Mathematics - 2014-06-12 (page 3 of 3)

Ilan Aizelman WS

17:01

anyone?

Studentmath

Yeah, it's the polar form @Ilan

Ilan Aizelman WS

@Studentmath How do you get to it?

Studentmath

Hrm, there are formulas in...

I think it's unit 5.

Ilan Aizelman WS

Yup, I'll go take a look.

Unit 6 btw.

Studentmath

Don't recall them by heart sadly, though I should.. Yes, unit 6.

Ilan Aizelman WS

17:09

I just forgot the formula, and the explanations in the book suck.

Studentmath

The thin unit.

It's a terrible book.

I think Algebra I and Algebra II have the worst books in the Open University.

The only course with worse format representation is Java. But that doesn't really have any books (well it has one in English which you are not allowed to use really..)

c c

@Studentmath how much did you find for P(Y≥5| X is even)? I've something ugly with lots of fraction, and very unsure of my method (11/12*20/49*(71/20)^5*20/29

Studentmath

Erm, let me open it up

Abouts 0.287

It wasn't something complete. I did it this way:

c c

ow, my result must even be >1 :))

Studentmath

If I define $Y_i$ to be the Y of the ith coin, we get..

Maybe you are forgetting the 1/6 for every single cross?

I mean

You get

c c

17:16

@Studentmath indeed

Studentmath

Was that?

I forgot it at first too :P

Then I got over 1, and realised I am doing something too wrong..

Anyhow you get:

$$\frac{P(Y\ge 5 \cap X=2)+P(Y\ge 5 \cap X=4)+P(Y\ge 5 \cap X=6)}{P(X=2)+P(X=4)+P(X=6)}$$

If I did it right..

So:

$$2*\frac{1}{6}*[P(Y_2 \ge 5)+P(Y_4 \ge 5)+ P(Y_6 \ge 5)]$$

Then you can solve it generally for $Y_i$, and you get the nice formula $(\frac{1}{i}-1)^4$ for each $P(Y_i \ge 5)$, and then you sum it up for 2, 4, 6 and get the 1/3. I think I did it right at least, getting the same results on your side?

c c

my method was surely wrong, let me check yours

Hippalectryon

17:34

@TedShifrin math.stackexchange.com/questions/831124/… any idea ?

Chris's sis

@robjohn just wow. I created a lot of marvellous stuff ... Ramanujan himself would love me :-)

c c

@Studentmath well I don't see where your nice formula comes from :), all I have is $P(Y_i=1)=\frac{1}{6*i}$, $P(Y_i=2)=(1-\sum P(Y_i=1))\frac{1}{6*i}=\frac{71}{120}\frac{1}{6*i}$, $P(Y_i=n)=(\frac{71}{120})^{n-1}\frac{1}{6*i}$ then doing $P(Y_i\ge 5)$ is a geometric sum

Studentmath

Also possible. I just adeed up (taking the 1/6 out) to:

$P(Y_i \ge 5)=1-P(Y_i \le 4)$

And Added up $P(Y_i=1)+...+P(Y_i=4)$ and 1 minus that result gave me that nice formula.

The 1/6 are all out, of course.

Eitherway it should get to 0.287, the probability. Abouts.

c c

17:49

of course the complementary, I wanted to do it also

Hippalectryon

@TedShifrin @G.T.R math.stackexchange.com/questions/832016/… silly thing uh

c c

@Studentmath but still are you sure of your $P(Y_i=2)$, $P(Y_i=3)$?

how do you calculate them?

ok

Studentmath

Sorry

meant

Geometric

As it is Geometric distribution after all, the number of tosses until success.

To be horrid, you get:

$$1-[i^{-1}+(1-i^{-1})*i^{-1}+(1-i^{-1})^2*i^{-1}+(1-i^{-1})^3*i^{-1}]$$

Which is $(i^{-1}-1)^4$

I hope.

G.T.R

@Hippalectryon classic olympiad problem

Hippalectryon

@G.T.R Any hint ?

@G.T.R math.stackexchange.com/a/832030/150347 sérieux, je suis pas super intelligent mais quand même faut pas pousser ....

