An exact simplification challenge - 85 (wild EllipticPi, GAMMA)
Vladimir Bondarenko
Here comes something different.
- Pi^(3/2)/(2*GAMMA(3/4)^2) + (3+sqrt(6*sqrt(3)-9))/
2*EllipticPi((1 + sqrt(2*sqrt(3) - 3))/2, 1/sqrt(2))
?
(.....do not waste your time with identify() ;)
Cheers,
Vladimir Bondarenko
Co-founder, CEO, Mathematical Director
http://www.cybertester.com/ Cyber Tester Ltd.
-------------------------------------------------------
"We must understand that technologies
like these are the way of the future."
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clicl...@freenet.de
Vladimir Bondarenko schrieb:
>
> - Pi^(3/2)/(2*GAMMA(3/4)^2) + (3+sqrt(6*sqrt(3)-9))/
> 2*EllipticPi((1 + sqrt(2*sqrt(3) - 3))/2, 1/sqrt(2))
>
> ?
>
It might be of significance that this can be rewritten as
- EllipticK(1/sqrt(2)) + (3+sqrt(6*sqrt(3)-9))/
2*EllipticPi((1 + sqrt(2*sqrt(3) - 3))/2, 1/sqrt(2))
although I don't see how.
Martin.
clicl...@freenet.de
Jean-Claude Arbaut schrieb:
>
> As you seem to like challenges, here is a one I was not able to simplify
> with a CAS, some years ago, though it was not too difficult by hand.
> Maybe it would be interesting to test with more modern software...
>
> What is the solid angle of the rectangle
> { x=a, 0<=y<=b, 0<=z<=c } a>0, b>0, c>0
> seen from the point O(0,0,0) ?
I think this is best treated as a problem of sperical trigonometry.
You want to know the sum of the areas of the spherical triangle
a1 = arctan(b/a), b1 = arctan(sqrt(b^2+c^2)/a), gamma1 = arctan(c/b)
and the spherical triangle
a2 = arctan(c/a), b2 = arctan(sqrt(b^2+c^2)/a), gamma2 = arctan(b/c).
If your CAS has procedures to determine unknown parameters of
spherical triangles from sufficient sets of known ones you are done.
Else you would have to turn to formulae in books on spherical
trigonometry. Somewhere I should have a nineteenth century book on
the subject.
Martin.
mma...@personal.ro
wrote:
> it yields, if I remember well, arctan(a*b/(c*sqrt(a^2+b^2+c^2)))
For such computations, Derive is the best.
INT(INT(a/(a^2+y^2+z^2)^(3/2),y,0,b),z,0,c)
ATAN(b*c/(a*SQRT(a^2+b^2+c^2)))
Mate
Pfenniger Daniel
INT(INT(1/(a^2+y^2+z^2),y,0,b),z,0,c)
i.e., the area divided by the squared distance to (0,0,0)?
Dan
mma...@personal.ro
wrote:
The solid angle equals
Int_S (r . n) / ||r||^3 dS
Here r = [a,y,z], n = [1,0,0] (the normal), r . n = a (the dot
product).
Mate
clicl...@freenet.de
mmat...@personal.ro schrieb:
> On Dec 16, 3:18 pm, Pfenniger Daniel <daniel.pfenni...@unige.ch>
> wrote:
> > mmat...@personal.ro wrote:
> > > On Dec 15, 9:26 pm, Jean-Claude Arbaut <jeanclaudearb...@orange.fr>
> > > wrote:
> > >> cliclic...@freenet.de wrote:
> > >>> Jean-Claude Arbaut schrieb:
> > >>>>
> > >>>> { x=a, 0<=y<=b, 0<=z<=c } a>0, b>0, c>0
> > >>>> seen from the point O(0,0,0) ?
> > >>>
> > >>> I think this is best treated as a problem of sperical trigonometry.
> > >>> You want to know the sum of the areas of the spherical triangle
> > >>> a1 = arctan(b/a), b1 = arctan(sqrt(b^2+c^2)/a), gamma1 = arctan(c/b)
> > >>> and the spherical triangle
> > >>> a2 = arctan(c/a), b2 = arctan(sqrt(b^2+c^2)/a), gamma2 = arctan(b/c).
> > >>> If your CAS has procedures to determine unknown parameters of
> > >>> spherical triangles from sufficient sets of known ones you are done.
