From 7096725a5c50f4369e2c6c85f1dd6210626abb86 Mon Sep 17 00:00:00 2001
From: Richard Kreckel
Date: Sun, 18 Nov 2001 20:10:27 +0000
Subject: [PATCH] * Change section about Square-free decomposition reflecting
the recent bugfix in the sqrfree() function.
---
doc/tutorial/ginac.texi | 18 +++++++++++-------
1 file changed, 11 insertions(+), 7 deletions(-)
diff --git a/doc/tutorial/ginac.texi b/doc/tutorial/ginac.texi
index c86114df..237aa3c4 100644
--- a/doc/tutorial/ginac.texi
+++ b/doc/tutorial/ginac.texi
@@ -3617,28 +3617,32 @@ GiNaC still lacks proper factorization support. Some form of
factorization is, however, easily implemented by noting that factors
appearing in a polynomial with power two or more also appear in the
derivative and hence can easily be found by computing the GCD of the
-original polynomial and its derivatives. Any system has an interface
-for this so called square-free factorization. So we provide one, too:
+original polynomial and its derivatives. Any decent system has an
+interface for this so called square-free factorization. So we provide
+one, too:
@example
ex sqrfree(const ex & a, const lst & l = lst());
@end example
-Here is an example that by the way illustrates how the result may depend
-on the order of differentiation:
+Here is an example that by the way illustrates how the exact form of the
+result may slightly depend on the order of differentiation, calling for
+some care with subsequent processing of the result:
@example
...
symbol x("x"), y("y");
- ex BiVarPol = expand(pow(x-2*y*x,3) * pow(x+y,2) * (x-y));
+ ex BiVarPol = expand(pow(2-2*y,3) * pow(1+x*y,2) * pow(x-2*y,2) * (x+y));
cout << sqrfree(BiVarPol, lst(x,y)) << endl;
- // -> (y+x)^2*(-1+6*y+8*y^3-12*y^2)*(y-x)*x^3
+ // -> 8*(1-y)^3*(y*x^2-2*y+x*(1-2*y^2))^2*(y+x)
cout << sqrfree(BiVarPol, lst(y,x)) << endl;
- // -> (1-2*y)^3*(y+x)^2*(-y+x)*x^3
+ // -> 8*(1-y)^3*(-y*x^2+2*y+x*(-1+2*y^2))^2*(y+x)
cout << sqrfree(BiVarPol) << endl;
// -> depending on luck, any of the above
...
@end example
+Note also, how factors with the same exponents are not fully factorized
+with this method.
@node Rational Expressions, Symbolic Differentiation, Polynomial Arithmetic, Methods and Functions
--
2.31.1