Mathematics - 2015-06-14 (page 1 of 3)

Ashley Davies

00:02

Oh I typed the original question wrong, my bad

Where I concluded that theta is arctan half, they concluded it's that or cot theta = 0 and they got two extra solutions from that (90 & -90)

anon

@AshleyDavies I cannot follow your work at all

okay, following a bit

Ashley Davies

I probably overcomplicated it a bit; essentially all I did was substitute in the identify cosec^2 x = 1 + cot^2 x then multiply throughout by tan^2 x after cancelling the +1 on both sides

anon

00:18

when you converted cos/sin to 1/tan (two times) you lost the possibility of cos=0

for instance, cos/sin=0 has 90 degrees as a solution, but 1/tan=0 has no solution

Ashley Davies

I think that makes sense - I thought the two were equal though?

(I'm not doubting you, just trying to get my head around it so I can catch similar mistakes in future)

anon

do you think 0/1 equals 1/(1/0) ?

you can't divide the top and bottom of a fraction by 0

you can't divide by 0 period

so dividing the top and bottom of cos/sin by cos is only valid when cos is not 0

Pedro Tamaroff

@anon You missed a comma there.

Ashley Davies

Ah, that makes sense, so my solution is only valid for theta =/= arccos 0 and that's why I'm missing them?

anon

@AshleyDavies is only valid for cos(theta)=/=0

Ashley Davies

00:28

Awesome, I understand it now! Thanks! :) I've been struggling to get to terms with this all year, I was just thinking of cot as cos/sin rather than 1/(sin/cos)

Sorry to be a pest!

Joe

Hi. Why isn't taylor quadriture used very often (or at all, as far as I can tell)? I found a vague reference to it being unstable somewhere, but in my use case it works very well.

I am fortunate enough that higher-order derivatives of the function I'm integrating are very (very) simple algebraic equations of the lower-order ones. Which I'm thinking might explain why my case is numerically stable.

Semiclassical

01:38

@Chris'ssistheartist: following up on that bessel summation question, i found a question which addresses a generalization of the original one, and moreover with an answer that states a closed form in terms of complete elliptic integrals

4

Q: There is a closed form for $\sum _{n=1}^{\infty }{\frac {{{\it J}_{0}\left(\,\alpha\,n\right)} {{\it J}_{0}\left(\,\beta\,n\right)}}{{n}^{2}}}$?

Using the method showed here proposed by Olivier Oloa with simplifications proposed by Anastasiya-Romanova, it is possible to prove that $$\sum _{n=1}^{\infty }{\frac {{{\it J}_{0}\left(\,2\,n\right)} {{\it J}_{0}\left(\,n\right)}}{{n}^{2}}}={\frac {5}{8}}+{\frac {1}{6}}\,{\pi }^{2}-4\,{\frac { ...

Semiclassical

01:51

taking the limit $\beta\to\alpha$ is a little tricky, though, since the complete elliptic integral of the first kind diverges at unit argument

i'm working right now to see if i can get mathematica to compute it

hi @bolbteppa

did you figure out k-theory? :)

bolbteppa

If you think of a module as a vector space where the 'scalars' are functions instead of numbers from a field, what is the intuition for a projective module?

Hey @Semiclassical yeah I am happy with K theory now!

On a vague level

Semiclassical

neat

bolbteppa

If you think of vector bundles as 'linear algebraic topology' then the theory of characteristic classes is the cohomology of vector bundles, i.e. a way to measure non-exactness, or twistiness... Then K theory is just talking about direct sums of vector bundles and analyzing the cohomology invariants associated to this structure,

Have you made any progress with classical integrability?

Semiclassical

a bit, but i haven't been focusing on it lately

i did get a bit of a handle on some of the machinery, at least

bolbteppa

Me neither, it's something I just periodically dip into

Semiclassical

01:59

for example, i now recognize where Toeplitz determinants can crop up

bolbteppa

I would now like to understand in quantum integrability how the whole trigonometric R matrix stuff in the Ising model is an example of representation theory of quantum groups

Semiclassical

yeah, um, have fun with that :)

bolbteppa

You too ;)

Semiclassical

i'm happy enough looking to understand how classical integrability is useful for quantum mechanics

bolbteppa

If you find easy dumbed down explanations of this stuff send them to me ;)

Semiclassical

02:00

by which i mostly just mean schrodinger differential/ difference equations and reductions thereof

will do

bolbteppa

Yeah

Semiclassical

right now i'm more trying to get some understanding of the Riemann-Hilbert problem

Enjoys Math

How do you define the direct product of modules using functions again instead of tuples?

