Σειρά με ζήτα

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Tolaso J Kos
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Εγγραφή: Κυρ Αύγ 05, 2012 10:09 pm
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Σειρά με ζήτα

#1

Μη αναγνωσμένη δημοσίευση από Tolaso J Kos » Πέμ Απρ 06, 2017 11:36 am

Ας δηλώσουμε με \zeta τη συνάρτηση ζήτα του Riemann. Δειχθήτω:
\displaystyle{\frac{1}{2\pi} {\rm Li}_2 \left ( e^{-2\pi} \right ) = \log 2\pi - 1 -\frac{5 \pi}{12} - \sum_{n=1}^{\infty} \frac{(-1)^n \zeta(2n)}{n \left ( 2n+1 \right )}} όπου {\rm Li}_2 είναι η διλογαριθμική συνάρτηση.


Η φαντασία είναι σημαντικότερη από τη γνώση !
\displaystyle{{\color{blue}\mathbf{Life=\int_{birth}^{death}\frac{happiness}{time}\Delta time} }}

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Άβαταρ μέλους
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Εγγραφή: Τετ Μάιος 20, 2009 9:14 am
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Re: Σειρά με ζήτα

#2

Μη αναγνωσμένη δημοσίευση από Σεραφείμ » Παρ Ιούλ 07, 2017 7:32 am

Tolaso J Kos έγραψε:Δειχθήτω: \displaystyle{\frac{1}{2\pi} {\rm Li}_2 \left ( e^{-2\pi} \right ) = \log 2\pi - 1 -\frac{5 \pi}{12} - \sum_{n=1}^{\infty} \frac{(-1)^n \zeta(2n)}{n \left ( 2n+1 \right )}}
Σχόλια:

\displaystyle{\sum\limits_{n = 1}^\infty  {\left( {\frac{{{{\left( { - 1} \right)}^n}}}{{n \cdot {m^{2n}}}}} \right)}  =  - \log \left( {\frac{{1 + {m^2}}}{{{m^2}}}} \right){\rm{   }}\left( * \right)} και \displaystyle{\sum\limits_{n = 1}^\infty  {\left( {\frac{{{{\left( { - 1} \right)}^n}}}{{\left( {2n + 1} \right) \cdot {m^{2n}}}}} \right)}  = 1 - m \cdot \arctan \left( {\frac{1}{m}} \right){\rm{   }}\left( {2*} \right)} (γνωστές σειρές).

\displaystyle{\Gamma \left( i \right)\Gamma \left( { - i} \right) = \frac{{2\pi }}{{{e^\pi } - {e^{ - \pi }}}}{\rm{   }}\left( {3*} \right)} διότι \displaystyle{\Gamma \left( i \right)\Gamma \left( { - i} \right) = \frac{{\Gamma \left( i \right)\left( { - i} \right)\Gamma \left( { - i} \right)}}{{ - i}} = \frac{{\Gamma \left( i \right)\Gamma \left( {1 - i} \right)}}{{ - i}} = \frac{\pi }{{ - i \cdot \sin \left( {\pi  \cdot i} \right)}} = \frac{{2\pi }}{{{e^\pi } - {e^{ - \pi }}}}}

\displaystyle{\mathop {\lim }\limits_{n \to \infty } \left( {\frac{{z \cdot \left( {1 + z} \right) \cdot .. \cdot \left( {m + z} \right)}}{{{{\left( {m + 1} \right)}^z}m!}}} \right) = \frac{1}{{\Gamma \left( z \right)}}{\rm{   }}\left( {4*} \right)} (εναλλακτικός ορισμός της συνάρτησης Gamma function)

\displaystyle{\sum\limits_{n = 1}^\infty  {\frac{1}{{{x^2} + {n^2}}}}  =  - \frac{1}{{2{x^2}}} + \frac{\pi }{{2x}} \cdot \frac{{{e^{\pi x}} + {e^{ - \pi x}}}}{{{e^{\pi x}} - {e^{ - \pi x}}}}{\rm{   }}\left( {5*} \right)} (έχει αποδειχθεί αρκετές φορές στην σελίδα)

Στο θέμα μας

\displaystyle{\sum\limits_{n = 1}^\infty  {\frac{{{{\left( { - 1} \right)}^n}\zeta \left( {2n} \right)}}{{n\left( {2n + 1} \right)}}}  = \sum\limits_{n = 1}^\infty  {\frac{{{{\left( { - 1} \right)}^n}}}{{n\left( {2n + 1} \right)}}\sum\limits_{m = 1}^\infty  {\frac{1}{{{m^{2n}}}}} }  = 2\sum\limits_{m = 1}^\infty  {\sum\limits_{n = 1}^\infty  {{{\left( { - 1} \right)}^n}\left( {\frac{1}{{2n \cdot {m^{2n}}}} - \frac{1}{{\left( {2n + 1} \right) \cdot {m^{2n}}}}} \right)} }  = }

