Mathematics - 2015-03-01 (page 3 of 6)

3:01 PM

Hi

According to http://dlmf.nist.gov/13.3.E22 the first derivative of the confluent hypergeometric function can be wrriten as:

$$\frac{d}{dz}\mathop{U}\nolimits\!\left(a,b,z\right)=-a\mathop{U}\nolimits%
\!\left(a+1,b+1,z\right),$$

my question is that; if the instead of $z$ we have $z^{-1}$, what will be the difference?

Hippalectryon

@TedShifrin To come back to the previous exercise, how do you get from $X=P\begin{bmatrix}0&a&b\\0&c&d\\0&0&e\end{bmatrix}P^{-1}$ that the space is two dimensional ?

Ted Shifrin

Huh, no @Hippa. You'll get a matrix of the form $\begin{bmatrix} 0 & 0 & 0\\ 0 & a & -b \\ 0 & b & a \end{bmatrix}$.

Hippalectryon

@TedShifrin Oh true, it can be done that way too, I had forgotten :D

I just trigonalized it

Silenttiffy

@TedShifrin I think your counterexample can't disprove: 3^{n} \equiv 3^{n \mod k} (\mod k) I tried many examples and it really seems to be true

evinda

@DanielFischer I have seen that it doesn't give the right result. Now I tried the following, that gives the right result?
int ipow(int base, int exp) {
    int result = 1;
    while (exp!=0) {
             result*=base;
             exp-=1;
    }
    return result;
}

Is it the most effective? Or could I improve something?

Ted Shifrin

3:05 PM

$k=3$, $n=4$ was my counterexample @Silettiffy

Hippalectryon

@TedShifrin Urm so how do we link $A$ to the solutions of $AX'=X$ ?

Ted Shifrin

Change basis to make $A$ in the form I just typed, @Hippa.

robjohn

@Chris'ssis: Does Catalan come up in that integral. A lot of approaches have parts that include Catalan, but perhaps it gets cancelled out by the other parts that I haven't gotten yet

Silenttiffy

$k=3$ and $n=4$ is not a counterexample for $3^4 mod 3 = 3^1 mod 3$

A Beautiful Mind

Today I went out for a walk and a haircut. I feel much better now.

Ted Shifrin

3:09 PM

hi @robjohn

hi Jasper ! Good :)

Daniel Fischer

@evinda For exp >= 0, while you don't have overflow, it gives the right result. It's the naive O(exp) algorithm, by a far cry not the most efficient.

robjohn

@TedShifrin hey there... we are in the middle of a heavy rain. We really need it.

Ted Shifrin

yes, I know ... I would rather not move to a fire zone.

user112495

I'm considering the interval $(0, 1)$ on $\mathbb{R}$ with the zariski topology. Under this topology, would I be correct in saying that it isn't closed as it neither has finitely many points nor is it all of $\mathbb{R}$? And hence it is not compact by the Heine-Borel theorem?

Hippalectryon

@TedShifrin $\begin{bmatrix}ae^{bt}\\ce^{dt}\\fe^{gt}\end{bmatrix}$ defines a plane ?

robjohn

3:11 PM

@TedShifrin fires were not too bad last summer. It is the drought that hurts.

Ted Shifrin

no, @Hippa. You're not doing this right at all.

Hippalectryon

:/

AX'=X would give us something of the form $\begin{bmatrix}x'\\y'\\z'\end{bmatrix}=\begin{bmatrix}0\\\alpha\\\beta\end{bmatrix}$

robjohn

@Hippalectryon definitely not, unless $b=d=g$ and then it would be a line

Ted Shifrin

@Hippa: You're confuzled.

robjohn

@Hippalectryon is $A$ a constant matrix?

Ted Shifrin

3:14 PM

yeah, @robjohn. He's looking at the usual $X'=AX$ ...

Hippalectryon

@robjohn yes, with $\det A=0$

Ted Shifrin

with $\det A = 0$

robjohn

@TedShifrin nice problem

user112495

and similarly, $\{0, 1\}$ would be a compact subset under this topolgy, wouldn't it?

Ted Shifrin

I thought you were looking at the open interval, @user112495

robjohn

3:16 PM

wrong delimiters, I think

user112495

@TedShifrin These are two different questions.

@TedShifrin I just want to see if I'm going about these the right way.

robjohn

@user112495 If those delimiters are correct, that would be compact under any topology

Ted Shifrin

@user112495: Any topological space is itself both open and closed.

