r/math • u/inherentlyawesome Homotopy Theory • 4d ago
Quick Questions: April 02, 2025
This recurring thread will be for questions that might not warrant their own thread. We would like to see more conceptual-based questions posted in this thread, rather than "what is the answer to this problem?". For example, here are some kinds of questions that we'd like to see in this thread:
- Can someone explain the concept of maпifolds to me?
- What are the applications of Represeпtation Theory?
- What's a good starter book for Numerical Aпalysis?
- What can I do to prepare for college/grad school/getting a job?
Including a brief description of your mathematical background and the context for your question can help others give you an appropriate answer. For example consider which subject your question is related to, or the things you already know or have tried.
0
1
u/username_is_alread- 19h ago
I want to show that the limit as k goes to infinity of (1/(k+1)) * (k/(k+1))^k is 0.
From looking at a graphing calculator, it looks like (k/(k+1))^k converges to 1/e, so if I can somehow establish that, I can conclude that my original limit is 0 * (1/e) = 0 since the terms of the sequence can be written as products of terms of sequences, one of which converges to 0, and the other to 1/e.
I know that one characterization of e^x is as the limit of the sequence {(1 + x/n)^n}, but I wasn't able to express (k/(k+1))^k directly in terms of that, though (k/(k+1))^k is lower bounded by (1 - 1/k)^k = ((k-1)/k)^k.
Based on a graphing calculator, it looks like ((k-1)/k)^k + 1/k upper bounds (k/(k+1))^k and also converges to 1/e, so I figured maybe I could try applying the squeeze theorem.
However, I've hit a wall in showing that ((k-1)/k)^k + 1/k indeed upper bounds (k/(k+1))^k.. If you simplify, it amounts to showing that k^(2k) <= (k+1)^k * ( (k-1)^k + k^(k-1) ), but I'm not sure how to prove that.
2
u/maffzlel PDE 19h ago
I know that one characterization of e^x is as the limit of the sequence {(1 + x/n)^n}, but I wasn't able to express (k/(k+1))^k directly in terms of that, though (k/(k+1))^k is lower bounded by (1 - 1/k)^k = ((k-1)/k)^k.
If your exponent was k+1 you'd be there. Can you rearrange or multiply and divide by a well chosen term?
1
u/username_is_alread- 18h ago
Ah, I think I see. I asked somewhere else and they pointed out that a much easier way to reach the conclusion I wanted was to simply note that you can simply apply squeeze theorem directly to (1/(k+1)) * (k/(k+1))^k with 1/(k+1) as the upper bound and 0 as the lower bound.
That being said, I think I see what you're saying. Since (k/(k+1))^k is the product of the (k+1)th term of the sequence that converges to e^-1 and k/(k+1) (whose limit is 1), we can take those limits separately, and then use those to get that the limit of (k/(k+1))^k is the product of the limits, so e^-1 * 1
1
u/maffzlel PDE 16h ago
Yes the squeeze theorem is more efficient but you were nearly there with your own method so I went with that direction
And yes that second paragraph is correct except its the product of the k+1 th term and k+1/k not k/k+1 but this makes no difference in the limit being 1/e
1
u/Alternative-Way4701 1d ago
If we have a 3x3 matrix A, with the first row all with 1's and the second and third row with zeros:
A =
(1 1 1
0 0 0
0 0 0)
So we just get ATA as a 3x3 matrix with ones. When I am calculating the eigen values of A, I get 1, 0 0(which is obvious), but when I am calculating the eigen values of ATA, I get (3,0,0), since the trace of the new matrix ATA is now 3, so it makes sense for them to sum to 3. Does the theorem(Eigen values of A are lamda, so the eigen values of ATA and AAT are lamda squared) apply only if A has independent rows? I am not able to properly understand the concept of eigen values. Any help would be appreciated here, thank you very much :).
3
u/bear_of_bears 19h ago
You saw that AT A has an eigenvalue 3 with eigenvector (1,1,1)T . If you take A(1,1,1)T then you get (3,0,0)T which is sqrt(3) times the length of (1,1,1)T . So it is true that multiplying A times this particular vector scales it by sqrt(3), just that there is also a rotation so it isn't an eigenvector. This is getting at the idea of singular value decomposition. (1,1,1)T is a singular vector of A with singular value sqrt(3). In general it is true that the singular values of any matrix A are the square roots of the eigenvalues of AT A.
