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sci / sci.math.symbolic / Re: radicalSolve() in FriCAS is pathetic

SubjectAuthor
* radicalSolve() in FriCAS is patheticclicliclic@freenet.de
+* Re: radicalSolve() in FriCAS is patheticclicliclic@freenet.de
|`* Re: radicalSolve() in FriCAS is patheticclicliclic@freenet.de
| `- Re: radicalSolve() in FriCAS is patheticValentin Zapod
`- Re: radicalSolve() in FriCAS is patheticclicliclic@freenet.de

1
Subject: radicalSolve() in FriCAS is pathetic
From: clicliclic@freenet.d
Newsgroups: sci.math.symbolic
Organization: Killfiles, Unlimited
Date: Thu, 30 Nov 2023 11:40 UTC
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From: nobody@nowhere.invalid (clicliclic@freenet.de)
Newsgroups: sci.math.symbolic
Subject: radicalSolve() in FriCAS is pathetic
Date: Thu, 30 Nov 2023 12:40:33 +0100
Organization: Killfiles, Unlimited
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Oh dear! In FriCAS 1.3.9 try:

radicalSolve(64*z^4 + 64*z^3 + 32*z^2 - 8*z + 1)

and then try:

ii := rootOf(ii^2+1)
factor(64*z^4 + 64*z^3 + 32*z^2 - 8*z + 1, [ii])

to see what is wrong with radicalSolve(). (The integer quartic splits
into gaussian-integer quadratics.)

Also, unparse() in FriCAS refuses to convert the result from
radicalSolve() of the type List(Equation(Expression(Integer))) back
to InputForm. Why shouldn't this be be allowed?!? (The prettyprinted
ASCII monster is appended below.)

Derive 6.10 factors the integer polynomial right away as:

FACTOR(64*z^4 + 64*z^3 + 32*z^2 - 8*z + 1, Complex)
= 64*(z - SQRT(2)/4 + 1/4 + #i*(1/4 - SQRT(2)/4))
*(z + SQRT(2)/4 + 1/4 + #i*(SQRT(2)/4 + 1/4))
*(z + SQRT(2)/4 + 1/4 - #i*(SQRT(2)/4 + 1/4))
*(z - SQRT(2)/4 + 1/4 + #i*(SQRT(2)/4 - 1/4))

This polynomial arises in integral 5.66 (#401) from the Timofeev test
suite:

INT(TAN(x)/(SQRT(TAN(x)) - 1)^2, x)
= - x/2 + 1/(1 - SQRT(TAN(x))) + 1/2*LN(COS(x))
+ LN(1 - SQRT(TAN(x)))
+ 1/SQRT(2)*LN((1 + TAN(x) + SQRT(2)*SQRT(TAN(x)))/SEC(x))
- 1/SQRT(2)*ATAN(1 + SQRT(2)*SQRT(TAN(x)))
+ 1/SQRT(2)*ATAN(1 - SQRT(2)*SQRT(TAN(x)))
= - x/2 + 1/(1 - SQRT(TAN(x))) + 1/2*LN(COS(x))
+ LN(1 - SQRT(TAN(x)))
+ 1/SQRT(2)*ATANH((1 + TAN(x))/(SQRT(2)*SQRT(TAN(x))))
- 1/SQRT(2)*ATAN(1 + SQRT(2)*SQRT(TAN(x)))
+ 1/SQRT(2)*ATAN(1 - SQRT(2)*SQRT(TAN(x)))

which was discussed in the <sci.math.symbolic> thread
"Boooooooooooooooooooooooooo" from April to June 2016. Note that the
one and only radical extension needed to express the antiderivative is
SQRT(2).

No doubt, radicalSolve() in FriCAS needs to be overhauled.

Martin.

