Confusion about Complex Continued Fraction












10












$begingroup$


I faced some problems when I tried to evaluate the following continued fraction
$$cfrac{1}{i+cfrac{1}{i+cfrac{1}{i+cfrac{1}{i+cfrac{1}{i+cfrac{1}{i+cdots}}}}}} $$
The common trick as usual is to set
$$x= cfrac{1}{i+cfrac{1}{i+cfrac{1}{i+cfrac{1}{i+cfrac{1}{i+cfrac{1}{i+cdots}}}}}} $$
Then, I have
$$x=frac{1}{i+x}$$
Using the quadratic formula, I get
$$x=frac{-ipmsqrt{3}}{2}$$
There are two solutions for $x$, but I don't know which one is correct.
For any real continued fraction, for example
$$cfrac{1}{1+cfrac{1}{1+cfrac{1}{1+cfrac{1}{1+cfrac{1}{1+cfrac{1}{1+cdots}}}}}} $$
it is clear to see that the value is $frac{-1+sqrt{5}}{2}$ instead of $frac{-1-sqrt{5}}{2}$.



But for the complex continued fraction, I have difficulty distinguishing the correct solution.










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    10












    $begingroup$


    I faced some problems when I tried to evaluate the following continued fraction
    $$cfrac{1}{i+cfrac{1}{i+cfrac{1}{i+cfrac{1}{i+cfrac{1}{i+cfrac{1}{i+cdots}}}}}} $$
    The common trick as usual is to set
    $$x= cfrac{1}{i+cfrac{1}{i+cfrac{1}{i+cfrac{1}{i+cfrac{1}{i+cfrac{1}{i+cdots}}}}}} $$
    Then, I have
    $$x=frac{1}{i+x}$$
    Using the quadratic formula, I get
    $$x=frac{-ipmsqrt{3}}{2}$$
    There are two solutions for $x$, but I don't know which one is correct.
    For any real continued fraction, for example
    $$cfrac{1}{1+cfrac{1}{1+cfrac{1}{1+cfrac{1}{1+cfrac{1}{1+cfrac{1}{1+cdots}}}}}} $$
    it is clear to see that the value is $frac{-1+sqrt{5}}{2}$ instead of $frac{-1-sqrt{5}}{2}$.



    But for the complex continued fraction, I have difficulty distinguishing the correct solution.










    share|cite|improve this question









    $endgroup$















      10












      10








      10


      1



      $begingroup$


      I faced some problems when I tried to evaluate the following continued fraction
      $$cfrac{1}{i+cfrac{1}{i+cfrac{1}{i+cfrac{1}{i+cfrac{1}{i+cfrac{1}{i+cdots}}}}}} $$
      The common trick as usual is to set
      $$x= cfrac{1}{i+cfrac{1}{i+cfrac{1}{i+cfrac{1}{i+cfrac{1}{i+cfrac{1}{i+cdots}}}}}} $$
      Then, I have
      $$x=frac{1}{i+x}$$
      Using the quadratic formula, I get
      $$x=frac{-ipmsqrt{3}}{2}$$
      There are two solutions for $x$, but I don't know which one is correct.
      For any real continued fraction, for example
      $$cfrac{1}{1+cfrac{1}{1+cfrac{1}{1+cfrac{1}{1+cfrac{1}{1+cfrac{1}{1+cdots}}}}}} $$
      it is clear to see that the value is $frac{-1+sqrt{5}}{2}$ instead of $frac{-1-sqrt{5}}{2}$.



      But for the complex continued fraction, I have difficulty distinguishing the correct solution.










      share|cite|improve this question









      $endgroup$




      I faced some problems when I tried to evaluate the following continued fraction
      $$cfrac{1}{i+cfrac{1}{i+cfrac{1}{i+cfrac{1}{i+cfrac{1}{i+cfrac{1}{i+cdots}}}}}} $$
      The common trick as usual is to set
      $$x= cfrac{1}{i+cfrac{1}{i+cfrac{1}{i+cfrac{1}{i+cfrac{1}{i+cfrac{1}{i+cdots}}}}}} $$
      Then, I have
      $$x=frac{1}{i+x}$$
      Using the quadratic formula, I get
      $$x=frac{-ipmsqrt{3}}{2}$$
      There are two solutions for $x$, but I don't know which one is correct.
      For any real continued fraction, for example
      $$cfrac{1}{1+cfrac{1}{1+cfrac{1}{1+cfrac{1}{1+cfrac{1}{1+cfrac{1}{1+cdots}}}}}} $$
      it is clear to see that the value is $frac{-1+sqrt{5}}{2}$ instead of $frac{-1-sqrt{5}}{2}$.



