Mathematics

12:23 AM

that's a bit vague, if there is a valid (& suitably smooth) one dimensional parameterisation of the constraint space, then minimising the resulting one dimensional function will be equivalent in the appropriate sense.

leslie townes

yeah. maybe some ways of 'solving' g(x,y) = k don't immediately give you that, but the principle isn't wrong.

copper.hat

if i ever show up in somebody's email box i will be in bits...

Ted Shifrin

@copper.hat The mistake students make is that they consider the “substituted” function on the wrong domain. They forget that often the domain has boundary and don’t check boundary points.

This is pretty much what Mike was talking about.

copper.hat

12:38 AM

i see

Ted Shifrin

Obviously, if you parametrize a circle by $\theta$ you’ll be fine, but they often do $y=\pm\sqrt{1-x^2}$ and miss endpoints (not to mention differentiability issues). But a function like $y^2+x$ will work fine unless you forget that $x\in [-1,1]$.

copper.hat

i stand suitable elaborated

Ted Shifrin

1:03 AM

I put that there for anyone else who might be interested :)

leslie townes

munchkin's new thing is to accuse one of us of having a 'tantrum,' which at least in my wife's case sometimes does produce anger.

Ajay

They're not wrong. After all, kids don't lie.

leslie townes

yes

Ajay

However, I do suppose they will reach the age when they lie like politicians do.

Ted Shifrin

1:31 AM

The tantrum produces anger or the accusation does?

leslie townes

the accusation.

munchkin is very good at presenting alternate realities in which authority figures are having tantrums and 'not listening.'

Ted Shifrin

We need to work on munchkin’s (or your) syntax.

She has all sorts of Trompian characteristics at the ripe age of 3.

leslie townes

4!

Ted Shifrin

Oh, then normal.

Can’t wait to see 7.

leslie townes

it's certainly going to be some kind of journey.

Ted Shifrin

1:42 AM

Do you have a child psychiatrist on speed-dial?

leslie townes

nah. we don't want her to be punished for having too many good ideas.

Ted Shifrin

One of those “good ideas” is to enroll her parents in an insane asylum.

leslie townes

1:57 AM

uncle ted is no fun at all.

Ted Shifrin

Nope. Plenty of people here agree with that.

I was about to link my latest response toma homework post, but my iPad refuses.

one potato two potato

3:03 AM

0

Q: Calculating an induced map $\phi_*:H_i(M_2,\Bbb Z)\to H_i(M_2,\Bbb Z)$

Let $M_2$ be a closed orientable genus $2$ surface and $\phi:M_2\to M_2$ be a homeomorphism given by rotation of angle $\pi$ by the $x$-axis. Calculate $\phi_*:H_i(M_2,\Bbb Z)\to H_i(M_2,\Bbb Z)$ for $i =0,1,2$ with respect to a basis of $H_i(M_2,\Bbb Z)$ for each $i$. (Hint: use $\phi$-invarian...

algebraic-topology

Don't know what to do

one potato two potato

5:29 AM

https://math.stackexchange.com/q/4576167/668308
Reposting because of the downvote

Ted Shifrin

5:41 AM

Do not repost. Edit the original post.

And ordering people not to downvote is pretty bad form, too.

BAYMAX

5:55 AM

Suppose if all eigenvalues of a square matrix $A$ are less than one in absolute value, then that implies $\det(I-A) > 0$

Is this true?

copper.hat

6:11 AM

Choose an 8x8 matrix with eigenvalue $i \tan {\pi \over 8}$.

Koro

Showing that [0,1] is not Frechet compact as a subspace of $R_l$ (that is R with lower limit topology):

Consider the set S={1-1/n, n is greater than 1}. 1 is not a limit point of S because {1} is a neighborhood of 1 that contains no element of [0,1].

And it is clear to me that any number outside [0,1] can't be a limit point of [0,1].

one potato two potato

7:25 AM

@TedShifrin I'm not ordering. I usually interpret it as a sign to delete. If someone wants me to revise the post, that person usually leaves a comment.

robjohn

7:53 AM

@copper.hat However, if the matrix is real...

one potato two potato

9:42 AM

I draw everything one by one (the rotation process) and the CW structure does not change at all (obviously..?). Does that imply the induced homomorphism on homology is an identity map?

one potato two potato

9:54 AM

Ah no

Yes it should be like this.

Suddenly everything makes sense

Koro

10:17 AM

In a T1 space, why is limit point compactness equivalent to countable compactness?

Suppose that X is countably compact. Let A be an infinite subset of X. Choose a countable subset {a_i} out of A. {X-a_i} is an open cover of X so there exist finitely many indices 1,2,...,k such that X-a_i cover X.

Atleast one of X-a_i, 1<=i<=k contains infinitely many elements of a_n.

Let the index be k WLOG and call the elements of a_n as $a_{n_k}$'s. Claim: a_k is a limit point of $ a_{n_k}$

Let U be an open set containing a_k. U has to be of the form X-F, where F is finite set that doesn't contain a_k. Clearly X-F contains some $a_{n_k} $.

QED.

But I have a doubt here: is it true that in a T_1 space, every open set has to be of the form X- F, where F is finite set? In T_1, finite sets are closed.

For the other direction: Let U_i be an open cover of X. Assume on the contrary that there is no finite subcover. Choose x_k in X- $ \cup_{i\le k} U_i$.

Let x be the limit point of {x_n}. x is some U_m. U_m doesn't contain u_i with i>m-1.

So U_m intersects x_n at finitely many points.

