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How to compute the coefficient of a generating function?


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0












$begingroup$


I would like to find a closed formula for the coefficients of the generating function
$$f(x)=-{frac {{x}^{4}+6,{x}^{3}-2,{x}^{2}+6,x+1}{ left( x+1 right)
left( {x}^{2}+1 right) left( x-1 right) ^{3}}} .$$



I am trying to do the following.
begin{align}
f(x)=sum_{i,j,k_1,k_2,k_3=0}^{infty} (-x)^i (-x^2)^j x^{k_1+k_2+k_3}(1+6x-2x^2+6x^3+x^4).
end{align}



The expression is very complicated. Are there some simpler method (or some software) to find a closed formula for the coefficients? Thank you very much.






combinatorics generating-functions







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    Tried partial fractions? $$-{x^4+6x^3-2x^2+6x+1over(x+1)(x^2+1)(x-1)^3}={Aover x+1}+{Bx+Cover x^2+1}+{Dover x-1}+{Eover(x-1)^2}+{Fover(x-1)^3}$$
    $endgroup$
    – Gerry Myerson
    14 hours ago
















0












$begingroup$


I would like to find a closed formula for the coefficients of the generating function
$$f(x)=-{frac {{x}^{4}+6,{x}^{3}-2,{x}^{2}+6,x+1}{ left( x+1 right)
left( {x}^{2}+1 right) left( x-1 right) ^{3}}} .$$



I am trying to do the following.
begin{align}
f(x)=sum_{i,j,k_1,k_2,k_3=0}^{infty} (-x)^i (-x^2)^j x^{k_1+k_2+k_3}(1+6x-2x^2+6x^3+x^4).
end{align}



The expression is very complicated. Are there some simpler method (or some software) to find a closed formula for the coefficients? Thank you very much.






combinatorics generating-functions







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    Tried partial fractions? $$-{x^4+6x^3-2x^2+6x+1over(x+1)(x^2+1)(x-1)^3}={Aover x+1}+{Bx+Cover x^2+1}+{Dover x-1}+{Eover(x-1)^2}+{Fover(x-1)^3}$$
    $endgroup$
    – Gerry Myerson
    14 hours ago














0












0








0





$begingroup$


I would like to find a closed formula for the coefficients of the generating function
$$f(x)=-{frac {{x}^{4}+6,{x}^{3}-2,{x}^{2}+6,x+1}{ left( x+1 right)
left( {x}^{2}+1 right) left( x-1 right) ^{3}}} .$$



I am trying to do the following.
begin{align}
f(x)=sum_{i,j,k_1,k_2,k_3=0}^{infty} (-x)^i (-x^2)^j x^{k_1+k_2+k_3}(1+6x-2x^2+6x^3+x^4).
end{align}



The expression is very complicated. Are there some simpler method (or some software) to find a closed formula for the coefficients? Thank you very much.






combinatorics generating-functions







share|cite|improve this question











$endgroup$




I would like to find a closed formula for the coefficients of the generating function
$$f(x)=-{frac {{x}^{4}+6,{x}^{3}-2,{x}^{2}+6,x+1}{ left( x+1 right)
left( {x}^{2}+1 right) left( x-1 right) ^{3}}} .$$



I am trying to do the following.
begin{align}
f(x)=sum_{i,j,k_1,k_2,k_3=0}^{infty} (-x)^i (-x^2)^j x^{k_1+k_2+k_3}(1+6x-2x^2+6x^3+x^4).
end{align}



The expression is very complicated. Are there some simpler method (or some software) to find a closed formula for the coefficients? Thank you very much.






combinatorics generating-functions




combinatorics generating-functions






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 14 hours ago









Rócherz

2,8562821




2,8562821










asked 14 hours ago









LJRLJR

6,63141849




6,63141849








  • 3




    $begingroup$
    Tried partial fractions? $$-{x^4+6x^3-2x^2+6x+1over(x+1)(x^2+1)(x-1)^3}={Aover x+1}+{Bx+Cover x^2+1}+{Dover x-1}+{Eover(x-1)^2}+{Fover(x-1)^3}$$
    $endgroup$
    – Gerry Myerson
    14 hours ago