1
u/Alternative-Way4701 19h ago
Interesting, I never thought about it like that. In hindsight, yeah you're right to make that kind of comparison since when we do A = USigmaVT this Sigma is a diagonal matrix with the square root of the eigen values of ATA.
3
u/Pristine-Two2706 1d ago
(Eigen values of A are lamda, so the eigen values of ATA and AAT are lamda squared) apply only if A has independent rows
This only applies if A is normal, meaning (for real matrices) AT A = AAT.
0
u/Alternative-Way4701 19h ago
Hmm, okay! So if A has to be of rank n if it has order n. Is this what you mean by normal?
2
u/Pristine-Two2706 7h ago
I encourage you to read the entirety of my comment, rather than the first 7 words.
1
u/lucy_tatterhood Combinatorics 12h ago
No, normal means what the comment says it means. It has nothing to do with rank.
1
u/ecorda98 1d ago
What is the formula to find the diameter of a cone with the height and volume given? I know that the formula to find the cone’s volume is V = 1/3 * r²h but I’m not sure what to rearrange in order to find height
1
1
u/-building_ 1d ago
I noticed this about 1÷7. Does that mean something? Are there other numbers in which that occurs? I tested 1 to 20, and none other than 7 worked.
1
2
u/Syrak Theoretical Computer Science 1d ago
Multiplying by powers of 2 then by powers of 1/100 is the same as multiplying by powers of 1/50. So you're really computing the sum
7/50 + 7/502 + 7/503 + 7/504 + ...
Factor:
7 (1/50 + 1/502 + ...)
The sum of powers of X (= 1/50), when X < 1, equals X/(1-X):
7 ((1/50)/(1 - 1/50)) = 7 (1/49) = 1/7
This calculation really relies on 50 = 1 + 72
For any n, if you multiply n by powers of (1/(1 + n2)) and sum them, the result will equal 1/n. When n=7, we have 1+n2 = 50 which can be decomposed as a product of powers of 2 and 10.
1
u/HaHaLaughNowPls 2d ago edited 2d ago
I found a different long hand version of the choose function. Has anyone seen it before, if it would have any applications, and also how it relates to the original formula? The formula I found was prod as n goes from 1 to x of [(x-n+1)/n], and this formula is equal to xCn.
Edit: Sorry, I wrote x+n-1 originally
1
u/Langtons_Ant123 2d ago
I don't quite understand that formula--I think you might have made a mistake in writing it down. "n" is used in "xCn", which seems to imply that you're thinking of some fixed number n, but then you let n vary inside the product. (It's a bit like saying: what's the sum, from n = 1 to n = n, of 2n ?) If you say "the product as i goes from 1 to n of ((x - n + 1)/n)" (note the changed signs in the numerator), then that's right--it's a slight variation on one of the standard ways of writing binomial coefficients.
We have n choose k = n!/(n - k)! k! -- that's the most common form. The numerator, n! = n * (n-1) * ... * (n - k + 1) * (n - k) * (n - k - 1) * ... * 1 is divisible by the (n-k)! = (n-k) * (n - k - 1) * ... * 1 in the denominator, so we can cancel those and get n choose k = (n * (n - 1) * ... * (n - k + 1))/(k * (k-1) * ... * 1).
This is probably the second most common form: we usually abbreviate n * (n - 1) * ... * (n - k + 1) as the "falling factorial" (n)_k and write n choose k = (n)_k / k! Then we can rearrange a little: rewrite (n * (n-1) * ... * (n - k + 1))/(k * ... * 1) as (n/1) * ((n - 1)/2) * ... * ((n - k + 1)/k), then rewrite that as ((n - 1 + 1)/1) * ((n - 2 + 1)/2) * ... * ((n - k + 1)/k). But this is just the product, from i = 1 to i = k, of ((n - i + 1)/i). (Or, in your notation, the product from i= 1 to i = n of ((x - i + 1)/i).)