      But for the complex continued fraction, I have difficulty distinguishing the correct solution.







      complex-numbers continued-fractions






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      share|cite|improve this question











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      share|cite|improve this question










      asked 16 hours ago









      LarryLarry

      2,52331131




      2,52331131






















          3 Answers
          3






          active

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          25












          $begingroup$

          Well, Vasily has put his finger on one problem, but I would like to point out a much more serious one.



          We write a continued fraction to get a number that is the limit of the convergents, that is, of the expressions that you get when you cut your continued fraction off, to be a finite c.f.



          The first convergent is $frac1i$, no problem, but the second is
          $$frac1{i+frac1i}=frac10,,$$ a most unfortunate development. My recommendation would be to go on to other complex continued fractions, where the partial denominators are rather larger than $i$.






          share|cite|improve this answer









          $endgroup$









          • 4




            $begingroup$
            [+1] very thorough remark.
            $endgroup$
            – Jean Marie
            13 hours ago










          • $begingroup$
            I'm not sure what you mean by your last sentence. What about $[x;x,x,cdots]$ where $x = sqrt{2}i$?
            $endgroup$
            – user21820
            7 mins ago










          • $begingroup$
            I didn’t mean much. But $sqrt2i$ is not what I had in mind by “rather larger”. Indeed, even $x=2i$ seems to give an uninteresting limiting value.
            $endgroup$
            – Lubin
            34 secs ago



















          20












          $begingroup$

          The problem with this expression is that function $f(x)=(i+x)^{-1}$ is a tricky one:



          $$
          f(f(x)) = frac{1}{i+frac{1}{i+x}}=frac{i+x}{-1+ix+1}=-i+frac1x,\
          f(f(f(x))) = frac{1}{i+(-i+1/x)}=x.
          $$



          So every number generates the orbit of length 3, except for two numbers you have found, which are fixed points. Thus, the series of nesting functions $f$ does not converge if you haven't already started with a fixed point. So the question as it is stated has no sense: you cannot assign any number to this expression.






          share|cite|improve this answer









          $endgroup$





















            2












            $begingroup$

            For future complex continued fractions...



            For a continued fraction to converge, the sequence of convergents (values finite initial segments of the partial fraction expression) must converge to a particular complex number, that is, to a specific argument and magnitude. For the real continued fraction you mention, the algebraic expression allows the argument to be either $0$ (containing "$+sqrt{5}$") or $pi$ (containing "$-sqrt{5}$"). The same thing can be done with the complex version. Your two expressions have arguments $-pi/6$ and $7pi/6$. If your continued fraction had any hope of converging (which, as others have shown, it does not) it converges to something with a specific argument. You generically hope that one of the algebraic solutions you have has an argument matching the limit argument of the convergents. Finding that limit argument usually requires another computation, typically solving another recurrence.






            share|cite|improve this answer









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              3 Answers
              3






              active

              oldest

              votes








              3 Answers
              3






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes









              25












              $begingroup$

              Well, Vasily has put his finger on one problem, but I would like to point out a much more serious one.



              We write a continued fraction to get a number that is the limit of the convergents, that is, of the expressions that you get when you cut your continued fraction off, to be a finite c.f.



              The first convergent is $frac1i$, no problem, but the second is
              $$frac1{i+frac1i}=frac10,,$$ a most unfortunate development. My recommendation would be to go on to other complex continued fractions, where the partial denominators are rather larger than $i$.






              share|cite|improve this answer









              $endgroup$









              • 4




                $begingroup$
                [+1] very thorough remark.
                $endgroup$
                – Jean Marie
                13 hours ago










              • $begingroup$
                I'm not sure what you mean by your last sentence. What about $[x;x,x,cdots]$ where $x = sqrt{2}i$?
                $endgroup$
                – user21820
                7 mins ago










              • $begingroup$
                I didn’t mean much. But $sqrt2i$ is not what I had in mind by “rather larger”. Indeed, even $x=2i$ seems to give an uninteresting limiting value.
                $endgroup$
                – Lubin
                34 secs ago
















              25












              $begingroup$

              Well, Vasily has put his finger on one problem, but I would like to point out a much more serious one.



              We write a continued fraction to get a number that is the limit of the convergents, that is, of the expressions that you get when you cut your continued fraction off, to be a finite c.f.



              The first convergent is $frac1i$, no problem, but the second is
              $$frac1{i+frac1i}=frac10,,$$ a most unfortunate development. My recommendation would be to go on to other complex continued fractions, where the partial denominators are rather larger than $i$.






              share|cite|improve this answer









              $endgroup$









              • 4




                $begingroup$
                [+1] very thorough remark.
                $endgroup$
                – Jean Marie
                13 hours ago










              • $begingroup$
                I'm not sure what you mean by your last sentence. What about $[x;x,x,cdots]$ where $x = sqrt{2}i$?
                $endgroup$
                – user21820
                7 mins ago










              • $begingroup$
                I didn’t mean much. But $sqrt2i$ is not what I had in mind by “rather larger”. Indeed, even $x=2i$ seems to give an uninteresting limiting value.
                $endgroup$
                – Lubin
                34 secs ago














              25












              25








              25





              $begingroup$

              Well, Vasily has put his finger on one problem, but I would like to point out a much more serious one.



