Mathematics - 2012-01-16

N3buchadnezzar

00:48

Hi!

Srivatsan

01:25

@robjohn Yes. I was just back.

robjohn

@Srivatsan was it good?

Srivatsan

Hm, never thought my understanding of probability would be so lacking... :-/

robjohn

In what way?

Srivatsan

@robjohn Yes, it was... good... Kind of... =)

@robjohn I cannot follow this answer at all. :-)

@robjohn: I am asking this only to remind myself and you (and not to hurry you or anything): Did you have a chance to look at the asymptotics of the error in the $L^1$ approx. question?

robjohn

@Srivatsan The error, not the function? I think it was $2^{-n}$ times the error in the $L^1[-1,1]$ case.

I haven't worked it out, but isn't the error in the $[0,1]$ case documented?

Srivatsan

01:42

@robjohn "documented" -- Where is it documented?

Does David give an estimate for the error? I don't think so.

@robjohn Did you mean the $L^{\infty}$ error...?

@robjohn Yes, you guys solved the function part, no? :=)

robjohn

@Srivatsan yes, but I thought that there was literature on this in the $L^1$ case.

If I can't find anything, then I should find the norms :-)

Srivatsan

@robjohn Someone commented that this should be a standard problem and should be in any book on approximation theory. But sadly, no one suggested any literature... :=)

robjohn

I have been doing a short bit of research and all of what I have seen has been $L^\infty$

Srivatsan

@robjohn Yep, $L^1$ seems to be the ignored step-child.

robjohn

01:58

@Srivatsan Well, then I should work out the norms for the $n^{\rm{th}}$ degree case

Srivatsan

You mean the $L^1$ norm of the polynomial you both wrote down? [The Chebyshev polynomial of the Xth kind.]

robjohn

@Srivatsan Yeah.

David Speyer still has not replied to my comment on his answer.

Srivatsan

=)

robjohn

Chapter 6.4 talks a bit about the $L^1$ case

I don't think it computes the norms.

Srivatsan

One second. I will be back.

robjohn

02:09

I gotta go anyway to take Lilly to the park.

bbl

Srivatsan

@robjohn Have fun

Sivaraman

05:08

@Srivatsan Are you here?

Srivatsan

05:20

@Sivaraman I am at school now

Martin Sleziak

07:02

@Srivatsan I am not sure, either. That tag could be useful.

Srivatsan

Yes, I felt so too. Let's just keep it then.

Martin Sleziak

Not sure if someone visits that room, but you could post that message in Jury Duty, so that it would be more visible.

If you find it important to get some opinions from other people.

Anyway, the usual people for tag talk were robjohn, J.M. and t.b.

I guess robjohn would have written something, if he wanted to. The other two are not here.

Srivatsan

@MartinSleziak Two of them are not seen in chat these days... :/

Martin Sleziak

Exactly.

But as you wanted my opinion, I'm ok with keeping the tag.

Martin Sleziak

07:15

I've noticed that you add a lot new stuff to your answer using Stolz-Cesaro, Srivatsan. (The last time I saw, there was method 1 only.

Method 2 is very similar to the proof of Stolz-Cesaro in general, right?

And that generalised Cesàro theorem is a special case of Stolz-Cesaro, isn't it?

Hmm... not sure about the last think I wrote, perhaps I was too quick with that conclusion.

Srivatsan

@MartinSleziak I did say so in the beginning of my answer. :) "however, note that if we unroll the proof of Stolz-Cesàro, the resulting proof is essentially the same as the second approach." =)

Martin Sleziak

I should have read more carefully.

Srivatsan

@MartinSleziak True. ("This theorem can be proved by mimicking the above two approaches, so I will skip the proof." ;))

Martin Sleziak

Nice answer.

I've upvoted already first version.

Srivatsan

Ah, Thanks.

Martin Sleziak

07:23

Ok, I guess I'll have to abstain from MSE today. (I've came because of the notifications in my inbox.)

Have a nice and productive day!

Srivatsan

@MartinSleziak Ah, thanks for that, Martin.

Later. Bye!

Asaf Karagila

07:46

Good morning to you, folks.

Srivatsan

08:18

Morning, Asaf, Kannappan.

