Basically we're making $f_n$ a rectangle centered around $x=a$, taller and thinner as $n$ grows, but of constant "area" $\int f_n^2=1$

hi

i am wondering why a line is closed

@Hippalectryon why $\varepsilon$ ?

That way for any function $h\in X$, we have $\lim\int_0^1 f_n(t)^2h(t)^2=h(a)^2$

@MikeMiller and continue like that down the $\mathbb{R}^k$'s What sort of properties does $\psi$ have? Like if I compose $\psi^1 \circ d_\psi^1$ will that be the same as $d^1 \circ\psi^2$?

@Vrouvrou It's just so that we're in an interval around $a$, getting smaller and smaller as $n$ tends to $\infty$

@user19405892 By definition. What is the complementary of a line ? Or, you can also use the sequential definition

if a line doesn't contain the limit point $\infty$ in $\mathbb{R}$ then how is it closed

@user19405892 A line in $\Bbb{R}$ is $\Bbb{R}$. Did you mean a segment ?

no i mean a line, but the definition of closed is that it contains all of its limit points. Infinity is a limit point of $\mathbb{R}$ but not in $\mathbb{R}$

@user19405892 $\infty\notin\Bbb{R}$

It's not a limit point, since it's not in the space itself

@Hippalectryon what is a?

Oh, so when defining closedness you must define a larger space?

@Vrouvrou A point at which $g$ reaches its $\sup$ : $g(a)=\max g(x)$

@user19405892 Well if you say that $A$ is closed in $B$, it's equivalent to saying that $A$ contains all its limit points that are in $B$. Otherwise it doesn't make much sense.

@user19405892 Saying that something tends to a number and that something tends to infinity are not two similar definitions. Only the notations really are similar

@JuanSebastianLozano I would have to think some things out. You should try to think about them!

Always assume the forms are compactly supported (zero outside a bounded set) so that you can actually integrate them.

@Hippalectryon i found $||Tf_n||_2=\sqrt{\int_{a-\frac\varepsilon{n}}^{a+\frac\varepsilon{n}} g^2(x) dx}$ how to continue

@Vrouvrou Oh damn :( I'm very sorry I might have made a mistake. Give me a sec.

ok

@Vrouvrou Ok please correct me if I'm wrong, but can we not say the following : Using the scalar product $f\cdot h=\int_0^1 f(t)h(t)dt$, by the cauchy-schwarz inequalty $||Tf||^2=f^2\cdot g^2\le (\int f^4)(\int g^4)$. The equality occurs iff $f,g$ are coolinear. Hence $||T||=\int g^4$.

@MikeMiller okay, I will think about this some more, thanks.

Cauchy-Schwarz: $\int fg \leq (\int f^2)^{1/2} (\nt g^2)^{1/2}$

@Hippalectryon

Yep, hence $||Tf||^2=(\int f^2 g^2)^2\le \int f^4\int g^4$

so $\int f^2 g^2 \leq (\int f^4)^{1/2})\int g^4)^{1/2}$

Wait what ? (let me rewrite it properly)

ok

$||Tf||^2=\int (f^2) (g^2)\le\sqrt{\int (f^2)^2\int (g^2)^2}$ hence $||T||=\sqrt\int f^4\int g^4$. I don't see where I got it wrong on this one.

$||Tf||=\sqrt{\int f^2 g^2}$

Oh so there's an additional square root

yes

@Vrouvrou (last edit messed up) So we have $||Tf||^2\le\sqrt{\int f^4\int g^4}$ thus as before $||T||=\sqrt{\int g^4}$

but how to find equality ?

@Hippalectryon

@Vrouvrou Cauchy schwarz says there's equality iff the two functions are coolinear. So $||Tf||^2\le\sqrt{\int f^4\int g^4}\rightarrow \max||Tf||^2=\sqrt{\int g^4\int g^4}$

why f and g are coolinear ?

What do you mean ? $f,g$ aren't necessarily coolinear. But the maximum of $\int f^2 g^2$ is obtained when $f,g$ are coolinear.

so if f and g are not coolinear?

@Vrouvrou But $||T||=\max||Tf||$, so we only care about the max. Which reminds me that I forgot about the condition $||f||=1$, so it's actually $||T||=\sqrt{\int g^4}/\int g^2$

i don't understand

@Vrouvrou We're searching for $\sup||Tf||,||f||=1$, right ? Cauchy schwarz tells us that $\forall f,||Tf||^2\le\sqrt{\int f^4\int g^4}$. Good so far ?

For quadratic polynomials, the discriminant can tell you whether the roots are complex or not.

Is there a general case text for determining whether any n-degree polynomial will have complex roots?

@Owatch By complex, you mean non real ? :-)

Yes.

Should I always refer to as non-real?

My bad.

