Example 1. Find the
rational approximation
for
over the interval [-1,1].
1 (c). Use
Mathematica's built in MiniMaxApproximation procedure.
Solution 1 (c).
Solve for the formula for
.
Plot graphs of the function and its minimax rational approximation over the interval [-1,1]. But we will draw the graphs over [-2,2].
![[Graphics:../Images/RationalApproxMod_gr_91.gif]](../Images/RationalApproxMod_gr_91.gif)
Find the error over the interval [-1,1].
![[Graphics:../Images/RationalApproxMod_gr_94.gif]](../Images/RationalApproxMod_gr_94.gif)
Comparison with the Taylor approximation.
There were 5 coefficients to determine for the rational
approximation, and a Maclaurin polynomial of degree 4 requires 5
coefficients.
Compare with the error in a
degree Maclaurin polynomial over the interval
.
![[Graphics:../Images/RationalApproxMod_gr_99.gif]](../Images/RationalApproxMod_gr_99.gif)
![[Graphics:../Images/RationalApproxMod_gr_102.gif]](../Images/RationalApproxMod_gr_102.gif)
We can determine how much smaller the error is for the MiniMax rational approximation.
Just what is this MiniMax rational approximation minimizing ?
Plot the quotient
![[Graphics:../Images/RationalApproxMod_gr_108.gif]](../Images/RationalApproxMod_gr_108.gif)
![[Graphics:../Images/RationalApproxMod_gr_111.gif]](../Images/RationalApproxMod_gr_111.gif)
So we see that the MiniMax rational approximation is minimizing the "relative error." The nodes are selected in an iterative process so that the maximum of the absolute value of the relative error is minimum, and the extrema occur at six points in this example.
(c) John H. Mathews 2004