Mathematics - 2012-11-30 (page 2 of 6)

2:00 AM

I'm a bit confused

@PeterTamaroff sorry, I was slow in answering. I'm not used to chat rooms (you all "talk" so rapidly) and I write too much!

cassandra0

$$t_n = 2 t_{n-1} + (-1)^n$$

let's solve this

Peter Tamaroff

@Argon I mean $$t_{2n}=2t_{2n-1}+1$

Argon

Ok

cassandra0

yes I like that idea

Peter Tamaroff

2:00 AM

@cassandra0 Another easier way could be telescopy, maybe,

cassandra0

that will removed the $(-1)^n$

user19161

@PeterTamaroff Yes, I know, but what you said is meaningless.

amWhy

@WillHunting It's getting close to Dec 4th

user19161

@amWhy Well, yeah, but there won't be an earthquake or anything, just a mild tremor perhaps. =)

Peter Tamaroff

@WillHunting Well...

cassandra0

2:02 AM

0

Q: What is the meaning of Chebychev's result and why is PNT stronger?

I saw the proof by Chebychev that there are constants $c_1,c_2$: $$c_1 \frac{\log(x)}{x} < \pi(x) < c_2 \frac{\log(x)}{x}$$ and the Prime Number Theorem states $$\lim_{x \to \infty} \frac{\pi(x)}{\log(x)/x} = 1$$ I don't understand what exactly PNT is telling us, and why is it s...

I wanted to you if this is good?

do you want me to add any remarks

Argon

@cassandra0 How does one attack the recursion?

cassandra0

to the question

user19161

@PeterTamaroff See, we are talking about whether or not E is dense in X. I don't see how isolated points come into the picture. Yes, I do know what isolated points are.

cassandra0

@Argon, I'm just trying to do Peters idea fist of all

Peter Tamaroff

@Argon I have an idea.

Argon

2:02 AM

@PeterTamaroff Ok

cassandra0

so $$t_{n+2} = 2 (2 t_n + (-1)^n) + (-1)^{n+1}$$

Peter Tamaroff

@Argon Generating functions

Yeah!

Argon

@cassandra0 Ok, yes

cassandra0

Hence $$t_{2(m+1)} = 4 t_{2m} + 2 - 1$$

Argon

Ok :)

I see where this is going now...

I like it!

user19161

2:06 AM

So @amwhy which is the month, 5 or 9?

Peter Tamaroff

The GF of $t_n$ is $$T(x)=\frac {t_0}{1-2x}+\frac{1}{(1-x)(1-2x)}$$

Argon

GF?

cassandra0

$$t_{2(m+1)+1} = 4 t_{2m+1} - 1$$

that's the odds and evens separated out

Argon

Ok

Peter Tamaroff

@Argon generating function

cassandra0

2:09 AM

I think it goes 1, 4+1, 4^2+4+1, 4^3+4^2+4+1, ...

so would that be $4^{n+1}-1$

for the even part

Argon

@PeterTamaroff I think that is too fancy for my purposes

@cassandra0 Ok

Peter Tamaroff

@Argon It is uuusome

Argon

@PeterTamaroff Baoss

I like generating functions too

cassandra0

the next bit is like $4^3 - 4^2 + 4 - 1$

I don't know the closed form for that

Argon

@cassandra0 I think I do

It's a geometric series

cassandra0

2:10 AM

oh yeah so it is

Argon

Yay!

@PeterTamaroff How did you think I should do it w/ GFs?

cassandra0

you know what strikes me as odd here

@cassandra0 $2n+1$?

lol

Hahaha!

2:12 AM

well yes sort o

it's just funny there's two cases

I guess you might 3 cases if there was something like $(\sqrt[3]{-1})^n$ in it

Argon

Bleh

user19161

@Argon What is GF, girlfriend? Aaron is a ladies man!

Argon

4 mins ago, by Peter Tamaroff

@Argon generating function

Peter Tamaroff

@Argon Just consider $$t_n=t_{n-1}+(-1)^{n}$$ and sum over $n=1$ to $\infty$

Argon

@WillHunting

Peter Tamaroff

2:13 AM

Who cares about convergence anyways?

