Hey!

Venís mañana a la fac?

@MarianoSuárez-Alvarez Si, claro!

@MarianoSuárez-Alvarez Estudio quimica a la mañana y al mediodia me voy para alla.

ok :-)

@JasperLoy What?

@JasperLoy Tell me the model!

@robjohn This is ghost town.

@PeterTamaroff It is.

When I left to take Lilly for a walk we had 50% more people here

@robjohn You mean the $\sqrt{}$ one?

@PeterTamaroff yes

@robjohn Well, we all feel lazy at times! =)

Though yours is much better.

You can take $\epsilon'=\epsilon\sqrt s$... and you're done.

I went with the usual stuff...

@robjohn Why did you delete it?

@PeterTamaroff I didn't read his question, only the title. He gave my argument and then said he didn't know what to do next :-(

@robjohn Oh, then tell him to do that what I said!

@PeterTamaroff Obviously, his question was on a far more basic level.

@robjohn Than what?

@PeterTamaroff Than my answer was aimed at.

@PeterTamaroff Okay, I fixed it up.

@robjohn +1

@PeterTamaroff I fixed the last line of Pragabhava's answer.

@robjohn Quite the slip!

I think most readers would figure it out.

@robjohn I'm trying to prove that if $x$ is convex in some closed interval, and diff @ a then $f(x)\geq f'(a)(x-a)+f(a)$ for every $x$ in the interval.

The geometric interpretation is pretty striaghtforward.

Now, assume $x>a$.

This is equivalent to proving

$$\frac{{f\left( x \right) - f\left( a \right)}}{{x - a}} \geqslant f'\left( a \right)$$

@PeterTamaroff I think that should follow from the proof of Jensen that I have somewhere.

Assume $x<a$. Then this is equivalent to prving $$\frac{{f\left( x \right) - f\left( a \right)}}{{x - a}} \leqslant f'\left( a \right)$$

@robjohn I have a proof of Jensen.

whim.org/nebula/math/convex.html

@robjohn Oh, dear! Not ASCII again!

@PeterTamaroff :-)

@robjohn Oh, measure space? Out of my league.

@PeterTamaroff Oh, it's all formal stuff at this point. All easy

@robjohn See this

Search " just be quiet for a sec and see this" and see from there

@PeterTamaroff You didn't want to use power series?

@robjohn ?

@PeterTamaroff For $\dfrac1{1+x^2}$

@robjohn Well, no. Many things have to be addressed before integrating termwise.

@PeterTamaroff All you need to address is the remainder term

@robjohn You mean proving $\int_0^x R_n(t) dt \to 0$?

@PeterTamaroff yep

@robjohn True. But new methods are interesting, merely because they are new.

in either case, you only get convergence of the sum for $|x|<1$

@robjohn Yeah... bummer.

@PeterTamaroff No, I agree that the more methods, the merrier.

@robjohn Today I was surprised to find out (and prove) that if $f$ is monotonous on $[a,b]$ then it is integrable in $[a,b]$-

@PeterTamaroff well it is continuous on all but a countable set.

So it is Riemann integrable.

@robjohn ?

@robjohn (I can picture what you mean, just keep it coming)

@robjohn The idea is that if the $f(y)$ are (totally) ordered, then $f$ is continuous almost everywhere?

@PeterTamaroff A monotonic function is continuous on all but a countable set and a function is Riemann integrable iff its set of discontinuities has measure $0$

@robjohn "A monotonic function is continuous on all but a countable set " is that hard to prove?

@PeterTamaroff the sum of the gaps of the discontinuities is finite. A finite sum of positive values is countable

And does it have anything to do with $\mathbb R$ being totally ordered and the fact it is complete?

@PeterTamaroff I think it is simpler than that

@robjohn When you say gaps, you mean "$f(a^+)-f(a^-)$"?

@PeterTamaroff yes

@robjohn How do you sum a potentially uncountable number of them? Don't you mean total variation?

@peoplepower The sum is finite, so there cannot be an uncountable number that are not $0$

@robjohn Today I tried to do an exercise on Spivak where he sketches a proof to show that if $f$ is integrable on$[a,b]$ then it is continuous on infinitely many points of $[a,b]$

@robjohn The "sum" is not a sum when there are potentially uncountably many discontinuities.

