1:00 PM

but basically I understand the result of prime number theorem now so that was really nice

@robjohn Say $f$ is integrable over $[a,b]$

@cassandra0 Oh! I thought you were talking about Argon. sorry

I was trying to explain a basic things about generating functions to Argon but I don't think I did it well

My lemma is that if $P$ is a partition such that $U(f,P)-L(f,P)<(b-a)/m$ for some $m$ natural, then there must exist ad index $i$ for which $M_i-m_i<1/m$
I'm not sure about my proof

1:02 PM
it's quite hard with all the infinite series when you only write out first 3 terms

@PeterTamaroff What are $M_i$ and $m_i$?

@robjohn $\sup f$ and $\inf f$ over $[t_{i-1},t_i]$
THe thing is as follows

@PeterTamaroff Take the signum function over [-1,1]

@robjohn ?
Suppose for the sake of contradiction that for evert $1\leq i \leq n$ we have $M_i-m_i\geq 1/m$
Then $$\frac 1 m (b-a)=\frac 1 m \sum_{i=1}^n (t_i-t_{i-1})$$

@PeterTamaroff on some interval $M_i-m_i=2$ or $1$

1:07 PM
$$\sum_{i=1}^n \frac 1 m (t_i-t_{i-1})\geq \sum_{i=1}^n(f(t_i)-f(t_{i-1})) (t_i-t_{i-1})=U(f,P)-L(f,P)<(b-a)/m$$
@robjohn Well, but there are many points where $M_i-m_i=0$ right?

@PeterTamaroff Oh, you are looking for just one interval that is shallow?

@robjohn Yes. FOr some index $i$

@PeterTamaroff That is trivial.

@robjohn ¬¬
@robjohn Is my "Proof" ok?
The thing is I have a $\geq$ in the middle of a $<$ argument
I mean, I get $A<blah\geq blah <A$

@PeterTamaroff yes, if I am reading it right
@PeterTamaroff that is not good

1:13 PM
@robjohn Why is it "trivial"?
@robjohn Yes, I know...
=(

I don't see how there can be a contradiction using that sequence of inequalities.

@peoplepower That is why I said it is not good

$$\frac 1 m (b-a)=\sum_{i=1}^n \frac 1 m (t_i-t_{i-1})\geq \sum_{i=1}^n(M_i-m_i)(t_i-t_{i-1})=U(f,P)-L(f,P)<(b-a)/m$$
THat is what I wrote

@PeterTamaroff Yes, but you need to break that into two inequalities, one that you derived, and one that is supposed to be true

@robjohn Oh, OK.

1:16 PM
I can see what you mean, but your notation is not good

@robjohn Yeah.

do you have a favorite proof of betrands postulate?

$U(f,P)-L(f,P)<(b-a)/m$ by hypothesis
And I seem to arrive to $U(f,P)-L(f,P)\leq (b-a)/m$
That's no good.

\begin{align} U(f,P)-L(f,P) &=\sum_i(M_i-m_i)(x_i-x_{i-1})\\ &\ge\frac1m(b-a) \end{align}
but by hypothesis $U(f,P)-L(f,P)<\frac1m(b-a)$

Oh, duh.
@robjohn =)
I was thinking it right, writing it wrong.

1:21 PM
@PeterTamaroff trivial?

@robjohn Nah.

@PeterTamaroff Okay, then I retract my comment about triviality :-)

@robjohn I don't like the word trivial that much.

@PeterTamaroff Find a subgroup of a group. The answer to this trivial problem is one of the trivial subgroups. :)

hello
trivial just means that it comes without exploiting deeper properties

1:26 PM
@cassandra0 Not really.
(Methinks)
Trivial, for me, means dead obvious
And most things aren't.
Trivial is very subjective.
Our bag of trivialities gets filled up with time. Surely rob's is bigger than mine.
(And that should not be taken out of context.)

:)
* $\nu(x) = \sum_{p \text{ prime} \le x} \log(p)$
* $\psi(x) = \nu(x) + \nu(x^{\frac{1}{2}}) + \nu(x^{\frac{1}{3}}) + \ldots$
* $\log([x]!) = \psi(x) + \psi(x/2) + \psi(x/3) + \ldots$

@robjohn

@PeterTamaroff yes?

I'm showing that if $f$ is R integrable over $[a,b]$ then the set of "continuities" of $f$ is dense over $[a,b]$

@PeterTamaroff Don't you want to show that the set of discontinuities is a set of measure 0?

