Mathematics - 2018-01-20 (page 2 of 2)

Daniel Fischer is a well known genius

hi Meow, Balarka, Leaky

Hey @Ted

hi ted

can you guide me on an integral

well, integral problem

hi @TedShifrin

6:12 PM

Like what?

would you be interested in the link

it was like

hey guys

hi @Sha

hi @Ted long time no chat

6:13 PM

Mostly you've been hiding, Sha :)

prove $\int_0^x \frac{\sin t}{t+1} dt > 0$ for all $x > 0$

So what seems to be the problem, @Meow?

haha yea that is true, I've learned to just do my homework during the problem class sessions @Ted

We're glad not to be needed, Sha.

okay so first, am i right in saying that i need to find a partition for every $[0,x]$ whose lower riemann sum is $>0$?

6:14 PM

Hi @Ted

anyhow, does anyone have an idea what kind of derivative Spivak used on the rhs?

because they say they take $D_j$, but it doesn't look like it

No, @Meow. You don't want to go back to explicit computation with the definition.

Hi, demonic @Alessandro

(this is the full context)

6:15 PM

then what do i want to do?

It's right, Sha. He's doing implicit differentiation with respect to $x_j$.

Think about the graph of the function, @Meow, and see if that gives you any ideas.

@Ted o okay, let me look that up

i have

it looks like you start with a large positive sliver, then a slightly smaller negative sliver

Don't look anything up, @Sha. Just take the derivative of the equation, using the chain rule!

then go back to a slightly smaller positive, and so on

so it seems the net area is always > 0 but thats not rigorous at all

6:16 PM

Hello @TedShifrin

OK, so can you argue it is positive for $x\in [0,\pi]$? Then what about $x\in [\pi,2\pi]$? Etc.

of course

OK, @Meow, so somewhere you need to do something rigorous with an estimate.

@TedShifrin but how can a partial derivative be expressed as a sum? and I don't see how they use the chain rule, because first you take $D_{n+a}$ and then $D_j$?

o wait hm

Can you prove the negative area on $[\pi,2\pi]$ is less in absolute value than the positive area on $[0,\pi]$?

That's precisely the chain rule, @Sha :)

6:17 PM

$\sin t \geq 0$ in $[0, \pi]$, and $t + 1 \geq 1 > 0$ so $\sin t / (t + 1) > 0$ in $[0,\pi]$

You're missing parens.

But how do we prove what I just asked?

uhh

We need precise estimates.

well i was thinking

@Ted oooh I think I see it now!

6:19 PM

@Sha: I find that it helps to write out the chain rule in matrix form.

I didn't realise that $D_jg^\alpha(x)$ would yield the identity matrix for $j\leq m$

Oh, right.

$\sin t = - \sin (t + \pi)$

BTW, some of my video lectures on this stuff might help you ...

me? @ted

6:19 PM

sin is a periodic function, so the values in [0, pi] are upto absolute values not different from [pi, 2pi]. What changes is the denominator...

Sure, @Meow. But I asked absolute value, right?

Yes, @Sha.

but $t + 1 < $t + \pi + 1$

SSShhh, Balarka.

thats waht i was gonna say you goob

ah right, well thanks for the tip:p

6:20 PM

I didn't reveal much, and Meow got it

its the same thing the denominator is just larger

OK, that'll do it, or you can do something more global, Meow.

For example, $\int_0^\pi \frac{\sin t}{t+1}\,dt \ge \int_0^\pi \frac{\sin t}{\pi +1}\,dt$. Indeed, $>$.

is that supposed to be $\pi$ to $2\pi$

Nope.

oh didt read the $\pi$

6:21 PM

I'm gonna leave you to do the next stuff.

So how do we get it for all $x>0$?

also why just $\pi + 1$ in the denominator

Why just?

