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Topic: logistic equation (Read 536 times) |
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Jean-Jacques
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logistic equation
« on: Jan 15th, 2004, 1:12am » |
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Hello,
I would like to implement a logistic equation, after several attempts I still can't do it even if it appears simple.
This is the logistic equation that i try to implement:
Xn+1=RXn(1-Xn)
where X = original account
where R = growth rate
Xn = where n is 1 after the first iteration, 2 after the second etc....
Your help is welcome !
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Jean-Jacques
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Re: logistic equation
« Reply #1 on: Jan 15th, 2004, 1:36am » |
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this is the most ugly code you've ever seen about logistic equation.
I'm sure that i must use loops for this but i don't know how !
Very ugly logistic equation:
float x = 0.1;
float rate = 2;
float X0 = 0;
float X1 = 0;
float X2 = 0;
float X3 = 0;
X0=(rate*x)*(1-x);
println(X0);
X1=(rate*X0)*(1-X0);
println(X1);
X2=(rate*X1)*(1-X1);
println(X2);
X3=(rate*X2)*(1-X2);
println(X3);
If someone could help me to give a structure to this code ?
If someone could help me to avoid the repetitions into this code ?
: )
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Bijeoma
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Re: logistic equation
« Reply #2 on: Jan 15th, 2004, 3:20am » |
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//n = number of iterations
int n = 4;
//x0, x1, x2, xn...
float[] values = new float[n];
float rate = 2;
//x
float init_val = 0.1
values[0] = (rate * init_val) * ( 1 - init_val);
for( int i = 1; i < n + 1; i++ ) {
values[i] = (rate * values[i-1]) * (1 - values[i-1]);
println(values[i];
}
i just woke up i and im brain dead on how to insert code... i also didnt bother testing the code. but i think it should work.
bryan
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| « Last Edit: Jan 15th, 2004, 3:25am by Bijeoma » |
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Jean-Jacques
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Re: logistic equation
« Reply #3 on: Jan 15th, 2004, 11:20am » |
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thank you for your help.
In order to post some code "it works like an html tag - you have to put a [ code ] at the begining, and a [/ code ] at the end (without the spaces before and after the word "code")."
I tried your code,i've added a ";" and a ")" but it stills doesn't work.
I'm trying to debug it but i'm not sure to succeed.
This is the new code version:
Code:
//n = number of iterations
int n = 4;
//x0, x1, x2, xn...
float[] values = new float[n];
float rate = 2;
//x
float init_val = 0.1 ;
values[0] = (rate * init_val) * ( 1 - init_val);
for( int i = 1; i < n + 1; i++ ) {
values[i] = (rate * values[i-1]) * (1 - values[i-1]);
println(values[i]);
}
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I obtain this error message:
java.lang.ArrayIndexOutOfBoundsException: 4
at Temporary_6985_838.draw(Temporary_6985_838.java:15)
at BApplet.nextFrame(BApplet.java:407)
at BApplet.run(BApplet.java:369)
at java.lang.Thread.run(Thread.java:536)
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| « Last Edit: Jan 15th, 2004, 11:58am by Jean-Jacques » |
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toxi
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Re: logistic equation
« Reply #4 on: Jan 15th, 2004, 12:04pm » |
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if you only have "n" elements in your array and knowing array indexes start at 0, the "for" loop should terminate at "n" not at "n+1"... so write the line like that and it should work:
Code:| for( int i = 1; i < n; i++ ) { |
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http://toxi.co.uk/
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Jean-Jacques
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Re: logistic equation
« Reply #5 on: Jan 15th, 2004, 12:21pm » |
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You're right Toxi now it works, this is the working code:
Code:
//This code is an attempt to implement a simple logistic equation
//based upon http://sprott.physics.wisc.edu/sa.htm
//As a newbie i did not succed to write this code on my own,
//Bijeoma and Toxi helped me a lot thank you guys !
//january,15,2004 Duclaux Jean Jacques
//n = number of iterations
int n = 4;
//x0, x1, x2, xn...
float[] values = new float[n];
//Here you define the rate
float rate = 2;
//x this is the initial value
float init_val = 0.1 ;
//calculate for X0
values[0] = (rate * init_val) * ( 1 - init_val);
//print X0 value
println(values[0]);
//start the loop and calculate for i= 1 to n
for( int i = 1; i < n; i++ ) {
//print i
println(i);
//this is the logistic equation
values[i] = (rate * values[i-1]) * (1 - values[i-1]);
//print values for i= 1 to 4
println(values[i]);
}
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