Mathematics - 2017-06-28 (page 2 of 7)

@ALannister One (probably silly) observation is that you can assume without loss of generality that $f$ is monic over Z/p. (the first term isn't 0 mod p prime, so it has a multiplicative inverse over Z/p which you can multiply f by.)

Dodsy

2:40 AM

How are you semi

hey listen to this

here's a putnam question

"find a formula to find the sum of the first $n$ odd numbers"

Chris Nguyen

Is the product of two convergent series always convergent?

Dodsy

isn't the formula literally just $n^2$

Chris Nguyen

Yes

Dodsy

how can that be a putnam question?

that's too easy.

Chris Nguyen

idk

is putnam a college math competition?

Dodsy

2:46 AM

yes

Chris Nguyen

Can you answer my question?

Dodsy

oh idk :C I am just an amateur, sorry.

Chris Nguyen

oh its ok

What grade are you in

Dodsy

I am going into University next year :)

Chris Nguyen

what school

Dodsy

2:48 AM

Canadian school.

Oh

yeah.

Not U.S.

Nah, I felt like staying in Canada :P

Chris Nguyen

I'm studying for a national math competition in 2 weeks

Dodsy

2:48 AM

oh nice

Chris Nguyen

thats going to take place in two weeks

Dodsy

are you a high school student?

Chris Nguyen

yes

Dodsy

Well, I hope you do well!

Chris Nguyen

Did you take the amcs in high school

Dodsy

2:49 AM

No, I was severely unmotivated and directionless in high school.

I didn't even like math :)

Chris Nguyen

OH

oops i meant oh

and now you like math?

Dodsy

It's been a long journey, friend.

I went to "college" (different in Canada than the states)

and didn't like it

I dropped out

and then found an engineering book

started liking engineering

took distance education courses and started to love mathematics.

and now here I am.

Chris Nguyen

oh i see

Dodsy

I know shockingly less than I should, but I am motivated and passionate and I think if I stick to it I can hopefully become a good mathematician :)

Chris Nguyen

good luck :)

Dodsy

2:53 AM

Well it wasn't perfect, but I am on the right path now.

And now I am going to the University that was down the road from the college I dropped out of.

Pretty radical.

and if it turns out that I am not good at math, I can always become an engineer! ;)

Chris Nguyen

I thought engineers were good at math

Dodsy

Eh, it's mostly a joke. But engineering is more computation.

Semiclassical

Depends on the kind of math.

Chris Nguyen

Semiclassical can you answer my question

Semiclassical

Product of two convergent series?

Chris Nguyen

2:55 AM

yes

Semiclassical

My gut says yes, but that isn't a proof.

Chris Nguyen

thats what i thought!

apparently its not

Semiclassical

huh.

Not sure why you needed me, then :P

Chris Nguyen

I just looked at the solution

I just wanted to know if that was an obvious question or not

Semiclassical

This bit on the Wiki page for convergent series seems relevant: en.wikipedia.org/wiki/…

Chris Nguyen

2:57 AM

Semiclassical are you in hs

Semiclassical

looool no

Chris Nguyen

in college?

Semiclassical

I'm at the end of grad school. (Which is formally defined as "any point in time when you know you want to get the hell out of grad school")

Chris Nguyen

Math major?

Semiclassical

math+physics for my undergrad

Dodsy

2:58 AM

He double majored I believe.

Semiclassical

Physics PhD program.

Chris Nguyen

What school?

Semiclassical

University of Minnesota for grad.

Convergence and Merten's theorem

Chris Nguyen

undergrad?

Semiclassical

also in Minnesota.

I've stayed pretty local.

Chris Nguyen

3:00 AM

Did you do putnam?

Semiclassical

Nah.

There were some regional contests but it wasn't a big deal where I was.

Chris Nguyen

oh

Semiclassical

I did like to attempt the problems in the math journals, though.

Chris Nguyen

Did you do the amcs in hs

Semiclassical

I should take another look at that.

I don't think so? But that was over a decade ago, so it's not within memory now.

Chris Nguyen

3:02 AM

lol i feel you

I'm still in hs :(

Semiclassical

gotcha.

kinda surprised to see you doing convergent series, though.

Is that for coursework, or for a math exam?

Chris Nguyen

Really?

I'm a rising senior I just finished calculus but I'm studying for a math competition in july

Semiclassical

Granted, my recollection of high school calculus is also dated.

Dodsy

I never even learned integrals.