Hippalectryon

@TedShifrin Urm I'd be most confused indeed if we didn't get $\begin{bmatrix}x'\\y'\\z'\end{bmatrix}=\begin{bmatrix}0\\\alpha y\\\beta z\end{bmatrix}$

Ted Shifrin

No, @Hippa, not in the complex eigenvalue case. In the real eigenvalue case, yes.

No, no, either way.

You're missing $\alpha y$, $\beta z$.

Hippalectryon

3:18 PM

I just did $\begin{bmatrix} 0 & 0 & 0\\ 0 & a & -b \\ 0 & b & a \end{bmatrix}\begin{bmatrix}x'\\y'\\z'\end{bmatrix}$

ooh

Typo q_q

user112495

@TedShifrin @robjohn I'm looking at the Zariski topology on the set $\mathbb{R}$. I'm then looking at the subsets $(0, 1)$ and $\{0, 1\}$ to determine whether or not they're compact and/ or closed.

Ted Shifrin

Ohhh, I misunderstood the first one, @user112495, sorry.

The second one is both. The first one is neither.

Hippalectryon

@TedShifrin So, the corrected one is ok ?

Ted Shifrin

yes, @Hippa.

Now think about solving it.

user112495

@TedShifrin Are the reasons I gave above the correct way of getting to these answers?

Hippalectryon

3:20 PM

@TedShifrin Doesn't that give us the exponentials I wrote above ? @TedShifrin

$\begin{bmatrix}a\\ce^{dt}\\fe^{gt}\end{bmatrix}$

Ted Shifrin

hell no, @Hippa :P

Hippalectryon

Well $\begin{bmatrix}x'\\y'\\z'\end{bmatrix}=\begin{bmatrix}0\\\alpha y\\\beta z\end{bmatrix}\Rightarrow\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}c\\ne^{‌\alpha t}\\me^{\beta t}\end{bmatrix}$

Ted Shifrin

yes, @Hippa, in the case that $\alpha$ and $\beta$ are real. ... No, you have your constants of integration wrong.

Hippalectryon

Oh yeah wrong ones

evinda

@DanielFischer We could use the following algorithm, right?

Alg(int n, int k){
    if (k==0) return 1;
    result=Alg(n, floor(k/2));
    result=result*result;
    if (k==1 mod 2) result=result*n;
    return result;
}

@DanielFischer Its time complexity is $T(k)=T(floor(k/2))+d$
a=1, b=2, f(n)=d
$n^{\log_b a}=1$
So $f(n)=\Theta(n^{\log_b a})$, so from the second case of the Master theorem, we have that $T(k)=\Theta(\lg k)$, right?

Daniel Fischer

3:24 PM

@evinda Have you followed the code to understand it and implemented and tested to check that it apparently works?

Ted Shifrin

Think about solving $u'=\lambda u$ when $\lambda = a+bi\in\Bbb C$, @Hippa.

Daniel Fischer

@evinda Yes, complexity is $O(\log k)$.

Balarka Sen

@Ted!

Ted Shifrin

hi @Balarka

Sayan Chattopadhyay

Hi @BalarkaSen

Hi professor @TedShifrin

David Wheeler

3:25 PM

@user112495 I wouldn't use Heine-Borel....

Ted Shifrin

hi @Sayan

user112495

@DavidWheeler really? But doesn't the result follow straight away from it? Or am I using it incorrectly?

David Wheeler

I would argue this way: suppose $\mathscr{O}$ is an open cover of (0,1), pick any element of $\mathscr{O}$ that has a non-null intersection with (0,1)

Hippalectryon

@TedShifrin But why are complex numbers involved here ? $A,X$ are real matrices. We got rid of the complex eigenvals by using the modified $A$ matrix.

David Wheeler

Said set misses at most finitely many elements of (0,1)

Ted Shifrin

3:28 PM

because we were talking about the complex eigenvalue case, @Hippa.

You have to solve $y'=ay-bz$, $z'=by+az$.

David Wheeler

So pick one set from $\mathscr{O}$ that covers each of those points-Voila! a finite open subcover.

Sayan Chattopadhyay

@BalarkaSen how will I write the proof of something like this....
If $A-B={\nullset}$
A is a subset of B.....
I know the idea but how do I write the proof

Balarka Sen

A - B is the collection of all elements in A that is not in B

David Wheeler

$A - B = \emptyset$ means there is no element of $B$ that is NOT an element of $A$.

