@buddhababe It seems you got both good and bad advice earlier. L'Hôpital's rule doesn't work in multivariable. The point is that you do know from single variable that $$\lim_{u\to 0}\frac{\sin u - u}{u^2} = 0,$$ either by Taylor polynomials (my preference) or by L'Hôpital. What happens if you substitute $u=2(x+y)$? (Note that $(x+y)^2\le (|x|+|y|)^2\le \sqrt{x^2+y^2}$ for $|x|$ and $|y|$ small enough.)

Hello @ted

morning @Ted

and @Jasper

Is there someone?

good night, @Mike, and @Jasper, for that matter.

@Valery: Offhand, it seems false to me.

Hmm, maybe not.

@MikeMiller @TedShifrin I was talking about my 30 years of my life with my mum just now. I feel that none of it makes sense. No words can describe what I feel now.

This is why you need to meet with a professional, @Jasper.

@TedShifrin So, you can't say certain?

I'm now believing it, @Valery. Do you have some differentiability hypotheses?

@TedShifrin No, that is why I need someone who understands me. But I will meet with the pros too.

@TedShifrin Well I'd guess f(x) is differentiable. Is that what you asking?

But not even you understand you, @Jasper. :)

Guess? @Valery Can we guess twice differentiable?

@TedShifrin I think ultimately, I am the best candidate. Ted, one of my greatest regrets in life is not going to Cambridge.

@TedShifrin I just assume since it's convex, it should be twice differentiable in the interval of interest

Meanwhile, if I just draw it in pictures it seems false

Or I am insane

Well, I thought that too, @Valery, but now I don't have a picture. What's yours?

@TedShifrin You wouldn't believe ...

Here is it

It seems correct. A convex function must be continuous. We can replace $f$ with $f+c$ for any constant $c$. So let $c=-\min(f)$, which definitely exists.

Roughly of course

Well sorry not the best picture

Yeah, that gives a positive integral. As I commented, vertical shifts don't affect the integral. Also, it might be easier to think on $[-\pi,\pi]$, rather than on $[0,2\pi]$.

Yeap

really

it makes sense

$\int_0^{2\pi} x^2\cos x \,dx = 4\pi > 0$ . :)

Ah, I was wrong about thinking on $[-\pi,\pi]$. Then it becomes false.

$\int_{-\pi}^{\pi} x^2\cos x\,dx = -4\pi$. So it follows that the result is false, and your intuition is right.

Hmm ... maybe not.

Yeah I know

I'm lost

LOL

this is funny

lol

I think it was Chris

stop laughing at me, mr eyeglasses :D

who said that

Well, @Chris'ssis knows all sorts of tricky stuff with integrals. I've never thought about this before.

I admit I should know a proof ...

@ted I don't think I will understand @Chris'ssis book.

Seems that cosine has a "valley" in that region in that valley the function should be less that on the sides

But then the contribution of the negative part may turn greater than the positive one

Yeah, we can't translate, because $\cos(x\pm\pi) =-\cos x$ ...

Nor I, @Jasper. But she thinks we're all dumber than a high school student.

@Valery: However, what I said about vertical translation is correct. $\int_0^{2\pi}[f(x)+c]\cos x\,dx = \int_0^{2\pi} f(x)\cos x\,dx$, of course.

@Ted: you might like this quote from a book by Kirby.

"...then $M_i$ has 'corners' where the $k_i$-handle was attached, but the phrase 'corners can be smoothed' has been a phrase that I have heard for 30 years, and this is not the place to explain it."

I saw your quote, @Mike.

Oh.

I know many high school students that know more math than I do :(

Same thing with smoothing corners of piecewise smooth curves ... which does show up in differential topology, etc.

Ok vertical translation is clear

But I think of the things relatively

I actually wrote out an argument for doing Gauss-Bonnet with a conical point, by smoothing appropriately.

Yes. It's just the colorful language that I found amusing.

I can't see WHY the negative contribution should be smaller

or I guess you mean a singularity in the metric that looks like a cone...

I still remember when Rob couldn't be bothered worrying about Chern class $+1$ versus Chern class $-1$ for the Hopf/tautological bundle on $S^2$. That was late 70's. I worried about such things, being an algebraic geometer type. We had some humorous discussions. :)

I have trouble with sign issues like that sometimes. It's a weakness I'm allowing myself to have for now.

