Mathematics - 2016-02-23 (page 3 of 3)

Mike Miller

16:05

I guess I've been implicitly assuming that $f$ is unbounded as you go to infinity. :P You win.

Secretly the problem is caused because there are "critical points at infinity".

Balarka Sen

:D

Yeah.

Mike Miller

Or maybe not secretly.

Now assume it's unbounded in both directions at infinity.

Yeesh.

Akiva Weinberger

Wow, it seems they've updated the mobile version of the chat!

Mike Miller

16:31

@BalarkaSen: Though this one, I think, finally gives the desired result, let's rephrase 4) to "Provided $f$ accumulates on the critical set"

err, $\gamma$

Jessy Cat

0

Q: Proof of existence and uniqueness of the exponential function using ODEs

In our lecture notes for our complex analysis class, we were given the following theorem: Theorem: There exists a unique complex function $f$ such that $f(z)$ is a single valued function $f(z) \in \mathbb{R}$ whenever $z \in \mathbb{R}$ and $f(1)=e$. $\forall z_{1},z_{2} \in \mathb...

What I'm having trouble with now is what I mentioned in my last comment at the bottom of the page.

I don't know if the guy who answered me is ever going to get back to me with an answer to my follow-up question.

sharaf zaman

16:50

@s.harp you gave me awesome example for my doubt but now the another doubt pops up, so anyody please help.. i know $\lim_{N\to \infty}\sum_n^N f_n(N) \neq \sum_n^\infty \lim_{N\to \infty}f_n(N)$ but my doubt is

i got it that we should sum firsat the apply the limit

but then how can i apply limit here $\lim_{x\to \infty} (x+x^2+x^3)$

Jessy Cat

anybody want to answer my question, please do. I have to go to class now.

Balarka Sen

@MikeMiller Actually, I am not sure anymore where compactness of the critical set is required for 1). Given $f : \Bbb R^n \to \Bbb R$ is unbounded as I head to infinity, and that $\lim_{N \to \infty} f(\gamma(N))$ is finite, $\gamma(t)$ must always be finite, i.e., image of $\gamma$ must be bounded. As $(f \circ \gamma)'(t) = \nabla f(\gamma(t)) \cdot \gamma'(t) = - \| \nabla f(\gamma(t))\|^2 \leq 0$, $f$ decreases monotonically along the image of $\gamma$. But $f \circ \gamma$ must also have bounded image, so $f$ cannot decreasing along $\gamma$ without bound.

I mean, we are essentially looking at a compact set $K \subset \Bbb R^n$ where the image of $g$ lives ('cause it has to be bounded). And then the bit of the critical set that lies in $K$ is automatically compact, so we don't really need any compactness assumption, do we?

s.harp

@sharafzaman if you want you can reformulate the expression as $x^3+x^2+x=x^3(1+1/x+1/x^2)$ for $x\neq0$. Then you can see that if eg $x>1$ the 3 little sums are always in the inteval $[1,3]$ whereas the $x^3$ term grows without bound

Mike Miller

You're making me work too hard, @Balarka!

But, yes. Once you have unbounded at infinity, you're right, you've just given a correct proof that you must accumulate to the critical set.

Let's modify appropriately.

sharaf zaman

@s.harp the can i apply limit after that

Mike Miller

17:02

3) Find an example where you get arbitrarily close to the critical set, but do not accumulate to it.

That's what I was gunning for in 3).

This is not possible if the critical set is compact.

Balarka Sen

@MikeMiller Phew, ok. I was trying to be very careful so that I don't say anything silly, good to know I haven't so far.

Mike Miller

Drop the unboundedness assumption for (3)-(4), you've solved (5), you still need to do (1) and (6).

(My mistake came because in the example I usually think about, the critical set is compact and finite energy is equivalent to accumulating to it, but this is because I can reformulate the notion of being on the critical set in this special case.)

Balarka Sen

OK, thanks, noted.

s.harp

@sharafzaman I don't know what you mean with "apply the limit"

Balarka Sen

@MikeMiller I can't quite understand the statement of (3), but do you mean that "Find an example where given the gradient flow line $\gamma$, for any $\epsilon > 0$ there is a $T > 0$ such that $d(\gamma(t), A) < \epsilon$ whenever $t > T$, but for every $x \in A$ there is an nbhd $U_x$ of $x$ which is disjoint from the image of $\gamma$"?

