Solution to $x_i=a(1+sum_{ine j}frac{b_ix_j}{1+b_ix_j})$












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I'm dealing with a set of variables $x_i$ and following equations:



$forall i$ $x_i=a(1+sum_{ine j}frac{b_ix_j}{1+b_ix_j})$



Wolfram mathematica's NSolve could help, but it's not very accurate and is very limited in capacity for the number of $x_i$s, like 8.



I'm aiming for 20-100 $x_i$s so, therefore, looking for other ways like an algorithm to find $x_i$s. Any suggestion?










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    0














    I'm dealing with a set of variables $x_i$ and following equations:



    $forall i$ $x_i=a(1+sum_{ine j}frac{b_ix_j}{1+b_ix_j})$



    Wolfram mathematica's NSolve could help, but it's not very accurate and is very limited in capacity for the number of $x_i$s, like 8.



    I'm aiming for 20-100 $x_i$s so, therefore, looking for other ways like an algorithm to find $x_i$s. Any suggestion?










    share|cite|improve this question



























      0












      0








      0







      I'm dealing with a set of variables $x_i$ and following equations:



      $forall i$ $x_i=a(1+sum_{ine j}frac{b_ix_j}{1+b_ix_j})$



      Wolfram mathematica's NSolve could help, but it's not very accurate and is very limited in capacity for the number of $x_i$s, like 8.



      I'm aiming for 20-100 $x_i$s so, therefore, looking for other ways like an algorithm to find $x_i$s. Any suggestion?










      share|cite|improve this question















      I'm dealing with a set of variables $x_i$ and following equations:



      $forall i$ $x_i=a(1+sum_{ine j}frac{b_ix_j}{1+b_ix_j})$



      Wolfram mathematica's NSolve could help, but it's not very accurate and is very limited in capacity for the number of $x_i$s, like 8.



      I'm aiming for 20-100 $x_i$s so, therefore, looking for other ways like an algorithm to find $x_i$s. Any suggestion?







      algorithms






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      share|cite|improve this question













      share|cite|improve this question




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      edited Dec 1 at 11:09

























      asked Dec 1 at 4:22









      Anais

      286




      286






















          1 Answer
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          3














          Adding and subtracting $dfrac{x_i}{1+x_i}$ inside the parenthesis gives



          $$x_i = a(1 + sum_{j}frac{x_j}{1+x_j} - frac{x_i}{1+x_i})$$



          The sum does not depend on $i$; it is a constant. Can you continue?






          share|cite|improve this answer

















          • 1




            Sum is not a constant and depends on all $x_i$s.
            – Anais
            Dec 1 at 5:34










          • @Parisa It is a constant in a sense that it is the same in all equations in the system. Solve any for $x_i$, and calibrate to satisfy the constant. Another look at the problem is to assume that all $x_i$ are equal.
            – user58697
            Dec 1 at 5:39












          • My mistake, I attempted to make the problem simpler by omitting some elements. Will change it to the original which cannot have identical $x_i$s
            – Anais
            Dec 1 at 5:47











          Your Answer





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          1 Answer
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          1 Answer
          1






          active

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          active

          oldest

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          active

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          3














          Adding and subtracting $dfrac{x_i}{1+x_i}$ inside the parenthesis gives



          $$x_i = a(1 + sum_{j}frac{x_j}{1+x_j} - frac{x_i}{1+x_i})$$



          The sum does not depend on $i$; it is a constant. Can you continue?






          share|cite|improve this answer

















          • 1




            Sum is not a constant and depends on all $x_i$s.
            – Anais
            Dec 1 at 5:34










          • @Parisa It is a constant in a sense that it is the same in all equations in the system. Solve any for $x_i$, and calibrate to satisfy the constant. Another look at the problem is to assume that all $x_i$ are equal.
            – user58697
            Dec 1 at 5:39












          • My mistake, I attempted to make the problem simpler by omitting some elements. Will change it to the original which cannot have identical $x_i$s
            – Anais
            Dec 1 at 5:47
















          3














          Adding and subtracting $dfrac{x_i}{1+x_i}$ inside the parenthesis gives



          $$x_i = a(1 + sum_{j}frac{x_j}{1+x_j} - frac{x_i}{1+x_i})$$



          The sum does not depend on $i$; it is a constant. Can you continue?






          share|cite|improve this answer

















          • 1




            Sum is not a constant and depends on all $x_i$s.
            – Anais
            Dec 1 at 5:34










          • @Parisa It is a constant in a sense that it is the same in all equations in the system. Solve any for $x_i$, and calibrate to satisfy the constant. Another look at the problem is to assume that all $x_i$ are equal.
            – user58697
            Dec 1 at 5:39












          • My mistake, I attempted to make the problem simpler by omitting some elements. Will change it to the original which cannot have identical $x_i$s
            – Anais
            Dec 1 at 5:47














          3












          3








          3






          Adding and subtracting $dfrac{x_i}{1+x_i}$ inside the parenthesis gives



          $$x_i = a(1 + sum_{j}frac{x_j}{1+x_j} - frac{x_i}{1+x_i})$$



          The sum does not depend on $i$; it is a constant. Can you continue?






          share|cite|improve this answer












          Adding and subtracting $dfrac{x_i}{1+x_i}$ inside the parenthesis gives



          $$x_i = a(1 + sum_{j}frac{x_j}{1+x_j} - frac{x_i}{1+x_i})$$



          The sum does not depend on $i$; it is a constant. Can you continue?







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 1 at 5:22









          user58697

          1,829512




          1,829512








          • 1




            Sum is not a constant and depends on all $x_i$s.
            – Anais
            Dec 1 at 5:34










          • @Parisa It is a constant in a sense that it is the same in all equations in the system. Solve any for $x_i$, and calibrate to satisfy the constant. Another look at the problem is to assume that all $x_i$ are equal.
            – user58697
            Dec 1 at 5:39












          • My mistake, I attempted to make the problem simpler by omitting some elements. Will change it to the original which cannot have identical $x_i$s
            – Anais
            Dec 1 at 5:47














          • 1




            Sum is not a constant and depends on all $x_i$s.
            – Anais
            Dec 1 at 5:34










          • @Parisa It is a constant in a sense that it is the same in all equations in the system. Solve any for $x_i$, and calibrate to satisfy the constant. Another look at the problem is to assume that all $x_i$ are equal.
            – user58697
            Dec 1 at 5:39












          • My mistake, I attempted to make the problem simpler by omitting some elements. Will change it to the original which cannot have identical $x_i$s
            – Anais
            Dec 1 at 5:47








          1




          1




          Sum is not a constant and depends on all $x_i$s.
          – Anais
          Dec 1 at 5:34




          Sum is not a constant and depends on all $x_i$s.
          – Anais
          Dec 1 at 5:34












          @Parisa It is a constant in a sense that it is the same in all equations in the system. Solve any for $x_i$, and calibrate to satisfy the constant. Another look at the problem is to assume that all $x_i$ are equal.
          – user58697
          Dec 1 at 5:39






          @Parisa It is a constant in a sense that it is the same in all equations in the system. Solve any for $x_i$, and calibrate to satisfy the constant. Another look at the problem is to assume that all $x_i$ are equal.
          – user58697
          Dec 1 at 5:39














          My mistake, I attempted to make the problem simpler by omitting some elements. Will change it to the original which cannot have identical $x_i$s
          – Anais
          Dec 1 at 5:47




          My mistake, I attempted to make the problem simpler by omitting some elements. Will change it to the original which cannot have identical $x_i$s
          – Anais
          Dec 1 at 5:47


















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