power series expansion in two variables












0














Let $p,q in (0,1)$ sucht that $p + q = 1$. For $x_{1},x_{2} in (-1,1)$ define the function $g(x_{1},x_{2})$ by
$$
g(x_{1},x_{2}) = frac{qx_{1} + p^{2}x_{1}x_{2}}{1 - pqx_{1}x_{2}}.
$$
Now $g$ is the generating function of some probability distribution $q(n_{1},n_{2})$ for $n_{1},n_{2} geq 0$,i.e.
$$
sum_{n_{1},n_{2}=0}^{infty}q(n_{1},n_{2})x_{1}^{n_{1}}x_{2}^{n_{2}} = g(x_{1},x_{2}).
$$
How can I expand $g$, so I can determine $q$.






real-analysis calculus multivariable-calculus power-series







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    0














    Let $p,q in (0,1)$ sucht that $p + q = 1$. For $x_{1},x_{2} in (-1,1)$ define the function $g(x_{1},x_{2})$ by
    $$
    g(x_{1},x_{2}) = frac{qx_{1} + p^{2}x_{1}x_{2}}{1 - pqx_{1}x_{2}}.
    $$
    Now $g$ is the generating function of some probability distribution $q(n_{1},n_{2})$ for $n_{1},n_{2} geq 0$,i.e.
    $$
    sum_{n_{1},n_{2}=0}^{infty}q(n_{1},n_{2})x_{1}^{n_{1}}x_{2}^{n_{2}} = g(x_{1},x_{2}).
    $$
    How can I expand $g$, so I can determine $q$.






    real-analysis calculus multivariable-calculus power-series







    share|cite|improve this question



























      0












      0








      0


      1





      Let $p,q in (0,1)$ sucht that $p + q = 1$. For $x_{1},x_{2} in (-1,1)$ define the function $g(x_{1},x_{2})$ by
      $$
      g(x_{1},x_{2}) = frac{qx_{1} + p^{2}x_{1}x_{2}}{1 - pqx_{1}x_{2}}.
      $$
      Now $g$ is the generating function of some probability distribution $q(n_{1},n_{2})$ for $n_{1},n_{2} geq 0$,i.e.
      $$
      sum_{n_{1},n_{2}=0}^{infty}q(n_{1},n_{2})x_{1}^{n_{1}}x_{2}^{n_{2}} = g(x_{1},x_{2}).
      $$
      How can I expand $g$, so I can determine $q$.






      real-analysis calculus multivariable-calculus power-series







      share|cite|improve this question















      Let $p,q in (0,1)$ sucht that $p + q = 1$. For $x_{1},x_{2} in (-1,1)$ define the function $g(x_{1},x_{2})$ by
      $$
      g(x_{1},x_{2}) = frac{qx_{1} + p^{2}x_{1}x_{2}}{1 - pqx_{1}x_{2}}.
      $$
      Now $g$ is the generating function of some probability distribution $q(n_{1},n_{2})$ for $n_{1},n_{2} geq 0$,i.e.
      $$
      sum_{n_{1},n_{2}=0}^{infty}q(n_{1},n_{2})x_{1}^{n_{1}}x_{2}^{n_{2}} = g(x_{1},x_{2}).
      $$
      How can I expand $g$, so I can determine $q$.






