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Interesting.
The way I approached this problem for the single digit case was:
Reduce everything mod 10. You can see that the second matrix and the product matrix are identical. At first glance you might think the first matrix should act as an identity matrix, which is not entirely correct.
Reducing mod 10, you can solve the problem for the first column, and then for the second column. So you see that it's impossible to solve the problem unless the matrix mod 10 has an eigenvalue of 1. To test for this, you simply subtract 1 from the main diagonal and verify that the determinant is a multiple of 10. If that does not happen you can never satisfy the equation. That cuts a lot of possibilities for the first matrix.
To solve for the second matrix, we only need to scan all eigenvectors corresponding to the eigenvalue 1. In principle, you could just find one eigenvector v and generate v, 2v, 3v, ..., 9v. For each of these you check if the mod 10 identity carries over to digit concatenation with respect to either the first or the second column. If it does you can generate all solutions combinatorially.
Finding an eigenvector, though, is a bit hard. The "right" way to do it would be to apply the Chinese remainder theorem and look at the equations mod 2 and 5. The problem is that sometimes degeneracies occur. For example, if you have an equation 2x + 4y = 0 mod 10, this equation is always true mod 2.
In the end, it was so annoying to deal with that I just brute forced all eigenvectors because I was lazy.
That should help you with the two-digit case, though, but I think using CRT is much more annoying because you have to work on base 100, and you have to look at equations mod 4 and 25, which are powers of primes, not simple prime numbers.
P.S: I also figured out why there are so many matrices whose square obeys the rule. It's because of the Cayley-Hamilton theorem. If the characteristic equation is x^2-11x = 0, the matrix will satisfy X^2 = 11X
So, any 1-digit 2x2 matrix which is singular and has a trace of 11 has the desired property.
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Recently I came across this post from a math social media page:
I thought it was pretty interesting, and thought there would be other examples of matrices satisfying a similar property. It's barely possible to simply brute force all pairs of 2x2 matrices and search for the ones whose product is just digit concatenation, but out of boredom, I actually managed to come up with an algorithm to optimize this search using number theoretic concepts.
Anyway, now I have a list of all 2x2 matrices whose product is obtained by just concatenating their base 10 digits.
One thing that's very cool is that in some pairs the matrices are actually the same. Anyway, the problem is: come up with an algorithm, better than plain brute force, to find two single digit 2x2 matrices whose product is obtained by concatenating the base 10 digits of each element.
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One thing that is really cool, and which is not obvious from Fractal's solution, is that the arc between the tangency point at x=0 and the other one is exactly 60 degrees.
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One very straightforward way to see why nuclear reactions release so much energy is to simply look at the measurement units.
Chemical reaction energies are usually of the order of a few electron-volt, while in nuclear physics the reaction energy is usually measured in mega electron-volts, which are one million times greater. The reason for this depends a lot on how far you want to go to explain the strong interaction, but ultimately it boils down to measuring stuff, even at the level of particle physics, where you study QCD and other stuff. Ultimately, you have to fit some parameters to make the experimental measurement come out.
Perhaps the simplest explanation is de Broglie wave-length and relativity. We know that momentum is inversely proportional to the length of an object. In a crude approximation, we can assume that relativistic energy is proportional to momentum, and we find that energy is inversely proportional to length.
From this, since the nucleus is around a million times smaller than an atom, you would expect that reactions that involve nuclei to be a million times more energetic than chemical ones, which involve the atom as a whole.
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OmnipotentEntity wrote:
Binding energy generally goes the other way. It manifests as a mass defect.
Yup, the binding energy in QCD is different from the one in traditional nuclear physics. It happens to create mass because of some peculiarities that only happen in relativistic field thories.
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This is more like a shitpost, but interesting nonetheless. I came across this image:
It turns out that the apparent paradox of three Magnemites having a mass 10 times greater than an isolated one is possible if we take into account quantum mechanics and special relativity. Intuitively, it can be viewed as follows. When two systems interact, they can have a binding energy as a consequence of this interaction. It's because of this energy that molecules do not break apart, for example. At the same time, the relativistic mass-energy equivalence E=mc^2 tells us that mass and energy are more or less the same thing, so if we have a lot of binding energy, it can manifest itself as mass.
The classic example is the QCD binding energy, which makes the proton much heavier than its constituent quarks, for example.
