Mathematics - 2014-01-19 (page 2 of 5)

7:04 AM

@Nick What device? Wheel?

@BalarkaSen: Yes, you need something to attach to your cart. What is the first aspect of that shape that you think is important?

Balarka Sen

Uh, not sure.

usukidoll

I'm doomed :(

Balarka Sen

@usukidoll Just wait up for Buggs.

Nick

@Balarka: Ok then, I propose a cuboid. Do you have any objections?

Balarka Sen

7:16 AM

@Nick ? Cuboid? What do you think that is useful?

Nick

@usukidoll: Why not try main? I pretty sure they could help.

usukidoll

:/

Nick

@BalarkaSen: Or a cube? Why don't you think it's useful? Your bullock could pull the cart and the cube wheels will offer less resistance to the pull than if it didn't have them. Could I have a more efficient shape?

Balarka Sen

@Nick Ah I see the reason. Not the most efficient one though, I propose oval (can you guess why?)

Nick

@Balarka: A smoother ride and even less resistance?

Balarka Sen

7:21 AM

@Balarka : No, it doesn't minimize both, alas. The cart can go faster though.

Nick

Ah yes, it can.

But it's be a bumpy ride though

What do we do about that?

Balarka Sen

Yes, IIIIT'S GONNNA BEE A BUMPY RIDEEE!!!! Nothing can be done about that.

Nick

@Balarka: :D Ok let's consider that you're not in India but in some grand place where they have nice smooth roads.

It's still going to be a bumpy ride if you're using ovals.

Balarka Sen

@Nick Well said. Yes, we need smooth streets first.

Nick

Why is it bumpy even with smooth streets?

Why was the square bumpy?

Why was the triangle bumpy?

Balarka Sen

7:24 AM

They are not smooth -- obvious.

Nick

What is this commonality between shapes that make them a bumpy wheel?

Yes and what is smoothness?

Balarka Sen

Ahem... good question.

I am not sure.

Need to think about them.

Nick

One thing common about the shape is that the distance from the centre of it to it's edge is not always constant.

If I need a really smooth ride on a smooth road, I need a shape which has a constant width.

@BalarkaSen: Can you find me such a shape?

Balarka Sen

@Nick Circle.

Sphere.

Nick

@BalarkaSen: and we have the reasons why wheels and ball bearings are the shapes they are :D

Balarka Sen

7:29 AM

Interesting.

Nick

Balarka: Now, comes the hard part.

Give me one more shape like that.

Balarka Sen

4-d sphere?

Nick

Nope, you're just stating polydimensional equivalents of the 2D circle.

usukidoll

cries

Balarka Sen

@Nick : Not sure then.

Nick

7:32 AM

@usukidoll: "Your lack of faith is disturbing" - Darth Vader

usukidoll

>:O

Nick

@BalarkaSen: It's a very mind boggling question isn't it? Another shape with constant width.

Balarka Sen

@Nick Yes, I am just wondering how can you think of one...

Mike

hopefully that's not spoilers to what you're trying to do @Nick

Nick

@mike: Indeed it is Mike. Indeed it is.

Balarka Sen

7:33 AM

@Mike : Wonderful!

Mike

Then it's gone :D

Oh noo

Nick

XD

Mike

@Balarka you didn't see that!

Nick

XD mike. It's ok. You told him at the right time.

Reuleaux triangle

A Reuleaux triangle is the simplest and best known Reuleaux polygon. It is a curve of constant width, meaning that the separation of two parallel lines tangent to the curve is independent of their orientation. Because all diameters are the same, the Reuleaux triangle is one answer to the question "Other than a circle, what shape can a manhole cover be made so that it cannot fall down through the hole?" The term derives from Franz Reuleaux, a 19th-century German engineer who did pioneering work on ways that machines translate one type of motion into another, although the concept was known b...

Balarka Sen

It maximizes both smoothness and speed... or does it?

@Nick Really, so much to learn.

Nick

7:35 AM

no, physically it's not THAT great compared to a sphere but it's pretty close. Close enough to use.

Balarka Sen

@Nick : Speed is more than sphere, actually, I think.

Mike

doesn't the circle actually optimize both of those

Balarka Sen

...Depending on the gravitational force.

@Mike : I don't think so.