The second form is useful because you can use it even if you replace n with something that isn't an integer. Non-integer factorials are tricky to make sense of, but the falling factorial (x)_k = x * (x - 1) * ... * (x - k + 1) makes sense no matter what x is. These "generalized binomial coefficients" (x)_k / k!, where x is any complex number, are used in the generalized binomial theorem: (1 + x)a = sum from k = 0 to infinity of xk * (a)_k / k!.
1
u/HaHaLaughNowPls 2d ago
Oh yeah I see what you mean, I have the formula saved on desmos and I probably just remembered it wrong. I can send the link if it would help understand more. I think it may have been that where it says x at the top of the product it should have been so it would instead be x choose k but I can't exactly remember.
1
u/lucy_tatterhood Combinatorics 2d ago edited 2d ago
This formula doesn't really make sense as written: you say it equals xCn but there are different values of n involved. The product you've written is actually equal to (2x-1)Cx, but possibly you meant to write something slightly different? There is a formula for xCn that looks similar to this; see Wikipedia.
1
1
u/HaHaLaughNowPls 2d ago
I meant to write x-n+1 if that removes any confusion
1
u/lucy_tatterhood Combinatorics 2d ago
As written (product going from 1 to x), this just makes it always equal 1. To match that formula on Wikipedia, the product would be from 1 to n and the variable inside would have a different name.
1
u/HaHaLaughNowPls 2d ago
https://www.desmos.com/calculator/ibm3ufulsy Yeah, this was the formula I had yesterday. I think the reason I was confused is because I wasn't originally trying to find another way to write the choose function, I was just wondering how I could figure out the number of combinations you could have if you had n items and could pick as many of those items as you want. I eventually realised you could just use the sum of xCn from n=1 to x but yeah. Sorry if what I'm saying isn't that clear
2
u/johnlee3013 Applied Math 2d ago
Suppose I have a semi-metric d(x,y), defined on N discrete points, expressed as a matrix D_ij = d(x_i, x_j). (Semi-metric is a distance function that do not necessarily respect the triangle inequality, but is otherwise a metric). Is there a way to tell how close it is to a Euclidean metric?
That is, is there a constructive algorithm (could be a heuristic or approximation), to select N points {y_i} in Rm (you get to choose m, but a smaller m is preferable), such that the matrix D'_ij = d2(y_i, y_j), where d2 is the L2 norm in Rm, is as close to D as possible? ("close" can be measured in either Frobenius norm or any nontrivial norm you like)
I asked a related question here a few weeks ago, and I was pointed to the Lindenstrauss lemma, but I think it doesn't cover my case, as Lindenstrauss assumes that d is already Euclidean in some high dimensional space, and m is fixed.
3
u/bear_of_bears 16h ago
There is a criterion to determine whether your matrix D is exactly Euclidean in m dimensions. See this page: https://en.m.wikipedia.org/wiki/Euclidean_distance_matrix#Properties
Under "Characterizations," the boxed theorem. Basically, you apply a simple formula to the entries of D to get another matrix G, and it's Euclidean in m dimensions if and only if G is positive semidefinite with rank at most m.
In your situation, this means you can compute G and then try to find a "close by" positive semidefinite matrix of low rank. Singular value decomposition is the usual way to do this. I am informed by Wikipedia that this technique (along with variations) is called "multidimensional scaling" by statisticians. See: https://en.m.wikipedia.org/wiki/Multidimensional_scaling
Under "Details" it is clear that they are considering the same problem as you. The rest of the page is a great illustration of how statistics is a different field from math, for better or for worse.
1
u/Aggressive_Sink_7796 2d ago
In Conway's Game of Life, is there some kind of expression which let's us calculate the state of cell nxn (in, say, a grid of NxN with a known initial state) without actually evolving the states?
If not, maybe some way that let's us not calculate ALL of the intermediate states?
1
u/Dizzy-Reality4346 2d ago
I'm junior high g9 anything hard I need to be prepared for the next school year
1
3
u/shad0wstreak 2d ago
Have more people considered the possibility of a set theory built on quantum logic besides Gaisi Takeuti and Masanao Ozawa?
1
u/NumberBrix 2d ago
Is there an equation that characterizes all composite numbers? Let me explain better, is there an equation that is satisfied when the (or one of the) independent variable(s) is a composite number?
