Mathematics - 2015-02-28 (page 1 of 3)

12:02 AM

@DanielFischer The functions are:

$$\frac{n}{\lg n} , \ \ n^{\lg n} ,\ \ (\sqrt{2})^{\lg n}=\sqrt{n}, \ \ n^2, \ \ n!, \ \ (\lg n)! ,\ \ \left( \frac{3}{2} \right)^n ,\ \ n^3 ,\ \ \lg^2 n ,\ \ \lg(n!) ,\ \ 2^{2^n}, \ \ n^{\frac{1}{\lg n}}=2, \ \ \ln{\ln n}, \ \ e^{\log_{10} n }, \ \ n \cdot 2^n, \ \ n^{\lg{\lg n}}, \ \ \ln n, \ \ 1 ,\ \ 2^{\lg n}, \ \ (\lg n)^{\lg n}, \ \ e^n ,\ \ 4^{\lg n}=n^2, \ \ (n+1)!, \ \ \sqrt{\lg n}, \ \ \lg{\lg{\lg n}}, \ \ 2^{\sqrt{2 \lg n}}, \ \ n, \ \ 2^n ,\ \ n \lg n, \ \ 2^{2^{n+1}}$$

Maximilian

For another problem I got stuck again as well:P

I did cross multiplication and got $1+sin^2\Theta=cos^2\Theta$ but thats just one part of one side of an equation already

and on the other I got $-sin^2\Theta=cos^2\Theta$

I think I did something wrong

Definitely did those wrong

Maximilian

12:25 AM

and a third time on another problem:/ Im getting no progress

$sin(x)-sin(x)cos^2(x)+sin(x)cos(x)-cos^2(x)(sin(x)cos(x))$

evinda

@Axoren Does it hold that -1=o(n) ?

Maximilian

Anyone know what I can do for this one?

Axoren

@evinda Does it hold that every positive $c, n_0$ results in $-1 < cn$ for $n > n_0$?

@Maximilian $1 = \sin^2\Theta + \cos^2\Theta$

evinda

@Axoren Yes... but then we cannot add at the inequality f(n), since it could also be negative... can we?

Axoren

@evinda I don't follow.

What inequality and what can also be negative?

evinda

12:31 AM

We have that $\forall c>0, \exists n_0(c) \in \mathbb{N}$ such that $\forall n \geq n_0: 0 \leq g(n)<cf(n)$.
But do we that f isn't negative and so that the following is allowed?
$f(n) \leq f(n)+g(n) < cf(n)+f(n)=(c+1)f(n), \forall n \geq n_0$. @Axoren

Axoren

@evinda Little-o doesn't have a positivity restriction.

It's simply an upperbound.

evinda

@Axoren So we cannot add f(n) at the inequality, can we?

Maximilian

@Axoren I know that, but Im not seeing that in the problem?

Axoren

@evinda $\forall n \geq n_0: g(n)<cf(n)$

You keep including a lower-bound of 0.

Maximilian

Or at least the last one I posted

Axoren

12:34 AM

@Maximilian $1+\sin^2\Theta=\cos^2\Theta$

Replace $1$ with that identity and you can solve it fairly easily.

Your most recent one is a mess that involves far more identities

Maximilian

Oh ok, ya I know..

Axoren

How did you even get that last term?

evinda

@Axoren And since $0 \leq g(n)$ and $g(n)<cf(n)$ we have that $cf(n) \geq 0 \Rightarrow f(n) \geq0$, right?

Axoren

I'm impressed if that actually came from a real-world problem.

Maximilian

Lol

Axoren

12:36 AM

@evinda are we still talking about $g(x) = -1$?

Maximilian

The original problem is $\frac{tan x}{1+cos x}+\frac{sin x}{1-cos x}=cot x +sec(x)csc(x)$

Axoren

Oh, I think I get your question now.

evinda

@Axoren For the general case, for example $g(x) = -1$..

Maximilian

for the right side I got it down to sin^2(x)

Axoren

@evinda You can perform algebraic steps to any inequality

As long as you apply it to all sides in a way that preserves the relation.

Maximilian

12:38 AM

and then on the left side is all the issue. I got the last term from combining the denominators to get $1-cos^2x$

Axoren

Multiplying by positive $c$ and adding positive $n_0$ will preserve the relations.

evinda

@Axoren But if f isn't positive we aren't sure that the signs will remain the same, are we?

