I mean $m^2 + m + n$

Also, $n$ is smaller than (...)

Hullo, @anon

@BalarkaSen May I disturb you for some time?

@Sawarnik Sure. Not for long though, I need to go soon.

What is it?

@anon Find a $\Bbb C$-rational transformation from $E : y^2 = ax^4 + bx^3 + cx^2 + d$ to $E' : y^2 = ax^3 + bx^2 + cx + d$.

Oh sorry, I as lost a bit.

Internet connection?

No, phone call. I wanted to ask that which book studying introductory calculus would be better?

Stewart, I have.

I have no idea. I studies from an infamous vague book.

Have Spivak if you really want any.

Does it make any real difference?

Well, Spivak is not available here, I saw that on Flipkart.

I have a nudge-nudge-wink-wink infamous illegal site to offer you then ;D

There are lots of books there.

For free.

which one?

i mean ehich site

Concentrate here. A quick look, and I'll delete it.

Okay?

Ya done

Good.

And by the way, I am concerned that I am racing a bit too fast on calculus.

For example, I am not going on a proof based approach. That really concerns me

@BalarkaSen I hope you have altered that opinion.

@Sawarnik Nothing new. I did that too. Just concentrate yourself in analysis if you ever jump in there.

@robjohn I have already.

@Sawarnik Just skim through the proofs and do exercises. That's all.

Yes, I am doing that only. But when do I know a rigorous approach would be better.

When you come to the part in analysis when you don't understand things at all. Go back and search through the book. That's what happened to me.

Thanks.

No problem. Have you found Spivak there?

Btw you are studying some undergraduate courses?

@Sawarnik I am not sure what you mean by 'studying'. I will take one, if you mean that, by the way, in IISc

I mean that you must be on something, some things that you self study.

ya i found it :)

Yes.

@Sawarnik Good luck!

You'll find much more there than you can think of.

I barely give this little secret to anyone.

=D

Thanks:)

I am following you, wait.

@BalarkaSen This is not secret but it's illegal:-)

Some day, I hope I could be a prodigy like you.

@SamiBenRomdhane Indeed.

@Sawarnik I am surely not a prodigy.

@BalarkaSen Then what do we mean by a prodigy?

Anyways, you are as I know, and bye :)

@Pedro How so?

Nice timing

Yeah.

Well, I am just saying Hungeford has a very clear exposition.

Plus it has some category theory.

Which I should get to know a bit.

:D

@Mike

Now I know what a product and a coproduct means.

@Mike you can let the quartic be a general one, if you wish.

There still exists one.

Find it!

What does it mean @pedro?

@Pedro I liked that, but even moreso I liked knowing why we call about twenty different objects 'free'

Okay, I meant $\Bbb C$-algebraic than $\Bbb C$-rational.

Nevermind.

Restated version : Find a $\Bbb C$-algebraic transformation between $E : y^2 = ax^4 + bx^3 + cx^2 + dx + e$ and $E' : y^2 = ax^3 + bx^2 + cx + d$

@Mike

@Charlie Well, you have a bunch of objects $A_i$, we call $P$ another object together with maps $\pi_i:P\to A_i$ a product of the $A_i$s if for any object $B$ and any set of maps $\psi_i:B\to A_i$ there is a unique map $\varphi:B\to P$ for which $\pi_i\circ \varphi=\psi_i$.

The coproduct is the same but with the things to the other side.

@Mike isn't it a interesting problem?

@BalarkaSen Do you have a solution?

Yes.

@BalarkaSen I haven't solved the problem, but discovered a few things. If $n$ is not prime, then $n=p^2$ or $n=pq$, being $p$ and $q$ primes.

If $n\gt 3$

@IanMateus What problem?

And I used it to relate $y^2 = x^4 + 1$ to Legendre normal form

@IanMateus Have you tried my hint?

@BalarkaSen yes, I'm trying it

Do you want another hint?