@peoplepower How many can be bigger than $1/n$ for each $n$?

@robjohn Yes, that is Spivaks sketch!

@robjohn Could you help me?

@PeterTamaroff I can try

@robjohn Well, that's good enough.

@robjohn So, suppose $f$ is integrable on $[a,b]$-

@robjohn Yes, a sum on a restricted set of the discontinuities, which is necessarily finite. I see, thank you.

Given a partition $P$ with $U(f,P)-L(f,P)<b-a$, it follows there must be an index $i$ for which $M_i-m_i<1$

(Assume $M_i-m_i\geq 1$ for all $i$ and arrive to a contradiction)

What are $M$ and $m$?

@robjohn Oh, sorry $M_i=\sup\limits_{[t_{i-1},t_i]}f(x)$

$m_i=\inf\limits_{[t_{i-1},t_i]}f(x)$

okay

Now, if $i$ is not $=1$ or $=n$; we can be sure there is $a_1$ and $b_1$ with $a<a_1<b_1<b$ such that $\sup\limits_{[a_1,b_1]}f(x)-\inf\limits_{[a_1,b_1]}f(x)<1$ since we can just choose that $t_{i-1}$ nd $t_i$

And Spivak says an easy artifice fixes things if $i=1$ or $n$

Then he asks to do the same with $a_1<a_2<b_2<b_1$ but showing this time $\sup-\inf<1/2$

Then $a_3,b_3$ and $<1/3$

and so on to get $<1/n$

Then use the fitted intervals theorem to find a point where $f$ is continuous.

I guess one can do something similar with $<(b-a)/n$; right??

I didn't think how to fit the points inside each other though, maybe refine the initial $P$ or something...

@PeterTamaroff I believe so

@robjohn OK =P I'll get some sleep now

@PeterTamaroff At least I won't be here alone :-)

@robjohn Bye byes =)

@PeterTamaroff laters

if the method of functional iteration is used on $F(x)=x^2+x-2$ and produces a convergent sequence of positive numbers, what is the limit of that sequence and what was the starting point?

Ok, so we're given that we can use functional iteration, the limit exits for the iteration and even nicer they're all positive numbers for the sequence.

So we assume $\lim_{n\to\infty}x_n=\lim_{n\to\infty}x_{n+1}=\lim_{n\to\infty}F(x_n)=F(\lim_{n\‌​to\infty}x_n)=x$

So, $x=x^2+x-2\implies x^2-2=0\implies x=\sqrt{2}$, hence the limit of the sequence must be $\sqrt{2}$.

So, how do we find the starting point?

I'm a little duped on that one

plugging some values in we see $x_0=0$ gives us $-2$. Thus this doesn't work because all values are positive.

@arete why can't the limit be $-\sqrt{2}$?

@JonasTeuwen !

@robjohn Hi!

@arete The iteration is not stable, so the only way you will get something converging to a limit is if you start at $\sqrt{2}$ or $-\sqrt{2}$.

@JonasTeuwen how goes?

@robjohn Just woke up!

@JonasTeuwen but it's 12:30 AM! ;-)

Oh 8-). Good time to wake up.

This guy has a stolen name! or is it really common?

@Charlie At night, theorems sit around the campfire and tell stories about Gauss.

$ {}$

!!

2 hours later...

BASIC TOPOLOGY is not easy.

@anon Is Rudin's Principles of Mathematical Analysis your textbook?

For what? I skipped analysis.

Eh? AP is for high school classes. I took calc I-III, Lin Alg and Diff Equ in HS but not an official analysis class. My classes this semester are Calc III (I didn't get college credit in HS), Abstract Algebra I, and Elementary Topology

@JasperLoy Well, you're smart.

@JasperLoy Oh. well, I haven't even read that! I'm just worrying about the exercises of basic topology.

@JasperLoy And I do not think the proofs of these theorems are easy enough.

@JasperLoy Do you know Demidovich's problemset on mathematical analysis?

@JasperLoy Can you prove $e^x\ge1+x$ only by definition of $e=\lim_{n\to\infty}(1+1/n)^n$ and some elementary methods?

