Is there such a thing as Dirichlet series for trigonometric functions like cosine and sine?

Why am I getting so much chatter about using different symbols recently? first it's $\sigma_p$ and now $\Phi$

I define what I mean by the symbols. There are too few symbols not to have overlap.

Tradition?

@gmlime If you are still around, I have some recommendations

@OldJohn yep, thansk

For algebraic number theory, you might try Cassels and Frolich, or the book by Weil (it is called "Basic Number Theory", but is definitely hard)

@gmlime For either of those, you probabl;y need more algabra than you need for Ireland and Rosen (one of my favourite books on NT)

@robjohn Feynman wrote about using his own, more meaningful symbols, in math and physics :-)

@gmlime But if you want to spent as little money as possible, have you looked at the notes by Milne - all free online

I am thinking of asking if Dilip would please change his name. I know another Dilip and his usage might confuse me...

@OldJohn Thanks! I have the phonebook!

@OldJohn I'll look into both, I'll check out the google/amazon preview for Frolich. I'd love to get through a touch book with some grunt work.

@MattN. I am. Here.

touch=tough

@JonasTeuwen Excellent - hope you enjoy it :)

@gmlime Have you looked at the Milne notes - they go pretty advanced

@OldJohn It would be rather masochistic :-). I should learn it the easy way first and then go for the phone book!

@MattN Yes it is Parseval/Plancherel, however you would like to call it.

@JonasTeuwen Yep - I think it might be a book more useful for looking up particular topics - I would hate to think of anyone ever working through it linearly :)

@OldJohn Yep.

@OldJohn @JasperLoy seems to like that... but I am sure he will change his mind after he looked into something easier 8-)).

@JonasTeuwen I'm sure!

@JonasTeuwen Yay, bro, thank you!! : ) So there's nothing more to it?

I dig Fourier analysis I think.

Although it's a bit early to say something like that.

@MattN. I don't think so. Or I might miss something. It is a standard functional analysis theorem. Like in the chapter on Hilbert spaces. Probably chapter 0 or 1.

Very cool, thank you bro : )

@MattN. You better.

@JonasTeuwen : )

Ok, I'm going to prove the theorem now. See you later : )

(Might come back later with more Fourier analysis questions : ))

@MattN. Sure.

Lets see if this works: $\text{The Dirichlet function }\lim_{m\to\infty}\,\lim_{n\to\infty}\,\cos^{2n}(m!\pi x)\text{ is equal to one at the rational points }x\text{ and to zero at the irrational points.}$

@gmlime Hey your irrational points are in the comments!!!

They are truly acting irrationally :-D

I tried not to be too sarcastic.

@robjohn It does not sound sarcastic at all. Try harder next time.

I should have changed the variable name to $\text{Dilip}$

@robjohn Yeah.

@robjohn I could have been much more pointed, if I had want to be ;)

@OldJohn I am still considering asking Dilip to change his name :-)

@robjohn yes - to Phi or Omega, I hope :)

Hi guys

@robjohn I have no idea why a "retired professor..." would say such a thing. Who would be confused?

@gmlime Ahhhh.. the irrational points are coming at our (gr)avatars!!!

@Mats Granvik: I think I got that wrong.

@JasperLoy Just saw this message. Is it sarcastic? 8-).

Is there such a thing as a (sort of set) where it matters how many times a number is in it?

I wanted to use this to definde the lcm and the gcd

@ChuckFernández Multiset?

@ChuckFernández here

@OldJohn The least he could do is upvote the answer. The vote that I have is from JM, I think. It's like, "I don't like the plates you served my food on, but now that you've changed them, I wasn't going to tip you anyway."

Is there a notion of substracting multisets? ike in normal sets(the complement)?

@robjohn Yould could get some satisfaction by downvoting his question, maybe :)

@ChuckFernández Not sure - I never use multisets - I can usually get by with sequences, vectors etc and avoid them :)

Do you know how i could define the lcm of a number?

i mean the gcd, by using the prime factorization?

@ChuckFernández Just use the factorisations of $a$ and $b$ and use the minimum of each exponent?

what is a caftorization?

Wait, i got it, thanks, you meant factorization right?

@ChuckFernández yep - there is a standard formula for gcd using min as the exponent for each prime factor

Would OS X 10.8 mess with the printer settings again like 10.7 did? (did not work anymore...)

