# How to perform simple linear interpolation on a data set

With the following data set, what is the best way to interpolate the data for each time.

Time    X    Y
0      10    15
...
...
24     28    17
...
...
49     9     14

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Best way in what sense? There are different methods and each has its own dis/advantages. –  Gigili Aug 1 '12 at 8:51
Now I think about it more, I guess just a linear straight line fit is best for my usage. As I know my objects position at 0 and at 24, and need to estimate as close as possible where it is at any other time. in this instance it is a cube structure rotating slightly off-axis. –  davivid Aug 1 '12 at 9:03

You can use Newton's divided differences interpolation polynomial which is easy to use and if you add a new point to the set, you don't have to calculate everything again.

So you'll have a table with four columns, $x_i, y_i$ and divided differences where:

$$f\left[x_0,x_1,\dots, x_n\right]=\dfrac{f\left[x_0,x_1,\dots, x_{n-1}\right]-f\left[x_1,\dots, x_n\right]}{x_0-x_n}$$

Then, for example, you have:

$x_0=10, y_0=15, x_1=28,y_1=17$: $$f[x_0,x_1]=\dfrac{y_0-y_1}{x_0-x_1}=\dfrac{15-17}{10-28}=0.11$$

$$f[x_1,x_2]=\dfrac{17-14}{28-9}=0.15$$

$$f[x_0,x_1,x_2]=\dfrac{0.11-0.15}{10-9}$$

Thus, the interpolating polynomial is:

$$p(x)=f_0+(x-x_0)f[x+0,x_1]+\dots+(x-x)\dots(x-x_{n-1})f[x+0,\dots,x_n]$$

And it's easy to take it from here.

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