Newton's Method in Matlab

I am trying to create a function that implements Newton's Method to solve the equation . I know from the past few questions that my zero should be close to x = 2.6357 when my initial guess x0 = 1. Any sort of advice would be helpful because at this point I do not produce any output in the first code and then I get 0.4109 from the second.

**Function 1:
function [y] = Newton3(x0)
    a = @(x) exp(2*sin(x)) - x;
    b = @(x) (2 * exp(2*sin(x))* cos(x)) - 1;
    tol = 10^12;
    x(1) = x0 - (a(x0) / b(x0));
    er(1) = abs(x(1) - x0);
    k = 2;
    while (er(k-1) > tol) && (k <= 50)
        x(k) = x(k-1) - (a(x(k-1)) / b(x(k-1)));
        er(k) = abs(x(k) - x(k-1));
        k = k + 1;
        y = x(k);
    end
end




**Function 2:
function [r] = Newton(x0)
    a = @(x) exp(2*sin(x)) - x;
    b = @(x) (2 * exp(2*sin(x))* cos(x)) - 1;
    tol = 10^-12;
    x = x0;
    for k = 1:50
        y = x0;
        x = y - (a(x) / b(x));
        if abs(a(k)) <= tol
            break
        end
    end
    r = x;
end

Answer 1

[John D'Errico]

First, consider if you are trying to solve the wrong problem.

In your question, you state it as e^2*sin(x) - x = 0

However, in your code, you write exp(2*sin(x)) - x. You do realize there is a difference? What I don't know is if you have miswritten your question, or is it your code?

fun = @(x) exp(2)*sin(x) - x;
fplot(fun,[0,3])
yline(0);
fzero(fun,3)
ans =
       2.7589

So the function you claim to want to solve has a zero around x==2.76, which is inconsistent with your claim of where the root lies. Next, look at the function you wrote code for:

fun = @(x) exp(2*sin(x)) - x;
fplot(fun,[-5,3])
yline(0);

Indeed, this does seem to have a root near 2.6357. So, just possibly, you really do want to solve the problem exp(2*sin(x))-x==0.

fzero(fun,3)
ans =
       2.6357

But now, look at the plot! What happens when you start Newton's method at x==1? THINK! Where will the first iteration go? On which side of that hump is x==1?

The point is, Newton's method tries to drive the functino to zero. But if you start the iterations BELOW x==1.5028, which direction will Newton's method push you?

fminbnd(@(x) -fun(x),1,3)
ans =
       1.5028
       

Think about the meaning of the iterations of Newton's method. What is the goal? What will happen?

[Sarah Johnson]

The equation in the question is incorrect, the code is the one to solve. I tried to fix it and this form won't let me raise 2sin(x), it only does the 2 no matter what I've tried. However the code is correct.

[Matt J]

I have fixed the equation rendering.

Answer 2

[Matt J]

Did you check whether the while loop is ever executed, even once? I don't think it is.