Semiclassical

which does show up in babelon's book, albeit not in a way that i can really get a lot out of

but i've got a few books from the library which should help

Enjoys Math

The set of all functions $\omega : I \to \cup_i M_i$ where $I$ is the index set over which the product is being taken. Or something...

Mary Star

02:08

Hello @robjohn !! Are you familiar with Cryptography and especially with the voting scheme?

bolbteppa

@Semiclassical yeah I had the exact same problem, tried to find out what R-H actually said, found Babelon no help, still have no idea, but just now I think I found something good ams.org/notices/200311/fea-its.pdf

Semiclassical

yeah, that's what i found too

and i've got a book on Painleve transcendents via RHP with him as an author, so that should be a strong source (readable or not)

i also grabbed percy deift's book on RHP applied to orthogonal polynomials and random matrix theory.

so hopefully between all of those i'll be able to triangulate something

bolbteppa

"the presence of the spectral parameter in the Lax pair brings the tools of complex analysis into the problem, and this ultimately transforms the original problem of solving a system of differential equations into the question of reconstructing an analytic function from the known structure of its singularities.

In turn, this question (almost) always can be reformulated as a Riemann-Hilbert problem of finding an analytic function (generally matrix valued) from a prescribed jump condition across a curve."

Semiclassical

right

bolbteppa

So the spectral parameter is the time-independent eigenvalue

Semiclassical

02:12

what confuses me a bit is that one gets into deformations which are iso monodromic rather that iso spectral

bolbteppa

"In general, the Riemann-Hilbert formalism provides a representation in terms of the solutions of certain linear singular integral equations, which in turn can be related to the theory of infinite-dimensional Grassmannians and holomorphic vector bundles."
Amazing, grassmannians which we were talking about last time, and randomly vector bundles from the K theory I just mentioned :p

Semiclassical

yep

bolbteppa

Well, monodromy is what happens to the solutions of, say a linear 2nd order ode, when you pass around a singularity of the ode

Semiclassical

which doesn't actually shock me much, seeing as one of the main motivations of this from the physics side is to do work with some simple types of topological insulators/superconductors in 1D

aka the kind of thing which K-theory is used to classify :)

bolbteppa

I had something on this before

Semiclassical

02:14

yeah, i do grok monodromy

bolbteppa

Yeah, you can see an insulator as like a base space plus something :p

I wish I knew this stuff better, it's cool stuff

Semiclassical

yeah. my research partner has been familiarizing himself with the whole classification, though from the perspective of time-reversal / particle-hole / chiral symmetries rather than K-theory

bolbteppa

Intuitively, my guess is that it's ultimately just like solving a non-linear PDE using an integral equation (the Gelfand-Marchenko-Levitan integral equation) which has singularities thus requiring an analysis of the monodromy of the solutions, thus solutions with similar monodromy are homotopic or something, and isospectrality is the condition for integrability

Semiclassical

mmkay

the thing i'm going to try to use as a foothold on this is the orthogonal polynomials aspect

since i (mostly) understand how that shows up

plus it'll come in handy if i ever want to do random matrix theory

Samuel Yusim

@EnjoysMath you can define it as the unique module $M$ (up to isomorphism) with the following property: if $\pi_i: M \to M_i$ is a surjective homomorphism for each $i$ then for any set of homomorphisms $f_i : N \to M_i$ there's a unique homomorphism $f: N \to M$ such that $f_i = f \circ \pi_i$ for all $i$.

bolbteppa

02:21

Well, I do know the ultimate jist of this theory comes from looking at dL/dt = [B,L] where you assume L is assumed to have regular singularities, so from this you can rewrite it as a fuschian system as in the first equation here http://www.ucl.ac.uk/~ucahrha/LTCC/CA-LTCC-2012-5.pdf
so since this is a non-linear pde with a lax form it can be solved by the inverse scattering integral equation

