Mathematics - 2016-06-02 (page 1 of 2)

user19405892

00:23

hi

Abhishek Bhatia

00:37

@Axoren This makes concept of random variables very clear. Can you please elucidate more on the expectation function.

Forever Mozart

01:18

the site died

user3502615

what theorem states that

if $\gamma$ and $\gamma_0$ are closed curves which satisfy $\gamma \sim_G \gamma_0$ for some simple connected subset $G$ which a function $f$ is holomorphic, then $$\oint_\gamma f(z) dz= \oint_{\gamma_0}f(z)dz$$

Akiva Weinberger

02:08

@MikeMiller Nice to know you're safe and sound

Mike Miller

02:29

Thanks.

Akiva Weinberger

02:41

Traditionally, when a Jew hears that someone has died, they're supposed to say baruch dayan emet ("Blessed is the True Judge", i.e., God). It's essentially the Hebrew version of "RIP". I always found that interesting.

So, um, baruch dayan emet.

Mike Miller

I didn't know the victim, but I appreciate the thought. It all feels a bit distant to me, since I'm around 1500 miles away. I found out about it because it was on TV at lunch.

There is a picture of the police having my roommate (and others) kneeling down while patrolling them with assault rifles.

Semiclassical

it's both personal and distant at the same time, I suppose. you can place it into context of what you know, but you weren't actually around for it.

user243301

07:49

All users, it is possible determine the max(distances between barycentres) of two triangles with non empty intersection of their interior (I say the interior set of each triangles) having those triangles the same perimeter?

user 1618033

08:22

@Semiclassical see here

0

A: An interesting inequality $\sum_{k=1}^n \frac{1}{n+k}<\frac{\sqrt{2}}{2}, \ n\ge1$

(Alternatively) By using Root-Mean Square-Arithmetic Mean-Geometric we get that $$\sum_{k=1}^{n}\frac{1}{n+k}<\sqrt{n\sum_{k=1}^{n}\frac{1}{(n+k)^2}}<\sqrt{n\sum_{k=1}^{n}\frac{1}{(n+k)(n+k-1)}}=\frac{1}{\sqrt{2}}.$$

Agawa001

08:38

@ForeverMozart consider all se corners dead, mathematics.se will never die

wb m. robjohn

user 1618033

@Agawa001 even the universe itself is going to die.

Agawa001

@user1618033 the universe is built on maths, and when it dies, mathematics dictates the beginning of another universe cycle

user 1618033

@Agawa001 I wouldn't know to tell that, all is too mysterious about the beginning and the end of the universe.

Agawa001

08:53

besides , it is summer, the hidden reason behind the low activity, people are less interested in posing math questions in summer-holydays, you ll see the difference after 3 months

user 1618033

Well, it's not bad to go out and get some tan. :D Even on the beach or in other parts, doing some math is enjoyable.

Agawa001

yes of course, but i hate sunlight !

user 1618033

Why?

Agawa001

every normal person sees sunlight disturbing , no ? it perturbs the vision, makes u sweat, and i dont need to look as choco-vanilla ice-cream

user 1618033

:-)))

It's said, according to some studies, that a little bit of sun exposure is linked to a better mood.

Agawa001

09:04

also i cant think properly under sun-heat, thus i must cut my hair once upon 15 days.

@user1618033 that is not here, because in africa, sun attacks (reminds me of the movie mars attacks)

user 1618033

imdb.com/title/tt0116996

:D

Agawa001

funniest movie i have seen, even better than star wars

user 1618033

@Agawa001 star wars? I couldn't watch the last one more than 10 min or so.

Agawa001

i think star wars is overrated

user 1618033

Yeap.

Agawa001

09:15

@user1618033 when $\frac{1}{n+k}$ is smaller than $n*(\frac{1}{n+k})^2$ ?

user 1618033

@Agawa001 take a look here to see what I applied

artofproblemsolving.com/wiki/…

@Agawa001 not sure what your point is

When $$\frac{1}{n+k}<\frac{n}{(n+k)^2}$$ holds?

First note that it can be reduced to

$$1<\frac{n}{n+k}$$

Agawa001

i should have said $\sqrt{n}\frac{1}{n+k}$

user 1618033

Since $k$ is a positive integer one gets a contradiction.

Agawa001

@user1618033 ok ok i misunderstood , we know that $a+b+c...>\sqrt{a^2+b^2+c^2...}$ how come the second end multiplied by $\sqrt{n}$ becomes bigger ?

(reading schwarz ineq meanwhile)

user 1618033

09:30

@Agawa001 In the link I gave you you may find 2 proofs for the inequality I used.