@ChuckFernández see equation 6 on this page

thanks

nice

@skullpatrol No problem!

lol

Yep. Is that statement hard to grasp?

They probably mean that there is so much water it can be dangerous, but the sign is too small for that. Perhaps they could make some Fermatesque remark: Sign too small for warning.

What else would be on the road during rain?

Your car, some worms,...

That drivers would have to be cautious about?

@OldJohn Thanks again, I'm trying to get to the a point where I can understand cutting edge research and the extremely difficult theories that each topic has with at most two books. For number theory I'd love to be able to verifiy the Shimura-Taniyam-Weil conjecture and the Fermats last proof.

@gmlime What level are you at so far with number theory? Have you done any graduate level courses?

I'm going to start a Ph.D. in two months time and I still can't understand some research done in the 80s, never mind cutting-edge research!

I was a physicist a few years ago , but only had a minor in applied math which wasn't proof based. I'm doing this self study because it is very enjoyable when I am able to solve the problems and finally understand the material.

@ZhenLin That is why your are going to do a PhD program right?

but but but

@ZhenLin I wouldn't worry about that

But... but... but... what?

@gmlime Once a physicist, always a physicist :-D

there's a thesis to be written!

If you already knew it you could skip that step.

Sure, but you have what 3 years?

In Cambridge.

(which is cutting-edge research, by definition, no?)

So about 1 year to figure stuff out and 2 years to write the thesis.

aaaaaah

panic

breath deeply

@ZhenLin Are you serious? :-). You are fully aware that that is the normal way of doing things right?

@gmlime I would expect to spend several years getting to the cutting edge then - and a lot more than just a few books

I've been told 3 years is ... rushed ...

Yes, it probably is, but not that bad.

If they thought you could not do it, they would not have hired you anyway.

^good point

That's probably true. Maybe. I know some students who took 4 years...

Four years is actually the common thing to do in many other countries.

Anyway, you will be fine. Just enjoy the time and suffering.

@OldJohn I'm compensating by specialization, Algebraic Number Theory, with a side of category theory and after that I'll have to figure out a good topological course. I'm also doing this to stay sharp since I don't really get to use my education in my day job.

So I hear. But I only have funding for 3 years. I think I read somewhere that they don't charge fees past the third year...

@ZhenLin Well, that is more than three years away. No need to worry about it right now 8-).

The situation could be way different in three years, so that would be lost energy.

@gmlime If you have a day job, then it is really going to take several years - despite specialisation

I suppose. Perhaps if I narrow my view a little I will be able to catch up with the research better...

I salute the students in algebraic number theory and algebraic geometry. They seem to have so much to learn, even without proofs!

back later - real life calls

later

Hi Jonas =)

Hi.

@ZhenLin Don't worry, you will be fine. You seem much more competent than many PhD students I know (but they are not in Cambridge, but some in Oxford).

Thanks! But seeming is easier than being...

Sure. Good luck. Do something where you cannot think that much else if you worry too much.

Good morning

hi

Morning Gnintendo, and later folks

later

I'm going to work soon...I'm working a 9 hour shift

:(

No offense intended :-D

The quotation is from Einstein.

Recall he worked at the patent office while writing his 1905 paper.

"Recall"? That is common knowledge in the commonest form.

even for a commoner

True, having to work is common knowledge in the commonest form even for a commoner.

I'm glad we're all on the same page.

Me too :-)

grammar twitch

Bizarre. I prefer to stay in my office, door closed and lights off 8-).

In front of the computer.

8-P

"fraction fields of a domain of integrity" :o

what a remarkable phrase

integral domain?

almost surely

odd

@skullpatrol Is there a deeper reason for most of your questions being removed?

@MichaelGreinecker work + play + keeping your mouth shut = success

@ZhenLin "domain of integrity" is an older term for various types of integral domains. See this question for some history.

interesting

indeed

If we agree to use the phrase "integral coefficients" then we should also agree to using "coefficients of integrity", no?

Hi all!

hi

is there any discrete math freaks that can help me understand a few things better? I am studying for a final friday

are there*

@treehau5 Freaks come out at Night

How hard is a simple induction?

@robjohn I would reply: "Nope.".

@JonasTeuwen I could ask if he needs his nose wiped.

@robjohn :P.

