What Bo means is that if you execute
a =: irrational_number
and then convert the approximation a to rational, the denominator will
be large.
Not necessarily true: first of all, x: uses tolerant comparison:
x: 1.000000000000001
1
so to look at all the bits of a you need
(x:!.0) 1.000000000000001
900719925474100r900719925474099
[and note: that was a perfectly respectable rational number, but it
comes out with a large denominator]
Moreover,
x: 1.000000000000000000014973485793 NB. It's an irrational number,
I just got tired of typing
1
Henry Rich
On 6/14/2018 10:12 AM, William Tanksley, Jr wrote:
I have no idea why you're saying that. Irrationals don't have denominators;
they cannot be written as fractions. And no floating point numbers are
irrationals.
On Thu, Jun 14, 2018 at 12:18 AM 'Bo Jacoby' via Programming <
[email protected]> wrote:
If a floating point number (a) , is irrational, then (x:a) has a
denominator greater than 1e10.
Some rational and irrational numbers are:
x:%%:>:i.5
1 676982533219r957397879968 9943229281777r17222178307344 1r2
8728368286235r19517224820674
The denominators are:
1 957397879968 17222178307344 2 19517224820674
1 9.57398e11 1.72222e13 2 1.95172e13
So the problem of identifying irrational numbers is reduced to the problem
of finding the denominator in x:a
/Bo.
Den 23:42 onsdag den 13. juni 2018 skrev Linda Alvord <
[email protected]>:
I wrote a definition with only one argument.
Linda
Get Outlook for Android<https://aka.ms/ghei36>
________________________________
From: Programming <[email protected]> on behalf of
Raul Miller <[email protected]>
Sent: Wednesday, June 13, 2018 4:36:31 PM
To: Programming forum
Subject: Re: [Jprogramming] Finding Irrational Numbers
So, something like this?
fe=:4 :0
p=. 9!:10''
vi=. _.
ve=. y
try.
9!:11 x
ve=. ":&.>y
9!:11]15
vi=. ":&.>y
catch.
end.
9!:11 p
ve=vi
)
Thanks,
--
Raul
On Wed, Jun 13, 2018 at 4:22 PM Skip Cave <[email protected]> wrote:
So I'm thinking something like this, to find exact numbers in a floating
point array:
pps 15 NB. Set the print precision.
Create a mixed floating point vector b:
]b=:(,3? 10^1+i.4),%a=:1+i.20
0 2 7 36 93 57 808 875 559 4674 8635 6196 1 0.5 0.333333333333333 0.25
0.2
0.166666666666667 0.142857142857143 0.125 0.111111111111111 0.1
0.0909090909090909 0.0833333333333333 0.0769230769230769
0.0714285714285714
0.0666666666666667 0.0625 0.0588235294117...
Create a "find exact" verb 'fe' where x=the number of digits max that can
be exact, and y is the vector to be tested (boolean result):
fe=. 4 : 'x> ># each ": each <"1,.y'
]c=.9 fe b
1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 0 0 1 0 1 0 0 0 0 0 1 0 0 0 1
Show exact:
c#b
0 2 7 36 93 57 808 875 559 4674 8635 6196 1 0.5 0.25 0.2 0.125 0.1 0.0625
0.05
Show inexact:
(-.c)#b
0.333333333333333 0.166666666666667 0.142857142857143 0.111111111111111
0.0909090909090909 0.0833333333333333 0.0769230769230769
0.0714285714285714
0.0666666666666667 0.0588235294117647 0.0555555555555556
0.0526315789473684
#b
32
#c#b
20
I'm sure this scheme has some holes, but it is a reasonable first cut.
Skip
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