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An exact simplification challenge - 68 (Sqrt, Log)
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Vladimir Bondarenko
Guest






PostPosted: Thu Jul 24, 2008 4:51 am    Post subject: An exact simplification challenge - 68 (Sqrt, Log) Reply with quote

Hello,

(48 + Sqrt[3] ArcCosh[2642885282] - 11 Sqrt[3] Log[2] -
10 Sqrt[3] Log[7] + Log[(2 - Sqrt[3])^(3 Sqrt[-22 + 10
I Sqrt[3]])] + Log[(2 + Sqrt[3])^(15 I)] + Sqrt[3] Log[
578509309952 (7 + 4 Sqrt[3])])/144

?

Best wishes,

Vladimir Bondarenko

Co-founder, CEO, Mathematical Director

http://www.cybertester.com/ Cyber Tester, LLC

-------------------------------------------------------

"We must understand that technologies
like these are the way of the future."

-------------------------------------------------------
Back to top
Guest







PostPosted: Thu Jul 24, 2008 1:09 pm    Post subject: Re: An exact simplification challenge - 68 (Sqrt, Log) Reply with quote

Vladimir Bondarenko schrieb:
[quote]
(48 + Sqrt[3] ArcCosh[2642885282] - 11 Sqrt[3] Log[2] -
10 Sqrt[3] Log[7] + Log[(2 - Sqrt[3])^(3 Sqrt[-22 + 10
I Sqrt[3]])] + Log[(2 + Sqrt[3])^(15 I)] + Sqrt[3] Log[
578509309952 (7 + 4 Sqrt[3])])/144

?
[/quote]
Instantaneous result on Derive 6.10: SQRT(3)*LN(SQRT(3)+2)/9 + 1/3.

(I didn>t participate in the Meijer G challenges since Derive has zero
intrinsic knowledge of Meijer G and hypergeometric pFq functions.)

Martin.
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Vladimir Bondarenko
Guest






PostPosted: Thu Jul 24, 2008 3:05 pm    Post subject: Re: An exact simplification challenge - 68 (Sqrt, Log) Reply with quote

On Jul 24, 6:09 am, cliclic...@freenet.de wrote:
[quote]Vladimir Bondarenko schrieb:



(48 + Sqrt[3] ArcCosh[2642885282] - 11 Sqrt[3] Log[2] -
10 Sqrt[3] Log[7]  + Log[(2 - Sqrt[3])^(3 Sqrt[-22 + 10
I Sqrt[3]])] + Log[(2 + Sqrt[3])^(15 I)] + Sqrt[3] Log[
578509309952 (7 + 4 Sqrt[3])])/144

                        ?

Instantaneous result on Derive 6.10: SQRT(3)*LN(SQRT(3)+2)/9 + 1/3.

(I didn>t participate in the Meijer G challenges since Derive has zero
intrinsic knowledge of Meijer G and hypergeometric pFq functions.)

Martin.
[/quote]
I know that Derive 6.10 cracks this easily (and this is pleasant
for me, so that it was not in vain that I invested my time into
Derive improvement), while Maple 12 and Mathematica 6.0.3 not,
but I expected that someone, maybe you, a Derive fan like me, will
announce this :)