Maximilian

and then I multiplied the denominator to the fraction to make it the top part only and then I got the end

Axoren

Multiplication by a negative doesn't always preserve the relation.

But adding a negative always does.

$x - c \le y - c$, for example.

evinda

@Axoren A ok... But at the definition we have this: $\forall n \geq n_0: 0 \leq g(n)<cf(n)$
so why do we allo negative functions?

Axoren

12:40 AM

Is that the definition you were given in class, because I don't have that one in front of me.

I have one that does not provide a lower-bound in little-o

But it makes sense that you may have been given one like that in class.

For time-complexity, we lack the technology to perform and finish operation before we start them.

So a lower bound of 0 is acceptable.

At the most, we can expect things to already be done, netting a BigO of O(0)

But accessing the result of a finished thing is still BigO O(1)

evinda

Yes, dropbox.com/s/d1ucdvokxy4o0na/… at page 50... @Axoren

So in this case we are sure that f(n) is positive, right?

Axoren

It wouldn't make sense within the context to expect a negative function.

So you can be confident that it's positive.

Maximilian

For $\frac{sec^4\Theta-tan^4\Theta}{sec^2\Theta+tan^2\Theta}=sec^2\Theta-tah^2\Theta‌$ I got $1=sin^4\Theta$

I don't think thats correct

Axoren

I can't tell you can be sure, that's your instructor's interpretation of the book's questions.

Maximilian

Well for the questions I'm only to verify that each trigonometric equation is an identity

Axoren

12:47 AM

@Maximilian my last message was aimed at @evinda

Maximilian

Oh lol

Axoren

As for you, let me see.

Maximilian

So far all of the questions had both sides equal, so they're identities, and I don't have the answers so I don't know if they are all supposed to be identities or I just put that its not?

Axoren

$\sec^2 = \frac 1 {cos^2}$
$\tan^2 = \frac {sin^2} {cos^2}$

My tan may be upside down

Maximilian

No its right

Axoren

12:49 AM

Oh, wait, is that the full equation?

That's right because of difference of squares.

Maximilian

Yup

Semiclassical

hello all

Axoren

How did you get $1 = \sin^4$ from that?

Maximilian

I think I mixed up something at the end so its not $1=sin^4\Theta$, hmmm

Axoren

@Semiclassical yo.

Maximilian

12:51 AM

I mixed it up and I think I put part of the left side I was working on just on the other side and messed it up:P

Semiclassical

anyone know much (read: anything) about projective representations? one of the physics seminars i saw today used the phrase 'projective symmetry group' a lot, and i suspect that's something which is familiar on the math side

Maximilian

Ya, I simplified the left part, I separated the sec and tan and then simplified them and accidentally set them to equal at the end

Oops

evinda

Thanks for you help @Axoren :)

Maximilian

So I think I have now $\frac{1}{cos^2\Theta+cos^2\Theta sin^2\Theta}+\frac{sin^4\Theta}{cos^2\Theta+cos^2\Theta sin^2\Theta}$

Axoren

@evinda No problem. Good luck with your studies.

evinda

12:54 AM

Thank you!!! I wish you good luck too!!! :)
I will go to sleep now... Good night!!!

Axoren

Good night

Maximilian

So I have $\frac{1}{cos^2\Theta+cos^2\Theta sin^2\Theta}+\frac{sin^4\Theta}{cos^2\Theta+cos^2\Theta sin^2\Theta}$=$\frac{1}{sin^2\Theta}-\frac{sin^2\Theta}{cos^2\Theta}$, seem correct?

then on the right side I can do $sin^2\Theta-cos^2\Theta=1$ and $1-sin^2\Theta=cos^2\Theta$?

Maximilian

1:31 AM

Its not my day for math

Stuck on the 4th problem in a row

$\frac{4cosx}{1+cos^2x} = 4\frac{cosx}{sin^2x}$

$1+cos^2x=sin^2x$, so maybe I did the other side wrong and the 4 should only go on top

Ok I got it

Maximilian

2:04 AM

Is anyone here?

Kaj Hansen

3:25 AM

@Maximilian, the identity is $\sin^2(\theta) + \cos^2(\theta) = 1$.

Committing to a challenge

Are all interior points of a metric space also limit points. E.g. Does $x\in E^O \imples x\in E'$?

Kaj Hansen

Yes @Committingtoachallenge

Committing to a challenge

@KajHansen Awesome, what about in general topology?