Or something not too far from that anyway

Semiclassical

yeah

i know they also describe it as a kind of complex WKB method, which is intriguing

especially since the first paper i was a co-author on was about semiclassical quantization of non-self-adjoint periodic potentials :)

aka the kind of thing which is properly derived from WKB

bolbteppa

I can see orthogonal polynomials as somehow coming out of the solution to this setup, and the whole random matrix stuff as coming out of the path integral formulation of this set-up, which obviously links to orthogonal polynomials since both relate to the solution to this problem set up, all vague though

Semiclassical

yeah. if we're flying right now, it's by the seat of our pants :P

bolbteppa

That sounds cool, the wiki en.wikipedia.org/wiki/… seems to say you deform a problem so that you can only then apply stationary phase to it

Semiclassical

yeah

i really dealt more with the output of WKB rather than the guts of it, though

namely taking classical action integrals, requiring them to be quantized, and checking that that gave the correct energy levels (within the approximation, of course)

Bohr-Sommerfeld quantization

bolbteppa

02:27

Yeah

Semiclassical

which did give some fun pictures

arxiv.org/abs/1303.6386 is the preprint

for example, figure 12 is kind've fun :P

Karim Mansour

@Semiclassical

hi

Semiclassical

hi

Karim Mansour

had you know 90 in mechanics exam I made algebra mistake would have gotten 100 otherwise :S

1 moment let me show you my exam

Semiclassical

ah, algebra. i can usually forgive myself for algebra errors in a scenario like that, as long as i remember to check my result to make sure i didn't do something obviously wrong

(like wrong units, or results waaaay too big, etc.)

Karim Mansour

02:37

Semiclassical

that's what you showed before, no?

Karim Mansour

Semiclassical

glad i don't have to do those anymore. just too tedious for my taste

Karim Mansour

no not those

yeah

Clarinetist

02:56

Looks like multivariable calc

Or physics?

Balarka Sen

bah, I am up way too early.

Karim Mansour

physics

mechanics 3rd level

@Clarinetist

Clarinetist

I think I'll just stick to probability measures :P

Balarka Sen

03:13

@MikeMiller This guy says Munkres uses covering spaces to prove the lemma. I don't recall this at all. Can't we just construct the homotopy $H : S^1 \times I \to X$ by taking $H(\gamma, t)$ to be the $t$-th intermediate of the path-homotopy of $\gamma$ to the constant loop?

Mike Miller

considering $f_*(\text{id}) = f$ it's a tautology that $f_*$ is the trivial homomorphism iff $f$ is null-homotopic

Balarka Sen

right

and what I say works, right?

Mike Miller

can't really parse what you're saying

not sure what there is to say

Balarka Sen

I mean, any loop $\gamma : S^1 \to S^1$, when composed with $h : S^1 \to X$, can be path-homotoped to the zero loop. Let $\{f_t\}$ denote such a path homotopy, with $f_0 = h \circ \gamma, f_1 = e_0$

Set $H : S^1 \times I \to X$ to be $H(-, t) = f_t$

Semiclassical

sigh. trying to track down an error in an answer based on numerical evidence alone is frustrating me

Mike Miller

03:17

yes, I agree

Semiclassical

math.stackexchange.com/a/940612/137524 doesn't appear to quite work numerically, but it's very close

Balarka Sen

ok, good. i'll post that as an answer, then.

Semiclassical

so presumably there's some little typo. but finding it is ugh

Mike Miller

seems to me Qiaochu's comment should suffi e

don't understand why it wasn't resolved then

Balarka Sen

Munkres doesn't use S^1 in the domain of the loops, I guess that's the origin of his problem.

Mike Miller

03:19

then the comments after that should have resolved it :p

Balarka Sen

and Munkres mucks up all his proof using rigorous calculations, as always :(

Mike Miller

but yes, i encourage posting a full answer with all this

Samuel Yusim

I was thinking about how you usually go about showing something like $(\Bbb Z / 10\Bbb Z)\otimes_{\Bbb Z} (\Bbb Z / 12\Bbb Z) \cong (\Bbb Z / 2\Bbb Z)$. In particular, to show $1 \otimes 1 \neq 0$ you usually look at the bilinear map $f: (\Bbb Z / 10\Bbb Z) \times (\Bbb Z / 12\Bbb Z) \to (\Bbb Z / 2 \Bbb Z)$ given by $(a, b) \mapsto ab$, but is there a way to see that $1 \otimes 1 \neq 0$ without looking "outside of the tensor product"?