PS: The result from radicalSolve(64*z^4 + 64*z^3 + 32*z^2 - 8*z + 1) in
FriCAS 1.3.9 is:

[
z
=
-
+--+
\|13
*
ROOT
+------------------+2
+-+ +---+ 3| +---+ +-+
(68 \|3 - 18 \|- 1 )\|18 \|- 1 \|3 + 35
+
+------------------+
+-+ +---+ 3| +---+
+-+ +-+
(208 \|3 - 234 \|- 1 )\|18 \|- 1 \|3 + 35 -
676 \|3
*
ROOT
+------------------+2
+---+ +-+ 3| +---+ +-+
(- 18 \|- 1 \|3 + 35)\|18 \|- 1 \|3 + 35
+
+------------------+
3| +---+ +-+
169 \|18 \|- 1 \|3 + 35 - 169
+
+------------------+2
+---+ +-+ 3| +---+ +-+
(702 \|- 1 \|3 - 1365)\|18 \|- 1 \|3 + 35
+
+------------------+
3| +---+ +-+
- 6591 \|18 \|- 1 \|3 + 35 - 13182
+
+-+
13 \|3
*
ROOT
+------------------+2
+---+ +-+ 3| +---+ +-+
(- 18 \|- 1 \|3 + 35)\|18 \|- 1 \|3 + 35
+
+------------------+
3| +---+ +-+
169 \|18 \|- 1 \|3 + 35 - 169
+
- 507
/
2028
,

z
=
+--+
\|13
*
ROOT
+------------------+2
+-+ +---+ 3| +---+ +-+
(68 \|3 - 18 \|- 1 )\|18 \|- 1 \|3 + 35
+
+------------------+
+-+ +---+ 3| +---+
+-+ +-+
(208 \|3 - 234 \|- 1 )\|18 \|- 1 \|3 + 35 - 676
\|3
*
ROOT
+------------------+2
+---+ +-+ 3| +---+ +-+
(- 18 \|- 1 \|3 + 35)\|18 \|- 1 \|3 + 35
+
+------------------+
3| +---+ +-+
169 \|18 \|- 1 \|3 + 35 - 169
+
+------------------+2
+---+ +-+ 3| +---+ +-+
(702 \|- 1 \|3 - 1365)\|18 \|- 1 \|3 + 35
+
+------------------+
3| +---+ +-+
- 6591 \|18 \|- 1 \|3 + 35 - 13182
+
+-+
13 \|3
*
ROOT
+------------------+2
+---+ +-+ 3| +---+ +-+
(- 18 \|- 1 \|3 + 35)\|18 \|- 1 \|3 + 35
+
+------------------+
3| +---+ +-+
169 \|18 \|- 1 \|3 + 35 - 169
+
- 507
/
2028
,

z
=
-
+--+
\|13
*
ROOT
+------------------+2
+-+ +---+ 3| +---+ +-+
(- 68 \|3 + 18 \|- 1 )\|18 \|- 1 \|3 + 35
+
+------------------+
+-+ +---+ 3| +---+
+-+ +-+
(- 208 \|3 + 234 \|- 1 )\|18 \|- 1 \|3 + 35 +
676 \|3
*
ROOT
+------------------+2
+---+ +-+ 3| +---+ +-+
(- 18 \|- 1 \|3 + 35)\|18 \|- 1 \|3 + 35
+
+------------------+
3| +---+ +-+
169 \|18 \|- 1 \|3 + 35 - 169
+
+------------------+2
+---+ +-+ 3| +---+ +-+
(702 \|- 1 \|3 - 1365)\|18 \|- 1 \|3 + 35
+
+------------------+
3| +---+ +-+
- 6591 \|18 \|- 1 \|3 + 35 - 13182
+
-
+-+
13 \|3
*
ROOT
+------------------+2
+---+ +-+ 3| +---+ +-+
(- 18 \|- 1 \|3 + 35)\|18 \|- 1 \|3 + 35
+
+------------------+
3| +---+ +-+
169 \|18 \|- 1 \|3 + 35 - 169
+
- 507
/
2028
,

z
=
+--+
\|13
*
ROOT
+------------------+2
+-+ +---+ 3| +---+ +-+
(- 68 \|3 + 18 \|- 1 )\|18 \|- 1 \|3 + 35
+
+------------------+
+-+ +---+ 3| +---+
+-+ +-+
(- 208 \|3 + 234 \|- 1 )\|18 \|- 1 \|3 + 35 +
676 \|3
*
ROOT
+------------------+2
+---+ +-+ 3| +---+ +-+
(- 18 \|- 1 \|3 + 35)\|18 \|- 1 \|3 + 35
+
+------------------+
3| +---+ +-+
169 \|18 \|- 1 \|3 + 35 - 169
+
+------------------+2
+---+ +-+ 3| +---+ +-+
(702 \|- 1 \|3 - 1365)\|18 \|- 1 \|3 + 35
+
+------------------+
3| +---+ +-+
- 6591 \|18 \|- 1 \|3 + 35 - 13182
+
-
+-+
13 \|3
*
ROOT
+------------------+2
+---+ +-+ 3| +---+ +-+
(- 18 \|- 1 \|3 + 35)\|18 \|- 1 \|3 + 35
+
+------------------+
3| +---+ +-+
169 \|18 \|- 1 \|3 + 35 - 169
+
- 507
/
2028
]