Balarka Sen

if A is a subset of B, then every element of A is in B

so there you go.

user112495

3:32 PM

@DavidWheeler Sorry, could you explain why it misses at most finitely many elements of (0, 1)?

David Wheeler

@user112495 Zariski topology

user112495

@DavidWheeler Oh yeah, :p. Thanks!

Hippalectryon

@TedShifrin That'd give us like $y=e^{at}(m\cos(bt)+n\sin(bt))$

Ted Shifrin

OK, @Hippa, ça marche.

Balarka Sen

@TedShifrin I have worked out a few exercises of Hatcher and I'll finish the rest of them after I go through chapter 2.1 and 2.2 again in detail to acquire a few geometric ideas.

Hippalectryon

3:33 PM

And... that defines a plane ? (with z)

Balarka Sen

@SayanChattopadhyay do it yourself.

Ted Shifrin

yes, @Hippa

David Wheeler

Suppose $x \in A$. Either $x \in B$, or $x \not\in B$

Hippalectryon

@TedShifrin Ok thanks

Sayan Chattopadhyay

Fine @BalarkaSen

David Wheeler

3:34 PM

If $x \not \in B$, then $x \in A - B$, contradiction. Hence $x \in B$.

Balarka Sen

@David! glares

Ted Shifrin

has no idea what is going on

David Wheeler

I know, the law of excluded middle isn't accepted by everyone. But let's pretend, OK?

Balarka Sen

:P

@TedShifrin Oh, and did I tell you that I finally understood the snake map?

Ted Shifrin

geometrically, you mean, @Balarka?

Balarka Sen

3:36 PM

yup

Ted Shifrin

yippee

David Wheeler

For noisy abelian categories we have the rattlesnake lemma

Balarka Sen

it was after a bit of help from prof though. i told mike already, let me find the discussion.

Semiclassical

if i remember the picture you gave, balarka, it (roughly speaking) amounted to viewing a sphere with handles from the 'perspective' of one of the handles?

Balarka Sen

here

Ted Shifrin

3:38 PM

I'll look later, @Balarka

Balarka Sen

okay.

essentially it sends a relative cycle of H_n(X, A) to the boundary of it, which is a cycle of H_{n-1}(A)

@Semiclassical i don't know what you mean.

Semiclassical

well, a sphere with handles = a torus of some genus

Ted Shifrin

yes, @Balarka ... now do Mayer-Vietoris :)

user112495

Under the same topology, $\{0, 1\}$ isn't connected is it? As we can decompose it into $A=\{0\}, B=\{1\}$

Balarka Sen

yeah, essentially you look at the genus g surface as g torii squashed

Semiclassical

3:40 PM

though i guess that analogy doesn't work quite right, since that boundary has the handle 'inside' it

Balarka Sen

then look at the last punctured torus you get by chopping the end off

user112495

wait, they they aren't closed. Ignore that

Semiclassical

right.

Balarka Sen

a relative cycle all over the first piece is sent by the snake map inside the boundary of the disc removed from \Sigma_g-1, i.e, the boundary of the first piece

this is a cycle in the second piece

but in turn is a boundary, so has homology class 0

Semiclassical

i guess the analogy i have mentally in mind is: take a coffee mug, and draw a circle around the handle (not along the loop, but so as to enclose the handle)

Balarka Sen

3:41 PM

yeah. that's right.

@TedShifrin i "believe" mayer-vietoris. it's essentially a long version of van kampen

but i'll read the proof.

Semiclassical

i imagine you worked out how to make that geometry-story more precise. i haven't worked with homology stuff lately, and when i did it was hardly at a rigorous level

user112495

hmm. $\mathbb{R}$ isn't a metric space in this case either.

Ted Shifrin

no pushouts of non-abelian groups, @Balarka

user112495

Can I have a hint for showing this?

Ted Shifrin

homologically, it all comes down to a short exact sequence of complexes, just like in the relative case

Balarka Sen

3:43 PM

yeah, well, i haven't read it.

evinda

@DanielFischer At each call of Alg(j,i) we calculate the value $j^{\floor{\frac{i}{2}}}$.
When $j=n, i=k$ we have the desired result.

Right?

Balarka Sen

@user112495 in which case?

Semiclassical

(which in retrospect is silly---understanding relative homology groups would probably sharpen my understanding of picard-lefschetz transformations of cycles on riemann surface)

Balarka Sen

zariski topology? it's not hausdorff

Ted Shifrin

or you can think about integrals and periods, @Semiclassical

Semiclassical

3:44 PM

that's what i've mostly done, yeah

Balarka Sen

definitely, @Semiclassical. it's frustrating at first because of the huge amount of algebra, but gets very interesting when you see the topology.

