Mathematics - 2013-01-30 (page 1 of 3)

Q: The Mysterious Kalva (Looking for data on John Scholes and kalva.demon.co.uk)

@pre-kidney I'm assuming you're talking about this question

0

When I started off with contest math, the site to go to was John Scholes' kalva.demon.co.uk; without fail, his succinct solutions to challenging problems inspired me. Anyhow, the site only exists in archives now, and other forums are much better learning sources for aspiring problem solvers. So...

MJD

I think this might be my favorite se.math comment ever: math.stackexchange.com/questions/288614/…

It is certainly one of the drier ones.

I (obviously) wasn't involved in closing this question (as I'm nowhere near that rep level). However, the question you asked was fairly open-ended. How would you determine a correct answer? (is there such thing as a wrong answer to that question?)

On SE, we like questions that are "answerable," so they have a "right" answer. Questions that are simply matters of opinion, or have no wrong answer, typically aren't recommended.

If you phrased your question in the form of, "What books has John Scholes written?" or similar, it might get a better reception. However, even that would be a fai…

Miles Alden

Is anyone around who can help me understand the math behind finding the normal vector of a plane? I'm working off this and there are just some symbols I don't understand ----> gamedev.stackexchange.com/questions/27785/…

3:16 AM

@MilesAlden I can help...

What symbols do you not understand?

@math101 Did you recently change your profile pic? For some reason, it's not displaying on my computer (Some image hosting sites are blocked on my network)... i was just wondering if the missing pictures I'm (not) seeing are because of that.

@MilesAlden This may actually ping your network acct, idk. Just saying again that I could probably explain the normal vector stuff to you...

Miles Alden

3:54 AM

Thanks anorton, I had another chat going on the same questions and also had to put my daughter to bed. Thanks for answering though!

4:13 AM

@anorton I changed my profile pic a lil over a week ago

Hi guys/gals.

Hey @OrangeHarvester

Are you by any chance related to Gustavo?

@math101 No, no . I'm not. He is in Brazil, I am in India. :-)

@math101 Wassup? How was your day?

It was not bad :) I am the laziest kid eva. I cant get myself to focus :\

@math101 Doing a job AND a course is tough. Hell, I find it difficult to just study, sans any job. :-)

4:18 AM

@OrangeHarvester the job is stress free. But it just wastes a lot of my precious time which I cld use for studying

@math101 Yes.

I can take off from work if necessary. I think I will do that on Thursday.

Hmm.

mhmm.

hahaha

oops, I thought you did not want to answer, so I deleted it. :P

4:22 AM

I will answer it just thinking what to say cuz its public

Its okay. I would rather not have an answer, than an abridged one which might give an incorrect idea. :-)

I chose to do it this way

Okay. Nice. :-)

I dont pay for my studies but I work for my dad and ya

I see you are studying abstract algebra too.

4:31 AM

Nope. I will be studying it next semester

Okay.

I took a course called Transition to Higher Mathematics and it covered a chapter or 2 on Abstract algebra

Oh. I see.

I am not that great at math if you cant tell :\

Well, I suck, so.... :-)

4:34 AM

I am gonna be heading into finances \ business after i finish with this stuff

Okay. When will you finish?

I hope to finish by September

Okay, I thought you were a full time student, 4 year course and all.

Well I did one year at a local college and then it will be 2 years online

so that would be 3 years of studying

Okay. You studied all in math? Or did you take different courses?

4:39 AM

Well my major is math

So a chunk of my courses are math

OK I am heading to bed early. I am soo tired and cant get myself to focus. Gnite

Good night.

5:25 AM

sup

<3

Not much

wat about you?

just got outta chem lab

cyclopeptane!

hehe. nice. What was the assignment?

determine the unknown liquid

by obtaining its mass, boiling point, and solubility

okay. I liked organic chemistry in high school. Morrison Boyd really brings the subject to life.

5:36 AM

@OrangeHarvester you're my age, where do you go to school ?

@RustynYazdanpour I graduated in electronics communication engineering from VNIT, India. I am now trying to get into graduate school in mathematics.

same.

Same?

i'm trying to get into grad school at the moment as well.

I have to study for the subject test.

Okay. Mathematics?

5:42 AM

yeppers

@RustynYazdanpour Ahh cool. Where are you from?

I am from Idaho

have you seen Andres Caicedo on this website?

