Tutorial 5 of 5: additional tips - samhocevar/lolremez GitHub Wiki

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What to do when the solver doesn’t converge?

It may happen that the solver doesn’t converge. There are many possible reasons for that:

• the function being approximated is not smooth enough; try to increase the polynomial degree, or reduce the approximation range.
• the solver cycles between a finite number (usually 3 or 4) of solutions; in this case, experiments indicate that you may safely choose the solution with the smaller error.

Floating point precision

The Remez algorithm performs high precision computations to find its solutions. As a result, the polynomial coefficients found are printed with far more decimals than the usual double or float numbers can store. So when you write:

float a = 8.333017134192478312472752663154642556843e-3;

The exact value of a is actually:

8.333017118275165557861328125e-3
         ^-- this decimal changes!

See “fixing lower-order parameters” for both an explanation of why this can be a problem, and a method to reduce the introduced error.

When not to use Remez?

There are cases when you should expect the Remez algorithm to potentially perform badly:

• when the function is not continuous (for instance, a step function)
• when the function is not differentiable (for instance, the $x+|x|$ function)
• sometimes when the function is not smooth
• sometimes even when the function is not analytic

There are cases where you should not try to use the Remez algorithm at all:

• when the source range is not finite, e.g. $[0,+\infty]$
• when the destination range is not finite, e.g. $\log(x)$ tends to $-\infty$ near 0
• when the function to approximate has an infinite derivative at a point contained in or near the approximation range, e.g. $\sqrt{x}$ or $\sqrt[3]{x}$ in 0

What if I want to use Remez anyway?

If you need to approximate a function $f(x)$ over $[a,+\infty]$ and for some reason you want to use the Remez exchange algorithm, you can still through a change of variable: $y = 1/x$. The function to approximate becomes $f(1/y)$ and the new range is $[0,1/a]$ (see “changing variables” for how to deal with $1/x$ in 0). The minimax polynomial will use $1/x$ as its variable; please be aware that computing $1/x$ at runtime may be expensive.

Conclusion

Please report any trouble you may have had with this document to [email protected]. You may then return to the [wiki:doc/maths/remez Remez exchange documentation].