Solved – Difference between non-linear curve fitting and interpolation

curve fittinginterpolationregression

I understand the difference between linear curve fitting and interpolation. In interpolation, the targeted function should pass through all given data points whereas in linear curve fitting we find the general trend of dependent variable. The cost function could be the distance between them.

If we keep on going with same sense of cost function, are not in case of interpolation the difference between data points zero ?
Should we not call interpolation as non-linear curve fitting ?
If not then what is the difference

Best Answer

In interpolation, the targeted function should pass through all given data points whereas in linear curve fitting we find the general trend of dependent variable.

Not at all! Interpolation is

a method of constructing new data points within the range of a discrete set of known data points.

You can use many different methods for interpolation including linear interpolation and polynomial, or spline curves.

When you are fitting curve to the data it is up to you to decide how close do you want it to fit the data. Nothing stops you from choosing the curve that perfectly fits to your data.

We are talking about interpolation when you use the fitted curve to re-create, or guess, the unobserved datapoints.

Related Solutions

Solved – Determining best fitting curve fitting function out of linear, exponential, and logarithmic functions

You might want to check out the free software called Eureqa. It has the specific aim of automating the process of finding both the functional form and the parameters of a given functional relationship.
If you are comparing models, with different numbers of parameters, you will generally want to use a measure of fit that penalises models with more parameters. There is a rich literature on which fit measure is most appropriate for model comparison, and issues get more complicated when the models are not nested. I'd be interested to hear what others think is the most suitable model comparison index given your scenario (as a side point, there was recently a discussion on my blog about model comparison indices in the context of comparing models for curve fitting).
From my experience, non-linear regression models are used for reasons beyond pure statistical fit to the given data:
1. Non-linear models make more plausible predictions outside the range of the data
2. Non-linear models require fewer parameters for equivalent fit
3. Non-linear regression models are often applied in domains where there is substantial prior research and theory guiding model selection.

Solved – How to interpolate (spline, LOESS, etc.) with a mix of critical and non-critical points

You may want to use weighted loess, as in:

a = data.frame(x=1:10, y=c(2,5,3,4,5,9,7,8,9,10))
fit1 = loess(y~x, a)  # unweighted loess fit
fit2 = loess(y~x, a, weights = c(0.1,1,0.1,0.1,0.1,0.1,1,0.1,0.1,0.1))
plot(1:10, predict(fit1), col="blue", type="l", ylim=c(1,10)) # unweighted fit
lines(1:10, predict(fit3), col="darkgreen") # weighted fit
points(y~x, a, col="brown", type="b") # all points 
points(y~x, a[c(2,7),], col="red") # points of high importance

plot

One issue with this approach is that weighted regression assumes that weights are known exactly, while this is very rarely the case. Some ideas on estimating weights for the case of weighted linear regression can be found here. Quite often however even a vague idea on differences in weights may work better than nothing. For example, here even though there was no guarantee that the loess fitted line will go exactly through the "important" points, this was very nearly the case with a rather arbitrary choice of weights.

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Related Solutions

Solved – Determining best fitting curve fitting function out of linear, exponential, and logarithmic functions

Solved – How to interpolate (spline, LOESS, etc.) with a mix of critical and non-critical points

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