How to plot Fit functions?

Question

If I do something like this

f[x_]:=x^2

Plot[f[x],{x,1,10}]

Mathematica plots the function $f(x)=x^2$, as expected. However, if I do something like this instead

data=Table[x^2,{x,1,10}]

f[x_]:=Fit[data,{1,x},x]

Plot[f[x],{x,1,10}]

Mathematica throws me some weird errors such as -4.9998 is not a valid variable. What is going on here?

Answer 1

The definition of your function f[x_] is the problem. Here is the corrected version:

data = Table[x^2, {x, 1, 10}]

(* ==> {1, 4, 9, 16, 25, 36, 49, 64, 81, 100} *)

f[x_] = Fit[data, {1, x}, x]

(* ==> -22. + 11. x *)

Plot[f[x], {x, 1, 10}]

The mistake was to define the function with SetDelayed. In that case, the symbol x on the right-hand side of the definition will be given the value with which the function is called, whereas Fit expects x to be a symbol and not a numeric value.

Answer 2

As an alternative, you can use FindFit[] instead of Fit[] for defining your regression function:

data = Table[x^2, {x, 1, 10}];
f = Function[x, a + b x] /. FindFit[data, a + b x, {a, b}, x];

Plot[f[x], {x, 1, 10}] ~Show~ ListPlot[data]

This hinges on the fact that FindFit[] returns the parameters (as replacement rules) as opposed to the regression function itself, making substitution into the regression model easy.

---

Mr. Wizard reminds me that variable localization might be necessary, especially if this is being done in the midst of a long session, where the variables may have had previous un-Clear[]ed assignments. One way to go about it is to use so-called "formal symbols" which are guaranteed to never have taken previous values, since you can's assign to them.

Here's how to go about it in this case:

f = Function[\[FormalX], \[FormalA] + \[FormalB] \[FormalX]] /.
   FindFit[data, \[FormalA] + \[FormalB] \[FormalX], {\[FormalA], \[FormalB]}, \[FormalX]]

(It looks somewhat complicated here, since the SE interface does not have the proper font faces for the characters that denote formal symbols. It looks better when pasted into Mathematica.)

Answer 3

An example:

data = {{0, 1}, {1, 0}, {3, 2}, {5, 4}}

f[x_] = Fit[data, {1, x}, x]

Plot[f[x], {x, 0, 10}]

Please note that I used = instead of :=. The right hand side needs to evaluate before being assigned to f.

Answer 4

Also

data = Table[x^2, {x, 1, 10}];
f1 = Fit[data, {1, x}, x];
f2 = Fit[data, {1, x^2}, x];
Show[Plot[{f1, f2}, {x, 1, 10}], ListPlot@data]

Edit

Note that the following gives you the same result

data = Table[x^2, {x, 1, 10}];
f1 = Fit[data, {1, #}, x] & /@ {x, x x}
Show[Plot[f1, {x, 1, 10}], ListPlot@data]

Answer 5

At present none of the answers given localize variables during the definition of f to protect against global assignments. For actual use I strongly suggest you do:

x = 0; (*example left-over assignment*)

data = {{0, 1}, {1, 0}, {3, 2}, {5, 4}}

Block[{x},
  f[x_] = Fit[data, {1, x}, x]
]

Plot[f[x], {x, 0, 10}]

Or:

f[\[FormalX]_] = Fit[data, {1, \[FormalX]}, \[FormalX]]

Plot[f[x], {x, 0, 10}]

---

I also wrote a function localSet to answer a prior question which you could use:

x = 0; (*example left-over assignment*)

data = {{0, 1}, {1, 0}, {3, 2}, {5, 4}}

localSet[ f[x_], Fit[data, {1, x}, x] ]

DownValues[f]

{HoldPattern[f[x_]] :> 0.186441 + 0.694915 x}