Better. I got 123 points today. Roughly half yours (if I do some division mistakes), I see.

I honestly WTF'd at that last comment.

I don't really understand the notion of polynomial

Polynomials were first defined by Karp in order define what is NP-complete.

@Srivatsan Thank you, I suppose ;-)

@AsafKaragila I guess I should thank you for the answer. =)

You're welcome, of course.

:-)

I wonder, how did you get to that answer recently anyway?

For example 3|n^3 + 2n + 3

but if you consider "n^3 + 2n + 3" as a polynomial in Z/3Z it's NOT equal to 0

despite the fact that it equals 0 for every n

@AsafKaragila Mostly browsing around. That answer caught my attention because that funny comment had come up in our conversation before. (In fact, I might have first gone there for Dylan's answer: when I first reached there, I didn't know it was that question.)

Why are polynomials something different than a subset of functions?

that's what I don't get

Polynomials are the best. =) No clue why.

what's the benefit of this definition

Definition of polynomials?

@Srivatsan I see. Oh well. If you have any follow up questions do let me know.

- sure, thanks for the offer.

Mark as offensive, so it gets deleted.

why not just wait for the post to get deleted

Because it would get deleted faster if people are active about it.

If something has like 5 or so flags of spam/offensive then it gets deleted automatically.

you're talking about the comment only, though?

No, it was an offensive "answer".

but there's no way to mark an answer as offensive

oh do you mean "it is not welcome in our community"

Flag it under "it is not welcomed in our community".

yeah, I did that

I did that too.

I guessed that his friends were trolling through his user, as he said, but little do I care. If you cannot trust your friends you shouldn't leave your user open.

Why does JM keep writing answers in comments

shoudl post it as an answer so it can be accepted!

Either way, I am going to sleep. I hope the world ends and I don't have to wake up and go to that never ending algebra class! :-)

@AsafKaragila I hope you get uninterrupted dreams of being forced to sit through never-ending algebra classes.

Sleep tight.

What do we do now? math.stackexchange.com/q/98076/13425

nothing?

Well, either rollback or delete. Something has to be done, but I don't know what's the best route. So I will leave it at that.

let the moderators deal with it

it's been flagged

Yes, that's my stand as well.

Thanks for flagging - if you did.

don't know if you were interested in it particularly, but I asked the polynomial question and Qiaochu raised an interesting point

Yes, I did see it. I had a similar point in mind actually. It feels nicer if Qiaochu writes it though. =)

basically even if distinct polynomials coincide sometimes, they give give distinct functions over a big enough field extension

In general, it seems like a good idea to separate out the function view and formal view in abstract algebra -- here's a different case of that: math.stackexchange.com/questions/29432/….

That reminds me of the theorem about power series

they converge only in a circle

that was a bit surprise

@QED "In any case the basic ideas are quite simple if you merely take off your analyst hat and, instead, put on your algebraist or combinatorist hat." -- That sentence is nice. It still does not explain why we define formal polynomials that way though.

yeah

@Srivatsan: It looks like we're starting a conversation in the comments, so I thought I would move it to chat.

I was thinking we were more like ending our conversation. :)

@MikeSpivey Hi, nice to see you after a long time.

Maybe that's true. :) I suppose, since we've agreed that there won't be a bounty, that we probably should delete the comments under the answer.

So: I guess it's going to be a wee bit awkward for one for us. So I am ok with your not wanting my bounty. Are we agreed on that?

@MikeSpivey Right. We could do that now, just a second.

@Srivatsan Yes, I've been busier the past couple of months and haven't been on chat much.

I haven't been on the main site that much the past few weeks, either.

@MikeSpivey Ah, ok. Busy is also good, I suppose.

BTW, next step: Can you edit out the bounty announcement from the question? Or should I do it?

@Srivatsan Sure, I will edit it out now.

Thanks. It did feel a little strange for me -- silently pulling down an offer. :=)

@Srivatsan I hope I didn't run over you too much there. I just felt a little bad about you giving up 300 rep for me answering my own question. :)

No, not a problem. =)

Finally: I will leave a comment later saying there was a bounty here. I have to think up something for that.

