Joined: 12/28/2013
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Location: Rio de Janeiro, Brasil
I believe if we take a = b = c = l/2 + t, for a very small t, it's not hard to see it's impossible to construct the pyramid, unless there is something I understood wrong. The top vertex would need to be in the intersection of 3 regions that don't intersect at all.
With a = b = c, that's the simplest case, as I understand it, seeing as the three sides of the pyramid will be identical. The top vertex would be above the dead center (centroid it's called?) of the base.
At least, if you take a triangular pyramid to mean a pyramid made out of four triangles (http://mathworld.wolfram.com/TriangularPyramid.html). How do you understand it Bruno?
Joined: 12/28/2013
Posts: 396
Location: Rio de Janeiro, Brasil
My point was that the top vertex being above the centroid requires t to not be too small. If a, b and c are just slightly bigger than l/2, I believe the construction is impossible
I see what you mean. If a=b=c, then they need to be > l/2cos(30°) (about 0,577l). If you choose a = 0, then b and c have to be at least l.
So if a, b, c > 0,577l, then it's possible for sure, but if one of them is less, the others have to compensate.
Prove or disprove:
Let a, b, c be real numbers. Suppose that l is a real number satisfying l > max(|a-b|,|b-c|,|c-a|) and l < min(a+b,b+c,c+a), so that it's possible to form a triangle having one of the sides measuring l, and any two numbers from a, b and c. Then, in these conditions, it's always possible to construct a triangular pyramid having an equilateral triangle of side l as basis, and having a, b and c as the lengths of the remaining edges.
If false, try to come up with a condition that allows the pyramid with equilateral basis to be constructed.
Ptolemy's inequality on the tetrahedron (=triangular pyramid) with equilateral triangle base of side l and other edges a,b,c gives al+bl>cl, which gives a+b>c. Note that it is equality only if the four points lie on a circle in a plane, which is not the case here. Likewise b+c>a and c+a>b. If any one of these inequalities does not hold, then there exists no such tetrahedron. E.g. For a=1, b=1, c=2.5, l=1.75, it is not possible, even though the triangle inequality on all faces is satisfied.
Note that a+b>c, b+c>a, c+a>b is still not sufficient; for example, as explained in the posts above, l=1, a=b=c=0.55 still does not give a tetrahedron even though all necessary conditions given so far are satisfied.
Searching online reveals a paper giving a necessary and sufficient condition: Six given edge lengths determine a tetrahedron iff the triangle inequality holds for all faces and its Cayley-Menger determinant is greater than 0. Using the last formula on the Wikipedia page and substituting A=a2, B=b2, C=c2, L=l2 gives:
4ABC+(A+B-L)(B+C-L)(A+C-L)-A(B+C-L)2-B(A+C-L)2-C(A+B-L)2>0
or, after expanding and writing as a polynomial in L:
-L3+(A+B+C)L2+(AB+AC+BC-A2-B2-C2)L>0
Since L must be positive, divide by -L to get:
L2-(A+B+C)L+(A2+B2+C2-AB-AC-BC)<0
Now A2+B2+C2-AB-AC-BC=(1/2)((A-B)2+(B-C)2+(A-C)2)>0, and A+B+C is positive, so solving x2-(A+B+C)x+(AB+AC+BC-A2-B2-C2)=0 gives us positive solutions for x. We get:
x=(1/2) (A+B+C ± sqrt(3)sqrt(2AB+2AC+2BC-A2-B2-C2))
But note that 2AB+2AC+2BC-A2-B2-C2 = (a+b+c)(-a+b+c)(a-b+c)(a+b-c). And so this gives us:
x=(1/2) (a2+b2+c2 ± sqrt(3)sqrt((a+b+c)(-a+b+c)(a-b+c)(a+b-c)))
Thus:
(1/2) (a2+b2+c2 - sqrt(3)sqrt((a+b+c)(-a+b+c)(a-b+c)(a+b-c))) < L < (1/2) (a2+b2+c2 + sqrt(3)sqrt((a+b+c)(-a+b+c)(a-b+c)(a+b-c)))
and since this interval is in the positive reals, this gives for l:
One more thing: how do we know that (a+b+c)(-a+b+c)(a-b+c)(a+b-c) must be positive? Well, that just follows from a+b>c, b+c>a, c+a>b which I derived from Ptolemy's inequality above. So the necessary and sufficient conditions that allows the equilateral base to be constructed are l > max(|a-b|,|b-c|,|c-a|), l < min(a+b,b+c,c+a), a+b>c, b+c>a, c+a>b, and the bound on l above.
