Solve $x^2 - frac{1}{ln(x) - 1} + frac{1}{x^2(ln(x)-1)} = 0$ [closed]












0












$begingroup$


How would I find the roots of this function?



begin{equation}
x^2 - frac{1}{ln(x) - 1} + frac{1}{x^2(ln(x)-1)} = 0
end{equation}



I don't really know how i would tackle this problem because we're working with natural logarithms. Does anyone have an idea?






logarithms roots







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closed as off-topic by RRL, Cesareo, José Carlos Santos, mrtaurho, Leucippus Jan 20 at 1:47


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – RRL, Cesareo, José Carlos Santos, mrtaurho, Leucippus

If this question can be reworded to fit the rules in the help center, please edit the question.





















    0












    $begingroup$


    How would I find the roots of this function?



    begin{equation}
    x^2 - frac{1}{ln(x) - 1} + frac{1}{x^2(ln(x)-1)} = 0
    end{equation}



    I don't really know how i would tackle this problem because we're working with natural logarithms. Does anyone have an idea?






    logarithms roots







    share|cite|improve this question









    $endgroup$



    closed as off-topic by RRL, Cesareo, José Carlos Santos, mrtaurho, Leucippus Jan 20 at 1:47


    This question appears to be off-topic. The users who voted to close gave this specific reason:


    • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – RRL, Cesareo, José Carlos Santos, mrtaurho, Leucippus

    If this question can be reworded to fit the rules in the help center, please edit the question.



















      0












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      0





      $begingroup$


      How would I find the roots of this function?



      begin{equation}
      x^2 - frac{1}{ln(x) - 1} + frac{1}{x^2(ln(x)-1)} = 0
      end{equation}



      I don't really know how i would tackle this problem because we're working with natural logarithms. Does anyone have an idea?






      logarithms roots







      share|cite|improve this question









      $endgroup$




      How would I find the roots of this function?



      begin{equation}
      x^2 - frac{1}{ln(x) - 1} + frac{1}{x^2(ln(x)-1)} = 0
      end{equation}



      I don't really know how i would tackle this problem because we're working with natural logarithms. Does anyone have an idea?






      logarithms roots




      logarithms roots






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      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Jan 18 at 17:08









      ViktorViktor

      16310




      16310




      closed as off-topic by RRL, Cesareo, José Carlos Santos, mrtaurho, Leucippus Jan 20 at 1:47


      This question appears to be off-topic. The users who voted to close gave this specific reason:


      • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – RRL, Cesareo, José Carlos Santos, mrtaurho, Leucippus

      If this question can be reworded to fit the rules in the help center, please edit the question.







      closed as off-topic by RRL, Cesareo, José Carlos Santos, mrtaurho, Leucippus Jan 20 at 1:47


      This question appears to be off-topic. The users who voted to close gave this specific reason:


      • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – RRL, Cesareo, José Carlos Santos, mrtaurho, Leucippus

      If this question can be reworded to fit the rules in the help center, please edit the question.






















          3 Answers
          3






          active

          oldest

          votes


















          1












          $begingroup$

          If you plot the fuction, you see solutions around $x=1$ and $x=3$ from which you can start Newton method.



          You can have approximations using series expansions.



          Around $x=0$,
          $$x^2 - frac{1}{ln(x) - 1} + frac{1}{x^2(ln(x)-1)} = 1+4 (x-1)+Oleft((x-1)^3right)$$ and then the solution is close to
          $$x=frac{3}4$$



          Around $x=3$, the expansion would be
          $$left(9-frac{8}{9 (log (3)-1)}right)+frac{2 (x-3) left(86+81 log ^2(3)-163
          log (3)right)}{27 (log (3)-1)^2}+Oleft((x-3)^2right)$$           and the the solution is close to
          $$x=frac{3 left(83+81 log ^2(3)-156 log (3)right)}{2 left(86+81 log ^2(3)-163
          log (3)right)}approx 3.00039$$



          Edit



          With regard to the first root, you could have better and better rational approximations building the $[1,n]$ Padé approximant. This would give
          $$left(
          begin{array}{ccc}
          n & x_{(n)} & x_{(n)} approx \
          1 & frac{3}{4} & 0.750000 \
          2 & frac{25}{33} & 0.757576 \
          3 & frac{1247}{1643} & 0.758977 \
          4 & frac{5179}{6822} & 0.759162 \
          5 & frac{11947}{15737} & 0.759166 \
          6 & frac{1339361}{1764260} & 0.759163
          end{array}
          right)$$






          share|cite|improve this answer











          $endgroup$





















            1












            $begingroup$

            Hint: By a numeric method we find $$xapprox 0.75916167335311382164$$ or $$xapprox 3.00039175573599420937$$






            share|cite|improve this answer









            $endgroup$









            • 1




              $begingroup$
              I've used WolframAlpha which also gave this sollution, but i was wondering if there was an analytic solution for this because solving this equation was necessary to solve a part of a question on my exam.
              $endgroup$
              – Viktor
              Jan 18 at 17:16






