4 ms·

Using tau would uglify euler's equation, wouldn't it?

kqr 10y ago
You be the judge of that. Compare e^(iπ) = -1 to e^(iτ) = 1. The whole business of re-writing the identity as e^(iπ) + 1 = 0 is nothing but a hack to get around the weirdness of π as a constant.
- kej 10y ago
  Writing it equal to 0 isn't a hack, it's a common method of understanding a function. You factor polynomials by setting them equal to 0, for example. In the case of Euler's identity, what we're really asking is "what values of x make e^(ix) + 1 = 0 true?" and the answer is "every multiple of π". Using τ instead hides half of the answers.
  - kazinator 10y ago
    You mean "every multiple of 2π". The value -1 only comes around once per revolution of the unit circle!
  - kazinator 10y ago
    You mean "every multiple of 2π, shifted by π". The value -1 only comes around once per revolution of the unit circle! The solutions to e^(ix) + 1 = 0 are { π, 3π, 5π, ... } Whereas the solutions to e^(ix) - 1 = 0 are: { 0, 2π, 4π, 3π, ... } i.e. { 0, τ, 2τ, 3τ, ... } Also, note that when we set a polynomial to zero, the roots appear subtracted on the opposite side from the independent variable: (x - r0)(x - r1)...(x - rn) = 0 In the Tau-oriented Euler formula written homogeneously, there is a vague analogy to this since we're similarly subtracting that 1: e^(ix) - 1 = 0
    - kej 10y ago
      Poor phrasing and math-before-coffee on my part. The point I was going for is that e^(ix) has real values for each multiple of π. That is mathematically interesting, and is obscured if you use τ instead.
      - kazinator 10y ago
        Yes; similarly, that a sinusoidal crosses zero twice in each period is obscured by a focus on the period itself.
Steuard 10y ago
e^(i tau) - 1 = 0 Seems pretty comparable to me: this one elegantly demonstrates the full periodicity of complex exponentials, while the original elegantly demonstrates how complex exponentials can produce pure negative numbers.

kqr 10y ago

You be the judge of that. Compare e^(iπ) = -1 to e^(iτ) = 1. The whole business of re-writing the identity as e^(iπ) + 1 = 0 is nothing but a hack to get around the weirdness of π as a constant.

kej 10y ago
Writing it equal to 0 isn't a hack, it's a common method of understanding a function. You factor polynomials by setting them equal to 0, for example. In the case of Euler's identity, what we're really asking is "what values of x make e^(ix) + 1 = 0 true?" and the answer is "every multiple of π". Using τ instead hides half of the answers.
- kazinator 10y ago
  You mean "every multiple of 2π". The value -1 only comes around once per revolution of the unit circle!
- kazinator 10y ago
  You mean "every multiple of 2π, shifted by π". The value -1 only comes around once per revolution of the unit circle! The solutions to e^(ix) + 1 = 0 are { π, 3π, 5π, ... } Whereas the solutions to e^(ix) - 1 = 0 are: { 0, 2π, 4π, 3π, ... } i.e. { 0, τ, 2τ, 3τ, ... } Also, note that when we set a polynomial to zero, the roots appear subtracted on the opposite side from the independent variable: (x - r0)(x - r1)...(x - rn) = 0 In the Tau-oriented Euler formula written homogeneously, there is a vague analogy to this since we're similarly subtracting that 1: e^(ix) - 1 = 0
  - kej 10y ago
    Poor phrasing and math-before-coffee on my part. The point I was going for is that e^(ix) has real values for each multiple of π. That is mathematically interesting, and is obscured if you use τ instead.
    - kazinator 10y ago
      Yes; similarly, that a sinusoidal crosses zero twice in each period is obscured by a focus on the period itself.

Steuard 10y ago

e^(i tau) - 1 = 0 Seems pretty comparable to me: this one elegantly demonstrates the full periodicity of complex exponentials, while the original elegantly demonstrates how complex exponentials can produce pure negative numbers.