@skullpatrol I'm back.

I had a dentist appointment before work.

well I guess I am having a problem conceptualizing certain binary relations if anyone would be willing to give me a few pointers

9 hour shift I'm gonna have to start soon

I can't write papers while working! What is this

@robjohn It could be worse, he could have asked you to change the name of the variable p to something else because its not prime.

like a binary relation which can be explicitly written out as a subset I can find whether it is transitive, symmetric, reflexive, ect ect. I have difficulty when my instructor asks me questions like for xRy, x,y are elements of the set of all integers, x=/= y

or x = ymod3

ect

@treehau5 I might be able to help with this stuff for 15 min or so

@BillDubuque I should be thankful for what I get.

@MichaelGreinecker okay thanks man I will be here just ping when you have free time

I gained a badge for reading the FAQ? eyeroll

@treehau5 I have time now. What do you want to know?

ok

so I can memorize why he tells us the answer I just have a hard time conceptualizing or visualizing that

@treehau5 Visualizing equivalence relations?

when the sets are written out I can "see" it, but I guess when he gives us a relation like xRy, x = y+1 and he explains why it has certain properties, I get it at the time, but I couldn't for example, go back and figure out why for myself because I guess I just dont have that ability to conceptualize an entire vast set like that

@robjohn Show me the steps! Show me the steps! is surely the MSE version of Jerry Maguire. Maybe you should post an animation of dominoes falling....

I guess I need a much firmer grasp of just exactly what it means to be reflexive, irreflexive, transitive, symmetric, anti-symmetric, or asymmetric

@treehau5 Maybe start with a concrete example. Then we can look how you can come up with the answer yourself.

ok, does LaTeX work in here?

@treehau5 Kind of, there is a script robjohn has developed. Ican certainly read it.

ok well Ill just type it out. so the only examples I have are one he has already gone over in class, so Ill just vary it a little bit

ok

xRy, x,y are elements of the set of all ints, x cannot equal y

so he asks: is it reflexive? irreflexive? symmetric? anti-symmetric? transitive?

ints?

integers

sorry, im a programmer at heart lol

@Gnintendo "Badges"? we don't need no stinking "badges."

ok, so xRy if and only if x and y are different integers

yes

so where do I begin to answer those if I don't have anything to "look" at

I cant possiblity write out the *entire set. so there must be a conceptual way of solving

which is what I am not grasping

With the definitions. Start with reflexive and irreflexive. How are they defined?

do you want how I understand it, or how my book/teacher defines it?

hi @Gigili

Hi.

@treehau5 yes, and yoou need that to hold for every element. Does it apply, to the case a=-23 here?

so I understand it as, if you have an element in a relation, for it to be reflexive you have to have that element and itself in the relation

the book says: "A relation R on the set A is called reflexive if (a,a) exists R for every element a exists A."

@treehau5 yes, and yoou need that to hold for every element. Does it apply, to the case a=-23 here?

no because if x cannot equal y you cant have -23 and -23

that wouldnt be in the relation

@treehau5 perfect. you have just proven that the relation is not reflexive by showing that -23 is counterexample to the claim that zRz for every integer z. Let's look at irreflexivity

yeah it is irreflexive because theres never a case where it can be reflexive

@frank f(n)=f(n-1)^3/f(n-2)

@treehau5 your idea is right, but the formulation is not quite right. Reflexive means that zRz for all z and irreflexive means not(zRz) for all z. So the formulation "no case where it is reflexive" doesn't quite fit. But you are right, it is irreflexive.

Try symmerty now.

well it can be symmetric because the opposite is true for y. if you have x =/= y, then y=/= x either

(1,2) (2,1) is valid, and (15, 500) (500,15) is valid

But is it always the case? Is it always the case that when y is differnet from z, then z is also different from y?

yes because if it wasn't it wouldn't be in the relation

so the order doesnt matter between the ordered pairs, they will never equal each other

Exactly, so the relation is not symmetric. Is it anti-symmetric?

ok now I am confused. I would figure it would be symmetric

Indeed, but is it anti-symmetric too? The relation "=" is both symmetric and anti-symmetric.

ohh

There's a homotopy theory for set theory... :o

Don't get anxious, take a look at the definition of anti-symmetry.

ok in the book it says A relation R on a set A such that for all a,b, in A, if (a,b) exists in R, and (b,a) exists in R, then a=b is called antisymmetric

This is again a statement of the form "for all". It is wrong if you can find a single case for which it does not hold. So pick an example of two integers x,y such that both xRy and $yRx$ hold. Is it true that x=y?

no x cannot equal y

Yes, so the relation is not anti-symmetric either. Asymmetry should also be easy. Let's try transitivity.

