uncool

The point of the problem is to use induction and the properties of determinants to find the answer, not to find each determinant separately.

Depends on what you mean. For any angle A, sin(A) = (e^(i*A) - e^(-i*A))/2i, which is an expression of the sine using exponents, but I'm guessing that's not what you're going for (in part, because you seem to have an objection to numerically finding the value). It sounds like you are trying to define sines in terms of radicals of rational numbers. There are uncountably many angles, and only countably many expressions using radicals. So we'd have to restrict ourselves to some countable subset of angles. You chose to look at pi/4; that suggests using only rational multiples of pi. And with that restriction, the answer is yes - sin(pi*(a/b)) can always be expressed in terms of radicals. In fact, the expression I gave above can count: sin(pi*(a/b)) = (1^(a/2b) - 1^(-a/2b))/(2*(-1^(1/2))). That expression...isn't really helpful in finding the value of sin, but I'm pretty sure it can be converted to an expression that is.

No matter the other 394 digits, nor the other 600 digits of the larger number, no, it can't be the smallest (non-trivial)-factor. Because the quotient would be around 10^207.

I think you are thinking of "normal vectors", which are not cotangent vectors, no. For one thing, normal vectors depend on how you have embedded your manifold into a larger Euclidean space, whereas tangent and cotangent vectors do not.

Yes (even better: that the map is a local diffeomorphism; the difference being whether there is *some* map or whether it is *this* map), but that is a very weak statement; any two smooth manifolds of the same dimension are locally diffeomorphic - because locally, the differential topology is that of Euclidean n-dimensional space. The term I think you want is that they are locally isometric (or more specifically, that the "wrapping map" is a local isometry), that is, that the metric, or the geometry, matches.

Because the map is not invertible - only locally invertible. When you wrap the sheet around the cylinder, it will cover itself (infinitely many times, even). Also, "diffeomorphism" is a term properly from differential topology - the metric plays no role in it (e.g. a "smoothed" coffee cup is diffeomorphic to a donut). The term "isometry" is much stronger - literally "same metric".

I think you mean the opposite: the spaces are not diffeomorphic, but they are locally isometric.

You are missing the logic behind the idea of taking limits, which is to avoid actually dividing by zero. The fact that you seem to think that the process is "magical" is a strong indicator of why.

you're missing the point taeto is making. Let me put it quite simply: no, the process of taking the derivative never involves a division by 0. Period.

...y = 1 is a horizontal line, not a vertical one; f(x) = 1 is a function.

...yes, it does. The definition of the limit of a function specifically excludes the value of the function of the limiting value from being relevant. "For all epsilon greater than 0, there exists a delta greater than 0 such that for any 0 < |x - a| < delta, |f(x) - L| < epsilon"

Please point out where you explained any actual division by 0, which is the specific distinction taeto is making. ...no, it isn't.

Except in neither case is there any actual division by 0.

...Mordred, I've said something like this to you before (about other people than Conjurer), but I think you are trying to tell Conjurer things at much too high a level.

The answer...is a bit complicated. From the other thread, I'm guessing that your level is around a freshman or sophomore undergraduate with a new interest in math. At that level, and for a few more years, that is the approximate idea: learn the definitions, understand the postulates (there's less of a difference between those two than you might think), play around with them a bit, see if you can figure out patterns for yourself, figure out or read theorems, learn the proofs, etc. In time, you will be able to figure out some of the standard proofs for yourself. (One of my favorite/hated lines from Munkres, chapter 4: "Why do we call the Urysohn lemma a ‘deep’ theorem? Because its proof involves a really original idea, which the previous proofs did not. Perhaps we can explain what we mean this way: By and large, one would expect that if one went through this book and deleted all the proofs we have given up to now and then handed the book to a bright student who had not studied topology, that student ought to be able to go through the book and work out the proofs independently.") To a large extent, the "game" is "Here are our assumptions; what are our implications?" When you have gotten comfortable with this, usually around the time you graduate from a math or math-related major, things start to change a bit. By that time, you stop simply accepting definitions and start asking "Why was this definition chosen, rather than that one?" The "game" shifts; definitions become more fluid, although rigorous proofs stay, in some sense, the center. You may or may not have heard of Terry Tao; he's a very famous mathematician. He explains what I said above in much more detail here: https://terrytao.wordpress.com/career-advice/theres-more-to-mathematics-than-rigour-and-proofs/ For now, it is a good idea to focus on learning the definitions and thinking in the "undergraduate" way. Getting to the "graduate" way takes a lot of investment and time that you haven't had the chance to put in yet, and knowledge that you haven't had a chance to learn yet. But you are well on your way through the first transition, from the sounds of it.

I'm not sure I understand what you are trying to say. What's the difference between "rounding to the nearest real number" and truncation? The result on the hyperreals is also exact. The "standard" function is specifically defined to give a result in the real numbers, not the hyperreal numbers. There are two correspondences I can think of, but the one I think you are referring to is that the reals are a subset of the hyperreals (i.e. there is a natural embedding of the reals in the hyperreals). This isn't "super advanced algebraic structure theory"; it's just...a fact. I'm not sure what you think needs enlightening.

I am "looking for" someone who cares whether what they say is correct. You apparently don't. I don't simply trust what people say, especially when I specifically know better. When we say you are wrong, consider the possibility that it is because you are wrong. To put it quickly: We are saying you are wrong because you are wrong. We are not correcting you because you seem to not be looking for nor appreciating correction.

...no. That's not even close to what I said. I don't even know how you read that from what I said. If anything, I'd say the problem is that you've talked too much. Specifically, you've confidently made assertions that are clearly wrong, and otherwise made statements that fail to demonstrate a deeper understanding of the things you discuss, when such a demonstration is clearly appropriate. For example: in this case, when challenged on your understanding of phase space, you superficially described uses of it. The challenge was about what phase space is. You eventually provided it, but only by copying and pasting from Wikipedia, without showing that you understood what you copied and pasted.

Because your descriptions are asserted with much more confidence than is warranted. As well as very often being wrong.

...this description, if anything, only confirms what Strange said: you have no idea what "phase space" means.

No, they do not. Hyperreal is not simply "not real"; it is a specific extension of the reals. One of the major properties is that it is an ordered field extension; that is, it allows addition, subtraction, multiplication, and division (except division by 0), and that these operations act nicely with respect to "Which of these is bigger". One result from that is that all squares are positive (as x^2 = (-x)^2, and one of x and -x is positive, and the product of two positive things must be positive). So if i were a hyperreal, then -1 would be positive - which it can't be. That said, it is possible to construct the hypercomplex numbers, in the same way the complex numbers are constructed from the real numbers.

What's the difference, in your opinion? I'm not sure what "it" refers to here, but both hyperreals and dual numbers are rigorous (though the "dual number" has some trouble with properly defining higher derivatives); a version of "dual numbers" is actually used in algebraic geometry. That shouldn't be especially surprising; the hyperreals aren't really meant to be a fundamental change in outcome, only in outlook. As such, most formal processes should look essentially the same.

Formally, no. Informally, spot the fuck on. And of course, with the new structures, you get new patterns and relationships, which need new structures, which ...

Among many, many other reasons, because I have some basic knowledge from quantum mechanics: photons are bosons, not fermions. What does this mean? This is not a good argument.

This is false.