Eufisky - The lost book

无穷套根式的一些特殊情形（Nested Radical）

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

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

得

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

参考资料

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