Mathematics - 2017-03-08 (page 5 of 8)

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12:01 PM

@DHMO can I just ask though, when a questions says: Show that ...., does that mean I should prove it? or give an example?

How would I 'show' what I have just written in the above paragraph?

dimension... basis of Ax... how should I know

Anonymous

An unrelated problem: How to find the numbers of sets of positive numbers ${a,b,c}$ such that abc =2^{4}3^{5}5^{2}7^{3} ? I can't see any simple method... @DHMO @MickLH

stack exchange chat is inserting a line break in your latex and screwing the render for me

Anonymous

@MickLH Yeah, I removed the dollar sign

@MickLH use spaces in the code

Anonymous

12:12 PM

Now ok?

So that's just a product?

Trevör Anne Denise

Hello everyone, is there anybody how speaks french ? :)

Anonymous

@MickLH Yes abc=2^{4}3^{5}5^{2}7^{3}. I need to find number of sets {a,b,c}.

Anonymous

a, b and c are positive numbers

Anonymous

natural numbers

Anonymous

12:14 PM

What would be the shortest way of solving it?

@TrevörAnneDenise je parle un peu de francais

@Mystic there are $4+5+2+3 = 14$ prime factors (including duplicates), enumerate all possible ways to split 14 items into 3 groups.

Trevör Anne Denise

@DHMO Hello, so actually I have a question about this (and there is a little bit of french in it)

Anonymous

@MickLH What ? 14 prime factors ?

(also include the cases with some sets empty, 1 is still a natural number)

Trevör Anne Denise

12:16 PM

user image

Anonymous

I can see only 2,3,5 and 7 are prime numbers..

@Mystic yes, $abc$ has 14 prime factors counting duplicates

Anonymous

I think you mean divisors

Trevör Anne Denise

I am wondering if what comes after "Ou bien" is necessary or if this is just some other way to prove it...

Anonymous

@MickLH Umm, so how do you make groups of 3? I've done it when all objects are different.

Trevör Anne Denise

12:18 PM

user image

@TrevörAnneDenise which language do you want me to speak in?

Trevör Anne Denise

@DHMO The one you prefer, I am okay with both I just wanted someone who could understand the demonstration !

on ne sait pas la valeur de $\epsilon$

sa valeur est arbitraire

c'est possible que a+e<b

c'est aussi possible que a+e>b

donc on a besoin de considerer les deux cas

Trevör Anne Denise

@DHMO Oh je vois, merci beaucoup !

@Mystic IIRC it's 16C2

Anonymous

12:21 PM

@MickLH Eh? How ?

@TrevörAnneDenise de rien

Hi and bye

@Mystic ${{n+r-1}\choose{r-1}}$ ways to distribute $n$ items across $r$ groups

I'm trying to find a reference :/

but it's 430am and I promised to sleep by 4am, and I also promised to eat but I didn't do that either lol

Anonymous

@MickLH That is the number of ways to distribute n indistinguishable objects among r boxes... Here they are not all same objects...

It doesn't matter if you have duplicates, it's still a unique number

Anonymous

12:25 PM

@MickLH I am not talking about duplicates. I am saying that 2,3,5 are different digits...

Anonymous

They aren't "indistinguishable"

Anonymous

That is not the right formula in this situation

It's not the right formula, but not for the reason you said

Anonymous

Although I think I can modify it a bit

The issue is that it might be over counting "different pairs" of the same two numbers, if you get what I mean

I'd have to find a god damn reference, good luck

Anonymous

12:28 PM

I don't really get you. Anyway goodnight!

Anonymous

I guess 4:30 is too late to sleep :P

I mean that you have 4 different 2s, but you don't want to count $2_0$ as unique from $2_1$

either way, just find the right combinatoric bullshit to plug the 14 and the 3 into

Anonymous

@MickLH Yes, say that. I need to apply it for each of the digits separately.

Anonymous

I have 4 2's to put in three boxes

Anonymous

and so on for 3 5 and 7

12:30 PM

Give me like 30 minutes to eat

(I get angry and stupid when hunger kicks in, I can feel it)

Anonymous

$$\binom{4+2}{2} \binom{5+2}{2} \binom{2+2}{2} \binom{3+2}{2}$$ <----this might be the way

Anonymous

Ah, now I need to remove some cases

Anonymous

Because a set can't contain duplicates

depends what duplicates, two of the same factor is necessary (ie, $2^1 \neq 2^2$)

@BalarkaSen what do you call two topologies when you can impose equivalent classes to its element so that they become the same?