Semiclassical

and monodromy etc

user112495

@BalarkaSen The set $\mathbb{R}$ equipped with the zariski topology. I'm trying to determine whether or not $\{0, 1\}$ is a connected subset of $\mathbb{R}$

Daniel Fischer

@evinda Can you prove it?

David Wheeler

yeah, it's hard to separate points, because neighborhoods are "big"

Semiclassical

3:45 PM

nod. knowing a bit about algebraic topology from the level of concepts is a useful thing

Balarka Sen

@Semiclassical well, you mean an algebraic proof (because that geometry-story is essentially quite precise)?

@user112495 zariski topology is not metrizable. hausdorffness gets void.

open sets are huge.

Ted Shifrin

@Balarka still hates non-Hausdorff :D

Semiclassical

no. i just mean that i haven't heard about the snake map till now, so i'm not in a position to 'truly' appreciate the geometric picture you gave

evinda

@DanielFischer Should I prove it using the invariant that result is at each step of the form $j^{\floor{\frac{i}{2}}}$ ?

David Wheeler

it is T1, though, so it gets good download speeds.

Semiclassical

3:46 PM

i can see the intuition, but i haven't thought it through

Balarka Sen

@Semiclassical oh. well, look up snake lemma.

Daniel Fischer

@evinda Sounds like a good plan.

Balarka Sen

@TedShifrin yeah. stupid, really. and zariski topology is not really a good topology.

Semiclassical

lol, that's what i was just looking up in another chrome tab

Balarka Sen

just a fancy language to handle affine schemes

and i am not really in the position to appreciate schemes yet

Semiclassical

3:48 PM

schemes are one of those words i don't even try to understand

Ted Shifrin

natural to look at level sets of functions as your closed sets, @Balarka

David Wheeler

@BalarkaSen "good" is a relative term-some things are good for certain tasks

Semiclassical

mostly b/c i can't see any reason i'd care about them from an applied math perspective

whereas stuff like homology matters to me b/c i care about integrals on riemann surfaces

Silenttiffy

3^{n} \equiv 3^{n \mod 10^k} (\mod 10^k)— how can I show that this is true for k>1?

Balarka Sen

sheaf cohomology, @Semiclassical

and sheaves are related to schemes

Semiclassical

3:50 PM

yes, well, i haven't heard a good explanation for why i should care about sheaves either :P

Balarka Sen

great, now i am speaking about stuff i have no idea about.

Ted Shifrin

natural from an analytic viewpoint in terms of gluing local solutions together ... what's the obstruction to their making a global one, @Semiclassical ?

Semiclassical

i can well believe you'd want those concepts if you wanted to have a properly rigorous discussion. i just don't care about rigor all that much

(physicist)

Ted Shifrin

but that's where monodromy comes from ... duh

Balarka Sen

@Ted I dropped atiyah-macdonald because it felt dry. but i will have to revise my algebra before getting into cohomology, and might as well study a bit of comalg too [i read a bit about the idea of cohomology from hatcher and they seem to be getting at something like graded rings from cup products]

user112495

3:52 PM

The closure of $(0, 1)$ would be $\mathbb{R}$ wouldn't it?

Ted Shifrin

@Balarka: I've told you many times that some things should wait until you're 18.

user112495

under the zariski topology on $\mathbb{R}$

Balarka Sen

@TedShifrin i should wait until 18 before reading cohomology?!

that'd be too late, i have to study multivariable calculus too.

Ted Shifrin

I'm a big believer that one should understand differential forms ... they're a very concrete version of cohomology.

Semiclassical

eh. the way i know monodromy is from drawing cycles in the complex plane, analytically continuing some parameter through a cycle to deform branch cuts/poles, and then looking at how the cycles have changed

Balarka Sen

3:53 PM

@TedShifrin yes, and for that i need differential geometry. that's why i'd be studying mult calc.

Ted Shifrin

just showed my class yesterday in essence why $H^1(\Bbb R^2-\{0\}) \cong \Bbb Z$.

David Wheeler

@TedShifrin Because race cars?

user112495

And that must mean that $(0, 1) \cup (2, 3)$ isn't a connected subset of $\mathbb{R}$ then?

Ted Shifrin

huh?