Agawa001

ok

user 1618033

@Agawa001 this is what I used ^^^

$x_1=1/(n+1), x_2=1/(n+2), ..., x_n=1/(n+n)$

Agawa001

now it is clearer

user 1618033

If you bring $n$ from the left side to the right side, then you have $n$ *$1/\sqrt(n)$

and since $n=\sqrt{n}*\sqrt{n}$, you get $\sqrt{n}$ in front

Agawa001

yes yes i implied that

@user1618033 how is yur book going ?

user 1618033

09:39

@Agawa001 All is fine, thanks, except that there is a delay that does not depend on me. All that is related to me is perfectly as planned.

Agawa001

good to hear

that is the result of the very extremely hard work isnt it ?

user 1618033

@Agawa001 Sometimes people expect you to tell them a story about a fascinating mind that allowed you to get some results, but things are pretty simple actually: very hard work, very hard work, and so on.

@Agawa001 very hard work is my strongest trait.

Agawa001

i ll be off this server too, i need it to do something non virtually productive

user 1618033

OK

Mary Star

09:55

Hello!!

We have that $R$ is a commutative ring.

Suppose that $0\rightarrow A\rightarrow B\overset{f}{\rightarrow} C\rightarrow 0$ and $0\rightarrow C\overset{g}\rightarrow D\rightarrow E\rightarrow 0$ are exact sequences.

Then $0\rightarrow A\rightarrow B\overset{gf}{\rightarrow} D\rightarrow E\rightarrow 0$ is also exact.

I want to show that each exact sequence can be arised by short exact sequences as above.

Can we show that by induction?

user125535

11:09

anyone here

Balarka Sen

Hi @Huy

Huy

hi @BalarkaSen

Balarka Sen

You were wrong that I wouldn't like stat :P

Huy

=(

good for you I guess :)

user147690

12:02

Who are the AG people of this room?

user147690

We don't have an AG person do we?

Mike Miller

there are no researchers in algebraic geometry here, no, but I'm guessing that's not what you're looking for

user147690

It isn't

user147690

Do you know much AG @MikeMiller?

Mike Miller

nope

"ask; don't ask to ask"

user147690

12:12

I don't have a singular question to ask, and there are no AG people so I can't ask anything :P

Balarka Sen

I know the basics. But I probably wouldn't be able to answer whatever you want to ask.

user147690

I wanted to ask a bunch of questions, some about the canonical divisor

Balarka Sen

OK, I don't know anything about divisors.

Mike Miller

Yes, but someone might still know how to answer your question. In any case this conversation was of no value; that's why one asks the question before worrying about whether or not someone is going to answer.

I'm gonna go.

Balarka Sen

Byes.

user147690

12:15

Cya @MikeMiller

user147690

The result is that I can't ask any questions about this level of AG :P, so of value for future me

user147690

Interestingly Weil divisors have been mentioned once in this chat before.

user147690

Frank and Ted might be the guys to go to

Semiclassical

12:46

@user1618033 that's a nice way. so all one needs is either AM-GM or Cauchy-Schwartz as a tool

Danu

13:00

@MikeMiller I've got an elementary proof of that thing I asked about some days ago. Wanna hear it?

Mike Miller

13:16

I don't know what that is, but sure.

Balarka Sen

The word problems in my physics textbook are ambiguous as hell.

Danu

13:31

@MikeMiller Distinguishing those two Hopf bundles

You gave some ideas for a proof, but this is more elementary/working directly with the definitions

and it in the end derives a contradiction with all maps $\pi_1(S^3)\to \pi_1(S^1)$ being trivial (as promised)

Assume you have an isomorphism $\phi$. Because it preserves fibers, we have $\phi(w)=w\mu(w)$ for some $\mu(w)\in S^1$. This defines a map $\mu: S^3\to S^1$ by $w\mapsto \mu(w)=w^{-1}\phi(w)$

It satisfies $\mu(w\lambda)=\mu(w)\lambda^{-2}$

Fix $w_0\in S^3$ and define $F:S^1\to S^3\to S^1$ by $\lambda\mapsto w_0\lambda\mapsto \mu(w_0)\lambda=\mu(w_0)\lambda^{-2}$

This is a self-map of $S^1$ with degree -2, but it factors through $S^3$. Contradiction. @MikeMiller

^The thing this is showing is that the actions $S^3\curvearrowleft S^1$ given by $w\mapsto w\lambda^{\pm 1}$ do not yield equivalent $S^1$-bundles over $S^2$.

Semiclassical

@BalarkaSen Example?

Hiroto Takahashi

Hi everyone

Someone know what is the original paper of the Smith Normal Form?

Semiclassical

13:46

if you trust wikipedia's refs on their page on the Smith normal form, then this is the relevant historical paper

they also link to here for Smith's collected works, citing pages 367-409

Hiroto Takahashi

@Semiclassical Thanks!

