If $x_1=sqrt 2$ and $x_{n+1}=(sqrt2)^{x_n}$ then sequence $x_n$ converges to 2.












0












$begingroup$


If $x_1=sqrt 2$ and $x_{n+1}=(sqrt2)^{x_n}$ then show that sequence $x_n$ converges to 2.



I know this sequence is monotonically increasing. But how to prove it converges to 2?



The sequence is bounded above is already answered here. And I know that$sqrt 2 ^l=l$ has only solution 2. But how to prove formally? Thanks.






real-analysis sequences-and-series







share|cite|improve this question









$endgroup$












  • $begingroup$
    You can use the monotone convergence theorem
    $endgroup$
    – Dr. Sonnhard Graubner
    Dec 10 '18 at 7:48










  • $begingroup$
    @Dr. Sonnhard Graubner but how to prove 2 is least upper bound?
    $endgroup$
    – ramanujan
    Dec 10 '18 at 7:53










  • $begingroup$
    Taking $log$, then taking the limit $n to infty$.
    $endgroup$
    – xbh
    Dec 10 '18 at 7:53










  • $begingroup$
    You can use induction to show that $$x_n<2$$
    $endgroup$
    – Dr. Sonnhard Graubner
    Dec 10 '18 at 7:54






  • 1




    $begingroup$
    $l=4$ is also a solution
    $endgroup$
    – Masacroso
    Dec 10 '18 at 8:22
















0












$begingroup$


If $x_1=sqrt 2$ and $x_{n+1}=(sqrt2)^{x_n}$ then show that sequence $x_n$ converges to 2.



I know this sequence is monotonically increasing. But how to prove it converges to 2?



The sequence is bounded above is already answered here. And I know that$sqrt 2 ^l=l$ has only solution 2. But how to prove formally? Thanks.






real-analysis sequences-and-series







share|cite|improve this question









$endgroup$












  • $begingroup$
    You can use the monotone convergence theorem
    $endgroup$
    – Dr. Sonnhard Graubner
    Dec 10 '18 at 7:48










  • $begingroup$
    @Dr. Sonnhard Graubner but how to prove 2 is least upper bound?
    $endgroup$
    – ramanujan
    Dec 10 '18 at 7:53










  • $begingroup$
    Taking $log$, then taking the limit $n to infty$.
    $endgroup$
    – xbh
    Dec 10 '18 at 7:53










  • $begingroup$
    You can use induction to show that $$x_n<2$$
    $endgroup$
    – Dr. Sonnhard Graubner
    Dec 10 '18 at 7:54






  • 1




    $begingroup$
    $l=4$ is also a solution
    $endgroup$
    – Masacroso
    Dec 10 '18 at 8:22














0












0








0





$begingroup$


If $x_1=sqrt 2$ and $x_{n+1}=(sqrt2)^{x_n}$ then show that sequence $x_n$ converges to 2.



I know this sequence is monotonically increasing. But how to prove it converges to 2?



The sequence is bounded above is already answered here. And I know that$sqrt 2 ^l=l$ has only solution 2. But how to prove formally? Thanks.






real-analysis sequences-and-series







share|cite|improve this question









$endgroup$




If $x_1=sqrt 2$ and $x_{n+1}=(sqrt2)^{x_n}$ then show that sequence $x_n$ converges to 2.



I know this sequence is monotonically increasing. But how to prove it converges to 2?



The sequence is bounded above is already answered here. And I know that$sqrt 2 ^l=l$ has only solution 2. But how to prove formally? Thanks.






real-analysis sequences-and-series




real-analysis sequences-and-series






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 10 '18 at 7:40









ramanujanramanujan

714713




714713












  • $begingroup$
    You can use the monotone convergence theorem
    $endgroup$
    – Dr. Sonnhard Graubner
    Dec 10 '18 at 7:48










  • $begingroup$
    @Dr. Sonnhard Graubner but how to prove 2 is least upper bound?
    $endgroup$
    – ramanujan
    Dec 10 '18 at 7:53










  • $begingroup$
    Taking $log$, then taking the limit $n to infty$.
    $endgroup$
    – xbh
    Dec 10 '18 at 7:53










  • $begingroup$
    You can use induction to show that $$x_n<2$$
    $endgroup$
    – Dr. Sonnhard Graubner
    Dec 10 '18 at 7:54






  • 1




    $begingroup$
    $l=4$ is also a solution
    $endgroup$
    – Masacroso
    Dec 10 '18 at 8:22


















  • $begingroup$
    You can use the monotone convergence theorem
    $endgroup$
    – Dr. Sonnhard Graubner
    Dec 10 '18 at 7:48