Kannappan Sampath

Good Morning, Srivatsan

See this. Does this require notation

Srivatsan

@KannappanSampath Think so.

robjohn

08:51

Good morning!

@MartinSleziak Is there something I should be looking at?

Srivatsan

@robjohn We were discussing this.

robjohn

roots-of-unity?

Srivatsan

(That comment of mine goes on for the next 3-4 lines.)

@robjohn Yes.

robjohn

I think it is a good tag.

Srivatsan

OK. Then we need to populate it. (Eric will get the Taxonomist badge now. :=))

robjohn

08:56

I see no cyclotomic

So we should have something like it

Srivatsan

@robjohn Sounds like a good tag, but I am not completely sure.

robjohn

@Srivatsan Do you have expressible concerns?

Srivatsan

Over a 125 hits.

robjohn

or just uncertainty?

Srivatsan

@robjohn No, nothing of that sort. // Since you ask: yes, just uncertainty.

robjohn

09:01

cyclotomic and roots-of-unity should be related or one should be a synonym

we don't need both, but I mean the topics are related.

Srivatsan

I will pass on giving my judgement on that. // On an administrative note, there is nothing called "related tags", is there?

robjohn

other than synonyms, I don't think so.

Also, I see that you've added a bounty to the AM-GM-HM Triplets question.

Srivatsan

@robjohn Yes. =)

robjohn

and that you've accepted my answer :-)

Srivatsan

@robjohn Yes, again. :)

robjohn

09:05

along with that it got 3 more upvotes.

probably because it got percolated back to the top of the list.

Srivatsan

@robjohn Yes, it cannot be coincidence, I think. I got 3 upvotes too.

+2 each for David and the other answer.

robjohn

I noticed that during the Holiday break, there were a lot fewer upvotes. After term started again, things started going again.

Srivatsan

I cannot give away my bounty yet. Apparently I must wait for 24 hours. It's my rep; what's their problem?

robjohn

I need to read David Speyer's answer a bit closer.

Srivatsan

@robjohn It's also a neat idea.

robjohn

09:13

Yes, his case of "all equal but one" is the case $\lambda=\frac{n-1}{n}$ of $\lambda=\frac{1}{n}$ from my answer.

Srivatsan

@robjohn Hm, either that or $\lambda = \frac1n$.

robjohn

@Srivatsan yes they are the border cases in the picture

one edge is $\lambda=\frac{n-1}{n}$ the other is $\lambda=\frac{1}{n}$

Srivatsan

I think that for completeness, David's answer must also contain a separate analysis for minimising the geometric mean subject to fixing AM and HM. He skipped this - that should turn out to be analogous though.

@robjohn Yes. In other words, one corresponds to maximising the GM (subject to the constraints); the other corresponds to minimising.

robjohn

Yes.

Srivatsan

I like the way he does the induction.

robjohn

09:17

there are critical points at all $\lambda=k/n$ for $k=1\dots n-1$, but the others are interior critical points.

Srivatsan

@robjohn That detail does not arise in his answer - that's my guess at least.

robjohn

That is the reason for the monotonicity that took me so long. That shows that the only $\lambda$ we need to consider are the max and min

Srivatsan

Let me try to explain why it does not matter to him; let's see if I get it right.

Unlike you, he does not tackle the problem for $n$ variables directly. He splits the proof into two parts:
- an optimisation step, where you prove stuff for $n=3$. (This is like the base step of the induction also.)
- an induction step, where he uses the previous part to prove stuff for general $n$.

In the first step, you are given three weights, and a target (weighted) AM and a target (weighted) HM. Your objective is to find three numbers whose weighted AM/HM matches the given value, and whose GM is maximum/minimum. What he is able to show is that the GM is maximised/minimised when two of three numbers are equal.

Since $n=3$, there are no interior critical points, which is nice... =)

Did I make sense till now?

yes? no? maybe? ;)

robjohn

okay, I think so, I have not finished digesting his answer.

Srivatsan

@robjohn Fine. I will leave you alone then :-)

robjohn

09:31

I see that he does work for $n=3$ and then more work for other $n$

Srivatsan

@robjohn Yes. Larger n is handled through induction.

...in such a way that whenever he is optimising something there are only three variables involved; so the interior critical points are sort of absent for his purposes.