Now so you would not sit without a challenge, I propose you to
simplify this

SQRT(3)*LN(SQRT(3)/2-1/2)/72-SQRT(3)*LN(3*SQRT(3)-5)/18+5*SQRT~
(3)*LN(SQRT(3)+1)/72+1/3-#i*ATAN((COS(-10*SQRT(3)*ACOT(SQRT(3)~
/9)-15*LN(14*SQRT(3)+14))*(COS(-6*SQRT(3)*ACOT(SQRT(3)/9)-3*LN~
(SQRT(3)/2-1/2))*(COS(18*LN(14*SQRT(3)-14)-4*SQRT(3)*ACOT(SQRT~
(3)/9))*(SIN(51*LN(2))*COS(33*LN(7))+COS(51*LN(2))*SIN(33*LN(7~
)))+SIN(18*LN(14*SQRT(3)-14)-4*SQRT(3)*ACOT(SQRT(3)/9))*(SIN(5~
1*LN(2))*SIN(33*LN(7))-COS(51*LN(2))*COS(33*LN(7))))-SIN(-6*SQ~
RT(3)*ACOT(SQRT(3)/9)-3*LN(SQRT(3)/2-1/2))*(COS(18*LN(14*SQRT(~
3)-14)-4*SQRT(3)*ACOT(SQRT(3)/9))*(COS(51*LN(2))*COS(33*LN(7))~
-SIN(51*LN(2))*SIN(33*LN(7)))+SIN(18*LN(14*SQRT(3)-14)-4*SQRT(~
3)*ACOT(SQRT(3)/9))*(SIN(51*LN(2))*COS(33*LN(7))+COS(51*LN(2))~
*SIN(33*LN(7)))))+SIN(-10*SQRT(3)*ACOT(SQRT(3)/9)-15*LN(14*SQR~
T(3)+14))*(COS(-6*SQRT(3)*ACOT(SQRT(3)/9)-3*LN(SQRT(3)/2-1/2))~
*(COS(18*LN(14*SQRT(3)-14)-4*SQRT(3)*ACOT(SQRT(3)/9))*(COS(51*~
LN(2))*COS(33*LN(7))-SIN(51*LN(2))*SIN(33*LN(7)))+SIN(18*LN(14~
*SQRT(3)-14)-4*SQRT(3)*ACOT(SQRT(3)/9))*(SIN(51*LN(2))*COS(33*~
LN(7))+COS(51*LN(2))*SIN(33*LN(7))))+SIN(-6*SQRT(3)*ACOT(SQRT(~
3)/9)-3*LN(SQRT(3)/2-1/2))*(COS(18*LN(14*SQRT(3)-14)-4*SQRT(3)~
*ACOT(SQRT(3)/9))*(SIN(51*LN(2))*COS(33*LN(7))+COS(51*LN(2))*S~
IN(33*LN(7)))+SIN(18*LN(14*SQRT(3)-14)-4*SQRT(3)*ACOT(SQRT(3)/~
9))*(SIN(51*LN(2))*SIN(33*LN(7))-COS(51*LN(2))*COS(33*LN(7))))~
))/(COS(-10*SQRT(3)*ACOT(SQRT(3)/9)-15*LN(14*SQRT(3)+14))*(COS~
(-6*SQRT(3)*ACOT(SQRT(3)/9)-3*LN(SQRT(3)/2-1/2))*(COS(18*LN(14~
*SQRT(3)-14)-4*SQRT(3)*ACOT(SQRT(3)/9))*(COS(51*LN(2))*COS(33*~
LN(7))-SIN(51*LN(2))*SIN(33*LN(7)))+SIN(18*LN(14*SQRT(3)-14)-4~
*SQRT(3)*ACOT(SQRT(3)/9))*(SIN(51*LN(2))*COS(33*LN(7))+COS(51*~
LN(2))*SIN(33*LN(7))))+SIN(-6*SQRT(3)*ACOT(SQRT(3)/9)-3*LN(SQR~
T(3)/2-1/2))*(COS(18*LN(14*SQRT(3)-14)-4*SQRT(3)*ACOT(SQRT(3)/~
9))*(SIN(51*LN(2))*COS(33*LN(7))+COS(51*LN(2))*SIN(33*LN(7)))+~
SIN(18*LN(14*SQRT(3)-14)-4*SQRT(3)*ACOT(SQRT(3)/9))*(SIN(51*LN~
(2))*SIN(33*LN(7))-COS(51*LN(2))*COS(33*LN(7)))))+SIN(-10*SQRT~
(3)*ACOT(SQRT(3)/9)-15*LN(14*SQRT(3)+14))*(SIN(-6*SQRT(3)*ACOT~
(SQRT(3)/9)-3*LN(SQRT(3)/2-1/2))*(COS(18*LN(14*SQRT(3)-14)-4*S~
QRT(3)*ACOT(SQRT(3)/9))*(COS(51*LN(2))*COS(33*LN(7))-SIN(51*LN~
(2))*SIN(33*LN(7)))+SIN(18*LN(14*SQRT(3)-14)-4*SQRT(3)*ACOT(SQ~
RT(3)/9))*(SIN(51*LN(2))*COS(33*LN(7))+COS(51*LN(2))*SIN(33*LN~
(7))))-COS(-6*SQRT(3)*ACOT(SQRT(3)/9)-3*LN(SQRT(3)/2-1/2))*(CO~
S(18*LN(14*SQRT(3)-14)-4*SQRT(3)*ACOT(SQRT(3)/9))*(SIN(51*LN(2~
))*COS(33*LN(7))+COS(51*LN(2))*SIN(33*LN(7)))+SIN(18*LN(14*SQR~
T(3)-14)-4*SQRT(3)*ACOT(SQRT(3)/9))*(SIN(51*LN(2))*SIN(33*LN(7~
))-COS(51*LN(2))*COS(33*LN(7)))))))/72

:-)

Best wishes,

Vladimir Bondarenko

Co-founder, CEO, Mathematical Director

http://www.cybertester.com/ Cyber Tester, LLC

-------------------------------------------------------

"We must understand that technologies
like these are the way of the future."

