Mathematics - 2013-02-28 (page 1 of 3)

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Charlie

00:20

@skull @argon hello

Ethan

dtyjghjnthcmij65rufvbn5r6tfg

Peter Tamaroff

´089´347rgp´0ownhep´ngt{pwej

Ethan

eyhs5wuyshfdhr6ufjg

Charlie

I thought that too @ethan

Ethan

lyftg7i5e6djtyi76rikhg

skullpatrol

00:23

Hi @Charlie

Charlie

@skullpatrol how are you?

skullpatrol

@Charlie Fine thanks, how are you?

Charlie

@skullpatrol good

Peter Tamaroff

I need @anon or @peoplepower

skullpatrol

$\Huge\text{STOP WITH THE (removed)'s}$

please.

skullpatrol

00:45

14 mins ago, by skullpatrol

$\Huge\text{STOP WITH THE (removed)'s}$

please

5 mins ago, by skullpatrol

14 mins ago, by skullpatrol

$\Huge\text{STOP WITH THE (removed)'s}$

robjohn

@Argon I used a different contour to integrate $\int_0^\infty\frac{e^{ix}}{e^{2\pi x}-1}\mathrm{d}x$. I used the entire upper imaginary axis with small half circles missing $ik$ for all positive integer $k$ and a quarter circle missing $0$.

Then take the imaginary part.

The answer I got was $\dfrac{3-e}{4(e-1)}$

skullpatrol

01:10

I think I scared him away >8(

Peter Tamaroff

@skullpatrol Good!

skullpatrol

...or maybe it was robjohn's imaginary axis ;-)

Peter Tamaroff

@amWhy

Argon

01:42

@robjohn You never cease to amaze me, this is correct

Argon

01:53

@robjohn How did you handle the integral that travels up the imaginary axis?

lyj

02:45

@Alex Youcis sorry I didn't have the chance to respond yesterday - I'm wondering in particular about the Matrix Analysis REU at William and Mary and how it compares to others

Peter Tamaroff

03:08

@Sanchez

robjohn

03:28

user image

$$
\mathrm{Res}\left(\frac{e^{iz}\,\mathrm{d}z}{e^{2\pi z}-1}\right)=\frac{e^{iz}}{2\pi}\quad\text{at }z=ik
$$
$$
\begin{align}
\mathrm{Im}\left(\color{#00A000}{\int_0^\infty\frac{e^{it}\,\mathrm{d}t}{e^{2\pi t}-1}}\right)
&=\mathrm{Im}\left(\color{#0000FF}{\int_0^\infty\frac{e^{-t}\,\mathrm{d}it}{e^{2\pi it}-1}}\right)\\
&+\mathrm{Im}\left(\color{#C00000}{\frac142\pi i\frac1{2\pi}}\right)
+\mathrm{Im}\left(\color{#C00000}{\frac122\pi i\frac{e^{-1}}{2\pi}}\right)
+\mathrm{Im}\left(\color{#C00000}{\frac122\pi i\frac{e^{-2}}{2\pi}}\right)

anon

@robjohn who's paying you to do that?

robjohn

@anon That should link back to Argon's question

Peter Tamaroff

@anon master of the sopas

robjohn

@anon I linked the LaTeX as well

Peter Tamaroff

@robjohn maybe you can help here

@anon That was a local reference, see here

anon

03:38

yes, I found that out

Peter Tamaroff

Today I made you soup
But not any soup
Behold the soup of soups
The soup that slaps every other soup
The ultimate soup
the mothership of the soups
Behold, the master of the soups!

@anon How you doin'?

anon

aight

Sanchez

hi @Peter

Peter Tamaroff

@Sanchez @anon I happen to be looking at $G$ finite, $|G:H|=n$; then there exists a subgroup in $H$ with $|G:J|\mid n!$

I have seen a solution using a homomoprhism of $G$ to $G/H$

By mapping $g$ to $xH\mapsto (gx)H$

anon

you mean an action of G on the coset space G/H?