Nice solution, Fractal. I also used the determinant. I'm still wondering if we can derive this formula geometrically, though. The big square root inside is proportional to the area of the triangle of sides a,b,c. I have no idea how to carry it out, though.
I also have to confess something about this next one, I haven't proved it! But if you understand the combinatorics, and run some cases on a computer, an obvious pattern emerges.
You are the owner of a casino and obviously want to rip people off using math. You have come up with a brand new game for your clients.
The game works like this. The dealer deals an amount N>=2 of cards face down in a straight line (the content of the cards doesn't matter). The player and the dealer take alternating turns, with the player starting. At any point, the person who's playing can either:
(a) Remove exactly two consecutive cards from any point in the interval.
(b) Remove a single card, provided that it's isolated from the rest of the pack.
The game goes on like this until it's someone's turn and there are no cards left. In that case, the person who has the turn loses.
To add complexity to the game, the number N of cards initially dealt is random. Given an amount of cards, assuming the player and the dealer both play perfectly, the outcome depends only on the initial number N. For perfect skilled dealers and players, what is the fair payoff for this game? (i.e., if the player loses all the money when losing, how many dollars for every dollar bet do you return to the player when he wins to break even?)
The game works like this. The dealer deals an amount N>=2 of cards face down in a straight line (the content of the cards doesn't matter). The player and the dealer take alternating turns, with the player starting. At any point, the person who's playing can either:
(a) Remove exactly two consecutive cards from any point in the interval.
(b) Remove a single card, provided that it's isolated from the rest of the pack.
The game goes on like this until it's someone's turn and there are no cards left. In that case, the person who has the turn loses.
I'll just cut straight to the answer since it would otherwise take a giant post to explain every little thing here:
This game is the octal game ".17". This OEIS sequence gives the nim values (or "Sprague-Grundy values") for this game; starting N=0:
0,1,1,0,2,1,3,0,1,1,3,2,2,3,4,1,5,3,2,2,3,1,1,0,3,1,2,0,1,1,4,4,2,6,
4,1,1,0,2,1,3,0,1,1,3,2,2,3,4,4,5,7,2,2,3,1,1,0,3,1,2,0,1,1,4,4,3,6,
4,1,1,0,2,1,3,0,1,1,3,2,2,3,4,4,5,7,2,2,3,1,1,0,3,1,2,0,1,1,4,4,3,6,
...
The starting position of N cards is a win for the player (who has the first move) if and only if its nim value is nonzero. Since it is periodic with period 34, and only 4 values in each period are zero, assuming that there is no limit to how large N can be, this gives a winning probability of 30/34 = 15/17. So a fair payoff would be 2/15 a dollar for every dollar bet.
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PS: Every position in an impartial game under normal play convention (last to play wins) can be assigned a nim value; if the nim value is n, there exist moves to positions with nim value 0, 1, 2, ..., n-1, but no moves to positions with nim value n. Hence positions can be thought of as heaps in a slightly-modified Nim game. Positions that can be divided into independent subpositions have a nim value equal to the nim sum (binary XOR sum) of the nim values of all its subpositions.
The nim value of a position can be calculated as: a position where no move is possible has nim value 0, and for all other positions, the nim value is the smallest non-negative integer not appearing as the nim value of any position it can move to. Using this, it is possible to calculate the nim values for the octal game ".17", and it turns out to be periodic with period 34 and just 4 exceptional values; this can be verified computationally.
