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Compare and contrast the methods of interpolation and extrapolation in mathematical terms.

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Interpolation is the process of estimating or predicting the value `y = f(x_3)` of an unknown but observable function`f` at a point `x_3` between two points `x_1` and `x_2` (`x_2>x_1`) where the value of the function can be/has been observed at `x_1` and `x_2` .

For example, linear interpolation is where we estimate or interpolate the value of `f` at `x_3`, `y =f(x_3)` using the formula `y = ((x_2-x_3)/(x_2-x_1))f(x_1) + ((x_3-x_1)/(x_2-x_1))f(x_2)`

By doing this, we make the assumption that `f` is linear over the interval `[x_1,x_2]` . If this seems unreasonable in the context, we would use some sort of non-linear interpolation eg quadratic interpolation.

Extrapolation is the process of estimating or predicting the value `y` of an unknown but observable function `f` at a point `x_3` outside of the range of points where the value of`f` can be/has been observed. If `x_1` is the minimum point at which the value of`f` has been observed and `x_2` is the maximum point, then `x_3` is outside of this range (either `x_3 < x_1`  or `x_3>x_2`). We use the form for `f` that we have fitted on the range `[x_1,x_2]` to estimate or extrapolate the value `y =f(x_3)`. If we have fitted `f` such that it is linear on the range `[x_1,x_2]` , so that `f(x) = (f(x_2)(x-x_1)-f(x_1)(x-x_2))/(x_2-x_1)`

then we extend this line out to `x_3` and linearly extrapolate ` ``y` as `f(x_3)`. The formula can be seen to be the same as that for interpolation except that in this case `x_3` is outside of the range `[x_1,x_2]` rather than inside it.

Answer (in short):

Interpolation is predicting the value y of a function f at a point `x_3` inside the range `[x_1,x_2]` where the values `f(x_1)` and `f(x_2)` have been observed.

Extrapolation is predicting the value of y when it is outside of the range `[x_1,x_2]`  defined by points `x_1` and `x_2`  where the values `f(x_1)` and `f(x_2)` have been observed.

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