  • 3




    $begingroup$
    Tried partial fractions? $$-{x^4+6x^3-2x^2+6x+1over(x+1)(x^2+1)(x-1)^3}={Aover x+1}+{Bx+Cover x^2+1}+{Dover x-1}+{Eover(x-1)^2}+{Fover(x-1)^3}$$
    $endgroup$
    – Gerry Myerson
    14 hours ago








3




3




$begingroup$
Tried partial fractions? $$-{x^4+6x^3-2x^2+6x+1over(x+1)(x^2+1)(x-1)^3}={Aover x+1}+{Bx+Cover x^2+1}+{Dover x-1}+{Eover(x-1)^2}+{Fover(x-1)^3}$$
$endgroup$
– Gerry Myerson
14 hours ago




$begingroup$
Tried partial fractions? $$-{x^4+6x^3-2x^2+6x+1over(x+1)(x^2+1)(x-1)^3}={Aover x+1}+{Bx+Cover x^2+1}+{Dover x-1}+{Eover(x-1)^2}+{Fover(x-1)^3}$$
$endgroup$
– Gerry Myerson
14 hours ago










1 Answer
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$begingroup$

$$-{frac {{x}^{4}+6,{x}^{3}-2,{x}^{2}+6,x+1}{ left( x+1 right)
left( {x}^{2}+1 right) left( x-1 right) ^{3}}} =frac{0.25}{1-x}-frac{0.75}{1+x}+frac{x}{1+x^2}-frac{1.5}{(1-x)^2}+frac{3}{(1-x)^3}$$
then use the following series and differentiate it two times to replace all terms to series
$$frac{1}{1-x}=sum_{n=0}^{infty}x^n$$






share|cite|improve this answer









$endgroup$













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    1 Answer
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    active

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    2












    $begingroup$

    $$-{frac {{x}^{4}+6,{x}^{3}-2,{x}^{2}+6,x+1}{ left( x+1 right)
    left( {x}^{2}+1 right) left( x-1 right) ^{3}}} =frac{0.25}{1-x}-frac{0.75}{1+x}+frac{x}{1+x^2}-frac{1.5}{(1-x)^2}+frac{3}{(1-x)^3}$$
    then use the following series and differentiate it two times to replace all terms to series
    $$frac{1}{1-x}=sum_{n=0}^{infty}x^n$$






    share|cite|improve this answer









    $endgroup$


















      2












      $begingroup$

      $$-{frac {{x}^{4}+6,{x}^{3}-2,{x}^{2}+6,x+1}{ left( x+1 right)
      left( {x}^{2}+1 right) left( x-1 right) ^{3}}} =frac{0.25}{1-x}-frac{0.75}{1+x}+frac{x}{1+x^2}-frac{1.5}{(1-x)^2}+frac{3}{(1-x)^3}$$
      then use the following series and differentiate it two times to replace all terms to series
      $$frac{1}{1-x}=sum_{n=0}^{infty}x^n$$






      share|cite|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        $$-{frac {{x}^{4}+6,{x}^{3}-2,{x}^{2}+6,x+1}{ left( x+1 right)
        left( {x}^{2}+1 right) left( x-1 right) ^{3}}} =frac{0.25}{1-x}-frac{0.75}{1+x}+frac{x}{1+x^2}-frac{1.5}{(1-x)^2}+frac{3}{(1-x)^3}$$
        then use the following series and differentiate it two times to replace all terms to series
        $$frac{1}{1-x}=sum_{n=0}^{infty}x^n$$






        share|cite|improve this answer









        $endgroup$



        $$-{frac {{x}^{4}+6,{x}^{3}-2,{x}^{2}+6,x+1}{ left( x+1 right)
        left( {x}^{2}+1 right) left( x-1 right) ^{3}}} =frac{0.25}{1-x}-frac{0.75}{1+x}+frac{x}{1+x^2}-frac{1.5}{(1-x)^2}+frac{3}{(1-x)^3}$$
        then use the following series and differentiate it two times to replace all terms to series
        $$frac{1}{1-x}=sum_{n=0}^{infty}x^n$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered 14 hours ago









        E.H.EE.H.E

        15.8k11968




        15.8k11968






























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