Peter Tamaroff

03:44

Sorry $\operatorname{sym}(G/H)$

anon

yes

Peter Tamaroff

I missed the "sym"

Sanchez

What is $H$?

Can't you just take $J = H$?

Oh okay, you probably mean normal subgroup

anon

he wants it normal probably

Peter Tamaroff

Yes normal.

Missed that too =)

anon

03:45

(kernels are normal)

Peter Tamaroff

So we take $\ker \phi$

Where $\phi$ is the said homomorphism

This basically uses the first isomoprhism theorem and $|\operatorname{sym} G/H|=n!$

anon

mmhmm

Sanchez

Yes.

Peter Tamaroff

Is there any other path?

Sanchez

Not sure, but what you said is a very natural approach to me.

Peter Tamaroff

03:47

The author suggested to look at the action by right translations on the coset space

this happens to have kernel $\cap_{i=1}^n x_i^{-1}Hx_i$ where $G=\cup x_iH$

Sanchez

it wouldn't be so different from the solution you already have, if you are just looking at a different group action

anon

if you want it to be a left action (this allows us to construct a homomorphism G->Sym(G/H) rather than an antihomomorphism), you're going to have to define $g:Hx\mapsto Hxg^{-1}$

Peter Tamaroff

No no, I meant the action $g(x_iH)=(gx_i)H$

So left translations

anon

the kernel of both actions is the normal core of H. my previous comment was in reference to your talk about the author's suggestion of using right actions, though.

Peter Tamaroff

@anon Yes, he says left, I rememberred wrongly.

How would one prove that $$\bigcap_{i=1}^n x_i^{-1}Hx_i$$ is the solution we seek?

Because that is as far as I got

anon

03:50

Note that's the same as $\bigcap_{g\in G} gHg^{-1}$

Peter Tamaroff

I really didn't think about using homomorphisms.

@anon Oh, right.

That is the largest normal subgroup contained in $H$.

anon

Also observe Stab(gH) under the left action is $gHg^{-1}$; the kernel of the homomorphism $G\to{\rm Sym}(X)$ (for any action) will always be the intersection of every stabilizer.

Alex Youcis

@lyj Which others? I know someone that did William and Mary, and I did SMALL.

Peter Tamaroff

@anon Right. I just chose to write $\operatorname{stab}\bar x_i$ as $H^{x_i}$ because it hints normality immeadiately.

Here $\bar{x_i}=x_iH$

@anon I didn't know that last fact.

@anon Do you think it is natural to think about the "induced" homomorphism $g\mapsto (xH\mapsto (gx)H)$ for this kind of problems?

anon

oh yes, very

Peter Tamaroff

03:57

When I first started the problem, I immediately thought of stabilizers and noted the solution had to involve them

But since this section was about group actions, I mostly thought about the orbit stab and related theorems, not homomoprhisms.

@anon OK

anon

having a group act on its subgroups is the most immediately available action you can equip the group with, since you already have said subgroups and don't have to invent other sets for it to act on

furthermore, the failure of coset spaces G/H to be groups when H is not normal may be displeasing, but with a good memory we can return to the idea later in our study of group theory and resurrect it: a group can still act on its cosets via left multiplication, even if multiplication of cosets isn't well-defined

Peter Tamaroff

@anon Right.

@anon I am a little clueless about this one though: "Let the partition associated with a conjugacy class be $(n_1,\dots,n_q)$ where $$n_1=\dots =n_{q_1}>n_{q_1+1}=\dots n_{q_1+q_2}>n_{q_1+q_2+1}=\dots$$ Show that the number of elements in this conjugacy class is $$\frac{n!}{\prod q_i!\prod n_k}$$

O.o

anon

what you are showing is that every finite index subgroup contains a finite index normal subgroup. this can be used to show, as I was told in an answer to one of my questions, that the profinite completion of a group is the the category of groups is isomorphic to the profinite complection in the category of G-spaces

@PeterTamaroff are you talking about a symmetric group?