That may be a stupid question, I haven't done any topology outside of metric spaces so far

Kaj Hansen

So limit point $x$ of a subset $S \subset X$ for a general top space $X$ is such that every open set containing $x$ also contains at least one other point in $S$.

Committing to a challenge

That is the same definition I am working off for metric spaces which is good

Except my $X$ is a metric space

E.g. Rudin PMA Chapter 2

Thanks @Kaj

Kaj Hansen

3:31 AM

Metric spaces make things easier to think about

You can make it match up with general top spaces by noting that, if $x \in S \subset X$ is in the interior of $S$, then there exists an open set containing $x$ that is a subset of $S$.

Committing to a challenge

Yep I think Rudin uses that also. It seems Rudin actually works with the actual general topology definitions(which I thought may be the case, based on the blog entry by @JulianRachman matching up with my understanding)

Kaj Hansen

IIRC, Rudin works only with the metric topology in PMA.

Which is fine for real analysis

Committing to a challenge

He does, but he defines limit points and interior points exactly as you have defined above(as far as I can tell and [open set = neighbourhood])

Kaj Hansen

3:46 AM

Yes indeed

Committing to a challenge

I actually found the chapter really pleasant, but I am moving sluggishly through the exercises

Kaj Hansen

Chap 2 was enjoyable for me as well

Committing to a challenge

I am working on 9d) Prove that the complement of $E^O$ is the closure of the complement of $E$, and I am having trouble, but I think I am almost there

Kaj Hansen

As a warning: Some of those in #9 were false.

I can't remember if it was Part D or another part

Committing to a challenge

Whattt

Oh probably e and f

They ask if things are true, rather than saying to prove it

Kaj Hansen

3:50 AM

mhmm

I can pull out my previous write-ups if you get truly stuck.

Committing to a challenge

$E^O$ is always open, $E$ is open iff $E^O=E$, If $G\subset E$ and $G$ is open, $G\subset E^O$

They are all true?

Kaj Hansen

Looks right

for all of them

Committing to a challenge

I'm a little surprised you don't have a blog xD, especially with having some youtube videos up

Kaj Hansen

The YouTube vids were inspired by me trying to drudge through the literature on that exact subject and finding it difficult just because they'd try to describe some crazy graph and all sorts of colorings on the edges without any visuals, which is problematic because I found that meaning would get lost underneath formality and ink.

Committing to a challenge

@KajHansen Fair enough, I guess I should have worked that out from the 'internets first' part

Kaj Hansen

3:53 AM

And that too. I know some people learn better from having graphical and audio cues than just plain text.

For the most part, though, what I know about math in general is already well-known and well-understood.

A lot of the stuff I'm doing in topology, e.g., can be found in all corners of the internet

Balarka Sen

Hi guys.

Kaj Hansen

hey hey

The Substitute

In real variables, the solution to $f''(x)=4f(x)$ is the collection of functions $ae^{2x}+be^{-2x}$. Would this solution also hold if $f$ is maps the complex plane to itself?

sorry, off topic

Committing to a challenge

@Kaj Fair enough, I guess mine is just to self-motivate, but if you haven't got any problems with motivation they would be pointless at a non-research level

Balarka Sen

cayley cayley

Committing to a challenge

3:55 AM

@BalarkaSen How are you?

Balarka Sen

speaking of which, have you got the isometry problem, @Kaj?

that every group can be realized as isometry group of some metric space?

Kaj Hansen

Not yet, but I don't have to until Wednesday @BalarkaSen if I choose to. There are loads of interesting problems on that set.

Julian Rachman

Hello! @commitingtothechallenge thank you for mentioning my blog and notes. What chapter 2 were you guys referring to @Kaj

Balarka Sen

fine, @Committingtoachallenge. what about you?

Kaj Hansen

@JulianRachman, of Rudin's Principles of Mathematical Analysis

Committing to a challenge

3:56 AM

@JulianRachman Rudin's Principles

Balarka Sen

@KajHansen yeah, which is confusing.

Committing to a challenge

@BalarkaSen Alright, I start my courses in 2 days

Balarka Sen

can't even select 7 of them.

Kaj Hansen

What's confusing @BalarkaSen ?

Ultimately I have to choose 14 total over a 2-week span.

Balarka Sen

choose them well

Kaj Hansen

3:57 AM

One of the ultrametric implications I found interesting: Two open balls in an ultrametric space are either disjoint or one is contained in the other.

