資源簡介
新陳代謝GM(1,1),還不錯,
代碼片段和文件信息
function?[?c?a?b?y]?=?GM_1_1oxcdx(x0x)
%師兄的原始GM新陳代謝?x0=[4.44?5.89?8.06?8.74?9.63?10.74];
format?long
n?=?length(x0);
m=?length(x0);
y=0;?
while(x>=m)
????for?k=1:n
????????x1(k)=sum(x0(1:k));%生成x1序列
????end
????%?x1;
????for?k=2:n
????????z1(k-1)=(x1(k)+x1(k-1))/2;%生成背景值z1序列
????end
????Y=x0(2:n)‘;
????B=[-z1‘?ones(n-11)];
????N=inv(B‘*B)*B‘*Y;
????a=roundn(N(11)-6)
????b=roundn(N(21)-6)%ab正確說明z1構造成功過
????b_chu_a=
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