Mike Miller

17:15

Sure.

sharaf zaman

@s.harp i was evaluating a question and i got the limit $$\lim_{n\to \infty}[ \frac{n(n+1)(2n+1)}{6n^3}x+\frac{n(n+1)(2n+1)}{6n^3}]$$
and i just took the higher power of n out then i just substituted the value of limit at places of n and i got $$\frac{x}{3}+\frac{1}{3}$$ and answer is correct but i wondered in that $\lim_{N\to \infty}\sum_n^N f_n(N) \neq \sum_n^\infty \lim_{N\to \infty}f_n(N)$ case i didn't get my answer but here i got how?

Balarka Sen

Thanks.

Mambo

@BalarkaSen Hello

s.harp

@sharafzaman if you pull the $1/n^3$ into the products, you get expressions like $1(1+1/n)(2+1/n)/6 x +1(1+1/n)(2+1/n)/6$. You can use standard statements of convergent sequences to calculate such expressions. (see en.wikipedia.org/wiki/Limit_of_a_sequence#Properties )

sharaf zaman

@s.harp ya! this mean $\frac{1}{n}$ will tend to zero so i will get my answer, correct?

Huy

17:35

hey @Soham

how's it going

Balarka Sen

@MikeMiller Right, I agree with you that (3) is impossible if critical set is compact. An example can be obtained by looking at a function $f : \Bbb R \to \Bbb R$ with critical points at $1/n$'s, $n$ integer, I believe. If I can make $\gamma$ pass through $0$, I am done, because that's the limit point. Making $f$ $C^\infty$ at the origin needs care though, but I think it's possible.

Eh, sorry, there's something wrong.

Mike Miller

You can't possibly do this in $\Bbb R$ for topological reasons.

Clarinetist

@Huy Could you give me a hint on how to get started? I have two matrices, say $A$ and $B$, with the same number of rows. If their column spaces are equal, how can I show that their projection matrices must be equal? Geometrically, this makes sense, but I don't see how to prove this using matrix algebra

Mike Miller

A gradient flow line in $\Bbb R$ that doesn't go to infinity accumulates somewhere and has a limit because, well, obviously.

(But also you can prove it.)

Huy

@Clarinetist: what do you mean by their projection matrices, precisely?

Clarinetist

17:46

@Huy E.g., $P_A = A(A^TA)^{-}A^T$

Huy

is this the generalized inverse stuff again? :D

Clarinetist

Yep

Balarka Sen

What I was thinking is in the positive half of $\Bbb R$ I can make $f$ have local max at $1/n$'s of gradually decreasing height and on the non-positive half make $f$ zero. Start $\gamma$ at $0$. Then it stays at $0$, which has distance $0$ from $\{1/n:n \in \Bbb N\}$ but accumulates nowhere on the critical set.

But I doubt this $f$ would be smooth.

Mike Miller

If $f'$ is continuous, and $f'$ is zero at $1/n$, then...

Huy

@Clarinetist: any useful properties of the projection matrix you can use?

Balarka Sen

17:48

Right.

It cannot possibly be even $C^1$.

Clarinetist

@Huy Symmetry, idempotency... umm

@Huy Rank of projection of $A$ is equal to rank of $A$ which is equal to trace

Semiclassical

@Clarinetist in what sense is that a projection matrix? without some other properties of $A$, i don't see how it'd be idempotent

Clarinetist

@Semiclassical Idk, that's just how we define it in stats. It projects whatever we put after it to the column space of $A$ in that case

Semiclassical

weird

Huy

@Semiclassical: take a 1-dimensional subspace and a unit vector $v$ spanning it. then, you can project to that space with $P_v = v v^T$. if you have a $k$-dimensional space, take a matrix $A$ with columns being an ONB of that space. then, $P = A A^T$ does the job. however, if you just take a matrix $A$ with columns forming a basis, you need to normalize. that's what the $(A^T A)^{-1}$ does. if you think vectors, you have $(v^T v)^{-1} = \| v \|^{-2}$.

