Mathematics - 2022-11-27

one potato two potato

06:24

peep

user 7269591

not a

> 8.999... hours

copper.hat

06:49

$-{1 \over 12}$

one potato two potato

@copper.hat About this... the proof using some algebraic manipulation on $S= 1+2+3+\cdots$ is not valid, right?

copper.hat

well, there is some context in which is has justification.

but i get a bit miffed at the amount of attention that attention grabbing stuff gets :-)

one potato two potato

This. We can't just add those we don't know the convergence.

copper.hat

some nice answers here math.stackexchange.com/questions/39802/…

one potato two potato

Yes I know that but the series expression of zeta function is valid only for $Re(s)>1$

copper.hat

06:56

indeed

one potato two potato

The result is by analytic continuation so strictly speaking, it's nonsense.

Especially the 'proof' in the image I uploaded is nonsense.

copper.hat

it has meaning within a very tight context, but people just leap off and announce these things because they are such noise generators

i pretend to be more miffed than i am, of course

user 7269591

Make Math Great Again.

copper.hat

i live in albany, ca, a left leaning sort of place, i was joking with some friends about getting a make albany great again hat, but the idea was quickly shot down

for a real attention getter math.stackexchange.com/a/152788/27978

2

user 7269591

07:17

coolio

Koro

08:21

Suppose that in R^2, I take a loop A based at (0,0). Suppose that there is a hole at (0,1). Inside region of A does not contain (0,0) inside it. Now there is another loop B at (0,0) whose inside contains (0,1). Then intuitively there is no path homotopy between A and B as continuous deformation would mean crossing (0,1) once.

one potato two potato

09:45

Why $(1-\zeta^2)^{-1/2} = i(\zeta^2-1)^{-1/2}$ when $\zeta>1$? $(1-\zeta^2)^{-1/2} = (1-\zeta)^{-1/2}(1+\zeta)^{-1/2}$.
\begin{align*}
(1-\zeta)^{-1/2} & = e^{-{1\over 2}\log(1-\zeta)}\\
& = e^{-{1\over 2}(\ln |1-\zeta|+i\arg(1-\zeta))}\\
& = e^{-{1\over 2}(\ln|\zeta-1|+i\arg(\zeta-1)+i\pi)}\\
& = e^{-{\pi i\over 2}}e^{-{1\over 2}(\ln|\zeta-1|+i\arg(\zeta-1))}\\
& = -i(\zeta-1)^{-1/2}.\\
\end{align*}

Koro

10:55

The fundamental groups of convex subspaces of R^n are the identity groups as convex subspaces are path connected.

leslie townes

the conclusion sounds right to me. there's a cheap generalization to star-shaped subsets of R^n. i wonder about the justification.

Koro

Ah, that's an exercise problem. I'll get there soon.

leslie townes

S^1 is path connected and its fundamental group is nontrivial, of course. the thing about convexity is it lets you homotope loops to constant loops in a particularly simple way.

and for a star shaped set, at least for one basepoint you can do that, and then use path connectedness to get it for other basepoints.

Koro

yeah. linearly

some say rectilinearly

leslie townes

i don't do that, if only because i want the number of meanings of the word 'linearly' and its variants to be minimized.

Koro

11:11

Leslie, for star convex, is the following justification correct?: Suppose A is star convex. Then there is a point p in A such every line segment connecting p to other points in A lies completely inside A. So take any x in A, connect it to p by a line segment. So we have a path in A from p to x. It follows that fundamental group of A at x is isomorphic to as that at p.

QED.

"It follows that..." is because of the theorem that "If f is a path in X from a to b, then $\pi_1(X,a)$ is isomorphic to $\pi_1(X,b)$."

leslie townes

if i were in a class i would annotate that with pin citations to the results i am using (e.g. the theorem you cite).

also, if the idea is to show that pi_1 is trivial, i'd include something about the straight line homotopies you can use at p.

Koro

Oh right. I'll use that now.

leslie townes

so what's on your plate right now? algebraic topology, elsewhere it seems ring/module theory, what else?

