hi @Daminark @BalarkaSen

@orlp I wonder, though. I see the downvotes. Did you make those? (I wouldn't have if I were you; they're good faith efforts to answer your question. Unless they were definitely, glaringly wrong I wouldn't downvote.)

@Wildcard they're not

@orlp Okay, good. :)

but it's also against SE policy to 'cancel' downvotes with upvotes

I only upvoted one (for a clever protocol), and I commented on it to that effect, because even that one I am not convinced the numerical answer is correct.

@orlp I didn't.

so while I don't want their answers to sit at -1

they didn't seem clear or convincing enough to upvote

@orlp I think that's too vague. I would define "cancel" as "to upvote without reading the content, simply because it has a downvote on it."

@orlp If reading them helped you at all in sifting through the problem, you are justified in upvoting. That's not the same as "canceling" a downvote.

But I agree that they are certainly not clear/convincing.

hi @Adeek

Yo @Adeek

(sniped)

Hiiii.

@orlp I don't think anybody controls the way you vote.

@Daminark Nifty, sick, LOL.

@BalarkaSen It is way past your bedtime, LOL.

Lmao

i rekt my sleep

again

By the way, I deleted my ELU and MSE accounts, I am using my ELL account now, LOL.

Oh hello @Faust LOL

MMorning

anyone done complex analysis?

i got an A+ in complex calculus but my analysis is self taught so im alittle worried about taking it

I have not seen a course called complex calculus anywhere.

But I guess it's possible, of course.

Nor have I, Jasper. Just sayin'.

Hello @TedShifrin LOL

hmm ok

Hi @Ted

hi Leaky

So how was the first day of classes, Faust?

morning leaky

@Faust I am guessing that the complex calculus course teaches you a lot of computations without the proofs, and the complex analysis one teaches you the proofs.

really good @Jasper no we did rigours proofs of most things

so far all my profs seem awesome!

@BalarkaSen This is far from newsworthy. :( You excel at this.

@Faust LOL. Then I have no idea what the course teaches.

That's great, Faust.

Now you don't need me :D

we only did like epsilon and delta proofs like once each section

Wait. What're you taking?

but i wasn very good at them despite that they appear alot easier than over the reals

i wanna take

They're the same no matter where they are :P

math.uvic.ca/courses/2017f/math401/a01/Course_outline_401.pdf

@orlp I've made some further comments on the answers. In particular:

Can one prove that all quartic polynomial in $\Bbb R[x]$ is reducible without resorting to complex numbers?

Oh, you took the first complex analysis already? I like Gamelin's book. That should be great stuff.

all the topics i find really intresting in that course outline

@Leaky: Do we know that the only irreducibles in $\Bbb R[x]$ are linears and quadratics?

i know about the first 300 or so pages appaerntly of Gamelins book according to my prof

@TedShifrin How do we know that?

It's a good book. If you do, you're in good shape.

I need the fundamental theorem of algebra for that, @Leaky.

So, yes, I'm resorting to $\Bbb C$.

so complex number is a must?

I think so.

well thats what he claims math 301 covered and i got an A+ in it but i have no analysis pre-req hes offered to waive it im just trying to decide if i can get by

then how does the fundamental theorem imply that?

If I hadn't gotten cancer, I was going to teach our graduate complex analysis course using Gamelin. It's a decent book.

You definitely need real analysis basics.

hmm\

i have self taught skillz but not sure how skillful they are

@TedShifrin You got cancer? I didn't know that. Sorry to hear that. I hope you recover soon.

Definitely need solid uniform convergence skills.

i see

@Jasper: I am fine, thanks. I am now almost 6 years out from surgery.

@TedShifrin Oh, LOL. I thought it was a recent thing.

No, I didn't teach that grad course because I wasn't sure how I'd be and so I assigned it to someone else.

I shouldn't have mentioned the C word.

i can manage the definition and maybe even prove certain things are but only by leaning on theorems maybe i just watch it too many interesting topics not to go.

If you haven't done solid uniform convergence proofs, you'd better practice those, Faust.

Heya @MikeM.

Really sorry to hear about the C lost my grandpa to it :(

but it was one of those untreatable kinds

The C word is the most vulgar word in the English language, LOL.

what is a metric?

nvm

d(a,b)

@Faust Sorry to hear that.

