How to solve $alpha,beta$ in $frac{x}{1-x-x^2}=frac{x}{(1-alpha x)(1-beta x)}$












0












$begingroup$


I'm trying to understand a procedure within solving a generating function problem:



$$G(x) = dfrac{x}{1-x-x^2}=dfrac{x}{(1-alpha x)(1-beta x)}$$



I understand that the reason for this step is to utilize partial fraction s.t.



$$ =dfrac{a}{(1-alpha x)} + dfrac{b}{(1-beta x)}$$



I don't know why turning $(1-x-x^2)$ into $(1-alpha x)(1-beta x)$ is possible, and is there some formula for this case solving $alpha$ and $beta$? The method I'm familiar with is



$$(1-x-x^2)=-(x-r_1)(x-r_2)$$



then I can apply the formula $$r_i=frac{-bpmsqrt{b^2-4ac}}{2a}$$ but I don't know how to deal with the case $(1-alpha x)(1-beta x)$. I need a general method/formula to solve $alpha$ and $beta$...






algebra-precalculus







share|cite|improve this question











$endgroup$












  • $begingroup$
    To utilize en.wikipedia.org/wiki/Binomial_series
    $endgroup$
    – lab bhattacharjee
    Dec 21 '18 at 7:18






  • 2




    $begingroup$
    It's just factorisation. Always works over the complex numbers.
    $endgroup$
    – Lord Shark the Unknown
    Dec 21 '18 at 7:20










  • $begingroup$
    @LordSharktheUnknown So this is not related to testing whether $x=0$ is a root (my current idea)?
    $endgroup$
    – user7813604
    Dec 21 '18 at 7:21






  • 2




    $begingroup$
    If you have $1-x-x^2=-(x-r_1)(x-r_2)$ then $r_1r_2=1$ and then $1-x-x^2=(1-x/r_2)(1-x/r_1)$.
    $endgroup$
    – Lord Shark the Unknown
    Dec 21 '18 at 7:31










  • $begingroup$
    @LordSharktheUnknown: Oh so since $r_1r_2=1implies r_1,r_2not=0$ so I can divide it by $r_1cdot r_2$?
    $endgroup$
    – user7813604
    Dec 21 '18 at 7:33
















0












$begingroup$


I'm trying to understand a procedure within solving a generating function problem:



$$G(x) = dfrac{x}{1-x-x^2}=dfrac{x}{(1-alpha x)(1-beta x)}$$



I understand that the reason for this step is to utilize partial fraction s.t.



$$ =dfrac{a}{(1-alpha x)} + dfrac{b}{(1-beta x)}$$



I don't know why turning $(1-x-x^2)$ into $(1-alpha x)(1-beta x)$ is possible, and is there some formula for this case solving $alpha$ and $beta$? The method I'm familiar with is



$$(1-x-x^2)=-(x-r_1)(x-r_2)$$



then I can apply the formula $$r_i=frac{-bpmsqrt{b^2-4ac}}{2a}$$ but I don't know how to deal with the case $(1-alpha x)(1-beta x)$. I need a general method/formula to solve $alpha$ and $beta$...






algebra-precalculus







share|cite|improve this question











$endgroup$












  • $begingroup$
    To utilize en.wikipedia.org/wiki/Binomial_series
    $endgroup$
    – lab bhattacharjee
    Dec 21 '18 at 7:18






  • 2




    $begingroup$
    It's just factorisation. Always works over the complex numbers.
    $endgroup$
    – Lord Shark the Unknown
    Dec 21 '18 at 7:20










  • $begingroup$
    @LordSharktheUnknown So this is not related to testing whether $x=0$ is a root (my current idea)?
    $endgroup$
    – user7813604
    Dec 21 '18 at 7:21






  • 2




    $begingroup$
    If you have $1-x-x^2=-(x-r_1)(x-r_2)$ then $r_1r_2=1$ and then $1-x-x^2=(1-x/r_2)(1-x/r_1)$.
    $endgroup$
    – Lord Shark the Unknown
    Dec 21 '18 at 7:31










  • $begingroup$
    @LordSharktheUnknown: Oh so since $r_1r_2=1implies r_1,r_2not=0$ so I can divide it by $r_1cdot r_2$?
    $endgroup$
    – user7813604
    Dec 21 '18 at 7:33














