How to evaluate $$\sum_{\substack{k_1,\ \cdots,\ k_n \in \mathbb{N}_0 \\ k_1\ne\ \cdots\ \ne k_n}}\frac1{m^{k_1+\cdots k_n}}$$ I have that this is a geometric series with weights $a_k$. $a_k$ should be the number of distinct weak compositions of length n of k. So it should be $a_k=\binom{k+n-1}{n-1}$ right? However if I put this back into the geo series and use the generating function I get an answer but it is inconsistent with simple cases of $m$ and $n$.

basically...

@TedShifrin Hello.

$a_k$ is the number of n-tuples whose non-negative integral elements add to k.

Am I being an idiot here?

@Alizter But you're asking the thingies to be different.

I don't follow.

Yes.

Also, isn't that possibly divergent?

So you can switch two $k_1<k_2<\cdots <k_n$ plus a weight.

@PedroTamaroff No. It converges for simple cases so I guess it is not hard to show.

Oh, you can only have one occurence of $0$.

Sorry.

It converges.

hmm the lowest power will be $n(n-1)/2$

0, 1, ..., n-1

I think $a_k$ is just the second stars and bars theorem

<---- Losted

@TedShifrin Hello.

heya

@TedShifrin I am trying to count how many times a certain expoenent of $m$ will come up.

So it will be written as $\sum a_k/m^k$ where $a_k$ is the number of times it comes up

I don't know a formula for the number of partitions into distinct summands.

@TedShifrin Maybe this will help

@PedroTamaroff Do you follow?

I taught that stuff in probability ... But it doesn't count the summands distinct.

@TedShifrin I think non-distinct ones are called partitions and distinct ones are called compositions

there are $2^{n-1}$ compositions of $n$

however I need to include 0

No, you have a fixed length. Your $n$ is fixed. Stop reusing the same letter.

Yes I made a mess. Step over the puddle.

hmm I think I found the mistake

I need a formula for the number of distinct n-tuples with distinct elements that add to k.

Ok @PedroTamaroff and @TedShifrin my question has been restated into a more concise form above

@Alizter Yes, I see what your question is.

So, you want to count the number of solutions of $k_1+\cdots+k_n=k$ where each $k_i$ is distinct.

yes

Now, consider $k_1<\cdots <k_n$.

This can be reordered int $n!$ ways.

We have $k_i\in \Bbb N_0$ as well

Yes.

@PedroTamaroff Yup.

So you should count the number of solutions to $k_1+\cdots+k_n=k$, $k_1<\cdots <k_n$ and multiply by $n!$.

Can you do this?

Is this just n-length weak partitions of k?

You tell me.

hmm k-n!

no

@Alizter Note that you're allowing $k_i=0$.

This is easily fixed by adding $1$.

@PedroTamaroff Adding 1 to what?

@Alizter To everything.

$k_1'+\cdots+k_n'=k+n$.

so we add n to the equations?

ah

Now you want $k_1'<\cdots <k_n'$ and $k_n'>0$.

Can you count them?

having trouble with that part

Hint Losing the commas, there are $r_i>0$ such that $k_1+r_1=k_2$, $k_2+r_2=k_3,\ldots,k_{n-1}+r_{n-1}=k_n$.

So you can write your equality as?

$k_1+r_1+\cdots+r_{n-1}$

For example, in the case $n=4$, you get $4k_1+3r_1+2r_2+r_3=k$.

oh

Relabeling, you get $4r_0+3r_1+2r_2+r_3=k$. And now the $r_i$ are arbitrary positive thingies.

And now you really want generating functions, I'd guess.

How can we count these?

@Alizter Well, it is the coefficient of $x^k$ in $\frac{1}{(1-x)(1-x^2)(1-x^3)(1-x^4)}$

Seems ugly as hell.

Actually it is not even that.

Let me see if I can fix this.

