About the Fibonacci “continued fractions”











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Fibonacci used a different version of continued fraction which I'm curious about. It seems the notation is equivalent to
$$[a_1,a_2,a_3,cdots]=frac{1}{a_1}+frac{1}{a_1a_2}+frac{1}{a_1a_2a_3}+cdots,$$
where $a_k$ are positive integers and (I believe) it is required that $1 le a_1 le a_2 le a_3 le cdots .$
The reason for my belief is that if we drop the nondecreasing assumption uniqueness is lost.



What I'd like to see is a proof (or reference to one) that each positive real has a unique such expression. Or just a reference to this type of "continued fraction" which I could follow up on.










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  • 1




    What you refer to is an Engel expansion and the attribution to Fibonacci is unsupported.
    – Somos
    8 hours ago










  • @Somos -- the link you give refers to Fibonacci's fractions, which I also saw in Burton's "Intro to Number Theory" book.
    – coffeemath
    4 hours ago










  • @Somos The exact title is "Elementary Number Theory" for the Burton book. He only mentions it briefly as an introduction to the modern version, see chapter 15 (section 15.2] I hadn't heard of Engel expansions (or maybe that would have stopped my even asking the question). But thanks much for that reference, which I can now follow up on.
    – coffeemath
    2 hours ago















up vote
1
down vote

favorite












Fibonacci used a different version of continued fraction which I'm curious about. It seems the notation is equivalent to
$$[a_1,a_2,a_3,cdots]=frac{1}{a_1}+frac{1}{a_1a_2}+frac{1}{a_1a_2a_3}+cdots,$$
where $a_k$ are positive integers and (I believe) it is required that $1 le a_1 le a_2 le a_3 le cdots .$
The reason for my belief is that if we drop the nondecreasing assumption uniqueness is lost.



What I'd like to see is a proof (or reference to one) that each positive real has a unique such expression. Or just a reference to this type of "continued fraction" which I could follow up on.










share|cite|improve this question


















  • 1




    What you refer to is an Engel expansion and the attribution to Fibonacci is unsupported.
    – Somos
    8 hours ago










  • @Somos -- the link you give refers to Fibonacci's fractions, which I also saw in Burton's "Intro to Number Theory" book.
    – coffeemath
    4 hours ago










  • @Somos The exact title is "Elementary Number Theory" for the Burton book. He only mentions it briefly as an introduction to the modern version, see chapter 15 (section 15.2] I hadn't heard of Engel expansions (or maybe that would have stopped my even asking the question). But thanks much for that reference, which I can now follow up on.
    – coffeemath
    2 hours ago













up vote
1
down vote

favorite









up vote
1
down vote

favorite











Fibonacci used a different version of continued fraction which I'm curious about. It seems the notation is equivalent to
$$[a_1,a_2,a_3,cdots]=frac{1}{a_1}+frac{1}{a_1a_2}+frac{1}{a_1a_2a_3}+cdots,$$
where $a_k$ are positive integers and (I believe) it is required that $1 le a_1 le a_2 le a_3 le cdots .$
The reason for my belief is that if we drop the nondecreasing assumption uniqueness is lost.



What I'd like to see is a proof (or reference to one) that each positive real has a unique such expression. Or just a reference to this type of "continued fraction" which I could follow up on.










share|cite|improve this question













Fibonacci used a different version of continued fraction which I'm curious about. It seems the notation is equivalent to
$$[a_1,a_2,a_3,cdots]=frac{1}{a_1}+frac{1}{a_1a_2}+frac{1}{a_1a_2a_3}+cdots,$$
where $a_k$ are positive integers and (I believe) it is required that $1 le a_1 le a_2 le a_3 le cdots .$
The reason for my belief is that if we drop the nondecreasing assumption uniqueness is lost.



What I'd like to see is a proof (or reference to one) that each positive real has a unique such expression. Or just a reference to this type of "continued fraction" which I could follow up on.







elementary-number-theory






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asked 10 hours ago









coffeemath

1,8551313




1,8551313








  • 1




    What you refer to is an Engel expansion and the attribution to Fibonacci is unsupported.
    – Somos
    8 hours ago










  • @Somos -- the link you give refers to Fibonacci's fractions, which I also saw in Burton's "Intro to Number Theory" book.
    – coffeemath
    4 hours ago










  • @Somos The exact title is "Elementary Number Theory" for the Burton book. He only mentions it briefly as an introduction to the modern version, see chapter 15 (section 15.2] I hadn't heard of Engel expansions (or maybe that would have stopped my even asking the question). But thanks much for that reference, which I can now follow up on.
    – coffeemath
    2 hours ago














  • 1




    What you refer to is an Engel expansion and the attribution to Fibonacci is unsupported.
    – Somos
    8 hours ago










  • @Somos -- the link you give refers to Fibonacci's fractions, which I also saw in Burton's "Intro to Number Theory" book.
    – coffeemath
    4 hours ago










  • @Somos The exact title is "Elementary Number Theory" for the Burton book. He only mentions it briefly as an introduction to the modern version, see chapter 15 (section 15.2] I hadn't heard of Engel expansions (or maybe that would have stopped my even asking the question). But thanks much for that reference, which I can now follow up on.
    – coffeemath
    2 hours ago








1




1




What you refer to is an Engel expansion and the attribution to Fibonacci is unsupported.
– Somos
8 hours ago