So, I know people posed this challenge as a joke, but science can explain the higher Magneton mass. In the Pokemon universe, the magnemites probably have a "Magnemite charge" which introduces a binding energy between them, and gives rise to this extra mass when they combine to form a Magneton :D
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It's been a long time since I had a look at analytical mechanics, but I suppose it's because if you differentiate a holonomic constraint, which is always something like f(x,y) = c, if I remember correctly, you end up with an exact differential equation?
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Warp wrote:
It's also not very useful to engage in trickery like "we define the function smartass(a, b) to be the exact perimeter of an ellipse with radiuses a and b. Therefore the closed-form expression for the perimeter is smartass(a, b)." This would just be a useless circular (hah!) definition.
It's because of stuff like this that my disposition to reply to any of your questions decreases exponentially over time.
If you bothered to look up the references I sent you, you'd see that the elliptic function, which is the answer to your question, is not a smart-ass function, but has many interesting properties, see for example the Jacobi elliptic functions, which are linked to geometry and have very similar properties to trigonometric ones. If there are doubts about its usefulness, they are also linked to elliptic curves, which form the basis of cryptographic algorithms that allow you to post stuff like this.
Warp wrote:
I don't think it's very useful to start nitpicking about the exact definition, and just go with, for example, what's listed at the wikipedia article.
The thing is: it is useful, much more useful than set in stone some limited form of expressions. In the words of Grothendieck, "The introduction of the digit 0 or the group concept was general nonsense too, and mathematics was more or less stagnating for thousands of years because nobody was around to take such childish steps..."
For example, for centuries people wanted to find a "closed-form" expression for prime numbers. The difficulty of the problem was just because what they understood by "closed-form" (polynomials, exponentials, etc.) could never work out. And it turns out it is rather simple to generate prime numbers if you do not limit yourself to these kinds of expressions. So, the useless thing, in retrospect, was to use the wrong tools to solve the problem.
Here's something that would be much more productive: why not come up with problems that you think are useful, solve them and explain to people how to do it? As far as I know, staying on the sidelines saying something is useless is the easiest thing in the world.
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Warp wrote:
Matt Parker in his latest video points out how, perhaps a bit surprisingly, there is no closed-form expression for the perimeter of an ellipse (at least not for the general case).
That got me thinking: Are there ellipses (other than the circle) where the major radius, the minor radius and the perimeter are all expressible with closed-form expressions?
(In this context "closed-form expression" excludes integrals.)
I would be careful to say "closed form expression" because this term can be a bit biased. For example, if you solve a problem and find sin(x) as an answer, most people would consider this a closed-form expression, until you actually go calculate it. How do you calculate sin(x) in general? You need to have an infinite series, integral representation, or the limit of an expression to do it, just like you need for more complicated expressions.
When people usually say a problem has a "closed-form" solution, what they mean by that is that the general answer has so many interesting algebraic properties that makes it very simple to calculate, and in the example you provided, the answer is given by elliptic functions, which satisfy many "nice" properties like trigonometric functions do, so I would consider it as a closed form expression.
By "nice" I mean more precisely that they are closely linked to modular forms, which can solve problems in many distinct areas of math. A lot of fancy research mathematics is just fancy ways to manipulate modular forms to show that they have all the structure needed to describe another algebraic structure.
If I take your question to mean, for which ellipses the perimeter is an algebraic number, the answer was found at the end of the 19-th century, and beginning of the 20-th. They happen at singular moduli. This term is very old and you're unlikely to find recent references about this.
Modern mathematicians usually study the j-invariant first. It turns out that the j-function is algebraic whenever its argument is an imaginary quadratic complex number, essentially because of a phenomenon called complex multiplication, and these singular moduli are all a consequence of that.
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I have two similar routes to solving it.
The first is to notice that (1 + k/n²) ~ ek/n² when n goes to infinity. There are two ways to notice this. The first one is to realize that (1+k/n²)^n² converges to e. So, by taking the 1/n² power of both sides we get the result. The other way is to simply take the first order Taylor expansion of ek/n² to see that it's indeed (1 + k/n²).
This way, we reduce the product to prod ek/n² for k from 1 to n-1. This is just en(n-1)/2n², which tends to sqrt(e) as n goes to infinity.
