The same inequality, but two equivalent ways to express it

[mcastillo356]

TL;DR Summary

I've been wondering why $0\leq{\theta}<2\pi$ is equal to $-\pi<\theta'\leq{\pi}$, but no way. I've been given the right answer, but still don't know how to move from one to another.

I've been given this answer: to move from negative to positive angles, $-\theta'=2\pi-\theta$; and to move from positive to negative angles, $\theta'=\theta-2\pi$. But my question is if there is any way to calculate it in a sequence of inequalities' steps.
If I am being cumbersome, forgive me.
Greetings

 

[fresh_42]

Imagine a unit circle. If you measure angles by $0\leq \theta < 2\pi$ then you start at $(1,0)$ and walk around once. If you measure angles by $-\pi\leq \theta' < \pi$ then you start at $(-1,0)$ and walk around once. The appropriate set up depends on the question asked, the way you want to scale your figures.

If you measure the miles between Chicago and Los Angeles, then do you say mile $0$ is in Chicago, or is it in New York and Chicago would have mile $789$? It is up to you and there is no right and wrong. At best there it's convenient and inconvenient.

 

[PeroK]

Summary:: I've been wondering why $0\leq{\theta}<2\pi$ is equal to $-\pi<\theta'\leq{\pi}$, but no way. I've been given the right answer, but still don't know how to move from one to another.

I've been given this answer: to move from negative to positive angles, $-\theta'=2\pi-\theta$; and to move from positive to negative angles, $\theta'=\theta-2\pi$. But my question is if there is any way to calculate it in a sequence of inequalities' steps.
If I am being cumbersome, forgive me.
Greetings

These two intervals are not "equal", but they are "equivalent", in the sense that you can choose either as a half-open interval of length $2\pi$. And thus serve as a domain for the angle around one complete circle.

You can find the transformation by considering $\theta' = a\theta + b$, for some constants $a, b$.

Hint: the transformation is almost, but not quite, simply $\theta' = \theta - \pi$.

 

[jedishrfu]

another way to look at it:

In both cases from a unit circle perspective, the endpoints are the same angle and so there's no need to include that angle twice in the set of angles defined by the semi open interval.

 

[mcastillo356]

Imagine a unit circle. If you measure angles by $0\leq \theta < 2\pi$ then you start at $(1,0)$ and walk around once. If you measure angles by $-\pi\leq \theta' < \pi$ then you start at $(-1,0)$ and walk around once. The appropriate set up depends on the question asked, the way you want to scale your figures.

If you measure the miles between Chicago and Los Angeles, then do you say mile $0$ is in Chicago, or is it in New York and Chicago would have mile $789$? It is up to you and there is no right and wrong. At best there it's convenient and inconvenient.

Fine, fresh_42, I did not manage to see it so graphically, until you've told it.

These two intervals are not "equal", but they are "equivalent", in the sense that you can choose either as a half-open interval of length $2\pi$. And thus serve as a domain for the angle around one complete circle.

You can find the transformation by considering $\theta' = a\theta + b$, for some constants $a, b$.

Hint: the transformation is almost, but not quite, simply $\theta' = \theta - \pi$.

The hint is definite. I feel quite dummy. Perfect.

another way to look at it:

In both cases from a unit circle perspective, the endpoints are the same angle and so there's no need to include that angle twice in the set of angles defined by the semi open interval.

That's the way I look it: why must I consider $-\pi\leq \theta' < \pi$ interval, instead of $0\leq \theta < 2\pi$?; it's because I must avoid integer multiples of $2\pi$. This question is understood in the context of complex numbers, just to calculate the main phase, the main argument, of any complex number. But I still don't know what is all about: $2\pi$ doesn't appear in $0\leq \theta < 2\pi$.

 

[jbriggs444]

That's the way I look it: why must I consider $-\pi\leq \theta' < \pi$ interval, instead of $0\leq \theta < 2\pi$?; it's because I must avoid integer multiples of $2\pi$. This question is understood in the context of complex numbers, just to calculate the main phase, the main argument, of any complex number. But I still don't know what is all about: $2\pi$ doesn't appear in $0\leq \theta < 2\pi$.

Ummm, zero is an integer multiple of $2\pi$.

 

[PeroK]

Fine, fresh_42, I did not manage to see it so graphically, until you've told it.The hint is definite. I feel quite dummy. Perfect.That's the way I look it: why must I consider $-\pi\leq \theta' < \pi$ interval, instead of $0\leq \theta < 2\pi$?; it's because I must avoid integer multiples of $2\pi$. This question is understood in the context of complex numbers, just to calculate the main phase, the main argument, of any complex number. But I still don't know what is all about: $2\pi$ doesn't appear in $0\leq \theta < 2\pi$.

