| T.R | Title | User | Personal
Name | Date | Lines |
|---|
| 486.1 | | ENGINE::ROTH | | Fri May 09 1986 19:15 | 30 |
| In any continued fraction that repeats you'll get an expression of the
form x = [i j k l m n ... + 1/x]; if you expand any expression of this
type out, you always have x = T(x) = (a*x + b)/(c*x + d), and the
solution for the fixed point of this transform is never worse than a
quadratic. You can also turn this around and solve any quadratic
by expressing it in terms of a regular continued fraction.
For example, you can always get a root of N by picking the nearest
integer root and expressing a recurrance, as in the SQRT(79) example.
Quadratic surds are the only continued fractions that have periodic
expansions.
Consider a filament in the plane stretched from the origin to a point
far away, with an irrational slope. If you pull it away from the origin
it will meet lattice points that are successive convergents of the
CF expansion of its slope.
Continued fraction type expansions occur in some other interesting
contexts. For example, if you want to color in a regular tiling
of the non-Euclidean plane, you can express it in terms of combinations
of 'inversions' with respect to a circle, and rotations about a fixed
point; these operations can effectively replicate copies of some
region throughout the plane, and basically generate the group of all
motions possible. Writing down the sequence of transforms shows that
they are just like a continued fraction, with inversion corresponding
to taking a reciprocal, etc, and you can get a group theoretic/geometric
analog to number theory concepts.
- Jim
|
| 486.2 | solution to sqrt(79)'s continued fraction . . . | 7480::KOSTAS | Kostas G. Gavrielidis <o.o> | Sat May 24 1986 18:29 | 85 |
| re. .0
This is the solution to the question:
What is the sqrt(79)'s continued fraction ?
Since the greatest inter of [sqrt(70)] = 8 and
(sqrt(79) - 8) * (sqrt(79) + 8 ) = 15
we have
sqrt(79) = 8 + (sqrt(79) - 8)
15
= 8 + ------------
8 + sqrt(79)
1
= 8 + ---------------
8 + sqrt(79)
------------
15
1
= 8 + ----------------
sqrt(79) - 8
1 + ------------
15
1
= 8 + ----------------
1
1 + ------------
8 + sqrt(79)
1
= 8 + ----------------
1
1 + ------------
8 + sqrt(79)
1
= 8 + --------------------------
1
1 + ----------------------
15
8 + (8 + ------------)
8 + sqrt(79)
1
= 8 + --------------------------
1
1 + ----------------------
15
16 + (------------)
8 + sqrt(79)
= . . .
_____
= <8, 1, 16>
which in your notation is:
sqrt(79) = [8 1 16 1 16 1 16 . . .]
I would like to remind anyone interested in the continued fractions
to also look up on this notes file note # 483.*
Enjoy,
Kostas G.
<><><><><>
|
| 486.3 | solution to [2 1 1 4 1 1 4 1 1 4 . . .] problem | 7480::KOSTAS | Kostas G. Gavrielidis <o.o> | Sat May 24 1986 19:43 | 135 |
| re. 486.0
This is a reply to a problem raised in this notes file node# 486.0
and is related to continued fractions problems.
The problem was this:
"Can we work backwards ? For instance, what is the closed
form for [2 1 1 4 1 1 4 1 1 4 . . .]."
The answer is:
Yes. And the closed form for [2 1 1 4 1 1 4 1 1 4 . . .] follows.
Since I have used a different notation in the problems related to
continued fraction in the note # 483.0, I will try to satisfy both
of us ( Eric and Kostas).
So,
let
x = [2 1 1 4 1 1 4 1 1 4 . . .]
or
_____ 1
x = <2 1 1 4> = 2 + ------------------------
1
1 + --------------------
1
1 + ----------------
1
4 + ------------
1 + . . .
then
1
x - 2 = ------------------------
1
1 + --------------------
1
1 + ----------------
1
4 + ------------
1 + . . .