@PedroTamaroff here

@pedro Hmmmm mm thanks

No, let me struggle a bit

Stumped, @Mike? =D

@BalarkaSen the least prime dividing $m^2+m+n$ is $\lt \sqrt{2n}+\frac 18$. Not sure what to do it with it, I'll think about it

$$\int_{-1}^1\int_{-\sqrt{1-y^2}}^{\sqrt{1-y^2}}c \,dx\,dy=1,$$ where $c$ is a constant. Then, $c=\dfrac1\pi.$ Am I right? @robjohn?

@Sush it seems so

$\displaystyle\int_{-\sqrt{1-y^2}}^{\sqrt{1-y^2}}\,dx=2\sqrt{1-y^2}$, so you just have to find $2\int_{-1}^1\sqrt{1-y^2}\,dy$

which is $\pi$

So how is $$2\int_{-1}^1\sqrt{1-y^2}\,dy=\pi?$$

Think of a half circle

@Sush Substitute $y=\sin(\theta)$

@IanMateus, @robjohn, thank you very much. I will have to think about this more, and have to take time.

I want to learn how to do 3d drawing using 2d lines to construct cubes in a 3d space...suppose i draw a line in 3d space, suppose from (5,5,5) to (10,10,10). Suppose my eye is at position (x,y,z) and i can move it and rotate it. How would i map this line to a flat surface? I don't know if this question makes sense, but i am hoping someone knows a resource for me to start learning about basic math concepts involving this

Do you mean using computer programs or drawing by hand @mathguy ?

Also, a question. I need to show that given the ordinals a, b, and a<b. I have to show that Ua=Ub iff b=a+1, and a is not a following number (i.e. it doesn't have a direct predecessor). I am completely lost.

Any ideas, anyone?

@Studentmath I want to use it for computers, but either way, the same math concepts should apply. I know that we have to define the degrees of vision that the eye has. The equation for the 3d line will always be the same no matter what, but the equation for the 2d representation for htat line will keep changing if the camera keeps changing

@IanMateus You'd want to express the bound of the prime factor in terms of $m$.

That'd be easier.

@Sush Arcsin.

@BalarkaSen interesting, I didn't know a y^2=quartic could be transformed into a y^2=cubic

@anon Indeed it is.

It's hard, but it can be done.

And in some cases the transformation would be also $\Bbb C$-rational.

what do C-rational and C-algebraic mean btw?

I'm assuming given by rational or algebraic expressions

algebraic over C, rational over C.

$\sqrt{2} + i\sqrt{3}$ is C-algebraic and $2/3i$ is C-rational.

we're talking about functions right?

Yes. Just adjoin $x$.

I am very confused. with regards to your examples, what's the difference between C-algebraic and C-rational? second, your examples are numbers, but you are using algebraic and rational to describe functions.

@anon This is why one does not consider $$\int_{\text{blah}}^{\text{blah'}} \frac1{\sqrt{P(x)}}$$ is not considered to be hyperelliptic if $\text{deg}P = 4$

@anon C-rational implies both imaginary part and real part are Q-rational. Similar for the algebraic. About functions, I mean en.wikipedia.org/wiki/Rational_function and en.wikipedia.org/wiki/Algebraic_function

that sounds like lame terminology, haven't heard of it before.

it is used frequently

I have always heard "K-algebraic" to mean "algebraic over K." you are using C-algebraic to mean what I would call Q-algebraic.

C-algebraic mean algebraic over C.

what you would call Q[i]-algebraic, i think

@Mike i is algebraic over Q...

err

oh. i don't define things that way. i see.

@BalarkaSen every complex number is algebraic over C

$e + i\pi$

that is algebraic over C

oh, okay, i see the point.

$e+i\pi$ is the root of the polynomial $x-(e+i\pi)\in\Bbb C[x]$

I mean $\bar {\Bbb Q}$-algebraic then

Sorry.