Ok, I guess that does it then =).

Your solution looks nice, by the way. I have only skimmed through it. It will take me several passes for me to get through it.

@Srivatsan Thanks. I really did think about that problem for a long time, and it's satisfying to have found an answer I'm happy with.

By the way, thank you for all you did to promote my question. :)

@MikeSpivey You are welcome, of course.

Points are to give back, right?

That leaves me with some spare points then. I'll see what to spend it on. =)

So what has been keeping you busy these days?

@Srivatsan Various things... working on some of my own problems, family visiting, general holiday busy-ness.

Is this you in the picture?

@Srivatsan No, I was assigned that picture at random when I signed up for WordPress.

What? That's a random picture. Quite interesting.

Yes, I liked it so I haven't bothered to change it.

Oh, my wife and kids just arrived home so I have to go. Catch you later!

See you later. Bye!

It's funny that the linear recurrences can define polynomials and exponentials, but there's nothing in between

When you say it like that, it is indeed funny. "Between polynomials and exponentials" is a fascinating topic.

This is sort-of relevant: en.wikipedia.org/wiki/Grigorchuk_group, although I don't really understand it.

Now that's weird!

"The group G is finitely generated but not finitely presentable"

that makes a bit of sense

It seems I have missed Mike Spivey. It has been a while since I have seen him here.

@robjohn - After a week of meditation, i still can't find a reliable or clear intuitive reference for a proof of multivariable changing of variables formula. Please enlight a way for me to get it, at least give me the most important concept during the proof

@Victor I have tried to do that, but without a knowledge of matrices and determinants, which you said you did not have, I can't really see a way to do what you want.

@robjohn - Isn't that you could just expand it? Maybe for 4 variable case?

I don't understand what you mean by "expand it". Expand what?

@robjohn - The determinant...

especially the Jocobian

@Victor Sure, I could expand the determinant, but that wouldn't give you any more of an intuition as to what is going on. You really need to learn about linear algebra (especially matrices and determinants) before tackling multivariable calculus.

Of course the determinant of a $4\times4$ matrix is a sum of $24$ $4^{\rm{th}}$ degree terms. more confusing that enlightening, IMO.

@robjohn - I think i am able to enpand it, but why does some terms are negetive(the most improtant component)?

So, how does mathmatician knows that that formula is correct?

@Victor, Can you formally state the "change of variables" theorem you want to prove?

@Victor That has to do with the fact that when looked at as an n-linear form of the columns (or rows) of a matrix the determinant is an alternating form.

@robjohn - If i already knows the intuition for 2 and 3 variables case, does that help?

@Victor One proves the formula from the fact that the determinant is an alternating n-linear function of the rows (or columns) of a matrix and that the determinant of the identity matrix is 1.

@robjohn - How do you show it? i maybe able to read it

@Victor you get that intuition from visualizing the geometry of the situation. The geometry of 4 dimensions is harder to visualize.

@robjohn - Different dimentional case is very different?

Example: 4 and 5?

@Victor - do you understand my question?

@Victor do you remember the last time we spoke about this? You were having trouble because you were not able to generalize your intuition from 3 dimensions to 4.

@Victor The abstraction of matrices allows us to view all dimensions pretty much on even footing.

@Victor ?

@robjohn - i think if i could overcome the 4 dimension case, i may understand all cases, also, what is the prelimitaries on definition of 4 dimension other than linear algebra

@Victor do you understand that any vector in $\mathbb{R}^n$ can be written as a linear combination of the standard basis vectors?

@QED - I think i am not sophisticate enough to make a formal statement

@robjohn - I know 3 variables case have 3 vectors from a point and so on...

@Victor, You need to be able to do that if you want to understand and prove it

@QED - Can you write that for the 4 variable cases?

@Victor a standard basis vector looks like $\begin{bmatrix}0\\0\\\vdots\\0\\1\\0\\\vdots\\0\\0\end{bmatrix}$. all $0$s, except a single $1$.