Hi all

Hint: $e^x\ge(1+x/n)^n$.

@FrankScience Can you use newton's binomial theorem?

@JasperLoy I never understood the confusion between L'hopital and l'hospital. I still don't know which is correct or even why there are two different names.

@JayeshBadwaik It's an interesting problem in Demidovich's. It's just an exercise after the section about the limit of sequence and the definition of $e$.

@FrankScience Okay. I mean, do you know the generalized binomial theorem.

@JayeshBadwaik BTW, I finished the chapter 2 of Rudin's but there are some problems I cannot solve out.

@OldJohn hi

@FrankScience Okay, ask me. May be even I cannot solve. ;-) And then we can share our lack of understanding.

@JayeshBadwaik Hi there

@JayeshBadwaik Unfortunately, other people told me how to attack these problems and I got the way.

@FrankScience ?? okay, so you know how, but the devil is in the details?

@JasperLoy This is fortunately not relevant to me: I took AP classes in HS, and my teachers were very competent and rigorous for those courses. (Also: the only AP tests I took were in math, and most of the math I've learned, I've learned before seeing it in a classroom anyway.) I've tutored friends and family, however, in AP classes with utterly horrible teachers / materials, and I despaired because I felt like I would have to teach the whole course in order to do the subjects proper justice.

@JayeshBadwaik Eh, I cannot get your idea clearly. I mean that I could not solve them alone but somebody gave some hint and then I solved.

@FrankScience Okay, I thought you are still stuck on some problems even after people have told you many hints and direction (basically not being able to get the details right.)

@JayeshBadwaik Now I'll come back to the classical analysis. Next week, I'll try to prove these theorems again and test whether I truly understand these conceptions.

@FrankScience good.

@JonasTeuwen I thought about that inner product at work earlier, and quickly found myself a stumbling block: how is the inner product of two pure tensors of different ranks defined? I assumed $0$, but then $$\Phi_n=\sum_{k=1}^n\alpha_k(\psi_1^k\otimes\psi_2^k\otimes\cdots\otimes\psi_k^‌​k)$$ as a sequence for $n=1,2,\cdots$ (and $\{\psi_i^k\}$ orthonormal) has $$\|\Phi_n-\Phi_m\|^2\le \sum_{l=\min\{n,m\}}^\infty|\alpha_l|^2,$$ and so $(\Phi_n)$ is a Cauchy sequence if $(\alpha_k)\in\ell^2$

Mm. Why would you need to define that? If you tensor two Hilbert spaces $H_1$ and $H_2$, can't you define the inner product on the tensor product as the product of the inner products? Then you would like a linear extension.

I mean, why mention the rank.

I guess my issue is not with tensor products but with direct sums (because we're putting tensors of different ranks together). if $V,W$ come equipped with an inner product, what is the inner product on $V\oplus W$?

@JonasTeuwen Hi - found some more wonderful lute music

Would you rather define the tensor product as all the finite rank operators from the dual of $H_1$ to $H_2$?

@OldJohn Hi :-).

@anon Pointwise?

so if $v\in V$ and $w\in W$, then $\langle v,w\rangle =0$ in $V\oplus W$ no?

@JonasTeuwen did you look at that readme, you think its any good?

So you have $(v, w) \in V \oplus W$, right. So, you get $\langle v, 0 \rangle + \langle 0, w \rangle$. So, yes $0$.

@JonasTeuwen Then the sequence I gave is Cauchy but its limit is not in the direct sum of tensor products, right?

(which would mean $\displaystyle\bigoplus_{n=1}^\infty\bigotimes_{k=1}^n V_k$ with the inner product you mention is not complete)

Wait - let me think about that.

@JayeshBadwaik No, I slept.

@JonasTeuwen okay.

garden needs me - back later

@anon So you are claiming your element is non-zero but has non-zero norm?

@JonasTeuwen If by element you mean the limit of the sequence, yes. The norm should be just be the $2$-norm of the sequence of coefficients $(a_l)$, right?

I am still amazed by what gets a lot of votes and what gets none!

@anon Yes.

First one thing: Only define the inner product on an orthonormal basis (rest by linearity).

right

That should like kinda make it non-zero. But only finitely many of those will be non-zero.