Peter Tamaroff

@anon That last thing is pretty meaningless to me, for the time being =P

@anon Yes.

anon

It's a combinatorics question: ask yourself how many ways you can fit $n$ items into $q$ bags, where $q_1$ bags are color 1, $q_2$ bags are color 2, etc.

you can distinguish bags of different colors but not those that are the same color. (the bags represent disjoint cycles in a permutation's disjoint cycle representation, and the colors represent the size of said cycles). note that in a symmetric group the conjugacy class of an element is precisely all those permutations that have the same shape of disjoint cycle decomposition

user19161

04:09

Just got a new pair of shoes.

robjohn

@PeterTamaroff are those too much?

Peter Tamaroff

@anon I see. And put into algebraic terms we're looking at how many ways a permutations can have its elements permutated among $q$ disjoint cycles, where there are $q_i$ cycles of each order i.e. $n_{q_i}$-cycle?

@robjohn No clue on equicontinuity! =P

anon

right

Peter Tamaroff

@anon Hallelujah!

robjohn

@PeterTamaroff It's an easy concept. It is like uniform continuity, but across all the functions.

anon

04:15

all_the_functions

Peter Tamaroff

@anon Ah?

@robjohn I know, but I never really read about it. =)

@anon OK, so we start of with $n!$ permutations. By "quoting off" all $n_i$ cycles together, we end up with $n!/\prod n_i$.

anon

You can permute cycles of the same size amongst each other arbitrarily (this accounts for the $q_i!$s in the denominator) and you can cycle the elements within each particular cycle (this accounts for the $n_k$s in the denominator), and you will still be have the disjoint cycle representation of the same permutation. If that's what you mean.

Peter Tamaroff

Then, looking at the $n_1=\dots=n_{q_1}$ cycles, we see that we have $q_i!$ ways of permuting them.

anon

The number of elements in a conjugacy class (of say x) is equal to the number of orbits of a x under conjugation, which by orbit-stabilizer is equal to the index of x's centralizer in Sn. It is in fact possible to explicitly describe the centralizer C(x) of an element, as a wreath product of these cyclings within cycles against the permutations of cycles of given length - I describe this here http://math.stackexchange.com/a/208821/11763

(The centralizer and normalizer of a single element are identical btw.)

Peter Tamaroff

Hmm, I think I might need some combinatorics classes.

anon

04:28

hmm, seems I incorporated inconsistent letter choices from two different versions of the answer I was drafting. oh well.

Peter Tamaroff

@anon I hate when that happens.

I made a Rock of Ages playlist

Gustavo Bandeira

@anon For what operations the set $\mathbb{C}$ is not closed?

anon

what?

Peter Tamaroff

@GustavoBandeira If something is an operation on a set $A$, it is "closed" by definition.

(I know what you mean, just being pedantic.)

Gustavo Bandeira

I'm trying to reason something about this:

Closure (mathematics)

A set has closure under an operation if performance of that operation on members of the set always produces a member of the same set. For example, the real numbers are closed under subtraction, but the natural numbers are not: 3 and 8 are both natural numbers, but the result of 3 − 8 is not a natural number. Another example is the set containing only the number zero, which is a closed set under multiplication. Similarly, a set is said to be closed under a collection of operations if it is closed under each of the operations individually. A set that is closed under an operation or ...

Tyler Bailey

04:32

howdy

anon

Pick any set $X$ containing $\bf C$ as a proper subset, and let $f:X^\eta\to X$ (for any cardinal $\eta$) be an $\eta$-ary operation for which the image of $\bf C$ has nontrivial intersection with $X\setminus\bf C$. Then $\bf C$ is not closed under the operation $f$.

Gustavo Bandeira

Real numbers are closed under subtraction, are there operations for which the complex numbers aren't closed?

anon

That's like asking, "are there any sets that strictly contain the complex numbers?"

Gustavo Bandeira

And this is something dumb to ask I supose?

@PeterTamaroff =D

anon

Would you ever ask if there are any sets strictly containing C? If so, why, and if not, why?

Gustavo Bandeira

04:35

Hmn. Actually I'm really not sure on what I'm asking - but I guess you got the idea.