Argon

@PeterTamaroff Ok

:)

By sum do you mean $t_0+t_1+\cdots$

user19161

@peter I hereby declare you as Potato.

cassandra0

$$G = \sum_{i} t_i x^i$$

Peter Tamaroff

@Argon No, I mean $\sum t_n x^n$

Argon

@PeterTamaroff Classic power series. Ok :)

Peter Tamaroff

2:15 AM

@WillHunting [ERROR 435 - TRY AGAIN ]

Argon

@WillHunting If you try again, you will probably get error 436

cassandra0

no one is answeing my question

on the website

Argon

@cassandra0 What was it?

cassandra0

I linked here it a amoment ago

user19161

@peter is very naughty today. He has said a number of naughty things today. I don't know what he is thinking, but I think he is probably wrong.

Argon

2:16 AM

@WillHunting How could @Peter ever be wrong?

user19161

@Argon Well, I have not told you what I think he is thinking, so yes, he could be partially wrong.

Peter Tamaroff

@WillHunting Spit your blabbery

user19161

But since I don't even know what Peter wants to say, I won't say anything. QED.

cassandra0

@amWhy, did you see my question?

@Argon,

$$
{\rm li} (x) = \gamma + \ln \ln x + \sqrt{x} \sum_{n=1}^\infty \frac{ (-1)^{n-1} (\ln x)^n} {n! \, 2^{n-1}} \sum_{k=0}^{\lfloor (n-1)/2 \rfloor} \frac{1}{2k+1} . $$

spotted that from wikipedia

Argon

@cassandra0 SO COOL!

cassandra0

2:19 AM

I thought you might like it :D

Argon

I love the $\ln \ln x$ :)

cassandra0

I'm looking up li because it's another version of the PNT

Argon

@PeterTamaroff Do you think I should do something like this: en.wikipedia.org/wiki/…

Peter Tamaroff

@Argon Yes.

Argon

@cassandra0 Yup, and a better approximation then $\frac{x}{\log x}$

cassandra0

2:20 AM

why is it better?

by the way I think it was Gauss who originally conjectured pi ~ li

amWhy

@WillHunting I used my real (first) name when I posted a "congrats" answer to Brian M Scott. I'm getting bold!

Argon

@PeterTamaroff I'm a bit confused as to why $f=xf+x^2 f+x$

amWhy

@cassandra0 No, I didn't. How far to scroll up?

cassandra0

@amWhy, it's here

Peter Tamaroff

@Argon Fibonacci!

cassandra0

2:23 AM

@Argon, I want to explain that

Argon

@cassandra0 Shoot!

user19161

@amWhy You wrote it wrongly. As far as I know, the first letter should be in majuscule and not minuscule form. =)

amWhy

@cassandra0 Ahhh! yes, I see it.

@WillHunting Hahahaa...

cassandra0

Let $$G = \sum_{i=0}^\infty c_i x^i$$ any generating function at all, and we should set all the negative $c_{-k} = 0$ (for later).

user19161

@amWhy I was trying to sound pretentious. I could have just said capital letters and small letters. =)

Peter Tamaroff

2:24 AM

@Argon Man

amWhy

@Argon nice answer re: modular arithmetic 9 upvotes!

cassandra0

$$x^k G = \sum_{i=0}^\infty c_{i-k} x^i$$

amWhy

@WillHunting You sounded pretentious, so it worked!

Argon

@amWhy ?

@PeterTamaroff ?

user19161

@amWhy Yay!

amWhy

2:24 AM

@Argon amWhy what?

cassandra0

@Argon, notice that I changed the index instead of changing the exponent - of course these are equivalent but this is the key.

Argon

@amWhy " nice answer re: modular arithmetic 9 upvotes!"?

@cassandra0 Ok

amWhy

@Argon Ohhh, I'm sorry, I clicked on the wrong post. I meant to post to cassandra0

Argon

@amWhy Oh, hahaha

cassandra0

Therefore if we have a linear relation between coefficients e.g. $c_{i} - c_{i-1} - c_{i-2} = 0$ for the fibonacci numbers we get a polynomial relation for $G$: $(1 - x - x^2)G = 1 + x$

$1$ and $1 \cdot x$ are the initial conditions.

user19161

2:27 AM

@amWhy That's called victim of the moving target.

