Mathematics - 2019-09-13

Subhasis Biswas

00:07

Won't it be $A_{ij}$, when $i \neq r$?

maths student

07:49

0

Q: Total function Problem

$$ \begin{array}{l}{\text { Consider the following function } f \text { defined over } \mathcal{R} \rightarrow \mathcal{R}} \\ {\qquad f(x)=\left\{\begin{array}{ll}{\log (x) \cdot \cos (x)} & {x>0} \\ {0} & {x \leq 0}\end{array}\right.}\end{array} $$ $$ \begin{array}{l}{\text { (a) Determine whe...

1010011010

10:14

I'm a bit tangled up as to whether the distributivity rule of limits takes precedent over apply the delta function

I have a product that can be factored out to have a delta function as one of the factors

The other factors permit their individual limits, and I'm confused as to whether I can split the products or not

So for example I have $\prod_i f_i g_i h_i$ with all smooth functions and variables, everything is real and analytic, yadda yadda. Normally I would allow them to be split as $\prod_i f_i \prod_i g_i \prod_i h_i$ but then one of the products is just the trivial product (?!)

How would I interpret that?

So to make it clear, if I take my limit, I can take the limit over each factor individually. One of the products resolves to be $\exp[A*\delta(x-\lambda_i)]$ where $\lambda_i$ and an indeterminate residuel parameter of the compound product above

Actually let's take $\lambda$ defined in the reals to make it "easy"

Mike Miller

Unfortunately I have no idea what you're trying to say. You want to plug $\prod_i f_i g_i h_i$ into a delta function?

Or one of the $h_i$ is a delta function?

And what do limits have to do with this? It would be nice if there was a very clear specific question.

1010011010

Sorry

Okay, so I have a product of factors over which I'm taking a limit

If I write $\prod_i f_i g_i h_i$ as $\prod_i f_i \prod_i g_i \prod_i h_i$, then take the limit over each of the individual products, one of the products becomes a Dirac delta

But I believe that the limit transition has been made at that point, and the delta function will not restrict the form of the other products

Luyw

10:55

Hello! On factoring polynomials, is it intuition that told us that if $a$ is a root then $P(x)=(x-a)Q(x)$? Or did we derive it algebraically?

Rithaniel

11:14

It is derived algebraically. I'd detail exactly how, but I don't know the full approach offhand.

I recall that the approach requires a definition of quotient/remainder division of polynomials, though.

ÍgjøgnumMeg

11:34

@1010011010 if $X - a \mid P(X)$ then it's obvious. Suppose $a$ is a root of $P$ but $X-a \nmid P(X)$. Then $P(X) = (X-a)Q(X) + r(X)$ with $\deg r < \deg(X-a) = 1$ so $r(X)$ is constant, say $k$. Then since $P(a) = 0$ you have $P(a) = (a - a)Q(a) + k$ so that $k = 0$, contradiction.

Secret

Integral f inverse

►

schn

15:32

How would one, using a delta-epsilon proof, prove that the limit of $\frac{2xy}{1+x^2+y^2}$ equals 0 as $(x,y)\to0$?

Mary Star

Hello!!

Does someone of you have an idea about my question:

0

Q: Laplace equation : Separation of variables

We have the laplac problem in $(0,1)^2$ with the boundary conditions $$u(x,0)=f(x) , \ u(x,1)=0, \ \ x\in [0,1] \\ u(0,y)=u_x(1,y)=0, \ \ y\in [0,1]$$ Using the separtaion of variables we get the following two problems $$X''(x)=\lambda X(x)=0, X(0)=X'(1)=0 \\ Y''(y)-\lambda Y(y)=0, \ \ Y(1)=0$$ F...

ordinary-differential-equations pde

The East Wind

16:05

can someone help me with 2 b)

Ted Shifrin

16:16

@schn Start by using $|x|\le \sqrt{x^2+y^2}$ and the same for $|y|$.

@TheEastWind This is implicit differentiation. You need to write down the chain rule for differentiating $F(x,y) = (f(x,y,u(x,y),v(x,y)),g(x,y,u(x,y),v(x,y))$.

yuvraj singh

16:38

Hi everyone can i ask a question

The East Wind

@TedShifrin I need help with that..

