| Title: | Mathematics at DEC |
| Moderator: | RUSURE::EDP |
| Created: | Mon Feb 03 1986 |
| Last Modified: | Fri Jun 06 1997 |
| Last Successful Update: | Fri Jun 06 1997 |
| Number of topics: | 2083 |
| Total number of notes: | 14613 |
Well,
This note will not create anything new, just some problems to
be solved on the area of finite continued fractions.
Some definitions fist:
A finite continued fraction is an expression of the
form:
1
a + -----------------------------
1 1
a + -------------------------
2 1
a + --------------------- (*)
3 .
.
.
1
+ --------------
1
a + -------
k-1 a
k
where the a are neal or complex numbers and only a is
i 1
permitted to be zero.
For our purposes, we require that the a , which are called the
i
terms or partial quotients of the continued fractions, be rational
numbers.
Notations used to designate the continued fraction in (*) include:
<a , a , ... , a >
1 2 k
[a , a , ... , a ]
1 2 k
{a , a , ... , a }
1 2 k
we will use the first of these notations.
Example 1:
----------
<-2,3,4,2> is numerically equal to the rational number -49/29
since, the usual procedure for simplifying fractions
1
<-2,3,4,2> = -2 + -------------
1
3 + ---------
1
4 + -----
2
1
= -2 + ---------
2
3 + -----
9
9
= -2 + ------
29
49
= - ------
29
Problems:
---------
1. Find the comtinued fractions of the following in the
<a , a , ... , a > notation for
1 2 k
1.1: 0
1.2: 4/3
1.3: 3/520
1.4: 17711/28657
1.5: SQRT(2)
1.6: 20/13
1.7: -3.31
2. What is the fraction of the following:
2.1: <1,2,3,4>
2.2: <-3,2,1>
2.3: <3,3,3,3,3>
Enjoy,
Kostas G.
<><><><><>
| T.R | Title | User | Personal Name | Date | Lines |
|---|---|---|---|---|---|
| 483.1 | I got one of 'em | CAD::PRENTICE | Edward G Prentice HLO2-2/G13 225-4061 | Wed May 21 1986 10:06 | 10 |
re: .0
1.5: SQRT(2) <1, 2, 2, 2, ...>
Give me some time, I'll get a few more. It's interesting that this irrational
number has such a nice representation in this format. Anyone know how to
generate PI in a continued fraction? /egp
| |||||
| 483.2 | please see 486.0 | SIERRA::OSMAN | and silos to fill before I feep, and silos to fill before I feep | Wed May 21 1986 11:57 | 5 |
Please see my note at 486.0.
/Eric
(hi Ed!)
| |||||
| 483.3 | missing steps to the sqrt(2) problem | 7481::KOSTAS | Kostas G. Gavrielidis <o.o> | Sat May 24 1986 11:27 | 48 |
re. .1
Since you just gave the soloution to the problem: "Find the continued
fraction of SQRT(2) in the <a , a ,..., a > notation"
1 2 k
I have included the missing steps in this reply.
sqrt(2) = 1 + (sqrt(2) - 1)
1
= 1 + --------------------
1 + sqrt(2)
1
= 1 + --------------------
2 + (sqrt(2) - 1)
1
= 1 + --------------------
1
2 + ----------------
1 + sqrt(2)
1
= 1 + --------------------
1
2 + ----------------
2 + (sqrt(2) -1)
= . . .
= <1, 2, 2, 2, ...>
_
= <1, 2 >
Enjoy,
Kostas G.
| |||||
| 483.4 | quadratic irrationals and periodic cont fractions | 7481::KOSTAS | Kostas G. Gavrielidis <o.o> | Sat May 24 1986 12:08 | 53 |
Well,
some more info and problems.
Definition: If n is a positive integer which is not a square,
then sqrt(n) is a quadratic irrational and
__________________________
sqrt(n) = <a , a , a , . . . , a , 2a >
1 2 3 k 1
for some integer k.