  • $begingroup$
    @Dr. Sonnhard Graubner but how to prove 2 is least upper bound?
    $endgroup$
    – ramanujan
    Dec 10 '18 at 7:53










  • $begingroup$
    Taking $log$, then taking the limit $n to infty$.
    $endgroup$
    – xbh
    Dec 10 '18 at 7:53










  • $begingroup$
    You can use induction to show that $$x_n<2$$
    $endgroup$
    – Dr. Sonnhard Graubner
    Dec 10 '18 at 7:54






  • 1




    $begingroup$
    $l=4$ is also a solution
    $endgroup$
    – Masacroso
    Dec 10 '18 at 8:22
















$begingroup$
You can use the monotone convergence theorem
$endgroup$
– Dr. Sonnhard Graubner
Dec 10 '18 at 7:48




$begingroup$
You can use the monotone convergence theorem
$endgroup$
– Dr. Sonnhard Graubner
Dec 10 '18 at 7:48












$begingroup$
@Dr. Sonnhard Graubner but how to prove 2 is least upper bound?
$endgroup$
– ramanujan
Dec 10 '18 at 7:53




$begingroup$
@Dr. Sonnhard Graubner but how to prove 2 is least upper bound?
$endgroup$
– ramanujan
Dec 10 '18 at 7:53












$begingroup$
Taking $log$, then taking the limit $n to infty$.
$endgroup$
– xbh
Dec 10 '18 at 7:53




$begingroup$
Taking $log$, then taking the limit $n to infty$.
$endgroup$
– xbh
Dec 10 '18 at 7:53












$begingroup$
You can use induction to show that $$x_n<2$$
$endgroup$
– Dr. Sonnhard Graubner
Dec 10 '18 at 7:54




$begingroup$
You can use induction to show that $$x_n<2$$
$endgroup$
– Dr. Sonnhard Graubner
Dec 10 '18 at 7:54




1




1




$begingroup$
$l=4$ is also a solution
$endgroup$
– Masacroso
Dec 10 '18 at 8:22




$begingroup$
$l=4$ is also a solution
$endgroup$
– Masacroso
Dec 10 '18 at 8:22










3 Answers
3






active

oldest

votes


















1












$begingroup$

Here is a formal way considering $f(x) = left( sqrt{2}right)^x$ and using MVP:




  • On $[1,2]$ we have $0 < f'(x)= ln{sqrt{2}}left( sqrt{2}right)^x leq 2 ln{sqrt{2}} < 2 cdot frac{2}{5}= frac{4}{5}$
    $$0 leq 2 - x_{n+1} = left( sqrt{2}right)^2 - left( sqrt{2}right)^{x_n} = f'(xi_n)(2-x_n) < frac{4}{5}(2-x_n)$$


It follows
$$0 leq 2 - x_{n+1} < left(frac{4}{5}right)^n (2-x_1)stackrel{n to infty}{longrightarrow} 0 Rightarrow boxed{lim_{n to infty}x_n = 2}$$



To sum this up:





  • $f$ maps $[1,2]$ into $[1,2]$ and, hence, has a fixpoint there.

  • The fixpoint is unique as $|f'(x)| leq q < 1$ on $[1,2]$.

  • For any starting value $x_1 in [1,2]$ the iteration $x_{n+1} = f(x_n)$ will converge to the fixpoint which is the solution of $x = f(x)$ on $[1,2]Leftrightarrow x=2$






share|cite|improve this answer











$endgroup$





















    1












    $begingroup$

    To find a candidate for the limit you have the equation



    $$x=(sqrt 2)^xtag1$$



    for some possible limit $x:=lim_{ntoinfty}x_n$. This gives, assuming $x>0$, the equivalent equation $x^{1/x}=sqrt 2$, what have the possible solutions $2$ and $4$, what can be seen by the study of the function $x^{1/x}$.



    Thus, if you shows that the sequence is monotone and bounded above by $2$ you are done.



    REMARK: you dont need to show that $2$ is the least upper bound, just that it is an upper bound, because the unique possible solutions are $2$ and $4$, and by the monotony of the sequence we knows that the sequence converges if it is bounded.






    share|cite|improve this answer











    $endgroup$





















      1












      $begingroup$

      As the exponential is a growing function,



      $$x<2implies sqrt2^{,x}<sqrt2^{,2}=2$$ and the sequence starting from $x=sqrt2<2$ is bounded above.






      share|cite|improve this answer









      $endgroup$













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        3 Answers
        3






        active

        oldest

        votes








        3 Answers
        3






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        1












        $begingroup$

        Here is a formal way considering $f(x) = left( sqrt{2}right)^x$ and using MVP:




