Solving a nonpolynomial equation












0












$begingroup$


How can we solve the following equation for $x$ without a calculator/computer? ($m$ and $n$ are positive integers)



$m^{n}(1-frac{x}{m})(1-frac{x}{m-1})(1-frac{x}{m-2})=1.$



I tried to estimate $1-frac{x}{m}$ by $e^{-frac{x}{m}}$ and do the same for the rest, but I got stuck.






algebra-precalculus numerical-methods







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$endgroup$








  • 2




    $begingroup$
    Is $n$ equal to the number of factors? That is, are there other than this 3-factor instance? Or in other words, what is $n$, is it arbitrary?
    $endgroup$
    – LutzL
    Jan 9 at 19:42












  • $begingroup$
    You have added the tag numerical-methods. Do you really need a formula for the exact solution or will you settle for a reliable algorithm for computing the possible values of $x$ as function of $m$ and $n$? It seems to me $x in {m-1,m-2,m}$ are good approximations of the roots when $n$ is large and the right hand side of $1$ is relatively insignificant. I expect the case of small $n$ to be handled by solving a sequence of problems starting with a large value of $n$.
    $endgroup$
    – Carl Christian
    Jan 10 at 10:04


















0












$begingroup$


How can we solve the following equation for $x$ without a calculator/computer? ($m$ and $n$ are positive integers)



$m^{n}(1-frac{x}{m})(1-frac{x}{m-1})(1-frac{x}{m-2})=1.$



I tried to estimate $1-frac{x}{m}$ by $e^{-frac{x}{m}}$ and do the same for the rest, but I got stuck.






algebra-precalculus numerical-methods







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    Is $n$ equal to the number of factors? That is, are there other than this 3-factor instance? Or in other words, what is $n$, is it arbitrary?
    $endgroup$
    – LutzL
    Jan 9 at 19:42












  • $begingroup$
    You have added the tag numerical-methods. Do you really need a formula for the exact solution or will you settle for a reliable algorithm for computing the possible values of $x$ as function of $m$ and $n$? It seems to me $x in {m-1,m-2,m}$ are good approximations of the roots when $n$ is large and the right hand side of $1$ is relatively insignificant. I expect the case of small $n$ to be handled by solving a sequence of problems starting with a large value of $n$.
    $endgroup$
    – Carl Christian
    Jan 10 at 10:04
















0












0








0





$begingroup$


How can we solve the following equation for $x$ without a calculator/computer? ($m$ and $n$ are positive integers)



$m^{n}(1-frac{x}{m})(1-frac{x}{m-1})(1-frac{x}{m-2})=1.$



I tried to estimate $1-frac{x}{m}$ by $e^{-frac{x}{m}}$ and do the same for the rest, but I got stuck.






algebra-precalculus numerical-methods







share|cite|improve this question











$endgroup$




How can we solve the following equation for $x$ without a calculator/computer? ($m$ and $n$ are positive integers)



$m^{n}(1-frac{x}{m})(1-frac{x}{m-1})(1-frac{x}{m-2})=1.$



I tried to estimate $1-frac{x}{m}$ by $e^{-frac{x}{m}}$ and do the same for the rest, but I got stuck.






algebra-precalculus numerical-methods




algebra-precalculus numerical-methods






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 10 at 5:44









Dylan

14.2k31127




14.2k31127










asked Jan 9 at 19:35









KenKen

43028




43028








  • 2




    $begingroup$
    Is $n$ equal to the number of factors? That is, are there other than this 3-factor instance? Or in other words, what is $n$, is it arbitrary?
    $endgroup$
    – LutzL
    Jan 9 at 19:42












  • $begingroup$
    You have added the tag numerical-methods. Do you really need a formula for the exact solution or will you settle for a reliable algorithm for computing the possible values of $x$ as function of $m$ and $n$? It seems to me $x in {m-1,m-2,m}$ are good approximations of the roots when $n$ is large and the right hand side of $1$ is relatively insignificant. I expect the case of small $n$ to be handled by solving a sequence of problems starting with a large value of $n$.
    $endgroup$
    – Carl Christian
    Jan 10 at 10:04
