$t \leq \pi$ on that domain of integral.

oh

You ninja that into the denominator

(Cool trick @Ted)

6:24 PM

well

Meow's getting to the part of the book where I wrote more of the problems :P

@TedShifrin s'il vous plait est ce que vous connaissez ça : $\ker(g)\subset\ker(h)$ donc il existe $\beta\in\mathbb{R}, h=\beta g$

you also have that $\int_{2\pi}^{3\pi} (\sin t) / (t + 1) dt > -\int_{3\pi}^{4\pi} (\sin t) / (t + 1) dt$

Tout à fait pas en générale, Vrouvrou.

@Meow: Presumably, you can write a rigorous proof now?

and $\int_{2n\pi}^{(2n+1)\pi} (\sin t) / (t + 1) dt > -\int_{(2n+1)\pi}^{2(n+1)\pi} (\sin t) / (t + 1) dt$

fori nteger $n$

6:26 PM

Modulo typos.

well it seems like

regardez ici @TedShifrin s'il vous plait, ya juste une erreur c'est $\ker(g)\subset \ker(f-\tilde{f})$

we'll always have $I > 0$ (integral $I$) when $x = 2\pi n$ for int $n$

because of the uhh

integral sum theorem or what youd call it

Right, @Meow.

Justement, @Vrouvrou. C'est que ce sont des fonctions linéaires $X\to\Bbb R$.

@TedShifrin comment? je ne comprends pas

6:32 PM

$\dim X/\ker(g) = 1$.

Si tu prends $x\notin\ker(g)$, ça te donne un $\lambda$. Pourquoi savons-nous que ce $\lambda$ marche pour tout $x$?

If $M$ is a topological manifold, then thinking of it as embedded as the diagonal of $M \times M$, you could study a small neighborhood $U$ of the diagonal and there is a natural retraction $p : U \to M$ which is actually locally trivial in the sense that $p$ is topologically conjugate to the projection $V_x \times \Bbb R^n \to V_x$ for any $x \in M$. I think this is what they call the tangent microbundle of $M$.

I've never studied microbundles, Balarka.

then i notice that if you have $x < \pi$, $\int_{(2n + 1)\pi}^{(2n + 1)\pi + x} f$ must be greater than $\int_{(2n+1)\pi}^{2(n+1)\pi} f$

notice i got lazy and just wrote $f$

@TedShifrin pour chaque?

@Leaky: I'm rusty on my idiomatic French math. But note that in English we can say both "for all" and "for every."

6:38 PM

@TedShifrin Me neither. I'm just guessing, at this point. I think the point is this guy has a "classifying map" $f : M \to B \text{Homeo}_0(\Bbb R^n)$ where $\text{Homeo}_0(\Bbb R^n)$ is the space of germs of homeomorphisms at $0$. I'm guessing $M$ has a smooth structure iff $f$ lifts to a map $\tilde{f} : M \to B \text{Diffeo}_0(\Bbb R^n)$.

@Vrouvrou which expression is used in french?

and for $x < \pi$, $\int_{2n\pi}^{(2n + x)\pi}f > 0$

what i wrote was true i think but this is more important

so do you want me to synthesize everything i wrote so far on a piece of paper and send it to you, or do you see how it all conects

I think you understand the right approach, @Meow. There are more interesting problems as you continue.

Howdy, DogAteMy.

@TedShifrin pourquoi dim (X\setminus \ker (g))=1$ ? je ne comprends pas votre raisonnement

Pas $\setminus$ !!

Il existe un vecteur $x_0\in X$ tel que pour chaque $x\in X$ on ait $x=cx_0 + v$ pour $c\in\Bbb R$ et $v\in\ker(g)$.

6:55 PM

je ne connais pas cette conséquence

Il faut que tu l'établisses.

Je dois m'en aller ... See you all later.

Does the archimedian property hold in any topologly ? if not does it hold in any metrisizable topology?

thats definitely how u spell that yes...

Tobias Kildetoft

@Faust The archimedian property does not make sense with just a topology

Ich verstehe Französisch nicht.

what about one with a metric on it

Tobias Kildetoft

not sure what sort of formulation of it you could do with just a metric

6:59 PM

hmm i see

Anderson Felipe Viveiros

7:16 PM

If $f:U\subset \Bbb R^m\to \Bbb R^n$ is continuous, $U$ open, and $\lim_{x\to x_0}f(x)$ does exist for all $x_0\in \overline{U}$ (closure of $U$), then is the natural extension $\overline{f}:\overline{U}\to \Bbb R^n$ automatically continuous?

