CODEWITHSUNDEEP
polynomial-mathscilab
javaniewbie
javaniewbie

Reputation: 23

Evaluation of a polynomial in SciLab

Could you please help me find a mistake in my code? I need to calculate the polynomial, with parameters the vector and the value of the variable. I am not allowed to use the power sign (^) I have got this code but it doesn't work and I don't see what I am doing wrong.

function f=veelTermEvaluatie(V,x)
        f=V(1);
        for i=2:length(V)
            for j=0:i-1
                if j=0 then x=x
                    else x=x*x
                end
                f=V(i)*x
            end
        end
    endfunction

    V=[1 2 3 4 5 6]
    x=3

Upvotes: 1

Views: 2430

Answers (3)

Stéphane Mottelet
Stéphane Mottelet

Reputation: 3004

Scilab has already a horner function so you don't need to reinvent the wheel. For example, to evaluate the polynomial 1+2*x+4*x^2+8*x^3 at x=-1,0,1 you can proceed as follows:

p = poly([1 2 4 8],'x','coeff')
pval = horner(p,[-1 0 1])

which yields the following output

--> p = poly([1 2 4 8],'x','coeff')
 p  = 
            2    3
   1 +2x +4x  +8x

--> pval = horner(p,[-1 0 1])
 pval  = 

  -5.   1.   15.

Upvotes: 1

Lutz Lehmann
Lutz Lehmann

Reputation: 25972

What you should do is implement the Horner scheme.

f=V(n)
for i from n-1 down to 0 do
    f = f*x
    f = f+V(i)
end for
return f

In the question, you return the array of values V(i)*x^(2^i), in the previous answer, the array of evaluated terms V(i)*x^i is returned, but the value of the polynomial is the sum over those terms.

Please elaborate on the input format. Is V an array with index range 1:n? What is the intended relation between index and degree?

Upvotes: 0

spoorcc
spoorcc

Reputation: 2955

I first refactored your code, to remove the j=0 situation, since this does not change anything since x=x it can be skipped by starting at j=1. I also added some debug printing :

function f=veelTermEvaluatie(V,x)
    f=V(1);

    for i=2:length(V)

        printf("\nI: %d --> ", i)            
        for j=1:i-1                
            x = x * x
            printf("j: %d, x=%d ",j,x)
        end

        f=V(i)*x
    end

    return f
endfunction

After that it became clear that you multiply the same x, each and every time:

I: 2 --> j: 1, x=9 
I: 3 --> j: 1, x=81 j: 2, x=6561 
I: 4 --> j: 1, x=43046721 j: 2, x=-501334399 j: 3, x=0 
I: 5 --> j: 1, x=0 j: 2, x=0 j: 3, x=0 j: 4, x=Inf 
I: 6 --> j: 1, x=Inf j: 2, x=Inf j: 3, x=Inf j: 4, x=Inf j: 5, x=Inf

What also happens is that you update the value of f to the highest term, disregarding any other terms. I think your meaning is to return all terms

So you should create a local x which is reset for every term.

Working example

function f=veelTermEvaluatie(V,x)

    //Initialize the result to be a copy of the input
    f=V;

    // For each element in the input (except the first)
    for i=2:length(V)

        printf("\nI: %d --> ", i);
        //Initialize a local x variable to the global x
        new_x = x;

        for j=1:i-1
            // Update only the local x                 
            new_x = new_x * x;
            printf("j: %d, x=%d ",j,x);
        end

        // Multiply the value in the array with the local x value
        f(i)=V(i)*new_x;
    end

    return f
endfunction

V=[1 2 3 4 5 6]
x=3

result = veelTermEvaluatie(V,x)

disp(result)
disp(sum(result))

Upvotes: 1

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