Well, @Valery, if you do the case of a quadratic polynomial, it's clear that it works out. But then why couldn't we make it grow much faster in the final interval $[3\pi/2,2\pi]$ to make the negative contribution of $\cos$ much more serious?

Topologists just reverse orientation whenever they feel like it, @Mike, but you can't do that in algebraic geometry.

The key thing to understand in this game is why blowing up a point in a complex surface is connect summing with $\overline{\Bbb CP^2}$, I suppose.

@TedShifrin That's what I think. It seems not to work with an arbitrary function for god sake

well, when you're working with 4-manifolds, one usually wants to classify things with orientation... and if you're working with framed links etc you can't just flip orientations unless you also flip the things that depend on the orientation

@TedShifrin No Ted I think it's actually right

Yeah, I do too.

I don't have a proof yet. And I have to go teach ... Have to prove Stokes's Theorem on manifolds today, so I need to think for a few seconds before class.

See you all later ...

Bye

Bye, @Ted.

@ᴇʏᴇs

@MikeMiller Are you happy?

Yes.

I think I never was really happy.

I think most people in this world live very unhappy lives.

What makes you think that?

I am too naive for you.

question to @mike or other algebraic topologists. I'm working on hatcher 2.2.30, For the mapping torus $T_f$ of a map $f:X\to X$, we constructed in Example 2.48 a long exact sequence $\dots\to H_n(X)\xrightarrow{1-f_*} H_n(X)\to H_n(T_f)\to H_{n-1}(X)\to\dots$. Use this to compute the homology of the mapping tori of the following maps:

I feel confident with the first few, since we are dealing with spheres and reflections on the sphere, so the degree is easily calculated,. I am a bit confused as to how to get the degree on the examples with $S^1\times S^1$

(c) The map $S^1\times S^1\to S^1\times S^1$ that is the identity on one factor and a reflection on the other.

Are there people familiar with integrals?

What do you mean by how to get the degree, @JMoravitz? The degree of the map itself?

Or the degree of the map coming from an inclusion of a 2-cell and then collapsing other cells etc?

Why is this \int_{0}^{2\pi}f(x)\cos x dx positive if f(x) is convex?

Yes, @mikemiller in order to find the kernel and image of the map $1-f_*$ in the exact sequence above for the purposes of calculating $H_n(T_f)$

I guess Hatcher hasn't defined what the degree of a map is when the spaces involved aren't both spheres.

If he did, I hadn't noticed it.

Can you be precise about what map you're trying to compute the degree of? Your yes above wasn't very precise. (Sorry :))

As I understand $H_2(S^1\times S^1)\approxeq \mathbb{Z}$, and the map $(1-f_*)$ from $H_2(S^1\times S^1)\to H_2(S^1\times S^1)$ is determined by where the generator gets mapped

OK, so you meant what that map is.

I guess most of the things I would point out are tools you don't quite have yet. Maybe the best thing to do here is to calculate the map via cellular homology.

The answer you're going to get is $f_* = -1$.

Would you mind going into a bit more detail? For $S^1\times S^1$, as I understand, we can describe it as a filled in square with opposite edges identified and all corners identified. I.e. there is one 2-cell, two 1-cells, and one 0-cell.

What's convenient is that the map you describe is cellular - it maps $k$-cells to $k$-cells. In this case, it maps the 2-cell to its negative. That's why the map $f_*$ is $-1$.

More rigorously, the maps on cellular homology are defined by looking at the long exact sequence $H_n(X_n, X_{n-1})$, and using your $f$ as a chain map to $H_n(Y_n, Y_{n-1})$. In other words, you see precisely where $f$ sends each chain in $X_n$, and pass to homology in the lower sequence.

Here 'passing to homology' isn't really happening, because none of the chains are boundaries.

I feel like that might be incomprehensible, but I have to go. If you can't make heads or tails of it, I'm just pointing out that you define the induced map on cellular homology the exact same way you do in singular homology - by looking at the chain groups. The value of cellular homology here is that the chain groups are nice and small.

Its cool. You at least gave me places to continue looking in the book.

I appreciate your time. Thank you.

Sure, it's no problem.

That would be 1 million USD.