Anonymous

12:34 PM

@MickLH I mean that {2,2,999} can't be a set...I need to remove those cases...

@Mystic I don't understand the 999, but {2,2,...} is fine

Anonymous

@MickLH {a,b,c}={x,x,y} isn't permissible in sets...

Anonymous

Two elements of a set can't be same

Anonymous

Atleast at my level...

oh I took {...} to be the prime factors of only a (or only b / c)

Anonymous

12:37 PM

@DHMO any ideas?

Anonymous

^

no

Anonymous

@DHMO Should I ask it on the main site?

Anonymous

Or will it be too silly ? =P

@Mystic yes and with your attempt

Anonymous

12:38 PM

I see

Anonymous

ok

Hi hi

Compulsive Mathurbator

@Mystic That would indeed give you the number of tuples (a,b,c). But isolating the individual sets {a,b,c} is harder.

Anonymous

@CompulsiveMathurbator Indeed. Any ideas from your side?

Compulsive Mathurbator

All I can think of is case by case, but that's gonna be painfull

Anonymous

12:45 PM

I guess this must have been discussed on the main site before

That's sorta where I'm at with it too, if I had to solve it for work I'd just work them out one at a time with conditional probabilities lol

that's not even the right term, sheesh

Compulsive Mathurbator

@Mystic At best because only the exponent on seven is a multiple of three, we could find the number of duplicates with any one pair, subtract the duplicates where seven is distributed evenly, divide by three and add them back at the end.

Anonymous

Yeah, but still there is a lot of problem =P

Anonymous

Doing it for each number is terrible

Anonymous

0

Q: How to find the numbers of sets of positive numbers $\{a,b,c\}$ such that $(a)(b)(c) =2^{4}3^{5}5^{2}7^{3}$ ?

How to find the numbers of sets of positive numbers $\{a,b,c\}$ such that $(a)(b)(c) =2^{4}3^{5}5^{2}7^{3}$ ? $$\binom{4+2}{2} \binom{5+2}{2} \binom{2+2}{2} \binom{3+2}{2}$$ This would give the number of tuples (a,b,c). But isolating the individual sets ${a,b,c}$ is harder as there would be d...

combinatorics elementary-number-theory

Anonymous

12:56 PM

Asked it on main site

Anonymous

Let's wait and see what happens

Compulsive Mathurbator

If you favourite a question do you get notified of any answers?

@CompulsiveMathurbator I tried it and I don't get notified

so I don't know what it is for

Anonymous

@CompulsiveMathurbator I don't think so. You need to check your profile...

it's for when you need a digital gold star sticker

Compulsive Mathurbator

1:00 PM

So it's just another upvote button then. (Glances at open tabs with unanswered questions)

Anonymous

@CompulsiveMathurbator They are saved on the favorites section on your profile

Anonymous

So that you don't lose them

Compulsive Mathurbator

@Mystic Ah thanks

@DHMO that's terribly phrased. do you have quotient topology in mind?

@BalarkaSen for example, consider the topology on N with basis {{0,1},{2,3},...}

it's quite similar to the discrete topology

Anonymous

1:04 PM

Combinatorics is tougher than all the calculus I have done =P

Anonymous

I used to think otherwise before =P

Calculus is polished.

Anonymous

Yeah, calculus has standard methods for most things

Anonymous

Even integration

@DHMO you can quotient that to get the discrete topology

Anonymous

1:05 PM

has standard tricks

@BalarkaSen oh thanks

what do we call two topological spaces which can be identified by taking a quotient topology?

