continued fractions help

[richardt]

Greetings Members:

Would someone kindly show me or point me toward proving that "every non-terminating continued fraction (CF) converges to some real number"?

I can show that a CF converges to a rational number if and only if it terminates (i.e., has finitely many coefficients, a_i). Moreover, I believe I can show that every irrational number has a non-terminating CF representation. But I still need to show that [a₀, a₁, a₂, ...] = a₀ + 1/(a₁ + 1/(a₂ + 1/(a₃ +... does indeed converge to some real number, r.

My thinking is to show the sequence of convergents, x_m, converges, i.e., {x₀, x₁, x₂, ...} = {[a₀], [a₀, a₁], [a₀, a₁, a₂], ...}. But how?

Thank you in advance.

Rich

 

[HallsofIvy]

Certainly, \(\displaystyle a_0+ 1/a_1\) lies in the interval \(\displaystyle ([a_0- 1/|a_1|, a_0+ 1/|a_1)|\), \(\displaystyle a_0+ 1/(a_1+ 1/a_2)\) lies in the interval \(\displaystyle (a_0+ 1/(a_1- 1/|a_2|), a_0+ 1/(a_1+ 1/|a_2|))\), etc. Show that those are a sequence of nested intervals whose length goes to 0.

 

[richardt]

now that is slick. And I do thank you for not giving it all to me. I believe i can make something of your suggestion.

Thanks much!
Rich

 

[richardt]

HallsofIvy: I am stuck. I can see intuitively that if u = a₀ + 1/(a₁ + 1/(a₂ + ... + 1/(a_n-1 + 1/a_n)))...) and v = a₀ + 1/(a₁ + 1/(a₂ + ... + 1/(a_n-1 - 1/a_n)))...), that u-v goes to zero as n goes to infinity, but how do I show this rigorously? Is is a less than unwieldy way in which to subtract these algebraically?

Thanks.
Rich

 

[richardt]

Can anyone answer my last inquiry? Thank you.

 

[stapel]

I can see intuitively that if u = a₀ + 1/(a₁ + 1/(a₂ + ... + 1/(a_n-1 + 1/a_n)))...) and v = a₀ + 1/(a₁ + 1/(a₂ + ... + 1/(a_n-1 - 1/a_n)))...), that u-v goes to zero as n goes to infinity, but how do I show this rigorously? Is is a less than unwieldy way in which to subtract these algebraically?

For clarity, I've formatted the above continued fractions.

. . . . .\(\displaystyle u\, =\, a_0\, +\, \cfrac{1}{a_1\, +\, \cfrac{1}{a_2\, +\, \cfrac{1}{a_3\, +\, \cfrac{\ldots}{a_{n-1}\, +\, \frac{1}{a_n}}}}}\)

. . . . .\(\displaystyle v\, =\, a_0\, +\, \cfrac{1}{a_1\, +\, \cfrac{1}{a_2\, +\, \cfrac{1}{a_3\, +\, \cfrac{\ldots}{a_{n-1}\, -\, \frac{1}{a_n}}}}}\)

If I've understood correctly, the only difference in the two fractions is in the sign between the final two terms at the very "bottom" of the fractions.

 

[daon2]

I thought about this problem a bit, and the solution is "obvious" when you simplify the fractions. Treating \(\displaystyle a_0, a_1,..., a_n\) as variables, the difference is on the order of a rational function having "degree" \(\displaystyle -n^2\), or probably more accurately \(\displaystyle -(n-1)^2-1\). The only issue here is that the general simplification will be quite large and ugly.

For example (and I used wolfram)

\(\displaystyle \dfrac{1}{b + \dfrac{1}{c+\dfrac{1}{d+\frac{1}{e}}}} - \dfrac{1}{b + \dfrac{1}{c+\dfrac{1}{d-\frac{1}{e}}}} = \dfrac{-2 e}{(-1-b c+b e+d e+b c d e) (1+b c+b e+d e+b c d e)} \approx \dfrac{-1}{(bcd)^2e}\)

edit: I should clarify. If all but finitely many \(\displaystyle a_i\) are equal to 1, this approximation wont help, you'll need to rely on the number of terms in those factors in the denominator.

Also, see chapter 8 of this textbook: http://www.math.binghamton.edu/dennis/478.f07/EleAna.pdf

The solution is there, it is just a little complicated how the study of continued fractions begins. I've looked at other sources and they are similar.

 

[richardt]

Thank you kindly. I am quite certain I can make something of this.

Rich B.