Mathematics - 2016-05-24 (page 1 of 2)

7:26 AM

Can we get some quick extra flags on the answer to math.stackexchange.com/questions/1797673/… to get it removed faster?

@TobiasKildetoft If this is intended seriously, maybe c-r-u-d-e would be also a reasonable room for such requests.

@MartinSleziak I thought this room had more people so was likely to get there faster (I feel like an answer like that needs to become invisible as fast as possible).

And I think that the correct way to deal with this is to case a close vote. Then it get into close votes review queue and it will go fast from there.

@MartinSleziak Not the question, the answer

D'oh! You linked to the question.

7:33 AM

@MartinSleziak ahh, yeah, that was the faster one to link (didn't even consider linking directly to the answer)

Still, I think that moderators have a lot to do. If a community can deal with something without moderator intervention, it is always good.

So I voted to delete that answer. You can add your vote. IIRC only three votes are needed to delete.

I think it takes 20k rep to vote to delete answers

You have already flagged it as "not an answer'?

I do not think that adding more flags speeds up the process. It just adds to the number of flags moderators have to clear.

It is different with spam flags - enough spam flags lead to deletion. But this is not spam - by definition.

@MartinSleziak I thought I had flagged it as "rude or offensive" (quite frankly, I think the answerer is just begging for a suspension by leaving such an answer).

Ok, so it was flagged. If my understanding of flagging system is correct, with the exception of spam flags, it will have to wait for a moderator.

7:40 AM

ahh, I see

I should also say that I appreciate that you try to lead the OP in the comments and use your time to help the OP both understand the problem and improve their question.

I have mentioned this in the other room. Somebody may notice it there. (Mainly high rep users, which can vote to delete, frequent that room. And also some mods can be seen from time to time. Especially Daniel Fischer is often there.)

BTW only 5 users in the main chatroom. This is an unusually low number.

Thanks (also somewhat worrisome, the answer has been upvoted, though I have a feeling that might be a sockpuppet since the guy made a rude comment on the question telling me to fuck off (or something like that) and that was immediately upvoted).

@TobiasKildetoft I see that the comment is no longer there. Did you have time to flag it before it was deleted?

BTW I am afraid this comment might be misleading for the OP:

Re-read the definition of symmetric. It cannot be shown by example (though it can be shown not to hold by example if that is the case). — Tobias Kildetoft 3 mins ago

$\gcd(x,y)>1$ $\Leftrightarrow$ $\gcd(y,x)>1$. (So the relation is symmetric, unless I misunderstood something.)

@MartinSleziak Ahh, good point.

7:57 AM

BTW the answer is deleted now: "This answer was marked as spam or offensive and is therefore not shown." So either some users flagged it as spam. Or I was wrong and "rude or abusive" flags can lead to deletion of a post, too.

@DanielFischer I am not sure whether you came here in connection with the flagged answer. In any case, the comment mentioned here (now deleted) might also need moderators' attention.

Indeed, I was wrong: "How does the Spam flag differ from the Offensive flag? In terms of getting the post deleted, there is no functional difference aside from separate counts - 3/6 of either will be sufficient to delete." meta.stackexchange.com/questions/58032/…

Daniel Fischer

@MartinSleziak This is one of the tabs I always have open, I drop in here whenever I start my browser. Thanks for the tags, by the way. Indeed "rude or abusive" flags also lead to deletion of a post. If a post gets six flags of type "spam" or "rude or abusive" combined, it is deleted and locked. If a moderator flags as one of those, it's immediately deleted and locked. The matter has been taken care of before I started my browser, for the moment, I don't think any further action is required.

8:18 AM

Is a vector space of finite dimention is a complete space

?

@Vrouvrou being complete requires more structure than being just a vector space

@Vrouvrou You probably mean normed space (so that it makes sense to speak about completeness). If this is what you are asking, the answer is yes.

yes it is a normed space

and is a complete normed space of finite dimention is localy compact ?

