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Vladimir Bondarenko
Posted: Tue Jun 05, 2007 11:28 pm
Guest
Hello analytic calculations fans,

Is there a simplifier to come up with (and display.... this is
added for Sir M Wink a sequence of CAS commands to compress this
328 bytes long dilogarithms involving expression

-1/12*Pi^2-ln(3+2*2^(1/2))*ln(-1-2^(1/2))-dilog(2*(2^(1/2)-1)/
(-3+2*2^(1/2)))+ln(3+2*2^(1/2))*ln(2^(1/2)-1)+ln(3+2*2^(1/2))^
2-4*I*ln(2^(1/2)-1)*Pi-1/2*ln(3-2*2^(1/2))^2+I*ln(3+2*2^(1/2))
*Pi+4*dilog(2^(1/2)-1)+4*ln(2^(1/2)-1)^2-4*ln(1+2^(1/2))*ln(2-
2^(1/2))+4*ln(1+2^(1/2))*ln(2^(1/2)-1)+I*ln(3-2*2^(1/2))*Pi-4*
I*ln(1+2^(1/2))*Pi

into 15 bytes ?

Best wishes,

Vladimir Bondarenko

VM and GEMM architect
Co-founder, CEO, Mathematical Director

http://www.cybertester.com/ Cyber Tester, LLC
http://maple.bug-list.org/ Maple Bugs Encyclopaedia
http://www.CAS-testing.org/ CAS Testing
Back to top
dimitris
Posted: Wed Jun 06, 2007 4:45 am
Guest
Hello Vladimir.

The expression in Mathematica notation is

In[418]:=
o= -(Pi^2/12) + 4*dilog[-1 + Sqrt[2]] - dilog[(2*(-1 + Sqrt[2]))/(-3 +
2*Sqrt[2])] + I*Pi*Log[3 - 2*Sqrt[2]] -
(1/2)*Log[3 - 2*Sqrt[2]]^2 - 4*I*Pi*Log[-1 + Sqrt[2]] + 4*Log[-1 +
Sqrt[2]]^2 - 4*I*Pi*Log[1 + Sqrt[2]] -
4*Log[2 - Sqrt[2]]*Log[1 + Sqrt[2]] + 4*Log[-1 + Sqrt[2]]*Log[1 +
Sqrt[2]] + I*Pi*Log[3 + 2*Sqrt[2]] +
Log[-1 + Sqrt[2]]*Log[3 + 2*Sqrt[2]] - (I*Pi + Log[1 +
Sqrt[2]])*Log[3 + 2*Sqrt[2]] + Log[3 + 2*Sqrt[2]]^2;

The function dilog is not implementated in Mathematica.
However, taking the integral represantation, we have

In[420]:=
rul=Integrate[Log[t]/(1 - t), {t, 1, x}, GenerateConditions -> False]

Out[420]=
PolyLog[2, 1 - x]

Applying this rule to the original expression we get

In[421]:=
om = o /. dilog[x_] -> rul

Out[421]=
-(Pi^2/12) + I*Pi*Log[3 - 2*Sqrt[2]] - (1/2)*Log[3 - 2*Sqrt[2]]^2 -
4*I*Pi*Log[-1 + Sqrt[2]] + 4*Log[-1 + Sqrt[2]]^2 -
4*I*Pi*Log[1 + Sqrt[2]] - 4*Log[2 - Sqrt[2]]*Log[1 + Sqrt[2]] +
4*Log[-1 + Sqrt[2]]*Log[1 + Sqrt[2]] +
I*Pi*Log[3 + 2*Sqrt[2]] + Log[-1 + Sqrt[2]]*Log[3 + 2*Sqrt[2]] -
(I*Pi + Log[1 + Sqrt[2]])*Log[3 + 2*Sqrt[2]] +
Log[3 + 2*Sqrt[2]]^2 + 4*PolyLog[2, 2 - Sqrt[2]] - PolyLog[2, 1 -
(2*(-1 + Sqrt[2]))/(-3 + 2*Sqrt[2])]

Let's check it!