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Subject: Re: radicalSolve() in FriCAS is pathetic
From: clicliclic@freenet.d
Newsgroups: sci.math.symbolic
Organization: Killfiles, Unlimited
Date: Thu, 21 Dec 2023 17:16 UTC
References: 1
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From: nobody@nowhere.invalid (clicliclic@freenet.de)
Newsgroups: sci.math.symbolic
Subject: Re: radicalSolve() in FriCAS is pathetic
Date: Thu, 21 Dec 2023 18:16:17 +0100
Organization: Killfiles, Unlimited
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"clicliclic@freenet.de" schrieb:
>
> Oh dear! In FriCAS 1.3.9 try:
>
> radicalSolve(64*z^4 + 64*z^3 + 32*z^2 - 8*z + 1)
>
> and then try:
>
> ii := rootOf(ii^2+1)
> factor(64*z^4 + 64*z^3 + 32*z^2 - 8*z + 1, [ii])
>
> to see what is wrong with radicalSolve(). (The integer quartic splits
> into gaussian-integer quadratics.)
>
> Also, unparse() in FriCAS refuses to convert the result from
> radicalSolve() of the type List(Equation(Expression(Integer))) back
> to InputForm. Why shouldn't this be be allowed?!? (The prettyprinted
> ASCII monster is appended below.)
>
> Derive 6.10 factors the integer polynomial right away as:
>
> FACTOR(64*z^4 + 64*z^3 + 32*z^2 - 8*z + 1, Complex)
> = 64*(z - SQRT(2)/4 + 1/4 + #i*(1/4 - SQRT(2)/4))
> *(z + SQRT(2)/4 + 1/4 + #i*(SQRT(2)/4 + 1/4))
> *(z + SQRT(2)/4 + 1/4 - #i*(SQRT(2)/4 + 1/4))
> *(z - SQRT(2)/4 + 1/4 + #i*(SQRT(2)/4 - 1/4))
>
> This polynomial arises in integral 5.66 (#401) from the Timofeev test
> suite:
>
> INT(TAN(x)/(SQRT(TAN(x)) - 1)^2, x)
> = - x/2 + 1/(1 - SQRT(TAN(x))) + 1/2*LN(COS(x))
> + LN(1 - SQRT(TAN(x)))
> + 1/SQRT(2)*LN((1 + TAN(x) + SQRT(2)*SQRT(TAN(x)))/SEC(x))
> - 1/SQRT(2)*ATAN(1 + SQRT(2)*SQRT(TAN(x)))
> + 1/SQRT(2)*ATAN(1 - SQRT(2)*SQRT(TAN(x)))
> = - x/2 + 1/(1 - SQRT(TAN(x))) + 1/2*LN(COS(x))
> + LN(1 - SQRT(TAN(x)))
> + 1/SQRT(2)*ATANH((1 + TAN(x))/(SQRT(2)*SQRT(TAN(x))))
> - 1/SQRT(2)*ATAN(1 + SQRT(2)*SQRT(TAN(x)))
> + 1/SQRT(2)*ATAN(1 - SQRT(2)*SQRT(TAN(x)))
>
> which was discussed in the <sci.math.symbolic> thread
> "Boooooooooooooooooooooooooo" from April to June 2016. Note that the
> one and only radical extension needed to express the antiderivative is
> SQRT(2).
>
> No doubt, radicalSolve() in FriCAS needs to be overhauled.
>

According to old notes of mine, a simple way to split any suitable
quartic:

a + b*x + c*x^2 + d*x^3 + e*x^4

into quadratics is by factorizing its resolvent:

