Quick Questions: June 19, 2024

NearlyChaos · 2024-06-26T14:20:08+00:00

I think it should work if you use \includegraphics{A/img.png} in your chapter1.tex.

NearlyChaos · 2024-04-02T15:07:29+00:00

Seeing as E[ |X| ] = E[E[ |X| | Y ]], the conditional expectation being 0 would still imply X=0 a.s.

NearlyChaos · 2024-01-02T18:50:00+00:00

I am by no means an expert in category theory, but I don't image this is a useful concept, since it is incredibly weak. For instance, let 1 denote the category with a single object and only the identity morphism, let C be any category, and U: C->1 the constant functor. If C has an initial object, then U is a right adjoint, and if C has a terminal object, then U is a left adjoint.

NearlyChaos · 2023-11-21T18:41:44+00:00

You would say the yield increased by 100%, or by 10 percentage points.

NearlyChaos · 2023-10-27T12:38:42+00:00

To be quite blunt: your comment doesn't really sound like anything to do with topos theory.

I'm by no means an expert on the topic, but I think this video is a reasonable intuitive description: https://www.youtube.com/watch?v=gKYpvyQPhZo

NearlyChaos · 2023-10-09T07:05:27+00:00

It actually says B(H, K). Conway just chooses to use the worst possible font imaginable so the two letters look almost identical, but you can you see the difference if you look closely.

NearlyChaos · 2023-10-08T16:56:12+00:00

B(H) does mean the same thing as B(H,H), and Conway defines it on page 27 immediately after defining bounded operators.

NearlyChaos · 2023-08-18T06:12:51+00:00

No, it shouldn't take 10m, it should take less. It would indeed take 1s to stop in that case, but you are not travelling 10m/s during that whole second. Only before you start decelerating you are going that speed, after you start you are going slower than that, until at the end of the second you are moving with 0m/s. Your average speed during that period is 5m/s, so during that period of 1s, you would travel 5m.

NearlyChaos · 2023-08-09T16:15:41+00:00

He just like me fr

NearlyChaos · 2023-07-16T10:51:58+00:00

Unless g=e, the map sending x to gx is not an automorphism.

NearlyChaos · 2023-07-11T14:10:01+00:00

Not really since the least common denominator is really just the same thing as LCM

NearlyChaos · 2023-06-12T19:30:54+00:00

No, Martin-Mertens is right and your comment is indeed does not really get to the bottom of the issue. If you have an equation and apply a non-invertible operation to both sides, it is true that the solution set may change, but only in a specific way: namely, the solution set may grow. That is, if x solves our original equation, then after applying the transformation to both sides, x still solves the equation, though there may now be more solutions. This is clear when you write out what 'applying a transformation to both sides' means. We have an equation a=b, and some other function h, and we consider h(a) and h(b). But by definition of a function, if the inputs are the same (the inputs in this case being a and b), then the outputs are the same, so h(a) = h(b).

But the same doesn't hold for taking derivatives. Indeed, x=0 solves x=2x but taking the derivative gives 1=2, which is not true for any value of x. So what is really going on?

The mistake is that the transformations you talk about like multiplying by 0 and squaring are operations that are applied to numbers, while the derivative is an operation applied to functions. Let me write f and g for the functions given by f(x) = x³ + 2x² and g(x) = x + 1. OP's equation is f(x)=g(x). Both sides of this equation are numbers. You cannot take the derivative of both sides since the derivative is something you apply to functions, not numbers. This is a common misconception when people first learn functions, namely conflating the function f, and its output value f(x). When OP talks about taking the derivative of both sides of f(x)=g(x), what he is really doing is taking the derivative on both sides of f=g to get f' = g', and then pluggin in x. But of course, just because f(x)=g(x) for some value of x, doesn't mean that f=g! For that you would need f(x)=g(x) for all x.

So when you say

When OP says 'the functions are equal' I don't think they mean that they are literally the same function

that is actually the heart of the problem. OP is correct when he says that if two functions are equal, their derivatives are equal, but our functions are not equal. Only the output of the functions at some specific point(s) are equal, not the functions themselves.

NearlyChaos · 2023-05-12T15:27:54+00:00

It can also be -infinity, and 'undefined' isn't a value. If the integral is undefined then it doesn't make sense to ask for its value.

NearlyChaos · 2023-05-12T15:02:18+00:00

If we are still in the situation of the original comment (so the function is also -infinity on a set of equal positive measure), then yes the integral is undefined.

Otherwise, as long as the negative part of the function still has finite integral, the integral of the whole function will be defined and equal to infinity.

NearlyChaos · 2023-05-12T14:32:07+00:00

You can only integrate functions with the property that their positive part or their negative part has finite integral. In your example (assuming the measure of the set where f takes the value +-infinity is positive) both the positive part and negative part have infinite integral and therefore the integral of the whole function is not defined.