I don't know what it is you want to prove

@Victor: I have to leave for a while. I will see later if you are around and we can perhaps do more to prove the formula for a determinant. Think on what I have said so far.

@robjohn - So how do you make a definition(geometrically) or lemma to prove the 4D case

@robjohn - I wonder why a math professor is that busy at night

@robjohn - Thanks

@robjohn - Are you there?

@Srivatsan you might like that, actually. Chapter VIII of de la Harpe's book is entirely devoted to it. The group has an easy enough description as a group acting on trees and also in terms of finite state automata. This handbook chapter is also quite nice.

@tb oh! Thanks.

If I ask you how you know this stuff, would it make sense? [I am asking this because many a time, I have been told that such a question doesn't make sense and is just funny. I think they are all wrong.]

Is this something all or almost all math people would have seen?

@Srivatsan Long story... I was very interested in geometric group theory when I was working on my diploma and at the beginning of my thesis. The Grigorchuk group is one of the major sources of counterexamples for many natural conjectures. The first conference I attended was in honor of Grigorchuk's 50th birthday.

@tb Oh ok.

@Srivatsan I guess not, but people interested in infinite group theory, probably yes.

Just clarifying.

The question on growth is a very nice one. I think. Given a finitely generated group, what is the asymptotic growth rate of the size of $n$-balls in the group.

Do you know [about] Gromov's theorem? I don't know (and not looking for an explanation now: that would be at best a bed-time story to me right now). I have seen Terry write about it a lot, hence I ask.

Sure. (March 31)

(that was before the proof via harmonic analysis by Kleiner)

Heh

(and before Tao's blog, too :))

@tb In which year was this conference(?)?

2005

It was more like a reading seminar.

I don't understand this comment math.stackexchange.com/questions/98345/…

"You can think of polynomials as functions" didn't I just give a counter-example?

@tb Seeing this comment made me realise I subconsciously revealed how I am secretly on a first-name basis with Terry Tao. =) In fact, it's so secret that he doesn't know this yet...

@Victor First of all, I am not currently a math professor. Second of all, why should a math professor not have a life at night?

@robjohn Not "life at night"; "busy at night". How can a math professor not read messages properly? =)

@Srivatsan :-p

Perhaps I was being too harsh. Perhaps not :-)

I felt like saying something as much. I decided that would be out of place for me.

I don't think it would have been out of place.

@QED I think that's the way people thought about polynomials before the algebraic revolution around the beginning of the 20th century. It looks like a valid point.

Clear answer, that one was.

@tb what is virtual nilpotency?

@robjohn a group is virtually (P) if it has a finite index subgroup having property (P)

okay, sounds reasonable.

that's why I'm not particularly interested in finite groups -- they're virtually trivial!

(standard joke, I admit)

So (to myself, mostly): it's a group satisfying (P) + a little bit extra.

haha

@Srivatsan well, $1/n^{\rm{th}}$ of a group has the property

@robjohn Are you saying I should've said a group satisfying (P) times a little bit extra? =)

so the little bit extra could be quite a bit more than the part satisfying (P)

@Srivatsan that would be better :-)

[I was going to give a lame example to show that I understand, but nevermind.]

What? four people upvoted this nonsense?

hey all.

Oh great wise ones on high, please link me to a simple explanation of where the exponential decay function comes from.

Wait, what!? The great wise ones are on high? =)

They're on something.

@KorganRivera Exponential decay function? What is that?

@Srivatsan, you know, along the lines of ce^-(ƛ t)

Oh, that's just the exponential function. I guess decay refers to the fact the function monotone decreases to zero as $t \to \infty$.

Do you know about radioactive decay?

That's a big question :)

But yes, I understand how the function can be used to measure the amount of a material with a given half-life after some amount of time.

I just want to know how this function is derived from scratch.

First: have you seen this Wikipedia article on radioactive decay?

I read the one on exponential decay, let me check through this one..

The explanation has a "let's rearrange the terms" step, and they use dy/dt notation, and their step is to 'multiply' by dt. I'm trying to avoid the whole infinitesimal thing.