-------------------------------------------------------
Back to top
Guest







PostPosted: Thu Jul 24, 2008 3:43 pm    Post subject: Re: An exact simplification challenge - 68 (Sqrt, Log) Reply with quote

On 24 juil, 17:05, Vladimir Bondarenko <v...@cybertester.com> wrote:
[quote]On Jul 24, 6:09 am, cliclic...@freenet.de wrote:





Vladimir Bondarenko schrieb:

(48 + Sqrt[3] ArcCosh[2642885282] - 11 Sqrt[3] Log[2] -
10 Sqrt[3] Log[7]  + Log[(2 - Sqrt[3])^(3 Sqrt[-22 + 10
I Sqrt[3]])] + Log[(2 + Sqrt[3])^(15 I)] + Sqrt[3] Log[
578509309952 (7 + 4 Sqrt[3])])/144

                        ?

Instantaneous result on Derive 6.10: SQRT(3)*LN(SQRT(3)+2)/9 + 1/3.

(I didn>t participate in the Meijer G challenges since Derive has zero
intrinsic knowledge of Meijer G and hypergeometric pFq functions.)

Martin.

I know that Derive 6.10 cracks this easily (and this is pleasant
for me, so that it was not in vain that I invested my time into
Derive improvement), while Maple 12 and Mathematica 6.0.3 not,
but I expected that someone, maybe you, a Derive fan like me, will
announce this :)

Now so you would not sit without a challenge, I propose you to
simplify this

SQRT(3)*LN(SQRT(3)/2-1/2)/72-SQRT(3)*LN(3*SQRT(3)-5)/18+5*SQRT~
(3)*LN(SQRT(3)+1)/72+1/3-#i*ATAN((COS(-10*SQRT(3)*ACOT(SQRT(3)~
/9)-15*LN(14*SQRT(3)+14))*(COS(-6*SQRT(3)*ACOT(SQRT(3)/9)-3*LN~
(SQRT(3)/2-1/2))*(COS(18*LN(14*SQRT(3)-14)-4*SQRT(3)*ACOT(SQRT~
(3)/9))*(SIN(51*LN(2))*COS(33*LN(7))+COS(51*LN(2))*SIN(33*LN(7~
)))+SIN(18*LN(14*SQRT(3)-14)-4*SQRT(3)*ACOT(SQRT(3)/9))*(SIN(5~
1*LN(2))*SIN(33*LN(7))-COS(51*LN(2))*COS(33*LN(7))))-SIN(-6*SQ~
RT(3)*ACOT(SQRT(3)/9)-3*LN(SQRT(3)/2-1/2))*(COS(18*LN(14*SQRT(~
3)-14)-4*SQRT(3)*ACOT(SQRT(3)/9))*(COS(51*LN(2))*COS(33*LN(7))~
-SIN(51*LN(2))*SIN(33*LN(7)))+SIN(18*LN(14*SQRT(3)-14)-4*SQRT(~
3)*ACOT(SQRT(3)/9))*(SIN(51*LN(2))*COS(33*LN(7))+COS(51*LN(2))~
*SIN(33*LN(7)))))+SIN(-10*SQRT(3)*ACOT(SQRT(3)/9)-15*LN(14*SQR~
T(3)+14))*(COS(-6*SQRT(3)*ACOT(SQRT(3)/9)-3*LN(SQRT(3)/2-1/2))~
*(COS(18*LN(14*SQRT(3)-14)-4*SQRT(3)*ACOT(SQRT(3)/9))*(COS(51*~
LN(2))*COS(33*LN(7))-SIN(51*LN(2))*SIN(33*LN(7)))+SIN(18*LN(14~
*SQRT(3)-14)-4*SQRT(3)*ACOT(SQRT(3)/9))*(SIN(51*LN(2))*COS(33*~
LN(7))+COS(51*LN(2))*SIN(33*LN(7))))+SIN(-6*SQRT(3)*ACOT(SQRT(~
3)/9)-3*LN(SQRT(3)/2-1/2))*(COS(18*LN(14*SQRT(3)-14)-4*SQRT(3)~
*ACOT(SQRT(3)/9))*(SIN(51*LN(2))*COS(33*LN(7))+COS(51*LN(2))*S~
IN(33*LN(7)))+SIN(18*LN(14*SQRT(3)-14)-4*SQRT(3)*ACOT(SQRT(3)/~
9))*(SIN(51*LN(2))*SIN(33*LN(7))-COS(51*LN(2))*COS(33*LN(7))))~
))/(COS(-10*SQRT(3)*ACOT(SQRT(3)/9)-15*LN(14*SQRT(3)+14))*(COS~
(-6*SQRT(3)*ACOT(SQRT(3)/9)-3*LN(SQRT(3)/2-1/2))*(COS(18*LN(14~
*SQRT(3)-14)-4*SQRT(3)*ACOT(SQRT(3)/9))*(COS(51*LN(2))*COS(33*~
LN(7))-SIN(51*LN(2))*SIN(33*LN(7)))+SIN(18*LN(14*SQRT(3)-14)-4~
*SQRT(3)*ACOT(SQRT(3)/9))*(SIN(51*LN(2))*COS(33*LN(7))+COS(51*~
LN(2))*SIN(33*LN(7))))+SIN(-6*SQRT(3)*ACOT(SQRT(3)/9)-3*LN(SQR~
T(3)/2-1/2))*(COS(18*LN(14*SQRT(3)-14)-4*SQRT(3)*ACOT(SQRT(3)/~
9))*(SIN(51*LN(2))*COS(33*LN(7))+COS(51*LN(2))*SIN(33*LN(7)))+~
SIN(18*LN(14*SQRT(3)-14)-4*SQRT(3)*ACOT(SQRT(3)/9))*(SIN(51*LN~
(2))*SIN(33*LN(7))-COS(51*LN(2))*COS(33*LN(7)))))+SIN(-10*SQRT~
(3)*ACOT(SQRT(3)/9)-15*LN(14*SQRT(3)+14))*(SIN(-6*SQRT(3)*ACOT~
(SQRT(3)/9)-3*LN(SQRT(3)/2-1/2))*(COS(18*LN(14*SQRT(3)-14)-4*S~
QRT(3)*ACOT(SQRT(3)/9))*(COS(51*LN(2))*COS(33*LN(7))-SIN(51*LN~
(2))*SIN(33*LN(7)))+SIN(18*LN(14*SQRT(3)-14)-4*SQRT(3)*ACOT(SQ~
RT(3)/9))*(SIN(51*LN(2))*COS(33*LN(7))+COS(51*LN(2))*SIN(33*LN~
(7))))-COS(-6*SQRT(3)*ACOT(SQRT(3)/9)-3*LN(SQRT(3)/2-1/2))*(CO~
S(18*LN(14*SQRT(3)-14)-4*SQRT(3)*ACOT(SQRT(3)/9))*(SIN(51*LN(2~
))*COS(33*LN(7))+COS(51*LN(2))*SIN(33*LN(7)))+SIN(18*LN(14*SQR~
T(3)-14)-4*SQRT(3)*ACOT(SQRT(3)/9))*(SIN(51*LN(2))*SIN(33*LN(7~
))-COS(51*LN(2))*COS(33*LN(7)))))))/72

