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Re: Making a function from the output of the Fit

  • To: mathgroup at smc.vnet.net
  • Subject: [mg121554] Re: Making a function from the output of the Fit
  • From: Daniel Lichtblau <danl at wolfram.com>
  • Date: Mon, 19 Sep 2011 07:07:05 -0400 (EDT)
  • Delivered-to: l-mathgroup@mail-archive0.wolfram.com

----- Original Message -----
> From: "Just A Stranger" <forpeopleidontknow at gmail.com>
> To: mathgroup at smc.vnet.net
> Sent: Saturday, September 17, 2011 5:25:53 AM
> Subject: Making a function from the output of the Fit function.
> Hello,
> 
> So I am trying to use the Fit ffunction to make another function.
> 
> So I want to do something like this:
> 
> f[x_] := Fit[ data, {1,x}, x]
> 
> But I want f[x] to be the function for the line, a+bx. But as written
> above
> it does not have this result of course (I know it's because whatever
> gets
> passed is used as an argument for the fit function, so passing a
> number
> doesn't work, and passing a variable simply isn't what I want to do)
> So how
> can I use the fit function to get
> f[x_] := a+bx (where a and b are the appropriate constants. )
> 
> My jerry rigged solution is to just execute Fit[ data, {1,x}, x] then
> copy
> and past the output to define a function. But this tacky solution just
> won't
> do :)
> 
> Thank you for your help mathgroup.

Might be easier to use FindFit.

In[49]:= data = 
  Table[{j, 3*j - 2}, {j, 20}] + RandomReal[{-.1, .1}, {20, 2}];

Now one can do

In[58]:= a + b*x /. FindFit[data, a + b*x, {a, b}, x]
Out[58]= -1.939582811115561 + 2.994503289542548 x

or

In[61]:= f[x_] = a + b*x /. FindFit[data, a + b*x, {a, b}, x]
Out[61]= -1.939582811115561 + 2.994503289542548 x

Many years ago I used the term "jerry rigged" in some in-house email. I was informed that it is not proper phraseology. It seems that "jury rig" and "jerry build" are quite different in meaning.

Several years after that incident, I noticed "jerry rigged" in a James Ellroy novel (White Jazz, I think it was). It seems that the term has entered the language as a crossover of sorts.

http://www.wordcourt.com/archives.php?show04-03-10

http://en.wikipedia.org/wiki/Jury_rig

http://www.answerbag.com/q_view/9245

Daniel Lichtblau
Wolfram Research







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