You are determining probabilities using a visual aid. Which visual aid would be most effective for finding the compound probability of picking a blue marble out of a hat with multiple colors of marbles and rolling a prime number on a die?

@Rithaniel Thanks a lot!

No problem

@RyanUnger Is that big?

@J.Doe maybe, maybe not. But it’s the first time I’ve seen the stuff balarka has been talking about in the wild

What does the notation $M_{2^\infty}(\Bbb C)$ mean?

@AlessandroCodenotti There are a few posts with this notation.

So maybe looking at them might help. There are also 2 results on MO.

This notation is certainly unknown to me.

It seems to be some kind of direct limit of $M_{2^n}$.

Thanks, the book mentioned in the first answer on MO seems promising

Hi all, can I ask a category theory question here?

@James You can try - but there is no guarantee that somebody will know the answer. (There is also category theory chatroom - but it has virtually no activity.)

As a side note: List of chatrooms on meta.

Ok so I'm trying to understand the definition of a Comma category (en.wikipedia.org/wiki/Comma_category). I'm confused because I don't understand why it's necessary for the diagram in the article to commute (i.e. why T(g) o h = h' o S(f) )

If you look at some special cases, for example slice category, then this condition looks quite natural, doesn't it?

In the case of slice category you want to have morphisms between "morphisms ending in A_*".

Without the condition the the diagram commute, you would get all morphisms between A and A' and the relationship between them and pi_A, pi_{A'} would be lost.

For other special cases mentioned in the WP article you linked (coslice category, arrow category), the requirement that we want the diagram to commute seems rather natural, too.

I suppose that's the best I can do. (I did not work with comma categories, although I have seen the definition before.) Maybe somebody else will be able to explain this better or have some better insight.

Your answer makes a lot of sense

I still want to see that omitting that condition somehow breaks a law and prevents the slice category from actually being a category somehow

In order to justify it

I don't think it prevents it from being category. It would just be less useful. (At least for some reasonable choices of the functors S, T.)

In the case of slice category, of you omit this condition, then you get category isomorphic to C.

Oh really?

Hello everyone, I have a function that maps from Lie algebra se(3) to R (not sure if it's a necessary detail). I need to linearise the function using Taylor expansion. How can I verify if my derivative is correct or not? I guess one way to check can be to see if the linear approximation is close to the actual value, but I'm not sure how close should it actually be.

Sorry if it's off-topic, I really hope someone can help me out as it's been driving me nuts for a few days now

@James No, my mistake. This would be true only $A_*$ is terminal objects, i.e., if there is only one morphism into $A_*$ from any object.

Ah yes that makes sense

In this case, there is bijection between objects in $\mathcal C$ and $(\mathcal C\downarrow A)$.

And the morphisms in the slice category correspond to the morphism in the original category.

Let’s say I have a monoid on a class of partial functions with a “combine” operation $f(x) \diamond g(x) = \left\{\begin{array}{l}f(x) \text{ when } f(x) \text{ is defined} \\g(x) \text{ otherwise}\end{array}\right$. How do you call a property that distinguishes the operation from $f(x) \diamond g(x) = \left\{\begin{array}{l}g(x) \text{ when } g(x) \text{ is defined} \\f(x) \text{ otherwise}\end{array}\right$?

I was searching for a “direction” and “bias” but was not able to find any concrete jargon.

I feel like this is too trivial for a question but can still open one.

@JanTojnar You need to have something after right to get that rendered correctly.

This should work:

$f(x) \diamond g(x) = \left\{\begin{array}{l}f(x) \text{ when } f(x) \text{ is defined} \\g(x) \text{ otherwise}\end{array}\right.$

$f(x) \diamond g(x) = \left\{\begin{array}{l}g(x) \text{ when } g(x) \text{ is defined} \\f(x) \text{ otherwise}\end{array}\right.$

All I added was the dot after \right.

For instructions how to render MathJax(TeX) in chat see this post on meta or go directly to robjohn's website. (Sorry for the digression from your problem towards TeX-nical issues.)

Consider a polynomial $f(\lambda)$ in $\lambda$. "If it takes value zero in an open interval on $\mathbb{R}$, the polynomial is identically zero." This is a result which we often. Can anyone give me an idea for proving this?

And, if the polynomial takes a value zero for some countable values of $\lambda$. Then what can we say in this regard?

It will be also helpful if someone can direct me towards the answers which have been already answered in the site.

This follows from the fact that a degree $n$ polynomial has at most $n$ roots. To prove this, exploit the correspondence between the roots of a polynomial and its linear factors.