Semiclassical

I guess there may have been some series stuff for that, though. I can't cast my mind that far back.

Dodsy

3:03 AM

I probably won't even be ready for putnam after first year though.

Chris Nguyen

Does every university have putnam

Semiclassical

An example of two conditionally convergent series whose product isn't convergent, btw, is apparently $1-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}-\cdots$ with itself.

Depends what you mean by that.

If you mean whether every university has people who attempt Putnam, I doubt it.

Certainly at the college level it's not true.

Dodsy

You should be able to request it in your math department

though it's only for U.S and Canada students, I believe.

Semiclassical

Whether someone at a given university could take it if they want, is another question and one I have no insight on.

Dodsy

the school I am going to meets once a week to discuss putnam

Semiclassical

3:06 AM

Oh, neat.

Chris Nguyen

After this competition I have to start college apps kms

Semiclassical

The FAQ for Putnam is here: maa.org/math-competitions/putnam-competition

Dodsy

hm

are you going into grade 12?

that seems early to begin applying..

or late.

Chris Nguyen

I'm going to grade 12 yes

Semiclassical

Looks like in principle anyone at a US/Canadian college can take the exam, but it requires a bit of setup on the part of your university/college

But I doubt that's much of a hurdle.

Dodsy

3:08 AM

yeah if you really want to do it you should be able to.

Chris Nguyen

I don't like this competition I'm studying for. Nothing like the AMCs

Semiclassical

I think I'll look up the latest list of problems in the AMM, see if there's any that seem both interesting and tractable.

Chris Nguyen

Whats AMM

Semiclassical

(It's behind a paywall, but hey university access)

American Mathematical Monthly

Chris Nguyen

oh

Semiclassical

3:12 AM

maa.org/press/periodicals/american-mathematical-monthly

Chris Nguyen

What if I qualify/ win a bunch of math awards after I already applied to college

Semiclassical

Could not tell you.

Chris Nguyen

qualify for AIME

Can I update colleges?

Semiclassical

Again, could not tell you.

Chris Nguyen

its ok

Is there like a math competition room here or something

Semiclassical

3:14 AM

That's a good question. Not as far as I know, though.

But that's something which other people might know, so---starred!

hmm, this is a neat-looking question in the latest AMM:

For $x\in(0,1)$, show that $$\prod_{n=1}^\infty (1-x^n)\geq \exp\left(\frac12-\frac1{2(1-x^2)}\right)$$

Chris Nguyen

Guys plz helphttp://www.mualphatheta.org/useruploads/files/mu_sequences_&_series_mu_alpha‌_theta_national_2015.pdf

oops

mualphatheta.org/useruploads/files/pasttests/…

question 14?

oh shoot wrong one

mualphatheta.org/useruploads/files/…

question 14 on this one

Semiclassical

$\displaystyle \lim_{n\to\infty}\sum_{i=1}^{n^2}\dfrac{e^{i/n}}{ne^n}=?$

Chris Nguyen

yes

Semiclassical

Well, first think I notice is that the 1/ne^n factor can be brought outside of the sum.

for the remaining series, my thinking is to write out the terms and see if I recognize anything

so that's $1+e^{1/n}+e^{2/n}+\cdots+e^{n^2/n}$

Which looks to be a geometric series.

Once that's summed in closed form and simplified, all that's left is a limit which should be straightforward enough.

Chris Nguyen

Ok thnx let me try

Semiclassical

3:23 AM

mmkay

Hrm. This seems more tedious than it should be.

Chris Nguyen

what do you mean?

Semiclassical

Well, I'm doing the hand calculations for what the sum would be.

And the limit just seems painful.

Doesn't mean it's not the right approach, of course.

Chris Nguyen

How would you do the limit

Semiclassical

Not sure, tbh.

Chris Nguyen

Can you explain the solution to me then? Its kinda vague

mualphatheta.org/useruploads/files/…

Semiclassical

3:29 AM

oh hey, I wondered if that was the approach to use instead.

Chris Nguyen

oh really

Semiclassical

So basically they solve it by making it equivalent to a Riemann sum.

Chris Nguyen

How can you recognize that so quickly

Semiclassical

Well, it helps that they've got the solution :P

Chris Nguyen

Can you explain the solution?

Semiclassical

3:30 AM

And in general when you go from a sum directly to an integral it's because of recognizing it as a Riemann sum.

Well, do you know how a Riemann sum relates to an integral?

Chris Nguyen

yes

How did they get from the 2nd to third line

Semiclassical

Yeah, that's the important one.

