无穷套根式的一些特殊情形(Nested Radical) - Eufisky - The lost book
$\sum{\frac{n}{{{e^{2\pi n}} - 1}}}$型的级数求解
Euler积分的一个延伸

无穷套根式的一些特殊情形(Nested Radical)

Eufisky posted @ 2014年7月25日 02:45 in 数学分析 with tags 无穷套根式 Ramanujan , 1808 阅读

情形一:(Vieta)\[\frac{2}{\pi } = \sqrt {\frac{1}{2}} \sqrt {\frac{1}{2} + \frac{1}{2}\sqrt {\frac{1}{2}} } \sqrt {\frac{1}{2} + \frac{1}{2}\sqrt {\frac{1}{2} + \frac{1}{2}\sqrt {\frac{1}{2}} } }  \cdots .\]

情形二:\[x = \sqrt[n]{{\left( {1 - q} \right){x^n} + q{x^{n - 1}}\sqrt[n]{{\left( {1 - q} \right){x^n} + q{x^{n - 1}}\sqrt[n]{ \cdots }}}}}.\]

有以下几种特殊情形:

\begin{align*}\frac{{b + \sqrt {{b^2} + 4a} }}{2} &= \sqrt {a + b\sqrt {a + b\sqrt {a + b\sqrt  \cdots  } } } \left( {n = 2,q = 1 - \frac{a}{{{x^2}}},x = \frac{b}{q}} \right)\\x &= \sqrt[n]{{{x^{n - 1}}\sqrt[n]{{{x^{n - 1}}\sqrt[n]{{{x^{n - 1}}\sqrt[n]{ \cdots }}}}}}}\left( {q = 1} \right)\\x &= \sqrt {x\sqrt {x\sqrt {x\sqrt {x\sqrt  \cdots  } } } } \left( {q = 1,n = 2} \right).\end{align*}
情形二:由以上情形可推得:\[{q^{\frac{{{n^k} - 1}}{{n - 1}}}}{x^{{n^j}}} = \sqrt[n]{{{q^{\frac{{{n^{k + 1}} - n}}{{n - 1}}}}\left( {1 - q} \right){x^{{n^{j + 1}}}} + \sqrt[n]{{{q^{\frac{{{n^{k + 2}} - n}}{{n - 1}}}}\left( {1 - q} \right){x^{{n^{j + 2}}}} + \sqrt[n]{ \cdots }}}}}.\]
特殊情形有:
\[\sqrt 2  = \sqrt {\frac{2}{{{2^{{2^0}}}}} + \sqrt {\frac{2}{{{2^{{2^1}}}}} + \sqrt {\frac{2}{{{2^{{2^2}}}}} + \sqrt {\frac{2}{{{2^{{2^3}}}}} + \sqrt {\frac{2}{{{2^{{2^4}}}}} +  \cdots } } } } } \left( {q = \frac{1}{2},n = 2,x = 1,k =  - 1} \right).\]
情形三:\[\sqrt[{n - 1}]{x} = \sqrt[n]{{x\sqrt[n]{{x\sqrt[n]{{x \cdots }}}}}}.\]
由于
\[\left\{ \begin{array}{l}1 + \frac{1}{n} + \frac{1}{{{n^2}}} +  \cdots  = 1 + \frac{1}{{n - 1}}\\\frac{1}{n} + \frac{1}{{{n^2}}} + \frac{1}{{{n^3}}} +  \cdots  = \frac{1}{{n - 1}}\\\frac{1}{n}\left( {1 + \frac{1}{n}\left( {1 + \frac{1}{n}\left( {1 +  \cdots } \right)} \right)} \right) = \frac{1}{{n - 1}}\end{array} \right.\left( {n \ge 2,n \in {N_ + }} \right).\]
令$n=3$,我们有\[\sqrt x  = \sqrt[3]{{x\sqrt[3]{{x\sqrt[3]{{x \cdots }}}}}}.\]
情形四: (Ramanujan)
\[x + n + a = \sqrt {ax + {{\left( {n + a} \right)}^2} + x\sqrt {a\left( {x + n} \right) + {{\left( {n + a} \right)}^2} + \left( {x + n} \right)\sqrt {a\left( {x + 2n} \right) + {{\left( {n + a} \right)}^2} + \left( {x + 2n} \right)\sqrt  \cdots }}}.\]
特殊地,有\[x + 1 = \sqrt {1 + x\sqrt {1 + \left( {x + 1} \right)\sqrt {1 + \left( {x + 2} \right)\sqrt {1 +  \cdots } } } } \left( {a = 0,n = 1} \right).\]由此有我们熟知的
\[3 = \sqrt {1 + 2\sqrt {1 + 3\sqrt {1 + 4\sqrt {1 + 5\sqrt  \cdots  } } } } \left( {a = 0,n = 1,x = 2} \right).\]
The justification of this process both in general and in the particular example of $\ln\sigma$, where $\sigma$ is Somos's quadratic recurrence constant in given by Vijayaraghavan (in Ramanujan 2000, p. 348).

情形五:

由下面两式

\[\left\{ \begin{array}{l}e = 1 + \frac{1}{{1!}} + \frac{1}{{2!}} + \frac{1}{{3!}} +  \cdots \\e = 1 + 1 + \frac{1}{2}\left( {1 + \frac{1}{3}\left( {1 + \frac{1}{4}\left( {1 + \frac{1}{5}\left( {1 +  \cdots } \right)} \right)}\right)} \right)\end{array} \right.\]

\[{x^{e - 2}} = \sqrt {x\sqrt[3]{{x\sqrt[4]{{x\sqrt[5]{{x \cdots }}}}}}}. \]

参考资料

[1] http://mathworld.wolfram.com/NestedRadical.html#eqn13


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