Locally compact means that every point has a compact neighborhood, right?

For this, it suffices if you know that the closed unit ball in finite dimensional normed space is compact.

In fact, this characterizes finite dimensional normed spaces: Closed unit ball if a normed space $X$ is compact if and only if $X$ is finite dimensional.

yes i know it but the fact that it is complete imply that it is localy compact

8:22 AM

Being not a native speaker I always hesitate whether to say finitely dimensional or finite dimensional.

@MartinSleziak definitely not finitely (whether to put a space or a hyphen I am always less sure of)

There are many infinitely dimensional normed space which are complete. But since closed unit ball is not compact, they are not locally compact. So complete $\not\Rightarrow$ locally compact.

thank you @MartinSleziak

Related posts on the main: A normed space is locally compact iff its closed unit ball is compact.

Compactness, Local Compactness, Completeness

The latter has counterexamples for metric spaces. (Although it seems that you are mainly interested in normed spaces, so this post is less useful for you.)

I almost fell into the regex trap there, spending more time figuring out to make a regex for a replacement than the replacement would take by hand (replacing the notation $(\cdots )_{\nabla}$ with a different type of brackets).

Mary Star

11:20 AM

Hello!!

Could someone of you take a look at my question: http://math.stackexchange.com/questions/1793918/each-exact-sequence-can-be-arised-by-short-exact-sequences ?

@MaryStar Why did you link the newer version which is a duplicate of an older version (also by you) which is answered?

Mary Star

At the newer question I want to show that it holds in general...
At the beginning I asked it also at the older version but they told me to ask a new question for that... @TobiasKildetoft

Evinda

11:54 AM

Hi!!! It doesn't hold that $ \int_{-\infty}^{+\infty} \int_{-\infty}^{+\infty} |Du| dx_1 dy_2 = \left( \int_{-\infty}^{+\infty}|Du| dy_1 \right)\left( \int_{-\infty}^{+\infty}|Du| dy_2 \right)$.

Does it ?

Q: The largest product of two n-digit numbers which is palindrome

Is Du something specific here?

Because in general this does not hold. For example, take a function which is $2$ on $[0,1] \times [0,1]$ and 0 elsewhere.

Evinda

@SteamyRoot $Du=(u_{x_1}, u_{x_2}, \dots, x_n)$ @SteamyRoot

Secret

12:13 PM

My opinion based on my experience on all 3 math programs

Sidarth

1:20 PM

Hey...can someone explain this?
why is Aij(Xi +Yj) != AijXi +AijYj ??
in tensors...

N3buchadnezzar

1:49 PM

1

Project Euler: 4 is stated as follows: Largest palindrome product A palindromic number reads the same both ways. The largest palindrome made from the product of two 2-digit numbers is 9009 = 91 × 99. Find the largest palindrome made from the product of two 3-digit numbers. No...

Heh

2:31 PM

hi

is it possible that (a mod 5 and b mod 3) and (c mod 5 and d mod 3) where a,b,c,d are not equal could be the same number?

by "and" do you mean "plus"?

no

then I have no idea at all what you mean

i mean (a mod 5 and b mod 3) = (c mod 5 and d mod 3)

a mod 5 and b mod 3 is a number

or a class

ok sorry

let me phrase it differently

we know that a mod 5 and b mod 3 is a number modulo 15, but is it possible that there exist other different numbers c and d such that c mod 5 and d mod 3 is the same number modulo 15?

ah, you're speaking in Chinese Remainder Theorem terms

2:38 PM

yes, that's right

i don't see how. once you've got a number mod 15, the remainders mod 5 and mod 3 are determined

at most, you could have something like a and a+5. but those are the same class.

right

that leads to this question

Let $n$ be an integer that is not divisible by any square greater than $1$. Denote by $x_m$ the last digit of the number $x^m$ in the number system with base $n$. Prove that the sequence $x_m$ is periodic with period $t$ independent of $x$ and find the minimal period $t$ in terms of $n$.