So they want to interpret $\displaystyle \frac{1}{n}\sum_{i=1}^{n^2} e^{i/n-n}$ as an integral.

Chris Nguyen

ok

Semiclassical

By comparison, the Riemann sum definition of an integral is

$$\int_a^b f(x)\,dx= \sum_{i=1}^m f(x_m)\frac{1}{m-1}$$ if I'm remembering right?

hrm.

no, I'm remembering it wrong.

$$\int_a^b f(x)\,dx = \sum_{i=1}^m f(x_m^*)(x_i-x_{i-1})$$ where $[a,b]$ can be partitioned using the $x_m^*$'s

Chris Nguyen

ok

Semiclassical

3:37 AM

But I'll be honest, I'm not see it immediately. What they've written seems plausible, but I never really felt like this kind of thing was my strong suit.

hmmm

hrm. I guess I really don't see it right now.

They have to be doing a Riemann sum somehow, but how they're doing it is far from obvious to me.

Chris Nguyen

ikr

Semiclassical

I guess I'll default for the moment to the obvious step of writing out the first few terms and see if inspiration strikes:

$$\frac1n e^{1/n-n}+\frac1n e^{2/n-n}+\cdots +\frac1n e^{n^2/n-n}$$

Zee

Is 3 hours of math reading/solving a day too little?

Chris Nguyen

no

Semiclassical

Depends what you're doing.

Zee

3:43 AM

For example , first year gradute math

Semiclassical

hmm

Chris Nguyen

why is everyone here in college lol

Zee

Can't tell if am lazy...

Semiclassical

that seems too low? but I've successfully repressed the memory of my first year of grad school.

Zee

Oh man am in trouble

Chris Nguyen

3:46 AM

How are they multiplying by 1/n when the limits of integration are from 0 to negative infinity

Zee

Three hours and am exhausted

Chris Nguyen

tbh during the summer especially before competitions I just do math all day

Semiclassical

@ChrisNguyen yeah, I'm perplexed by that too.

I mean, it seems right insofar as n is going out to infinity

but I can't piece it together

@Zee Work with others, it helps a lot.

Seriously, that's the only way to not go insane during first year.

Chris Nguyen

shouldn't the limits of integration be from 1 to 0 if they are multiplying by 1/n

Zee

@ChrisNguyen do you actually sit down and and think technically all day??

I find that unrelatable

Semiclassical

3:48 AM

@ChrisNguyen Well, bear in mind that the $i$-independent part includes the factor of $e^{-n}$.

Chris Nguyen

I guess I don't know how grad school works :P

Semiclassical

So they're not just multiplying by 1/n.

Zee

But it's not about math per say, just the ability to focus for hours

Am not sure anybody can focus on anything HARD for more than a few hours tbh

Chris Nguyen

I can actually when I'm doing hard problems

Semiclassical

@Zee I'll point out that I'm also not quite the right person to ask, since, y'know , Physics not Math PhD.

So other math grad students may have better insight.

Chris Nguyen

3:50 AM

@ Semiclassical Oh yea I forgot

Zee

Ya, I need more self discipline

Chris Nguyen

@ Zee I'm only in hs

@Semiclassical So they are actually mutliplying by $$\frac{1}{ne^n}$$

Semiclassical

Right. At least that's how I think it should be understood?

Zee

@ChrisNguyen maybe it makes a difference, I didn't gradute HS :p

Chris Nguyen

I still don't see how they got to the integral

Semiclassical

3:55 AM

At this point a lot of my self-discipline has just been exhausted.

Zee

@Semiclassical sounds to me like you need a vacation

Semiclassical

why do you think I want to get out of grad school?

Chris Nguyen

My vacation is a math competition in New York :P

Zee

Well you can travel during the summer semi

Go to Thailand or Argentina

Semiclassical

Eh. Depends what you're doing.

Zee

3:57 AM

Aware me

Semiclassical

This particular summer I'm trying to get my exit plan figured out.

So travelling wouldn't be a great idea.

Zee

Do it while traveling, or travel for a couple weeks, trust me, two weeks and you will know what the heck you need to do

Semiclassical

there's some truth to that.

Zee

Has nothing to do with "finding yourself" it's all about just changing your perspective outwardly

And for a short period of time, till you can look at your life soberly

Chris Nguyen

Anyone see how they got the integral?

