关于Logistic模型的matlab代码t=0:51; %令1954年为初始年x=[60.2 61.5 62.8 64.6 66 67.2 66.2 65.9 67.3 69.1 70.4 72.5 74.5 76.3 78.5 80.7 83 85.2 87.1 89.2 90.9 92.4 93.7 95 96.259 97.5 98.705 100.1 101.654 103.008 104.357 105.851 107.5 109.

关于Logistic模型的matlab代码t=0:51; %令1954年为初始年x=[60.2 61.5 62.8 64.6 66 67.2 66.2 65.9 67.3 69.1 70.4 72.5 74.5 76.3 78.5 80.7 83 85.2 87.1 89.2 90.9 92.4 93.7 95 96.259 97.5 98.705 100.1 101.654 103.008 104.357 105.851 107.5 109.
关于Logistic模型的matlab代码
t=0:51; %令1954年为初始年
x=[60.2 61.5 62.8 64.6 66 67.2 66.2 65.9 67.3 69.1 70.4 72.5 74.5 76.3 78.5 80.7 83 85.2 87.1 89.2 90.9 92.4 93.7 95 96.259 97.5 98.705 100.1 101.654 103.008 104.357 105.851 107.5 109.3 111.026 112.704 114.333 115.823 117.171 118.517 119.85 121.121 122.389 123.626 124.761 125.786 126.743 127.627 128.453 129.227 129.988 130.756];
[c,d]=solve('c/(1+(c/60.2-1)*exp(-5*d))=67.2','c/(1+(c/60.2-1)*exp(-20*d))=90.9','c','d') ;%求初始参数
b0=[ 241.9598,0.02985]; %初始参数值
fun=inline('b(1)./(1+(b(1)/60.2-1).*exp(-b(2).*t))','b','t');
[b1,r1,j1]=nlinfit(t,x,fun,b0)
y= 180.9871./(1+( 180.9871/60.2-1).*exp( -0.0336.*t)); %非线性拟合的方程
plot(t,x,'*',t,y,'-or') %对原始数据与曲线拟合后的值作图
R1=r1.^2;
R2=(x-mean(x)).^2;
R=1-R1/R2 %可决系数
W=sum(abs(r1)) %残差绝对值之
那个初始参数值是怎么定的,还有fun函数是固定的吗?拟合函数的那些系数又是怎么回事.

关于Logistic模型的matlab代码t=0:51; %令1954年为初始年x=[60.2 61.5 62.8 64.6 66 67.2 66.2 65.9 67.3 69.1 70.4 72.5 74.5 76.3 78.5 80.7 83 85.2 87.1 89.2 90.9 92.4 93.7 95 96.259 97.5 98.705 100.1 101.654 103.008 104.357 105.851 107.5 109.
【1】初始值是这么定的：
>> [c,d]=solve('c/(1+(c/60.2-1)*exp(-5*d))=67.2','c/(1+(c/60.2-1)*exp(-20*d))=90.9','c','d')
c =
62.457687920920008140916437719297-1.6291275002135204129081793780683*i
62.457687920920008140916437719297+1.6291275002135204129081793780683*i
241.95986840768380162085441918688
d =
-.95675170588036288272009149214033e-1+.43712460153973020393407947582405*i
-.95675170588036288272009149214033e-1-.43712460153973020393407947582405*i
.29854891132649696390901254874996e-1
【2】fun的函数形式就是logistic函数.
【3】b1 =
180.9871 0.0336
可以改成这样,效果一样：
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