Yup, that's the solution. Interestingly, even though it makes no sense for the sequence to be periodic, by reading Conway's book where he treats these kind of games, he proves that if you compute the sequence long enough, the computation itself is proof that it'll keep periodic forever!
That's very interesting and comes from the definition of the mex function (which you use to calculate the nim numbers) when applied to a periodic array.
-------------------
Now, some trigonometry:
Prove that sin(10o)cos(20o)cos(40o) = 1/8
Wow, that was fast. Another solution is to multiply and divide by sin(20o), use sin(10o)=cos(80o) and use the double angle formula: sin(2x)=2sin(x)cos(x)
sin(20o)cos(20o)cos(40o)cos(80o)/sin(20o) =
1/2*sin(40o)cos(40o)cos(80o)/sin(20o) =
1/4*sin(80o)cos(80o)/sin(20o) = 1/8 * sin(160o)/sin(20o)
And since sin(160o)=sin(20o), it equals 1/8.
-----------------
Number theory:
Find all pairs of positive integers n and m, such that n/m + m/n is also an integer.
Interestingly, even though it makes no sense for the sequence to be periodic, by reading Conway's book where he treats these kind of games, he proves that if you compute the sequence long enough, the computation itself is proof that it'll keep periodic forever!
So you do know about Winning Ways for your Mathematical Plays. It's one of my favorite mathematical books (or rather, set of volumes); it covers not just impartial games, but the angel problem (among other various combinatorial problems), as well as Conway's Game of Life.
By the way, computation itself may be used to prove periodicity of nim sequences, but this is assuming of course that the nim sequence is periodic. It is conjectured that all octal games have periodic nim sequence, but that is not proven, and there are more than a few octal games whose periodicity has not been determined even when searching up to 233.
p4wn3r wrote:
Prove that sin(10o)cos(20o)cos(40o) = 1/8
p4wn3r wrote:
Another solution is to multiply and divide by sin(20o), use sin(10o)=cos(80o) and use the double angle formula: sin(2x)=2sin(x)cos(x)
You can also multiply and divide by cos(10o), use the double angle formula, and use cos(10o)=sin(80o). Same idea. Though I never knew there was a way to do this without using cubics (casus irreducibilis was my first thought for this question).
p4wn3r wrote:
Find all pairs of positive integers n and m, such that n/m + m/n is also an integer.
Suppose n,m is such a pair. Without loss of generality, assume n≥m. If n>1, then there exists a prime p dividing n. Since n/m + m/n = (n2+m2)/(mn), it follows that if p divides n, then p divides n2+m2, and therefore m, as well. Then we can write n=pc, m=pd, and so n/m + m/n = (pc)/(pd) + (pd)/(pc) = c/d + d/c, and so c,d is a smaller pair satisfying the above statement. So we can cancel common prime factors in n and m until it is no longer possible to do so, which is the case n=1. It is easily checked that n=m=1 satisfies the statement, and so solutions occur exactly when n=m.
Joined: 12/28/2013
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Does the following sum always converge, for all non-zero values of a and b?
What if n, in the denominator, is raised to c, 0<c<1? Can this sum diverge for small enough values of c?
That's a complicated question.
To answer the simple part: No, it might fail to converge.
First notice that the absolute values of the terms is 1/n. Therefore the series of the absolute values is the harmonic series, which diverges. From this, we conclude that the series, if it does converge, it does so conditionally.
To make it diverge, simply set a = 1, b = 2pi/3. Now look at the remainder of n when divided by 3:
* If it's 1, then the term is 1/n
* If it's 2, then the term is -1/n
* If it1's 0, then the term is 1/n
From this, you can rearrange it to a series with term:
1/(3n+1) - 1/(3n+2) + 1/(3n+3) = [(3n+2)(3n+3)-(3n+1)(3n+3)+(3n+1)(3n+2)]/(3n+1)(3n+2)(3n+3)
Now you have some messy algebra, but you get 9n² + something in the numerator and 27n³ + something in the denominator. From this, you can set up some inequalities like:
9n² + something > 8n² after some large n (that happens because a quadratic term grows faster than the rest).
