12:23 AM
@FernandoMartin

sup @Pedro

How you doin'?

chillin

nope

12:42 AM
@FernandoMartin did you see the linear problem I sent?

nope

@FernandoMartin given a matrix with strictly positive real entries, prove it has a positive real eigenvalue

let's see

as of three minutes ago the only proof I knew was topological

I was trying to do it elementarily
I failed
then I recalled I'd seen this proof in my topo class

12:54 AM
@FernandoMartin "Oh my God! I've never seen so much blood!"

I solved it this morning as an exercise in Bredon
which has great exercises so far

uh
it had something to do with some quadrant of the sphere having the fpp
right?

intersect the sphere with the first "orthant" (general word for quadrant, orthant)
then that intersection is homeo to the closed disk

that's right, I didn't recall the word

$x \mapsto Ax/\|Ax\|$ thus defines a map $D^{n-1} \rightarrow D^{n-1}$ and has a fixed point, thus $A$ has a positive real eigenvalue

12:56 AM
right

not hard to do when the book says "use the brouwer fixed point theorem to prove ..."
but i would have never come up with that if it didn't give me that hint

1:27 AM
@JasperLoy

1:39 AM
@LasperJoy

1 hour later…
3:05 AM
@TedShifrin Heya-.
@DanielFischer Maybe.
@TedShifrin Too.

3:26 AM
Anyone mind helping me out?

Depends. With homicide? Probably not.

who are you murdering

We're not murdering anyone!

I need to determine an equation of a line tangent to the curve at the given point: 9(x^2 + y^2)^2 = 100xy^2; (1, 3)
Implicit differentiation, of course.

OK, differentiate then.

3:28 AM
which i suck at
18x^2 + 18y^2
=
200y?

Chain rule, mister.

don't know how I put this into a chain rule format

Write things as 9(x^2+f(x)^2)^2=1000x f(x)^2
Now differentiate w.r.t. to x.
And use all the rules.

@PedroTamaroff Maybe you're not.

@PedroTamaroff yeah, I kinda suck at calc and derivatives altogether, that's why I came for help here

3:36 AM
@DemCodeLines Well, can you differentiate xf(x)^2 wrt to x?

@PedroTamaroff Well, first I'm not sure I understand what wrt stands for (I probably know, but can't figure it out just by looking at it right now). Second, that would be 2f(x), I think.

@DemCodeLines "with respect to"
@DemCodeLines You're not using the product rule nor the chain rule.
What is the derivative of f x g for f,g two functions? here "x" denotes "times".
If you check your notes, it is f g' + f' g
Above, you have a product of x and f(x)^2

so x f(x)'^2 + x' f(x)^2?

Almost there.
So (x f(x)^2)' = x' f(x)^2 + x (f(x)^2)'= f(x)^2+ x(f(x)^2)'.
What is the derivative of f(x)^2 ?

2f(x)

3:41 AM
Nope.
It is also a product.
f x f
So (f x f)'= f f' + f' f = 2 f f'
You can also use the chain rule.

This is all way too confusing to me. Calculus might not have been a good choice.

Before starting to diff, you need to identify what your function is. Is it a product? Is it a composition?
There are certain rules that tell you exactly how to deal with those things.
@DemCodeLines Just practice.

It's a product

Yes, x f(x)^2 is a product.
You can choose to see it as a product say (x f(x))(f(x)).
Then use the product rule on that to get (xf(x))' f(x) + xf(x) f'(x). But now (x f(x) ) is another product.
So again you use the product rule.
And so on.

so I'm thinking about writing something for the blog, and I'm thinking of writing on evidence of Zipf's Law in MSE: which basically says that the terms used happen to follow a power law in distribution. Does that sound interesting?
@DemCodeLines one step at a time - that's all

3:46 AM
@mixedmath Come again? =P

@PedroTamaroff But how would you answer the initial question then?