:-)

Best wishes,

Vladimir Bondarenko

Co-founder, CEO, Mathematical Director

http://www.cybertester.com/ Cyber Tester, LLC

-------------------------------------------------------

"We must understand that technologies
like these are the way of the future."

-------------------------------------------------------- Masquer le texte des messages précédents -

- Afficher le texte des messages précédents -
[/quote]
Bonsoir,

I do not ever agree with : ""We must *understand* that technologies
like these are the way of the future."
I believe that with these technologies a lot of sane people
will not even understand a word about anything belonging
to their world,

Alain
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Guest







PostPosted: Thu Jul 24, 2008 3:51 pm    Post subject: Re: An exact simplification challenge - 68 (Sqrt, Log) Reply with quote

Vladimir Bondarenko schrieb:
[quote]On Jul 24, 6:09?am, cliclic...@freenet.de wrote:
Vladimir Bondarenko schrieb:

(48 + Sqrt[3] ArcCosh[2642885282] - 11 Sqrt[3] Log[2] -
10 Sqrt[3] Log[7] ?+ Log[(2 - Sqrt[3])^(3 Sqrt[-22 + 10
I Sqrt[3]])] + Log[(2 + Sqrt[3])^(15 I)] + Sqrt[3] Log[
578509309952 (7 + 4 Sqrt[3])])/144

?

Instantaneous result on Derive 6.10: SQRT(3)*LN(SQRT(3)+2)/9 + 1/3.

(I didn>t participate in the Meijer G challenges since Derive has zero
intrinsic knowledge of Meijer G and hypergeometric pFq functions.)

