@LeakyNun

yes

what are we trying to prove?

@LeakyNun $f(x)=e^{ax}$

can you prove it now?

No

just let $a = \ln(f(1))$

@LeakyNun why?

$f(1)$ is arbitrary

$a$ is also arbitrary

How to proceed from f(x)=f^x(1) without knowing $f(x)=e^{ax}$ ?

f(1) is arbitrary

so you just let it be $b$

Hmm then?

then can you rewrite the equation you just gave me?

$f(x)=b^x$

can you see it now?

Aha. $b=e^a$ taking arbitrarily

yes

@LeakyNun it become e^a when its derivative condition is given right?

I don't understand what you are asking.

The arbitrarily thing becomes cumpolsary if given condition is f'(x)=f(x)

it has nothing to do with it.

$e^a$ is a constant.

$b^x = (e^a)^x = e^{ax}$

@LeakyNun $(A+B)^T=A^T+B^T$ ?

yes

If $\csc \theta - \cot \theta =2017$ then $\theta$ lies in which interval?

I got till $cot\theta>0$

This give $\theta$ is either in 1 or 3rd quadrant

why $\cot\theta>0$?

First I got $\cot\frac{\theta}{2}=2017$

why?

Then using double angle formula we can get $\cot \theta$

why $\cot\dfrac\theta2=2017$?

$\csc\theta -\cot\theta=\cot\dfrac\theta 2 $ wrong?

can you prove it?

@LeakyNun yes.

could you prove it?

$\dfrac{1}{\sin\theta}-\dfrac{\cos\theta}{\sin\theta}=\dfrac{1-\cos\theta}{sin\t‌​heta}=\dfrac{2\cos^2 \frac\theta 2}{2\sin\frac\theta 2 \cos\frac\theta 2}=\cot\frac\theta 2$

you need spaces

Is it clear now?

Don't worry of that one $\theta$

@LeakyNun

Can you prove $1-\cos\theta = 2\cos^2\dfrac\theta2$?

@LeakyNun yes.Should I prove here?

@BalarkaSen I have a setup where I know the $\Bbb Z_2$-cohomology has one generator in degree 3, which squares to the single generator in degree 6. I also know that $H^3(M;\Bbb Z)\cong H^6(M;\Bbb Z)\cong \Bbb Z_2$, and the copy of $\Bbb Z_2$ comes from an isomorphism with the $\Bbb Z_2$ cohomology groups. Can I somehow try to argue that the generator of $H^3(M;\Bbb Z)$ squares to the one in degree 6?

@Fawad yes, you should.

$\cos(a+b)=\cos a\cos b-\sin a \sin b$

Putting $a=b=\theta /2$

$\cos\theta=\cos^2 \frac{\theta}{2} -\sin^2\frac{\theta}{2}$

Since $\sin^2 a +\cos^2 a=1$

$\cos\theta=1-2\sin^2\theta$

Oops so @LeakyNun $1-cos\theta=2\sin^2\theta$

hi there, is there a 'positive' definition of divergence?

Hmm so expression is $\tan \theta/2$ right? @LeakyNun

@mrnovice just negate the epsilon-delta definition

@Fawad why?

So $\forall epsilon >0, \exists N\in\mathh{N}:n\geq N \implies |a_n-a|<\epsilon$

is the def of convergence

$\exists \epsilon:\forall N\in\mathbb{N}, n\geq N\implies |a_n-a|\geq \epsilon$

Is that the correct negation @LeakyNun?

I thought you were talking about real functions

turns out you're talking about sequences

the first equation is missing a quantifier, so so is the second equation.

