function Nov11()

% z is a fixed point of the iteration function G:R-->R if z = G(z)
% Here are the G functions for the four methods
% Am1: y_{n+1} is a fixed point of
%       G1(z) = = y_{n} + h*f(t_{n+1},z)
% Am2: y_{n+1} is a fixed point of
%       G2(z) = = y_{n} + (h/2)*(f(t_{n+1},z) + stuff)
% Am3: y_{n+1} is a fixed point of
%       G3(z) = = y_{n} + (h/24)*(f(t_{n+1},z) + stuff)
% Am4: y_{n+1} is a fixed point of
%       G4(z) = = y_{n} + (h/720)*(f(t_{n+1},z) + stuff)

%  What can guarantee the convergence of z_{k+1} = G(z_{k} ?
%  If the thing converges to z* then
%      z* - z(k+1} = z* - G(z_{k}) 
%                  = ((G(z)-G(z_{k})/(z*-z_{k}) ) (z*-z_{k})
%   So if |G'(z)| = mu < 0, we have
%       |z* - z_{k+1}| <= mu |z* - z{k}| = mu^(k+1)|z* - z_{0}|
%  If mu<1 we have convergence. For the above G functions, if tau is
% an upper bound for the y-partial f_{y}(t,y) we would like
%     AM1    h*tau      < 1
%     AM2   (h/2)*tau)  < 1
%     AM3   (h/24)*tau  < 1
%     AM4   (h/720)*tau < 1

    








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