Zee

4:00 AM

What integral

Semiclassical

Basically, we're trying to figure out the following step which a solution makes:

$$\lim_{n\to\infty}\frac{1}{n}\sum_{i=1}^{n^2} e^{i/n-n} = \int_{-\infty}^0 e^{x}\,dx.$$

It seems plausible as a "oh hey Riemann sum" thing, but

plausible isn't a proof :P

Zee

Man, it's been a while since I seen a specific integral

Chris Nguyen

Do you guys not do this stuff as math majors

Semiclassical

not so much, really. and if I did, it's been literally years

life isn't just math competitions :)

Chris Nguyen

True

Zee

4:06 AM

Depends on the course, I only seen specific integral in calculus but even then they were simple

Semiclassical

one thing I can see is that, if you write $j=n^2-i$, then the sum becomes $\frac{1}{n}\sum_{j=0}^{n^2-1}e^{-j/n}$

Zee

Math isn't math competitions...

Semiclassical

Not convinced that's helpful, though.

Chris Nguyen

I just need to do well on math competitions to get into college ya know

Zee

Nothing wrong with competitions just does not capture the poetic aspects of math

Semiclassical

4:09 AM

And really doesn't tell you much about research.

Chris Nguyen

I know

But how can I do research when I'm in hs

Zee

@ChrisNguyen let me tell you about a little secret in math, a second rate department might still have a few leading top mathematicians in their specific domain

Semiclassical

Therein lies the problem.

But, hey, that's why you go to college.

Chris Nguyen

I'm probably gonna do a math minor not major

Zee

@ChrisNguyen why???

You don't wanna be loved and admired?

Semiclassical

4:11 AM

Got something else in mind?

@Zee ...

you're kidding, right

Zee

Chicks love mathematcians bro

Chris Nguyen

I want to go to med school

Semiclassical

Can I live in your universe? Sounds like an interesting place.

Zee

Pffff docs are just glorified mechanics

Chris Nguyen

lol

Semiclassical

4:12 AM

naaaaah

mechanics don't end up with nearly as much debt :P

Chris Nguyen

LOL

Zee

Hahaha

if that's your calling it's your calling

But lots of people wanna be docs for reasons that will make them miserable

Chris Nguyen

like

Semiclassical

i'll say that med school seems like an absolute grinder

i say that with more ignorance than knowledge of the process, though

Zee

like going in to help people and then wishing that patient that never stops cursing you out would die so you would not have to deal with his crap

Chris Nguyen

4:15 AM

chill

Semiclassical

It is a pretty exhausting profession, yeah.

Zee

Am telling you what I heard from a friend of mine whose a nurse

You stop caring at all who lives and who dies until one day you break down couse the guy who died has a name close to yours

Semiclassical

Burnout isn't uncommon, no.

Zee

Am just giving you the awful stuff I heard , but it has its good moments too

After 12 years of debt

Become a scientist, they are like doctors but cooler

Semiclassical

That's one thing which grad school has over med school: You get paid as a TA.

i.e. you don't have to accumulate more debt on top of what you got from college.

Zee

4:19 AM

And you do research

Chris Nguyen

Can we just get back to the math q plz?

Works for me.

Lol he got shook

lol

The moral of what am saying is, know what you like, don't listen to what is "in" , just go for anything you love and pays the bills

Semiclassical

4:21 AM

The main thing I'm noticing right now is that $\int_0^{n^2}\frac{1}{n}e^{-x/n}\,dx=1-e^{-n}\to 1$ as $n\to\infty$.

But I'm not seeing how to quite connect that with $\frac{1}{n}\sum_{j=0}^{n^2-1}e^{-j/n}$

Zee

So the left side converges to 1?

Semiclassical

Sure, that's the final desired result.

I feel like it should almost be obvious :(

Zee

Idk why bring in the integral , isn't more pleasant to work with the series?

Semiclassical

well, the resulting limit isn't great.

Zee

Can you post the original question?

Semiclassical

4:26 AM

But it's probably worth giving the series solution in any case.

It's up there in the transcript somewhere...

solution 14 here: mualphatheta.org/useruploads/files/…

The problem is found elsewhere, but tbh it's pretty much self-explanatory from the setup of the solution.

Chris Nguyen

When I solved the integral I didn't get what you got

Zee

And you wanna prove the solution?

Chris Nguyen

@ Zee yes

Semiclassical

@ChrisNguyen Yeah, I'm doing a slightly modified route I think.

But really, $\int_{-\infty}^0 e^x\,dx$ and $\int_0^\infty e^{-x}\,dx$ aren't that different.