1:35 PM
@robjohn I don't really know how to show that. But it would be good. =)
How does one construct a measure?

'm having trouble seeing this for x=7. I actually get two different answers

@PeterTamaroff You can define the idea of a null set (set of measure 0) without really defining a measure

DEFINITION: A null set is...?

set of measure 0

1:37 PM
do you have definition of set of measure 0?
I'll explain it yf you like

@robjohn You Wikipediated me! How dare you!
@robjohn =)

@PeterTamaroff It can be contained in a union of intervals whose total length is arbitrarily small

@robjohn I see.
2
That is bliss.

Parsevals identity, why art thou so evil...

@N3buchadnezzar They are nice! Not evil,

1:42 PM
Maybe

1

The fourier coefficients for nonzero $n$ are $$\begin{eqnarray} \hat f(n) &=& \frac{1}{2 \pi} \int_0^{2 \pi} (\pi-t) e^{-i n t} dt \\ &=& \frac{1}{2} \int_0^{2 \pi} e^{-i n t} dt - \frac{1}{2 \pi} \int_0^{2 \pi} t e^{-i n t} dt \\ &=& - \frac{1}{2 \pi} \left(\left[t \frac... parseval identity So given$$ f(x) = \left\{ \begin{array}{ccc} (\pi - a)x & \text{if} & x\leq a \\
(\pi-x)a & \text{if} & x>a \end{array} \right. $$Then Parsevals identity when developing the series gives what$$ \frac{1}{2\pi} \int_0^{\pi} f(x)^2\mathrm{d}x = \sum_{n=1}^\infty a_n^2 + b_n^2 $$? =) @rob Is the FIT equivalent to completeness? 1:51 PM @PeterTamaroff FIT Fourier Inversion Theorem? @robjohn I meant Fitted intervals theorem Nested intervals theorem @PeterTamaroff what is that? Cantor's Yes. 1:52 PM Then yes, the space has to be complete but only the case of \Bbb R I mean if we take NIP as an axiom, we can deduce completeness NIP=Nested INterval Property @cassandra0 my favorite method of computing \zeta(2k) is here @PeterTamaroff I guess it depends on how you state the FIP (finite intersection property) @cassandra0 The fact that the generating function is pretty simple is nice. That's incredible! (1) is actually such a simple calculation @robjohn Let I_n be a sequence of closed intervals with the property that I_{i+1}\subseteq I_{i}. Then \bigcap_{n\in \Bbb N} I_n is nonempty. @PeterTamaroff I would require that the I_n have the finite intersection property 2:02 PM @PeterTamaroff does [n,\infty) count as a closed interval? @robjohn I'm stating it as a principle. For I_n subsets of \Bbb R @OldJohn Nope. Thats a ray! =D @PeterTamaroff just checking on your terminology :) @PeterTamaroff That's right, you have a decreasing chain of non empty sets... okay @OldJohn, do you have an insight on this? 33 mins ago, by cassandra0 * \nu(x) = \sum_{p \text{ prime} \le x} \log(p) * \psi(x) = \nu(x) + \nu(x^{\frac{1}{2}}) + \nu(x^{\frac{1}{3}}) + \ldots * \log([x]!) = \psi(x) + \psi(x/2) + \psi(x/3) + \ldots @cassandra0 looks vaguely familiar from Hardy and Wright ... what is the problem? 2:04 PM I just don't see why the last equation holds I've been trying examples and they do match up OK - will do a bit of checking ... wait don't go to the trouble it's ok I'll try more examples and see if I get anything 2:15 PM @cassandra0 what do you get for the value of \psi(10)? well \sqrt[4]{10} = 1.778.. so \psi(10) = \nu(10) + \nu(3.162277..) + \nu(2.1544) + 0 + 0 +\ldots  = (l(7)+l(5)+l(3)+l(2))+(l(3)+l(2))+l(2) 7.8320141... so ... 3\log 2 + 2\log + \log 5 + \log7? yes (better to leave it as logs, I think) interesting to collect like term like that 2:19 PM I am convinced it is true now - just need to justify the equivalence of the 2 sides must be something to do with contributions from prime powers dividing x!, I think Why don't we take exponentials and compare with \sum [n/p^k] which represents the greatest power of p dividing n!. @peoplepower yep - I was about to suggest the formula for the greatest power of p dividing n! :) and notice that p<x^{1/2} means the same as p^2<x etc. e^{\psi(x)} = \prod_{p^r \le x} p^{r} You can also look at e^{\psi(x)} as the product of p^{\alpha} (over p, now) where \alpha is maximal such that p^{\alpha}\leq x. Now, how do e^{\psi(x)} and e^{\psi(x/n)} relate, isolating a prime? oh I see the if p^{\beta_p} is in n then we'll have p^{\alpha-\beta_p} in the second one so it cuts out some primes 2:31 PM What do you mean by "is in n"? higest prime power dividing n Doesn't work. For instance 5<7 but 5>7/3. I think we are interested in the factorization of n with respect to p though. Since p^{\beta}\leq x/p implies p^{\beta+1}\leq x. yes that seems a good way to do it just prove it for every prime oh I thought of something consider the prime powers in x, then the prime powers in sqrt{x} are the just the same but with the indices halved e.g. if x = 2 * 3^2 * 5^3 then \sqrt{x} has 3 * 5 that's example is not really right but I think the statement is true 2:39 PM Is