CyclotomicField

The point of using affine spaces is to avoid defining an origin. While it's often useful to choose one for proofs I think in the definition it should be avoided.

$Mathematics$ Discussion on question by user3716267…

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CyclotomicField

Oct 22, 2024 13:59

I'd prefer to call everything Euler's theorem and let the reader figure out what's meant from context. Then it's like a minigame you get to play while you're doing the math.

$Mathematics$ Discussion on question by Elvis: Defi…

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CyclotomicField

Oct 6, 2024 02:48

Instead of choosing $(0,0)$ as the origin instead choose some other origin $(a,b)$ and then define your vectors as $v=(x,y)-(a,b)$. So there is nothing special about $(0,0)$ as an element of $\mathbb{R}^2$ and we can pick any element we like to be the origin. So to create a vector space we have to add some additional information, which is a choice of origin.

$Mathematics$ Discussion on question by Constantine…

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CyclotomicField

Sep 17, 2024 09:26

Voted to close as it's unclear what the question is. Please elaborate more completely as to what you're asking.

$Mathematics$ Discussion on question by Princess Mi…

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CyclotomicField

May 25, 2024 20:39

@PrincessMia addition of the natural numbers is derived from counting which is how we created addition. That's why this approach is taken, it's the way we all learned as children.

$Mathematics$ Discussion on question by Flummox: Ho…

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CyclotomicField

May 24, 2024 00:38

Do you have the fundamental theorem of calculus? Set $f(x) = \sum_{n=1}^\infty \frac{x^n}n$ then calculate $f'(x)$ to get a geometric series which you can solve. Then integrate to undo the differentiation.

$Mathematics$ Discussion on answer by Lauren S: Cir…

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CyclotomicField

May 24, 2024 00:37

@NicolasBourbaki then how should I have been helpful? Opinions without reasoning are useless. Explain how I could have done better.

CyclotomicField

May 24, 2024 00:37

@NicolasBourbaki what evidence can you provide that they weren't helpful? You've imagined it's so but I have no reason to believe you. I have double the points you have in half the time and it's almost all from answering questions and providing information. If you think you can be more helpful then get on my level.

CyclotomicField

May 24, 2024 00:37

@NicolasBourbaki that's correct, I can't counter your point. You're arbitrary criteria on an a subjective topic stands firm. Well done I guess.

CyclotomicField

May 24, 2024 00:37

@NicolasBourbaki there wasn't a discussion. You decided you wanted to fight instead of talk. I like answering question and I like providing more information to users. You apparently think that's a problem and it ruins the website for you. If you have anything constructive to add now's the time to mention it.

CyclotomicField

May 24, 2024 00:37

@NicolasBourbaki you've definitely made me understand that some people take comments extremely personally. I'll be sure to account for the fragility of the reader in the future.

CyclotomicField

May 24, 2024 00:37

@NicolasBourbaki if you refuse to read you're no position to know if I'm right or wrong. Do the work or stay ignorant.

CyclotomicField

May 24, 2024 00:37

@NicolasBourbaki your disingenuousness is apparent. If you're rather be bitter and petty then I can't stop you. I don't have to live with you, you do.

CyclotomicField

May 24, 2024 00:37

@NicolasBourbaki definition 3.1.2 is right there, just a few words after you stopped reading. The whole of chapter 2 is dedicated to the topological properties of the real numbers as a lead up to the chapter on continuity. The one-point compactification of the reals is given as an exercise and is purely topological. They use the topological definition to prove that continuous functions on a compact set are uniformly continuous. It's impossible to miss if you make even a trivial effort.

CyclotomicField

May 24, 2024 00:37

@NicolasBourbaki you asked for an example of the topological definition being used in an analysis text and when I provided you one you decided that reading a single sentence from a single paragraph somehow made that example disappear. If you want to have an honest discussion about which definition is easier I'm happy to but your dishonesty isn't compelling.

