5 ms·

I disagree, I think it's very much essential to the point. For one, 10^(i*pi) is some horrible complex number, which is what you'd expect when exponentiating r

by ilyagr 7y ago

I disagree, I think it's very much essential to the point. For one, 10^(i*pi) is some horrible complex number, which is what you'd expect when exponentiating random complex numbers. Exercise: figure out what it is from the rotation point of view. Secondly, usual definitions of e don't involve complex numbers at all. Nor do questions such as whether e is rational, algebraic, etc. Thirdly, if you go with this argument too far, you'll conclude that the value of pi is also not important. After all, circles must have some length, who cares what it is precisely?

jacobolus 7y ago
> 10^(iπ) is some horrible complex number Sure, you want to use a convenient coordinate system. But you already have 2 horrible numbers in the conventional coordinate system: the π and e. In the conventional (“natural”) coordinate system for logarithms, the coordinates represent powers of e and rotations by 1/2π turns, respectively. You have just decided that you’ll use those specific horrible numbers in the logarithmic coordinates, because e and π are what you get out if you choose a coordinate system for the logarithm of quotients-of-vectors where the derivative of log(x) at the point x = 1 is 1. This turns out to be convenient for simplifying differential equations and related formulas. Personally I like using a coordinate system where the “scale” axis of the logarithm uses units of doublings, and the “rotation” axis of the logarithm uses units of full turns. If I use that one when making a picture, I can easily tell you exactly what the value is (as a rational number) at every grid intersection. e.g. here is a plot of a Möbius transformation, https://raw.githubusercontent.com/jrus/images-for-observable/master/v.-*(z%2B1)%7C(z-1).b.jpg https://raw.githubusercontent.com/jrus/images-for-observable... Every contour at the most prominent level is either a doubling or 1/12 of a turn. (You’ll notice that when we zoom on any part of the picture, we see little rectangles rather than little squares; the “natural” coordinate system is a square one, which is also often convenient.)

by ilyagr 7y ago

jacobolus 7y ago

> 10^(iπ) is some horrible complex number Sure, you want to use a convenient coordinate system. But you already have 2 horrible numbers in the conventional coordinate system: the π and e. In the conventional (“natural”) coordinate system for logarithms, the coordinates represent powers of e and rotations by 1/2π turns, respectively. You have just decided that you’ll use those specific horrible numbers in the logarithmic coordinates, because e and π are what you get out if you choose a coordinate system for the logarithm of quotients-of-vectors where the derivative of log(x) at the point x = 1 is 1. This turns out to be convenient for simplifying differential equations and related formulas. Personally I like using a coordinate system where the “scale” axis of the logarithm uses units of doublings, and the “rotation” axis of the logarithm uses units of full turns. If I use that one when making a picture, I can easily tell you exactly what the value is (as a rational number) at every grid intersection. e.g. here is a plot of a Möbius transformation, https://raw.githubusercontent.com/jrus/images-for-observable/master/v.-*(z%2B1)%7C(z-1).b.jpg https://raw.githubusercontent.com/jrus/images-for-observable... Every contour at the most prominent level is either a doubling or 1/12 of a turn. (You’ll notice that when we zoom on any part of the picture, we see little rectangles rather than little squares; the “natural” coordinate system is a square one, which is also often convenient.)