Hey all. I am studying calculus 2 this summer and my professor has put forward a new criteria for getting an A and a quite unrealistic one at that. He said if using chatgpt or any assistance , we are able to create a new convergence test that has never been presented before (works even if it is applicable for even just a few types of series and fails for others) , we can get a good grade. Is it kinda possible that we do this? Has anybody got an idea. HELPPPP

p.s dont mind my english, its my 2nd language.

Bear with me, this is a maths question. A less whimsical way of asking it would be: "How to determine whether a set of points in a 2D plane is randomly distributed?"

So, when I'm out walking in places like the Lake District, I often see fields full of sheep, from elevated viewpoints where I can discern the spatial distribution of the sheep. Like [this one](https://www.geograph.org.uk/photo/570884), say.

The other day I got to wondering whether the sheep were randomly distributed. They certainly weren't all flicking together. But you know how, when you ask people to place points randomly in an area, they tend to make them all quite evenly spaced? I wasn't seeing that either. There were a few small groups of sheep, and areas where there were none, and it all looked like it might be genuinely random.

I thought that's quite interesting. You might think that sheep might prefer to be near other sheep, or might prefer not to be near other sheep, but if the distribution is genuinely random it suggests that they're simply indifferent.

Anyway, how to tell?

I can use knowledge of my position and the terrain to morph the photo into what it would look like from a vantage point perpendicularly above the field. I can assign (x,y) coordinates to each sheep and calculate the distances between each pair of sheep, or the distance of each sheep from its nearest neighbour, or anything like that.

But what then? What metric should I use, and how do I test for randomness?

Insertion Sort Algorithm:

The example with the solution:

This is wrong: If it actually belongs in position j, then k-j+1 comparisons will be used to move it, because there will be k-j+1 attempted iterations of the while loop...

Isn't it that there are k-j (not k-j+1) comparisons used to move the value a[k] to j position (it is correct that there are k-j+1 attempted iterations of the while loop)?

For example, if n=5 and we are inside the while loop and we want to move value in a[5] to a[1], we are going to need 4 comparisons (and 5 attempted iterations of the while loop).

So it looks like the solution (once again) computes the number of while loop iterations instead of the number of comparisons.

---

Of course, the rest of the solution is also wrong because we're making calculations with k-j+1 instead of k-j.

I don't know if this is the right sub(or flair), I don't know if this is math, physics or something else.

I asked my dad a question(it had something to do with points I think. I can't remember it right now) we then discussed and the answer i came to was that it depended on whether a vacuum has volume.

Imagine a separate dimension(or a pocket dimension), doesn't exist in this universe, that is empty has literally nothing does this space have a volume? And could you fit one vacuum in another?

I don't know if more context is needed if ot is just ask.

Again, I apologize if this is the wrong sub and thank you in advance.

I thought I couldn't do math for as long as I can remember. For context on my learning difficulty, I have no idea how to "do math" the way my classmates do theirs. I'm really slow and that's me being generous to myself. I see them do fraction problems in multilinear algebra and my brain.exe stops working, the symbols appear and reappear like magic to me. That changed recently when I figured out how to study on my own, I just realised how my brain works, that it sees shapes. How I do calculations is like animations moving on slides, differential is a shape losing energy, integral is a shape gaining energy, convex and concave is like walking on terrain while blind so I need to guess whether the next step will incline or decline, and so on and so forth.

I've done the algebra study, calculus I–III, analytic geometry, and the like, but now I'm working toward studying real analysis and people say it's very notation heavy and gets more abstract. The problem is my mind can only do up to 4th dimensional graphs to help me with math, manifold dynamics and the like. So far the book that has helped me most is Stewart, but even for the Calculus version, I needed other books with a geometry-first intuition. LLMs sometimes help me translate what an equation is actually doing into shapes I can understand, but as we know, I can't rely on them for everything due to user bias, etc. The question is any tips for studying real analysis? Any recommended books or courses? Thanks in advance.

TL;DR: My brain is one useless GeoGebra graph generator minus the immediate equation translation and I need tips for studying real analysis.