I studied under him

no, I have not. I will see.

and his contemporaries

Idaho is in which state?

5:43 AM

boise, ID

idaho is the state eh

okay.

Idaho is a state? hehe.

;)

this is my last semester, i finished the math program and the spanish program

i am taking all the lower level shit I put off

like all 100 levels

5:57 AM

triple major

BDub

What kind of classes do you have to take during your last semester, @Rustyn? Like general education stuff?

yeah triple major

I have to take general chem, physics with calculus, philosphy of morality and english literature

8:48 AM

@MichaelGreinecker: hi, how are you?

@Ilya I'm good. I just drank coffee and am alive now. How are you doing?

@MichaelGreinecker didn't do the coffee trick yet 6_6

@Ilya I see.

@MichaelGreinecker tried to look into "Stochastic relations" - need to fight with a language a bit

@Ilya It's a different world and culture I guess.

8:52 AM

@MichaelGreinecker well, I'm familiar with computer science papers since I'm doing (approximate) bisimulations

languages, huh (really need coffee)

@Ilya What do you use them for?

@Michael Are you familiar with model-checking?

@Ilya I'm familiarizn ing myself with wikipedia right now...

:)

the key idea is to do an algorithmic analysis of models, so each model-checking problem has two ingridients: 1. a model and 2. a property

the task it to verify whether the model satisfies the property

I got that.

8:56 AM

models are: Kripke structures, ODEs, SDEs, Markov Chains etc.

Sounds more familiar.

properties are usually expressed using some modal logic

e.g. with a formula like
$$
\mathsf X^t \alpha\vee \lozenge \beta
$$

Okay.

this one means that at time $t$ the trajectory visits the set $\alpha$, or eventually it visits the set $\beta$

for Markov Chains, of course, one is interested in a probability that such event happen (given an initial state/distribution)

And that's what you do?

8:58 AM

not yet :)

in finite state spaces all these problems can be solved automatically

for infinite state spaces, clearly - no

Which makes life fun, I guess.

so given a logic or a class of properties, a precision level, and a infinite model I'm interested in finding a finite approximation thereof

@Ilya I remember your questions about the distance of random paths induced by Markov kernels...

e.g. if $\mathscr C$ is the class of event I'm interested in and $\mathsf P$ is the probability measure of the original process, I'm looking for a finite process such that its measure $\tilde{\mathsf P}$ satisfies
$$
\sup_{\mathscr C}|\mathsf P(C) - \tilde{\mathsf P}(C)|\leq \varepsilon
$$

when I came to Delft, they were only doing this for $\mathscr C$ being a single event on the finite time horizon

so I worked in both directions: the infinite horizon case and the approximation over the whole class of events

@Ilya The approximation is over the whole $\sigma$-algebra?

9:04 AM

well, over a given horizon

@Ilya Sure, sure. Do you employ some kind of metrics for kernels?

of course - TV

now I'm looking into connections with the Prokhorov metric, as people who came from deterministic systems construct approximations using distances between trajectories, not measures

super-conservative results, you may guess

@Ilya Crauel has something in his book on this, right (note to myself: convince boss to order it).

@MichaelGreinecker which one? never heard of this guy

@Ilya "Random Probability Measures on Polish Spaces"

9:09 AM

@MichaelGreinecker something on this = ?..

on what?

@Ilya It does weak convergence theory for kernels. The motivation comes from stocastic dynamcal systems.

@MichaelGreinecker nice, I shall check it. The point is that I'm not that interesting in purely convergence results - rather in the inequalities

@Ilya Sure, sure. But the book treats stuff like metrizability and separability of weak-convergence of prob kernels and these ideas might be relevant.

sure

I know the feeling.

9:17 AM

@Michael btw, I've recently got a paper accepted on this topic - there I make a gentle but formal introduction (even acknowledged be reviewers), so if you're interested I can send you a preprint

@Ilya Yes, I'd like to see it.

@MichaelGreinecker I'll do that. Btw, I actually liked very much the ordering based on plays rather than on outcomes

@Ilya Thanx!

since outcomes seemed always to me an inessential part for the concept itself (now I'm talking about games)

and the concept of the measurability w.r.t. partition is very natural given a well-known theorem about the existence of a measurable map $Y = f(X)$ whenever $Y$ is $X$-measurable

which makes me think that measurability accomplishes two tasks:
1. the structural one: avoid dealing with weird sets
2. the informational one: relates level sets of measurable maps

I wonder, if these two features are only seemingly different

also relates to the question by Tim on which structures do measurable maps preserve/reflect

@Ilya Yes. Of course measurability with respect to partitions is not my invention. The connection between modeling information with partitions and $\sigma$-algebras is not that straightforward. There are some subtle issues.