I think

To prove that $x_m$ is periodic with period $t$ independent of $x$, we see that it is sufficient to find the residues modulo every prime factor of $n = a_1 a_2 a_3 \cdots$ where $a_1,a_2,\ldots$ are the prime factors of $n$ listed in ascending order. Then using Fermat's Little Theorem, $x^{a_i}\equiv x \pmod{a_i}$ and thus the periods of each of the modular congruences modulo $a_i$ is $a_i-1$.

Thus, since the residues of $x^m$ will be the same as the residues of $x^{m+t}$, modulo each prime $a_i$, period $t = \text{lcm}(a_1-1,a_2-1,\ldots) = \phi(n)$.

my question is is this minimal?

sorry, i don't know. i know basic modular arithmetic but i haven't done anything with it lately

3:11 PM

morning folks

good afternoon

how's things

morning @mike

3:26 PM

$$\sum \limits_{k=1}^{k=n} \mu(k) = 1 - \sum \limits_{a=2}^{a \leq n} 1 + \sum \limits_{a=2}^{a \leq \frac{n}{b}} \sum \limits_{b=2}^{b \leq \frac{n}{a}} 1 - \sum \limits_{a=2}^{a \leq \frac{n}{b \cdot c}} \sum \limits_{b=2}^{b \leq \frac{n}{a \cdot c}} \sum \limits_{c=2}^{c \leq \frac{n}{a \cdot b}} 1 ...+$$

Mats Granvik

3:40 PM

oeis.org/A255315

5:01 PM

Is anyone familiar with the number theoretic partition function and its derivatives?

And by derivatives, I don't mean differentials.

"ask; don't ask to ask"

That was my question.

I'm curious how many people are actually familiar with it.

Yes, someone is familiar.

The partition function such as the number of ways a number can be expressed as sums of different integers?

If so, then I know of it, but familiar depends on your definition of familiar. xD

you really didn't care about anything about the partition function? ok

Sawarnik

5:08 PM

@Mike Is it ok to differentiate f(x)=sinx*cosx as ln(y)=ln(sinx)+ln(cosx) and then chain rule?

Sorry, I'm back from a call.

Yeah, I did actually care about the partition function, @MikeMiller. I was just being silly.

that's what I assumed, but then you vanished :)

I'm trying to find out how to work with a variant of it, where I only consider partitions of at least size $k$.

Rather, partitions whose slices are at least size $k$.

You pray to the number theory gods and weep as they do not answer the calls of any man.

so you mean that each integer in the sum is at least $k$?

5:12 PM

Yes.

I'm coming into it from the idea of partitions on sets and their cardinalities, so forgive me if I forget the context is sums and numbers.

so you should be careful, because that's not the same thing

It's enough of the same thing, for the most part.

Operand order doesn't matter, like the slices in a partition. The whole must be the sum of its parts, just as the union of all slices must be the full set.

that's not true. there are two partitions of 3 in the number theory sense (3, 2+1); there are five partitions of the 3-element set (the trivial partition, three partitions into a set of 2 elements and 1 element, and one partition into three sets). if you know the set of partitions of n in the number theory sense of a particular size k you can recover both (the first is just the sum $\sum p_k$, the latter is $\sum \binom{n}{k-1} p_k$), but they're not really the same thing

Hmm, then I guess I'm looking for something more rigid than the partition function.

One that cares about order.

That's a shame. I need to dig deeper before continuing with this.

Thanks.

eh, actually my talk of recovering them is wrong, sorry.

5:18 PM

What exactly is your question, Ax?

so the situation is even worse than that.

@SAWblade How many ways are there to partition a set into slices of no smaller $k$ elements each?

On the outside, it sounds like Stars and Bars, but I don't know how many bars there are.

