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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

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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.

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  • $\begingroup$ Are you saying that interpolation and non-linear curve fitting are one and the same? I would distinguish them by noting that interpolation tends towards the tautological where non-linear curve fitting involves fitting a model of some type, e.g., polynomial or spline curves, as you note. But that's just my POV. $\endgroup$
    – user78229
    Commented Apr 1, 2016 at 11:59
  • $\begingroup$ @DJohnson Exactly the opposite: they are two different things :) Curves can be used for interpolation, as one of the many tools available. That's all. $\endgroup$
    – Tim
    Commented Apr 1, 2016 at 12:00
  • $\begingroup$ @Tim Your are answering to this Question: 'What is interpolation'. You just explained interpolation with another way, probably with an Engineering perspective where we get data from sensor whose resolution does not fulfill our requirements. I have implemented polynomial interpolation, $\endgroup$
    – TonyParker
    Commented Apr 1, 2016 at 12:31
  • $\begingroup$ @hmluqman I do not understand what do you mean..? $\endgroup$
    – Tim
    Commented Apr 1, 2016 at 12:32

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