@BalarkaSen

Balarka Sen

yeah.

Kaj Hansen

Kinda cool to imagine

Balarka Sen

think about the cantor set

Maximilian

How would you solve $\frac{cos\Theta+1}{tan^2\Theta}=\frac{cos\Theta}{sec\Theta-1}$

Balarka Sen

every two point is either too close or too far away

Committing to a challenge

3:59 AM

How do I make a large overbar?(in mathjax)? $\bar{(E^o)}$ looks terrible

Maximilian

Or atleast make each side equal

Julian Rachman

@Kaj @commitingtothechallenge ah I see

Kaj Hansen

$\text{ \overline{ stuff } }$

Like that?

Committing to a challenge

Yes, that's very good. Thank you

Kaj Hansen

:D

Julian Rachman

4:00 AM

I have one more prob in chap 2!

Committing to a challenge

Which one?

30 from PMA?

You are using Munkres for topo and PMA for real ana?

Julian Rachman

No. In munkres

Nono. I am done with analysis

Dave

what is the term for a number that can divide into two other numbers that both produce whole numbers for use of simplifying

Committing to a challenge

Did all of PMA?

Dave

im trying to word it correctly and i hate the way i worded it

Julian Rachman

4:02 AM

I am focusing on algebra and topology now

Balarka Sen

@KajHansen Pick up a metric space $X$.

Kaj Hansen

Ok, sure

Julian Rachman

@commitingtothechallenge I didn't. I went up to the point right when they start getting into too much detail of the "why's" and "how's" of calculus.

Maximilian

Anyone know?

Committing to a challenge

@JulianRachman Well that's pretty much why I am doing the book haha. I can mechanically do calculus problems, but I don't know why or how[the methods work]

Balarka Sen

4:05 AM

A map $\gamma : [0,1] \to X$ will be called a geodesic if for $\gamma(a) = x$ and $\gamma(b) = y$, supremum of $\{\sum_{i=1}^r d(\gamma(t_{i-1}), \gamma(t_i)) | a = t_0 < t_1 < ... < t_r = b, r \in \Bbb N\}$ is $d(x, y)$

Kaj Hansen

I am familiar with geodesics from differential geometry @BalarkaSen

Committing to a challenge

@Maximilian What are you solving for?

Theta?

Balarka Sen

this is a generalization of geodesics in metric spaces, @Kaj

Maximilian

The problems want me to verify that each trigonometric equation is an identity

So I just need both sides to be equal

Kaj Hansen

Cool cool. I got a taste of that working in $\mathbb{H}$ towards the end.

Julian Rachman

4:06 AM

@commitingtothechallenge well, ya. Topology doesn't require any analysis. But they said a foundation in the subject would be great. So I got my foundation.

Committing to a challenge

So you didn't find it interesting?

Just going straight for topology?

Balarka Sen

if for any two points $x, y \in X$, $x$ and $y$ can be joined by a geodesic, $X$ is called a geodesic metric space.

Committing to a challenge

@Maximilian Use trignometric identities to simplify that to another (known correct) trig identity?

Maximilian

ya

Kaj Hansen

I'm with you so far @BalarkaSen

Julian Rachman

4:07 AM

Nono. I believe that analysis is beautiful and that you should learn it for what it is and what it could bring you in the future.

Balarka Sen

example : a graph with metric induced from assigning a fixed length to the edges.

Julian Rachman

@commitingtothechallenge

^^

Committing to a challenge

@JulianRachman But you are primarily focusing on Topology?

Julian Rachman

Currently yes.

But I might end up going back

Balarka Sen

as every graph is a geodesic metric space, every finitely generated group is also a geodeisc metric space (cayley graph)

Julian Rachman

4:09 AM

After I learn everything I want to leaen

Kaj Hansen

Still with you @BalarkaSen

Balarka Sen

Right, now the fun begins.

If you have a geodesic triangle in your g. met. space $X$ (i.e., three points with each two joined by geodesics) then define this :

Julian Rachman

What subject are you guys talking about

Maximilian

I need help with that

Balarka Sen

a geodesic triangle $abc$ is $\delta$-slim if any of the three sides $ab$, say, is contained in the union of the \delta-nbhds of the other two sides, i.e., $ab \subseteq N_\delta(bc \cup ac)$

Kaj Hansen

4:13 AM

I'll have to think about that a little more in a sec @BalarkaSen . It's less obvious what's going on in my mind, and I'm trying to get something submitted within the hour.