Hatcher hatches, Miller mills, Jasper jasps.

hi @JasperLoy

That deserves a star.

there we go

You should not think about topology, you never listen to our advice, finish naive set theory and calculus first.

well i am doing calculus......

did naive set theory

And after that there are many more things to study before topology.

i was just looking through the book pdf and found that i understand some stuff

thats it

Sure, I understand the full stop at the end of every sentence.

well currently i am doing a very important part which i think is going to take me weeks or months.......differential equations

i just started it today

@JasperLoy how do you think differential equations will go?

@SayanChattopadhyay They will go with two legs.

but what if it is not a homosapien

@JasperLoy hi

You know.

I never understood multiplying matrices.

Mostly because of the randomization of the size of said result.

WHY IS IT THAT I KILL EVERY CHAT ROOM I ENTER

@DeltaEscher Multiplying matrices is equivalent to composing linear maps. You will learn in higher math.

I'm in pre-calc.

But it freaks me out when I learn about the complex reason of how the size of the resulting matrix is chosen.

I just don't get it.

Quick question to anyone. If I have a rhombus, under what transformation can it be made into a rectangle?

Affine, right?

@Delta if you have two matrices $A$ and $B$, and you take their product $AB$, it is only allowed if the number of columns of $A$ is the same as the number of rows of $B$, and the result $AB$ will have the same number of rows as $A$ and the same number of columns of $B$.

Hi @meer2kat

hi professor @TedShifrin

rehi @Jasper, goodnight, @Mike, hi, @JMoravitz

hi, @Sayan

@TedShifrin Hi.

@Valery: I no longer believe it. If you take piecewise linear functions, with increasing slope as you move to the right, this is "almost" convex, and one can get a negative integral.

Morning, @Ted.

I can't go home after work today, @Mike, so let's go get a drink :P

I'd say yes, but I think there are geographical problems.

Annoyingly.

@evinda Hello. Somewhat.

@Axoren: If one corner of the rhombus is at the origin, you can do it with linear. But, yes, affine.

@MikeMiller @TedShifrin If you create a good enough bijection, it would be like you're almost in the same place, wouldn't it?

No.

NO @Axoren. Bijections can be ridiculous.

@ValerySaharov When I talked to you I gave a link on chat with the @robjohn's proof.

@TedShifrin Thanks, I just wanted a sanity check.

@Axoren Do you know if this command: dg=@(y(:,i+1)) 1+h*(rho-1)*part(t(i+1),y(:,i+1)); is allowed in matlab? Or can't we give as argument of a function a vector?

@TedShifrin While injections are painful.

@Chris'ssis: With what hypotheses? I'm no longer believing it in generality.

Often, yes, @Jasper ...

@JasperLoy For you.

@TedShifrin $f(x)$ has to be convex.

I made a bane joke in MathSE Chat

I'm going to meme hell.

Strictly convex? How differentiable? I have a piecewise-linear counterexample, and it is (non-strictly) convex. But I can make tiny perturbations and make it convex, so I'm suspicious.

@Axoren No, I am not afraid of injections. I am afraid of life.

@evinda You're defining a lambda when you do @(anything)

I don't think you can use subindexing in the domain definition

@JasperLoy youtube.com/watch?v=w9wi0cPrU4U

@TedShifrin See the proof of @robjohn's here - chat.stackexchange.com/transcript/36/2015/1/29/12-17

Go a bit downwards on that page.

@evinda You use: @(x)

And then treat $x$ as a vector inside the lambda body

In your case, you pass in the value y(:, i+1) as the argument to the lambda function when you use it.

But when defining the lambda function, use the argument variable instead?

@Chris'ssis: I see a statement of it, but no proof.

@Axoren Which lambda function do you mean? :/

@evinda This one: @(y(:,i+1)) 1+h*(rho-1)*part(t(i+1),y(:,i+1))

@TedShifrin ^^^

Try $f(x)=\begin{cases} x, & 0\le x\le \pi/2 \\ 3x, & \pi/2<x<3\pi/2\\ 3.1x, & 3\pi/2\le x\le 2\pi\end{cases}.$ Gives negative integral.