What you refer to is an Engel expansion and the attribution to Fibonacci is unsupported.
– Somos
8 hours ago












@Somos -- the link you give refers to Fibonacci's fractions, which I also saw in Burton's "Intro to Number Theory" book.
– coffeemath
4 hours ago




@Somos -- the link you give refers to Fibonacci's fractions, which I also saw in Burton's "Intro to Number Theory" book.
– coffeemath
4 hours ago












@Somos The exact title is "Elementary Number Theory" for the Burton book. He only mentions it briefly as an introduction to the modern version, see chapter 15 (section 15.2] I hadn't heard of Engel expansions (or maybe that would have stopped my even asking the question). But thanks much for that reference, which I can now follow up on.
– coffeemath
2 hours ago




@Somos The exact title is "Elementary Number Theory" for the Burton book. He only mentions it briefly as an introduction to the modern version, see chapter 15 (section 15.2] I hadn't heard of Engel expansions (or maybe that would have stopped my even asking the question). But thanks much for that reference, which I can now follow up on.
– coffeemath
2 hours ago










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Let $x_1$ be a positive real number. Take $a_1>0$ minimal such that $frac{1}{a_1} < x_1$ and set $x_2 = a_1x_1 - 1$. Note that $x_2$ is positive and by the minimality of $a_1$ we have
$$
x_1 le frac{1}{a_1-1}
$$

which is equivalent to $x_2le x_1$. Now letting $a_2>0$ be minimal such that $frac{1}{a_2} < x_2$ we must have $a_2ge a_1$ and obtained
$$
x_1 = frac{1}{a_1} + frac{x_2}{a_1} = frac{1}{a_1} + frac{1}{a_1a_2} + frac{x_2a_2-1}{a_1a_2}.
$$

Repeating this, setting $x_{i+1} = x_i a_i-1$ in each step, should yield the desired unique expansion.






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    Let $x_1$ be a positive real number. Take $a_1>0$ minimal such that $frac{1}{a_1} < x_1$ and set $x_2 = a_1x_1 - 1$. Note that $x_2$ is positive and by the minimality of $a_1$ we have
    $$
    x_1 le frac{1}{a_1-1}
    $$

    which is equivalent to $x_2le x_1$. Now letting $a_2>0$ be minimal such that $frac{1}{a_2} < x_2$ we must have $a_2ge a_1$ and obtained
    $$
    x_1 = frac{1}{a_1} + frac{x_2}{a_1} = frac{1}{a_1} + frac{1}{a_1a_2} + frac{x_2a_2-1}{a_1a_2}.
    $$

    Repeating this, setting $x_{i+1} = x_i a_i-1$ in each step, should yield the desired unique expansion.






    share|cite|improve this answer

























      up vote
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      down vote



      accepted










      Let $x_1$ be a positive real number. Take $a_1>0$ minimal such that $frac{1}{a_1} < x_1$ and set $x_2 = a_1x_1 - 1$. Note that $x_2$ is positive and by the minimality of $a_1$ we have
      $$
      x_1 le frac{1}{a_1-1}
      $$

      which is equivalent to $x_2le x_1$. Now letting $a_2>0$ be minimal such that $frac{1}{a_2} < x_2$ we must have $a_2ge a_1$ and obtained
      $$
      x_1 = frac{1}{a_1} + frac{x_2}{a_1} = frac{1}{a_1} + frac{1}{a_1a_2} + frac{x_2a_2-1}{a_1a_2}.
      $$

      Repeating this, setting $x_{i+1} = x_i a_i-1$ in each step, should yield the desired unique expansion.






      share|cite|improve this answer























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        up vote
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        down vote



        accepted






        Let $x_1$ be a positive real number. Take $a_1>0$ minimal such that $frac{1}{a_1} < x_1$ and set $x_2 = a_1x_1 - 1$. Note that $x_2$ is positive and by the minimality of $a_1$ we have
        $$
        x_1 le frac{1}{a_1-1}
        $$

        which is equivalent to $x_2le x_1$. Now letting $a_2>0$ be minimal such that $frac{1}{a_2} < x_2$ we must have $a_2ge a_1$ and obtained
        $$
        x_1 = frac{1}{a_1} + frac{x_2}{a_1} = frac{1}{a_1} + frac{1}{a_1a_2} + frac{x_2a_2-1}{a_1a_2}.
        $$

        Repeating this, setting $x_{i+1} = x_i a_i-1$ in each step, should yield the desired unique expansion.






        share|cite|improve this answer












        Let $x_1$ be a positive real number. Take $a_1>0$ minimal such that $frac{1}{a_1} < x_1$ and set $x_2 = a_1x_1 - 1$. Note that $x_2$ is positive and by the minimality of $a_1$ we have
        $$
        x_1 le frac{1}{a_1-1}
        $$

        which is equivalent to $x_2le x_1$. Now letting $a_2>0$ be minimal such that $frac{1}{a_2} < x_2$ we must have $a_2ge a_1$ and obtained
        $$
        x_1 = frac{1}{a_1} + frac{x_2}{a_1} = frac{1}{a_1} + frac{1}{a_1a_2} + frac{x_2a_2-1}{a_1a_2}.
        $$

        Repeating this, setting $x_{i+1} = x_i a_i-1$ in each step, should yield the desired unique expansion.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered 10 hours ago









        Christoph

        11.2k1341




        11.2k1341






























             

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