Balarka Sen

17:56

@MikeMiller I am having dinner right now, but I will whip up an example after I finish.

Semiclassical

um, but it's $(A^T A)$ (no reciprocal) in the expression before

Huy

@Semiclassical: did you read my message right after that expression?

(also it's not just $A^T A$ in that expression)

Semiclassical

generalized inverse stuff?

Mike Miller

@BalarkaSen: You should probably prove what I said above: a gradient flow line on $\Bbb R$ that doesn't go off to infinity has a unique limit in the critical set. (It might be a stationary point, but that's fine.)

Huy

@Semiclassical: I think by $(A^T A)^-$, @Clarinetist meant to write $(A^T A)^{-1}$ which seemed more reasonable in this context

Semiclassical

17:59

ugh, i misread that expression worse than that

Clarinetist

@Huy $A^TA$ doesn't necessarily have to be invertible

Huy

yes, hence the generalized inverse

Semiclassical

i read it too quick and thought the minus was a subtraction

Huy

lol

Semiclassical

facepalm

Huy

17:59

that would have been an odd projection

Semiclassical

yep

regardless, that's a lot more sensible

Huy

given some space $V$, there are a lot of different projections projecting to $V$, right?

is there some way to restrict this to get a unique one? kernel?

Mike Miller

yes

the kernel determines the projection

Huy

kernel + image is enough, yes?

Mike Miller

kernel & image I should say

Huy

18:03

yeah, that's what I assumed

Semiclassical

by what properties is $A^-$ defined?

Clarinetist

@Semiclassical $AA^{-}A = A$

Evinda

Hi!!!

Suppose that we have a vector $v$ with $m$ non-zero components.
Also assume that $w$ is a vector with $2m$ non-zero components.
Why does it hold that $v+w$ has $m$ non-zero components?

Semiclassical

thanks

Clarinetist

@Evinda Without loss of generality, suppose $v, m \in \mathbb{R}^k$, $k > 2m$.

Huy

18:04

@Clarinetist: I think that is the argument I'd try to go for - showing kernel and image of the projections coincide. but maybe there's a better method.

Clarinetist

@Huy < hasn't heard of kernels in something like 3 years

Huy

get used to them

Evinda

@Clarinetist m is a positive real number, isn't it?

Semiclassical

not seeing how that can possibly be true. i mean, isn't $(1,0)+(1,1)=(2,1)$ a counterexample?

Clarinetist

@Semiclassical That's true. I didn't think about that. The zeros would have to be in the same coordinates

< needs to spend this summer reviewing linear algebra

Semiclassical

18:07

now, i could see it being true if it were "show that $v+w$ has at least $m$ non-zero components"

Huy

@Clarinetist: I thought you were studying topology so you could teach me some

Clarinetist

@Huy Sigh, I've decided to prioritize knocking out the quals I have in May

Huy

:P

priorities

Clarinetist

I'm getting a study group together from the local university for math stuff in the summer though

Huy

such a pain

Clarinetist

18:08

The power of reddit :)

Huy

the last study group on reddit I joined didn't work out well

Clarinetist

I can hope :P

Evinda

$v$ is a vector of weight $m$, so with m non-zero components. So all vectors in $C+v$ have weight at least $m$.
$w \in C$ is a vector of weight $2m$ and $v$ is a "subset" of $w$ then $v,w+v \in C+v$ are two vectors of weight exactly $m$. @Clarinetist @Semiclassical

Why does it hold the weight $w+v$ is exaclt m?

Semiclassical

what is $C$ here, the vector space?

Evinda

It is a linear code @Semiclassical

Mike Miller

18:11

@Huy: I am tempted to not blame teddit

Semiclassical

that's a rather huge part of the context you've omitted.

Mike Miller

the last study group I joined also failed

Huy

@MikeMiller: you joined a study group?

Evinda

@Semiclassical Oh sorry, is my question related to the fact that C is a linear code?

