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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

edited 14 hours ago

Rócherz

2,8562821

asked 14 hours ago

LJRLJR

6,63141849

$endgroup$

• 
  
  
  

  
  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
  

  
  
  
  

add a comment | 

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

edited 14 hours ago

Rócherz

2,8562821

asked 14 hours ago

LJRLJR

6,63141849

$endgroup$

• 
  
  
  

  
  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
  

  
  
  
  

add a comment | 

$begingroup$

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

edited 14 hours ago

Rócherz

2,8562821

asked 14 hours ago

LJRLJR

6,63141849

$endgroup$

share|cite|improve this question

edited 14 hours ago

Rócherz

2,8562821

asked 14 hours ago

LJRLJR

6,63141849

share|cite|improve this question

edited 14 hours ago

Rócherz

2,8562821

asked 14 hours ago

LJRLJR

6,63141849

edited 14 hours ago

Rócherz

2,8562821

edited 14 hours ago

Rócherz

2,8562821

asked 14 hours ago

LJRLJR

6,63141849

asked 14 hours ago

LJRLJR

6,63141849

1 Answer
1

active

oldest

votes

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

answered 14 hours ago

E.H.EE.H.E

15.8k11968

$endgroup$

add a comment | 

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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

answered 14 hours ago

E.H.EE.H.E

15.8k11968

$endgroup$

add a comment | 

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

answered 14 hours ago

E.H.EE.H.E

15.8k11968

$endgroup$

add a comment | 

$begingroup$

$$-{frac {{x}^{4}+6,{x}^{3}-2,{x}^{2}+6,x+1}{ left( x+1 right)
left( {x}^{2}+1 right) left( x-1 right) ^{3}}} =frac{0.25}{1-x}-frac{0.75}{1+x}+frac{x}{1+x^2}-frac{1.5}{(1-x)^2}+frac{3}{(1-x)^3}$$
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

answered 14 hours ago

E.H.EE.H.E

15.8k11968

$endgroup$

share|cite|improve this answer

answered 14 hours ago

E.H.EE.H.E

15.8k11968

answered 14 hours ago

E.H.EE.H.E

15.8k11968

answered 14 hours ago

E.H.EE.H.E

15.8k11968