@Bhargob Polynomials can only have finitely many roots. In fact, if the polynomial has degree $n$, there can be at most $n$ values of $\lambda$ with $f(\lambda)=0$. (Proof: If $f(c) = 0$, then $\lambda-c$ has to be a factor of $f$. Divide and continue ...)

Oops. I didn't notice that Thorgott already said this. Well, now you have it twice.

Thank you very much Sir. And thanks to Thorgott also.

Hi @Ted

Hi, demonic @Alessandro

Tempted to write up an identity theorem-esqe response to Bhargob's question

I looked that up, it reads like magic lol

You could also take $n=\deg f$ derivatives and find that $n!a_n = 0$ where $a_n$ is the leading coefficient.

@Karl: So you want to use Rolle's Theorem inductively? I'm fine with that.

Seems like proving a real analytic version of the identity theorem is a bit deeper than necessary.

I was simply thinking about the weaker statement where $f=0$ on an open interval.

Well, then obviously all derivatives vanish, so we're done. But he mentioned the countable case.

The idea I had in mind re identity theorem was to factor out $(\lambda-a)^k$ and use continuity of the polynomial left-over.

I guess I don't follow.

If we have a whole interval of zeroes, then obviously we should use calculus.

So is there any (natural) condition weaker than being polynomial that implies being zero as soon as it has an infinite number of zeroes? What if we require those zeroes to be within a bounded interval?

@TedShifrin If $f(x)=c_0+c_1(x-a)+\dots+c_n(x-a)^n$ and $f=0$ in a nbhd of $a$, then $c_0=c_1=\dots=c_k=0$ for a maximal $k$. Factor out $(x-a)^{k+1}$ to get $f(x)=(x-a)^{k+1}(c_{k+1}+\dots+c_n(x-a)^{n-k-1})$. The factor $(x-a)^{k+1}$ is nonzero for any $x\ne a$. The polynomial $c_{k+1}+\dots+c_n(x-a)^{n-k-1}$ is thus zero in a punctured nbhd of $a$.

Oh, so you're doing complete induction on degree?

@TobiasKildetoft for holomorphic functions having infinitely many zeroes in a bounded set implies being zero, but I suppose you already know that

@Tobias: Then it reduces to the Identity Principle. Right.

@TedShifrin This is just a contradiction (assuming to start that $f\ne 0$). $c_{k+1}\ne 0$ by hypothesis, but it would have to be zero given the continuity of polynomials.

@Alessandro: So, as Karl was suggesting, it applies to real analytic functions, as well. For smooth functions, of course, it's false.

I guess the argument that I learned for the identity theorem is not the standard one.

Oh, I see, @Karl.

Sure, the usual example of non-analytic smooth function is zero on an unbounded set even

Well, I was going to use the usual $e^{-1/x}\sin(1/x)$ to get a countable number of zeroes, @Alessandro.

@AlessandroCodenotti I had probably seen that result before, but I did not remember it.

I need to choose a branch of $f(z) = z^{-2/3}(1-z)^{-1/3}$ such that $f$ is holomorphic on $\mathbb C \setminus [0,1]$. The solution does the following:

"Imposing the restrictions $0 < \arg z < 2\pi$ and $0 < \arg(z-1) < 2\pi$, we obtain:

$z^{-2/3} = \text{algebra} = |z|^{-2/3} e^{-2/3 i \arg z}$ and

$(1-z)^{-1/3} = e^{-1/3} \log(1-z) = |1-z|^{-1/3} e^{-1/3 i \arg (1-z)} = |1-z|^{-1/3} e^{-i\pi/3}}e^{-1/3 i \arg(z-1)}}$

\color{red}{yo}

alright try 2

$(1-z)^{-1/3} = e^{-1/3 \log(1-z)} = |1-z|^{-1/3} e^{-i/ i \arg (1-z)} = |1-z|^{-1/3} \color{red}{ e^{-i\pi/3}e^{-i/3 \arg(z-1)}}$

Yay

thanks Karl

Complex analysis O__O

what's the solution doing there in the red part?

You want to analyze what happens to the function as you go around a little circle around $z=0$ and around a little circle at $z=1$ (this is like the usual keyhole path).

The point should be to make sure that going around both brings $f$ back to its original value.

Hm.

But why play with $z-1$ rather than $1-z$?

@TobiasKildetoft it's the identity principle, often it is phrased as "two holomorphic functions that agree on an open set are equal", but it's really enough for the set to have an accumulation point

Because we usually orient the circle around $z=1$ by considering $z-1$.

ah, right

@AlessandroCodenotti Ahh, I had somehow not put together in my mind that an infinite number of elements in a bounded interval will have an accumulation point

What's up, everyone? What's going in here?