David Wheeler

You know, "laps"

Ted Shifrin

3:54 PM

oh

Balarka Sen

:P

Semiclassical

my sense is that physics people have some sense of differential forms just from their background in stuff like EM and thermodynamics. but we're not very precise about it, and so would probably have a lot of misconceptions

Ted Shifrin

why isn't it connected, @user112495? Give me a separation.

Balarka Sen

poincare duality with the fact that R^2 - {0} def rets onto S^1.

easy.

waits for smacks

evinda

@DanielFischer How can we prove an invariant for a recursive algorithm?

Ted Shifrin

3:55 PM

Poincaré duality only works on compact, oriented manifolds, @Balarka

Balarka Sen

lol ok

David Wheeler

race cars do that, too, it's called "driving between the lines"

user112495

@TedShifrin Whoops. It just means I can't use this specific theorem to prove that it's connected.

Daniel Fischer

@evinda Induction?

Semiclassical

but, for example, $H^1(\Bbb R^2-\{0\}) \cong \Bbb Z$ in physics terms would probably be stated in terms of a current and the magnetic field/vector potential that can be created by it.

Ted Shifrin

3:56 PM

I haven't been paying attention, @user112495. I don't know to which theorem you refer.

user112495

@TedShifrinOh wait, i misread it. They are connected

Balarka Sen

@Semiclassical oh? mild interest

Ted Shifrin

yes, @Semiclassical, indeed. I showed them $d\theta$ and argued why it couldn't be exact.

Semiclassical

right

Ted Shifrin

We're just starting differential forms ... this is a freshman/sophomore class.

Balarka Sen

3:58 PM

now that i see so much connection to differential geometry, i wish i knew some analysis

anyway, back to interpreting homology geometrically

user112495

@TedShifrin If C and D are connected subspaces of a topological space T, and if $C \cap \bar D \neq \phi$, then $K = C \cup D$ is connected.

As the closure of $(0, 1)$ is $\mathbb{R}$, they have nonempty intersection, and so all necessary conditions hold.

Ted Shifrin

It doesn't say anything about closures in your theorem, @user112495

Semiclassical

one of my favorite uses of diff. forms, though, is in deriving maxwell identities in thermodynamics. usually one has to do it a bit tediously, but there's a cute way to get it directly

Ted Shifrin

And you'd first have to prove that $(0,1)$ is connected ... maybe you've already done that.

user112495

@TedShifrin Sorry, I'm using the notation $\bar D$ to denote the closure of D.

Ted Shifrin

4:00 PM

yes, but there wasn't any closure in your statement.

user112495

@TedShifrin Doesn't that follow as (0, 1) is an interval?

Ted Shifrin

I took a year of thermo in college, @Semiclassical. Loved it. What are you calling Maxwell identities?

wrong topology @user112495

user112495

@TedShifrin Which topologies does that hold for then?

Semiclassical

basically just what you get from taking mixed second partials of the various thermodynamic potentials

Ted Shifrin

The subspace/order topology from the usual topology on $\Bbb R$, @user112495

right, @semiclassical, that's what I thought. But that's really circular reasoning. $d^2 = 0$ is equivalent to mixed partials being 0.

Semiclassical

4:03 PM

well, that's not what i was using as 'derivation'

Ted Shifrin

oh?

Semiclassical

start from a statement of the thermodynamic identity as $dE=T\,dS-p\,dV$

Ted Shifrin

Yes

David Wheeler

if only we could make turning a continuous function-stupid calendar

user112495

@TedShifrin As we're under the zariski topology, I know that $(0, 1)$ is open, and that its closure is all of $\mathbb{R}$. How can I show it is connected?

Semiclassical

4:05 PM

then take the exterior derivative of both sides and use $d^2=0$ to get $dT\wedge dS=dp\wedge dV$

Ted Shifrin

No, @user112495, $(0,1)$ isn't open

Precisely, @semiclassical. You're using $d^2=0$.

Semiclassical

if that's what you're getting at in terms of circularity, i don't disagree. it's a formal derivation

Ted Shifrin

Yup.

It's the same proof the chemists give without mentioning differential forms and just manipulating differentials formally.

Semiclassical

well, there's one further formal wrinkle

user112495

@TedShifrin But I thought that under the zariski topology on $\mathbb{R}$, a subset of $\mathbb{R}$ is closed iff it consists of finitely many points or if it is all of $\mathbb{R}$?

