Is it possible to solve polynomials with degree larger equal 5 using formulas with transcendental numbers?












3












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Algebra tells us that it is not possible to solve a polynomial with degree larger equal 5 using formulas containing roots, multiplication etc.
But the use of transcendental numbers ist not allowed in these formulas.
Am i able to generally solve polynomials with degree larger equal 5 by also allowing the use of transcendental numbers in those formulas?



Thanks in advance!






polynomials transcendental-numbers







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    3












    $begingroup$


    Algebra tells us that it is not possible to solve a polynomial with degree larger equal 5 using formulas containing roots, multiplication etc.
    But the use of transcendental numbers ist not allowed in these formulas.
    Am i able to generally solve polynomials with degree larger equal 5 by also allowing the use of transcendental numbers in those formulas?



    Thanks in advance!






    polynomials transcendental-numbers







    share|cite|improve this question









    $endgroup$















      3












      3








      3


      1



      $begingroup$


      Algebra tells us that it is not possible to solve a polynomial with degree larger equal 5 using formulas containing roots, multiplication etc.
      But the use of transcendental numbers ist not allowed in these formulas.
      Am i able to generally solve polynomials with degree larger equal 5 by also allowing the use of transcendental numbers in those formulas?



      Thanks in advance!






      polynomials transcendental-numbers







      share|cite|improve this question









      $endgroup$




      Algebra tells us that it is not possible to solve a polynomial with degree larger equal 5 using formulas containing roots, multiplication etc.
      But the use of transcendental numbers ist not allowed in these formulas.
      Am i able to generally solve polynomials with degree larger equal 5 by also allowing the use of transcendental numbers in those formulas?



      Thanks in advance!






      polynomials transcendental-numbers




      polynomials transcendental-numbers






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      share|cite|improve this question











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      asked Jan 24 at 7:41









      Alexander ZornAlexander Zorn

      161




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          $begingroup$

          No. To be precise, the Abel-Ruffini theorem is valid over any field:




          Let $K$ be a field and $ngeq 5$. Then the roots of the polynomial $$x^n+a_{n-1}x^{n-1}+dots+a_0$$ with coefficients in the field $K(a_0,dots,a_{n-1})$ of rational functions cannot be expressed in terms of the coefficients $a_i$ and elements of the field $K$ using only the operations of addition, subtraction, multiplication, division, and extraction of $m$th roots for arbitrary positive integers $m$.




          In particular, taking $K=mathbb{C}$, this says that there is no general formula using radicals for the roots of a polynomial of degree $ngeq 5$, even if the formula is allowed to refer to whatever specific complex numbers you want. Specifically, the theorem tells you that any putative such formula will be incorrect when applied to a polynomial whose coefficients are algebraically independent over the subfield of $mathbb{C}$ generated by the complex numbers used in the formula.



          (It should be noted, though, that if you allow reference to transcendental numbers, then you can express any root of any one specific polynomial with coefficients in $mathbb{C}$ at a time, by a formula that works only for that polynomial. Indeed, if $r$ is a root of your polynomial, then either $r$ itself is transcendental so you can refer to it directly in the formula, or it is algebraic so you can find it by the formula $alpha-pi$ where $pi$ and $alpha=pi+r$ are both transcendental.)






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            $begingroup$

            No. To be precise, the Abel-Ruffini theorem is valid over any field:




            Let $K$ be a field and $ngeq 5$. Then the roots of the polynomial $$x^n+a_{n-1}x^{n-1}+dots+a_0$$ with coefficients in the field $K(a_0,dots,a_{n-1})$ of rational functions cannot be expressed in terms of the coefficients $a_i$ and elements of the field $K$ using only the operations of addition, subtraction, multiplication, division, and extraction of $m$th roots for arbitrary positive integers $m$.




            In particular, taking $K=mathbb{C}$, this says that there is no general formula using radicals for the roots of a polynomial of degree $ngeq 5$, even if the formula is allowed to refer to whatever specific complex numbers you want. Specifically, the theorem tells you that any putative such formula will be incorrect when applied to a polynomial whose coefficients are algebraically independent over the subfield of $mathbb{C}$ generated by the complex numbers used in the formula.