Semiclassical

hrm. or maybe i'm being careless about the domain of the elliptic integrals. blah

yep, careless on my part

Samuel Yusim

I guess more generally, what's the most trick-free possible approach to arguing that a particular tensor product of modules is nonzero?

I ask because the above question was given in a text I'm looking at right after the definition of the tensor product, and before any notion of bilinearity was mentioned

given what the text would expect, how can I do this? it seems harder than it should be for me

Balarka Sen

03:31

Done, @Mike.

Let's see if this resolves his confusions.

Rajesh D

1

Q: Countering Gibb's phenomenon, and approximating jumps with jumps in Fourier Analysis : An attempt and a question in this regard

Main Question (this section is fully self contained and does not need anything that comes further, for answering the question). Let $f:\mathbb{R}\to\mathbb{R}$, $f(t) = 0, t<0$. Let $f \in L^2(\mathbb{R})$ and is locally BV. Let its jump set be $\{x_i\},i=1,2,3,...,x_i>0$, and the corresponding...

real-analysis fourier-analysis harmonic-analysis fourier-transform

Mike Miller

which dwfinition of tensor product @SamuelYusim

Samuel Yusim

the quotient of the free module by all the stuff

Mike Miller

what's the stuff, can you remind me?

Anthony

Hey everyone, anyone have any thoughts on how to get rid of the (t+1)! in this :math.stackexchange.com/questions/1323183/…

Also hey @MikeMiller

Samuel Yusim

03:42

take the free $A$-module with generating set $M \times N$, then mod out by the submodule generated by the elements $(m_1,n) + (m_2,n) - (m_1 + m_2, n)$, $(m, n_1) + (m, n_2) - (m, n_1 + n_2)$, $a(m,n) - (am, n)$, and $a(m, n) - (m, an)$ for all $m, m_1, m_2 \in M$, $n, n_1, n_2 \in N$, and $a \in A$.

it's the bad-but-concrete definition

Mike Miller

I guess you can probably calculate precisely what that ideal is but I just fiddled on paper and did not have a good time

submodule* is

Balarka Sen

04:32

lol

2

Soham Chowdhury

05:05

@ShinKim look at my question history, I asked about that once on main.

half of the project is over :)

Enjoys Math

:0

::: math.stackexchange.com/q/1324601/26327

@TedShifrin

Remember me

06:28

Hey@Balarka

Balarka Sen

hi

Remember me

How to ask dumb questions?? hahaha

Balarka Sen

there's nothing funny about that

the flow-chart's what's funny.

Remember me

I do mean the answer

I was able to do questions of munkres without taking any help... Thanks to Simmons

@Balarka Can we take the product and quotient of two topological spaces

Balarka Sen

yes

Remember me

06:34

How I mean do I have to write ..
$\frac{X}{Y}$

Balarka Sen

there are well-defined notions of product and quotient of top. spaces out there. it's in Simmons, I think.

@Remember you have to define what quotients mean first before bothering about notations of quotient spaces :P

Remember me

Definitions

Hey@Skill

morning @Mike

skill patrol

Hi pal @Rememberme

How are you?

Remember me

07:03

@Balarka I had this question ...
let me define a set $B:\{[a,b) : a,b \in \Bbb{Q}\}$ . Does the topology generated by this set B different from the lower limit topology?

Soham Chowdhury

07:36

heh. chat is ded.

[sic]

Balarka Sen

yes, @Remember (why?)

hello @Soham

Soham Chowdhury

is Sayan doing Counterexamples or what?

what's up, B?

still sick?

Balarka Sen

nope, I'm alright now.

Soham Chowdhury

mmh, good.

Balarka Sen

trying to make up some notes of the Galois theory I know of to make sure I don't forget it.

Soham Chowdhury

07:40

cool.

Balarka Sen

finished with the darn project of yours?

Soham Chowdhury

tell me an interesting theorem (in the spirit of Borsuk-Ulam or Banach-Tarski) from Galois theory. other than the obvious quintics one.

no, still have ~15 pages to write.

and ~20 graphs.

i have 15 Uttarakhand tabs open and my cheap laptop doesn't like it. damn.