Being hasty

\bar{Q}-algebraic and Q-algebraic are the same; we always say the latter

But $\bar{\Bbb Q}$-rational is not the same as $\Bbb Q$-rational

true. note $\overline{\Bbb Q}$-rational is the same as $\overline{\Bbb Q}$-algebraic! (which is in turn the same as $\Bbb Q$-algebraic)

That wasn't my point

didn't seem like you were making one

@anon what do you think of this?

pings can be hyperlinks? O_O

i plan to use and abuse this

me too

me also.

@AlexanderGruber as user says, you mean rational functions not polynomials. I think the natural setting to prove it in is algebra but I will think combinatorially about it.

@AlexanderGruber I'd think there exists one.

@AlexanderGruber Can you give me a class of polynomials with $L(3, 2)$ groups?

In particular whats the smallest coefficient (is abs values) polynomial that has $L(3, 2)$ galois?

@BalarkaSen what does that mean?

What does what mean?

the group of a polynomial means the galois group of its splitting field

@anon right.

or not

i mean what is an L(3,2) group

@AlexanderGruber the fact the expression has a lot of minus signs means there's probably not a proof involving just pure counting. generating functions with positive coefficients can be used to mimic combinatorial proofs algebraically, but gfs generalize to allow more exotic interactions. I suppose things with - signs might track something like "uncounting," akin to negative numbers generalizing positive.

@AlexanderGruber see Fano plane

And when I mean polynomial, I mean septics.

@AlexanderGruber GL(2, 3), if this notation helps you

Or PSL(2, 7), by exceptional iso

Are anyone here familiar with any recent development of the constant problem?

Hey @Charlie

Idiot wind blowin' everytime ya move your mouth

@PedroTamaroff You're an analysis guy, you might enjoy David's answer here.

Hi. I've got a stupid latex question. How do I make a theorem environment inside a question?

@anon @AlexanderGruber Here's an interesting question : Say we have a function $f$ holomorphic such that $f(X)^5 + f(X) + X = 0$. Is it possible to prove $f$ must be modular?

@Henrique you mean a question on MSE? just use the dollar signs for inline math and double dollar signs for centered display equations. else you can use \begin{array}s and whatnot

I tried that, but it didn't work.

@Henrique just use a mix of text and math as necessary

Oh, so that's not a feature?

no, most of the document formatting things are not present, because online posts are not documents

ok

it's just mathjax

all right, thanks

@ccorn

Huh! success!

Is this lema correct: an ordinal does not have a direct predecessor iff it is of the form w*n?

@anon do you think a generating function proof would be possible here?

@BalarkaSen i see. i might be able to but i'd have to think about it. i don't do much inverse galois theory.

@AlexanderGruber Where?

I like EXGFs.

GFs, too.

In fact a little more than EXs.

What does 'applied' mean in sciences?

It means "has a practical use"

An adjective to get money.

@skullpatrol Practical sounds more appropriate

Attracts investors, like flies.

@PedroTamaroff Are you talking to me :D

?

Yes.

Engineering is also called applied science.

@saadtaame Everything can be potentially applied.

@PedroTamaroff But somethings more than others. I think that everything becomes applied eventually.

The parts of math that can be used in physics and engineering are applied math.

So differential equations, Fourier stuff

sure

Guys, what's that inequality for approximating a summation by integrals?

@skullpatrol @PedroTamaroff

@PedroTamaroff on my lagrange polynomial question

@PedroTamaroff @skullpatrol

@anon It should be with regard to and not with regards to.

Of course

@Charlie The cat looks very grumpy, lol.

Meow

I like scratching too

Hi @charlie

Hi @skull

How are you? @charlie

hi @Charlie, @skull, @Jasper ... and all other grumps.