> > >>> Else you would have to turn to formulae in books on spherical
> > >>> trigonometry. Somewhere I should have a nineteenth century book on
> > >>> the subject.
> > >>
> > >> Ok. I started from the integral :-) After simplifications
> > >> it yields, if I remember well, arctan(a*b/(c*sqrt(a^2+b^2+c^2)))
> > >
> > > For such computations, Derive is the best.
> > > INT(INT(a/(a^2+y^2+z^2)^(3/2),y,0,b),z,0,c)
> > > ATAN(b*c/(a*SQRT(a^2+b^2+c^2)))
> >
> > INT(INT(1/(a^2+y^2+z^2),y,0,b),z,0,c)
> > i.e., the area divided by the squared distance to (0,0,0)?
>
> Int_S (r . n) / ||r||^3 dS
> Here r = [a,y,z], n = [1,0,0] (the normal), r . n = a (the dot
> product).
>
I think Mate's approach is to be preferred over mine: It doesn't take
much more brains to restate the problem, while Integrators are easier
to find than Spherical Trigonometers nowadays.
Is there an implication that some Computer Algebra Systems have
problems with the double integral INT(INT(a/(a^2+y^2+z^2)^(3/2),y,
0,b),z,0,c) ?
Martin.
mma...@personal.ro
> Is there an implication that some Computer Algebra Systems have
> problems with the double integral INT(INT(a/(a^2+y^2+z^2)^(3/2),y,
> 0,b),z,0,c) ?
>
Maple cannot compute it without a consistent help from the user.
For Mathematica the situation seems to be the same.
Mate
Daniel Lichtblau
What I get, in Mathematica, using a couple of assumption sets (which
might qualify as user help).
In[1]:= InputForm[Integrate[a/(a^2+y^2+z^2)^(3/2), {y,0,b}, {z,0,c},
Assumptions->{a>0,b>0,c>0}]]
Out[1]//InputForm= ArcTan[(b*c)/(a*Sqrt[a^2 + b^2 + c^2])]
In[2]:= InputForm[Integrate[a/(a^2+y^2+z^2)^(3/2), {y,0,b}, {z,0,c},
Assumptions->Element[{a,b,c},Reals]]]
Out[2]//InputForm=
a*Abs[c]*If[(Sqrt[-a^2 - c^2]/b != 0 && Re[Sqrt[-a^2 - c^2]/b] >= 0)
||
Re[Sqrt[-a^2 - c^2]/b] <= -1 || NotElement[Sqrt[-a^2 - c^2]/b,
Reals],
ArcTan[(b*c)/(a*Sqrt[a^2 + b^2 + c^2])]/(a*c),
Integrate[b/((a^2 + b^2*y^2)*Sqrt[a^2 + c^2 + b^2*y^2]), {y, 0, 1},
Assumptions -> Element[a | b | c, Reals] &&
!((Sqrt[-a^2 - c^2]/b != 0 && Re[Sqrt[-a^2 - c^2]/b] >= 0) ||
Re[Sqrt[-a^2 - c^2]/b] <= -1 || NotElement[Sqrt[-a^2 - c^2]/b,
Reals])]]
Daniel Lichtblau
Wolfram Research
Waldek Hebisch
FriCAS refuses to compute the integral because of possible
singularities. If you tell it that there are no singularities
then it gives the result:
(9) -> integrate(integrate(a/(a^2+y^2+z^2)^(3/2), y=0..b, "noPole"), z=0..c, "noPole")
+------------+
| 2 2 2 2 2
c\|c + b + a - c - a a
(9) atan(--------------------------) + atan(-)
a b b
Type: Union(f1: OrderedCompletion Expression Integer,...)
I would expect other systems to at least need assumption that a, b
and c are real.
--
Waldek Hebisch
heb...@math.uni.wroc.pl
clicl...@freenet.de
Daniel Lichtblau schrieb:
> On Dec 16, 3:33?pm, mmat...@personal.ro wrote:
> > On Dec 16, 10:23 pm, cliclic...@freenet.de wrote:
> >
> > > Is there an implication that some Computer Algebra Systems have
> > > problems with the double integral INT(INT(a/(a^2+y^2+z^2)^(3/2),y,
> > > 0,b),z,0,c) ?