27n³ + something <28n> 8/28n, so your term is larger than an harmonic one. Since the harmonic series diverges, so does your series.
Since it already diverges for n in the denominator, changing it to nc, for 0<c<1 only gives you a larger term, so this series can also diverge.
------------
Now, for the general question of when it diverges, and when it converges. If b/2pi is a rational number, then the exponents are eventually periodic, you can grab all terms in the period, and if you have an equal amount of terms with + and - signs, you can prove it's smaller than something times 1/n², which converges. Therefore, the series converges. If there are different quantities of signs, it will diverge.
Now, if b/2pi is irrational, the sequence of signs is not periodic. However, we have the equidistribution theorem to help. Basically, as n goes to infinity, the probability of hitting a point in the interval [0,2pi[ is uniform. From there, you expect a*sin(bn) to behave like an arcsine distribution. From this, you can find using integrals the probability of the floor values. If it's 1/2 for even/odd values, it should converge.
The problem is to prove this rigorously, the switch from a series to a continuous probability distribution is very complicated, because you have to make the error terms explicit and be creative with the bounds to show that the error terms do not spoil the asymptotic behavior. This reminds me a lot of proofs in analytic number theory, I wonder if this exercise was inspired by that.
About the series with n^c, when the signs alternate (+-+-+-...), it's the same as the Dirichlet eta function, which does converge for all c>0. The proof is more elaborate, because it involves some complex bounds. See here.
Riddler Classic this week has an interesting question. I'll paraphrase it below:
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You need to fairly choose one of N people to be the winner, using a coin. If N=2, then you just flip a fair coin to decide the winner; heads for one person winning and tails for the other. If N=4, then flip the coin twice; the outcomes are HH, HT, TH, TT with 1/4 probability each, and just assign each outcome to each person.
However, if N=3, it is not possible to assign the outcomes of a finite number of flips to the three people to fairly choose one of them with probability 1/3; for k flips, every outcome has probability 1/2^k, so no matter which outcomes you take, you can never get their probabilities to add to 1/3.
However, you can choose to use an unfair coin (probability p of landing heads, 1-p of landing tails) and select any p between 0 and 1 that you wish.
The question is: Is it possible to choose such a p so that you can flip the coin a finite number of times and assign the outcomes to fairly choose one of three people, and if so, how?
How would the question change if there were five people (N=5)? Or for the case of N people in general?
(Feel free to use WolframAlpha; I used it to help with this problem.)
Here's what I came up with.
Toss the unfair coin 4 times. If it lands all heads or all tails, choose person A.
Now, for the other possibilities, notice that there are 4 ways of getting 1 H and 3 T's, 6 ways of getting 2 H's and 2 T's and 4 ways of getting 3 H's and 1 T. For each of these, choose half and assign to person B, and the other half assign to person C.
With this, you guarantee that the probability of B being selected is the same as C. So, if you make the probability of choosing A 1/3, the probability of choosing the other two will be 1/3 for each one also. It turns out that the probability of choosing A is simply p^4 + (1-p)^4.
Therefore, solving p^4+(1-p)^4 = 1/3, which does give a number between 0 and 1, you can make your method work.
Now, for general N, it's easy to see if you solve for N prime, you solved for general N. For prime N, you should flip the coin a number of times m such that the m-th row in Pascal's triangle (except the extremes) are divisible by N-1. For that number of flips, you can put N-1 people in the combinations of these rows, and assign the last one to the extremes.
Divisibility patterns in Pascal's triangle are complicated, but iirc you do get an infinite number of m for which all the non-extreme values in the row are multiples of a given number. From that you have to solve p^m + (1-p)^m = 1/N.