Julian Rachman

@Maximilian do you know your trig identities?

Maximilian

ya

Julian Rachman

Then just use that.

Maximilian

Well some, I think there are more but they haven't been mentioned yet, I could be wrong

Balarka Sen

Oh sure. You can think about geodeisc triangles as deflated triangles, @Kaj.

Maximilian

4:14 AM

well I do but I don't get the correct answer

Julian Rachman

@Kaj @Balarka what are you guys talking about?

Maximilian

Im trying it again but ill get something way off

Balarka Sen

Definition : if every geodesic triangle in a g. met. space $X$ is $\delta$-slim for some $\delta$, call $X$ hyperbolic

there you go, @Kaj. i'll let you think about it for a while.

Kaj Hansen

@BalarkaSen, like more of a simplex whose nodes are defined by the intersection of 3 geodesics?

Balarka Sen

i'll have to go soon.

@KajHansen think about triangles in $\Bbb H$

Kaj Hansen

4:15 AM

Sure thing!

Julian Rachman

@Kaj @Balarka ^^

Balarka Sen

@Julian we are talking about one of Gromov's works.

about hyperbolicity.

Julian Rachman

What field of math?

Balarka Sen

geometric group theory.

Kaj Hansen

@Balarka et. al. I recall a quote that was along the lines of "Proving things is the easy part. Figuring out what to prove is difficult".

Does anyone know who said this? I think it was Riemann, trying to find a source.

Julian Rachman

4:16 AM

Oh. Geez. Sadly I can't compete with that.

Balarka Sen

Um, no idea.

anyway, when you've given this a thought, prove that $\Bbb H$ is a hyperbolic geodesic metric space.

you know the metric., so it shouldn't be hard

one of gromov's breakthrough idea was to define hyperbolic groups using this idea, which was the beginning of modern geometric group theory.

Maximilian

What would $\frac{cos^3\Theta+cos^2\Theta}{sin^2\Theta}$ be? $\frac{cos^2\Theta}{sin^2\Theta}$ would be $cot^2\Theta$ but I don't know about $\frac{cos^3\Theta}{sin^2\Theta}$

Balarka Sen

i have got to go

Maximilian

4:35 AM

:[

Julian Rachman

@Kaj hi

Maximilian

In another problem I got it to $\frac{1}{cos^6\Theta}-\frac{sin^6\Theta}{cos^6\Theta} = \frac{1}{cos^2\Theta}-\frac{sin^2\Theta}{cos^2\Theta}$

Is there something that I am missing to make these =

Oops, I think I see it for this one

I probably solved it and erased it on accident and doing it again

Kaj Hansen

5:01 AM

Hey there @JulianRachman

Julian Rachman

5:19 AM

@Kaj how's it going? What you working on?

Maximilian

How can I get $tanx+sinxcosx=cotx+secxcscx$

Well they are already = in the problem, but how can I make 1 side look like the other? If i make it the denominator of 1, ill get the other but if i do that then don't I need to do it to the other side? that would be counter productive

Semiclassical

@Maximilian: For $x=\pi/4$, the LHS is $1+1/2=3/2$ and the RHS is $1+2=3$.

so that identity isn't valid.

Maximilian

Im very confused

Hmm

The problem is $\frac{tanx}{1+cosx}+\frac{sin}{1-cosx}=cotx+secxcscx$

and I have no idea how to get it to equal each other

anon

5:53 AM

@Maximilian what do you do to add fractions together?

Semiclassical

as an alternative to fractions, you can multiply both sides by the two denominators. that clears the fractions, and has the advantage that $(1+\cos x)(1-\cos x)=1-\cos^2 x=\sin^2 x$

for comparison with my remark above, note that $\sin^2 x=1/2$ for $x=\pi/4$---precisely the factor missing in the RHS above

Maximilian

Because thats the problem in the book

@anon

so I multiply both sides by (1+cosx)(1-cosx)? that was probably my mistake, I only did the left side

Ok.. that makes sense. Ill try that

anon

@Maximilian in fact you can start from the LHS and obtain the RHS with absolutely no foresight about what it will end up being

just multiply top/bottom of the first fraction by 1+cos and the second by 1-cos

see what happens when you do that

Maximilian

Thats what I did already

and that gave me $tanx+sinxcosx=cotx+secxcscx$

anon

so... you already solved the problem?