> >
> > Maple cannot compute it without a consistent help from the user.
> > For Mathematica the situation seems to be the same.
>
> might qualify as user help).
>
> In[1]:= InputForm[Integrate[a/(a^2+y^2+z^2)^(3/2), {y,0,b}, {z,0,c},
> Assumptions->{a>0,b>0,c>0}]]
>
> Out[1]//InputForm= ArcTan[(b*c)/(a*Sqrt[a^2 + b^2 + c^2])]
>
> In[2]:= InputForm[Integrate[a/(a^2+y^2+z^2)^(3/2), {y,0,b}, {z,0,c},
> Assumptions->Element[{a,b,c},Reals]]]
>
> Out[2]//InputForm=
> a*Abs[c]*If[(Sqrt[-a^2 - c^2]/b != 0 && Re[Sqrt[-a^2 - c^2]/b] >= 0)
> ||
> Re[Sqrt[-a^2 - c^2]/b] <= -1 || NotElement[Sqrt[-a^2 - c^2]/b,
> Reals],
> ArcTan[(b*c)/(a*Sqrt[a^2 + b^2 + c^2])]/(a*c),
> Integrate[b/((a^2 + b^2*y^2)*Sqrt[a^2 + c^2 + b^2*y^2]), {y, 0, 1},
> Assumptions -> Element[a | b | c, Reals] &&
> !((Sqrt[-a^2 - c^2]/b != 0 && Re[Sqrt[-a^2 - c^2]/b] >= 0) ||
> Re[Sqrt[-a^2 - c^2]/b] <= -1 || NotElement[Sqrt[-a^2 - c^2]/b,
> Reals])]]
>
For the requested Element[{a,b,c},Reals]], the condition on the result
boils down to a^2+c^2 != 0 (it remains unclear what the condition
means for b=0) and the result becomes
Sign[c]*ArcTan[(b*c)/(a*Sqrt[a^2 + b^2 + c^2])].
The Sign[c] is clearly wrong; the original integral changes its sign
when c<0 (as it does when b<0).
Martin.
clicl...@freenet.de
Waldek Hebisch schrieb:
> mma...@personal.ro wrote:
> > On Dec 16, 10:23 pm, cliclic...@freenet.de wrote:
> >
> > > Is there an implication that some Computer Algebra Systems have
> > > problems with the double integral INT(INT(a/(a^2+y^2+z^2)^(3/2),y,
> > > 0,b),z,0,c) ?
> >
> > Maple cannot compute it without a consistent help from the user.
> > For Mathematica the situation seems to be the same.
>
> singularities. If you tell it that there are no singularities
> then it gives the result:
>
> (9) -> integrate(integrate(a/(a^2+y^2+z^2)^(3/2), y=0..b, "noPole"), z=0..c, "noPole")
>
> +------------+
> | 2 2 2 2 2
> c\|c + b + a - c - a a
> (9) atan(--------------------------) + atan(-)
> a b b
> Type: Union(f1: OrderedCompletion Expression Integer,...)
>
> I would expect other systems to at least need assumption that a, b
> and c are real.
>
This result is not equivalent to ATAN(b*c/(a*SQRT(a^2 + b^2 + c^2))),
and therefore wrong, as confirmed by numerical approximation for
specific values a,b,c. Are the FriCAS antiderivatives perhaps
discontinuous?
Martin.
Waldek Hebisch
>
> Waldek Hebisch schrieb:
> >
> > FriCAS refuses to compute the integral because of possible
> > singularities. If you tell it that there are no singularities
> > then it gives the result:
> >
> > (9) -> integrate(integrate(a/(a^2+y^2+z^2)^(3/2), y=0..b, "noPole"), z=0..c, "noPole")
> >
> > +------------+
> > | 2 2 2 2 2
> > c\|c + b + a - c - a a
> > (9) atan(--------------------------) + atan(-)
> > a b b
> > Type: Union(f1: OrderedCompletion Expression Integer,...)
> >
> > I would expect other systems to at least need assumption that a, b
> > and c are real.
> >
>
> This result is not equivalent to ATAN(b*c/(a*SQRT(a^2 + b^2 + c^2))),
> and therefore wrong, as confirmed by numerical approximation for
> specific values a,b,c. Are the FriCAS antiderivatives perhaps
> discontinuous?