Nick

Yeah, you're right but without proper manufacture, it's kind of unstable to use.

Circles are easier to make, don't you think?

Balarka Sen

In the triangle above, when you get to the vertices, the wheel falls fast and speedily, and thus accelerates the speed.

Nick

7:37 AM

Balarka Sen

@Nick : Yeah, circs are easier to make.

Nick

@BalarkaSen: and hence instability.

Balarka Sen

@Nick : Good one.

Nick

@BalarkaSen: Notice how the front wheel of that bike is a pentagonal curve.

That's the Reuleaux pentagon. Not exactly the best wheel (again difficult to make). But in the bike it helps increase stability at slower speeds.

Balarka Sen

@Nick Yes, I got that.

Question : Is there a particular reuleaux polygon that maximizes that optimizes smoothness and speed both in a smooth street on the equator of earth?

I conjecture : Reuleaux pentagon.

Mike

7:43 AM

why do you conjecture such

Balarka Sen

@Mike Apparent observation. The real proof would take extensive calculations.

Applying acceleration formulas, gravitational pull, etc.

How's the problem?

Nick

@BalarkaSen: ... First find the moment of inertia. Specifically, the radius of gyration.

Balarka Sen

@Nick Physics is not my thing. I'll leave the hard work to the others =D.

Nick

XD It's more math than physics and it ain't too hard either.

and why the equator of the earth?

Balarka Sen

@Nick : To invariant the gravitational pull.

Nick

7:49 AM

The curvature of the earth is absolutely negligible.

Oh!

nice.

Considering ofcourse that you neglect the existence of that minute error due to not so uniform density of the earth.

Balarka Sen

@Nick Yes, infinitesimal errors should be neglected but minimized ; ).

Nick

@BalarkaSen: That reminds me.. kind of off-topic but when exactly are the products of two irrational numbers rational.

Balarka Sen

$\frac{1}{\sqrt{2}}*\sqrt{2} = 1$

Nick

XD

usukidoll

<_<

Balarka Sen

7:56 AM

You need more than irrationality (or transcendence) to show that a * b is not rational.

Nick

@BalarkaSen: I mean remember how I asked if when the circumference is rational, the radius is irrational.

In that case what is this peculiar value that when multiplied by pi gives me a rational number.

Don't tell me it always has to be a multiple of 1/pi

Balarka Sen

@Nick : it has to be linearly dependent of $1/\pi$, yes.

Nick

Oh

well,that's kind of a bummer.

But totally coherent... so yay!

@BalarkaSen: Also, how do I generally represent the digits of a number. Can I write a two digit number as ab (Meaning 10a+b)

Balarka Sen

$\sum_{i \leq n} a_i 10^i$

Nick

oooooooooooooooooooh!

Balarka Sen

8:05 AM

Off topic : I cannot prove that $e^\pi$ is not in the algebraic closure of the field generated by hypergeometric functions over algebraic values. My approach is to consider the Ramnujan theta and invert it. Can this be done? I mean is the inverted Ramanujan theta in $\bar A$?

Nick

You can try it. If it works, then you haven answer(Doesn't necessarily have to be right, though). If it doesn't work, then go to main, it's not going to hurt your rep.

Balarka Sen

@Nick I have already squeezed it as a comment in here

This question needs MO instead of MSE (but I always would like to try the latter before the former).

Nick

Actually wait till the peak answering hours to ask the question;.

If it really deserves to be on MO, then the community will migrate it there.

usukidoll

mo?

Balarka Sen

I just installed robjohn's javascript to chat LaTeX.

@usukidoll MathOverflow. Useful for research questions.

usukidoll

8:19 AM

mathoverflow stackexchange

Nick

@usukidoll:yup.

Balarka Sen

Testing LaTeX $1, 2, 3, 4$, $\text{etc.}$... $\text{Gal}(\mathbb{Q(\sqrt{2}, \sqrt{3})}/\mathbb{Q})$

Nick

You mean testing $ \LaTeX$

Balarka Sen

LaTeX works!!!

Okay, testing $\TeX$, $\LaTeX$.

usukidoll

$a_n$

Balarka Sen

8:22 AM

Yep, I see that.