Semiclassical

4:06 PM

what i really like is that you can get all four versions out of that one expression without referring to the potentials

Ted Shifrin

Right, @user112495. So is the complement of $(0,1)$ a finite set?

user112495

@TedShifrin and $(0, 1)$ is neither of these.

Semiclassical

namely, by expanding each of the four one-forms in terms of (say) $S$ and $V$, $S$ and $p$, etc.

user112495

@TedShifrin No, it's not.

Ted Shifrin

AGH, @user112495. You are laboring under the misconception that any given set must be either open or closed. WRONG.

Semiclassical

4:07 PM

including with the correct signs by taking account of the wedge product being antisymmetric

Balarka Sen

@Ted :P

Ted Shifrin

yes, @Semiclassical, I know :)

Semiclassical

heh, fair enough. i just like it as a way of quickly obtaining the correct expressions, which is a bit tedious otherwise

granted, yes, it's just formalism. but it's a useful formalism for a lazy physicist :P

user112495

@TedShifrin Oh, so it's not closed. And as it's complement isn't closed, it isn't open either.

Ted Shifrin

right

Semiclassical

4:11 PM

i suspect there are more full-blown applications of diff. forms in thermodynamics, say for how one understands stuff like heat capacities and cycles

but it's not something i've labored with myself. (i tend to use stat mech much more than thermo in the stuff i do)

user112495

@TedShifrin How do I go from here to determine whether or not it's connected?

Ted Shifrin

show you can't separate?

back later

David Wheeler

@user112495 What does a closed set look like in ZT?

user112495

@DavidWheeler It is either the whole set we are looking in, or it contains only finitely many points.

David Wheeler

Ok, so suppose $(0,1) = U + V$, where $U,V$ are closed.

Wouldn't that mean $(0,1)$ was finite?

Semiclassical

4:16 PM

ugh. i really should start recreating the mathematica stuff i was doing yesterday. computer crashes are such a pain when you're sloppy about saving :P

user112495

@DavidWheeler Oh, so we can't write (0, 1) as a union of two finite sets, as that will only have finitely many points. Thus either U or V must be $\mathbb{R}$. But this cannot be the case. So (0, 1) must not be connected.

@DavidWheeler So can I use the same argument to show that $(0, 1) \cup (2, 3)$ isn't connected?

David Wheeler

In ZT, it's easier to work with closed sets.

Sayan Chattopadhyay

How do you write in LaTeX the big intersection

David Wheeler

\bigcap

evinda

@DanielFischer So is it right as follows?
So is it right as follows?

- Base Case: For $k=0$, the algorithm returns $1$, so it is correct since result=$n^{\lfloor \frac{k}{2} \rfloor}=n^0=1$.
- Induction Hypothesis: We suppose that $\forall 0 \leq j \leq k: result=n^{\lfloor{ \frac{j}{2} \rfloor }}$
- Induction Step: Suppose that the second argument of the algorithm is $k+1$.
Then result=pow($n, \lfloor \frac{k+1}{2} \rfloor$) $\overset{\text{ Induction hypothesis}}{=} n^{\lfloor \frac{k+1}{2} \rfloor}$

David Wheeler

4:21 PM

@user112495 You've shown $(0,1)$ CANNOT be written as the disjoint union of 2 closed sets-that means it IS connected.

The same argument shows the union is also connected. It seems bizarre, because ZT is bizarre.

user112495

@DavidWheeler Thanks!

David Wheeler

@user112495 Uh, wait a minute....

user112495

@DavidWheeler Wait, what?

David Wheeler

We're still good on (0,1), but not on the union

Because for connectedness, we have to check in the relative ZT.

user112495

@DavidWheeler What do you mean by that?

David Wheeler

4:28 PM

For example, $(-\infty,1)\cup(1,\infty)$ is open in ZT

So, in the relative ZT, we have for $A = (0,1) \cup (2,3)$, that $[(-\infty,1) \cup (1,\infty)] \cap A$ is open

user112495

@DavidWheeler Why do we need to check the relative ZT? I haven't come across this before.

David Wheeler

Nevermind, I'm over-thinking it

user112495

@DavidWheeler So does the original argument work?

David Wheeler

Yeah

Even in the relative topo, any open set intersect $A$, will still miss only finitely many points of either interval.

A Beautiful Mind

@Chris'ssis What are you doing now?

Chris's sis

4:56 PM

@ABeautifulMind Just back from jogging (actually, very hard training). How about you?