Maximilian

6:05 AM

No, thats what I get but I need both sides to equal each other

I have to use the identities to make both sides =

so I mean, they are but I need to simplify it down until I get there

Semiclassical

you lost the bottom terms

i.e., if you multiply top and bottom by 1+cos, then you need to simplify both numerator and denominator

Maximilian

And the closest I have is that both sides are inverted, now I'm multiplying the right side by (1-cos^2x) which is the denominators of the left side

ya i did

anon

@Maximilian wait, when you write that equation, are you saying you started with the LHS and changed it and that's it? you aren't saying you figured out the whole problem?

Maximilian

Ya

anon

don't operate on the equation, operate solely on the LHS and obtain the RHS

that's more natural

to show A=B, show A=X=Y=Z=...=B.

Maximilian

6:08 AM

The whole problem is $\frac{tanx}{1+cosx}+\frac{sinx}{1-cosx}=cotx+secxcscx$

anon

so, you know the original LHS equals tan+sincos. now you need to show tan+sincos equals cot+sec*csc.

@Maximilian yes, I know

Maximilian

so thats in the book, as the problem and I use the identities to make them = but I haven't gotten there

anon

the problem isn't to start with the thing you're trying to show. that's borderline circular reasoning.

infinitesimal

Which book @Maximilian?

anon

@infinitesimal doesn't matter

Maximilian

6:09 AM

lol

Semiclassical

what anon is suggesting is to take the $\frac{\tan x}{1+\cos x}$ term, and multiply by $\frac{1-\cos x}{1-\cos x}$

and do similarly for the second term. then they'll both have the same denominator and you can directly add them

anon

@Semiclassical no, don't directly add them

Maximilian

Yes, I multiplied $tanx$ by $1-cosx$ and multiplied $sinx$ by $1+cosx$ which is the opposites? Did I do it wrong? If you multiply by both to each other one cancels out so I had $tanx(1-cosx)+sinx(1-cosx)$

anon

@Maximilian if you multiply tan/(1+cos) by (1-cos)/(1-cos) you should get tan(1-cos)/(1-cos^2)

Semiclassical

@anon i meant in the sense of them having common denominators, so that the numerators can be directly added. sloppy wording on my part

anon

6:12 AM

@Semiclassical I know what you mean, and I'm saying not to do it

keep them separate for a bit longer

Maximilian

Ok... let me try that then, I got rid of the denominators

anon

I didn't want you to get rid of denominators

I want 1-cos^2 to be replaced by sin^2

Maximilian

ok let me do it

lots of erasing though :P

anon

also I replace tan with (sin/cos)

Maximilian

Ok, I got $\frac{tanx=tanxcosx}{1-cos^2x}$ for just the first part

the tan, then ill do the same to the sin x

anon

6:18 AM

$$\frac{\frac{\sin}{\cos}(1-\cos)}{\sin^2}+\frac{\sin(1+\cos)}{\sin^2}$$

that's what I get after replacing $1-\cos^2$ in the denominators with $\sin^2$

Maximilian

Ok, ya I got that now

anon

some sin's can be cancelled now...

Maximilian

So I have $\frac{tanx-sinx}{sin^2x}+\frac{sinx+sinxcosx}{sin^2x}$

anon

I would have written $$\frac{1-\color{Blue}{\cos}}{\cos\sin}+\frac{\color{Blue}{1}+\cos}{\sin} $$

Maximilian

Errrrr

How did you do that? I see on the right one you can cross out a sin, but the left?

Oh, multiply by cos then it cancels on top, but is in the bottom then also cross out a sin

wouldn't $1-cos+1+cos=2$?

cross out the $cos$?

Hmmm... I need 1 to be on top of cos and sin but need cos on top of sin

anon

6:33 AM

$$\frac{1}{\cos\sin}-\color{Blue}{\frac{\cos}{\cos\sin}}+\color{Blue}{\frac{1}{ \sin}}+\frac{\cos}{\sin} $$

Maximilian

ya, i put that but mine is off a bit:P

$secxcscx-cscx+cscx-cotx$ so I just did 1 thing wrong, need the cot + and done

and yes i see your work does it... very strange that out of all the problems, this is just like way different then them all by a lot

atleast to me:P

ah, I put a 1-cos on the top, ok.. done

Thank you so much

Ill try to see if anything I learned on this one works for the others I had trouble with... Im missing very simple things

Zelos Malum

6:55 AM

dear lord =O

Committing to a challenge

8:57 AM

In Rudin Chapter 3 - Convergent sequences, he starts referring to bounded sequences without defining them. Is this still following the definition from metric space topology?

