tag:
красивые_формулы_и_тождества
§
[Рамануджан]
\[1 + \frac{1}{{1 \cdot 3}} + \frac{1}{{1 \cdot 3 \cdot 5}} + \frac{1}{{1 \cdot 3 \cdot 5 \cdot 7}} + \frac{1}{{1 \cdot 3 \cdot 5 \cdot 7 \cdot 9}} + \cdot \cdot \cdot + \frac{1}{{1 + \frac{1}{{1 + \frac{2}{{1 + \frac{3}{{1 + \frac{4}{{1 + ...}}}}}}}}}} = \sqrt {\frac{{\pi \cdot e}}{2}} .\]
Примечание
\[\begin{array}{l} S = \sum\limits_{k = 1}^{ + \infty } {\frac{1}{{\left( {2k - 1} \right)!!}}} = \sqrt {\frac{{\pi \cdot e}}{2}} \cdot \left( {1 - \operatorname{erfc} \left( {\frac{{\sqrt 2 }}{2}} \right)} \right) \hfill \\ F = \frac{1}{{1 + \frac{1}{{1 + \frac{2}{{1 + \frac{3}{{1 + ...}}}}}}}} = \sqrt {\frac{{\pi \cdot e}}{2}} \cdot \operatorname{erfc} \left( {\frac{{\sqrt 2 }}{2}} \right) \hfill \\ \operatorname{erfc} x = 1 - \operatorname{erf} x \hfill \\ \operatorname{erf} x = \frac{2}{{\sqrt \pi }}\int\limits_0^x {{e^{ - {t^2}}}dt} \hfill \\ \end{array}\]
1410.
\[\begin{array}{l}
\sqrt[3]{{\cos \frac{{2\pi }}{7}}} + \sqrt[3]{{\cos \frac{{4\pi }}{7}}} + \sqrt[3]{{\cos \frac{{8\pi }}{7}}} = \sqrt[3]{{\frac{{5 - 3\sqrt[3]{7}}}{2}}} \hfill \\
\sqrt[3]{{\cos \frac{{2\pi }}{9}}} + \sqrt[3]{{\cos \frac{{4\pi }}{9}}} + \sqrt[3]{{\cos \frac{{8\pi }}{9}}} = \sqrt[3]{{\frac{{3\sqrt[3]{9} - 6}}{2}}} \hfill \\
\end{array}\]
1728.
\[{\text{Докажите}}{\text{, что }}\int\limits_0^{ + \infty } {\frac{{\sin x}}{{\sinh x}}dx} = \frac{\pi }{2}\tanh \frac{\pi }{2}.\]
1840.
\[{\text{Докажите}}{\text{, что }}\int\limits_0^{\frac{\pi }{2}} {\frac{{dx}}{{{{\sin }^{10}}x + {{\cos }^{10}}x}} = \sqrt 5 \pi .} \]
1858.
\[\int\limits_0^{\frac{\pi }{2}} {\left( {\prod\limits_{k = 1}^8 {\sin \left( {kx} \right)} } \right)dx} = \frac{\pi }{{256}}.\]
1888.
\[\begin{array}{l}
\left[ {{\text{Рамануджан}}} \right] \hfill \\
\sqrt[3]{{\sqrt[3]{2} - 1}} = \sqrt[3]{{\frac{1}{9}}} - \sqrt[3]{{\frac{2}{9}}} + \sqrt[3]{{\frac{4}{9}}} \hfill \\
\end{array}\]
1890.
\[\begin{array}{l}
\left[ {{\text{Рамануджан}}} \right] \hfill \\
\sqrt[4]{{\frac{{3 + 2\sqrt[4]{5}}}{{3 - 2\sqrt[4]{5}}}}} = \frac{{\sqrt[4]{5} + 1}}{{\sqrt[4]{5} - 1}} \hfill \\
\end{array}\]
1904.
\[\begin{array}{l}
{\text{Докажите тождество:}} \hfill \\
\sqrt[{{2^n}}]{{\frac{{2 + \sqrt 2 }}{{2 - \sqrt 2 }}}} + \sqrt[{{2^n}}]{{\frac{{2 - \sqrt 2 }}{{2 + \sqrt 2 }}}} = \underbrace {\sqrt {2 + \sqrt {2 + \sqrt {2 + ... + \sqrt {2 + 2\sqrt 2 } } } } }_{n{\text{ корней}}}. \hfill \\
\end{array}\]
1935.
\[\frac{{{\pi ^2}}}{6} = 1 + \frac{1}{{1 + \frac{1}{{1 + \frac{1}{{\frac{1}{2} + \frac{1}{{\frac{1}{2} + \frac{1}{{\frac{1}{3} + \frac{1}{{\frac{1}{3} + ...}}}}}}}}}}}} = \left[ {1;1,1,\frac{1}{2},\frac{1}{2},\frac{1}{3},\frac{1}{3},\frac{1}{4},\frac{1}{4},\frac{1}{5},\frac{1}{5}...} \right]\]
1948.
\[\begin{array}{l}
{\text{а) }}\sqrt {x + \frac{1}{2}\sqrt {x + \frac{1}{4}\sqrt {x + \frac{1}{8}\sqrt {x + \frac{1}{{16}}\sqrt {x + ...} } } } } = \frac{1}{4} + \sqrt x ; \hfill \\
{\text{б) }}\sqrt {x - \frac{1}{2}\sqrt {x - \frac{1}{4}\sqrt {x - \frac{1}{8}\sqrt {x - \frac{1}{{16}}\sqrt {x - ...} } } } } = - \frac{1}{4} + \sqrt x . \hfill \\
\end{array}\]
1954.
\[\frac{{{\pi ^2}}}{4} = 2 + \frac{1}{{1 + \frac{1}{2} + \frac{1}{{\frac{1}{2} + \frac{1}{3} + \frac{1}{{\frac{1}{3} + \frac{1}{4} + \frac{1}{{\frac{1}{4} + \frac{1}{5} + \frac{1}{{...}}}}}}}}}}\]
1955.
\[\frac{{{\pi ^2}}}{3} = 3 + \frac{1}{{\frac{2}{1} + \frac{2}{2} + \frac{1}{{\frac{2}{2} + \frac{2}{3} + \frac{1}{{\frac{2}{3} + \frac{2}{4} + \frac{1}{{\frac{2}{4} + \frac{2}{5} + \frac{1}{{...}}}}}}}}}}\]
2065.
\[{\text{Докажите}}{\text{, что }}\prod\limits_{k = 1}^{2n} {\Gamma \left( {\frac{k}{{2n + 1}}} \right)} = \frac{{{{\left( {2\pi } \right)}^n}}}{{\sqrt {2n + 1} }}.\]
2126.
\[{\text{Найдите }}x \in \mathbb{N}.\]
$$\frac{1}{{\sqrt[3]{{\cos \frac{{2\pi }}{9}}}}} + \frac{1}{{\sqrt[3]{{\cos \frac{{4\pi }}{9}}}}} + \frac{1}{{\sqrt[3]{{\cos \frac{{8\pi }}{9}}}}} = \sqrt[3]{{x\left( {\sqrt[3]{9} - 1} \right)}}$$