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Re: Natural numbers are uniquely defined
Posted:
May 27, 2009 4:55 AM
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WM wrote:
> On 26 Mai, 21:49, Virgil <virg...@nowhere.com> wrote:
>
> > > If we define:
> >
> > > 1 is a natural number
> > > and
> > > with n also n+1 is a natural number
> > > and
> > > N is the smallest set that satisfies both conditions
> >
> > > then N is uniquely specified.
> > > Of course there can be different models for N and there can be
> > > different names for the elements of N. But that does not matter. The
> > > natural numbers do not enter mathematics because someone "defines"
> > > them, names them, or makes models of them, but because the natural
> > > numbers are simply existing and mathematics is built upon them.
>
> > But according to WM, no such thing as N can exist.
> > So WM wold throw out the naturals on which so much is built.
>
> The question is not whether the complete set of all naturals exists.
> That question alrady is nearly as ridiculous as any affirmative
> answer.
>
> The question is whether we could inform someone who does not yet know,
> what we understand by the sequence of natural numbers.
>
> In order to answer this question, we need not wait until SETI gets
> contact. We can answer it in every first class of every elementary
> school. Of course we can inform any child with average intelligence
> what we understand by this sequence 1, 2, 3, ...
>
> Only logicians seem to see problems where no problems are. (As some
> kind of compensation they see no problems where problems are.)
>
> Of course this sequence 1, 2, 3, ... is uniquely defined, because it
> is possible to inform any intelligent being about it. There is no the
> slightest difference between the idea of N and the idea of a sonata or
> pi.
>
> Regards, WM
>
> On 26 Mai, 23:09, Virgil <virg...@nowhere.com> wrote:
> > In article
> > <b54a16b3-914d-47eb-bd89-967cc10d2...@u10g2000vbd.googlegroups.com>,
> >
> >
> >
> >
> >
> > WM <mueck...@rz.fh-augsburg.de> wrote:
> > > On 25 Mai, 17:40, WM <mueck...@rz.fh-augsburg.de> wrote:
> > > > ... classical logic was abstracted from the mathematics of finite sets
> > > > and their subsets .... Forgetful of this limited origin, one
> > > > afterwards mistook that logic for something above and prior to all
> > > > mathematics, and finally applied it, without justification, to the
> > > > mathematics of infinite sets. [H. Weyl].
> >
> > > > One example is this. For infinite sets with a linear ordering ³=<² the
> > > > implication
> > > > En Am: m =< n ==> Am En m =< n (1)
> > > > is held correct, but not the equivalence
> > > > En Am: m =< n <==> Am En m =< n (2)
> > > > For any finite set with linear ordering however, (2) is true. The
> > > > reason is that all elements of a finite set are subject to
> > > > investigation. Therefore, if in ³completed², i.e., ³actual² infinity,
> > > > as used in set theory, all elements are available, then also (2)
> > > > should be used. (2) can only be false, if potentially infinite sets
> > > > are considered, i.e., non-static sets which are never complete but
> > > > allow that elements can be added.
> >
> > But non-static sets do not exist in any mathematical set theory.
>
> They do not exist in what is commonly called ste theory and what is
> eternally false mathematics.
>
> Regards, WM
WM: I offer this text as further proof of polynomial time
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Watch List What is a Watch List? Thread Subject: An exact
simplification challenge - 91 (eerie PolyGamma) Subject: An exact
simplification challenge - 91 (eerie PolyGamma) From: Vladimir
Bondarenko Date: 25 May, 2009 03:23:31 Message: 1 of 6 Reply to this
message Add author to My Watch List View original format Flag as
spam 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 BondarenkoCo-founder, CEO,
Mathematical Directorhttp://www.cybertester.com/ Cyber Tester
Ltd.----------------------------------------------------------"We must
understand that technologieslike these are the way of the
future."----------------------------------------------------------
Subject: An exact simplification challenge - 91 (eerie PolyGamma)
From: Martin Musatov Date: 25 May, 2009 03:36:14 Message: 2 of 6
Reply to this message Add author to My Watch List View original
format Flag as spam Vladimir Bondarenko wrote:> 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.">>
----------------------------------------------------------People, did
I or did I not call it: "eerie effectiveness". Googlesearch sci.math:
"Musatov" and "Eerie". Want to keep going? I enjoyit. It keeps getting
more gratifying each passing day, like puttingpennies into a piggy
bank. Subject: An exact simplification challenge - 91 (eerie
PolyGamma) From: Martin Musatov Date: 25 May, 2009 03:40:46 Message: 3
of 6 Reply to this message Add author to My Watch List View
original format Flag as spam (C)2009:Here you go folks, call me
the information prophet what youare witnessing is simple P=NP
mathematics over large data sets:"Eerie" isn't it?...The
"Effectiveness"...Martin Musatov wrote:> Vladimir Bondarenko wrote:> >
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."> >> >
----------------------------------------------------------> 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 MusatovFounderMeAmI.org Subject: An
exact simplification challenge - 91 (eerie PolyGamma) From:
clicliclic@freenet.de Date: 25 May, 2009 18:32:02 Message: 4 of 6
Reply to this message Add author to My Watch List View original
format Flag as spam 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)3.698588915Martin. Subject: An exact simplification challenge - 91
(eerie PolyGamma) From: Vladimir Bondarenko Date: 25 May, 2009
19:41:14 Message: 5 of 6 Reply to this message Add author to My
Watch List View original format Flag as spam On May 25, 9:32 pm,
cliclic...@freenet.de wrote:> 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.Great!I'd only offer a
bit simpler answer1-pi^2*CSCH(pi)^2+4*LOG(2)3.698588915But there's
still something about this andmany other challenges untold :)
Cheers,Vladimir BondarenkoCo-founder, CEO, Mathematical
Directorhttp://www.cybertester.com/ Cyber Tester
Ltd.----------------------------------------------------------"We must
understand that technologieslike these are the way of the
future."----------------------------------------------------------
Subject: An exact simplification challenge - 91 (eerie PolyGamma)
From: clicliclic@freenet.de Date: 25 May, 2009 21:34:03 Message: 6 of
6 Reply to this message Add author to My Watch List View original
format Flag as spam Vladimir Bondarenko schrieb:> On May 25, 9:32
pm, cliclic...@freenet.de wrote:> > 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> >>> 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)) + 1which is
equivalent to your expression since LI(-1,z) = z/(1-z)^2.Can Maple or
Mathematica perform the conversion automatically?Martin.Tags for this
Thread Everyone's Tags: spam(2) Add a New Tag: Separated by
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Applied By Date/Time spam per isakson 25 May, 2009 00:25:24 spam
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