              We write a continued fraction to get a number that is the limit of the convergents, that is, of the expressions that you get when you cut your continued fraction off, to be a finite c.f.



              The first convergent is $frac1i$, no problem, but the second is
              $$frac1{i+frac1i}=frac10,,$$ a most unfortunate development. My recommendation would be to go on to other complex continued fractions, where the partial denominators are rather larger than $i$.






              share|cite|improve this answer









              $endgroup$



              Well, Vasily has put his finger on one problem, but I would like to point out a much more serious one.



              We write a continued fraction to get a number that is the limit of the convergents, that is, of the expressions that you get when you cut your continued fraction off, to be a finite c.f.



              The first convergent is $frac1i$, no problem, but the second is
              $$frac1{i+frac1i}=frac10,,$$ a most unfortunate development. My recommendation would be to go on to other complex continued fractions, where the partial denominators are rather larger than $i$.







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered 14 hours ago









              LubinLubin

              45k44687




              45k44687








              • 4




                $begingroup$
                [+1] very thorough remark.
                $endgroup$
                – Jean Marie
                13 hours ago










              • $begingroup$
                I'm not sure what you mean by your last sentence. What about $[x;x,x,cdots]$ where $x = sqrt{2}i$?
                $endgroup$
                – user21820
                7 mins ago










              • $begingroup$
                I didn’t mean much. But $sqrt2i$ is not what I had in mind by “rather larger”. Indeed, even $x=2i$ seems to give an uninteresting limiting value.
                $endgroup$
                – Lubin
                34 secs ago














              • 4




                $begingroup$
                [+1] very thorough remark.
                $endgroup$
                – Jean Marie
                13 hours ago










              • $begingroup$
                I'm not sure what you mean by your last sentence. What about $[x;x,x,cdots]$ where $x = sqrt{2}i$?
                $endgroup$
                – user21820
                7 mins ago










              • $begingroup$
                I didn’t mean much. But $sqrt2i$ is not what I had in mind by “rather larger”. Indeed, even $x=2i$ seems to give an uninteresting limiting value.
                $endgroup$
                – Lubin
                34 secs ago








              4




              4




              $begingroup$
              [+1] very thorough remark.
              $endgroup$
              – Jean Marie
              13 hours ago




              $begingroup$
              [+1] very thorough remark.
              $endgroup$
              – Jean Marie
              13 hours ago












              $begingroup$
              I'm not sure what you mean by your last sentence. What about $[x;x,x,cdots]$ where $x = sqrt{2}i$?
              $endgroup$
              – user21820
              7 mins ago




              $begingroup$
              I'm not sure what you mean by your last sentence. What about $[x;x,x,cdots]$ where $x = sqrt{2}i$?
              $endgroup$
              – user21820
              7 mins ago












              $begingroup$
              I didn’t mean much. But $sqrt2i$ is not what I had in mind by “rather larger”. Indeed, even $x=2i$ seems to give an uninteresting limiting value.
              $endgroup$
              – Lubin
              34 secs ago




              $begingroup$
              I didn’t mean much. But $sqrt2i$ is not what I had in mind by “rather larger”. Indeed, even $x=2i$ seems to give an uninteresting limiting value.
              $endgroup$
              – Lubin
              34 secs ago











              20












              $begingroup$

              The problem with this expression is that function $f(x)=(i+x)^{-1}$ is a tricky one:



              $$
              f(f(x)) = frac{1}{i+frac{1}{i+x}}=frac{i+x}{-1+ix+1}=-i+frac1x,\
              f(f(f(x))) = frac{1}{i+(-i+1/x)}=x.
              $$



              So every number generates the orbit of length 3, except for two numbers you have found, which are fixed points. Thus, the series of nesting functions $f$ does not converge if you haven't already started with a fixed point. So the question as it is stated has no sense: you cannot assign any number to this expression.






              share|cite|improve this answer









              $endgroup$


















                20












                $begingroup$

                The problem with this expression is that function $f(x)=(i+x)^{-1}$ is a tricky one:



                $$
                f(f(x)) = frac{1}{i+frac{1}{i+x}}=frac{i+x}{-1+ix+1}=-i+frac1x,\
                f(f(f(x))) = frac{1}{i+(-i+1/x)}=x.
                $$



                So every number generates the orbit of length 3, except for two numbers you have found, which are fixed points. Thus, the series of nesting functions $f$ does not converge if you haven't already started with a fixed point. So the question as it is stated has no sense: you cannot assign any number to this expression.






