Solve x in Periodic Continued Fraction Expansion

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In summary, the conversation discusses finding the real positive number x with periodic continued fraction expansion [0;a,b,a,b,...], where a and b are positive integers. The notation is explained and the quadratic equation x^2+bx-a=0 is used to solve for x. An additional question about computing 1/α-a is discussed, and the method of finding the solution by repeatedly performing operations on α until it is obtained again is mentioned. The conversation concludes with appreciation for the simple solution presented.
  • #1
*FaerieLight*
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Homework Statement



Find the real positive number x with periodic continued fraction expansion [0;a,b,a,b,...] where a and b are positive integers.

Homework Equations





The Attempt at a Solution



I have determined that x definitely involves a root.
 
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  • #2
It's been a while since I have done these kinds of problems but I think the notation "[0;a,b,a,b,...]" means a continued fraction of the form
[tex]x= \frac{a}{b+ \frac{a}{b+ \frac{a}{b+\cdot\cdot\cdot}}}[/tex]

If that is correct, then that is the same as
[tex]x= \frac{a}{b+ x}[/tex]
since the "continued" fraction in the denominator is still just "x".

Of course, that is the same as the quadratic equation [itex]x(b+ x)= x^2+ bx= a[/itex] or [itex]x^2+ bx- a= 0[/itex]. Use the quadratic formula to solve that.
 
  • #3
If [tex]\alpha=[0;a,b,a,b,...] [/tex], what do you get by computing

[tex]\frac{1}{\alpha} -a?[/tex]

Basically you want to continue performing operations on [tex]\alpha[/tex] until you get [tex]\alpha[/tex] back. The resulting equation will always be quadratic.

Edit: HallsofIvy, not quite, the continued fractions in the [tex][a_0;a_1,\ldots][/tex] notation are the so-called simple continued fractions, which always have unit numerators. So

[tex][0;a,b,a,b,...] = 0 + \frac{1}{a+\frac{1}{b+\frac{1}{a+\cdots}}}[/tex]

Your method still works for this of course, the result is just a bit more complicated.
 
  • #4
Thanks HallsofIvy!

I was taught that [0;a,b,a,b,a,...] denotes
[tex]\frac{1}{a+\frac{1}{b+\frac{1}{a+...}}}[/tex]
The quadratic then becomes ax2+abx-b=0

Thank you very much! I actually looked at the continued fractions expansions for [tex]\sqrt{n}[/tex] for n=1,2,3,...,20 and tried to find patterns. I did find some interesting ones, but I just couldn't manage to use them to work out this problem. I suspected the solution was a simple one, and indeed thank you very much for showing me this beautifully simple solution.
 
  • #5
Thanks to you too, fzero!
 

Related to Solve x in Periodic Continued Fraction Expansion

1. What is a periodic continued fraction expansion?

A periodic continued fraction expansion is a way to represent a real number as a fraction using an infinite repeating pattern of integers. This is often used to approximate irrational numbers.

2. How do I solve for x in a periodic continued fraction expansion?

To solve for x in a periodic continued fraction expansion, you can use a formula called the Euler's identity, which involves manipulating the fraction's coefficients and using the quadratic formula. Alternatively, you can use a calculator or computer program to approximate the solution.

3. What are the applications of periodic continued fraction expansions?

Periodic continued fraction expansions have various applications in mathematics, including in number theory, cryptography, and approximation of irrational numbers. They also have practical applications in fields such as engineering and computer science.

4. Can a periodic continued fraction expansion represent any real number?

No, a periodic continued fraction expansion can only represent irrational numbers and some rational numbers. Some real numbers, such as pi and e, have non-repeating decimals and cannot be fully represented by a periodic continued fraction expansion.

5. Are there any drawbacks to using periodic continued fraction expansions?

One potential drawback of using periodic continued fraction expansions is that they can be computationally intensive and require a large number of iterations to find an accurate approximation. Additionally, they may not always provide the most precise representation of a real number compared to other methods.

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