On 12/1/23 05:51, Waldek Hebisch wrote:
On Tue, Nov 28, 2023 at 07:02:16PM +0800, Qian Yun wrote:
I have examined integrals where rootSum is over degree 4, from a
subset of Nasser's test.
Martin Weltz (clicliclic) points out to example from Timofeev:
integrate(tan(x)/(sqrt(tan(x)) - 1)^2, x)
where root sum is over roots of
a^5 - (1/2)*a^3 - (5/8)*a^2 + (9/64)*a - 1/64
Is this integral example wrong? "internalIntegrate" shows a different
rootSum.
So it looks like that this "Riboo real functions" method is effective
for rootSum over polynomial with FRAC(INT) coefficients.
As for POLY(INT) coefficients, 'countRealRoots' fails and we have to
do case by case treatment, and for some cases, we have to assume
different signs (branches), giving a list of possible forms as result.
If you are working on the code, please continue, so our efforts will
not be duplicated.
- Qian
which factors as
(54) -> factor(a^5 - (1/2)*a^3 - (5/8)*a^2 + (9/64)*a - 1/64)
4 3 1 2 1 1
(54) (a - 1)(a + a + - a - - a + --)
2 8 64
Type: Factored(Polynomial(Fraction(Integer)))
The quartic gives complicated roots when passed to radicalSolve.
'rsimp2.input' which I posted some time ago when applied kernel-by-kernel
can express roots in terms of sqrt(-1) and sqrt(2), but ATM it is
not clear to me if this is good method. I would prefer to get
simple answer in more direct way.
Equation for real parts of the roots splits into 3 factors:
[[flag = "prime", factor = 8 u + 4 u + 1, exponent = 2],
2
[flag = "prime", factor = 16 u + 8 u - 1, exponent = 2],
2
[flag = "prime", factor = 16 u + 8 u + 3, exponent = 2]]
and 'countRealRoots' finds out that the second one is the
correct factor. This leads to simple expressions for real
parts:
(28) -> radicalSolve(fl(2).factor)
+-+ +-+
- \|2 - 1 \|2 - 1
(28) [u = ----------, u = --------]
4 4
Type: List(Equation(Expression(Integer)))
and complex parts. So here resolvent works nicely.
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