Note:
The bar over some of the terms in the expression
__________________________
<a , a , a , . . . , a , 2a >
1 2 3 k 1
means that the continued fraction is extended to infinite many
terms by the endless recurrence of the terms beneath the bar.
Problems:
1. Find the periodic continued fraction representation for
each quadratic irrational for 2 < n < 19.
_
Note: For n = 2 we know that this is <1, 2>
2. Find the irrational number respresented by the infinite
simple continued fractions:
_
(2a): < 3 >
____
(2b): < -4, 2, 5>
_______
(2C): < -4, 2, 3, 1>.
Enjoy,
Kostas G.
<><><><><>
| |||||
| 483.5 | solution to x =<2, 1, 1, 4> problem | 7480::KOSTAS | Kostas G. Gavrielidis <o.o> | Sat May 24 1986 20:44 | 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.
<><><><><>
| |||||
| 483.6 | are you sure? | CLT::GILBERT | Juggler of Noterdom | Sun May 25 1986 15:00 | 2 |
Hmmm. I'd have expected [2 1 1 4 1 1 4 1 1 4 . . .] to be positive, while -3 � sqrt(2)/2 is certainly negative. | |||||
| 483.7 | CLT::GILBERT | Juggler of Noterdom | Sun May 25 1986 15:03 | 3 | |
Can someone provide a method for *directly* dividing a continued
fraction by two? (by direct, I mean without first converting it
to the real number).
| |||||
| 483.8 | corrected solution to .0 | THEBUS::KOSTAS | Kostas G. Gavrielidis <o.o> | Sat May 31 1986 21:51 | 131 |
re. 483.6, 486.0
This is the corrected solution to .0
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 - 2) (2x + 5) = x + 3
2
2x + 5x - 4x - 10 = x + 3
2
2x + 0x - 13 = 0
2
2x = 13
2
x = 13/2
x = + sqrt(13/2) or x = - sqrt(13/2)
x = + sqrt(6.5) or x = - sqrt(6.5)
x = 2.5495098 or x = - 2.5495098
we can conclude that
_______
x = <2, 1, 1, 4> = + sqrt(6.5)
or
[2 1 1 4 1 1 4 1 1 4 . . .] = +sqrt( 6.5)
Enjoy,
Kostas G.
<><><><><>
[ End of note ]
| |||||
| 483.9 | solution to problems 1 and 2 of .0 | THEBUS::KOSTAS | Kostas G. Gavrielidis <o.o> | Wed Jun 04 1986 22:27 | 26 |
re. .0
These are solution to the problems # 1, and # 2 of 483.0
Solutions
for:
1.1: 0 = <0> = <-1, 1>
1.2: 4/3 = <1, 3> = <1, 2, 1>
1.3: 3/520 = <0, 173, 3>
1.4: 17711/28657 = <0, {20 1's}, 2>
1.5: sqrt(2) = <1, 2, 2, 2, ... >
1.6: 20/13 = <1, 1, 1, 6>
1.7: -3.31 = <-4, 1, 2, 4, 2, 3>
2.1: <1, 2, 3, 4> = 43/30
2.2: <-3, 2, 1> = -8/3
2.3: <3, 3, 3, 3, 3> = 360/109
Enjoy,
Kostas G.
| |||||
| 483.10 | solutions to problems 2a, 2b and 2c of 483.4 | THEBUS::KOSTAS | Kostas G. Gavrielidis <o.o> | Wed Jun 04 1986 22:32 | 24 |
re. .4
solutions to problems of 483.4
for:
_ 3 + sqrt(13)
(2a): <3> = ------------
2
____ -13 + sqrt(35)
(2b): <-4, 2, 5> = --------------
2
_______ -31 - sqrt(21)
(2c): <-4, 2, 3, 1> = --------------
10
Enjoy,
Kostas G.
| |||||