Is the positive root of the equation $x^{x^x}=2$, $x=1.47668433…$ a transcendental number?












22














I can prove using the Gelfond–Schneider theorem that the positive root of the equation $x^{x^x}=2$, $x=1.47668433...$ is an irrational number. Is it possible to prove it is transcendental?






irrational-numbers transcendental-numbers tetration transcendental-equations transcendence-theory







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  • 2




    I couldn't find a solution but from all the things I tried, the only one that made me feel like I might be on the right way was noticing that $left(x^xright)^{x^x}=x^{xx^x}=left(x^{x^x}right)^x=2^x$ which, if you assume $x$ is irrational algebraic, is transcendental.
    – xavierm02
    Apr 26 '13 at 22:52






  • 1




    I can't help at all with your question, but I'm curious how you used Gelfond-Schneider to prove $x$ is irrational. If $x$ and $x^x$ both happen to be rational, what's the problem?
    – Jason DeVito
    Apr 26 '13 at 23:13






  • 5




    For detailed proof see Marshall, Ash J., and Tan, Yiren, "A rational number of the form $a^a$ with $a$ irrational", Mathematical Gazette 96, March 2012, pp. 106-109.
    – Vladimir Reshetnikov
    Apr 27 '13 at 18:12
















22














I can prove using the Gelfond–Schneider theorem that the positive root of the equation $x^{x^x}=2$, $x=1.47668433...$ is an irrational number. Is it possible to prove it is transcendental?






irrational-numbers transcendental-numbers tetration transcendental-equations transcendence-theory







share|cite|improve this question


















  • 2




    I couldn't find a solution but from all the things I tried, the only one that made me feel like I might be on the right way was noticing that $left(x^xright)^{x^x}=x^{xx^x}=left(x^{x^x}right)^x=2^x$ which, if you assume $x$ is irrational algebraic, is transcendental.
    – xavierm02
    Apr 26 '13 at 22:52






  • 1




    I can't help at all with your question, but I'm curious how you used Gelfond-Schneider to prove $x$ is irrational. If $x$ and $x^x$ both happen to be rational, what's the problem?
    – Jason DeVito
    Apr 26 '13 at 23:13






  • 5




    For detailed proof see Marshall, Ash J., and Tan, Yiren, "A rational number of the form $a^a$ with $a$ irrational", Mathematical Gazette 96, March 2012, pp. 106-109.
    – Vladimir Reshetnikov
    Apr 27 '13 at 18:12














22












22








22


21





I can prove using the Gelfond–Schneider theorem that the positive root of the equation $x^{x^x}=2$, $x=1.47668433...$ is an irrational number. Is it possible to prove it is transcendental?






irrational-numbers transcendental-numbers tetration transcendental-equations transcendence-theory







share|cite|improve this question













I can prove using the Gelfond–Schneider theorem that the positive root of the equation $x^{x^x}=2$, $x=1.47668433...$ is an irrational number. Is it possible to prove it is transcendental?






irrational-numbers transcendental-numbers tetration transcendental-equations transcendence-theory




irrational-numbers transcendental-numbers tetration transcendental-equations transcendence-theory






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asked Apr 26 '13 at 21:23









Vladimir ReshetnikovVladimir Reshetnikov

24.2k4119231




24.2k4119231








  • 2




    I couldn't find a solution but from all the things I tried, the only one that made me feel like I might be on the right way was noticing that $left(x^xright)^{x^x}=x^{xx^x}=left(x^{x^x}right)^x=2^x$ which, if you assume $x$ is irrational algebraic, is transcendental.
    – xavierm02
    Apr 26 '13 at 22:52






  • 1




    I can't help at all with your question, but I'm curious how you used Gelfond-Schneider to prove $x$ is irrational. If $x$ and $x^x$ both happen to be rational, what's the problem?
    – Jason DeVito
    Apr 26 '13 at 23:13






  • 5




    For detailed proof see Marshall, Ash J., and Tan, Yiren, "A rational number of the form $a^a$ with $a$ irrational", Mathematical Gazette 96, March 2012, pp. 106-109.
    – Vladimir Reshetnikov
    Apr 27 '13 at 18:12