  • 2




    $begingroup$
    Is $n$ equal to the number of factors? That is, are there other than this 3-factor instance? Or in other words, what is $n$, is it arbitrary?
    $endgroup$
    – LutzL
    Jan 9 at 19:42












  • $begingroup$
    You have added the tag numerical-methods. Do you really need a formula for the exact solution or will you settle for a reliable algorithm for computing the possible values of $x$ as function of $m$ and $n$? It seems to me $x in {m-1,m-2,m}$ are good approximations of the roots when $n$ is large and the right hand side of $1$ is relatively insignificant. I expect the case of small $n$ to be handled by solving a sequence of problems starting with a large value of $n$.
    $endgroup$
    – Carl Christian
    Jan 10 at 10:04










2




2




$begingroup$
Is $n$ equal to the number of factors? That is, are there other than this 3-factor instance? Or in other words, what is $n$, is it arbitrary?
$endgroup$
– LutzL
Jan 9 at 19:42






$begingroup$
Is $n$ equal to the number of factors? That is, are there other than this 3-factor instance? Or in other words, what is $n$, is it arbitrary?
$endgroup$
– LutzL
Jan 9 at 19:42














$begingroup$
You have added the tag numerical-methods. Do you really need a formula for the exact solution or will you settle for a reliable algorithm for computing the possible values of $x$ as function of $m$ and $n$? It seems to me $x in {m-1,m-2,m}$ are good approximations of the roots when $n$ is large and the right hand side of $1$ is relatively insignificant. I expect the case of small $n$ to be handled by solving a sequence of problems starting with a large value of $n$.
$endgroup$
– Carl Christian
Jan 10 at 10:04






$begingroup$
You have added the tag numerical-methods. Do you really need a formula for the exact solution or will you settle for a reliable algorithm for computing the possible values of $x$ as function of $m$ and $n$? It seems to me $x in {m-1,m-2,m}$ are good approximations of the roots when $n$ is large and the right hand side of $1$ is relatively insignificant. I expect the case of small $n$ to be handled by solving a sequence of problems starting with a large value of $n$.
$endgroup$
– Carl Christian
Jan 10 at 10:04












2 Answers
2






active

oldest

votes


















0












$begingroup$

Computer algebra gives the solution:



$$x = -frac{sqrt[3]{27 m^{2 n}+54 m^{2 n-2}-81 m^{2 n-1}+sqrt{left(27 m^{2 n}+54 m^{2 n-2}-81 m^{2 n-1}right)^2-108
m^{6 n-6}}} m^{1-n}}{3 sqrt[3]{2}}-frac{sqrt[3]{2} m^{n-1}}{sqrt[3]{27 m^{2 n}+54 m^{2 n-2}-81 m^{2
n-1}+sqrt{left(27 m^{2 n}+54 m^{2 n-2}-81 m^{2 n-1}right)^2-108 m^{6 n-6}}}}+m-1$$



which suggests that you'll not easily find this solution by pen and paper.






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    Hint: Expanding all we obtain
    $$m^3 n^n-3 m^2 n^n x-3 m^2 n^n-m^2+3 m n^n x^2+6 m n^n
    x+2 m n^n+3 m-n^n x^3-3 n^n x^2-2 n^n x-2=0$$
    this is a cubic equation in $x$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Thank you for your comment. So you mean we can use the formula for a cubic equation, right?
      $endgroup$
      – Ken
      Jan 9 at 19:46










    • $begingroup$
      Yes you can, and good luck!
      $endgroup$
      – Dr. Sonnhard Graubner
      Jan 9 at 19:48






    • 1




      $begingroup$
      This is not an answer. It would better be given as a comment on the question.
      $endgroup$
      – robjohn♦
      Jan 9 at 19:58










    • $begingroup$
      But the Computer Algebra solution is an answer? Senseless your coment!
      $endgroup$
      – Dr. Sonnhard Graubner
      Jan 9 at 20:02












    Your Answer





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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