@robjohn Hah, frieze groups.

@JasperLoy If you follow the links back, you get to here

ok so I understand transitivity as the mathematical way of showing hypothetical syllogism

the book defines : A relation R on a set A is called transitive if whenever (a,b) exists R, and (b,c) exists R, then (a,c) exists R, for all a,b,c, in A

@J.M. is it a cyclic group? That is, do we end up where we started?

so using an easy example of 1 2 3,

(1,2), (2,3), (1,3)

For this example, it works. Maybe you can try a different one. Note that a,b,c in the definition do not have to be all distinct.

@robjohn It's periodic...

@J.M. The tango touches the dance floor at just one point...

hmm

so you are saying I can have a b and c all equal 1?

symmetry groups of periodic things can be noncyclic if they have noncommuting symmetries

Oh, no!!! a basis of open disco balls!

also, while products of cyclic groups are cyclic in the finite case, I don't think that's true in general

heh

@treehau5 In general, yes. But of course 1R1 does not hold. But you can have a=c.

oh okay

so (1,2) and (2,1), (1,1) but 1,1 can't be in there

My rep is a full house of 3's over 4's

Exactly. Hence?

its not transitive

Exactly.

ok well that was a pretty easy example I get really tripped up when he does a "divides" relation or a modulus relation

It is a hyperbole? Oh. Nice :-).

@JasperLoy Why should it be disliked because of that?

@JasperLoy Hmm, but sometimes it is just better because it adds some structure.

Can be a good insight.

@JasperLoy It's called fun with numbers.

@JasperLoy Huh? But metric spaces are more general!

@treehau5 Consider the following rule: If a relation is defined so that to elements are in the relation iff and only if the are equal with respect to some property, then the relation is an equivalence relation: reflexive, symmetric transitive. This applies for example to the modulus case. Try to reprove that the relation "divides" between natural numbers is transitive.

@J.M.: that is such a plain gravatar. Are you feeling all right?

@JasperLoy Not necessarily. One can eventually conclude something about the structure which would basically say it is a Banach space already.

@robjohn I'm not quite done with coloring my next feature, so consider it a placeholder... :)

@J.M. Okay :-) I didn't mean to be pushy, just curious.

We also have the other way around :-).

We would have gradient flows in Hilbert spaces (easy).

So we would now want to generalize to Banach spaces... turns out that does not work and we need a metric space.

@MichaelGreinecker I am not understanding the question

Then you will get it rephrased in the EVI (Evolution Variational Inequality).

@robjohn s'okay. I needed to molt, but I wasn't satisfied with my last few coloring attempts...

@JasperLoy I do not care how other people do it. I do it the way I think is the best. If I turn out to be wrong: fine.

(Who knew consistently producing quality art was so hard? :D )

so if a divides b and b divides c, a can divide c as well

@J.M. Renoir, Gaugin, Picasso, ... ?

@treehau5 Can you prove it formally?

I am not :-(.

well I see that as always being true

@JM You would be the man if you could actually color your formulas that way you would see a troll face.

@OldJohn Yes, that was rhetorical. ;) See, even those guys have bad days...

@JonasTeuwen I am surprised that you did not chime in on this question.

@JonasTeuwen I'll think about it, but I somehow find the "troll face" repulsive.

This is crowded!

@J.M. Sorry - I have a bad habit of mis-interpreting things literally :)

@J.M. Something else is fine too. Van Gogh?

@treehau5 Can you prove it formally? Note that the natural number a divides the natural number b if and only if there is a natural number n such that a*n=b.

@JasperLoy Nope.

@robjohn The main page makes me feel misanthropic.

@treehau5 Use some equations.

if a divides b, and b divides c, that means b= a * some number k, and c = b* some number j, you could say c=a(k*j)

@JonasTeuwen I don't feel like cutting my earlobe just yet. ;)

@PeterTamaroff "active" is the word :)

@JM Need some bipolarity?

