Mathematics - 2012-11-03 (page 2 of 4)

Greg Ros

2:00 PM

Can't I then turn it into Grandi series?

Johan Larsson

@Nimza sry about that, had promised to pick gf up but had forgot

Greg Ros

Hmm. I guess I could say that if A = -1 + 1 - 1 + 1... then it could either be the series {-1 + 1} or the series {(-1)^k}. Can't I?

Frank Science

It's not converge absolutely.

Therefore the interchanging is sometimes illicit.

In fact you can obtain an arrangement such that $\liminf x_n=A$ and $\limsup x_n=B$ for all $A\le B$.

jdoe

hello

Nimza

@jdoe hi

ガベージコレクタ

2:13 PM

@skullpatrol: If division by zero is undefined, $\tan 90^\circ$ is also undefined?

skullpatrol

@ガベージコレクタ Correct.

N3buchadnezzar

Indeed

ガベージコレクタ

Oops! There was a problem updating your profile:
Display name may only be changed once every 30 days; you may change again in 42 minutes.

I will change my display name soon. Any idea?

skullpatrol

@ガベージコレクタ skeletor

ガベージコレクタ

@skullpatrol: I will change to Casper Joy or Casper Toy, does it looks similar to someone?

Frank Science

2:20 PM

Even though it were similar to somebody, it would be no sin of you to do that.

jdoe

Hello

skullpatrol

@ガベージコレクタ He doesn't like you.

Johan Larsson

hej

jdoe

What mathematics is going on

Mats Granvik

tjenare

Frank Science

2:21 PM

Any mathematics.

ガベージコレクタ

@skullpatrol Who?

jdoe

I've got 4 different topics to study and I can't decide which one!

skullpatrol

@ガベージコレクタ You know who.

Frank Science

Goodbye, everybody.

ガベージコレクタ

@skullpatrol IC.

@FrankScience Welcome to Math.SeX.

skullpatrol

2:24 PM

@ガベージコレクタ Did you read the link?

ガベージコレクタ

@skullpatrol: Yes.

skullpatrol

@ガベージコレクタ Any questions?

ガベージコレクタ

@skullpatrol No. Clear enough.

skullpatrol

@ガベージコレクタ So what is your answer?

ガベージコレクタ

@skullpatrol Almost the same as the link explains.

skullpatrol

2:27 PM

@ガベージコレクタ Which is?

jdoe

ガベージコレクタ

@skullpatrol Your link above.

@jdoe What is the context of this photo?

skullpatrol

@ガベージコレクタ I asked for your answer.

ガベージコレクタ

@skullpatrol: First answer is about our intuition, i.e., we cannot divide a pie by zero number of people. Second answer is almost identical to math.stackexchange.com/a/26452/26975

skullpatrol

You have to make it make sense to you.

ガベージコレクタ

2:31 PM

@skullpatrol: Do you have mathematics certificates?

Greg Ros

What do you think about summation techniques like cesaro summation? How should I put this. You can often do algebraic manipulations that seem to make sense, but aren't supported using classic summation of infinite series.

It's kind of confusing me, mentally.

It would seem to make sense that if S = 1 - 1 + 1 - ⋯, then S = 1 - S ⇒ S = 1/2

And in terms of classical summation, it's putting your cart before the horse. But it still seems to have meaning.

jdoe

@GregRos, In general if S is not a well defined number then you should not expect algebraic manipulation of S to provide anything of value

ガベージコレクタ

Everybody here, I have a question about Math education.

jdoe

@GregRos, in specific cases (like the Euler zeta function) you might have something else in mind that guides or justifies the algebraic manipulation

ガベージコレクタ

how can we categorize the math subjects in senior high school?

Greg Ros

2:36 PM

Well, this manipulation does make sense, and the cesaro sum of 1-1 is 0.5. So it would seem like, when I'm saying S = 1 - 1 + 1... S is actually the cesaro sum of the sequence.

skullpatrol

@ガベージコレクタ Algebra, Geometry, Pre-Calculus

jdoe

but that's not the cesaro summation

I suppose you could just say "For the next three pages "..." means Cesaro summation"

ガベージコレクタ

@skullpatrol Logarithm belongs to algebra?

jdoe

but that's just obtuse, use sigma or whatever notation everyone else uses for it

skullpatrol

@ガベージコレクタ Yes.