And hence the space will not be complete.

Right.

Okay, fine. Then complete it!

Take.

$$\overline{\bigoplus_{n = 1}^\infty \bigotimes_{k = 1}^n V_k}.$$

Could we then as well kick out the completion in the tensor product? Guess not.

Well, only finitely many Hilbert spaces so yes.

@anon But I bet that quotienting the stuff out and completing it is the same result.

quotienting by what stuff?

Let things be equivalent if their difference is orthogonal?

Or norm $0$.

Like making the $L^p$ space by saying all stuff a.e. is the same.

ah, quotienting by elements with norm 0. I don't see that that is equivalent to the completion. fun stuff. I am so tired.

@anon No, I mean by only considering this on an orthonormal set and extending would give the same result as quotienting. And both need to be completed.

oh right

More stuff. Say $D = t\partial_t$. What would be the combinatorical proof for $D t^n = t^n D + n t^n$? Induction drains my mental energy.

so, like, product+power rule in the algebraic setting

Yes.

Mm, yes like $x = t$ and $y = \partial_t$ and $x, y$ the non-commutative variables in a Weyl algebra.

I think the first place to start would be a combinatorial interpertation of $\partial_t$

But my "objects" are $t^n$?

then a combinatorial interpretation of multiplication by $t^n$.

I'd say that is like removing a letter from a $n$-letter word.

and then put them together. presumably all wrapped in generatingfunction stuffs

gfology.

@JonasTeuwen sort of, but we get a coefficient in front. so it's more like removing letter #1, plus the result of removing letter number #2, etc. (when all of the letters are just $t$)

Yeah.

So the bunch of things you get from removing like one of the letters.

In an alphabet of... two letters?

Kinda, must think about it. Now go off to work. A professor retired. And this implies cake. And I didn't eat and I like cake so off to the cake. Be right back!

seems more like one letter to me, unless we want to conflate letters with operations on letters

That's a pretty lousy word if it only has one letter.

mmm cake

Yeah. Brb!

Shouldn't those contest questions, which stirred up things at meta, be reopened now? The contest is already over...?

Bye, all!

@MartinSleziak Hi there, how are you?

Hi Matt!

Fine, thanks for asking. Quite busy, as always.

See you all later!

Spending less time here in chatroom did not seem to help. I always find a way to procrastinate.

Bye!

Heh, I don't know, but I find this comment to be a bit over-the-top. Perhaps Hagen is used to working with matrices over fields with finite characteristic. However, I usually work with matrices over $\mathbb{R}$.

@robjohn But it seems that he commented, but he did not downvote. It's much nicer that what many of people do - they downvote and they don't leave any explanation.

@MartinSleziak Indeed, I replied once but I don't want to inflame anything, so I am not going to say anything negative. I might just say "thank you for commenting" :-)

@robjohn I did not understand this comment of yours. You say same can be done with real matrices, but then the only real value of cube root of 1 is 1, so its an identity matrix.

@JayeshBadwaik Did you see my answer using real matrices to represent the cube root of 1?

@robjohn Yes, okay like that. I thought of that too. But then I thought there might be something else I am missing.

@JayeshBadwaik Nope, that is all I was saying.

@robjohn okay.

Good eve.

@robjohn Hey, how is it going? =)

@N3buchadnezzar knda slow today, how about you?

Same

Differential equations are mean to me

@anon The quiffy is that $\partial_t$ is not an object itself in this interpretation but something that eats objects and spits out others 8-(.

Perhaps we should see it as $\partial_t 1$.

Yes, that works.

Hi everyone!

Does anyone know much about prime number generation?

Hi @robj

@RajeshD Hey there, how are things going?

@KristopherIves you want to generate a list of prime numbers?

hello

@Nimza hi.

okay.

@JayeshBadwaik good evening :)

@Nimza good evening to you too.