E.g: (X is a metric space) $E\subset X$ is bounded if there is a real number M and a point $q\in X$ such that $d(p,q)\lt M$ for all $p\in E$

bolbteppa

9:23 AM

@Committingtoachallenge he says on the first page of Ch. 3 that the first 3 sections are defined w.r.t. a metric space and since the range of the sequence is in the metric space one can ask whether it is a bounded set right?

Sayan Chattopadhyay

10:02 AM

Hi guys

user159870

10:25 AM

Hi!

@DanielFischer Can you look at this math.stackexchange.com/questions/1168439/… and tell me if it is correct?

user112495

12:54 PM

I am trying to show that if F is a finite set and (F, d) is a metric space, then the induced topology is the discrete topology.

I think if I can show that singletons are open in finite sets, then the result will follow by the union property of metric spaces.

But I'm having trouble showing that singletons are open (i'm not even sure if this is true).

Actually, don't worry. I think I've got it.

Daniel Fischer

@user112495 Good. How confident are you that your way is correct?

user112495

@DanielFischer I said that as we're in a finite set, for each point we can take a value that is the minimum distance between that point and any other point in the set. We can the construct an open ball with radius equal to this distance. As the ball is open, the only point in this ball is the point we took the ball about.

@DanielFischer And so all singletons are open. As (F, d) is a metric space, we know that the union of any collection of open sets is open and so we have the discrete topology.

Daniel Fischer

@user112495 Good.

An alternative: Metric spaces are Hausdorff, hence singletons are closed. A finite union of closed sets is closed, so all subsets are closed. So all subsets are open (since the complement of each is closed).

Saal Hardali

1:12 PM

Hey everyone! Is there someone here who's familiar with rudin's functional analysis and can tell me if i can read it without too much prior knowledge of hilbert spaces?

Daniel Fischer

@SaalHardali You can.

Saal Hardali

The table of contents gives the impression that it has a very broad perspective.

I liked his real and complex analysis very much and am now going over it again before going forward.

Daniel Fischer

Yes. One of the few where one can learn a bit about topological vector spaces in general.

Saal Hardali

The other books i looked at were really hilbert space focused with a bit of banach spaces that's why i worry about prerequisites.

@DanielFischer thanks for the advice!

Daniel Fischer

@SaalHardali There's a tendency that books on functional analysis focus on the applications to PDEs, where Hilbert spaces are used a lot. Rudin's book is not focused on that, so he treats more general stuff. That means on the other hand that he covers less of the Hilbert space theory than some others do.

Saal Hardali

1:24 PM

@DanielFischer Ah, sounds perfect for me. I'm only learning it because i like the theory anyway... plus, Rudin writes beautifully

Daniel Fischer

@SaalHardali Then you'll enjoy it. And learn a lot from it.

Saal Hardali

@DanielFischer thx!

evinda

1:50 PM

Hello @DanielFischer !!! Could I ask you something? I want to determine if $\lg^k(n) $ is O/ o / Ω / ω, Θ of $n^{\epsilon}, \epsilon>0, k \geq 1$. I proved that $\lim_{n \to +\infty} \frac{\lg^k(n)}{n^{\epsilon}}=0$. This means that $lg^{k}(n)$ is $o(n^{\epsilon})$. Can we say the following?
Since $lg^{k}(n)$ is $o(n^{\epsilon})$ we conclude that $lg^{k}(n)$ is $O(n^{\epsilon})$ and that $lg^{k}(n)$ isn't $\omega(n^{\epsilon})$ and $lg^{k}(n)$ isn't $\Omega(n^{\epsilon})$. Since $lg^{k}(n)$ isn't $\Omega(n^{\epsilon})$ it cannot hold that $\lg^{k}(n)$ is $\Theta(n^{\epsilon})$.

Daniel Fischer

@evinda Yes.

evinda

@DanielFischer Thank you :)

evinda

2:06 PM

@DanielFischer Is the use of the limit the only way to determine if $\sqrt{n}$ is O/ o / Ω / ω, Θ of $n^{\sin n}$ ?