  • 2




    I couldn't find a solution but from all the things I tried, the only one that made me feel like I might be on the right way was noticing that $left(x^xright)^{x^x}=x^{xx^x}=left(x^{x^x}right)^x=2^x$ which, if you assume $x$ is irrational algebraic, is transcendental.
    – xavierm02
    Apr 26 '13 at 22:52






  • 1




    I can't help at all with your question, but I'm curious how you used Gelfond-Schneider to prove $x$ is irrational. If $x$ and $x^x$ both happen to be rational, what's the problem?
    – Jason DeVito
    Apr 26 '13 at 23:13






  • 5




    For detailed proof see Marshall, Ash J., and Tan, Yiren, "A rational number of the form $a^a$ with $a$ irrational", Mathematical Gazette 96, March 2012, pp. 106-109.
    – Vladimir Reshetnikov
    Apr 27 '13 at 18:12








2




2




I couldn't find a solution but from all the things I tried, the only one that made me feel like I might be on the right way was noticing that $left(x^xright)^{x^x}=x^{xx^x}=left(x^{x^x}right)^x=2^x$ which, if you assume $x$ is irrational algebraic, is transcendental.
– xavierm02
Apr 26 '13 at 22:52




I couldn't find a solution but from all the things I tried, the only one that made me feel like I might be on the right way was noticing that $left(x^xright)^{x^x}=x^{xx^x}=left(x^{x^x}right)^x=2^x$ which, if you assume $x$ is irrational algebraic, is transcendental.
– xavierm02
Apr 26 '13 at 22:52




1




1




I can't help at all with your question, but I'm curious how you used Gelfond-Schneider to prove $x$ is irrational. If $x$ and $x^x$ both happen to be rational, what's the problem?
– Jason DeVito
Apr 26 '13 at 23:13




I can't help at all with your question, but I'm curious how you used Gelfond-Schneider to prove $x$ is irrational. If $x$ and $x^x$ both happen to be rational, what's the problem?
– Jason DeVito
Apr 26 '13 at 23:13




5




5




For detailed proof see Marshall, Ash J., and Tan, Yiren, "A rational number of the form $a^a$ with $a$ irrational", Mathematical Gazette 96, March 2012, pp. 106-109.
– Vladimir Reshetnikov
Apr 27 '13 at 18:12




For detailed proof see Marshall, Ash J., and Tan, Yiren, "A rational number of the form $a^a$ with $a$ irrational", Mathematical Gazette 96, March 2012, pp. 106-109.
– Vladimir Reshetnikov
Apr 27 '13 at 18:12










2 Answers
2






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20














I believe, it is a known open problem. Ditto for ${^3 x}=3$, ${^3 x}=4$, ${^3 x}=5$ (left superscript denotes tetration).






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    0














    In addition to the other answer and comments, the following might prove to be useful. Define the sequence $s_n$ as follows: $s_0:=1$ and for all positive integers $ngeq 1$ let $s_n:=2^{frac{1}{2}s_{n-1}}$. We have



    $$
    s_1=sqrt{2},; s_2=sqrt{2}^{sqrt{2}}, s_3=sqrt{2}^{sqrt{2}^{sqrt{2}}},ldots
    $$



    The limit of this sequence is



    $$
    lim_{ntoinfty}s_n=sqrt 2^{{{sqrt 2}^{{sqrt 2}^{ldots}}}}=2.
    $$



    The solution of the power tower $x^{x^{{x}^{ldots}}}=2$ is therefore $x=sqrt 2=1.41421356237309ldots$ which is irrational but not transcendental.






    share|cite|improve this answer





















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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      20














      I believe, it is a known open problem. Ditto for ${^3 x}=3$, ${^3 x}=4$, ${^3 x}=5$ (left superscript denotes tetration).






      share|cite|improve this answer


























        20














        I believe, it is a known open problem. Ditto for ${^3 x}=3$, ${^3 x}=4$, ${^3 x}=5$ (left superscript denotes tetration).






        share|cite|improve this answer
























          20












          20








          20






          I believe, it is a known open problem. Ditto for ${^3 x}=3$, ${^3 x}=4$, ${^3 x}=5$ (left superscript denotes tetration).






          share|cite|improve this answer












          I believe, it is a known open problem. Ditto for ${^3 x}=3$, ${^3 x}=4$, ${^3 x}=5$ (left superscript denotes tetration).