Clarinetist

what's a linear code

Semiclassical

18:13

seeing as the claim you gave above is clearly false according to the counterexample I gave above? yes, i'd say that it must be pretty important

Evinda

It's a linear subspace of $\mathbb{F}_q^n$. @Clarinetist

Huy

probably just some $X^n$ for some countable set $X$

Mike Miller

yeah but then we all stopped studying

Clarinetist

*sees abstract algebra*
*runs*

Evinda

@Semiclassical Why does the statement hold for linear codes?

Semiclassical

18:16

considering I have no idea how linear codes work, i couldn't tell you.

i'd start with understanding why all vectors in $C+v$ have weight at least $m$, since that clearly fails to be true if $-v\in C$

Vrouvrou

Hello; I want to prove that this set $D=\{x\in E, d(x,A)>d(x,B)\}$ is open

Semiclassical

@clarinetist what confuses me is that the projector is written as $A(A^T A)^- A^T$ rather than $A^T A(A^T A)^-$

Vrouvrou

I choose $r= d(x,A)-d(x,B)$ bur i can't prove that $B(x,r)\subset B$

Semiclassical

i can see how the latter expression is idempotent, but that doesn't seem right for the first

Vrouvrou

can you help me @Huy

Semiclassical

18:22

or the reverse order for that second expression. same difference.

unless $A^T$ is itself invertible, but that seems bizarre in a generalized-inverse context

Clarinetist

@Semiclassical $$A(A^TA)^{-}A^TA(A^TA)^{-}A^T = A(A^TA)^{-}A^T$$
This is because by definition of $(A^TA)^{-}$, $$A^TA(A^TA)^{-}A^TA = A^TA$$ and there's a result that says that $$AB = AC \Longleftrightarrow A^TAB = A^TAC$$

Semiclassical

hmm. so $A^T$ is effectively 'invertible' in that specific context.

Jessy Cat

0

Q: Proof of existence and uniqueness of the exponential function using ODEs

In our lecture notes for our complex analysis class, we were given the following theorem: Theorem: There exists a unique complex function $f$ such that $f(z)$ is a single valued function $f(z) \in \mathbb{R}$ whenever $z \in \mathbb{R}$ and $f(1)=e$. $\forall z_{1},z_{2} \in \mathb...

complex-analysis differential-equations derivatives complex-numbers exponential-function

Clarinetist

@Semiclassical "Invertible" yes. To see why this is true, $\Longrightarrow$ is obvious. For $\Longleftarrow$, you get $A^TA(B-C) = 0$ or $(B-C)^TA^TA(B-C) = 0$, which implies that $[A(B-C)]^TA(B-C) = 0$, and there's a result that says that this implies that $A(B-C) = 0$.

Semiclassical

that sounds right.

Clarinetist

18:38

@Huy @Semiclassical I found the proof in an obscure book I have

Semiclassical

nice

Clarinetist

The idea is to prove two things: let $A$ be $n \times p$ and $B$ be $n \times q$ with $C(B) \subset C(A)$. Then show: (1) $P_A B = B$ and $B^{\prime}P_A = B^{\prime}$, and (2) $P_A P_B = P_B P_A = P_B$

Balarka Sen

18:55

Back.

Vrouvrou

hello, @BalarkaSen

I want to prove that this set $D=\{x\in E, d(x,A)>d(x,B)\}$ is open

I choose $r= d(x,A)-d(x,B)$ but i can't prove that $B(x,r)\subset B$

have you an idea please

@DanielFischer have you an idea ?

Jessy Cat

19:13

People keep posting answers and comments and then deleting them!!

If anybody in here can please, please help me with the last part of the question I posted, I would be so incredibly grateful!

@PVAL, I think you posted something before.

Balarka Sen

@MikeMiller Getting the easy ones done: 6) Assume $\gamma$ is finite length but does not have a limit. Then there's a sequence $\{t_k\}$ in $[0, \infty)$ such that $\{\gamma(t_k)\}$ is not convergent.
It's also not Cauchy, as $\Bbb R^n$ is complete. So there's an $\epsilon$ such that $\|\gamma(t_{k+1}) - \gamma(t_k)\| \geq \epsilon$ regardless of whatever $k$ I choose. Then $\int_{t_k}^{t_{k+1}} \|\gamma'(u)\| du \geq \epsilon$ for all $k$, hence $\int_0^{t_k} \|g'(u)\| du \geq k \epsilon$. As $k \to \infty$, left side is the length while right side blows up to $\infty$: impossible situation.