Pranav Arora

18:36

Hello @Chris'ssis!

Wolfram Alpha is giving strange results for the identity you showed me today. This one: $$Li_2(e^x)+Li_2(e^{-x})=\frac{\pi^2}{3}-\frac{x^2}{2}$$

For $x=1$, W|A gives $$-\frac{\pi^2}{6}-\frac{1}{2}(1+i\pi)^2$$ but as per your identity, the answer should be $$\frac{\pi^2}{3}-\frac{1}{2}$$

confused

c c

@Studentmath thanks for that training, I find like you $P(Y_i=k)= P(Y_i\ne k-1)*\frac{1}{i} = \frac{1}{6}(\frac{i-1}{i})^{k-1}\frac{1}{i}$ and $P(Y_i\ge 5)=\frac{1}{6}(\frac{i-1}{i})^{4}$

r9m

@PranavArora do you mind me asking where are you from .. ? :-)

Pranav Arora

India :)

r9m

I can see that ... which state ? :)

Pranav Arora

Rajasthan!

Oh, so you are from Chennai, nice to meet you. :)

r9m

18:47

@PranavArora nice to meet you too :)

c c

you have around 100/1e9 chances to know each other

Pranav Arora

Judging by your age, you are in some college at the moment, right?

r9m

@PranavArora ya ... right :)

Pranav Arora

Do you mind telling me which one? :)

Hippalectryon

Is there a limit to the number of questions that can be asked per day ?

Pranav Arora

18:49

Wow!

That's great!

r9m

are you in college too ?

Pranav Arora

No, I passed out of high school last year and will take admission in some college this year. :)

r9m

ah .. I see :D ..

Pranav Arora

I wish I could study at CMI or ISI :(

r9m

why frown ?

Pranav Arora

18:52

I tried for ISI this year but didn't get selected, I am probably going to BITS-Pilani this year but I wanted to do a degree in Math from CMI or ISI

r9m

@PranavArora well .. I didn't get selected in ISI the first year and I didn't sit for cmi exam (wasnt aware that there was something called the cmi :P) .. so I dropped a the year and tried again next year .. :)

Pranav Arora

I didn't even know about CMI or ISI in my first attempt and this year, I forgot about the CMI exam. :P

r9m

ah .. Ic ic .. :|

Pranav Arora

What do you learn at CMI? I am really curious about it. :)

or what are you currently studying there? :)

r9m

@PranavArora well its a math-computer sc course (BSc) course

I just finished the 2nd year

Pranav Arora

19:02

I guess you got admission in an early age, it really makes me feel stupid for taking so much time to get into a college :P

r9m

@PranavArora when you put it like that I feel stupid ... :| I follow you on I&S .. you write nice solutions :) :D ..

Studentmath

@cc thanks for double checking! It was a real nice one.. Any tips for studying before the test?

Pranav Arora

@r9m: Integrals and Series? Never saw you there and thanks. :)

r9m

@PranavArora na .. I don't participate there (I just read the posts :) .. I like reading the site a lot :-) )

Chris's sis

@PranavArora maybe you'd like to expand $$-\frac{\pi^2}{6}-\frac{1}{2}(1+i\pi)^2$$

Pranav Arora

19:09

@r9m: Ah ok and you too post great solutions on MSE, I like reading them. :)

@Chris'ssis: But the imaginary part still remains.

Chris's sis

@PranavArora and?

Pranav Arora

Erm...it doesn't match the answer which the identity gives.

The identity gives: $$\frac{\pi^2}{3}-\frac{1}{2}$$

r9m

@PranavArora lol :P .. I don't believe I have written many nice solutions ..

Chris's sis

@PranavArora Figure out what that does mean.

Pranav Arora

@Chris'ssis: I don't think I will be able to figure out a reason, dilogarithms is completely a new thing to me. Can you please give a hint? :)

user44840

19:14

hey everyone,

anybody from the UK here?

r9m

although I am fond of a few solutions due to me :P laughing stock

Jasper Loy

@robjohn I am now the black square.

