Mathematics - 2014-04-12 (page 1 of 3)

Pedro Tamaroff

01:08

@KarlKronenfeld Kaaaaaaaaaaaaaaaaaaaaaaaaarl.

Karl Kronenfeld

@PedroTamaroff yello

Pedro Tamaroff

@KarlKronenfeld Hows you?

Karl Kronenfeld

@PedroTamaroff Good. And you?

Pedro Tamaroff

@KarlKronenfeld Can't complain.

Try to check some stuff about direct limits of modules.

A bit bothersome.

Karl Kronenfeld

god damnit, I just installed some antivirus software that came with a whole bunch of adware.

Pedro Tamaroff

01:21

@KarlKronenfeld what antivirus?

Karl Kronenfeld

@PedroTamaroff avast, I was surprised since it is extremely popular (yes I downloaded it from the main site)

Pedro Tamaroff

Weird. It has good rep.

@KarlKronenfeld Suppose $\Gamma$ is a directed set, and $\varphi_{ij}$ is a set of morphisms for $M_i$ with the properties for the direct limit.

I take $M=\bigsqcup_{i\in\Gamma} M_i/\sim$ where $x_i\sim x_j$ if we can find $k\geqslant i,j$ with $\varphi_{ik}(x_i)=\varphi_{jk}(x_j)$.

Now, one defines sum of classes as $$\overline{x_i}+\overline{x_j}=\overline{\varphi_{ik}(x_i)+\varphi_{jk}(x_j)}$$

Where $k\geqslant i,j$.

It is clear this is independent of $k$.

I have to show it is indenpendent of $x_i$ and $x_j$ too yes?

Karl Kronenfeld

yes

Pedro Tamaroff

Right, that's what I was doing.

Not too complicated though.

Mike

hello.

Pedro Tamaroff

01:28

@Mike Hey.

Mike

hella hello

Pedro Tamaroff

@FernandoMartin

did you get questiuns

in linalg

Fernando Martin

yup

bunch of them

none about dual spaces though

Mike

>@FernandoMartin

>helping students

Fernando Martin

>implying I answered the questions

Karl Kronenfeld

01:37

>implying answering questions is the same as helping

Pedro Tamaroff

>greentext sucks

Karl Kronenfeld

no @pedro

>greentext

>so rad

Karl Kronenfeld

02:08

lmfao, the OP accepted my answer for "how many sides does a dot have?"

user127001

lol

Fernando Martin

lel

Pedro Tamaroff

@FernandoMartin @KarlKronenfeld

How do you understand the direct limit?

That is if someone were to ask you informally what it is.

What would you say?

Fernando Martin

02:26

suppose your arrows were all injective

then the direct limit would be the smallest object containing the objects in your diagrams

in some sense

in general that doesn't happen but the idea is more or less the same

Pedro Tamaroff

@FernandoMartin Yes, I thought about it that way.

As some gluing process.

Fernando Martin

The thing you told me about with matrices

The $K_1$ thing

Pedro Tamaroff

@FernandoMartin Right, yes.

Fernando Martin

Sounds like a direct limit

Pedro Tamaroff

It is, yes.

An easy one!

Karl Kronenfeld

02:27

@PedroTamaroff The objects $M_i$ are representatives of the the direct limit of that system.

Pedro Tamaroff

@KarlKronenfeld Come again?

@FernandoMartin I actually think of the $\varphi_{ij}$ as embeddings of the $M_i$ into each other. I cannot think of an example where they are not injective.

I mean, of course there is.

But one that is "something", not something trivial.

Fernando Martin

The example we saw today of germs of functions had non-injective arrows

Karl Kronenfeld

@PedroTamaroff You can either interact with the direct limit $M$, which is certainly doable, or you can interact with the various $M_i$, yielding results about $M$.

Pedro Tamaroff

@FernandoMartin Ah, yes. =P

@FernandoMartin The "inverse limit" goes down, instead of up, right?

Fernando Martin

define "down"

Pedro Tamaroff

02:30

NE

VER

MI

ND.

Fernando Martin

Inverse limits are exactly the same, except your universal object is a co-cone instead of a cone

Pedro Tamaroff

I don't know what those things are. ¬¬

Well, for starters.

