Sequence from a generating function?












2














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







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    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







    share|cite|improve this question

























      2












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      2







      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







      share|cite|improve this question













      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






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      asked Nov 18 at 15:32









      Mittal G

      1,188515




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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 $a to infty.$






          share|cite|improve this answer























          • 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











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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 $a to infty.$






          share|cite|improve this answer























          • 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
















          2














          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 $a to infty.$






          share|cite|improve this answer























          • 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














          2












          2








          2






          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 $a to infty.$






          share|cite|improve this answer














          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 $a to infty.$







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Nov 18 at 16:20

























          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


















          • 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


















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