@Ted It went well, I got a neutral reaction as you suggested due to her inexperience with the stuff I was working with. She told me she'll ask around and see if anyone can get me in contact with someone who has experience with special functions

@teadawg: Cool. And your taste may well change as you learn more mathematics.

@Ted: Did you see the question I asked you on facebook?

@Mike. Yeah. Didn't feel like dealing with it there. Special functions still show up with harmonic analysis and representation theory, I think ...

And when you use dg0, you say: dg0(y(:, i+1))

Wasn't looking for much more than what you just said. Thanks.

Is the inverse of a shear transform the opposite shear transform? Or am I being too hopeful?

I just REALLY don't want to notate this as $Shear^{-1}_{\lambda, Horizontal}$

Or... hell, how would I notate this.

Which version of matlab are you using?

@Axoren MATLAB 7.7.0 (R2008b)

Wow, that's pretty old. Were lambdas included in that version of matlab?

@Ted I feel great, I'm excited about the possibilities in my future :D

@Axoren How can I check it... I don't really know what you mean with lambdas :/

@TedShifrin related to one of your previous commments, it's not at all what you believe since here I met some of the most clever persons I've ever met. The thing is that I have a different attitude towards mathematics, and yes, I think I'm boundless and I can reach any peak.

@evinda This is a lambda expression for a function: @(x) x

That function is f(x) = x.

I don't want to engage in that discussion, @Chris'ssis. I am trying to understand why my counterexample is not a counterexample. Clever or not.

@Axoren But doing it like that we cannot name the function, right?

@evinda Yes, be we can always give a lambda expression a name later, which is the benefit of it.

We can pass it around like a value, the same way we do with numbers.

And once we bind a variable to that value, that variable is a function.

@Axoren So should I try it?

@TedShifrin Yeah, but it's good to know that I appreciate people, but the fact that I believe in myself so strongly doesn't mean the opposite. If you don't wanna engage in these discussions then be more careful. I feel the need to say something ... ;)

@evinda Try the identity function definition I just gave you. If that one fails to compile, you don't have lambdas in your version of matlab.

f = @(x) x

@TedShifrin OK, let me see ...

One can believe in oneself without being overbearing and repeating oneself constantly, @Chris'ssis. But I understand you are in a different life and different setting.

Yeah, I have worked out robjohn's proof, and it's correct. But I don't see what's wrong with my example.

@evinda You need to store the lambda expression in a variable to use it, otherwise it's just a value. h wasn't bound to it.

@evinda However, it seems that lambdas are in your version since it didn't complain about the lambda's definition

@Axoren How could I do this?

@evinda h = @(x) x

@TedShifrin I didn't catch it, I don't think. What was your counterexample?

@evinda Yes, lambdas are in your version, there must be something wrong with your original lambda definition.

@Axoren: chat.stackexchange.com/transcript/36?m=20817951#20817951

@TedShifrin What does a different life and different setting have to do with the attitude to the mathematics? If you ever felt I was repeting myself then I did it with a noble purpose. I remind you the double integral that you weren't sure about it exists. So repeting myself was good for you, you took benefit of it and realized you were wrong. :-)

@evinda Are you passing the lambda into a function where a value is expected?

$$\Large \text{Repetitio est mater studiorum.}$$

@Axoren I define g like that: g=@(x) x-y(:,i)+h*(func(t(i),y(i))+(1-theta)*func(t(i+1),x)); but I dind't get a warning message... So does the problem lie on the function part?

@evinda it seems that the part function is being passed a lambda expression instead of a value.

Because it's crashing on a comparison with a matrix and a lambda.

Which doesn't make sense usually.

Hey @Chris'ssis!

@robjohn: I'm totally perplexed. Computing the integral $\int_0^{2\pi} f(x)\cos x\,dx$ (admittedly, lazily, with Mathematica) for chat.stackexchange.com/transcript/36?m=20817951#20817951 gives approximately $-0.57$. AGH. This appears not to be positive.

@JasperLoy Hey. Still sleepy, I felt bad today, the whole day. Too less sleep. :-)

@TedShifrin Are you using your hand-drawn function?

@Chris'ssis What happened?

@JasperLoy Less sleep as I said.

It should get even more negative if you use an even more negative convex function.

It wasn't my question, @Axoren. I'm using the function I linked.

Vertical shifts are irrelevant, @Axoren.