            • 1




              $begingroup$
              Either the person who made the exam made a mistake, or solving the equation explicitly was not required.
              $endgroup$
              – Robert Israel
              Jan 18 at 17:23












            • $begingroup$
              I've asked the full question in a new thread you can find here: math.stackexchange.com/questions/3078569/… the question is about differential equations.
              $endgroup$
              – Viktor
              Jan 18 at 17:56





















            0












            $begingroup$

            I usually do a quick plot like Plot[x^2-1/(Log[x]-1)+1/(x^2(Log[x]-1)),{x,-10,+10}] which shows the above roots quickly and that there are no others.






            share|cite|improve this answer









            $endgroup$




















              3 Answers
              3






              active

              oldest

              votes








              3 Answers
              3






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes









              1












              $begingroup$

              If you plot the fuction, you see solutions around $x=1$ and $x=3$ from which you can start Newton method.



              You can have approximations using series expansions.



              Around $x=0$,
              $$x^2 - frac{1}{ln(x) - 1} + frac{1}{x^2(ln(x)-1)} = 1+4 (x-1)+Oleft((x-1)^3right)$$ and then the solution is close to
              $$x=frac{3}4$$



              Around $x=3$, the expansion would be
              $$left(9-frac{8}{9 (log (3)-1)}right)+frac{2 (x-3) left(86+81 log ^2(3)-163
              log (3)right)}{27 (log (3)-1)^2}+Oleft((x-3)^2right)$$               and the the solution is close to
              $$x=frac{3 left(83+81 log ^2(3)-156 log (3)right)}{2 left(86+81 log ^2(3)-163
              log (3)right)}approx 3.00039$$



              Edit



              With regard to the first root, you could have better and better rational approximations building the $[1,n]$ Padé approximant. This would give
              $$left(
              begin{array}{ccc}
              n & x_{(n)} & x_{(n)} approx \
              1 & frac{3}{4} & 0.750000 \
              2 & frac{25}{33} & 0.757576 \
              3 & frac{1247}{1643} & 0.758977 \
              4 & frac{5179}{6822} & 0.759162 \
              5 & frac{11947}{15737} & 0.759166 \
              6 & frac{1339361}{1764260} & 0.759163
              end{array}
              right)$$






              share|cite|improve this answer











              $endgroup$


















                1












                $begingroup$

                If you plot the fuction, you see solutions around $x=1$ and $x=3$ from which you can start Newton method.



                You can have approximations using series expansions.



                Around $x=0$,
                $$x^2 - frac{1}{ln(x) - 1} + frac{1}{x^2(ln(x)-1)} = 1+4 (x-1)+Oleft((x-1)^3right)$$ and then the solution is close to
                $$x=frac{3}4$$



                Around $x=3$, the expansion would be
                $$left(9-frac{8}{9 (log (3)-1)}right)+frac{2 (x-3) left(86+81 log ^2(3)-163
                log (3)right)}{27 (log (3)-1)^2}+Oleft((x-3)^2right)$$                 and the the solution is close to
                $$x=frac{3 left(83+81 log ^2(3)-156 log (3)right)}{2 left(86+81 log ^2(3)-163
                log (3)right)}approx 3.00039$$



                Edit



                With regard to the first root, you could have better and better rational approximations building the $[1,n]$ Padé approximant. This would give
                $$left(
                begin{array}{ccc}
                n & x_{(n)} & x_{(n)} approx \
                1 & frac{3}{4} & 0.750000 \
                2 & frac{25}{33} & 0.757576 \
                3 & frac{1247}{1643} & 0.758977 \
                4 & frac{5179}{6822} & 0.759162 \
                5 & frac{11947}{15737} & 0.759166 \
                6 & frac{1339361}{1764260} & 0.759163
                end{array}
                right)$$






                share|cite|improve this answer











                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  If you plot the fuction, you see solutions around $x=1$ and $x=3$ from which you can start Newton method.



                  You can have approximations using series expansions.