Tyler Bailey

It reminds me of that thing John Lennon said: "I don't believe in Sets, I only believe in Math"

Man, the math stackexchange is so nice. I went over to the chemistry stackexchange and the users there seem like real jerks.

Gustavo Bandeira

@TylerBailey XD

Peter Tamaroff

@anon Detail: I think you missed that you can fit $n_i$ items correspondingly. That is what $\prod n_i$ accounts for, right?

The $\prod q_i!$ stand for the indistinguishable colours.

Gustavo Bandeira

@anon Thanks for the answer.

anon

@PeterTamaroff The $\prod n_i$ says that we can "cycle" the numbers already inside a given bag. For instance, (123)=(231)=(312)

15 mins ago, by anon

You can permute cycles of the same size amongst each other arbitrarily (this accounts for the $q_i!$s in the denominator) and you can cycle the elements within each particular cycle (this accounts for the $n_k$s in the denominator), and you will still be have the disjoint cycle representation of the same permutation. If that's what you mean.

Peter Tamaroff

04:40

@anon And that reduces the amount of elements precisely because some permutations produce the same effect, i.e, those that contain in this case, the same 3-cycle (123) in some part of their decomposition?

anon

I would phrase it as "some representations produce the same permutation," but I think you get the idea.

since we're first counting representations of permutations, then dividing by the multiplicity by which representations of permutations overcount the actual permutations, in order to (in the final analysis) count how many permutations there are of a given cycle type / conjugacy class

I am using the term "representation" informally of course.

So (123)(4657) and (7465)(231) are both representations of the same permutation

Peter Tamaroff

@anon Right.

Peter Tamaroff

04:56

@anon Consdier $2+3$

$$\eqalign{
& \left( {12} \right)\left( {345} \right) \cr
& \left( {13} \right)\left( {245} \right) \cr
& \left( {14} \right)\left( {235} \right) \cr
& \left( {15} \right)\left( {234} \right) \cr
& \left( {23} \right)\left( {145} \right) \cr
& \left( {24} \right)\left( {135} \right) \cr
& \left( {25} \right)\left( {134} \right) \cr
& \left( {34} \right)\left( {125} \right) \cr
& \left( {35} \right)\left( {124} \right) \cr
& \left( {45} \right)\left( {123} \right) \cr} $$

But according to the formula, I should get $10$ more. @anon

Oh, wait.

I missed changing the order of two elements in the 3 cycle

anon

there are not one but two ways of forming a three-cycle out of three numbers

Peter Tamaroff

@anon Right.

OK, I got the $\prod n_j$ role. Moving on.

OK, the $\prod q_i!$ accounts for the fact that for example,. $(12)(35)(46)$ is the same as $(35)(12)(46)$ or $(12)(46)(35)$?, @anon

My combinatorics are almost non existent, sorry.

anon

right

Peter Tamaroff

OK, I have 3/13 done =)

Maybe this formula helps me with "Determine the representatives of the conjugacy classes in $S_5$ and the number of elements in each class. Use this information to show that the only normal subgroups of $S_5$ are the trivial ones and $A_5$"

Ethan

05:14

@anon If I have a function $F(x)=\sum_{n\leq x}f(n)$, what are some ways I can alter F(x) to make it continuous?

@Sanchez

Sanchez

What is your purpose?

Ethan

what do you mean?

Sanchez

What are you trying to do? (with the continuity)

In any case, there is something called smoothing.

Essentially, you consider a function of the form $h(x) = \sum f(n) g(n/x)$

See example 3 here: tricki.org/article/Smoothing_sums

The idea is that, if I take $g$ to be the indicator function of $[0,1]$, then my corresponding $h$ is same as your $F$.

You can thus approximate $g$ by something continuous, smooth, etc.