Chris's sis

@PranavArora There you're interested in the real part only.

@r9m what solutions? :-)

Pranav Arora

@Chris'ssis: Yep, looks like it, thank you! :)

robjohn

@JasperLoy No more blue square?

Chris's sis

19:18

@robjohn it's coming ...

Jasper Loy

@robjohn Not for now, depends on my mood.

Pranav Arora

BTW, by any chance, does the other method involves the use of differentiation under the integral symbol? Just guessing...

robjohn

@Chris'ssis The alternating sum?

@Chris'ssis I started on the alternating sum, but I am currently proctoring an exam at UCLA. I won't get back to it until this evening.

Chris's sis

$$\sum_{n=1}^{\infty} \left(\zeta(2) - 1 - \frac{1}{2^2}-\cdots - \frac{1}{n^2} \right)^2= 3\zeta(3)-\zeta(2)^2$$

robjohn

@Chris'ssis Oh, that one. I should look into that. It is easier.

Chris's sis

19:20

@robjohn newly created

@robjohn I want to also attend the cubic variant.

robjohn

@Chris'ssis You mean $\zeta(3)-\dots$?

Chris's sis

@robjohn I mean what I typed.

robjohn

@Chris'ssis Oh, I just saw the square. That should be similar to the previous one we were working on

Chris's sis

@robjohn Yeah, to a certain extent. :D

robjohn

@Chris'ssis the cubic would be harder

Chris's sis

19:22

@robjohn But much nicer ...

robjohn

@Chris'ssis how so?

G.T.R

@r9m you may find this interesting math.stackexchange.com/questions/832071/f21f2-leq-1-implies-f-0

Chris's sis

@robjohn harder=nicer :-)

r9m

@Chris'ssis hmm .. raise that to higher powers ! :D

Chris's sis

@r9m I do that! :D

r9m

19:24

@Chris'ssis a few inequality solutions ... although they are nothing special :-)

Chris's sis

@robjohn Moreover, I'm also interested in (this should be awesome) $$\sum_{n=1}^{\infty} (-1)^{n+1} \left(\zeta(2) - 1 - \frac{1}{2^2}-\cdots - \frac{1}{n^2} \right)^2$$

@robjohn I'm going to develop many such series, really many ...

The trivial variant gives us $$\sum_{n=1}^{\infty} (-1)^{n+1} \left(\zeta(2) - 1 - \frac{1}{2^2}-\cdots - \frac{1}{n^2} \right)=\frac{\zeta(2)}{4}$$

N3buchadnezzar

Soccer

brazil

Chris's sis

@N3buchadnezzar It happens now?

N3buchadnezzar

y

G.T.R

@N3buchadnezzar when will the game start ?

N3buchadnezzar

19:38

20 minutes

r9m

23 mins from now :)

G.T.R

@DanielFischer interested in the world cup ?

Balarka Sen

Goodie. We also have Schanuel for $\Bbb {\bar Q} [[X]]$

Daniel Fischer

@G.T.R Only if England wins. Football is a club sport.

Chris's sis

Let me develop a new series ...

G.T.R

19:45

@Daniel England ? I don't get it :P

Chris's sis

$$\sum_{n=1}^{\infty} (-1)^{n+1} \left(\zeta(2)\zeta(3) - \left(1 + \frac{1}{2^2}+\cdots + \frac{1}{n^2}\right)\left(1 + \frac{1}{2^3}+\cdots + \frac{1}{n^3}\right) \right)$$

Daniel Fischer

@G.T.R What's the problem? I like England. (I also like France, but in Football, England is my favourite.)

G.T.R

@Daniel no problem with England (except for the accent) but what do you mean by club sport ?

Balarka Sen

@Chris'ssis Evaluate $$\sum_{n=1}^\infty \frac1{(3n+1)(3n+2)(3n+3)}$$

Daniel Fischer

@G.T.R Manchester United, Girondins Bordeaux, Werder Bremen: clubs.

Hippalectryon

19:53

@robjohn :D

@robjohn epic new avatar

N3buchadnezzar

it's started

Chris's sis

@BalarkaSen I did it a long time ago ...