@PedroTamaroff so, MSE has a blog, and I'm thinking of writing something for it.
There's this heuristic law called "Zipf's Law". Let's say that in MSE, the first most common word is "the", the second is "is", the third is "math", and so on

and further, that "the" happened 1000 times, "is" happened 500, and "math" happened 300, and so on
it just so happens that these relative rates, 1000 to 500 to 300 follow a pattern (heuristically) in many things
words in books, in common use, etc.
it just so happens that they follow a pattern on MSE too, a sort of power law

f(x)f(x)' + f(x)'f(x) = 2f(x)f(x)'

3:50 AM
in particular, if you plot the frequency (1000, 500, 300, ...) against the order (1st, 2nd, 3rd), you get a predicted curve (up to some parameter)

9(x^2 + 2f(x)f(x)') = 200xf(x)f(x)'
Plug in (1,3)
9x^2 + 18f(x)f(x)' = 200xf(x)'f(x)
sigh

Hi can someone create a room and teach me something real quick please?
Or can we do it in here (i don't care)

@PsychOPhobiA what do you want to learn?

Something on calculus.

@PedroTamaroff But yeah. Does that sound interesting?
@PsychOPhobiA Something quick?

3:52 AM
No, it can be patient.

@mixedmath like a linguistic benford's law?

@mixedmath Maybe starting with the basics?

@MikeMiller yes, exactly
@PsychOPhobiA I'll do it, I'm just trying to decide whether to do it here or in a different chat

I know nothing about it so I'd be interested
when I get off my butt I'm going to write something about Nevanlinna's 5-value theorem

@PsychOPhobiA let's do it here, but with constant pings, I think
@PsychOPhobiA so what's up?

3:54 AM
@PedroTamaroff don't leave me!!!!

@mixedmath Well some friends told me some stuff but said I needed to know a really good math.

@PsychOPhobiA is there a reason you can't post a question on the main site?

@MikeMiller ah, that's what I needed to know - I'll proceed. I've actually already run the data, and it follows through
@PsychOPhobiA ok, sure

I think if one person benefits, it's worth writing
in my case, I think I'll benefit from writing mine up :)

@MikeMiller Yes, I don't want to post a question because it would be hard to explain. Probably to long
@MikeMiller Something like "I need to learn calculus" may not work.

3:57 AM
@PsychOPhobiA I can teach you something. Jalapeños ain't good for doggies.

@PedroTamaroff Okay. I am trying to rember what I need to know

@PsychOPhobiA well, unless mixedmath is a lot more patient than I (entirely possible) someone here isn't likely to be willing to teach you all of calculus...

Something that has to do with angles and matrices

On a tip similar to Pedro's, you should use volumizing shampoo to make your hair properly poofy

@MikeMiller ah, I know how that feels

3:58 AM
@mixedmath do you know the theorem?

"Something, something, something... dark side."
@DemCodeLines I'm here.

@PedroTamaroff What can you possibly teach
?

@MikeMiller not yet, but perhaps I will after I read your post ;[

@PedroTamaroff wut

@PsychOPhobiA I am pretty good at fraktur typography.

4:00 AM
brb

Is that of any interest?

@mixedmath ah, I'm really only motivating the theorem rather than proving it (not possible, given length and background constraints), so I suppose I'll refrain from stating it for now

0

Find equation of a line tangent to the curve at the given point: determine an equation of a line tangent to the curve at the given point: 9($x^2$ + $y^2$)$^2$ = 100x$y^2$; (1, 3) How would I go about solving this. I know it requires implicit differentiation. I have been having a lot of trouble...

There you guys go
Get some free points

@MikeMiller Interesting thing: if $G$ is a finite group and we form $k[G]$, the center of the algebra has dimension $[G:Z(G)]$, that number of conjugacy classes in $G$.

if you say so boss

4:03 AM
@DemCodeLines That's not the point. Someone can give you a nice colorful uppity answer, but you need to read your notes and learn the rules, and use the and reuse them till they are easy to remember and come off naturally.
@MikeMiller In fact a sum $\sum \alpha_x x$ is central if the $\alpha_x$ are constant over the conjugacy classes. That's basically the important observation.
And $k[G]$ is bialgebra.

where are you learning this?
Rotman?