\displaystyle{ = 2\sum\limits_{m = 1}^\infty  {\left( {\sum\limits_{n = 1}^\infty  {\left( {\frac{{{{\left( { - 1} \right)}^n}}}{{2n \cdot {m^{2n}}}}} \right) - \sum\limits_{n = 1}^\infty  {\left( {\frac{{{{\left( { - 1} \right)}^n}}}{{\left( {2n + 1} \right) \cdot {m^{2n}}}}} \right)} } } \right)} \mathop { =  =  = }\limits_{\left( {2*} \right)}^{\left( * \right)} } \displaystyle{\sum\limits_{n = 1}^\infty  {\left( { - \log \left( {\frac{{1 + {m^2}}}{{{m^2}}}} \right) + 2\left( {1 - m \cdot \arctan \left( {\frac{1}{m}} \right)} \right)} \right)} }

Όμως \displaystyle{\sum\limits_{n = 1}^\infty  {\log \left( {\frac{{1 + {m^2}}}{{{m^2}}}} \right) = \log \left( {\prod\limits_{n = 1}^\infty  {\frac{{\left( {m + i} \right)\left( {m - i} \right)}}{{{m^2}}}} } \right)}  = } \displaystyle{\log \left( {\mathop {\lim }\limits_{n \to \infty } \left( {\frac{{i \cdot \left( {1 + i} \right) \cdot .. \cdot \left( {m + i} \right)}}{{m!}} \cdot \frac{{ - i \cdot \left( {1 - i} \right) \cdot .. \cdot \left( {m - i} \right)}}{{m!}}} \right)} \right) = }

\displaystyle{ = \log \left( {\mathop {\lim }\limits_{n \to \infty } \left( {\frac{{i \cdot \left( {1 + i} \right) \cdot .. \cdot \left( {m + i} \right)}}{{{{\left( {m + 1} \right)}^i}m!}} \cdot \frac{{ - i \cdot \left( {1 - i} \right) \cdot .. \cdot \left( {m - i} \right)}}{{{{\left( {m + 1} \right)}^{ - i}}m!}}} \right)} \right)\mathop { =  = }\limits^{\left( {4*} \right)} } \displaystyle{ - \log \left( {\Gamma \left( i \right)\Gamma \left( { - i} \right)} \right)\mathop { =  = }\limits^{\left( {3*} \right)}  - \log \frac{{2\pi }}{{{e^\pi } - {e^{ - \pi }}}}}

Επομένως \displaystyle{{\sum\limits_{n = 1}^\infty  {\frac{{{{\left( { - 1} \right)}^n}\zeta \left( {2n} \right)}}{{n\left( {2n + 1} \right)}}}  = \log \frac{{2\pi }}{{{e^\pi } - {e^{ - \pi }}}} + 2\sum\limits_{n = 1}^\infty  {\left( {1 - m \cdot \arctan \left( {\frac{1}{m}} \right)} \right)} }}

Επίσης \displaystyle{\sum\limits_{n = 1}^\infty  {\left( {1 - n \cdot \arctan \left( {\frac{1}{n}} \right)} \right)}  = \sum\limits_{n = 1}^\infty  {n\left( {\frac{1}{n} - \arctan \left( {\frac{1}{n}} \right)} \right)}  = } \displaystyle{\sum\limits_{n = 1}^\infty  {n\left( {\frac{1}{n} - \sum\limits_{k = 0}^\infty  {\frac{{{{\left( { - 1} \right)}^k}}}{{\left( {2k + 1} \right) \cdot {n^{2k + 1}}}}} } \right)}  = \sum\limits_{n = 1}^\infty  {\sum\limits_{k = 1}^\infty  {\frac{{{{\left( { - 1} \right)}^{k - 1}}}}{{\left( {2k + 1} \right) \cdot {n^{2k}}}}} }  = }

\displaystyle{ = \sum\limits_{n = 1}^\infty  {\sum\limits_{k = 1}^\infty  {\frac{{{{\left( { - 1} \right)}^{k - 1}}}}{{{n^{2k}}}}\int\limits_0^1 {{x^{2k}}dx} } }  = \sum\limits_{n = 1}^\infty  {\int\limits_0^1 {\sum\limits_{k = 1}^\infty  {\left( {{{\left( { - 1} \right)}^{k - 1}}{{\left( {\frac{{{x^2}}}{{{n^2}}}} \right)}^k}} \right)dx} } }  = } \displaystyle{\sum\limits_{n = 1}^\infty  {\int\limits_0^1 {\frac{{{x^2}}}{{{x^2} + {n^2}}}dx} }  = \int\limits_0^1 {{x^2}\left( {\sum\limits_{n = 1}^\infty  {\frac{1}{{{x^2} + {n^2}}}} } \right)dx} \mathop { =  =  = }\limits^{\left( {5*} \right)} }

