@ethan are you sure? Wolfram alpha gives me 0.9999998

@Khromonkey I didn't check the algebra, just the first part of your answer

If you did all the algebra correct, it should be correct

@PeterTamaroff kill me now

why did you multiply it

by 12

at the begining?

@Khromonkey Why?

because im one of the cool kids

no, because he had 12 as his first term instead of 1

@Khromonkey, then it shouldn't be n+1

$(x+x^2...x^n)=x(x^n-1)/(x-1)$

you put n+1

it should be n

but that isnt what i want

i want a+ax+ax^2....+ax^n

where a is 12

@ethan

@Khromonkey oh.

anyways the question asked for sequence not series

yes, and it took me a lot of time to write that also

I can only answer about 1 in every 100 questions so its more of a big deal for me

wait, I think I did answer the correct question

you mean I should use the word progression instead of series?

Awesome.

@JacobBlack

@JonasTeuwen what? is?

I DON'T KNOW.

is a banana awesome?

why not?

yes

wait need to go

can we go back to this?

Hello all!

@JacobBlack why do you ask?

@JacobBlack Ohhhhh...sorry if intruding...

@JacobBlack okay...I apologize habitually...as you probably noticed by now...

Habit I need to break...

Does anyone know if files can be uploaded to a post?

If $f(s)$ and $g(s)$ are absolutely convergent Dirichlet series for $s>1$, and $\frac{f(s)}{g(s)$, can be bounded between two constants, does $\lim_{s \to \ 1}\frac{f(s)}{g(s)}\$ exist?

@JacobBlack Yes, I can upload images...I'm talking about a non-image file.

Never mind...it's probably pointless...

@JacobBlack Perhaps it is pointless, too late, damage done...Anyway...I might look into Google docs to see what sorts of files I can upload. No, I'm not wanting to post any doc or image in chat.

@JacobBlack My file of an answer I typed out to edit an answer to a question...while waiting in case the OP needed more of a nudge than my initial "hints"...

@MJD Haven't seen you for awhile! Hello!

@PeterTamaroff now I am :-)

@PeterTamaroff I apologize, I have been busy on meta and my pings are silent. I have tried to make them audible again.

I found a Bill Gates like guy: youtube.com/…

@robjohn No problem, rob.

I hope my first comment on your answer didn't come out as harsh.

@PeterTamaroff No, I had obviously jumped the gun and assumed what your question was

@robjohn What do you think about my work? It took a while, but I managed to do it on my own =)

@PeterTamaroff It looks good, if a bit roundabout

@robjohn What does roundabout mean?

@PeterTamaroff not as direct as might be :-)

Can anyone help with a modular arithmetic question?

@user62753 ask it :-)

@robjohn Oh, well. The thing is it the direct reflection of a "progressive" solution.

I thought about it and worked on it when the idea came to mind

@GarbageCollector hi garb!

@PeterTamaroff Sure, things are much neater when you rewrite them

@Charlie dictionary.reference.com/browse/garb

@Argon Oh!

:)

@JacobBlack it is not amazing when we consider it will often be at the top of the search results for "define [blah]"

because of all places to look for a definition, dictionary.com sounds like the go-to site?

nope

shit

@amWhy Good evening. I didn't even realize I was here until I saw your message.

@MJD Oh...that happens to me all the time...when I "wake up" my computer...

Yes, that's just what happened here.

Did you change your username a few months ago?

Quick question... is this a convex function? I think it is, but I'm not 100% sure... math.stackexchange.com/questions/309707/…

@anon Looks like we can arbitrarily intersect categories; which explains Pete L Clarke's comment.

@MJD No, I've been amWhy for a long time, well over a year plus

@peoplepower I am referring to the deleted post of Asaf, actually.

