Is it just me, or does it look like black holes hovering over the planets?
I guess that is to be expected, if a planet with black background is used.
Pseudo-scientific question: Could black holes have anything to do with something being inverted?
A black hole is just an object with so much dense mass that its Schwarzschild radius[1^] extends outside of its body. Everything has a Schwarzschild radius.
For anyone else who's mind went blank trying to work out just what sort of "inverse" can be applied to a circle. It's interesting math and the diagrams are worth checking out, some very nice use of interactivity to assist with understanding the nature of the function.
But definitely not the kind of inverse I was expecting.
I went to "A function that's equal everywhere not the circle you define in the function", some sort of f(x,y) = !f(circle) by way some sort of algebraic geometry math. Then I was trying to work out if it meant something else... then I loaded it and was genuinely surprised to find its a much more specific kind of inverse that never even occurred to me.
This article is interesting but not rigorous I think.
* The inverse of a geometric shape makes no sense. We only inverse operations.
* aa^-1 = 1 only if you consider the multiplication over reals.
* 1/0 is not equal to infinity.
Because the article is interesting but some people might be put off by the first few sentences, I suggest to had a disclaimer that this article lean on edutainment to the detriment of rigorous mathematics.
Your assertions are simply not true in the context of complex analysis. It is common to use "inverse" to refer to the multiplicative inverse as shorthand (though potentially confusing). a a^-1 = 1 is absolutely and uncontroversially applicable to any complex number. It is common and natural to extend to complex plane to include a single point at infinity (known as the extended complex plane, see e.g. https://mathworld.wolfram.com/ExtendedComplexPlane.html ). When you are working in the extended complex plane, 1/0 does equal infinity.
It depends on your definitions and which mathematical objects you are working with. The notions in the blog post are not something the author invented themselves.
This is it. Inverse is a property of functions or other relational operators, not "static" individual objects. You need a direction in order to invert it.
As long as, C*Cinv = I, where C is the circle, Cinv is the inverse of circle, and I the identity. You're right. C, I, and * are entirely up in the air.
In general for inversion, we have object A (argument) and object I (identity) and a function F of two arguments, so we have equations: `F(A, X) == I, F(A, I) == I, F(X, I) == I, A != I, A != X`, where A, I, and X are objects in the same category, i.e. they must be circles `(x² + y² == r²)`.
If F is defined as `ra•rx`, then `ri == 1`, and inverse will be `rx = 1/ra`.
If F is defined as `ra + rx`, then `ri == 0`, and inverse will be `rx = 0 - ra`, where negative radius means hole.
If F is defined as `ra²•rx²`, then `ri == 1²`, and inverse will be `rx = sqrt(1/ra²)`.
If F is defined as `ra² + rx²`, then `ri == 0²`, and inverse will be `rx = sqrt(0 - ra²)`.
The dymaxion projection of the globe is one of my favorites and is essentially what an unbroken singular peel/shell of an orange would look like, centered on the North Pole
I interpreted the question myself from another angle: a circle is a function where every f(x) is an equal linear distance to an arbitrary fixed point z. So, the "inverse" to this function could be a function where every f(x) must have a different linear distance to z.
Yep, this would have the same effect. Both of these define all of the points that do not describe the circle.
However, as someone said above, f() is the inverse of g() if g(f(x)) = x. When put into practice, this means that the inverse is the reflection of the original function over y = x.
However, there's one problem with looking at the problem this way: A circle is NOT a function. Therefore, it does not have an inverse as we are thinking of it. A circle can be described by two functions, and both of these inverses combine to form the same circle. So, the inverse of a circle is (sort of) itself.
Readit News
https://www.adamponting.com/inside-out/
Related: Not Knot, 1991 Thurston-ish short film about knots and knot complements - "the space where the knot isn't".
https://www.youtube.com/watch?v=zd_HGjH7QZo
https://en.wikipedia.org/wiki/Not_Knot
https://en.wikipedia.org/wiki/Knot_complement
[1^]: https://en.wikipedia.org/wiki/Schwarzschild_radius
http://bl.ocks.org/KKostya/6075142http://bl.ocks.org/KKostya/6066548
Draw rectangle/circle on the right pane
But definitely not the kind of inverse I was expecting.
I went to "A function that's equal everywhere not the circle you define in the function", some sort of f(x,y) = !f(circle) by way some sort of algebraic geometry math. Then I was trying to work out if it meant something else... then I loaded it and was genuinely surprised to find its a much more specific kind of inverse that never even occurred to me.
* The inverse of a geometric shape makes no sense. We only inverse operations.
* aa^-1 = 1 only if you consider the multiplication over reals.
* 1/0 is not equal to infinity.
Because the article is interesting but some people might be put off by the first few sentences, I suggest to had a disclaimer that this article lean on edutainment to the detriment of rigorous mathematics.
For example you can find Riemann Sphere in wikipedia - https://en.wikipedia.org/wiki/Riemann_sphere
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If F is defined as `ra•rx`, then `ri == 1`, and inverse will be `rx = 1/ra`.
If F is defined as `ra + rx`, then `ri == 0`, and inverse will be `rx = 0 - ra`, where negative radius means hole.
If F is defined as `ra²•rx²`, then `ri == 1²`, and inverse will be `rx = sqrt(1/ra²)`.
If F is defined as `ra² + rx²`, then `ri == 0²`, and inverse will be `rx = sqrt(0 - ra²)`.
And so on.
Basically: Given a function `f`, then the function `g` is the inverse of `f` if `g(f(x)) = x`.
Dead Comment
Molehill: reciprocal of all points in a circle on the complex plane.
thinking of elephants stomping waves many miles... experience points to yes
However, as someone said above, f() is the inverse of g() if g(f(x)) = x. When put into practice, this means that the inverse is the reflection of the original function over y = x.
However, there's one problem with looking at the problem this way: A circle is NOT a function. Therefore, it does not have an inverse as we are thinking of it. A circle can be described by two functions, and both of these inverses combine to form the same circle. So, the inverse of a circle is (sort of) itself.
Dead Comment