I need help with the secant method. Need to find the root of Wilkinson polynomial

Im trying to use the secant method to prove that the root lies [20,22]. The equation that I'm trying to solve is

p(t) = (t-1)*(t-2)*...*(t-20) - (10^-8)*t^19

my code is as follows:

format long
my_tol = 1e-16
w =@(t,n) prod(t-n)-(10.^-8)*t.^19
t(1) = input('Enter first guess:  ')
t(2) = input('Enter second guess:  ')
iter = 0;
n = 1;
for i = 3:23
    t(i) = t(i-1)-(w(t(i-1),n))*((t(i-1)-t(i-2))/(w(t(i-1),n))-w(t(i-2)));
    iter = iter + 1;
    n = n + 1;
    if abs(t(i)-t(i-1))/t(i) < my_tol
        wilkroot = t(i)
        break
    end 
end

I'm not going to lie, i am terrible at this. This code looks like it should work but i get this error

*Error using @(t,n)prod(t-n)-(10.^-8)*t.^19
Not enough input arguments.

Error in test (line 9)
    t(i) = t(i-1)-(w(t(i-1),n))*((t(i-1)-t(i-2))/(w(t(i-1),n))-w(t(i-2)));*

Can anyone help me with this?? I'm losing my f****** mind

[John D'Errico]

It is much easier to read IF you flag as code those parts of your post that are code. In fact, simply preceding those lines with a double space will suffice.

Answer 1

[Guillaume]

You've defined w as a function of two variables, but the last part of your t calculation on line 9 calls w with only one argument, t(i-2). Maybe you meant:

t(i) = t(i-1)-(w(t(i-1),n))*((t(i-1)-t(i-2))/(w(t(i-1),n))-w(t(i-2), n));

Answer 2

[John D'Errico]

Anyway, why do you think that your function w(t,n) evaluates that polynomial?

t is a vector of length 2, end points of your interval apparently. n is a scalar, that seems to change with iteration?

I think you need to first figure out how to write a function that will CORRECTLY evaluate that polynomial.

Your polynomial has a fixed order, so why make it a function of n at all? Next, you need to be careful trying to use tools like prod when t is a vector. How does MATLAB know what to take the product over? READ THE HELP FOR PROD.

For example, IF you know that t will always be a row vector, then you MIGHT do something like this:

w = @(t) prod(bsxfun(@minus,t,(1:20)'),1) - (10^-8)*t.^19;

Note how I've changed w. See that the last term uses the .^ operator, instead of the ^ operator. There is a difference, and it is an important one when t is a vector.

I've also made w not a function of n. Why would you have it that way anyway?

And, if you wanted to make w insensitive to whether t is a row or column vector, this would work.

w = @(t) prod(bsxfun(@minus,t(:).',(1:20)'),1) - (10^-8)*t(:).'.^19;

Having done that, what does it take to prove a root lies in an interval? So if your interval of interest is [20,22], what does w([20 22]) tell you? Do you really need to do much more than that if your goal is to show at least one root lies in an interval?

[Tony Montgomery]

Forgive me I am VERY new to this and i really don't know what i'm doing. I'm trying to figure it out as i go trying to learn code as i go. I thought i had to iterate n and t, but im assuming by your answer i do not. I will rewrite the function as you did and run it.

[John D'Errico]

The polynomial is not a function of n. Somehow maybe you were confusing n with the numbers 1:20 in the polynomial. So rewrite it. Solve the small problem of determining there is a root in that interval.

The solve the secant method iteration. By the way, I'm not sure why you were using a var named n and iter, then incrementing both.