Uniqueness is clear.

(I should have mentioned I chose $\{t_k\}$ so that $t_k \to \infty$, sorry).

Vrouvrou

19:41

Someone help me please

Mambo

19:54

@Vrourou What is E?

Vrouvrou

@Mambo it is a metric space $(E,d)$

Mambo

Are A and B any two subsets?

Vrouvrou

yes

Vrouvrou

20:12

@Mambo have you an idea on a good choice of $r$ ?

Mambo

@Vrouvrou Do you know anything about the function $f(x) = d(x,A)$, where $A$ is any subset of $E$

Vrouvrou

it is continuous

Mambo

Can you prove it?

Vrouvrou

by using that $|d(x,A)-d(y,A)|<d(x,y)$

Jessy Cat

Help please.

Mambo

20:17

Do you notice something more?

something more than continuity?

Vrouvrou

@Mambo no, like what ? it is Lipschitz

Mambo

What is the definition of continuity?

Vrouvrou

in $\mathbb{R}$ ?

Mambo

In metric spaces

Say you have a function $f:(X,d) \to (Y,p)$

When will be $f$ continuous?

@JessyCat What kind of help do you need?

Jessy Cat

0

Q: Proof of existence and uniqueness of the exponential function using ODEs

In our lecture notes for our complex analysis class, we were given the following theorem: Theorem: There exists a unique complex function $f$ such that $f(z)$ is a single valued function $f(z) \in \mathbb{R}$ whenever $z \in \mathbb{R}$ and $f(1)=e$. $\forall z_{1},z_{2} \in \mathb...

complex-analysis differential-equations derivatives complex-numbers exponential-function

I need help getting the very last part done

Vrouvrou

20:23

$\forall \varepsilone>0, \exists \delta>0, \forall x\in E, d(x,a)<\delta\Rightarroe d(f(x),f(a))<\varepsilone$ @Mambo

but i don't see the relation with the choice of $r$

Mambo

and $\delta$ depends on?

Vrouvrou

f from $(E,d)$ to $(E,d)$

Mambo

okay

Vrouvrou

on $\varepsilone$

Mambo

For the function $f:\mathbb{R} \to \mathbb{R} x \to x^2$, Will you get such a delta working for all $x \in \Bbb R$

Vrouvrou

20:27

yes and then ?

Mambo

Take $\varepsilon = 1$ give me $\delta$

Overly Excessive

HELP PLEASE

user19405892

hi

Vrouvrou

@Mambo in what this help me to find $r$ ?

user19405892

say we have O at the origin and A,B,C, G points in space

what is the condition that A,B,C, G lie on the same plane?

Mambo

20:30

Do you know what continuity does to open sets?

user19405892

I am told we must have aOA+bOB+cOC = OG

Vrouvrou

the pre-image of open sets is open

@Mambo

but i don't knoe to use this

i want to find $r$

Mambo

Consider the function $f(x) = d(x,A) - d(x,B)$

Vrouvrou

it is continuous yes

Mambo

What is $f^{-1}((0,\infty))$?

Vrouvrou

20:32

equal $D$

but i have to prove that D is open usins $B(x,r)\subset D$

i have to find r

@Mambo

Overly Excessive

HELP PLEASE

Mambo

Let $X = A\cup B ; A,B $ are disjoint closed sets.

What is your desired set in this case?

Vrouvrou

you are with me @Mambo?

Mambo

yes

Overly Excessive

PLEASE ADVICE

Mambo

20:43

By the way, your definition of continuity said above is not correct. With that definition you cannot give the $\delta$ I asked for

Vrouvrou

i don't understand how the continuity give me r

Mambo

Take $\varepsilon = 1$ and find the corresponding $\delta$

$\delta$ is $r$.

Overly Excessive

PLEASE DO THE NEEDFUL

Mambo

Whatever you have stated as definition of continuity is actually the definition of uniform continuity

Vrouvrou

i don't know how to do @Mambo

i think that $r=\frac12 [d(x,A)-d(x,B)]$ works

Mambo

21:00

Can you prove it?