Since m can get arbitrarily large, you should always be able to find a p between 0 and 1 satisfying that, so that construction should always work.
Joined: 12/28/2013
Posts: 396
Location: Rio de Janeiro, Brasil
p4wn3r wrote:
Now, for general N, it's easy to see if you solve for N prime, you solved for general N.
I don't find it that obvious. It's easy to see that by solving for all primes you solve for powers of primes, but how about products of different primes? Not that this maters if for all N there is a Pascal Triangle row with the property you conjectured. Looking at small values of m, though, I couldn't find a row with all elements multiples of 4.
Also, nice work on the summation problem I posted. I hadn't considered picking b so that b/(2pi) is rational. The most challenging scenario is when it isn't, but after seeing your input I don't know if I'll adventure myself into trying to prove anything here.
Now, for general N, it's easy to see if you solve for N prime, you solved for general N.
I don't find it that obvious. It's easy to see that by solving for all primes you solve for powers of primes, but how about products of different primes? Not that this maters if for all N there is a Pascal Triangle row with the property you conjectured. Looking at small values of m, though, I couldn't find a row with all elements multiples of 4.
Well, if you have N = pq, then you can make p groups of people, each having q. Then you perform the process for p, to choose the group which survives. Then, you change the process to select the q people in the group. But now, that I think of it, this new process might require you to change p, so this would not work.
About Pascal's triangle, it seems that you're right. I made a program to compute it mod 4 and did not find a row for the first 100 iterations. Maybe that property only works for prime n, I don't remember.
Anyway, I still think you can adapt this method to work. The equation p^m + (1-p)^m = 1/N can always be solved for large enough m. That's because the maximum is at p={0,1}, which is simply 1, and the minimum is at p=1/2, which is a very small number if m is very large.
So, even if the rows in Pascal's triangle are not all divisible, you can pick a huge m, and set person A at the extremes, and for every row you put A on m mod (N-1) combinations. If m is large enough, these new combinations should be very small, and the equation still solvable.
For example, if N=5, looking at Pascal's triangle mod 4, at row 8:
8: 1 0 0 0 2 0 0 0 1
So, the equation now is this, which also works to get a fair probability for everyone.
Therefore, solving p^4+(1-p)^4 = 1/3, which does give a number between 0 and 1, you can make your method work.
One of the solutions is p = 1/2 - sqrt(12*sqrt(6)-27)/6 ≈ 0.24213; you can solve it by substituting p = 1/2 + x and expanding to get 2x4 + 3x2 - 5/24 = 0, which can be solved easily as a quadratic in x2.
p4wn3r wrote:
About Pascal's triangle, it seems that you're right. I made a program to compute it mod 4 and did not find a row for the first 100 iterations. Maybe that property only works for prime n, I don't remember.
It's not possible for all elements of a row of Pascal's triangle (other than the ends) to be a multiple of 4. The ith row of Pascal's triangle sums to 2i, a multiple of 4. Since the two 1's at the ends only add up to 2, there has to be at least one other number in the row that isn't a multiple of 4.
p4wn3r wrote:
So, even if the rows in Pascal's triangle are not all divisible, you can pick a huge m, and set person A at the extremes, and for every row you put A on m mod (N-1) combinations. If m is large enough, these new combinations should be very small, and the equation still solvable.
Good! This is what I eventually came up with when trying to show it is possible for all N. (I got tripped up on "m mod (N-1) combinations" for a while; I eventually figured out that you meant "m combinations of 'mod (N-1)'", though I think it should be m+1 rather than m.)
I use C(m,i) to mean the binomial coefficient m choose i. Assign, for each i between 0 and m inclusive, C(m,i) mod (N-1) of the C(m,i) outcomes of probability pi(1-p)m-i on A; everything else can be divided evenly among the N-1 others. Then the probability of A winning is:
f(p)=sum[0≤i≤m] (C(m,i) mod (N-1))*pi(1-p)m-i.
