function [num,temp,alpha] = method_Fibonacci(a0,b0,L,epsilon)
% epsilon = 1e-9;
ak =a0;
bk =b0;
%%  找到 满足 >= temp的第一个 斐波那契数
temp = (bk-ak)/L;
f = [1 1];
n = 1;
while f(n)< temp
    f(n+2)=f(n)+f(n+1);
    n = n+1;
end
%%
k = 0;
num=1;
temp(1) = phi(0.5*(ak+bk));
i = 2;
while bk-ak>= L
    lamudak = ak +(bk-ak)* (f(n-k-1)/f(n-k+1));
    muk     = ak +(bk-ak)* (f(n-k)/f(n-k+1));
    if phi(lamudak)<phi(muk)
        ak = ak;
        bk = muk;
    else
        ak = lamudak;
        bk = bk;
    end
    k=k+1;
    num = num+1;
    temp(i) = phi(0.5*(ak+bk));
    disp(['这是Fibonacci--->第', num2str(num),'次迭代,当前目标函数最优值为',num2str(temp(i))])
    i=i+1;
end
alpha = (ak+bk)/2;
end