@rob I found Didier's comment here amusing. :-)

Everyone seems to get "proof" and "proof sketch" mixed up.. :P

[Sorry, I didn't want to leave this "opportunity" to bring it up. Not trying to be mean... :)]

robjohn

09:49

@Srivatsan Yeah, Didier gets tough on my proofs and enthusiastic about proof sketches :-)

Most proofs are really sketches really. Adding all the details would make even the simplest proofs unwieldy. The details that can be left out are dependent on the audience.

Ilya

10:57

is anybody here?

robjohn

Ah, I am but working in a different window :-)

Ilya

ok

I'm just thinking: I have two models for iid short-term interest rate: $\max(0,\mathscr N(0.05,\sigma^2))$ and bimomially distributed interest rate over $0,0.01,\dots,0.1$

which one is more realistic, how do you think? (I know that both of them are barely realistic)

robjohn

11:23

@Ilya realistic in what sense?

Matt

With bacon or not?

Can someone explain this comment to me?

robjohn

I cannot speak for Ilya, but I assume that it refers to having the answer handed on a platter; I assume that Ilya is just wondering whether a platter with or without bacon :-)

Matt

Thanks, robjohn.

robjohn

11:39

of course, the platter with bacon is not kosher

Matt

I'm oblivious to religious beliefs.

robjohn

I consider kosher foods to be a culinary concern, but I guess it does have its basis in Judaism.

However, now that you mention it, I might remove that comment.

Matt

Why? It's neutral and therefore not offensive.

@robjohn I have yet to try kosher food.

robjohn

You probably have had Kosher food.

It is not a special preparation, but a list of foods that are not kosher

as far as I understand

Matt

Oh! I had it all wrong. I thought meat was kosher if the animal was butchered in a particular way!

My coffee here is probably kosher. If water and coffee beans are kosher.

robjohn

11:47

There are many considerations for what is kosher food. Some are what it is, some are how it is prepared.

But I think your coffee is indeed kosher :-)

Matt

: )

robjohn

@Ilya: is my assessment of your comment at all close?

Matt

12:00

Does this really need Chebychev and Borel-Cantelli to prove pointwise convergence a.e.?

greut

13:54

hey… I'm working on some Fourier series and cannot figure out how the sin(k*pi), cos(k*pi) and transfomed into (-1)^k and such…

if anyone's here is super keen… much appreciated.

en.wikipedia.org/wiki/…

like the example #1 from wikipedia.

Srivatsan

14:18

@rob Do you think this picture helps in an article about kosher foods? =)

I find it quite amusing.

robjohn

I think the title is odd.

the food I can see as kosher...

Srivatsan

Which title is odd?

robjohn

but why classify the consumer?

"Jew with kosher food.jpg"

Srivatsan

@robjohn Oh man, you're paying too much attention. Nevermind. =)

@robjohn But he is Jewish; I swear! Please believe me. :-)

robjohn

@Srivatsan :-) you can tell?

Srivatsan

14:23

@robjohn Oh no, not really. I was just kidding.

robjohn

@Srivatsan I figured. I get the impression that we joke with each other a lot and don't realize it.

Srivatsan

"whatsamatter, not a fan of Johnny Cash?"

@robjohn Yes, in fact, I also go over the top at times to make it plain that I am kidding. I wouldn't be surprised that the other person does not realise it even with all that. It's the textual medium after all.

People are quick with their downvotes.

Matt

14:38

@AsafKaragila Are you around?

Jonas Teuwen

15:16

The ARBO guy just came by (that is the person that checks if your work environment is suitable for you) and I need a whole new desk and chair!

They don't even seem to have those available... Strange, more people are 2m.

Srivatsan

@JonasTeuwen "More people are 2m" -- what does this mean?

Jonas Teuwen

More people have a length of 2 meters and more.

Certainly in The Netherlands!

Srivatsan

height? Are you serious? 2m is like 6'8''..

Jonas Teuwen

Yes.

Srivatsan

That is tall. =) What's your height? I suppose this is not too private an information.

Jonas Teuwen

15:20

My height? 1,99m.

I feel a bit encumbered that they now have to change this room because all the desks are connected.