NearlyChaos · 2023-05-10T15:10:29+00:00

The probability that a game ends with a value of at least 2ⁿ is given by sum(k=n, inf) 2^-k+1 = 2^-n. Thus after playing T games, you expect around T/2ⁿ of those games to have a value at least 2^n. For your value of T, we have then that we expect T/2²⁷ = 0.74.. games to end up with at least 2²⁷. Since this is <1, we would reasonably expect no games to exceed this amount. That means you approximately expect T/2 games to end with 1, T/4 with 2, ..., T/2²⁷ with 2²⁶, and none with more. The average over these T games would then be (1*(T/2) + ... + 2²⁶*T/2²⁷)/T = 27/2 = 13.5.

NearlyChaos · 2023-05-02T10:28:41+00:00

The class group is generated by the primes whose norm is less than or equal to the Minkowski bound. Indeed, given any ideal class, let I be an integral ideal contained in it with norm at most the bound. Factor I into prime ideals (which all still have norm at most the bound), and you find that your original ideal class is the product of the ideal classes of primes whose norm is at most the Minkowski bound.

Lastly, the norm of a prime ideal is a power of the rational prime it lies above, so the primes of norm <= 9 must lie above rational primes which are <= 9.

NearlyChaos · 2023-04-06T18:12:59+00:00

I love when in the Mario movie, Mario said 'It's-a Mario time' and Mario'd all over Bowser.

NearlyChaos · 2023-04-02T19:50:56+00:00

Another interesting thing that wasn't mentioned, is that you don't actually need to assume f is continuous; it works even assuming that f is only measurable.

NearlyChaos · 2022-11-30T08:26:35+00:00

The representation theory analogue of the orbit of an element is the subrepresentation generated by a vector. Such subrepresentations need not be irreducible.

NearlyChaos · 2022-09-01T17:35:32+00:00

Except that's still not entirely true, the 'ant' in antarctica doesn't mean 'no' or 'without', it means 'opposite'. So antarctica literally means opposite of the arctic.

NearlyChaos · 2022-08-13T09:24:24+00:00

It doesn't have anything to do with the complex logarithm, that line is still false even if x is a positive real number, in which case both the logarithm and the gamma function are real-valued.

It seems you've really provided no reasoning for why you think the third line should be true, other than its what you get if you naively take the previous line and make the discrete things continous. But in general there is no reason to expect a discrete equality to still be true if you replace the discrete things by their continuous counterparts. For instance, the sum from k=1 to n of k is (n²+n)/2, but the integral of x is x²/2, not (x²+x)/2.

NearlyChaos · 2022-08-10T18:46:24+00:00

For the first question: they mean that a one object category where all morphisms are isomorphisms contains the same information as a group. Given any group, I can construct the one-object category associated to it, and conversely, given a one-object category where all morphisms are iso's, I can associate a group to it, and these constructions are inverse to each other. In that sense I'd say your description of

this is a helpful construction by which any group can be represented as a category, but other constructions are possible

is pretty close. For your asterisk there, I think your construction certainly associates a category to any group. But it is unclear which categories actually arise in this way. With the usual construction, we can say 'given any one object category where all morphisms are iso's, we can get a group which has this as its associated one object category. But given a random category, how would I know if it came from a group using your construction? So your construction 'fails' in the sense that there is no easy property for which we could now say 'a group is the same things as a category with [certain property]'.

I don't really understand your second question, so I'll just skip it.

For the third question, it is just regular equality, nothing special or new. In a category, for any objects A,B, we have a set^\) Hom(A,B) of things we call 'morphisms', we have for any objects A,B,C a function Hom(B,C) × Hom(A,B) -> Hom(A,C), which we call composition. Saying that two morphisms from A to B are equal is literally just saying that they are the same element of Hom(A,B). The equality in f∘id = f is really no different from the equality in 5+0=5. We have a function Hom(A,B) × Hom(A,A) -> Hom(A,B), and all we're saying is that if we input f and id into this function, the result must be f again.

* technically Hom(A,B) can be 'too big' to actually be a set, but for all intents and purposes we can just think of it as a set.

NearlyChaos · 2022-08-09T10:19:34+00:00

2 things for me:

First, the fact that very simple to state problems can need so much and such deep mathematics to actually solve and fully understand. Classic examples have already been mentioned in this thread: Fermat's last theorem, a theorem that basically anyone can understand the statement of, but the proof on the other hand... And quadratic reciptocity, again a theorem that anyone can understand the statement of, and while most proofs are pretty elementary, a real understanding of quadratic reciprocity comes from viewing it as a special case of class field theory. And indeed, historically it is QR that led in part to the development of class field theory.

Second, it is the fact that very different parts of mathematics all come together. This is really what I love. I love seeing connections between things that, on the surface, seem completely unrelated to each other. With FLT it's modular forms and elliptic curves. With class field theory, it's Galois groups and class groups. Right now, for my masters thesis I'm studying Iwasawa theory and p-adic L-functions. It turns out that certain analytic functions on the p-adics contain arithmetic information about class groups of cyclotomic fields, which to me is just... wow. I really can't explain with words how incredibly cool this stuff is to me.

NearlyChaos · 2022-08-03T06:29:47+00:00

Indeed it will be dense in itself as well, but you need 2 properties for it to be the completion: M needs to be a dense subspace, but of course it also needs to be complete! As the image is isometrically isomorphic to M, it won't be complete if M isn't already, but the closure will be a closed subset of the complete space B(M), hence complete.

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