that relevant? I'm not seeing how we can use it yet. it helps me understand \psi better, but I don't know a way to use it Indeed if the p component of e^{\psi(x/n)} is p^{\alpha} and that of n is p^\beta then that of e^{\psi(x)} is p^{\alpha+\beta}. Thus, find n such that p^1 is the highest power less than or equal to x/n. Only problem is that we can take n+1 and get a different answer at the end. :( Ithink I was wrong about \sqrt{x}, but instead (I think OldJohn siad this) that if p \le x^{1/r} we would have r \log(p) in \psi(x) one for each the \nu that p still lives in Yes. @peoplepower If A is countable, does it have measure 0? I mean lebesgue measure, 2:49 PM @PeterTamaroff yes @OldJohn How does that follow?7 If A is countable, we have A listed as a_1, a_2, \dots just cover teh countable set in intervals that decrease with order \epsilon/n^2 @OldJohn OK. then cover with intervals ... what @cassandra0 said :) 2:50 PM We take \epsilon/2^n nbhds? @PeterTamaroff that would do that works too So$$\sum_{n=0}^\infty \epsilon /2^n=\epsilon$$or something of the sort? yep so, for example, the rationals in the unit interval can be covered by intervals of total length as small as you wish @OldJohn I was telling robjohn I showed the set of continuities of a riemann integrable function over some closed interval is dense there, Now I want to show the set of discontinuities is countable. viz, it has measure zero or to show it is a null set 2:53 PM You could also take a sequence of finite sets A_n whose union is the set in question. Then use the formula m(\bigcup A_n)\leq \sum m(A_n)=0. @peoplepower Hmm... I know nothing about set theory. I just know little random stuff But I can get through some simple definitions like that of null set This follows from m being a subadditive function. is argon here? @PeterTamaroff sounds plausible - but I can't recall the conditions for a function to be Riemann integrable and I have to go - got to get ready for a new kitchen :((( @OldJohn Super sad smiley? 2:57 PM @PeterTamaroff yeah - clearing absolutely everything out of the old kitchen is exceedingly boring but it must be done - later all! bye Todays integral: Prove that$$I = \int_{-\infty}^\infty \frac{\pi}{2} - \arctan(x^2)\,\mathrm{d}x = \pi \sqrt{2} $$@N3buchadnezzar I think I already did that. Once... I have to sleep sometime today, why not at 10am? ;) @peoplepower It is 12am here 2:59 PM @PeterTamaroff By parts until you lost your sanity or ? ;) @N3buchadnezzar I can't recall now. @cassandra0 Sorry I did not help you with the entirety of your question. Hopefully you will see what to do from here. It's from a proof of PNT right? I still don't see the equality but I'll keep going @PeterTamaroff The times look so similar, but really they are so far! Math <3 Exam tommorow :D 3:08 PM could we say$$n! = (\prod_{p \le n} p)(\prod_{p^2 \le n} p)(\prod_{p^3 \le n} p)\cdots$$seems not hi Argon @cassandra0 Hi sorry I didn't the generating function well yesterday :( @Argon See todays integral a few posts up =) you guys do realize I can't do a single one of these integrals @N3buchadnezzar I see it. So far I have$$\int_{-\infty}^\infty \int_{x^2}^1 \frac{1}{y^2+1} \,dy \,dx$$though I may be off the mark 3:18 PM wow @Argon Clever, but I do not think fubini applies here, since the area is not closed. @N3buchadnezzar Ok, I will try something else @Argon Is not the expression you wrote slightly off? It gives med \frac{\pi}{4} - \arctan(x^2) which clearly diverges.$$\int_{-\infty}^\infty \int_{x^2}^\infty \frac{1}{y^2+1} \,dy \,dx$$I think I figured it out @Argon Rewrite \arctan(x^2) as a taylor series, evaluated at infinity Eg 3:27 PM @N3buchadnezzar I had thought of that too But I didn't like the \pi/2 Because of symetry we have I will try it though @Argon You already have a \pi/2 in the integral... They cancel! Nice. What do we have then?$$ \int_{-\infty}^{\infty} \ldots = 2 \int_0^\infty \ldots \arctan(x) = \frac{\pi}{2} - \frac{1}{x} + \frac{1}{3x^3} - \frac{1}{5x^5} + \ldots$$You do the rest ^^ 3:35 PM I just can't see this identity at all @cassandra0 Well the taylorseries to \arctan(x) you know right? sorry I mean the prime numbers one Ooooh, sorry. Please do not ask me! Ask John or Rob! @Argon The x=0 messes things up... @N3buchadnezzar x=0? Where? @Argon  \displaystyle \int_{x=0}^{x = \infty} (Please do not hurt me for using this definitition) 3:42 PM @N3buchadnezzar Hahaha :) as x tends to zero the whole thing blows up @N3buchadnezzar That would be a problem. 