Spoken like a true algebraist, @Tobias :P

Okay so this works for $\mathbb C \setminus [0,+\infty)$ and then he says we show that the limit coming from the positive imaginary direction and the negative imaginary direction agree

on $(1,+\infty)$

makes sense

I would just check that the changes in argument as I go around the little circles cancel out.

I think it's confusing to think about branch cuts to infinity when I want to remove just $[0,1]$.

That said, I am not thinking about the particular question at hand.

To prove that every open subset $U$ of $\mathbb{R}^n$ is a union of open balls without using choice would it suffy to use $U = \bigcup \{B_R(x),\, x \in U\}$, where $R$ is the maximum of all balls that lie in $U$ with center at $x$: $R = \mbox{sup} \{r \in \mathbb{R}_>0 | B_r(x) \subset U\}.$ That's assuming that for every point we have a finite supremum, but otherwise $U$ would equal $R^n$.

just take some ball

you don't need choice

Not sure what you mean by the changes in argument cancelling out

@RyanUnger you do

No wonder why it's called "complex" analysis

:D

because there can be infintite amount of points

As I go counterclockwise around a little circle at $0$, $z^{2/3}$ changes by a factor of $e^{4\pi i/3}$. As I go counterclockwise around a little circle at $1$, $(z-1)^{1/3}$ changes by a factor of $e^{2\pi i/3}$.

Ah

So the product doesn't change.

Yours is a variant of this.

@famesyasd If you really wanted to carry out your construction, use $R/2$ rather than $R$. Note, however, that it only appears to need the axiom of choice to do the usual argument. For, just write $\bigcup\{B:\text{B is an open ball and $B\subset U$}\}$. Then, prove in the usual way that $U$ is that union.

I like your perspective better but theyre more or less the same

Thanks Ted

Heya @ÉricoMeloSilva

you're welcome, @GFauxPas.

hi

Doesn't the series $$u - a u' + a^2 u'' - \dots$$ diverge for $|a| >1$?

For $u(x)$ as a continuous function and $a$ as a number

You mean $u(x)$ is infinitely differentiable? And all those derivatives are just some fixed numbers? I have no idea what this means.

@AbhasKumarSinha There are certainly continuous functions $u$ where that is a fine definition of a function.

we out here

$$ \int e^{\frac xa} u(x) \, \mathrm dx = a e^{\frac xa} \left ( au(x) - a^2\dfrac{\mathrm du(x)}{\mathrm dx} + a^3\dfrac{\mathrm d^2u(x)}{\mathrm dx^2} - \dots \right ) \quad \dots\tag {*} $$

This is the problem, the integral exists for $|a|>1$ but not RHS

Whoa. Start by putting limits on the integrals and explaining what you're doing.

@TedShifrin Definite integrals. Clearly, the RHS is a series made by doing By parts again and again

See this question, in case you are confused: math.stackexchange.com/questions/3323330/…

This is too sloppy for me.

@TedShifrin Sloppy?

Yes, sloppy.

huh?!

@KarlKronenfeld feels bad not having enough experience with topology, that's genius, man

@AbhasKumarSinha Integration by parts how many times exactly?

@J.Doe Infinite

How do you define $\frac{d^\infty f}{dx^{\infty}}?$

@J.Doe I won't it's meaningless

can an arc length be infinite on a finite interval. (yes)

@Ultradark no...

@AbhasKumarSinha what if you integrate from a to b and the graph is convex and tends to infinity at a and b?

@AbhasKumarSinha let $I = \int e^{x/a}\sin{x}\,{dx}$, and $J = \int e^{x/a}\cos{x}$. Find $I+J$ and $I-J$.

$$I = \int \limits_{a}^{b} \sqrt{1+y'^2} \, dx$$ is always finite for a arc.

@Ultradark Sorry, please refer to your last text, you mentioned 'arc' which I assume it to be a part of a circle which is always supposed to be finite under a given interval.

@J.Doe I know how to evaluate the integral, just wondering why the series doesn't works for $|a|>1$...

@AbhasKumarSinha How do you even define the series?

@TobiasKildetoft Is it needed?

is it needed to define it? Yes, of course it is.

what we'll do by defining it here?

I am not sure what you mean. You wrote up an infinite sum. There is no such thing a priori, so you need to define what you mean by it.

My bad @AbhasKumarSinha I didn't realise. What you did inferring that oscillating sum is 0 is textbook definition of dodgy though. Couldn't you argue that it's also sin(x) by re-bracketing after the first term?

2:59

No, 259 is not a prime number. The list of all positive divisors (i.e., the list of all integers that divide 259) is as follows: 1, 7, 37, 259. To be 259 a prime number, it would have been required that 259 has only two divisors, i.e., itself and 1.

LOL, looks prime to me!