4*a*c*e - a*d^2 - b^2*e + e*(b*d - 4*a*e)*y - c*e^2*y^2 + e^3*y^3

and expressing the quadratics in terms of any one resolvent root y as:

1/(4*e)*(e*y - e*(d*y - 2*b)/SQRT(d^2 + 4*e*(e*y - c))
+ (d - SQRT(d^2 + 4*e*(e*y - c)))*x + 2*e*x^2)
*(e*y + e*(d*y - 2*b)/SQRT(d^2 + 4*e*(e*y - c))
+ (d + SQRT(d^2 + 4*e*(e*y - c)))*x + 2*e*x^2)

which applies if d^2 + 4*e*(e*y - c) /= 0 and doesn't involve a, or as:

1/(4*e)*(e*y - SQRT(e*(e*y^2 - 4*a))
+ (d - e*(d*y - 2*b)/SQRT(e*(e*y^2 - 4*a)))*x + 2*e*x^2)
*(e*y + SQRT(e*(e*y^2 - 4*a))
+ (d + e*(d*y - 2*b)/SQRT(e*(e*y^2 - 4*a)))*x + 2*e*x^2)

which applies if e*y^2 - 4*a /= 0 and doesn't involve c. For d^2 +
4*e*(e*y - c) = 0 one can use:

1/(4*e)*(e*y - SQRT(e*(e*y^2 - 4*a)) + d*x + 2*e*x^2)
*(e*y + SQRT(e*(e*y^2 - 4*a)) + d*x + 2*e*x^2)

while for e*y^2 - 4*a = 0 one finds:

1/(4*e)*(e*y + (d - SQRT(d^2 + 4*e*(e*y - c)))*x + 2*e*x^2)
*(e*y + (d + SQRT(d^2 + 4*e*(e*y - c)))*x + 2*e*x^2)

As no cube roots can appear, this way of splitting a quartic is
advantageous whenever its resolvent possesses at least one rational
root. Indeed, the resolvent of 64*z^4 + 64*z^3 + 32*z^2 - 8*z + 1
simply is:

(4*y - 3)*(4*y + 1)*y

Using all three resolvent roots in turn yields three different ways of
splitting the quartic; if the quartic is real, at least one pair of
factors is real as well.

Martin.