Balarka Sen

@Chris'ssis I want to know the closed form.

G.T.R

@Daniel I see

Chris's sis

@BalarkaSen it's $$\frac{1}{2}\left(\frac{\log(3)}{2}+\frac{\pi \sqrt{3}}{6}-\frac{1}{3}\right)$$ There is a famous formula for computing these forms.

Balarka Sen

19:57

@Chris'ssis That's pretty odd. Have you made sure numerically? Please do.

Chris's sis

@BalarkaSen Yeah.

Balarka Sen

Thanks. And that's very very odd. You just don't know how much I am amused by that result.

Chris's sis

@BalarkaSen Why? :-)

Balarka Sen

@Chris'ssis Here

Chris's sis

@BalarkaSen I don't get ....

Balarka Sen

19:59

@Chris'ssis ?

Chris's sis

@BalarkaSen did you enter that link?

Balarka Sen

OK, I linked you to the wrong thing.

Chris's sis

:D

Balarka Sen

Here and convert to PDF through here

Hippalectryon

Lol that's part of reddit's math census

Balarka Sen

20:02

They nearly proves the folklore that $\log(3)$ and $\pi$ is algebraically independent then, @Chris'ssis, if I both trust their result and yours.

Hippalectryon

Maths => ^

Parth Kohli

@Hippalectryon Which sub was this posted in?

Hippalectryon

@ParthKohli reddit.com/r/math/comments/1tlz4i/…

Chris's sis

@BalarkaSen Interesting ...

Balarka Sen

@Chris'ssis It is.

David Kirby

20:03

@Balarka Sen, I got $\frac{1}{12} \left(\pi \sqrt{3}-2-\log (27)\right)$ for the above series...

Parth Kohli

@Hippalectryon Wow, I frequent /r/math and it's usually better than this...

Balarka Sen

@DavidKirby Doesn't matter to me. Having log(3) and pi in the expression is all I want

N3buchadnezzar

croatia 1 : brazil 0

Chris's sis

@robjohn I think the greatest achievement is not to compute these series, but to wisely apply them to different difficult integrals and series. This is the real marvellous thing in my opinion. They must be applied! :-)

Hippalectryon

@N3buchadnezzar Would be so funny is croatia won

N3buchadnezzar

20:14

y

G.T.R

@Hippalectryon some gamblers would be mad

Hippalectryon

@G.T.R especially their winner predicting turtle xD

N3buchadnezzar

1:150 ratio or something

I could have betted a dollar and won a bazillion

Cortizol

@DavideGiraudo Because we don't have PM here, I will ask you here (I hope you will see this): do you know some "good" book for solving differential equations over distributions? I mean, I am interested in examples in general, because I can't find much of them on internet (my English is bad, I know)

Hippalectryon

@N3buchadnezzar a brazillion :D

N3buchadnezzar

20:18

^^

That can't be much since they spent all their money preparing for the tournament

Hippalectryon

@G.T.R il y a une limite au nombre de question posées / jour ?

G.T.R

@Hippalectryon yes

Hippalectryon

@G.T.R combien ?

G.T.R

@Hippalectryon I can't find it. @robjohn I believe there is a limited daily question quota ?

Alex

anyone know of any text that explains why even though there is symplectic structure for the cotangent bundle, the tangent bundle does not have a symplectic structure?

Daniel Fischer

20:30

@Hippalectryon six questions par jour (24h)

Hippalectryon

@DanielFischer ok merci

Daniel Fischer

de rien

G.T.R

@N3buchadnezzar done

N3buchadnezzar

yeah

Hippalectryon

Why are there so few people on chem SE :c

N3buchadnezzar

20:38

@Hippalectryon Maybe questions there get a bad reaction...

2

Hippalectryon

@N3buchadnezzar good one xD

G.T.R

CHIMIE KHRAAAAAAAAAAAAAAAS

Hippalectryon

@G.T.R ?