I know that Derive 6.10 cracks this easily (and this is pleasant
for me, so that it was not in vain that I invested my time into
Derive improvement), while Maple 12 and Mathematica 6.0.3 not,
but I expected that someone, maybe you, a Derive fan like me, will
announce this :)

Now so you would not sit without a challenge, I propose you to
simplify this

SQRT(3)*LN(SQRT(3)/2-1/2)/72-SQRT(3)*LN(3*SQRT(3)-5)/18+5*SQRT~
(3)*LN(SQRT(3)+1)/72+1/3-#i*ATAN((COS(-10*SQRT(3)*ACOT(SQRT(3)~
/9)-15*LN(14*SQRT(3)+14))*(COS(-6*SQRT(3)*ACOT(SQRT(3)/9)-3*LN~
(SQRT(3)/2-1/2))*(COS(18*LN(14*SQRT(3)-14)-4*SQRT(3)*ACOT(SQRT~
(3)/9))*(SIN(51*LN(2))*COS(33*LN(7))+COS(51*LN(2))*SIN(33*LN(7~
)))+SIN(18*LN(14*SQRT(3)-14)-4*SQRT(3)*ACOT(SQRT(3)/9))*(SIN(5~
1*LN(2))*SIN(33*LN(7))-COS(51*LN(2))*COS(33*LN(7))))-SIN(-6*SQ~
RT(3)*ACOT(SQRT(3)/9)-3*LN(SQRT(3)/2-1/2))*(COS(18*LN(14*SQRT(~
3)-14)-4*SQRT(3)*ACOT(SQRT(3)/9))*(COS(51*LN(2))*COS(33*LN(7))~
-SIN(51*LN(2))*SIN(33*LN(7)))+SIN(18*LN(14*SQRT(3)-14)-4*SQRT(~
3)*ACOT(SQRT(3)/9))*(SIN(51*LN(2))*COS(33*LN(7))+COS(51*LN(2))~
*SIN(33*LN(7)))))+SIN(-10*SQRT(3)*ACOT(SQRT(3)/9)-15*LN(14*SQR~
T(3)+14))*(COS(-6*SQRT(3)*ACOT(SQRT(3)/9)-3*LN(SQRT(3)/2-1/2))~
*(COS(18*LN(14*SQRT(3)-14)-4*SQRT(3)*ACOT(SQRT(3)/9))*(COS(51*~
LN(2))*COS(33*LN(7))-SIN(51*LN(2))*SIN(33*LN(7)))+SIN(18*LN(14~
*SQRT(3)-14)-4*SQRT(3)*ACOT(SQRT(3)/9))*(SIN(51*LN(2))*COS(33*~
LN(7))+COS(51*LN(2))*SIN(33*LN(7))))+SIN(-6*SQRT(3)*ACOT(SQRT(~
3)/9)-3*LN(SQRT(3)/2-1/2))*(COS(18*LN(14*SQRT(3)-14)-4*SQRT(3)~
*ACOT(SQRT(3)/9))*(SIN(51*LN(2))*COS(33*LN(7))+COS(51*LN(2))*S~
IN(33*LN(7)))+SIN(18*LN(14*SQRT(3)-14)-4*SQRT(3)*ACOT(SQRT(3)/~
9))*(SIN(51*LN(2))*SIN(33*LN(7))-COS(51*LN(2))*COS(33*LN(7))))~
))/(COS(-10*SQRT(3)*ACOT(SQRT(3)/9)-15*LN(14*SQRT(3)+14))*(COS~
(-6*SQRT(3)*ACOT(SQRT(3)/9)-3*LN(SQRT(3)/2-1/2))*(COS(18*LN(14~
*SQRT(3)-14)-4*SQRT(3)*ACOT(SQRT(3)/9))*(COS(51*LN(2))*COS(33*~
LN(7))-SIN(51*LN(2))*SIN(33*LN(7)))+SIN(18*LN(14*SQRT(3)-14)-4~
*SQRT(3)*ACOT(SQRT(3)/9))*(SIN(51*LN(2))*COS(33*LN(7))+COS(51*~
LN(2))*SIN(33*LN(7))))+SIN(-6*SQRT(3)*ACOT(SQRT(3)/9)-3*LN(SQR~
T(3)/2-1/2))*(COS(18*LN(14*SQRT(3)-14)-4*SQRT(3)*ACOT(SQRT(3)/~
9))*(SIN(51*LN(2))*COS(33*LN(7))+COS(51*LN(2))*SIN(33*LN(7)))+~
SIN(18*LN(14*SQRT(3)-14)-4*SQRT(3)*ACOT(SQRT(3)/9))*(SIN(51*LN~
(2))*SIN(33*LN(7))-COS(51*LN(2))*COS(33*LN(7)))))+SIN(-10*SQRT~
(3)*ACOT(SQRT(3)/9)-15*LN(14*SQRT(3)+14))*(SIN(-6*SQRT(3)*ACOT~
(SQRT(3)/9)-3*LN(SQRT(3)/2-1/2))*(COS(18*LN(14*SQRT(3)-14)-4*S~
QRT(3)*ACOT(SQRT(3)/9))*(COS(51*LN(2))*COS(33*LN(7))-SIN(51*LN~
(2))*SIN(33*LN(7)))+SIN(18*LN(14*SQRT(3)-14)-4*SQRT(3)*ACOT(SQ~
RT(3)/9))*(SIN(51*LN(2))*COS(33*LN(7))+COS(51*LN(2))*SIN(33*LN~
(7))))-COS(-6*SQRT(3)*ACOT(SQRT(3)/9)-3*LN(SQRT(3)/2-1/2))*(CO~
S(18*LN(14*SQRT(3)-14)-4*SQRT(3)*ACOT(SQRT(3)/9))*(SIN(51*LN(2~
))*COS(33*LN(7))+COS(51*LN(2))*SIN(33*LN(7)))+SIN(18*LN(14*SQR~
T(3)-14)-4*SQRT(3)*ACOT(SQRT(3)/9))*(SIN(51*LN(2))*SIN(33*LN(7~
))-COS(51*LN(2))*COS(33*LN(7)))))))/72