0












0








0





$begingroup$


I'm trying to understand a procedure within solving a generating function problem:



$$G(x) = dfrac{x}{1-x-x^2}=dfrac{x}{(1-alpha x)(1-beta x)}$$



I understand that the reason for this step is to utilize partial fraction s.t.



$$ =dfrac{a}{(1-alpha x)} + dfrac{b}{(1-beta x)}$$



I don't know why turning $(1-x-x^2)$ into $(1-alpha x)(1-beta x)$ is possible, and is there some formula for this case solving $alpha$ and $beta$? The method I'm familiar with is



$$(1-x-x^2)=-(x-r_1)(x-r_2)$$



then I can apply the formula $$r_i=frac{-bpmsqrt{b^2-4ac}}{2a}$$ but I don't know how to deal with the case $(1-alpha x)(1-beta x)$. I need a general method/formula to solve $alpha$ and $beta$...






algebra-precalculus







share|cite|improve this question











$endgroup$




I'm trying to understand a procedure within solving a generating function problem:



$$G(x) = dfrac{x}{1-x-x^2}=dfrac{x}{(1-alpha x)(1-beta x)}$$



I understand that the reason for this step is to utilize partial fraction s.t.



$$ =dfrac{a}{(1-alpha x)} + dfrac{b}{(1-beta x)}$$



I don't know why turning $(1-x-x^2)$ into $(1-alpha x)(1-beta x)$ is possible, and is there some formula for this case solving $alpha$ and $beta$? The method I'm familiar with is



$$(1-x-x^2)=-(x-r_1)(x-r_2)$$



then I can apply the formula $$r_i=frac{-bpmsqrt{b^2-4ac}}{2a}$$ but I don't know how to deal with the case $(1-alpha x)(1-beta x)$. I need a general method/formula to solve $alpha$ and $beta$...






algebra-precalculus




algebra-precalculus






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 21 '18 at 8:45









Lorenzo B.

1,8402520




1,8402520










asked Dec 21 '18 at 7:16









user7813604user7813604

15912




15912












  • $begingroup$
    To utilize en.wikipedia.org/wiki/Binomial_series
    $endgroup$
    – lab bhattacharjee
    Dec 21 '18 at 7:18






  • 2




    $begingroup$
    It's just factorisation. Always works over the complex numbers.
    $endgroup$
    – Lord Shark the Unknown
    Dec 21 '18 at 7:20










  • $begingroup$
    @LordSharktheUnknown So this is not related to testing whether $x=0$ is a root (my current idea)?
    $endgroup$
    – user7813604
    Dec 21 '18 at 7:21






  • 2




    $begingroup$
    If you have $1-x-x^2=-(x-r_1)(x-r_2)$ then $r_1r_2=1$ and then $1-x-x^2=(1-x/r_2)(1-x/r_1)$.
    $endgroup$
    – Lord Shark the Unknown
    Dec 21 '18 at 7:31










  • $begingroup$
    @LordSharktheUnknown: Oh so since $r_1r_2=1implies r_1,r_2not=0$ so I can divide it by $r_1cdot r_2$?
    $endgroup$
    – user7813604
    Dec 21 '18 at 7:33


















  • $begingroup$
    To utilize en.wikipedia.org/wiki/Binomial_series
    $endgroup$
    – lab bhattacharjee
    Dec 21 '18 at 7:18






  • 2




    $begingroup$
    It's just factorisation. Always works over the complex numbers.
    $endgroup$
    – Lord Shark the Unknown
    Dec 21 '18 at 7:20










  • $begingroup$
    @LordSharktheUnknown So this is not related to testing whether $x=0$ is a root (my current idea)?
    $endgroup$
    – user7813604
    Dec 21 '18 at 7:21






  • 2




    $begingroup$
    If you have $1-x-x^2=-(x-r_1)(x-r_2)$ then $r_1r_2=1$ and then $1-x-x^2=(1-x/r_2)(1-x/r_1)$.
    $endgroup$
    – Lord Shark the Unknown
    Dec 21 '18 at 7:31










  • $begingroup$
    @LordSharktheUnknown: Oh so since $r_1r_2=1implies r_1,r_2not=0$ so I can divide it by $r_1cdot r_2$?
    $endgroup$
    – user7813604
    Dec 21 '18 at 7:33
