Ethan

Well I am trying to evulate $$\lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^nF(\frac{k}{n})$$ but if I can construct a slightly different function say $F_2(x)$ that is continuous and still has the same limit, when substituted into the aforementioned expression, I would have $$\lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^nF(\frac{k}{n})=\lim_{n\to\infty}\frac{‌1}{n}\sum_{k=1}^nF_2(\frac{k}{n})=\int_{0}^1F_2(x) dx$$

Which I can evaluate more easly

Sanchez

If your $F$ is integrable on [0,1], then your limit holds.

Ethan

05:22

what do you mean

Sanchez

I'm not sure if you really need something not even integrable.

So when you say $\int_0^1 F_2(x)dx$, one need to specify what it means.

And for what kind of $F_2$ you can do it.

Continuous functions are one such example, but there is a more general notion of integrability of functions.

Ethan

?

Sanchez

For example, it includes functions that are only discontinuous at a few points.

I'm not sure what $F$ you are having in mind, but if they are at least integrable, then you don't need an approximation by continuous functions.

Ethan

$f(n)$ is an integer function

$$F(x)=\sum_{n\leq x} f(n)$$

Sanchez

okay, then it's integrable I think.

Ethan

05:24

u think?

Sanchez

yes.

Ethan

how do I know?

wait

Sanchez

By the way

Ethan

the problem is @sanchez

Sanchez

you probably mean something else.

When you integrate $\int_0^1 F(x)dx$

Ethan

05:25

I plan to perform integration by substutition to simply the integral

but I cant do that if F(x) isn't continuous

Sanchez

Your $x$ are all smaller than 1, so your sum is pretty much vacuous.

Ethan

@Sanchez my bad

It should be

Sanchez

No, substitution works not only for continuous functions.

Ethan

how?

how would you derive it

with out assuming

Sanchez

@Ethan, just find the relevant results in an analysis book.

You were reading baby Rudin right? there should be something there.

I can't possibly go into the whole theory of Riemann/Lebesgue integral here.

Ethan

05:27

ok

Peter Tamaroff

@anon I got the representatives and the cards.

Of the conjugacy classes of $S_5$

anon

well, I am off to sleep

Peter Tamaroff

@anon Oh, OK. I will sleep after I'm done with this one. Sleep well

$$\displaylines{
\left| {\left( {12345} \right)} \right| = 24 \cr
\left| {\left( {1345} \right)} \right| = 30 \cr
\left| {\left( {345} \right)} \right| = 20 \cr
\left| {\left( {12} \right)\left( {345} \right)} \right| = 20 \cr
\left| {\left( {12} \right)\left( {34} \right)} \right| = 15 \cr
\left| {\left( {12} \right)} \right| = 10 \cr
\left| {{\rm{id}}} \right| = 1 \cr} $$

@Sanchez Do you think you can help?

Sanchez

What's the problem?

Peter Tamaroff

With the above information I have to show the only nontrivial normal subgroup of $S_5$ is $A_5$

Sanchez

05:35

the numbers are size of conjgagy class?

Peter Tamaroff

Yes.

And the element is a representative.

Sanchez

well i guess you can only do brute force

Peter Tamaroff

$A_5$ has $60$ elements.

Sanchez

I would just try random stuff

The key would be normal subgroup = union of conjugacy class

Peter Tamaroff

I still don't know how that info helps me.

Sanchez

05:37

I would just play with the divisibility conditions then

Peter Tamaroff

@Sanchez Oh, OK.

Sanchez

For example, if you have a proper normal subgroup, it can't contain a 2-cycle.

Peter Tamaroff

@Sanchez Why?

Sanchez

Let's say it contains a (12)(345)

Oh, because the 2-cycles would generate the whole symmetric group.

Peter Tamaroff

OK.

@Sanchez Right.

Sanchez

05:38

Every permutation is a product of 2 cycles (transpositions)

Peter Tamaroff

@Sanchez Yes.

Sanchez

Let's call that proper normal subgroup $H$. If it contains $(12)(345)$, then it contains 20 + 1 (id) elements at least.

Peter Tamaroff

Yes.

Sanchez

Now a normal subgroup is union of conjugacy class, so the order of $H$ might be 1 + 20 + sum of other numbers on your list

I don't think 120 has a factor that ends with 1, except 1.