Ted Shifrin

@TheEastWind: Do you know how to use matrices to write the chain rule? That will make your life easier.

The East Wind

@TedShifrin suppose we have f(x,y,u,v) where u and v are functions of x and y. What is delf/delx in this case

yeah jacobians and all that

@TedShifrin This is the main doubt i have in chain rule of multiple variables. How to apply it when the variables themselves depend on one another

Ted Shifrin

The simplest case of implicit differentiation is to consider $f(x,y)=0$ defining $y=y(x)$. Then $F(x)=f(x,y(x))=0$ for all $x$. Take the derivative carefully. You get $F'(x)=0=\partial f/\partial x + \partial f/\partial y \cdot dy/dx$.

This tells you that $dy/dx = -\dfrac{\partial f/\partial x}{\partial f/\partial y}$.

The East Wind

@TedShifrin ok

Ted Shifrin

16:45

You want to do the same thing with more variables. Using matrices is easiest, as I said.

Basically, you do the same argument, treating $x$ and $y$ in that last example as vectors (and replacing the division with multiplication (on the left) by the inverse matrix).

The East Wind

But the problem arises when extra variables that we included are themselves functions of the already present variables, like in the above case. This confuses me.

Ted Shifrin

No, what I just did is a simple case of that same phenomenon.

Write $\vec x = (x,y)$, $\vec u = (u,v)$, and $\vec F(\vec x,\vec u) = (f(\vec x,\vec u),g(\vec x,\vec u))$.

You're starting with $\vec F(\vec x,\vec u) = \vec 0$ defines $\vec u = \vec u(\vec x)$.

Set $\Phi(\vec x) = \vec F(\vec x,\vec u(\vec x))$ and $\Phi(\vec x) = \vec 0$ for all $\vec x$.

Now differentiate, just as before, using the chain rule.

Mike Miller

The grand bargain we made with engineering departments --- get teaching in exchange for writing the curriculum in an order suited to them --- is a cancer

There is absolutely no pedagogical reason to teach lin alg after multivar differentiation. It's abominable

Ted Shifrin

I presume the rationale was that their students needed material X in a greater hurry than material Y?

What lin alg is in the course?

Mike Miller

yes, and the chain rule, as you just said...

Ted Shifrin

16:50

I don't know any "engineering/science-style" calculus course that does the math "right" in this stuff.

Ultradark

I'm confused about something. With an algebraic variety, how many polynomials are simultaneously being solved?

The East Wind

@TedShifrin can you write this down for me. I have an answer and to want to check whether its correct.

Ted Shifrin

Why don't you say what you have?

The East Wind

i have : delf/delx + (delf/delu)*(delu/delx) + (delf/delv)*(delv/delx) = 0

Mike Miller

damn, I do enough office hours, no need to watch more... see ya

Ted Shifrin

16:54

LOL, bye, @MikeM

The East Wind

Am i correct?

Alessandro Codenotti

Hi everyone

Ted Shifrin

Yes, that's right, but you have to do $f$ and $g$ both, as well as the $y$ partials and then work with determinants. This is why I suggested writing vectors and the chain rule using matrices.

Ultradark

HI

Ted Shifrin

hi, demonic @Alessandro

The point, @TheEastWind, is that you can think of the derivative matrix $D\vec F(\vec x,\vec u)$ as a block matrix $\begin{bmatrix} \frac{\partial\vec F}{\partial\vec x} & \frac{\partial\vec F}{\partial\vec u}\end{bmatrix}$.

This is a $2\times 4$ matrix that I'm visualizing in terms of two $2\times 2$ blocks. Play with that until you understand it.

Ultradark

16:59

what is $p_1(x,y)=p_2(x,y)$ I know that there are two polynomials being equated

Ted Shifrin

What do you mean "what is"?

Ultradark

I mean is it an algebraic curve?

I think it is because the polynomials are bivariate

Ted Shifrin

You're not setting them both $=0$? You're just setting them equal to one another? Then this is one polynomial equation set equal to $0$.

Ultradark

sure

but since it's bivariate it would be a curve right?

instead of a point?

Ted Shifrin

Yes.

Unless the polynomials are stupid.

Working over $\Bbb R$, the curve might degenerate to a bunch of points, though. Over $\Bbb C$ that won't happen.

Ultradark

17:05

and is there a name for a finite set of disjoint pairs of such polynomials?