$begingroup$
To utilize en.wikipedia.org/wiki/Binomial_series
$endgroup$
– lab bhattacharjee
Dec 21 '18 at 7:18




$begingroup$
To utilize en.wikipedia.org/wiki/Binomial_series
$endgroup$
– lab bhattacharjee
Dec 21 '18 at 7:18




2




2




$begingroup$
It's just factorisation. Always works over the complex numbers.
$endgroup$
– Lord Shark the Unknown
Dec 21 '18 at 7:20




$begingroup$
It's just factorisation. Always works over the complex numbers.
$endgroup$
– Lord Shark the Unknown
Dec 21 '18 at 7:20












$begingroup$
@LordSharktheUnknown So this is not related to testing whether $x=0$ is a root (my current idea)?
$endgroup$
– user7813604
Dec 21 '18 at 7:21




$begingroup$
@LordSharktheUnknown So this is not related to testing whether $x=0$ is a root (my current idea)?
$endgroup$
– user7813604
Dec 21 '18 at 7:21




2




2




$begingroup$
If you have $1-x-x^2=-(x-r_1)(x-r_2)$ then $r_1r_2=1$ and then $1-x-x^2=(1-x/r_2)(1-x/r_1)$.
$endgroup$
– Lord Shark the Unknown
Dec 21 '18 at 7:31




$begingroup$
If you have $1-x-x^2=-(x-r_1)(x-r_2)$ then $r_1r_2=1$ and then $1-x-x^2=(1-x/r_2)(1-x/r_1)$.
$endgroup$
– Lord Shark the Unknown
Dec 21 '18 at 7:31












$begingroup$
@LordSharktheUnknown: Oh so since $r_1r_2=1implies r_1,r_2not=0$ so I can divide it by $r_1cdot r_2$?
$endgroup$
– user7813604
Dec 21 '18 at 7:33




$begingroup$
@LordSharktheUnknown: Oh so since $r_1r_2=1implies r_1,r_2not=0$ so I can divide it by $r_1cdot r_2$?
$endgroup$
– user7813604
Dec 21 '18 at 7:33










4 Answers
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Finally I understand @Lord Shark the Unknown's comment, it's



$$(1-x-x^2)=-(x-r_1)(x-r_2)$$



then solving for $r_1,r_2=dfrac{-1pmsqrt5}{2},$ and since $r_1r_2=-1$, simplify to



$$=dfrac{1}{r_1r_2}(x-r_1)(x-r_2)=(dfrac{x}{r_1}-1)(dfrac{x}{r_2}-1)$$



And since $dfrac{1}{r_1}=-r_2$ and $dfrac{1}{r_2}=-r_1$,



begin{align}&=(-r_2x-1)(-r_1x-1)\
&=(dfrac{1+sqrt5}{2}x-1)(dfrac{1-sqrt5}{2}x-1)end{align}



(But why $1+sqrt5over2$ is so strange...?)






share|cite|improve this answer











$endgroup$





















    1












    $begingroup$

    If you can factor a quadratic trinomial as



    $$(x-alpha')(x-beta')$$ then you can also solve for



    $$(alpha x-1)(beta x-1).$$



    Because



    $$(alpha x-1)(beta x-1)=alphabetaleft(x-frac1alpharight)left(x-frac1betaright)$$



    and the roots in the second factorization are the inverses of those in the first one.



    In other words,



    $$alpha,beta=frac1{alpha'},frac1{beta'}=frac{2a}{-bpmsqrt{b^2-4ac}}=frac{-bpmsqrt{b^2-4ac}}{2c}.$$






    share|cite|improve this answer











    $endgroup$





















      0












      $begingroup$

      Since $1-x-x^2=0$ has two real roots, which are$$alpha=frac{-1+sqrt{5}}{2},beta=frac{-1-sqrt{5}}{2},$$We can denote$$1-x-x^2=-(x-alpha)(x-beta).$$Thus, we may assume $$frac{x}{1-x-x^2}=frac{A}{x-alpha}+frac{B}{x-beta}.tag1$$
      Multiply $x-alpha$ to the both sides of $(1)$ and then let $x=alpha$. We have$$A=frac{alpha}{beta-alpha}.$$Likewise, multiply $x-beta$ to the both sides of $(1)$ and then let $x=beta$. We have$$B=frac{beta}{alpha-beta}.$$As a result,$$frac{x}{1-x-x^2}=frac{alpha}{(beta-alpha)(x-alpha)}+frac{beta}{(alpha-beta)(x-beta)}.$$






      share|cite|improve this answer









      $endgroup$





















        0












        $begingroup$

        Hint: we have $$1-x-x^2=1-(alpha+beta)x+alphabeta x^2$$therefore $$begin{cases}alpha+beta=1\alphabeta=-1end{cases}$$can you finish now?






