– Mittal G
Nov 18 at 16:19
add a comment |
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