OK, so, $f=f(0)+f'(0)x+f''(0)x^2/2+o(x^2)$

This is à propos of what?

and we know $f(0),f(h)<M_0$ and $f''(0),f''(h)<M_2$

Oh, that one. Note the domain is $(0,\infty)$.

and we want to show that $f'(0)<2\sqrt{M_0M_2}$

@TedShifrin Why would that matter? Shift it over

The hint is to expand in a Taylor polynomial centered at $x$.

Plus trigonometry, dot products in space, basic set theory, arithmetic in Z, basic number theory, rotation in the plane, vectors in space.

@TedShifrin That's all.

OK, I'll shift it over to being defined on $(-x,\infty)$ so that I can work with $0$, I like $0$

Weird mixture, @Mahmoud. How many more years before university?

1 @TedShifrin

LOL, DogAteMy: You're going to get us both totally confuzled.

1 after this year, Mahmoud?

@TedShifrin Yup

What do you study next year?

First time seeing complex numbers,

More arithmetic in Z,

And Calculus,

@Mahmoud in the integers? Don't you mean modular arithmetic?

@Sophie Sorry ._.

Operations on matrices,

Probability

OK ... so it's a weird mixture. I cannot understand how you have not seen geometric series.

@TedShifrin I will see them this year.

Ah, OK.

Do they make you write some proofs?

But I see the word sequence rather than series.

@TedShifrin A lot, all the time, in all problems.

Prove that ...

OK, that's good preparation if you want to study mathematics in the university.

several years ago I had a teacher who called $\{x|a<x<b\}$ an open interval and $\{x|a\leq x \leq b\}$ a closed interval. It wasn't until several years later when I understood the look on his face when I asked if the empty set was open or closed

Did I suggest to you already that you look at Spivak's calculus book? Is that where you got this problem about $f(x) = (ax+b)/(cx+d)$?

@Sophie: I hope he just said yes.

yes I should have asked XOR

Then he should have said no :D

@TedShifrin No, we were studying this function, and since we didn't cover derivatives yet, the teacher said that we'll use the discriminant. (without talking about the derivative)

Discriminant of what?

You mean determinant?

I'm confused.

Yes.

We didn't really use the word determinant (No Linear Algebra).

so the teacher was discussing it in terms of the $2\times 2$ matrix?

So .. I don't understand.

He wasn't because we're not supposed to know anything about Matrices I just saw it in the Internet.

If you really want to get a bit ahead in math in a constructive way, I would very much recommend Spivak's Calculus ... I dunno if you can find it underhandedly on the internet. But working through that book, you will learn serious mathematics and calculus as you go.

@TedShifrin Finally ! :D

cough libgeb cough

I do not know such things.

@TedShifrin Yes ! Thank you !

What things ?

What Sophie just said.

I suspect some of my books are in that unknown place, too.

cough ?

Oh, nvm.

@TedShifrin oh yes they are. Everything is

libgen, sorry. libgen.io to be precise

This site can’t be reached :I

Funny that they have an old, old, old (2008) version of my diff geo text, which is frequently updated and available for free on my website. So someone should replace that with the 2016 version :P ... They've got 2 of 4 books.

@TedShifrin Shall I jump the Vector-Spaces chapters ? :)

What vector spaces chapters?

I said Calculus ...

Yes the book you said.

There's no vector spaces.

I'm talking about the book that's called Calculus. NOT Calculus on Manifolds.

Let $a,b\in\mathbb{Z}\setminus\{0\}$ and $d\in\mathbb{N}$ be the smallest number for which $l,k\in\mathbb{Z}$ exist, such that $d=l\cdot a+k\cdot b$. How can you show by using $\text{mod}$ that $d$ divides $a$ and $b$?

@TedShifrin It occurs to me just now that $(ff')'=f'^2+ff''$. I don't think it's relevant to the problem but I'm writing it anyway

Oh no, meh, I got Advanced Calculus of LYNN H. LOOMIS :/ @TedShifrin

@NaCl use the Euclidean algorithm on $(a,b)$

That sort of thing is useful for differential equations sometimes, DogAteMy.

No, no, that's no good, Mahmoud.

In any case, I have no idea how to solve the problem

I'm lucky :D quora.com/…

DogAteMy: You're supposed to consider for $h>0$ that $f(x+h) = f(x) + f'(x)h + \dots$, etc. Then solve for $f'(x)$ and try to give estimates.