@Faust Have you done a course on metric spaces?

no

i have read about them apperntly

also have a book on them

@Faust: I'm glad to try to help you, but I think this course may be a bit ambitious. But see how it goes.

maybe avoid the complex analysis but they have reimans zeta functions and reimann surfaces and all kinds of intresting things

Oh, I know. I loved teaching that course.

Riemann is the name, though Reimann is also a name, LOL.

not fair and i dont think my uni will offer it next year so i wont get o take it

@Jasper: Faust doesn't spell too well. Don't pick on him.

i didnt even realize that he spelt the two diffrently till you pointed it out lol.

@TedShifrin Ah yes, I know. I thought I would mention because both happen to be names.

@jasper which one do i want?

I'm going to watch tennis, but keep me posted, Faust. :)

ttyl

@Faust Well, it's a Riemann surface.

lol kk

so not how it sounds

Dietmar and Deitmar are also both names, LOL.

Hi @Ted

Guys help? I don't remember how to factor... And now I need it for my SATs and Calc II class...

I don't ever remember being very good at it, it wasn't very intuitive

Is there a more mathematical way of doing it rather than trying to come up with factors that multiply to things?

(i) ac-method if a quadratic can be factored with integers, (ii) quadratic formula if it can't, (iii) factoring by grouping if there's four terms and it's nice, otherwise for less nice but still elementary higher degree polynomials it's (iv) rational roots theorem + synthetic division. (v) plus there lots of special case formulas (which are just shortcuts), like difference of squares, difference of cubes, sum of cubes, perfect square trinomials

I was slightly scared, now I'm slightly more scared....

Currently googling ac-method

are you only trying to factor simple quadratics? if so, then only (i) and (ii) applies to you.

I remember doing the quadratic formula to avoid factoring in the past

Yeah, simple quadratics. I'm hoping it stays that way.

I teach (i), (ii), (iii) and (v) in intermediate algebra, and (iv) in college algebra

I took linear algebra last year, thought I was done with that

I guess algebra has no end :/

now find the pair that adds to the middle coefficient, -1. that pair is 3,-4. therefore we can split the middle term -1x into 3x and -4x, after which we factor by grouping

6x^2+3x -4x-2

2x^2+3x−27

So I list all the pairs that multiply to -54?

yep, and find the pair that adds to +3.

9, -6

(x-9)(x+6)?

no

do what I said

split the middle term 3x into 9x and -6x, then factor by grouping

2x^2+9x -6x-27

= ...

do you know what I mean?

Factor... by grouping? I don't understand this step

group the first two terms together, 2x^2+9x, and the second two terms together, -6x-27

what can you factor out of 2x^2+9x?

x?

right, 2x^2+9x=x(2x+9). now what can you factor out of the second group, -6x-27?

3

-3

so -6x-27 = -3(2x+9)

factor out (2x+9) to (2x+9)(x-3)

yes

What happens if I have no C term?

then C=0

in any case, ax^2+bx = x(ax+b), clearly

oh yeah, your right.

if you want to use the ac-method it would look like this

the pair that multiplies to ac=0 and adds to b is the pair b,0

so we have ax^2+bx+0x+0

that would be unnecessary

Wish I was taught this method back in elementary school so I didn't just ignore and forget factoring.

Thank you.

mmhmm

If we plug in x=10000, then 2x^2+3x−27 becomes 200029973 and (2x+9)(x-3) becomes (20009)*(9997)

Sure enough, (20009)*(9997)= 200029973

Or, alternatively, desmos.com/calculator

@danu I just bought a bottle of this thewhiskeywash.com/reviews/… bottled at 61% abv. Did I do good?

Why did the chicken cross the Möbius strip?

NFI

To form a trivial chicken bundle on the Mobius strip

please define this bundle of chickens

can everyone see the "OBJ" in boxes?

I'm sorry, the number you have dialed is imaginary. Please multiply by i and try again.

i think there's a mathjax bug

@JennaSloan to get to the same side

any1 there??

"The common chord of the circle x^2 + y^2 = 5 and the parabola 6y = 5x^2 + 7x will pass through the point(s) ; "

i) (1,2) (ii) (4,4) (iii)(-2,1) (iv) (9,-6)

one straightforward way to do this is to solve the equations with one another

That however gives us a biquadratic equation

with two imaginary and two real roots

its a bit time taking

i am looking for a shorter way, if there is any

Thanks!!!

by inspection (ii) and (iv) are too big to be solutions to x^2+y^2=5. plugging in the other two are easy.

too big??

