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An exact simplification challenge - 91 (eerie PolyGamma)

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Vladimir Bondarenko

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May 24, 2009, 11:23:31 PM5/24/09
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Hello,

- PolyGamma[0, 1 + I]
- PolyGamma[0, 1 - I]
- PolyGamma[0, 3/2 + I]
- PolyGamma[0, 3/2 - I]
+ 2 PolyGamma[0, 2 + 2 I]
+ 2 PolyGamma[0, 2 - 2 I]
+ (1/2 + I) PolyGamma[1, 1 + I]
+ (1/2 - I) PolyGamma[1, 1 - I]
- (1/2 + I) PolyGamma[1, 3/2 + I]
- (1/2 - I) PolyGamma[1, 3/2 - I]
+ (2 + 4 I) PolyGamma[1, 2 + 2 I]
+ (2 - 4 I) PolyGamma[1, 2 - 2 I]
- 2 Re[I PolyGamma[1, 1 + I]]
+ 2 Re[I PolyGamma[1, 1 - I]]

?

Folks, please give not just the answer but the processing.

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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Martin Musatov

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May 24, 2009, 11:36:14 PM5/24/09
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People, did I or did I not call it: "eerie effectiveness". Google
search sci.math: "Musatov" and "Eerie". Want to keep going? I enjoy
it. It keeps getting more gratifying each passing day, like putting
pennies into a piggy bank.

Martin Musatov

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May 24, 2009, 11:40:46 PM5/24/09
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(C)2009:Here you go folks, call me the information prophet what you
are witnessing is simple P=NP mathematics over large data sets:
"Eerie" isn't it?
...The "Effectiveness"...
Martin Musatov
Founder
MeAmI.org

clicl...@freenet.de

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May 25, 2009, 2:32:02 PM5/25/09
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Vladimir Bondarenko schrieb:

>
> - PolyGamma[0, 1 + I]
> - PolyGamma[0, 1 - I]
> - PolyGamma[0, 3/2 + I]
> - PolyGamma[0, 3/2 - I]
> + 2 PolyGamma[0, 2 + 2 I]
> + 2 PolyGamma[0, 2 - 2 I]
> + (1/2 + I) PolyGamma[1, 1 + I]
> + (1/2 - I) PolyGamma[1, 1 - I]
> - (1/2 + I) PolyGamma[1, 3/2 + I]
> - (1/2 - I) PolyGamma[1, 3/2 - I]
> + (2 + 4 I) PolyGamma[1, 2 + 2 I]
> + (2 - 4 I) PolyGamma[1, 2 - 2 I]
> - 2 Re[I PolyGamma[1, 1 + I]]
> + 2 Re[I PolyGamma[1, 1 - I]]
>
> ?
>

This simple challenge seems to be specifically made for Derive 6.10:

-POLYGAMMA(0,1+#i)-POLYGAMMA(0,1-#i)-POLYGAMMA(0,3/2+#i)-POLYGAM~
MA(0,3/2-#i)+2*POLYGAMMA(0,2+2*#i)+2*POLYGAMMA(0,2-2*#i)+(1/2+#i~
)*POLYGAMMA(1,1+#i)+(1/2-#i)*POLYGAMMA(1,1-#i)-(1/2+#i)*POLYGAMM~
A(1,3/2+#i)-(1/2-#i)*POLYGAMMA(1,3/2-#i)+(2+4*#i)*POLYGAMMA(1,2+~
2*#i)+(2-4*#i)*POLYGAMMA(1,2-2*#i)-2*RE(#i*POLYGAMMA(1,1+#i))+2*~
RE(#i*POLYGAMMA(1,1-#i))

" ... is automatically rewritten to ... "