So the "other" stuff you need add in, must contain at least either a 15 or a 24.

Peter Tamaroff

@Sanchez It has 2,3,4,5,10,12,15,20,30,60 I think.

@Sanchez Right.

Oh, and 6

Sanchez

05:40

Let's start with 15. Then it contains a (12)(345) and a (12)(34). So by multiplying inverse of the latter to the former, you would get a 2-cycle.

Peter Tamaroff

And 8

Sanchez

By normality, you get all the 2-cycles, thus the whole group, contradicting $H$ is proper.

So the "other" stuff can't be 15.

Peter Tamaroff

I see.

I'll go to sleep.

Sanchez

Let's say it's 24. Then you get 1 + 20 + 24 = 45, and you need to add other stuff to it too.

Peter Tamaroff

I didn't sleep well today, and its giving me a head ache.

Sanchez

05:42

But the only factor of 120 greater than 45 is 60, and 120, and you see that there's no way you can get 60.

Peter Tamaroff

@Sanchez Right.

Sanchez

So it can't be 24 either.

Peter Tamaroff

I see where this is going.

Sanchez

Hmm, have a good sleep then :)

Hopefully you see the idea.

Peter Tamaroff

Thanks.

Ethan

05:42

@Sanchez

Sanchez

I don't see cleaner ways to do this.

@Ethan?

Ethan

Can you help me, I think I have made the problem more precise now

Or can you atleast point me in the right direction

Sanchez Im trying to evaluate $$\lim_{n\to\infty}\frac{1}{n}\sum_{k\leq n} \ln(\frac{k}{n})F(\frac{n}{k})$$ My idea was to construct continuous $F_2(x)$, so that $$\lim_{n\to\infty}\frac{1}{n}\sum_{k\leq n} \ln(\frac{k}{n})F(\frac{k}{n})=\lim_{n\to\infty}\frac{1}{n}\sum_{k\leq n} \ln(\frac{k}{n})F_2(\frac{k}{n})=\int_{0}^1\ln(x)F_2(\frac{1}{x})\ dx=\int_{1}^\infty\frac{\ln(x)F_2(x)}{-x^2} \ dx$$

Sanchez

hm?

Ethan

Which I can evaluate by differentiating a Mellin transform that I can obtain in terms of $F_2(x)$, namely if I have $$\psi(s)=\int_{1}^\infty\frac{F_2(x)}{x^{s+1}}$$
I can get $$\lim_{s\to+1}\frac{d}{ds}\psi(s)=\int_{1}^\infty\frac{\ln(x)F_2(x)}{-x^2} \ dx$$
that is assuming I can swap the order of the limits

Sanchez

If your $F$ is that partial sum thing, then you don't need to construct a continuous $F_2$

Ethan

05:44

why not? are you sure?

Sanchez

As long as your integral converges

Peter Tamaroff

@Ethan We already told you the second equality might not be true.

Ethan

@PeterTamaroff I said assuming I can swap limits

Peter Tamaroff

Also, you want $\frac nk$ not $\frac kn$ with $F$ and $F_2$

Ethan

my bad ur right

Peter Tamaroff

05:45

@Ethan I mean the $\lim\sum =\int$

Ethan

messed up there

@PeterTamaroff sanchez is saying it is true

Sanchez

@Peter, what are you saying is not true?

@Ethan, what is "it"?

Ethan

$$\lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^n\ln(\frac{k}{n})F(\frac{n}{k})=\int_{0‌}^1\ln(x)F(\frac{1}{x})\ dx $$, where $$F(x)=\sum_{n\leq x} f(n)$$ where f(n) is an integer function

Peter Tamaroff

@Sanchez That $\lim \frac 1 n \sum_{k=1}^n f(k/n)$ limits to the integral, without assuming the integral exists.

I'm off.

Sanchez

Ah okay.

@Ethan, Peter is right.

You need apriori the integral to converge, then you can say the limit holds.

Peter Tamaroff

05:48

Ethan: "the wise man builds his house upon the rock"

3

Remember that.

