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Obliczyć wartość iloczynu \(\displaystyle{ \prod_{k=4}^{\infty} \left( 1 - \left( \frac{3}{k} \right)^3\right) . }\)

\(\displaystyle{ \prod_{n=4}^{ \infty } \left( 1- \frac{27}{n^3} \right) = \prod_{n=4}^{ \infty } \left( 1- \frac{3}{n} \right) \left( 1- \frac{\omega_{1}}{n} \right) \left( 1- \frac{\omega_{2}}{n} \right) }\)

gdzie trójka i te omegi to pierwiastki zespolone z \(\displaystyle{ 27}\)

\(\displaystyle{ \omega_{1}=- \frac{3}{2} \left( 1+i \sqrt{3} \right) }\)

\(\displaystyle{ \omega_{2}=- \frac{3}{2} \left( 1-i \sqrt{3} \right) }\)


tu trzeba wytoczyć cięższe działa...

\(\displaystyle{ \prod_{n=4}^{ \infty }\left( 1- \frac{3}{n} \right) \left( 1- \frac{\omega_{1}}{n} \right) \left( 1- \frac{\omega_{2}}{n} \right)=\prod_{n=4}^{ \infty } \left( 1- \frac{3}{n} \right) \left( 1- \frac{\omega_{1}}{n} \right) \left( 1- \frac{\omega_{2}}{n} \right) \cdot e^{ \frac{3}{n} } \cdot e^{ \frac{\omega_{1}}{n} } \cdot e^{ \frac{\omega_{2}}{n} }= }\)

bo:

\(\displaystyle{ 3+\omega_{1}+\omega_{2}=0}\)


\(\displaystyle{ \prod_{n=4}^{ \infty } \left[ \left( 1- \frac{\omega_{1}}{n} \right) \cdot e^{ \frac{\omega_{1}}{n} } \right] \cdot \prod_{n=4}^{ \infty } \left[ \left( 1- \frac{\omega_{2}}{n} \right) \cdot e^{ \frac{\omega_{2}}{n} } \right] =}\)

\(\displaystyle{ \frac{\prod_{n=1}^{ \infty } \left[ \left( 1- \frac{\omega_{1}}{n} \right) \cdot e^{ \frac{\omega_{1}}{n} } \right] \cdot \prod_{n=1}^{ \infty } \left[ \left( 1- \frac{\omega_{2}}{n} \right) \cdot e^{ \frac{\omega_{2}}{n} } \right] }{e^{ \frac{\omega_{1}}{1} } \cdot e^{ \frac{\omega_{1}}{2} } \cdot e^{ \frac{\omega_{1}}{3} } \cdot e^{ \frac{\omega_{2}}{1} } \cdot e^{ \frac{\omega_{2}}{2} } \cdot e^{ \frac{\omega_{2}}{3} } \cdot \left( 1- \frac{\omega_{1}}{1} \right) \cdot \left( 1- \frac{\omega_{1}}{2} \right)\cdot \left( 1- \frac{\omega_{1}}{3} \right)\cdot \left( 1- \frac{\omega_{2}}{1} \right)\cdot \left( 1- \frac{\omega_{2}}{2} \right)\cdot \left( 1- \frac{\omega_{2}}{3} \right)}}\)

\(\displaystyle{ \omega_{1}+\omega_{2}=-3}\)

\(\displaystyle{ \omega_{1} \cdot \omega_{2}=9}\)

więc cały ten dół się uprości do:

\(\displaystyle{ e^{- \frac{11}{2} } \cdot \frac{741}{4} }\)

teraz zastosujemy wzór Weierstrassa dla iloczynów nieskończonych:

\(\displaystyle{ \frac{1}{\Gamma(-z)} =- \frac{z}{e^{\gamma z}} \prod_{n=1}^{ \infty } \left( 1- \frac{z}{n} \right) e^{ \frac{z}{n} }}\)

lub:

\(\displaystyle{ \prod_{n=1}^{ \infty } \left( 1- \frac{z}{n} \right) e^{ \frac{z}{n} }=- \frac{e^{\gamma z}}{z\Gamma(-z)} }\)

teraz za z podstawmy obie omegi i otrzymamy:

\(\displaystyle{ \prod_{n=1}^{ \infty } \left( 1- \frac{\omega_{1}}{n} \right) e^{ \frac{\omega_{1}}{n} }=- \frac{e^{\gamma \cdot \omega_{1}}}{\omega_{1}\Gamma(-\omega_{1})} }\)

\(\displaystyle{ \prod_{n=1}^{ \infty } \left( 1- \frac{\omega_{2}}{n} \right) e^{ \frac{\omega_{2}}{n} }=- \frac{e^{\gamma \cdot \omega_{2}}}{\omega_{2}\Gamma(-\omega_{2})} }\)



po wymnożeniu otrzymamy:

\(\displaystyle{ \frac{e^{\gamma \cdot \omega_{1}}}{\omega_{1}\Gamma(-\omega_{1})} \cdot \frac{e^{\gamma \cdot \omega_{2}}}{\omega_{2}\Gamma(-\omega_{2})}= \frac{e^{-3\gamma}}{9\Gamma\left( \frac{3}{2} +i \frac{3 \sqrt{3} }{2} \right) \cdot \Gamma\left( \frac{3}{2} -i \frac{3 \sqrt{3} }{2} \right) } = }\)

korzystając ze wzoru:

\(\displaystyle{ \Gamma\left( 1+ \frac{1}{2} +i \frac{3 \sqrt{3} }{2} \right)=\left( \frac{1}{2} +i \frac{3 \sqrt{3} }{2} \right) \cdot \Gamma\left( \frac{1}{2} +i \frac{3 \sqrt{3} }{2} \right)}\)

\(\displaystyle{ \Gamma\left( 1+ \frac{1}{2} -i \frac{3 \sqrt{3} }{2} \right)=\left( \frac{1}{2} -i \frac{3 \sqrt{3} }{2} \right) \cdot \Gamma\left( \frac{1}{2} -i \frac{3 \sqrt{3} }{2} \right)}\)

wymnażając to wszystko otrzymamy:

\(\displaystyle{ 7 \cdot \Gamma\left( \frac{1}{2} +i \frac{3 \sqrt{3} }{2} \right) \cdot \Gamma\left( \frac{1}{2} -i \frac{3 \sqrt{3} }{2} \right)= \frac{7\pi}{\sin\pi\left( \frac{1}{2}+i \frac{3 \sqrt{3} }{2} \right) }= \frac{7\pi}{\cosh \frac{3 \sqrt{3} \pi }{2} } }\)

co da nam:

\(\displaystyle{ \prod_{n=4}^{ \infty } \left( 1- \frac{\omega_{1}}{n} \right)e^{\frac{\omega_{1}}{n}} \left( 1- \frac{\omega_{1}}{n} \right)e^{\frac{\omega_{2}}{n}}= \frac{e^{-3\gamma+5,5} \cdot \cosh \frac{3 \sqrt{3}\pi }{2}}{63\pi \cdot \frac{741}{4} } }\)


zostaje nam do policzenia:

\(\displaystyle{ \prod_{n=4}^{ \infty } \left( 1- \frac{3}{n} \right)e^{ \frac{3}{n} } }\)

W tym wypadku wzór Weierstrassa za bardzo nie zadziała więc spróbujmy inaczej:

(*) \(\displaystyle{ \prod_{n=4}^{ \infty } \left( 1- \frac{3}{n} \right)e^{ \frac{3}{n} }=e^{\lim_{N \to\infty} \sum_{n=4}^{N}\left[ \ln\left( 1- \frac{3}{n} \right) + \frac{3}{n} \right] }=e^{\lim_{N\to\infty} \left[ \sum_{n=4}^{N}\ln\left( 1- \frac{3}{n} \right) + 3\sum_{n=4}^{N}\frac{1}{n} \right] }}\)

pobawmy się samą górą:

\(\displaystyle{ \sum_{n=4}^{N}\ln\left( 1- \frac{3}{n} \right) + 3\sum_{n=4}^{N}\frac{1}{n}=\ln\left( 6\left( N-3)!\right) \right)-\ln N! +3\left( \sum_{n=1}^{N} \frac{1}{n}-1- \frac{1}{2}- \frac{1}{3} \right) }\)

po podstawieniu do (*) otrzymamy:

\(\displaystyle{ \lim_{N\to\infty} e^{\ln 6} e^{-4}e^{- \frac{3}{2} } e^{ \sum_{n=1}^{N}\left( \frac{1}{n} -\ln N\right) } e^{ \sum_{n=1}^{N}\left( \frac{1}{n} -\ln (N-1)\right) }e^{ \sum_{n=1}^{N}\left( \frac{1}{n} -\ln (N-2)\right) }=}\)

\(\displaystyle{ e^{\ln 6} e^{-4}e^{- \frac{3}{2} } \lim_{N\to\infty} e^{ \sum_{n=1}^{N}\left( \frac{1}{n} -\ln N\right) }e^{ \sum_{n=1}^{N-1}\left( \frac{1}{n} -\ln (N-1)\right) }e^{ \sum_{n=1}^{N-2}\left( \frac{1}{n} -\ln (N-2)\right) }e^{ \frac{1}{N} +\frac{1}{N-1}+\frac{1}{N} }}\)

Biorąc pod uwagę, że:

\(\displaystyle{ e^{ \sum_{n=1}^{N}\left( \frac{1}{n} -\ln N\right) } \rightarrow e^{\gamma}}\)

\(\displaystyle{ e^{ \frac{1}{N} +\frac{1}{N-1}+\frac{1}{N} } \rightarrow 1}\)

otrzymamy:

\(\displaystyle{ \prod_{n=4}^{ \infty } \left( 1- \frac{3}{n} \right)e^{ \frac{3}{n} } =6e^{3\gamma-5,5}}\)

ostatecznie otrzymamy:

\(\displaystyle{ \prod_{n=4}^{ \infty } \left( 1- \frac{27}{n^3} \right) = \frac{4e^{5,5-3\gamma}\cosh \frac{3 \sqrt{3}\pi }{2} \cdot 6e^{-5,5+3\gamma}}{46683\pi} = \frac{8\cosh \frac{3 \sqrt{3}\pi }{2}}{15561\pi} }\)

...

gdzie:

\(\displaystyle{ \gamma=\lim_{n \to \infty}\left[ H_{n}-\ln(n+1)\right] }\)