Say my set was cardinality $10$ and $k = 2$, then would $\{ 4,6 \}$ be a proper partition of this form?

just sum over the possible numbers of bars

from 0 to the floor of n/k

@SAWblade No, but $\{[1, 4]_{\mathbb N}, [5, 10]_{\mathbb N}\}$ would be.

5:21 PM

I'm really not understanding what you mean then, Ax. xD

$4$ and $6$ as elements are not sets, they can't be slices.

Well. They can be sets, if you really hate simple discourse on the matter.

But they're not sets, to keep things simple.

Ah, I get what you mean I think?

Does $[1,4]_{\mathbb{N}}$ have 4 elements in it?

it's just notation; that means {1,2,3,4}

Mmk.

Ah, I see where your problem becomes tricky.

A partition $P$ of a set $S$ is a set such that $\bigcup_{p \in P} = S$ and $P_i \cap P_j = \varnothing$ for two different slices $P_i, P_j$.

5:24 PM

I am aware of what a partition is, thank you. xD

probably better to phrase it as "is a collection of subsets" than "is a set"

Just in case.

Collection of subsets gets even nastier, if I allow collections larger than sets.

If it's a set, it's got nicer, tamer properties.

your sets $S$ are finite so... i wouldn't worry about that so much

That aside, I can't use the partition function as written.

If you call $P(n,k)$ the number of ways to write $n$ as the sum of $k$ integers, then what you're looking for seems to be $\sum _{i=1}^{\lfloor \frac{n}{k} \rfloor} P(2n-nk,i)$. :0

5:28 PM

Let me pick that apart slowly. It's a lot to chew on at once.

It's not the whole solution, which is where your nastiness comes in.

What is the interpretation of each summand?

Oh wait I think I might have it, gimme a sec.

Yeah I think your overall solution would be $$1 + \sum _{i=2}^{\lfloor \frac{n}{k} \rfloor} \binom{n}{i} P(2n-nk,i)$$

The formula is all well and good, but could you tell me how to interpret the summand you've chosen?

Floor of $\frac{n}{k}$? :0

5:34 PM

No. $a + b = c$, $a$ and $b$ are called summands.

So, in a sum, $\sum a_n$

$a_n$ are the summands

Sure, just gimme one second. :)

Thanks.

Ah, yeah, that's wrong. xD

I do not feel like I am in good hands here, @SAWblade

I never said you were! :0

5:40 PM

Potentially because you're named after something sharp and rapidly spinny.

All confidence in me is most assuredly fault on your part. xD

But I fundamentally misunderstood a part of your question and get it now. xD

That's fine. I fundamentally misunderstood a lot of your answers. :P

In the meantime, I'm trying to follow the transcript back and figure out what spawned Undo's 6-star quote.

What's his quote? :0

It looks like things escalated out of nowhere, potentially with a lot of deleted messages.

On the right, where it shows starred messages.

I see nothing by Undo. :0

5:45 PM

That's spooky.

god you have nerd sniped the hell out of me with this question

Hanan N.

my question is why the number $1$ in the integral becomes $T$?

shouldn't it be T/2

?

$\int dT = T$

Hanan N.

6:02 PM

@SAWblade Sorry but i don't get it, could you please elaborate?

The integral of $1$ with respect to $dT$ is always $T$. The reason it's not $\frac{T}{2}$ is because the $\frac{1}{2}$ is pulled out of the integral and combined with the other $\frac{1}{2}$ present outside of the integral to create the $\frac{1}{4}$ we see in the far right.

I hope I'm not interrupting anything, but I have a question that no one has ever really answered for me (or at least not given me a satisfactory answer)

Why can't infinity be treated as a number?

you have to be more specific about what you mean by "treated as a number" to get an answer

Well, the only time I've seen the infinity symbol written down is in infinite sums or in limits

but why do we need to say "limit as n approaches infinity"

and not just "infinity"

ah, so that's a well-defined thing I can answer. Let's not do series, let's do limits of functions. People often say "what is $\lim_{x \to \infty} f(x)$?" And, I assume, you'd like to respond "It's $f(\infty)$", yeah?

