Mathematics - 2017-04-04 (page 2 of 3)

3:06 PM

On a partially related note, for a given function $f:\Bbb C\rightarrow\Bbb C$, one can define several $F$ such that $F^\prime(z)=f(z)$, correct?

Danu

@LegionMammal978 Do you know what the derivative of a constant is? :)

Balarka Sen

Upto a constant (like Danu is telling you) it is, however, unique.

mrnovice

hi there, would someone be willing to help me with a quick math query?

Akiva Weinberger

@LegionMammal978 I assume that simplifies to $\sqrt{x^2}$?

without going through it

where $x^2$ is rendered as $\exp(2\log(x))$ and $\sqrt x$ is rendered as $\exp(\log(x)/2)$

and where division ($x/y$) is rendered as $\exp(\log(x)-\log(y))$, etc. etc.

@MeowMix What's the problem?

Meow Mix

3:24 PM

Given three points in the plane, find the point which minimizes the sum of the distance from each point to that point

mrnovice

I need to find a sequence such that $a_{n}$ converges, $b_{n}$ is bounded, and $a_{n}b_{n}$ diverges. I chose $a_{n}=\frac{1}{n}$ and $b_{n}=(-1)^n\cdot n$. Is that correct?

Meow Mix

By the way, In $\Bbb R^3$, the set of points such that that sum is fixed would make an ellipsoid, right?

Akiva Weinberger

@mrnovice Your $b_n$ doesn't look bounded to me

Balarka Sen

I don't think so. Ellipsoid is set of points with fixed total distance from two points, right?

Foci.

Meow Mix

Oh, I tried to generalize it wrong-ly

Akiva Weinberger

3:26 PM

@BalarkaSen That would specifically be an ellipsoid of revolution, though, wouldn't it?

Balarka Sen

I think in general it's gonna be a cubic.

Meow Mix

What would 3 points make then? 1 point makes a circle, 2 makes an ellipse

mrnovice

@Akiva Weinberger I thought it was bounded since $|b_{n}| =1$?

Akiva Weinberger

It would have rotational symmetry around the line through the points.

Meow Mix

line?

Balarka Sen

3:26 PM

@AkivaWeinberger Don't remember what that is.

Yeah, I suppose.

Akiva Weinberger

@mrnovice $|b_n|=|(-1)^n\cdot n|=|(-1)^n|\cdot|n|=1\cdot|n|=|n|$

@BalarkaSen I just meant an ellipsoid that's also a solid of revolution.

Balarka Sen

@MeowMix Line through the foci.

mrnovice

@AkivaWeinberger Ah yes my mistake, thanks

Balarka Sen

@Akiva Sure. Didn't remember things other than that were called an ellipsoid too.

Meow Mix

@AkivaWeinberger Here's my analysis question, also

Balarka Sen

3:28 PM

But of course cross sections can be ellipse with varying eccentricity

Akiva Weinberger

$(x/a)^2+(y/b)^2+(z/c)^2=1$

Meow Mix

Assume you know $f(x)$ is continuous, but you're only given it on some interval, minus finitely many singletons

Akiva Weinberger

except in this case, two of $a,b,c$ are equal

Balarka Sen

Gotcha.

Meow Mix

Can you find out the function?

Similarly, say you're given the function on $[0,1] \setminus \{1, 1/2, 1/3, \dots \}$

Can you uniquely determine it on $[0,1]$?

Balarka Sen

3:30 PM

It need not extend to a continuous function on $[0, 1]$. Are you assuming it does?

Meow Mix

I was just asking.

Like, it seems if say, you're given $f(x)=x^2$ on $[-1,1]$ except at $x=0$

Balarka Sen

? I am asking you what your hypothesis is. $f$ is continuous on $[0, 1] - S$ where $S$ is a bunch of points. Is $f$ also continuous on $[0, 1]$?

Otherwise it won't extend!

Meow Mix

Oh, sorry. Yes.