Koro

Constant map c:I--> A, defined by c(t)= p for all t in I is a path. I want to show that path homotopy class of c, i.e., [c] is the only element in $\pi_1(A, p)$. This step requires straight lines as you said.

Leslie, we have algebraic topology next semester. This semester, we have Topology + introduction to algebraic topology, measure theory, abstract algebra, linear algebra, calculus of several variables.

one potato two potato

10

Q: Can any harmonic function on $\{z:0<|z|<1\}$ be extended to $z=0$?

After looking at this question for quite some time, I've asked a couple of other students, and they also couldn't seem to come up with an answer. This is from an old qualifying exam at our university. Let $u$ be a harmonic function bounded on the set $\{z:0 < |z| < 1\}$. Can it always be defin...

leslie townes

11:20

koro: that seems like fun. with the exception of measure theory and calculus of several variables (puke emoji)

one potato two potato

I didn't know you liked general topology

Koro

suppose that [f] is in $\pi_1(A, p)$. Consider F(x,t):= (1-t)f(x)+ t p. (this lies in A as A is star shaped). This F is path homotopy between f and c. So [f]=[c]. This concludes the proof.

leslie townes

oh, i'll puke at that too. good point.

Koro

Leslie: I find several variables calculus also very interesting. It's just that there is so much syllabus and very less time to cover.

Next semester we also have complex analysis and number theory.

leslie townes

complex analysis can be frighteningly difficult.

it was the one class in grad school where not only was i lost, but i thought, 'i can't understand how anyone could understand this.'

Koro

11:31

I had complex analysis during my UG also.

leslie townes

well, i hope for your sake it is just a redo of that.

Koro

let's see.

one potato two potato

@Koro UG means undergraduate? You're a graduate student?

Kumar Shuvam

I have a doubt

Umm so, there is this question which has been bugging me for a while.. f(x)= x-(1/x).. we can see its derivative is positve in its domain that means function is strictly increasing but still it is not injective why?

leslie townes

it's injective on some domains, but not others.

what domain are you considering f as a function on?

note that for a function defined only on some open subset of R, f'(x) > 0 for all x does not actually imply that f is strictly increasing. it implies only that if is increasing on any connected subset of the domain of f.

Kumar Shuvam

11:47

I'm taking R-{0} as domain

leslie townes

the attempt at a proof that f is strictly increasing goes, okay, consider a < b. then f(b) - f(a) is by the mean value theorem f'(c) (b - a) for some c in (a,b). that requires f to be defined on (a,b).

in particular it doesn't work if the domain of f doesn't include the interval (a,b).

so here, f(-1) = 0 and f(1) = 0 but there is no c in between with f(1) - f(-1) = f'(c) (1 - (-1)).

as one example.

Koro

@onepotatotwopotato yes UG means undegraduate. I'm doing masters now.

Kumar Shuvam

Ok I got it, So when does it helps to check if derivative of a function is increasing to find whether a function is injective?

leslie townes

if the domain is an interval, knowing that the function is increasing will tell you that it is injective.

i should rephrase that. if the domain of a function is an interval, the positivity of the derivative will tell you that the function is injective.

any strictly increasing function whose domain is a subset of R is injective. but when the domain is not connected, you can't deduce "strictly increasing" from a derivative condition.

Raffallo

12:03

anyone is able to help me with this?
https://math.stackexchange.com/questions/4584985/estimate-variable-using-curve-formula

Kumar Shuvam

Thanks a lot you were a big help :)

one potato two potato

12:49

1

Q: $\int_{D^2}f^*\omega = 0$ for any smooth map $f$ that maps $\partial D^2\to S^1$

Consider the differential $2$-from $\omega = {dx\wedge dy\over x^2+y^2}$ on $X =\Bbb R^2-\{0\}$. Prove that, for the unit disk $D^2$ and for any smooth map $f:D^2\to X$ which sends the boundray circle to the unit circle $S^1$, $$\int_{D^2}f^*(\omega) =0.$$ Since $df^*(\omega) = f^*(d\omega) =0$...

differential-geometry solution-verification smooth-manifolds

Koro

Suppose that p is point of $S^1$. Why is $(S^1\times \{p\})\cup (\{p\}\times S^1)$ a figure 8 ?

one potato two potato

I thought it was not a very difficult question but now I would say it's not very easy (to me).

