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Chebyshev's zeros do Newton interpolation under interpolation nodes

function newton_divided_difference( f, min_x, max_x, n )
%UNTITLED Use Newton difference quotient method to interpolate
% f Is the original function ,min_x max_x The interval is defined ,n Is polynomial degree

% step 1: seek x,f,f[1],f[2]...f[n] x = min_x : (max_x-min_x)/n : max_x; for i = 1:1:n+1      y(i) = f(x(i));  end  a=[x',y'];  for i = 1:1:n      for j = 1:1:(n-i+1)          a(j,i+2) = (a(j+1,i+1)-a(j,i+1))/(a(j+i,1)-a(j,1));      end  end    % step 2: Draw the original function image  figure(1);  STEP = 0.001;  x = min_x : STEP : max_x;  for i = 1:1:((max_x-min_x)/STEP+1)      y_1(i) = f(x(i));  end  plot(x,y_1,'r')    % step 3: Draw an interpolation function image  hold on  for i = 1:1:((max_x-min_x)/STEP+1)      y_2(i) = a(1,n+2);      for j = n:-1:1          y_2(i) = y_2(i) * (x(i) - a(j,1)) + a(1,j+1);      end  end  plot(x,y_2,'b')  hold off    % step 4: Draw error image  figure(2);  y_error = y_1 - y_2;  plot(x,y_error)  

end

How can I use the Newton interpolation polynomial with Chebyshev as the zero point ?

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