4^2+4^2=5

you can tell the left side is too big

doesnt this mean that we have to find the equation of the common chord

and then check if the points lie anywhere on it

not necessarily on the circle/inside it

ah

I was not familiar with the term common chord

coincidentally though, the answer is 1 and 3

but i think we first have to find the equation of the common chord

you can verify easily that (i) and (iii) are solutions, then use them to write down the equation of the secant line thru them, then verify if (ii) and (iv) are not on that line

the vertex of the parabola is inside the circle (something you can verify by hand) which guarantees there should be only two points of intersection

hmm yea thats a way. plug in 1 and 3 and check that they are lying in both the parabola and circle

@Typhon That's the "Object Replacement Character"; it was probably copy-pasted from somewhere.

and then write the equation and check for other points

but lets suppose we dont have options

what then??

then it looks like you solve the degree 4.

you can use rational roots theorem to get two solutions that way. (and I mentioned above we can verify there are only two solutions by hand.)

@Typhon as tiresome as I find questions like the one you linked

"Forget it, Jake. It's Chinatown."

(which is to say, people use this site to do homework and there's no easy/simple way to change that)

not sure why exactly it's there

@Semiclassical I sometimes use this chat to "guide" me with certain HW problems, and oc some hints are overly helpful

sure.

I more refer to cases on the main site where people seemingly don't put any effort in

("I'm totally lost on this problem" doesn't count in my book)

oh right. but at least they get called out most of the time (and still too often complacently answered in full)

even the calling out feels futile after a while.

I just noticed Typhon's rather blunt comment on that Q, lol

Wanted to share this: rlv.zcache.com/svc/…

Great tee design for a wonderful organization

Hi @Brody

Heya @Balarka

How are you?

Quite well. How's life on your end?

Glad to hear! It's been a while

Feels just ok, but no real complaints

Indeed

Doing any math recently?

Doing my uni's undergrad algebra course, but nothing too remarkable

I'm curious what your latest topic of interest is

Undergrad algebra can be interesting

My latest topic of interest are dank memes

Last I heard it was ultrafilters

Until you clarified 'dank' I was second guessing if there's actually a math term called memes

Nah i don't think i was ever interested in ultrafilters

lol

There was an actual 10% doubt haha

Yo

Don't get too caught up in (dank) memes

I was interested in ultrafilters

You'll never come out

for the 10 minutes i tried to read Terry Tao's post on them :)

@Brody I'm already too far gone

specifically, this post: terrytao.wordpress.com/2007/06/25/…

and he has a paragraph where he describes ultrafilters in voting theory language, and as a fan of that I found it appealing

I see

Oh no MC Mental Sandstorm?

@Balarka is that my fault?

@BalarkaSen that's honestly great

"We have now thoroughly discussed non-principal ultrafilters, interpreting them as voting systems which can extract a consistent series of decisions out of a countable number of independent voters."

@Daminark WIKID

that's a fun sentence.

I'm tempted to make that my bio on something

Hi chat

Hi @Eric

Sup

Glb @EricSilva

Same @Daminark

How's the Ricci stuff going?

hi peeps :-)

P good @Daminark lots of fun calculations

Is there a term/phrasing to describe a magma that satisfies a given set of axioms but does not require additional structure? Something like "at most" or "maximally" ___

I know it's imprecise but wondering if there's some informal jargon for this

@EricSilva So suppose $M$ is a complete Riemannian manifold and $p \in M$ is a point. The critical locus of $\exp_p$ is exactly the conjugate locus $C(p)$ of the exponential map - the intuition I have was this:

Take a "polar chart" on $T_p M$ with lines $t \cdot v$ coming out of the origin for $t \in [0, 1]$ where $v := v(s)$ is a curve in $T_pM$ with $v(0) = v$ and $v'(0) = w$ for $s \in [-\epsilon, \epsilon]$ and we exponentiate that in $M$ below to get a parametrized surface $f(t, s) = (d\exp_p) tv(s)$ below. $q = \exp_p v$ is conjugate to $p$ if $d/ds f(1, 0) = 0$

or rather, fails additional axioms, so it is appropriately "at most" in some context

So usually one would expect that this literally means the geodesic $\gamma(t) = \exp_p tv(0)$ merges with a bunch of nearby geodesics at $q = \gamma(1)$, so that the "spread" of the geodesics near $q$ is 0.

