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> You cannot represent an n-dimensional point in 2D Here's a 3d point in 2d with a discrete dimension and a continuous one: (x1, x2, x3) Or if you want it wi
by 6gvONxR4sf7o 6y ago
> You cannot represent an n-dimensional point in 2D
Here's a 3d point in 2d with a discrete dimension and a continuous one:
(x1, x2, x3)
Or if you want it with a continuous horizontal dimension, using dirac deltas makes it trivial, but you wanted it in a finite space, so we can use all sorts of other sets of orthogonal functions. I'll use sinusiods. To represent any 3D point (x1, x2, x3) as a function that takes a finite amount of 2D space, define
g(x1, x2, x3) = f, where f is the following function from R to R:
f(x) = x1 * sin(x) + x2 * sin(2x) + x3 * sin(3x).
So that's representing any point in R^3 in a finite 2D space, with x and y=f(x), just like in your original construction with x=frequency and y=time and z=luminosity=f(frequency, time).
For the reverse, to make it a bijection, make use of the fact that the integral from negative pi to pi of 1/pi sin(m x) sin(n x) is 1 if m == n and 0 otherwise.
h(f) = (
integral 1/pi sin(x) f(x),
integral 1/pi sin(2 x) f(x),
integral 1/pi sin(3 x) f(x)
) = (x1, x2, x3)
g and h are inverses, so we have a bijection. Crucially, the bijection is between R^3 and a very particular set of functions from R to R.
With the set of sin(i x) functions for different i, you can make a bijection between any R^N and a particular set of functions from R to R, so the fact that there's an (x, y=f(x)) representation doesn't matter.
And this why your original construction doesn't work. You didn't make a function from the set of frequency distributions to R^3. You made a function from the set of frequency distributions to the set of functions from R^2 to R. The set of functions from R^2 -> R is an infinite dimensional set, which is a very well established fact. See [0] for a bunch of different explanations of this fact.
[0] https://math.stackexchange.com/questions/2599827/proof-that-vector-space-ca-b-of-all-functions-from-a-b-to-mathbbr-i https://math.stackexchange.com/questions/2599827/proof-that-...
- sudosysgen 6y agoSorry, I didn't word it correctly. I'll reformulate it like this: you can represent any colour spectrum, which is in fact a 1 variable function, by a relationship of two continuous dimensions, such as frequency and amplitude. The example I gave was essentially the Fourier transform of the function; the result of the Fourier transform is encoded as a spectrogram, and the color spectrum over time is the result of the square. If you want to represent any given colour spectrum as a set of real numbers, of course that isn't possible, because a colour spectrum can only be described as either a 2D space or a function. However, that's not my assertion, my assertion is that you only need 3 dimensions in order to describe the human perception of color, and that you do not need an infinite number of dimensions in order to represent a colour spectrum.
- 6gvONxR4sf7o 6y agoSure, I'd agree that our perception is (probably close enough to) 3D, but what does that have to do with what you were writing before? Your arguments for the three dimensionality don't depend on the fact that we're humans talking instead of, say, mantis shrimp whose color perception needs over a dozen dimensional representation.
- sudosysgen 6y agoI was writing before that you don't need an infinite amount of dimensions to represent a color spectrum. Just that you need to move past a point into a 2-dimensional space instead, which our brains actually do because that's how we perceive sound, as a distribution of frequencies instead of as discrete real values.