>
BOTH formulas are discontinuous arond a = 0. Could you give
values for which the formulas disagree? AFAICS formulas are
equivalent modulo branches of atan, and for a few values I
tried they give me the same result.
--
Waldek Hebisch
heb...@math.uni.wroc.pl
clicl...@freenet.de
Waldek Hebisch schrieb:
I was mistaken and stand corrected (some late-night transcription
error I guess). For real a,b,c, the FriCAS evaluation
ATAN((c*SQRT(a^2+b^2+c^2)-a^2-c^2)/(a*b)) + ATAN(a/b)
is _fully_ equivalent to
ATAN(b*c/(a*SQRT(a^2+b^2+c^2))).
Martin.
Daniel Lichtblau
We need to force Integrate to figure out provisos in both stages
of the integration; the default behavior is to avoid this for all
but the outermost stage.
In[11]:= InputForm[Integrate[a/(a^2+y^2+z^2)^(3/2), {y,0,b}, {z,0,c},
Assumptions->Element[{a,b,c},Reals], GenerateConditions->True]]
Out[11]//InputForm=
a*If[c >= 0, c*If[(NotElement[Sqrt[-a^2 - c^2]/b, Reals] ||
Re[Sqrt[-a^2 - c^2]/b] <= -1 || (Sqrt[-a^2 - c^2]/b != 0 &&
(Re[Sqrt[-a^2 - c^2]/b] >= 1 || Re[Sqrt[-a^2 - c^2]/b] == 0)))
&&
((Element[a/b, Reals] && Im[a/b] >= 0) || Im[a/b] >= 1 ||
Im[a/b] <= -1 || Re[a/b] != 0), ArcTan[(b*c)/(a*Sqrt[a^2 + b^2 +
c^2])]/
(a*c), Integrate[b/((a^2 + b^2*y^2)*Sqrt[a^2 + c^2 + b^2*y^2]),
{y, 0, 1}, Assumptions -> c >= 0 &&
!((NotElement[Sqrt[-a^2 - c^2]/b, Reals] || Re[Sqrt[-a^2 -
c^2]/b] <=
-1 || (Sqrt[-a^2 - c^2]/b != 0 && (Re[Sqrt[-a^2 - c^2]/b]
>= 1 ||
Re[Sqrt[-a^2 - c^2]/b] == 0))) && ((Element[a/b, Reals]
&&
Im[a/b] >= 0) || Im[a/b] >= 1 || Im[a/b] <= -1 ||
Re[a/b] != 0))]], Integrate[Integrate[(a^2 + y^2 + z^2)^
(-3/2),
{z, 0, c}, Assumptions -> Element[a | b | c, Reals] && c < 0 &&
((b < 0 && b < y && y < 0) || (b > 0 && b > y && y > 0))], {y,
0, b},
Assumptions -> c < 0]]
Best I can tell from eyeballing this, the integrated case no longer
has anything
dependent on Sign[c] (or Sign[b]).
Daniel Lichtblau
Wolfram Research
clicl...@freenet.de
cliclic...@freenet.de schrieb:
By the way, Derive 6.10 (for real variables, which is the default
assumption) cannot simplify the difference to zero, or transform one
expression into the other; we have to wait for Derive 7.01 for this:
ATAN((c*SQRT(a^2+b^2+c^2)-a^2-c^2)/(a*b))+ATAN(a/b)
" ATAN(x)+ATAN(y) -> pi*SIGN(x+y)/2-ATAN((1-x*y)/(x+y)) "
pi*SIGN(c*(SQRT(a^2+b^2+c^2)-c)/(a*b))/2-ATAN(a*SQRT(a^2+b^2+c^2~
)/(b*c))
" If x^2>y^2, SIGN(x+y) -> SIGN(x) "
pi*SIGN(c*SQRT(a^2+b^2+c^2)/(a*b))/2-ATAN(a*SQRT(a^2+b^2+c^2)/(b~
*c))
" (???) "
pi*SIGN(c/(a*b))/2-ATAN(a*SQRT(a^2+b^2+c^2)/(b*c))
" 1/SIGN(x) -> SIGN(x) "
pi*SIGN(a*c/b)/2-ATAN(a*SQRT(a^2+b^2+c^2)/(b*c))
" 1/SIGN(x) -> SIGN(x) "
pi*SIGN(a*b*c)/2-ATAN(a*SQRT(a^2+b^2+c^2)/(b*c))
" ATAN(x) -> pi*SIGN(x)/2-ATAN(1/x) "
pi*SIGN(a*b*c)/2-(pi*SIGN(a*SQRT(a^2+b^2+c^2)/(b*c))/2-ATAN(b*c/~
(a*SQRT(a^2+b^2+c^2))))
" (???) "
pi*SIGN(a*b*c)/2-pi*SIGN(a/(b*c))/2+ATAN(b*c/(a*SQRT(a^2+b^2+c^2~
)))
" 1/SIGN(x) -> SIGN(x) "
pi*SIGN(a*b*c)/2-pi*SIGN(a*b/c)/2+ATAN(b*c/(a*SQRT(a^2+b^2+c^2)))
" 1/SIGN(x) -> SIGN(x) "
pi*SIGN(a*b*c)/2-pi*SIGN(a*b*c)/2+ATAN(b*c/(a*SQRT(a^2+b^2+c^2)))
" (one final step) "
ATAN(b*c/(a*SQRT(a^2+b^2+c^2)))
The string for one rule makes no sense and has been replaced by (???).