Ethan

what?

Peter Tamaroff

Get good rocks before building the house.

Now think it in terms on learning maths.

Ethan

@Sanchez If the integral converges, then they are equal?

Sanchez

For proper integral, yes.

Ethan

@Sanchez Can I show you a quick proof of the pnt, I think I can construct based apon some of what you said, it feels to good to be true

its short

Sanchez

05:51

You can try, but any short proof is likely to be wrong

Ethan

thats why I think somthing is wrong

How do I write a dirichlet convolution in latex?

*?

Sanchez

That works

Ethan

sec typing it up

Ethan

06:07

@Sanchez are you their

Sanchez

go on

Ethan

ok here it is

Sanchez

err..

Ethan

$$\psi(n)=\sum_{k=1}^n(\mu*log)(k)=\sum_{k=1}^n\ln(k)\sum_{j\leq \frac{n}{k}}\mu(j)=\sum_{k=1}^n\ln(k)M(\frac{n}{k})$$
$$1=\sum_{k=1}^n 1(n)=\sum_{k=1}^n(\mu*1)(k)=\sum_{k=1}^n\sum_{j\leq\frac{n}{k}}\mu(j)=\sum_{k=1}^nM(\frac{n}{k})$$
$$\psi(n)-\ln(n)=\sum_{k=1}^n\ln(k)M(\frac{n}{k})-\ln(n)M(\frac{n}{k})=\sum_{k=1}^n\ln(\frac{k}{n})M(\frac{n}{k})$$

$$\lim_{n\to\infty}\frac{\psi(n)}{n}-\frac{\ln(n)}{n}=\lim_{n\to\infty}\frac{\psi(n)}{n}=\lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^n\ln(\frac{k}{n})M(\frac{n}{k})$$
$$=\int_{0}^1\ln(x)M(\frac{1}{x}) \ dx=\int_{1}^\infty-\frac{\ln(x)M(x)}{x^{2}}\ dx$$
So we get, $$\lim_{n\to\infty}\frac{\psi(n)}{n}=\int_{1}^\infty-\frac{\ln(x)M(x)}{x^{2}}\ dx$$But we also have by the differentiation of a Mellin transform involving the mertens function that for all $Re(s)>1$,

$$-\frac{\zeta'(s)}{s\zeta(s)^2}-\frac{1}{s^2\zeta(s)}=\int_{1}^\infty\frac{-\ln(x)M(x)}{x^{s+1}}$$
So that $$1=\lim_{s\to +1}\frac{\zeta'(s)}{s\zeta(s)^2}-\frac{1}{s^2\zeta(s)}=\lim_{s\to 1}\int_{1}^\infty\frac{-\ln(x)M(x)}{x^{s+1}}=\int_{1}^\infty\frac{-\ln(x)M(x)}{x^{2}}$$

That is assuming we can interchange the limits on that last part
So $$1=\int_{1}^\infty\frac{-\ln(x)M(x)}{x^{2}}=\lim_{n\to\infty}\frac{\psi(n)}{n}$$

@Sanchez I made the assumption we could interchange the limits, other then that though, what isn't justifyable or is wrong?

Sanchez

What is $\psi(n)$? again?

Ethan

06:16

the summatory function for the vonmangoldt function

Sanchez

How do you know limit of $\psi(n)/n$ exists?

Ethan

because the integral exists?

the statement $\psi(n)\sim n$ is equivilent to $\pi(n)\sim \frac{n}{\ln(n)}$

Sanchez

Why do you think the integral exists?

Ethan

because its equal to one, assuming we can interchange the limits

Sanchez

Sure, if it exists

Ethan

06:19

if one assumes the limits are interchangeble, then it is equal to 1

Sanchez

I think the problem would be the interchange of the limit at last.

Ethan

@Sanchez If I make the assumption I can interchange the limits and that integral, then I would have the integral is equal to one?

Sanchez

Yes, but how do you know you can interchange the limit?

