Talk:Risch algorithm

Constant undecidability

I'm removing the following statement from the page because I'm not sure it's true:

The Risch decision procedure is not formally an algorithm because it requires an oracle that decides whether a constant expression is zero, a problem shown by Daniel Richardson to be undecidable.

According to MathWorld, Richardson's theorem states:

Let $R$ be the class of expressions generated by

1. The rational numbers and the two real numbers $\pi$ and $\ln 2$,
2. The variable x,
3. The operations of addition, multiplication, and composition, and
4. The sine, exponential, and absolute value functions.

Then if $E$ in $R$, the predicate $E=0$ is recursively undecidable.

This is based on a simple algebraic system that includes, among other things, an absolute value function. This implies an ordered field, while Risch integration is typically (always?) done over an unordered field. And this seems to make a big difference! My logic goes as follows:

1. Any constant expression involving nothing but rational numbers and the standard field operators (+ - * /) should be decidable. In sums and differences, multiply denominators by common multiples to make them equal, then combine fractions, throw away the denominator, and test for zero on the numerator. Multiplication is even simpler — just multiple the numerators and denominators seperately, and for division just interchange numerator and denominator.
2. Any simple algebraic extension over a field can be reduced using Euclidean long division to testing if a remainder in the underlying field is zero.
3. Any simple transcendental extension over a field requires all coefficients to be identically zero — each is testable in the underlying field.

Therefore, any algebraic system that consists purely of the rationals plus a finite number of algebraic and/or transcendental extensions should be decidably testable for equality to zero. This is the world of the Risch algorithm.

And the presence of the absolute value function makes all the difference for the Richardson theorem, right?

Baccala@freesoft.org 05:49, 14 January 2006 (UTC)