@PedroTamaroff Yes, I understand that, but I am stuck on the last question and having to spend an hour or two mastering the concept before I can finish it is a little tedious. Plus, it's a little late in the night here too.
Not that I won't learn, I don't feel like failing tests, so i have to learn eventually

@MikeMiller THe first is from my course. The second I scanned over in Jacobson's BAII.
@DemCodeLines Then go to sleep and do it properly tomorrow.
You'll gain nothing from spoonfeeding.
People like to spoonfeed but most aren't good at changing smelly diapers... that's where you don't want to find yourself.
My point is: don't learn the bad way, you'll catch up bad habits and mistaken ideas that will be very hard to get off your brain.

@PedroTamaroff You're right and it's very frustrating.

@DemCodeLines Chugga chugga chugga chugga...

4:11 AM
@DemCodeLines bwahaha
(evil laugh)

@PsychOPhobiA If you are trying to "learn calculus" for the first time, I might be so bold as to recommend a post I wrote for some of my students once on calculus at an introductory but math-literate level

@mixedmath need the http

yep
@MikeMiller thank you!

"I can't feel mah legs!"
"Bubba, they ain't there!"

4:13 AM
:D
I saw Forrest Gump on ABCFamily once
that is not even remotely a family movie

@DemCodeLines did you go away?

@MikeMiller Love it when school director leaves house and Forrest is "UH! UH! UH!"

@DemCodeLines anyway, your homework grade is probably miniscule, as compared to the effect of the tests. it's very important that you do the homework in order to pass the course, but the reason it's important is that it helps you understand the concepts. focus more on the reason you do things in your computations rather than the computations themselves.

@mixedmath No, I took @PedroTamaroff's advice and decided to properly do it tomorrow.
Meanwhile, I bookmarked the link about intro to calc you shared above ;)

so do the homework, but instead of focusing on getting it right for turning it, focus on understanding what's going on :)

4:15 AM
oh, boo. Someone gave him a complete answer on his question.

user105491
@mixedmath Is it me or Rene? Who are you referring to?

@mixedmath Yup. I definitely have had moments where I wished I had tried a little harder at learning the concepts.

@SanathDevalapurkar Rene. I noticed you posted and immediately came in here to see what was up
I appreciate that - that's honest

user105491
@mixedmath I've never been in this chat room before. What do you guys do in here?

@mixedmath I know, still, I'll probably just go and do it out myself

4:18 AM
@DemCodeLines that's the spirit!

@SanathDevalapurkar You're too young. This is a dark place.

@DemCodeLines And I hope you will.

The chat is dark and full of terror.

@mixedmath No really, I will. The last thing I want is to fail a test. That will destroy my grade.

user105491
@PedroTamaroff It is? Are you joking or s/t? :-)

4:19 AM
@SanathDevalapurkar hmm. Well, sometimes math. I'm usually more of a passive observer here. I came in here today to pitch a blog idea

From my experience in other chat rooms, the people in them usually talk about everything but the topic that the room was created for. I'm guessing the trend applies to this room too.

that's correct

@SanathDevalapurkar I'm a male and it is 1:20 a.m.

user105491
@PedroTamaroff And you're on M.SE? Wow, I got to appreciate your love of math! :-)

@SanathDevalapurkar Well, it is Monday.
Were it Friday... we wouldn't be talking.

4:22 AM
sounds like Pedro's got a bad case of the mondays

user105491
@PedroTamaroff Yeah, I understand how you feel about fridays. :-)

@PedroTamaroff it's 1:20am? Where do you live?!

a small island in the middle of the atlantic sea, @DemCodeLines

4:26 AM
@DemCodeLines Look at a map and surf through the meridians.
@MikeMiller Heh, BAII has a typo: $\otimes$ when they mean $\oplus$.
@SanathDevalapurkar Do you know a bit of group theory?

@PedroTamaroff oh no!

user105491
@PedroTamaroff Yeah, I do.

user105491
@PedroTamaroff Why?

@SanathDevalapurkar OK. Here's a problem for you.

user105491

4:30 AM
Prove that $\Bbb Q$ has no (nontrivial) subgroup of finite index.
That is, seen as a additive group.
In fact, prove that it does have nontrivial subgroups of finite index when seen as a multiplicative group.
Hint $\Bbb Q^\times \simeq (\Bbb Z/2\Bbb Z)\oplus \bigoplus\limits_{i\geqslant 1}\Bbb Z$
Hint $\Bbb Q^+$ is divisible.

user105491
@PedroTamaroff Let me prove the first one first. Suppose H is a subgroup of Q. Then, [Q:H]=n=|Q/H|. For every q\in Q, n(q+H)=H. Hence, nq+H=H, i.e., nq\in H. Thus, nQ\subseteq H. However, nQ=Q, hence Q=H.