-DIGAMMA(3/2-#i)-DIGAMMA(3/2+#i)+2*DIGAMMA(2-2*#i)+2*DIGAMMA(2+2~
*#i)-DIGAMMA(1-#i)-DIGAMMA(1+#i)-ZETA(2,1/2-#i)/2-ZETA(2,1/2+#i)~
/2+2*ZETA(2,1-2*#i)+2*ZETA(2,1+2*#i)+ZETA(2,1-#i)/2+ZETA(2,1+#i)~
/2+#i*(-CONJ(ZETA(2,1-#i))+CONJ(ZETA(2,1+#i))+ZETA(2,1/2-#i)-ZET~
A(2,1/2+#i)-4*ZETA(2,1-2*#i)+4*ZETA(2,1+2*#i))

" judicious substitution of the helper functions ... "

mpsi(z,m):=1/m*SUM(DIGAMMA((z+k)/m),k,0,m-1)+LN(m)

mzeta(s,z,m):=1/m^s*SUM(ZETA(s,(z+k)/m),k,0,m-1)

" ... along with manual help for CONJ(ZETA) ... "

-DIGAMMA(3/2-#i)-DIGAMMA(3/2+#i)+2*mpsi(2-2*#i,2)+2*mpsi(2+2*#i,~
2)-DIGAMMA(1-#i)-DIGAMMA(1+#i)-ZETA(2,1/2-#i)/2-ZETA(2,1/2+#i)/2~
+2*mzeta(2,1-2*#i,2)+2*mzeta(2,1+2*#i,2)+ZETA(2,1-#i)/2+ZETA(2,1~
+#i)/2+#i*(-ZETA(CONJ(2),CONJ(1-#i))+ZETA(CONJ(2),CONJ(1+#i))+ZE~
TA(2,1/2-#i)-ZETA(2,1/2+#i)-4*mzeta(2,1-2*#i,2)+4*mzeta(2,1+2*#i~
,2))

ZETA(2,1-#i)+ZETA(2,1+#i)+4*LN(2)

2*RE(ZETA(2,1+#i))+4*LN(2)

3.698588915

Martin.

Vladimir Bondarenko

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May 25, 2009, 3:41:14 PM5/25/09
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Great!

I'd only offer a bit simpler answer

1-pi^2*CSCH(pi)^2+4*LOG(2)

3.698588915

But there's still something about this and
many other challenges untold :)

clicl...@freenet.de

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May 25, 2009, 5:34:03 PM5/25/09
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Vladimir Bondarenko schrieb:
> Great!
>
> I'd only offer a bit simpler answer
>
> 1-pi^2*CSCH(pi)^2+4*LOG(2)
>
> 3.698588915
>

Sorry, I didn't try to "elementarize" the ZETA's.

ZETA(2, 1+#i) + ZETA(2, 1-#i) =
ZETA(2, 1+#i) + ZETA(2, -#i) + 1 =
-(2 pi)^2 LI(-1, #e^(-2 pi)) + 1

which is equivalent to your expression since LI(-1,z) = z/(1-z)^2.

Can Maple or Mathematica perform the conversion automatically?

Martin.

Nasser Abbasi

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May 27, 2009, 9:52:20 PM5/27/09
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<clicl...@freenet.de> wrote in message
news:1597e1ba-faaa-458e...@h28g2000yqd.googlegroups.com...

Matrin,

Just for fun, I entered this in Mathematica 7, and after 3 days (!) running
it returned with the result below:

In[26]:= FullSimplify[-PolyGamma[0, 1 + I] - PolyGamma[0, 1 - I] -

PolyGamma[0, 3/2 + I] -

PolyGamma[0, 3/2 - I] + 2*PolyGamma[0, 2 + 2*I] + 2*PolyGamma[0, 2 - 2*I]
+
(1/2 + I)*PolyGamma[1, 1 + I] + (1/2 - I)*PolyGamma[1, 1 - I] -
(1/2 + I)*PolyGamma[1, 3/2 + I] - (1/2 - I)*PolyGamma[1, 3/2 - I] +
(2 + 4*I)*PolyGamma[1, 2 + 2*I] + (2 - 4*I)*PolyGamma[1, 2 - 2*I] -
2*Re[I*PolyGamma[1, 1 + I]] + 2*Re[I*PolyGamma[1, 1 - I]]]