I'm fine @skull

Hi Professor @tedshifrin

Hi @ted can I ask you something?

you just did, @Charlie :D

Howdy @Alex and dormant @Pedro (no, I haven't had time to read your email yet)

Has anyone ever called you Theo? @ted

a few, also a few call me Théo (French)

Good.

May I apologize for my remark last time we talked @tedshifrin?

I've forgotten, @skull.

:-)

Eu já falei pra vocês deixarem de maresia, eu quero ver o Pedro na coreografia

stop that, @Charlie

Stop what @ted

ranting in Portuguese

But it's so nice

It is rude.

No one likes Portuguese here!

Only Leo

Do you like my new avatar @charlie?

I don't understand it ... it's not that I dislike it.

@skull it's funny

:D

@charlie plug.dj is a neat site to listen to music

@Charlie It saves time searching through youtube.

I have lot of time

Kidding

:-)

It looks like the youtube generation is setting up new communities.

Yeah

@charlie did you watch any of the game?

Nope

why would anyone outside the US (not to mention plenty of us in the US) care one whit about the game?

Hi, quick question. (somewhat physics) I have a mass m1 = 12kg, and a mass m2. They are balanced on a bar with L = 4m, and mass 6KG.

twiddla.com/1486454

@skullpatrol What is the game?

Drawn like that, essentially I take the (mass of the left side of the bar * distance to fulcrum) + m1 = (mass of right side of bar * distance to fulcrum)+m2

Is that right?

I don't know any physics, lol.

@JasperLoy, it's fine, I was just asking here because it's very simple, perhaps someone else can comment?

torques (moments) must cancel, @Link: so you must have $m_1d_1 = m_2d_2$.

@TedShifrin, does it have to be torque, or can I just use the mass instead?

Since if torque's cancel, masses on both sides are equal right?

Well, the mass * distance is.

@Charlie What is the game?

oh, the bar has mass? You have to take it into account.

@jasper that football thing

hola

mr @Mike !

Hola @mike como estás?

Hi Ms Mike!

@TedShifrin I am. Basically I have the following. (m1+mrod*distance) = m2(mrod*distance)

Is that pretty much right?

@Charlie no hablo espanol, amigo

but I'm alright

I don't know what that means, @Link. You have parentheses wrong, and I don't know what mrod means.

ahem @Mike

@Ted I'm a slow typer!

But hello.

LOL, only when you're smoking are you slow :D

A low blow... now I have to get juice on you for comeback purposes.

Does mike smoke?

LOL ... you know I'm a low sort, @Mike.

Is Michael your actual name @mike?

@mike amiga*

Amigos Para Siempre

Sarah Brightman looked really sexy in that video, lol.

@Jasper Mostly.

@Ted I bet I can take that out of context somehow.

Of course you can, @Mike.

@Mike: Have you been in touch with Jacob?

:)

@TedShifrin mrod is mass of the rod, and I meant this: (m1+mrod*distance) = (m2+mrod*distance)

@Ted He talked to me about meeting up at the open house, I said sure.

Jacob Black is hot.

No, @Link. You have to take the portion of the rod to one side and compute its moment. And same for the other portion of the rod.

Ah, cool, @Mike. Then my job is done :P

Who? @Jasper

@Ted Ayup. Still no emails from anyone else. :P

@TedShifrin en.wikipedia.org/wiki/Jacob_Black here

Ah, @Jasper. Hot and sinister.

It's still early, @Mike.

I see you did not watch Twilight, lol.

Nope. I watch very little popular TV.

It's a movie.

Rather, 5 movies.

oh, well, see, I'm ill-informed.

@Ted And I'm impatient :)

Movies are a waste of time IMO

Well, duh @Mike.

@skullpatrol IMO=International Math Olympiad, lol.

Lol

also = in my opinion ...

ok, time for me to eat dinner and start grading 30+ diff geo papers

Later

Yes, no chatting for you, lol.

Have a good meal

see ya ... thanks, @Charlie.

Say hi to @Pedro if he ever shows up.