@anon Oh, no, I need rep. :)

Hi

hello, hero of time, wind, and courage

Know anything about kinetic energy?

maybe

So, if I have a rotating disk

it has kniteic enrgy right?

yes

So if disk stops rotating

does not mean that there is no kinteic energy?

an object at rest has 0 kinetic energy

thanks

@Charlie Hi charlie! :-)

doom wii

My 8yo asked how glasses help you see, so I explained it. Then I told her about how when she was three, she asked why I had to wear glasses, and when I replied "Because my eyeballs are the wrong shape," instead of saying "Oh!", for the first time in her life she laughed, gave me a skeptical look, and said "Naww!"

daww

@anon Mister.

tell me it's algebra related

HAHAHAHAH, it is!

Did today's calculus get to you?

@anon Man. You there?

yes. question?

@anon Yes.

Let $K$ be normal in $G$

$H$ be a subgroup of $G$ containg $K$.

Let $\bar G=G/K$ and $\bar H=H/K$

We then have the "natural" homomorphisms $g\mapsto gK$ of $G$ with $\bar G$ and $gK\mapsto (gK)\bar H$ of $\bar G$ with $\bar G/\bar H$

So we have the natural homomorphism $g\mapsto (gK)\bar H$

I want to find the kernel of this last homomorphism.

the kernel of $G\to G/K\to(G/K)/(H/K)$ is $H$

Yes, I know.

I want to prove it.

Wait

So $H$ is supposed normal in $G$?

@peoplepower Yes, my bad. Though else we would be talking crap.

Indeed.

The kernel is the set of $g$ that get mapped to the unit of $\bar G/\bar H$

That is $(gK)\bar H=K$; yes?

well, the preimage of e_((G/K)/(H/K)) is H even if H isn't normal

$(gK)\bar{H}=K\bar{H}$, ie $gK(H/K)=K(H/K)$

note gK=Kg since K is normal

The problem is that you simply cannot define a group $(G/K)/(H/K)$ when $H$ is not normal in $G$.

If I'm not misunderstanding, we map $g$ to the collection of cosets $(gK) \bar H$ which is actually a collection of products of cosets $(gK)(hK);h\in H$

@peoplepower I know.

So if this happens to be $K\bar H$ we must have $g\in H$.

@PeterTamaroff I was talking to anon, sorry.

And conversely.

[the 3rd isomorphism theorem holds for G-spaces, ie even if quotients are not groups :) ]

OK, if I'm not getting things wrong, two cosets $gK$ and $g'K$ are equivalent modulo $H/K$ if $gg'=h$ for some $h\in H$

And the unit is $\bar H$.

New longest answer for me: math.stackexchange.com/a/309754/23353

That one took about an hour and a half...

I think there must be some kind of tool to write all expression in terms of nand/xnor since those two are the most commonly use gates everywhere.

There probably is...

if not, we should write one in haskell.

I think those gates are common because you can express all boolean statements with just one of those gates.

Yes.

Why Haskell, just wondering...

Also, because it is very easy to construct a NAND gate. (Two transistors in series and you are done.)

@anorton Haskell is the best for parsing trees. Also, I am falling in love with it. Also, I am doing some exercises in it to build my mojo.

Hmm... I should try to learn that sometime...

@anon Could you enunciate the first isomorphism theorem?

Homomorphic image is isomorphic to domain / kernel; $\phi(G)\cong G/\ker\phi$.

I think what I just discussed is the third iso, but it is the first according to Jacobson.

@anon OK.

@anon and he seems to call that the "Fundamental Theorem of Homomorphisms".

Oh, no sorry.

He leaves FIT 1 has a corollary to the FTHom.

@anon I think he adds something which is not stated in wikipedia, for example.

He says: Let $\eta$ be an epimorphism of $G$ onto $G'$ and let $\Lambda$ be the set of subgroups of $G$ containing $\ker \eta$. Then the map $H\mapsto \eta(H)$ of $\Lambda$ gives a bijection between $\Lambda$ and the complete setof subgroups of $G'$. $H$ is normal in $G$ iff $\eta(H)$ is normal in $G'$ and in this case $gH\mapsto \eta(g) \eta(H)$ is an isomorphism of $G/H$ with $G'/\eta(H)$

@anon Thanks for the reminder. I actually did derive these in such generality at some point.