Martin.
Vladimir Bondarenko
Hint:
= Pi*sqrt(...)
clicl...@freenet.de
cliclic...@freenet.de schrieb:
Writing the elliptic integrals in their trigonometric normal form and
substituting t=cos(phi) I arrive at this hypergeometric
representation:
(GAMMA(1/4)^2*((2*SQRT(2)*3^(3/4)+3*SQRT(3)+3*12^(1/4)+9)*HYPERG~
EOM([1,1/4],[3/4],2*SQRT(2)*3^(3/4)+4*SQRT(3)+4*12^(1/4)+7)-2)-4~
*(11*SQRT(2)*3^(3/4)+21*SQRT(3)+18*12^(1/4)+33)*GAMMA(3/4)^2*HYP~
ERGEOM([1,3/4],[5/4],2*SQRT(2)*3^(3/4)+4*SQRT(3)+4*12^(1/4)+7))/~
(8*SQRT(pi))
Systems with adequate knowledge might find a way to simplify this.
Martin.
Vladimir Bondarenko
M> Systems with adequate knowledge might find
M> a way to simplify this.
"with adequate knowledge"
Sounds like the systems from your definition
are neither Maple 12 nor Mathematica 7.
;)
clicl...@freenet.de
Vladimir Bondarenko schrieb:
> On Dec 19, 8:29?pm, cliclic...@freenet.de wrote:
> >
> > Writing the elliptic integrals in their trigonometric normal form and
> > substituting t=cos(phi) I arrive at this hypergeometric
> > representation:
> >
> > (GAMMA(1/4)^2*((2*SQRT(2)*3^(3/4)+3*SQRT(3)+3*12^(1/4)+9)*HYPERG~
> > EOM([1,1/4],[3/4],2*SQRT(2)*3^(3/4)+4*SQRT(3)+4*12^(1/4)+7)-2)-4~
> > *(11*SQRT(2)*3^(3/4)+21*SQRT(3)+18*12^(1/4)+33)*GAMMA(3/4)^2*HYP~
> > ERGEOM([1,3/4],[5/4],2*SQRT(2)*3^(3/4)+4*SQRT(3)+4*12^(1/4)+7))/~
> > (8*SQRT(pi))
> >
> > Systems with adequate knowledge might find a way to simplify this.
> >
>
> M> a way to simplify this.
>
> "with adequate knowledge"
>
> Sounds like the systems from your definition
> are neither Maple 12 nor Mathematica 7.
>
> ;)
Wasn't meant to sound like this. It was said from a Derive
perspective. Derive isn't adequately equipped for the simplification
of this hypergeometric expression. Other systems may be, but I simply
don't know what transformations are available to them, or what
transformations would be needed here.
Martin.
Vladimir Bondarenko
None of Simplify, FullSimplify, FunctionExpand, simplify(),
convert() works out directly in Mathematica 7 and Maple 12.
