Note that your equality \(e^{i\tau} = 1\) implies only that \(e^{i\pi} = \pm1\),... | Hacker News

Hacker Newsnew | past | comments | ask | show | jobs | submit

		JadeNB on June 28, 2010 \| parent \| context \| favorite \| on: No, really, pi is wrong: The Tau Manifesto (Tau Da... Note that your equality \(e^{i\tau} = 1\) implies only that \(e^{i\pi} = \pm1\), hence is strictly weaker than Euler's equality (which picks a sign). I think that you might have meant \(e^{i\tau/2} = -1\). (Is there a way to do LaTeX properly here?)

acangiano on June 28, 2010 [–]

I was actually reporting the equality as expressed by the manifesto. See here: http://tauday.com/#sec:euler_s_formula

Guidelines | FAQ | Lists | API | Security | Legal | Apply to YC | Contact