@PeterTamaroff called the lattice isomorphism theorem (note that the bijective correspondence is in fact a lattice homomorphism, but you don't need to know that)

@anon That is not the "first isomorphism theorem"? Or any "X th isomorphism theorem"?

sometimes, lattice is known as 4th or correspondence isomorphism theorem

Oh, well.

I will read it a few times.

@anon Man,.

@MarianoSuárez-Alvarez ?

two types of questions annoy me almost automatically, the «explain to the layman» and the «physical interpretation» ones... Explain to the layman what hypercohomology in the Nisnievich site is useful for? What is the physical interpretation of the fundamental theorem of arithmetic?

sigh

hello, Peter!

@MarianoSuárez-Alvarez HAHAHAHHAHAA

@MarianoSuárez-Alvarez El otro dia pase por Ciudad! Ya me inscribi =)

Bien :D

qué materias?

@MarianoSuárez-Alvarez Las materias todavia no estan listas (para inscripcion) pero inevitablemente sera algebra y analisis I

A la mañana.

ah

Creo que el 25 las suben y es todo on line

ok

@MarianoSuárez-Alvarez Pase por tu oficina para saludar pero no estabas, fue el lunes

What is hypercohomology?

@MarianoSuárez-Alvarez I think «physical interpretation of the tensor product» is arguably reasonable though; my intuition has been that we may view vector spaces as possible formal superpositions of pure states of a system, and tensor products as the vector space of the composite system.

but he knows that

Aaaaaaand switiching back to english

he wants the «abstract meaning»

it's like looking for the meaning of good seasoning

@Sanchez, the extension f cohomology to complexes. The cool new guys just say cohomology nowadays

@anon I am trying to prove that last theorem I wrote. Now, inyectivity of $H\mapsto \eta(H)$ wasn't hard. I have to prove it is surjective.

say, if X is a space and F a sheaf on X, then $H^*(X,F)$ is the chomology of F; if C is a complex of sheaves, then one sometimes wants the hypercohomology of C, $\mathbb H^*(X,C)$, which is more complicated

I don't have an intuition for these things - the wikipedia entry confuses me, as I don't understand what a quasi-isomorphism from a complex to a complex of injectives does.

@anon So I want to show that to every subgroup $X'$ of $G'$ there corresponds a n $H$ with $\ker \eta \subset H$ such that $X'=\eta(H)$.

@mariano, Is there a concrete example that I can work on? I am familiar with sheaf cohomology.

for example, take a complex C of sheaves on a space

then one can compute $H^0(X,C^p)$ for each p

and since $H^0(X,\mathord-)$ is a functor, we get a complex $H^0(X,C)$

usually, that is not very interesting

because we want to compute global sections and take into account the differential of C at the same time

that's what hypercohomology des

if you have a map of complexes $C\to D$ which induces an isomorphism in cohomology

since complexes are generally interesting only up to isomorphisms in cohomology, that is a problem

So just wanna make sure, What is the hypercohomology of a complex? Do we get a complex or just an object?

and in your example, is the sequence $\{H^0(X,C^p\}$ with the differentials exactly $H^0(X,C)$?

2nd question: yes

OK, that means we get a complex in return.

no

$H^0(X,-)$ is a functor from sheaves to abelian groups

if you apply it to a complex of sheaves, you get a complex of groups

OK, I should say a sequence of a group in this case.

That's what I mean.