You are over-thinking this. Imagine you have a tennis court, which is 24m long, and you want to mark it out every metre. You have a choice:

1) You could start making one baseline $0m$, then mark a line every metre, with the net being at $12m$ and the second baseline at $24m$. In summary, you have maked out the court from $0-24m$.

Or

2) You could mark the net as $0m$ and mark the lines on one side from $1m$ to $12m$ towards one baseline, then from $-1m$ to $-12m$ towards the other baseline. In summary, you have marked out the court from $-12m$ to $12m$.

Conceptually, that is all there is to it. With a circle, of course, you imagine the two baselines touching. In the first case $0m \equiv 24m$ and in the second case $-12m \equiv +12m$.

 

[mcastillo356]

Sorry, jbriggs444, I meant multiples of $2\pi$, $\dots, \theta-4\pi, \theta -2\pi, \theta, \theta+2\pi, \theta+4\pi, \dots$.
PeroK, thank you! The analogy is simply perfect.
Trouble solved. Greetings!

 

[PeroK]

Sorry, jbriggs444, I meant multiples of $2\pi$, $\dots, \theta-4\pi, \theta -2\pi, \theta, \theta+2\pi, \theta+4\pi, \dots$.
PeroK, thank you! The analogy is simply perfect.
Trouble solved. Greetings!

I'm not sure I understand. If you mean $2\pi \le \theta < 4\pi$, then there is no advantage of that instead of simply $0 \le \theta < 2\pi$. In general, you would expect $0$ to lie in your interval: either at the beginning or in the middle.

There's no right or wrong here, only what should seem natural and convenient.

 

[mcastillo356]

Hi PeroK! If $w=a+bi\neq{0}$, then every value of $\mbox{arg}(w)$ differ in a integer multiple of $2\pi$. The symbol $\mbox{arg}(w)$ its not just a number, its a set of numbers. When we write $\mbox{arg}(w)=\theta$, we mean the set of $\mbox{arg}(w)$ includes every number like $\theta+2k\pi$, $k\in{\mathbb{K}}$.
Sometimes is convenient to restrict $\theta=\mbox{arg}(w)$ to an interval of range $2\pi$, this is, the interval $0\leq \theta < 2\pi$, or $-\pi\leq \theta' < \pi$. This way, the complex numbers different to zero will have a unique phase or argument.
Does it make sense now?

 

[jbriggs444]

The phrasing that I was taught for this sort of thing is that we have an "equivalence class" of values that each differ from one another by various integer multiples of $2\pi$. Instead of a set of values, we have a set of equivalence classes of values.

But because it is clumsy to talk about equivalence classes, we choose to denote each equivalence class by a particular member, an "exemplar", of that class.

The problem at hand seems to be finding a convenient set of exemplars and knowing when two sets of exemplars are identical when understood as two sets of equivalence classes.

 

[PeroK]

Hi PeroK! If $w=a+bi\neq{0}$, then every value of $\mbox{arg}(w)$ differ in a integer multiple of $2\pi$. The symbol $\mbox{arg}(w)$ its not just a number, its a set of numbers. When we write $\mbox{arg}(w)=\theta$, we mean the set of $\mbox{arg}(w)$ includes every number like $\theta+2k\pi$, $k\in{\mathbb{K}}$.
Sometimes is convenient to restrict $\theta=\mbox{arg}(w)$ to an interval of range $2\pi$, this is, the interval $0\leq \theta < 2\pi$, or $-\pi\leq \theta' < \pi$. This way, the complex numbers different to zero will have a unique phase or argument.
Does it make sense now?

Yes, that's all there is to it. It's not complicated.

 

[mcastillo356]

jbriggs444, Hi, for every angle in the interval $0\leq \theta < 2\pi$ (or equivalently $-\pi\leq \theta' < \pi$), I am going to guess: each angle of the first circumference might be an equivalence class. But I know nothing more.

 

[fresh_42]

jbriggs444, Hi, for every angle in the interval $0\leq \theta < 2\pi$ (or equivalently $-\pi\leq \theta' < \pi$), I am going to guess: each angle of the first circumference might be an equivalence class. But I know nothing more.

The model of equivalence classes is a bit far fetched in this context in my opinion, especially as you seem to think about trigonometric functions, which are defined on the entire real number line, and equivalence classes are nowhere even near. Of course one can define $\alpha \sim \beta :\Longleftrightarrow \alpha-\beta \in (2\pi)\cdot\mathbb{Z}$, which is an equivalence relation, and factor the real number line along these classes to get the unit circle: $\mathbb{R}/\sim \;\cong S^1$, but I doubt this is helpful at this level of discussion.

 

[mcastillo356]

Hi fresh_42, yes, this level (my level) is very low