6:10 PM

yeah

So there are two reasons you can't do that. 1) If I hand you a well-known function, there might or might not be a natural way to define it at infinity. What is $\cos(\infty)$? Now, you could define this arbitrarily, but it wouldn't be super helpful. 2) Limits are about the journey, not the destination! Even if you had a notion of $f(\infty)$, you're more interested in the values near infinity, not at it.

This is the same reason when we take $\lim_{x \to 0} f(x)$ we can't naively just plug in $f(0)$ in every setting. (Think of the case where $f(x) = 0$ when $x \neq 0$ but $f(0)=1$. In this case, the limit is zero, but $f(0) = 1$.)

I see what you mean

especially about the cosine of infinity

Now it turns out (this won't show up in your calculus class) that there is a way of saying that a function is defined as a map $\Bbb R \cup \{-\infty,\infty\}$ to itself (so that points can take on infinite values, and that infinity takes on values). And I can make sense of what continuity means for these sorts of functions. Well, we have a general rule for normal functions - if $f$ is continuous, then $\lim_{x \to a} f(x) = f(a)$.

Lol I didn't get a word of that

In our "extended numbers" and "extended functions", this is still true. So it turns out that if $f$ is "continuous at infinity" (a reasonable thing to demand), then $f(\infty)$ is the same thing as $\lim_{x \to \infty} f(x)$. So we may as well use the latter. And if the latter doesn't exist? Then there's no way to choose what $f(\infty)$ is so that $f$ is continuous there.

Food for thought. I'm probably not going to elucidate it. If someone else wants to, they're free to.

6:15 PM

I think I get it. Not 100%, but I get the concept. Thanks

is LaTeX working on this chat for you?

See LaTeX in chat on the top right

Thanks man!

you might also find this helpful

72

Q: Is infinity a number?

Is infinity a number? Why or why not? Some commentary: I've found that this is an incredibly simple question to ask — where I grew up, it was a popular argument starter in elementary school — but a difficult one to answer in an intelligent manner. I'm hoping to see a combination of strong citati...

definition infinity

Thanks

Ilan Aizelman WS

Would love some help here: math.stackexchange.com/questions/1798418/…

6:23 PM

Ha I would help, but I have no idea what your question means. You would think that after AP calc, I could do these things, but I can't

It's a linear algebra question. There's a lot of math beyond calculus! Linear algenbra is one of the first things one needs to learn after/while learning multivariable calculus.

Mats Granvik

@Axoren oeis.org/A160096

I wish my school offered more math.

Because I freaking love math, but it's hard to learn a lot of things when they aren't taught in school

I assume you're in high school? Two things I tend to think a high school student excited about math could take a look at is number theory (I remember being very excited by Niven's book, but there are other books people like a lot) or Spivak's calculus book, which goes back through calculus again but from a mathematically rigorous perspective - no handwaving, all of the notions are actually defined, etc.

s.harp

6:38 PM

For calculus better read dieudonne `_´ (just kidding, its a cool book but may warrant already existing mathematical maturity)

7:25 PM

How do I write Euclid's 5th postulate in logical form?? If two lines are intersected by a transversal in such a way that the sum of the degree measures of the two interior angles on one side of the transversal is less than 180 ◦ , then the two lines meet on that side of the transversal.

I guess it depends on how exactly you've set up things, but

I wonder if Hilbert's axioms do this or if they just get around it.

I'm just trying to figure out how to convert sentences into logic. It's not always very clear for me as sometimes it appears that there are many ways to do it

For every pair of lines and another line (the "transversal"), and given a side of the transversal, if the transversal intersects the first two lines and the sum of the angles on the given side is less than 180 degrees, then there exists a point on both of the first two lines on the given side of the transversal

I think

I think "the given side" is a pretty serious sticking point, Akiva.