If $i\leqslant j$, we go from $A_j\to A_i$.

Instead of from $A_i\to A_j$.

Fernando Martin

nah

Pedro Tamaroff

That's what wiki says.

Fernando Martin

Oh, ok, you're asking the poset to be co-directed

or co-filtrant

Mike

02:36

But anyone can edit wikipedia.

Fernando Martin

@Pedro: a cone over a diagram is just an object $C$ with morphisms from each object in the diagram to $C$ such that everything in sight commutes

the limit of a diagram is just the universal cone

Pedro Tamaroff

@FernandoMartin "limit of a diagram"?

PLOP

I give up.

Fernando Martin

diagram = functor = poset

Pedro Tamaroff

NOPE.

Fernando Martin

suppose you have a poset like this: * -> * <- *

Pedro Tamaroff

02:39

I have given up.

I have to sleep.

Classes tomorrow morning.

=/

Fernando Martin

a functor from that poset to any other category is just 3 objects, and 2 morphisms

A->B<-C

that's ok

Pedro Tamaroff

@FernandoMartin Aha.

Yes, I understood that thing you explained.

Of thinking of posets as cats.

Fernando Martin

Well, then a functor that has a poset as domain

can be thought of just as a BIG commutative diagram

over the codomain of the functor

Pedro Tamaroff

Yes.

Fernando Martin

that's why it's ok to say the limit of a diagram

instead of limit of the functor, or limit of the directed system

Pedro Tamaroff

02:41

@FernandoMartin But I don't know what the limit of a functor is.

Fernando Martin

the direct limit brah

what we saw in class today

Pedro Tamaroff

But not all functors are of that form, yes?

Fernando Martin

we saw the construction for functors $F:\Gamma\rightarrow \text{R-mod}$

if $\Gamma$ is directed, they are

Pedro Tamaroff

Wait.

Oh, again I forgot.

Fernando Martin

no

$\Gamma$

where $\Gamma$ is one directed set

think of $F$ as just one big commutative diagram of $R$-modules

Pedro Tamaroff

02:43

Yes.

Mike

I like cats

Pedro Tamaroff

How does catdog poop.

Fernando Martin

The construction we saw today works for nice categories

not just for $R$-mods

you just need $\Gamma$ to be directed

Pedro Tamaroff

@FernandoMartin OK. I go now.

I leave that question up.

Fernando Martin

K, see you

Pedro Tamaroff

02:44

How does catdog poop.

Mike

03:07

don't ask

MickLH

@PedroTamaroff Childhood: Demolished

usukidoll

03:34

openstudy.com/study#/updates/5348ad17e4b01a3618780e2f

user127001

03:53

I got trouble on OpenStudy once

usukidoll

for?

user127001

I posted solutions for every question I answered lol

usukidoll

XD

Will Hunting

04:39

Hi @EnjoysMath!

Enjoys Math

Oh... my family :D home again home again jiggidy jig... goooood evening, JF!

blade runner reference

usukidoll

enjoys....come to me! :D

Will Hunting

What?

user127001

Hi @JasperLoy and @EnjoysMath

usukidoll

openstudy.com/study#/updates/5348ad17e4b01a3618780e2f errr just want enjoys to look at this T_T

Will Hunting

04:41

@user127001 Hi Bart you can't ping me with that now that I have changed my username

Enjoys Math

Sawadi krup to all yehs

usukidoll

I feel like combining theorems, lemmas, and colloraries

user127001

Oh Sorry @JasperLoy

usukidoll

@enjoysmath D:

Will Hunting

@usukidoll You misspelled corollaries

Enjoys Math

04:42

@usukidoll what are you trying to prove again?

usukidoll

Question: A number is said to be algebraic if it is the root (zero) of a polynomial with integer coefficients. Show that the set of algebraic numbers is countable.

see the url for attempts and such..

openstudy.com/study#/updates/5348ad17e4b01a3618780e2f

Will Hunting

This is trivial

Enjoys Math

You showed $\Bbb{Z}[x]$ is countable, and there is a mapping $a \to \Bbb{Z}[x]$ that's injective, and $a$ an algebraic integer, therefore there's a bijection onto a subset of $\Bbb{Z}[x]$. Prove that that means the alg numbers are countable.

wait... may not be injective

usukidoll

that's my issue do I have this right or what or lol or dsfjkfjwelrjlkejwelkjrl

I have so much theorems thrown at my face XD

Enjoys Math

You know that if $S \subset S'$ and $S'$ countable, then $S$ is also?

usukidoll

04:46

errrrrrrumm yeah I guess... S is contained in S'

so if S' is countable so is S?