9:30 AM

Hi folks. My class has gone over logic and I've been exploring the homework. I just barely got comfortable with some logic and was starting to enjoy it somewhat, then I ran into this:

Show that (p^(~q)) logically implies ~(p^q).

In this case, what is "logically implies" supposed to mean?

Do I set them in an equality and try to get them to match, or what?

@ilya Sorry, my connection went weird. I'm not sure Tim's question can really be answered.

@MichaelGreinecker mmm, any of them? :) at least of the last ones

I guess, you're asked to show that
$$
p\wedge (\neg q) \rightarrow \neg(p\wedge q)
$$
in a sence that whatever values of $p$ and $q$ you put there, the formula is true

@Ilya It might work in some nice subcategory. Like taking standard Borel spaces as objects, but factorization does not work in general afaik.

@Ilya Oh ok, so I put it like that then attempt to reduce it to a smaller form and make sure it's true for all values in a truth table?

@Yep yes, you can use e.g. De Morgan rule

9:36 AM

@Ilya Thank you very much! That helps a bunch.

@Yep nice :)

@Michael: sent, notice the name of t.b. in the Ack. Section

@Ilya Sent to printer.

the main result there is Lemma 1, otherwise it's just an introduction of a new (for guys in my field) framework, discussion on the well-posedness of the problem (never been given before) and refinement of known results to the new framework

@Iya That looks great. Only the CS-typical amount of abbreviaion scares me a little. ;-)

@MichaelGreinecker :p

I really tried my best and was polishing this paper a lot :) at the same time, I submitted a much weaker (technically) and less polished paper where I do also control synthesis - and reviews very much more enthusiastic.

So, I don't have doubts what's appreciated in my field

9:51 AM

@Ilya So the official focus is more on CS than Probability?

Is there anybody familiar with Lebesgue's theorem about the necessary and sufficient condition of Riemann integrability?

@Frank Discontinuities form a set of measure zero.

@MichaelGreinecker How many ways do you know to prove that?

@Frank None, but I can look it up.

@MichaelGreinecker Hi. Have a minute for forcing?

9:57 AM

@matt Sure.

Suppose $f$ is bounded. $f\in\mathfrak R\iff\forall\epsilon,\eta>0$, there's a partition $P$ such that $\sum_{M_k-m_k\ge\eta}(x_k-x_{k-1})<\epsilon$.

Cohen's forcing?

This is a very dumb question and I'm sorry, but I seem to have forgotten my algebra. If anyone has a quick second, what is the property by which $$3^{nlog_{9}n}$$ is equal to $$n^{n/2}$$?

@MichaelGreinecker Aces. In Farah&T. on page 1, : "Observe that for countable $\mathcal D$ a $\mathcal D$-generic filter always exists; it can be constructed by an easy induction." -- I think I know the inductive proof. But why does it have to be countable? Does transfinite induction not work?

Or better: why can't we apply transfinite induction to prove it for arbitrary $\mathcal D$?

$3^{n\log_9n}=3^{\log_3n\cdot n/2}=n^{n/2}$

And you got from log_9 to log_3 using change of base, which produced the 2?

10:05 AM

$\log_{b^m}n=(log_b n)/m$

@matt I am on it.

A special case of $\dfrac{\log_b n}{\log_b m}=\log_m n$.

Thank you @FrankScience. I must be quite rusty as I seem to have completely forgotten that.

And here I was about to start taking the derivative. Simplicity always wins. ;)

@MichaelGreinecker in a way

@MichaelGreinecker You can find a proof for the countable case in Halbeisen on page 264, proposition 13.1.. I need a holiday.

10:14 AM

@Matt Actually, the issue should be what additional properties the filter should have. The whole poset is a filter that trivially intersects every dense set...

Daniil

Hi!

@Ilya In a way?

@Daniil Hi, long time no see.

@MichaelGreinecker well, there is a community doing this stuff which is partly from CS, partly from Systems and Control theory

Daniil

Indeed :)

10:16 AM

@Ilya I see, I see.

@MichaelGreinecker Very well observed. But nothing in Halbeisen seems to exclude this trivial case.