"one is the quotient of another"

there is no one word

one doesn't have to be the quotient of another

consider {{0,1,2},{3,4,5},...} and {{0,1},{2,3},...}

well you were pretty vague with that. there is no word for such things.

you can go from one to the other by a zig-zag path of quotients

yes that's what I mean

Akiva Weinberger

1:10 PM

I suppose we could let $A\sim B$ mean $A$ is a quotient of $B$, and then consider the equivalence relation generated by $\sim$

Well, hm, that would have everything equivalent to a point (and thus to each other). That's not very interesting.

yup

@Mystic I got a result for the drunk man but I used my own mathematics that nobody else might accept :P

well, so your idea is gone to shit

When I try to translate it into the standard way, it gives me a limit I don't know how to evaluate

Alessandro Codenotti

Hi chat

1:15 PM

Hi @Alessandro

Compulsive Mathurbator

@AlessandroCodenotti Hey

@Mystic $$\Pr(\text{ded}) = \frac{p (p^{\infty} - (1-p)^{\infty})}{p (p^{\infty} + (1-p)^{\infty}) - (1-p)^{\infty} }$$

Alessandro Codenotti

Today's AT lecture was 100% algebraic

What did you prove

Alessandro Codenotti

Nothing in particular, we saw the algebra necessary for Van Kampen's theorem to make sense (free groups, abelianizations and group presentations)

1:18 PM

aha

what's $\pi_1(S^1 \vee S^1)$?

(btw if anyone can evaluate that on the reals as a limit, I'd appreciate knowing how)

Draw the shortest curve across as many cities in US as possible

(You wouldn't be able to prove this without van Kampen, or the essential ideas - which Akiva told you actually - but you can make a good guess)

Compulsive Mathurbator

@MickLH Do you mean $$ Lim_{n \rightarrow \infty} \frac{p (p^{n} - (1-p)^{n})}{p (p^{n} + (1-p)^{n}) - (1-p)^{n} }$$

Alessandro Codenotti

The free group with 2 generators I'd say?

1:20 PM

@CompulsiveMathurbator effectively, yes

Ya

The real hard part is to prove that $ab$ and $ba$ (a and b are loop going around the two S^1's resp, based at wedge point) are not homotopic.

@Alessandro So, $\pi_1(X \vee Y)$ in general is isomorphic to $\pi_1(X) * \pi_1(Y)$ (everything's a CW complex, otherwise you get stuff like cones on earrings blah blah)

Alessandro Codenotti

I'll find out soon, we should see Van Kampen's theorem on Monday

Typo'd. Fixed.

Alessandro Codenotti

Yup I noticed

From the above, do you know what $\pi_1(T^2 - D)$ is, where $D$ is an open disk on a chart in the torus?

So a "punctured torus"

Anonymous

1:27 PM

@MickLH What is that ? :O Beyond my comprehension!

It's the probability that our drunk friend dies!

Anonymous

Explanation ?

it's 1

ez

I fixed the recurrence equation and re-solved it

Anonymous

@MickLH :P I haven't learnt recurrence

1:28 PM

@BalarkaSen apparently only when $p \geq \frac{1}{2}$, though

Anonymous

Will learn it by the end of this year

Akiva Weinberger

What's the question?

Anonymous

@MickLH right

Anonymous

2 hours ago, by Mystic

"A drunk man stands with a cliff one step to his left. He takes steps randomly left and right. Each step has probability $p$ of going left and probability $q=1−p$ of going right. Each step is the same size. If allowed to randomly step indefinitely, what is the probability that the drunkard falls off the cliff? " Do you know any method to solve this elegantly without using generating functions ? Once again my method was ridiculously long. @DHMO

Akiva Weinberger

Also, don't kill drunkards, drunkards are people too

1:29 PM

Also I used my weird infinitesimal notation, so that might throw a wrench in it, you can just evaluate it as a limit if you give $p$ a value first

Alessandro Codenotti

@BalarkaSen it deformation retracts to a wedge of 2 circles, right?

@Alessandro Yup.

Akiva Weinberger

The notation which, as previously mentioned, overloads symbols like $0$ and $\infty$ with more than one meaning

and hence runs the risk of giving you the wrong answer

Anonymous

@AkivaWeinberger Drunkards kill themselves. We don't need to kill them =P

So it's again $F_2$

1:31 PM

@AkivaWeinberger Not if you do it strictly correctly :P

I moved the meaning of the old $0$ over to $\varnothing$

Alessandro Codenotti

Ah, well, there's only a 2-cell so it retracts to the 1-skeleton after puncturing

Akiva Weinberger

@MickLH That amounts to trying to remember "what you mean" in your head

Writing things down is always superior.

Maybe by whatever you assumed?