Out[26]= (1/2)*(1 - 2*I*Conjugate[PolyGamma[1, 1 - I]] +
2*I*Conjugate[PolyGamma[1, 1 + I]] - Pi^2*Csch[Pi]^2 - 2*PolyGamma[0, 1 -
I] -
2*PolyGamma[0, 1 + I] - 2*PolyGamma[0, 3/2 - I] - 2*PolyGamma[0, 3/2 + I]
+
4*PolyGamma[0, 2 - 2*I] + 4*PolyGamma[0, 2 + 2*I] -
(1 - 2*I)*PolyGamma[1, 3/2 - I] - (1 + 2*I)*PolyGamma[1, 3/2 + I] +
(4 - 8*I)*PolyGamma[1, 2 - 2*I] + (4 + 8*I)*PolyGamma[1, 2 + 2*I])

That is all the simplification one gets.

of course, numerically it is the same result as you have.

In[23]:= N[%]
Out[23]= 3.698588915485305 + 0.*I
In[24]:= Chop[%]
In[25]:= 3.698588915485305

--Nasser


Martin Musatov

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May 27, 2009, 11:36:55 PM5/27/09
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Thank you
And does the above progress in theory a proof p=np?
Martin

clicl...@freenet.de

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May 28, 2009, 12:47:47 PM5/28/09
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Nasser Abbasi schrieb:

>
> Just for fun, I entered this in Mathematica 7, and after 3 days (!) running
> it returned with the result below:
>
> In[26]:= FullSimplify[-PolyGamma[0, 1 + I] - PolyGamma[0, 1 - I] -
> PolyGamma[0, 3/2 + I] -
> PolyGamma[0, 3/2 - I] + 2*PolyGamma[0, 2 + 2*I] + 2*PolyGamma[0, 2 - 2*I]
> +
> (1/2 + I)*PolyGamma[1, 1 + I] + (1/2 - I)*PolyGamma[1, 1 - I] -
> (1/2 + I)*PolyGamma[1, 3/2 + I] - (1/2 - I)*PolyGamma[1, 3/2 - I] +
> (2 + 4*I)*PolyGamma[1, 2 + 2*I] + (2 - 4*I)*PolyGamma[1, 2 - 2*I] -
> 2*Re[I*PolyGamma[1, 1 + I]] + 2*Re[I*PolyGamma[1, 1 - I]]]
>
> Out[26]= (1/2)*(1 - 2*I*Conjugate[PolyGamma[1, 1 - I]] +
> 2*I*Conjugate[PolyGamma[1, 1 + I]] - Pi^2*Csch[Pi]^2 - 2*PolyGamma[0, 1 -
> I] -
> 2*PolyGamma[0, 1 + I] - 2*PolyGamma[0, 3/2 - I] - 2*PolyGamma[0, 3/2 + I]
> +
> 4*PolyGamma[0, 2 - 2*I] + 4*PolyGamma[0, 2 + 2*I] -
> (1 - 2*I)*PolyGamma[1, 3/2 - I] - (1 + 2*I)*PolyGamma[1, 3/2 + I] +
> (4 - 8*I)*PolyGamma[1, 2 - 2*I] + (4 + 8*I)*PolyGamma[1, 2 + 2*I])
>
> That is all the simplification one gets.
>
> of course, numerically it is the same result as you have.
>
> In[23]:= N[%]
> Out[23]= 3.698588915485305 + 0.*I
> In[24]:= Chop[%]
> In[25]:= 3.698588915485305
>

I see - and you could have heard me chuckling there - the three days
must have been THE reason why Vladimir posed this particular
challenge.

Strangely enough the Mathematica result contains Pi^2 Csch[Pi]^2 = (2
Pi)^2 PolyLog[-1, E^(-2 Pi)], but otherwise it is 10 PolyGammas out to
12 PolyGammas in.

Martin.

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