Akiva Weinberger

If it extends, it extends uniquely, if that's what you're asking

Balarka Sen

Right.

Meow Mix

3:31 PM

But what about the case listed above?

Balarka Sen

Still true.

Meow Mix

Does it still hold?

Is there some criterion that needs to hold here?

Balarka Sen

Extend it on each $\{1, 1/2, \cdots, 1/n\}$ uniquely, for all $n$.

Akiva Weinberger

As long as the set of points where you don't know it has "empty interior" (each point is the limit of things outside of the set), you can extend it

Meow Mix

Criterion for the set we're removing, I mean.

Oh, empty interior, that's right.

Akiva Weinberger

3:33 PM

So, like, $S=\{1,1/2,1/3,\dots\}$ has empty interior in $[0,1]$, Since each point in $S$ can be approached by things in $[0,1]\setminus S$.

Same for $T:=S\cup\{0\}$.

Meow Mix

So like, every point in our set has to be a limit point in $R - S$?

Or am I interpreting this wrong.

Akiva Weinberger

@MeowMix Yeah

Meow Mix

Cool :P

In $\Bbb C$ I'm pretty sure the requirement of holomorphicity is really strict

Akiva Weinberger

Although, here's a question: Suppose $f$ is continuous and defined on $[0,1]\cup\{2\}$.

Balarka Sen

Yes, it is.

Akiva Weinberger

3:35 PM

And we only know its values on $[0,1]$.

Meow Mix

Can't you like define it (Assuming it's holomorphic on a disk) on the boundary of a disk and get it defined for the whole disk?

Akiva Weinberger

Can we uniquely recover the original function?

Meow Mix

I'm pretty sure not.

Akiva Weinberger

Right

Meow Mix

Oh, wait.

Yeah

You can't

Akiva Weinberger

3:36 PM

The function will be continuous no matter what you define $f(2)$ to be

This means that $\{2\}$ (which has empty interior in, say, $\Bbb R$) doesn't have empty interior in $[0,1]\cup\{2\}$.

So we care what the surrounding space is.

Balarka Sen

@MeowMix Again, is that what you meant to ask? $f$ can be holomorphic inside the disk and still have singularities on the boundary of the disk.

Meow Mix

By holomorphic on, I mean, the whole closed disk

Balarka Sen

$1/(z - 1)$, and the disk is the interior of the unit disk.

Meow Mix

Like, the disk is closed.

Balarka Sen

@MeowMix That doesn't make sense, sorry. Holomorphicity is defined only on the interior.

Meow Mix

3:39 PM

Oh, you're right.

Balarka Sen

But you probably mean it doesn't have singularities on the boundary?

Meow Mix

Yeah

Akiva Weinberger

Holomorphic on the interior and defined/nonsingular on the closure

You can use Cauchy's formula to recover the values on the interior if you only know the values on the boundary, right?

assuming it's holomorphic

Balarka Sen

For sure.

Meow Mix

Yeah, that was the question I asked as well.

Balarka Sen

3:41 PM

I think if you're defined on the boundary you can analytically continue it on a bigger disk containing the closed disk.

Akiva Weinberger

This doesn't mean that every continuous function defined on $S^1$ can be extended to a homolorphic function on $D^2$

Meow Mix

Kind of like analytic continuation, because analytic and holomorphic are the same thing in $\Bbb C$ right?

Balarka Sen

@AkivaWeinberger Not true. $\bar{z}$.

Akiva Weinberger

I said it doesn't mean it can be extended

Meow Mix

@AkivaWeinberger $S^1$ is the unit circle and $D^2$ is the unit disk right?

Akiva Weinberger

3:42 PM

Yeah

Meow Mix

By the way, isn't $\bar{z}$ not holomorphic or sommething?

Balarka Sen

@Akiva You originally started with "Doesn't", which was a bit confusing. I agree.

Akiva Weinberger

@MeowMix Right

Balarka Sen

However, you can give a pretty strong condition on your function on $S^1$ (harmonic) and the extension problem is a deep problem in analysis. Also known as the Laplace boundary value problem.