3
u/GMSPokemanz Analysis 2d ago
The MRDP theorem implies there is a finite collection of Diophantine equations with a parameter x that can all be satisfied if and only if x is composite.
https://en.m.wikipedia.org/wiki/Formula_for_primes gives a similar example for primes.
1
u/NumberBrix 2d ago
Thanks Pokemanz! I wrote this one:
cos(2πy/x)+cos(2πx)=2
For a preprint and I wanted to know if by any chance you had come across a similar one. In the domain 1 < x < y, x, y ∈ R>1 is satisfied only when y is composite.
3
u/Langtons_Ant123 2d ago edited 2d ago
What kinds of equations do you allow? The obvious answer is that a is composite if and only if the system of equations "a = bc, b ≠ 1, c ≠ 1" has a solution where b, c are natural numbers. But maybe you don't want to allow "not-equal" constraints like "b ≠ 1"--maybe you want, for example, a single polynomial equation which is satisfied iff one of the variables is composite.
In that case, the MRDP theorem says that, for any computable (and more generally, recursively enumerable) set S of natural numbers, there exists a Diophantine equation (i.e. polynomial equation with integer coefficients) f(x_1, ..., x_n, y) = 0 which has solutions if and only y is in S. Such an equation (in 26 variables) has been constructed for the set of primes, i.e. a Diophantine equation f(x_1, ..., x_26, y) = 0 which is solvable iff y is a prime (under the restriction that y is positive). I don't know of an explicit construction for the composite numbers, but maybe you could modify this one to find it, and in any case the MRDP theorem guarantees its existence.
1
u/NumberBrix 2d ago edited 2d ago
Thank you for the answer Ant123. For a preprint of mine I wrote this equation:
cos(2πy/x)+cos(2πx)=2
Which, in the domain 1 < x < y, x, y ∈ R>1 is satisfied only when y is composite. I wanted to know if by any chance you had come across a similar equation.
2
u/Economist294 2d ago
I want to know how binomial distribution applies to a continuum of trials. Suppose the probability of success is $p$ and failure is $1-p$. What is the probability that $x \%$ of the trials is a success.
1
u/GMSPokemanz Analysis 2d ago
By the law of large numbers, with probability 1 the amount of trials that will be a success will converge to 100p%.
2
u/Interesting_Bag1700 2d ago
Is addition unique in peano arithmétique? As in, is there any other operation (°) that satisfies these 2 properties For all a:a°0=a For all a:a°b=S(a°b) And also the peano axioms
1
u/-building_ 2d ago
2 to the power of a prime number is a number that always ends with 2 or 8 (except number 4). Is there some significance to this?
2
u/dogdiarrhea Dynamical Systems 2d ago
I think it’s just the case that 2 to the power of an odd number ends with 2 or 8 and every prime number is odd (except 2).
1
0
u/WhateverDood03 3d ago
Why is a division operation equal to a fraction where the dividend is the numerator and the divisor is the denominator?
(I'm talking early high school math in college so please explain it to me as though I'm a beginner. Thanks for reading.)
3
u/AcellOfllSpades 3d ago
What is a division? When we write a÷b, what do we mean? (I'll use ÷ for a division in this post, and / for a fraction.)
At an intuitive level, we mean "I have a objects, and I want to split them among b people evenly. How much does each person get?"
So say I have 7 cakes, and I want to split them among 3 people. I can just cut each cake into 3 pieces, and give the first person the first piece of each cake, the second person the second piece of each, and the third person the third piece of each. Then each person gets seven pieces. Each piece is a third of a cake, so each person gets seven thirds. That's what "7/3" means.
At a higher level, we understand division as "the thing that undoes multiplication". a÷b is "whatever number you can multiply by b, to get a".
And the number a/b fits that perfectly: if we multiply it by b, we do indeed get a. So a÷b is a/b, then!
0
u/ecorda98 3d ago
How to add a specific math symbol on a ti-84 plus calculator?
Trying to do an equation (finding side length of a cube with volume) and it requires this math symbol (³√). I don’t know how to add it on my calculator (it’s a ti-84 plus). I tried doing the squared button but all it leaves me is 2√
1
u/AcellOfllSpades 3d ago
I don't think it has a key for it. But the cube root is the same as just raising to the 1/3 power, so you can do that.