Daniel Fischer

@evinda There's always more than one way. Well, almost always.

evinda

@DanielFischer What is the best way to determine if $\sqrt{n}$ is O/ o / Ω / ω, Θ of $n^{\sin n}$ ?

Daniel Fischer

@evinda Immediately seeing that it isn't and then showing that. For example by noting that among every seven consecutive integers, there is at least one with $\sin n > \sin 1$ and at least one with $\sin n < -\sin 1$.

Ted Shifrin

hi, @DanielF

hi @evinda

evinda

Hello @TedShifrin :)

Ted Shifrin

2:14 PM

Sadly, @DanielF, I caused a poor OP nightmares ...

evinda

@DanielFischer Do we see this, using the graph? Is there a way, without the use of the graph?

Daniel Fischer

@TedShifrin Oh. Can you make up for it by offering him a dinner invitation should he ever come near your home?

Ted Shifrin

LOL, that might be construed as improper :D

Daniel Fischer

@evinda You must know something about the behaviour of $\sin$ in order to be able to answer the question. Where you get that knowledge from, whether from looking at the graph, manipulations of power series or whatever, is unimportant.

evinda

@DanielFischer So can we also use the fact that $ \sin x=\sum_{n=0}^{\infty} (-1)^n \frac{x^{2n+1}}{(2n+1)!}$ ?

Ted Shifrin

2:22 PM

Seems unhelpful to me :D

Daniel Fischer

@evinda Yes, but getting the needed results from that requires tedious computations or some clever tricks.

evinda

So would it be easier to use $\limsup \sin n$ and $\liminf \sin n$ ? @DanielFischer

Daniel Fischer

@evinda That depends. Rigorously proving that $\liminf \sin n = -1$ and $\limsup \sin n = +1$ is also not entirely trivial. If you can just use these facts, fine. If not, it's easier to prove something weaker that also suffices.

evinda

@DanielFischer I don't think that the proof is needed, since it is from the course Algorithms and Complexity. But I am not sure.. What's the easiest way to prove something weaker?

Daniel Fischer

@evinda Since the circumference of the unit circle is $2\pi < 7$, among every seven successive integers, there is at least one with $\lvert n - (2k+1/2)\pi\rvert < 1/2$ for some $k\in \mathbb{Z}$, and ditto one with $\lvert n - (2k-1/2)\pi\rvert < 1/2$ for some $k\in\mathbb{Z}$.

evinda

2:37 PM

@DanielFischer Could you explain me an other way? :/

Ted Shifrin

That's the best way, @evinda. Pigeonhole principle.

evinda

@TedShifrin It is from the course Algorithms and Complexity from the CS department, that's what I thought that an other way would be better.. :/ @TedShifrin

@DanielFischer What operations could we make with the Taylor series?

Daniel Fischer

@evinda Don't go that route. That was just mentioned as an example illustrating that various ways are possible, some not so well-suited as others.

Ted Shifrin

Bis später :)

evinda

@TedShifrin Bis dann :)

A Beautiful Mind

2:47 PM

Hi.

evinda

@DanielFischer Ok.. How can we then use $\limsup \sin n$ and $\liminf \sin n$ ?

A Beautiful Mind

@ted Are you a good cook?

ʙᴀᴅ ᴀᴛ ᴍᴀᴛʜ

Hi @JasperLoy

A Beautiful Mind

3:02 PM

@ʙᴀᴅᴀᴛᴍᴀᴛʜ Hey Bart. I am now in the depths of darkness. I hope there is a light at the end of the tunnel.

ʙᴀᴅ ᴀᴛ ᴍᴀᴛʜ

@JasperLoy I hope so too

evinda

@DanielFischer Could we also do the following?
We suppose that $\sqrt{n}=o(n^{\sin n})$.
This means that $ \forall c>0, \exists n_0 \in \mathbb{N}$ such that $\sqrt{n}< c n^{\sin n}, \forall n \geq n_0$.

For $ n=2 \pi $:
$ \sqrt{n}< c n^{\sin n} \Rightarrow \sqrt{2 \pi}< c (2 \pi)^{\sin{2 \pi}} \Rightarrow \sqrt{2 \pi}<c (2\pi)^0=c$

So $c> \sqrt{2 \pi} , \forall c>0 $, contradiction.
So we conclude that $ \sqrt{n} \neq o(n^{\sin n}) $, so also that $ \sqrt{n} \neq O(n^{\sin n})$.