For a sequence of real numbers $a_n\to a \iff$, then my first line

is that what you mean?

no

you're missing a for-all quantifier

for which part

Hi

@mrnovice it should be $\forall \varepsilon >0, \exists N\in\mathbb{N}: \forall n\geq N: |a_n-a|<\varepsilon$, or:

@LeakyNun [i.sstatic.net/BT5Vr.jpg](like this)

$\forall \varepsilon >0, \exists N\in\mathbb{N}: \forall n: n \geq N \implies |a_n-a|< \varepsilon$ @mrnovice

@Fawad ok continue

@Danu Yeah, this should just be naturality of the cup product under the coefficient change map $H^*(-, \Bbb Z) \to H^*(-, \Bbb Z/2)$.

@LeakyNun so $\tan\theta<0$ which means $\theta$ is either in 2nd or 4th quadrant

when did we say anything about $\tan\theta$?

Hi, @MichaelAlbanese, @PaulPlummer.

Hi @BalarkaSen

Haven't seen you around for a while.

On the site or in chat?

@LeakyNun $\tan\theta=\dfrac{2\tan\frac\theta 2}{1-\tan^2\theta}$

On chat. I don't follow the main site anymore, but I have noticed some of your answers.

@Fawad alright, continue

can you say anything about the sign of $\csc\theta$?

Yeah, I don't chat often.

I go through phases.

What do you mean that you don't follow the main site?

@BalarkaSen Yea?

I don't keep up with what goes around there except for some occasional serving through questions/answers.

I wanted to say so too

Let me think :p

you are not answering my question

Don't know about its sign

@Danu If I am not super-dumb, this should be super-obvious once you think in terms of cochains.

@Fawad use the original equation they gave you

I'm worried because the map goes the other way @BalarkaSen. I don't know the cup product in $\Bbb Z$.

I konw it in $\Bbb Z_2$ instead

If I knew it in $\Bbb Z$, for sure

Which map? It's induced from $\Bbb Z \to \Bbb Z_2$, that's the right direction.

Oh, I see what you're worried about. But it's an isomorphism in groups.

So you can just invert it.

Yeah?

I was hoping so too

but I'm confused :D

Ok. So csc is $2017+\frac{2(2017)}{1-{2017}^2}$

Well, you know it's an isomorphism in groups, yup?

Yea, but why is the cup product natural under that kind of thing

This isn't reduction of coefficients anymore

Work it out cochain level.

This is easier than the argument that cup product is natural under maps $H^*(X) \to H^*(Y)$ induced from $X \to Y$

@LeakyNun what do you say it's sign will be? I don't think I have to calculate 2017 square in exam

You don't need to

You are given $\csc\theta - \cot\theta = 2017$

and you know that $\tan\theta < 0$

Yes

What can you say about $\csc\theta$ then?

+ve

So it is 2nd quadrant

yes

@Danu The point is simply that multiplication in $\Bbb Z$ is natural with multiplication in $\Bbb Z/2$ under the quotient map $\Bbb Z \to \Bbb Z/2$, precisely because it's a ring homomorphism.

Anybody do Ruby?

@BalarkaSen: Do you know much about homology spheres?

@MichaelAlbanese I know a tiny bit, but not much.

You might know the answer to my most recent question then: math.stackexchange.com/q/2275637/39599

I was assuming Mike Miller was going to answer it within minutes (which seems to be the case for my questions these days).

Haha.

Let me look.

@MichaelA Oh, cute question. So I know the only homology 3-sphere with finite fundamental group is Poincare, so dim = 3 is not the best place to look. Hmm.

Ah, you mentioned that below.

@BalarkaSen Okay, I see. I'm really overstressed right now and so I can't really get any thinking done. The point is that, because it is a ring homomorphism, the evaluation of a representing $\Bbb Z$-cochain on the front face of a chain, multiplied by the evaluation on the back chain, must reduce mod 2 to the same thing as what I get when I first mod-2 then multiply.

Do you think a good student already knows the class material in advance?

No.

@JingWang: No.

@Danu Right, exactly.

I'm thinking. I wonder if I truly ""know"" a lot of examples of homology spheres beyond dimension 3.

@Danu say you're beggining undergrad and already know the material for the advance upper courses, that would not make you a good student?