Semiclassical

np

Balarka Sen

@Semiclassical Here's a rather silly example off the top of my head. A steamer goes north with velocity $v$. Smoke is coming out of the chimney making an angle of 30 degrees (in the opposite direction) with east. If wind is flowing from west to east, what's its velocity? I initially thought the steamer was moving with velocity $u$ and hence smoke was moving with velocity $-u$, in which case I have $u + w = v$ and $u - w$ becomes the velocity of smoke ($w$ being velocity of wind).

But after calculation I got an answer which doesn't match my textbook so I realized that the question assumes that wind has no effect on the steamer (perhaps because it's moving on water, not wind) :/

Semiclassical

well, they tell you it's going north with velocity $v$, not that it's being pushed in that direction

if you were to take the wind into account as a force acting on the steamer, then what that'd imply is that the steamer is in fact pushing in a direction that partially opposes the wind

but that's inferred precisely from the fact that they tell you what direction it in fact moves

Balarka Sen

I am not good at interpreting these inherent assumptions.

Semiclassical

13:56

Fair enough. I think the key point is that, if they tell you it moves north with velocity $v$, that it means for you assume that it's moving with that as constant velocity.

which is to say, take it quite literally

But I have a hard time divorcing myself from how I'd unpack that problem, which is itself founded upon prior experience.

Mike Miller

@Danu: Oh, that's clear. My proof from the other day is an unnaturally complicated way of saying that.

Danu

@MikeMiller OK :)

Balarka Sen

@Semiclassical But in some problems, they do not actually mention that [steamer/aeroplane/etc] is pushed by the wind (as in, do not explicitly use the word pushed) while they mean it. How do I figure out what it's meant, then?

e.g., if the steamer was an aeroplane I am pretty sure I should have interpreted the problem as I did initially in this case.

Semiclassical

no, you still wouldn't. you're told that it has a certain velocity

now, I think that it should have said constant velocity

Balarka Sen

@Semiclassical Hm.

Semiclassical

14:01

Imagine it like this. You're watching this scene play out, and you see the steamer moving north at a velocity $v$.

Now, imagine that you instead saw an aeroplane (probably a glider, heh) moving north at that same velocity.

Mike Miller

@Danu: Sorry, by "that's clear" I meant "that's a very clear proof", not "that's obvious".

Danu

@MikeMiller I realized, don't worry :)

Semiclassical

Certainly the setting has changed, and so presumably the forces experienced by the steamer are different than the aeroplane.

Balarka Sen

Mhm

Semiclassical

but the motion is not ambiguous

Here's how I'd phrase the problem, just to be clear.

Akiva Weinberger

14:06

A steamer is a type of boat, right?

Semiclassical

"From a helicopter hovering above a river, you see a steamer moving north at a constant velocity of $v$. The smoke coming out of the chimney makes an angle of 30 degrees (in the opposite direction) with east. If you assume that the wind is flowing from west to east, what wind velocity would you report?"

right @akiva

Balarka Sen

@Semiclassical Oh, that's an interesting phrasing.

Semiclassical

I'd say it's the same problem, but written in a way that is more contextual rather than abstract.

I mean, one should be able to make it abstract/symbolic. But to avoid ambiguity, it helps to be concrete in formulation.

Balarka Sen

Right, that makes sense.

Semiclassical

Of course, that relies on the person writing the problem.

The issue isn't that physics word problems are inherently ambiguous, but rather than we're not always as good at writing problems as we think.

Balarka Sen

14:12

heh

Evinda

If $u \in W^{3,p}(\mathbb{R}^{+})$ how can we construct the catoptric extension $Eu$ of $u$ in $\mathbb{R}$ (reflection) such that $Eu \in W^{3,p}(\mathbb{R})$ ?

Do you have an idea?

Semiclassical

@Balarka So I am sympathetic to what you're bothered by.

Balarka Sen

It's probably just that I don't know how to think like a physics person would do. I am faced by similar although I must say less bothersome problems when I try to learn something new even in math. The way of thinking is probably what I need to get used to.

Semiclassical

Sure. It's the old f(x,y)=x^2+y^2 thing.

Balarka Sen

=P

Semiclassical

14:17

which I did actually pick up from one of our profs talking about physics education research, so that brings it full circle

The use of the word 'context' earlier was also an instance of that, actually. "Context-rich problems" are a big thing in the physics education research here.

A word problem that's not contextual kind've misses the point.

One way to take advantage of that, perhaps, is to take a problem that lacks context and imagine how you'd add it.

i.e. what I did in saying that you observe it moving north, rather than saying "it moves north" in passive fashion.