$\mathbb H^0(X,-)$ is a functor from complexes to groups

OK, I meant to say a complex of abelian groups. Just wanna sure it's a complex, not a single object (abelian group)

(if the complex you give it as argument happens to be a complex C concentrated at degree zero, which one usually identifies with a sheaf $C_0$, then $\mathbb H(X,C)=H(X,C_0)$)

hypercohomology is needed for more or less the same reason that homotopical constructions are needed inhomotopy theory

say (X,A) and (Y,B) are pairs of spaces which are homotopically equivalent

then it is not true, in general, that X/A and Y/B are homotopically equivalent

to get that, one needs to use instead "homotopy quotients"

Oh

namely, replace X/A by $X \cup CA$

(ie, attach a cone to $X$ on top of $A$)

and what difference does it make?

is X/A not homotopy equivalent to $X \cup CA$?

that the 2nd construction is comptabible with homotopical equivalences

@Mariano, is $X/A$ not homotopy equivalent to $X \cup CA$? Is there any example for this?

If (X,A) has the homotopy extension property, yes

Crap. How should I think about these things?

My intuition has been, say you start with $X \cup CA$, then all the $A$ part can be shrinked to the vertex of the cone, and that should give me $X/A$. What's wrong with this intuition?

the intution is correct :-)

X\cup CA is what is called a homotopy colimit

a canonical example is:

I'm not sure what it is, but if the intuition is correct, then where does HEP appear in the intuition?

consider the diagram $*\leftarrow S^{n-1} \rightarrow D^n$

(wth the second arrow the inclusion)

the colimit if that diagram is $S^{n+1}$

You mean $S^n$?

but that diagram is homotopy equivalent to $*\leftarrow S^{n-1}\rightarrow *$, becuse $D^n$ is contractible

yeah

and the colimit of the 2nd diagram is just a point

the two colimits are not homotopically equivalent

Sorry, what diagram is this?

hm?

I'm not sure what your notation means - I would guess that it is $S^{n-1} \to D^n$ with $S^{n-1}$ shrinks to a point?

But then I'm confused about the second one. Do you mind giving me a definition?

definition of what?

The meaning of a diagram $* \leftarrow * \rightarrow *$

it is just a diagram in the category of top spaces

ohoh

Got it. To solve $X'=\eta(H')$ we use $H'=\bigcup_{g'\in X'}\eta^{-1}(g')$

every such diagram has a colimit

And $\ker \eta\subset H'$ is immediate. One just checks $H'\leq G$ @anon

the colimit of $X\leftarrow Y\rightarrow Z$ is the result of gluing $X$ and $Z$ "along" $Y$

@PeterTamaroff shhh

there is a sensible notion of homotopy equivalence of diagrams

@skullpatrol Ah?

ah, so colimit means pushout products?

the two diagrams I draw are homotopy equivaent, yet their colomits are not

heh, yeah: for that diagram shape, yes :-)

Ah i see. I never remember what colimit/limit are, and can only deal with direct sum/product, fiber/pushout products etc...

So back to $X \cup CA$ and $X/A$. Are you claiming that they are two colimits of one thing?

there is an homotopy push out which for the diagram wth the $X$ $Y$ and $Z$ would be starting with $X\cup Y\times I\cup Z$, and attach one side of $Y\times I$ to $X$ and the other to $Z$ using the maps

which I suppose is $* \leftarrow A \rightarrow X$?

X/A is the colimit of that diagram

the otther is the homotopy colimit of the diagram

notice that $X\cup CA$ is just $X\cup A\times I\cup *$ with attaching as I described

I see. Two questions.

a similar situation occurs for quotients

suppose $f:X\to X$ is an homeo, then you want to study $X/f$

if $g:X\to X$ is another homeo, and $f$ and $g$ are homotopicall equivalent, generally $X/f$ and $Y/g$ are very different

2. no

1. it doesn't explain why $C/A$ does not work

examples should convince you of that :D

another: let $t:\mathbb R\to\mathbb R$ be translation by $1$, so that $\mathbb R/t\cong S^1$.

now the trivial map $*\to *$ is homotopy equivalent to $r:\mathbb R\to\mathbb R$, yet the quotient of $*$ by that map is not $S^1$ nor has that homotopy type, being just a point

we are taking a limit how (of a diagram which has one vertex and one loop)

and to get something homotopy-invariant we should take a homotopy limit

there is a whole industry made out of this :D