@Chris'ssis Get more sleep then.

I'm assuming that Fourier series/transforms will make much more sense when I'm a bit more familiar with complex analysis...

Um, yes, @teadawg :P

@JMoravitz Thanks. Our school uses the most confusing math program.

@JasperLoy I try to finish another solution to a problem from my book.

@Chris'ssis I think you should try to sleep.

@JasperLoy I also had some tutoring, but it was fine. It's too early now, 3, 4, hours laters.

@evinda I don't think you need to change any of that function, just be more careful about what you pass to it.

@TedShifrin Oh damn, my chatjax bookmark broke and I wasn't looking at your post. For some reason, someone's handdrawing was the only thing that appeared.

My message got only one star.

oh ... I reloaded the link in a message to robjohn a few lines up, @Axoren.

@TedShifrin It was all tech difficulties on my end. I see it now.

@Axoren Should the result of part be a vector or a real number?

@DeltaEscher I see you are in this room too.

You CS people aren't supposed to have tech difficulties, @Axoren :D

@JasperLoy hey

@TedShifrin Our technical difficulties are normally other people's code :P

@meer2kat How strong is your faith in God?

@JasperLoy Strong enough?

@evinda You tell me. I have no idea what part is supposed to do within your context.

Right now, it returns a number as defined.

It's times like this I regret not studying rhombuses when I was given the opportunity.

@Ted That's a shame, Fourier analysis looks fascinating... I should know by now that higher mathematics builds upon itself in a very specific manner

Yes, specifically, using its hands.

@TedShifrin I don't think your function is convex.

Specifically at the boundaries of 3x and 3.1x, you can draw a chord under the curve. (Wait, maybe not)

hello

@Axoren I want to implement a numerical method. For $\theta=0$, we have $g(y^{n+1})=y^{n+1}-y^n-hf(t^{n+1},y^{n+1})$. In order to find an approximation of $y^{n+1}$, we use Newton's method: $y_{k+1}^{n+1}=y_k^{n+1}-\frac{g(y_k^{n+1})}{g'(y_k^{n+1})}$

@evinda What is the notation $y^j_i$?

Who loves differential equations?

I don't believe you, @Axoren. As long as the slopes increase to the right, it's convex.

@TedShifrin My edit agrees, there's no chord to be drawn under the curve there.

I just realised John Nash just won the Abel Prize, LOL.

I don't know if it is allowed for me to ask here or not but ys wong and I need help on this differential equation question if any of you guys are interested. I have an equation but probably not in the smallest order. math.stackexchange.com/questions/1213133/…

@randomgirl I love girls, not differential equations, lol.

@TedShifrin Your function isn't convex, actually.

@Jasper ... Cool your jets, hotshot

The points above the curve don't form a convex set.

It's easier to see this when looking at the boundary of $x$ and $3x$

Girls are probably easy than most differential equations. :p

@TedShifrin I could still be wrong, though. I don't normally look at discontinuous functions' convexity.

@evinda I'm not familiar enough with ODEs to be very constructive there.

@TedShifrin I'm going to get this thing drawn and then give you a counter example for it's convexity.

Otherwise, I'm just a bumbling lunatic.

@Axoren: Thanks, you told me what's wrong. I didn't write the right formula. It was supposed to have been continuous. Duh. I'm a dope. Thanks.

@TedShifrin No problem

@robjohn: Ignore my earlier post. I didn't write the right formulas to glue together the lines properly. Cool proof.

My message got only two stars.

@evinda it might be the semicolon

The error messages in later matlab versions are more specific

Morning people

@Axoren Good point.

$$\lim_{n\to\infty} \frac{1}{n!}\int_0^n t^n e^{-t} \ dt $$

@Axoren Now I have an infinite loop..

This is disgusting to me.

Can someone offer better notation?

I need to prove that a quadrilateral has congruent sides iff it has congruent angles. Any idea how to prove it?

@meer2kat I'd probably do that by contradiction.

Assume a quadrilateral has congruent sides but doesn't have congruent angles.

so just use a rhombus?

hi.. I have been learning about matrix multiplication over a finite field. When is this used in applications?

Is this something physicists/engineers/computer scientists actually need to do?