Hi everyone! I could use some help. I am in the middle of reading Halmos's lectures on Boolean algebras. He makes a distinction between ∧, ∨, and ' meaning "and", "or", and "not" as representing the Boolean object but meet, join, and complement as functions in the Boolean algebra. However, in the lectures, he doesn't seem to explain the distinction or assumes the reader is already familiar. I don't think I am. At the same time, I'm thinking about union (∪), intersection (∩), and difference (∖). Are all of these supposed to represent similar operations but in a certain order of operation? So, if A is all Boolean objects, B is all Boolean algebras on those objects, and C is all finite set theoretic constructions on A and B, then I can think of it as something like C(B(A)))? Thanks for any assistance!

I'm in a discord server with some friends and we have a text /command game that we all play. /work every 10 minutes earns you money, /collect once a day gets you a paycheck dependant on your job, etc.

With the money, you can buy items and/or gamble. One of the games is chicken fights. You buy a chicken, place a bet, and then typing /cockfight generates a win or loss, either taking your money or doubling it- pretty simple. Well, its absolutely smoking me and I feel like it shouldn't be.

The chicken has a 49% win rate to start, every win increases the win rate of the next fight by 1% (so 49%, then 50%, then 51%, etc.). when you lose, the probability resets back to 49%.

Mathematically, I thought this would equal out to like a 51-52% win rate. But i am getting crushed. See my 3 bouts below.

43 wins - 66 losses

46 wins - 53 losses

3 wins - 11 losses

Is my sample size too small? Are the odds worse than what I estmated? Am i just super unlucky?

Does Anyone know what the exact estimated win rate should be for the equation mentioned?

(edit: i have no idea how i made that bad of a mistake in the title)

So today i'm trying to figure out homothety which i already did but theres a problem, i cant use it while solving the homework (well its not homework but i dont know what else to call it) here is the problem i couldnt solve, at least with homothety,

Given ABCD trapezoid, AB is the biger bottom. AD=CD=1; AB=3, angle A = 2 times angle B; find BC and angle ABC

i'm on pc and i can't take proof of my work, thanks to everyone in advance!

Insertion Sort Algorithm:

The example with the solution:

The penultimate sentence of the solution a. doesn't imply the algebra that follows.

It says: Thus the maximum number of comparisions for a given value of k is k-1.

What should follow is this:

When k=2, the number of comparisions is k-1 = 2-1 = 1.
When k=3, the number of comparisions is k-1 = 3-1 = 2.
When k=4, the number of comparisions is k-1 = 4-1 = 3.
                    .
                    .
                    .
When k=n, the number of comparisions is k-1 = n-1.

By summing all the comparisions for each k from 2 to n, we get:

1 + 2 + 3 + ... + n-1 = (n-1)(n-1+1)/2 [by the sum of the first n-1 integers] 
                      = n(n-1)/2
                      = (1/2)n(n-1)
                      = (1/2)(n^2 - n)
                      = (1/2)n^2 - (1/2)n

---
Is my observation correct?

It seems so incredibly intuitive? Like I've been trying to prove it by showing that if there weren't twin prime numbers at all, then you couldn't uniquely factor all of the imaginable composite numbers. You could still "create" an infinite number of composite numbers, and uniquely factor them with all existing numbers, but you just couldn't create ALL of them.

Like for example imagine the set of primes (2,3,7,11,17...)

You could still make an infinite number of composite numbers, you just couldn't factor ALL of the POSSIBLE ones. Again, the FTA doesn't break, and so there's no "issue" really, logically it still makes sense.

But even if you just add 1 more twin prime to the set (2,3,5,7,11,17...) you get more composite numbers than you had before. And then you just keep adding twin primes to the set of numbers and if you ever miss one then there will be composite numbers you couldn't "make up" or create. There would still be infinitely many, and the FTA wouldn't break, you'd just have... Less... Like 99.9999% instead of 100% but OBVIOUSLY the number line is 100% of the composite numbers + all the primes and so you have to add infinitely many twin primes.

That seems reasonable enough but why is a proof so hard?

There are 64 green dots on a circle. Choose one dot and color it red. Repeat the following steps until there is only one green dot left.