Greg Ros

2:38 PM

That's not what I mean. I mean, the cesaro sum of {(-1)^k} is 0.5. When I say, "assuming the sum of the series exists and it is S", and then I write

Johan Larsson

whhat belongs to pre-calculus?

Greg Ros

S = 1 - S ⇒ S = 0.5, it seems like I've just calculated the cesaro sum.

ガベージコレクタ

@skullpatrol: The link abcte.org/teach/exam-preparation/mathematics/standards seems to be different from your taxonomy.

Greg Ros

To put it another way, if I proved the series is cesaro summable, I could do this manipulation, and S would be the cesaro sum. The many ways in which you can define summation of this sort confuse me.

jdoe

I suppose you have calculated the Cesaro sum on the condition that it exists

Johan Larsson

2:40 PM

trig E geometry?

skullpatrol

@ガベージコレクタ You asked for senior high school, no?

ガベージコレクタ

@skullpatrol Yes.

jdoe

@GregRos, do all summation methods coincide in the cases when both are defined?

Greg Ros

I don't know about all of them. Only those I know.

But it seems like the sum of an infinite series should be one thing.

skullpatrol

@JohanLarsson Pre-calculus has analytic geometry....

jdoe

2:42 PM

analytic continuations are unique

maybe we could define "summation method" in terms of continuations

Greg Ros

Are there functions which cannot be integrated with one definition, but cannot be integrated in another?

Hmm. I guess if we wanted the summation form that corresponds to integration it would be classical summation.

jdoe

Wait, Given 1 - 1 + 1 - ⋯ Cesaro summable, why is S = 1 - S?

because -S = -1 + 1 - 1 + ⋯ is Cesaro summable and so 1 + -S = S, fine

There are perhaps summation methods and algebraic manipulations such that some part of the proof which seems algebraically fine actually turns a summable sequence into a non-summable one

in which case you could end up with a wrong answer

skullpatrol

@ガベージコレクタ My taxonomy were the general subjects covered in senior high school.

jdoe

What I mean is: It's not enough to just assume something summable and then do algebra with it, you need to check that each step is again summable

ガベージコレクタ

@skullpatrol OK.

Greg Ros

2:48 PM

Of course, I'm not making a formal argument. I'm just thinking about what summation means.

jdoe

which answers the question about which summation methods coincide: (At least) all those for which the algebraic manipulations transform summable sequences into summable sequences, (and they must exist of course)

Greg Ros

Hmm. It is interesting that from a purely algebraic point of view, you could use any of a dozen summation methods.

So they're all valid in the algebraic sense.

skullpatrol

hi @Charlie

ガベージコレクタ

We have a method to calculate a square root by hand. Is there a method to calculate trigonometric function by hand for any angle in radian?

Greg Ros

Series expansion

jdoe

2:51 PM

There are many algorithms to compute trigonometric functions

ガベージコレクタ

@jdoe Which one can be done by a human?

Greg Ros

Series expansion is the one that requires least thought.

jdoe

@ガベージコレクタ, The best for a human would be a big table of values and then ways to reflect and combine them to cover almost all the whole circle

Greg Ros

best for a human would be using a calculator :)

jdoe

e.g. double angle and sum identities

skullpatrol

2:55 PM

hi @anon

Greg Ros

Is there such a thing as a broadest form of series summation, where its set of summable infinite series are a superset of all others?

anon

no, because summability methods can give contradictory answers

also hi skull

Greg Ros

Really? Can you give an example?

ガベージコレクタ

I just changed my display name, but the change has not taken affect in this chat room.

anon

@GregRos I do not have one memorized, am googling.

Greg Ros

3:02 PM

thanks :)

skullpatrol

@ガベージコレクタ You said: "First answer is about our intuition, i.e., we cannot divide a pie by zero number of people." Isn't the pie just left undivided?