Yet another way is to take logs. If we call the limit L, we have
ln L = sum_k ln(1+k/n²)
We can expand ln(1+k/n²) = k/n² + o(1/n²)
The error terms are all o(1/n²) because k<n. Now, if we sum n terms, the o(1/n²) terms become o(1/n), and we have
ln L = sum(k)/n² + o(1/n) = n(n-1)/2n² + o(1/n)
Since the o(1/n) terms go to 0 we have ln L = 1/2, and L = sqrt(e)
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I had no idea of where I'd put this sort of rant, but this thread is the one that seemed most appropriate.
Recently, I've started working on finance, and I was extremely surprised by the enormous amount of people that all of a sudden decided to ask me for advice on how to get jobs like that, I never imagined that they had such an interest.
In any case, I've kept some contacts and still participate in several groups involving non-profit causes, a very curious one involves a chess club I attended since childhood. They have a WhatsApp group with several high-profile people, including the president of the state federation and even some IM's and GM's. It's quite a mess, usually, with the rated players constantly spamming the group with their streams/courses, or people making accusations against people who manage the federations or claims that someone beat them online using a chess engine.
The company where I work sponsors some social causes, and I was helping them get these funds, because if the government approves a project, the company gets nice tax deductions, so in the end, everyone is happy.
Today there was some discussion about the current government not doing anything to help culture, and at some point I told them that I had worked at the government and it is completely dysfunctional, because of the blind hierarchy and the impossibility of firing anyone for doing something wrong, and it was useless to discuss politics there, it would be much more productive to leverage online platforms to popularize the sport.
At this point, one of the admins changed the group to admin posts only, and said "chess only, no politics" and unlocked it two minutes later. Anyway, since I had discovered recently how crazed people are for money, I decided to try something to see how they would react.
I simply said something like "well, if you read what I wrote, you'd see that I was telling people not to discuss politics. no problem, in case you need sponsorships, I'll be sure to remember your reaction". That caused an admin to lock the group again, only for two minutes later someone senior to him unlock it and do damage control saying how many players they are helping, that they are organizing lots of tournaments, and that no sport survives without sponsorships.
Anyway, I decided to drop this anecdote here as a friendly reminder about how people who work on "noble causes" wanting to "change the world" are willing to sell their soul for money just like everybody else...
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It's not very hard to see why it should be huge.
We're basically looking for crossings of the graph x with tan(x). For large values of x, we can approximate the graph of tan(x) by vertical lines at (n+1/2)*pi. So, we want (n+1/2)*pi to be pretty close to an integer. So, essentially, we need to compute rational approximations to pi, hoping we hit a prime number. The thing is, the better the approximation is, the larger the denominator, and the smaller the probability of the number being prime, since it falls with 1/log(n). But because log(n) grows very slowly, it's not unlikely that you will find one.
To calculate this number, one probably needs to use some assymptotics from trigonometry and number theory to make what I said rigorous, until you can prune the search enough to hit the prime number. It looks like a lot of work, but doable.
(Incidentally, the recent advances in number theory, like the proofs that there are infinitely many pairs of consecutive primes with gap smaller than a finite quantity, weak Goldbach, the proof of the Duffin-Schaeffer conjecture, etc., all came from clever combinatorics arguments that made rigorous many statistical estimates. Although the most spectacular application is to number theory, they all work in settings like dynamical systems, and hypergraphs, which do not encode any number theoretic information at all!)
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FractalFusion wrote:
Do you like the harmonic series?
Of course. The reason I tend to post problems like this, though, is that I seem to get a lot more engagement from you guys with these a la Euler questions, which involve algebraic manipulation of expressions. I like them, too. But they become rather simple once you have enough experience, so these days I am tending more towards more abstract stuff.
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An interesting fact is that Bolsonaro's popularity has actually risen quite a lot, even when he constantly denies the pandemic. But if you actually go about your daily life, it's not very hard to understand that.
The number of deaths, at least those evaluated here, includes anyone who test positive for COVID. I am fairly certain of that, because I know someone who died from a stroke, but because he had tested positive, he was counted as a COVID death, and there were sever restrictions on the mourning procedures. I am now working from home in my parents' house, there's a retirement home nearby where COVID has killed five elderly people, and I also had close contact with two people who tested positive. To be honest, I'm not into doing a test myself because of whatever crap the government orders me to do if they find antibodies.
This situation is very weird. One of the reasons that there are few vaccine denialists in my country is because serious diseases like polio, measles, etc, were only eradicated at the end of the 90's, because of the extensive vaccination campaigns. I still remember people forming long lines in health clinics to get their children vaccinated.