        • On $[1,2]$ we have $0 < f'(x)= ln{sqrt{2}}left( sqrt{2}right)^x leq 2 ln{sqrt{2}} < 2 cdot frac{2}{5}= frac{4}{5}$
          $$0 leq 2 - x_{n+1} = left( sqrt{2}right)^2 - left( sqrt{2}right)^{x_n} = f'(xi_n)(2-x_n) < frac{4}{5}(2-x_n)$$


        It follows
        $$0 leq 2 - x_{n+1} < left(frac{4}{5}right)^n (2-x_1)stackrel{n to infty}{longrightarrow} 0 Rightarrow boxed{lim_{n to infty}x_n = 2}$$



        To sum this up:





        • $f$ maps $[1,2]$ into $[1,2]$ and, hence, has a fixpoint there.

        • The fixpoint is unique as $|f'(x)| leq q < 1$ on $[1,2]$.

        • For any starting value $x_1 in [1,2]$ the iteration $x_{n+1} = f(x_n)$ will converge to the fixpoint which is the solution of $x = f(x)$ on $[1,2]Leftrightarrow x=2$






        share|cite|improve this answer











        $endgroup$


















          1












          $begingroup$

          Here is a formal way considering $f(x) = left( sqrt{2}right)^x$ and using MVP:




          • On $[1,2]$ we have $0 < f'(x)= ln{sqrt{2}}left( sqrt{2}right)^x leq 2 ln{sqrt{2}} < 2 cdot frac{2}{5}= frac{4}{5}$
            $$0 leq 2 - x_{n+1} = left( sqrt{2}right)^2 - left( sqrt{2}right)^{x_n} = f'(xi_n)(2-x_n) < frac{4}{5}(2-x_n)$$


          It follows
          $$0 leq 2 - x_{n+1} < left(frac{4}{5}right)^n (2-x_1)stackrel{n to infty}{longrightarrow} 0 Rightarrow boxed{lim_{n to infty}x_n = 2}$$



          To sum this up:





          • $f$ maps $[1,2]$ into $[1,2]$ and, hence, has a fixpoint there.

          • The fixpoint is unique as $|f'(x)| leq q < 1$ on $[1,2]$.

          • For any starting value $x_1 in [1,2]$ the iteration $x_{n+1} = f(x_n)$ will converge to the fixpoint which is the solution of $x = f(x)$ on $[1,2]Leftrightarrow x=2$






          share|cite|improve this answer











          $endgroup$
















            1












            1








            1





            $begingroup$

            Here is a formal way considering $f(x) = left( sqrt{2}right)^x$ and using MVP:




            • On $[1,2]$ we have $0 < f'(x)= ln{sqrt{2}}left( sqrt{2}right)^x leq 2 ln{sqrt{2}} < 2 cdot frac{2}{5}= frac{4}{5}$
              $$0 leq 2 - x_{n+1} = left( sqrt{2}right)^2 - left( sqrt{2}right)^{x_n} = f'(xi_n)(2-x_n) < frac{4}{5}(2-x_n)$$


            It follows
            $$0 leq 2 - x_{n+1} < left(frac{4}{5}right)^n (2-x_1)stackrel{n to infty}{longrightarrow} 0 Rightarrow boxed{lim_{n to infty}x_n = 2}$$



            To sum this up:





            • $f$ maps $[1,2]$ into $[1,2]$ and, hence, has a fixpoint there.

            • The fixpoint is unique as $|f'(x)| leq q < 1$ on $[1,2]$.

            • For any starting value $x_1 in [1,2]$ the iteration $x_{n+1} = f(x_n)$ will converge to the fixpoint which is the solution of $x = f(x)$ on $[1,2]Leftrightarrow x=2$






            share|cite|improve this answer











            $endgroup$



            Here is a formal way considering $f(x) = left( sqrt{2}right)^x$ and using MVP:




            • On $[1,2]$ we have $0 < f'(x)= ln{sqrt{2}}left( sqrt{2}right)^x leq 2 ln{sqrt{2}} < 2 cdot frac{2}{5}= frac{4}{5}$
              $$0 leq 2 - x_{n+1} = left( sqrt{2}right)^2 - left( sqrt{2}right)^{x_n} = f'(xi_n)(2-x_n) < frac{4}{5}(2-x_n)$$


            It follows
            $$0 leq 2 - x_{n+1} < left(frac{4}{5}right)^n (2-x_1)stackrel{n to infty}{longrightarrow} 0 Rightarrow boxed{lim_{n to infty}x_n = 2}$$



            To sum this up:





            • $f$ maps $[1,2]$ into $[1,2]$ and, hence, has a fixpoint there.

            • The fixpoint is unique as $|f'(x)| leq q < 1$ on $[1,2]$.

