I want to decompose the rational function P ( s ) Q ( s ) = ∏ i = 1 m ( s + a i ) ∏ i = 1 n ( s + b i ) where a i &gt; 0 for every i = 1 , … , m b i &gt; 0 for every i = 1 , … , n and n &gt; m.In other words, I'm looking for coefficients x j , j = 1 … , n such that P ( s ) Q ( s ) = x 1 s + b 1 + ⋯ + x n s + b n ⟺ ∑ j = 1 n x j ∏ i = 1 i ≠ j n ( s + b i ) = ∏ i = 1 m ( s + a i )Making some attempts with Mathematica for low values of m and n it seems that the x j are given by x j = ∏ i = 1 m ( a i − b j ) ∏ i = 1 i ≠ j n ( b i − b j ) So my questions are:1) Can I say beforehand that there exist unique such x j s?2) How can I prove that x j has in general (as it seems) the form above?

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I want to decompose the rational function                                                  P          (          s          )                                                                  Q          (          s          )                      =                                                              ∏                          i              =              1                        m                    (          s          +                      a            i                    )                                                                              ∏                          i              =              1                        n                    (          s          +                      b            i                    )                    where       a    i    &amp;gt;  0 for every   i  =  1  ,  …  ,  m       b    i    &amp;gt;  0 for every   i  =  1  ,  …  ,  n and   n  &amp;gt;  m.In other words, I&#039;m looking for coefficients       x    j  ,   j  =  1  …  ,  n such that                                                  P          (          s          )                                                                  Q          (          s          )                      =                                                              x            1                                                                            s          +                      b            1                                +  ⋯  +                                                              x            n                                                                            s          +                      b            n                                  ⟺        ∑          j      =      1        n        x    j        ∏                                                      i              =              1                                                          i              ≠              j                                            n    (  s  +      b    i    )  =      ∏          i      =      1        m    (  s  +      a    i    )Making some attempts with Mathematica for low values of   m and   n it seems that the       x    j   are given by      x    j    =                                                              ∏                          i              =              1                        m                    (                      a            i                    −                      b            j                    )                                                                              ∏                                                                                                      i                      =                      1                                                                                                  i                      ≠                      j                                                                                            n                    (                      b            i                    −                      b            j                    )                    So my questions are:1) Can I say beforehand that there exist unique such       x    j  s?2) How can I prove that       x    j   has in general (as it seems) the form above?

aletantas1x · Accepted Answer

These are real (or complex) polynomials? This is known as &quot;partial fractions&quot;, perhaps first seen by the typical student in elementary calculus courses.(1) Suposing that your   P  (  s  )      /    Q  (  s  ) in lowest terms, the necessary and sufficient condition that there exists       x    1    ,  …  ,      x    n   such that                                                  P          (          s          )                                                                  Q          (          s          )                      =                                                              x            1                                                                            s          +                      b            1                                +  ⋯  +                                                              x            n                                                                            s          +                      b            n                              Is:   m  &amp;lt;  n and no repeats among       b    1    ,      b    2    ,  …  ,      b    n  .(2) If such       x    j   exist, they are unique.Let&#039;s show       x    1   is unique. Suppose                                                              x            1                                                                            s          +                      b            1                                +  ⋯  +                                                              x            n                                                                            s          +                      b            n                                =                                                              u            1                                                                            s          +                      b            1                                +  ⋯  +                                                              u            n                                                                            s          +                      b            n                                .Multiply by   s  +      b    1  , then take the limit as s→−b1. Result:       x    1    =      u    1  . Uniqueness for the others is the same.(3) They have the form you describe.Let&#039;s prove it for       x    1  . We have                                                              x            1                                                                            s          +                      b            1                                +  ⋯  +                                                              x            n                                                                            s          +                      b            n                                =                                                              ∏                          i              =              1                        m                    (          s          +                      a            i                    )                                                                              ∏                          i              =              1                        n                    (          s          +                      b            i                    )                    Multiply by   s  +      b    1   to get      x    1    +  (  s  +      b    1    )      (                                                                                x              2                                                                                                  s            +                          b              2                                            +    ⋯    +                                                                                x              n                                                                                                  s            +                          b              n                                            )    =                                                              ∏                          i              =              1                        m                    (          s          +                      a            i                    )                                                                              ∏                          i              =              2                        n                    (          s          +                      b            i                    )                    Take the limit as   s  →  −      b    1   [remembering that       b    j    ≠      b    1   for all other       b    j   ]      x    1    +  0  =                                                              ∏                          i              =              1                        m                    (          −                      b            1                    +                      a            i                    )                                                                              ∏                          i              =              2                        n                    (          −                      b            1                    +                      b            i                    )                    Does that match your claimed formula?

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