Not by how I actually function

If I don't write it down, I forget it

Akiva Weinberger

@MickLH I don't know why you can't just write $h$ and put a $\lim_{h\to0}$ in front. It's essentially the same thing.

Or keep it at $\varnothing$ and write $\lim_{\varnothing\to0}$, I suppose

I usually do that at the end, but I tried that and I can't evaluate the limit

1:33 PM

@AlessandroCodenotti Right. Alternatively you can see it off from the square description of the torus.

Akiva Weinberger

(Or the hexagon description!)

I think it needs to be split into cases

Hello!

Alessandro Codenotti

Right, the interior of the square is the 2-cell and the border is the 1-skeleton

If you delete a disk off from the interior of the square everything deformation retracts to the boundary of the square, which identifies to S^1 v S^1

@Alessandro Yeah.

Akiva Weinberger

1:34 PM

Həi @user21820

I've a strange request; if anyone here has a few minutes to spend, can you check this post and confirm my judgement that it is irreparably logically flawed and help to delete it? If not then never mind.

For those with the reputation, you can check this other post to see that he is a crank who likes to prove Fermat's Last Theorem in a simple way.

@Alessandro So, now. Think of the double torus as two copies of $T^2 - D$ glued along $\partial D$.

By double torus I mean the genus 2 surface.

baller

Can you similarly "guess" what is $\pi_1(\Sigma_2)$ from that description?

Compulsive Mathurbator

No scratch that

I messed up :(

1:37 PM

really, it looked plausible

Compulsive Mathurbator

But this looks like a plausible limit $$ Lim_{n \rightarrow \infty} \frac{1- \left( \frac{1}{p} -1 \right)^n }{ 1- \left( \frac{1}{p} -1 \right)^{n+1}} $$

After that L'Hopital's rule gives you $$ \frac{p}{1-p}$$ For $ p \le \frac{1}{2} $

Anonymous

@CompulsiveMathurbator There are 2 answers now

Anonymous

Check

Compulsive Mathurbator

Inequality on the other side

Alessandro Codenotti

@BalarkaSen hmm, that doesn't look homotopy equivalent to a wedge of tori

1:40 PM

@CompulsiveMathurbator I think you've cracked it!

it definitely isn't

Compulsive Mathurbator

@MickLH for one case

Alessandro Codenotti

Because the circle they're glued along is not contractible

Mhm.

Anonymous

0

A: How to find the numbers of sets of positive numbers $\{a,b,c\}$ such that $(a)(b)(c) =2^{4}3^{5}5^{2}7^{3}$?

Let $S = \{(a,b,c) \colon abc = 2^4 3^5 5^2 7^3\}$. Let's split it into few smaller sets. $S_1 = \{(a,b,c) \in S \colon a \neq b;\, b \neq c;\, a \neq c\}$, $S_2 = \{(a,b,c) \in S \colon a=b;\, b \neq c$ and let $S_3, S_4$ be similar to $S_2$ only with equal pairs $a=c$ and $b=c$. It's easy to no...

Alessandro Codenotti

1:42 PM

Ok, I'll think about it, I'm supposed to be following this probability lecture right now

@CompulsiveMathurbator well here's the thing, we know that at $p=\frac{1}{2}$ the probability is 1, and that the probably can only increase in response to the underlying cause becoming more probable

Find the solution to the following lhcc recurrence:
a_{n} = 30 a_{n-1} - 225 a_{n-2} \text{ for } n \geq 3 with initial conditions a_1 = 210, a_2 = 6300.

Akiva Weinberger

@BalarkaSen Remind me what $\Sigma_2$ is?

Oh, genus two surface

You need to understand how the two copies of $\pi_1(T^2 - D)$ are "glued along" in the fundamental group of $\Sigma_2$.

@AkivaWeinberger Yeah

Grumble at my internet

do you use wifi?

Compulsive Mathurbator

1:44 PM

@MickLH True that. Makes sense from the limit as well, both the numerator and denominator approach 1.

extra grumbles

@skill Yes.