Meow Mix

Why?

Balarka Sen

3:43 PM

Cauchy-Riemann

Akiva Weinberger

Try differentiating it with the definition

(or that^^)

Meow Mix

Aren't you just flipping the plane over the real axis?

Akiva Weinberger

You know how "differentiable" on $\Bbb R$ means "if you zoom in, it'll look like a line"?

On $\Bbb C$, it's essentially the same thing; "if you zoom in, it'll look like a linear function"

Linear functions are like $z\mapsto az+b$.

There's no place on $\bar z$ where you can zoom in and have it look like something of the form $az+b$.

Balarka Sen

@MeowMix Yes, but that means the derivative (at 0, say) along the imaginary axis is negative and the derivative along the real axis is positive.

Akiva Weinberger

It "flips orientation" in a way that $az+b$ does not, if that makes sense.

Balarka Sen

3:45 PM

Whereas holomorphicity means you have to have all derivatives from all directions the same.

Meow Mix

That's extremely weird...

I mean, it makes sense.

Balarka Sen

It's a strong condition. (x, y) mapsto (x, -y) is differentiable as a multivariable function, but is not holomorphic.

You essentially want everything to be "like" polynomials like Akiva said

That's why complex analysis is so rigid

Meow Mix

Oh, and that's related to analytic-ness?

Balarka Sen

You can say that

Akiva Weinberger

On the other hand, if I have this right, things that are differentiable as multivariable functions will look like something like $az+b\bar z+c$ when you zoom in.

Which is more general.

Balarka Sen

3:49 PM

Nice perspective. It's right, of course.

Akiva Weinberger

And this is related to the concept of Wirtinger derivatives, which are written like $\dfrac\partial{\partial z}$ and $\dfrac\partial{\partial\bar z}$.

Which I assume would give you $a$ and $b$, respectively (from the $az+b\bar z+c$ thing).

Sha Vuklia

4:20 PM

Let $E=\{(x,y,z\in\mathbb R^3:x>0,y>0,z>0\}$ and $f\colon E\to\mathbb R$ given by $f(x,y,z)=x^yy^zz^x$. Compute the partial derivatives of $f$.
It suffices to compute only the first derivative of $f$. I started as follows:
\begin{align}D_1f(\vec x)&=\lim_{h\to0}\frac{f(\vec x-h\vec e_1)-f(\vec x)}{h}\\
&=\lim_{h\to0}\frac{((x-h)^y,y^z,z^{x-h})-(x^y,y^z,z^x)}{h}\\
&=\lim_{h\to0}\left(\frac{(x-h)^y-x^y}{h},0,\frac{z^{x-h}-z^x}{h}\right).\end{align}
However, I don't know if I can continue from here on. Is there anything elke I could do?

Or should I maybe not use the definition and work with "regular" differentiation?

oh wait, I see I messed up badly too

I should have written:
\begin{align}
D_1f(\vec x)&=\lim_{h\to0}\frac{f(\vec x-h\vec e_1)-f(\vec x)}{h}\\
&=\lim_{h\to0}\frac{(x-h)^yy^zx^{x-h}-x^yy^zz^x}{h}.
\end{align}

I'm guessing the answer is this:
$$D_1f(\vec x)=\frac{\partial f(\vec x)}{\partial x}=yx^{y-1}y^zz^x+x^yy^zz^x\ln x.$$

Adeek

4:46 PM

hi @arctictern do you know a good book along with Gelfand homological algebra as reference ?

for homological algebra ?

Secret

5:01 PM

Everything is easier with matrices

Akiva Weinberger

@Secret Isn't that mixed derivatives (of a multivariable function), rather than Taylor series?

Balarka Sen

You use them for multivariable Taylor series

Secret

Well, that matrix just makes it easier to keep track of what derivatives I have computed for the taylor series

I tend to miscount things when they are presented horizontally

This is why I don't like polynomial expansion

Adeek

5:17 PM

@MikeMiller I was wondering can we define homology theory in terms of non-abelian groups ?