0
u/ecorda98 3d ago
What should I input in the calculator then to solve the problem? /gen
1
u/AcellOfllSpades 3d ago
Something like [^] [1] [/] [3], IIRC? It's been a while since I used a TI-84.
1
1
u/sqnicx 3d ago edited 3d ago
Let A be an algebraic algebra over a finite field F. Let E be the algebraic closure of F and consider the scalar extension of A over F, A⊗E. Let B=A⊗E. Take a bilinear form f:AxA→F such that F(x,x-1)=0 for all invertible x in A. Can you extend f to a bilinear form g:BxB→E so that g(y,y-1)=0 for all invertible g in B? From what I've researched so far I think there may be some restrictions over the characteristic of F.
I tried to define g as g(∑ai⊗𝛼i,∑bi⊗𝛽i)=∑𝛼i𝛽if(ai,bi) but could not succeed to show that g(y,y-1)=0 for all invertible g in B.
3
u/777upper 3d ago
Is it possible to prove that an axiomatic system has no equivalent system with fewer axioms? Has that been done before to a well known one?
3
u/ilovereposts69 3d ago edited 3d ago
If your axiomatic system has a finite number of axioms, then it can be formulated through a single axiom by concatenating all of them with "and".
Some systems (like ZFC and PA) have infinitely many axioms, and some of them can be proven to not be expressible in finitely many such axioms.
So just looking at the number of axioms, you can only really classify things into finitely axiomitizable and non-finitely axiomatizable systems.
You could also look at a given set of axioms and ask whether the system changes when you remove some of them. In some cases it always changes, meaning that none of your axioms follow from the others (like certain formalizations of the axioms of groups), but this isn't as easy for non-finitely axiomatized systems, because if you have a schema ranging over all formulas, removing a single axiom from that schema won't change your system at all.
2
u/777upper 3d ago
Is there no way to retrieve the effective number of axioms from that concatenation of axioms?
1
u/Esther_fpqc Algebraic Geometry 3d ago
I guess it classifies exactly like sets then, I could create an uncountable amount of axioms that can't be reduced (e.g. the theory of ℝ-algebras)
1
u/SouBPC 3d ago
hey guys, i heard a maths question of a boss offering bonus to its employees such that each employee can pick any bonus between 0 to 1 million dollars. however the condition is that if one third of the avg of bonuses picked by all employees is more than 333k $ than none of the employees get any bonus, so what should each employee choose to get maximum money. This question doesnt has those exact values and i am looking for the original question. Can anyone help me to find me the og one?
2
u/GMSPokemanz Analysis 3d ago
I suspect you mean https://en.m.wikipedia.org/wiki/Guess_2/3_of_the_average
1
u/ChobotsRobot 3d ago
Is this correct?
0.084 × 0.143 × 0.0025 × 0.003 × 0.05 equals 1/222,000,222.
?
1
u/AcellOfllSpades 3d ago
Not exactly equal, but close enough for all practical purposes.
1
u/whatkindofred 3d ago
I wonder though if there are any practical applications in which this approximation is more useful than the exact solution of 9009/2000000000000.
0
u/Independent-Ad-4712 4d ago edited 4d ago
Why does the shape of the eye and the cornea seem equally round from the front and from the side? If you look at the circle on the eyeball slightly from the side, it should be an oval, not a circle, right? But it is a circle. How is this possible from a geometric point of view?
2
u/edderiofer Algebraic Topology 4d ago
But it is circle.
It isn't a circle. You can determine this by Googling "face in profile" and zooming in.
1
u/Independent-Ad-4712 4d ago
I said that it seems like a circle if you look at it from the side a little, not completely from the side. Unfortunately, I can't send some pictures of it what I mean
3
u/edderiofer Algebraic Topology 4d ago
Yes, a circle viewed from the side a little, looks like a circle viewed from the front. I don't know what's so surprising about that.
1
u/OGOJI 3h ago
Forgive me for my psychological question: when you say you understand a proof does it mostly feel like you can "directly see it all at once" like a=a, or is it more of trust in a linear process like "at t1 I checked that p1 was valid, then t2 p2 was valid etc.." along with a kinda vague feeling that it all makes sense?