Daminark

Henlo

Ted Shifrin

No.

Hi, Demonark.

Ultradark

$p_1=p_2$ and $p_3=p_4$ ...

Ted Shifrin

Are you imposing all those equations at once?

Ultradark

no, then it would be a variety i believe

Ted Shifrin

17:07

Any system of equations (or one equation) defines a variety.

Daminark

In contrast to what would happen if you just had one

Ugh sniped

Ted Shifrin

No.

Ultradark

no

I just felt like saying no

Daminark

And so you did

Ted Shifrin

One equation still defines (typically a codimension 1) variety.

Daminark

17:08

Oh I was being sarcastic

Ultradark

why didn't they tell me that any system of equations defines a variety when i learned about them

Ted Shifrin

Because that's the language of algebraic geometry, not high school algebra.

Ultradark

fair point

Ted Shifrin

Why didn't they tell you that you were doing commutative ring theory in elementary school when you learned arithmetic?

I mean, geez ...

Daminark

"Carrying is a 2-cocycle..."

Ted Shifrin

17:10

LOL

Daminark

That would be a sight to see

But yeah how's everything going with you guys? Any fun math?

Ultradark

so each equation

Ted Shifrin

I'm retired from fun math, Demonark.

Ultradark

is a 1 dim. variety

Alessandro Codenotti

At some point they'll start using integral separated scheme of finite type instead of variety, never trust mathematicians and their language

Ted Shifrin

17:13

An equation is not a variety. Learn to use the language correctly.

Ultradark

a polynomial set to 0 defines a 1 dim. variety?

Daminark

Lol in AG we now upgraded from algebraic subsets of k^n to a "makeshift definition" of a variety which isn't a priori embedded

Codimension 1

Ted Shifrin

Hmm, to me varieties come with an embedding. Otherwise they're complex manifolds or analytic spaces.

Alessandro Codenotti

"A thing which is isomorphic to an algebraic subset of $k^n$"?

Ultradark

Co-dimension okay

Alessandro Codenotti

17:15

This definition is nice because $\Bbb C\setminus\{0\}$ is a variety now @Dami

Ted Shifrin

Demonark: I'm doing fun cooking today and tomorrow. That will substitute.

Daminark

But yeah now we have a set X and an algebra R of functions X->k such that every maximal ideal in R is the kernel of "evaluate at x" for some x

Alessandro Codenotti

Oh, the expert is here, I'll stop talking about AG before I say something stupid :P Hi @loch

Ted Shifrin

Hi, @loch.

loch

Hi all

Ultradark

17:15

but what if you have a quadra-variate polynomial set equal to $0$

loch

i'm just lurking

Daminark

Hey loch what's up?

Ultradark

it's still of codimension 1?

Ted Shifrin

@Ultra: You keep using statistics language. Math people don't say "...variate."

Alessandro Codenotti

@Ultradark Its zero set is a codimension 1 variety embedded in $\Bbb C^4$

Ultradark

17:16

what dimensional varieties are usually studied?

Daminark

In principle any dimension

Dimensions 1 and 2 seem easiest to say things about

Ultradark

okay, I guess what I'm trying to say is, are codimension 1 varieties embedded in $C^n$ trivial or non-trivial?

Daminark

E.g. Bezout, 27 lines in a cubic surface, etc

Ultradark

I'm thinking they are pretty non-trivial

Daminark

So people I think more often work with projective varieties but modulo that yeah hypersurfaces aren't quite trivial to study I imagine

Ultradark

17:20

like the twisted cubic?

Daminark

I don't know what that is exactly

Ultradark

It is generally considered to be the simplest example of a projective variety that is not linear or a hypersurface,

Daminark

Oh it's degree 3 Veronese of P^1 in P^3

Ted Shifrin

Demonark: $(t,t^2,t^3)$ in $\Bbb A^3$.

More interestingly, if you look at it in $\Bbb P^3$ it's a basic example of something that is not a complete intersection. Try as hard as you want, you cannot give it as the intersection of just two surfaces.

You need $3$ equations to define it. $2$ will always give you something extra.