@BalarkaSen Not in dimension 4, since there the intersection form controls homeomorphism type. :-)

@JingWeng Not necessarily, no. It does make the chances of you being a good student larger (by quite a lot).

True, but in this case I want odd dimensions. 5 is a good place to try.

@Danu How do you motivate yourself to study advance material from a class?

Doing personal mathematics doesn't seem to have the same motivation as doing assigned mathematics.

@MichaelAlbanese I think this can be turned into a group theoretic question. By Kervaire, any group $G$ which has finite presentation, $H_1 = H_2 = 0$ appears as $\pi_1$ of a homology 5-sphere.

If you can produce such a group such that $H^6(G; \Bbb Z/p) \neq 0, \Bbb Z/p$ for some prime $p$, then by this it cannot act freely on $S^5$.

I'm just patching up what I know. I am not sure if this is a useful approach.

I think this should be doable, though. I am pretty sure I can produce a group like that.

That sounds interesting.

It's topologically a little uninspiring though.

@MikeMiller might know an easier way to get an example.

It looks like the awesome @ShaVuklia is not around today. Who I will greet today?

@Waiting you're just in the wrong room

@Danu Oh, I see!

@SimplyBeautifulArt how is it going?

I didn't see @robjohn either (for some long time).

@Secret You're around too! What are you working on now?

no plan for maths tonight. I am currently procrastinating by playing computer games.

@Secret No computer game here can beat the pleasure of doing mathematical research.

@Waiting Do you have trouble do school mathematics?

@JingWeng I'm not at school anymore.

@Secret (just for curiosity) how would you translate "kui5 bei2 tiu4 zai2 sik9 zyu6" to English?

@Waiting was that ever a problem?

@LeakyNun Sounds like "some female is being hooked up by a guy".

@Secret not "hooked up"

@JingWeng I've always managed to be successful about anything when learning along. Learning in school was never for me, it never compared with learning things alone. I excel when I learn alone in my own style.

I mean, e.g. the girl does everything the boy asks, in a relationship @Secret

ah yes

@Waiting So learning in school was sometimes demotivating?

@Waiting I'm coding

But the code isn't fully working yet

@BalarkaSen Cool thanks

@Secret so how would you translate it?

@Akiva Indeed, now I'll remember :p

@JingWeng Yeah, you can say that. I mean often in school people make comparisons, like who is the best in class and so on. I mean in school you don't learn how to beat yourself and excel, and focus only on that, on trying to be better than yourself every day, sometimes I felt it was about beating the one next to you. Then it's about beauty and miracles of science you can discover alone, and say wow ... repeatedly.

Hi chat

Hi @Astyx

@SimplyBeautifulArt I see, nice.

:-)

@LeakyNun If it is not hook up, then it has to be "the guy is in a relationship with the girl". I am not sure if there's a better way to translate it

@Secret it's not as simple as the guy is in a relationship, but that the relationship is being dominated by the guy

i.e. sik9 zyu6

@Waiting Do you know how to get yourself to study the school math?

Oh hi @Waiting sorry, yea I was indeed in the other room :P

how are you?

@ShaVuklia Not that bad. It seemed weird I didn't see you around. :P How is it going?

Does anyone here remember what $\operatorname{Sq}^i(a^j)$ is?

@BalarkaSen ?

@JingWeng I don't know if I can give you the best piece of advice, but I can tell you that in general one needs very hard work and a deep love, passion for doing math, daily. Don't get discouraged, never. Be highly optimitics even when confronting the hard-to-understand concepts.

Steenrod square maybe but I don't know it, @MichaelAlbanese

haha, yea I'm always on the chat XD well, I'm doing great actually. I just finished oscillations, and now I can leave that for a while. I'm going to be doing electromagnetism for the next days, and I'll make a start with quantum! I've kind of abandoned maths temporarily, just like I didn't do physics for 5 months straight :P anyhow, I don't even know what to say actually @Waiting

Sorry, I know it is the Steenrod square, but it should reduce to $ka^{i+j}$ for some constant $k$.