I was looking for a scenario where that doesn't happen. I guess it suffices to look at examples where $f(1, s) = \exp_p v(s)$ looks like a $y = x^3$ curve in the base $M$?

So that we end up having a critical point at $s = 0$ anyway, but it doesn't have the "merging" behavior

@EricSilva Noice

@Brody "at most" sounds good.

Provided you give plenty of context.

@Balarka I had a problem about this at some point and I recall that my solution was a deformed 2 sphere that I got by multiplying the standard metric by some ad hoc function

Hm, that probably makes sense.

Right. Thanks @skullpatrol

The idea was to force it so that it was only conjugate along non minimal geodesics or something

Idr it very well though

What I said is probably nonsense, my brain is fried from calculating a bunch of derivatives earlier

hehe, no worries

np @Brody :-)

I think it's doable by hand; I just wondered if there's a natural example of this situation

@Eric Were they that bad?

$$\sum_{n=1}^{k}\frac{H_n}{n} = H_k+ \sum_{n=1}^k\sum_{m=n-2}^1\frac{1}{n(n-m)}+H_{k,2}$$

@Daminark no, not particularly, but I wanted to work out by hand what happens to all the usual geometric quantities under Ricci flow and it took a while to bash out all the formulae

Ah

@Balarka yeah idk abt natural but I bashed it out on a homework set with something stupid so it's definitely not hard, if you have a natural one hmu

I picked up Fulton and Harris in advance for fall @Daminark, p excited to learn something I know 0 about

@EricSilva I think I see what the picture of the exponential map should be like in that case; I'm more or less content with that unless I happen to find an example in a manifold I know and adore

Yeah I've got FH and Serre

what's FH on again?

Rep theory

Intersection theory?

I've heard a lot of good things recently abt when Benson teaches that class

ah ok

wrong book

Why did that autocorrect happen??

So I'm p excited

lol, autorekt

Yeah he hasn't taught that in a while, it was all AG, so yeah this'll be a fun time

P sad I can't take ag

At least you'll do it 4th year!

Are they doing algebraic curves this spring?

Idk but in the spring I'm probably taking qm, harmonic analysis w schlag, complex with Marianna and sosc

Second quarter grad algebra is commutative + algebraic geometry

So I have no space

Ah cool, I'll probably take Benson's geometric literacy course fourth year, where he teaches "algebraic geometry for non-algebraic geometers"

Quote from him?

Yes

Lel

André Sat in on It and told me it was good and that I should take it lol

4th year I haven't thought about much. Algebra for sure, logic if I don't do it this year, and some kind of algebraic topology. Plus that computational/metric geometry class at TTIC

I just know I'll be tying up loose ends/doing random weird shit

I'm toying with linguistics too

zazzle.com/pythagoras_theory_math_t_shirt-235251323966018309 ugh

I know there's a scholar of old Norse coming to uchicago next year so I'll probably take a bunch of classes from that dude

Sick

I heard rumors from medievalists that there would be a few courses on the Icelandic sagas which I would definitely take

Unless they conflicted with math ofc

Lmao

What are you taking in fall @Damunark?

Definitely algebra and bio

Core bio or topic?

Core bio, though I've done a quarter of multiscale and they're letting me retroactively call it a topic

Thats nice

Yeah for sure. Beyond that I'm thinking possibly logic, possibly analysis

what is bio, is it a maths discipline?

Might try for AT (with Danny) or taking rep outright

Biology @Secret

Ah I see

Check out biostatistics

Ah, now I see why multiscale had something to do with biology

Yeah it's called multiscale modeling of biological systems

Wasn't fond tbh

biostats always caught my eye in the course catalogs

of course UGA has an entire prefix devoted to biostatistics

MS and PhD programs included

Oh wow

being in a Georgia college that's not very comprehensive, it's always amusing reading UGA's various programs

it's not that funny but I still laugh at their Poultry Science Ph.D.