Hello

@teadawg1337 are you available by any chance?

hi @ted @owatch

Hey

I need some help with this :

gyazo.com/797a4c5600213819e5f139a35e2888a6

I know that the canonic way is $a(x-\alpha)+\beta$ with $\alpha = \frac{5+8}{2}$, but I can't figure out what $a$ or $\beta$ is.

$\int{\frac{x^{2}+x+1}{(x^{2}+1})^{2}}dx$

@Owatch Was that answer for me ?

Oh no.

I cannot help you, I don't know what canonic even is. I just solve integrals.

Wait, I will look.

anyone?

@Ramanewbie it's x^2-13x+40. I don't understand. What are you struggling with?

nvm

@Owatch canonical form I mean

You did have x intercepts though, it should have been fairly easy.

Also, you had minimum.

@Chris'ssis I prove that is $\frac12$ somewhere on the site. I think I even compute a few terms of an asymptotic expansion.

I was searching for paper but it looks like the Meercat solved it.

I wrote my integral wrong damnit

@Owatch it's not hard lol. (x-5)(x-8). Putting it in standard form from there should be fairly streaightforward @raman

@robjohn That would be great. Anyway, if you find the link let me know. That limit is more than a simple limit. It's related to another limit I attended a long time ago.

@TedShifrin Thanks :-)

$\int{\frac{x^{2}+x+1}{(x^{2}+1)^{2}}}dx$

the question math.stackexchange.com/questions/624894/… has no REAL answers . it would be nice if sb would take a look at it.

@meer2kat thank you but can you explain how you got the expression ?

How should I set up my integral?

@Ramanewbie Factoring the equation will give you the x intercepts

@Owatch what equation ?

@Ramanewbie Sure. I second degree polynomial's equation is found by its x-intercepts, in this case x=5 and x=8. The equation of a second degree polynomial is written y=(x-r)(x-s), where r and s are the x-intercepts such that y=(x-5)(x-8). Distribute from there and you have your equation.

You don't have one, but you can see it is a parabola, which have an equation of

i gotchu @owatch

$a^{2}+bx+c$

Thanks Meercat

I have a picture of a meercat oh my phone I think.

fun :)

@meer2kat @Owatch thank you. I would have never thought I would have been so simple... I don't really need the answer, but how would you proceed if you don't have any intersection with the x axis ?

imgur.com/y21CnVC

I will collapse the link to not take too much room.

Anyone got suggestions for my integral? I can split it up into Bx+C + Dx + E

But the denominators are the same and it doesn't help solve the problem.

@Owatch is this related wityh maths ?

@Ramanewbie If there is no intersection with the x axis it involves using i as an imaginary number. do you really want to learn it?

@Ramanewbie Yes.

@Owatch what integral?

@meer2kat Sure I would like, but I really know nothing about it. All I know is that ther's a number $i = \sqrt{-1}$, with a real part and an imaginal part...

@Owatch ohhh pretty. give me a few minutes to look at it

@Ramanewbie this should help a little. let me know if you have questions after reading it algebra.com/algebra/homework/quadratic/…

@meer2kat thanks !

@owatch separate it into three integrals to start

It's integration by partial fractions though

You generally don't separate it?

yeah sorry hang on

@Chris'ssis Here is the answer. I can add more terms if needed. I computed six terms for a post on sci.math.

@robjohn Good, I saw that! :-)

there's a starting point

Uh.

what?

How did it split it?

This is after it recombined

I would need to split it, then solve for A and B

Well, at least it is useful

For the denominators

yeah youre on your own for that part; i haven't messed with partial fractions in almost 2 years

I just don't know what it did before that first step. Typically, given this problem, you want to factor the denominator as follows: $\int{\frac{x^{2}+x+1}{(x^{2}+1)(x^{2}+1)}}dx$ = $\frac{Ax+B}{x^{2}+1} + \frac{Bx+C}{x^{2}+1}$

@BalarkaSen

here?

However, the problem is that it suddenly becomes useless to me.

Because $\frac{(Ax+B)(x^{2}+1)}{x^{2}+1} + \frac{(Cx+D)(x^{2}+1)}{x^{2}+1}$ becomes

$(Ax+B) + (Cx+D)$

In the integral earlier I wrote Bx+C rather than Cx+D but ignore that, I cannot edit now

i'm going to look this up really quick. give me a minute