I can't remember what $k$ is.

Some binomial coefficient.

Ah, I'm afraid I can't help with that.

I never learnt these

It's probably $\binom{j}{i}$.

@ShaVuklia Well, after so much work, maybe you should also need to take a break, to watch a good movie. You deserve that. :P

honestly, doing this feels like a break for me :P when I have do be social and do sports and talk to people, I feel like I'm working hard. this is very relaxing actually @Waiting

I'm also thinking to watch a movie, not decided yet about the right one.

hm, too bad I can't advise you on that:(

@ShaVuklia No worry! I also love to do sport, I actually enjoy pretty much going jogging in the evening. It's very relaxing.

I've heard they were making a second Blade runner movie

Cool

yea, that's really good:) @Waiting

Works for $i = 1, 2$ at least.

I hope they don't mess it up

I should watch Blade Runner 1

@Astyx Release date is october afaik

You definitely should @Balarka

Also read the book (Do androids dream of electric sheep?)

@ShaVuklia Still at uni now? :P It's Friday anyway ...

lol I didn't even go to the uni. i literally did not even attend a single class this week, if I remember correctly. or maybe just one

but today I worked at home

usually i do go the uni :P

@ShaVuklia Oh, then I misunderstood you when you told me this week (not sure though) you are at uni.

no you understood me well! i go to the uni to work there, but I don't attend class:P sorry that was a bit confusing

@ShaVuklia I also love learning at home. :P

@Astyx there's a trailer already

@ShaVuklia Ah, I see. Gotcha! :P

@SteamyRoot Ah Philip K Dick. Thanks, noted.

Yeah, it looks promising @Alessandro

I have pretty much literally whatever by Philip K Dick and J G Ballard on my reading list

And I trust that Scott, as executive producer will keep the director (who is a good one by the way) from doing weird stuff

Hell if I'd ever get my ass off and read anything

Also the cast looks good

@Waiting :)

I'm always a bit scared of "late" sequels... it'll probably be a good movie in its own right, but whether it's a good sequel to the original...

@ShaVuklia I'll return a bit later. I wanna finish an article. :P Take care!

haha I was about to say the same!

I'm going to buy some food

good luck with the article!

@ShaVuklia haha :D

@ShaVuklia Thanks! ;)

For those playing at home, $\operatorname{Sq}^i(a^j) = \binom{j}{i}a^{i+j}$.

@Waiting do you know some technqiues to get over discouragement ?

@JingWeng discouragement? Of what sort?

@JingWeng It's about the mindset. Look, there is a simple question: ff thrown in the middle of the jungle, what would you do? Some would answer they wouldn't survive. Some will answer they will do their best to survive. What would you answer?

I tell you what I would like to answer: to become the king of the jungle. You need a powerful mindset to succeed.

haha :D

Life is very tough, you have to be tougher, stronger than anything you face.

@Waiting Would you be disappointed if you fail?

To become king of the jungle.

@AkivaWeinberger Never, because I did all my best to get where I wanna get. Maximum pleasure doing and believing in my aims.

I would never give up.

That's a good mindset, I think. To have higher aims than expectations.

Yeap.

I have to go now.

Hello @BalarkaSen

If you have a linear operator $A$, does the value $\min_{||x||=1}||Ax||$ have a name ?

(in finite dimension)

@Astyx minimum magnitude of range of unit sphere?

Hai

Yeah I guess, but I mean, like with $|||A|||=\max_{||x||=1}||Ax||$ being the matrix triple norm of $||A||$, I feel like you'd find some use for nonsingular matrices in taking that value into account (for instance for continuity arguments)

Hi @Daminark

How are you ?

Everything's alright, how about you?

What percentage of 1 meter is 2.54cm? How do I quickly find that out without much calculation.

I'm good, enjoying my three days holidays :)

Glad to see some new faces in this room

$2,54/100 *100$

And I don't actually know of this, I'm not sure I've ever seen the minimum used before

=2.54%