            • For any starting value $x_1 in [1,2]$ the iteration $x_{n+1} = f(x_n)$ will converge to the fixpoint which is the solution of $x = f(x)$ on $[1,2]Leftrightarrow x=2$







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Dec 10 '18 at 8:25

























            answered Dec 10 '18 at 8:19









            trancelocationtrancelocation

            10.3k1722




            10.3k1722























                1












                $begingroup$

                To find a candidate for the limit you have the equation



                $$x=(sqrt 2)^xtag1$$



                for some possible limit $x:=lim_{ntoinfty}x_n$. This gives, assuming $x>0$, the equivalent equation $x^{1/x}=sqrt 2$, what have the possible solutions $2$ and $4$, what can be seen by the study of the function $x^{1/x}$.



                Thus, if you shows that the sequence is monotone and bounded above by $2$ you are done.



                REMARK: you dont need to show that $2$ is the least upper bound, just that it is an upper bound, because the unique possible solutions are $2$ and $4$, and by the monotony of the sequence we knows that the sequence converges if it is bounded.






                share|cite|improve this answer











                $endgroup$


















                  1












                  $begingroup$

                  To find a candidate for the limit you have the equation



                  $$x=(sqrt 2)^xtag1$$



                  for some possible limit $x:=lim_{ntoinfty}x_n$. This gives, assuming $x>0$, the equivalent equation $x^{1/x}=sqrt 2$, what have the possible solutions $2$ and $4$, what can be seen by the study of the function $x^{1/x}$.



                  Thus, if you shows that the sequence is monotone and bounded above by $2$ you are done.



                  REMARK: you dont need to show that $2$ is the least upper bound, just that it is an upper bound, because the unique possible solutions are $2$ and $4$, and by the monotony of the sequence we knows that the sequence converges if it is bounded.






                  share|cite|improve this answer











                  $endgroup$
















                    1












                    1








                    1





                    $begingroup$

                    To find a candidate for the limit you have the equation



                    $$x=(sqrt 2)^xtag1$$



                    for some possible limit $x:=lim_{ntoinfty}x_n$. This gives, assuming $x>0$, the equivalent equation $x^{1/x}=sqrt 2$, what have the possible solutions $2$ and $4$, what can be seen by the study of the function $x^{1/x}$.



                    Thus, if you shows that the sequence is monotone and bounded above by $2$ you are done.



                    REMARK: you dont need to show that $2$ is the least upper bound, just that it is an upper bound, because the unique possible solutions are $2$ and $4$, and by the monotony of the sequence we knows that the sequence converges if it is bounded.






                    share|cite|improve this answer











                    $endgroup$



                    To find a candidate for the limit you have the equation



                    $$x=(sqrt 2)^xtag1$$



                    for some possible limit $x:=lim_{ntoinfty}x_n$. This gives, assuming $x>0$, the equivalent equation $x^{1/x}=sqrt 2$, what have the possible solutions $2$ and $4$, what can be seen by the study of the function $x^{1/x}$.



                    Thus, if you shows that the sequence is monotone and bounded above by $2$ you are done.



                    REMARK: you dont need to show that $2$ is the least upper bound, just that it is an upper bound, because the unique possible solutions are $2$ and $4$, and by the monotony of the sequence we knows that the sequence converges if it is bounded.







                    share|cite|improve this answer














                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited Dec 10 '18 at 8:26

























                    answered Dec 10 '18 at 8:19









                    MasacrosoMasacroso

                    13k41746




                    13k41746























                        1












                        $begingroup$

                        As the exponential is a growing function,



                        $$x<2implies sqrt2^{,x}<sqrt2^{,2}=2$$ and the sequence starting from $x=sqrt2<2$ is bounded above.






                        share|cite|improve this answer









                        $endgroup$


















                          1












                          $begingroup$

                          As the exponential is a growing function,



                          $$x<2implies sqrt2^{,x}<sqrt2^{,2}=2$$ and the sequence starting from $x=sqrt2<2$ is bounded above.






                          share|cite|improve this answer









                          $endgroup$
















                            1












                            1








                            1





                            $begingroup$

                            As the exponential is a growing function,



                            $$x<2implies sqrt2^{,x}<sqrt2^{,2}=2$$ and the sequence starting from $x=sqrt2<2$ is bounded above.






                            share|cite|improve this answer









                            $endgroup$



                            As the exponential is a growing function,



                            $$x<2implies sqrt2^{,x}<sqrt2^{,2}=2$$ and the sequence starting from $x=sqrt2<2$ is bounded above.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Dec 10 '18 at 8:38









                            Yves DaoustYves Daoust

                            125k671223




                            125k671223






























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