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

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David C. Ullrich

Posts: 3,156
Registered: 12/13/04
Re: Fractious Fractions
Posted: Jun 3, 2013 2:22 PM
  Click to see the message monospaced in plain text Plain Text   Click to reply to this topic Reply

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 (*).






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