Balarka Sen

Right

bolbteppa

Think I finally understand how to explain the origins of Riemann-Roch to a child, any thoughts/ideas on/about what I'm about to paste(?):

Danu

^I'm calling BS :P

Evinda

Hi @bolbteppa
Are you familiar with sobolev spaces?

bolbteppa

A plane curve $F(x,y) = 0$ of degree $n$ has $1 + 2 + \dots + n + (n+1) = \frac{1}{2}(n+1)(n+2)$ coefficients. Since $F$ generates the same curve as $cF(x,y) = 0$, there are $\frac{1}{2}(n+1)(n+2) - 1 = \frac{1}{2}n(n+3)$ independent coefficients. Thus $ \frac{1}{2}n(n+3)$ points in the plane determine an algebraic curve. If the curve has $d$ double points, then assume a curve $G$ of degree $n$ passes through these $d$ double points.

The intersections of our two $n$ degree curves $F$ & $G$ occurs, by Bezout, at $n^2$ points, but equating $F(x,y) = G(x,y) = 0$ we can eliminate one further constant in $H(x,y) = F(x,y) - G(x,y) = 0$ so that $H$ is determined by $\frac{1}{2}n(n+3) - 1$ points. In other words, out of $n^2$ points, $\frac{1}{2}n(n+3) - 1$ determine the curve, while $\frac{1}{2}(n-1)(n-2)$ are left dependent, since $n^2 = \frac{1}{2}n(n+3) - 1 + \frac{1}{2}(n-1)(n-2)$.

Therefore a curve can have, at most, $\frac{1}{2}(n-1)(n-2)$ double points. Thus, for any curve with $d$ double points, define the genus/deficiency $g = \frac{1}{2}(n-1)(n-2) - d$ as the difference between the number of independent points and the number of double points. Riemann-Roch arises from the idea that this number remains invariant under birational transformations.

Danu

14:25

"child" :P

bolbteppa

Okay apart from Bezout :p

Danu

Your kid will crack under the pressure ;)

bolbteppa

@Evinda I can only really wave my hands about Sobolev spaces, I'll do my best if you have a basic question

Evinda

0

Q: Construct extension of function

If $u \in W^{3,p}(\mathbb{R}^{+})$ how can we construct the catoptric extension $\overline{u}$ of $u$ in $\mathbb{R}$ (reflection) such that $\overline{u} \in W^{3,p}(\mathbb{R})$ ?

This is my question

bolbteppa

runs far away and curls up into a ball

Balarka Sen

14:29

Bezout is quite intuitive.

bolbteppa

Honestly I don't even believe Bezout, it seems like a trick to make something uncomfortable true so we get what we want, being stupid but can't get over that yet :p

Semiclassical

Bezout is a nice example of: That which is 'usually' true for curves in the plane is 'always true' for curves in the projective plane.

(that's a lazy formulation, yes. bah humbug)

Balarka Sen

The literal Bezout is almost never true for curves in affine plane.

Semiclassical

Yeah, 'usually true' is not the right way to put it :(

there's a general principle there, I think, but I don't know how to say it

Balarka Sen

(e.g., since when did two distinct circles ever intersect at 4 points in the plane?)

bolbteppa

14:35

The difference between a mathematician and a physicist is that a physicist will only understand Bezout using flashlights :p

Balarka Sen

You can reformulate as "given degree p and q curves on the plane, # of intersection pts is always <= pq".

Semiclassical

yeah. i mean, a generic conic + a circle can work

So going from the affine plane to the projective plane promotes that upper bound into an equality.

Balarka Sen

Right.

bolbteppa

In Ueno's Algebraic Geometry he really nicely motivates why you should do algebraic geometry using commutative (quotient) rings with the whole flashlight ellipse = hyperbola = circle idea

Semiclassical

I think something like that last statement is what I want as a heuristic

No idea how to make that rigorous, though.

Though I guess one cheap version is: "Counting is easier in the projective plane than the affine plane." :p

Balarka Sen

14:40

Maybe you want to prove that pick two curves generically from the moduli space of all curves on the plane of degree p and q respectively they intersect in pq points. That seems too good to be true, and I don't want to believe it.

Semiclassical

though i suppose even that has some counterexample, sigh.

Balarka Sen

*picking

By generically I mean you pick from a Zariski open set (which are huge)

@Semiclassical The key point is that intersection theory on projective spaces is easier as they are compact. Affine spaces are noncompact.

Semiclassical

Good point.

Balarka Sen

In noncompact world you can always do something bad "off in the infinity" to make intersections bad and transversality fail horribly.

bolbteppa

Picture two concentric non-intersecting circles in the plane around the origin, you looking at them amounts to aiming a flashlight right at the center of the plane. If you now tilt your flashlight to an angle, you can intuitively see that eventually the circles will intersect at 4 points in some tilted plane in space, I know Bezout is basically just generalizing this idea, but I don't see it for crazier curves.