Srivatsan

I am rethinking my plans of trying to settle in Europe now... =)

Jonas Teuwen

Don't worry, I'm surely among the tallest of the Europeans :-).

Srivatsan

That's reassuring. :D

Jonas Teuwen

I don't think they will be much taller than most people from the US, maybe the Dutch can be slightly taller.

Srivatsan

@JonasTeuwen Interesting. I don't quite feel too short here, so it may not be that bad then. [From time to time, I do see some people who are too tall for their own good. =)]

Jonas Teuwen

15:28

I surely am too tall for my own good.

Matt

1.99 m?! Wowzers!

I'm 1.75 m. I think that's about average for people around here.

Is that reassuring, @Srivatsan?

Jonas Teuwen

The average male height in The Netherlands is 1.84m.

Srivatsan

I would be about the average then =) (After Jonas' edit: slightly below average..)

Matt

I think Dutch people are above average tall among Europeans.

Jonas Teuwen

The tallest in the world!

Srivatsan

15:42

Alright. I have to go now. See you, Matt and Jonas.

Matt

See you later!

Jonas Teuwen

See you.

Zhen Lin

16:33

Computational Science should really be Scientific Computation...

Martin Sleziak

16:52

@Matt Do you have to avoid using Axiom of Choice in you injection question for some reason?

Matt

@MartinSleziak No, I think I can use AC.

Martin Sleziak

Then why not put $f(\alpha)$ arbitrary element of $R\setminus \{f(\gamma); \gamma<\alpha\}$?

Matt

But I'm not sure. I haven't really grasped the concept of assuming or not assuming AC.

Martin Sleziak

That set is clearly non-empty, by cardinality argument.

Matt

@MartinSleziak I'm still thinking about why my $f$ doesn't work! If you write an answer I'll certainly up vote and possibly accept especially given that Andres just apologised and said that he's going to think over his answer again.

Martin Sleziak

16:57

Well, this is too trivial for an answer. I'll post it as a comment. But I wanted to be clear, whether it is ok to use AC.

Matt

I'm still confused about Andres' answer. He seems to be saying that I cannot have an injective function and then again that I can.

Martin Sleziak

He is talking about the question whether you can show that without AC.

Matt

@MartinSleziak I'm grateful either way : )

Martin Sleziak

If you have a look at the first version of his answer, he thought at first that you're trying to define something that will, in addition, be strictly increasing.

When I look in the pdf-file you linked: "By the Well Ordering Principle (Theorem 0.3 in Folland), we can assume that R is well ordered."

So that text obviously assumes AC.

Matt

@MartinSleziak I'm not sure the function I defined is not strictly increasing...

Martin Sleziak

17:03

To be honest I am little confused by your definition.

Matt

Which bit?

I map $\alpha \in \omega$ to $n = |\alpha|$.

And what I'm trying to do for the limit ordinal case is to map the elements in $\beta$ to decimal places. So $\beta$ maps to some $r$ in $(0,1)$.

Martin Sleziak

But there are many choices how to order $\beta$ into a sequence.

I mean you write $\beta = \sup \{ \alpha_i \mid \alpha_i < \beta , i \in \mathbb{N} \}$.

This is dependent on the choice of map $i\mapsto\alpha_i$, which goes from $\mathbb N$ to $\beta$.

Matt

I thought that the $\alpha$ in $\beta$ are already ordered by $\in$.

Martin Sleziak

Yes, they are.

But the order type is $\beta$, not $\omega$.

Matt

I think this is where I went wrong!

Martin Sleziak

17:08

This is what AC means, when he writes that it is not clear, whether it is well-defined.

Matt

Hold on. No. The construction still works, it just needs to be written in a different way.

Martin Sleziak

Also it is clear that the series converges?

@MartinSleziak Here AC stands for Andres Caicedo, not Axiom of Choice.

Matt

: D

Martin Sleziak

Ok, I'll check tomorrow whether you were able to modify your construction or whether AC writes something interesting there.

Matt

Thanks!

Martin Sleziak

17:11

A friend of mine, who studies in Switzerland (Geneva), is coming for a visit this evening...

See you later!

Matt

See you later!

Have fun! : )

Asaf Karagila

18:05

@Matt Yeah.

anon

My dad's worked as a programmer for half his life but when I mentioned the local library had linux he didn't know what that was.. How is that even possible?