$\LaTeX$ works here.

user105491
@PedroTamaroff For the multiplicative group problem, you do mean that Q/\{0\} has nontrivial subgroups of finite index, right?

@SanathDevalapurkar That's what $\Bbb Q^\times$ denotes, yes.
@SanathDevalapurkar You're thinking?

user105491
@PedroTamaroff Yeah; give me a few minutes - I'm trying.

user105491
4:45 AM
@PedroTamaroff I don't know. How do you prove it?

@SanathDevalapurkar Did you see my hint?

@Bobthezealot Hi Bob! (I'm here too, and still interested)
oh, bye @Bobthezealot

Can you prove that isomorphism?

ok
hope this is clear enough

user105491
4:51 AM
@PedroTamaroff Probably not right now; it's almost 2200 here, and it's a weekday. I've got school tomorrow, so I hope it's OK if I chat tomorrow (possibly with a proof).

@Bobthezealot out of curiosity, how did the winners from middle, high, and the world champion compare?

@SanathDevalapurkar Hint $\pm p_1^{\pm \alpha_1}p_2^{\pm \alpha_2}p_3^{\pm \alpha_3}\cdots$

what do you mean?

user105491
@PedroTamaroff Thanks for the hint; I'll try proving it tomorrow at school. Thanks for the enlightening conversation, but it's time for me to say goodbye. Until tomorrow, my fellow mathematician!

4:59 AM
@mixedmath do you read any other math blogs?

@Bobthezealot is the high school champion better than you in a meaningful way (or vice versa?)

no, each division is diffrent

@MikeMiller yeah. I read Terry Tao, Tim Gowers, Jordan Ellenberg, the AMS grad student blog, and sometimes click randomly on Terry Tao's blogroll for posts

@MikeMiller I do.

(since his blog is one of the best blogrolls that I know of, other than mathblogging.org)

5:01 AM
@mixedmath I once read Gowers' blog. Didn't enjoy the nontexification. =P

@PedroTamaroff what do you mean? he uses TeX

I imagine he's a great writer though

@MikeMiller I didn't until I met him, and I really like both him and his writing

Oh, SHOOT. Math.
@mixedmath Hm... not always.

5:03 AM
@PedroTamaroff that could be. I don't remember. I probably wouldn't read nonTeX math either
I've grown soft

@PedroTamaroff oh gross. And he even went through the effort of making subscripts and superscripts

Gowers' nontechnical writing is great, but his technical stuff is always a bit too dense for me (technically self-contained, but clearly not meant for the combinatorial neophyte)

@MikeMiller I remember he wrote on Taylor's Theorem once, and I felt very surprised at how hard it was for me to really read something that I understand so well
it's not that I don't get it - I do - but the presentation is just so different than how I present it, and different than how I think of it

I once read someone on this site (perhaps Georges E.?) who said that the best sign of understanding something is that everyone else's presentation of it seems wrong to you

5:10 AM
@MikeMiller ah, that's funny. I sort of like that

@MikeMiller Interesting.

Man I clearly didn't properly learn Taylor's theorem! The only statement I remember is that an analytic function is locally expressed by its Taylor series
which appears to not be the theorem at all!

@Chris'ssis when you wake up .. you don't wanna miss this AWESOME thing :D .. some one save our souls :P :D ;)

@MikeMiller Do you read any math blogs?

@mixedmath I try to keep up with terry tao's and tim gowers' as well; and when I can understand the Secret Blogging Seminar it's incredible.