About this pandemic? I can say the hospitals are not nearly as chaotic as they were during some Dengue outbreaks we had in the past. I still cannot understand how this disease can need strict lockdown actions, while at the same time be very difficult to detect in your daily routine (except, of course, through the media, which has been airing COVID numbers nonstop for the last five months).
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Warp wrote:
If the counter-argument is "you can't replace all the infinite strings with finite ones in a unique way" you'll have to explain why. The answer cannot be "because the set of all infinite strings is uncountable" because that's precisely what we are trying to prove, and we cannot assume what we are trying to prove.
For the record, I did explain why. It's not wrong to do that mapping, it just does not lead to any meaningful proof of the uncountability of the reals, precisely because by doing it you throw away the information necessary to prove it.
Your argument is essentially doing this: someone says "hey, there exist two matrices A and B such that AB is different from BA!", and you reply, "Wait a second, if I take the determinant, it's always true that det(AB)=det(A)*det(B), and since swapping the two factors give the same number, I don't see how we could have AB different from BA".
There's nothing wrong with this statement, but you simply showed that a line of proof can never succeed, because by restricting your view to the determinant you throw away all the information that's relevant to prove that matrices do not commute. It does not mean that other proof strategies are wrong.
Warp wrote:
So if we do that, we are replacing the original list of infinite strings with a list of finite strings, and the only thing that the diagonalization does is to tell us "there are no infinite strings in the list", which is self-evident to the point of being tautological. Of course there aren't, because we replaced them with finite lists in the first place. The only thing that the "proof" is now saying is, in essence, "if you replace all infinite strings with finite strings, there will be no infinite strings."
OK, so now I will ask you to consider something (not accept, just consider). People have been studying set theory formally for more than a century. In the mean time, many other fields, like theoretical computer science, rely on the diagonalization argument extensively, and we have some foundational results, like the undecidability of the Halting problem, are proven in a very similar manner. It's possible to prove the uncountability of the reals without it, as Cantor did, before he came up with the diagonalization proof.
Now, consider something before you put large quotes around the proof word: what is more probable? That the collective minds of mathematicians of various generations spanning hundreds of years is wrong? Or that you are the one who's not understanding something?
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Warp wrote:
1/5 is in the list. It's just that it has been "remapped" to eg. the value 16 (using the Calkin-Wilf sequence). If we are just trying to demonstrate that there are numbers that do not exist in the set of rational numbers, this shouldn't make any difference.
If the counter-argument is "but you can't do that with irrational numbers like sqrt(2) or pi", then you are already assuming that irrational numbers are not countable and thus cannot be listed, and thus cannot be "remapped" to the natural numbers. So you are already assuming what you are trying to prove (ie. that the set is uncountable and thus unindexable), which makes the argument circular and contradictory.
The problem is that this "argument" is not Cantor's argument at all.
Cantor's argument is: represent a real number by its decimal expansion, and use this algorithm to contruct another real number which is not on the list. The key for it to work is that the reals can be represented by an infinite decimal expansion. Sets which do not have this property can be countable.
Instead, what you're doing is, you are reassigning every real number to a natural number, which unlike the reals, must have a terminating digit expansion, and apply something analogous to the diagonal argument there. Of course it does not work, because if it did, you would prove that any infinite set is uncountable, you have not used any property of the reals that distinguish them from an arbitrary infinite set.
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Warp wrote:
r57shell wrote:
Its basic usage is poof that real numbers from 0 to 1 are uncountable.
I think there's a contradiction in the argument, then.
That's because the argument starts with the assumption that you can list all the numbers between 0 and 1 in some order (because if you couldn't, then it would be impossible to "choose the first digit of the first number, the second digit of the second number, and so on".)
If you can list all the numbers in the set as a list, that means that you can assign a natural number to it: Its position on the list.
I don't see any reason why you couldn't list the numbers in their base-2 increasing order... which then results in the "diagonal" number being 1111111... (or 0.111111... which would just be 1), which doesn't really tell us anything.
(Most particularly, if you were to list the numbers in the order 0.000..., 0.100... 0.010..., 0.110... and so on, and you included 1.000... in the list, the "diagonal number" would give you 0.111111... which is just 1.0. A number that's already in the list. Thus the diagonal argument actually fails to give a new number in this case.)