[/quote]
A bit harder, but still easy - it needs some juggling with the
simplification settings:

[Trigonometry:=Collect, Logarithm:=Collect]
1/3-SQRT(3)*LN((2-SQRT(3))^(1/9))
Logarithm:=Auto
1/3-SQRT(3)*LN(2-SQRT(3))/9

Martin.

PS: But Derive (under my command) failed to fully simplify your
monster challenge

SQRT(6)*SQRT(-SQRT(16*pi^2*LN(35*SQRT(2)+20*SQRT(6)+28*SQRT(3)+49)^2-
pi^2*LN(~
35*SQRT(2)+20*SQRT(6)+28*SQRT(3)+49)*LN(3967654484264881281965698789310658220~
572212595135191362121484088486083360*SQRT(2)+32395763207048357347710948248549~
50687594346152507722273751834468297053056*SQRT(3)+229072638454175499028284157~
8360115154299405761187724567354303143178465280*SQRT(6)+5611110782457822952710~
[...........................................................................]
2*pi^2)+LN(2)*LN(1853020188851841)*(144*LN(2)^2+47*pi^2)+2304*LN(2)^4+2209*pi~
^2*LN(2)^2)-
LN(SQRT(3)-1)*LN(184141500186093409798577801761927946149371551404~
77097961418915840*SQRT(3)+318942434104267658870956964797945820754099234494920~
25397756297216)-48*LN(SQRT(3)-1)^2+LN(3)*LN(447127149604363042816-25814898018~
6069073920*SQRT(3))-48*LN(SQRT(3)+1)^2+12*LN(3)^2+48*LN(2)^2)/12