@ABeautifulMind Well, I begin to feel more and more the mental stress, it affects my mood to a certain extent. I think I'll need a break from doing math, a longer one.

@robjohn didn't you start with the one with $\sin(2x)$?

robjohn

@Chris'ssis I can see how to do that one, I think. It is just a sub and IBP

Chris's sis

@robjohn You mean you have a solution or?

OK.

Semiclassical

i still haven't done the triple integral one :/

right now i'm actually trying to figure out the simplest way to compute the following integral (which i know is a combination of complete elliptic integrals, but i'd appreciate a less tedious calculation)

Chris's sis

@Semiclassical No hurry with that one.

Semiclassical

$$\int_0^{2\pi} \sqrt{\frac{\lambda+e^{-i x}}{\lambda+e^{i x}}}\,dx$$

for $\lambda>0$

A Beautiful Mind

5:07 PM

@Chris'ssis I went out for a walk and haircut today.

Chris's sis

@ABeautifulMind That's good. You arranged yourself to be beautiful, right? :-)

(maybe are you about to date a girl?:D)

A Beautiful Mind

@Chris'ssis I have made some plans to solve 99 per cent of my OCD in April. I hope I succeed. However, the other 1 per cent might take another year or so.

@Chris'ssis Nope, I won't date anyone until I am totally well. =)

Chris's sis

@ABeautifulMind What's the plan? How would you do that?

A Beautiful Mind

@Chris'ssis Hard to describe here. It's a very long story. But I might upset a lot of people next month, but I think it will be fine.

Sayan Chattopadhyay

Hi jasper @ABeautifulMind

Hi @Chris'ssis

Chris's sis

5:11 PM

@SayanChattopadhyay Hi

Ramanewbie

hi @sayan

A Beautiful Mind

@SayanChattopadhyay Hi. You can buy books online from Amazon, either amazon.in in India or amazon.com in US to ship to India.

Chris's sis

@Semiclassical Interesting integral.

Sayan Chattopadhyay

Hi @Ramanewbie

Hippalectryon

@SayanChattopadhyay o/

Sayan Chattopadhyay

5:13 PM

Thanks jasper

I didn't get u @Hippalectryon

Semiclassical

i thought so too. mathematica can't compute it, amusingly, or at least not until one tosses out the odd part of it (one can integrate over $[-\pi,\pi]$ instead since the integrand is periodic)

Ramanewbie

@hippa how did you arrive so fast ??

Hippalectryon

@Ramanewbie teleportation

A Googler

"Rarely if ever expressible as a ratio of integers." - what does this mean?

Ramanewbie

@hippa "Hey, people, you see he's mad !" lol

Hippalectryon

5:14 PM

@SayanChattopadhyay o/ = hello

A Beautiful Mind

@AGoogler Rarely rational. Pun intended.

Semiclassical

irrational

Ramanewbie

@SayanChattopadhyay fr.urbandictionary.com/define.php?term=o%2F

A Googler

lol

Semiclassical

(and even when mathematica does give it in general, it's not in the simplest form possible. mathematica is kind of stupid when it comes to complete elliptic integrals)

Sayan Chattopadhyay

5:16 PM

I feel I am too oldfashioned

A Googler

doe "general discussion" include things completely unrelated to math?

*does

A Beautiful Mind

@AGoogler Yes.

Chris's sis

@robjohn You mean you use sub one time and then IBP and you're done? If so, your way is far far more brilliant than mine. At any rate, I admit I might miss some obvious ways though.

A Googler

great

So um , I wanna learn programming , where should i start? I've been suggested python

Hippalectryon

Python is good

Lua too

Basic, also. There are tons of Basics out there

A Googler

5:20 PM

i learned a bit of basic in school

Chris's sis

I found it!

73

Q: Math and mental fatigue

Just a soft-question that has been bugging me for a long time: How does one deal with mental fatigue when studying math? I am interested in Mathematics, but when studying say Galois Theory and Analysis intensely after around one and a half hours, my brain starts to get foggy and my mental condi...

A Googler

ok so , how should i learn python? many people are suggesting to learn by doing , but what do i do?

Chris's sis

@robjohn how do you deal with the metal fatigue from doing math? I don't afford to take breaks until I publish my book ...

Hippalectryon

@AGoogler You could follow some tutorials online

Chris's sis

Actually it's a question for all: how do you deal with the metal fatigue from doing math?

Hippalectryon

5:22 PM

There are some good websites out there

@Chris'ssis Chocolate

robjohn

@Chris'ssis you need to take breaks from anything taxing.

