Question

Let \(n\) be some natural number and let \(\theta\) and \(\phi \) be in the interval \([0,2\pi)\).
Suppose \(\theta\) and \(n\phi\) are equivalent angles (congruent modulo \(2\pi\)).
Find the general solution for \(\phi\).

Answer 1

This was an interesting problem I thought about last night.

Answer 2

I'm wondering if I'm understanding the question properly, so bear with me.

Suppose \(\theta=\dfrac{\pi}{2}\), then \(\phi\) can be any of \(\dfrac{\pi}{2},~\dfrac{\pi}{4},~\dfrac{\pi}{6},~\cdots\), or \(\phi=\left\{\dfrac{\pi}{2n}\right\}_{n=1}^\infty\)

Is the question asking for the general form of \(\phi\), as in a set of angles, for any \(\theta\) in the interval?

Answer 3

The solution for \(\phi\) as a set of angles.

Answer 4

Basically, there are multiple solutions.

Answer 5

\(\theta \mod 2\pi = n \phi \mod 2\pi \)?

Answer 6

Okay, so if \(\theta\) was \(\dfrac{\pi}{2}\), my solution for \(\phi\) is correct?

Answer 7

first thing im thinking about is a set of prime numbers would be involved in n, for theta=nphi, and then +2*k*pi to that

Answer 8

@SithsAndGiggles Let me look at your solution for a moment, I don't quite get it.

Answer 9

@wio sure take your time, I'm just trying to wrap my mind around the question.

Answer 10

\(\theta = n \phi+m 2\pi\)

Answer 11

mmm diophantine equation

Answer 12

@SithsAndGiggles Okay, there is a problem with you solution. First of all, \(n\) is a constant natural number, it is not a free variable but you are using it as one. Second of all, it is missing solutions, I believe.

Answer 13

That would be \(\theta = \pi/2\) and \(n=2\) solutions.

Answer 14

if \(\theta =\frac{ \pi }{2}\)
then \(n\phi\)= \(\theta +m2\pi\)

Answer 15

One other thing I want to add. I want a unique set of solutions. So I don't want repeats. If you create a free variable for expressing solutions (as is necessary), then give that variable some bounds.

Answer 16

So you're saying for \(\theta=\dfrac{\pi}{2}\), you have \(\phi=\dfrac{\pi}{4}\) and \(\dfrac{5\pi}{4}\) as solutions? Okay, I see...

I used the bracketed sequence notation in place of a set to account of for any value of \(n\). What I meant was that \(\phi=\dfrac{\pi}{4}\) was the only solution in the case that \(n=2\), which is clearly not a complete solution since \(2\cdot\dfrac{5\pi}{4}=\dfrac{5\pi}{2}\equiv\dfrac{\pi}{2}\text{ mod }2\pi\).

Okay, I got it now. Interesting question indeed.

As a conjecture, I'd guess that there are \(n\) solutions for a given \(n\).

Answer 17

No, only when \(n=2\) are those the solutions.

Answer 18

what about like

Answer 19

So you need to generalize to the \(n\)th case.

Answer 20

Right, let's see what the others can come up with :P

Answer 21

I have a solution in my head already. I wanna see what others do.

Answer 22

cases where its not possible to meet in 1 rotation or 2
theta = pi/5 and
phi = pi/7
n*pi/7=pi/5+2kpi

Answer 23

n*phi=Theta + 2k pi
solutions to this are from diophantine

Answer 24

Are you saying the general solution can't be expressed as a function?

Answer 25

mm well , we can solve diophantine like this ?
x=theta
y=n phi
x=y mod 360 mmm too much variables , but we have interval [0,360) for x , y

Answer 26

The thing is, these angles don't have to be rational or anything. They can be real numbers.

Answer 27

ik mm im still thinking of it maby mod thingy wont work

Answer 28

maybe*

Answer 29

The thing about\[
\phi = \frac{\theta +2\pi k}{n}
\]is how to limit it to unique solutions.

Answer 30

When \(k = n\), we have: \[
\phi = \frac{\theta + 2\pi n}{n} = \frac\theta n+2\pi
\]So at this point, the solutions have begun to repeat.

Answer 31

When \(k=0\) we get the more trivial solution: \[
\phi = \frac \theta n
\]This one will always apply.

Answer 32

\(k\) can't be negative, otherwise we get a negative \(\phi\).

Answer 33

I've started seeing that kind of pattern. Keeping to my example with \(\theta=\dfrac{\pi}{2}\), I've been considering one \(n\) at a time:
\[\begin{matrix}\underline{n}&&&\underline{\phi}\\
1&&&\frac{\pi}{2}\\
2&&&\frac{\pi}{4},\frac{5\pi}{4}\\
3&&&\frac{\pi}{6},\frac{5\pi}{6},\frac{9\pi}{6}\\
4&&&\frac{\pi}{8},\frac{5\pi}{8},\frac{9\pi}{8},\frac{13\pi}{8}
\end{matrix}\]
If you were to plot these angles, you'd see the plane divided equally into \(n\) parts.

Now, notice that
\[\frac{\pi}{2}\equiv\frac{5\pi}{2}\equiv\frac{9\pi}{2}\equiv\frac{13\pi}{2}\equiv\cdots\equiv\frac{(4k+1)\pi}{2}\equiv\frac{\pi}{2}\text{ mod }2\pi\]
for \(k=0,1,2,...\).

So judging by that, I would expect
\[\phi=\frac{\pi}{10},\frac{5\pi}{10},\frac{9\pi}{10},\frac{13\pi}{10},\frac{17\pi}{10}\]

I haven't checked for other values of \(\theta\) yet, but I'm convinced the pattern (or something similar) will hold.

Answer 34

its 3:49 AM lol cant open my eyes O.O
so sith just need to check other values of theta
mmm cant wait to see wio general solution :-\

Answer 35

Well, first thing I did consider was just \(n=2\). You have two solutions:

So you have \(\theta /2\) and you have \(2\pi -\theta/2\)