Subject: Re: radicalSolve() in FriCAS is pathetic
From: clicliclic@freenet.d
Newsgroups: sci.math.symbolic
Organization: Killfiles, Unlimited
Date: Thu, 11 Jan 2024 12:13 UTC
References: 1 2
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From: nobody@nowhere.invalid (clicliclic@freenet.de)
Newsgroups: sci.math.symbolic
Subject: Re: radicalSolve() in FriCAS is pathetic
Date: Thu, 11 Jan 2024 13:13:03 +0100
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"clicliclic@freenet.de" schrieb:
>
> "clicliclic@freenet.de" schrieb:
> >
> > Oh dear! In FriCAS 1.3.9 try:
> >
> > radicalSolve(64*z^4 + 64*z^3 + 32*z^2 - 8*z + 1)
> >
> > and then try:
> >
> > ii := rootOf(ii^2+1)
> > factor(64*z^4 + 64*z^3 + 32*z^2 - 8*z + 1, [ii])
> >
> > to see what is wrong with radicalSolve(). (The integer quartic
> > splits into gaussian-integer quadratics.)
> >
> > Also, unparse() in FriCAS refuses to convert the result from
> > radicalSolve() of the type List(Equation(Expression(Integer))) back
> > to InputForm. Why shouldn't this be be allowed?!? (The prettyprinted
> > ASCII monster is appended below.)
> >
> > Derive 6.10 factors the integer polynomial right away as:
> >
> > FACTOR(64*z^4 + 64*z^3 + 32*z^2 - 8*z + 1, Complex)
> > = 64*(z - SQRT(2)/4 + 1/4 + #i*(1/4 - SQRT(2)/4))
> > *(z + SQRT(2)/4 + 1/4 + #i*(SQRT(2)/4 + 1/4))
> > *(z + SQRT(2)/4 + 1/4 - #i*(SQRT(2)/4 + 1/4))
> > *(z - SQRT(2)/4 + 1/4 + #i*(SQRT(2)/4 - 1/4))
> >
> > This polynomial arises in integral 5.66 (#401) from the Timofeev
> > test suite:
> >
> > INT(TAN(x)/(SQRT(TAN(x)) - 1)^2, x)
> > = - x/2 + 1/(1 - SQRT(TAN(x))) + 1/2*LN(COS(x))
> > + LN(1 - SQRT(TAN(x)))
> > + 1/SQRT(2)*LN((1 + TAN(x) + SQRT(2)*SQRT(TAN(x)))/SEC(x))
> > - 1/SQRT(2)*ATAN(1 + SQRT(2)*SQRT(TAN(x)))
> > + 1/SQRT(2)*ATAN(1 - SQRT(2)*SQRT(TAN(x)))
> > = - x/2 + 1/(1 - SQRT(TAN(x))) + 1/2*LN(COS(x))
> > + LN(1 - SQRT(TAN(x)))
> > + 1/SQRT(2)*ATANH((1 + TAN(x))/(SQRT(2)*SQRT(TAN(x))))
> > - 1/SQRT(2)*ATAN(1 + SQRT(2)*SQRT(TAN(x)))
> > + 1/SQRT(2)*ATAN(1 - SQRT(2)*SQRT(TAN(x)))
> >
> > which was discussed in the <sci.math.symbolic> thread
> > "Boooooooooooooooooooooooooo" from April to June 2016. Note that the
> > one and only radical extension needed to express the antiderivative
> > is SQRT(2).
> >
> > No doubt, radicalSolve() in FriCAS needs to be overhauled.
> >
>
> According to old notes of mine, a simple way to split any suitable
> quartic:
>
> a + b*x + c*x^2 + d*x^3 + e*x^4
>
> into quadratics is by factorizing its resolvent:
>
> 4*a*c*e - a*d^2 - b^2*e + e*(b*d - 4*a*e)*y - c*e^2*y^2 + e^3*y^3
>
> and expressing the quadratics in terms of any one resolvent root y as:
>
> 1/(4*e)*(e*y - e*(d*y - 2*b)/SQRT(d^2 + 4*e*(e*y - c))
> + (d - SQRT(d^2 + 4*e*(e*y - c)))*x + 2*e*x^2)
> *(e*y + e*(d*y - 2*b)/SQRT(d^2 + 4*e*(e*y - c))
> + (d + SQRT(d^2 + 4*e*(e*y - c)))*x + 2*e*x^2)
>
> which applies if d^2 + 4*e*(e*y - c) /= 0 and doesn't involve a, or
> as:
>
> 1/(4*e)*(e*y - SQRT(e*(e*y^2 - 4*a))
> + (d - e*(d*y - 2*b)/SQRT(e*(e*y^2 - 4*a)))*x + 2*e*x^2)
> *(e*y + SQRT(e*(e*y^2 - 4*a))
> + (d + e*(d*y - 2*b)/SQRT(e*(e*y^2 - 4*a)))*x + 2*e*x^2)
>
> which applies if e*y^2 - 4*a /= 0 and doesn't involve c. For d^2 +
> 4*e*(e*y - c) = 0 one can use:
>
> 1/(4*e)*(e*y - SQRT(e*(e*y^2 - 4*a)) + d*x + 2*e*x^2)
> *(e*y + SQRT(e*(e*y^2 - 4*a)) + d*x + 2*e*x^2)
>
> while for e*y^2 - 4*a = 0 one finds:
>
> 1/(4*e)*(e*y + (d - SQRT(d^2 + 4*e*(e*y - c)))*x + 2*e*x^2)
> *(e*y + (d + SQRT(d^2 + 4*e*(e*y - c)))*x + 2*e*x^2)
>
> As no cube roots can appear, this way of splitting a quartic is
> advantageous whenever its resolvent possesses at least one rational
> root. Indeed, the resolvent of 64*z^4 + 64*z^3 + 32*z^2 - 8*z + 1
> simply is:
>
> (4*y - 3)*(4*y + 1)*y
>
> Using all three resolvent roots in turn yields three different ways of
> splitting the quartic; if the quartic is real, at least one pair of
> factors is real as well.
>

In version 3.1.10 of FriCAS, unparse() can convert the result from
radicalSolve() of the type List(Equation(Expression(Integer))) back
to InputForm:

radicalSolve(64*z^4 + 64*z^3 + 32*z^2 - 8*z + 1)