Balarka Sen

14:45

I think there is a version using cohomology with compact supports through.

Semiclassical

'Noncompact = that which can go wrong will go wrong' :p

Mike Miller

@BalarkaSen how could that possibly be true? cohomology with compact supports is zero outside of dimension $n$ on $\Bbb A^n$.

i guess $2n$ if we're working complex

Balarka Sen

I dunno, I know nothing about that. I vaguely remember I was told one does intersection theory in the noncompact world with cohomology with compact support.

Mike Miller

sure, that's trueish, but there's still no nontrivial intersection theory in $\Bbb A^n$

Balarka Sen

I can believe that.

A^n is contractible, so topologically there should be nothing. I don't know if one can do something algebraically.

Mike Miller

14:49

n

bolbteppa

Okay I can believe Bezout now I think :p

Balarka Sen

Intersection of a bundle of $m$ disjoint lines transversely with a bundle of $n$ disjoint lines is $mn$ many points. That's how I visualize Bezout.

Ops, I should get back to work instead of talking about Bezout now.

Mike Miller

:)

bolbteppa

I think that bundle of lines comes from the flashlight :p

Steven Stewart-Gallus

16:38

Under Coulomb's law $\vec{E} = \int_C \frac{k \rho}{r^2} \hat{r} \mathrm{d} V $ but I'm having trouble figuring out the functional derivative with respect to charge.

skill patrol

16:59

wb @robjohn

Owatch

17:22

I've got a question.

Would you find it acceptable that a hypothetical calculator refuses to accept input like: $x^{-1}$

But must take it as: $x^{(-1)}$

(But will accept non-negative powers without parentheses fine: $x^{1}$)

robjohn

@skillpatrol thanks. It was a nice long weekend

skill patrol

:-)

Agawa001

17:54

@skillpatrol hey hey hey

skill patrol

Hey there @Agawa001

Evinda

18:52

Why does the equality hold? Why isn't it as follows?

$\int_{\mathbb{R}^n} | D|u|^{\gamma}| dx= \int_{\mathbb{R}^n}\gamma |D |u|^{\gamma-1}|$

Agawa001

@Evinda still on exams ?

Evinda

I haven't written yet my first exam @Agawa001
I am studying for this

What's with you?

Agawa001

@Evinda no, it had been one year long since my last studies

Evinda

@Agawa001 Have you graduated?

Agawa001

a white year

wait do they say this in english ?

Evinda

18:58

I don't know... What does it mean? @Agawa001
Have you finished your master?

Agawa001

a white year, like a legal suspension from studies

Evinda

Why?

Agawa001

a voluntary i mean

because i have got tired

i wasnt been doing my desired field, so i held it for a year

Semiclassical

This is a pretty dumb question, but i like this answer: math.stackexchange.com/a/1810038/137524

Agawa001

it sucks when u study things u dislike, i prefer wasting my time answering SE questions

@Semiclassical do u mean the way round ?

Semiclassical

19:05

nope. dumb question, fun answer.

though that goes to taste, of course

@Agawa001 I can't disagree with that statement without being entirely hypocritical.

Balarka Sen

@Semiclassical Curious to hear how you'd interpret this problem. An object is dropped downwards with velocity 14 m/s and it hits the ground after 2 seconds. It then bounces off the ground with the same velocity it hit it with and goes higher up perpendicular to the ground. A second later an object is dropped again from the same height as the previous. In what time will both the objects hit each other?

There is an ambiguity regarding what "a second later" means. A second after the first object hits the ground, or a second after the first object is dropped?

Agawa001

@Semiclassical this question isnt dump

Semiclassical

Well, considering there's no way to mathematically prove that a given finite sequence of digits is coming from a truly random distribution...

All you can prove is that certain things are exceedingly unlikely. That provides a practical standard, but one needs to specify such a threshold.

Balarka Sen

@Semiclassical There is a probability for it to appear though.

Agawa001

this depends which criterion decides a randomness of a fractional digits of a number

Balarka Sen

19:11

Sometimes the probability is closer to 1, in which case it's more likely, sometimes it's closed to 0, in which case it's less so.

Semiclassical

Sure. But it's a probability. It's like testing a coin for bias by flipping it a bunch of times. You don't use such a test to 'prove' the coin is biased. All you do is assert that there's a very low probability that, were it fair, that a certain amount of bias would occur.

Balarka Sen

Sure, I agree.

Semiclassical

Hence why the question as posed is a bit silly. (I shouldn't have said dumb.)

@BalarkaSen I'd interpret that as "a second after the first object hits the ground."