CyclotomicField

May 24, 2024 00:37

@NicolasBourbaki cherry-picking one paragraph won't rewrite the book. It's a major theme in the text and nothing you say will change that. You're clearly intent on being angry over nothing so I'll leave you to it.

CyclotomicField

May 24, 2024 00:37

@NicolasBourbaki "Introduction to Analysis" by Douglass does. The simplifications brought by the topological definition are a major theme of the text. They're rather passionate about it and I found their reasoning compelling. Epsilon-delta proofs have a reputation for being difficult and that's been my experience with them as well. I don't think I'm alone in that either.

CyclotomicField

May 24, 2024 00:37

@NicolasBourbaki the point of my comment is to provide more information so they can further explore the subject. The topological definition of continuity is simpler than the epsilon-delta definition and I wasn't aware that it's considered more advanced. I'd also note that the OP answered their own question so rehashing what they already know seems less useful than providing them with an easier approach. It's exactly what I would want when I asked a question or post an answer. Whatever motivations you've hallucinated I have other than to be helpful are entirely of your own imagining.

CyclotomicField

May 24, 2024 00:37

The epsilon-delta method is only applicable to metric spaces. The topological definition applies to any topological space and reduces to the epsilon-delta definition for metric spaces. en.wikipedia.org/wiki/…

$Mathematics$ Discussion on question by ummg: Techn…

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CyclotomicField

Mar 4, 2024 16:53

@Trebor the "trust me bro" line of reasoning isn't sufficient. If you have a mathematical reason to exclude it then provide it. Until then I'm unconvinced.

CyclotomicField

Mar 4, 2024 16:53

@freakish here is a problem given where they ask if the empty set is a ball but they explicitly state that the metric space is non-empty. I would suggest they did because they have to. The author of this problem didn't exclude the empty metric or the empty ball. I would suggest because they didn't have to. Allowing the empty metric to have an empty ball doesn't cause any complications that aren't inherited from the empty metric. math.stackexchange.com/questions/1638820/…

CyclotomicField

Mar 4, 2024 16:53

@freakish your position just doesn't hold up to scrutiny. You keep claiming there is utility to it but aren't able to provide a single example of that utility. Until you can your reasoning is unconvincing.

CyclotomicField

Mar 4, 2024 16:53

@freakish balls can only be empty if the metric is empty. You don't have to state the ball is non-empty in a metric space that has elements. All the complaints you have about the empty ball apply to the empty metric in exactly the same way. We gain nothing useful by extending metrics to the empty set yet we do. Including the empty metric and excluding the empty ball is internally inconsistent. Do both or do neither but this mix-and-match approach is arbitrary and confusing.

CyclotomicField

Mar 4, 2024 16:53

@freakish I find arbitrarily excluding it confusing, not the other way around. If there is no compelling reason to do so why single it out for exclusion?

CyclotomicField

Mar 4, 2024 16:53

@ummg the empty set is open in every metric space, but it's only a ball in the empty metric space. I'm not convinced there are any problems by allowing this.

CyclotomicField

Mar 4, 2024 16:53

@freakish what paradoxes and inconsistencies do you mean? What causes the problem if you allow the empty set to be a ball in the empty metric space?

CyclotomicField

Mar 4, 2024 16:53

@freakish Metric spaces come with a metric and that's also defined using the nonexistent elements of $M$. If we can define the distance then we can define the ball. I feel like if you allow one you're allowing the other.

CyclotomicField

Mar 4, 2024 16:53

I don't see a problem with a ball in $A$ being empty. Topologically the empty set is open and closed so $A$ would be contained in every open and closed ball. The ball is defined as a subset of the elements in the metric space and the empty set is the only set that meets that criteria. $A$ would then be bounded since it's contained in every ball.