[equation(z,((-1)*(((((-18)*((-1)/3)^(1/2)+68)*((54*((-1)/3)^(1/2)+35)^(1/3))^2+((-234)*((-1)/3)^(1/2)+208)*(54*((-1)/3)^(1/2)+35)^(1/3)+(-676))*((((-54)*((-1)/3)^(1/2)+35)*((54*((-1)/3)^(1/2)+35)^(1/3))^2+169*(54*((-1)/3)^(1/2)+35)^(1/3)+(-169))/3)^(1/2)+((702*((-1)/3)^(1/2)+(-455))*((54*((-1)/3)^(1/2)+35)^(1/3))^2+(-2197)*(54*((-1)/3)^(1/2)+35)^(1/3)+(-4394)))/39)^(1/2)+(((((-54)*((-1)/3)^(1/2)+35)*((54*((-1)/3)^(1/2)+35)^(1/3))^2+169*(54*((-1)/3)^(1/2)+35)^(1/3)+(-169))/3)^(1/2)+(-13)))/52),equation(z,((((((-18)*((-1)/3)^(1/2)+68)*((54*((-1)/3)^(1/2)+35)^(1/3))^2+((-234)*((-1)/3)^(1/2)+208)*(54*((-1)/3)^(1/2)+35)^(1/3)+(-676))*((((-54)*((-1)/3)^(1/2)+35)*((54*((-1)/3)^(1/2)+35)^(1/3))^2+169*(54*((-1)/3)^(1/2)+35)^(1/3)+(-169))/3)^(1/2)+((702*((-1)/3)^(1/2)+(-455))*((54*((-1)/3)^(1/2)+35)^(1/3))^2+(-2197)*(54*((-1)/3)^(1/2)+35)^(1/3)+(-4394)))/39)^(1/2)+(((((-54)*((-1)/3)^(1/2)+35)*((54*((-1)/3)^(1/2)+35)^(1/3))^2+169*(54*((-1)/3)^(1/2)+35)^(1/3)+(-169))/3)^(1/2)+(-13)))/52),equation(z,((-1)*((((18*((-1)/3)^(1/2)+(-68))*((54*((-1)/3)^(1/2)+35)^(1/3))^2+(234*((-1)/3)^(1/2)+(-208))*(54*((-1)/3)^(1/2)+35)^(1/3)+676)*((((-54)*((-1)/3)^(1/2)+35)*((54*((-1)/3)^(1/2)+35)^(1/3))^2+169*(
54*((-1)/3)^(1/2)+35)^(1/3)+(-169))/3)^(1/2)+((702*((-1)/3)^(1/2)+(-455))*((54*((-1)/3)^(1/2)+35)^(1/3))^2+(-2197)*(54*((-1)/3)^(1/2)+35)^(1/3)+(-4394)))/39)^(1/2)+((-1)*((((-54)*((-1)/3)^(1/2)+35)*((54*((-1)/3)^(1/2)+35)^(1/3))^2+169*(54*((-1)/3)^(1/2)+35)^(1/3)+(-169))/3)^(1/2)+(-13)))/52),equation(z,(((((18*((-1)/3)^(1/2)+(-68))*((54*((-1)/3)^(1/2)+35)^(1/3))^2+(234*((-1)/3)^(1/2)+(-208))*(54*((-1)/3)^(1/2)+35)^(1/3)+676)*((((-54)*((-1)/3)^(1/2)+35)*((54*((-1)/3)^(1/2)+35)^(1/3))^2+169*(54*((-1)/3)^(1/2)+35)^(1/3)+(-169))/3)^(1/2)+((702*((-1)/3)^(1/2)+(-455))*((54*((-1)/3)^(1/2)+35)^(1/3))^2+(-2197)*(54*((-1)/3)^(1/2)+35)^(1/3)+(-4394)))/39)^(1/2)+((-1)*((((-54)*((-1)/3)^(1/2)+35)*((54*((-1)/3)^(1/2)+35)^(1/3))^2+169*(54*((-1)/3)^(1/2)+35)^(1/3)+(-169))/3)^(1/2)+(-13)))/52)]

but the result involves unnecessary cube roots as before.

Martin.