Daminark

Oh that's kinda sick

Ted Shifrin

17:25

Yes, complete intersections are very nice and hard to come by, in general. All sorts of nice things happen for them (if you think back to diff top, you get things like the normal bundle is the sum of the normal bundles of the hypersurfaces, etc.). Of course, I omitted to say that you have transverse intersections of the hypersurfaces.

Demonark: You should be able to write down equations in $\Bbb P^3$ and see you need 3 of them. As you learn AG you can mull over how you'd prove it's impossible to do it with just 2.

Ultradark

could you just take a codim. 1 variety and let another codim. 1 variety "intersect" it to create a codim. 2 variety?

wouldn't this basically amount to adding more polynomials to the system

ÍgjøgnumMeg

Yooo

Ted Shifrin

Happy ultimate day, @ÍgjøgnumMeg.

ÍgjøgnumMeg

Thank you @Ted !

Alessandro Codenotti

Hi @ÍgjøgnumMeg

ÍgjøgnumMeg

17:37

It's oooover

Hey @Alessandro

Daminark

Wait what's going on? Also yo igjo

ÍgjøgnumMeg

@Dami hey :) It was my final day at work today

hence ultimate

lol

Daminark

Ah congrats fam

ÍgjøgnumMeg

Cheers bruh

Ted Shifrin

No second round of goodbye presents? :P

ÍgjøgnumMeg

17:38

Alas, my team is but a small one

with its own office

Ted Shifrin

Demonark: You working on those equations for the twisted cubic? :)

ÍgjøgnumMeg

so we don't socialise with the rest of the staff much :P

Alessandro Codenotti

When will you move to Germany?

ÍgjøgnumMeg

17 days I thin

Ted Shifrin

I miss the days when we had 3 or 4 Parisians in here.

ÍgjøgnumMeg

17:39

k

Just reading about the j-invariant of an elliptic curve, super spooky

Daminark

Right now I'm walking outside but once I sit down and find some chalk

Oh speaking of elliptic curves I should prob try to bother someone about the reading course

ÍgjøgnumMeg

@Dami I picked up another number theory course this semester lol

Daminark

A couple of us independently asked for one and when we met up and decided on elliptic curves (assuming he thought it was a good idea) one of us emailed him

ÍgjøgnumMeg

nice :)

Daminark

But no response yet

Eyy, what on?

ÍgjøgnumMeg

17:43

construction of L-functions

Ted Shifrin

ah, that's the analytic side.

ÍgjøgnumMeg

Hey speaking of nt

MatheinBoulomenos

Hey everyone

Ultradark

hi

Ted Shifrin

Hi/bye @mathein

Daminark

17:45

Yo Mathein whaddup

ÍgjøgnumMeg

Hey @Mathein

schn

18:49

How would one go about determining the limit of $\frac{xy}{x^2+|y|}$ as $(x,y)\to0$?

Rithaniel

Depends on the path you take when approaching 0

schn

The limit is 0 regardless of the path, no?

Ted Shifrin

Try coming in along $y=x$.

Oh, nope.

Rithaniel

Well, potentially, but how you go about determining the limit depends on the path you take.

Ted Shifrin

If the limit is really $0$, we need to show this without doing any paths at all.

schn

18:54

Right, so is there any algebraic manipulation?

Ted Shifrin

Aha, it's too simple.

Think about $|y|+x^2\ge |y|$.

Oh oh, @ÍgjøgnumMeg is back. He must be drinking to celebrate his completion of one chapter of his life.

ÍgjøgnumMeg

:D I have a friend coming to visit for celebration in fact

Ted Shifrin

I'm not as dumb as I look :)

ÍgjøgnumMeg

Heh

:P

schn

@TedShifrin That would yield $\frac{xy}{|y|}$, correct?

Ted Shifrin

19:00

What do you mean yield? Be precise.

schn

$\frac{xy}{x^2+|y|} \leq \frac{xy}{|y|}$?

Ted Shifrin

There you go. Good. Except you might want to work with absolute values of everything.

That's easier for squeeze arguments, because then $0$ is always on the bottom.

schn

Okay. But if the squeeze theorem shows that the absolute value of the original expression converges to 0, how does that prove it for without the absolute value of it?

Ted Shifrin

Stop and think about what you just asked.

schn

Absolute convergence implies convergence?

Ted Shifrin

19:11

Way too complicated. We're not doing infinite series here.

