Hello, @AlexClark

nothing, merely skimming through the algebraic topology questions

having fun with complex analysis?

I have to restudy complex analysis thoroughly at some point of time.

@AlexClark only if there are some nontrivial problems :P

there are a few problems here and there which I can't answer, but mostly because I haven't studied cohomology/homotopy. I usually stop by the problems I can't solve and try to pick up the tricks from the answers.

the only nontrivial question I answered here got something like 2 upvotes, and the most trivial answer got 8. sigh.

which Neumann problem?

@AlexClark basics, but forgotten a lot.

I never did the real complex analytic stuff (conformal mappings, etc). amongst the nontrivials, I recall residues because that's the only thing I have practically applied somewhere (proof of PNT).

I have no idea.

right, I didn't do that stuff.

I'll take your word for it :) I'm skeptic about branches I don't know about (which is not good, I admit).

ps, @AlexClark, have you guys done residues yet?

ugh

you might try to read up a proof of prime number theorem from somewhere, then, if you are interested.

are you, by any chance, taking analytic number theory next semester?

ah, i see

must be a tight schedule.

ah, i see.

@AlexClark Are you just talking about a PDE with a Neumann boundary condition?

Hello

@AlexClark @AlexClark That looks right to me, its been a long time since I did any PDEs though

@AlexClark Does the linearity partial derivatives not just immediately imply your result? I may be being stupid though (very long day now).

this phrase seems to have got viral

"what are you working on/thinking about"

Well I checked out a couple of books, and looking at what if anything inspires me to work on them more. I am looking at "Geometric Topology in Dimensions 2 and 3" by Moise @AlexClark

it's old alright, but i am seeing too much of it lately

geometric topology in dimension 2 must be sad

i like it, just saying that everyone seems to be using it :P

Not 100% sure yet, but it is one of the books I am thinking about keeping around. Frankly I should probabably just buckle down with one or two books, but I just can't seem to get myself to do that. My focus is not that great, and don't like just doing problem after problem, and doing that feels monotonous. (Part of the reason I started the blog, so that I could do projects, and then right stuff on them) @AlexClark

@BalarkaSen: What are you working on?

nothing, at this moment.

And what are you thinking about, @BalarkaSen?

@BalarkaSen What work are you doing? :P

sigh

Pretty sure that sense of completion is just an illusion :D

no, i think he means although we think we have finished some book, we really haven't.

haha, I like how you keep track of stuff by plotting your study functions

yes, better study than planning to study

Because "finishing" a text book is not really completing something , or necessarily something worth completing. It is like turning on and off the light switch 10 times

Obsessing to do every problem and read every page is typically not worthwhile, and you probably don't actually get what you want out of it, unless in the end all you want is to finish a book.

yet, it has been helpful for me to read everything in some certain chapter paragraph-wise, but i guess that's just because Hatcher hides his jwells in between the important theorems and proofs.

I don't think so, I don't really know @AlexClark

@AlexClark I really don't know what's the difference between geometric topology is 2-dimensions and manifold topology in 2-dimensions. That being said, it's about classifying 2-manifolds.

I presume you know what a manifold is?

oh, it's nothing very complicated. an n-manifold is just a topological space such that each point on it has an open nbhd homeomorphic to R^n

some people include hausdorff and second-countable just to make life easier

yes.

so for example, a sphere is a 2-manifold (take open nbhd to be the open half containing that point).

@AlexClark good question. there are 2-manifolds which cannot be embedded in R^3, funnily.

it is locally so. pick any point in S^2. consider a small open nbhd around that point.

isn't that nbhd hoemomorphic to R^2?

How about the projection of a neigh onto the plane @AlexClark

well, draw a small disk lying on $S^2$ around $x \in S^2$. consider the homeomorphism by throwing rays from the center of $S^2$ and sliding the disk through that line as long as it doesn't "lie flat"

so you have mapped your small nbhd to a disk in your tangent plane $T_x$ touching $x \in S^2$

that is, something homeomorphic to $\Bbb R^2$

almost, @AlexClark, but I throw my rays from the center instead

Well that is what it is doing @AlexClark

It is warping it flat

yeah.

(and maybe some "scaling")

for your question, it's hard to imagine a manifold that can't be embedded in $\Bbb R^3$ but the answer to it is klein bottle.

every proof of this i know is hard.

you'll learn about them if you do armstrong thoroughly

especially his chapters on quotient topology

anyway, the whole point of manifold topology in 2-dimensions is about classifying 2-manifolds upto homeomorphisms.

@AlexClark i have forgotten.

page 65.