Subject: Re: radicalSolve() in FriCAS is pathetic
From: clicliclic@freenet.d
Newsgroups: sci.math.symbolic
Organization: Killfiles, Unlimited
Date: Sat, 27 Jan 2024 17:01 UTC
References: 1
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From: nobody@nowhere.invalid (clicliclic@freenet.de)
Newsgroups: sci.math.symbolic
Subject: Re: radicalSolve() in FriCAS is pathetic
Date: Sat, 27 Jan 2024 18:01:27 +0100
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"clicliclic@freenet.de" schrieb:
>
> [...]
>
> This polynomial arises in integral 5.66 (#401) from the Timofeev test
> suite:
>
> INT(TAN(x)/(SQRT(TAN(x)) - 1)^2, x)
> = - x/2 + 1/(1 - SQRT(TAN(x))) + 1/2*LN(COS(x))
> + LN(1 - SQRT(TAN(x)))
> + 1/SQRT(2)*LN((1 + TAN(x) + SQRT(2)*SQRT(TAN(x)))/SEC(x))
> - 1/SQRT(2)*ATAN(1 + SQRT(2)*SQRT(TAN(x)))
> + 1/SQRT(2)*ATAN(1 - SQRT(2)*SQRT(TAN(x)))
> = - x/2 + 1/(1 - SQRT(TAN(x))) + 1/2*LN(COS(x))
> + LN(1 - SQRT(TAN(x)))
> + 1/SQRT(2)*ATANH((1 + TAN(x))/(SQRT(2)*SQRT(TAN(x))))
> - 1/SQRT(2)*ATAN(1 + SQRT(2)*SQRT(TAN(x)))
> + 1/SQRT(2)*ATAN(1 - SQRT(2)*SQRT(TAN(x)))
>
> which was discussed in the <sci.math.symbolic> thread
> "Boooooooooooooooooooooooooo" from April to June 2016. Note that the
> one and only radical extension needed to express the antiderivative is
> SQRT(2).
>

In version 3.1.10 of FriCAS, the integral:

integrate(tan(x)/(sqrt(tan(x)) - 1)^2, x)

evaluates to:

(((2^(1/2)+(-1))*tan(x)+((-1)*2^(1/2)+1))*log(2^(1/2)*tan(x)^(1/2)+(tan(x)+1))+((4*tan(x)+(-4))*log(tan(x)^(1/2)+(-1))+((((-1)*2^(1/2)+(-1))*tan(x)+(2^(1/2)+1))*log((-1)*2^(1/2)*tan(x)^(1/2)+(tan(x)+1))+((((-2)*2^(1/2)+2)*tan(x)+(2*2^(1/2)+(-2)))*atan(2^(1/2)*tan(x)^(1/2)+1)+((((-2)*2^(1/2)+(-2))*tan(x)+(2*2^(1/2)+2))*atan(2^(1/2)*tan(x)^(1/2)+(-1))+((-4)*tan(x)^(1/2)+(-4)))))))/(4*tan(x)+(-4))

which can be restated as:

(SQRT(TAN(x)) + 1)/(1 - TAN(x)) + LN(SQRT(TAN(x)) - 1)
+ (SQRT(2) - 1)/4*LN(TAN(x) + SQRT(2)*SQRT(TAN(x)) + 1)
- (SQRT(2) + 1)/4*LN(TAN(x) - SQRT(2)*SQRT(TAN(x)) + 1)
- (SQRT(2) - 1)/2*ATAN(SQRT(2)*SQRT(TAN(x)) + 1)
- (SQRT(2) + 1)/2*ATAN(SQRT(2)*SQRT(TAN(x)) - 1)

and thus constitutes an interesting alternative to the solutions above.

Martin.

Subject: Re: radicalSolve() in FriCAS is pathetic
From: Valentin Zapod
Newsgroups: sci.math.symbolic
Organization: A noiseless patient Spider
Date: Wed, 28 Feb 2024 05:51 UTC
References: 1 2 3
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From: val.zapod.vz@gmail.com (Valentin Zapod)
Newsgroups: sci.math.symbolic
Subject: Re: radicalSolve() in FriCAS is pathetic
Date: Wed, 28 Feb 2024 08:51:42 +0300
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Meh, at least you are solving degree 2 correctly. You do, right?

https://cnrs.hal.science/hal-04116310

1

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