Finding real root of Newton's cubic

f = @(x) (x^5)-(2*x)-5;

dff = @(x) (3*x^2)-2;

x(0) = 1;

itr_num = 10;

TOL = 1E-6;

x = zeros(itr_num+1, 1);

x(1) = x^0;

val_n = 2;

n_fin = (itr_n <= itr_num+1)

while (val_n <= itr_num+1)

fe = f(x)*(val_n-1);

dffe = dff(x(val_n-1));

xval_n = (x(val_n-1) - (fe/dffe));

if (abs(fe) <= TOL)

n_fin = val_n;

break;

end

val_n = val_n+1;

end

fprintf('The approximate root is %f',x(val_n))

figure('color','white')

plot(0:n_fin-1,x(1:n_fin,'0-'));

title('f(x) = (x^3)-(2*x)-5');

xlabel('Iteration', 'Fontsize',16);

ylabel('value');

Now when I run the code it says 'Array indices must be positive integers or logical values'. How do I fix this as it doesn't give a specific line.

Answer 1

[James Tursa]

The third line:

x(0) = 1;

You have a 0 index. You need to start your indexing with 1, not 0.

You've got other issues with your code also. E.g., take these lines:

x = zeros(itr_num+1, 1);
x(1) = x^0;

Given that you have now defined x as a vector of 0's, what were you expecting the x^0 calculation to give you? Does it make sense to do that calculation and then assign the result to the x(1) element?

And then look at this line:

fe = f(x)*(val_n-1);

Here you don't index into x, but pass the whole vector into your f( ) function. That isn't going to do what you want either, and will need to be fixed.

And take a look at your function and derivative definitions:

f = @(x) (x^5)-(2*x)-5;
dff = @(x) (3*x^2)-2;

Does dff really look like the derivative of f to you?

[Corey Butler]

Okay the derivative is 5x^4 - 2 .

I'm new to Matlab. Could you point me in the right direction?

[James Tursa]

I've given you several pointers already. We are waiting for you to implement them and then get back to us on your progress and where you are still having problems.

Answer 2

[Corey Butler]

f = @(x) (x^5)-(2*x)-5;

dff = @(x) (3*x^2)-2;

x(1) = 1;

itr_num = 10;

TOL = 1E-6;

x = zeros(itr_num+1, 1);

val_n = 2;

n_fin = (itr_n <= itr_num+1)

while (val_n <= itr_num+1)

fe = f(x)(val_n-1);

dffe = dff(x(val_n-1));

xval_n = (x(val_n-1) - (fe/dffe));

if (abs(fe) <= TOL)

n_fin = val_n;

break;

end

val_n = val_n+1;

end

fprintf('The approximate root is %f',x(val_n))

figure('color','white')

plot(0:n_fin-1,x(1:n_fin,'0-'));

title('f(x) = (x^3)-(2*x)-5');

xlabel('Iteration', 'Fontsize',16);

ylabel('value');