MATLAB: Finding the first 50 roots

roots

The below code was posted by Matt Fig as a response in a thread last fall. I have a similar function that I need to find the first 50 roots. How would you modify the code to accommodate for the change?
function [z] = find_roots()
x = 0.01:0.01:(20*pi/2)-0.01;
plot(x,tan(x)+5.6*x),
grid on
f = @(x) tan(x)+5.6*x;
N = 250; % Fineness of the search, bigger is finer
z = zeros(1,N);
for jj = 1:30
    for n=1:N
        try
            z(n) = fzero(f,jj+[(n-1)/N n/N]);
            if abs(f(z(n)))>1e-8
                z(n) = 0;
            else
                fprintf('Found a root at: %.5f\n',z(n))
            end
        catch
        end
    end
end
x = 0:pi/50:10*pi;
y = f(x);
plot(z,zeros(1,length(z)),'o',x,y,'-')
line([0 10*pi],[0 0],'color','black')
axis([0 10*pi -1000.0 3000.0])
z = sort(z(z~=0)); % Return only the needed values

Best Answer

I think the idea there was to incrementally search for the roots. If you need a certain number of roots, you could use a WHILE loop that evaluates on the root counter. It often helps to know some details of the function in question. Can you post it?
Here is how I would modify the function.
function [z] = find_roots() 
f = @(x) x.*cot(x)+99;
NR = 50;  % The number of roots to find.
N = 150; % Fineness of the search, bigger is finer z = zeros(1,N);
rcnt = 1; % The root counter
cnt = 0;  % The loop counter
while rcnt<=NR 
    cnt = cnt + 1;
    for n = 1:N 
        try z(rcnt) = fzero(f,cnt+[(n-1)/N n/N]); 
            if abs(f(z(rcnt)))>1e-8 
                z(rcnt) = nan; % FZERO was messing with us
            else
                fprintf('Found a root at: %.5f\n',z(rcnt)) 
                rcnt = rcnt + 1;
                break % Put here only if root spacing is >1!
            end
        catch
        end
    end
end
x = 0:pi/50:max(z); 
plot(x,f(x),'-',z,zeros(1,length(z)),'o') 
line([0 max(z)],[0 0],'color','black') 
axis([0 max(z) -1000.0 3000.0]) 
z = z(~isnan(z)); % Return only the needed values
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