Hard Riddle #1

@Nomen-Nonatur
I was talking about your words below my quoted text. Why do you act as a laymen as if you don't know anything?
Plus, I never gave my definition of fun.
I just gave a statement and you probably assumed of possible definition according to me which isn't the case.

@Akbar2thegreat said in #41:

Plus, I never gave my definition of fun.
I just gave a statement and you probably assumed of possible definition according to me which isn't the case.

CAUTION: the following will contain some big words which are not easy to understand. Parental oversight is strongly suggested.

I will try to explain it in a way most specimen of the species "homo" will be able to understand. I beg your pardon in advance if you are not among the intended audience, this is meant as a "best effort":

You said (in #30):

[...] this forum was meant to be fun and not for intellectuals[...].

By saying so you implied "fun" and "for intellectuals" being opposites. If, for instance, you say "This is white, not black." you also imply that "black" and "white" are opposites or, at the very least, mutually exclusive. Otherwise the sentence would make no sense at all. For instance: "This is white, not hard." would make no sense, because "white" and "hard" are not opposites at all. Something can be hard AND white at the same time, it can be neither or have one of the attributes without having the other.

So, in fact you implied that "fun" was something which anything "for intellectuals" can't be. Why this is so that something appealing to ones intellect isn't - and CAN'T - be fun for you is something you should speculate about yourself. It would be rude to intrude in this process of self-reflection.

@ChessMathNerd @InkyDarkBird I posted a solution in #9 and I elaborated on it in #28, and I am pretty sure it is correct. In particular, I give an exact expression for the distance covered by the robot at any time t (starting from t=1 for convenience).

@Nomen-Nonatur I think if we assume the rubber stays circular at all time then the expansion has to happen everywhere uniformly?

@polylogarithmique said in #43:

@Nomen-Nonatur I think if we assume the rubber stays circular at all time the the expansion has to happen everywhere uniformly?

Again: this may or may not be the case and if it is then see above. Adding magically 1m of material at some point is about as realistic as an infinitely strechable rubber band, and my point was merely to explicitly state it in posing the question rather than having this implied quality of the band be assumed by the reader.

This is math after all, we deal with epsilons all the time and exactness is advisable when dealing with infinites.

@Nomen-Nonatur said in #44:

Again: this may or may not be the case and if it is then see above. Adding magically 1m of material at some point is about as realistic as a an infinitely strechable rubber band, and my point was merely to explicitly state it in posing the question rather than having this implied quality of the band be assumed by the reader.
No but I mean, if you add one meter of material in one single point of the circle without changing the metric elsewhere, what you obtain is no longer a circle I think?

Though I must admit I haven't thought carefully about the question.

@polylogarithmique said in #45:

No but I mean, if you add one meter of material in one single point of the circle without changing the metric elsewhere, what you obtain is no longer a circle I think?

It is possible to form a perfect circle from 1 m of material. Why shouldn't it be possible to form a perfect circle from 2 m of the same material? Just say "the insertion takes place in a manner so that a perfect circle results" is all it takes, no? I mean, we are dealing with some imaginary material with imaginary properties here, why not add one more property?

@Nomen-Nonatur you are right. My reasoning was the following : you already have a circle of circumference 1m. It has a certain curvature.
Now add 1m of material in a single point, without changing the metric elsewhere. Then apart where you added the material, the curvature stays the same, and since the curvature of a 2m circle must be least that the curvature of a 1m circle, the thing you obtain is no longer a circle.

The reason it doesn't work is you can actually change the curvature without changing the metric.

@Nomen-Nonatur
You are on wrong track if you think that one thing can either be same or opposite of a particular thing just like 'not liking' doesn't necessarily mean 'hating'. There is third thing which most homosapiens (including you) miss. Better ask a logician how to use logic correctly.

The rubber band is growing in both directions, so divide the speed in two.
The robot is heading in one direction. Which equals to the speed that was divided in two.
So the robot is on the X and unable to move forward.

It's like a carpet that's getting pulled away from your feet and you are landing on the X every time you jump forward.

The universe is expanding and we are still on the planet. When a shuttle leaves the earth, it can still come back to the earth even if the earth is moving. Everything is connected together on that rubber band called the universe.

@Nomen-Nonatur So, as a mathematician, your previous comments have gotten me thinking about more exact ways to measure the expansion of the circle. There are two ways that I have been thinking. The first type involves the circle changing center, like starting with the circle x^2 + (y+1)^2 = 1 and then transitioning the center down to keep the top at the origin, so like x^2 + (y+2)^2 = 4. Yeah I know the circumference is wrong but you get the point. In that case the expansion is not uniform, in other words, the circle expands the most at the point at it's base, along the y axis. We could use some normal line analysis of some kind to determine an equation representing the expansion. This gets pretty complex because of the circular nature, but I think it could be done. This is how I was imagining it in my solution earlier.
The second way, which might be what most people have been thinking, is to have uniform expansion, always centered at zero, from x^2 + y^2 = 1 -> x^2 + y^2 = 2, etc. In this case, at time t, the circle has circumference 1+t, and at time t+1, circumference 2+t, so the rate of expansion at each point is the same, namely (2+t)/(1+t). That would simplify matters some I guess.