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Topic:
Fractious Fractions
Replies:
10
Last Post:
Jun 6, 2013 7:14 AM




Re: Fractious Fractions
Posted:
Jun 3, 2013 2:22 PM


On Sun, 2 Jun 2013 18:53:42 0700, William Elliot <marsh@panix.com> wrote:
Since you love to lecture people on how to write posts, someone should point out that an informative subject line is better than a cute one. And much better than something you _think_ is cute!
Somewhere you should mention that you're talking about _continued fractions_, for the benefit of readers who aren't familiar with these things.
>Let a = (aj)_j be a sequence of positive integers. > >Define [a1] = a1; [a1,a2] = a1 + 1/a2; [a1,a2,a3] = a1 + 1/(a2 + 1/a3); > [a1,.. aj] = a1 + 1/(a2 + 1/(a3 +..+ 1/aj))..) > >Shome use the more visual notation > a1 + 1/a2+ 1/a3+ ..+1/aj >for [a1,.. aj]. > >Let cj = [a1,.. aj]. > >Assume j < k < n. How does one show > ck  c_n < cj  c_n?
It's enough to show this:
(i) The numbers c_{j+1}  c_j are alternating in sign and decreasing in absolute value.
(You have to figure out why what you want follows from (i)).
Now (i) follows from (ii):
(ii) The numbers c_{j+1}  c_j alternate in sign. If j is odd then c_{j+2} > c_j, while if j is even then c_{j+2} < c_j.
To prove (ii): Let d_j = [a_2,...,a_j], so that
(*) c_j = 1 + 1/d_j.
By induction you can assume (iii):
(iii) The numbers d_{j+1}  d_j alternate in sign. If j is odd then d_{j+2} < d_j, while if j is even then d_{j+2} > d_j.
And then it's not hard to show that (ii) follows from (iii) and (*).