@MichaelGreinecker You should've written "ic ic" : )

@Matt I usually do.

Because Ilya does, too : )

if you write one more I, you will have the name of a bank.

ICICI. :P

@Matt I guess one needs countability because one can be assured only that finite sets have lower-or upper bounds (depending on which one picks).

But I have to talk to some folks in rl now. Later folks!

10:21 AM

See you later! Nice talking, will try to work this out.

@MichaelGreinecker see you! sorry, been away for a couple of minutes

10:50 AM

Where should I look to learn about how to simplify sigma notations like the following? $$\sum\limits_{i=1}^n\sum\limits_{j=1}^i\sum\limits_{k=i+j}^{2n}1$$

1:42 PM

Look here: http://en.wikipedia.org/wiki/Summation#Some_summations_of_polynomial_expressions

$$\sum_{i=1}^n\sum_{j=1}^i\sum_{k=i+j}^{2n} 1 = \sum_{i=1}^n\sum_{j=1}^i\left(-(i+j) + \sum_{k=1}^{2n} 1\right)$$

@anorton you've broken the silence...

@Robjohn sorry... :P

I didn't know you were in here...

@anorton That would have kept you from posting?

@robjohn No. But I thought everyone had left the room, and was surprised to see a response so fast

@anorton Ah, makes sense.

skullpatrol

1:47 PM

>8(

We are just zombies.

@skullpatrol what?

Good grief! everyone came out of the woodwork...

@skullpatrol That should ping me, but no...

4

:)

@anorton Unless @robjohn knows of more powerful methods, the only ones I know is solving the stuff simplifying the inner most summation and moving outwards, unless there is some very obvious connection I can see, which may make me interchange the order of summation or something.

1:51 PM

@OrangeHarvester
True. I was responding to Yep's post--I can simplify the triple sum

Ahh, your and Yep's Gravatar are too similar. I thought you were reposting.

@orange Ah. ok. I should probably change my gravatar, but I kinda like the geometric nature of it...

@anorton Hmm. Even Ilya's gravatar is very similar. :P I think, there is something wrong with the automatic generated gravatar's. We need better algo's to generate them, so that we can have sufficiently different ones.

Mats Granvik

2:14 PM

@draks Thanks for the link to your question, looks interesting. I am not sure what you have done, but you seem to have some interesting expression for the prime zeta function linking the logarithmic integral function and the Moebius function.

@draks... Not sure how to ping you. Maybe this works.

The ellipsis directly after the name does not ping people, AFAIK.
@MatsGranvik...

user19161

2:34 PM

@robjohn Wow, that got 3 stars! Today is free star day!

2

2:53 PM

hey guys, stack isn't letting me post a discrete math proof question. anyone free for a second wanna help me out?

That is weird

I know

pastebin.com/BEhkbjc1

Anyone free?

hmm you cant prove with an actual problem you need to prove using a general statement

Unless its a counter example

actually that's the wrong problem

one sec let me post you the right one

Thank you, Jason

Here is the problem pastebin.com/30H32m7T

user19161

Wait the original problem, are you sure you know how to do it?

2:58 PM

gyazo.com/f75712a97c3e1938daca780023eb5969 there is table 1

the one i just posted?

i solved the one i posted earlier

i'm just having trouble on this last one

user19161

Are you sure you solved the first one?

The first one you misunderstood what A-C means

user19161

Because if the solution there is yours, then you don't quite understand set operations.

where did i post a solution?

oh on the reddit?

yeah that was the wrong answer

user19161

@jtm22 But do you know how to do the first problem now?

3:02 PM

yeah

the first problem i figured it out

it's this other one that i'm having trouble with

user19161

OK, and was the original solution there yours?

no, i was working with someone

but we figured it out now

user19161

Ah OK. Anyway you are now trying to prove that set intersection is associative.

user19161

It's really straightforward. To prove that two sets are equal, just prove that each is a subset of the other.

can we do that by example?

user19161

3:04 PM

To prove that X is a subset of Y, show that each element of X is also an element of Y.

user19161

@jtm22 No, you need to give a general proof. Examples are only for illustration or intuition.

Damn...

Umm

err

user19161

Have you done any proofs before?

we don't do formal proofs, per se but we do do them

user19161

Oh well, if you had proven correctly the first problem, the second one is much easier.