I think we can use Ferrers diagrams to simply things.

for k+n=4

there is one

How do you read them?

@PedroTamaroff I think it just means 3 +1

Right, but some people read them differently.

At any rate, I thought you didn't allow zeros.

Suppose $a_k$ counts the number of solutions to $k_1+k_2+k_3+k_4=k$, $0<k_1<\cdots <k_4$.

new $a_k$ or old $a_k$?

this would just be the old $a_k/n!$

Let me think about it for a second.

@Alizter I think your best bet is to use the bijection I suggested.

That is, look at solutions of $r_1+2r_2+\cdots+nr_n=k$.

Where the $r_i$ are nonzero.

ok

As you noted, the first nonzero term starts at $1+\cdots+n=\binom n2$

yup

Hello @alizter, lol. Could you explain the current A Level system to me? There are two subjects, Math and Further Math, and you can get A or A* for each, is that right? Is there another math subject? Is there something called an S paper? I don't get it.

So if two dice are rolled, the probability of the sum being 10+ is $\frac{7}{36}$. And if you roll two dice three times, the probability of at least one roll having a sum of 10+ is still $\frac{7}{36}$, right?

@JasperLoy STEP paper for certain uni entry.

@JasperLoy Hi

18 modules. You pick c1-c4 + fp1 default and rest the other modules for further maths

@JasperLoy Basically this

@PedroTamaroff Why is this problem harder than it looks?

@Alizter It is not a nice problem.

=D

@JasperLoy Do you change the color of your avatar based on your mood?

@Alizter It sounds complicated. Here in local bookstores, I see books covering Pure Math 1,2,3, Mechanics 1,2 and Probability and Statistics 1,2. Is that the old syllabus? Is that only 1 math subject?

I think so.

@user130018 Yes, I do. But I choose blue because I like it.

@PedroTamaroff Maybe if we ignore $k_1<k_2...$

and just count the partitions

@Alizter Why'd you ignore that?

@Alizter So the list of books you gave me that day covers all 18 modules?

Is any closed proper subset if S_1 simply connected? I beleive it is since if given any closed curve on this subset, we can contract it to a point. However, a friend of mine said that it isn't simply connected because we can't contract the whole space to an entire point. Who's correct?

So if two dice are rolled, the probability of the sum being 10+ is 7/36 . And if you roll two dice three times, the probability of at least one roll having a sum of 10+ is still 7/36 , right?

@user130018 You should change your username and your email address to things more easily readable.

@TheSubstitute You said proper.

$S_1$ is not a proper closed subset of itself.

Yes, I was constructing an example of a nested sequence of closed sets, each of which is simply connected and whose union is not. The union in my example happens to be S_1

@Alizter So you want the coefficient of $x^{k+n(n-1)/2}$ in $\prod_{k=1}^n (1-x^k)^{-1}$

Is that a satisfactory answer?

=P

@Alizter Wow, it looks very complicated, thanks.

@PedroTamaroff Eek.

@alizter I think my country no longer offers Further Math, any reason why it is a crap place to live, lol.

I need a formula for the number of distinct n-tuples with non-negative distinct integral elements that add to k.

God this problem is hard.

This is the kind of thing Erdos would think about

@alizter Thank you for telling me about A Level Math. It has opened my eyes. You should not sleep too late.

@JasperLoy How can I sleep with this bastard of a problem?

Don't curse

@Alizter I think perhaps I should share some of my life problems with you over email in future. You are a wise kid.

@user130018 bastard is not a curse.

My eyes

I feel so unhappy living where I am now. So many of the things I consider important in life are not here, because people choose to remove them or ban them on purpose. Sick, sick, sick.

@user130018 Forgive me if it is a curse. I grew up with it not being that bad. Internet says otherwise.

I thought it was on the same level as idiot

If they can't say it in a Disney movie, then it's probably bad

@JasperLoy At least you don't get chewing gum on your shoe.