5:25 AM
ah, I haven't read the Secret Blogging Seminar in a long time

they also haven't posted much in a long time :)
Scott Carnahan appears to have posted something a couple weeks ago I should read
I learned fairly recently (couple months ago) that a great way to learn some new math is to go to MO once a week and click the week tab - because of the number of upvotes, these will generally be applicable to/understandable by a wide audience, and usually will be really cool

@MikeMiller oh, that sounds like a good plan. I keep on intending to read more MO. I also feel like I should be spending more time asking/answering on MO, as I'm trying to become a research mathematician
but... it's just so hard...

haha yeah
usually I don't get a whole lot out of my weekly checks except for a few interesting new things (since so much I don't understand), but those new things are often great
of course, as a wee child, I have more time to check stuff like that

in Tagging, 47 secs ago, by Martin Sleziak
There is a new tag called . What is the purpose of this tag? Should it be synonym of ? (AFAIK precalculus is an American term, so maybe some comments on this familiar with American education system could be useful.)

@MartinSleziak Agreed it should.

5:34 AM
@MartinSleziak I don't think it should exist at all. If anything, it should be a synonym of algebra-precalculus.

@MartinSleziak I agree with Mike

ok, I'll post a request on meta
Thanks for the feedback!

@mixedmath I asked you once, but what do you work on?

@MikeMiller I work in analytic number theory on automorphic forms, Maass forms, and mutliple Dirichlet series usually; recently, I happen to be working on explicit asymptotics arising from number theory in function fields
but my dissertation (assuming it goes along in the same direction it's currently going), is on GL(3) automorphic forms
and I'd be happy to explain anything more - just point me where

5:51 AM
morally, what's the goal?
(I don't think I'd get much out of the details, so I guess I should go in the opposite direction!)

@MikeMiller that's a nice question
these are functions that have zeta functions (or L-functions) as data in some way, like as coefficients or as complex residues at poles. I study bounds on these functions, which translate to bounds on the data
if we were super duper and understood all powers of the bound I'm looking at, we'd have the Riemann Hypothesis and the Lindelhof Hypothesis
(no chance - don't hold your breath)

ah :)
interesting, thanks!

anyhow, I'm off to sleep
good night all!

night!

2 hours later…
8:21 AM
Greetings
@robjohn it's interesting to see that Mma goes crazy when computing that series. Just look at the result of Sum[(Zeta[2] - HarmonicNumber[n, 2]) (Zeta[3] - HarmonicNumber[n, 3]), {n, 1, Infinity}]
Numerically it's about $-0.128523$. However, the partial sums are correctly computed.

Greetings my friend :-)

@skullpatrol Hi :-). How are you doing?

Fine thanks how are you?

How was the interview?

8:25 AM
@skullpatrol I wasn't called yet ... (I only had that interview by phone).

Did you tell them what they wanted to hear?

@skullpatrol Yeah, excepting the salary part.

:-)
@robjohn Mathematica does some horrible mistakes when dealing with double, triple Euler sums.
@robjohn this is another one Sum[HarmonicNumber[n, 2] HarmonicNumber[n, 3] - Zeta[2] Zeta[3], {n,
1, Infinity}]
It says the series evaluates to -(Pi^4/36) + (17*Pi^6)/5760 + (1/4)*(-4 + Pi^2)*Zeta[3]
that is completely wrong.
@N3buchadnezzar can you test the series above in Maple? Mathematica fails to compute it correctly.

@Chris'ssis To compute your series, use the Hurwitz Zeta function. Not that it can compute your series in closed form, but it gets the numerical value quickly.

8:33 AM
$$\sum_{n=1}^\infty \sum_{k=1}^\infty \frac{1}{n^2}\frac{1}{k^2} - \zeta(2)\zeta(3)$$ ?

@robjohn Can you show me the code pls? I think you also showed me soemthing like that in the past, but I forgot it ...

@Chris'ssis Yeah, I can not really read mathematica code

@N3buchadnezzar $$\sum_{n=1}^{\infty} (H _n^{(2)} H_n^{(3)}- \zeta(2)\zeta(3))$$

Zeta[n]-HarmonicNumber[k,n]=HurwitzZeta[n,k+1]

No closed form

8:43 AM
@robjohn Thanks, I already did it.

@Chris'ssis I don't know if you can get a good form for that using the Hurwitz Zeta.