Koro

I don't see how it's a union of two circles touching at one point.

Why is either of the sets in the union is a subset of the circle in the first place?

one potato two potato

@Koro Embed that into a torus $S^1\times S^1$.

Koro

why? sorry, I am new to this.

if looks like union of a circle in xy plane with a circle in yz plane that is passing through the diameter of the first circle.

In no way, it looks like 8.

one potato two potato

12:56

@Koro I guess you have a copy of Munkres topology. See section 53 (covering space) example 5.

Koro

yes, that's where I am coming from.

"B_0 is a union of two circles " why?

B_0 is the set that I wrote above.

Thorgott

13:21

$S^1\times\{p\}$ is a circle, $\{p\}\times S^1$ is a circle, their intersection is $\{p\}\times\{p\}=\{(p,p)\}$

thus, it's two circles intersecting in a point

Koro

I understood that now. :-)

Thorgott

if you picture $S^1\times S^1$ as torus like it has been suggested, this is the union of a longitudinal and meridian circle

Koro

I had asked this in General Topology room and got the answer.

ILikeMathematics

> A function of degree $n$ has at most $n - 1$ extrema.

This is wrong, right? It could have at most $n + 1$ extrema. It could have $n - 1$ inner extrema, but there is the possibility of the domain not being $\mathbb{R}$ which gives two more edge extrema.

monoidaltransform

13:39

0

Q: Bijective fiber bundle map is a bundle isomorphism proof attempt

We are working in the continuous category. Definition 1: A bundle map between two fiber bundles $\pi_1:E_1\rightarrow B$, $\pi_2:E_2\rightarrow B$ is a continuous map $f:E_1\rightarrow E_2$ such that $\pi_2\circ f = \pi_1$ Definition 2: A bundle map $f:E_1\rightarrow E_2$ between two fiber bundle...

differential-geometry solution-verification fiber-bundles

Thorgott

this is not true

ever space is a fiber bundle over the one-point space in a unique way and then this would imply that continuous bijections are homeomorphisms

one potato two potato

14:04

I know there is a branch cut of $\sqrt{1-z^2}$ that makes the function analytic on $\Bbb C-[-1,1]$. Is there a branch cut that makes the function analytic on $\Bbb C-((-\infty,-1]\cup [1,\infty))$?

copper.hat

14:31

The downvote dribble continues.

user123234

14:42

is there a way to show that a state is recurrent if I have the transition matrix and a measure which is invariant under the transition matrix?

Koro

15:02

$S^1$ is not simply connected.

Because its fundamental group at (1,0) is isomorphic to $\mathbb Z$.

But that's strange. I remember a discussion here a few days ago about $S^n$ being simply connected.

I suppose the discussion was for $n\ge 2$.

Thorgott

yes, precisely

the $n$-sphere has connectivity $n-1$, which is $\ge1$ precisely when $n\ge2$

user123234

1

Q: How do I show that all the states are recurrent in this Markov chain?

Let me consider a Markov chain with transition matrix given by $$P_{0,1} =1, P_{i,i-1}=1-p, P_{i,i+1}=p$$for $i\geq 1$ and $p\in (0,1)$. I need to prove that for $p<\frac{1}{2}$ all states are recurrent. We have got the hint from our Prof that we need to find a measure which is invariant for th...

probability probability-theory markov-chains stochastic-calculus

my last question referrs to this one here.

Koro

15:30

thank @Thorgott.

Koro

16:25

Suppose that Y is path connected. [I,Y]= the set of homotopy classes of maps from I into Y. It is to be shown that [I,Y] is singleton.

I tried to do it like this: Fix a b in Y. Let c: I--> Y be the constant map mapping I to b. So [c] is in [I,Y]. Now let f:I--> Y be continuous. By path connectedness, for every x in I, there exists a path $p_x$ such that $p_x(0)=b, p_x(1)=f(x)$. Now consider $F:I\times I\to Y$ defined by $F(x,t)= p_x(t)$. $F(x,0)= b, F(x,1)=f(x)$. It only remains to now show that F is continuous.