In[425]:=
Chop[N[om,30]]

Out[425]=
0.776819399895695981500574710089

In Maple

evalf(-1/12*Pi^2-ln(3+2*2^(1/2))*ln(-1-2^(1/2))-dilog(2*(2^(1/2)-1)/
Quote:
(-3+2*2^(1/2)))+ln(3+2*2^(1/2))*ln(2^(1/2)-1)+ln(3+2*2^(1/2))^
2-4*I*ln(2^(1/2)-1)*Pi-1/2*ln(3-2*2^(1/2))^2+I*ln(3+2*2^(1/2))
*Pi+4*dilog(2^(1/2)-1)+4*ln(2^(1/2)-1)^2-4*ln(1+2^(1/2))*ln(2-
2^(1/2))+4*ln(1+2^(1/2))*ln(2^(1/2)-1)+I*ln(3-2*2^(1/2))*Pi-4*
I*ln(1+2^(1/2))*Pi,30);
0.77681939989569598150057471008 + 0. I


Now, copy/paste this estimation in Plouffer's Inverter
we get
http://bootes.math.uqam.ca/cgi-bin/ipcgi/lookup.pl?Submit=GO+&number=0.776819399895695981500574710089&lookup_type=simple

1/ln(exp(1))*log(1+sqrt(2))^2

with

In[427]:=
LeafCount[(1/Log[Exp[1]])*Log[1 + Sqrt[2]]^2]

Out[427]=
10

Dimitris







/ Vladimir Bondarenko :
Quote:
Hello analytic calculations fans,

Is there a simplifier to come up with (and display.... this is
added for Sir M Wink a sequence of CAS commands to compress this
328 bytes long dilogarithms involving expression

-1/12*Pi^2-ln(3+2*2^(1/2))*ln(-1-2^(1/2))-dilog(2*(2^(1/2)-1)/
(-3+2*2^(1/2)))+ln(3+2*2^(1/2))*ln(2^(1/2)-1)+ln(3+2*2^(1/2))^
2-4*I*ln(2^(1/2)-1)*Pi-1/2*ln(3-2*2^(1/2))^2+I*ln(3+2*2^(1/2))
*Pi+4*dilog(2^(1/2)-1)+4*ln(2^(1/2)-1)^2-4*ln(1+2^(1/2))*ln(2-
2^(1/2))+4*ln(1+2^(1/2))*ln(2^(1/2)-1)+I*ln(3-2*2^(1/2))*Pi-4*
I*ln(1+2^(1/2))*Pi

into 15 bytes ?

Best wishes,

Vladimir Bondarenko

VM and GEMM architect
Co-founder, CEO, Mathematical Director

http://www.cybertester.com/ Cyber Tester, LLC
http://maple.bug-list.org/ Maple Bugs Encyclopaedia
http://www.CAS-testing.org/ CAS Testing
Back to top
dimitris
Posted: Wed Jun 06, 2007 4:51 am
Guest
I guess however this not exactly what Vladimir has in his mind!
And moreover Plouffer's inverter is not panachea!

So let's make an attempt to simplify indeed the expression (Mma 5.2 is
used).

In[427]:=
om

Out[427]=
-(Pi^2/12) + I*Pi*Log[3 - 2*Sqrt[2]] - (1/2)*Log[3 - 2*Sqrt[2]]^2 -
4*I*Pi*Log[-1 + Sqrt[2]] + 4*Log[-1 + Sqrt[2]]^2 -
4*I*Pi*Log[1 + Sqrt[2]] - 4*Log[2 - Sqrt[2]]*Log[1 + Sqrt[2]] +
4*Log[-1 + Sqrt[2]]*Log[1 + Sqrt[2]] +
I*Pi*Log[3 + 2*Sqrt[2]] + Log[-1 + Sqrt[2]]*Log[3 + 2*Sqrt[2]] -
(I*Pi + Log[1 + Sqrt[2]])*Log[3 + 2*Sqrt[2]] +
Log[3 + 2*Sqrt[2]]^2 + 4*PolyLog[2, 2 - Sqrt[2]] - PolyLog[2, 1 -
(2*(-1 + Sqrt[2]))/(-3 + 2*Sqrt[2])]

A quick attempt gives

In[431]:=
FullSimplify[om]

Out[431]=
-(Pi^2/12) - ArcSinh[1]*Log[68 - 48*Sqrt[2]] + I*Pi*Log[3 - 2*Sqrt[2]]
- (1/2)*Log[3 - 2*Sqrt[2]]^2 +
4*PolyLog[2, 2 - Sqrt[2]] - PolyLog[2, 3 + 2*Sqrt[2]]

with

In[433]:=
LeafCount[e]

Out[433]=
79

which is much better than

In[434]:=
LeafCount[om]

Out[434]=
226

Dimitris

/ dimitris :
Quote:
Hello Vladimir.