Mike Miller

You can define $H_1(X;G)$ when $G$ is non-abelian. You can't do $H_k$ for $k>1$.

Can you? I guess I know you can define $H^1(X;G)$.

Adeek

why can't we do it for higher homologies ?

Mike Miller

What happens if you try?

Adeek

oh I see we get this issue with it not being necessarily a group

hm is there some variation for homology theory for non-commutative things ?

It seems that a non-commutative homology theory would be very useful

Akiva Weinberger

5:36 PM

I remember you guys telling me once that it had to be an abelian category

(not that I fully know what those are)

Balarka Sen

it's essentially a category where you can take derived functors. you should read the construction of Ext and Tor in Hatcher

chapter 3

Adeek

@AkivaWeinberger Abelian category you can think of it as generalization of abelian objects modules,abelian groups,etc

@AkivaWeinberger I don't recommend hatcher

Try Methods of homological Algebra by YU.I.Manin

Balarka Sen

he doesn't need to learn all of homological algebra to understand abelian categories, assuming he's not going to get to technicalities in the first dip

"generalization of abelian modules, abelian groups" is also a rather uninspiring description

Adeek

no there is section on abelian category on that book which is short

It covers it better than Hatcher

Balarka Sen

Hatcher doesn't cover abelian category....

i just told him to learn the defn of Ext and Tor from Hatcher.

Adeek

5:43 PM

oh okay

LegionMammal978

@Danu Sorry, the question I had in mind was: How would one determine a specific value of the antiderivative (ignoring the constant of integration for now) on the complex plane? It can be done easily enough on the real line with $F(x)=\int_a^xf(x)\,\mathrm dx$, etc. but I can't figure out any of this contour craziness.

Balarka Sen

@LegionMammal978 It's the same idea in complex analysis. If $f$ is holomorphic on a neighborhood of $0$, you write $F(a) = \int_\gamma f(z) dz$ where $\gamma$ is an oriented path going from $0$ to $a$ given by moving from $0$ to $\Re[a]$ along the real axis, and then moving from $\Re[a]$ to $a$ along the imaginary axis.

You can prove $F'(z) = f(z)$ by using the Goursat's theorem (integral of hol. function along triangles are zero) - which is a version of Cauchy's theorem.

LegionMammal978

@BalarkaSen "Holomorphic on a neighborhood of $0$?" Sorry, never been real good with this terminology

Balarka Sen

Look in Stein-Shakarchi for details.

@LegionMammal978 Holomorphic on a open disk containing $0$. Just assume holomorphic on $\Bbb C$ if you want

LegionMammal978

Also, what are you denoting by $\Re[a]$?

Balarka Sen

5:52 PM

Real part....

LegionMammal978

Oh, okay :P

Balarka Sen

I have to go for now. See ya

LegionMammal978

Oh, okay, "holomorphic" basically means "differentiable"

Meow Mix

But be careful

Like @Balarka said, it's much stronger of a condition than, say, multivariable differentiability

LegionMammal978

As in...?

Mike Miller

5:57 PM

@Adeek When you write down $\partial^2$ and your coefficients are nonabelian, $\partial^2 \neq 0$. So you're screwed.

LegionMammal978

Huh, it also ensures that it's analytic

Will only really be taking these for elementary (or, more accurately, Liouvillian) functions anyway in my use case

Adeek

yeah I see @MikeMiller

Vrouvrou

i have a function $f:\mathbb{R}_+\rightarrow \mathbb{R}_+$ increasing such that $f(0)=0$ and for all $s,t\in\mathbb{R)_+, f(s+t)\leq f(s)+f(t)$ please if $f(x)=0$ can we deduce that $x=0$ ?

LegionMammal978

@Vrouvrou Isn't $0\not\in\Bbb R_+$?