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          4 Answers
          4






          active

          oldest

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          4 Answers
          4






          active

          oldest

          votes









          active

          oldest

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          active

          oldest

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          1












          $begingroup$

          Finally I understand @Lord Shark the Unknown's comment, it's



          $$(1-x-x^2)=-(x-r_1)(x-r_2)$$



          then solving for $r_1,r_2=dfrac{-1pmsqrt5}{2},$ and since $r_1r_2=-1$, simplify to



          $$=dfrac{1}{r_1r_2}(x-r_1)(x-r_2)=(dfrac{x}{r_1}-1)(dfrac{x}{r_2}-1)$$



          And since $dfrac{1}{r_1}=-r_2$ and $dfrac{1}{r_2}=-r_1$,



          begin{align}&=(-r_2x-1)(-r_1x-1)\
          &=(dfrac{1+sqrt5}{2}x-1)(dfrac{1-sqrt5}{2}x-1)end{align}



          (But why $1+sqrt5over2$ is so strange...?)






          share|cite|improve this answer











          $endgroup$


















            1












            $begingroup$

            Finally I understand @Lord Shark the Unknown's comment, it's



            $$(1-x-x^2)=-(x-r_1)(x-r_2)$$



            then solving for $r_1,r_2=dfrac{-1pmsqrt5}{2},$ and since $r_1r_2=-1$, simplify to



            $$=dfrac{1}{r_1r_2}(x-r_1)(x-r_2)=(dfrac{x}{r_1}-1)(dfrac{x}{r_2}-1)$$



            And since $dfrac{1}{r_1}=-r_2$ and $dfrac{1}{r_2}=-r_1$,



            begin{align}&=(-r_2x-1)(-r_1x-1)\
            &=(dfrac{1+sqrt5}{2}x-1)(dfrac{1-sqrt5}{2}x-1)end{align}



            (But why $1+sqrt5over2$ is so strange...?)






            share|cite|improve this answer











            $endgroup$
















              1












              1








              1





              $begingroup$

              Finally I understand @Lord Shark the Unknown's comment, it's



              $$(1-x-x^2)=-(x-r_1)(x-r_2)$$



              then solving for $r_1,r_2=dfrac{-1pmsqrt5}{2},$ and since $r_1r_2=-1$, simplify to



              $$=dfrac{1}{r_1r_2}(x-r_1)(x-r_2)=(dfrac{x}{r_1}-1)(dfrac{x}{r_2}-1)$$



              And since $dfrac{1}{r_1}=-r_2$ and $dfrac{1}{r_2}=-r_1$,



              begin{align}&=(-r_2x-1)(-r_1x-1)\
              &=(dfrac{1+sqrt5}{2}x-1)(dfrac{1-sqrt5}{2}x-1)end{align}



              (But why $1+sqrt5over2$ is so strange...?)






              share|cite|improve this answer











              $endgroup$



              Finally I understand @Lord Shark the Unknown's comment, it's



              $$(1-x-x^2)=-(x-r_1)(x-r_2)$$



              then solving for $r_1,r_2=dfrac{-1pmsqrt5}{2},$ and since $r_1r_2=-1$, simplify to



              $$=dfrac{1}{r_1r_2}(x-r_1)(x-r_2)=(dfrac{x}{r_1}-1)(dfrac{x}{r_2}-1)$$



              And since $dfrac{1}{r_1}=-r_2$ and $dfrac{1}{r_2}=-r_1$,



              begin{align}&=(-r_2x-1)(-r_1x-1)\
              &=(dfrac{1+sqrt5}{2}x-1)(dfrac{1-sqrt5}{2}x-1)end{align}



              (But why $1+sqrt5over2$ is so strange...?)







              share|cite|improve this answer














              share|cite|improve this answer



              share|cite|improve this answer








              answered Dec 21 '18 at 8:04


























              community wiki





              user7813604
























                  1












                  $begingroup$

                  If you can factor a quadratic trinomial as



                  $$(x-alpha')(x-beta')$$ then you can also solve for



                  $$(alpha x-1)(beta x-1).$$



                  Because



                  $$(alpha x-1)(beta x-1)=alphabetaleft(x-frac1alpharight)left(x-frac1betaright)$$



                  and the roots in the second factorization are the inverses of those in the first one.