@Owatch Well all reals are complex numbers :P

Ah well, I don't know how to phrase my question then.

For instance, my calculator will tell me it cannot find any real roots if I input an equation like:

$x^{2} + 1 = 0$

@Owatch Well it's a bit of a hack but if there's an odd number of sign changes then it has a real root

like, $x^n-x^{n-m}=0$ always has a positive root. Same for $\alpha x^a-\beta x^b-\gamma x^c+x^d-x^e=0,a>b>c>d>e$ etc

@Owatch Well if it's just second degree it's easy ...

Oh wow.

Yes of course.

I have a calculator that computes for nth degree.

(Not the one I'm using)

And I want it to avoid complex solutions because I don't want to deal with them right now.

Kind of like emulating what my Ti calculator does when it encounters it.

Not all n-th degree equations are solvable though

Does it always give exact results ??

It uses a numerical approach?

Oh, that makes sense. Doesn't it just say it doesn't have roots when it can't find any ?

So it uses an iterative algorithm to get close.

I right now, go up to 100 iterations max or up to the point where it is close enough to the solution.

I'd be interested in knowing the answer to your question, I suggest you ask it on main

It outputs zero everywhere for complex solutions.

I could work through the example by hand and find out why.

Might take me a bit though.

It worked okay for 6th degree polynomials so far.

Which is nice.

But that was with a online generator with easy roots. Oh well. I may well ask then!

Which method do you use ? Newton ?

No.

I use the QR algorithm.

With something called the Wilkinson shift.

It's on slide 11 here: terminus.sdsu.edu/SDSU/Math543_s2008/Lectures/22/…

Newtons method had divergence problems I didn't want to deal with.

Yeah that's true, but it's much much simpler than QR

I needed to guess an initial estimate close to the actual roots, which I wasn't about to do. So I ended up looking for an alternative approach. And this is one of them.

it is possible to use randomly generated roots and continually try computing the solution (While abandoning cycling or diverging attempts), but that seems wasteful to me.

There are approaches like Hill-climb, local beam search or something I can do for that. But it's like playing a game of hot and cold.

Hill-climb works very well usually (well, if it's what I think, i.e. gradient descent)

I love neural networks :>

Well it is a method where you begin to choose better and better potential solutions based on a heuristic I think.

While abandoning lesser ones.

Oh. Gradient descent should work very well too, since polynomials are very smooth

I have no experience with neural networks, so IDK.

@Owatch Tbh I have little experience with them :-) I just find them very interesting, but I don't have enough free time right now to study them further

Well, I am currently in an AI-course.

It's not my choosing, just part of the program. But apparently its a bit of a held-up field.

As in, AI isn't getting anywhere right now.

I just read stuff on the internet :P but if you have interesting notes you're ready to share I'd gladly take them

Hi, can anyone explain a problem regarding Laurent series?

Well just based off what I've learned in lectures. The approaches right now to AI involve two different ideas. The first is to create machines that think intelligently. And the other is to make machines that act intelligently, or exhibit behavior that humans need intelligence to do. And the second one is where most existing AI fall into right now

@Paradox101 depends on the problem :-)

So they're not really intelligent at all. The problem for the first approach is that nobody exactly understands what makes humans intelligent to begin with, so it is difficult to create anything without having that solved first.

@Owatch You don't have any online notes like that pdf before by any chance ?

Which branches into a field called cognitive-science.

I do have PDF's related to stuff I am researching on my own. But the lecture slides I use are locked behind the schools login.

(I login and use them over BB)

Aw too bad :(

On a totally unrelated note, how's the planets app doing ? :-D

That is done.

Was published last August or so.

It did very poorly at first. Spiked in September or October with about 3k downloads, then in December my license expired so they pulled it from the App Store.

And that is the genesis of it.

Well, it lived its life :D

I'm busy with something else right now, while I'm not under attack from school. We have four finals a year, and mid-terms too. So It's always a crisis with no time for my own projects. (Or not as much)

Hopefully in the summer I can get more done.

@Owatch Yeah I understand ._. I just finished 2 full weeks of exams, 8 hours of exam a day

@Hippalectryon ok. I was a bit confused while finding the laurent expansion for $f(z)=\frac{b-a}{(z-a)(z-b)}$ in the domain $|z|<|a|$. Converting it to partial fractions i get $f(z)=\frac{-1}{z-a} + \frac{1}{z-b}$. When I draw the region, the singularity at point $b$ is outside the region. Finding the expansion for singularity at $a$ seems simple enough but how should I proceed for the one at $b$?

What?

8 Hours a day. Why?

@Owatch Because 2 subjects per day, 4 hours each on average

Where are you going to school that has you go through 8 hours of exams each day for two weeks.

@Owatch It's a bit special :P en.wikipedia.org/wiki/…

So you must be taking 20 different courses or so?