Sush

$k^2-k-2(n-1)\leq0$ implies $k\leq\dfrac{1+\sqrt{1+8(n-1)}}{2}$ how is that possible? I know that if $k^2-k-2(n-1)=0$, then $k=\dfrac{1+\sqrt{1+8(n-1)}}{2}$ or $k=\dfrac{1-\sqrt{1+8(n-1)}}{2}$

Nick

$ \text{Look textbook font}$

usukidoll

save my math soul lol

Balarka Sen

@Sush : Complete the square in a similar style as when you do in solving the quadratic.

usukidoll

all of these things in the exercise are just logic laws...

Balarka Sen

8:23 AM

Okay, having fun with $\LaTeX$,... a lot.

usukidoll

how to increase the font in latex

Nick

@usukidoll: \large, \Large are pretty useful.

usukidoll

http://assets.openstudy.com/updates/attachments/52da4347e4b0b71c4f8dd63d-usukidoll-1390035831646-scan1401170001.jpg
just realized that a is negation b and c are demorgans d and e are distributive f was something and g is contrapositive... so b and c are one liner proofs. the same goes for d and e

Nick

@BalarkaSen: try formatting! It's very fun.

usukidoll

$\large A$

ok suppose someone has bad vision and can't see that is there an extra large?

no such thing nevermind

Balarka Sen

8:25 AM

$\bigg A$

Nick

\Huge

Balarka Sen

Oh, yeah, I remember that.

Nick

$\Huge { \text{I'm a tough little baby and I dance like a man}}$

usukidoll

$\huge thanks because I can't see small font$

:O

now space?

$ \huge X \space dcs$

Nick

Use \text{} to format text

usukidoll

8:27 AM

oh there

Nick

@usukidoll: There's a zoom feature on your browser, ya know.

usukidoll

openstudy.com/study#/updates/52db5f51e4b05a53debcfc5a

Balarka Sen

8:44 AM

Hullo, @N3buchadnezzar

Nick

9:10 AM

@Balarka: Someone just pointed something out to me. I think i screwed up in basic differentiation. $\frac{d}{dx} \frac{1}{1-x}$ = $\frac{d}{dx} (1-x)^{-1}$ = $-1(1-x)^{-2}$

Balarka Sen

@Nick : You forgot to use the chain rule. $\frac{d}{dx} ( 1 - x)^{-1} = \left [\frac{d}{d(1-x)} (1-x)^{-1} \right ]\cdot \left [\frac{d}{dx} (1 - x) \right ]$

Nick

Aww, I can't believe I mistook a mistake and thought I made a mistake which I indeed did do because I overlooked that but still made a mistake.

... I hate it when I can't understand myself.

Balarka Sen

...

Nick

@BalarkaSen: How do you not ever screw up like that?

You must have some method?

Balarka Sen

@Nick Loads of times.

Huy

9:23 AM

Can anyone verify a proof of a consequence of the closed graph theorem for me?

Balarka Sen

...Although not frequently, I agree.

Nick

@BalarkaSen: Can you give me some advice on how to decrease the frequence of my idiocy?

Balarka Sen

@Nick Depends on what kinds of screwing-ups you do.

(Off topic) Have anyone considered constants related infinite sum of multidimensional integrals of algebraic functions before? Is there any reference?

Nick

@BalarkaSen: Just general things that are easy to spot by most people. Like mistaking the geometry of a sp3 hybridized molecule to be square planar rather than tetrahedral.

Balarka Sen

@Nick I am not sure then. My mistakes are usually rather hard to spot, as well as huge.

(I am not proud of it, though)

Nick

9:31 AM

@BalarkaSen: I sometimes even make mistakes while making mistakes and often if I'm lucky, I get the right answer. (I'm not proud of it either)

Balarka Sen

(off topic continued) The ring of periods $\mathbb{P}$ in already on that ring. But there are constant that is not a period but still in this ring. Example : $e^\pi$ (!).

@Nick : The general idea is two-sided checks, which I often do at my cases (not sure it is applicable to your kind of screw-ups, though).

(off topic going on...) $e^\pi$ is a doable candidate because it can represented as infinite sum of areas of even dimensional unit balls.

Wait, no, it's not obvious whether $\mathbb{P}$ is a subset -- I have said infinite sum of multidimensional integrals of algebraic functions.

Nick

9:52 AM

I just found the most coolest video

►

it's sherlock and the doctor!