1 sec This is the expansion at infinity, right? @Argon Indeed @N3buchadnezzar what are you guys doing? @JayeshBadwaik$$ \int_{\mathbb{R}} \frac{\pi}{2} - \arctan(x^2)\,\mathrm{d}x = \pi \sqrt{2}$$3:48 PM Is Fubini really futile here:$$\int_{-\infty}^\infty \int_{x^2}^\infty \frac{1}{y^2+1} \,dy \,dx? @Argon don't we have to show the integral of the absolute value is finite in the first place? hi @amWhy @N3buchadnezzar Use \begin{align} \frac{\pi}{2} - \arctan(x^2) = \operatorname{arccot} (x^2) \end{align} \arcot What is it? \operatorname{arccot} Use \operatorname{} Yeah I thought arccot would be defined for sure? ;-) Anyway, after this substitute x^2 = \cot \theta 4:13 PM The answer seems to be \frac{\pi}{\sqrt{2}} @JayeshBadwaik Clever Indeed @N3buchadnezzar Well, every dawg has his day. :-) What's another fun one?\int_0^1 \log(t) \log(1-t)\,\mathrm{d}t$$? ^^ 4:24 PM Hmmm... Symmetrical Compare it to$$ y = \sqrt{1/4 - (t-1/2)^2} $$^^ @JayeshBadwaik Sir, @PeterTamaroff Dawg, @JayeshBadwaik Me has a question,. @PeterTamaroff Speaketh,. 4:29 PM @JayeshBadwaik Let f:V\to V be a linear transform, and dim V=2 okay. Then f is nilpotent if and only if f^n=0 One direction is trivial by defintion Consdier the other. Say \dim V=n and f is nilpotent okay. And you want to prove that f^n = 0? then there exists an s\in \Bbb N suh that f^s=0 @JayeshBadwaik Yes. I'm advised to think if the inclusions Ker(f^i)\subseteq Ker(f^{i+1}) are strict or not @PeterTamaroff I am still not getting you, what does f is nilpotent imply other than f^k(v)=0 for some k? 4:35 PM @JayeshBadwaik Nothing more than that I have proven that f can't be neither mono neither epimorphism @PeterTamaroff Then isn't it just a matter of definition? If you go into matrices, then a non-zero matrix with determinant zero is nilpotent. @JayeshBadwaik Well, the thing is to prove that k=n If \dim V=n and f is nilpotent, then f^n=0 @PeterTamaroff No, k \leq n where k is the \inf of all such k_i @JayeshBadwaik ? Well, yeah Hehehehe @JayeshBadwaik I mean that n is one of the solutions "solutions" So the thing would go @PeterTamaroff This is strict I think. 4:38 PM Let \dim V=n and f nilpotent. Then there exists k such that f^k=0,. Consider the set K=\{k:f^k=0\}. By the well ordering principle K has a least element k'. We wish to prove k'\leq n @PeterTamaroff okay, I am inclined to kick in linear independence at this stage. how would I show \nu(x) - \nu(x^{\frac{1}{2}}) + \nu(x^{\frac{1}{3}}) - \nu(x^{\frac{1}{4}}) + \ldots \le \nu(x)? @JayeshBadwaik I'm all eyes. @cassandra0 You should study from some book... Apostol's Analytic number theory is good, I guess. strangely, I have not found an answer to this question by a quick search is the Cantor space separable? I just want to prove this 4:42 PM I saw that, Peter @cassandra0 And what theory do you know that you think will do? @Ilya =P btw, added you on facebook ??? @Ilya Me? 4:43 PM aha check out your requests @PeterTamaroff I suppose it might be easier to show the stuff for matrix, and then use isomorphism of any vector space of dimension n to F^n @Ilya sorry. Victim of the moving chat. np :) I got it each - \nu(x^{\frac{1}{n}}) + \nu(x^{\frac{1}{n+1}}) is \ge 0. @Ilya Cantor space is countable in itself? Countable union of finite sets? And it is dense. 4:46 PM @JayeshBadwaik How so? @JayeshBadwaik the first thing about the Cantor set is that it is uncountable. Even in itself @JayeshBadwaik I mean, this proof, not the isomorphism @Ilya I don't see it @PeterTamaroff Cantor set is \{0,1\}^\mathbb N @Ilya I know, I mean your FB request!!! this is getting messy"! @Ilya, that's not the Cantor set I have in mind 4:47 PM @cassandra0 What is yours? It must surely be homeomorphic @Ilya Hehe, I went strolling. Sorry (again!). If the Cantor set consists of all base 3 "decimals" with only 0 and 2, then isn't the collection of all such "decimals" which terminate a countable dense subset? @OldJohn dense sunset - that's shall be romantic :) @cassandra0 Well, that is it, yes. 4:48 PM :) @Ilya Heheh I must be in a romantic mood :) @OldJohn, I managed to understand this log[x]! thing @OldJohn seems so, thanks 4:49 PM @PeterTamaroff I had proved it sometime back, it is in Artin. Let me recall. @cassandra0 well done! - I bet it was something to do with showing that the highest power of p dividing both sides was the same @OldJohn Yes, nice. :-) yes exactly that what happened was I was staring at 10 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 the prime 7 occurs once because it's in the bigger half, but the prime 5 occurs twice because it's in the lower half and then you just do that sum it over powers and 3 occurs three times because it is in the lower third etc. yes exactly that and then 3^2 is in the upper half so you get 3^4 4:51 PM - and an extra one because 3^2 < 10 yep$$\sum_m \sum_n \sum_{p^m \le \frac{x}{n}} 1
gives the order of each prime
and it's easy to turn the claim into this algebraically

