How to re-write one fraction as two others.I have the two following fractions. A B x α + 1 and C D x α + β The form i want E F x α + β + 1 I was thinking to do partial fractions or potentially doing a Laplace or Fourier transform, then rearranging and transforming back. How could one possibly get the two fractions into one fraction of this form. A,B,C,D,E and F can be number or possible a factorial if we go down the laplace transform route, as can α,β

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How to re-write one fraction as two others.I have the two following fractions.            A              B                  x                      α            +            1                              and            C              D                  x                      α            +            β                              The form i want            E              F                  x                      α            +            β            +            1                              I was thinking to do partial fractions or potentially doing a Laplace or Fourier transform, then rearranging and transforming back. How could one possibly get the two fractions into one fraction of this form. A,B,C,D,E and F can be number or possible a factorial if we go down the laplace transform route, as can α,β

taweirrvb · Accepted Answer

I&#039;m assuming, as you seem to have implied, that    A  ,  B  ,  C  ,  D  ,  E  ,  F are arbitrary, and, in particular, that E and F can be defined as some combination of   A  ,  B  ,  C  ,  D      A          B              x                  α          +          1                      =      A          B              x                  α          +          1                      ⋅            x      β              x      β        =            A              x        β                    B              x                  α          +          1                            x        β              =            A              x        β                    B              x                  α          +          β          +          1                    And, similarly,      C          D              x                  α          +          β                      =            C      x              D              x                  α          +          β          +          1                    So you can have            A              x        β                    B              x                  α          +          β          +          1                      −            C      x              D              x                  α          +          β          +          1                      ⋅      x          β      −      1        =            A              x        β                    B              x                  α          +          β          +          1                      −            C              x                  β                            D              x                  α          +          β          +          1                      =            A      D      −      B      C              B      D              x                  α          +          β          +          1                    and thus, we can let   E  =  A  D  −  B  C and   F  =  B  D to get      E          F              x                  α          +          β          +          1                      =      (          A              B                  x                      α            +            1                                )    −      (          C              D                  x                      α            +            β                                )        x          β      −      1      Edit:This also shows that you cannot get appropriate values of E,F from a linear combination. Suppose, for the sake of contradiction, that there are constants M,N such that  M      (          A              B                  x                      α            +            1                                )    +  N      (          C              D                  x                      α            +            β                                )    =            E      1                      F        1                    x                  α          +          β          +          1                    for some        E    1    ,      F    1   Then we have            E              /                    E        1                    F              /                    F        1                  (    M          (              A                  B                      x                          α              +              1                                          )        +    N          (              C                  D                      x                          α              +              β                                          )        )    =            E              /                    E        1                    F              /                    F        1                        E      1                      F        1                    x                  α          +          β          +          1                      =      E          F              x                  α          +          β          +          1                    For simplicity, let   ϵ  =            E              /                    E        1                    F              /                    F        1            . Then,                     ϵ        M                  (                      A                          B                              x                                  α                  +                  1                                                              )                +        ϵ        N                  (                      C                          D                              x                                  α                  +                  β                                                              )                                    =                  (                      A                          B                              x                                  α                  +                  1                                                              )                −                  (                      C                          D                              x                                  α                  +                  β                                                              )                          x                      β            −            1                                              ϵ        M                  (                      A                          B                              x                                  α                  +                  1                                                              )                                    =                  (                      A                          B                              x                                  α                  +                  1                                                              )                −                  (                      C                          D                              x                                  α                  +                  β                                                              )                          (                      x                          β              −              1                                +          ϵ          N          )                                    ϵ        M                            =        1        −                  (                                    B              C                                      A              D                              x                                  β                  −                  1                                                              )                          (                      x                          β              −              1                                +          ϵ          N          )                                    ϵ        M                            =        1        −                              ϵ            B            C            N                                A            D                          x                              β                −                1                                                                        M                            =                  1          ϵ                −                              B            C            N                                A            D                          x                              β                −                1                                                        And this contradicts the assumption that M is constant.

Spencer Lutz · Accepted Answer

Assuming that I properly understood (which is not sure, I must confess) : let  P  =      A          B              x                  α          +          1                      Q  =      C          D              x                  α          +          β                    So      P                  1                  α          +          1                      =            (            A    B                      )                            1                  α          +          1                          1    x        P                  1                  α          +          1                      Q  =            (            A    B                      )                            1                  α          +          1                                (            C    D              )            1          x              α        +        β        +        1              =            E              F                  x                      α            +            β            +            1                              So            E      F        =            (            A    B                      )                            1                  α          +          1                                (            C    D              )      You can choose any value you want for E or F and compute the other one from the above equation.

How to re-write one fraction as two others. I have the two following fractions. <mstyle displa

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