                  Around $x=0$,
                  $$x^2 - frac{1}{ln(x) - 1} + frac{1}{x^2(ln(x)-1)} = 1+4 (x-1)+Oleft((x-1)^3right)$$ and then the solution is close to
                  $$x=frac{3}4$$



                  Around $x=3$, the expansion would be
                  $$left(9-frac{8}{9 (log (3)-1)}right)+frac{2 (x-3) left(86+81 log ^2(3)-163
                  log (3)right)}{27 (log (3)-1)^2}+Oleft((x-3)^2right)$$                   and the the solution is close to
                  $$x=frac{3 left(83+81 log ^2(3)-156 log (3)right)}{2 left(86+81 log ^2(3)-163
                  log (3)right)}approx 3.00039$$



                  Edit



                  With regard to the first root, you could have better and better rational approximations building the $[1,n]$ Padé approximant. This would give
                  $$left(
                  begin{array}{ccc}
                  n & x_{(n)} & x_{(n)} approx \
                  1 & frac{3}{4} & 0.750000 \
                  2 & frac{25}{33} & 0.757576 \
                  3 & frac{1247}{1643} & 0.758977 \
                  4 & frac{5179}{6822} & 0.759162 \
                  5 & frac{11947}{15737} & 0.759166 \
                  6 & frac{1339361}{1764260} & 0.759163
                  end{array}
                  right)$$






                  share|cite|improve this answer











                  $endgroup$



                  If you plot the fuction, you see solutions around $x=1$ and $x=3$ from which you can start Newton method.



                  You can have approximations using series expansions.



                  Around $x=0$,
                  $$x^2 - frac{1}{ln(x) - 1} + frac{1}{x^2(ln(x)-1)} = 1+4 (x-1)+Oleft((x-1)^3right)$$ and then the solution is close to
                  $$x=frac{3}4$$



                  Around $x=3$, the expansion would be
                  $$left(9-frac{8}{9 (log (3)-1)}right)+frac{2 (x-3) left(86+81 log ^2(3)-163
                  log (3)right)}{27 (log (3)-1)^2}+Oleft((x-3)^2right)$$                   and the the solution is close to
                  $$x=frac{3 left(83+81 log ^2(3)-156 log (3)right)}{2 left(86+81 log ^2(3)-163
                  log (3)right)}approx 3.00039$$



                  Edit



                  With regard to the first root, you could have better and better rational approximations building the $[1,n]$ Padé approximant. This would give
                  $$left(
                  begin{array}{ccc}
                  n & x_{(n)} & x_{(n)} approx \
                  1 & frac{3}{4} & 0.750000 \
                  2 & frac{25}{33} & 0.757576 \
                  3 & frac{1247}{1643} & 0.758977 \
                  4 & frac{5179}{6822} & 0.759162 \
                  5 & frac{11947}{15737} & 0.759166 \
                  6 & frac{1339361}{1764260} & 0.759163
                  end{array}
                  right)$$







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Jan 19 at 8:40

























                  answered Jan 19 at 7:21









                  Claude LeiboviciClaude Leibovici

                  126k1158135




                  126k1158135























                      1












                      $begingroup$

                      Hint: By a numeric method we find $$xapprox 0.75916167335311382164$$ or $$xapprox 3.00039175573599420937$$






                      share|cite|improve this answer









                      $endgroup$









                      • 1




                        $begingroup$
                        I've used WolframAlpha which also gave this sollution, but i was wondering if there was an analytic solution for this because solving this equation was necessary to solve a part of a question on my exam.
                        $endgroup$
                        – Viktor
                        Jan 18 at 17:16






                      • 1




                        $begingroup$
                        Either the person who made the exam made a mistake, or solving the equation explicitly was not required.
                        $endgroup$
                        – Robert Israel
                        Jan 18 at 17:23












                      • $begingroup$
                        I've asked the full question in a new thread you can find here: math.stackexchange.com/questions/3078569/… the question is about differential equations.
                        $endgroup$
                        – Viktor
                        Jan 18 at 17:56


















                      1












                      $begingroup$

                      Hint: By a numeric method we find $$xapprox 0.75916167335311382164$$ or $$xapprox 3.00039175573599420937$$






                      share|cite|improve this answer









                      $endgroup$









                      • 1




                        $begingroup$
                        I've used WolframAlpha which also gave this sollution, but i was wondering if there was an analytic solution for this because solving this equation was necessary to solve a part of a question on my exam.
                        $endgroup$
                        – Viktor
                        Jan 18 at 17:16






                      • 1




                        $begingroup$
                        Either the person who made the exam made a mistake, or solving the equation explicitly was not required.
                        $endgroup$
                        – Robert Israel
                        Jan 18 at 17:23












                      • $begingroup$
                        I've asked the full question in a new thread you can find here: math.stackexchange.com/questions/3078569/… the question is about differential equations.
                        $endgroup$
                        – Viktor
                        Jan 18 at 17:56
