            (It should be noted, though, that if you allow reference to transcendental numbers, then you can express any root of any one specific polynomial with coefficients in $mathbb{C}$ at a time, by a formula that works only for that polynomial. Indeed, if $r$ is a root of your polynomial, then either $r$ itself is transcendental so you can refer to it directly in the formula, or it is algebraic so you can find it by the formula $alpha-pi$ where $pi$ and $alpha=pi+r$ are both transcendental.)






            share|cite|improve this answer











            $endgroup$


















              4












              $begingroup$

              No. To be precise, the Abel-Ruffini theorem is valid over any field:




              Let $K$ be a field and $ngeq 5$. Then the roots of the polynomial $$x^n+a_{n-1}x^{n-1}+dots+a_0$$ with coefficients in the field $K(a_0,dots,a_{n-1})$ of rational functions cannot be expressed in terms of the coefficients $a_i$ and elements of the field $K$ using only the operations of addition, subtraction, multiplication, division, and extraction of $m$th roots for arbitrary positive integers $m$.




              In particular, taking $K=mathbb{C}$, this says that there is no general formula using radicals for the roots of a polynomial of degree $ngeq 5$, even if the formula is allowed to refer to whatever specific complex numbers you want. Specifically, the theorem tells you that any putative such formula will be incorrect when applied to a polynomial whose coefficients are algebraically independent over the subfield of $mathbb{C}$ generated by the complex numbers used in the formula.



              (It should be noted, though, that if you allow reference to transcendental numbers, then you can express any root of any one specific polynomial with coefficients in $mathbb{C}$ at a time, by a formula that works only for that polynomial. Indeed, if $r$ is a root of your polynomial, then either $r$ itself is transcendental so you can refer to it directly in the formula, or it is algebraic so you can find it by the formula $alpha-pi$ where $pi$ and $alpha=pi+r$ are both transcendental.)






              share|cite|improve this answer











              $endgroup$
















                4












                4








                4





                $begingroup$

                No. To be precise, the Abel-Ruffini theorem is valid over any field:




                Let $K$ be a field and $ngeq 5$. Then the roots of the polynomial $$x^n+a_{n-1}x^{n-1}+dots+a_0$$ with coefficients in the field $K(a_0,dots,a_{n-1})$ of rational functions cannot be expressed in terms of the coefficients $a_i$ and elements of the field $K$ using only the operations of addition, subtraction, multiplication, division, and extraction of $m$th roots for arbitrary positive integers $m$.




                In particular, taking $K=mathbb{C}$, this says that there is no general formula using radicals for the roots of a polynomial of degree $ngeq 5$, even if the formula is allowed to refer to whatever specific complex numbers you want. Specifically, the theorem tells you that any putative such formula will be incorrect when applied to a polynomial whose coefficients are algebraically independent over the subfield of $mathbb{C}$ generated by the complex numbers used in the formula.



                (It should be noted, though, that if you allow reference to transcendental numbers, then you can express any root of any one specific polynomial with coefficients in $mathbb{C}$ at a time, by a formula that works only for that polynomial. Indeed, if $r$ is a root of your polynomial, then either $r$ itself is transcendental so you can refer to it directly in the formula, or it is algebraic so you can find it by the formula $alpha-pi$ where $pi$ and $alpha=pi+r$ are both transcendental.)






                share|cite|improve this answer











                $endgroup$



                No. To be precise, the Abel-Ruffini theorem is valid over any field:




                Let $K$ be a field and $ngeq 5$. Then the roots of the polynomial $$x^n+a_{n-1}x^{n-1}+dots+a_0$$ with coefficients in the field $K(a_0,dots,a_{n-1})$ of rational functions cannot be expressed in terms of the coefficients $a_i$ and elements of the field $K$ using only the operations of addition, subtraction, multiplication, division, and extraction of $m$th roots for arbitrary positive integers $m$.




                In particular, taking $K=mathbb{C}$, this says that there is no general formula using radicals for the roots of a polynomial of degree $ngeq 5$, even if the formula is allowed to refer to whatever specific complex numbers you want. Specifically, the theorem tells you that any putative such formula will be incorrect when applied to a polynomial whose coefficients are algebraically independent over the subfield of $mathbb{C}$ generated by the complex numbers used in the formula.



                (It should be noted, though, that if you allow reference to transcendental numbers, then you can express any root of any one specific polynomial with coefficients in $mathbb{C}$ at a time, by a formula that works only for that polynomial. Indeed, if $r$ is a root of your polynomial, then either $r$ itself is transcendental so you can refer to it directly in the formula, or it is algebraic so you can find it by the formula $alpha-pi$ where $pi$ and $alpha=pi+r$ are both transcendental.)







                share|cite|improve this answer














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                share|cite|improve this answer








                edited Jan 24 at 8:40

























                answered Jan 24 at 8:32









                Eric WofseyEric Wofsey

                189k14216347




                189k14216347






























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