Vrouvrou

i don't know using continuity

as you tel me

Mike Miller

@Balarka More straight forwardly: The integral from 0 to infinity of $\grad f$ exists and is finite.

Jessy Cat

I figured it out 8-)

Mambo

Take any $y \in B(x,r)$ and compute $d(y,A) - d(y,B)$ by plugging in $d(x,A)$ and $d(x,B)$

Vrouvrou

ok thank you i found it

Balarka Sen

21:08

@MikeMiller Right.

Hi @TedShifrin.

Mambo

@Vrourou Did you understand what was the problem with the definition of continuity?

Vrouvrou

no

Mambo

Can you recall again?

Vrouvrou

i use $|d(x,A)-d(y,A)|<d(x,y)$ and $|d(x,B)-d(y,B)|<d(x,y)$

Mambo

Yes

@Vrouvrou The definition of continuity?

Vrouvrou

21:14

$\forall \varepsilone>0, \exists \delta>0, \forall x\in E, d(x,a)<\delta\Rightarroe d(f(x),f(a))<\varepsilone$

Balarka Sen

@MikeMiller About 1), lots of curves with finite energy but not finite length. E.g. $f(x) = 1/x$ on $[1, \infty)$. $\int_1^\infty f(x)^2 dx = 1$, whereas $\int_1^\infty f(x) dx$ isn't finite. I can extend $f$ to a smooth curve on $[0, \infty)$ by hand: glue a curve $g$ on $[0, 1]$ to the end of $f$, so that all the derivatives of $g$ at $g(1)$ is equal to the derivatives of $f$ at $f(1)$.

Mambo

@Vrouvrou Do you mean If you fix an $\varepsilon$ then $\forall x \in E$ same $\delta$ works?

Balarka Sen

(Sorry, I forgot to say that the curve is $\gamma(x) = \log(x)$ on $[1, \infty)$. That has finite energy, but not finite length).

I am a bit skeptic that finite length implies finite energy though. Antiderivative of $f(x) = 1/\sqrt{x}$ on $(0, 1]$ has finite length: $\int_0^1 f(x)dx = 2$. But $\int_0^1 f(x)^2dx$ doesn't converge, so anti-derivative of $f$ doesn't have finite energy. Can't this happen with $[0, \infty)$ instead of $(0, 1]$?

user19405892

Hi

say we have O at the origin and A,B,C, G points in space
what is the condition that A,B,C, G lie on the same plane?

I am told we must have aOA+bOB+cOC = OG

is this right?

does this have to do with linear combinations?

Mike Miller

21:36

@Balarka: No. Do it!

PVAL

22:18

What's a good policy on unnecessary logical shorthand?

user19405892

hi

Daniel Fischer

@PVAL Avoidance?

PVAL

@DanielFischer grading of

Daniel Fischer

@PVAL You're grading and the students are still in the phase of "if it isn't all symbols it ain't maths"? Tell them that words are awesome for readability.

PVAL

@DanielFischer Apparently a professor (not the one I'm grading under) told them that "Real Math" was filled with logical shorthand. Arxiv was open on my computer so I clicked a random article.

Daniel Fischer

22:32

@PVAL Real maths uses a balanced mix of symbols and words.

9

user19405892

can anyone help with my question?

PVAL

These symbols are usually not logical shorthand.

Ted Shifrin

23:13

@PVAL @DanielF: When I took point-set topology from Munkres, he forbade us to use any symbols (other than implies, element of, etc.) ... He insisted that we write words and complete sentences.

I've always taught my students the same way. I find it impenetrable to look at a line (or worse, a page) of symbols.

PVAL

23:26

@Ted I was at some point taught to not use the logical symbol for implies.

I have regressed though.

Ted Shifrin

I use $\implies$, not the right arrow.

PVAL

I think I only use that for bidirectional proofs so it is easier to denote the direction (as saying something like the "if direction" or "forward direction" can get confusing otherwise.

Ted Shifrin

Well, good for you. I still write "if then" if I'm in prose mode, but use the implies symbol if there are some lines of math formulas ... I think ... :)

PVAL

Well I haven't had to do analysis in a while..

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