Balarka Sen

There are probably statistical ways to decide if a variable is indeed a random variable which I do not know though. But I am aware that there is no mathematical way to decide it.

"given large enough data", etc.

Semiclassical

It's the difference between "can I conclude it with absolute certainty" and "would it be reasonable for me to accept it"

Balarka Sen

19:16

@Semiclassical That's what I thought.

The book does it with "a second after the first object is dropped" :/

Semiclassical

pffffft

That's just badly written.

Balarka Sen

I know right.

Agawa001

@Semiclassical hypothetically pi is forseen to be normal in base 10 which doesnt imply binary base

Semiclassical

sure. but absent a proof of either, the only way to check it is statistically

Balarka Sen

Curious: how does one actually check - statistically - if a data comes from a random variable?

Semiclassical

19:22

well, since one is working with $\pi$ in base 2, it's basically a sequence of coin flips

Balarka Sen

By data I mean an infinite set. So you can truncate it to a set of cardinality n and do something statistical there and say that happens as limit goes to n or something.

Semiclassical

so any method you'd use to test a coin for bias would apply.

Clement C.

Is the point system in Math.SE currently down?

Balarka Sen

@Semiclassical Hm, good point. And I can test coin for bias by just looking at # of heads/# of data set as # of data set --> \infty.

If that's 1/2, I am good.

Clement C.

This will sound very petty (warning), but I got upvoted 7 times for a question, which was rather surprising to me (math.stackexchange.com/questions/1809965/…); yet the reputation tab and score give me +31 for that... which seems like a score (+30+1) I had never encountered

Semiclassical

19:23

right. or looking for runs of heads/tails/etc.

any downvotes? @ClementC.

Clement C.

Nope, apparently none

Semiclassical

weird.

could be that you downvoted an answer that's since been removed

Clement C.

Oh, true.

Semiclassical

or that a downvote was removed, etc.

still more than a little weird

Agawa001

@Semiclassical exactly

but that should sum up to 36

Clement C.

19:26

Oh, well. Whatever.

Semiclassical

The not-an-answer I'd give to the randomness question, btw, would go like this.

Clement C.

(thanks, though)

Semiclassical

In order to prove that the binary sequence generated from $\pi$ is random, an infinite sequence of digits must be tested so that the probability goes to zero or one.

But this takes an infinite amount of time. Hence you are, in fact, already dead and in mathematician hell.

Evinda

@Agawa001 Aha ok

Agawa001

hmmmm

An interesting new method was recently proposed by David Bailey, Peter Borwein and Simon Plouffe. It can compute the Nth hexadecimal digit of Pi efficiently without the previous N-1 digits. The method is based on the formula:

pi = sum_(i = 0)^oo (1 16^i) ((4 8i + 1) - (2 8i + 4) - (1 8i + 5) - (1 8i + 6))

in O(N) time and O(log N) space. (See references.)

The following 160 character C program, written by Dik T. Winter at CWI, computes pi to 800 decimal digits.

int a=10000,b,c=2800,d,e,f[2801],g;main(){for(;b-c;)f[b++]=a/5;

as quoted from cs.uwaterloo.ca/~alopez-o/math-faq/mathtext/node12.html

Semiclassical

19:36

that algorithm is weird

Agawa001

yes C can be both obfuscated and readable for its reader as jscript is

Semiclassical

eh, i didn't even mean at the level of readability. the fact that such a method exists at all is plenty strange.

Evinda

https://i.sstatic.net/kP1be.jpg

Hi @robjohn
Why does the equality hold? Why isn't it as follows?

$\int_{\mathbb{R}^n} | D|u|^{\gamma}| dx= \int_{\mathbb{R}^n}\gamma |D |u|^{\gamma-1}|$

Balarka Sen

yeah I heard of existence of such a method

it is kind of weird

Agawa001

19:58

@BalarkaSen i m in accord with you because since pi digits arent even proven to be infinite, the algorithm must return 0 with extremely big i, i think it doesnt

Semiclassical

What.

The digits of pi are certainly infinite in any base. If they weren't, its expansion would terminate in that base and it would therefore be rational.

user3502615

^

Is it ok if i e-mail a professor in a university which I don't attend? (I'm not in college yet)

I have a question

Agawa001

@Semiclassical no i meant digits arent proven to appear infinitely, so it takes a finite distance for 0 to appear infinitely

Semiclassical

you mean, that it's not known if (for base 10) the digits 0-9 all appear infinitely often in $\pi$?

Balarka Sen

That is certainly believable.

Semiclassical

20:06

anyways, I don't doubt that the algorithm is true. I'm sure it's been proven. It's just that it's weird.

Agawa001

or 1

see here : Pi = 3.1415926.....01001000100001000001... the researchers captured a very long stream of 1 and 0 at the end, where they remarked the other digits barely found

bolbteppa

@Evinda Just the chain rule no?

user3502615

> at the end \

there is no end!