Vrouvrou

$\mathbb{R}_+=[0,+\infty[$ and $\mathbb{R}_+^*=]0,+\infty[$

LegionMammal978

6:06 PM

ah, okay

Is $f$ strictly increasing, or no?

Vrouvrou

no just increasing

LegionMammal978

In that case, wouldn't $f(x)=0$ for all $x$ be a simple counterexample?

Vrouvrou

yes

LegionMammal978

So no, we cannot in general deduce that $x=0$

Vrouvrou

there is a question suppose that there existe $s>0$ such that $f(s)=0$ how to schow that $f$ vanish on $\mathbb{R}_+$

@LegionMammal978 please if we suppose that $f$ does not vanish if f(x)=0 can we deduce that x=0 ?

someone here ?

?????

@AlessandroCodenotti hello

trilolil

6:29 PM

Hello everyone

this is something I have been stuck with for a while
I want to calculate the covariance
not in advance, but as the data is coming in...
So if at T1 x1 enters the system cov1 should be calculated
at T2 x2 cov2 should be calculated (without having to save in memory x1!)
some sort of moving covariance
Any suggestions?

LegionMammal978

Hmm

I figured out an O(1) moving standard deviation once

You just have to save the sum of the elements and the sum of their squares

Adeek

@MikeMiller have you read homological algebra by Gelfand before ?

Methods of homological algebra *

trilolil

@LegionMammal978 yes but I only have 1 element at a time

and need to have a covariance immediately

LegionMammal978

@trilolil O(0) isn't necessarily possible

trilolil

Sorry?

LegionMammal978

6:39 PM

As in, you can't calculate anything "immediately"

Mike Miller

6:50 PM

@Adeek Nope.

Vrouvrou

hello, can someone help me on my question ?

trilolil

7:05 PM

vrom vrom

CausingUnderflowsEverywhere

7:28 PM

immediately as in atomicly?

what?

Hi @Meow

Hi @Balarka

yawn I give up trying to numerically approximate values of the complex antiderivative

Nothing I'm doing is working

Vrouvrou

hello

CausingUnderflowsEverywhere

7:46 PM

hi balarka

hi meow

Meow Mix

@BalarkaSen Did you see the drawing of homer simpson using Fourier Series?

And they showed it as like circles rotating around circles rotating around circles rotating around ...

quallenjäger

Hi anyone familiar with spinor bundles?

Balarka Sen

what's homer simpson

Meow Mix

Ptolemy and Homer (Simpson)

►

quallenjäger

Meow Mix are you really middle school student

Meow Mix

7:49 PM

Yes.

quallenjäger

impressive

Balarka Sen

@MeowMix neat

i guess

Meow Mix

@BalarkaSen I might try creating one of those lol

All I'd have to do is just draw it out as one continuous curve, create functions for the x and y then approximate them with fourier series?

Balarka Sen

something like that

Meow Mix

Is there a theorem analogous to Weierstrass Approximation for Fourier Series?

Like, fourier series are dense in the set of continuous functions on some interval?

Balarka Sen

7:55 PM

Yes. Sufficiently nice periodic functions are approximated (in some appropriate norm) by partial sums of their Fourier series.

It's a subtle topic, so I don't know much about it.

I think it's analogous to the proof of Weierstrass using approximate identities for C^1 functions on the circle with uniform norm.

Meow Mix

Interesting.

Balarka Sen

@MeowMix There's a chapter in Ted's notes, and a whole lecture he devotes to them, where he described this as a sort of orthogonal projection thing on towers of subspaces spanned by $1,\sin(kx), \cos(kx)$ (n = 1, 2, ..., N), in the vector space of C^1 functions on [0, 1] say.

It explains the intuition behind why it "should" happen nicely.

Alessandro Codenotti

Every function in $L^2(-\pi,\pi)$ has a converging Fourier series

Balarka Sen

@Alessandro There are C^0 counterexamples, IIRC?

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$Mathematics$ Mathematics