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












0












$begingroup$


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







share|cite|improve this question











$endgroup$

















    0












    $begingroup$


    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







    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      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







      share|cite|improve this question











      $endgroup$




      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




      algorithms






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 1 '18 at 11:09







      Anais

















      asked Dec 1 '18 at 4:22









      AnaisAnais

      336




      336






















          1 Answer
          1






          active

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          3












          $begingroup$

          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









          $endgroup$









          • 1




            $begingroup$
            Sum is not a constant and depends on all $x_i$s.
            $endgroup$
            – Anais
            Dec 1 '18 at 5:34










          • $begingroup$
            @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.
            $endgroup$
            – user58697
            Dec 1 '18 at 5:39












          • $begingroup$
            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
            $endgroup$
            – Anais
            Dec 1 '18 at 5:47











          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









          3












          $begingroup$

          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









          $endgroup$









          • 1




            $begingroup$
            Sum is not a constant and depends on all $x_i$s.
            $endgroup$
            – Anais
            Dec 1 '18 at 5:34










          • $begingroup$
            @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.
            $endgroup$
            – user58697
            Dec 1 '18 at 5:39












          • $begingroup$
            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
            $endgroup$
            – Anais
            Dec 1 '18 at 5:47
















          3












          $begingroup$

          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









          $endgroup$









          • 1




            $begingroup$
            Sum is not a constant and depends on all $x_i$s.
            $endgroup$
            – Anais
            Dec 1 '18 at 5:34










          • $begingroup$
            @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.
            $endgroup$
            – user58697
            Dec 1 '18 at 5:39












          • $begingroup$
            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
            $endgroup$
            – Anais
            Dec 1 '18 at 5:47














          3












          3








          3





          $begingroup$

          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









          $endgroup$



          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 '18 at 5:22









          user58697user58697

          1,829512




          1,829512








          • 1




            $begingroup$
            Sum is not a constant and depends on all $x_i$s.
            $endgroup$
            – Anais
            Dec 1 '18 at 5:34










          • $begingroup$
            @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.
            $endgroup$
            – user58697
            Dec 1 '18 at 5:39












          • $begingroup$
            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
            $endgroup$
            – Anais
            Dec 1 '18 at 5:47














          • 1




            $begingroup$
            Sum is not a constant and depends on all $x_i$s.
            $endgroup$
            – Anais
            Dec 1 '18 at 5:34










          • $begingroup$
            @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.
            $endgroup$
            – user58697
            Dec 1 '18 at 5:39












          • $begingroup$
            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
            $endgroup$
            – Anais
            Dec 1 '18 at 5:47








          1




          1




          $begingroup$
          Sum is not a constant and depends on all $x_i$s.
          $endgroup$
          – Anais
          Dec 1 '18 at 5:34




          $begingroup$
          Sum is not a constant and depends on all $x_i$s.
          $endgroup$
          – Anais
          Dec 1 '18 at 5:34












          $begingroup$
          @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.
          $endgroup$
          – user58697
          Dec 1 '18 at 5:39






          $begingroup$
          @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.
          $endgroup$
          – user58697
          Dec 1 '18 at 5:39














          $begingroup$
          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
          $endgroup$
          – Anais
          Dec 1 '18 at 5:47




          $begingroup$
          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
          $endgroup$
          – Anais
          Dec 1 '18 at 5:47


















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