"identification spaces" that's what you want to know thoroughly for most of algebraic topology.

you don't need to know anything else.

fact that might intrigue you, btw : any closed, compact, orientable (where you have a well-defined notion of "inside" and "outside") 2-manifold is homeomorphic to either a sphere, or a torus, or a double torus, etc...

yes, @AlexClar, armstrong is very good at doing a lot of stuff with ranting as less as possible and providing the best geometric intuition for each notion if possible.

@AlexClark yes.

ta-ta.

i have to go too.

See yah

$$\sum_{n=1}^{\infty} (-1)^{n+1}\frac{\displaystyle 1+\frac{1}{2}-\frac{1}{3}-\frac{1}{4}+\cdots \pm\frac{1}{n}}{n^2}$$

@r9m ^^^ (of course I can nicely write the numerator but that one is a part of the problem)

The version I wanna add to my book though is this one

$$\sum_{n=1}^{\infty} \left(\frac{\displaystyle 1-\frac{1}{2}+\cdots +(-1)^{n+1}\frac{1}{n}}{n}\right)^2$$

I removed some terms, there were too many.

@robjohn the version above looks amazing. It's a kind of sister of the Au-Yeung series excepting that there one had the alternating terms.

Marvellous!

@TedShifrin How are you doing? BTW, how would you tackle the last series? Not a full proof, of course, but a tiny starting point only.

@Chris'ssis The one I am sort of interested in is $$\sum_{n=1}^\infty\sum_{k=n+1}^\infty\frac{(-1)^k}{k\,n}$$

or even with $n^p$ in the denominator. That converges for $p\gt0$

@robjohn There are many things you can do there. You mean you cannot calculate it?

@Chris'ssis I have not tried. I just thought it would be interesting since it converges for a larger range of $p$. Does it have a nice closed form?

@robjohn Obviously.

I have a feeling I may have computed it in the past (without the $p$)

@robjohn $$\frac{\log ^2(2)}{2}-\frac{\pi ^2}{12}$$

@robjohn This is just $$\sum_{n\ge1}\frac{(-1)^n H_n}{n}$$

@Chris'ssis yeah... that does look familiar. I think I computed that over a year ago. I will have to look at my notes

@Chris'ssis Ah, yes $A(1,1)$

@robjohn I'm afraid I did a mistake though. Sorry.

here is where it was.

@robjohn I know all I have to do ... (just to check something)

@robjohn I think we can do somehting very interesting ...

$$\sum_{n=1}^{\infty} (-1)^{n-1} \frac{1}{n+x}=\log(2)-\psi(x)+\psi(x/2)+1/x$$

@Chris'ssis The sum I give above is actually $$\sum_{n=1}^\infty\frac{(-1)^nH_{n-1}}{n}$$

Then I multiply both sides by $1/x$ that yields

$$\sum_{n=1}^{\infty} (-1)^{n-1} \frac{1}{x(n+x)}=\log(2)/x-\psi(x)/x+\psi(x/2)/x+1/x^2$$

Then I replace $x$ by $k$

$$\sum_{n=1}^{\infty} (-1)^{n-1} \frac{1}{k(n+k)}=\log(2)/k-\psi(k)/k+\psi(k/2)/k+1/k^2$$

Then I multiply both sides by $(-1)^{k+1}$ and thus

$$\sum_{n=1}^{\infty} (-1)^{n+k} \frac{1}{k(n+k)}=(-1)^{k+1}\log(2)/k-(-1)^{k+1}\psi(k)/k+(-1)^{k+1}\psi(k/2)/k+(‌​-1)^{k+1}/k^2$$

@robjohn I see.

At this point I use my research I used for computing the previous series I posted on channel and get the closed form.

@robjohn ^ (related to the series that was connected to the integrals you asked me about these days)

Q.E.D. (since all reduces to computing $\sum_{k=1}^{\infty} (-1)^{k+1}(\psi(k/2)/k-\psi(k)/k)$)

@Chris'ssis The integral I posted was to compute the sum $$\sum_{k=1}^\infty(-1)^{k-1}\frac{H_{2k}}{k}$$ I believe.

@robjohn Really? It can be brought to such a nice form? That's nice!

I didn't check that integral (entirely), but if it can be expressed like that it's really nice.

$$\sum_{k=1}^\infty(-1)^{k-1}\frac{H_{2k}}{k}=2\sum_{k=1}^\infty(-1)^{k-1}\frac{‌​H_{2k}}{2k}$$ and then use $i$ in the generating function of the harmonic number.

Ah, no... It was not that. it was $$\sum_{k=1}^\infty\frac{H_{2k}}{k^3}$$ Now that I recall, and that is related to $A(1,3)$

@robjohn That is not really bad if you know the generating function of $H_n/n^3 x^n$

Well, better say

$$8\sum_{k=1}^\infty\frac{H_{2k}}{(2k)^3}$$ and hence the sum expresses as a sum of $2$ special values in the generating function (divided by 2).