Waldek Hebisch
>
> cliclic...@freenet.de schrieb:
> >
> > I was mistaken and stand corrected (some late-night transcription
> > error I guess). For real a,b,c, the FriCAS evaluation
> >
> > ATAN((c*SQRT(a^2+b^2+c^2)-a^2-c^2)/(a*b)) + ATAN(a/b)
> >
> > is _fully_ equivalent to
> >
> > ATAN(b*c/(a*SQRT(a^2+b^2+c^2))).
> >
>
> By the way, Derive 6.10 (for real variables, which is the default
> assumption) cannot simplify the difference to zero, or transform one
> expression into the other; we have to wait for Derive 7.01 for this:
In FriCAS normalize function can simplify the difference to 0:
(7) -> ex1 := atan((c*sqrt(a^2+b^2+c^2)-a^2-c^2)/(a*b)) + atan(a/b)
+------------+
| 2 2 2 2 2
c\|c + b + a - c - a a
(7) atan(--------------------------) + atan(-)
a b b
Type: Expression(Integer)
(8) -> ex2 := atan(b*c/(a*sqrt(a^2+b^2+c^2)))
b c
(8) atan(----------------)
+------------+
| 2 2 2
a\|c + b + a
Type: Expression(Integer)
(9) -> normalize(ex1 - ex2)
(9) 0
Type: Expression(Integer)
However, this does not neccesserly mean that two corresponding
numerical function are equal: normalize may treat as equal two
functions having different branch cuts.
--
Waldek Hebisch
heb...@math.uni.wroc.pl
clicl...@freenet.de
Vladimir Bondarenko schrieb:
> On Dec 20, 12:45?am, cliclic...@freenet.de wrote:
> > Vladimir Bondarenko schrieb:
> >
> > > On Dec 19, 8:29?pm, cliclic...@freenet.de wrote:
> >
> > > > Writing the elliptic integrals in their trigonometric normal form and
> > > > substituting t=cos(phi) I arrive at this hypergeometric
> > > > representation:
> >
> > > > (GAMMA(1/4)^2*((2*SQRT(2)*3^(3/4)+3*SQRT(3)+3*12^(1/4)+9)*HYPERG~
> > > > EOM([1,1/4],[3/4],2*SQRT(2)*3^(3/4)+4*SQRT(3)+4*12^(1/4)+7)-2)-4~
> > > > *(11*SQRT(2)*3^(3/4)+21*SQRT(3)+18*12^(1/4)+33)*GAMMA(3/4)^2*HYP~
> > > > ERGEOM([1,3/4],[5/4],2*SQRT(2)*3^(3/4)+4*SQRT(3)+4*12^(1/4)+7))/~
> > > > (8*SQRT(pi))
> >
> > > > Systems with adequate knowledge might find a way to simplify this.
> >
> > > M> Systems with adequate knowledge might find
> > > M> a way to simplify this.
> >
> > > "with adequate knowledge"
> >
> > > Sounds like the systems from your definition
> > > are neither Maple 12 nor Mathematica 7.
> >
> > > ;)
> >
> > Wasn't meant to sound like this. It was said from a Derive
> > perspective. Derive isn't adequately equipped for the simplification
> > of this hypergeometric expression. Other systems may be, but I simply
> > don't know what transformations are available to them, or what
> > transformations would be needed here.
> >
>
> convert() works out directly in Mathematica 7 and Maple 12.
Hmmm. You seem to be discouraging people from trying this
hypergeometric route, which suggests that you know an explicit
reduction path for your challenge.
I have another suggestion left: Complete elliptic integrals of the 3rd
kind can always be expressed in terms of incomplete and complete
integrals of the 1st and 2nd kinds with moduli k and sqrt(1-k^2).
Since we have k = sqrt(1-k^2) = 1/sqrt(2), the complete integrals
reduce again to gamma functions. I haven't done the transformation, so
I don't know if the incomplete integrals can be fused and/or reduced
to something elementary.
But I would have thought that hypergeometric functions are easier to
deal with than incomplete elliptic integrals of the 1st and 2nd kinds.
Martin.
Vladimir Bondarenko
Probably, some help should be provided...
Hint:
- Pi^(3/2)/(2*GAMMA(3/4)^2) + (3+sqrt(6*sqrt(3)-9))/
2*EllipticPi((1 + sqrt(2*sqrt(3) - 3))/2, 1/sqrt(2))
=
Pi*sqrt(2+sqrt(3)+sqrt(7+38*sqrt(3)/9))