ガベージコレクタ

@skullpatrol I ate it alone. :-)

Greg Ros

I've been taught that you can perform certain algebraic operations on infinite sums only in some cases. For example, you can't do something like $∑(-1)^{k} = ∑(1 - 1) = 0$. However, it seems to me that whatever manipulation you do results in a different series. It's not that some manipulations are legal. It's that all manipulations result in different series, and some of those series don't converge, or converge to a different value.

Does that make sense?

anon

you might be interested in ch 9 of Classical and Modern Methods in Summability, somewhat available on google books

skullpatrol

@GarbageCollector That would be the pie divided by 1 person, no?

anon

3:08 PM

I know what you're saying greg

jdoe

yeah, a manipulation is either admitted or not depending on what summation method you're considering

N3buchadnezzar

@GarbageCollector I was able to read that instantaneously

Garbage Collector

@skullpatrol by zero people if possible.

Greg Ros

No, manipulations are fine, but they always result in different series. The question is, is this new series convergent or not? Some manipulations are, let's say, "closed" in terms of convergence within a given summation method.

jdoe

not just "is it convergent" but is it equal

Garbage Collector

3:09 PM

@N3buchadnezzar Good.

jdoe

both of ∑(-1)^{k}, ∑(1 - 1) are cesaro convergent

skullpatrol

@GarbageCollector I'm saying it is possible to divide a pie by 0 people, leave it undivided.

Greg Ros

@skullpatrol that's one person.

anon

perhaps you're confusing dividing n times with dividing by n

Greg Ros

I think the question whether two series are equal is tricky. It's kind of like equality and equivalence in matrices. If you have $a_{1} + a_{2} + ...$ then turning it into $a_{2} + a_{1} + ...$ seems to result in an equivalent series that is different but probably converges to the same sum as the previous one. Though you might be able to define summations where this isn't the case. This idea seems more elegant to me than saying some manipulations are illegal.

skullpatrol

3:17 PM

hi @OldJohn is it half-time?

jdoe

Surely any "summation method" should not care about finitely many terms

Old John

@skullpatrol No - all finished - "we" won :)

skullpatrol

@OldJohn What was the score?

jdoe

(I mean take a finite initial segment and replace it with any other finite initial segment with the same sum)

Garbage Collector

@skullpatrol We have different perspective. :-)

Greg Ros

3:18 PM

Oh. I was sort of thinking of flipping e.g. 1-2, 3-4, 5-6.

Old John

@skullpatrol 2-1, but United missed chances, and ought to have won 5-1

Greg Ros

like transposing the elements in a predictable way

Jayesh Badwaik

@OldJohn and so many offisdes.

Old John

@JayeshBadwaik too many, yes

Jayesh Badwaik

@OldJohn but good to see such an awesome attacking game. Missed that for quite some time now.

Old John

3:19 PM

@JayeshBadwaik Yep - makes for better viewing

Greg Ros

In fact, the notation $a_{1} + a_{2} + ⋯$ seems kind of deceitful, since summation of an infinite series doesn't actually mean performing the addition operation an infinite number of times.

skullpatrol

@OldJohn Do you watch any American sports?

jdoe

that's a good point

Old John

@skullpatrol No - I watch very little sport, really - just a bit of football and maybe tennis

@GregRos Just think of the notation as shorthand for a limit

Greg Ros

If you say that $$-1 + 1 - 1 + ⋯≠ (-1 + 1) + (-1 + 1) ⋯$$ that already means your addition operator isn't associative, so it's not an addition operator at all.

@OldJohn I understand the definition, I'm just kind of pondering about different summation methods and notation.

Old John

3:24 PM

we use a lot of notations in maths which really ought to be straightened out - mainly for historical reasons. Notations of derivatives in calculus are a case in point.

Greg Ros

What do you mean, notation of derivatives? Isn't the form $$df/dx$$ fairly non-ambiguous?

jdoe

yeah so let's say like

instead of sigma_{n = 1}^infinity a_n

cesaro_{n = 1}^infinity a_n

abel_{n = 1}^infinity a_n

etc.

Old John

@GregRos well - it is not really a quotient, is it - it is really a limit

Jayesh Badwaik

@GregRos You have to write of which variables is $f$ a function of for instance.