                  In other words,



                  $$alpha,beta=frac1{alpha'},frac1{beta'}=frac{2a}{-bpmsqrt{b^2-4ac}}=frac{-bpmsqrt{b^2-4ac}}{2c}.$$






                  share|cite|improve this answer











                  $endgroup$


















                    1












                    $begingroup$

                    If you can factor a quadratic trinomial as



                    $$(x-alpha')(x-beta')$$ then you can also solve for



                    $$(alpha x-1)(beta x-1).$$



                    Because



                    $$(alpha x-1)(beta x-1)=alphabetaleft(x-frac1alpharight)left(x-frac1betaright)$$



                    and the roots in the second factorization are the inverses of those in the first one.



                    In other words,



                    $$alpha,beta=frac1{alpha'},frac1{beta'}=frac{2a}{-bpmsqrt{b^2-4ac}}=frac{-bpmsqrt{b^2-4ac}}{2c}.$$






                    share|cite|improve this answer











                    $endgroup$
















                      1












                      1








                      1





                      $begingroup$

                      If you can factor a quadratic trinomial as



                      $$(x-alpha')(x-beta')$$ then you can also solve for



                      $$(alpha x-1)(beta x-1).$$



                      Because



                      $$(alpha x-1)(beta x-1)=alphabetaleft(x-frac1alpharight)left(x-frac1betaright)$$



                      and the roots in the second factorization are the inverses of those in the first one.



                      In other words,



                      $$alpha,beta=frac1{alpha'},frac1{beta'}=frac{2a}{-bpmsqrt{b^2-4ac}}=frac{-bpmsqrt{b^2-4ac}}{2c}.$$






                      share|cite|improve this answer











                      $endgroup$



                      If you can factor a quadratic trinomial as



                      $$(x-alpha')(x-beta')$$ then you can also solve for



                      $$(alpha x-1)(beta x-1).$$



                      Because



                      $$(alpha x-1)(beta x-1)=alphabetaleft(x-frac1alpharight)left(x-frac1betaright)$$



                      and the roots in the second factorization are the inverses of those in the first one.



                      In other words,



                      $$alpha,beta=frac1{alpha'},frac1{beta'}=frac{2a}{-bpmsqrt{b^2-4ac}}=frac{-bpmsqrt{b^2-4ac}}{2c}.$$







                      share|cite|improve this answer














                      share|cite|improve this answer



                      share|cite|improve this answer








                      edited Dec 21 '18 at 8:29

























                      answered Dec 21 '18 at 8:23









                      Yves DaoustYves Daoust

                      128k674227




                      128k674227























                          0












                          $begingroup$

                          Since $1-x-x^2=0$ has two real roots, which are$$alpha=frac{-1+sqrt{5}}{2},beta=frac{-1-sqrt{5}}{2},$$We can denote$$1-x-x^2=-(x-alpha)(x-beta).$$Thus, we may assume $$frac{x}{1-x-x^2}=frac{A}{x-alpha}+frac{B}{x-beta}.tag1$$
                          Multiply $x-alpha$ to the both sides of $(1)$ and then let $x=alpha$. We have$$A=frac{alpha}{beta-alpha}.$$Likewise, multiply $x-beta$ to the both sides of $(1)$ and then let $x=beta$. We have$$B=frac{beta}{alpha-beta}.$$As a result,$$frac{x}{1-x-x^2}=frac{alpha}{(beta-alpha)(x-alpha)}+frac{beta}{(alpha-beta)(x-beta)}.$$