Zhen Lin

Linux is just over 20 years old.

It's kinda new in the grand scheme of things.

Daniil

19:11

Hi everyone!

I started reading "Godel. Escher. Bach." today; pretty rad book

robjohn

19:22

@anon he didn't know what Linux was?

robjohn

19:33

Finally, I got an upvote on my even Fibonacci answer. The OP hasn't been back since the initial "flurry" of answers. That's what I get for not seeing the question for a couple of weeks.

Kannappan Sampath

@robjohn Can we discuss a graph theory concept

robjohn

@KannappanSampath I am not very up on graph theory, but I can try.

MaX

Hey guys!

Kannappan Sampath

How to draw the dual graph of a graph with a bridge?

@MaX Hi

Daniil

@KannappanSampath what do you mean, "with a bridge"?

Kannappan Sampath

19:41

Say I have a loop at $A$ and I have a square on the vertices BCDE and an edge connecting A and B. I term AB a bridge.

MaX

Hey Kannappan.

robjohn

@KannappanSampath by dual, you mean replacing edges with vertices and vice-versa?

Kannappan Sampath

No, I mean replacing vertices with faces and faces with vertices, and keeping the no. of edges the same.

@robjohn

robjohn

Doh, of course. Yes. (V+F-E is constant)

Daniil

Yes, so every planar graph has a dual

right?

Kannappan Sampath

19:50

Yeah, true.

And, planar connected graphs are isomorphic to their double dual.

And funny thing is two isomorphic graphs can have different dual!

Is anyone still working on it? @robjohn @Daniil

Daniil

Sorry, what is the question again?

robjohn

@KannappanSampath I am making a drawing.

Kannappan Sampath

Thank You, @robjohn

@Daniil I would like to know about the dual of a graph.

Daniil

Any particular graph?

Kannappan Sampath

Yes, a graph with a loop at A and a square on four vertices BCDE, with an edge btn A and B.

Daniil

20:06

If a graph has a loop, then a dual should have a loop too, right?

Kannappan Sampath

No.

A dual of a graph with a loop will only have a bridge.

Daniil

And the bridge produces a loop?

(in dual graph)

Kannappan Sampath

Yes.

(But this is where I think things are funny!)

Daniil

Is this the answer: imgur.com/VMb0i.jpg ?

robjohn

how about i.sstatic.net/SjtEx.png

Daniil

20:14

@robjohn looks handy and nice :D

Kannappan Sampath

But, now draw its double dual.

@robjohn There's a little problem with that.

robjohn

@KannappanSampath I wouldn't be surprised :-)

Daniil

@robjohn not sure where the vertices are, frankly :S

robjohn

vertices are dots, hopefully color-coded

Daniil

but I see only two dots

Kannappan Sampath

20:17

double dual has a vertex of degree 1, but my original graph with which we started, to which this double dual must be isomorphic to has no vertex of deg 1?

robjohn

well, I guess there should be two dots at either end of the edge between faces A and B, representing the outer face (white)

Daniil

Hm, I don't see any vertex of deg 1

Most likely I made a mistake then

Kannappan Sampath

BTW, @robjohn how did you upload your hand drawn picture, was it through a scanner or other means?

robjohn

@KannappanSampath hand-drawn? Daniil uploaded the hand-drawn image.

Daniil

I uploaded my hand-draw picture, and I just snapped my paper on an iPod

Kannappan Sampath

20:22

Oh I mistook the links for on and the same ones. Sorry!!

Daniil

I've also drawn a double-dual (in hatching): imgur.com/PyWXH.jpg

Sorry it looks like crap

Kannappan Sampath

Finding it extremely hard, @Daniil. Please draw distinct ones for the graph, its dual, its double dual, it helps!

@robjohn not following your graph. There should be three vertices right.

Daniil

@KannappanSampath Sorry, I am kinda nodding off right now,

robjohn

@KannappanSampath Mine is probably messed up. There were two faces in graph 1 so I assumed there would only be 2 vertices in graph 2.

Daniil

I am going to go to the bed, but I am surely going to read the transcript :)

@robjohn but there were 3 faces

Kannappan Sampath

20:28

@Daniil bye! Catch you later.

robjohn

@Daniil are you talking about the surrounding face?

