3 ms·
It's real (see my blog post about it: http://math-blog.com/2010/06/28/forget-pi-here-comes-tau/ http://math-blog.com/2010/06/28/forget-pi-here-comes-tau/).
by acangiano 16y ago
It's real (see my blog post about it: http://math-blog.com/2010/06/28/forget-pi-here-comes-tau/ http://math-blog.com/2010/06/28/forget-pi-here-comes-tau/).
- JadeNB 16y agoNote 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 16y agoI was actually reporting the equality as expressed by the manifesto. See here: http://tauday.com/#sec:euler_s_formula http://tauday.com/#sec:euler_s_formula