5 ms·

No, we measure angles in radians so that d^4/dx^4 (sin x) = sin x.

by programjames 2y ago

cperciva 2y ago
No, we measure angles in radians so that e^(ix) = cos x + i sin x.
- nine_k 2y ago
  But this does not depend in the unit.
  - deleted 2y ago
    [deleted]
  - JadeNB 2y ago
    > But this does not depend in the unit. It does! e^z, defined as the series \sum_{n = 0}^\infty z^n/n!, can only be a function of a dimensionless number z. sin(z) and cos(z), defined as power series, technically also work this way. And that's OK, because angles are dimensionless: a radian is just C/(2πr), where C is the circumference of a circle of radius r. But it is sometimes convenient to pick your favorite number of radians, like π/180 of them, and call that a degree, and then to say that sin(x degrees) is the same as sin(xπ/180 radians). With this convention, where the left-hand side of e^(ix) = sin(x) + icos(x) is a function of a dimensionless variable, and the right-hand side can be viewed as a function of a dimensioned argument only in the sense written above, it really is the case that the equation written is true, but the equation e^(ix) = sin(x degrees) + icos(x degrees) is false. (On the other hand, you could make the case that e^(ix) is really a function of an angle, where its value is the complex number that lies on the unit circle at that angle. Then you do recover a "dimensioned" version of e^(ix) = sin(x) + i*cos(x) that's valid even if you measure angles in degrees.)
  - kazinator 2y ago
    If we use a degrees version of sin and cos (call them sind and cosd), then we cannot have e on the left side without a conversion factor. (iπx/180) e = cosd x + i sind x ix π/180 -> e = cosd x + i sind x π/180 -> let f = e ix f = cosd x + i sind x Probem is, f doesn't have nice properties like: d x x - f /= f dx There is something uniquely special about the unit circle, and about using the unscaled distance around the unit circle as the measure of the angle.
    - kazinator 2y ago
      Radians have the property that if we step x by some tiny amount δ, then the cos/sin coordinates will move by that same distance around the unit circle: |[cos(x+δ) + i sin(x+δ)] - [cos(x) + i sin(x)]| = δ This is also related to how we can estimate sin(x) = x for small values next to zero, if using radians. In radians, the derivative sin'(x) is cos(x), and cos'(x) is -sin(x). Derviation just shifts the waveform left by ninety degrees. In units other than radians, we get wacky constant terms that change at each step. That's related to how e^x is its own derivative.

No, we measure angles in radians so that d^4/dx^4 (sin x) = sin x.

cperciva 2y ago

No, we measure angles in radians so that e^(ix) = cos x + i sin x.

nine_k 2y ago
But this does not depend in the unit.
- deleted 2y ago
  [deleted]
- JadeNB 2y ago
  > But this does not depend in the unit. It does! e^z, defined as the series \sum_{n = 0}^\infty z^n/n!, can only be a function of a dimensionless number z. sin(z) and cos(z), defined as power series, technically also work this way. And that's OK, because angles are dimensionless: a radian is just C/(2πr), where C is the circumference of a circle of radius r. But it is sometimes convenient to pick your favorite number of radians, like π/180 of them, and call that a degree, and then to say that sin(x degrees) is the same as sin(xπ/180 radians). With this convention, where the left-hand side of e^(ix) = sin(x) + icos(x) is a function of a dimensionless variable, and the right-hand side can be viewed as a function of a dimensioned argument only in the sense written above, it really is the case that the equation written is true, but the equation e^(ix) = sin(x degrees) + icos(x degrees) is false. (On the other hand, you could make the case that e^(ix) is really a function of an angle, where its value is the complex number that lies on the unit circle at that angle. Then you do recover a "dimensioned" version of e^(ix) = sin(x) + i*cos(x) that's valid even if you measure angles in degrees.)
- kazinator 2y ago
  If we use a degrees version of sin and cos (call them sind and cosd), then we cannot have e on the left side without a conversion factor. (iπx/180) e = cosd x + i sind x ix π/180 -> e = cosd x + i sind x π/180 -> let f = e ix f = cosd x + i sind x Probem is, f doesn't have nice properties like: d x x - f /= f dx There is something uniquely special about the unit circle, and about using the unscaled distance around the unit circle as the measure of the angle.
  - kazinator 2y ago
    Radians have the property that if we step x by some tiny amount δ, then the cos/sin coordinates will move by that same distance around the unit circle: |[cos(x+δ) + i sin(x+δ)] - [cos(x) + i sin(x)]| = δ This is also related to how we can estimate sin(x) = x for small values next to zero, if using radians. In radians, the derivative sin'(x) is cos(x), and cos'(x) is -sin(x). Derviation just shifts the waveform left by ninety degrees. In units other than radians, we get wacky constant terms that change at each step. That's related to how e^x is its own derivative.