Such limits exist with $x\to x_0$ with $x\in U$. But for $\overline{f}$, I need $x$ going through $\overline{U}$, which has more points (theoretically) than $U$... That's my doubt

i think the answer is probally yes, only because of the limit condition of f(x) you state

but you dont know alot about what you land in it still may not be closed

unless u is also bounded

but it is not so useful i think because the way you have seem to of stated it is if U is closed the map on the closure of U is also continuous

but my understanding of analysis is thrid rate so i could be completely wrong

7:35 PM

@Daminark any good troll's today?

I never troll!

philmcole

8:26 PM

Just to reassure myself a zero can have a multiplicity of $n \ge 1$ but $n=0$ doesn't make sense?!

8:56 PM

context?

zero could make sense it could just mean that its not a root

Trey

9:12 PM

how can I find the radius of convergence of a serie?

David Reed

ratio test or root test

be careful the way you apply it though, as it winds up being inverted from normal (i.e. you wind up with $|a_k/a_{k+1}|$ instead of $|a_{k+1}/a_k|$

philmcole

@Faust The multiplicity of a root of a polynomial. So a root with multiplicity 0 is just no root you mean?

@Trey Wikipedia describes it quite well: en.wikipedia.org/wiki/…

9:36 PM

@philmcole it depends how you define it but if it is a root its multiplicity must be at least 1

unless your in some strange algerbra space like Z(x)

philmcole

9:57 PM

@Faust Thanks, that's what I thought also.

Q: Are functions converging to the same value locally equivalent?

10:07 PM

0

Suppose we have two functions $g(x)$ and $f(x)$ such that the left hand limit as x approaches 1 exists. Let us now define a continuous sequence of g or f as a function $h(x) = f(x + (1-x)a)$ where $a \in [0,1)$ does there necessarily exist continuous sequences for f and g that are equal on $[0,...

Hi, Determined all bases with : $\text{gcd}(10101,1001001)>1$.

where did all you rep go?

@Faust : do you ask that to me ?

no of the turtle

what base are you numbers given in?

there are the base 2,3,4... example : 11=3 in base 2

10:18 PM

thats why im slightly confused by the question

is 10101 given in base 10 or is that the representation in every base?

find the bases b, with gcd(10101,1001001)>1 where 10101 and 1001001 is two number in base b.

interesting question i have no idea, other than b< 10101

try writting out the prime power decomposition of each number in base 10

its likely the largest prime will give you an upper bound on the base

of the smaller number

10:57 PM

Hey @Akiva

Hi

You fancy some classical music?

Josh A - Another One

►

youtube.com/watch?v=c2_VdzHQWNs A Beethoven classic

Naturally Ted comes when I'm in shitposting mode. Hello!

the answer : artofproblemsolving.com/community/…

Ted might just disappear for good.

11:00 PM

Hey @Ted

Rehi.

I'll have to go to bed in a minute or so

"or so"

:P

Let's see how long that'll be :P

(Really though do go to bed at the right time, I'm a shit influence wrt sleep schedules)

11:02 PM

I think that's a minute. Cya!

Oh wow aight, peace!

So @Ted how've things been going with the aops?

If I ping him, what are the odds he'll come back

An 80% chance

11:17 PM

@MatheinBoulomenos what's the point of category theory, seriously?

Is it possible to hit you in the head in such a way that you (a) go to sleep, or rather get knocked unconscious, and (b) do not suffer lasting brain damage? @BalarkaSen

@LeakyNun No, the whole point is that the point isn't the points, it's the arrows between the points

interesting

11:32 PM

@BalarkaSen 80% of the times it works 100% of the times

howdy

If $\varphi : A \to B$ is a homomorphism between rings with unity, must a prime ideal in $A$ be the preimage of a prime ideal in $B$?