      real-analysis calculus multivariable-calculus power-series




      real-analysis calculus multivariable-calculus power-series






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 4 at 22:57







      love_math

















      asked Jan 4 at 19:54









      love_mathlove_math

      1545




      1545






















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          Since the constraints that you give result into $$|pq x_1 x_2|<1$$
          then you can apply the geometric series to the denominator and the expansion is simply
          $$
          eqalign{
          & left( {q,x_1 + p^2 x_1 x_2 } right)sumlimits_{0 le k} {left( {pq,x_1 x_2 } right)^{,k} } = cr
          & = sumlimits_{0 le k} {p^{,k} q^{,k + 1} ,x_1 ^{,k + 1} x_2 ^{,k} + p^{,k + 2} q^{,k} ,x_1 ^{,k + 1} x_2 ^{,k + 1} } = cr
          & = sumlimits_{0 le k} {left( {p^{,k} q^{,k + 1} ,x_2 ^{,k} + p^{,k + 2} q^{,k} ,x_2 ^{,k + 1} } right)x_1 ^{,k + 1} } = cr
          & = sumlimits_{0 le n_1 } {sumlimits_{0 le n_2 } {cleft( {n_1 ,n_2 } right)x_1 ^{,n_{,1} } x_2 ^{,n_{,2} } } } cr
          & quad quad quad Downarrow cr
          & cleft( {n_1 ,n_2 } right) = left[ {1 le n_1 } right]left( {left[ {n_2 = n_1 - 1} right]left( {p^{,n_1 - 1} q^{,n_1 } } right) + left[ {n_2 = n_1 } right]p^{,n_1 + 1} q^{,n_1 - 1} } right) cr}
          $$



          where $[P]$ denotes the Iverson bracket
          $$
          left[ P right] = left{ {begin{array}{*{20}c}
          1 & {P = TRUE} \
          0 & {P = FALSE} \
          end{array} } right.
          $$






          share|cite|improve this answer























          • The above function is the generating function of some probability distribution $q(n_{1},n_{2})$, i.e. $sum_{n_{1},n_{2}=0}^{infty}q(n_{1},n_{2})x_{1}^{n_{1}}x_{2}^{n_{2}} = g(x_{1},x_{2})$. So I want to expand it to determine $q(n_{1},n_{2})$. I think your answer does not make sense in my special situation
            – love_math
            Jan 4 at 22:55












          • @love_math: I expanded the sum , hope now is clear what the coefficients of the double expansion will be.
            – G Cab
            Jan 5 at 1:35











          Your Answer





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          1 Answer
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          0














          Since the constraints that you give result into $$|pq x_1 x_2|<1$$
          then you can apply the geometric series to the denominator and the expansion is simply
          $$
          eqalign{
          & left( {q,x_1 + p^2 x_1 x_2 } right)sumlimits_{0 le k} {left( {pq,x_1 x_2 } right)^{,k} } = cr
          & = sumlimits_{0 le k} {p^{,k} q^{,k + 1} ,x_1 ^{,k + 1} x_2 ^{,k} + p^{,k + 2} q^{,k} ,x_1 ^{,k + 1} x_2 ^{,k + 1} } = cr
          & = sumlimits_{0 le k} {left( {p^{,k} q^{,k + 1} ,x_2 ^{,k} + p^{,k + 2} q^{,k} ,x_2 ^{,k + 1} } right)x_1 ^{,k + 1} } = cr
          & = sumlimits_{0 le n_1 } {sumlimits_{0 le n_2 } {cleft( {n_1 ,n_2 } right)x_1 ^{,n_{,1} } x_2 ^{,n_{,2} } } } cr
          & quad quad quad Downarrow cr
          & cleft( {n_1 ,n_2 } right) = left[ {1 le n_1 } right]left( {left[ {n_2 = n_1 - 1} right]left( {p^{,n_1 - 1} q^{,n_1 } } right) + left[ {n_2 = n_1 } right]p^{,n_1 + 1} q^{,n_1 - 1} } right) cr}
          $$



          where $[P]$ denotes the Iverson bracket
          $$
          left[ P right] = left{ {begin{array}{*{20}c}
          1 & {P = TRUE} \
          0 & {P = FALSE} \
          end{array} } right.
          $$






          share|cite|improve this answer























          • The above function is the generating function of some probability distribution $q(n_{1},n_{2})$, i.e. $sum_{n_{1},n_{2}=0}^{infty}q(n_{1},n_{2})x_{1}^{n_{1}}x_{2}^{n_{2}} = g(x_{1},x_{2})$. So I want to expand it to determine $q(n_{1},n_{2})$. I think your answer does not make sense in my special situation
            – love_math
            Jan 4 at 22:55












          • @love_math: I expanded the sum , hope now is clear what the coefficients of the double expansion will be.
            – G Cab
            Jan 5 at 1:35
















