Take the 2-minute tour ×
Mathematica Stack Exchange is a question and answer site for users of Mathematica. It's 100% free, no registration required.

This is what I have done so far. The points are of angles-of-deviation and wavelengths from my physics lab experiment.

V = ListPlot[
  {{589.3, 53.50}, {435.8, 56.02}, {535.4, 54.52}, {546.1, 
    53.57}, {577.0, 53.13}, {690.9, 52.10}, {402.6, 55.72}, {447.2, 
    55.02}, {492.2, 54.53}, {501.6, 54.83}, {587.6, 53.03}, {667.8, 
    52.83}, {706.5, 52.05}
   }, PlotRange -> {{400, 710}, {50, 60}}, 
  AxesLabel -> { wavelength, deviation}, PlotStyle -> Red, 
  GridLines -> Automatic]

this is what i have done so far.

I am trying to find the best fit line of these points. I know it is some type of exponential function. But I can not figure out how to do it. Please help me out.

share|improve this question
    
The relation appears to be linear. Why not use regression? –  David Carraher Nov 12 '13 at 2:35
1  
Which is the dependent and which is the independent variable (or are both independent)? Regression of y with respect to x is not the same as regression of x with respect to y. –  geordie Nov 12 '13 at 4:20
    
@geordie. Very good point. That makes me suspect that the (linear) Fit that I used below works differently from regression. –  David Carraher Nov 12 '13 at 5:17
    
@geordie If the experiment is to calibrate one set of measurements to the other, as taya suggests in a comment to dwa, then the standard measure would presumably correspond the dependent variable. –  David Carraher Nov 12 '13 at 5:27
    
@DavidCarraher. Given the scatter in both x and y, it seems reasonable to assume there is an uncertainty associated with each measurement - in which case a simple linear regression (least squares, etc.) will produce an unreliable fit. Better methods are available using robust stats. Then again, perhaps I'm over thinking this... –  geordie Nov 13 '13 at 1:06

2 Answers 2

Given the distribution of the data, you should use linear model, unless you have compelling theoretical reasons for believing the model is not linear.

data = {{589.3, 53.50}, {435.8, 56.02}, {535.4, 54.52}, {546.1, 53.57}, {577.0, 53.13},{690.9, 52.10}, {402.6, 55.72}, {447.2, 55.02}, {492.2, 54.53}, {501.6, 54.83}, {587.6,53.03}, {667.8, 52.83}, {706.5, 52.05}};
Show[
  ListPlot[data, PlotRange -> {{400, 710}, {50, 60}}, AxesLabel -> {"wavelength", "deviation"}, 
    PlotStyle -> Red, GridLines -> Automatic], 
  Plot[Evaluate@Fit[data, {1, x}, x], {x, 400, 710}]
     ]

linear

share|improve this answer
1  
The Dungeon Master told them it's an exponential. That's what I call a compelling experimental reason. –  belisarius Nov 12 '13 at 3:01
1  
@belisarius hits the evil dungeon master with his comment of snarkiness +1? –  bobthechemist Nov 12 '13 at 3:13
    
@bobthechemist It was with the hammer of cynicism (+1 to hit, +3 damage) –  belisarius Nov 12 '13 at 3:27

It's straightforward, and can get gleaned from the documentation.

After Sort@data,

nfit = NonlinearModelFit[data, a Exp[-t/\[Tau]], {a, \[Tau]}, t]
nfit["AdjustedRSquared"]
Show[{V,
  Plot[nfit[t], {t, 400, 710}]
  }
 ]

will compare your experimental data to an exponential fit. Documentation on NonlinearModelFit gets you a bunch of diagnostics.

Having said that, after looking at \[Tau], why not fit a linear function?

share|improve this answer
    
Thank you so much. –  taya Nov 12 '13 at 1:42
    
in the lab hand out, the professor gave the example graph that the calibration curve should resemble a curve line,,, which i guess it some kind of exponential... –  taya Nov 12 '13 at 1:45
    
I see no reason to sort the data. Am I missing something. –  David Carraher Nov 12 '13 at 2:53
    
No compelling reason, since NonlinearModelFit appears to handle data 'as is'. –  dwa Nov 12 '13 at 3:28

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.