Oh yes that is quite prestigious.

After two years or so, we have very important exams to get into whichever school we want. More specifically, each school is part of one of 5 "exam bank". Each exam bank has its own written and oral exam (but the orals are only for those who did well enough in the written part)

I did the first three "exam banks" so that took be 2 weeks and 1 day

Finally over ._.

Now I have to study for the orals haha

@Paradox101 $|b|>|a|$ ?

What do they ask you there?

I imagine it is mathematics?

I mean, it must have a designed purpose over the written part)

Probably to get rid of the math robots.

@Owatch Each exam bank has roughly (with some variations) 2 math exams, two physics exams, ne chemistry exam, one French exam, one English exam. (for the written part)

@Hippalectryon yes $|b|>|a|$. Would the series be zero in this case?

@Paradox101 The series isn't defined for $|z|>|a|$. So neither does the expansion.

But this is specialized towards a certain area is it not, given the bias towards sciences in your exams?

@Hippalectryon so for $f(z)=\frac{-1}{z-a}$ part I can find the Laurent expansion but as it's not defined for $f(z)= \frac{1}{z-b}$ part, it'll be zero?

@Owatch The coefficients are not the same everywhere, clearly. For instance I forgot to mention there's also a computer science exam, with a coefficient around 2 compared to for instance 30 for both math exams. And as you can see, we do have more "scicen" exams

How does it affect other students by the way?

Do people just quit?

@Owatch Those who quit quit during the first year, although I don't know many who quit around me. That being said, I have the chance to be in one of the best prep school so it's not totally a representative view

@Paradox101 Just not defined, not zero

How many students take exams with you?

@Owatch Overall each year we're about... 2000 in my section maybe ? (PC for Physics Chemistry. There are also MP for Maths Physics and PSI for Physics and Engineering Science)

@Hippalectryon wait if it's not zero and not defined will I just consider $f(z)=\frac{-1}{z-a}$ part and consider that my answer? And why won't it be zero?

But there are more than that if you cound the two last exam banks. (they're kind of ranked by difficulty, the first are the most difficult) @Owatch

@Paradox101 The Laurent series is undefined outside the domain of convergence, in the same way that $\log$ is undefined over negative reals (well, the real log at least)

I've heard about this before, Its very admirable. I would probably get annihilated.

I'm happy I do not have to endure those exam hours.

I have not much to complain about I suppose. My exams are 3h long, and usually spaced out over a week.

@Owatch Hah don't say that :-) I'm sure you'd do great. It's just an... unusual system. I like it a lot actually because our classes are kind of great, and in the end we learn tons of stuff, there isn't really anything like that system in the world elsewhere

No I would not. I am a math robot.

For instance (correct me if i'm wrong, I don't know that much about your system either) your exams are meant to be "doable". Our exams are "competitive exams" : the goal is not to do everything perfectly, it's to do it better than the others

It's not unusual to have exams that can't be finished, and often the ones with full scores haven't done everything right

Our exams have a cutoff mark and you need to pass it. That is the only requirement.

Well actually it is't quite true.

You need to pass overall. The exam has a cutoff where if you fail, you fail entirely. But if you pass, you might still fail if you did not do other work to a sufficient degree.

yeah but for instance you have a fixed notation, usually fixed in advance. Our grades are harmonized to fit a given grades distribution. Also if you do a question that nearly no one else has done, you'll get more points.

@Hippalectryon ok and for the $f(z)=\frac{-1}{z-a}$ part I'll find the taylor series right?

It's really all about ranking people

@Paradox101 Yeah. Just don't forget to add $\frac1{a-b}$ in the constant term.

It sounds extraordinary tough to me.

But it has produced very good scientists.

@Owatch Well it all depends what you're aiming for. I'm aiming for the top school out there, so yeah it's definitely hard... but just getting in some mid grade (which by the way even if it might sounds a bit bad is actually already really cool) school is really doable. You have to work seriously though.

@Owatch Do you do any chemistry by any chance ? I've just finished translating our chem exam from the second exam bank in English for someone else

@Hippalectryon so then I'll get summation from zero to infinity of $\frac{z^n}{a^{n-1}}$ for the first bit right?

I did very little of it.

So basically: no

About a year.

Oh nevermind then @Owatch

I'm going to go shower since it is late, then probably go to bed. Thank you for helping me, and talking!

@Owatch No problem :-) I'll probably be here tomorrow anyway

Good night

Night!

@Paradox101 Why not $\frac{z^n}{a^{n+1}}$ ?

$\frac{-1}{a-z}=\frac{1}{a}\frac{1}{1-(z/a)}$

@Hippalectryon oh sorry I was supposed to write an addition sign

Ok then :-)

but then with the $f(z)=\frac{1}{z-b}$ part I just state that it's outside the domain?