so you have to convince yourself that this all works for any prime
it is a bit like some of the arguments with p-adic numbers

@PeterTamaroff I did it like this $\det( A^n) = (\det A)^n$

yes we are taking the p-adic valuation of $e^{\log[x]!}$

@JayeshBadwaik I have no determinants yet!

4:53 PM
@cassandra0 yes - I guess we are :)

This is CH3, and dets is CH5, I think...
=D

so now I got halfway through page 1

@N3buchadnezzar - there is a question on main which covers the thing I was talking about the other day: $x$ is its won power series :)
@cassandra0 I suspect that it might not be too hard after you have mastered the first bit :)

@PeterTamaroff Hmm. Without determinants, you will get into some lengthy argument of linear independence. I am not sure how. Let $L b_i = a_1i b_1 + a_2i b_2 + \cdots + a_ni b_n$ where $b$ are the basis of course. Something like that, which are basically hypervectors. Let me think.

@PeterTamaroff Pitt T. you are, right?

4:56 PM
@Ilya Yes. Tamaroff

@PeterTamaroff then I sent you request

@Ilya I am /ptamaroff
@Ilya How would Tamaroff be spelled?
In cyrilic

Тамаров

@Ilya Oh, OK.
And Peter is what?
Piotr?

why is log([x]!) - 2 log([x/2]!) <= log(Gamma(x+1)) - 2 log(Gamma(x/2+1/2))?

5:02 PM
@PeterTamaroff Пётр
Ё = IO
like in [ur] withour [r]

@Ilya Oh, OK. That's a $\Pi$!
@Ilya Yes, sounds like spanish.
@Ilya I am actually Peter Nicholas. Some Tsar names going on. Not that I'm a tsarist, I swear!

I am, so don't worry :)

@Ilya In what sense? Isn't tsarism something "dead". I mean, politically.
I was just joking, also. I really don't have any position

ah it's log convex

it is, which mean I'm not a politician

5:08 PM
these halves are really difficult :/

bbl guys, I'm leaving

later!