Akiva Weinberger

This one is different in that the intersection is not simply connected

@CompulsiveMathurbator @Mystic
$$
\Pr(\text{ded})=
\begin{cases}
\frac{p}{1-p} & p \leq \frac{1}{2} ,\\
1 & p \geq \frac{1}{2}
\end{cases}
$$

thanks for pointing out L'Hopital's rule, @CompulsiveMathurbator, I'll hopefully remember to see if it applies next time I get stuck like that

Compulsive Mathurbator

@MickLH Mind telling us how you got the original expression as well?

@Mystic Second one looks good. The first is just the idea of the second but no actual calculations imo

Anonymous

@CompulsiveMathurbator true

Anonymous

1:56 PM

I need to read that again to get it fully

Anonymous

I'm too tired now

Anonymous

Bye all =)

Anonymous

See you tomorrow!

cya

I wrote a (wrong) recurrence for the probability that he dies at each step, at first it was flawed because it modeled him as dying at any step left, so I made it also compute the next step in terms of itself to account for him needing to get back to the center before a step to the left would be fatal. With that it had gained a 2nd recursive term and so I had some unknowns, so I filled in one variable by forcing the probability to 1 if he made any leftward progress

then I filled in the other by assuming that if he walked forever, he never died, and so I forced the element at infinity to probability zero and that solved the rest

@Mystic night

@CompulsiveMathurbator Oh, and so that expression was the element corresponding to not having made any steps yet (which also corresponds to having made an equal number of steps right and then left)

(my dirty secret: I skipped the hard work and used computer algebra on the recurrence)

2:06 PM

Hi i was going to make a post about this but it seems more appropriate to put it in the chat, i'm currently a second year undergraduate going into my third year, and for my project next year i'm stuck on deciding whether to do one about solar sails or one on green functions, i cant seem to find any of the mathematics behind solar sails, so if anyone knows any good books on the subject i would like to know about them

@CompulsiveMathurbator In symbols: It was $F[0]$ where $F[-1]=1, F[\infty]=0, F[n]=p F[n-1] + (1-p) F[n+1]$

This was the raw (wrong) intuition: chat.stackexchange.com/transcript/message/35896521#35896521

hi chat

Compulsive Mathurbator

@MickLH I get it now. Thanks

@Semiclassical Hey

hi @Semiclassical

Any good math this morning?

2:15 PM

The only substantial problem with the first attempt was that it was indexed by (steps right + steps left), instead of (steps right - steps left), so it couldn't recurse onto itself properly, but that was a minor adjustment

@MickLH As you go farther into actual applications, you'll find that becomes less of a dirty secret and more of a "well, of course you do that."

@Semiclassical Lol I know, I've pretty much exclusively used computer algebra since the get go

By the time I was doing grad-level electrodynamics, for instance, there were certain HW problems where a chunk of my work was just "Here's the integral I typed into Mathematica and here's what I got as output."

I learned computer algebra before I learned calculus, so I'm really just more comfortable with it than paper

I mean, I'd include a printout of the Mathematica input/output itself in my HW.

2:18 PM

I just say "dirty secret" for giggles :P

That still usually required some written work, though, to show how I maneuvered the problem into that form and possibly some more to get it into the desired form.

Fair enough :P

I actually love computer algebra and I'd be significantly less interested in math without it

because fun problems like that one earlier would become fucking tedious

Compulsive Mathurbator

That's why we invented the damned things isn’t it.

Also if it weren't for Mathematica, then I would never have pulled off some of my favorite designs IRL

Life's too short to do everything by hand.

4

2:24 PM

true

Though there's a counterargument that if you rely on PC's too much then you lose some of the skills you picked up in order to do stuff without it.

I found a compromise: I usually use a weaker CAS that isn't very smart but just automates the most tedious parts

I just resort to Mathematica when I'm stumped

sometimes you need to learn how to un-stump yourself

no PC can teach you that

that's true, but at least a hint is nice :P

sure hints are nice, but they can make you lazy

imho

2:32 PM

Depends on what kind of lazy you mean.

I swear I just take the hint, not the whole answer! One day I fed Mathematica a terrible integral and it introduced me to the hypergeometric functions. I spent at least a month studying those.

I'd explain what I mean by that, but I don't want to. (Now -that's- bad laziness.)

yup

well said

Compulsive Mathurbator

@Semiclassical I'll explain the other type later. :p

lol

Ah, procrastination humor.