1
= ------------------------
1
1 + --------------------
1
1 + ----------------
4 + (x - 2 )
1
= ------------------------
1
1 + --------------------
1
1 + ----------------
(x + 2 )
1
= ------------------------
1
1 + --------------------
(x + 3)
------------
(x - 2 )
1
= ------------------------
2x + 5
-----------
(x + 3)
(x + 3)
= -----------
2x + 5
Thus
(x + 4)(2x + 5) = x + 3
2
2x + 5x + 8x + 20 = x + 3
2
2x + 12x + 17 = 0
2
-12 + or - sqrt( 12 - 4 * 2 * 17)
x = ----------------------------------
4
-12 + or - sqrt(8)
= -----------------------
4
since
[(-6 + sqrt(8))/2] = -2
and
[(-6 - sqrt(8))/2] = -4
we can conclude that
_______ -6 - sqrt(8)
x = <2, 1, 1, 4> = ---------------
2
or
-6 - sqrt(8)
[2 1 1 4 1 1 4 1 1 4 . . .] = ---------------
2
Enjoy,
Kostas G.
<><><><><>
|
| 486.4 | ... but it's wrong ... | METOO::YARBROUGH | | Tue May 27 1986 12:50 | 4 |
| And this solution, like the one in the other note, is wrong. There
is a typo early in the calculation.
MAPLE does this sort of thing very nicely (and correctly).
|
| 486.5 | Correct solution of <2, 1, 1, 4> is on note 483.8 | THEBUS::KOSTAS | Kostas G. Gavrielidis <o.o> | Sat May 31 1986 20:59 | 13 |
| re. .4
I have fixed the bug please read note 483.8 for the corrected solution.
I would like to see the output of MAPLE. I have never used this
tool before, so could you provide some examples or a reference to
where I can get info on this MAPLE tool.
Thanks,
Kostas G.
|
| 486.6 | See 439 | METOO::YARBROUGH | | Mon Jun 02 1986 08:52 | 1 |
| see note 439 for info on MAPLE.
|
| 486.7 | some of these look a lot alike | METOO::YARBROUGH | | Mon Jun 02 1986 08:59 | 3 |
| I may have miscounted 1's in an earlier reply. The continued fraction
for 6.5^.5 is [2,1,1,4,1,1,4,...], while
for 7^.5 it is [2,1,1,1,4,1,1,1,4...].
|
| 486.8 | Find f(n) = [2,{n 1's, 4}] | ROXIE::OSMAN | and silos to fill before I feep, and silos to fill before I feep | Mon Jun 02 1986 10:41 | 10 |
|
o.k., so [2,{1,1,4}] is sqr(6.5) and
[2,{1,1,1,4}] is sqr(7).
What about [2,{n 1's, 4}] is f(n) ? Can anyone find a correct
function ?
/Eric
|
| 486.9 | looks like the interval [6^.5, 8^.5] = <2,{n 1's,4}> | THEBUS::KOSTAS | Kostas G. Gavrielidis <o.o> | Tue Jun 03 1986 08:41 | 40 |
| Well,
what we have as of now:
____
6^.5 = <2, 2, 4>
_______
6.5^.5 = <2, 1, 1, 4> and 6.5^.5 = 2.5495098
-5 + (433)^.5 _____________
-------------- = <2, 1, 1, 1, 1, 4>
6
and (-5 + 433^.5)/6 = 2.6347753
__________
7^.5 = <2, 1, 1, 1, 4> and 7^.5 = 2.6457513
8^.5 = <2, 1, 4> and 8^.5 = 2.8284271
This looks like we have the interval
from: > 6^.5
to: 8^.5
(6^.5, 8^.5]
Still no function f(n) available yet.
Enjoy,
Kostas G.
|
| 486.10 | More data on this particular class of C.f.'s | METOO::YARBROUGH | | Tue Jun 03 1986 10:03 | 15 |
| There appears to be a calculation error in the previous reply. Let
K(n) be the continued fraction with 2 as first term and n 1's preceding
a 4 as the repeating cycle after that. Then
K(0) = sqr(5)
K(1) = sqr(8)=2sqr(2)
K(2) = sqr(26)/2
K(3) = sqr(7)
K(4) = sqr(170)/5
K(5) = sqr(110)/4
K(6) = sqr(1157)/13
K(7) = 4sqr(21)/7
K(8) = sqr(7922)/34
(I leave the rest as an exercise for the reader :-))
Thanks, MAPLE.
|
| 486.11 | a.k.a. fie, "Osman's Constant" | AURORA::HALLYB | Free the quarks! | Tue Jun 03 1986 11:23 | 7 |
| Lynn, would you or MAPLE happen to have decimal expansions of the
preceding values?