                      1












                      1








                      1





                      $begingroup$

                      Hint: By a numeric method we find $$xapprox 0.75916167335311382164$$ or $$xapprox 3.00039175573599420937$$






                      share|cite|improve this answer









                      $endgroup$



                      Hint: By a numeric method we find $$xapprox 0.75916167335311382164$$ or $$xapprox 3.00039175573599420937$$







                      share|cite|improve this answer












                      share|cite|improve this answer



                      share|cite|improve this answer










                      answered Jan 18 at 17:13









                      Dr. Sonnhard GraubnerDr. Sonnhard Graubner

                      79.2k42867




                      79.2k42867








                      • 1




                        $begingroup$
                        I've used WolframAlpha which also gave this sollution, but i was wondering if there was an analytic solution for this because solving this equation was necessary to solve a part of a question on my exam.
                        $endgroup$
                        – Viktor
                        Jan 18 at 17:16






                      • 1




                        $begingroup$
                        Either the person who made the exam made a mistake, or solving the equation explicitly was not required.
                        $endgroup$
                        – Robert Israel
                        Jan 18 at 17:23












                      • $begingroup$
                        I've asked the full question in a new thread you can find here: math.stackexchange.com/questions/3078569/… the question is about differential equations.
                        $endgroup$
                        – Viktor
                        Jan 18 at 17:56
















                      • 1




                        $begingroup$
                        I've used WolframAlpha which also gave this sollution, but i was wondering if there was an analytic solution for this because solving this equation was necessary to solve a part of a question on my exam.
                        $endgroup$
                        – Viktor
                        Jan 18 at 17:16






                      • 1




                        $begingroup$
                        Either the person who made the exam made a mistake, or solving the equation explicitly was not required.
                        $endgroup$
                        – Robert Israel
                        Jan 18 at 17:23












                      • $begingroup$
                        I've asked the full question in a new thread you can find here: math.stackexchange.com/questions/3078569/… the question is about differential equations.
                        $endgroup$
                        – Viktor
                        Jan 18 at 17:56










                      1




                      1




                      $begingroup$
                      I've used WolframAlpha which also gave this sollution, but i was wondering if there was an analytic solution for this because solving this equation was necessary to solve a part of a question on my exam.
                      $endgroup$
                      – Viktor
                      Jan 18 at 17:16




                      $begingroup$
                      I've used WolframAlpha which also gave this sollution, but i was wondering if there was an analytic solution for this because solving this equation was necessary to solve a part of a question on my exam.
                      $endgroup$
                      – Viktor
                      Jan 18 at 17:16




                      1




                      1




                      $begingroup$
                      Either the person who made the exam made a mistake, or solving the equation explicitly was not required.
                      $endgroup$
                      – Robert Israel
                      Jan 18 at 17:23






                      $begingroup$
                      Either the person who made the exam made a mistake, or solving the equation explicitly was not required.
                      $endgroup$
                      – Robert Israel
                      Jan 18 at 17:23














                      $begingroup$
                      I've asked the full question in a new thread you can find here: math.stackexchange.com/questions/3078569/… the question is about differential equations.
                      $endgroup$
                      – Viktor
                      Jan 18 at 17:56






                      $begingroup$
                      I've asked the full question in a new thread you can find here: math.stackexchange.com/questions/3078569/… the question is about differential equations.
                      $endgroup$
                      – Viktor
                      Jan 18 at 17:56













                      0












                      $begingroup$

                      I usually do a quick plot like Plot[x^2-1/(Log[x]-1)+1/(x^2(Log[x]-1)),{x,-10,+10}] which shows the above roots quickly and that there are no others.






                      share|cite|improve this answer









                      $endgroup$


















                        0












                        $begingroup$

                        I usually do a quick plot like Plot[x^2-1/(Log[x]-1)+1/(x^2(Log[x]-1)),{x,-10,+10}] which shows the above roots quickly and that there are no others.






                        share|cite|improve this answer









                        $endgroup$
















                          0












                          0








                          0





                          $begingroup$

                          I usually do a quick plot like Plot[x^2-1/(Log[x]-1)+1/(x^2(Log[x]-1)),{x,-10,+10}] which shows the above roots quickly and that there are no others.






                          share|cite|improve this answer









                          $endgroup$



                          I usually do a quick plot like Plot[x^2-1/(Log[x]-1)+1/(x^2(Log[x]-1)),{x,-10,+10}] which shows the above roots quickly and that there are no others.







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Jan 18 at 19:28









                          Lorenz H MenkeLorenz H Menke

                          9611




                          9611















                              -

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