The expression in Mathematica notation is

In[418]:=
o= -(Pi^2/12) + 4*dilog[-1 + Sqrt[2]] - dilog[(2*(-1 + Sqrt[2]))/(-3 +
2*Sqrt[2])] + I*Pi*Log[3 - 2*Sqrt[2]] -
(1/2)*Log[3 - 2*Sqrt[2]]^2 - 4*I*Pi*Log[-1 + Sqrt[2]] + 4*Log[-1 +
Sqrt[2]]^2 - 4*I*Pi*Log[1 + Sqrt[2]] -
4*Log[2 - Sqrt[2]]*Log[1 + Sqrt[2]] + 4*Log[-1 + Sqrt[2]]*Log[1 +
Sqrt[2]] + I*Pi*Log[3 + 2*Sqrt[2]] +
Log[-1 + Sqrt[2]]*Log[3 + 2*Sqrt[2]] - (I*Pi + Log[1 +
Sqrt[2]])*Log[3 + 2*Sqrt[2]] + Log[3 + 2*Sqrt[2]]^2;

The function dilog is not implementated in Mathematica.
However, taking the integral represantation, we have

In[420]:=
rul=Integrate[Log[t]/(1 - t), {t, 1, x}, GenerateConditions -> False]

Out[420]=
PolyLog[2, 1 - x]

Applying this rule to the original expression we get

In[421]:=
om = o /. dilog[x_] -> rul

Out[421]=
-(Pi^2/12) + I*Pi*Log[3 - 2*Sqrt[2]] - (1/2)*Log[3 - 2*Sqrt[2]]^2 -
4*I*Pi*Log[-1 + Sqrt[2]] + 4*Log[-1 + Sqrt[2]]^2 -
4*I*Pi*Log[1 + Sqrt[2]] - 4*Log[2 - Sqrt[2]]*Log[1 + Sqrt[2]] +
4*Log[-1 + Sqrt[2]]*Log[1 + Sqrt[2]] +
I*Pi*Log[3 + 2*Sqrt[2]] + Log[-1 + Sqrt[2]]*Log[3 + 2*Sqrt[2]] -
(I*Pi + Log[1 + Sqrt[2]])*Log[3 + 2*Sqrt[2]] +
Log[3 + 2*Sqrt[2]]^2 + 4*PolyLog[2, 2 - Sqrt[2]] - PolyLog[2, 1 -
(2*(-1 + Sqrt[2]))/(-3 + 2*Sqrt[2])]

Let's check it!

In[425]:=
Chop[N[om,30]]

Out[425]=
0.776819399895695981500574710089

In Maple

evalf(-1/12*Pi^2-ln(3+2*2^(1/2))*ln(-1-2^(1/2))-dilog(2*(2^(1/2)-1)/
(-3+2*2^(1/2)))+ln(3+2*2^(1/2))*ln(2^(1/2)-1)+ln(3+2*2^(1/2))^
2-4*I*ln(2^(1/2)-1)*Pi-1/2*ln(3-2*2^(1/2))^2+I*ln(3+2*2^(1/2))
*Pi+4*dilog(2^(1/2)-1)+4*ln(2^(1/2)-1)^2-4*ln(1+2^(1/2))*ln(2-
2^(1/2))+4*ln(1+2^(1/2))*ln(2^(1/2)-1)+I*ln(3-2*2^(1/2))*Pi-4*
I*ln(1+2^(1/2))*Pi,30);
0.77681939989569598150057471008 + 0. I

Now, copy/paste this estimation in Plouffer's Inverter
we get
http://bootes.math.uqam.ca/cgi-bin/ipcgi/lookup.pl?Submit=GO+&number=0.776819399895695981500574710089&lookup_type=simple