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    active

    oldest

    votes









    0












    $begingroup$

    Computer algebra gives the solution:



    $$x = -frac{sqrt[3]{27 m^{2 n}+54 m^{2 n-2}-81 m^{2 n-1}+sqrt{left(27 m^{2 n}+54 m^{2 n-2}-81 m^{2 n-1}right)^2-108
    m^{6 n-6}}} m^{1-n}}{3 sqrt[3]{2}}-frac{sqrt[3]{2} m^{n-1}}{sqrt[3]{27 m^{2 n}+54 m^{2 n-2}-81 m^{2
    n-1}+sqrt{left(27 m^{2 n}+54 m^{2 n-2}-81 m^{2 n-1}right)^2-108 m^{6 n-6}}}}+m-1$$



    which suggests that you'll not easily find this solution by pen and paper.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      Computer algebra gives the solution:



      $$x = -frac{sqrt[3]{27 m^{2 n}+54 m^{2 n-2}-81 m^{2 n-1}+sqrt{left(27 m^{2 n}+54 m^{2 n-2}-81 m^{2 n-1}right)^2-108
      m^{6 n-6}}} m^{1-n}}{3 sqrt[3]{2}}-frac{sqrt[3]{2} m^{n-1}}{sqrt[3]{27 m^{2 n}+54 m^{2 n-2}-81 m^{2
      n-1}+sqrt{left(27 m^{2 n}+54 m^{2 n-2}-81 m^{2 n-1}right)^2-108 m^{6 n-6}}}}+m-1$$



      which suggests that you'll not easily find this solution by pen and paper.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        Computer algebra gives the solution:



        $$x = -frac{sqrt[3]{27 m^{2 n}+54 m^{2 n-2}-81 m^{2 n-1}+sqrt{left(27 m^{2 n}+54 m^{2 n-2}-81 m^{2 n-1}right)^2-108
        m^{6 n-6}}} m^{1-n}}{3 sqrt[3]{2}}-frac{sqrt[3]{2} m^{n-1}}{sqrt[3]{27 m^{2 n}+54 m^{2 n-2}-81 m^{2
        n-1}+sqrt{left(27 m^{2 n}+54 m^{2 n-2}-81 m^{2 n-1}right)^2-108 m^{6 n-6}}}}+m-1$$



        which suggests that you'll not easily find this solution by pen and paper.






        share|cite|improve this answer









        $endgroup$



        Computer algebra gives the solution:



        $$x = -frac{sqrt[3]{27 m^{2 n}+54 m^{2 n-2}-81 m^{2 n-1}+sqrt{left(27 m^{2 n}+54 m^{2 n-2}-81 m^{2 n-1}right)^2-108
        m^{6 n-6}}} m^{1-n}}{3 sqrt[3]{2}}-frac{sqrt[3]{2} m^{n-1}}{sqrt[3]{27 m^{2 n}+54 m^{2 n-2}-81 m^{2
        n-1}+sqrt{left(27 m^{2 n}+54 m^{2 n-2}-81 m^{2 n-1}right)^2-108 m^{6 n-6}}}}+m-1$$



        which suggests that you'll not easily find this solution by pen and paper.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 9 at 19:43









        David G. StorkDavid G. Stork

        11.9k41735




        11.9k41735























            0












            $begingroup$

            Hint: Expanding all we obtain
            $$m^3 n^n-3 m^2 n^n x-3 m^2 n^n-m^2+3 m n^n x^2+6 m n^n
            x+2 m n^n+3 m-n^n x^3-3 n^n x^2-2 n^n x-2=0$$
            this is a cubic equation in $x$






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Thank you for your comment. So you mean we can use the formula for a cubic equation, right?
              $endgroup$
              – Ken
              Jan 9 at 19:46










            • $begingroup$
              Yes you can, and good luck!
              $endgroup$
              – Dr. Sonnhard Graubner
              Jan 9 at 19:48






            • 1




              $begingroup$
              This is not an answer. It would better be given as a comment on the question.
              $endgroup$
              – robjohn♦
              Jan 9 at 19:58










            • $begingroup$
              But the Computer Algebra solution is an answer? Senseless your coment!
              $endgroup$
              – Dr. Sonnhard Graubner
              Jan 9 at 20:02
















            0












            $begingroup$

            Hint: Expanding all we obtain
            $$m^3 n^n-3 m^2 n^n x-3 m^2 n^n-m^2+3 m n^n x^2+6 m n^n
            x+2 m n^n+3 m-n^n x^3-3 n^n x^2-2 n^n x-2=0$$
            this is a cubic equation in $x$






