function []=hw6_2_12(n)

E=zeros(n);
for i=1:n
    for j=1:n
        if (i==j)
            E(i,j)=i*(n-i+1);
        elseif (j>i)
            E(i,j)=E(i,j-1)-i;
        else
            E(i,j)=E(j,i);
        end
    end
end
E=1/(n+1)*E;

evals=sort(eig(E));

% avoid a for loop by making k a vector, the only catch is that you have
% to make a vector of ones the same size then divide the two vectors element by
% element using the ./ operator
k=[1:n]';
lambda=sort(ones(n,1)./(2-2*cos(pi*k/(n+1))));

disp('    approx    exact')
disp([evals,lambda])
disp('distance between')
disp(norm(lambda-evals))