@Alizter I am not the "ungrateful" type. I am the "perfectionist" type who wants the best of everything for humanity.

@PedroTamaroff I just had an idea

If I manually work out the number of ways

@Alizter Maybe I will get a copy of all the edexcel math books in future, perhaps.

see if OEIS knows it :P

Is there anyone taking IB math here? Just want to find out more about it.

@user130018 What is the name of the qualification or exam you take at the end of high school in the US?

@JasperLoy No such thing

@JasperLoy Just pass your classes and you get a diploma

@JasperLoy Sometimes there are state-mandated exams for each subject

@user130018 OK, I am just trying to compare the different education systems around the world, especially with regard to pre-university math, which I don't know about.

Anyone have a good continuous function that isn't lipschitz on the unit interval?

Question: I'm working through the details of showing $\widehat{f'}(\xi) = 2\pi i \xi \widehat f(\xi)$ and I use integration by parts to get $\lim_{N \to \infty} \bigg[ f(x)e^{-2\pi i \xi x} \bigg]_{-N}^N + 2\pi i \xi \widehat f(\xi)$. I'm trying to show that the middle term goes to zero but I'm not seeing it?

@Alizter what's the problem you're trying to solve?

non-negative i.e. including 0?

And do you allow repetition of elements?

Also what do you mean by integral elements?

Elements are nonnegative integers that are distinct

For example 3

3+0

2+1

1+2

0+3

so that is 4

of length 2

Okay.

Will try to think about it, if I have anything useful in mind will share

@PedroTamaroff I think I am close.

@Alizter To what?

an answer

It would be easier if it was with repetitions possible

But no idea how to go without repetitions.

With repititions it's equivalent to $k$ lines and $n-1$ breaks, which is then $(n+k-1)!/[k!(n-1)!]$

ye

Hey @Studentmath

@Ted!

How's the course going on?

I am running out of hair to pull.

Hi @TedShifrin

Hi @user130018 ... I preferred mr eyeglasses ;)

Haha, I never officially went by that username though :p

You were mr. Eyeglasses?! Wondered where you disapeared

Exam quite disappointing :( Some of these students don't belong in college ... :(

No, that was my name for you :)

@TedShifrin I feel bad for you..even though you tell them explicitly the questions on the exam, they still don't get it? Maybe they thought it was reverse psychology trying to trick them lol

@Studentmath They were eyebrows, not eyeglasses

lol @MikeMiller

No, I didn't tell them everything, but every question was easier than most homework.

@Mike I believe we already had that discussion

@Studentmath Yes, and we all agreed they were eyebrows

@Mike I will take your word for it. I don't recall where I left my keys couple of minutes ago, so...

I didn't care, and I still don't :)

@Ted that's disappointing indeed..

At least the A students leave the course with some extra knowledge/experience?

I have to memorize a half-page long monologue for my acting class by tomorrow

I agree, @Studentmath. It's very disappointing that Ted doesn't care about the great eyebrows debate

I made a question today.

Here.

@TedShifrin @MikeMiller

Wonder if Mariano or Pete may answer.

@Mike it could change the world of mathematics

@Studentmath: I told them a computer needs to be rebooted on average once a day, then asked for a reasonable random variable expressing the number of reboots in a day. About 1/4 got it ...

What is $a$, @PedroTamaroff? Also, only assholes write $M[a^{-1}]$ for the localization of a module at anything.

lollll

That's really disappointing

The As and Bs are fine. But almost 2/3 the grades were C, D, F.

Gah

What's the distribution of the grades like? ;)

I will complain about your English @Pedro: "which I don't know if it was translated"?

I failed to get one of my 'students' to pass to the higher maths class in School.. It was rather heartbreaking, we worked for weeks

@TedShifrin Oh, yes. Yikes.

:)

I don't know how to re-write that.

@MikeMiller It is nice notation.

No, it's really not. You're not adjoining anything for a module.