As mentioned before, f(0)=1 and f(1/2) = sum[0≤i≤m] (C(m,i) mod (N-1))*(1/2)m < (m+1)(N-2)(1/2)m. So to get 1/N as a possible output of f(p), we need f(1/2)<1/N so we can use the intermediate value theorem. So we just need f(1/2)<(m+1)(N-2)(1/2)m<1/N, or (m+1)(1/2)m<1/(N(N-2)), and this can clearly be satisfied for large enough m, since (m+1)(1/2)m goes to 0 as m goes to infinity.
p4wn3r wrote:
So, the equation now is this, which also works to get a fair probability for everyone.
This equation p8+2p4(1-p)4+(1-p)8 = 1/5 works out pretty nicely. The left-hand side is just (p4+(1-p)4)2, so after taking square roots, you can solve it similarly to p4+(1-p)4 = 1/3 above.
Note that for N=5, 8 flips is not the minimum number of flips required; it is possible to do it in 6 flips since the polynomial f(p) above with m=6 and N=5 can be made equal to 1/5. Though it would be a lot harder to solve even with the p = 1/2 + x substitution (you'd have a cubic in x2 now), especially if you don't have WolframAlpha.
I think it's time for an "abstract nonsense" exercise.
A category consists of a collection of objects, and for each pair of objects, a set of morphisms between them. The collection of objects of a category C is often denoted obj(C), but the collection is usually also denoted by C. If A,B∈C, then the set of morphisms from A to B is denoted Mor(A,B).
A morphism is often written f:A→B, and A is said to be the source of f, and B the target of f. Morphisms in a category have the notion of composition. There is a composition ◦: Mor(B,C)×Mor(A,B)→Mor(A,C), and if f ∈ Mor(A,B) and g∈Mor(B,C), then their composition is denoted g◦f. Morphisms are associative: f◦(g◦h) = (f◦g)◦h
For each object A∈C, there is always an identity morphism idA:A→A, such that when you (left- or right-)compose a morphism with the identity, you get the same morphism.
We also have a notion of isomorphism between two objects of a category (a morphism f:A→B such that there exists some - necessarily unique - morphism g:B→A, where f◦g and g◦f are the identity on B and A respectively), and a notion of automorphism of an object (an isomorphism of the object with itself).
Now, the problem:
If A is an object in a category C, show that the invertible elements of Mor(A,A) form a group, the so-called automorphism group of A, denoted by Aut(A). Show that two isomorphic objects have isomorphic automorphism groups.
A category consists of a collection of objects, and for each pair of objects, a set of morphisms between them.
I never really understood what category theory is supposed to be. Isn't a morphism just a "function", and an object just a "set"? Now I suppose the difference is that a function is a map from a set to another, whereas object and morphism are mathematical things that no one can make any assumptions about; the morphisms from A to B are literally defined by the elements comprising Mor(A,B) and nothing else.
OK, I oversimplified a bit. There are some rules you have to follow, and p4wn3r already described them above. Also, composition has to be associative.
I'll give this question a try.
So consider the morphisms from A to A, defined solely by the elements comprising Mor(A,A). I'm assuming that a morphism f in Mor(A,A) is invertible if there exists a morphism h in Mor(A,A) such that h◦f=f◦h=idA; h is unique and denoted f-1. So to show that invertible morphisms form a group, three of the properties are automatically satisfied (associativity, identity, invertibility) so all we need is to show closure. If f and g are invertible, then so is g◦f, since
(g◦f)◦(f-1◦g-1) = g◦(f◦f-1)◦g-1 = g◦idA◦g-1 = g◦g-1 = idA
and the other way around, likewise. So invertible morphisms of Mor(A,A) form a group denoted Aut(A).