So, -from a physical point of view- the defining characteristic of the exponential function is the "first order growth or decay". That is, if $X$ is the amount of some substance, and if the rate of change of $X$ is directly proportional to how much amount is currently present (i.e., the derivative of $X$ is proportional to $X$ itself), then you end up with such a differential equation; the solution happens to the exponential function.

I guess I can work it out from there, but if anyone has a link to something else I'd like to see it.

It's not really infinitesimal thing: it's a differential equation. You can solve it rigorously as well.

Here's how to do it:

$$ \frac{dX}{dt} = - \lambda t .$$

Rearrange it to: $$ \frac{dX}{dt} + \lambda t = 0. $$

Now the magic step: multiply by $e^{\lambda t}$. It is the "integrating factor" for the differential equation. Then an application of the product rule will simplify it to:

Is everything ok till now? Can you go from here?

I thought the first step would have been $\frac{dX}{dt}=-ƛX$

I don't understand; isn't that what I wrote?

no you said it's equal to -ƛt

btw, mathjax isn't rendering on this page, should it?

Ah, sorry. Both my first and second equations are wrong. They should be $\lambda X$, as you say.

ok.

See this page for MathJax support in chat.

Ok this is what I understand so far. So, the proportion of our substance reduces as a proportion of its initial 'weight' over time. So I can say dX/dt = -ƛX. After that, I'm stumped. I suppose I find a solution for X by integrating?

thx for the link.

@KorganRivera I wrote down the solution - except for the last step. Of course, correct the first two equations.

k thx.

Yeah that didn't help, but thanks :) 'Magic step' threw me off. I'm sure it's not that complicated, I'll find a solution somewhere.

Do you want me to try this once more - this time splitting the magic step into two?

If you have the patience, I would appreciate it. Can you start from where my understanding ends?

Your understanding ends within you; how would I know where it ends? =)

Because I told you.

4th post before this one.

..of mine.

OK. The third step is to multiply by the integrating factor of this differential equation, namely $e^{\lambda t}$.

Ok, but where does that come from?

OK: do you know about the exponential function? Or is all this an indirect way of asking us to explain that?

[I'm not scolding you or anything; I don't know if my tone suggests that. I'm not sure what you know and what you wish to know through chat.]

No no I know. You're trying to help. I'm not sure if I know enough intuitively about the exponential function to help me here. I thought I did.

Let's carry on with this path to see if it clicks for me.

so I have $\frac{dX}{dt}e^{\lambda t} = -\lambda X e^{\lambda t}$

@KorganRivera Most of the times, the discomfort is related to "how do I manipulate this expression or solve this function", not what that function "is" or what the expression "means".

If your trouble is the latter, I cannot help you that much; since to me also, some bits appear out of the blue.

I understand. :)

Let's carry on.

Ok. Let's agree that multiplying by $e^{\lambda t}$ is the right thing to do. I can give some more intuition - but later. Not now.

Ok, now what?

So: $$ e^{\lambda t} \frac{dX}{dt} + \lambda X e^{\lambda t} = 0 .$$

ok

Now it's time to recognise patterns: in fact, this is why we multiplied by $e^{\lambda t}$ in the first place.

Right now, I'm seeing a massive loss of spirits. What do you see? :)

Let's call $e^{\lambda t}= Y$ temporarily. Then what is $Y'$?

$\lambda e^{\lambda t}$

I take it that you mean $\lambda e^{\lambda t}$. So: in terms of this new function, we can rewrite the equation as: $Y \frac{dX}{dt} + X \frac{dY}{dt} = 0$.

yes

Does this look better?

oh yes. It does.

@KorganRivera Do you want to take over and complete the solution?

I want you to drag me there like a corpse.

@KorganRivera You don't even want to try?

uh.

Well, that is not quite what I meant. Because both the terms involve X.

Yeah I just saw that :(

Integrating factor

@KorganRivera HINT: The expression $Y \frac{dX}{dt} + X \frac{dY}{dt}$ should remind you of some specific thing.

oh it's the product rule.

@KorganRivera So, what is the next step - after applying the product rule (actually, you should apply the product rule in "reverse" =))?

