The <<>> in the formatted output is a shorthand to indicate that there is an expression there that is too large to fit in the formatted version.

If you execute

InputForm[nlm]

you will see that everything you expect (and more) is there.

For plotting your expression is correct. Note that Mathematica chooses not to display all of the plot's y-range since the "interesting" part of the function would then note be visible. To see the entire y-range you can set PlotRange->All as in

Plot[nlm[x], {x, 0, 18000}, PlotRange -> All]

In the problem as stated, the best fit for b appears to be around 0.00008. For some reason NonlinearModelFit does not find it. If we rewrite the model slightly, so a good fit for b is about 0.8, it does well.

nlm = NonlinearModelFit[data, 
   1 - ((1/(1 + (2 b x/4.0952977))^3)) /. b -> b/10000, {b}, x];

Show[ListPlot[data], Plot[nlm[x], {x, 0, 18000}]]

Hi David. Are you sure? I used b->b/10000, so in your Out[85] there is still an expression which contains b. And that is the expression defining the model, with parameter b.

The goodness-of-fit is many times in the eye of the beholder. As a statistician I see that the first five values have positive residuals and the last three have negative residuals. That suggests that the model does not fit well (either the proposed curve form doesn't fit or the assumption of additive independent random errors is not supported or both). Also, R-squared is not so meaningful when describing the strength of the relationship of curves forced through the origin. In addition, I’d argue that the point {0,0} is likely NOT a data point as the curve form has to go through the origin.

A better measure of goodness is the root mean square error which is in the same units as the dependent variable. Given the undesirable residuals you might consider a slightly more complex model where the exponent "3" is replaced with a parameter.

data = {{1800, 0.258738807}, {3600, 0.362822881}, {5400, 
    0.473489839}, {7200, 0.54841429}, {9000, 0.614915391}, {10800, 
    0.638559079}, {14400, 0.68690743}, {18000, 0.730710491}};

(* Fit the model using the data *)
nlm = NonlinearModelFit[data, 1 - ((1/(1 + (b (x/1000)))^k)), {b, k}, 
  x]

(* Parameter estimates *)
nlm["BestFitParameters"]

(* Root mean square error *)
nlm["EstimatedVariance"]^0.5

(* Show the fit *)
Show[Plot[nlm[x], {x, 0, 18000}, PlotRange -> All], ListPlot[data]]

with output

{b -> 0.225117, k -> 0.818872}
0.0140814

I've also explicitly divided the independent variable by 1,000 as standardizing independent variables (either by subtracting the mean and dividing by the standard deviation or subtracting the minimum and dividing by the range) many times fixes numerical instabilities.

Hi David,

Thanks. That worked. The actual data is:

data = {{0, 0}, {1800, 0.258738807}, {3600, 0.362822881}, {5400, 0.473489839}, {7200, 0.54841429}, {9000, 0.614915391}, {10800, 0.638559079}, {14400, 0.68690743}, {18000, 0.730710491}};

I wrote the equation wrong. The actual equation is:

nlm = NonlinearModelFit[data, 1 - ((1/(1 + (2 b x/4.0952977))^3)), {b}, x].

Which gives an output of:

1 - 1/(1 + <<20>> x)^3

What do the <<>> mean?

For a plot of the graph is the following code correct:

Plot[nlm[x], {x, 0, 18000}]

Thanks,

Whoops! My bad! I need more coffee. Care to join me?

Yes your approach pushes the b into the region where whatever the Automatic Method is using will properly find it rather than send the search in the wrong direction.

I think Oregon is the better choice...

I don't understand why {b, 0.01} is incorrect for an initial guess for b?