My god VFX is cool.

Balarka Sen

10:05 AM

Someone in a forum named me "Mr. TNT" for Transcendental Number Theorist. Cool, eh?

does XD smiley refer to Xenophilleus Dudley? :P

Nick

@BalarkaSen: Firstly, Trinitrotoluene is hot. Very hot under the right circumstances.

Secondly, XD precedes harry potter.

Balarka Sen

@Nick No, that was Xenophilleus Lovegood. XL. XD.

@Nick Heh.

Nick

Yes, but Dudley is a character too, right?

Balarka Sen

Oh, right, right. Dudley Dursley.

Big D.

:D.

Okay, lots of spam talk, I have to go now : It's LOTR time.

Lord Of The Rings, I mean.

Nick

XD good luck

Balarka Sen

10:38 AM

...And back.

Nick

►

Huy

11:09 AM

Riesz representation theorem states that for a Hilbert space (over $\mathbb{R}$), for every $\ell \in H^*$ there exists a unique $y \in H$ with $$\forall x \in H: \ell(x) = \langle y, x \rangle_H.$$ In our lecture notes, the proof goes as follows: W.l.o.g. $\ell \neq 0$ and $\| \ell \|_{H^*}=1$. Then, we have a sequence $(y_k)_{k \in \mathbb{N}} \subset H$ such that $\| y_k \|_H = 1$ and $\ell(y_k) \to \| \ell \|_{H^*} = 1$ as $k \to \infty$. We start by proving that this is a Cauchy sequence.

However, I am wondering: Why can we assume $\| y_k \|_H = 1$? According to the definition of $\| \cdot \|_{H^*}$, isn't $\| y_k \|_H \leq 1$ the only thing we can assume?

Mike

if $||y_k|| < 1$ then multiply it by a suitable constant to have norm one

by linearity of $\ell$, $\frac{|\ell(y_k)|}{||y_k||} = |\ell(\text{rescaled }y_k)|$

Huy

@Mike: Assume $\| y_k \| = \frac{1}{2}$ for all $k$. You propose that we instead look at the sequence $(2 y_k)_k$. But then, $\ell(2 y_k) = 2 \ell (y_k) \to 2$?

Mike

i'm not sure what you're objecting to

Huy

The sequence doesn't converge to $\| \ell \|$ anymore as we would like it to, or am I wrong?

Mike

ah

the point is that if $||y_k|| < 1$, then $|\ell(y_k)|$ is going to be strictly less than $|\ell(\text{rescaled})|$

so that if the former converges $||\ell||$ so must the latter (since we're taking the sup)

Huy

11:18 AM

Oh, I see now.

Thanks a lot.

Mike

not a problem, and pardon my terrible notation

Balarka Sen

Pondering over some old question on transcendental number theory.

Hullo, @DanielFischer

Daniel Fischer

11:36 AM

Hello, @Balarka.

Huy

Lax-Milgram's theorem states that for a continuous bilinear form $a: H \times H \to \mathbb{R}$ with $$|a(x,y)| \leq \Lambda \|x \|_H \| y \|_H$$ and for some $\lambda > 0$ $$a(x,x) \geq \lambda \| x \|_H^2$$ there is a continuous bijection $A \in L(H)$ with $$a(x,y) = \langle Ax, y \rangle_H$$ and $\| A \|_{L(H)} \leq \Lambda$, $\| A^{-1} \|_{L(H)} \leq \lambda^{-1}$.

To prove this we apply Riesz' representation theorem on $\ell_x: y \mapsto a(x,y)$. This gives us the existance of $Ax = J^{-1} \ell_x \in H$, where $J$ is the isometric isomorphism given by Riesz' theorem. However, isn't $J…

Mike

no - $J^{-1}\ell_x = x$ when $\ell_x = \langle \cdot, x\rangle$

but there's no reason to believe that $a(x,y)$ is of that form

Huy

Oh, of course. The notation kind of confused me there. :s

Mike

11:51 AM

yeah... especially considering I normally see written $\ell_x := \langle \cdot, x\rangle$

or maybe $\langle x, \cdot \rangle$ but who cares, our codomain is the reals

Huy

12:06 PM

@Mike: I am slightly confused: Are all $L^p$ spaces Hilbert? I know that $L^2$ spaces are, what about the rest?