Skip one green dot clockwise from the current dot and move to the next green dot. Color the dot you just visited red and connect it with a line to the previous dot.

No three lines intersect at the same point. How many intersection points are there, other than the vertices?

Ok, phenomenologically I understand it physically with the visual example of rotating the plate/cup (or a belt or something in general) with your hand, but structurally, why does that happen with the geometric configuration — the fact that your arm gets twisted/entangled (or whatever reference is rotating it)? And what is the analog-geometric structural reason why you need two 360° rotations for it to return to the same geometric orientation? Do you understand what I'm asking? I mean, could someone explain conceptually, describing step by step the analogic-geometric mechanism of why, geometrically, that structural geometric-configurative fact ends up happening in the spatial arrangement? Because that it happens, it happens — it's a fact, you can do it in real life — but the explanation of why this geometric structural behavior occurs in the disposition, I don't understand.

I'm working my way through a logic textbook and the first chapter's on set theory. The exercise is to prove that "For any sets A and B, if there is a one-to-one correspondence from A to B, then there is also a one-to-one correspondence from B to A" - likely it wants me to use some theorems proven in previous exercises, but since I wanted the challenge I tried to prove it without explicit recourse to those theorems about inverses (I do still make use of inverses in the proof). Can someone check to see if my proof is fallacious or misses something? Thank you.

Exercise: Prove that for any sets A and B, if there is a one-to-one correspondence from A to B, then there is also a one-to-one correspondence from B to A

Label the one-to-one correspondence/bijection from A to B f. Again, f: A→B is bijective. Since f is bijective, f is surjective (onto). So, for each b∈B there is some a∈A such that f(a)=b. Now, by the Axiom of Choice, there is a function h: B→A such that f(h(b))=b for every b∈B. Now suppose x∈B and y∈B and h(x)=h(y). Then f(hx)=x and f(hy)=y and, by Leibniz's Law, f(hy)=x [substituting hy for hx in f(hx)=x, justified by hx=hy] and thus x=y [substituting x for f(hy) in f(hy)=y, justified by f(hy)=x]. Thus for any b∈B and b'∈B if h(b)=h(b') then b=b'. And therefore h is injective/one-to-one. Suppose a∈A, then f(a)∈B and so f(h(fa))=fa. Because f is bijective, f is injective, so h(fa)=a. Thus for all a∈A, h(fa)=a. Since fa∈B, for each a∈A there is some b∈B such that h(b)=a. So, h is surjective/onto. Because h is injective and surjective, h is bijective. Therefore, for any sets A and B, if there is a one-to-one correspondence from A to B, then there is also a one-to-one correspondence from B to A.

For contrast, here's the proof I assume the textbook wanted because of its position in the exercises. Because f is bijective, f has an inverse (already proven). Call the inverse h. Since h is an inverse, it itself has an inverse function (f). Since every function that has an inverse is a bijection (already proven) and h has an inverse, h is a bijection. Thus for any sets A and B, if there is a one-to-one correspondence from A to B, then there is also a one-to-one correspondence from B to A.

Is there a consensus on using LaTeX for graduate school applications in math? Is it recommended or considered overkill? I’d probably be using it for my CV and other written material.

Currently self-learning Calculus BC through Khan Academy, I am right now at Unit 7 Introducing separable equations.

https://www.khanacademy.org/math/ap-calculus-bc/bc-differential-equations-new/bc-7-6/v/separable-differential-equations-introduction

Integration symbol will be referred as S()dx.

As shown at 3:30, Sal manipulates the equation

y*dy = -xe^-x^2 dx into

S(y)dy = S(-xe^-x^2) dx

I don't get it, why turn

-xe^-x^2 dx

into S(-xe^-x^2) dx

and not S(-xe^-x^2 dx)dx (it does seem very hard though, but thats besides the point)

I'd imagine an example such as x^3 be integrated as S(x^3)dx, so why does Sal drop the inner dx in S(-xe^-x^2 dx)dx?

let S(A)dx be an integral, then A should be the "height" and dx, the "length", shouldn't the height be -xe^-x^2 dx if we integrate the former?