N3buchadnezzar

2

Hippalectryon

haha

google.fr/… :O google has improved

robjohn

20:52

93

Q: The Complete Rate-Limiting Guide

I noticed that I can only perform certain actions such as commenting a finite number of times in a given period of time. Obviously, rate limiting is in place to prevent accidental misuse or intentional abuse of certain features. Where else is rate limiting applied on Stack Exchange sites, and wh...

Hippalectryon

'maximum of 50 questions per 30 days' awwww :c

2

G.T.R

Damn I only have 2 questions left

Balarka Sen

@Hippa What's with the naem erauqs?

Hippalectryon

@BalarkaSen ? when have i mentioned the sq mean ?

Balarka Sen

Robjohn is the mean square.

Hippalectryon

20:58

ooh

I just felt like doing it :)

N3buchadnezzar

|-|

Daniel Fischer

@G.T.R Unless you have deleted a lot, you have an easy 20+ questions left.

Balarka Sen

@robjohn I guess we ban people who impersonates a mod, right?

smirks

Hippalectryon

@BalarkaSen @rob is a mod ?

Balarka Sen

Indeed, he is.

Hippalectryon

21:01

ooh

Then i'm a Dom ? :D

G.T.R

@DanielFischer Damn I read 30 instead of 50. Some sleep needed

Balarka Sen

Haha, you are, but you are going to get yourself end up being banned if you impersonate him, @Hippa

Chris's sis

Someone should save me ... in my house there is a flood of series :D

Hippalectryon

@robjohn Even if i wanted to impersonate him i couldn't i'm not smart enough :)

@G.T.R youtube.com/watch?v=gklZVCo_MuE 14:25 quel accent magnifique :3

G.T.R

@Hippalectryon normal

Hippalectryon

21:14

@G.T.R c'est moche que le seul présentateur que j'ai vu faire une faute d'anglais à l'E3 soit français :/

@G.T.R tu connais des gens à Pasteur ?

G.T.R

oui mon père était là

Hippalectryon

@G.T.R je veux dire, des personnes en prépa là bas

G.T.R

non

Hippalectryon

Ah ok

Chris's sis

Does your Mathematica answer back to this? N[Integrate[(E^(y + z) y z PolyLog[2, E^(y + z)])/((-1 + E^y) (-1 + E^z) (-1 + E^(y + z))), {y, 0, Infinity}, {z, 0, Infinity}], 10]?

G.T.R

21:29

Not after 5 minutes

Hippalectryon

@Chris'ssis It's eating up my RAM, but that's all :P

Chris's sis

@Hippalectryon :-)

Ted Shifrin

Is RAM tasty ce soir? :D

Hippalectryon

hmm hmm RAM

Ted Shifrin

Un drôle de repas, @Hippa :)

N3buchadnezzar

21:36

1-2

Hippalectryon

@TedShifrin ça me changera de me gaver de chocolat :D

Ted Shifrin

Tu grossis un peu? :D

Hippalectryon

@TedShifrin même pas :)

@G.T.R math.stackexchange.com/questions/832268/… ça te dis qqch ?

Antonio Vargas

21:55

@Hippalectryon, if you don't know the Laplace method then I don't know if an answer using it would be helpful, but if you want I can outline one.

Hippalectryon

@AntonioVargas it's ok @mookid's link gave me the answer

Antonio Vargas

@Hippalectryon alright, cool

Hippalectryon

:)

Antonio Vargas

Are you an admirer of robjohn's? ;)

Hippalectryon

'You can answer your own answer in two days' aww

@AntonioVargas yeah he gave such a beautiful answer to my question :D

Antonio Vargas

22:02

@Hippalectryon He does have some awesome answers.

Balarka Sen

Hello professor @TedShifrin

Welcome back again, @AntonioVargas

Antonio Vargas

@BalarkaSen Hello again

Balarka Sen

@N3buchadnezzar A similar pun can be made up, now I think about it. Why are there so few question in [elliptic-curves] these days? A : The question get a bad reduction sometimes.

Ted Shifrin

22:18

Hi @BalarkaSen

Not really a pun. Not even humor. Maybe just more ellipsis than before.

Balarka Sen

sigh

Ted Shifrin

LOL

Isn't this a school day?

Balarka Sen

Yep. But there are people dying of heat.