Yeah, @Studentmath, it's frustrating. Most of my students aren't good enough to be as casual as they've been.

"As I don't know if it's been translated"?

@MikeMiller $R[a^{-1}]$ is $R[x]/(ax-1)$. $M[a^{-1}]$ is $R[a^{-1}]\otimes_R M$.

I like that notation.

Yes. Do you not notice a difference between the two?

@TedShifrin OK.

It's good for rings, not for modules.

@MikeMiller One can define $M[x]$

That lack of $ at the start led to a nice chain-reaction there

Ah, agreed, @Mike

It is $R[x]\otimes_R M$.

@TedShifrin Yes, that's part of it. He can't expect to pass just by attending and looking at the homework questions, he needs to sit down and do some exercises

But you're adjoining in coefficients, not in the module, @Pedro

More than that for my students, @Studentmath

@TedShifrin It's the same to me. In $M[a^{-1}]$, you have polinomials elements of the form $\sum r_i a^{-i}m_i$. I don't think it is a sin to use that notation.

@PedroTamaroff, I challenge you to a duel over your notation.

@alizter Please go to bed, lol.

@JasperLoy Impossibru

@Ted They passed calculus, right?

I will fight you to the death if you continue using it.

@Pedro. Suppose $V=\Bbb Q^n$. I don't like $V[\sqrt2]$.

@Alizter I have half a mind to get those 18 books but they are terribly expensive, especially with shipping from the UK!

@JasperLoy There are joint ones that are cheaper in the long run

however my school pays for most of it

textbooks are a scam

@TedShifrin But that is not related to what I'm doing here.

You're adjoining inverses, not arbitrary elements.

Yes, @Studentmath, but some of them couldn't integrate $(1+x)^{-2}$. Sigh.

@PedroTamaroff Do you think it is worth asking that question on the main?

@Alizter What do you mean by joint ones?

@JasperLoy You can find non-official c1+c2 books

@Alizter Dunno. Try.

I also have digital copies of the books

@TedShifrin Which class couldn't integrate that?

...

some of them anyway

@Ted you kidding me, right?

It's logic used in high-school calculus

A few of my probability students :(

I ain't good in integrals, but still.. this is a bit silly

Maybe they're better at Lesbesgue integrals

LOL, forget that, mr eyeglasses.

There's a business major who's outperforming most of the math majors ...

Anyhow, enough.

@alizter I want to ask you. Are there any other books other than the 18 you showed me that are also in the same series? It seems there are none, but I want to be sure before I get a complete set.

Oh Prof. @Ted, I think I've managed the problem I was working on!

@JasperLoy Well this is for the exam board Edexcel. You can try other exam boards but I am unsure of their books.

I used in the solution something similliar to what I sent to you, but simplified it

@Alizter I know. So the 18 is complete for Edexcel, right?

@TedShifrin My real analysis class has a business major in it and he's also taken honors calculus which uses Spivak

@JasperLoy right.

@Alizter problem is the number of partitions having repetitions seems to change without any nice formula between different $n$s and $k$s

@Studentmath I am writing the question to ask on the main currently

Curious if there is a nice formula

@Alizter Too bad the Bostock and Chandler books are no longer in print. But I think Pure Math 1 and 2 is going to be published by OUP, since OUP has taken over Nelson and Thornes. However, they don't list Applied Math 1 and 2 on their website. The strange thing is that even the Edexcel website does not contain all 18 books.

I'm back to bed, it's 4:40 here.. g'night all!

I have posted the question

It is almost 3am

@Alizter Tmr is 1 Nov. I will start running then. I have not run for 10 years.

@JasperLoy Oh, have fun!

@Alizter Where do you want to go for graduate school after undergraduate?

@JasperLoy Let's get into undergraduate first

I have no idea about grad school

@Alizter The Edexcel site is terribly confusing, lol.

@JasperLoy Tell me about it.

