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Here's another good reason to think in turns: it turns Euler's formula from this Eldritch Terror: e^(ix) = cos(x) + isin(x) into something you can kinda

Here's another good reason to think in turns: it turns Euler's formula from this Eldritch Terror: e^(i*x) = cos(x) + i*sin(x) into something you can kinda understand by staring at the complex plane: -1^(2x) = cost(x) + i*sint(x) Credit to justinpombrio for this: https://news.ycombinator.com/item?id=32986869 https://news.ycombinator.com/item?id=32986869

judofyr 2mo ago
I'm confused. How is this simpler? Is there something in (-1)^(2x) that can easily understood by staring at the complex plane? It seems mostly that you've gotten rid of "e", but one of the goals of Euler's formula IMO is to explain what "e^(i …)" means so I'm not sure how this variant is useful.
- voidmain 2mo ago
  I'll defend i^4x since I like it better. (cost x, sint x) is a point on the unit circle x turns counterclockwise from (1,0). cost x + i sint x is a point in the complex plane x turns counterclockwise from 1. Now look at integer powers of i, a point in the complex plane 1/4 turn from 1: i^0 = 1 (0 turns from 1) i^1 = i (1/4 turn from 1) i^2 = -1 (2/4 turn from 1) i^3 = -i (3/4 turn from 1) and we define complex exponentiation such that, for all real x, i^x = cost (x/4) + i sint (x/4) (x/4 turn from 1)
lefra 2mo ago
Now define exponentiation by a non-integer.
WCSTombs 2mo ago
Sorry but this is pretty bogus. (-1)^x is only well defined when x is an integer. This is generally the case for r^x whenever r isn't a positive real number. For example, when x = 0.5, r has two distinct square roots. Sure, you can choose one of them arbitrarily and declare it to be the value of r^0.5 (and math libraries typically do this), but there's unfortunately no good way to make this arbitrary choice consistently for all values of r simultaneously.
lioeters 2mo ago
Funny how the so-called eldritch terror is another face of what is widely considered one of the most beautiful equations in mathematics, Euler's identity that unites five fundamental constants. e^(i*pi)+1 = 0 ..which is a result of the more general formula. e^(i*x) = cos(x) + i*sin(x) Pi is hiding there in the sin and cos functions implicitly, because the unit radian is defined by 2*pi. In comparison, the version you mentioned that takes x in "turns". -1^(2x) = cost(x) + i*sint(x) It got rid of pi and e, which already seems a win for simplicity. i is still there for the imaginary component, or y in the complex plane. So the need for pi was removed thanks to the "turn", defined by 1 as the whole circle or cycle. Multiplying -1 to itself every half turn makes it an alternating series of 1 and -1.. Weird, but it is visually clear to understand, without involving e. Though I still don't see where e went. Oh, this comment explains: > If we rearrange the products in the exponent we get 2πix πi2x ( πi ) 2x e -> e -> (e ) > Where e^(πi) is -1. That shows there is something to the turns units; we can express the analog of the Euler identity using exponentiation using a base and factor which are integers. Yeah I get it now, a "turn" acts like a dimensionless unit to the circle/cycle.
voidmain 2mo ago
How about i^4x? It makes it clearer which direction we are rotating (vs -1 which is a 0.5 turn rotation in either direction) and avoids the garden path confusion that could arise from it not being obvious from the left side of the equation that we are working in the complex plane at all.

Here's another good reason to think in turns: it turns Euler's formula from this Eldritch Terror: e^(i*x) = cos(x) + i*sin(x) into something you can kinda

judofyr 2mo ago

I'm confused. How is this simpler? Is there something in (-1)^(2x) that can easily understood by staring at the complex plane? It seems mostly that you've gotten rid of "e", but one of the goals of Euler's formula IMO is to explain what "e^(i …)" means so I'm not sure how this variant is useful.

voidmain 2mo ago
I'll defend i^4x since I like it better. (cost x, sint x) is a point on the unit circle x turns counterclockwise from (1,0). cost x + i sint x is a point in the complex plane x turns counterclockwise from 1. Now look at integer powers of i, a point in the complex plane 1/4 turn from 1: i^0 = 1 (0 turns from 1) i^1 = i (1/4 turn from 1) i^2 = -1 (2/4 turn from 1) i^3 = -i (3/4 turn from 1) and we define complex exponentiation such that, for all real x, i^x = cost (x/4) + i sint (x/4) (x/4 turn from 1)

lefra 2mo ago

Now define exponentiation by a non-integer.

WCSTombs 2mo ago

Sorry but this is pretty bogus. (-1)^x is only well defined when x is an integer. This is generally the case for r^x whenever r isn't a positive real number. For example, when x = 0.5, r has two distinct square roots. Sure, you can choose one of them arbitrarily and declare it to be the value of r^0.5 (and math libraries typically do this), but there's unfortunately no good way to make this arbitrary choice consistently for all values of r simultaneously.

lioeters 2mo ago

Funny how the so-called eldritch terror is another face of what is widely considered one of the most beautiful equations in mathematics, Euler's identity that unites five fundamental constants. e^(i*pi)+1 = 0 ..which is a result of the more general formula. e^(i*x) = cos(x) + i*sin(x) Pi is hiding there in the sin and cos functions implicitly, because the unit radian is defined by 2*pi. In comparison, the version you mentioned that takes x in "turns". -1^(2x) = cost(x) + i*sint(x) It got rid of pi and e, which already seems a win for simplicity. i is still there for the imaginary component, or y in the complex plane. So the need for pi was removed thanks to the "turn", defined by 1 as the whole circle or cycle. Multiplying -1 to itself every half turn makes it an alternating series of 1 and -1.. Weird, but it is visually clear to understand, without involving e. Though I still don't see where e went. Oh, this comment explains: > If we rearrange the products in the exponent we get 2πix πi2x ( πi ) 2x e -> e -> (e ) > Where e^(πi) is -1. That shows there is something to the turns units; we can express the analog of the Euler identity using exponentiation using a base and factor which are integers. Yeah I get it now, a "turn" acts like a dimensionless unit to the circle/cycle.

voidmain 2mo ago

How about i^4x? It makes it clearer which direction we are rotating (vs -1 which is a 0.5 turn rotation in either direction) and avoids the garden path confusion that could arise from it not being obvious from the left side of the equation that we are working in the complex plane at all.