Lukas Heger

@Koro no that's not true in general. it's not even true in metric spaces

Koro

10:33 AM

Let $\{x_1,...,x_{m-1}\}\cap U_m = $ non empty. But I don't see how to get any contradiction from here. It seems that limit point compactness is NOT equivalent to countable compactness in T_1.

@LukasHeger oh yes, thanks. Not true even in R, take N, the set of naturals for example.

one potato two potato

11:05 AM

0

A: Calculation of induced map on homology by self homeomorphism by rotation on genus $2$ surface

Using the above CW structure, on a chain level the map $\phi_*$ is given by \begin{cases} a\mapsto d\\ b\mapsto c\\ c\mapsto b\\ d\mapsto a\\ \end{cases} Since $a,b,c,d$ are the generators of $H_1(M_2)$ (consider cellular homology for example), we conclude $\phi_*:H_1(M_2)\to H_1(M_2)$ is given b...

Enjoy nonsensical pictures with cumbersome calculations

Koro

1:22 PM

0

Q: The integral $\int_A f$ exists $\iff$ the series $\sum_i \int_A \phi_i|f|$ converges.

The theorem 16.5 in Munkres' analysis on manifolds reads: Let $A$ be open in $\mathbb R^n$; let $f : A → ℝ$ be continuous. Let $\{ϕ_i\}$ be a partition of unity on $A$ having compact supports. The integral $∫_A f$ exists if and only if the series $\sum_i \int_A \phi_i|f|$ converges and in this ca...

real-analysis multivariable-calculus

one potato two potato

2:03 PM

@Thorgott Hello, could you take a look at my solution (my latest chat before this)? I wonder if it makes sense and (whether it's ok or not) if there's an easier way to solve it.

I think the calculation of $\phi_*:H_2\to H_2$ can be improved. I used $\Delta$-complex but it's quite complicated.

Ajay

2:24 PM

Is it okay for me to swear on here?

user193319

Let $F_n$ denote the $n$-th Fibonacci number. I am trying to show that if $n \ge 29$, then $F_n > (1.6)^n$. The hint is to prove this by contradiction and use the smallest counterexample. By counterexample, I take it to mean any $n \in \Bbb{N}$ such that $F_n \le (1.6)^n$. But isn't it true that $F_0 = 0 < 1 = (1.6)^0$, so $n=0$ would be the smallest counterexample? How does that even help? I don't see how to prove this by contradiction (or even induction).

Ajay

Well you can let your base case be n = 29

then work your way from there by induction

user193319

Yeah, I didn't get very far with induction. So, $F_{n+1} = F_n + F_{n-1} > F_n > (1.6)^n$. That's all I've come up with so far, but that inequality isn't good enough.

I don't think we can necessarily say $F_{n-1} > (1.6)^{n-1}$ because it's possible $n-1=28$ which is a counterexample.

@Ajay Do you know how to get the induction to work?

Ajay

Well, you need to use strong induction since it's a fibonacci sequence

Do you know how to do it?

user193319

Yes, I know strong induction.

Ajay

2:38 PM

Then simply apply it here.

user193319

Apply it where?

Ajay

Well, you can "let P(n) be the statement $n \geq 29$, then $F_n > (1.6)^n$"

Then the induction process should flow naturally

user193319

Yes, I am quite familiar with the fundamental idea behind (strong) induction.

How exactly does it flow naturally? I showed you what I tried and it doesn

...doesn't seem to flow naturally.

Ajay

Then let's try something a little easier

$f_{n+1} < \big( \frac{7}{4} \big)^n$ for all $n \geq 1$

We'll work our way up to your question

user193319

$P(n)$ should actually be $F_n > (1.6)^n$. Quantifiers shouldn't be mixed in.

Ajay

2:41 PM

Yes, I know. Doesn't really matter though, nothing formal here.

user193319

Well, you're wavering between formality and informality. I think formality is essential.

Ajay

Ok fine, i'lll be more formal from now on.

Anyways, let's do the induction problem now.

Ajay

2:55 PM

@user193319 your base case is not 0

it is 29

You first need to show that the 29th fibonacci number is indeed greater than $(1.6)^{29}$

Wait what...

$F_n = F_{n-1} + F_{n-2}$

so $F_{29} = F_{28} + F_{27}$

Which is greater than $(1.6)^{29}$

Now assume that P(k) is true.

SO what we need to show is that $F_{k+1} > (1.6)^{k+1}$

SInce we have that $F_{k+1} = F_{k} + F_{k+1}$

Our initial hypothesis says that $F_k > (1.6)^k$

So $F_{k+1} > (1.6)^k + F_{k-1}$

Ohhhh, but now we need a second base case

uhhhh i'm confused

ANy help?

Koro

3:31 PM

My question has been answered. I still need one clarification though.

2 hours ago, by Koro

0

Q: The integral $\int_A f$ exists $\iff$ the series $\sum_i \int_A \phi_i|f|$ converges.

The theorem 16.5 in Munkres' analysis on manifolds reads: Let $A$ be open in $\mathbb R^n$; let $f : A → ℝ$ be continuous. Let $\{ϕ_i\}$ be a partition of unity on $A$ having compact supports. The integral $∫_A f$ exists if and only if the series $\sum_i \int_A \phi_i|f|$ converges and in this ca...

real-analysis multivariable-calculus

Partition of unity is defined for $\scr A$, a collection of open sets. How should one interpret it in case it is defined for an open set A? It seems natural to me to treat that as follows: For every a in A, there is a ball $B_a$ lying completely in A. So take $\scr A=$ $\{B_a: a\in A\}$.

Transcript for

$Mathematics$ Mathematics