Anyways, going from the sum I just wrote out: $\frac1n \sum_{j=0}^{n^2-1} e^{-j/n}$ is a geometric sum and gives the closed form

$\dfrac{1}{n} \dfrac{e^{-n}-e^{-1/n}}{e^{-1/n}-1}$

...blah, I've done something wrong. That doesn't give the correct limit.

Chris Nguyen

wait howd you get that

all the e stuff

Semiclassical

4:32 AM

Well, if the form I wrote out was correct, that's a sum of the form $\sum_{j=0}^{m-1}x^j=1+x+\cdots+x^{m-1}$

But that's a geometric series which equals $\frac{x^m-1}{x-1}$

Chris Nguyen

So you are not going the rieman sum to integral route?

Semiclassical

Which actually means I was a bit wrong. Should be $e^{-n}-1$ on top.

Yeah, I want to see how well the direct approach works.

So (moving it around a bit to be more convenient) I want $\dfrac{1}{n} \dfrac{1-e^{-n}}{1-e^{-1/n}}$ as $n\to\infty$

This is not too hard to do, though. As $n\to\infty$, the top becomes 1.

On the bottom, the behavior of $e^{x}$ for small $x$ is $1+x+\cdots$.

Akiva Weinberger

Hi $\phantom{What's up}$

Chris Nguyen

Hi

Semiclassical

So $n(1-e^{-1/n})=n(1-(1-1/n+1/(2n^2)\cdots))=1-1/(2n)\to 1$ as $n\to\infty$

hence it goes to 1/1 = 1 as $n\to\infty$

I don't really like that approach, but it does get you to the right place.

Chris Nguyen

4:36 AM

I got lost

Semiclassical

...yeah, there's the other problem :P

Chris Nguyen

let me catch myself up

Semiclassical

So, here's the series solution starting from the problem statement.

okay, I'll hold on

Chris Nguyen

How did the numerator in the geometric sum

I thought a geometric sum was a(1-r^n)/(1-r)

r is the common ratio and a is the first term

Semiclassical

Let's be specific just so that I don't mix myself up:

$$a\dfrac{1-r^n}{1-r}=a+ar+ar^2+\cdots + ar^{n-1}$$

Right?

Chris Nguyen

4:43 AM

yes

Semiclassical

And, just to make sure we're on the same page, which summation are you trying to do? (I've got two versions written down, and while they're equivalent I'd rather avoid any confusion)

Chris Nguyen

the e^(1/n)+e^(2/n)+...+e^(n^2/n) summation

Semiclassical

okay.

So that's got $a=r=e^{1/n}$, and instead of $n$ it's $n^2$

So that'll sum to $e^{1/n}\dfrac{1-e^{n^2/n}}{1-e^{1/n}}$

Recalling that the original series was $\displaystyle \sum_{i=1}^n \dfrac{e^{i/n}}{ne^n}$

Chris Nguyen

Wait plz

howd you get the e^(n^2/n) term

Semiclassical

Remember, that's $r=e^{1/n}$ with the $n$ in the geometric series being $n^2$.

So the numerator should contain $1-r^{n^2}=1-(e^{1/n})^{n^2}=1-e^{n^2/n}=1-e^n$.

Chris Nguyen

4:52 AM

I'm still thinking sorry

ok i got it

Semiclassical

Mmkay.

Chris Nguyen

but what happened to the e^(1/n)

Semiclassical

If you're comparing to what I had earlier, note that what I was doing there wasn't the original series.

it was the one I got by taking $j=n^2-i$

So that one would look slightly different.

But we don't really need to do that here. We can just work with the original series.

So let me do that...

Leaky Nun

How do you guys pronounce $\ln$?

Semiclassical

lawn.

usually.

(sometimes L-N)

Zee

4:58 AM

L n

Lawn lol

Semiclassical

We just argued that $$\sum_{i=1}^{n^2}e^{i/n}=e^{1/n}\dfrac{1-e^n}{1-e^{1/n}}$$

Chris Nguyen

yes

Mithlesh Upadhyay

0

Q: How to decide $36th$ smallest element in max-heap tree of $100$ elements?

Consider a max heap tree with $100$ elements and a node from the same level is chosen randomly. What is the probability that it is the $36th$ smallest element______ . My attempt: Somewhere, it explained as: $P = 1/7∗0+1/7∗1/2+1/7∗1/4+1/7∗1/8+1/7∗1/16+1/7∗1/32+1/7∗1/37=0.087$ According ...