Mike

at most countable

usukidoll

okkk

but what about the polynomial.. root being 0

I mean by lemma 7.3.13 The set $Z[x]$ of all polynomial functions with integer coefficients is countable

Will Hunting

For each polynomial there is a finite number of roots

usukidoll

and then by prop 7.3.14 The number of distance roots of a polynomial function of degree n is at most n.

since I have a zero root then the function is at most zero right?

yeah I am trying to get the finite number of roots thing because I was reading something similiar and it was infinite

Will Hunting

There are countably many polynomials with integer coefficients

usukidoll

04:49

finitie... countable.... bijection $ N \rightarrow S$

nughhhhhhhhh!

then I got to get a surjection map

el oh el

Enjoys Math

Define $f : \Bbb{Z}[x] \to \text{alg nums} : p(x) \to \text {zeros of } p$ Then $f$ is onto. Any time $f : S \to S'$ is onto and $S$ is countable means that $S'$ is countable.

usukidoll

theorem 7.2.5 Let A be any set, and suppose that there exists a surjection f : N ->A. Then A is finite or countable
so let that A be the set of algebraic numbers... since by that theorem A can be any set
then that would be that a surjection f: N -> A exists From natural numbers -> set of algebraic numbers.

Enjoys Math

QED

usukidoll

ok do I need to show an example or is that it or errr or huh or what?!

Enjoys Math

crap that didn't work

usukidoll

04:52

all we need is a surjection mapping from natural numbers -> the set of algebraic numbers.

and then the define from your stuff @enjoysmath since it has the onto which is surjection

Enjoys Math

Something like that... good luck!

usukidoll

nooo don't run D: *drags @enjoysmath

I should write all of this on scratch and then re post or something maybe that would be a good idea because it's a bunch of theorems, a lemma, and collary sdjfsajfasdkfjsa;

Enjoys Math

Okay that will work. There's an injection of alg numbers into $S' = \{B : B = \text{zeros of } p, p \in \Bbb{Z}[x]$. That completes the proof

usukidoll

alright I just have to gather all of this and write it down...

Enjoys Math

So we have $|A| \leq |S'| = |\Bbb{Z}[x]|$

usukidoll

04:57

Cardinality !

how does that get involved?

Enjoys Math

I'm just compressing the existence of all those maps I mentioned into a cardinality statement

*the expression of the existence

usukidoll

what if I don't want to use cardinality? then would the proof be alright or crash?

Enjoys Math

To review. Define $f: \Bbb{Z}[x] \to \text{sets of alg numbers }$

in the obvious way, ie $f(p(x)) = \{ a : p(a) = 0\}$

usukidoll

oooh! I should also use the surjection and injection definitions in this too

Enjoys Math

Let $\im f = S'$

usukidoll

05:00

that's the surjection def you just used right?

lol

so anyway , you have used the surjection defintion right?

dang typoz

Enjoys Math

Yes

$f$ is a surjection onto $f(\Bbb{Z}[x])$ since any $f : A \to B$ is a surjection onto $f(A)$.

Then there exists an injection of $A = \text{alg numbers}$ into $f(\Bbb{Z})[x]$, by simply taking an alg number $a$ and choosing any set in $f(\Bbb{Z}[x])$ that contains $a$. Walah!

So you're left to prove that when you have $A \to B \leftarrow C$ and the first map is surjective, the second map injective, then $|A| \geq |C|$, but I haven't proved that yet, so not sure

usukidoll

D: D:

k k

Enjoys Math

you can replace the $|A|$ part with statements about countability and maps between sets

usukidoll

I have one more proof to correct... after dinner so would you still be available @enjoysmath ... it's about equivlaence relations and dual relation of R*

Enjoys Math

Since $A \to B$ is surj, there's a right inverse well-defined, which is injective by def and so you have a sequence of injections into $A$, which means there's a bijection of $C$ into a subset of $A$. That subset is countable so $C$ is countable.