Greg Ros

Does that matter? I already said I was differentiating by x.

jdoe

3:29 PM

yes it matters

Greg Ros

Hmm, in a practical sense, why?

jdoe

you can define a differentiation operator "D" such that for f : R -> R, Df is the derivative (another function R -> R)

but if f is a function then df/dx doesn't mean anything, df(x)/dx on the other hand is the derivative of f evaluated at x

Greg Ros

Hmm. I maybe summation of infinite series was invented under the idea that, while actually performing an infinite number of additions isn't logical, maybe we can define something that's sort of like adding an infinite number of terms, for cases that make sense. Similarly, while dividing by an infinitely small quantity doesn't really make sense, you can define something that's like it, and you get a derivative.

Well, people probably did it before they understood it doesn't make sense. But, urm, I have a point in there, somewhere.

jdoe

@GregRos, Do you know about Cauchy sequences?

Greg Ros

3:44 PM

Yeah. Don't think I've done them yet though, in school.

jdoe

@GregRos, "infinite number of operations isn't logical" is a very insightful observation which I think Cauchy sequences resolve in a purely finitary way

(In particular the fact that the diagonal of a cauchy sequence of cauchy sequences is a cauchy sequence)

Greg Ros

I'm not sure what you mean

Pilot

Hello people,I have a quick question after reading something that anon linked me to I came up with an idea that a function f:Q->R can never be continuous?

jdoe

For example, in the case of the most basic summation method. "Sum a_n converges" really means the sequnce a_0, a_0+a_1, a_0+a_1+a_2, a_0+a_1+a_2+a_3, ... is Cauchy

and Cesaro summation convergence is the statement that the sequence a_0, (a_0+a_1)/2, (a_0+a_1+a_2)/3, (a_0+a_1+a_2+a_3)/4, ... is Cauchy

Greg Ros

Is it different from saying that the sequence of partial sums has a finite limit?

jdoe

3:55 PM

it's equivalent to that

@Pilot, every function Q -> R is continuous,given the discrete topology on Q

Old John

@Pilot erm - isn't a constant function from Q to R always continuous?

Pilot

Oh ok you are right,I was focusing on its discontinuity on R

I was lookin for a function continuous on Q but discontinuous at some irational point

Greg Ros

Can't you say something like, f(x) = 0 if x ≠ √2 and f(x) = 1 otherwise? :)

It's certainly continuous in every point in Q. and it's certainly discontinuous in one irrational point.

Old John

@GregRos or : zero for all x less than $\sqrt{2}$ and one for all values above $\sqrt{2}$

Greg Ros

As long as your function is discontinuous in a countable number of points in ℝ while being continuous in ℚ, there isn't any problem.

Pilot

4:06 PM

@GregRos that was quite nice example.

Greg Ros

You could have like, f(x) = 1 for every x where there exists p ∈ ℙ such as that x = √p.

and f(x) = 0 otherwise

Pilot

Thank you

can I say f(x)=1 for all irrational x s but then a set of discontinuities of f would be irrational which is wrong

Greg Ros

A function that is continuous in all x ∈ ℚ and discontinuous in all x ∉ ℚ doesn't exist.

skullpatrol

hi @PeterTamaroff

jdoe

can a R -> R function only be discontinuous at countably many points?

Pilot

4:12 PM

@ jdoe no cause R has to be then meager set

Greg Ros

yes?

I gave two functions, one of which discontinuous in 1 point, and the other discontinuous in a countably infinite number of points.

You can also have a function discontinuous for every x ∈ ℚ.

You just can't have the opposite.

Peter Tamaroff

@jdoe Yes. Why not?

jdoe

example?

Greg Ros

What do you mean by countable number of points?

Countably infinite or just countable?

Pilot

f(x) = 1 for every x where there exists p ∈ ℙ such as that x = √p.,this one? so you take p from rationals only?

jdoe

4:16 PM

Countably infinite

Greg Ros

p is prime, so √p is irrational.

the above. Or this, en.wikipedia.org/wiki/Thomae%27s_function

Garbage Collector

@JasperLoy: Hi, I changed my display name.

user19161

@GarbageCollector Great, easier to ping now.