What does it mean for a quantity to approach $0$?

schn

19:22

So $|f(x,y)|\to 0 \iff f(x,y)\to 0$?

Semiclassical

the fact that this question from yesterday already has 11 votes perplexes me: math.stackexchange.com/q/3353508/137524

I mean, yes, the chain rule is a common source of confusion

But I’d have thought it had been covered to death on MSE by now

schn

@TedShifrin $f(x,y)\leq|f(x,y)|\leq0$ when $(x,y)\to0$. If there’s a lower bound for $f(x,y)$ that is 0, then I see how we could conclude that $f(x,y)$ also converges to 0.

Ted Shifrin

@schn: You haven't given me the definition of the statement $f(x,y)\to 0$.

This is actually quite important.

Hi, DogAteMy.

Akiva Weinberger

Ted Shifrin

@Semiclassic: People do not remotely search for their own question/answers.

Akiva Weinberger

19:33

This is so clever!^^

I dunno if my instructor wrote it himself or stole it from somewhere

Ted Shifrin

Probably stole it. He's obviously a very different flavor of mathematician from me, I've noticed. I wonder if he even covers things like multiple integrals and differential forms "properly."

Akiva Weinberger

The rest of the pset is stuff like "prove if $a_n\to a$ and $b_n\to b$ then $a_nb_n\to ab$"

"Here are some slightly wrong definitions of a limit (with quantifiers swtiched and whatnot), give examples that show these differ from a limit"

Ted Shifrin

Yeah, Spivak has a question like that in his book, and I don't think I ever assigned it. I mean, understanding order of quantifiers is important, but ...

Lunchtime. Bye.

Akiva Weinberger

Have fun

Akiva Weinberger

19:54

Puzzle: Construct a Cantor set embedded in the plane such that any line passing through its convex hull intersects it

ALannister

20:19

@Semiclassical I don't think that Cesareo's code is correct. I wrote it differently and am not getting his convergence.

none

Is anyone here familiar with rotation numbers?

I was wondering whether there is any good justification for giving a circle map rotation number $1$ as opposed to $0$ (the rotation number is taken mod $1$). There are cases in which it seems to make sense to do so. For instance, considering a continuously varying family of circle maps starting with $\rho(f_0) = 0$ and ending with $\rho(f_1) = 1$. Then every rotation number in between is attained, by continuity of $\rho$.

But technically $\rho(f_1) = 1 = 0 \mod 1$.

schn

22:05

Consider the following multivariable function: $\frac{e^{|\textbf{u}|^2}-1}{|\textbf{u}|^2+x^2y+y^2z}$ where $\textbf{u}=(x,y,z)$ and $|\textbf{u}|=\sqrt{x^2+y^2+z^2}$. How would one find its limit as $\textbf{u}\to0$?

If one approaches the origin from where $y=0$, the terms $x^2y+y^2z$ in the denominator equal 0. Substituting $|\textbf{u}|^2$ for $t$ yields the expression $\frac{e^t-1}{t}$, which has limit 1 as $\textbf{u}\to\textbf{0}$ and thus $t\to0$. So the limit should be 1 if it exists.

Ted Shifrin

OK, so what can you say about $x^2y+y^2z$ compared to $\|\mathbf u\|^2$? With proof?

schn

Do you mean there is a way to express $x^2y+y^2z$ in terms of $\textbf{u}$ or $\|\mathbf u\|^2$?

Ted Shifrin

No, how does its absolute value compare to $\|\mathbf u\|^2$ when $x,y,z$ are small?

Limits/analysis is all about estimates.

schn

22:23

Are you saying that $x^2y+y^2z$ are negligible when $x,y,z$ are small? But so would $\|\mathbf u\|^2$ be, no?

Ted Shifrin

Think about the quotient of the two.

What does it mean to say one quantity is "negligible" compared to another? The quotient should go to $0$.

Think about how you would do a beginning calculus limit like $\lim\limits_{x\to 0} \dfrac{3x+x^3}{5x+3x^2}$.

schn

22:36

@TedShifrin Alright, so $x^2y+y^2z$ is smaller compared to $x^2+y^2+z^2=\|\mathbf u\|^2$, and therefor it would be negligible?