$(f(x+h)-f(x)-f''(x)h^2/2+\dotsb)/h=f'(x)$?

OK, now give an upper bound on $|f'(x)|$ and play.

Here you want the error term with $f''(c)$ in it, BTW, no ... left.

@Sophie What do you mean by that? Saying that $a=t_0\cdot b+r_0$ for specific $t_0,r_0\in\mathbb{Z}$ doesn't seem to lead to a good road.

$(2M_0+M_2h)/h$??

@TedShifrin It isn't scary :D

Thanks for recommending this $:)$

@NaCl you know the good old Euclidean algorithm $(120, 35)=(15,35)=(15,5)=(0,5)=5$ for example

Lots of good stuff to read and lots of exercises, Mahmoud. (If you read the preface, you'll find I wrote a lot of the exercises.)

I'm kind of confused

@Sophie Yeah, I guess...

Wait .. You wrote ?

It looks almost right, DogAteMy.

Although I'm quite sure it should be $(120,35)=(35,15)=(15,5)=(0,5)=5$ using your notation

Oh, you lost another 2, DogAteMy.

Why? $f''(x)h^2/2$ contributes $M_2h/2$, and the error term contributes that much also.

$f''(c)h^2/2$

No?

Oh, you misinterpreted the error.

What's the error term then

$f(x+h) = f(x) + f'(x)h + f''(c)h^2/2$ for some $c$.

Here we're using the actual Taylor with Remainder formula.

Oh…

So $(2M_0+M_2h/2)/h$

@NaCl it's the same thing because $(a,b)=(b,a)$

Edit again, DogAteMy :)

But you're close. Now you should finish.

@Sophie So you're talking about $\text{gcd}$'s

ok

I need to prove that $d$ is actually the $\text{gcd}$ of $a$ and $b$

$(2M_0+M_2h^2/2)/h=M_0\dfrac2h+M_2\dfrac h2$

Better, DogAteMy :)

DogAteMy??

Me

I really don't see how I can proof anything with that

And that, after multiplying by $h/2$ and completing the square, shows that the minimum is the desired value

@NaCl: Divide $a$ by $d$. If you get a nonzero remainder $r$, you'll find a contradiction to the construction of $d$.

if two numbers have a common divisor then their sum also has that common divisor

@TedShifrin Thank you

DogAteMy: So there are some cool consequences to this. Like if $f$ and $f''$ go to $0$ at $\infty$, then ...

Or if one does and the other stays bounded, then ...

$f'$ must go to $0$ also?

Yup ... Not entirely obvious a priori.

@Ted: What did your ping say?

Prove that

Oh, it was relative to DogAteMy's comment over there >>>> that no one else could use your name. I said you might go on hiatus and disappear and then your name is available.

I'm shocked I actually remember. Maybe Alzheimers isn't too bad yet.

Prove that convex twice-differentiable functions have at most two roots.

So I try to make up a contradiction from $a=l_a\cdot d+r_a$

@DHMO: What have you tried?

@MikeMiller, Semi made a pretty picture

@TedShifrin i've constructed a terrible 4-case unrigorous proof.

Looking for something neater

Yuck. Way easier than that, @DHMO.

If you had three roots, what would have to happen?

(Take a grid of squares, and replace each square by a pair of parallel line segments that each join midpoints of sides; choose between the two choices randomly @MikeMiller)

Not like I spend much time here in the past weeks anyway.

(to get the picture)

how do I know @Ted

Basic calculus theorem, @DHMO.

Between any two roots of a differentiable function, what must happen?

@TedShifrin you would have to have two optima

well, critical points, yes.

so f' would have two roots

If $f''>0$ can that happen?

nope

Done.

because between the two roots there is also an optimum

for which f'' = 0

you mean a critical point of $f'$ ...

Hmmm.... $d=l\cdot a+k\cdot b=l\cdot(l_a\cdot d+r_a)+k\cdot b=l\cdot l_a\cdot d+k\cdot b+l\cdot r_a$

mean value theorem

@ted thanks

yeah, or $f''>0$ means that $f'$ is an increasing function.

It's a nice question.

@NaCl: Can you take the original equation and solve it for $r_a$?

What do you mean by original?

The first equation you typed out with $r_a$ in it ...

@AkivaWeinberger do you like wine and hamburger?

So $a=l_a\cdot d+r_a$ uuh... $r_a=a-l_a\cdot d$ see no problem here

Now substitute for $d$.

