这个是自动建模，模型是MA（2），系数就是ma1，ma2，下面的s.e应该是标准差最下的是一下评价准则的结果

请问写出来的函数是什么样的

可以的，有个forecast的包，可以直接确定ARIMA模型的阶数
data <- read.csv("book1.xlsx",sep=",")
data
y <- ts(data$y,start = 1978,frequency = 1)
y
par(mfrow = c(1,1))
plot(data$y)
hist(data$y)
yd <- diff(log(y),1)
yd
plot(yd)
acf(yd,lag = 24)
pacf(yd,lag = 24)
fit <- arima(yd,order = c(1,0,1),method = 'ML')
fit
fit1 <- arima(yd,order = c(1,0,0),method = 'ML')
fit1
tsdiag(fit)
predict(fit,n.ahead = 3)$pred
library(forecast)
data$y
fit <- auto.arima(y)
fit
model <- arima(y,order = c(1,1,1))
accuracy(model)
accuracy(fit)
accuracy(fit1)
accuracy(fit2)
model$aic
fit$aic
forecast(fit,5)
plot(forecast(fit,5))

> x<-ts(x,start=1988)
> plot(x)
>  for(i in 1:2)
+  {print(Box.test(x,lag=6*i))}

        Box-Pierce test

data:  x
X-squared = 62.543, df = 6, p-value = 1.368e-11


        Box-Pierce test

data:  x
X-squared = 70.871, df = 12, p-value = 2.2e-10

>  x.dif<-diff(x)
>  plot(x.dif)
> acf(x.dif)
> pacf(x.dif)
> y.dif<-diff(x.dif)
> plot(y.dif)
> acf(y.dif)
> pacf(y.dif)
> x.fit<-arima(x,order=c(1,2,2))
> x.fit

Call:
arima(x = x, order = c(1, 2, 2))

Coefficients:
         ar1      ma1     ma2
      0.2088  -1.3440  0.3441
s.e.  1.8221   1.8088  1.8016

sigma^2 estimated as 361.6:  log likelihood = -106.54,  aic = 221.08
> library(forecast)
> x.fore<-forecast(x.fit,h=3)
> x.fore
     Point Forecast    Lo 80    Hi 80    Lo 95    Hi 95
2014       360.1367 335.2747 384.9986 322.1136 398.1598
2015       357.2889 323.7881 390.7896 306.0538 408.5239
2016       354.6531 314.2735 395.0327 292.8979 416.4084
根据这个程序得到了未来三年的预测值，但是如何得到14年之前的拟合值

why do you need 14 years ago fitted value? you could get all the fitted value by using predict function.

用auto.arima