Semiclassical

sure. but that's consistent with the suspicion that pi is in fact a normal number, since that's just one particular sequence of 19 digits

it might feel weird to see it, but it's probably just the law of large numbers in action

Owatch

I've got a problem. I'm using a QR algorithm with the Wilkinson shift to calculate eigenvalues numerically, but when the Shift is zero, (Because one of the eigenvalues is in-fact zero), it cycles and can't find the others. Probably not something I can get help with here, but if anyone's done it before some input would be awesome.

Mike Miller

20:14

morning

Balarka Sen

OK, I gotta get some sleep.

Mike Miller

night

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

later

Obliv

How can I start this proof: $|GL_2(\mathbb{F}_p)|$ is $p^4 - p^3 - p^2 + p$ where $p$ is a prime number? I know that this is a group of $2 \times 2$ matrices from the field $\{\bar{0},\bar{1},\bar{2}...\overline{p-1}\}$

arctic tern

@Obliv better to think of the size as (p^2-1)(p^2-p).

p^2-1 is the number of ways of picking the first column vector, p^2-p counts the ways of picking the second after the first

Obliv

20:18

what do you mean?

oh I get it

But that doesn't give me information of if the determinant is 0. Then it's not in the group

arctic tern

pay closer attention

bolbteppa

Think of it as a space of matrices, use the matrix to count the number of elements

arctic tern

I didn't say p^2 times p^2. I said p^2-1 times p^2-p.

The first column can be any nonzero vector. Hence p^2-1. The second column can be any vector that's not any of the p possible multiples of the first column, hence p^2-p.

Obliv

does that make it impossible for the determinant to be 0?

actually it makes sense yeah

for 2x2 matrices anyway

bolbteppa

What is the definition of $GL$?

arctic tern

20:21

@bolbteppa group of invertible matrices

it's called the "general linear group"

bolbteppa

So how does that link to determinants?

arctic tern

a matrix is invertible iff its determinant is not zero

Obliv

$GL_n(F) = \{ A~|~det(A) \ne 0, n\times n$ matrices where the entries are from $F\}$

Evinda

@bolbteppa How?

I tried it but I get an other result

Obliv

thanks @arctictern

user 1618033

20:24

@Semiclassical yeah, that's why I didn't want to tell you how it works, it is just a matter of noticing something simple.

Semiclassical

gotcha.

user 1618033

@Semiclassical sure, this also means working on such problems to train your eye to easily identify them.

bolbteppa

So how many elements can you put into the first column of the matrix? Then how many in the second column? How many altogether?

arctic tern

@Evinda $\frac{d}{dx}\left( u^\gamma\right)$ equals $\gamma u^{\gamma-1}\frac{du}{dx}$. it does not equal $\gamma\frac{d u^{\gamma-1}}{dx}$ as you claim. (that's $n=1$ for simplicity)

user 1618033

@Semiclassical I don't say in vain work, work, work, ceaselessly. People here once thought I brag saying that or what not, but this is the only way to go, without a huge amount of work nothing good can be obtained in this area.

Semiclassical

20:26

Sure. It's supposed to be aspirational.

Obliv

uh @arctictern $\begin{bmatrix} \underline{?}_A & \underline{?}_B \\ \underline{?}_C & \underline{?}_D \end{bmatrix}$ is the correct line of reasoning that $\underline{?}_A$ has $(p-1)$ choices, $\underline{?}_B$ has $(p-2)$ choices and so on? Then you multiply them together and subtract any un-invertible matrices?

user 1618033

Bragging works when things happen like in movies, you read a book in no time and then you make great discoveries - however, this might only happen in movies.

arctic tern

@Obliv no.

Obliv

darn. Can you correct it then?

bolbteppa

In other words, a) how many distinct column vectors can you form in a finite field? b) Does the fact you are in $GL$ mean any of those are not possible as a column vector in a $GL$ matrix?, c) How many columns vectors can go in the second column of your matrix, given that you are in $GL$? (Hint, relate the 2nd column to the first)

arctic tern

20:29

9 mins ago, by arctic tern

The first column can be any nonzero vector. Hence p^2-1. The second column can be any vector that's not any of the p possible multiples of the first column, hence p^2-p.