Thanks in advance, I gotta sleep rn :3 ❤️

In the popular elementary probability problem where there is, say, 10 seniors and 5 freshmen, and you need to find the probability of 1 senior and 1 freshmen being picked from the group, why does the probability only account for unique combinations and disregard the order being picked? Does the problem assume that both students are picked at the same time?

If there is one reality where Bradley is selected first, and Caleb second, this feels quite causally distinct from the reality where Caleb is picked first, and Bradley second. It appears that there are two distinct sets of possible conditions that would bring the outcome of those two boys being picked. But in probability theory, they appear to only represent a single unique connection, and thus are counted as only one possible outcome.

Why is this the case? I can actually understand the problem if I do consider the two students as chosen simultaneously, like grabbing two marbles from a bag at once, but if one is chosen before the other, then it feels like the probability for seniors opens up way more. For n seniors there would be (n-1)^2 connections instead of half of that.

I’ve been sinking my teeth into the problem but only in my own head because I’ve only seen the problem mentioned various places and don’t remember where. But is it really as simple as the fact that I just disregard the notion of order in this problem to grasp its solution?

In physics some quantities are described by pseudovectors, namely the magnetic field B, though they are usually treated as vectors because there's usually no need to do transformations that would make the difference apparent. To calculate the Lorentz force we need to take the cross product between the velocity vector and the magnetic field pseudovector and get a vector as the result. Am I correct in saying that to do so we need to define a map that takes a vector and a pseudovector and produces a vector? Are there some properties that force us into a certain definition or are multiple definitions possible?

Why do we use n+1/2 for finding the median of a number,(what is the point of adding the 1?) whereas with a cumulative frequency diagram, we just do n/2? What does discrete and continuous data have to do with this if sometimes cumulative frequency (of whatever) can be discontinuous?

Something happening 98% of the time is the same as it not happening 1/50 times, whereas 99% is not happening 1/100 - so it's twice as good. 80% to 90% is 1/5 to 1/10, also twice as good. 50% to 75% is 1/2 to 1/4, again twice as good.

In all of the above I'm halving the gap between the initial value and 1, and by looking at the reverse in all cases it's twice as good (even going from 0% to 50% is twice as good looking at it in this way).

So in the title, if you go from 20% to 60% chance of something happening, looking at the reverse it's 4/5 of not happening to 2/5 not happening. Would this be a 2x improvement or is it 3x?

I'm noticing something amazing. When I:

• generate 1000 random numbers with mean=10 and STDEV=2,, and then I sort them ascendingly, and then take the accumulation curve in XLS (source code for those cells is of this format: =SUM($A$2:A2)/(SUM(A:A))

The graph generated looks very straight.

I was introduced to the PROBIT function last night. I'm wondering if there is any utility in doing what I've done.

If this curve has a positive or negative second derivative, does that mean anything, or is it ever used to describe these data points?

Ultimately, I'm interested in having something like a "Fast Fourier Transform for Segmenting Data Points". I'm trying to segment data points using mathematical filters just for my edification.

I play around with numbers pretty often. Yesterday I was playing around with the equation X^0.5 = x-1. (the square root of x = x-1). And I realized the answer is the golden ratio +1. I am guessing I am not the first one to make that realization. I was wondering if anyone can put some mathmatical logic to it.

I like to think about prime numbers a lot for fun and I've noticed a pretty easy pattern you can follow to visually represent the possible distribution of prime numbers by plotting a grid of all numbers with a consistent number of columns. The easiest pattern to use would be 10 columns, for base 10, and you can immediately see that all numbers greater than 10 that end in 0, 2, 4, 5, 6, or 8 are not prime, and any other number could be prime (not always obviously). The obvious reasoning is because those numbers are divisible by 2 or 5 which are the prime factors of the base you used. You can use this with any base system to do the same. 6 is a good example, and you can actually visually represent the proof for all prime numbers greater than 3 being representable as 6n ± 1 like below:

This is a cool way to look at primes in my opinion and it's very visually easy to follow, but for the life of me, I cannot find a name for this to use and I'd love to research practical uses for this or other proofs you can represent visually this way such as 6n ± 1. Could anyone help?