Laguerre interpolation algorithm, something's wrong with my implementation - c++

Question

This is a problem I have been struggling for a week, coming back just to give up after wasted hours...
I am supposed to find coefficents for the following Laguerre polynomial:
P0(x) = 1
P1(x) = 1 - x
Pn(x) = ((2n - 1 - x) / n) * P(n-1) - ((n - 1) / n) * P(n-2)
I believe there is an error in my implementation, because for some reason the coefficents I get seem way too big. This is the output this program generates:
a1 = -190.234
a2 = -295.833
a3 = 378.283
a4 = -939.537
a5 = 774.861
a6 = -400.612
Description of code (given below):
If you scroll the code down a little to the part where I declare array, you'll find given x's and y's.
The function polynomial just fills an array with values of said polynomial for certain x. It's a recursive function. I believe it works well, because I have checked the output values.
The gauss function finds coefficents by performing Gaussian elimination on output array. I think this is where the problems begin. I am wondering, if there's a mistake in this code or perhaps my method of veryfying results is bad? I am trying to verify them like that:
-190.234 * 1.5 ^ 5 - 295.833 * 1.5 ^ 4 ... - 400.612 = -3017,817625 =/= 2
Code:
#include "stdafx.h"
#include <conio.h>
#include <iostream>
#include <iomanip>
#include <math.h>
using namespace std;
double polynomial(int i, int j, double **tab)
{
double n = i;
double **array = tab;
double x = array[j][0];
if (i == 0) {
return 1;
} else if (i == 1) {
return 1 - x;
} else {
double minusone = polynomial(i - 1, j, array);
double minustwo = polynomial(i - 2, j, array);
double result = (((2.0 * n) - 1 - x) / n) * minusone - ((n - 1.0) / n) * minustwo;
return result;
}
}
int gauss(int n, double tab[6][7], double results[7])
{
double multiplier, divider;
for (int m = 0; m <= n; m++)
{
for (int i = m + 1; i <= n; i++)
{
multiplier = tab[i][m];
divider = tab[m][m];
if (divider == 0) {
return 1;
}
for (int j = m; j <= n; j++)
{
if (i == n) {
break;
}
tab[i][j] = (tab[m][j] * multiplier / divider) - tab[i][j];
}
for (int j = m; j <= n; j++) {
tab[i - 1][j] = tab[i - 1][j] / divider;
}
}
}
double s = 0;
results[n - 1] = tab[n - 1][n];
int y = 0;
for (int i = n-2; i >= 0; i--)
{
s = 0;
y++;
for (int x = 0; x < n; x++)
{
s = s + (tab[i][n - 1 - x] * results[n-(x + 1)]);
if (y == x + 1) {
break;
}
}
results[i] = tab[i][n] - s;
}
}
int _tmain(int argc, _TCHAR* argv[])
{
int num;
double **array;
array = new double*[5];
for (int i = 0; i <= 5; i++)
{
array[i] = new double[2];
}
//i 0 1 2 3 4 5
array[0][0] = 1.5; //xi 1.5 2 2.5 3.5 3.8 4.1
array[0][1] = 2; //yi 2 5 -1 0.5 3 7
array[1][0] = 2;
array[1][1] = 5;
array[2][0] = 2.5;
array[2][1] = -1;
array[3][0] = 3.5;
array[3][1] = 0.5;
array[4][0] = 3.8;
array[4][1] = 3;
array[5][0] = 4.1;
array[5][1] = 7;
double W[6][7]; //n + 1
for (int i = 0; i <= 5; i++)
{
for (int j = 0; j <= 5; j++)
{
W[i][j] = polynomial(j, i, array);
}
W[i][6] = array[i][1];
}
for (int i = 0; i <= 5; i++)
{
for (int j = 0; j <= 6; j++)
{
cout << W[i][j] << "\t";
}
cout << endl;
}
double results[6];
gauss(6, W, results);
for (int i = 0; i < 6; i++) {
cout << "a" << i + 1 << " = " << results[i] << endl;
}
_getch();
return 0;
}

Answer 1

I believe your interpretation of the recursive polynomial generation either needs revising or is a bit too clever for me.
given P[0][5] = {1,0,0,0,0,...}; P[1][5]={1,-1,0,0,0,...};
then P[2] is a*P[0] + convolution(P[1], { c, d });
where a = -((n - 1) / n)
c = (2n - 1)/n and d= - 1/n
This can be generalized: P[n] == a*P[n-2] + conv(P[n-1], { c,d });
In every step there is involved a polynomial multiplication with (c + d*x), which increases the degree by one (just by one...) and adding to P[n-1] multiplied with a scalar a.
Then most likely the interpolation factor x is in range [0..1].
(convolution means, that you should implement polynomial multiplication, which luckily is easy...)
[a,b,c,d]
* [e,f]
------------------
af,bf,cf,df +
ae,be,ce,de, 0 +
--------------------------
(= coefficients of the final polynomial)

Answer 2

The definition of P1(x) = x - 1 is not implemented as stated. You have 1 - x in the computation.
I did not look any further.