Newton Raphson Getting Wrong Root

I'm trying use the Newton Raphson method to produce a root of -3.4256, according to fzero(f,-3.5). I think I have the equations right but for some reason it gets -0.9428. Please help correct mistakes. Thanks!

f=@(x) x*tan(x)-1;
df = @(x) tan(x) + x*(tan(x)^2 - 1);
x1 =-5;
x2=-1.5;
x=x1;
for i=1:1:10
    x3=x-(f(x)/df(x));
    x=x3;
end
fprintf('Approx. Root is %.15f\n',x);

Answer 1

[Walter Roberson]

This is a general characteristic of Newton Raphson: because of the way it does the projections, you can easily end up projecting into a different "basin of attraction" than the one you are currently in.

Try this:

function x = do_nr(x1, f, df)

x=x1;
for i=1:1:10
    x3 = x - (f(x)/df(x));
    x=x3;
end

end

together with

f=@(x) x*tan(x)-1;
df = @(x) tan(x) + x*(tan(x)^2 - 1);

x1 = linspace(-5, -2.5, 500);

root_found = arrayfun( @(X1) do_nr(X1, f, df), x1);

plot(x1, root_found)

The values on the y axis will be the root located. Notice the scale -- and notice how bumpy the graph is.

The closer you get to a root, the further you can end up projecting, because the angle can be very steep.

[Farrah Rinehart]

The plot doesn't show for some reason. I typed it in as:

x1=linspace(-5,-1.5,500);
root_found=arrayfun(@(X1) do_nr(X1,f,df),x1);
plot(x1,root_found)
function x = do_nr(x1,f,df)
    x=x1;
    for i=1:1:10
        x3=x-(f(x)/df(x));
        x=x3;
    end
end

[Walter Roberson]

If you are using a release before R2016b you will need to use separate files for the two pieces.

[Farrah Rinehart]

I'm using R2017b.

Edit Found the plot, it just wasn't showing up like earlier. However it's not bumpy at all, although I do notice the scale.

[Walter Roberson]

On the scale of 10^10 the bumps are not showing up. If you were to change your range in the linspace() then you would change the size of that peak. You can also use the zoom tool to look at that range between -5 and -2