          0














          Since the constraints that you give result into $$|pq x_1 x_2|<1$$
          then you can apply the geometric series to the denominator and the expansion is simply
          $$
          eqalign{
          & left( {q,x_1 + p^2 x_1 x_2 } right)sumlimits_{0 le k} {left( {pq,x_1 x_2 } right)^{,k} } = cr
          & = sumlimits_{0 le k} {p^{,k} q^{,k + 1} ,x_1 ^{,k + 1} x_2 ^{,k} + p^{,k + 2} q^{,k} ,x_1 ^{,k + 1} x_2 ^{,k + 1} } = cr
          & = sumlimits_{0 le k} {left( {p^{,k} q^{,k + 1} ,x_2 ^{,k} + p^{,k + 2} q^{,k} ,x_2 ^{,k + 1} } right)x_1 ^{,k + 1} } = cr
          & = sumlimits_{0 le n_1 } {sumlimits_{0 le n_2 } {cleft( {n_1 ,n_2 } right)x_1 ^{,n_{,1} } x_2 ^{,n_{,2} } } } cr
          & quad quad quad Downarrow cr
          & cleft( {n_1 ,n_2 } right) = left[ {1 le n_1 } right]left( {left[ {n_2 = n_1 - 1} right]left( {p^{,n_1 - 1} q^{,n_1 } } right) + left[ {n_2 = n_1 } right]p^{,n_1 + 1} q^{,n_1 - 1} } right) cr}
          $$



          where $[P]$ denotes the Iverson bracket
          $$
          left[ P right] = left{ {begin{array}{*{20}c}
          1 & {P = TRUE} \
          0 & {P = FALSE} \
          end{array} } right.
          $$






          share|cite|improve this answer























          • The above function is the generating function of some probability distribution $q(n_{1},n_{2})$, i.e. $sum_{n_{1},n_{2}=0}^{infty}q(n_{1},n_{2})x_{1}^{n_{1}}x_{2}^{n_{2}} = g(x_{1},x_{2})$. So I want to expand it to determine $q(n_{1},n_{2})$. I think your answer does not make sense in my special situation
            – love_math
            Jan 4 at 22:55












          • @love_math: I expanded the sum , hope now is clear what the coefficients of the double expansion will be.
            – G Cab
            Jan 5 at 1:35














          0












          0








          0






          Since the constraints that you give result into $$|pq x_1 x_2|<1$$
          then you can apply the geometric series to the denominator and the expansion is simply
          $$
          eqalign{
          & left( {q,x_1 + p^2 x_1 x_2 } right)sumlimits_{0 le k} {left( {pq,x_1 x_2 } right)^{,k} } = cr
          & = sumlimits_{0 le k} {p^{,k} q^{,k + 1} ,x_1 ^{,k + 1} x_2 ^{,k} + p^{,k + 2} q^{,k} ,x_1 ^{,k + 1} x_2 ^{,k + 1} } = cr
          & = sumlimits_{0 le k} {left( {p^{,k} q^{,k + 1} ,x_2 ^{,k} + p^{,k + 2} q^{,k} ,x_2 ^{,k + 1} } right)x_1 ^{,k + 1} } = cr
          & = sumlimits_{0 le n_1 } {sumlimits_{0 le n_2 } {cleft( {n_1 ,n_2 } right)x_1 ^{,n_{,1} } x_2 ^{,n_{,2} } } } cr
          & quad quad quad Downarrow cr
          & cleft( {n_1 ,n_2 } right) = left[ {1 le n_1 } right]left( {left[ {n_2 = n_1 - 1} right]left( {p^{,n_1 - 1} q^{,n_1 } } right) + left[ {n_2 = n_1 } right]p^{,n_1 + 1} q^{,n_1 - 1} } right) cr}
          $$



          where $[P]$ denotes the Iverson bracket
          $$
          left[ P right] = left{ {begin{array}{*{20}c}
          1 & {P = TRUE} \
          0 & {P = FALSE} \
          end{array} } right.
          $$