A Googler

@Hippalectryon which do you recommend?

Chris's sis

@robjohn Yeah, I think so.

@Hippalectryon hmmm, it doesn't work anymore :-(

A Googler

is "too long for a comment" a valid reason for writing comments as answers?

Hippalectryon

@AGoogler It has been done before, but only if it's way over the limit. And do mention that it was intended to be a comment.

@AGoogler Have you ever leart programming before ? For other languages ?

robjohn

5:25 PM

@Chris'ssis breaks let me climb out of ratholes into which I might be descending

A Googler

@Hippalectryon a bit of c++ (old console version ) , some basic and very little javascript

Chris's sis

@robjohn Realizing I'm very irritable, tired, sleepy, there must be something wrong. I think you're right.

Hippalectryon

@AGoogler codecademy.com/en/tracks/python maybe ?

robjohn

@Chris'ssis sounds like time for a break and some food.

A Googler

@Hippalectryon i'll try thanks

Chris's sis

5:28 PM

@robjohn And, yes, I skipped many meals in the last period of time.

Hippalectryon

@Chris'ssis That can't do you any good

Ilya_Gazman

Hello people

Can anyone help me with a good book recommendation?

A Beautiful Mind

What book?

Chris's sis

@Hippalectryon The brain is better irrigated with blood when you eat less, so the performance is better.

Ilya_Gazman

@ABeautifulMind I posted a question about it

Silenttiffy

5:34 PM

Is it possible to simplify \sum_{i=1}^{k} \binom{k}{i} \cdot (-2^{k-i})^{i} \cdot (2^k-1)^{k- i} any further?

Hippalectryon

@Chris'ssis But disrupting your eating cycle is bad

Chris's sis

@Hippalectryon Yeap, I feel that now.

A Googler

@Chris'ssis Don't pressure yourself so much. No mathematician changes her/his eating habits for math.

Chris's sis

@AGoogler Thanks, that seems so.

@robjohn I'm referring to this one $$\int_0^{\pi/2}\frac{x\log{\sin{(x)}}}{\sin(2x)}\,dx$$

Semiclassical

my immediate intuition for that is to try to replace $x$ with some useful function $f(sx)$, and characterize that integral as the first derivative of $f$ at 0

Chris's sis

5:58 PM

Also note that

$$\int_0^{\pi/2}\frac{x\log{\sin{(x)}}}{\sin(2x)}\,dx=\int_0^{\pi/2}\frac{x\log{‌\tan{(x)}}}{\tan(x)}\,dx$$

Semiclassical

interesting. eyeballing that, it seems like that should amount to writing the second log as $\log \sin x-\log \cos x$ and doing an $x\to \pi/2-x$ reversal on the second term

except that mucks up the $x$ out in front. hrm.

Chris's sis

Integrating by parts again here $$\int_0^{\pi/2}\frac{x\log{‌\tan{(x)}}}{\tan(x)}\,dx$$ we get the way to go (all combined with beta function too).

Semiclassical

ahh, neat

Chris's sis

:-)

AvZ

6:32 PM

Finally hit 1k!

Random Variable

6:43 PM

@robjohn If you're on $W_{0}, W_{-1}$, or $W_{1}$ and you circle around $z = - \frac{1}{e}$ twice, do you end up where you started?

Semiclassical

@RandomVariable: you may have seen this already, but this MO answer looks relevant (including sources)

4

A: Equation between the two branches of the lambert w function

You should specify what do you mean by "equation". If you mean algebraic equation, then evidently there is none. Because an algebraic equation has finitely many solutions, and if there is an algebraic equation $F(W_0,W_{-1})=0$, analytic continuation will give you infinitely many solutions. This ...

robjohn

@RandomVariable IF you start on $\mathrm{W}_{-1}$ near $-\frac1e$ and go counter clockwise around $-\frac1e$ you will pass across the branch cut to $\mathrm{W}_0$. You can then go counter clockwise around $-\frac1e$ a full rotation then pass through the branch cut to $\mathrm{W}_1$. Then you can go counter clockwise a half a rotation and cross the branch cut onto $\mathrm{W}_{-1}$.

So if you are close to $-\frac1e$, you can rotate counter-clockwise $\pi$ on $\mathrm{W}_{-1}$, $2\pi$ on $\mathrm{W}_0$, then $\pi$ on $\mathrm{W}_1$, then start over again.

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$Mathematics$ Mathematics