There is no contradiction in the argument at all. The problem you allude to arises from the fact that the standard way to interpret a string of 0's and 1's as a real number between 0 and 1 is not a bijection, because some numbers have two representations. However, the set of numbers which do not have a unique binary representation is countable, and it's straightforward to fix the function to be a bijection. Take a look at the Wikipedia article. Nothing in this is new, by the way, it already appears on Cantor's article in the 19th century.
The reason that people don't mention this possibility of two distinct representations is that it only arises if you insist on constructing a bijection from the set of binary strings to an interval of real numbers, and if your goal is to simply prove that the reals are uncountable, that's not necessary.
For example, you could just understand the string as a base 10 representation instead of base 2. No problems relating to multiple decimal expansions happen, because no 9's appear. And you can prove that you cannot index the real numbers that have only 0's and 1's in their decimal expansion, and since this is a subset of the reals, you have no hope of indexing the larger set also.
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Warp wrote:
This got me thinking: It seems to be kind of taken for granted that the set of integers is the "smallest" possible infinite set. Why?
Would the question "is there an infinite set whose cardinality is strictly smaller than that of the integers?" be something completely ridiculous to ask? Is the answer self-evidently "no"? Would it be as ridiculous as asking if there's a smaller non-negative number than zero?
I tried reading about cardinality and aleph numbers, but I didn't really find something that would definitely state that there cannot be an infinite set which cardinality is strictly smaller than that of the integers. Maybe because it's such a self-evident thing that it isn't even worth mentioning? But is it?
"Taken for granted" is not the right word, it is proven.
Suppose there is a set A whose cardinality is smaller than N. That means one can find a function f: A -> N such that f is injective. Now, look at its image, clearly there is a bijection between A and Im(f), so they have the same cardinality.
If Im(f) is finite, then A is also finite. If Im(f) is infinite, because it's a non-empty subset of the natural numbers, then by the well-ordering principle, it has a least element. So, we can define recursively a function g: N -> Im(f) such that:
g(0) is the least element of A
g(1) is the least element of A - {g(0)}
g(2) is the least element of A - {g(0),g(1)}
...
Notice that this function is well-defined, since if A is infinite, subtracting any finite set from A results in a non-empty set. It's also easy to prove it's a bijection. So there is a bijection between Im(f) and N, and thus also from A to N. Because of this, we have proved that if a set has the cardinality at most that of N, then it's either finite or has the same cardinality as N.
If you neglect the axiom of choice, it's possible to construct a set S whose cardinality is not comparable to that of N (there's neither an injection from S to N or from N to S), but it's still true that N is the smallest infinite cardinal.
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I might as well post the solution to my problem already. In my opinion, it's the first result in category theory that's difficult to see without resorting to it, and helps abstract lots of proofs. The trick is, for very abstract statements like this, for them to be true, the statement has to follow from piecing together the functions in the problem statement, and quite often you can do this by only "chasing" the types.
I'm taking the yellow images from here:
(1) If we have two natural transformations n: F -> G and m: G -> H, where F,G,H: C -> D (that is, all of them are functors from category C to category D), show that we can compose them, and that composition has the structure of a monoid (it's associative and has an identity). This is the so-called vertical composition of natural transformations.
(Replace alpha with n, beta with m)
Given this diagram, how could we possibly construct the natural transformation? For each object X in obj(C), we must associate a morphism. We have two morphisms available, nX from the transformation n, and mX from the transformation m. If we write them as a commutative diagram:
The first square commutes by the naturality of n. The second commutes by the naturality of m. Thus, the whole diagram commutes. If we think of the composition mX o nX, it satisfies the requirements for a natural transformation. So, to build the vertical composition, we simply compose the morphisms. It clearly has an identity (take the identity morphism) and is associative, because morphism composition is associative.
(2) If we have a natural transformation a: F -> G, where F, G: C->D, and a functor H: D -> E, show that we can find a natural transformation from H o F to H o G, called the right whiskering of a with respect to H.
This time, we have a morphism and a functor. The key insight is that the morphism aX maps objects in the category D and we can apply the functor H to it to make it map objects in E. So, we apply H to the whole square of the natural transformation:
By looking at this square, it's clear from the definition that H o a defines a natural transformation from the functor H o F to the functor H o G.
(3) Similarly, show that if we have a natural transformation b: G -> H, where G,H: D -> E, and a functor F: C -> D, show that we can find a natural transformation from G o F to H o F, called the left whiskering of b with respect to F.