7:41 PM

Right

I think you need some set of axioms that deals with that. Euclid's weren't good enough, formally, for 20th century logicians, right?

It sounds good to me, but what do I know...

I mean, we never defined what a "side" of a line is

There are a set of axioms...

so it works for me

The first 4 of Euclid's postulates don't talk about sides of lines, noto

You can hopefully define it from them, but the definition isn't built into the system.

Oh I see what you mean, so you can't use Hilbert's axioms.

7:48 PM

Oh sure you can if you want. I only mentioned them in hopes that he would have some good formalism

Since Euclid's postulates are missing a few technical things.

but Hilbert's are complete enough to formally recover all of The Elements, iirc.

8:00 PM

I think I heard somewhere that, technically, Euclid even messed up in the proof of his very first proposition (constructing an equilateral triangle). He assumed without proof that the two circles intersect in two points, without trying to prove it solely from the axioms.

Or something.

Hello, i want to prove that $\ovrset{\circ}{S}=\emptyset$ where $S=S(x,r)=\{y\in E,||x-y||=r\}$ is it right to suppose that $\overset{\circ}{S}\neq \emptyset$ i.e., there exist $\varepsilon>0, B(y,\varepsilon)\subset S(x,r) $ then there exists $z$ such that $||y-z||<\vrepsilon\leq r $ and $ ||x-y||=r$ contradiction

Yeah, there are a lot of "completeness" issues like that, that Euclid wasn't thinking about. Euclid's system also allows for a model consisting of rational lines, or something like that.

Can someone help me ?

What's E and does over-circ mean interior points?

@Vrouvrou You're trying to prove the interior of a circle in a normed space E is empty?

8:03 PM

yes @SteamyRoot

is my proof is correct?

I think you have the right idea; it looks like you're trying to formalize "if a point is interior in the sphere then it has a full neighborhood living in the sphere, but something that neighborhood is closer to $y$ than $r$; contradiction."

It's not super clear to me why such a $z$ must exist, though.

Yeah... I'm trying to figure that bit out...

You have to use something about the full structure of the normed space, since the statement is false in subsets of normed spaces.

I f i suppose that $\overset{\circ}{S}\neq \emptyset$ then there exist $y\in \overset{\circ}{S}$

i.e there exists $\varepsilon>0$ $y\in B(y,\varepsilon)\subset S$

Exactly

But do we know enough about the space $E$ to continue?

8:11 PM

$E$ is a normed space

Hmmm... But suppose that you are using the trivial norm... what does that give?

what is trivial norm ?

In a vector space with a $0$, the trivial norm is defined as $\|0\| = 0$ and $\|x\| = 1 \forall x \neq 0$.

Perhaps the statement is true for this norm, but the process of finding a second point inside the ball $B(y,\epsilon)$ might not work?

i don't see the problem

hy B(y,\varepsilon)=\emptyset ?

$B(y,\epsilon) = \{y\}$

Well, if $\epsilon < 1$ that is

8:17 PM

so we can find $z\in B(y,\varepsilon)$

Steamy, I don't think that's a normed space

Oh, okay it is but it's not a normed TVS

feynnnn

Then you find $z = y$, but since $y$ was a point on $S(x,r)$, then $\|z-x\| = \|y-x\| = 1$.

I may be wrong on many things... It's been ages since I did any topology...

However I think the statement remains true, actually O.O; no point in the sphere is in its interior because every nontrivial ball around such a point contains the center of the (original) sphere, and this point isn't in the sphere.

Yes, I think the statement is true indeed.

so the statment is true but the method no

right

8:21 PM

I believe so

In many spaces the method should work, though... but in general, I have my doubts.

the problem is if z=y

No, actually this is not a norm, Steamy, I am now convinced.

Steven Stewart-Gallus

How would you notate a volume element with respect to velocity or $\mathrm{d} v_x \, \mathrm{d} v_y \, \mathrm{d} v_z $?

Norms have to have $|ax|=|a||x|$, and this doesn't do that.