Well at this point I have $\frac{d}{dt}(e^{\lambda t}\cdot x)=0$

Yes. So, the derivative of a particular function is zero. What could that function be?

I feel I ought to be very embarrassed at how slow my progress is right now. Just to assure you, I'm well aware of it :)

That function would be some constant.

The function $e^{\lambda t} X(t)$ is a constant; call this $C$. I made it $X(t)$ for emphasis.

Ok.

$e^{\lambda t}X(t) = C \implies X(t) = C e^{- \lambda t}$.

Ok that makes sense.

Thank you I appreciate you doing that. I'll look at this tomorrow morning and wonder what was so difficult about it :)

@Srivatsan Can I get a couple of minutes? :)

Where are you?

GHC - office

@Srivasan: that multiplying by $e^{\lambda t}$ is a very clever step. I wouldn't have figured that out in 50 years.

actually I would have, but I wouldn't have figured it out tonight.

chat.stackexchange.com aka the skeletal trebuchet.

..always throwing me bones, thanks folks please tip your waitresses, g'night!

@Korgan I don't know if that will ping @Srivatsan

-hi =)

@N3buchadnezzar Hi, it's so quiet in here ... that you can hear a pin drop before it hits the floor!!!

(3 hours later...)

I was just dropping by to ask a simple question, before going to my lectures ^^

@N3buchadnezzar How simple?

My question, is why are these points discarted? By my graph, it looks like the only point where the function is not smooth is when t=0. But my calculations says otherwise.

@N3buchadnezzar Maybe you can "ping" one of the professors ... if you set the denominator equal to 0 and solve for t ... what do you get?

Its unsolvable in terms of an explicit expression

Have you tried polar coordinates?

Hello @Matt

And *inside* research analysts it's too dark to see general convergence in topological spaces.

Hello Kannappan.

I'm a 4k user!!! How did that happen!?

You explained that $\omega + 1$ has a last element... :)

: )

@N3buchadnezzar The curve can fail to be smooth only if both dx/dt and dy/dt are zero.

Hi

@RahulNarain or if $\dfrac{\mathrm{d}x}{\mathrm{d}t}$ or $\dfrac{\mathrm{d}y}{\mathrm{d}t}$ are not smooth

heh, it appears I've missed Rahul.

Hm.

almost started with my coursework :D

@Daniil started? don't get ahead of yourself :-)

@robjohn I found the Algebra Bible

@Skullpatrol and what would that be?

@robjohn Algebra Bible

@Skullpatrol You mean Dummit and Foote

never heard of it :(

Is it a modern algebra text

@KannappanSampath The link says "Algebra II Bible"

@Skullpatrol My head is composed of mud! that's some classical algebra text.

But, good that I came to know of it!

@robjohn Today I came across a book in the library entitled "Is God a Mathematician" and now I find an Algebra Bible, do you think that this means that mathematics is a religion to some people Rob?

@Skullpatrol It has proofs of what it says and relies not on faith. I think that disqualifies it from being a religion.

@robjohn What did you think of the "Algebra II Bible?"

blackboardsinporn.blogspot.com/2010/09/5.html haha

(not very safe for work)

@Skullpatrol Are there only 6 chapters?

@robjohn I think that is just the study guide link.

@Skullpatrol yeah, but I only saw six chapters. How many chapters does "Algebra and Trigonometry Structure and Method" have, I wonder?

@robjohn 16

@robjohn Have you read "Is God a Mathematician?"

@Skullpatrol No

@Daniil This one at least has some non-trivial (and correct) math.

It also looks like something I would watch

@Daniil anything to learn, eh?

@Daniil Do you prefer "formulae" or"formulas?"

Latter.

@Daniil Latter.

@robjohn My 12 year-old nephew asked me "How can I be sure I'm right when I solve an equation?"

@Skullpatrol depends on the equation. Usually, you plug the values back into the equation and see if it works.

@robjohn That is what I told him, but he said "why is that true?"

That doesn't make sense. You are looking for values to make an equation true. That means if you plug those values into the equation, it is true. How more basically can you verify it?