The key there is that injection $\implies$ existence of bijection onto subset

maybe baby

You should also become very familiar with these two statements: for any map $f: A \to B$, $f$ is surjective $\iff$ there's a right inverse for $f$, and $f$ is injective $\iff$ there's a left inverse for $f$.

That's showing existence of a function $g: B \to A$ s.t. $f\circ g = id_B = $ the unique identity map on $B$, $id_B(b) = b : B \to B$.

(i.e. that's a necc. & suff. condition for surjectivity, then the analogous applies to injectivity)

You should def prove those two statements if you haven't already, since they're used EVERYWHERE!!!!!

Mike

05:16

heya, @StephenMontgomery-Smith

usukidoll

@enjoysmath

here's the other question

Let $R$ be a reflexive and transitive relation on a set $S$. Let $R*$ be the dual relation, $(a,b) \in R*$ if and only if $(b,a) \in R$. Show that $R \cap R*$ and $R \cup R*$ are equivalence relations.

$ R \rightarrow $ reflexive and transitive.
$(a,b) \in R* \leftrightarrow (b,a) \in R$. Set $T = R \cap R*$. Show that $T$ is an equivalence relation.

1. $(a,a) \in R$ and $ (a,a) \in R* \rightarrow (a,a) \in R \cap R*$

2. Suppose $(a,b) \in R \cap R* \rightarrow (a,b) \in R$ and $(b,a) \in R \rightarrow (b,a) \in R \cap R*$

SO #3 doesn't work if I'm trying to prove that $R \cup R*$ is an equivalence relation... hmmm

dang starz.. brb gonna do stuff

Sush

05:34

@usukidoll, how can i download the lecture videos of calculus ocw course ?

Enjoys Math

05:54

$(a,b) \in R\cup R^* \implies (b,a)$ is definitely in there.

symmetry

$(a,b), (b,c) \in R\cup R^* \implies (b,a), (c,b) \in R\cup R^*$ fuck I give up

usukidoll

heyyY! *drags @enjoysmath * you just have to tell me how the third part doesn't really work .. that like's symmetric and transitivity isn't it?

Mats Granvik

08:24

@GabrielR. I don't know what I mean by looking at the power series directly. I just had to say something. Knowing that sine and cosine are axis positions of a point on a circle is much closer to saying that sine and cosine are periodic. So you probably need to prove that the power series for sine and cosine are related to the circle in order to prove that they are periodic.

@GabrielR

Gabriel R.

@MatsGranvik Yes a geometric approach is possible, but it wouldn't be as rigorous as using power series only. @DanielFischer posted a solution yesterday

@MatsGranvik "From the power series, you immediately get Euler's formulae. Also, from the power series it is easy to check that $\exp$ is a homomorphism $\mathbb{C}\to \mathbb{C}\setminus\{0\}$. Since $\exp$ is a non-constant holomorphic function, you get that $\ker \exp$ is not trivial. From basic considerations, it is clear that $\ker \exp$ is infinite cyclic, and it remains to see that it is generated by $2\pi i$."

@MatsGranvik He also provided details about the last part of his reasoning :"$\ker \exp$ is a closed subgroup of $\mathbb{C}$, and contained in $i\mathbb{R}$. A closed subgroup of $\mathbb{R}$ is either all of $\mathbb{R}$ or cyclic. Since the kernel is discrete, it is cyclic."

Mats Granvik

ok

Enjoys Math

0

Q: Appropriateness of using an interval in $\Bbb{R}$ as the parameter to a continuous path in space $X$.

The fundamental group $\pi_1(X,x)$ is usually defined using continuous paths $p : [0,1] \to X$. But... (1) Can you use other spaces besides a closed interval of $\Bbb{R}$ in the usual topo? (2) How will the resulting fundamental group change if at all. (3) Do there exist spaces $X$ where a cl...

yeah, good luck with that

Alex Youcis

09:00

@EnjoysMath What would make you happy with that problem? You can think of p_1(X) as [S^1,X]. You can then consider other spaces H for which [H,X] has a natural group structure. The most natural example is if H is, a, well, H-space.