Now consider isomorphic objects A and B; that is, there exists h in Mor(A,B) such that there exists some g in Mor(B,A) with g◦h=idA and h◦g=idB; g is unique and denoted h-1. ("Isomorphic" here has no meaning beyond this; objects A and B are just objects with no further specifications.) Let f be an invertible morphism in Mor(A,A), and consider h◦f◦h-1. By the property of composition, this is a morphism of Mor(B,B). Is it invertible? It sure is, since h◦f-1◦h-1 is also a morphism in Mor(B,B), and h◦f◦h-1◦h◦f-1◦h-1 works out to idB (and the other way around, likewise). So h◦f◦h-1 is an invertible morphism in Mor(B,B).
Now that we have that, we now define the function α: Aut(A)→Aut(B), α(f)=h◦f◦h-1. We will prove that α is a group isomorphism (we are now getting back into the more familiar notion of isomorphism). First, α is one-to-one, since if α(f) = α(g), then h◦f◦h-1 = h◦g◦h-1, and by left composing by h-1 and right composing by h, we get f=g. Second, α is onto, since for any morphism g in Aut(B), h-1◦g◦h is in Aut(A) (for similar reasons as to why h◦f◦h-1 is in Aut(B)), and α(h-1◦g◦h) = h◦h-1◦g◦h◦h-1 = g. Finally, α is a homomorphism, since α(g◦f) = h◦g◦f◦h-1 = h◦g◦h-1◦h◦f◦h-1 = α(g)◦α(f). Therefore, α is a group isomorphism, and so Aut(A) and Aut(B) are isomorphic as groups.
...
OK, that did feel like a lot of "abstract nonsense". Yet it feels like something I went through over and over during my university studies.
A category consists of a collection of objects, and for each pair of objects, a set of morphisms between them.
I never really understood what category theory is supposed to be. Isn't a morphism just a "function", and an object just a "set"? Now I suppose the difference is that a function is a map from a set to another, whereas object and morphism are mathematical things that no one can make any assumptions about; the morphisms from A to B are literally defined by the elements comprising Mor(A,B) and nothing else.
Thanks for pointing out about associativity. I somehow forgot to copy that part.
The category of sets has more structure than a general category. The category of sets somehow has to encode the recursive enumerability of sets, so that you can have peano arithmetic and the cantor diagonal argument on it. There are multiple ways to define the category of sets, though, each with its own characteristics.
FractalFusion wrote:
OK, that did feel like a lot of "abstract nonsense". Yet it feels like something I went through over and over during my university studies.
The thing is, there's nothing in category theory that is "new" in category theory, because the whole point of it is to abstract things from other disciplines. Category theory is just language, but it's language with logic that you already know.
If you look at the development of mathematics, you can see that the greatest progress has been made when people realized that two distinct areas were actually "the same thing", and then started to apply techniques from one area to the other.
For example, the series
1 - 1/3 + 1/5 - 1/7 + ... = pi/4
The proof of this involves some seemingly uncorrelated areas of mathematics. It arises by looking at the 1/1+x² function and integrating it. From it, you have algebra (expanding it into a geometric series), analysis (integrating it term by term and showing convergence), and geometry (the result is the arctan function, which has a geometric interpretation). From this point of view, you can try to come up with more general objects, and if they describe simultaneously algebra, analysis and geometry, you expect to find more general identities. Category theory allows you to do this systematically, it's in practice very hard to do this in an ad hoc way if the mathematical objects are too complicated.
Historically, the first problem that gave widespread adoption to category theory was that of elliptic curves over finite fields. If elliptic curves are defined over infinite fields (or, more precisely, those of zero characteristic), it was well known that you could interpret elliptic curves by looking at Riemann surfaces, lattices, modular groups of complex functions, and binary quadratic forms in number theory.
Yet, no one could think of a similar analogy for those over finite fields. It took nearly two decades to understand that they encoded the topological properties of some geometric spaces, because the construction was so difficult. In the Grothendieck philosophy, if you can "look" at distinct objects at different ways, then they should be defined in the same way. That culminated in the development of the notion of a scheme. Its definition is obviously highly abstract, because it actually tries to abstract four or five areas of mathematics at the same time! But once you do understand the definitions, the proofs are kinda natural.