Wait that's a thing? Lol

Hi

Any quick eyeballing $x^{i}(1 - x^{i})$ for $ i =1,2,3$ are orthogonal to each other over $[0,1]$

@Daminark apparently. looks like it's motivated by local economics

bulletin.uga.edu/MajorsGeneral.aspx?MajorId=158

Wow

$$\int_0^1 x^i(1-x^i)dx= \left[\frac{x^{i+1}}{i+1}-\frac{x^{2i+1}}{2i+1}\right]_{0}^1 = \frac{i}{(i+1)(2i+1)}\neq 0$$

thus not orthogonal

$\int_{0}^{1} \phi_{i} \phi_{j} dx = \delta_{ij}$

for orthogonality

This is going on in two rooms at the same time :-)

@MartinSleziak do you know what happened to the math mods' office?

I think it was deleted.

Oh well, one less place to lurk :P

@skullpatrol It seems to be private rather than deleted

When you try chat.stackexchange.com/rooms/20352/math-mods-office you get: "The room you are attempting to access is private. If you believe that you should have access to this room, you can request access to it."

Thanks @MartinSleziak

Still -1 lurking location

doing basic math at 2 a.m. and I can't be confident bcuz of sleepy brain

$$\int_0^1 x^i(1-x^i)x^j(1-x^j)dx= \int_0^1 x^{i+j}-x^{2i+j}-x^{i+2j}+x^{2i+2j}dx$$

$$=\left[\frac{x^{i+j+1}}{i+j+1}-\frac{x^{2i+j+1}}{2i+j+1}-\frac{x^{i+2j+1}}{i+2‌​j+1}+\frac{x^{2i+2j+1}}{2i+2j+1}\right]_0^1 = \frac{1}{i+j+1}-\frac{1}{2i+j+1}-\frac{1}{i+2j+1}+\frac{1}{2i+2j+1}$$

$$= \frac{i}{(i+j+1)(2i+j+1)}-\frac{i}{(i+2j+1)(2i+2j+1)}=\frac{3ij+3j^2+2j}{(i+j+1)‌​(2i+j+1)(i+2j+1)(2i+2j+1)}\neq \delta_{ij}$$

[Chemistry] I need to redo all the calculations again because the error blow up

so far, I have both absolute sum $|x+y|$ and absolute difference $|x-y|$ commutative over $\Bbb R$, but not associative or admitting an identity element or invertibility

grr, this is like the 4th time I need to redo this batch of calculations!

@Daminark I only used to lurk there because one of my favourite mods was there :-)

(and he stopped coming here :(

there exists an identity element under $\max$

in $\Bbb R\cup\{-\infty,+\infty\}$ ;)

hi chat

hello

@Sharcoux Sure. So have you found the common combined distance from those points to the edge?

Ok, just to illustrate my issue, I made this : docs.google.com/drawings/d/…

@Sharcoux Ahh, great, so I misunderstood what you meant by left and right points.

ah ok, yes, I don't know the focal

Well, you should be able to find them from this. They are the points with the same combined distance to each of the three points you have. They are also on the line between the two end points

Hum... ok, so I guess I have to solve the equation, knowing that F1A+F2A = F1B+F2B = F1C+F2C ? Then, I can guess F1 and F2 that's what you mean ?

How do I deduce the small axis from F1 and F2 (F1 and F2 being the focal of course)

Oh, wait, looking at this, I think I have better : fr.wikipedia.org/wiki/Ellipse_(math%C3%A9matiques)#/media/…

Let's see. I can easily have the circle C2. I have M. I can have M2. Then, I can draw OM2 and then I can easily get M1

OM1 will be my small axis

it is always true that for an olomorphic function $f$ there exists another olomorphic function $F$ such that $F'=f$?

Holomorphic where?

You need a simply connected $\Omega$

@PVAL-inactive That sounds really great! I had a small taster of a Ledaig once, it was a nice smokey whisky that also had some nice flowery notes, sorta reminiscent of Highland Park. A lot better though :) Also it wasn't cask strength so I'll assume that's going to be even better!

Do you guys regularly go to Octoberfest?

Is $SL_{n-1}\mathbb{R} \times \mathbb{R}$ a reductive subgroup of $SL_{n}\mathbb{R}$?

via $(A,\lambda) \mapsto \begin{pmatrix}e^\lambda A & 0 \\ 0 & e^{-(n-1)\lambda}\end{pmatrix}$

hi echoes