Mike

Nope, they're not

I don't recall how the proof goes, though.

user112495

Let p(x) be a polynomial with positive leading coefficient with a_j>0 (a_j represents the polynomial coefficients in the series representation). Show that there exists a* in R such that P(x) + a = 0 if a<a*. How do I go about doing this?

Mike

Actually, I do

Use the parallelogram law of norms on a Hilbert space (that you inherit from the linearity of the inner product), and for $p \neq 2$, use it to find a counterexample

i.e. pick two elements whose norms do not satisfy that

What I don't recall is which elements it was.

user112495

How do I show it?

(the question I asked a few minutes ago?

Huy

@Mike: I think I've seen this somewhere done by splitting up $\Omega$ into two disjoint sets of measure $M$ and defining $f = 1/M$ on $M$ and $0$ everywhere else and $g$ the other way around.

And then solving for $p$ would give $p=2$.

Mike

12:19 PM

That seems right, the example I just came up with (for $\Bbb R$) was just the characteristic functions for $[0,1]$ and $[1,2]$, and the appropriate generalization of that argument would be yours

though I don't think $\Omega$ needs to be split into two sets - just pick two disjoint subsets of finite and equal measure and consider the characteristic functions on this set... but of course this is not guaranteed for all measures so perhaps we restrict to some nicer subset, probably regular measures

Chris's sis

Greetings

Balarka Sen

Hullo.

Chris's sis

@N3buchadnezzar I have a cookie for you. Are you around? :-)

Balarka Sen

Phew, it took some calculations to get the partial result on transcendence right...

Can someone tell me what's the link group for Whitehead link is?

Sush

Please someone answer this.

user112495

1:10 PM

Let p(x) be a polynomial with positive leading coefficient. Show that there exists a* in R such that P(x) + a = 0 if a<a*. How do I go about doing this?

Balarka Sen

@Nick Digits of Pi aren't random, actually. It's what we call "pseudorandomness" -- random but deterministic (Bailey–Borwein–Plouffe! Fun! Fun! Fun! ;D).

user112495

Please can someone help with my question?

Balarka Sen

@JasperLoy Hullo.

@user112495 what have you tried?

Jasper Loy

@BalarkaSen Hi, do you live near Nick?

Balarka Sen

@JasperLoy Nick presumably lives in Gujarat, whereas I am in west Bengal. So, no.

Daniel Fischer

1:12 PM

@user112495 Do you know the intermediate value theorem?

user112495

I think I may be getting a bit confused with the question. But I tihnk i've probably got to use the IVT but I can't really make any progress on it.

Jasper Loy

@BalarkaSen But both in India, so yes, lol.

Huy

@Mike: We have seen a theorem stating that if $M$ is seperable, then $A \subset M$ is as well. To prove this, doesn't $D \cap A$ where $\overline{D} = M$ suffice?

Balarka Sen

@JasperLoy Where do you live?

Jasper Loy

@BalarkaSen Singapore

Balarka Sen

1:13 PM

@JasperLoy Uh? I didn't know people can have name "Jasper" there? Or is it "Joy"? Joy Lasper -- yes, that'd fit well.

user112495

@BalarkaSen But I think i've probably got to use the IVT but I can't really make any progress on it.

@DanielFischer Yeah, I do know the IVT.

Daniel Fischer

@user112495 What does the sign of the leading coefficient tell you about the values of $p(x)$?

Jasper Loy

@BalarkaSen Why not? It's just an ordinary English name.

user112495

That p(x) tends to infinty as x tends to +_infinty

@DanielFischer
That p(x) tends to infinty as x tends to +_infinty

@DanielFischer That p(x) tends to infinty as x tends to +_infinty

Balarka Sen

@JasperLoy I see.

user112495

1:15 PM

Sorry for the triple post.

Daniel Fischer

@user112495 in Particular, they are positive for large $x > 0$.

user112495

So that can give us some value to use as an upper bound to use in the IVT?

Daniel Fischer

That doesn't change when you add a constant and look at $p(x)+a$.

@user112495 You don't need a value, you just need that one exists.

user112495

What do you mean by it doesn't change?

Daniel Fischer

@user112495 So now you need to find an $a^\ast$ such that for all $a < a^\ast$, you know that $p$ has at least one negative value.