Garbage Collector

@JasperLoy: I wanted to change to Casper Joy but @skullpatrol said you don't like me. :-)

user19161

@GarbageCollector You should not try to impersonate others.

Garbage Collector

4:18 PM

@JasperLoy Casper Joy sounds funny :-)

@JasperLoy: lets talk about mathematics

Greg Ros

I've always wondered how a function that goes:
$$f(x) = 0\space if\space x∈ ℝ, f(x) = ab\space for\space x = \frac{a}{b}$$ is in lowest terms

Old John

@JasperLoy Hmm - you are still catching up with me ...

user19161

@OldJohn Well, you are always ahead of me. =)

Old John

@JasperLoy probably for not much longer

Peter Tamaroff

@GregRos ?¿

You mean $\Bbb R\setminus \Bbb Q$

user19161

4:22 PM

@GregRos Does not make sense since rationals are reals.

Greg Ros

whoops

user19161

And note that real numbers are complex numbers in a sense.

Greg Ros

Well, I meant if x ∉ ℚ.

user19161

So to be precise one has to say for example non-real complex numbers.

user19161

In this case, it is just irrational vs rational.

Garbage Collector

4:23 PM

$$\Huge\text{ Garbage Collector}$$ just a test!

Greg Ros

Yeah, that's what I meant.

user19161

@GregRos What do you mean by lowest terms? That itself is not well-defined.

Greg Ros

Lowest terms is sufficiently well-defined? It's used in many places? en.wikipedia.org/wiki/Thomae%27s_function

Anyway, I think it's possible to understand the function from what I wrote. It was just idle musing. It doesn't have to be perfectly rigorous and formal.

Ed Gorcenski

Guten tag

Greg Ros

Ignore its existence if you like.

Old John

4:25 PM

@EdGorcenski sawubona

I don't speak German, so maybe you are OK with Zulu? :)

Pilot

Another question please: So I needed a continuous function f:Q->R and g:R->R s.t f(x)=g(x) for all rational x s. So if I say f(x)=0 and g(x)=0 for x in R/{sqrt2} and g(x)=1 at x=sqrt 2 .would I be right?

Greg Ros

what's g(x) for non-rational x's?

Old John

@EdGorcenski You beat me to the answer with the differentiable question just now :(

Ed Gorcenski

@OldJohn Hehe, I figured once that comment was made, I may as well answer my own question

Greg Ros

youtube.com/watch?v=BRRolKTlF6Q I really don't like this video -_-

Ed Gorcenski

4:30 PM

I don't know why I didn't think of that; this book keeps you on your toes.

Old John

@EdGorcenski Yep - why not :)

Greg Ros

It's so stupid.

jdoe

I'm confused about this fractional ideal thing

Pilot

@GregRos if it is any real number than sqrt2 then it is 0

Ed Gorcenski

Also, I don't speak German either, just some phrases here and there.

I also don't speak Zulu ;)

jdoe

4:31 PM

He defined fractional ideals for a number field K, then started talking about fractional ideals in O_K the ring of integers

Old John

@EdGorcenski same with me and Zulu ...

I can say hello and goodbye in Zulu - but nothing in between

@EdGorcenski Which Rudin is it from?

jdoe

could that just be a typo?

Greg Ros

Wow I want to punch this guy in the face.

user19161

@GregRos Stay calm dude.

user19161

No punching is allowed in this chat.

Ed Gorcenski

4:34 PM

@OldJohn PMA

Greg Ros

Yeah but did you see his face. He has this obnoxious smile.

And by his logic you can say that $$\frac{1}{0^{2}}=∞$$

Old John

@EdGorcenski OK, Never really used that one, but I struggled through about half of Real and Complex Analysis - found some tough problems in there

Ed Gorcenski

I do like this book

Old John

@EdGorcenski I think he is a great writer

Ed Gorcenski

I don't find that his proofs are super slick like some people said; I guess maybe I find that they're slick, but not diabolically so.

I do enjoy the book greatly.