Completely analogous, just write down the square for the natural transformation, and make it explicit that objects from D come from the functor F applied to an object in C:
From this diagram, it's clear that b o F defines a natural transformation from G o F to H o F.
(4) Finally, if we have two natural transformations a: F-> G and b: H -> K, where F,G : C -> D and H,K: D-> E, show that we can define the horizontal composition b o a : H o F -> K o G in two ways:
(a) Left whiskering followed by right whiskering
(b) Right whiskering followed by left whiskering
Also, show that the definitions (a) and (b) define the same natural transformation.
Just look at the types.
(a) We start with b from H to K, and forget the functor G. If we left whisker, we have the natural transfomation b o F from H o F to K o F. Now, if we start with a form F to G and ignore the functor H, we can right whisker and find the natural transformation K o a from K o F to K o G. Finally, we can vertically compose them to find:
(K o a) "o" (b o F): H o F -> K o G
Notice that here o denotes functor composition, while "o" denotes vertical composition of natural transformations, so I denote them differently, because they are different transformations.
(b) Do the same thing as (a). Start with a and forget the functor K. Right whisker to obtain H o a: H o F -> H o G.
Now, start with b and forget the functor F. Left whisker to obtain b o G: H o G -> K o G. Compose them vertically to find:
(b o G) "o" (H o A): H o F -> K o G
How do we prove these are the same thing? There are several commuting diagrams you can draw. I don't like the one on the slides I got the yellow images from. The one I like is:
You can think of it as a cube. Naturality and functoriality makes everything commute. All arrows from one square to the other are the functor applied to the morphism f, I didn't write it because it would be a mess.
The construction in (a) amounts to the path H(F(X)) -> K(F(X)) -> K(F(Y)) -> K(G(Y)), and the construction in (b) is simply H(F(X)) -> H(G(X)) -> H(G(Y)) -> K(G(Y)). Since the whole thing commute, these two things are equal.
Bonus: why are they called vertical and horizontal composition?
Compare the diagram in (1) to the diagram in (4). In (1), the transformations that are composed are drawn one above the other, while in (4), they are written next to each other. Because of that, (1) is vertical and (4) is horizontal. Of course, this is merely convention, and you can obviously draw things another way. However, these names have stuck.
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Ferret Warlord wrote:
Now clearly, actually understanding why this is the case is beyond me, but I do have to wonder, is this fact part of assigning the EM constant to Zeta(1)?
Pretty cool that you're exploring these things! One way to understand it is to simply plug the harmonic series into the definition of Ramanujan summation. You end up with the difference between an integral of 1/x, which gives ln n, and the n-th harmonic number, this is pretty much the definition of the constant.
To link with the zeta function, the idea is that it can be expanded as a Laurent series at s=1 like this:
zeta(s) = 1/(s-1) + gamma + O(|s-1|)
The first term obviously gives rise to a singularity at s=1, but there's a rigorous way of treating this pole as an "infinitesimal", it's pretty similar to the p-adic metric I talked about earlier. So, in this sense, you could write zeta(1) "=" gamma
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Warp wrote:
If the sum of the natural numbers is -1/12 according to these summation methods, why is the sum of the reciprocals of the natural numbers infinity?
(If you could keep the answer as close to arithmetic as possible, I would be grateful.)
Well, the method that sums the natural numbers to -1/12, Ramanujan summation, actually sums the reciprocals of the natural numbers to the Euler-Mascheroni constant.
Just so we're in the clear, I will not attempt to explain why only in terms of arithmetic for the following reason: there seems to be a widespread misconception that everything mathematicians do should be neatly expressible using numbers, addition, and multiplication.
This is completely false, it happens a lot that the methods used to derive results can be more interesting than the results themselves, so the more advanced results go completely beyond numbers and actually it's very hard to see why they're relevant if you express them in elementary terms, they just look like a garbled mess of propositions.
Incidentally, that's how I make exercises intended to be hard, like the integral I posted before. I start with some very abstract notions and specialize to more elementary things and see what the abstract notions imply. When I get something that's very had to see with elementary methods, I propose as a hard exercise.