Ethan

@Sanchez thats what I thought, anyway terry gives like a couple page elementry proof showing the statement $M(x)=o(x)$ is equivilent to the pnt

And I think showing it exists would require showing $M(x)=o(x)$

so, I guess its not that good

Sanchez

That's true.

So I think that's how you circumvent the difficulty

Ethan

06:22

what do you mean?

Sanchez

If you have $M(x) = o(x)$ you would be able to justify the last interchange of limit

You don't, and I don't see how you can justify it.

Ethan

Which is just as hard to prove as the pnt

Sanchez

Yes.

Ethan

lol

Sanchez

good try though

Ethan

06:24

I thought it was so clever, lol, oh well

Sanchez

haha

For something with content (PNT), it's hard to prove it by just transforming it to equivalent statements

Ethan

@Sanchez Are there some useful conditions that permit the interchanging of limits

Sanchez

limit and integral? Dominated convergence theorem.

You may also want to read about uniform convergence.

Ethan

yes I was trying to read that earlier today I don't understand any of the articles on any wiki sites

Sanchez

To apply DCT in your case would amount to show that the integral converges for the exponent in denominator to be 2.

i.e. you didn't get pass the difficulty

Ethan

06:33

If the integral exists, I can justify the switching of limits right?

Sanchez

the integral with $x^2$ in the denominator, yes.

Well actually not quite

You need the convergence of $\int_1^{\infty} \left| \frac{...}{x^2}\right|$

i.e. even stronger, you need the absolute value of the whole thing to converge

Ethan

@Sanchez have you read the erdos selberg elementry proof of the pnt or any variations of it?

Sanchez

no

Ethan

@Sanchez What should I study

I don't know enough

Sanchez

Are you reading baby rudin?

Ethan

06:40

no

should i?

Sanchez

For your interest right now, go through basic analysis

Ethan

i have an old version, should i get a newer 1

Sanchez

as in, you should know epsilon-delta definition of limit, basic theory of differentiation/Riemann integral

As you want. Whichever version is fine I guess.

Read the first 8-9 chapters or so.

Then read a book on complex analysis

and perhaps, Apostol's Introduction to analytic number theory.

Ethan

excited 4 that lol

Sanchez

If you are dead serious in math, you would need to learn other branches of math as well.

But if you only want to know enough to explore in the stuff you are doing, at least go through basic analysis/complex analysis.

Ethan

06:44

@Sanchez I have Rudin's principles of mathematical analysis second edition, is that too old? or is it ok?

Sanchez

That's okay.

There's no need to go for new editions in general.

Ethan

the copy right stamp says 1954, are there any new tools I might miss out learning that have been developed sense then? if I use this book

Sanchez

Unlikely I think.

The math you do now? Unlikely

Ethan

ye nvm that

Sanchez

But you are still young.

At least go to university, then you will have a better idea after a few years.

Sorry I don't know.

The advisors at your school would know better.

If you can go to UCLA that would be pretty decent in terms of math.

Ethan

06:53

@Sanchez did you ever study anything before math?

Sanchez

Not really.

Ethan

me neither lol

Sanchez

Anyway, good night @Ethan.

Gotta go now.

Ethan

@Sanchez ok, night

lyj

07:20

@Alex Youcis Cornell, IU, Clemson (still waiting on some but I'm interested in any information nonetheless)? I was shortlisted at small but most people accepted. could you ask your friend about his experience at william and mary? thanks!

Gustavo Bandeira

I watched Stargate in 98, I had like 8 or 9 years old. There's a scene in which the linguist draws the following figure in the sand:

user image

And then they take him to a private place, and a women comes in and undress herself.

In 98, I had no idea on the conection between both things... Now I do.

I want my innocence back.

Jonas Teuwen

07:52

Why?

Gustavo Bandeira

@JonasTeuwen Just kidding. =)

skullpatrol

End of the innocence

That^ is why...

Gustavo Bandeira

08:12

@skullpatrol =D

skullpatrol

@GustavoBandeira ;-)

Orange Harvester

08:28

@JonasTeuwen TikZ Question in case you are interested.

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