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- Noether
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[tex]\sqrt[2]{ \pi\sqrt[3]{ \pi\sqrt[4]{ \pi\sqrt[5]{\pi...}} } }[/tex]
Morsom sak.
La [tex]P = \sqrt[2]{ \pi\sqrt[3]{ \pi\sqrt[4]{ \pi\sqrt[5]{\pi...}}}}[/tex]
Da er
[tex]\ln(P) = \ln \left( \sqrt[2]{ \pi\sqrt[3]{ \pi\sqrt[4]{ \pi\sqrt[5]{\pi...}}}} \right) = \frac 1 2 \left( \ln( \pi) + \frac 1 3 \left(\ln(\pi) + \frac 1 4 \left( \ln(\pi) + \frac 1 5 \left( ... \right) \right) \right) \right) = \left( \sum _{n=2} ^\infty \frac{1}{n!}\right) \ln(\pi) \\ = (e-2) \ln(\pi)[/tex]
Dermed får vi at [tex]P = e^{(e-2) \ln(\pi)} = \pi ^{e-2}[/tex]
La [tex]P = \sqrt[2]{ \pi\sqrt[3]{ \pi\sqrt[4]{ \pi\sqrt[5]{\pi...}}}}[/tex]
Da er
[tex]\ln(P) = \ln \left( \sqrt[2]{ \pi\sqrt[3]{ \pi\sqrt[4]{ \pi\sqrt[5]{\pi...}}}} \right) = \frac 1 2 \left( \ln( \pi) + \frac 1 3 \left(\ln(\pi) + \frac 1 4 \left( \ln(\pi) + \frac 1 5 \left( ... \right) \right) \right) \right) = \left( \sum _{n=2} ^\infty \frac{1}{n!}\right) \ln(\pi) \\ = (e-2) \ln(\pi)[/tex]
Dermed får vi at [tex]P = e^{(e-2) \ln(\pi)} = \pi ^{e-2}[/tex]