1/ln(exp(1))*log(1+sqrt(2))^2

with

In[427]:=
LeafCount[(1/Log[Exp[1]])*Log[1 + Sqrt[2]]^2]

Out[427]=
10

Dimitris







/ Vladimir Bondarenko :
Hello analytic calculations fans,

Is there a simplifier to come up with (and display.... this is
added for Sir M Wink a sequence of CAS commands to compress this
328 bytes long dilogarithms involving expression

-1/12*Pi^2-ln(3+2*2^(1/2))*ln(-1-2^(1/2))-dilog(2*(2^(1/2)-1)/
(-3+2*2^(1/2)))+ln(3+2*2^(1/2))*ln(2^(1/2)-1)+ln(3+2*2^(1/2))^
2-4*I*ln(2^(1/2)-1)*Pi-1/2*ln(3-2*2^(1/2))^2+I*ln(3+2*2^(1/2))
*Pi+4*dilog(2^(1/2)-1)+4*ln(2^(1/2)-1)^2-4*ln(1+2^(1/2))*ln(2-
2^(1/2))+4*ln(1+2^(1/2))*ln(2^(1/2)-1)+I*ln(3-2*2^(1/2))*Pi-4*
I*ln(1+2^(1/2))*Pi

into 15 bytes ?

Best wishes,

Vladimir Bondarenko

VM and GEMM architect
Co-founder, CEO, Mathematical Director

http://www.cybertester.com/ Cyber Tester, LLC
http://maple.bug-list.org/ Maple Bugs Encyclopaedia
http://www.CAS-testing.org/ CAS Testing
Back to top
dimitris
Posted: Wed Jun 06, 2007 7:24 am
Guest
Hi again.

So, what is the simplified expression?

Trying really hard I could not simplify more than

-(Pi^2/12) - ArcSinh[1]*Log[68 - 48*Sqrt[2]] + I*Pi*Log[3 - 2*Sqrt[2]]
- (1/2)*Log[3 - 2*Sqrt[2]]^2 +
4*PolyLog[2, 2 - Sqrt[2]] - PolyLog[2, 3 + 2*Sqrt[2]]

My attempts include various identities of dilog

http://functions.wolfram.com/ZetaFunctionsandPolylogarithms/PolyLog2/

I even try Abel's identity (http://en.wikipedia.org/wiki/Dilogarithm)

I think someone must consult so some of these idenitities.
I don't think pure setting of built in functions of Mathematica
can simplify the expression.
But I don't know what Maple can do!

Anyway...

Is it Log[1 + Sqrt[2]]^2 how I found with Plouffe's inverted in the
first message?
Can you show us the steps of simplification?

Greetings from Greece
Dimitris




/ Vladimir Bondarenko :
Quote:
Hello analytic calculations fans,

Is there a simplifier to come up with (and display.... this is
added for Sir M Wink a sequence of CAS commands to compress this
328 bytes long dilogarithms involving expression

-1/12*Pi^2-ln(3+2*2^(1/2))*ln(-1-2^(1/2))-dilog(2*(2^(1/2)-1)/
(-3+2*2^(1/2)))+ln(3+2*2^(1/2))*ln(2^(1/2)-1)+ln(3+2*2^(1/2))^
2-4*I*ln(2^(1/2)-1)*Pi-1/2*ln(3-2*2^(1/2))^2+I*ln(3+2*2^(1/2))
*Pi+4*dilog(2^(1/2)-1)+4*ln(2^(1/2)-1)^2-4*ln(1+2^(1/2))*ln(2-
2^(1/2))+4*ln(1+2^(1/2))*ln(2^(1/2)-1)+I*ln(3-2*2^(1/2))*Pi-4*
I*ln(1+2^(1/2))*Pi

into 15 bytes ?

Best wishes,

Vladimir Bondarenko

VM and GEMM architect
Co-founder, CEO, Mathematical Director

http://www.cybertester.com/ Cyber Tester, LLC
http://maple.bug-list.org/ Maple Bugs Encyclopaedia
http://www.CAS-testing.org/ CAS Testing
Back to top
 
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