I think I am about to construct a number set superior to the real numbers.

Hey @PVAL, how's it going?

well sorta superior

depends on context

11:35 PM

Alright still getting back into the swing of things.

@Daminark how are you doing?

@LeakyNun Why can't you figure this out yourself?

@TheGreatDuck I'm doing alright, how about you?

@PVAL after winter break you mean?

Yeah

@PVAL-inactive I already can't figure that out, and now I need to figure out why I can't figure that out?

no, I can't figure out the reason behind my failure to figure that out myself

11:40 PM

@TheGreatDuck I can think of a few that are pretty cool, like the dual numbers, complex numbers, and surreals (though that's technically too big to be a set)

PearlSek

I found a sum of sines that makes a weird fractal

What are some examples of homomorphisms you know?

Nice @PearlSek

PearlSek

$$

@PVAL-inactive Z/mZ to Z/nZ, Z to Q, Z^2 to Z, Q to Q[sqrt2], R to C, C to R, etc...

11:41 PM

What's your grading/teaching schedule like now? Hopefully better than last semester

PearlSek

I suppose Z to Q provides the counterexample

Q: Are functions converging to the same value locally equivalent?

@AkivaWeinberger how about the set of all limits convergent and divergent?

their value

1

Suppose we have two functions $g(x)$ and $f(x)$ such that the left hand limit as x approaches 1 exists for both and they are both equal. Let us now define a continuous sequence of g or f as a function $h(x) = f(x + (1-x)a)$ where $a \in [0,1)$ Does there necessarily exist continuous sequences f...

real-analysis limits

well I have a construction assuming this question is correct.

@Daminark Just broke 50 pages on my paper.

I will be past 60 by next week.

@TheGreatDuck What, like having $\lim_{x\to\infty}x$ and $\lim_{x\to\infty}x+1$ as formally distinct values?

Trivial homomorphism

11:44 PM

@AlessandroCodenotti nice

@AkivaWeinberger bounded limits

I don't think theres a ring homomorphism C to R, but the inclusion Z into Q surely works.

like sin(x) as x goes to infinity

Duck: nifty

There's a lot of examples if you think about evaluation maps of polynomial rings F[x] \to F and things similar to that.

11:45 PM

it is needed in studies of time travel paradoxes, hypercomputation, and my step function integral algorithm

the last only makes sense in context of my paper

which is currently not on my profile due to trolls

XD

they tried to claim a pure math paper was a nazi manifesto

@Daminark I have no teaching responsibilities.

@TheGreatDuck Hm. Do you have an order on this set? If so, would $\lim\sin(x)$ be larger or smaller than zero?

which puts me under a lot of pressure to catch up.

@AkivaWeinberger I suspect it is multidimensional

Would $\lim(\sin(x)+1/x)$ be equivalent to it?

11:47 PM

Ah, well good luck!

@AkivaWeinberger depends on whether my questions proposed equivalency works

if so then one would prove wether or not that is equal

not assert it

@TheGreatDuck Actually, I think I understand your question, and if so the answer is "no"

Take $f(x)=(x-1)^2$ and $g(x)=-(x-1)^2$

oh

O.O

Then $h_f(x)=((1-a)(x-1))^2$ I think

FUUUUUUUUUU!

11:51 PM

The best you can say is that if $\varphi:A\to B$ is an injective homomorphism then you can think of $B$ as an $A$-algebra, if this algebra is integral then it satisfies the lying over property, that is if $p$ is prime in $A$ then $p=q\cap A$ for some prime ideal $q$ of $B$ (identifying $\varphi(A)$ and $A$)

well that sucks

@AkivaWeinberger last time I told him that he didn't define it formally, he threw a fit and said I'm rude to think he doesn't know what he's talking about

There are a few similar in flavour results that are called the going up and going down theorems

and that wasn't the last time he threw a fit when people told him he's wrong

so I don't see any point trying to convince him to formalize his definition

Gib rep plox ^_^

(Answer posted)