Obliv

what do you mean by nonzero vector?

user 1618033

@Semiclassical I didn't reach that stage of getting results. For all I got (in terms of results) I had to work extremely hard and much.

arctic tern

@Obliv any vector except the zero vector.

nonzero means not zero

Obliv

what is a vector?

like in this context

just an arrangement of entries?

arctic tern

an element of $\mathbb{F}_p^2$. it concerns me you don't know what the word vector means, but you can write things down like $\mathrm{GL}_2(\mathbb{F}_p)$.....

one applies matrices to column vectors, such an act is a linear transformation of the vector space

Obliv

20:31

isn't this the field $\mathbb{F}_p$?

does the 2 signify a 2x2 matrix?

arctic tern

the field is $\mathbb{F}_p$, the vector space is $\mathbb{F}_p^2$. just like $\mathbb{R}$ is a field, but $\mathbb{R}^2$ is the vector space of ordered pairs of reals.

user 1618033

Here is almost the midnight (and maybe I'm very sleepy and tired from the previous days with so much lack of sleep) and now I have to finish some part of a project, I have a damn deadline.

arctic tern

@Obliv are you learning about this stuff outside of a classroom?

Obliv

yes. I didn't take linear algebra yet but I'm going through an abstract algebra book. I'm going in order so I don't know anything about vector spaces

We went over some vectors in my physics class but that wasn't rigorously defined

arctic tern

you can't really understand the solution to this problem if you aren't conversant with some of the basic notations that appear before it.

user 1618033

20:34

Go to prepare some tea and I'm back to the hard work for some more hours (hope not too many).

Obliv

if I take it for granted that if each column isn't a multiple of any other column, that it is invertible, then I suppose it makes sense.

arctic tern

that's only true in the 2x2 case

Obliv

oh ok

arctic tern

for example, in the 3x3 case, you want the second column to not be a multiple of the first, and you want the third to not be in the linear span of the first two columns.

Obliv

I mean determinants are just another way of doing arithmetic so isn't there a way to arithmetically prove that inverses in 2x2 matrices exist only if the 2 columns aren't multiples of each other?

arctic tern

20:36

for 2x2 yes

@MikeMiller The hopf fibration $\Bbb S^3\to \Bbb S^2$ should be visualizable as a self-homotopy of the trivial based map $\Bbb S^2\to \Bbb S^2$, but I am having trouble seeing what the self-homotopy is. Ideas?

Obliv

$\begin{bmatrix} _A & _B \\ _A & _B \end{bmatrix}$ is it true that the entries of the $A$ column cannot be a multiple of any of the entries of the $B$ column, and vice versa? This means they are relatively prime with each other, then?

for the inverse to exist ***

arctic tern

@Obliv A few things: (i) you have already asked that, and we have already answered. (ii) yes: for 2x2 matrices, a matrix is invertible if and only if neither column is a scalar multiple of the other (iii) the phrase "relatively prime" does not apply here, but it is analogous.

Obliv

Just making sure I'm understanding correctly, visually speaking.

Mike Miller

@arctictern well, let's see what happens to the map when we restrict it to $S_t = \text{Im}(w)=t$, $t \in [-1,1]$. on $S_t$, which is homeomorphic to $S^2$ by the map $(z,w) \mapsto (z,\text{Re}(w))/(1-t^2)$. the map is still $(z,w) \mapsto z/w$. so we should invert these homeomorphisms and write down the relevant formulae to see what the "formula" for your self-homotopy is, then try to visualize that

i don't have access to paper right now or i'd have gone further with that

let's make that $w/z$ for convenience

meh, these formulae are not at all inspiring.

arctic tern

20:53

I don't quite understand what you're saying (how can $S_t$ equal a number and be homeo to S^2?), but I did work out some formulas from pairs of complex numbers and stereo projection etc. and like you say it was not inspiring.

Mike Miller

$S_t$ was meant to be the set of points on the sphere with $\text{Im}(w)=t$, i stuck that in at the end of writing the message, whence poorly written sentence. sorry.

arctic tern

Currently I'm trying to picture S^3 as R^3, partitioning it into S^1s as is done in those pictures on google image search for hobf fibration. To see that they're arranged in the form of an S^2, one can put a flat copy of D^2 bounded by the circle inside the nested tori: then every S^1 in space intersects said D^2 in exactly one point except its D^2's bounding circle, which quotients down to a single point (D^2/~ is ofc S^2).

Then one may instead partition R^3 into a bunch of spheres centered around points on the z-axis and intersecting the origin, except for the plane through the origin. We should project these spheres along the S^1 fibers onto the disk D^2 and imagine that as a continuous family of maps S^2->S^2.

problem is, I can't actually visualize it

Mike Miller

is the goal to have a picture, or something else?

arctic tern

goal is to describe the process in words

for instance: the identity map S^2->S^2 can be reinterpreted as a self-homotopy of the trivial based map S^1->S^2, accomplished by pulling a rubber band around a baseball.

Mike Miller

i think the thing i would probably do is find some formula for your favorite homotopy (either the previous one or this new one) and have mathematica show you how the image changes over time.

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$Mathematics$ Mathematics