@Chris'ssis The formula I got involved $\mathrm{Li}_3$

@Chris'ssis yes, that trick is used to get sums of every $n^{\text{th}}$ term of some series. Evaluating the generating function at the roots of unity

@robjohn I think you should not get trilogarithm though, but $\operatorname{Li}_2(1/4)$

@Chris'ssis I'll have to find my notes...

@robjohn You need the values in the generating function at $1$ and $-1$ only. In terms of polylogarithms, you only get $\operatorname{Li}_2(1/4)$, nothing more.

@robjohn OK

@Chris'ssis I was talking about the sums I mentioned above, not just this particular sum

@robjohn OK

@robjohn One of the greatest achievements to me (since I was thinking of that today) was to compute such integral by real analysis

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

@robjohn I'm very very happy for doing that. Even last night I dreamt about it. I actually have an entire research plan for such integrals. :-)

It's that kind of thing that makes you proud of your work and fully content, fulfilled.

Sure, there was a time when I wished I could do that. I'm glad such evaluation can be done in a very nice way by real analysis.

No bragging at all, but all lovers of integrals will be very glad to hear that. There are still integrals where I cannot employ real anaysis to get nice proofs, and complex analysis is needed.

This is eventually another part of the art, to make what seems impossible, possible (especially when referring to ugly integrals that apparently only work by complex analysis).

@robjohn when you have time, give it a try, it's incredibly amazing.

@Chris'ssis I assume you are taking the principal value.

@robjohn Yeah.

I have under work an integral that yields the fraction $$\frac{1+e^2}{1+\pi^2}$$ that is, I especially design it for that (and it's similar to the integral above).

Have you given the central extentions talk yet @AlexClark

Hi @DanielFischer

Hi @TimDavids

Hi @huy how's it going?

I'm fine, what about you?

@TimDavids Hi.

And hi @Huy.

Hi @DanielFischer

Is anyone here familiar with Euler's Method, in terms of Differential Equations?

I don't think this is supposed to be difficult, but I was hoping someone could help me verify I was doing the following problem right:

@DanielFischer Do you mabye understand the reasnoing why you can't extend Asifs answer in the post to the countably infinite case? Isn't the point of proof by induction? To show that it holds for $\mathbb{N}$?

Use Euler's Method with step size 0.2 to estimate $y(1)$, where $y(x)$ is a solution of the initial value problem: $y' = xy-x^{2}$, $y(0)=1$

@DanielFischer He is showing that the axiom of choice is not required for a finite index.

So, Euler's method is (As written in the book): $y_n = y_{n-1} + h*F(x_{n-1}, y_{n-1})$

So, since I am supposed to estimate $y(1)$, I said that $y_{1} = (1) + (0.2)*(1*0 - o^{2})$

Which is just 1

@Owatch $y_1$ is an approximation for $y(0.2)$. You need $y_5$.

@TimDavids Hmmmmm. Not sure I can explain it. But induction just gives you something for all individual finite $n$, it doesn't give you the same thing for the set of all finite $n$. You have "$n$ is finite" for all $n\in \omega$, but you don't get "$\omega$ is finite" by induction.

If we let for an ordinal $\alpha \leqslant \omega$ denote $P(\alpha)$ the property that there is a function $f_\alpha \colon \alpha \to \bigcup_{m\in\alpha} y_m$ such that $f_\alpha(k) \in y_k$ for all $k\in\alpha$, then induction gives you $P(n)$ for all $n\in \omega$, but not $P(\omega)$.

@DanielFischer Ah!

@DanielFischer I think I must be confused, what would I use for $x$ in the next case? It says $x_{n-1}$, but what would that be?

@Owatch $x_k = k\cdot 0.2$

Hi @TedS.

Hi @DanielF

Oh, thank you.

Hi mr eyeglasses

@DanielFischer That seems to make sense, but there is something else. See if the reasoning is correct which I will write now.

@DanielFischer Let me think about this for a second.

Hi @TedShifrin

I want a white board

@Owatch: Buy one.

Given a probem, do you have an idea how we could show that we cannot find a quicker algorithm than O(m^2) ?

@evinda: Assume there exists such an algorithm, then use it to solve a problem in a time it cannot be solved, for example.

@Huy There are given three vectors $V_1, V_2, V_3$ of dimension $m$, the elements of which are real numbers. Is there a quicker algorithm than $O(m^2)$ that determines if there are three numbers, one of each of the matrices $V_1, V_2$ and $V_3$, that have sum equal to $0$?

@Huy How could we do it for example in this case?

@evinda: Did you do any other matrix related algorithms?

