Why is the arima function giving odd answers I have a problem in interpreting what the arima function in R is doing.  I have the following code:
x <- 1000*.8^(0:100)
arima(x, order = c(1,0,0), include.mean = F)

The resulting coefficient is "0.9988".  But I would think the coefficient should be exactly "0.8", since x[t] = 0.8 * x[t-1].
I must be missing something that R is doing in processing the data.
Any help would be appreciated.
New Info I:
If I change the function to 
arima(x, order = c(1,0,0), include.mean = F, method="CSS")

then it solves for the correct coefficient of 0.8.
My problem that I am generally finding that the function arima with order = c(1,0,0) is often producing different results to:
lm(x[-1] ~ I(x[-length(x)]))

for all sorts of different time series that I am reviewing. 
New Info II:
Some of the comments and answers below are concerned that there is no random fluctuation in my data.  I did this to make the problem as simple as possible, but even if you add in random fluctuation, arima still produces the same wrong results in the default case.  To add insult to injury, arima will sometimes get the right answer if you change method = "CSS". This suggest that there is perhaps a computational issue with arima and not my misunderstanding the statistical model. Here is an extended set of two examples that highlight the problem (I ignore the intercept difference between lm and arima as these two items are not the same thing, but the coefficients should be.
set.seed(1)
# Example 2
x <- rep(1000,200)
for (i in 2:200) x[i]=x[i-1]*.8 + runif(1)*100 
plot(x,type="l")
arima(x, order = c(1,0,0))  #Incorrect answer:  coefficient = 0.9944
arima(x, order = c(1,0,0), method="CSS") # correct answer: coefficient = 0.7915
lm(x[-1] ~ I(x[-length(x)])) # correct answer: coefficient = 0.7914

# Example 3
x <- rep(0,200)
for (i in 2:200) x[i]=x[i-1]*.8 + runif(1)*.2 
plot(x,type="l")
arima(x, order = c(1,0,0))  #Incorrect answer:  coefficient = 0.8836
arima(x, order = c(1,0,0), method="CSS") # correct answer: coefficient = 0.8158
lm(x[-1] ~ I(x[-length(x)])) # correct answer: coefficient = 0.8158

 A: You need to read more about properties of an AR(1) model such as its mean and variance. So first see Example 1 in here. You can see that the mean of an AR(1) becomes zero if $\mu=0$, which is your case. Right? So this means that your time series should fluctuate around zero if it is coming from an AR(1) with a zero mean. Now lets look at your generated x
x <- 1000*.8^(0:100)
ts.plot(x)


As you can see, before time 20, it is not alternating around zero! Therefore, you should NOT expect to see a coefficient of .8 even if you fit it with an arima function. R does not check whether it makes sense to fit an AR(1) to your data or not. Basically, this is your job! To do it correctly, first you need to simulated an AR(1) time series of data as follow:
> z=arima.sim(n = 101, list(ar = c(0.8)))
> ts.plot(z)


As you see, right now it is now almost mean stationary. So let's try an arima function on z
> arima(z, order = c(1,0,0),include.mean =FALSE)
Series: z 
ARIMA(1,0,0) with zero mean     

Coefficients:
         ar1
      0.7922
s.e.  0.0612

sigma^2 estimated as 1.175:  log likelihood=-151.96
AIC=307.93   AICc=308.05   BIC=313.16

Now the AR coefficient is correctly estimated as 0.7922 ... Cheers!
