Convergence of the sequence $x_{k+1}=frac{1}{2}(x_k+frac{a}{x_k})$












3















Consider the sequence $x_{k+1}=frac{1}{2}(x_k+frac{a}{x_k}), agt 0, xinmathbb{R}$. Assume the sequence converges, what does it converge to?




I'm having trouble seeing how to start,



Any help would be appreciated



Thanks






real-analysis sequences-and-series







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    3















    Consider the sequence $x_{k+1}=frac{1}{2}(x_k+frac{a}{x_k}), agt 0, xinmathbb{R}$. Assume the sequence converges, what does it converge to?




    I'm having trouble seeing how to start,



    Any help would be appreciated



    Thanks






    real-analysis sequences-and-series







    share|cite|improve this question



























      3












      3








      3


      2






      Consider the sequence $x_{k+1}=frac{1}{2}(x_k+frac{a}{x_k}), agt 0, xinmathbb{R}$. Assume the sequence converges, what does it converge to?




      I'm having trouble seeing how to start,



      Any help would be appreciated



      Thanks






      real-analysis sequences-and-series







      share|cite|improve this question
















      Consider the sequence $x_{k+1}=frac{1}{2}(x_k+frac{a}{x_k}), agt 0, xinmathbb{R}$. Assume the sequence converges, what does it converge to?




      I'm having trouble seeing how to start,



      Any help would be appreciated



      Thanks






      real-analysis sequences-and-series




      real-analysis sequences-and-series






      share|cite|improve this question















      share|cite|improve this question













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      edited Nov 26 '18 at 7:21









      Jean-Claude Arbaut

      14.7k63464




      14.7k63464










      asked Mar 22 '13 at 17:47









      bobdylan

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          The number $x$ to which it converges should satisfy $x =frac12left(x+frac a xright)$. Solve that equation for $x$. (It may help to begin by multiplying both sides by $x$, thereby getting rid of the fraction.)






          share|cite|improve this answer





























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            If $x_n to g$, then, if $g ne 0$, we have $frac{1}{2}(x_n + frac{a}{x_n}) to frac{1}{2}(g+frac{a}{g})$. We also have $x_{n+1} to g$, so since $x_{n+1} = frac{1}{2}(x_n + frac{a}{x_n})$, we have that $g = frac{1}{2}(g + frac{a}{g})$.






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              This sequence has a closed form.
              First, if $x_0=sqrt{a}$, $x_n=sqrt{a}$ for all $n$, and the sequence is constant. We will thus assume $x_0nesqrt{a}$ in the following.



              Now, given $x_nnesqrt{a}$ for some $n$, we have
              $$x_{n+1}-sqrt{a}=frac{1}{2}left(x_n+ frac{a}{x_n} right) - sqrt{a}=frac{1}{2}left( sqrt{x_n} - frac{sqrt{a}}{sqrt{x_n}}right)^2 > 0$$
              Hence $x_n>sqrt{a}$ for all $n>0$. To simplify a bit, we can safely assume that $x_0 > sqrt{a}$ too.



              We can thus write $x_n = sqrt{a} coth{t_n}$. Then
              $$x_{n+1} = frac{1}{2}left( sqrt{a} coth{t_n} + frac{a}{sqrt{a} coth{t_n}} right) = frac{1}{2}sqrt{a} left( coth{t_n} + mathrm{th} t_nright) = sqrt{a} coth 2 t_n$$
              We have thus $t_{n+1}=2 t_n$, and $t_n= 2^n t_0$, then for $x_n$ :
              $$x_n=sqrt{a} coth left( 2^n arg coth frac{x_0}{sqrt{a}}right)$$
              And the value inside parentheses tends to infinity, so $lim(x_n)=sqrt{a}$.






              share|cite|improve this answer























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                3 Answers
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                3 Answers
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                active

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                active

                oldest

                votes









                2














                The number $x$ to which it converges should satisfy $x =frac12left(x+frac a xright)$. Solve that equation for $x$. (It may help to begin by multiplying both sides by $x$, thereby getting rid of the fraction.)






                share|cite|improve this answer


























                  2














                  The number $x$ to which it converges should satisfy $x =frac12left(x+frac a xright)$. Solve that equation for $x$. (It may help to begin by multiplying both sides by $x$, thereby getting rid of the fraction.)






                  share|cite|improve this answer
























                    2












                    2








                    2






                    The number $x$ to which it converges should satisfy $x =frac12left(x+frac a xright)$. Solve that equation for $x$. (It may help to begin by multiplying both sides by $x$, thereby getting rid of the fraction.)






                    share|cite|improve this answer












                    The number $x$ to which it converges should satisfy $x =frac12left(x+frac a xright)$. Solve that equation for $x$. (It may help to begin by multiplying both sides by $x$, thereby getting rid of the fraction.)