Alex Youcis

09:11

@EnjoysMath I accdidentally reversed the indices there, unfortunately.

@EnjoysMath I left a comment on your post :)

mirgee

Hi, how do you prove a series with negative terms diverges? When working with series with strictly positive terms it is easy, Cauchy, d'Alembert, Raabe, Abel, Gauss + you can always compare it. But all the tests I was taught for series is in the form conditions met->series converges (except for Gauss)...

Alex Youcis

@mirgee Does it have finitely many negative terms?

mirgee

No, I mean generally any number of negative and positive terms

Alex Youcis

09:27

@mirgee How are you defining the series? Are you taking limits simultaneously ($\lim_{n\to\infty}\sum_{-n}^n$) or independently?

mirgee

I am not sure what you mean, I see one limit... I am not separating anything, so simultaneusly I guess

Alex Youcis

@mirgee Then, you can just bring the terms all to one side, and make it just a sum of "positive" terms.

mirgee

What side? I can't leave out the negative terms, do I

Alex Youcis

@mirgee Just think about it as a sum of just positive terms. $\sum_{k=-n}^{n}a_k=a_0+\sum_{k=1}^{n}(a_k+a_{-k})$

mirgee

Alright, but when such a series diverges, it tells me nothing about the original series...

No, sorry, I see what you mean now! Thanks :)

Alex Youcis

09:37

@mirgee No problem. Be cognizant of the fact that this was under the assupmtion you defined the sum symmetrically like that. If you don't, things get more sticky.

mirgee

I see. The number of positive and negative terms must repeat periodically for such approach to work.

But that's not all is it

The limit of each individual term of the original series must converge to zero

Enjoys Math

09:56

Thanks @AlexYoucis

Hawk

@GabrielR. by Cauchy Condensation Test.

Enjoys Math

@AlexYoucis In other words they've already generalized there. But my third question is unanswered, I put that in a comment. I see though that the fundamental group is simply defined with respect to $\Bbb{R}$ and that whether or not a space has "paths" with more points than $\Bbb{R}$ doesn't matter.... or something

Hawk

@robjohn I now undestand...what you said is not so obvious is because, I asked that question to DanielFischer...

@ParthKohli any idea when SK will be coming?

@Sawarnik Let me know when you come...we have a plan to execute.

Sawarnik

10:12

@Hawk Here is he.

Hawk

The room we created is it there?

@Sawarnik Yes, I got it.

Harold Cavendish

Greetings. Can anyone help me with feature extraction from EEG signal? I am trying to utilise PCA, but it does not seem to work properly so I need validation.

Gabriel R.

10:28

@Hawk don't you think there's more simple approach ?

Hawk

@GabrielR. Yes, there is, but that is what I seemed to learn recently, so, that is what I am using...but I think Cauchy Condensation makes it quite easy to prove...what would be your way?

Gabriel R.

@Hawk $ln(n)<n-1$

Hawk

@GabrielR. How will you prove this?

Gabriel R.

Convexity of log, or derivatives

Hawk

Okay, will this extend the proof?

Gabriel R.

10:33

I meant concavity

Hawk

yes, I got that...

Gabriel R.

What do you mean?

Hawk

I mean that..you were meaning concavity...

I understood that...

Gabriel R.

No, what did you mean by "extend the proof"?

robjohn

@Hawk what I said about what?

Hawk

10:39

What I meant is that...Cauchy Condensation Test would prove this in the second or third line I suppose....

@robjohn You said that $\sum\limits_{n=2}^{\infty} \dfrac{1}{log n}$ wouldn't be so obvious...regarding Cauchy Condensation test.

robjohn

@Hawk $\sum\frac1{n\log(n)}$ almost converges

Hawk

@robjohn What is meant by 'almost' converges? Doesn't Cauchy Condensation Test prove that it is divergent?

robjohn

@Hawk yes, it diverges, but $\sum\frac1{n\log(n)^a}$ converges for any $a\gt1$.

Gabriel R.