@user112495 I mean that the values of $p(x)+a$ will still be positive for large $x$, whatever $a$ is. What "for large $x$" means depends on $a$, but not the fact.

Balarka Sen

1:19 PM

@JasperLoy My geographical sense limits me to thinking that singapore must be somewhere on Africa. =P

Jasper Loy

@BalarkaSen No, it is in Asia. Don't forget me when you win the Fields medal.

user112495

@Daniel Sorry, i'm still confused

@DanielFischer Sorry, i'm still confused.

Daniel Fischer

@user112495 What about in particular?

Balarka Sen

@JasperLoy I will (en.wikipedia.org/wiki/Vacuous_truth)

@JasperLoy Asia? Is that in Europe-Russia border?

Jasper Loy

@BalarkaSen No, it is on the southern tip of Malaysia, lol.

user112495

1:22 PM

@DanielFischer I think possibly just about what the question is asking. So I need to show that there exists an a* such that P(x)+a=0 has a root if a<a*. So do we just need to choose a* such that a_{2n}+a*<0?

Balarka Sen

@JasperLoy Where Indian revolution happened? (And they say my historical knowledge is weird. Hmph.)

Jasper Loy

@BalarkaSen I forgot my history, lol.

Daniel Fischer

@user112495 That choice will not always work.

Balarka Sen

@JasperLoy I believe Malaysia is somewhere on Paris, isn't it? Where WWI happened?

user112495

@DanielFischer Why not?

Jasper Loy

1:24 PM

@BalarkaSen What? Paris is in Europe!

Mats Granvik

@JasperLoy Have you been to the bird park in Singapore?

Jasper Loy

@MatsGranvik Maybe twenty years ago as a kid...

Daniel Fischer

@user112495 $$p(x) = x^4 + 3x^2 + 5$$

Balarka Sen

@JasperLoy Really? Then is Malaysia on squadron 36 in the uncharted space?

Mats Granvik

@JasperLoy I was there 30 years ago, when I was a kid.

Jasper Loy

1:26 PM

@BalarkaSen No idea, but it is also in Asia, lol.

Balarka Sen

lol.

Jasper Loy

@MatsGranvik I see. Anyway, I hope you solve Riemann Hypothesis soon!

Balarka Sen

Multidimensional intersection of alternate universes.

Mats Granvik

@JasperLoy me too.

Balarka Sen

@JasperLoy RH?

user112495

1:27 PM

@DanielFischer But then wouldn't just choose a* such that the constant +a*<0? So a*<-5

Jasper Loy

@BalarkaSen Well, he is very interested in the zeta function.

Daniel Fischer

@user112495 That would work.

Balarka Sen

@JasperLoy I see. Who would not be? It's one of the fascinating branches of mathematics.

Sush

@BalarkaSen, I think you are mixing me and Nick. I live in Gujarat, but as Nick said, he speaks Dravidian language, as he said on that day, so he might be from some southern Indian states, though not sure!

Balarka Sen

@Sush Perhaps.

Daniel Fischer

1:29 PM

@user112495 Since the condition is $a < a^\ast$, choosing $a^\ast = - a_0 = -p(0)$ is a possible choice.

Sush

So, let me ask Nick him/her self!

@Nick, Where do you live?

Balarka Sen

@Nick where do you live?

Done.

Lol.

Sush

@BalarkaSen, haha, my was Where and your is where!

user112495

@DanielFischer Ahh. Thanks. And so then by the IVT, we know there must be a point where p(x)+a=0?

Balarka Sen

I am a Legilimence, as I have said before.

A skilled one.

Jasper Loy

1:31 PM

I am going to see my psychiatrist tmr.

Balarka Sen

So beware.

@JasperLoy psychiatrist?

Sush

@BalarkaSen, are Sens Brahmin? Why are Sens so brilliant and progressive in general?

Jasper Loy

@BalarkaSen Yes, I have OCD, I told Nick.

Sush

@BalarkaSen I am undergrad in Economics and am fan of Amartya Sen.

Balarka Sen

@Sush I am not. I am rather lower class, but then do you believe these things? In math, everyone's equal and are mathematicians

@Sush Ah, I see.

