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
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.
--
Waldek Hebisch
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