ShoutOutAndCalculate

(A point was a measure zero.)

There doesn't need a point disappear on the trajectory, it's a trajectory. Also it's measure zero and it's not a pole so it doesn't matter(in physics).

A bijection function.

From $\mathbb{R}^2$ to $\mathbb{R}$

But it could be mapped by an isomorphic map. even open sets to open sets.

But as mentioned in the post, both coordinates and measure could be complete by a single test trajectory.

Or this one: en.wikipedia.org/wiki/Product_topology it's the simple product topology

@XanderHenderson en.wikipedia.org/wiki/One-dimensional_space

@XanderHenderson Could it been seen as a single one dimensional object along the particle's trajectory, or was $(x,y)$ necessary.

@XanderHenderson I've seen them before(a little bit), but the question was inspired by the compactification in string theory, i.e. with a Ricci-flat manifold, which was why I used the phrase "test particle". the space filling curves was not always a geodesic, but the one given in the post was. (which made it a lot of more interesting.)

@amWhy As for the format, math.stackexchange.com/questions/3572304/… looked the same.

I'm self learning and I tried google, but I'm not sure which subject does it belong to. It seemed to be an algebraic question, but I'm not sure there's a generic solution.

@amWhy I got "You can tell us where the question comes from." "Give full references." and "Give definitions."

There's literately no more details.

It listed as "This question needs details or clarity." However, I have put the entire question in the book there.

Hi! Could you help me to reopne this question please:math.stackexchange.com/questions/3576602/…

@dandavis the code was standard wire library and worked very well on Arduino. I tried other scripts online(which supposed to be working, but did not) as well. I mentioned that in case 8 and case 6.

@StarCat I think that might be the reason, as I have attempted the rest, somehow CP2102 has nasty behavior with Wire library and i2c bus, I might try the one with CH340 later. but why cp2102 didn't work?

@Juraj yes, see case 7 power supply. I tried multiple breadboard in case9 as well.

@Juraj I tried multiple ones as mentioned in case 2, they were all working well, the tape is just to protect the circuit and, if you don't trusted it, I've used multiple pre soldered chips as well.

@StarCat wait, v3 use the unstable CH340, could that be the reason why it actually worked? (could you check if you are using CH340 or CP2102)

@StarCat the scanner can get the address, online code in case 6 get 0(nothing). my code for gyro was fine if I don't move it too fast, but accelerator angle was wrong.

@StarCat I tried both frequency, but isn't V3 unofficial? (the one in graph was v2)

@StarCat how? could you try the script from case 6 please? (I have honestly tried everything you've told me to, plus 72 hours) what brand of ESP8266-12E "NodeMCU" are you using?(also, are you sure those value reading are correct?)

@StarCat I have tried that as well, in previous case 7 when adjusting the power supply. both 5V and 3.3V. Pull up resistors have been attempted from 0 to 100k

If you are beliving in and are going to write infinite bars and kets on a paper, fine, you can say that it seemed to be infinite to you, but even if you spend all your time wiritng a single states, you won't finish it and the symbol you wrote was sitll finite.

In sate vector formalism, the modern approach usually assume hilber space to be finite, which means it can not describe continuous orbit.

In functional represtation, if you can write spin in a funciton, then its probabilistic evoluation is determined, that's bascially wrong.

Funcitonal represtation is somewhat "determinsistic".

No, you are not getting what's the difference between them, like I said, what happend to spin?

are you answering the question?

You can't "draw some spherical graph of the continuous orbit" with state vector formalism.

No, hydrogen atom's electron orbit always use funciton, you are consusing two concepts.

states is more abstruct, wave is limited.

not really, there is a siginificant difference between wave and states. Think about spin, there is no function for it. On the other hand, what's the states for a hydrogon atom's sphere?

but bascially states vector use hilbert space, wave funciton use functional expension, and matrix use matrix.

I'm not sure how to say this

wave vector, or matrix

Not really, but I'm new to state vector formalism

hello?

?

@OfekGillon so the measurement of $|u_1>$ and $|u_2>$ collapse the states into $|A>$ and $|B>$ before the measurement? How could that be? I didn't measure perform the measurement $|A><A|+|B><B|$..

@OfekGillon what do you mean? $|\psi>=c_1 A+c_2 B$, en.wikipedia.org/wiki/Quantum_superposition#Theory It's either indistinguishable quantum particles of $A$ or $B$ or a single particle of superposition $A$ and $B$.

@OfekGillon Suppose a mixure of $N$ particles with $|c_1|^2 N$ of particle $A$. Or shot single particles with probability of obtian $A$ to be $|c_1|^2$.

@OfekGillon It's $|c_1|^2$ portion/probability of particle $A$ and $|c_2|^2$ for particle $B$. The measurement $|A><A|+|B><B|$ obtain either particle $A$ or $B$, yet, if one add them together, the secondary result for measurement $|u_1><u_1|+|u_2><u_2|$ is differed. (Acturally, if you add them together, the sate is no longer normalized)

@OfekGillon Yes. $|\psi>$ is a suposition of particle $|A>$ and $|B>$ normalized by $c_1,c_2$. Particle $|A>$ is normalized with respect to state vector $|u_1>,|u_2>$ by $a_1,a_2$. Particle $|B>$ is normalized with respect to state vector $|u_1>,|u_2>$ by $b_1,b_2$.