How do I find a root in a nonlinear equation?

my equation is

function y=f(x)

x = (-20:0.05:20); %-20<x<20

y=5*(x^2)*(sin(x))

I already have the graph of this function, but what would I write in the script (m file) to solve this equation for the root and have it plot on the same graph? I tried fzero, but I keep getting an error message back, so maybe I'm using it incorrectly?

Answer 1

[Star Strider]

One option:

x = (-20:0.05:20); %-20<x<20
y=5*(x.^2).*(sin(x));

yz = y.*circshift(y, [0 -1]);
yzi = find(yz < 0);
for k1 = 1:size(yzi,2)-1
    xzeros(k1) = interp1(y(yzi(k1):yzi(k1)+1),x(yzi(k1):yzi(k1)+1),0);
end

figure(1)
plot(x, y, '-m')
hold on
plot(xzeros, zeros(size(xzeros)), 'pb')
hold off
grid

The routine finds the indices near the zero-crossings by multiplying the y-values with a one-index circularly-shifted version of itself and then finds the values where these products are negative. It then uses interp1 and the intervals defined by these indices to find the actual values of the zeros, producing the vector ‘xzeros’ that are the x-coordinates of the zeros. It then plots them.

[Joseph Cheng]

you'd want to find(yz<=0) or you'll miss when x==0

[Star Strider]

True. The consecutive indices though provide enough information for interp1 to find the actual zeros.