khaled

using an algorithm to generate a picture is not useful, pictures we know are no pattern

but, chaos algorithm is good to simulate cosmic unreachable events ...

it's because we may give initial conditions (sometimes random),

because the random chaos simulation is the best choice for anonymous events,

but about what you mentioned of "repeating N times can go infinite",

it means a computer can run infinite, means you won't be able to wait until it stops

but like the non-terminating decimal numbers, you round up to some level

and when the N algorithm is running on a computer, it shows current results at the moment,

you study current to its past,

like a doctor who diagnoses a patient ...

that's all i can guess,

but in the end, this solution does not solve the main problem

saving space .. no,

0 = __

1 = {}

2 = {}{}

3 = {}{}{}

4 = {}{}{}{}

5 = {}{}{}{}{}

6 = {}{}{}{}{}{}

7 = {}{}{}{}{}{}{}

8 = {}{}{}{}{}{}{}{}

9 = {}{}{}{}{}{}{}{}{}

i think i should waste time on more beneficiary ...

i did read your post,

and if you really read my second post, you'd realize that i started a new idea ...

so we have, density 14.5 * 10³kg m³

and, Density = mass/volume

so, mass = density * volume ..and.. volume = mass / density

... mass = 14.5 * 10³ * volume ..&.. volume = mass / 14.5 * 10³

UNIT mass = 14.5 * 10³ * 1 kg ..for volume = 14.5 * 10³ m³

UNIT volume = 1 m³ / (14.5 * 10³) = 6.8965517241379310344827586206897e-5

UNIT volume (round) = 6.9

but the question is, how did you estimated the density in the first place,

.. what are you working on, explain your geometrical work space ...

geometric representation can let you notice possible relations easily,

geometric representation allows transformation,scaling,..etc

sometimes translating one formula into different field expand its enhancement scope,

it's true that if x = 3y + 5 then,

1. change sides

x = 3y + 5

y = x/3 - 5/3

2. validation

x = 3(x/3 - 5/3) + 5 = 3x/3 - 3*5/3 + 5 = x -5+5

y = (3y + 5)/3 - 5/3 = 3y/3 + 5/3 - 5/3 = y +5-5 = y

also, let's not forget the possibility of inserting the pi value as fraction into the formula,

for example,

formula: (X + 2[math]X^2{}[/math]) = r

Result = [math]r^2{}[/math] * pi

A: r^2 * (22/7)

B: (X + 2 X^2) * (X + 2 X^2) * (22/7) = (X^2 + 4 X^3 + 4 X^4) * (22/7)

....... = (22 X^2 / 7) + (88 X^3 / 7) + (88 X^4 / 7)

i guess that such process can give better solutions to particular formula

when we want to tell the computer "for example" to do

calculation like calculating the circle's area,

we may give the following instruction in different versions,

A: Result = [math]r^2{}[/math] * 3.14

B: Result = [math]r^2{}[/math] * 3.1416

C: Result = [math]r^2{}[/math] * 3.14159265

example, input: r = 100000

A: Result = 31400000000

B: Result = 31416000000

C: Result = 31415926500

but yet when we go with real numbers, it start to be worse ...

so the best solution would to be,

let the machine shows you the best it can do !

this means, let the computer do the division ...

and this would be my version of Circle Area Calc .. in C,

long double Circle_Area ( double r )
{
     double pi = (double)22.0/(double)7.0;

     long double Result =  r * r * pi;

     return Result;
}

irrational numbers, with continues non-repeating decimal representation,

we round the value for use,

here is a way to make this theory fits for real numbers,

i just thought of it,

n > 0

0 = ..{ }

1 = { { } }

2 = { { }, {{}} }

3 = { { }, {{},{{}}} }

4 = { { }, {{},{{},{{}}}} }

guess you notice that there is always a single empty set at the beginning of the global set

so, we can use it to define precision as follows,

note: i used ,{} to distinguish 0 from 1 ...

0.0 = { }

0.1 = { ,{} }

0.2 = { ,{},{} }

0.03 = { ,{ {},{},{} } }

0.004 = { ,{{ {},{},{},{} }}

example:

3.1235 = { { {}, { {},{}, { {},{},{}, { {},{},{},{},{} }}}}, {{},{{}}} }

................. DECIMAL ..................... INTEGER ....

so degree of precision is specified in levels of { }'s into the single set

PS, i should get a certification for this, you know

the integral logo is my vote,

<logo> [ .l` f(n) ]

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