@EricStucky why

the norm is given in general

8:23 PM

Let $x$ nonzero. If $|a|\neq 1$, then $|ax|=1$ by definition, but $|a||x|=|a|\neq 1$.

i dont understand

Try it on R^2.

Oh, right...

Look at $(1,0)$ and $(2,0)$.

The trivial norm is a division ring norm

My bad

8:25 PM

but E is given as a normed space with the norm ||.||

we don't know what is equal to ||.||

i don't understand the problem

Vrouv, I'm talking specifically about the trivial norm; it's not a vector space norm. (In that case we do know exactly what is ||.||)

Yeah, that norm won't work as a counterexample for the argument

as in this case it doesn't count as a norm

so the problem is with z=y?

Well, the problem is finding a $z \in B(y,\epsilon)$ that is not on $S(x,r)$.

I don't see why it has to exist, without more information about the space $E$.

but as $B(y,\varepsilon)\neq \emptyset $ so $z$ exist

in $B(y,\varepsilon)$

8:33 PM

Like, if $E = \mathbb{R}^n$, then it's really easy. But in general?

Sure, there is something in the ball, Vrouv.

But it could conceivably be that

everything in the ball is also in the sphere.

$B(y,\varepsilon)\subseteq \{p: ||p-x||=r\}$.

In this case, you couldn't get the contradiction.

So you have to show that this is impossible; in other words, you need to find a $z$ in the ball that is not in the sphere.

yes so one time we have $||y-z||<\varepsilon$ and in the other time $||x-y||=r$

yes... But how will you go from there to $\|x-z\| < r$?

you are right

my proof i in general fals

i do a mistake we have ||y-z||<\varepsilon and ||x-z||=r

@SteamyRoot

8:53 PM

@EricStucky Do you know of any simple true-but-unprovable statements in geometry? (Hilbert's axioms, I guess, if Euclid's aren't good enough)

These are guaranteed to exist by Gödel, I think

Wait, Hilbert's stuff isn't first-order

I don't, fwiw.

This thread seems relevant, Akiva.

Forever Mozart

the seed is strong what does that mean

?

Forever Mozart

9:09 PM

i am watching game of thrones from the beginning

can anyone explain to me how these basic examples of cycle decomposition notation are evaluated? $(1~2) \circ (1~3) = (1~3~2)$ and $(1~3) \circ (1~2) = (1~2~3)$? How is this read? In standard function composition $f\circ g$ is $f(g(x))$ so I imagine the first permutation permutes the latter permutation, to get $(1~2~3)$ but this is incorrect

Think of it as a function

What does it do to the number $1$?

sends it to 3

3 doesn't go anywhere so i thought it should be fixed

$((1~2)\circ(1~3))(1)=((1~2))(3)=3$

Might be nonstandard notation

yeah exactly how I see it

wait..

9:13 PM

And $3$ gets sent to $1$ and then $2$

and $2$ gets sent to $2$ and then $1$

So this is $(1~3~2)$

why does 3 get sent to 1?

$(1~3)$ is a cycle. It switches $1$ and $3$.

I thought it just meant $\sigma (1) = 3$ not vice versa

thats why i was confused. i see now though

It means both. It's a permutation (which is really just another way to say "bijection from itself to itself"), so we can't send $1$ and $3$ both to $3$.

And something needs to go to $1$.

Similarly, $(1~2~3~4)$ maps $1\mapsto2\mapsto3\mapsto4\mapsto1$.

couldn't you write the cycle as $(3~1~2)$ too , then?

9:17 PM

$(1~2~3)=(3~1~2)=(2~3~1)$, yeah

In $S_3$ (set of permutations of $\{1,2,3\}$), there are only two $3$-cycles, $(1~2~3)$ and $(1~3~2)$.

(And three $2$-cycles, plus the identity permutation that does nothing)

(So $6=3!$ permutations of $\{1,2,3\}$)