The result of the automatic simplification is recalcitrant:
SQRT(6)*SQRT(-SQRT(-pi^2*LN(2-
SQRT(3))*(12*LN(SQRT(2)+1)+LN(2))-4*pi^2*LN(SQR~
T(15*SQRT(2)+21)+2*SQRT(5*SQRT(2)+7))^2-144*LN(2)^2*LN(11*SQRT(3)-19)^2-4*pi^~
2*LN(12-6*SQRT(3))*LN(3*SQRT(3)-5)+144*LN(SQRT(3)-1)^4+576*LN(SQRT(3)+1)*LN(S~
QRT(3)-1)^3+4*LN(SQRT(3)-1)^2*(216*LN(SQRT(3)+1)^2-18*LN(3)^2+252*LN(2)^2+65*~
pi^2)+LN(SQRT(3)-1)*(576*LN(SQRT(3)+1)^3-16*LN(SQRT(3)+1)*(9*LN(3)^2+4*(36*LN~
(2)^2-5*pi^2))+32*pi^2*LN(3)-377*pi^2*LN(2))
+144*LN(SQRT(3)+1)^4-4*LN(SQRT(3)~
+1)^2*(18*LN(3)^2-72*LN(2)^2-25*pi^2)+5*pi^2*LN(SQRT(3)+1)*(4*LN(3)-47*LN(2))~
+9*pi^2*LN(SQRT(2)+1)^2+9*LN(3)^4+72*LN(2)^2*LN(3)^2-24*pi^2*LN(2)*LN(3)+144*~
LN(2)^4+138*pi^2*LN(2)^2)-12*LN(SQRT(3)-1)^2-24*LN(SQRT(3)+1)*LN(SQRT(3)-1)-1~
2*LN(SQRT(3)+1)^2+3*LN(3)^2+12*LN(2)^2)/6+SQRT(6)*SQRT(SQRT(-pi^2*LN(2-
SQRT(3~
))*(12*LN(SQRT(2)+1)+LN(2))-4*pi^2*LN(SQRT(15*SQRT(2)+21)+2*SQRT(5*SQRT(2)+7)~
)^2-144*LN(2)^2*LN(11*SQRT(3)-19)^2-4*pi^2*LN(12-6*SQRT(3))*LN(3*SQRT(3)-5)+1~
44*LN(SQRT(3)-1)^4+576*LN(SQRT(3)+1)*LN(SQRT(3)-1)^3+4*LN(SQRT(3)-1)^2*(216*L~
N(SQRT(3)+1)^2-18*LN(3)^2+252*LN(2)^2+65*pi^2)+LN(SQRT(3)-1)*(576*LN(SQRT(3)+~
1)^3-16*LN(SQRT(3)+1)*(9*LN(3)^2+4*(36*LN(2)^2-5*pi^2))
+32*pi^2*LN(3)-377*pi^~
2*LN(2))
+144*LN(SQRT(3)+1)^4-4*LN(SQRT(3)+1)^2*(18*LN(3)^2-72*LN(2)^2-25*pi^2~
)+5*pi^2*LN(SQRT(3)+1)*(4*LN(3)-47*LN(2))
+9*pi^2*LN(SQRT(2)+1)^2+9*LN(3)^4+72~
*LN(2)^2*LN(3)^2-24*pi^2*LN(2)*LN(3)+144*LN(2)^4+138*pi^2*LN(2)^2)-12*LN(SQRT~
(3)-1)^2-24*LN(SQRT(3)+1)*LN(SQRT(3)-1)-12*LN(SQRT(3)+1)^2+3*LN(3)^2+12*LN(2)~
^2)/6

Do you know how to proceed?
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David W. Cantrell
Guest






PostPosted: Thu Jul 24, 2008 8:46 pm    Post subject: Re: An exact simplification challenge - 68 (Sqrt, Log) Reply with quote

Vladimir Bondarenko <vb@cybertester.com> wrote:
[quote]On Jul 24, 6:09=A0am, cliclic...@freenet.de wrote:
Vladimir Bondarenko schrieb:



(48 + Sqrt[3] ArcCosh[2642885282] - 11 Sqrt[3] Log[2] -
10 Sqrt[3] Log[7] =A0+ Log[(2 - Sqrt[3])^(3 Sqrt[-22 + 10
I Sqrt[3]])] + Log[(2 + Sqrt[3])^(15 I)] + Sqrt[3] Log[
578509309952 (7 + 4 Sqrt[3])])/144

=A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 ?

Instantaneous result on Derive 6.10: SQRT(3)*LN(SQRT(3)+2)/9 + 1/3.

(I didn>t participate in the Meijer G challenges since Derive has zero
intrinsic knowledge of Meijer G and hypergeometric pFq functions.)

Martin.

I know that Derive 6.10 cracks this easily (and this is pleasant
for me, so that it was not in vain that I invested my time into
Derive improvement), while Maple 12 and Mathematica 6.0.3 not,
but I expected that someone, maybe you, a Derive fan like me, will
announce this :)