Huy

1:33 PM

@DanielFischer: For $A: D_A \subset X \to Y$ linear with $\overline{D_A} = X$, we defined its dual operator $A^*: D_{A^*} \subset Y^* \to X^*$ to have the domain $$D_{A^*} = \{y^* \in Y^* | \, \ell_{y^*}: D_A \ni x \mapsto y^*(Ax) \text{ is continuous}\}.$$ Is $\overline{D_{A^*}} = Y^*$?

Balarka Sen

@JasperLoy Oh.

Daniel Fischer

@user112495 Yes. Then $p(0)+a < 0$, and $p(x) + a > 0$ for large enough $x > 0$.

Mats Granvik

@JasperLoy I got my first computer in Singapore. I remember changing the source code to an ice-cream selling program. The user interface was only numbers and a few words.

Daniel Fischer

@Huy Not necessarily. There are operators with $D_{A^\ast} = \{0\}$.

Huy

@DanielFischer: Could you give me an example of such an operator?

Jasper Loy

1:35 PM

@MatsGranvik Hahaha.

Balarka Sen

@MatsGranvik good one.

Sush

@BalarkaSen, I just wanted to know, by the way where is Sodhepur? your school? Is it a suburb of Kolkata?

Daniel Fischer

@Huy Not explicitly off-hand, but take an operator whose graph is dense.

Balarka Sen

@Sush It's in north 24-parganas, panihati.

Sush

@BalarkaSen I see.

Balarka Sen

1:37 PM

@Sush I don't really live exactly on the middle of Kolkata, but I am sure it is not superb. Quite the other way.

Sush

@BalarkaSen, haha!

Kolkata has amazing institutes like ISI, where I am going to apply, @BalarkaSen

Balarka Sen

@Sush Unfortunately $\sf ISI$ has just started B-math, so I don't think I am going to apply there for math anytime soon.

But I heard that it was great on Eco.

I live right beside ISI, do you know that?

Sush

@BalarkaSen, yes it is the best in India for Econ.

@BalarkaSen, Oh amazing! Do you have access to ISI?

@BalarkaSen, ISI has world fame in Econ and Stats.

Balarka Sen

@Sush Not yet. A professor on Bangalore referred me to another professor who is currently at ISI and if I contact him, I should have. I haven't done that yet, since I hadn't had the need to (I usually find the papers I need by contacting the authors if very necessary).

@Sush Yes, it has quite a bit of fame in stats.

Sush

@BalarkaSen, bye, as my mother is calling me!

Balarka Sen

1:46 PM

@Sush Bye. I have to go too for I have got some reading to do. (pouff)

Huy

3:17 PM

Let $f(z) = p(z) + \sum_{m=1}^M c_m (\alpha_m - z)^{-k_m}$ be a rational function. The spectral mapping theorem states that $f(A) \in \mathrm{Gl}(X)$ iff $\forall z \in \sigma(A): f(z) \neq 0$ and (second statement) $\sigma(f(A)) = f(\sigma(A))$.

I have a small problem with the proof of $\implies$ - direction of the first statement. In our notes, this is proven by contraposition, i.e. let $\lambda \in \sigma(A)$ with $f(\lambda) = 0$. By polynomial division, we obtain a rational holomorphic $g$ with $f(z) = g(z) \cdot (z- \lambda)$ and therefore $f(A) = g(A) (A-\lambda) = (A-\lambda)g(A)$.…

skullpatrol

3:35 PM

hi @JayeshBadwaik how are you?

Balarka Sen

Hullo @Skull.

skullpatrol

:-)

Peter Sheldrick

4:51 PM

wow that 1+2+3+4..=-1/12 stuff really went viral

Huy

@PeterSheldrick: That YouTube video where $\sum_{k=0}^\infty (-1)^k = \frac{1}{2}$?

Peter Sheldrick

yep, and the social network stuff attached to that

skullpatrol

anyone who thinks positive integers can add up to a negative fraction definitely has a viral infection

Peter Sheldrick

boingboing.net/2014/01/17/why-the-sum-of-all-positive-in.html

i frist saw it on twitter

Huy

I didn't watch it yet but a friend (engineer) asked me about whether the proof was correct and I stopped when that divergent sum was "set" to be 1/2. Is the rest of it at least correct?

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$Mathematics$ Mathematics