Multiple solutions of partial fractions decomposition

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I'm learning about series, and there this bit about partial fractions decomposition that I want to ask.



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Am I right to assume that, there are multiple way to achieve the form $fracA1 - alpha_1x+fracB1 - alpha_2x$, i.e. there might be several solutions to $(1 - alpha_1x)(1 - alpha_2x) = 1 - 3x +x^2$, and hence to $A$ and $B$, but we choose the one most suited to simplify $H(x)$?



If this is true, is there some observation that can make choosing the right set of $alpha_1,alpha_2,A,B$ easier?




partial-fractions




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    up vote
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    down vote

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    I'm learning about series, and there this bit about partial fractions decomposition that I want to ask.



    enter image description here



    Am I right to assume that, there are multiple way to achieve the form $fracA1 - alpha_1x+fracB1 - alpha_2x$, i.e. there might be several solutions to $(1 - alpha_1x)(1 - alpha_2x) = 1 - 3x +x^2$, and hence to $A$ and $B$, but we choose the one most suited to simplify $H(x)$?



    If this is true, is there some observation that can make choosing the right set of $alpha_1,alpha_2,A,B$ easier?




    partial-fractions




    share|cite|improve this question





















      up vote
      1
      down vote

      favorite









      up vote
      1
      down vote

      favorite











      I'm learning about series, and there this bit about partial fractions decomposition that I want to ask.



      enter image description here



      Am I right to assume that, there are multiple way to achieve the form $fracA1 - alpha_1x+fracB1 - alpha_2x$, i.e. there might be several solutions to $(1 - alpha_1x)(1 - alpha_2x) = 1 - 3x +x^2$, and hence to $A$ and $B$, but we choose the one most suited to simplify $H(x)$?



      If this is true, is there some observation that can make choosing the right set of $alpha_1,alpha_2,A,B$ easier?




      partial-fractions




      share|cite|improve this question











      I'm learning about series, and there this bit about partial fractions decomposition that I want to ask.



      enter image description here



      Am I right to assume that, there are multiple way to achieve the form $fracA1 - alpha_1x+fracB1 - alpha_2x$, i.e. there might be several solutions to $(1 - alpha_1x)(1 - alpha_2x) = 1 - 3x +x^2$, and hence to $A$ and $B$, but we choose the one most suited to simplify $H(x)$?



      If this is true, is there some observation that can make choosing the right set of $alpha_1,alpha_2,A,B$ easier?





      partial-fractions





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      share|cite|improve this question




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      asked Jul 22 at 10:24









      Josh Ng

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          Your claim is not true. The partial fraction decomposition of any rational function is unique up to term order, as proven (for example) here, so there will always be only one possible solution for the parameters involved, again up to term order.






          share|cite|improve this answer





















          • I've just realized that the "different solutions" I got were in fact the same, but expressed in different forms (though clearly some form is better suited to deduce $h_n$ than other). But thanks for the recommended reading!
            – Josh Ng
            Jul 22 at 10:50










          Your Answer




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          1 Answer
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          active

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          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes








          up vote
          2
          down vote



          accepted










          Your claim is not true. The partial fraction decomposition of any rational function is unique up to term order, as proven (for example) here, so there will always be only one possible solution for the parameters involved, again up to term order.






          share|cite|improve this answer





















          • I've just realized that the "different solutions" I got were in fact the same, but expressed in different forms (though clearly some form is better suited to deduce $h_n$ than other). But thanks for the recommended reading!
            – Josh Ng
            Jul 22 at 10:50














          up vote
          2
          down vote



          accepted










          Your claim is not true. The partial fraction decomposition of any rational function is unique up to term order, as proven (for example) here, so there will always be only one possible solution for the parameters involved, again up to term order.






          share|cite|improve this answer





















          • I've just realized that the "different solutions" I got were in fact the same, but expressed in different forms (though clearly some form is better suited to deduce $h_n$ than other). But thanks for the recommended reading!
            – Josh Ng
            Jul 22 at 10:50












          up vote
          2
          down vote



          accepted







          up vote
          2
          down vote



          accepted






          Your claim is not true. The partial fraction decomposition of any rational function is unique up to term order, as proven (for example) here, so there will always be only one possible solution for the parameters involved, again up to term order.






          share|cite|improve this answer













          Your claim is not true. The partial fraction decomposition of any rational function is unique up to term order, as proven (for example) here, so there will always be only one possible solution for the parameters involved, again up to term order.







          share|cite|improve this answer













          share|cite|improve this answer



          share|cite|improve this answer











          answered Jul 22 at 10:36









          Parcly Taxel

          33.6k136588




          33.6k136588











          • I've just realized that the "different solutions" I got were in fact the same, but expressed in different forms (though clearly some form is better suited to deduce $h_n$ than other). But thanks for the recommended reading!
            – Josh Ng
            Jul 22 at 10:50
















          • I've just realized that the "different solutions" I got were in fact the same, but expressed in different forms (though clearly some form is better suited to deduce $h_n$ than other). But thanks for the recommended reading!
            – Josh Ng
            Jul 22 at 10:50















          I've just realized that the "different solutions" I got were in fact the same, but expressed in different forms (though clearly some form is better suited to deduce $h_n$ than other). But thanks for the recommended reading!
          – Josh Ng
          Jul 22 at 10:50




          I've just realized that the "different solutions" I got were in fact the same, but expressed in different forms (though clearly some form is better suited to deduce $h_n$ than other). But thanks for the recommended reading!
          – Josh Ng
          Jul 22 at 10:50












           

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