I am serious, 3 guys died in where I live. Simply sunstroke.

Ted Shifrin

Damn. I'm sorry.

Drink plenty of water.

Hippalectryon

Anyone dead of mathstroke yet ? :)

Ted Shifrin

22:23

I died long ago, @Hippa.

Balarka Sen

@TedShifrin Yes, thanks. I think India is becoming a place of no mathematics for the deathly heat and terrible mosquito.

Hippalectryon

@TedShifrin :O you sold your soul like @BalarkaSen ? :D

Balarka Sen

Cut it out @Hippa

Ted Shifrin

@Hippa: Je n'ai pas d'âme.

Balarka Sen

Isn't it bad enough to have 6 stars on that?

Hippalectryon

22:24

@BalarkaSen wow 6 stars :O

@TedShifrin ah je me disais aussi ;)

Balarka Sen

Great, rapid french. And I'd never even know if I am being sneered at in your conversations. Further, I'd never know if you both are planning to spray Hypozen all over the New York city.

N3buchadnezzar

@BalarkaSen ?

Balarka Sen

@N3buchadnezzar ?

N3buchadnezzar

?

Balarka Sen

I am not doing this.

What's with the '?'

Ted Shifrin

22:28

Nothing in French to do with you, @Balarka. Our insults are in English.

Hippalectryon

^

Ted Shifrin

Hi @N3

Balarka Sen

@TedShifrin Fine. But how can I be sure that you aren't planning to spray hypnozen?

Ted Shifrin

I don't know what that is.

Hippalectryon

@TedShifrin Quoique .. c'est pas bête l'hypnozen en fait ;)

Balarka Sen

22:30

@TedShifrin A chemical object used to hypnotize people, discovered by a Norwegian (ping @N3buchadnezzar) mad scientist Alexander Crag.

Don't take what I say too literally.

Ted Shifrin

Don't take it at all!

Balarka Sen

Take it litterely instead.

Hippalectryon

craigalexander.net hum hum xD

Balarka Sen

@Hippalectryon haha

Ted Shifrin

Drink water and go to school, @Balarka!

Hippalectryon

22:32

@TedShifrin bon alors on en commande combien de l'hypnozen ? :D

Ted Shifrin

Pas moi @Hippa.

Hippalectryon

@TedShifrin 100$m^3$ pour toute la room ça devrait aller non ?

xD

Balarka Sen

@TedShifrin I will drink water, but no school. I'll sit at home and do math instead.

Ted Shifrin

Ça ira, oui.

Dinnertime for me... Bubye, bubbas.

Hippalectryon

@TedShifrin continuous meal ? :D

Balarka Sen

22:34

ntinuous comeal, I believe.

Ted Shifrin

Continuous?

You're silly geese.

Hippalectryon

@TedShifrin Well, i love to buy fractal meal, the price is based on the ratio perimeter/area :D

Balarka Sen

@TedShifrin Do you recall the Galois theory problem you mentioned here?

Probably months ago?

OK, @Pedro is here. Nice. Give me a bunch of (elementary, please. I don't know big names) group theory problem to work my head around, @PedroTamaroff.

Hippalectryon

Well i'm off, have a good day/evening/night :)

Pedro Tamaroff

@BalarkaSen Suppose $G$ is a group that acts on a set, and that $H$ is a subgroup that acts transitively on this set. Then $G=H\,{\rm Stab}_G\; x$ for any $x\in X$.

Do you know what a stabilizer is?

Balarka Sen

22:41

$H \rm{Stab}_G$?

What does that mean?

@PedroTamaroff Yes.

Pedro Tamaroff

@BalarkaSen $AB=\{ab:a\in A,b\in B\}$-

Balarka Sen

OK. You should have defined that, you gave me a fright.

Recite the Frattini part please.

Pedro Tamaroff

Solve the first one.

Balarka Sen

OK. I am going to think about it.

r9m

@robjohn @Chris'ssis have you seen this ?! :-) ... I tried to apply the only trick I know for these sort of inequality and failed :( ..