3:07 PM

i'm just having difficulty understanding this one and planned to go in for office hours but it was cancelled on me last minute

user19161

So I am guessing that you did not get the first one right either.

no, i believe we did

want me to post it?

user19161

Yes.

pastebin.com/JzwtJzvv

Charlie

Meh

3:10 PM

just can't figure out the first one now

even if this one is right haha

any help if we work off this one?

what is the meh all abt @Charlie

Charlie

@math101 hi.

user19161

@jtm22 It looks OK except for the phrasing.

@Charlie Hi -_-

user19161

Just don't say X-Y consists of all elements of X but not Y.

user19161

3:12 PM

Rather say that X-Y consists of all elements in X but not in Y.

that sounds better

thanks for that advice

user19161

And the second proof is really way easier than the first.

user19161

Let x be in (A intersection B) intersection C.

user19161

Then x is in A intersection B and also in C.

Charlie

@JasonBourne hi Jasbear

user19161

3:14 PM

Hence x is in A and B and also C.

user19161

So it is in A and also B intersection C.

user19161

So it is in A intersection (B intersection C).

user19161

Do the other direction similarly and you are done. QED

the other direction similarly?

user19161

Yes I am telling you to do it yourself.

3:16 PM

isn't it just those lines though?

5 line proof? haa

hahah

user19161

Well, you know why the opposite directions look so similar?

user19161

Because this proof is really simple.

user19161

But sometimes, the opposite direction looks very different.

OH i didn't even realize

ok let me try now

wait i'm confused

i think you meant to do it opposite, but it's not?

user19161

3:18 PM

Confused? Good.

lol

user19161

If you are confused, it means you are thinking.

How does it deviate from ->

user19161

So the opposite inclusion goes like this dude.

user19161

Let x be in A intersection (B interesection C)

user19161

3:20 PM

SO what's the next line?

@JasonBourne Whats up?

user19161

@math101 Haha! Nothing's up. Everything's down.

Let x be in A intersection( B intersection C).
Then x is in B intersection C and also in A.

user19161

@Charlie Hey M.

user19161

@jtm22 Yes, go on.

3:22 PM

Hence x is in B and C and also in A?

user19161

Yes.

so then next..

So it is in B and (A intersection C) ?

user19161

Well that doesn't help you prove things does it?

user19161

You need to work towards the end result.

user19161

So you need to group A and B together.

user19161

3:29 PM

@jtm Are you here? LOL

@JasonBourne Not quite :P

user19161

Oh noes, you have been kidnapped!

user19161

@math101 Erm, how do you know? =)

yeah

i'm here

so

the next step would be

So it is in A and B intersection C?

rather

C and A intersection B

?

Ok I gotta go Jason. Speak to ya

user19161

3:31 PM

@math101 See you!

user19161

@jtm22 Yes.

So it is in (A intersection B ) intersection C

QED?

user19161

@jtm22 Yes! See, trivial.

thanks Jason!

user19161

And because of this we can write A intersect B intersect C without parentheses.

user19161

3:34 PM

There is no ambiguity.

user19161

Similarly we have associativity of addition of numbers.

user19161

So we can write 1+2+3 without caring about whether it is (1+2)+3 or 1+(2+3) @jtm22

Thanks a lot man

really help

$$
\begin{align}
\sum_{i=1}^n\sum_{j=1}^i\sum_{k=i+j}^{2n}1
&=\sum_{i=1}^n\sum_{j=1}^i(2n-i-j+1)\\
&=\sum_{i=1}^n\sum_{j=i}^{2i-1}(2n-j)\tag{$j\mapsto j-i+1$}\\
&=\tfrac12\sum_{i=1}^n(2n-i+1)(2n-i)-(2n-2i+1)(2n-2i)\\
&=\tfrac12\sum_{i=1}^n(2n-i+1)(2n-i)-(2n-2i+2)(2n-2i)+(2n-2i)\\
&=\tfrac12\sum_{i=1}^n(2n-i+1)(2n-i)-4(n-i+1)(n-i)+2(n-i)\\
&=\tfrac12\sum_{i=0}^{n-1}(n+i+1)(n+i)-4(i+1)i+2i\tag{$i\mapsto n-i$}\\[6pt]
&=\tfrac16(2n+1)2n(2n-1)-\tfrac16(n+1)n(n-1)-\tfrac46(n+1)n(n-1)+\tfrac12n(n-1)\\[12pt]