Solve $x^2+y^2=25$ and $x+y=7$. $x=3$ and $y=4$ is a solution, and we check that by $3^2+4^2=25$ and $3+4=7$. There may be other solutions, but at least that is a solution.

@robjohn He wanted to know why it works or what makes the method work.

Why what works? plugging values in to check? I don't understand what is confusing about that.

Yes, plugging values into check.

you want to solve $x^2-3x+2=0$. Saying that $x=1$ is a solution, is simply saying that $1^2-3\times1+2=0$.

Find out what he thinks solving an equation means. If he means something else, I would like to know.

That confuses me. The equation is true for some value of $x$ when you substitute that value of $x$ into the equation and the equation is true. $x$ is simply a placeholder for some value.

where the final equation is a transformation of the original equation.

Perhaps he is confused about the role of a variable.

Perhaps.

Oh, well, I was assuming you were plugging the values back into the original equation.

I was.

Then where is the confusion?

but he wanted to know where I got the value to plug in?

what is being asked when I ask for a solution of $x^2-3x+2=0$?

@Skullpatrol Didn't you show him how you solved the equation? Plugging back in is just the check.

I said I solved the equation. We only worked on linear equations.

I said to him

The check should be the test of the correctness of an answer.

I agree.

But kids like to ask why, why, why,...

Is he confused about why if $3x+2=11$ then $3x=11-2$?

no

but when we say x = 3, and sub in 3

he asks why

Is he asking why we check?

no, how do we know if the 3 is right?

Because we checked it by plugging in

I said because it satisfies the original equation

The transformations work both ways

with the exception of multiplying by 0

It doesn't matter that the transformations work both ways. You have plugged the values in to the original equation, and that shows that they solve the equation.

True.

If he is asking why we solve linear equations as we do, then it is necessary that the operations are reversible.

Yes, I think that he is asking that.

But if I want to solve $3x+2=11$, subtracting $2$ from both sides says that $3x=9$ dividing both sides by $3$ says that $x=3$. So we have said that if $3x+2=11$ then $x=3$. Without reversible operations, we have not said that if $x=3$ then $3x+2=11$, but that is why we check (even if the operations are reversible).

I wonder what's going on with the moderators. I have two pending flags on two separate posts one of which is waiting since yesterday.

In the case of $x^2-3x+2=0$, we usually complete the square to get $x^2-3x+\tfrac94=\tfrac14$ so that $(x-\tfrac32)^2=\tfrac14$

This is different. There is no single reversible operation that will give us $x$.

We can factor $(x-\tfrac32)^2-(\tfrac12)^2=(x-2)(x-1)=0$

Then we have that either $x=2$ or $x=1$, but we can't know what $x$ was; $2$ or $3$. We just know it is one or the other.

But we still check the answer by plugging in.

I have to go for a while. BBL

@robjohn In your first example Rob, if x being 3 is the value that satisfies the original equation then plugging it back in will give a true statement (provided no errors have been made), then have we said if x = 3, then 3x + 2 = 11?

Plugging in shows that if $x=3$ then $3x+2=11$

The derivation of the solution shows that if $3x+2=11$, then $x=3$

@robjohn But you said we have not said that if x=3 then 3x+2=11, but that is why we check (even if the operations are reversible).

@robjohn So why we check is to show the converse is true. That is why we check.

@Matt It sometimes takes some time to think about the flags, if they are borderline cases. Also I recall Mariano saying that sometimes he's not sure whether or not to accept the flag and leaves it to the other moderators. I'm certain it is the same with the others as well.

@Dylan Hi.

Hey Asaf.

How are the sets?

What's up?

Oh very little. Teaching in an hour. Reading more Deligne.

What do you teach?

It's this linear algebra/calculus of several variables hybrid course.

Who's in the audience?

It's awkward for the students, because they don't really like being precise but any course involving the abstract definition of a vector space is going to require that somehow.

First-year students.

Undergraduates.

Math? Engineering?

"Science"

Broadly speaking.

I don't think they're math majors, by and large.

I'm going to grab some coffee. Back in a bit.