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Mar 22 '13 at 17:50









                    Michael Hardy

                    1




                    1























                        2














                        If $x_n to g$, then, if $g ne 0$, we have $frac{1}{2}(x_n + frac{a}{x_n}) to frac{1}{2}(g+frac{a}{g})$. We also have $x_{n+1} to g$, so since $x_{n+1} = frac{1}{2}(x_n + frac{a}{x_n})$, we have that $g = frac{1}{2}(g + frac{a}{g})$.






                        share|cite|improve this answer


























                          2














                          If $x_n to g$, then, if $g ne 0$, we have $frac{1}{2}(x_n + frac{a}{x_n}) to frac{1}{2}(g+frac{a}{g})$. We also have $x_{n+1} to g$, so since $x_{n+1} = frac{1}{2}(x_n + frac{a}{x_n})$, we have that $g = frac{1}{2}(g + frac{a}{g})$.






                          share|cite|improve this answer
























                            2












                            2








                            2






                            If $x_n to g$, then, if $g ne 0$, we have $frac{1}{2}(x_n + frac{a}{x_n}) to frac{1}{2}(g+frac{a}{g})$. We also have $x_{n+1} to g$, so since $x_{n+1} = frac{1}{2}(x_n + frac{a}{x_n})$, we have that $g = frac{1}{2}(g + frac{a}{g})$.






                            share|cite|improve this answer












                            If $x_n to g$, then, if $g ne 0$, we have $frac{1}{2}(x_n + frac{a}{x_n}) to frac{1}{2}(g+frac{a}{g})$. We also have $x_{n+1} to g$, so since $x_{n+1} = frac{1}{2}(x_n + frac{a}{x_n})$, we have that $g = frac{1}{2}(g + frac{a}{g})$.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Mar 22 '13 at 17:51









                            xyzzyz

                            5,5361221




                            5,5361221























                                2














                                This sequence has a closed form.
                                First, if $x_0=sqrt{a}$, $x_n=sqrt{a}$ for all $n$, and the sequence is constant. We will thus assume $x_0nesqrt{a}$ in the following.



                                Now, given $x_nnesqrt{a}$ for some $n$, we have
                                $$x_{n+1}-sqrt{a}=frac{1}{2}left(x_n+ frac{a}{x_n} right) - sqrt{a}=frac{1}{2}left( sqrt{x_n} - frac{sqrt{a}}{sqrt{x_n}}right)^2 > 0$$
                                Hence $x_n>sqrt{a}$ for all $n>0$. To simplify a bit, we can safely assume that $x_0 > sqrt{a}$ too.



                                We can thus write $x_n = sqrt{a} coth{t_n}$. Then
                                $$x_{n+1} = frac{1}{2}left( sqrt{a} coth{t_n} + frac{a}{sqrt{a} coth{t_n}} right) = frac{1}{2}sqrt{a} left( coth{t_n} + mathrm{th} t_nright) = sqrt{a} coth 2 t_n$$
                                We have thus $t_{n+1}=2 t_n$, and $t_n= 2^n t_0$, then for $x_n$ :
                                $$x_n=sqrt{a} coth left( 2^n arg coth frac{x_0}{sqrt{a}}right)$$
                                And the value inside parentheses tends to infinity, so $lim(x_n)=sqrt{a}$.






                                share|cite|improve this answer




























                                  2














                                  This sequence has a closed form.
                                  First, if $x_0=sqrt{a}$, $x_n=sqrt{a}$ for all $n$, and the sequence is constant. We will thus assume $x_0nesqrt{a}$ in the following.