Kannappan Sampath

Yes. That is also counted for the vertex of the dual graph.

robjohn

@KannappanSampath then that is where I messed up. As I said, I have never really done graph theory :-)

Kannappan Sampath

didn't you do graphs during your undergrad days?

robjohn

@KannappanSampath not as a real subject. I just played around with them a bit on my own.

Kannappan Sampath

20:31

Oh fine. So I didn't get "real" answers as of now. sorry but that's what comes to my mind :(

robjohn

Here is what I meant: i.sstatic.net/iTuLS.png

when I said that the two ends of the edge connecting face A and face B needed to be a single vertex

the blue dot(s) represents the outer face

Kannappan Sampath

Now there are four vertices , so, the blue dots coincide, am I right?

robjohn

since the outer face shares an edge with itself, something has to be funky, I would think.

Yes, the blue dots are one.

I am most likely off in some fashion.

Kannappan Sampath

OK. But, I don't quite have the perspective to see what that would mean for other edges.

robjohn

Definitely, don't rely on my graph. :-)

Kannappan Sampath

20:40

I am frustrated.

Ilya

me too

who did delete my comment?

Kannappan Sampath

@Ilya which one?

Ilya

the one with bacon

robjohn

@Ilya which comment?

@Ilya we were discussing that last night...

@Ilya I know what happened...

Ilya

@robjohn sorry, I didn't join - was a bit busy

@robjohn tell me )

robjohn

20:44

One of the comments is removed when a question is closed, and there was none saying that the question was a duplicate, so your comment was chosen.

deleted by automation.

Ilya

One of the comments is removed when a question is closed why?

robjohn

because there is usually a comment saying "this question is a duplicate of ..." and that comment is removed since it itself is now redundant.

Somehow, whatever algorithm is used chose your comment as the one that told why the question should be closed.

Ilya

brrr

I know you and I trust you

robjohn

@Ilya dangerous thing to do :-)

Ilya

but that sounds ridiculous (and again, that's not a stone into your garden)

@robjohn I can live with it

robjohn

20:49

We can ask a mod...

@Mariano: do you know what the algorithm is that deletes the comment that supposedly tells why a question should be closed?

Zhen Lin

Some of those comments were automatically generated to begin with.

Ilya

@Rob: please, that comment does not cost too much - and it makes a little of sense only

just the way guy asked for the paper reminded me people buying cheeseburger king deal

robjohn

@Ilya I am interested about the algorithm anyway. I asked a question of the Cheshire Cat on Friday and it was ignored, so I don't think I am bothering Mariano. :-)

Ilya

@robjohn hohoho

robjohn

If I get an answer, great. If not, then no bother.

@Ilya: see this

Ilya

21:03

but there were no links in my comment

moreover, I saw it in its place even after the question was closed

robjohn

@Ilya Then my explanation does not apply. I have no idea why your comment was removed.

Srivatsan

22:05

I find it funny when the question begins with a "Sorry for asking such a stupid question."

robjohn

They should remove that and change the image to LaTeX

Srivatsan

Who is "they"? =)

robjohn

The OP

Srivatsan

(telling someone else) Burn! =)

robjohn

@Srivatsan for the comment I deleted? :-)

Srivatsan

22:11

@robjohn Yes, I couldn't reply to it. I think I clicked the previous message to reply.

robjohn

I didn't want it starred :-)

Srivatsan

@robjohn Hm. I don't see why not. But it's fine.

The scicomp.SE banner they put up -- how long will it loom on top of all the posts? It's quite annoying. What's up with the pitch black background? It grabs a lot of my attention; not very subtle.

2

Also: more seriously, why doesn't the banner not provide a link for us to link? I must open a new tab, copy-paste the name, click enter, ... I am not used to working so hard.

@rob: If I want to complain against the latter point, should I post it in meta.math? It's not really a question about MSE per se...

Asaf Karagila

@Srivatsan: If you bumped that post couldn't you have fixed the LaTeX bug in my answer caused by M<N?

Jonas Teuwen

The banner makes me think somebody died.

6

robjohn

@Srivatsan I was thinking about that. I find it annoying as well.