If the topic was something like: why is the golden ratio the solution to the equation x^2=x+1, then certainly I would stay close to arithmetic. But that's not what the topic is about, the topic is about why it might make sense to assign 1+2+3+4+... to the value -1/12, and specifically how blackpenredpen's three-line calculation fits into that. I answered a technical level that I think is appropriate for this topic. To appreciate it, you of course need to read a considerable portion of the literature and understand some definitions, which takes even some proofs to show they make sense, and a lot of knowledge to contextualize them.
Doing all of this looks intimidating at first, but all the time I am very surprised by smart people that manage to do it. If, of course, it looks superfluous or uninteresting, then the answer to the original question would look just as superfluous or uninteresting.
I thought it was pretty interesting, and thought there would be other examples of matrices satisfying a similar property. It's barely possible to simply brute force all pairs of 2x2 matrices and search for the ones whose product is just digit concatenation, but out of boredom, I actually managed to come up with an algorithm to optimize this search using number theoretic concepts.
Anyway, now I have a list of all 2x2 matrices whose product is obtained by just concatenating their base 10 digits.
One thing that's very cool is that in some pairs the matrices are actually the same. Anyway, the problem is: come up with an algorithm, better than plain brute force, to find two single digit 2x2 matrices whose product is obtained by concatenating the base 10 digits of each element.
It turns out that the apparent paradox of three Magnemites having a mass 10 times greater than an isolated one is possible if we take into account quantum mechanics and special relativity. Intuitively, it can be viewed as follows. When two systems interact, they can have a binding energy as a consequence of this interaction. It's because of this energy that molecules do not break apart, for example. At the same time, the relativistic mass-energy equivalence E=mc^2 tells us that mass and energy are more or less the same thing, so if we have a lot of binding energy, it can manifest itself as mass.
The classic example is the QCD binding energy, which makes the proton much heavier than its constituent quarks, for example.
So, I know people posed this challenge as a joke, but science can explain the higher Magneton mass. In the Pokemon universe, the magnemites probably have a "Magnemite charge" which introduces a binding energy between them, and gives rise to this extra mass when they combine to form a Magneton :D
(Replace alpha with n, beta with m)
Given this diagram, how could we possibly construct the natural transformation? For each object X in obj(C), we must associate a morphism. We have two morphisms available, nX from the transformation n, and mX from the transformation m. If we write them as a commutative diagram:
The first square commutes by the naturality of n. The second commutes by the naturality of m. Thus, the whole diagram commutes. If we think of the composition mX o nX, it satisfies the requirements for a natural transformation. So, to build the vertical composition, we simply compose the morphisms. It clearly has an identity (take the identity morphism) and is associative, because morphism composition is associative.
This time, we have a morphism and a functor. The key insight is that the morphism aX maps objects in the category D and we can apply the functor H to it to make it map objects in E. So, we apply H to the whole square of the natural transformation:
By looking at this square, it's clear from the definition that H o a defines a natural transformation from the functor H o F to the functor H o G.
Completely analogous, just write down the square for the natural transformation, and make it explicit that objects from D come from the functor F applied to an object in C:
From this diagram, it's clear that b o F defines a natural transformation from G o F to H o F.
Just look at the types.
(a) We start with b from H to K, and forget the functor G. If we left whisker, we have the natural transfomation b o F from H o F to K o F. Now, if we start with a form F to G and ignore the functor H, we can right whisker and find the natural transformation K o a from K o F to K o G. Finally, we can vertically compose them to find:
(K o a) "o" (b o F): H o F -> K o G
Notice that here o denotes functor composition, while "o" denotes vertical composition of natural transformations, so I denote them differently, because they are different transformations.
(b) Do the same thing as (a). Start with a and forget the functor K. Right whisker to obtain H o a: H o F -> H o G.
Now, start with b and forget the functor F. Left whisker to obtain b o G: H o G -> K o G. Compose them vertically to find:
(b o G) "o" (H o A): H o F -> K o G
How do we prove these are the same thing? There are several commuting diagrams you can draw. I don't like the one on the slides I got the yellow images from. The one I like is:
You can think of it as a cube. Naturality and functoriality makes everything commute. All arrows from one square to the other are the functor applied to the morphism f, I didn't write it because it would be a mess.
The construction in (a) amounts to the path H(F(X)) -> K(F(X)) -> K(F(Y)) -> K(G(Y)), and the construction in (b) is simply H(F(X)) -> H(G(X)) -> H(G(Y)) -> K(G(Y)). Since the whole thing commute, these two things are equal.