Continued fractions and nested radicals

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Discussion Overview

The discussion revolves around the representation of roots of integers as continued fractions, particularly focusing on square roots and higher-order roots. Participants explore methods for expressing these roots in continued fraction form and examine the limits of nested radicals.

Discussion Character

  • Exploratory
  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant questions whether any square root or integer root can be expressed as a continued fraction, providing an example with Sqrt[2].
  • Another participant mentions Mathematica's capability to convert rational or quadratic irrationals into continued fractions and shares a method for square roots of numbers that are one more than a square.
  • A different participant suggests that roots of the form (\sqrt{x} + a) / b can be represented by simple continued fractions, but notes that higher-order roots lead to more complex representations.
  • One participant challenges the applicability of the proposed form for Sqrt[3], suggesting that it results in an alternating sequence.
  • Another participant acknowledges the potential for repeating sequences in continued fractions for square roots with rational offsets and hints at a formula for conversion between forms.

Areas of Agreement / Disagreement

Participants express differing views on the generalizability of methods for representing roots as continued fractions. While some propose specific forms and methods, others question their validity or applicability to certain roots, indicating that the discussion remains unresolved.

Contextual Notes

Participants reference specific cases and examples, but the discussion lacks a consensus on a general method for all roots, particularly higher-order roots. There are also unresolved assumptions regarding the nature of sequences in continued fractions.

soandos
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is there a way to express any given root of an integer in a continued fraction? i.e. Sqrt[2] = 1 + 1/(2 + Sqrt[2] - 1) and the process can be continued infinitely to get a fraction that defines the radical with only integers.

so my question is can this kind of thing be done with any square root? any integer root?

next question is is there a way to determine the limit of the following:
Sqrt[x^0*a+Sqrt[x*a+Sqrt[x^2*a+Sqrt[x^3*a...]]]] or a similar form for a progression of numbers?
 
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Mathematica has a function that will take a rational or a quadratic irrational and turn it into a continued fraction.
I know a simple one for the square root of any number that is one more than a square. for example: Sqrt[17]-4 == [(Sqrt[17]-4)*(Sqrt[17]+4)]/Sqrt[17]+4
multiplying across the top one gets: 1/Sqrt[17]+4
to make the denominator the same as the original, have 1/Sqrt[17]-4+8, retaining the original value of the expression, and creating a loop back to the beginning. just add four at the very top.
what is the method for doing this manually for any square root?
 
anyone have any ideas?
 
Any root of the form ([tex]\sqrt{x}[/tex] +a ) / b can be represented by a simple continued fraction. Consider how you solve for X = a + [tex]\frac{1}{b+\frac{1}{b+\cdots}}[/tex] using the self similarity identity.

Going for higher order roots means more complex continued fractions, that either aren't self similar or have sequences in their numerators. I have no idea how to handle those in general, and I think its impossible for there to be a general method.
 
i am not sure that that form works. for Sqrt[3] i believe that b is an alternating sequence of 1,2
 
Yes, sorry, that's where the "a" came in. My bad typing.

You may get a repeating sequence, but it will be a finite repeat. It's just like the decimal expansion of 1/7, an infinite sequence with a repeating finite group. This covers all square roots with rational offsets.

I don't remember it now, but I came upon the formula to go from one to the other by solving (0; 1, 1, 1, 1, ... ) and (0; 2, 2, 2, ... ) and extending it to x. In the same way, solve (0; 1, 2, 1, 2, 1, 2,... ) and then generalize that. The algebra doesn't get that bad.
 

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