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Is there a way to fit a line similar to power trendline in Excel?

Something in this fashion: Fit[data, {1, x, x^(-n)}, x]

EDIT:

modelFit = NonlinearModelFit[data, a*x^n, {a, n}, x];

Show[
 ListPlot[data, PlotStyle -> Red],
 Plot[modelFit[x], {x, 10, 300}]
 ]

Gives a result which is a little off:

enter image description here

Data:

data = {{10, 0.229456252}, {11, 0.197084485}, {12, 0.190385018}, {13, 
0.167211837}, {14, 0.162024048}, {15, 0.146360843}, {16, 
0.141582714}, {17, 0.128658408}, {18, 0.12634757}, {19, 
0.115664973}, {20, 0.112934492}, {21, 0.103436493}, {22, 
0.101525578}, {23, 0.094280256}, {24, 0.093998465}, {25, 
0.087612133}, {26, 0.085961964}, {27, 0.081235224}, {28, 
0.079490311}, {29, 0.075953893}, {30, 0.073722495}, {31, 
0.070194375}, {32, 0.069373963}, {33, 0.066115294}, {34, 
0.064971653}, {35, 0.061982956}};
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Use the same function if you want the same result: something like NonlinearModelFit[data, a + b x + c x^n, {a,b,c,n}, x]; –  bill s May 5 '13 at 13:43
    
Out of curiosity, what does your data represent? –  Szabolcs May 5 '13 at 20:48
    
@Szabolcs, variance of scores in a computer boardgame. –  JavaCake May 5 '13 at 21:35
    
can you explain in what way the data is a little off? –  jens_bo May 6 '13 at 7:26
    
@jensen, im not sure if my screenshot is clear enough, but the line does not go through the first points in the upper bound of the graph. –  JavaCake May 6 '13 at 12:59

3 Answers 3

up vote 4 down vote accepted

Sure. Try e.g. data = Transpose[{Range[10], Range[10]^3}] and then use NonlinearModelFit[data, x^n, {n}, x]. As bill s said, read for more details Mathematicas help regarding NonlinearModelFit.

For your data you can use:

fitFkt = NonlinearModelFit[data, a*x^n, {a, n}, x]
fitFkt["ParameterConfidenceIntervalTable", ConfidenceLevel -> .95]

The result is the same as within Excel (y = 2.3409x^-1.018; you only forgot the factor a) as you can see from:

Show[ListPlot[data], Plot[fitFkt[x], {x, 10, 35}]]

Fit to data

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I have updated my post with an example im working on. The fitted line is a bit off. –  JavaCake May 5 '13 at 13:32
    
Could you please give us the data too? Then we could try to get a better fit. –  partial81 May 5 '13 at 14:04
    
Sure, i can give a small sample. Excel fits it perfectly to a power trendline. –  JavaCake May 5 '13 at 14:14
    
Thanks for the great example. What i dont understand is that my fitted line does not begin from 10 as my points. I need to enter something like {-2000,300,x} which seems ok. I have updated the example with my issue. –  JavaCake May 5 '13 at 20:22
1  
Ok, just to summarize: The fit is ok, just the plot not. Do you agree? That is because you plot over a big range. If you use Show[ListPlot[data, PlotStyle -> Red, PlotRange -> All], Plot[modelFit[x], {x, 10, 300}, PlotRange -> All]] you will see that the function fits to the data (without choosing such a big range as above). By the way: If you do not trust your fit, you can use something like Table[{i, modelFit[i]}, {i, 10, 30}]. If you compare the result with your data (or calculate the residual), you will see that the fit is well. –  partial81 May 6 '13 at 7:10

Plot automatically reduces the Plot range depending on the function that is plotted. Adding PlotRange->All shows the function completely in the Plot plot. Show uses the data range from the first plot that is shown, so you eventually need to add a PlotRange->All inside the Show command (depending on your data).

modelFit = NonlinearModelFit[data, a*x^n, {a, n}, x];

Show[
 ListPlot[data, PlotStyle -> Red], 
 Plot[modelFit[x], {x, 10, 300}, PlotRange->All]
]

enter image description here


An advice for plotting and fitting power laws:

Often it is better to take the logarithm of the function and the data, or to use a logarithmic scale (LogLogPlot or ListLogLogPlot).

Let's say you have $f(x)=a\cdot x^n$, taking the logarithm leads to: $$\log(f(x)) = \log(a) + n \cdot \log(x)$$ so you basically get a straight line if you plot $\log(f(x))$ vs. $\log(x)$ and the slope is the exponent $n$. This makes it easier to see deviations.

Show[ListPlot[Log@data, PlotStyle -> Red], 
  Plot[Log@modelFit[Exp@x], {x, 1, 4}], Frame -> True, 
  FrameLabel -> {"log(score)", "log(Var)"}, 
  BaseStyle -> {14, FontFamily -> "Helvetica"}]

enter image description here

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If the model is linear, you can use LinearModelFit. If the model is nonlinear, then NonlinearModelFit. To see how to apply these, check out the help (F1 on the word LinearModelFit`) where there are plenty of examples.

To mimic the function you've included, you could try something like:

NonlinearModelFit[data, a + b x + c x^n, {a,b,c,n}, x];

Though it also looks like an exponential decay, so you might want to see how that fits as well.

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