ラマヌジャンの初等的恒等式観賞会

$\sqrt{1+2\sqrt{1+3\sqrt{1+\cdots}}}=3.$

$\sqrt{6+2\sqrt{7+3\sqrt{8+\cdots}}}=4.$

$\displaystyle \sqrt[3]{\cos \frac{2}{7}\pi}+\sqrt[3]{\cos \frac{4}{7}\pi}+\sqrt[3]{\cos \frac{8}{7}\pi} = \sqrt[3]{\frac{5-3\sqrt[3]{7}}{2}}.$

$\displaystyle \sqrt[3]{\cos \frac{2}{9}\pi}+\sqrt[3]{\cos \frac{4}{9}\pi}+\sqrt[3]{\cos \frac{8}{9}\pi} = \sqrt[3]{\frac{3\sqrt[3]{9}-6}{2}}.$

$\sqrt[3]{3(\sqrt[3]{a^3+b^3}-a)(\sqrt[3]{a^3+b^3}-b)} = \sqrt[3]{(a+b)^2}-\sqrt[3]{a^2-ab+b^2}.$

$\displaystyle \sqrt{\sqrt[5]{\frac{1}{5}}+\sqrt[5]{\frac{4}{5}}} = \sqrt[5]{1+\sqrt[5]{2}+\sqrt[5]{8}} = \sqrt[5]{\frac{16}{125}}+\sqrt[5]{\frac{8}{125}}+\sqrt[5]{\frac{2}{125}}-\sqrt[5]{\frac{1}{125}}.$

$\displaystyle \sqrt[3]{\sqrt[5]{\frac{32}{5}}-\sqrt[5]{\frac{27}{5}}} = \sqrt[5]{\frac{1}{25}}+\sqrt[5]{\frac{3}{25}}-\sqrt[5]{\frac{9}{25}}.$

$\displaystyle \sqrt[4]{\frac{3+2\sqrt[4]{5}}{3-2\sqrt[4]{5}}} = \frac{\sqrt[4]{5}+1}{\sqrt[4]{5}-1}.$

$\sqrt{8-\sqrt{8+\sqrt{8-\cdots}}} = 1+2\sqrt{3}\sin 20^{\circ}.$

$\sqrt{11-2\sqrt{11+2\sqrt{11-\cdots}}} = 1+4\sin 10^{\circ}.$

$\sqrt{23-2\sqrt{23+2\sqrt{23+2\sqrt{23-\cdots}}}}=1+4\sqrt{3}\sin 20^{\circ}.$

$\displaystyle \sqrt[3]{\sqrt[3]{2}-1} = \sqrt[3]{\frac{1}{9}}-\sqrt[3]{\frac{2}{9}}+\sqrt[3]{\frac{4}{9}}.$

$\displaystyle \sqrt{\sqrt[3]{5}-\sqrt[3]{4}} = \frac{\sqrt[3]{2}+\sqrt[3]{20}-\sqrt[3]{25}}{3}.$

$\displaystyle \sqrt{\sqrt[3]{28}-\sqrt[3]{27}} = \frac{\sqrt[3]{98}-\sqrt[3]{28}-1}{3}.$

$\displaystyle \sqrt[6]{7\sqrt[3]{20}-19} = \sqrt[3]{\frac{5}{3}}-\sqrt[3]{\frac{2}{3}}.$

$\displaystyle \sqrt[8]{4\sqrt[3]{\frac{2}{3}}-5\sqrt[3]{\frac{1}{3}}} = \sqrt[3]{\frac{4}{9}}-\sqrt[3]{\frac{2}{9}}+\sqrt[3]{\frac{1}{9}}.$

$(6a^2-4ab+4b^2)^3=(3a^2+5ab-5b^2)^3+(4a^2-4ab+6b^2)^3+(5a^2-5ab-3b^2)^3.$

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数、特に整数に関する記事。

ラマヌジャンの初等的恒等式観賞会