Linear interpolation

The linear interpolation is certainly the simplest method of interpolation. For example, if we wish to determine F (2,5) whereas one knows the values of F (2) = 0,9093 and F (3) = 0,1411, this method consists in taking the average of the 2 values knowing that 2,5 is the medium of the 2 points. One obtains consequently $f\; (2,5)\; =\; \backslash \; frac\; \{0,9093+0,1411\}\; \{2\}\; =0,5252$.

More generally, the linear interpolation between 2 points ( X _has , _has there) and ( X _B , there _B ), the line of interpolation will have as an equation (three equivalent formulations):

$F\; (X)\; =\; \backslash \; frac\; \{y\_a\; there\; \_b\}\; \{x\_a-x\_b\}\; X\; +\; \backslash \; frac\; \{x\_a.y\_b-x\_b.y\_a\}\; \{x\_a-x\_b\}$

or (formula of Taylor-Young to the first order):

$F\; (X)\; =\; y\_a\; +\; (x-x\_a)\; \backslash \; frac\; \{y\_b\; there\; \_a\}\; \{x\_b-x\_a\}$

or:

$F\; (X)\; =\; \backslash \; frac\; \{x\_b-x\}\; \{x\_b-x\_a\}\; y\_a\; +\; \backslash \; frac\; \{x-x\_a\}\; \{x\_b-x\_a\}\; y\_b$

This method is fast and easy but it misses precision. Another disadvantage is that the function is not derivable at the point X _K .

The error in estimation shows that the linear interpolation is not very precise. If the function G which one wishes to interpolate is twice continuement derivable and if $x\; \backslash \; in$ then the interpolation error is given by:

$|F\; (X)\; -\; G\; (X)|\; \backslash \; C\; (x\_b-x\_a)\; ^2\; \backslash \; quad$ where $\backslash \; quad\; C\; =\; \backslash \; frac18\; \backslash \; max\_\; \{there\; \backslash \; in\}\; |G$ (there)|.