                share|cite|improve this answer









                $endgroup$
















                  20












                  20








                  20





                  $begingroup$

                  The problem with this expression is that function $f(x)=(i+x)^{-1}$ is a tricky one:



                  $$
                  f(f(x)) = frac{1}{i+frac{1}{i+x}}=frac{i+x}{-1+ix+1}=-i+frac1x,\
                  f(f(f(x))) = frac{1}{i+(-i+1/x)}=x.
                  $$



                  So every number generates the orbit of length 3, except for two numbers you have found, which are fixed points. Thus, the series of nesting functions $f$ does not converge if you haven't already started with a fixed point. So the question as it is stated has no sense: you cannot assign any number to this expression.






                  share|cite|improve this answer









                  $endgroup$



                  The problem with this expression is that function $f(x)=(i+x)^{-1}$ is a tricky one:



                  $$
                  f(f(x)) = frac{1}{i+frac{1}{i+x}}=frac{i+x}{-1+ix+1}=-i+frac1x,\
                  f(f(f(x))) = frac{1}{i+(-i+1/x)}=x.
                  $$



                  So every number generates the orbit of length 3, except for two numbers you have found, which are fixed points. Thus, the series of nesting functions $f$ does not converge if you haven't already started with a fixed point. So the question as it is stated has no sense: you cannot assign any number to this expression.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 15 hours ago









                  Vasily MitchVasily Mitch

                  2,5941312




                  2,5941312























                      2












                      $begingroup$

                      For future complex continued fractions...



                      For a continued fraction to converge, the sequence of convergents (values finite initial segments of the partial fraction expression) must converge to a particular complex number, that is, to a specific argument and magnitude. For the real continued fraction you mention, the algebraic expression allows the argument to be either $0$ (containing "$+sqrt{5}$") or $pi$ (containing "$-sqrt{5}$"). The same thing can be done with the complex version. Your two expressions have arguments $-pi/6$ and $7pi/6$. If your continued fraction had any hope of converging (which, as others have shown, it does not) it converges to something with a specific argument. You generically hope that one of the algebraic solutions you have has an argument matching the limit argument of the convergents. Finding that limit argument usually requires another computation, typically solving another recurrence.






                      share|cite|improve this answer









                      $endgroup$


















                        2












                        $begingroup$

                        For future complex continued fractions...



                        For a continued fraction to converge, the sequence of convergents (values finite initial segments of the partial fraction expression) must converge to a particular complex number, that is, to a specific argument and magnitude. For the real continued fraction you mention, the algebraic expression allows the argument to be either $0$ (containing "$+sqrt{5}$") or $pi$ (containing "$-sqrt{5}$"). The same thing can be done with the complex version. Your two expressions have arguments $-pi/6$ and $7pi/6$. If your continued fraction had any hope of converging (which, as others have shown, it does not) it converges to something with a specific argument. You generically hope that one of the algebraic solutions you have has an argument matching the limit argument of the convergents. Finding that limit argument usually requires another computation, typically solving another recurrence.






                        share|cite|improve this answer









                        $endgroup$
















                          2












                          2








                          2





                          $begingroup$

                          For future complex continued fractions...



                          For a continued fraction to converge, the sequence of convergents (values finite initial segments of the partial fraction expression) must converge to a particular complex number, that is, to a specific argument and magnitude. For the real continued fraction you mention, the algebraic expression allows the argument to be either $0$ (containing "$+sqrt{5}$") or $pi$ (containing "$-sqrt{5}$"). The same thing can be done with the complex version. Your two expressions have arguments $-pi/6$ and $7pi/6$. If your continued fraction had any hope of converging (which, as others have shown, it does not) it converges to something with a specific argument. You generically hope that one of the algebraic solutions you have has an argument matching the limit argument of the convergents. Finding that limit argument usually requires another computation, typically solving another recurrence.






                          share|cite|improve this answer









                          $endgroup$



                          For future complex continued fractions...



                          For a continued fraction to converge, the sequence of convergents (values finite initial segments of the partial fraction expression) must converge to a particular complex number, that is, to a specific argument and magnitude. For the real continued fraction you mention, the algebraic expression allows the argument to be either $0$ (containing "$+sqrt{5}$") or $pi$ (containing "$-sqrt{5}$"). The same thing can be done with the complex version. Your two expressions have arguments $-pi/6$ and $7pi/6$. If your continued fraction had any hope of converging (which, as others have shown, it does not) it converges to something with a specific argument. You generically hope that one of the algebraic solutions you have has an argument matching the limit argument of the convergents. Finding that limit argument usually requires another computation, typically solving another recurrence.







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered 10 hours ago









                          Eric TowersEric Towers

                          32.8k22370




                          32.8k22370






























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