Given that phi = [1,1,1,1,1,1,1,1,.....] = (1+sqrt(5))/2 ~= 1.618034,
I would conjecture that [2,1,1,1,1,...,4] = (3+sqrt(5))/2 in the limit.
John
|
| 486.12 | | CLT::GILBERT | On the road again... | Tue Jun 03 1986 20:06 | 29 |
|
Let z = [1,1,1,...,b,1,1,1,...,b,...] = [{1,1,1...,b}]
<- n 1s -> <- n 1s -> <- n 1s ->
n n-1
Let f(x) = 1 + 1/x, f�(x) = f(f(x)), and in general, f (x) = f(f (x)).
n
Now we see that z = f (b+1/z).
n x F(n+1) + F(n)
But f (x) = ---------------, where F(n) is the nth Fibonacci number
x F(n) + F(n-1)
(this results from a simple induction, and the formula for f(n)).
n (b+1/z) F(n+1) + F(n) (bz+1) F(n+1) + z F(n)
So, z = f (b+1/z) = --------------------- = ----------------------, or
(b+1/z) F(n) + F(n-1) (bz+1) F(n) + z F(n-1)
z� (b F(n) + F(n-1)) - z F(n+1) - F(n+1) = 0.
F(n+1) + sqrt( F(n+1) F(n+1) + 4 F(n+1) (b F(n) + F(n-1)) )
So, z = -----------------------------------------------------------.
2 (b F(n) + F(n-1))
Right so far?
|
| 486.13 | PI - continued fraction | CADSYS::PRENTICE | Ed 225-4061 HLO2-2/G13 (E13) | Tue Jun 23 1987 14:19 | 13 |
| > < Note 486.0 by SIERRA::OSMAN >
> -< The New Method (Square roots, Continued Fractions) >-
>
> Not all irrational numbers come out rational as continued fractions,
> however. In this sense, might might say PI is less rational than e.
>
> pi = [3 7 15 1 185 . . .]
Since the regular continued fraction for PI is irregular, Does anyone
have any ideas on an efficient means of calculating the terms of PI from
some other values in continued fraction format (or a way to produce the
terms directly)?
/egp
|
| 486.14 | That 5th term is off a bit | AKQJ10::YARBROUGH | Why is computing so labor intensive? | Tue Jun 23 1987 16:09 | 9 |
| MAPLE will produce the continued fractions for Pi out a ways. The first 20
terms are:
Pi ~ [3, 7, 15, 1, 292, 1, 1, 1, 2, 1, 3, 1, 14, 2, 1, 1, 2, 2, 2, 2, ...]
You can also get the convergents (i.e. rational approximations) as a side
effect.
Lynn Yarbrough
|
| 486.15 | | CLT::GILBERT | eager like a child | Wed Jun 24 1987 00:52 | 7 |
| Re .12, the last few equations should be ...
z� (b F(n) + F(n-1)) - z b F(n+1) - F(n+1) = 0.
b F(n+1) + sqrt( b� F(n+1) F(n+1) + 4 F(n+1) (b F(n) + F(n-1)) )
So, z = ----------------------------------------------------------------.
2 (b F(n) + F(n-1))
|
| 486.16 | | ENGINE::ROTH | | Wed Jun 24 1987 08:37 | 12 |
| Re continued fraction for PI; I think Gosper has the record for number
of terms calculated so far, millions of them, on a Symbolics workstation.
I don't know the method he used. However, I *just* got a neat new
recreational math book called PI and the AGM, by Borwien and Borwien.
Haven't really looked thru it yet, but it probably has an algorithm
for this.
This book is chock full of neat number theory, elliptic function theory,
and computational mathematics...
- Jim
|