                          share|cite|improve this answer









                          $endgroup$


















                            0












                            $begingroup$

                            Since $1-x-x^2=0$ has two real roots, which are$$alpha=frac{-1+sqrt{5}}{2},beta=frac{-1-sqrt{5}}{2},$$We can denote$$1-x-x^2=-(x-alpha)(x-beta).$$Thus, we may assume $$frac{x}{1-x-x^2}=frac{A}{x-alpha}+frac{B}{x-beta}.tag1$$
                            Multiply $x-alpha$ to the both sides of $(1)$ and then let $x=alpha$. We have$$A=frac{alpha}{beta-alpha}.$$Likewise, multiply $x-beta$ to the both sides of $(1)$ and then let $x=beta$. We have$$B=frac{beta}{alpha-beta}.$$As a result,$$frac{x}{1-x-x^2}=frac{alpha}{(beta-alpha)(x-alpha)}+frac{beta}{(alpha-beta)(x-beta)}.$$






                            share|cite|improve this answer









                            $endgroup$
















                              0












                              0








                              0





                              $begingroup$

                              Since $1-x-x^2=0$ has two real roots, which are$$alpha=frac{-1+sqrt{5}}{2},beta=frac{-1-sqrt{5}}{2},$$We can denote$$1-x-x^2=-(x-alpha)(x-beta).$$Thus, we may assume $$frac{x}{1-x-x^2}=frac{A}{x-alpha}+frac{B}{x-beta}.tag1$$
                              Multiply $x-alpha$ to the both sides of $(1)$ and then let $x=alpha$. We have$$A=frac{alpha}{beta-alpha}.$$Likewise, multiply $x-beta$ to the both sides of $(1)$ and then let $x=beta$. We have$$B=frac{beta}{alpha-beta}.$$As a result,$$frac{x}{1-x-x^2}=frac{alpha}{(beta-alpha)(x-alpha)}+frac{beta}{(alpha-beta)(x-beta)}.$$






                              share|cite|improve this answer









                              $endgroup$



                              Since $1-x-x^2=0$ has two real roots, which are$$alpha=frac{-1+sqrt{5}}{2},beta=frac{-1-sqrt{5}}{2},$$We can denote$$1-x-x^2=-(x-alpha)(x-beta).$$Thus, we may assume $$frac{x}{1-x-x^2}=frac{A}{x-alpha}+frac{B}{x-beta}.tag1$$
                              Multiply $x-alpha$ to the both sides of $(1)$ and then let $x=alpha$. We have$$A=frac{alpha}{beta-alpha}.$$Likewise, multiply $x-beta$ to the both sides of $(1)$ and then let $x=beta$. We have$$B=frac{beta}{alpha-beta}.$$As a result,$$frac{x}{1-x-x^2}=frac{alpha}{(beta-alpha)(x-alpha)}+frac{beta}{(alpha-beta)(x-beta)}.$$







                              share|cite|improve this answer












                              share|cite|improve this answer



                              share|cite|improve this answer










                              answered Dec 21 '18 at 8:09









                              mengdie1982mengdie1982

                              4,897618




                              4,897618























                                  0












                                  $begingroup$

                                  Hint: we have $$1-x-x^2=1-(alpha+beta)x+alphabeta x^2$$therefore $$begin{cases}alpha+beta=1\alphabeta=-1end{cases}$$can you finish now?






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                                  $endgroup$


















                                    0












                                    $begingroup$

                                    Hint: we have $$1-x-x^2=1-(alpha+beta)x+alphabeta x^2$$therefore $$begin{cases}alpha+beta=1\alphabeta=-1end{cases}$$can you finish now?






                                    share|cite|improve this answer









                                    $endgroup$
















                                      0












                                      0








                                      0





                                      $begingroup$

                                      Hint: we have $$1-x-x^2=1-(alpha+beta)x+alphabeta x^2$$therefore $$begin{cases}alpha+beta=1\alphabeta=-1end{cases}$$can you finish now?






                                      share|cite|improve this answer









                                      $endgroup$



                                      Hint: we have $$1-x-x^2=1-(alpha+beta)x+alphabeta x^2$$therefore $$begin{cases}alpha+beta=1\alphabeta=-1end{cases}$$can you finish now?







                                      share|cite|improve this answer












                                      share|cite|improve this answer



                                      share|cite|improve this answer










                                      answered Dec 21 '18 at 10:30









                                      Mostafa AyazMostafa Ayaz

                                      15.6k3939




                                      15.6k3939






























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