30
$\begingroup$

One out of the almost endless supply of identities discovered by Ramanujan is the following: $$ \sqrt[3]{\rule{0pt}{2ex}\sqrt[3]{2}-1} = \sqrt[3]{\frac19} - \sqrt[3]{\frac29} + \sqrt[3]{\frac49}, $$ which has the following interpretation in algebraic number theory: the fundamental unit $\sqrt[3]{2}-1$ of the pure cubic number field $K = {\mathbb Q}(\sqrt[3]{2})$ becomes a cube in the extension $L = K(\sqrt[3]{3})$.

Are there more examples of this kind in Ramanujan's work?

$\endgroup$
1
  • 2
    $\begingroup$ Dear Matt: $2^{1/3} - 1$ is a fundamental unit, so probably that is what Franz meant to write. $\endgroup$
    – BCnrd
    Oct 24, 2010 at 16:53

1 Answer 1

17
$\begingroup$

$$(7 \sqrt[3]{20} - 19)^{1/6} = \ \sqrt[3]{\frac{5}{3}} - \sqrt[3]{\frac{2}{3}},$$ $$\left( \frac{3 + 2 \sqrt[4]{5}}{3 - 2 \sqrt[4]{5}} \right)^{1/4}= \ \ \frac{\sqrt[4]{5} + 1}{\sqrt[4]{5} - 1},$$ $$\left(\sqrt[5]{\frac{1}{5}} + \sqrt[5]{\frac{4}{5}}\right)^{1/2} = \ \ (1 + \sqrt[5]{2} + \sqrt[5]{8})^{1/5} = \ \ \sqrt[5]{\frac{16}{125}} + \sqrt[5]{\frac{8}{125}} + \sqrt[5]{\frac{2}{125}} - \sqrt[5]{\frac{1}{125}},$$ and so on. Many of these were submitted by Ramanujan as problems to the Journal of the Indian Mathematical Society. See the following link: jims.ps for more precise references. Quote: "although Ramanujan never used the term unit, and probably did not formally know what a unit was, he evidently realized their fundamental properties. He then recognized that taking certain powers of units often led to elegant identities."

$\endgroup$
3
  • 1
    $\begingroup$ I would have gone with 'enery $$ $$ en.wikipedia.org/wiki/I'm_Henery_the_Eighth,_I_Am $\endgroup$
    – Will Jagy
    Oct 25, 2010 at 1:05
  • $\begingroup$ Not a real link, in an answer rather than a comment, even if the wikipedia link has extra punctuation, one can click on the hyperlink symbol and convince MO to make the entire character string into a genuine link. $\endgroup$
    – Will Jagy
    Oct 25, 2010 at 1:08
  • $\begingroup$ Excellent. I had no idea the song went back to 1910. $\endgroup$
    – Will Jagy
    Oct 25, 2010 at 1:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.