But it seems that F is not continuous. How to fix this?

Tempting to say that for any open set U in Y: $F^{-1}(U)= \cup_{x\in I} I\times p_x^{-1}(U)$.

But there is a trouble proving that RHS is contained in LHS, which makes me want to believe that F is not continuous.

Thorgott

you have chosen the paths $p_x$ completely independently of one another

there's no reason why this should yield something continuous in $x$

Koro

But if the equality were true, then F^{-1} will indeed be open.

Thorgott

another good indicator this cannot work is that you actually haven't used any property of the interval $I$

you could repeat your argument verbatim with any space $X$ in place of $I$ and surely not all homotopy sets $[X,Y]$ are singletons

Koro

I think you're right.

Then how to prove this?

Thorgott

I would do this in steps

first, show that any map I -> Y is homotopic to a constant map (this is where you have to use the particular structure of the space I). then, show that any two constant maps are homotopic (this only uses that Y is path-connected).

Koro

16:41

the first step is what I tried to do.

Thorgott

@Koro this equality is wrong, to be clear. the correct equality is $F^{-1}(U)=\bigcup_{x\in I}\{x\}\times p_x^{-1}(U)$

and drawing a picture should clarify that just because the "slices" of a set are open doesn't mean the set is open

Koro

I agree.

Koro

17:59

$\mathbb Q^c$ is $T_3$ and has a countable basis also.

Let $B_i$'s be a countable basis of R. Then $B_i\cap Q^c$'s give a countable basis of $\mathbb Q^c$.

Shootforthemoon

18:47

Hey everyone! Been a long time since my last visit here. Some of you might remember me asking bout the implcit function theorem :)

I remember reasoning with Thorgott and Rithaniel

I've got a new question on a completely different topic, written here math.stackexchange.com/questions/4586173/…. i wonder whether any of you has suggestions...

geocalc33

21:18

$\sum_{n \le x} (2n-1) + \sum_{n \le x} \frac{1}{2n} $ Is this correct notation?

I want to get this $(5, 1+1/2+3+1/4+5)$ for $x=5$ for example

copper.hat

it is not clear what the domain of summation is, what the pair means, and using $x$ for a discrete seummation seems a bit odd.

geocalc33

$n,x \in \Bbb N $

Ted Shifrin

21:41

No, it’s wrong. Try it out for yourself.

Watercrystal

So, I have a small question.

I'm currently designing a randomized algorithm. In each round, the algorithm samples $k$ i.i.d. 0-1-RVs, and I want to estimate the expected value of those RVs up to a multiplicative error of $1 + \varepsilon$, say.

AFAIK, the usual way to go about this is to use Chernoff bounds to determine the number of samples needed.

However, to do this, I need a lower bound on the expected value, which I do not have (or rather: I could give a very crude estimation which would not suffice to derive the result I need)

So my idea is: For any such RV $X$ we have $\E(X) = \Pr(X = 1)$ and $\Pr(X = 1) = 1 - \Pr(X = 0)$, and for my purposes it suffices to estimate either $\Pr(X = 1)$ or $\Pr(X= 0)$ well enough (rather than both).

I know that at least one of these probabilities is bounded from below by $1 / 2$, but I do not know which one of them.

Hence, I simply choose the one which the algorithm empirically found to have value $\geq 1 / 2$.

This, of course, could be wrong, and I need to account for this.

So I thought of the following: Suppose that the relative frequency of $i \in \{0, 1\}$ was at least $1 / 2$. If, say, $\Pr(X = i) < 1 / 4$, then the probability of choosing $i$ is very low, and since we see an absolute difference of $\geq 1 / 4$ we can apply Hoeffding's inequality to bound the probability of choosing $i$ in this case.

This, hopefully, yields that $\Pr(X = i) \geq 1 / 4$ with high probability if we chose it.

But this kind of feels wrong?

Also, I think I might be running into correlation issues if I use the same samples to both choose $i$ and estimate $\Pr(X = i)$.

But this could be fixed by taking independent samples for both tasks, I guess?

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