@DHMO Hamburgers: yes. Wine: very much no.

@AkivaWeinberger :(

@AkivaWeinberger then do you like wineburger?

I'll take DogAteMy's share, if it's good wine.

I must admit I've never tried the two together.

The first thing I learned to cook was Julia Child's recipe for hamburgers with a red wine sauce :)

@Weinberger wineburger

Yes, I know

aber Berger ist verschiedener als Burger :P

Ich nein spreach Deutsh

LOL

@ted i thought you speak french

Was

So, @DHMO?

@NaCl ¿Qué?

So you're trilingual @ted

There's an old joke:

My French once was as good as my English. My German was never quite so good. And I know some Latin and a bit of Russian.

"What do you call someone who speaks three languages? Trilingual;

@ted pentalingual!

Sooo $r_a=a-l\cdot l_a^2\cdot d-l_a\cdot k\cdot b-l_a\cdot l\cdot r_a$

We did this the other day, DogAteMy.

Oh, did we?

@TedShifrin Pfff, not even old Greek?

Sorry

No, no, @NaCl. Remember that $d = ka+\ell b$ or something.

Oh meant that $d$

The punchline is monolingual = American

Yup, we did that the other day.

@akiva I haven't heard it

Sorry, @Krijn: I'm an ignoramus.

@DHMO It ends, "What do you call someone who speaks one language? American."

what is the joke?

Or maybe the President of the US.

The sad commentary is that Americans are proud of being barely monolingual.

@akiva english also

and australian

Hi @DHMO

@TedShifrin It's good that you're no ignorabimus, they have no place in mathematics.

basically anyone whose mother tongue is english

The word you're looking for is "Anglophone"

@Mahmoud salaam

What did you mean by the piano ...

@TedShifrin do you think there are integers $x,y,n$ with $n>2$ and $x>1$ so that $x^n-2y^n=1$? There's some similarity with FLT but it's probably easier

In Europe many (most?) people travel to countries where other languages are spoken.

@DHMO Wa alayka salam !

@mahmoud what piano?

$r_a=a-l_a\cdot d=a-l_a\cdot(l\cdot a+k\cdot b)=a(1-l_a\cdot l)+l_a\cdot k\cdot b$

I have no idea, @Sophie. I'm as far from a number theorist as one can get.

OK, @NaCl ... I'm not checking details, but now what do you conclude?

@DHMO You wrote something about some piano and said lol

To me.

@TedShifrin $r_a=d$?

But $0\le r_a<d$, @NaCl.

He wrote "lol" after I "demonstrated" that I don't speech German

DogAteMy: In a few days I'll give you a few more questions :)

Should start sneaking in differential geometry :P

@DHMO LOL, you wrote exactly this ''@Mahmoud peano in kindargarten lol''

Hello :)

Hi.

I have a very quick question can I post it here ?

"Just ask; don't ask to ask."

Wow, cool!! However, can you enlighten me a bit more why $r_a=d$ follows? $a$ and $b$ are set, $l$ and $k$ can be chosen in any way from $\mathbb{Z}$, but they must make $d$ the smallest possible $d$. I don't really see how $1-l_a\cdot l$ and $l_a\cdot k$ are minimal... Or can we just say "Choose $l_a,l,k$ such that $a(1-l_a\cdot l)+l_a\cdot k\cdot b$ gets minimal, then it equals $d$"? But what if it isn't minimal?

But we chose $d$ to be the minimal such (positive) combination. Could you have a smaller one that's positive?

No, but $a(1-l_a\cdot l)+l_a\cdot k\cdot b$ could be bigger, couldn't it?

No that can't be!

No, $0\le r_a<d$.

haha, yes! I just saw it

So what do you conclude?

Thank you very much

We proved my contradiction, that no such $r_a$ exists, and thus $d\vert a$

Well, more precisely, $r_a=0$.

So this means that ...

@Mahmoud Ohh. The "Peano axioms", not "piano"

@AkivaWeinberger I don't think he meant that ..

The (or, "a") basis for all arithmetic, but too abstract to teach before teaching arithmetic

Yeah, sorry. I was inaccurate

I'm just helping you with good proof-writing, @NaCl.

Okay, so my problem is that I'm trying to prove that the limit of $x^n/ e^x$ when x tends toward infinity is always 0 without using L hospitals rule ? How can I do that ? Or is it possible to rearrange it as the limit of $x^n * e^x$ when x tends toward minus infinity ?

hello

Hi @meow