Balarka Sen

@SohamChowdhury not much low-brow applications I can think of out there.

of course, there are all kinds of impossibility results (trisecting angles/doubling cube/etc)

Remember me

@Balarka I was just thinking of topologies generated by sets which seem to be quite the same

Balarka Sen

high-brow applications would be the Kummer theory, say, to solve FLT for regular primes.

Remember me

07:46

Well I got to know of something very nice in topology.... seems very interesting and intuitive

Gluing....

Balarka Sen

everything in mathematics can be made intuitive, if you play with them a lot

Remember me

Yup I also thought of something ....
How can you get a sphere from a disk

Start gluing the ends to a point

Balarka Sen

Define gluing.

Remember me

Its defined using equivalence relations and quotient spaces

Balarka Sen

Well, write it down :P

Remember me

07:49

Equivalence relation I know

Balarka Sen

I mean, write down the definition of quotient topology.

Remember me

Quotient topology.. ahh dont seem so :p

Balarka Sen

And prove the quotient of a disk by it's boundary circle is a sphere.

Intuition is great, but it's not good to talk about intuitive things without knowing the right definitions.

Remember me

Oh my god @Balarka You are shelling me with questions :p

Well but I can say this much (now) Gluing is pretty intuitive

Balarka Sen

No, I am trying to get you to understand that you'll attain mathematical maturity only when you both have the intuition and the correct translation of intuition to rigorous mathematics in mind.

@Rememberme Even topological quantum field theory is intuitive, but it's useless if you don't know the definition :P

skill patrol

07:51

Define intuition.

Remember me

yes you are right @Balarka

lol@Skill

Intuition

Intuition, a phenomenon of the mind, describes the ability to acquire knowledge without inference or the use of reason. The word "intuition" comes from Latin verb intueri translated as consider or from late middle English word intuit, "to contemplate". Intuition is often interpreted with varied meaning from intuition being glimpses of greater knowledge to only a function of mind; however, processes by which and why they happen typically remain mostly unknown to the thinker, as opposed to the view of rational thinking. Intuition has been subject of discussion from ancient philosophy to moder...

I think this will suffice you @skill

skill patrol

Good start @Rememberme :-)

en.m.wikipedia.org/wiki/Mathematical_maturity

Chris's sis the artist

08:22

@Semiclassical Oooooooooooo, that is nice! I didn't think of such an amazing form.

Sunday here, a great day for hard work.

2

BBL

Soham Chowdhury

"A great day for hard work." That's a very nice way to put it. :)

3

Balarka Sen

08:43

whoa, what? homotopy equivalent aspherical manifold are homeomorphic?

Mike Miller

Conjecturally.

(Add the word closed, but that was implied. There's a modified, stricter version that's true for compact manifolds with boundary in small dimensions; I'm not sure if it conjecturally extends to all dimensions.)

Balarka Sen

oh, I see.

pretty strong conjecture.

Mike Miller

Known for dimensions at most 3. Known for manifolds with enough geometric structure to work with. Issue with higher dimensions, I think, is that there's just not enough relation between the topology and the geometry to get a handle on the problem.

Balarka Sen

Did they actually prove this for 3-manifolds, or did they just check through all the manifolds using the classification theorem (if that's even possible?)

Mike Miller

The correct statement of the theorem is "A homotopy equivalence $M \to N$ of aspherical closed manifolds is homotopic to a homeomorphism". This involves checking on the level of maps, not manifolds, so even if you wanted to "check a list" (not exactly feasible in the world of 3-manifolds) it's not going to prove the full theorem.

There is not a classification theorem like there is for 2-manifolds.

Balarka Sen

08:57

ahh, the corrected statement makes sense.

I don't know anything about 3-manifolds, so I am going to believe you about the classification theorem bit.

Mike Miller

All manifolds here are orientable. "Every 3-manifold has a unique prime decomposition. Every prime piece has a (unique minimal choice of, I think) decomposition along tori into pieces with a finite volume geometric structure. These geometric structures are fairly easily understood for 7 of the geometries, and the last piece is pretty rough."

Balarka Sen

@MikeMiller Decomposition in the sense of connected sums or what?

Anonymous

How would someone start with this problem?math.stackexchange.com/questions/1296342/…

Balarka Sen

And by geometric structure, you mean a geodesic metric space/Riemannian manifold structure?

Anonymous

I don't think it's always possible,as mentioned in the comments

Anonymous

09:10

@SohamChowdhury Someone,don't worry!