Looks to me as it either diverges, or converges to -12 or something >.<

@Chris'ssis are you sure that converges?
@Chris'ssis $H_n^{(2)}\sim\zeta(2)-\frac1n$ and $H_n^{(3)}\sim\zeta(3)-\frac1{2n^2}$

@robjohn Not really. That was my point too. But Mathematica says it converges and shows a closed form.

@Chris'ssis Numerically, it grows about 3 for each power of 10 in the partial sum.

8:52 AM
@robjohn I think this would be the proper form $$\sum_{n=1}^{\infty} \frac{1}{n}(H _n^{(2)} H_n^{(3)}- \zeta(2)\zeta(3))$$

@Chris'ssis that would converge

@Chris'ssis This is weird... I think the plot should be smoother... DiscretePlot[n^2(Zeta[3]-HarmonicNumber[n, 3]), {n,1000,100000,1000}, ImageSize -> 1000]

Got it.

@Chris'ssis This one looks good: DiscretePlot[n(Zeta[2]-HarmonicNumber[n, 2]), {n,1000,100000,1000}, ImageSize -> 1000]

9:07 AM
@robjohn Yeah, that one looks nice.

9:47 AM
I find several definitions, and some say that " A homothetic function is a monotonic transformation of a function which is homogeneous of degree 1. " while others do not require that "degree 1" restriction.

@Sush Sorry, I only know that a homothety is a map $x \mapsto r\cdot x$ for some $r > 0$. Never encountered "homothetic functions".

@DanielFischer, OK!

1 hour later…
10:58 AM
^ nvm

11:13 AM
Is Maths SE down ?
Nvm it's back up for me

11:33 AM
@DanielFischer, how does $\displaystyle \sum_k \binom r k x^k \sum_{n-k} \binom s {n - k} x^{n - k}=\displaystyle \sum_n \left({\sum_k \binom r k \binom s {n-k} }\right) x^n$ hold?

Switching order of summation.
Try this with arbitrary finite series

@BalarkaSen, what is "Switching order of summation?"

Just a big name. Try looking at arbitrary series, as I suggested.
$\sum a_n$ and $\sum b_n$ for example

OK!

@Sush Did you get it?

11:46 AM
@BalarkaSen, No!

OK. Consider $\left ( \sum a_n \right) \left ( \sum b_n\right )$. And consider $\sum a_n \left (\sum b_n \right )$

@Hippalectryon yo hippa, t'avais disparu ?
@BalarkaSen good afternoon

afternoon, @G.T.R

@G.T.R exactement :) - Lundi c'était férié et j'étais VRAIMENT fatigué donc je ne me suis pas connecté
@G.T.R Vous avez fini la RMS ?

caps, but not shown for readability issues would you guys stop talking in rapid French?

11:49 AM
@BalarkaSen :c Let's try russian :D

Heck.
"LOLOLOLLOLOL" as @G.T.R would say.

@Hippalectryon Nope, too much to do in there

@G.T.R My brother's teacher gave 10 exercises to search to each students
@G.T.R So before the orals they had a whole RMS tome with corrections
@G.T.R :D

@Hippalectryon too ambitious IMHO

@G.T.R Well they managed to do it :)
@G.T.R Some of the RMS's exercises are just big jokes
Some others are really hard

11:54 AM
@Hippalectryon I'm going to post a weird puzzling one on main to see if it is "a big joke"

@G.T.R Hehe

@BalarkaSen, still not getting that original question! I expanded first polynomial, but is that only way?

@Sush What's not to get? It's a rearrangement, right?
$$\sum_k \binom r k x^k \sum_{n} \binom s {n-k} x^{n - k}=\displaystyle \sum_n \left({\sum_k \binom r k \binom s {n-k} }\right) x^n$$
Both are finite sums, one is rearranged to get the other.

@BalarkaSen, how there $n-k$ becomes $n$?

For example, $(x_1 + x_2 + x_3) \cdot (y_1+y_2+y_3) = x_1(y_1+y_2+y_3) + x_2(y_1+y_2+y_3) + x_3(y_1+y_2+y_3)$
@Sush I thought that $n - k$ was a typo...?

11:58 AM
NO! @BalarkaSen, I am trying to understand this

@Sush I am not sure what the subscript means then.