Pedro Tamaroff

22:53

@BalarkaSen The Frattini argument follows from the above with $X$ the Sylow $p$-subgroups of $H$ for some $H$ normal in $G$ (else $G$, the whole group cannot act on the Sylow subgroups by conjugation). Since ${\rm stab}\; P=N_G(P)$, $G=H N_G(P)$. Although innocent looking, this is a very useful result.

For example, it shows that $\Phi(G)$ is a nilpotent subgroup whenever $G$ is finite.

Mike Miller

holy crazy

Pedro Tamaroff

Parth Kohli

@PedroTamaroff It is Friday here. Stay jealous.

Pedro Tamaroff

@MikeMiller Stanley has a great proof that ${\rm Hom}(\prod \Bbb Z,\Bbb Z)\sim \bigoplus \Bbb Z$.

Where all those are countably many copies.

Mike Miller

23:08

Isomorphism is $\cong$ bro.

Pedro Tamaroff

@MikeMiller Actually $\simeq$.

I never use two dashes.

sea turtles

hipster

@PedroTamaroff what is the proof?

Pedro Tamaroff

@seaturtles One first shows that if $f\in $ the Hom vanishes in $\bigoplus \Bbb Z$ then it vanishes identically.

One then shows that $f(e_k)$ must be zero for $k$ large enough.

sea turtles

what are $e_k$?

Pedro Tamaroff

This means that $f=\sum f(e_i)\pi_i$ a finite sum, since the difference vanishes on $\bigoplus Z$.

@seaturtles Canonical base for the direct sum.

sea turtles

23:19

@PedroTamaroff but if $f$ vanishes on the sum then it obviously vanishes on the canonical base, not just for k large enough?

oh, you're no longer using the hypothesis in that sentence I guess

Pedro Tamaroff

@seaturtles Sure, sure, the point is that one observes that, then observes that for an arbitrary $f$, $f-\sum f(e_i)\pi_i$ vanished on $\bigoplus\Bbb Z$.

And the sum makes sense, i.e. it is finite.

One concludes the injection $\bigoplus \Bbb Z\to {\rm Hom}(\prod \Bbb Z,\Bbb Z)$ which sends $\sum a_ie_i$ to $\sum a_i\pi_i$ is surjective.

This gives a proof that $\prod \Bbb Z$ is not a free $\Bbb Z$ module.

sea turtles

how does vanishing on $\oplus$ imply identically?

also, could have sworn that t.b. gave this proof in chat. good times.

Pedro Tamaroff

@seaturtles It's a dirty trick. Suppose $x=(x_i)\in\prod\Bbb Z$. Write $x_i=a_i2^i+b_i 3^i=y_i+z_i$.

Then $f(x)=f(y)+f(z)$.

By the vanishing in $\bigoplus\Bbb Z$, $f(y)\in\bigcap 2^i\Bbb Z=\{0\}$.

Same for the other.

That is, we can kill the first $k$ coords to get $2^k\mid f(y)$.

sea turtles

that's nice

inb4 mr turtles

Ted Shifrin

Mssrs @pedro and @seaturts

Pedro Tamaroff

23:23

Yeah.

Heehhe hello.

@seaturtles +1

Ted Shifrin

@Pedro $\simeq$ is homotopy equiv, not iso. ;)

Pedro Tamaroff

@TedShifrin Not in my books. =D

Ted Shifrin

Messed up Spanish books ;)

Pedro Tamaroff

23:41

@TedShifrin Two lines makes it too messy.

Ted Shifrin

LOL. We're not talking about your cooking habits!

Mike Miller

Or his drug habits

Ted Shifrin

Not funny @Mike

Mike Miller

>:(

Agreed that $\simeq$ is homotopy equivalent. Upsettingly enough Hatcher uses $\approx$ as isomorphic

Ted Shifrin

I use that for approx. eq. :)

Pedro Tamaroff

23:45

@TedShifrin Cooking?

Ted Shifrin

Messy @Pedro

Pedro Tamaroff

@TedShifrin Messy = Untidy.

Ted Shifrin

Or reckless, sloppy ... :D

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$Mathematics$ Mathematics