Java programming?

[Obsidian]

I was wondering if someone could help me with this program I'm working on. It's focusing on interpolation (for those that don't know what that means, it means that you're taking a set of points and creating a new set of data points}. I've got the input down, I'm just not sure how to program the method.

/**
* @(#)Interpolation.java
*
* Interpolation application
*
* @author
* @version 1.00 2007/5/5
*/

import java.io.*;
import java.lang.Double;

public class Interpolation {

public static void main(String[] args)throws Exception {

int N; // the number of points to be interpolated
int mew; // the number of x points to be inputed
double[] xinter; // the x-values of the interpolated points
double[] yinter; // the y-values of the interpolated points
double[] newx; // the new x-values inputed
double[] newy; // the new y-values outputed
String input = "";
BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
double[] lfct; // the L function
double[] yfct; // the Y function
double[] pfct; // the Lagrange polynomial value


System.out.println("Please enter the number of points to be interpolated");
input = in.readLine();
N = Integer.parseInt(input);

xinter = new double[N];
yinter = new double[N];


int sub = 0;

for (int i = 1; i <= N; i++){

System.out.println("Please enter x " + i);
input = in.readLine();
xinter[sub] = Double.parseDouble(input);

System.out.println("Please enter y " + i);
input = in.readLine();
yinter[sub] = Double.parseDouble(input);
sub++;
}

System.out.println("Please enter the number of x-values");
input = in.readLine();
mew = Integer.parseInt(input);

newx = new double[mew];

for (int j = 0; j < mew; j++){
System.out.println("Please enter the x(s):");
input = in.readLine();
newx[j]= Double.parseDouble(input);
}




}
}

I'm going to use Lagrange formula, but I can't for the life of me figure out how to code it. I understand the method pretty well, but when it gets down to coding, I'm lost. I have a general idea that it's going to involve a few loops, but yeah, if anyone can help me ASAP, it would be appreciated.

that's the general equation where the Yj refers to the Y coordinates that are going to be entered. The Xj will be used in the l(x) part of the equation, l(x) is the only part I'm really worried about. Once I get that, I'll be fine with the rest.

If you don't understand it, I'm welcome to explaining it.

Cheers,
Obsidian

 

[aptget]

Probably not the best forum to ask for programming help
Sorry, I could help you if this was in Python
Have you tried a java IRC channel?

 

[Kritish]

I don’t think anyone can help you with this one, come back when you need to figure out how to do a scan disk.

 

[forumgogo]

its really difficult for me ..



sharing free forum with you :www.forumgogo.com

 

[Spelunking Man]

Well, that's an easy problem. Wait here until I finish my PhD in a few years. D'oh!

 

[wiegeabo]

Just after a quick look, it seems that L(x) is a simple enough loop multipling yj to the result of lj(x).

Solving lj(x) may best be done by recursion or another loop, skipping the recursion/loop iteration when i = j. It would proably be easiest to make lj(x) another function containg the recursion/loop. That way, when you've got the Product down, you won't have to worry about it anymore and can just multiply the result of the function to yj in the Summation.