Old John

4:36 PM

but some people criticise PMA for doing all the topology definitions in one page

Ed Gorcenski

I prefer that approach. I looked at some of my older analysis books from undergrad

user19161

The proof of l'Hospital's rule is the shortest in Rudin.

user19161

Essentially he covered all cases at once.

Ed Gorcenski

And presenting topology stuff later makes the text schizophrenic

Old John

OK - have fun guys - back later

user19161

4:37 PM

Schizophrenia is often misunderstood for split personality.

user19161

They are two different things.

Greg Ros

Split personality is a fictionalized disorder for example.

Schizophrenia may make you believe cats are hunting you

Ed Gorcenski

I do have cats hunting me.

user19161

@GregRos Fictionalised? What do you mean? It does exist.

Ed Gorcenski

@JasperLoy Actually, it was recently found that the foundational literature on MPD was falsified by an overzealous psychologist, who was also having an affair with the patient.

Pilot

4:40 PM

Alright I hope my function was ok

anon

the foundational literature was the work of a single individual and a single patient?

Ed Gorcenski

his patient

Sybil (book)

Sybil is a 1973 book by Flora Rheta Schreiber about the treatment of Sybil Dorsett (a pseudonym for Shirley Ardell Mason) for dissociative identity disorder (then referred to as multiple personality disorder) by her psychoanalyst, Cornelia B. Wilbur. The book was made into two movies of the same name, once in 1976 and again as a television movie in 2007. Overview Mason is given the pseudonym "Sybil" by her therapist to protect her privacy. Originally in treatment for social anxiety and memory loss, after extended therapy involving amobarbital and hypnosis interviews, Sybil manifests ...

"The book is believed by Mark Pendergrast and Joan Acocella to have established the template for the later upsurge in the diagnoses of dissociative identity disorders."

Greg Ros

DID is real, what people refer to as 'split personality disorder' is fictionalized.

user19161

@EdGorcenski I have watched videos of these split personality patients before. And no, I don't have schiz or split.

anon

there is a difference between fictional and fictionalized

user19161

4:42 PM

Yes anon...

Ed Gorcenski

A 2006 study compared scholarly research and publications on DID and dissociative amnesia to other mental health conditions, such as anorexia nervosa, alcohol abuse and schizophrenia from 1984 to 2003. The results were found to be unusually distributed, with a very low level of publications in the 1980s followed by a significant rise that peaked in the mid-1990s and subsequently rapidly declined in the decade following.

Compared to 25 other diagnosis, the mid-90's "bubble" of publications regarding DID was unique. In the opinion of the authors of the review, the publication results suggest a period of "fashion" that waned, and that the two diagnoses "[did] not command widespread scientific acceptance".

anon

cool

but did does exist, I have seen him answer many questions on mse

Ed Gorcenski

That's from en.wikipedia.org/wiki/Dissociative_identity_disorder#History last paragraph; other text shows the historical link between DID and MPD/other names

user19161

@anon Haha, I was wondering yesterday if you=did.

Ed Gorcenski

The lack of scientific rigor in the psychiatric fields is mind-bogglingly terrifying.

anon

4:47 PM

too many variables ignored and unaccounted for

user19161

Hey @anon are you at work now?

anon

I work in a couple hours

I am unable to be online while at work

Ed Gorcenski

And the fragmentation of psychology into distinct "camps" is extremely harmful in general; I have submitted some research grants that included a behavioral psychologist (actually two, world class researchers both) as part of the team, and we just happened to draw a primary reviewer from a competing branch

Gigili

Hello, any Canadian here?

Garbage Collector

@JasperLoy : I sent you a new email, check it out!

Ed Gorcenski

4:48 PM

He eviscerated the proposal without regard to the scientific merit contained therein; in fact, he willfully ignored sections where we cited prior research to justify the proposal.

Greg Ros

Can you pretend to have factitious disorder?

user19161

@Gigili Mahnax and Mr Shiny are if you need any.

jdoe

@EdGorcenski, fascinating !!

user19161

@GarbageCollector What you sent me is not a secret, Herbert Voss is a well known figure.

Ed Gorcenski

The secondary reviewers gave us a solid average score (average of about 2, lower is better); primary reviewer gave us an average score of 7.

Garbage Collector

4:50 PM