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered May 4 '13 at 22:13









          Oksana GimmelOksana Gimmel

          2,83822535




          2,83822535























              0














              In addition to the other answer and comments, the following might prove to be useful. Define the sequence $s_n$ as follows: $s_0:=1$ and for all positive integers $ngeq 1$ let $s_n:=2^{frac{1}{2}s_{n-1}}$. We have



              $$
              s_1=sqrt{2},; s_2=sqrt{2}^{sqrt{2}}, s_3=sqrt{2}^{sqrt{2}^{sqrt{2}}},ldots
              $$



              The limit of this sequence is



              $$
              lim_{ntoinfty}s_n=sqrt 2^{{{sqrt 2}^{{sqrt 2}^{ldots}}}}=2.
              $$



              The solution of the power tower $x^{x^{{x}^{ldots}}}=2$ is therefore $x=sqrt 2=1.41421356237309ldots$ which is irrational but not transcendental.






              share|cite|improve this answer


























                0














                In addition to the other answer and comments, the following might prove to be useful. Define the sequence $s_n$ as follows: $s_0:=1$ and for all positive integers $ngeq 1$ let $s_n:=2^{frac{1}{2}s_{n-1}}$. We have



                $$
                s_1=sqrt{2},; s_2=sqrt{2}^{sqrt{2}}, s_3=sqrt{2}^{sqrt{2}^{sqrt{2}}},ldots
                $$



                The limit of this sequence is



                $$
                lim_{ntoinfty}s_n=sqrt 2^{{{sqrt 2}^{{sqrt 2}^{ldots}}}}=2.
                $$



                The solution of the power tower $x^{x^{{x}^{ldots}}}=2$ is therefore $x=sqrt 2=1.41421356237309ldots$ which is irrational but not transcendental.






                share|cite|improve this answer
























                  0












                  0








                  0






                  In addition to the other answer and comments, the following might prove to be useful. Define the sequence $s_n$ as follows: $s_0:=1$ and for all positive integers $ngeq 1$ let $s_n:=2^{frac{1}{2}s_{n-1}}$. We have



                  $$
                  s_1=sqrt{2},; s_2=sqrt{2}^{sqrt{2}}, s_3=sqrt{2}^{sqrt{2}^{sqrt{2}}},ldots
                  $$



                  The limit of this sequence is



                  $$
                  lim_{ntoinfty}s_n=sqrt 2^{{{sqrt 2}^{{sqrt 2}^{ldots}}}}=2.
                  $$



                  The solution of the power tower $x^{x^{{x}^{ldots}}}=2$ is therefore $x=sqrt 2=1.41421356237309ldots$ which is irrational but not transcendental.






                  share|cite|improve this answer












                  In addition to the other answer and comments, the following might prove to be useful. Define the sequence $s_n$ as follows: $s_0:=1$ and for all positive integers $ngeq 1$ let $s_n:=2^{frac{1}{2}s_{n-1}}$. We have



                  $$
                  s_1=sqrt{2},; s_2=sqrt{2}^{sqrt{2}}, s_3=sqrt{2}^{sqrt{2}^{sqrt{2}}},ldots
                  $$



                  The limit of this sequence is



                  $$
                  lim_{ntoinfty}s_n=sqrt 2^{{{sqrt 2}^{{sqrt 2}^{ldots}}}}=2.
                  $$



                  The solution of the power tower $x^{x^{{x}^{ldots}}}=2$ is therefore $x=sqrt 2=1.41421356237309ldots$ which is irrational but not transcendental.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 28 '18 at 13:35









                  KlangenKlangen

                  1,66211334




                  1,66211334






























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