\displaystyle{ = \frac{1}{2}\int\limits_0^1 {\left( { - 1 + \pi x\frac{{{e^{\pi x}} + {e^{ - \pi x}}}}{{{e^{\pi x}} - {e^{ - \pi x}}}}} \right)dx}  =  - \frac{1}{2} + \pi \int\limits_0^1 {x \cdot \frac{{{e^{\pi x}} + {e^{ - \pi x}}}}{{{e^{\pi x}} - {e^{ - \pi x}}}}dx}  = } \displaystyle{ - \frac{1}{2} + \frac{\pi }{2}\int\limits_0^1 {x \cdot \frac{{1 + {e^{ - 2\pi x}}}}{{1 - {e^{ - 2\pi x}}}}dx}  = }

\displaystyle{ =  - \frac{1}{2} + \frac{\pi }{2}\int\limits_0^1 {x \cdot \left( {1 + {e^{ - 2\pi x}}} \right)\sum\limits_{n = 0}^\infty  {{e^{ - 2n\pi x}}} dx}  =  - \frac{1}{2} + \frac{\pi }{2}\int\limits_0^1 {x\sum\limits_{n = 0}^\infty  {{e^{ - 2n\pi x}}} dx} } \displaystyle{ + \frac{\pi }{2}\int\limits_0^1 {x\sum\limits_{n = 0}^\infty  {{e^{ - 2\left( {n + 1} \right)\pi x}}} dx}  = }

\displaystyle{ =  - \frac{1}{2} + \frac{\pi }{2}\sum\limits_{n = 0}^\infty  {\int\limits_0^1 {x \cdot {e^{ - 2n\pi x}}dx} }  + \frac{\pi }{2}\sum\limits_{n = 0}^\infty  {\int\limits_0^1 {x \cdot {e^{ - 2\left( {n + 1} \right)\pi x}}dx} }  = } \displaystyle{ - \frac{1}{2} + \frac{\pi }{4} + \frac{\pi }{2}\sum\limits_{n = 1}^\infty  {\int\limits_0^1 {x \cdot {e^{ - 2n\pi x}}dx} }  + \frac{\pi }{2}\sum\limits_{n = 0}^\infty  {\int\limits_0^1 {x \cdot {e^{ - 2\left( {n + 1} \right)\pi x}}dx} }  \Rightarrow }

\displaystyle{ \Rightarrow \sum\limits_{n = 1}^\infty  {\left( {1 - n \cdot \arctan \left( {\frac{1}{n}} \right)} \right)}  =  - \frac{1}{2} + \frac{\pi }{4} + \pi \sum\limits_{n = 1}^\infty  {\left( { - \frac{{{e^{ - 2n\pi }}}}{{4{\pi ^2}{n^2}}} + \frac{1}{{4{\pi ^2}{n^2}}} - \frac{{{e^{ - 2n\pi }}}}{{2\pi n}}} \right)} } \displaystyle{ =  - \frac{1}{2} + \frac{\pi }{4} - \frac{1}{{4\pi }}\sum\limits_{n = 1}^\infty  {\left( {\frac{{{e^{ - 2n\pi }}}}{{{n^2}}}} \right)}  + }

\displaystyle{ + \frac{1}{{4\pi }}\sum\limits_{n = 1}^\infty  {\left( {\frac{1}{{{n^2}}}} \right)}  - \frac{1}{2}\sum\limits_{n = 1}^\infty  {\left( {\frac{{{e^{ - 2n\pi }}}}{n}} \right)}  =  - \frac{1}{2} + \frac{{7\pi }}{{24}} - \frac{1}{{4\pi }}L{i_2}\left( {{e^{ - 2\pi }}} \right) + \frac{1}{2}\log \left( {1 - {e^{ - 2\pi }}} \right)}

Τότε \displaystyle{\sum\limits_{n = 1}^\infty  {\frac{{{{\left( { - 1} \right)}^n}\zeta \left( {2n} \right)}}{{n\left( {2n + 1} \right)}}}  = \log \frac{{2\pi }}{{{e^\pi } - {e^{ - \pi }}}} + 2\left( { - \frac{1}{2} + \frac{{7\pi }}{{24}} - \frac{1}{{4\pi }}L{i_2}\left( {{e^{ - 2\pi }}} \right) + \frac{1}{2}\log \left( {1 - {e^{ - 2\pi }}} \right)} \right)} \displaystyle{ = \log 2\pi  - 1 - \frac{{5\pi }}{{12}} - \frac{1}{{2\pi }}L{i_2}\left( {{e^{ - 2\pi }}} \right)}

και τελικά \displaystyle{\frac{1}{{2\pi }}L{i_2}\left( {{e^{ - 2\pi }}} \right) = \log \left( {2\pi } \right) - 1 - \frac{{5\pi }}{{12}} - \sum\limits_{n = 1}^\infty  {\frac{{{{\left( { - 1} \right)}^n}\zeta \left( {2n} \right)}}{{n\left( {2n + 1} \right)}}} } :) :)




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