Anonymous

@BalarkaSen probably lives in a microscopic world,he seems to have an intuition for Quantum Mechanics.

Balarka Sen

09:29

@Ashwin it's a common misconception that anything with the word "quantum" in it means it's quantum mechanics.

While topological quantum field theory might originate from physics, it has little to do with it.

Abdou Abdou

@Rememberme intuition has strong relation with sixth sens , i think intuition is the appearance of sixth sens in its modestest form

in the times of non-spoken communication means , intuition was in its highest use among people via telepathy , until the invention of "language"

i think there was a description for it , memetics (Mimesis) -actually memes- , it means the simulation and figuring human behavior 's evolution by time and ages

Paul Plummer

For example the "quantum BS" that pseudo-scientific cranks use to explain their BS, like water has memory @BalarkaSen

3

Balarka Sen

09:44

lol

S.C.

Hello

Paul Plummer

I don't even know why I post questions this late, they are not going to get any answers...

Balarka Sen

@Paul How're you doing, btw?

S.C.

Just thinking: Can we have a continuous function f defined on X =(0,1) such that f(X)= [1/2,1)??

Abdou Abdou

i think noone could make a remarkable advancement in my question

Paul Plummer

09:47

I am doing okay, have been doing bits and pieces of a bunch of things (or as you would say learning nothing), how about yourself? @BalarkaSen

Balarka Sen

I was ill all the past week, but now I am back to doing multivariable calc.

S.C.

@BalarkaSen Hey Balarka you hail from India??

Abdou Abdou

i couldnt understand quite the formula of poisson , but I think it is just an approximation , I want the exact limit , do I ask for much ?

Mike Miller

You can embed a set of tori. Now just consider the disjoint union of the closures of the connected components of the complement of this set of tori. Aka, cut along them.

Balarka Sen

That's what it means to "decompose"?

Mike Miller

09:55

In the tori bit, yes. Unless you're asking about the prime decomposition, then it's a connect sum decomposition. (Which is essentially equivalent to this but with spheres.)

Balarka Sen

ah, makes sense.

Paul Plummer

@S.C. Think about a parabola

Well I am going to try to go to sleep now...

Mary Star

10:12

Hello @robjohn @ThomasAndrews !! Are you familiar with Cryptography and especially with the voting scheme?

Clash

Hey guys, what is this function supposed to do? $diag(A_1, A_2)$ thanks for the help! I thought the diag function would just return a matrix with $0$s everywhere except on the diagonal, but the comma is trying me on a loop

evinda

10:45

@Clash Hi!!! Do you mean in Matlab?
DIAG(V,K) when V is a vector with N components is a square matrix
of order N+ABS(K) with the elements of V on the K-th diagonal. K = 0
is the main diagonal, K > 0 is above the main diagonal and K < 0
is below the main diagonal.

DIAG(X,K) when X is a matrix is a column vector formed from
the elements of the K-th diagonal of X.

Is anyone familiar with differential equations? I have the following question:

1

Q: Why is the general solution of this form?

I found the following in my lecture notes: $$u_t=u_{xx}, x \in \mathbb{R}, t>0 \\ u(x,0)=f(x)$$ $$u(x,t)=X(x)T(t)$$ $$\Rightarrow \frac{T'(t)}{T(t)}=\frac{X''(x)}{X(x)}=-\lambda \in \mathbb{R}$$ $$X''(x)+\lambda X(x)=0, x \in \mathbb{R}$$ $$X \text{ bounded }$$ The characteristic equation is...

differential-equations heat-equation

Clash

Thanks @evinda, I just posted the question and realized it's a diagonal made out of blocks of matrices, so blkdiag did the trick on Matlab! Thanks!

evinda

@Clash You are welcome :)

Semiclassical

@PaulPlummer as a grad student in condensed-matter physics, that kind of thing is teeth-gnashingly annouying

just because QM is strange and hard to understand doesn't mean it predicts telekinesis or telepathy

evinda

Hi @Semiclassical
What's up?

Semiclassical

hi. up earlier than usual, and not really awake yet

evinda

10:57

@Semiclassical what time is it there?

Semiclassical

coming up on 6am

evinda

@Semiclassical Aha... Do you have exams now?

Semiclassical

nah. i'm past the course-taking stage of my PhD

evinda

@Semiclassical So are you in the research field now?