          share|cite|improve this answer














          Since the constraints that you give result into $$|pq x_1 x_2|<1$$
          then you can apply the geometric series to the denominator and the expansion is simply
          $$
          eqalign{
          & left( {q,x_1 + p^2 x_1 x_2 } right)sumlimits_{0 le k} {left( {pq,x_1 x_2 } right)^{,k} } = cr
          & = sumlimits_{0 le k} {p^{,k} q^{,k + 1} ,x_1 ^{,k + 1} x_2 ^{,k} + p^{,k + 2} q^{,k} ,x_1 ^{,k + 1} x_2 ^{,k + 1} } = cr
          & = sumlimits_{0 le k} {left( {p^{,k} q^{,k + 1} ,x_2 ^{,k} + p^{,k + 2} q^{,k} ,x_2 ^{,k + 1} } right)x_1 ^{,k + 1} } = cr
          & = sumlimits_{0 le n_1 } {sumlimits_{0 le n_2 } {cleft( {n_1 ,n_2 } right)x_1 ^{,n_{,1} } x_2 ^{,n_{,2} } } } cr
          & quad quad quad Downarrow cr
          & cleft( {n_1 ,n_2 } right) = left[ {1 le n_1 } right]left( {left[ {n_2 = n_1 - 1} right]left( {p^{,n_1 - 1} q^{,n_1 } } right) + left[ {n_2 = n_1 } right]p^{,n_1 + 1} q^{,n_1 - 1} } right) cr}
          $$



          where $[P]$ denotes the Iverson bracket
          $$
          left[ P right] = left{ {begin{array}{*{20}c}
          1 & {P = TRUE} \
          0 & {P = FALSE} \
          end{array} } right.
          $$







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 5 at 1:33

























          answered Jan 4 at 20:26









          G CabG Cab

          18.1k31237




          18.1k31237












          • The above function is the generating function of some probability distribution $q(n_{1},n_{2})$, i.e. $sum_{n_{1},n_{2}=0}^{infty}q(n_{1},n_{2})x_{1}^{n_{1}}x_{2}^{n_{2}} = g(x_{1},x_{2})$. So I want to expand it to determine $q(n_{1},n_{2})$. I think your answer does not make sense in my special situation
            – love_math
            Jan 4 at 22:55












          • @love_math: I expanded the sum , hope now is clear what the coefficients of the double expansion will be.
            – G Cab
            Jan 5 at 1:35


















          • The above function is the generating function of some probability distribution $q(n_{1},n_{2})$, i.e. $sum_{n_{1},n_{2}=0}^{infty}q(n_{1},n_{2})x_{1}^{n_{1}}x_{2}^{n_{2}} = g(x_{1},x_{2})$. So I want to expand it to determine $q(n_{1},n_{2})$. I think your answer does not make sense in my special situation
            – love_math
            Jan 4 at 22:55












          • @love_math: I expanded the sum , hope now is clear what the coefficients of the double expansion will be.
            – G Cab
            Jan 5 at 1:35
















          The above function is the generating function of some probability distribution $q(n_{1},n_{2})$, i.e. $sum_{n_{1},n_{2}=0}^{infty}q(n_{1},n_{2})x_{1}^{n_{1}}x_{2}^{n_{2}} = g(x_{1},x_{2})$. So I want to expand it to determine $q(n_{1},n_{2})$. I think your answer does not make sense in my special situation
          – love_math
          Jan 4 at 22:55






          The above function is the generating function of some probability distribution $q(n_{1},n_{2})$, i.e. $sum_{n_{1},n_{2}=0}^{infty}q(n_{1},n_{2})x_{1}^{n_{1}}x_{2}^{n_{2}} = g(x_{1},x_{2})$. So I want to expand it to determine $q(n_{1},n_{2})$. I think your answer does not make sense in my special situation
          – love_math
          Jan 4 at 22:55














          @love_math: I expanded the sum , hope now is clear what the coefficients of the double expansion will be.
          – G Cab
          Jan 5 at 1:35




          @love_math: I expanded the sum , hope now is clear what the coefficients of the double expansion will be.
          – G Cab
          Jan 5 at 1:35


















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