Jonas Teuwen

22:25

Well, there surely did but someone we know.

robjohn

@Srivatsan I don't know to whom one would complain.

Srivatsan

@AsafKaragila I have no clue which post you're talking about...

Now I know. I didn't notice it. (Had I noticed the obvious typesetting problem, wouldn't I have made the edit myself?)

Asaf Karagila

SHEBANG! One can add \newcommands at the end of the post.

Argh!!! IT TRICKED ME!!

@Srivatsan Maybe you're a jerk that way! :-)

Kannappan Sampath

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Q: Prove the proposition is a tautology

I have two homework questions that I've been struggling with. For the first I need to prove that $(p \lor q) \land (\lnot p \lor r) \to (q \lor r)$ is a tautology. I've tried two approaches. First I tried substituting other logically equivalent statements for the propositions on the LHS. Once ...

Srivatsan

@AsafKaragila Yes, that was a trick rhetorical question. The correct answer is "No, I wouldn't have made the edit myself so that Asaf looks uncool in everyone's eyes." :=)

Kannappan Sampath

22:33

Aren't the logical equivalences written like $\implies$ ??

Srivatsan

@KannappanSampath Since when does one-sided arrow mean an equivalence?

Kannappan Sampath

or $\Rightarrow$ instead of $\rightarrow$.

Srivatsan

@KannappanSampath Ah, I see what you mean. Logical equivalence can be written as $\leftrightarrow$ or $\iff$.

(or $\Leftrightarrow$..)

Similarly, one way implication is written $\rightarrow$ or $\Rightarrow$ or $\implies$.

The $\rightarrow$ and $\leftrightarrow$ is used in logic context usually. Not in general mathematical writing (for example, you may not find it in an analysis textbook).

Asaf Karagila

Logical implication is $\Rightarrow$ and material implication $\rightarrow$. The difference is that the former is a statement about propositions while the second is a connective between propositions (within propositional logic, that is).

Kannappan Sampath

@AsafKaragila Enlightened. This makes more sense in terms of the question I linked to.

@Srivatsan Thank You for listing out all possible ways of writiing these implications on TeX.

Srivatsan

22:40

It seems my interpretation is all wrong.. :) Nevermind.

Asaf Karagila

@Srivatsan Which one?

Srivatsan

@AsafKaragila Well, my explanation preceding yours. At least naive, if not outright wrong.

Kannappan Sampath

Ah, finally, the banner can take you to scicomp.SE

Srivatsan

Yes, it seems some other user has already requested that in the meta post...

Zhen Lin

@Asaf: Is it possible to construct a non-principal ultrafilter using only the ultrafilter theorem, rather than a cardinality argument?

Asaf Karagila

22:47

What do you mean?

Zhen Lin

Well, the claim that there are P(P(X)) ultrafilters immediately tells us that there are non-principal ultrafilters.

Asaf Karagila

Ah.

Zhen Lin

But suppose instead we assume that the conclusion of the ultrafilter theorem holds for the boolean algebra P(X). Can we construct a non-principal ultrafilter?

Asaf Karagila

The statement of the ultrafilter lemma says that every filter can be extended to an ultrafilter. Take the filter generated by the co-finite subsets and it has to be free.

I think that there is a model in which there are only countably many free ultrafilters over $\omega$.

Zhen Lin

Ah, you're using the fact that an ultrafilter is principal iff it contains a finite subset. I forgot about that.

Thanks!

Asaf Karagila

22:58

Sure.

Kannappan Sampath

23:19

What is happening to all those questions that I tag with counting? This tag is stripped of them without a trace, that is, there is no one who has edited the question after I did?

Srivatsan

23:30

@KannappanSampath [counting] has been made a synonym of [combinatorics]. So your counting tag would've been converted to combinatorics automatically.

Kannappan Sampath

23:41

@Srivatsan Oh, I see. Thank You. I was busy setting up a Google Reader account.

Srivatsan

@KannappanSampath ah ok.

Asaf Karagila

23:54

@ZhenLin I was wrong. As Joel observes if there is one free ultrafilter then there are continuum many of those.

Ray

http://math.stackexchange.com/questions/99631/how-do-i-differentiate-this
Anyone know how to solve it? I can't seem to find $\frac{d^2}{du^2}$

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