Semiclassical

i'm doing research, yeah

evinda

11:01

@Semiclassical In physics?

Semiclassical

yep

evinda

@Semiclassical Nice.... :)

Semiclassical

@Chris'ssistheartist i'm pretty tickled by it myself. though I wish he'd included the domain of applicability (i.e. what branch cuts have to be introduced)

that's absolutely crucial for not mis-understanding how the function behaves under analytic continuation

math110

@Chris'ssistheartist,@robjohn, can you see detai prove : Q-function equal complementary error function？

Anonymous

@BalarkaSen Got it,thanks!

Semiclassical

11:09

kay, back to sleep i think

Anonymous

@SohamChowdhury I saw this question of yours:math.stackexchange.com/questions/392374/…

Calculus

Someone please help with the integrals?

Anonymous

@SohamChowdhury Visual Group Theory by Nathan Carter fits the bill!

math110

Calculus

$$\int_{-\infty}^\infty \frac{x^2}{(x^2+1)^2(x^2+2x+2)} dx$$

Someone else with the beautiful integrals

Anonymous

11:12

I don't think such integrals are beautiful LOL

Calculus

You don't think that one is beautiful?

Anonymous

I vaguely understand them,maybe that's why.

Calculus

Is anyone good with integrals here?

math110

@Calculus,you can partial fraction expansion

Calculus

I suppose it looks like $\frac{1}{x^4}$

math110

11:14

Note $$\dfrac{x^2}{(x^2+1)^2(x^2+2x+2)}=\dfrac{2x-1}{5(x^2+1)^2}-\dfrac{2(3x+1)}{25(x‌^2+1)}+\dfrac{2(3x+7)}{25(x^2+2x+2)}$$

Calculus

How did you do that so quickly?

math110

also see :[wolframalpha.com/input/…

I use wolf computer

Calculus

The 5, and 25 normally aren't on the denominator, they were introduced via calculating ax+b etc

Chris's sis the artist

@Semiclassical I see. I didn't study the details yet. The experience in the area of elliptic functions is low.

math110

if you partial,then your integral is easy to find it

Calculus

11:17

@math110 Can you recap me on the numerator of partial fractions?

so I will have denominators with (x^2+1), (x^2+1)^2 and (x^2+2x+2)

And numerators are A, Bx+C, D respectively?

math110

yes

Calculus

Thank you

Chris's sis the artist

@robjohn are you aware of any function that has its first derivative as his function but rotated at 90-degrees? I was just asked that. Maybe I should think more of it. If it's easy, don't tell me the solution.

math110

You are welcome.，@Calculus

Chris's sis the artist

@math110 OK

math110

11:19

@Chris'ssistheartist,hello, you see this proof with my integral

Chris's sis the artist

@math110 Which proof?

math110

this reslut can find in wiki:[en.wikipedia.org/wiki/Error_function]

this indenity is very interesting,But I tried sometimes to prove,and at last,I Failed，and I google sometime,I only find this paper:[ W. Craig, A new, simple and exact result for calculating the probability of error for two-dimensional signal constellaions, Proc. 1991 IEEE ]

But I can't find proof on this paper

First,when I see this problem,It seem not easy,But when I try some Substitution( for example,\dfrac{z}{\sin{x}}=u),all can't solve it.

Calculus

12:02

hi

can someone help me with showing $$\lim_{R\to\infty} \int_{C_R} \frac{z^2+8z+7}{(z^2+4)(z^2+2z+2)} dz =0$$

transversed positvely

@MaryStar you can help? I saw you doing integrals

center at 0

Mary Star

@Calculus Is $C_R$ a circle with center at 0 and radius R?

Calculus

12:28

@MaryStar Yep

Shin Kim

@SohamChowdhury Okay, i checked it. i perfectly agree with Siminore's opinion. it is a bad notation.

it will be much much better if authors change their habit to using the standard notations. otherwise it will be ambiguous and confusing.

Chris's sis the artist

12:55

I'm watching Interstellar, back later :-)

Mary Star

@Calculus We have a path integral over the circle with radius R. So, use $\int_{C_R} f(\xi) \mathrm{d} \xi| \leq L(C_R) \max_{ z \in R} |f(z)|$, where $L$ is the length of the circle.

Transcript for

$Mathematics$ Mathematics