@Robjohn what's your proof for $1/(nln(n))$? I achieved it by summing integrals, but I'm looking for something more straightforward

robjohn

@GabrielR. You can use Cauchy Condensation, you can use Integral Test, there are others.

Hawk

10:46

@robjohn How is that?

robjohn

@Hawk Try Cauchy Condensation.

Hawk

@robjohn Okay, just a minute.

mirgee

Hi, please, how do I find if $\sum_{n=1}^\infty {(-a)^n \over {a^n + b^n}}$ converges or diverges?

robjohn

@mirgee when $b\gt a$

@mirgee I assume that $a,b\gt0$

mirgee

Yes

@robjohn How did you find out, please

Hawk

10:50

@robjohn I am getting $\sum \dfrac{1}{a\log 2}$ after Condensation...

robjohn

@Hawk You should get $\sum\frac1{(n\log(2))^a}$

Hawk

@robjohn Okay, just a min, let me check once more...

mirgee

@robjohn I mean, why it diverges for a \gt b

robjohn

@mirgee alternating series test. When $b\gt a$ the terms monotonically go to $0$

@mirgee if $a\ge b$, the terms don't go to $0$

Chris's sis

Greetings

mirgee

10:54

@robjohn Yes, but it doesn't imply divergence AFAIK

robjohn

@mirgee If the terms don't go to $0$ the series diverges. Term test

Hawk

@robjohn Isn't that for $\sum \dfrac{1}{n(\log n)^a}$?

robjohn

@Hawk that is what I wrote

mirgee

@robjohn Stupid me! Thanks.

Hawk

@robjohn Sorry, but I thought you wrote $\sum \dfrac{1}{n\log (n)^a}$

robjohn

10:57

@Hawk Yes, not $\sum\frac1{n\log(n^a)}$

Sawarnik

@ParthKohli Hi.

Trying your question now.

Hawk

@robjohn Okay, yes, $\sum \dfrac{1}{n(\log n)^a}$ converges.

I understand that.

robjohn

@Hawk Put the parentheses around the arguments and things work better.

$\sum\frac1{n\color{#C00000}{\log(n)}^a}$

that is the way precedence works

Hawk

@robjohn Yes, so it does...Is Cauchy Condensation test itself easy or I am feeling it to be easy?

@robjohn Now, I do not understand what you mean by that...

robjohn

The parentheses around the argument of a function bind to the function so the exponentiation comes after the log

Hawk

11:02

Okay. I get that now.

@robjohn Is there any situation where Cauchy Condensation does not work? I feel this is the easiest way to test convergence or divergence.

robjohn

@Hawk well it only works for monotonically decreasing series...

Hawk

@robjohn Not monotonically increasing series?

robjohn

@Hawk are you talking about the negative of a monotonically decreasing series or a really increasing positive series?

Hawk

@robjohn aren't both monotonically increasing series?

robjohn

@Hawk yes, but you must know the difference

Hawk

11:10

@robjohn Okay, please tell me the difference.

robjohn

@Hawk one tends to zero and the other doesn't

for starters

come on

Hawk

Yes, that I understand...

a little advanced than that?

robjohn

multiply by -1 to get a series for which the condensation test works.

Hawk

I think I do not understand clearly, can you please provide an example?

robjohn

@Hawk to keep from having to cover numerous cases with increasing negative series and decreasing positive series, the condensation test is formulated with positive decreasing series, and you are left to apply multiplication by $-1$ to handle the other case

Hawk

11:14

Oh, multiplying by -1 to make the series monotonically decreasing right?

robjohn

so when I say monotonic decreasing, I mean POSITIVE monotonic decreasing.

Hawk

Okay...

Yes, I understand very clearly now I think...

robjohn

The Cauchy condensation test as given in most texts (and wikipedia) assumes a positive decreasing series

Hawk

Yes, it does....

Sawarnik

..

Hawk

11:18

@robjohn so, when the monotonically decreasing series converges to 1 or some other positive value...there cauchy condensation doesn't work? is that correct?

robjohn

@Hawk it works, but the series diverges

@Hawk You mean the terms converge to 1

Hawk

@robjohn yes, that is what I mean.