@JonasTeuwen Why is that? and where do you get questions to answer?

@JonasTeuwen No thanks, I'm content with the mental disorders I already have. :)

@robjohn I did not answer them in a long time, I just look what people put here.

@treehau5 If $a\mid b$, then $b=k\times a$. If $b\mid c$ then $c=k'\times b$. Now we have that $c=k' (k\times a)$, so....

its transitive

(Why did I know somebody would star that... :D )

@treehau5 Correct. I have to leave now. I hope that was of a little help.

@treehau5 Right, so that $a\mid c$ becase $c=K \times a $; where $K=k\times k'$

@MichaelGreinecker yeah most definitely thanks brother

@J.M. stars make it dangerous to state some things in this room ...

I wonder which of those mentality thingies actually help in learning things.

@JasperLoy yep

@OldJohn Oh well, I'll leave it.

@JasperLoy agreed :P

Perhaps some onset schizophrenia can help (Nash?).

Natural LSD, yeah.

Did Erdős have any such disorders?

That is the nitpick disorder, a common comorbidity to the GND.

Hm, I bet not. Perhaps some personality thingie.

@JonasTeuwen Just when you think you already have enough voices in your head... :D

I mean disorder means that it actually has to interfere with your normal way of being.

@J.M. :D.

@JasperLoy I have to prove that $f:(X,2^X)\to (Y,\mathfrak I)$ is always continous, as is $f:(X,\mathfrak I)\to (Y,\{X,\varnothing\})$

@JasperLoy ...where X is any of the set {hysteria, OCD, ...}

GND is a form of OCD. You compulsively try to correct other peoples grammar. Even if it is correct already.

Female hysteria!

I think haloperidol will do splendid work in the treatment of GND.

As a start I suggest 20mg daily.

@J.M. I think you're a donut addict.

@JonasTeuwen The depot injection would be best. Lasts long and all that...

@PeterTamaroff Sure, I like tori. :)

Or at least, you have a torus fixation.

@J.M. And if that doesn't work I think adding some cytarabine will.

@JonasTeuwen ...that's for tumors.

@J.M. Yep.

@JonasTeuwen Hmm... :)

@JasperLoy Yes, sure.

@J.M. No time for whining when you have that stuff.

@JonasTeuwen Oh, certainly.

"So, now you know what feeling miserable really feels like eh?"

That's quite soft.

@JasperLoy That's sad. But those brain tumors tend to occur younger than older, I've found.

My mother her best friend has a tumour most neurosurgeons said: oh man. Then this experimental neurosurgeon said: retractive stereographical surgery!

And then she died anyway.

@JasperLoy Yeah, hypochondria. Very common case of brain "tumours".

Nope, I mean, lots of people think they have them. But it actually quite hard to miss if you really have them...

Perhaps not for small ones, but they will not do an MRI for that anyway.

Anyway, those thingies are quite agressive if I remember correctly. Between 0 and death there is often less than a year.

@JasperLoy Yeah, I imagine.

Well, apparently it is not trivial for Peter.

@JasperLoy But maybe some of you guys had something interesting to say.

@JonasTeuwen Very.

So if you get those: stop worrying.

You should do that anyway, by the way.

I love statistics.

Oh bloody monkey, the closest printer is like 40m away. Somebody might see me... 8-).

@JonasTeuwen Is that really so bad?

@OldJohn No, but I don't feel like that for the moment. 8-). I'll pick it up anyway. Guerilla style!

@JasperLoy Not hairy enough!

@JasperLoy for that, you're supposed to have hair even in the places not shown in polite company...

Hi everyone.

@MattN. Hi, one. :)

@J.M. I like your new avatar. : )

@MattN. Thanks. I went with a candy-ish Gravatar for the time being; my last coloring attempts on what was supposed to be this week's Gravatar went awry.

What's with the disorders here? I think most of the people in here are normal.

@MattN. ...whose definition of "normal"? ;)

Anyone else think that the mathematical connotation of "normal" being at right angles to all the others is a bit odd?

@J.M. Mine. : ) And since I'm normal, I can measure it well enough.

@OldJohn I'll admit not knowing the etymology for that...

@OldJohn It's at a right angle to the usual meaning of normal...

@J.M. Yeah - just looking for it myself

@MattN. yep

does google books have math books?

@ChuckFernández yes

do they have a decent collection?