Or maybe the other way around?

schn

22:48

$\frac{e^{|\textbf{u}|^2}-1}{|\textbf{u}|^2+x^2y+y^2z}\leq\frac{e^{|\textbf{u}|^2}-1}{|\textbf{u}|^2}$

Ted Shifrin

No, that's not right, because the denominator could have negative terms. How do you work out that single-variable problem correctly?

schn

$\dfrac{3x+x^3}{5x+3x^2}\leq\dfrac{3x+0}{5x+0}=\dfrac{3}{5}$, right?

Ted Shifrin

Nope. How in the world do you justify that?

Don't you remember learning how to do this at the beginning of calculus?

Even if the $\le$ were right, you don't have a lower bound of $3/5$ as well, so how does that justify a limit?

schn

23:04

You're right. $\dfrac{3x+x^3}{5x+3x^2}=\dfrac{x(3+x^2)}{x(5+3x)}\approx\dfrac{3+0}{5+0}$

Ted Shifrin

Right. Now how do you make the last step rigorous?

Well, of course, you take the limit as $x\to 0$. Note where the $x^2$ and $3x$ came from. They came from factoring out the $x$. Can you approach the multivariable problem in a similar fashion?

schn

Right, by the properties of limits I would justify the last step by the sum rule and the fact (which also is sometimes stated as a "rule") that $\lim\limits_{x\to 0} x=0$.

Regarding the multivariable problem, is there something that one can factor out in the numerator?

Ted Shifrin

No, but you already know a basic limit fact that you used. You already know $\lim\limits_{\mathbf u\to 0} \frac{e^{\|\mathbf u\|^2}-1}{\|\mathbf u\|^2}$. So how should you proceed?

schn

23:23

$\dfrac{e^{\|\mathbf u\|^2}-1}{\|\mathbf u\|^2+x^2y+y^2z}=(e^{\|\mathbf u\|^2}-1) \dfrac{1}{\|\mathbf u\|^2} \dfrac{1}{1+\frac{x^2y+y^2z}{\|\mathbf u\|^2}}$

Ted Shifrin

OK, although I would have put the first two terms together in the fraction you already understand. Now what do you want to do?

schn

Since these are continuous functions, one could simply plug in $x=0,y=0,z=0$ in the expression and together with the fact that $\lim\limits_{\mathbf u\to 0} \frac{e^{\|\mathbf u\|^2}-1}{\|\mathbf u\|^2}=1$ evaluate the limit. Yet in $\dfrac{x^2y+y^2z}{\|\mathbf u\|^2}$ it remains to determine whether the numerator or denominator is bigger...

Ted Shifrin

No, be very careful saying stuff like that. You can't plug in and get 0/0.

Analyzing $x^2y+y^2z$, etc., you should think about using the hint I gave you hours ago for the earlier problem.

schn

23:41

$\dfrac{x^2y+y^2z}{\|\mathbf u\|^2}=\dfrac{x^2(y+\frac{y^2z}{x^2})}{x^2(1+\frac{y^2+z^2}{x^2})}=\dfrac{y+\frac{y^2z}{x^2}}{1+\frac{y^2+z^2}{x^2}}$

Ted Shifrin

No, no, no.

You want to think about how to compare the numerator to $\|\mathbf u\|$.

schn

You mean which of $x^2y+y^2z$ and $\sqrt{x^2+y^2+z^2}$ is smaller?

Ted Shifrin

No, you should know which is smaller, but you need some precise estimate to do this stuff.

Just saying $2x$ is smaller than $3x$ doesn't tell you the limit of their quotient is $0$.

schn

What estimate are you thinking about?

Like $y=x$, and then see what the expression turns into?

Ted Shifrin

No, what can you say about $|x|,|y|,|z|$ specifically?

schn

23:52

$|x|\leq \sqrt{x^2+y^2+z^2}$, or something similar.

Ted Shifrin

OK, so use that.

schn

Alright, so that together with the triangle inequality (to separate the sum in the numerator) and the squeeze theorem should show that $\dfrac{x^2y+y^2z}{\|\mathbf u\|^2}$ equals 0 as $x,y,z\to0$?

Ted Shifrin

Not "equals," but "goes to." Yes.

I hope you try to understand these techniques and apply them :)

schn

Certainly! Thanks for your time and help.

Ted Shifrin

You're welcome.

Transcript for

$Mathematics$ Mathematics