I say

HAI

@pedro Ted said hi

oh, there he is. HI/BYE.

Speak of the rabbit.

You're late!

Once you reach 30, you are an old man.

(removed)

Thanks a lot, @Jasper.

@TedShifrin I am an old man too, it's OK.

You're old. I'm dead.

OK, outta here.

Hibye @TedShifrin

Hi Krispy Karl

If only we stayed at 20 years old for all eternity.

Ah, perhaps such a world does exist.

Hello, @Charlie, how goes it?

It. Would be crowded

The user I hate also uses "how goes it", lol.

Pretty good @karl and you?

Good thanks

@JasperLoy Ah cool.

How goes it @jasper?

@jasper time will passes slower if you enjoy life more

I have become quite evil, I hate people for small things...

Just love

@Charlie or if you get into a really fast rocket ;)

Hey @TedShifrin @KarlKronenfeld

@ian like c...

hey yo

How's it going?

good good. you?

Fine, thanks

I stopped saying fine, thanks long ago, because that is too cliche.

@jasper start saying "I'm good"

It will not happen if you don't say it.

Maybe you don't want it to happen. @jasper

Is a closed line integral over an analytic function in R^2 always zero?

people don't seem to appreciate my definition of $\sqrt{z^2-1}$...

@N3buchadnezzar of, not over

@anon Yea ofcourse, just a tad tired.

Hello everyone

@N3buchadnezzar if it doesn't enclose any singularities.

I know it is true in C, but does it hold for R^2 ?

And in a packed forum tonight, we might talk about recursion theory, Lp-Lq duality, and arithmetic geometry.

What about the children?!

@N3buchadnezzar How are you defining your line integrals in $\mathbb{R}^2$. You need the complex structure to even define an analytic function

Is the recursion theorist here?

well $ f = u + i v$, where u and v are real harmonic functions so

$\oint f = \oint u + i \oint v $

@N3buchadnezzar If you duplicate your complex integrals the same as is done in $\mathbb{C}$, then it should work. But you need the complex structure and the contour integrals.

ok thanks

complex functions are $\Bbb C\to\Bbb C$; are you considering $\oint {\bf f}\cdot d{\bf x}$ for functions ${\bf f}:\Bbb R^2\to\Bbb R^2$?

@N3buchadnezzar You can't just use a path integral, you need to use $\mathrm{d}z=\mathrm{d}x+i\,\mathrm{d}y$

@N3buchadnezzar yeah, what anon is saying is different than contour integration

doing $\oint fds$ for $f:\Bbb R^2\to\Bbb R$ just doesn't seem as analogous

the domain/codomain have different dimensions and "$fds$" is real scalar whereas $f(z)dz$ is complexified

then again, the $\oint$ I described yields a scalar, hrm

it's probably more analogous to something with $\times$

multiplication in $\Bbb C$ is understood rotationally, after all

@anon What's the discussion about?

oh, things

things are bad

@PedroTamaroff.

Taking full advantage of my new knowledge

@KarlKronenfeld Oh, I see.

thing within a thing

Certainly not?

I am not totally convinced that it's not true, but my opinion holds no water in this area of mathematics.

Someone who edits Wikipedia on? Look at this mess

@IanMateus Oh sorry, I did that.

How could you @karl desiccate such a holy online resource such as Wikipedia?

@KarlKronenfeld then you better fix it! >8(

NOW!!!

@Karl It looks great, thanks for those improvements.

@Mike Thanks. I need that sometimes.

I thought it would be a matter of just removing the \begin{align} crap.

Seems to be more than that.

I don't see anything wrong with it

Maybe because I'm on my phone

Hm... some pages were offline some hours ago

@Mike wikipedia is having trouble parsing the section Other Heegler Numbers perhaps your phone collapsed that section.

Nope, fine for me. Bizarre.

wtf

anyway you can see my ip if you look at edit history. Sure you want that info.