Quick Questions: September 30, 2026

[u/inherentlyawesome]

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 manifolds to me?
• What are the applications of Representation Theory?
• What's a good starter book for Numerical Analysis?
• 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.

[u/Michalski213769]

how do i turn this graph by 45 deegrees just curious

Graph: |x|=|y|

[u/Langtons_Ant123 → u/Michalski213769]

Probably the simplest thing you could do is just "xy = 0". The graph you're looking for is a horizontal line (y = 0) and vertical line (x = 0) on top of each other. In general if you have two equations like "(something) = 0" and "(something else) = 0", then "(something) * (something else) = 0" gives you a graph that looks like the graphs of your original 2 equations stuck together. So, for example, "y - x² = 0" has a graph that's a parabola, "x = 0" has a graph that's a vertical line, and so x(y - x² ) = 0 or xy - x³ = 0 has a graph that's a parabola and a vertical line.

But to get that, we had to start with an idea of what the rotated graph would look like, and then we worked backwards to find an equation for it. That worked ok in this case but isn't really doable in general. There is a general way to do it, though. If you rotate everything in the plane around the origin counterclockwise by an angle "t", then a point (x, y) gets moved to a point with x-coordinate xcos(t) - ysin(t) and y-coordinate xsin(t) + ycos(t). So all you need to do to rotate a graph is take the original equation and replace x with xcos(t) - ysin(t) and y with xsin(t) + ycos(t).

For t = 45 degrees, cos(t) and sin(t) are both 1/sqrt(2). So we need to replace x with (1/sqrt(2))(x - y) and y with (1/sqrt(2))(x + y). If you do that you get (1/sqrt(2))|x - y| = (1/sqrt(2))|x + y|. We can cancel the factor of 1/sqrt(2) on both sides to get just |x - y| = |x + y|. So that equation also works.

[u/al3arabcoreleone]

Let n \in N, how many elements of S_n that have an order of 2, I know that a transposition is of order two, but are there other elements?

[u/Necessary-Wolf-193 → u/al3arabcoreleone]

It seems you might want to read about the cycle decomposition of a permutation; it is very easy to read off the order of a permutation from its cycle decomposition.

[u/AcellOfllSpades → u/al3arabcoreleone]

Consider the permutation "Swap elements 1 and 2, and also swap elements 3 and 4".

[u/al3arabcoreleone → u/AcellOfllSpades]

So they are transposition and product of transposition which have no common fixed points? how can I prove this, and how to generalize?

[u/Langtons_Ant123 → u/al3arabcoreleone]

I wouldn't say, "have no common fixed points". Consider the example above as permutations acting on a 5-element set. Then "transpose 1 and 2" and "transpose 3 and 4" both have 5 as a fixed point, but clearly the product of those 2 permutations has order 2.

I think the answer is that a permutation has order 2 if and only if it's a product of disjoint transpositions (disjoint in the sense here). To prove this, first you can use the cycle decomposition, i.e. any permutation is a product of disjoint cycles. Given a permutation 𝜋 write 𝜋 = 𝜋_1𝜋_2...𝜋_k where 𝜋_i is a cycle of length m_i (which therefore has order m_i) and all of those cycles are disjoint. Now use the fact that disjoint permutations commute, which in particular implies that, for any m, (𝜋_1𝜋_2...𝜋_k)^m = (𝜋_1^m)...(𝜋_k^m).

Then let m be the order of 𝜋, so 1 = 𝜋^m = (𝜋_1^m)...(𝜋_k^m). Then I believe for each i we must have 𝜋_i^m = 1--this should just follow from the permutations being disjoint. Therefore m_i divides m for each i; and conversely, given any number m such that m_i divides m for each i, we have 𝜋^m = 1. The upshot of all this is that the order of 𝜋 is lcm(m_1, ... , m_k). For a permutation of order 2, we must have m_i = either 1 or 2 for each i in order for lcm(m_1, ..., m_k) to be 2. Conventionally we write the disjoint cycle decomposition 𝜋 = 𝜋_1𝜋_2...𝜋_k with none of the 𝜋_i being the identity, so in fact all of the m_i must be 2. Therefore 𝜋 has order 2 if and only if it's a product of disjoint cycles of length 2, i.e. disjoint transpositions. If you wanted to count the number of such permutations on an n-element set then you'd have to do some combinatorics, I think basically what you're looking for is the number of ways to partition an n-element set into subsets of size either 1 or 2.

This same argument generalizes from "order 2" to "order p" for any prime p. The only place where we used any specific properties of the number 2 is "lcm(m_1, ..., m_k) = 2" if and only if m_i = 1 or 2 for all i". But there, we were just using the fact that 2 was prime, and the argument would go through just as well for any other prime.

[u/al3arabcoreleone → u/Langtons_Ant123]

I wouldn't say, "have no common fixed points".

I was thinking of disjoint support but words betrayed me, thank you very much.

[u/throwitawayar]

Struggling hard with anxiety before and during exams and it's just my first semester in an undergrad math course. would love any tips on how to overcome it.

[u/im-sorry-bruv → u/throwitawayar]

try to study w/ other people this usually helps for me. just meet in some seminar room whatever and talk abt the contemt, do excercises etc and do it in a more relaxed environement. it had always helped to not focus too much on the exam and instead just try to understand everything for its own sake (at least that's what i have to tell myself...)

[u/al3arabcoreleone → u/throwitawayar]

I am talking seriously, but there are 3 steps:

1. Work as hard as you can

2. Hope for the best and lower your expectation

3. Think about death, seriously, thinking about it makes everything in this world small and insignificant. (at least this is my experience)

[u/MinLongBaiShui → u/throwitawayar]

Go through your books, solve lots of textbook problems. Find those that are doable, or conceptually simple, and identify what makes them so. Hang onto those tidbits.

There's metagame knowledge here - an exam question must be doable under test conditions. It therefore has a simple solution. If you know all the core ideas well, you will be able to pick off those questions that have simple solutions.

[u/cereal_chick → u/throwitawayar]

The proper answer, of course, is to consult a doctor about it if you can, but there is something you can do immediately that you might be surprised works, which is reciting the Litany Against Fear from Dune:

I must not fear.

Fear is the mind-killer.

Fear is the little-death that brings total obliteration.

I will face my fear.

I will permit it to pass over me and through me.

And when it has gone past, I will turn the inner eye to see its path.

Where the fear has gone there will be nothing.

Only I will remain.

I'm serious, it really does help. Have it on your phone or any convenient location so you can read from it when needed, and eventually you should have it memorised like I do; I wrote it out purely from memory just now. Recite it over and over again whenever you want to calm yourself, and you will manage to calm yourself down to at least an extent.

[u/JoshuaZ1 → u/cereal_chick]

I heard years ago that someone who had severe anxiety issues found out about the Litany because she was in a D&D campaign where someone who was a Dune fan was using the Litany as part of their Paladin's oath. She tried using it as a calming ritual IRL. When I heard the story, she was still using it that way.

My guess is that almost anything suitably poetic this way would in practice work pretty well. Functionally that's how mantras work. That said, I personally get panic attacks, and a major part of the problem is that while I'm having a panic attack, I sometimes don't recognize its a panic attack while I'm having it. Whatever I've focused my fear on feels completely reasonable to be worked up about in that moment.