@robjohn why will the series diverge?

robjohn

@Hawk the terms don't go to $0$

before or after condensation

Hawk

Does convergence always mean that it should converge to 0?

robjohn

@Hawk terms converging to $0$ is necessary, but not sufficient for the series to converge.

Hawk

11:24

@robjohn I do not understand this...

robjohn

@Hawk if the terms don't tend to $0$ the series does not converge however, if the terms tend to $0$, we don't know anything about the convergence of the series.

Hawk

@robjohn Yes, I understand now...

@robjohn and convergence will be confirmed only after some test...in this discussion...Cauchy Condensation...is that right?

robjohn

@Hawk Yes. Cauchy Condensation, and Dirichlet's Test (which covers the alternating series test) are the two main tests for convergence

The integral test covers most of the other cases

Hawk

@robjohn other cases?

@robjohn how to know when Cauchy Condensation won't work and when to apply Dirichlet's Test?

robjohn

Things that Cauchy condensation and Dirichlet do'nt cover

Sawarnik

11:29

..

Hawk

@robjohn are there such cases too? what are such cases? please tell me about it.

robjohn

@Hawk Cauchy's test is for monotonic decreasing and Dirichlet usually is used for series with sign changes

Hawk

@robjohn You mean alternating signs of terms?

robjohn

@Hawk they don't need to be alternating. for instance $\sum\frac{\sin(n)}{n}$

they don't alternate with any regular pattern

Hawk

@robjohn Okay...I understand...

robjohn

11:32

thogh they definitely change sign

Hawk

@robjohn Yes, they do...

@robjohn does the integral test cover all?

robjohn

@Hawk no, generally, the integral test applies to monotonically decreasing series

Hawk

@robjohn Is this too, positive monotonically decreasing?

Sawarnik

..

robjohn

@Hawk yes, or then we need to say positive decreasing or negative increasing. It is easier to cover one and leave it to the hopefully intelligent user to multiply by $-1$

Hawk

11:36

@robjohn how does it then cover the 'other cases'? It is of same league as Cauchy Condensation...isn't it?

robjohn

@Hawk yes, but not all POSITIVE decreasing sequences are easily handled by the condensation test

Hawk

@robjohn okay, please give one such example...

@robjohn I got one more question too, how could I tackle $\sum \dfrac{\sin n}{n}$ with Dirichlet's test...does it always satisfy $a_n \ge a_{n+1}>0$?

@Sawarnik You're posting '..' since then...you want to say something??

:D

Sawarnik

@Hawk No, I am just making sure that I am alive :D

..

Hawk

@Sawarnik What would kill you? You can always try pinching yourself... :D

Sawarnik

@Hawk No, I am telling you that I am alive :P

I just pinched, btw, won't try it again :D

Hawk

11:52

Robjohn leaving chat???!?!?!

never saw that before.

Chris's sis

A funny series :-) $$\sum_{n=1}^{\infty} \frac{\displaystyle \sin\left(\frac{n \pi}{2}\right)}{n^2}=C$$

Sawarnik

Greeting, greatest one!

Chris's sis

$C$- Catalan's constant

@Sawarnik Hello :-)

Sawarnik

You missed Rob just, btw.

Chris's sis

No problem. He'll be back later on. :-)

Mats Granvik

11:54

Do you know a formula such that regardless of what input you feed it, it will always converge to the same number?

Hawk

@Chris'ssis have you ever seen robjohn leaving this room? I saw this first time.

Chris's sis

@Hawk Yeah, once in a while.

Hawk

@Chris'ssis Okay, this is the first time..was a little shock for me. Never saw that before...

Though I am not here for very long time

Chris's sis

@Hawk shock? hehe, no worry for that.

Sawarnik

@Hawk He should be alive.

Hawk

11:57

@Chris'ssis No, he was teaching me a lot of things and concepts...I was overwhelmed...

Chris's sis

@Hawk He is very good at doing that. He has also taught me a lot of nice things.

Hawk

@Chris'ssis Yes, I know...great...

Sawarnik

@Hawk I pinched again :D

Mats Granvik

Can anyone give me an example of a formula that always converges to the same answer?

Transcript for

$Mathematics$ Mathematics