Now so you would not sit without a challenge, I propose you to
simplify this

SQRT(3)*LN(SQRT(3)/2-1/2)/72-SQRT(3)*LN(3*SQRT(3)-5)/18+5*SQRT~
(3)*LN(SQRT(3)+1)/72+1/3-#i*ATAN((COS(-10*SQRT(3)*ACOT(SQRT(3)~
/9)-15*LN(14*SQRT(3)+14))*(COS(-6*SQRT(3)*ACOT(SQRT(3)/9)-3*LN~
(SQRT(3)/2-1/2))*(COS(18*LN(14*SQRT(3)-14)-4*SQRT(3)*ACOT(SQRT~
(3)/9))*(SIN(51*LN(2))*COS(33*LN(7))+COS(51*LN(2))*SIN(33*LN(7~
)))+SIN(18*LN(14*SQRT(3)-14)-4*SQRT(3)*ACOT(SQRT(3)/9))*(SIN(5~
1*LN(2))*SIN(33*LN(7))-COS(51*LN(2))*COS(33*LN(7))))-SIN(-6*SQ~
RT(3)*ACOT(SQRT(3)/9)-3*LN(SQRT(3)/2-1/2))*(COS(18*LN(14*SQRT(~
3)-14)-4*SQRT(3)*ACOT(SQRT(3)/9))*(COS(51*LN(2))*COS(33*LN(7))~
-SIN(51*LN(2))*SIN(33*LN(7)))+SIN(18*LN(14*SQRT(3)-14)-4*SQRT(~
3)*ACOT(SQRT(3)/9))*(SIN(51*LN(2))*COS(33*LN(7))+COS(51*LN(2))~
*SIN(33*LN(7)))))+SIN(-10*SQRT(3)*ACOT(SQRT(3)/9)-15*LN(14*SQR~
T(3)+14))*(COS(-6*SQRT(3)*ACOT(SQRT(3)/9)-3*LN(SQRT(3)/2-1/2))~
*(COS(18*LN(14*SQRT(3)-14)-4*SQRT(3)*ACOT(SQRT(3)/9))*(COS(51*~
LN(2))*COS(33*LN(7))-SIN(51*LN(2))*SIN(33*LN(7)))+SIN(18*LN(14~
*SQRT(3)-14)-4*SQRT(3)*ACOT(SQRT(3)/9))*(SIN(51*LN(2))*COS(33*~
LN(7))+COS(51*LN(2))*SIN(33*LN(7))))+SIN(-6*SQRT(3)*ACOT(SQRT(~
3)/9)-3*LN(SQRT(3)/2-1/2))*(COS(18*LN(14*SQRT(3)-14)-4*SQRT(3)~
*ACOT(SQRT(3)/9))*(SIN(51*LN(2))*COS(33*LN(7))+COS(51*LN(2))*S~
IN(33*LN(7)))+SIN(18*LN(14*SQRT(3)-14)-4*SQRT(3)*ACOT(SQRT(3)/~
9))*(SIN(51*LN(2))*SIN(33*LN(7))-COS(51*LN(2))*COS(33*LN(7))))~
))/(COS(-10*SQRT(3)*ACOT(SQRT(3)/9)-15*LN(14*SQRT(3)+14))*(COS~
(-6*SQRT(3)*ACOT(SQRT(3)/9)-3*LN(SQRT(3)/2-1/2))*(COS(18*LN(14~
*SQRT(3)-14)-4*SQRT(3)*ACOT(SQRT(3)/9))*(COS(51*LN(2))*COS(33*~
LN(7))-SIN(51*LN(2))*SIN(33*LN(7)))+SIN(18*LN(14*SQRT(3)-14)-4~
*SQRT(3)*ACOT(SQRT(3)/9))*(SIN(51*LN(2))*COS(33*LN(7))+COS(51*~
LN(2))*SIN(33*LN(7))))+SIN(-6*SQRT(3)*ACOT(SQRT(3)/9)-3*LN(SQR~
T(3)/2-1/2))*(COS(18*LN(14*SQRT(3)-14)-4*SQRT(3)*ACOT(SQRT(3)/~
9))*(SIN(51*LN(2))*COS(33*LN(7))+COS(51*LN(2))*SIN(33*LN(7)))+~
SIN(18*LN(14*SQRT(3)-14)-4*SQRT(3)*ACOT(SQRT(3)/9))*(SIN(51*LN~
(2))*SIN(33*LN(7))-COS(51*LN(2))*COS(33*LN(7)))))+SIN(-10*SQRT~
(3)*ACOT(SQRT(3)/9)-15*LN(14*SQRT(3)+14))*(SIN(-6*SQRT(3)*ACOT~
(SQRT(3)/9)-3*LN(SQRT(3)/2-1/2))*(COS(18*LN(14*SQRT(3)-14)-4*S~
QRT(3)*ACOT(SQRT(3)/9))*(COS(51*LN(2))*COS(33*LN(7))-SIN(51*LN~
(2))*SIN(33*LN(7)))+SIN(18*LN(14*SQRT(3)-14)-4*SQRT(3)*ACOT(SQ~
RT(3)/9))*(SIN(51*LN(2))*COS(33*LN(7))+COS(51*LN(2))*SIN(33*LN~
(7))))-COS(-6*SQRT(3)*ACOT(SQRT(3)/9)-3*LN(SQRT(3)/2-1/2))*(CO~
S(18*LN(14*SQRT(3)-14)-4*SQRT(3)*ACOT(SQRT(3)/9))*(SIN(51*LN(2~
))*COS(33*LN(7))+COS(51*LN(2))*SIN(33*LN(7)))+SIN(18*LN(14*SQR~
T(3)-14)-4*SQRT(3)*ACOT(SQRT(3)/9))*(SIN(51*LN(2))*SIN(33*LN(7~
))-COS(51*LN(2))*COS(33*LN(7)))))))/72
[/quote]
Your new expression simplifies to the same thing as the original expression
in this thread.

David
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