2:37 PM

let's procrastinate on humor

Now is that humor about procrastination, or humor for the purpose of procrastination?

Compulsive Mathurbator

@MickLH We can cross that bridge when we come to it.

Either/or.

Nobody said it can't be both?

My funniest anecdote about procrastination is that, for a behavioral econ/psych course, I wrote a paper on the subject.

2:38 PM

drum roll...

It'd be even more fun it you wanted to write it but didn't

In which I led off by specifically noting that I'd delayed starting the paper until the last minute.

...and then got it in a day late.

Compulsive Mathurbator

@Semiclassical For research purposes of course

Of course.

Compulsive Mathurbator

@Semiclassical I'd give you a perfect score for dedication.

2:39 PM

Everything I've done before/since then is just field testing.

Compulsive Mathurbator

And that's why I'm not in academia

Man, I miss @CompulsiveMathurbator, it's too bad he pissed off the academia gestapo

What always surprises me is how often math that I thought I'd be able to forget about forever has shown up in one way or another.

never forget math stuff!

lol

Right now the piece of old-becomes-new math is "irregular singular points."

2:47 PM

semi classical

what is a pole

and removable singularity and essenstial singularity

$f(z)=1/z$ is the obvious example of a (simple) pole.

any math knowledge you have is either a standard enhancement to all your other math knowledge, or it's a new enhancement that's waiting for you to discover it

well said mickLH

The way to think of the various singularities is basically: By what factor do I need to multiply to get rid of the singularity?

For a simple pole, you just multiply by a linear factor e.g. $f(z)=1/z\implies z f(z)=1$ is well-behaved (analytic) at $z=0$.

hmm and what is the order of a zero

What i dont quite get is those 2 quetions , order of zeros and difference between pole and essenstial singularity

2:50 PM

how many times can you divide and still have the zero left :P

Zeros and poles are pretty much inverses

if we take 1/z^3

what is the order of 0 here?

The order of the zero is basically: How many times can I divide by $z$ without creating a singularity?

shouldnt we multiply by z here?

For $f(z)=z$, I can only divide once so it's a zero of order 1.

Sure. I was just completing the description.

Can you give me a good example ?

2:51 PM

How many times do I have to multiply $f(z)=1/z^3$ by $z$ in order to get rid of the singularity?

well 3 times makes it constant

Right. So it's a triple pole.

hmm

why cant we do anything about essential singularity

Suppose you'd had $f(z)=\frac1{z^3}\frac1{z-1}$.

a second thinking

2:52 PM

Would that change the order of the pole?

@KasmirKhaan Because the Laurent series has infinitely many terms of the form 1/z^n for n = 1, 2, 3, ... around an essential singularity.

Eg exp(1/z) at z = 0

hmm can we multiply by (z-1 ?

The problem with an essential singularity is that it's, well, essential: No matter how many times you multiply by $z$, an essential singularity at zero will remain a singularity.

@KasmirKhaan You would if you wanted to study the pole at $z=1$.

But suppose we're still just looking at the pole at $z=0$.

@BalarkaSen yes my teacher took that example also e^ 1/z)

well we can multiply by z^4

or wait z^3 is enough

Yeah, z^3 is still enough.

2:55 PM

we get 1/ (z-1)

ah :D thanks guys :D

The point is that while 1/(z-1) isn't analytic at z=1, it's still analytic at z=0.

So it doesn't affect the order of the pole at z=0.

hmm right :D

but for e^( 1/z^2)

By contrast, if I did $f(z)=\frac{\sin z}{z^3}$, I'd note that $\sin z$ has a (simple) zero at $z=0$

cant we use other ways ?like multiply by e^z^2 ?

No. That'd completely miss the point.

2:57 PM

1) e^(z^2)*e^(1/z^2) = e^(z^2+1/z^2) != 1.

so we only allowd a linear factor ?

sinz / z^3 = sinz /z * 1/z^2 = 1/z^2

Well, there were two issues. One is that multiplying by e^(z^2) wouldn't actually get rid of the essential singularity.

as an aside, (sin z / z)=1 only makes sense in the vicinity of z=0.

So I wouldn't write that as an equality.

Yes yes but you know what i mean =p

Sure. But probably better to do sin z ~ z.

does $\frac{\sin z}{z}$ ever occur outside of a sinc function?

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