:D

why

Why?

Just does, but it isn't. Intriguing!

It

looked

I always used to get stuck on 51, but that's 3*17

Then 57 is not prime following 51

but 53 is

57 is G

?

G

Grothendieck's prime sorry

Oh, he made a mistake or did he do it on purpose?

mistake

lol, Abstract math vs arithmetic in the integers, two separate things

People are amazed that I can't do arithmetic very well, but I just use a calculator for that.

@Ultradark prime factorization is another NP-complete problem such that if you found a polynomial time algo, then P = NP

But most strongly believe P != NP

Which is likely

@ShineOnYouCrazyDiamond hamiltonian path problem

is another

@Ultradark yes true

I was actually working on that for the past 3 days

Want to hear my conclusion / best approach?

I believe that the H path problem

what does resolve mean

is hard for even a lattice!

@Ultradark yeah you're probably right

@TobiasKildetoft can you help me?

but it doesn't hurt to analyze it algebraically, as that may lead to a P != NP proof

@ShineOnYouCrazyDiamond I want to hear your conclusion and best approach

@Ultradark room?

He probably didn't care what number it was. It's like when people call unidentified corpses J. Doe. I never take it personally.

@ShineOnYouCrazyDiamond What? Do you have a reference for prime factorizationg being NP-complete?

@Vrouvrou With what?

I thought NP-complete means you can verify it in Polynomial Time?

yes room

@ShineOnYouCrazyDiamond No, that is just being NP

it's probs NP hard

@TobiasKildetoft thank you, I was wrong on that

I heard Kummer couldn't do basic arithmetic though.

WHY?"^

Because abstract math is infinitely more beautiful than mundane whole numbers

@Ultradark Why do you expect that? Isn't the fastest factorization algorithms faster than the fastest travelling salesman for example?

@Ultradark anyway, we should make a room as not to spam here

@TobiasKildetoft how to delete the imagine part

ehh, what?

Don't delete your imagine part, you'll need that ;)

@Ultradark join the room! :D

I do not join the room

The room joins me

Write out everything in a form where it is clear what the imaginary part is first if you need a real solution

@TedShifrin Suppose now I consider a power series which is taking value zero for a countable (say for $\mathbb{N}$) number of points. Thus it imply that the power series is also zero?

@TobiasKildetoft $y=c_1 \cos(\sqrt{2}x)+i c_1 \sin(\sqrt{2}x)+c_2\cos(-\sqrt{2}x)+ic_2\sin(-\sqrt{2}x)+c_3\cos(x)+ic_3\sin(x)+c_4\cos(-x)+ic_4\sin(-x)$

Ok, and now you apply the usual rules for signs in those

what is this rule?

cos(-x)=cos(x), etc

One of my friend gave an counter example of power series of sinx which is zero at $n \pi$. Ofcourse that is not identically zero

@RyanUnger

@Vrouvrou You should be able to take it from there

@TobiasKildetoft i found a crazy integral that evaluates to 1

@TobiasKildetoft but the imagine part is sin how i must delete sin ?

@Vrouvrou By choosing the constants properly

@TobiasKildetoft but the same constants are in sin and cos

Please actually write up what the imaginary part of that thing is

$c_1\sin(\sqrt{2}x)-c_2\sin(\sqrt{2}x) +c_3\sin(x)-c_4\sin(x)$

so this must be equal 0

@TobiasKildetoft

Right

that is $c_1-c_2=c_4-c_3$

@TobiasKildetoft

And those being 0

yes so c_1=c_2 and c_4=c_3

this is the methods?

but like this there is no sin in the solution

@TobiasKildetoft

any one know

group theory?

i know what a group is

whats your question

@Bhargob This is not the correct statement. You need a countable set of points that have a limit point somewhere. For example, $\sin x$ vanishes at every multiple of $2\pi$. ... At any rate, you need to say convergent power series and be talking about what happens in its domain of convergence.

@TedShifrin hello, sorry I think you can help me in O.d.e

I have this solution

whaatt

lolol

Hello all

i think im gonna be sick, see you tomorrow

Gute Besserung!

@TedShifrin I have this solution of ode I want to find the real solution without an imaginer part

good luck with that lol

:D

@Vrouvrou: It really is not polite to ping random people in the room with questions that might not be of interest to them. I do have one comment on your question: Ordinarily in this situation, if you start with a real differential equation, the real and imaginary parts of your solution should give (usually linearly independent, I believe) solutions, and a general linear combination of them gives the general real solution.

Consider $y''+y=0$ with solutions $e^{ix}$ and $e^{-ix}$. The general real solution is a linear combination of the real part $\cos x$ and the imaginary part $\pm \sin x$.