That's what I'd write.

Ok and if the domain is $|z|>|b|$ instead, then $f(z)=\frac{-1}{z-a}$ will be undefined and $f(z)=\frac{1}{z-b}$ will be expanded into taylor series? @Hippalectryon

Well since one term is undefined, the whole expression is undefined. Basically, the series doesn't converge for $|z|>|a|$

@MikeMiller for the diagram to commute in the way I want it is sufficient to show that your mapping takes an exact form on $\mathbb{R}^n$ to an exact form on $\mathbb{R}^{n-1}$, but I don't actually think this is true.

so $|z|>|b|$ will have no expansion at all? @Hippalectryon

Well $|z|>a$ has no expansion, so if $|b|>|a|$ then it follows that $|z|>b$ has no expansion indeed.

ok thanks a lot for all your help! ^_^

:-D

How to simplify $P(A \land (1-P(B))$ to something simpler?

Also, if it were true, then it would be pretty cool because (if we consider each cochain a category) then it would be a functor between cochain complexes.

@Link Depends what A,B are

@Hippalectryon It's a probability class, so P(A) is chance of A happening iirc.

it would be a forgetful full functor, but not faithful, I think?

@Link I mean, is there any link between A and B ? (no pun intended)

I know nothing about catagory theory, actually, so this might be really wrong.

@Hippalectryon No, from what I know, there is no link between A and B. That is we were not told if they are independent or not.

@Link Hm but something looks really wrong here. A is an event. (1-P(B)) is a number. Writing $A\wedge (1-P(B))$ doesn't make any sense to me

@Hippalectryon This is part of a proof. See: math.stackexchange.com/questions/1771883/…

@Link I don't see it written anywhere in that page though

@JuanSebastianLozano To convince me, you'll need to give me a counterexample.

@Hippalectryon Because that proof was flawed, and this is part of a different try.

@Link So, how exactly did you arrive to this line ? Because it's clearly flawed.

@MikeMiller convince you that that is sufficient, or convince you that my sufficient condition isn't true?

Convince me that exact forms don't map to exact forms.

You should not think of cochain complexes as categories. The sort of morphism you describe is called a morphism of chain complexes.

@Hippalectryon Give me a minute then to type it all out.

Sure

Why can't they be thought of as categories? But okay, that does describe it better.

I mean, uh, I guess you could. You get nothing out of it.

Other than to use cool words like "faithful and forgetful functor"? ;P

Let me think of a counterexample

Oh. Hmm, my basic premise is flawed.

Welp.

@MikeMiller I actually don't know any examples of compactly supported forms at all

@Link >:-D

@Hippalectryon Haha, thanks. Do you know a way to uh, simplify:

$P(A \land B)$ ?

It's hardly simplifyable since it's very compact

You can expand it though

Well, expand then.

Since the expansion I know of is for $P(A \lor B)$

Well you can think of it this way : Let's break A,B down into elementary events. Say our space $\Omega$ is comprised of elementary events : $\Omega=(\omega_1,\dots,\omega_z)$. Then $A=(\omega_1,\dots,\omega_n,\omega_{n+1},\dots,\omega_m) ,B=(\omega_{n+1},\dots,\omega_m,\omega_{m+1},\dots,\omega_p)$

Okay

By definition for $i\neq j$, the events $\omega_i$ and $\omega_j$ are independant

Thus $P(A\wedge B)=P(\omega_1,\dots,\omega_m)P(\omega_{m+1},\dots,\omega_p)$

what a day

sheesh

But when I did that for my first proof, the people said I couldn't assume that since wasn't specified.

oh wait, nevermind.. carry on

@Link Ugh slight mistake in my "thus" though. Should be $P(A\wedge B)=P(w_{n+1},\dots,\omega_m)$. Right ?

?

I think that makes more sense

Likewise, $\neg B=(\omega_1,\dots,\omega_n,\omega_{p+1},\dots,\omega_z)$, thus $P(A\wedge \neg B)=P(\omega_1,\dots,\omega_n)$

@MikeMiller I was able to come up with one example of a compactly supported form $e^{\frac{1}{x^2-1}}dx+e^{\frac{1}{y^2-1}}dy$, but if it is exact (I can't show that it is), then so is its integral.

I can't find any other examples of compactly supported exact forms

Hmm, okay

@Link Thus $P(A\wedge B)+P(A\wedge\neg B)=P(\omega_{n+1},\dots,\omega_m)+P(\omega_1,\dots,\omega_n)= P(\omega_1,\dots,\omega_m)=P(A)$

Alright then.

@Link It's a special case of a broader theorem : mathworld.wolfram.com/TotalProbabilityTheorem.html

Do you know where I might find examples?

@MikeMiller Now I think it works...

@PVAL oh