@Huy Multiplication of matrices

I'm not sure about this specific case but what I said would be the "typical" way to prove such a statement, at least it was in our course on algorithms & complexity.

@Huy I also have no idea how we could do it in this case..

@evinda: I lack the typical matrix problems and their runtime to solve it.

@evinda: Are you certain there is no way to do it in a quicker way? Not that I know of any, but I can't think of a useful different algorithm to reduce "finding three numers that have sum equal to 0" to.

I am attempting to verify that $y = -tcost-t$ is a solution of the initial value problem $t*\frac{dy}{dt} = y + t^{2}sint$, where $y(\pi) = 0$.

I have obtained $\frac{dy}{dt}$

@Huy I am not sure.. Maybe there is a quicker one..

And I have plugged it into the initial value problem.

I have also plugged y in as well, giving me: $t(tsint-cost) = (-tcost-t)+t^{2}sint$

But I'm not sure if I'm doing this right.

@Owatch: Can you tell me your result for $\frac{dy}{dt}$ and how you obtained it?

$y = -tcost-t$, so $\frac{dy}{dt} = -t(-sint)+cost(-1)-1$

Forgot -1 up earlier

$\frac{dy}{dt} = tsint-cost-1$

Ok. Now you should quickly see that your $y = -t \cos t - t$ is indeed a solution of your IVP.

$t(tsint-cost-1) = (-tcost-t)+t^{2}sint$

@evinda: Where is the question from?

@Owatch: Yes.

I don't know... The prof gave it to us... @Huy

@evinda: And is it homework or?

@Huy Yes, it is..

There's not really a solution.

If I simplify, I get $t^{2}sint = t^{2}sint$

@Owatch: So?

I don't see how this is a solution.

@Owatch: You just showed that $y = -t \cos t - t$ does indeed satisfy your initial value problem.

How?

Because it still produces a zero?

@evinda: Whether or not there is an algorithm to solve 3SUM in $O(n^{2-\epsilon})$ seems to be an open problem. And I think if you can solve 3SUMx3, you can solve 3SUM linearly. Seems like an odd homework problem, but not my area at all, so I have no idea.

@Owatch: Is $x = -1$ a solution of the equation $x^2 = 1$?

Yes.

@Owatch: How can one verify that?

Because squaring -1 gives you 1.

@Owatch: Ok. And is $x=-1$ a solution of $x^2+1 = 2$?

Sure.

@Owatch: Why?

Because squaring -1 gives you 1, and adding 1 to that gives you 2.

@Owatch: Correct. Now go back to your example. Give yourself a minute or two, and if you still don't get it, I'll try to lead you to it in a different way.

No, it makes sense.

Great! :)

@Owatch: You have one thing left to verify though, just in case you forgot. It must satisfy the initial value!

So..

See if I get zero, if pi is substituted for x?

@Owatch: You also need to show $y(\pi) = 0$.

Using $y = -tcost-t$ you mean?

@Owatch: Yes.

@Owatch: You want to show $y = -t \cos t - t$ is a solution to a given initial value problem. Such a problem consists of two equations that must be satisfied. You have only shown your $y = \dots$ satisfies one of the two.

T = 0, or $\pi$

@Owatch: $\pi$.

Well, it is in there, so it works out I guess.

Good.

Cool, now I have to show that $y = \frac{lnx+c}{x}$ is a solution of $x^{2}y' + xy =1$

I guess I will differentiate y, then put in terms of the other.

$x^{2}y+xy=1$

Like before.

$\frac{dy}{dx} = \frac{-lnx+c}{x^{2}}$

$x^{2}(\frac{-lnx+c}{x^{2}}) + x(\frac{lnx+c}{x}) = 1$

$-ln(x)+lnx + c = 1$

Hello. A quick terminology question: in graph theory, does edge line on a cycle or in a cycle? Is it the same with forest?

@robjohn I also created this version that is awesome

Hi, a question: does anyone study / has anyone studied (undergraduate) mathematics at Cambridge or Oxford? I'm thinking about applying there and would like to know what it is like, especially 1) whether I should pick Cambridge or Oxford and 2) how both schools compare to US schools like MIT, Stanford, Berkeley or Harvard. Any information that could help me make a decision would be greatly appreciated! :)

I don't know anything about those, but if you aspire to attend the community college in my local county, I can give you a rundown there. :)

...I might just have to pass on that :O

Going for the safe schools I guess, huh.

heh...

@SMF: All nothing compared to my school! :)

I have a feeling I did not successfully show that every member of the family of functions is a solution of the differential equation given.

$-ln(x)+ln(x) = 1$, doesn't seem to satisfy $x^{2}y' + xy = 1$

Shouldn't I get $1 = 1$?

Goodnight @MikeM: Here's a good one for you.