4 ms·
e^(tau*i/2)+1=0 ??? Ewwww, yuck.
by drblast 16y ago
e^(tau*i/2)+1=0 ???
Ewwww, yuck.
- dgritsko 16y agoDid you get all the way down to here? http://tauday.com/#sec:euler_s_formula http://tauday.com/#sec:euler_s_formula e^(i*tau) = 1 Much nicer that way.
- rudd 16y agoI actually disagree that that is nicer. e^(ipi) = -1, which means you can square both sides to come to the also factual statement, e^(itau) = 1. However, if you only knew the latter you'd be wondering whether e^(i*pi) would be 1 or -1.
- dgritsko 16y agoBy "much nicer", I was merely attempting to point out that drblast's formula: e^(tau*i/2)+1=0 could be stated as: e^(i*tau) = 1 As in the article. Whether that's intrinsically "better" than Euler's Identity is a different question, but not what I was discussing here.
- pyrtsa 16y agoI actually disagree with that logic as well. If you just knew that e^(i * pi) = -1, then how would you know, whether halving that exponent would yield +i or -i? Considering that, we'd need to state Euler's identity as: e^(i * pi/2) = i, but again we'd know nothing about fractions of that exponent. And there we go! The point is, Euler's identity is a nice property of the definition of complex numbers, but in itself, not too generic. That's why OP's form, e^(i*tau) = 1, would do just fine. Edit: fixed asterisks.
- deleted 16y ago[deleted]
- gecko 16y agoI thought he addressed that in the paper rather well. Did you disagree with that section of the paper, or were you ignoring it?
- JeremyBanks 16y agoThere's a section about that: http://tauday.com/#sec:euler_s_formula http://tauday.com/#sec:euler_s_formula e^(tau * i) = 1 e^(pi * i) = -1 Which looks better?