$Mathematics$ Discussion on question by Haridasa: O…

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CyclotomicField

Feb 2, 2024 14:00

@Haridasa the rules are the axioms of a field and parenthesis always have priority. That allows you to distinguish between $2x^2$ and $(2x) ^2$. That's all the rules.

CyclotomicField

Feb 2, 2024 14:00

PEMDAS isn't how we do it in math. Instead we use the axioms of a field. This avoids all the problems that come with PEMDAS. mathworld.wolfram.com/FieldAxioms.html

$Mathematics$ Discussion on question by user107952:…

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CyclotomicField

Jan 14, 2024 02:16

I consider Euler the greatest mathematician of all time and many of his more exploratory results were not rigorous or even true. However they led to new directions and research and there was lasting value in his errors.

$Mathematics$ Discussion on question by Anacardium:…

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CyclotomicField

Jan 8, 2024 12:46

All complex reflections can be written as an identity matrix with a single entry changed to an $n$-th root of unity like you suggested earlier as long as you select the right basis. So circulant matrices and other representations are excluded for dimensional reasons as you observed. The isotypic components are the building blocks for an irreducible representation. Vaguely, think of them as the primes in a prime decomposition and the weights as the powers. It's not that easy but it's those sort of vibes.

CyclotomicField

Jan 8, 2024 12:46

It is a representation but it's not the minimal one when working over the complex numbers. The rotations can be generated by an primitive $n$-th root of unity my favorite being $e^{\frac{2i\pi}n}$. If you want an infinite cyclic group just pick an irrational multiple of $2i\pi$ instead. I actually prefer to call $\mathbb{C}$ the circle numbers for this reason as they encode cyclic properties in a natural way. The character is the trace of a complex representation and can be useful when classifying simple groups.

CyclotomicField

Jan 8, 2024 12:46

That representation works as well but $\zeta I$ would have $\zeta$ along all the diagonal entries. All you need is scalar multiplication without using matrices at all.

CyclotomicField

Jan 8, 2024 12:46

CyclotomicField

Jan 8, 2024 12:46

A complex representation can be one dimensional for cyclic groups. We don't have leave $\mathbb{C}$ at all. Note that real representations don't enjoy this property and you do need cyclic matrices in that case.

CyclotomicField

Jan 8, 2024 12:46

The representation doesn't have to be a permutation matrix. Consider a primitive $n$-th root of unity $\zeta$ and the matrices $\zeta^k I$ with $I$ the identity.

$Mathematics$ Discussion on question by Mohaboko: F…

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CyclotomicField

Dec 10, 2023 16:51

@Mohaboko ok just don't forget the absolute value. It's mandatory when you calculate areas.

CyclotomicField

Dec 10, 2023 16:51

@Mohaboko sure you can. It works both ways because you MUST take the absolute value.

CyclotomicField

Dec 10, 2023 16:51

@Mohaboko it's not wrong you just oriented the curve differently. Again, it's like $\int_a^b = -\int_b^a$. If you go in the other direction it will be the other sign.

CyclotomicField

Dec 10, 2023 16:51

@Mohaboko you already solve it. You just didn't take the absolute value. That's all you have to do. How many times do you need to be told?

CyclotomicField

Dec 10, 2023 16:51

@Randall Green's theorem isn't about area, it's about integrals. Integrals can be interpreted as an area under certain conditions but they aren't areas by design. Even trivial cases can be reoriented to give a negative instead of a positive every single time. That's a feature, not a bug.

CyclotomicField

Dec 10, 2023 16:51

@Randall the area enclosed is identical regardless of how you parameterize the boundary and reflections preserve area. That's why we have a convention for orientations, it's arbitrary.

CyclotomicField

Dec 10, 2023 16:51

You're getting a negative value because integrals can be negative. Note that $\int_a^bf(x) dx = -\int_b^a f(x) dx$ but they would both measure the same area between $a$ and $b$. The orientation doesn't matter and you must ALWAYS take the absolute value for area. There are no exceptions.