@Alizter First, when I navigate to X, it does not list all 18 books you listed. Second, when I navigate to Y, it starts to list those that are published by OUP instead. So, all different books, and all incomplete, lol.

@JasperLoy It is all a mess.

These books I showed you are written by the exam board.

@Alizter They are preventing themselves from making money from me.

Also make sure it is A level

@JasperLoy They are a charity XD

@Alizter I will trust you that the 18 is a complete set.

Money goes to the publisher pretty much

@JasperLoy Edexcel simply does not provide any more exams.

There are 18

@Alizter Good! My friend is currently in Oxford, about to return after PhD. Maybe I will ask him to get it for me, lol.

@JasperLoy What did he do it in?

@Alizter Physics. He did undergrad physics in Princeton, US. He is my best friend, and one of the wisest people on earth. I will ask him for help to solve my life problems when he returns.

@JasperLoy My Uncle is doing a PhD in physics

@Alizter He's really smart, got government scholarships to go to both Princeton and Oxford.

@Alizter I have PHD=Permanent Head Damage.

@Alizter since the numbers are distinct, you can arrange a solution in increasing order .. $k_1 < k_2 \cdots < k_n$, and successive elements in this list differ by atleast one, so make the change of variable $k_{i+1} - k_i = x_i + 1$, for $n-1 \ge i \ge 1$, then you are counting number of non-negative integer solutions .. does that sound correct ? :O

@r9m Yes, I already did that with him. =)

What happened, did it work? I can't remember. Also is it friday today?

@PedroTamaroff ah ! I just got here and saw the link .. so I didn't know =) Cool !!

Oh cool, @Studentmath. BTW, I assume you're awake early morning there ... Oh, I emailed you the exam.

@Pedro: Looks like you got a nice answer. Algebro-geometric example.

@TedShifrin Yes, I commented. He uses some jargon I am not familiar with.

But I'm loving this result.

It seems quite intuitive to me, although I usually think the word "intuitive" carries no real meaning.

It usually means "I personally believe I understand this result beyond the technicalities and can give some examples to convince people it is true."

@alizter Sorry to bother you again. I was looking at the Cambridge A level Math books on CUP and OUP websites. How come there seems to be only Math and no Further Math? I don't get it.

Link me

@Alizter education.cambridge.org/as/subject/mathematics/…

This is not edexcel

@Alizter So you are saying edexcel offers more than cambridge currently in the uk?

cambridge is just the publisher

@Alizter OK, so what about Cambridge as an examining board? Where to find those books?

@JasperLoy cambridge has further maths

Its all a mess

Good night all. I am starting to shut my eyes.

@Alizter See you in your dreams.

Hi

Hi

Hi

:)

:)

@IceBoy How are u doing :)

@FreeMind how are u :)

Fine thanks, how are you pal?

Fine :D

$F(\theta)=\sin \theta \int_{-L}^{L}e^{-ikz \cos \theta}f(z)\,dz$ where $k>0$ and $\theta \in R$, in addition, $f$ is defined in $|z| \le L$

Question number 1 :

For which function $g:[0,\pi]\rightarrow R$ one can find $f:[-L,L]\rightarrow R$ , $f \approx g$

?

@TheArtist Thank you, I'm fine.

@robjohn @PedroTamaroff ^^

Dizzifying

@JessyCat :)

@FreeMind, sup?

@JessyCat Fine, and you?

@TheArtist BTW, how are u ? :D

@FreeMind im fine too :)

@FreeMind I don't see what you are asking here and how it relates to the $F$ defined two lines above.

@robjohn @Chris'ssis have you seen an elementary evaluation of the series $\displaystyle \sum\limits_{n=1}^{\infty} \frac{H_n^2}{n^4}$ ? :-)

Second last assignment done, now one final assignment

@r9m is that $H_n^2$ or $H_n^{(2)}$?

Noone would want to pirate textbooks, but where would one find a large cache of textbooks to download anyway?