I hate cake :P

@JasperLoy replied

@Jayesh Feel free to flame me, I'm not a math expert, but I've been playing with an equation to generate the next prime numbers given all the previous primes as input. The basic equation is a sum of squares with another argument in the summing series. I'm analyzing the properties of this last argument in the series, because if I can model it the formula generates primes

I have messed with numbers up to 10,000 to test

For example, sqrt( 2*2 + 3*3 + 12) = 5 - Where 12 is the parameter I am attempting to better understand / learn

If you graph it you'll see the last part of that equation becomes less important as well, as it gets closer to guessing prime numbers as they get larger

There is also interesting things that happen around the "epoch" of 0, 1, and 2 and how to apply this explanation to getting the first prime (2) from no previous primes

Just curious if anyone a.) Has something to add or b.) Call me stupid and show me where I'm being dumb

@OldJohn, hi!

how are you doing?

@KristopherIves Could you give more examples of the sum and square-root you are talking about?

For any prime number, take all the previous primes and sum the squares

So for 11

11 = sqrt( 2*2 + 3*3 + 5*5 + 7*7 + x)

or (c) wonder why these sums of squares have anything more to do with prime numbers than any other formulation of multivariate functions. (this is my initial reaction.) unless something interesting numerically turns up - beyond just asymptotic bounds because those can probably be explained with some analytic NT machinery - I see no reason to expect this investigation to bear fruit.

@Steenrod Hi - I'm fine, thanks - and you?

@anon I agree there are lots of ways to add things up and many are not that important

What keeps me interested is how modelable X

@anon I would agree there

Well, disregarding primes, does anyone know a solution for x ?

@Old John,I am fine as well.School exams just got over and they have put my real analysis on the backburner.

@Steenrod Never mind - you will have plenty of time for analysis, I'm sure

@OldJohn, hopefully so.When I returned to stackexchange, I was suprised to see some heat on meta regarding contest questions.How did it all happen?

@KristopherIves You can perhaps get a simple asymptotic bound on your parameter by applying the formulas here for $s=-2$ with $p_n\sim n\log n$ (and I'm sure there are more precise bounds for the latter).

Also note we can use LaTeX in chat (called "chatjax"); see the panel on the right. (In case you're interested.)

@Steenrod somebody posted a question from a contest which was still open and then there was a debate whether the question would be closed if the moderator of that contest contacted us to take it down or something.

Not sure exactly - there have apparently been rumblings about people using MSE to "cheat" with contest questions, and some of the moderators don'r seem to be able to agree on the right way to deal with them

@anon Thanks I'll give that a read

I am not a huge fan of contest questions really, so not been following the argument very closely

@anon how do I get it to highlight latex in chat on StackExchange ?

clarify the terms "it" and "highlight"

but it would be nice if the high-rep users were not arguing with each other (for the sake of the site)

I guess I'll try this super legit javascript paste

@OldJohn,@Jayesh,thanks.I guess you are right.

"it" was Latex (previous pronoun) and highlighting was Layman for parse

@KristopherIves Have you seen the link to Chatjax - on the left of this screen?

Yeah I got it figured out but was suprised it was a bookmarklet paste

@OldJohn, good day(rather evening).I better get going.

@Steenrod OK - have fun!

I think it would be nice if some modern video games allowed the player to customize the texture package instead of having to hack it on roms.

@anon Hehe. What are you playing? and what did you hack?

just old nintendo stuff, mario 64 and ooc etc. I might look into gcn stuff on dolphin, like sunshine or galaxy.

so far I haven't hacked anything but I'm reading up on it.

I only like a few dozen console games, but I'm okay with playing them on the computer if I get cool extras for the inconvenience.

@anon ohh, what do you use? Nestopia?

not that old (except for smrpg)

@JasperLoy :D

@JasperLoy no, that was intentional.

@anon ohh.

@JasperLoy hmm.

@Charlie Wassup??

@JayeshBadwaik everything fine .is just a little hot in here.and you?

@Charlie I'm good. Its raining here, quiet heavily.

@JasperLoy hehe no.it's really sunny.

@JayeshBadwaik oh...:D

why telescoping series has this name?

excuse me...

@Charlie one of the meanings of telescope is "capable of being extended or compressed by the use of parts that slide over one another"

a telescopic series can be "compressed" or "collapsed" into few terms by canceling of one term "from another.

@JayeshBadwaik hmmmm

you must have heard of telescopic jacks used in cars.