It's very difficult to explain the huge amount of abstraction in modern mathematics. Like Fractal said, it's necessary to go through many different proofs and actually see that many are the same concept repeated over and over again. If someone has never done this, it does look like "abstract nonsense".
I like to make an analogy with software engineering. When you have a very complex problem to solve, you can actually just write a bunch of code, and solve it. Most likely, though, you'll end up with something that's very difficult to understand separately, and people will take a lot of time to analyze it. Another way is to write an API to solve some very general abstract problems, and go slowly making it more concrete until you solve the problem you have. That second approach usually works much better. That has something to do with the inventor's paradox.
In the words of Grothendieck:
Grothendieck wrote:
I can illustrate the ... approach with the ... image of a nut to be opened. The first analogy that came to my mind is of immersing the nut in some softening liquid, and why not simply water? From time to time you rub so the liquid penetrates better, and otherwise you let time pass. The shell becomes more flexible through weeks and months — when the time is ripe, hand pressure is enough, the shell opens like a perfectly ripened avocado!
A different image came to me a few weeks ago. The unknown thing to be known appeared to me as some stretch of earth or hard marl, resisting penetration ... the sea advances insensibly in silence, nothing seems to happen, nothing moves, the water is so far off you hardly hear it ... yet finally it surrounds the resistant substance.
It's usually said in popular circles that modern mathematics is too useless, but in my view, it has advanced a lot more recently than some hyped up fields, like materials science, and biotech. It seems that every year they promise some miraculous thing that after ten years turns out to not work.
In contrast, I see that people highly trained in mathematics in enterprise environments are increasingly able to take more tasks, and more efficiently, at the same time. There's an anecdote that physicists that joined the development of the radar in WW2 were outperforming people that were electrical engineers their whole life, just like today you see startups of 15 people scaring huge corporations. It all has to do because these smaller companies could understand some abstract parts of their business that their competitors still refuse to see.
The sine of the difference of two angles obeys this equality:
sin(A-B) = sin(A)*cos(B) - sin(B)*cos(A)
I can't help but to notice the similarity to the formula for rotating a 2D point around the origin by a given angle. If you have a point (x, y) and you want to rotate it around the origin by an angle A, you can multiply it by a rotation matrix, which results in the formula:
newX = x*cos(A) - y*sin(A)newY = x*sin(A) + y*cos(A)
I wonder if this similarity is just coincidental.
Indeed they are related. Here's one way to see it. Consider the point lying on the unit circle that makes an angle -B with the positive x axis (measured anticlockwise). Its coordinates will be (cos(-B), sin(-B)) = (cosB, -sinB). If you now rotate this point anticlockwise by A, it will move to a point that makes an angle A-B with the positive x axis, and the y coordinate of this final point should be sin(A-B) (by the definition of sin). This should be equal to the newY given by your formula, where the old point is (cosB, -sinB), so newY = cosB*sinA + (-sinB)*cosA, which is the same as sin(A-B).
I suppose the main point is that a 2D rotation by A-B is the same as the composition of two rotations, one by A and one by -B. You can look at sin(A+B) similarly and even the formulas for cosine by considering the x components instead.
One more thing: how do we know that (a+b+c)(-a+b+c)(a-b+c)(a+b-c) must be positive? Well, that just follows from a+b>c, b+c>a, c+a>b which I derived from Ptolemy's inequality above. So the necessary and sufficient conditions that allows the equilateral base to be constructed are l > max(|a-b|,|b-c|,|c-a|), l < min(a+b,b+c,c+a), a+b>c, b+c>a, c+a>b, and the bound on l above.
What if n, in the denominator, is raised to c, 0<c<1? Can this sum diverge for small enough values of c?