                                  Now, given $x_nnesqrt{a}$ for some $n$, we have
                                  $$x_{n+1}-sqrt{a}=frac{1}{2}left(x_n+ frac{a}{x_n} right) - sqrt{a}=frac{1}{2}left( sqrt{x_n} - frac{sqrt{a}}{sqrt{x_n}}right)^2 > 0$$
                                  Hence $x_n>sqrt{a}$ for all $n>0$. To simplify a bit, we can safely assume that $x_0 > sqrt{a}$ too.



                                  We can thus write $x_n = sqrt{a} coth{t_n}$. Then
                                  $$x_{n+1} = frac{1}{2}left( sqrt{a} coth{t_n} + frac{a}{sqrt{a} coth{t_n}} right) = frac{1}{2}sqrt{a} left( coth{t_n} + mathrm{th} t_nright) = sqrt{a} coth 2 t_n$$
                                  We have thus $t_{n+1}=2 t_n$, and $t_n= 2^n t_0$, then for $x_n$ :
                                  $$x_n=sqrt{a} coth left( 2^n arg coth frac{x_0}{sqrt{a}}right)$$
                                  And the value inside parentheses tends to infinity, so $lim(x_n)=sqrt{a}$.






                                  share|cite|improve this answer


























                                    2












                                    2








                                    2






                                    This sequence has a closed form.
                                    First, if $x_0=sqrt{a}$, $x_n=sqrt{a}$ for all $n$, and the sequence is constant. We will thus assume $x_0nesqrt{a}$ in the following.



                                    Now, given $x_nnesqrt{a}$ for some $n$, we have
                                    $$x_{n+1}-sqrt{a}=frac{1}{2}left(x_n+ frac{a}{x_n} right) - sqrt{a}=frac{1}{2}left( sqrt{x_n} - frac{sqrt{a}}{sqrt{x_n}}right)^2 > 0$$
                                    Hence $x_n>sqrt{a}$ for all $n>0$. To simplify a bit, we can safely assume that $x_0 > sqrt{a}$ too.



                                    We can thus write $x_n = sqrt{a} coth{t_n}$. Then
                                    $$x_{n+1} = frac{1}{2}left( sqrt{a} coth{t_n} + frac{a}{sqrt{a} coth{t_n}} right) = frac{1}{2}sqrt{a} left( coth{t_n} + mathrm{th} t_nright) = sqrt{a} coth 2 t_n$$
                                    We have thus $t_{n+1}=2 t_n$, and $t_n= 2^n t_0$, then for $x_n$ :
                                    $$x_n=sqrt{a} coth left( 2^n arg coth frac{x_0}{sqrt{a}}right)$$
                                    And the value inside parentheses tends to infinity, so $lim(x_n)=sqrt{a}$.






                                    share|cite|improve this answer














                                    This sequence has a closed form.
                                    First, if $x_0=sqrt{a}$, $x_n=sqrt{a}$ for all $n$, and the sequence is constant. We will thus assume $x_0nesqrt{a}$ in the following.



                                    Now, given $x_nnesqrt{a}$ for some $n$, we have
                                    $$x_{n+1}-sqrt{a}=frac{1}{2}left(x_n+ frac{a}{x_n} right) - sqrt{a}=frac{1}{2}left( sqrt{x_n} - frac{sqrt{a}}{sqrt{x_n}}right)^2 > 0$$
                                    Hence $x_n>sqrt{a}$ for all $n>0$. To simplify a bit, we can safely assume that $x_0 > sqrt{a}$ too.



                                    We can thus write $x_n = sqrt{a} coth{t_n}$. Then
                                    $$x_{n+1} = frac{1}{2}left( sqrt{a} coth{t_n} + frac{a}{sqrt{a} coth{t_n}} right) = frac{1}{2}sqrt{a} left( coth{t_n} + mathrm{th} t_nright) = sqrt{a} coth 2 t_n$$
                                    We have thus $t_{n+1}=2 t_n$, and $t_n= 2^n t_0$, then for $x_n$ :
                                    $$x_n=sqrt{a} coth left( 2^n arg coth frac{x_0}{sqrt{a}}right)$$
                                    And the value inside parentheses tends to infinity, so $lim(x_n)=sqrt{a}$.







                                    share|cite|improve this answer














                                    share|cite|improve this answer



                                    share|cite|improve this answer








                                    edited Nov 25 '18 at 22:13

























                                    answered Mar 22 '13 at 18:18









                                    Jean-Claude Arbaut

                                    14.7k63464




                                    14.7k63464






























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