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Thank you for your comment. So you mean we can use the formula for a cubic equation, right?
              $endgroup$
              – Ken
              Jan 9 at 19:46










            • $begingroup$
              Yes you can, and good luck!
              $endgroup$
              – Dr. Sonnhard Graubner
              Jan 9 at 19:48






            • 1




              $begingroup$
              This is not an answer. It would better be given as a comment on the question.
              $endgroup$
              – robjohn♦
              Jan 9 at 19:58










            • $begingroup$
              But the Computer Algebra solution is an answer? Senseless your coment!
              $endgroup$
              – Dr. Sonnhard Graubner
              Jan 9 at 20:02














            0












            0








            0





            $begingroup$

            Hint: Expanding all we obtain
            $$m^3 n^n-3 m^2 n^n x-3 m^2 n^n-m^2+3 m n^n x^2+6 m n^n
            x+2 m n^n+3 m-n^n x^3-3 n^n x^2-2 n^n x-2=0$$
            this is a cubic equation in $x$






            share|cite|improve this answer









            $endgroup$



            Hint: Expanding all we obtain
            $$m^3 n^n-3 m^2 n^n x-3 m^2 n^n-m^2+3 m n^n x^2+6 m n^n
            x+2 m n^n+3 m-n^n x^3-3 n^n x^2-2 n^n x-2=0$$
            this is a cubic equation in $x$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Jan 9 at 19:43









            Dr. Sonnhard GraubnerDr. Sonnhard Graubner

            78.5k42867




            78.5k42867












            • $begingroup$
              Thank you for your comment. So you mean we can use the formula for a cubic equation, right?
              $endgroup$
              – Ken
              Jan 9 at 19:46










            • $begingroup$
              Yes you can, and good luck!
              $endgroup$
              – Dr. Sonnhard Graubner
              Jan 9 at 19:48






            • 1




              $begingroup$
              This is not an answer. It would better be given as a comment on the question.
              $endgroup$
              – robjohn♦
              Jan 9 at 19:58










            • $begingroup$
              But the Computer Algebra solution is an answer? Senseless your coment!
              $endgroup$
              – Dr. Sonnhard Graubner
              Jan 9 at 20:02


















            • $begingroup$
              Thank you for your comment. So you mean we can use the formula for a cubic equation, right?
              $endgroup$
              – Ken
              Jan 9 at 19:46










            • $begingroup$
              Yes you can, and good luck!
              $endgroup$
              – Dr. Sonnhard Graubner
              Jan 9 at 19:48






            • 1




              $begingroup$
              This is not an answer. It would better be given as a comment on the question.
              $endgroup$
              – robjohn♦
              Jan 9 at 19:58










            • $begingroup$
              But the Computer Algebra solution is an answer? Senseless your coment!
              $endgroup$
              – Dr. Sonnhard Graubner
              Jan 9 at 20:02
















            $begingroup$
            Thank you for your comment. So you mean we can use the formula for a cubic equation, right?
            $endgroup$
            – Ken
            Jan 9 at 19:46




            $begingroup$
            Thank you for your comment. So you mean we can use the formula for a cubic equation, right?
            $endgroup$
            – Ken
            Jan 9 at 19:46












            $begingroup$
            Yes you can, and good luck!
            $endgroup$
            – Dr. Sonnhard Graubner
            Jan 9 at 19:48




            $begingroup$
            Yes you can, and good luck!
            $endgroup$
            – Dr. Sonnhard Graubner
            Jan 9 at 19:48




            1




            1




            $begingroup$
            This is not an answer. It would better be given as a comment on the question.
            $endgroup$
            – robjohn♦
            Jan 9 at 19:58




            $begingroup$
            This is not an answer. It would better be given as a comment on the question.
            $endgroup$
            – robjohn♦
            Jan 9 at 19:58












            $begingroup$
            But the Computer Algebra solution is an answer? Senseless your coment!
            $endgroup$
            – Dr. Sonnhard Graubner
            Jan 9 at 20:02




            $begingroup$
            But the Computer Algebra solution is an answer? Senseless your coment!
            $endgroup$
            – Dr. Sonnhard Graubner
            Jan 9 at 20:02


















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