Finding Coefficients in Bspline is making a problem ?

% Define the x array
N = 100; % Example N value, adjust as needed
x = linspace(0, 5, N + 2); % Example x values, adjust as needed
% Initialize the a array
a = zeros(1, N + 2);
% Compute the coefficients a_i
for i = 1:(N + 2)
    product = 1;
    for j = 1:(N + 2)
        if j ~= i
            product = product * (x(i) - x(j));
        end
    end
    a(i) = 1 / product;
end
% Display the coefficients
disp('Coefficients a_i:');
disp(a);

This is my code but the values of this coefficient are coming in negative is my code right?THIS IS THE ACTUAL FORMULA TO SOLVE

Answer 1

[sai charan sampara]

Hello Rohitasya,

The code you have written is accuarte for calculating the coeffecients.The following code achieves the same result as yours:

N = 100;
x = linspace(0, 5, N + 2);
b = zeros(1, N + 2);
for i=1:(N+2)
    x_new=x(i)-x;
    x_new(i)=1;
    product=prod(x_new);
    b(i)=1/product;
end
disp(b)

  Columns 1 through 18

   -0.0000    0.0000   -0.0000    0.0000   -0.0000    0.0000   -0.0000    0.0000   -0.0000    0.0000   -0.0000    0.0000   -0.0000    0.0000   -0.0000    0.0000   -0.0000    0.0000

  Columns 19 through 36

   -0.0000    0.0000   -0.0000    0.0000   -0.0000    0.0000   -0.0000    0.0000   -0.0001    0.0002   -0.0005    0.0013   -0.0031    0.0070   -0.0154    0.0322   -0.0643    0.1231

  Columns 37 through 54

   -0.2257    0.3966   -0.6679    1.0790   -1.6724    2.4882   -3.5546    4.8772   -6.4290    8.1434   -9.9137   11.6012  -13.0513   14.1168  -14.6814   14.6814  -14.1168   13.0513

  Columns 55 through 72

  -11.6012    9.9137   -8.1434    6.4290   -4.8772    3.5546   -2.4882    1.6724   -1.0790    0.6679   -0.3966    0.2257   -0.1231    0.0643   -0.0322    0.0154   -0.0070    0.0031

  Columns 73 through 90

   -0.0013    0.0005   -0.0002    0.0001   -0.0000    0.0000   -0.0000    0.0000   -0.0000    0.0000   -0.0000    0.0000   -0.0000    0.0000   -0.0000    0.0000   -0.0000    0.0000

  Columns 91 through 102

   -0.0000    0.0000   -0.0000    0.0000   -0.0000    0.0000   -0.0000    0.0000   -0.0000    0.0000   -0.0000    0.0000

Since "x" is an array of increasing values,for any iteration "i", the values of "x_new" in index positions 1 to "i-1" will be posiitve, at "i" the value of "x_new" is 1 and for index "i+1" to "N+2" the values are negative. So when calculating the product you are esssentially taking the product of "i" positive numbers and "N+2-i" negative numbers. So for even values of "i" the product is positive and for odd values it is negative. Hence the array "a" in your case has alternate negative and positive values. The same is shown in the code below:

for i=1:(N+2)
    x_new=x(i)-x;
    x_new(i)=1;
    product=prod(x_new);
    b(i)=1/product;
    if(i==25 ||i==26)
        disp("Value of i: "+i)
        x_new
        disp("Number of negative terms: "+sum(x_new<0))
        disp("Product of the terms: "+product)
    end
end

Value of i: 25

x_new = 1x102

1.1881 1.1386 1.0891 1.0396 0.9901 0.9406 0.8911 0.8416 0.7921 0.7426 0.6931 0.6436 0.5941 0.5446 0.4950 0.4455 0.3960 0.3465 0.2970 0.2475 0.1980 0.1485 0.0990 0.0495 1.0000 -0.0495 -0.0990 -0.1485 -0.1980 -0.2475

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Number of negative terms: 77

Product of the terms: -130056.8999

Value of i: 26

x_new = 1x102

1.2376 1.1881 1.1386 1.0891 1.0396 0.9901 0.9406 0.8911 0.8416 0.7921 0.7426 0.6931 0.6436 0.5941 0.5446 0.4950 0.4455 0.3960 0.3465 0.2970 0.2475 0.1980 0.1485 0.0990 0.0495 1.0000 -0.0495 -0.0990 -0.1485 -0.1980

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Number of negative terms: 76

Product of the terms: 42226.2662

If you are expecting the coefficients to be of different behaviour then it is purely based on the data "x".