Peter Tamaroff

amWhy

@WillHunting That was it, exactly!

user19161

@PeterTamaroff Pedro's handwriting looks as beautiful as Pedro!

cassandra0

@amWhy, oh that's funny I don't know why they like that calculation so much

Argon

@PeterTamaroff You have neat writing

cassandra0

2:28 AM

@Argon, so is anything I said unclear

Argon

@cassandra0 A bit

cassandra0

We got this for fibonacci numbers $$G = \frac{1 + x}{1 - x - x^2}$$

$$G = 1 + 2x + 3x^2 + 5x^3 + 8x^4 + 13x^5 + 21x^6 + 34x^7 + 55x^8 + 89x^9 + ...$$

Argon

How is that polynomial relation obtained?

amWhy

@cassandra0 Well, it seemed to be a hit!

cassandra0

It's based on $$x^k G = \sum_{i=0}^\infty c_{i-k} x^i$$

robjohn

2:30 AM

@PeterTamaroff Hey there!

Peter Tamaroff

@Argon One learns how to survive as time goes by

@robjohn Man.

Argon

@PeterTamaroff You should "read" mine, haha

Peter Tamaroff

@robjohn I aced my linear algebra test.

Hehehehe

robjohn

@PeterTamaroff Good job! Congratulations!

cassandra0

If $c_{i} - c_{i+1} - c_{i+2} = 0$ then $$0 = \sum_{i=0}^\infty (c_{i} - c_{i+1} - c_{i+2}) x^i = \sum_{i=0}^\infty c_{i} x^i - \sum_{i=0}^\infty c_{i+1} x^i - \sum_{i=0}^\infty c_{i+2} x^i$$ $$= (1 - x - x^2)\sum_{i=0}^\infty c_{i} x^i$$

Argon

2:31 AM

@PeterTamaroff YAY!

user19161

@PeterTamaroff Everyone is man to you.

Argon

@cassandra0 Which is 0?

robjohn

@PeterTamaroff I broke 40K today :-)

Peter Tamaroff

@robjohn Yeah.

user19161

@robjohn Thanks to someone. =)

Peter Tamaroff

2:33 AM

@WillHunting Well, I'm uncomfortable talking to unicorns.

cassandra0

@Argon, basically this^ although ttechnically it's $c_{i} - c_{i+1} - c_{i+2} = 0$ only for $i \ge 2$ there are two initial conditions c_0 and c_1 we have to define

robjohn

@WillHunting Couldn't have done it otherwise, I'm sure ;-)

cassandra0

that's where the 1 + x came from

Argon

I am very confused

cassandra0

sorry :(

user19161

2:34 AM

@Argon So am I. I am always confused.

cassandra0

$$x^k \sum_{i=0}^\infty c_{i} x^i = \sum_{i=0}^\infty c_{i-k} x^i$$

do you understand this?

this is where it all comes from

Argon

@WillHunting You are a Jambul

@cassandra0 Yes

I understand that bit

Peter Tamaroff

@cassandra0 Shouldn't you be changing some indices in the summation too?=

cassandra0

@PeterTamaroff, I actually reindexed the sum which is why we have $c_{i-k}$

Peter Tamaroff

@Argon OH NO! That is wrong. It should be $-x/(1+x)$ not $1/(1+x)$

cassandra0

2:36 AM

@Argon, so for example $$(5 - 3x^2) \sum_{i=0}^\infty c_{i} x^i = \sum_{i=0}^\infty (5 c_i - 3 c_{i-2}) x^i$$

Peter Tamaroff

@robjohn I have a silly question,

user19161

Hmm, if I can get 100 every day in December, I should reach 10k this year. But that would be very taxing...

Argon

@cassandra0 Ok

Peter Tamaroff

Suppose $\{f_n\}$ are bounded over $[a,b]$ and $f_n\to f$ uniformly. THen $f$ is bounded.

anon

How do I specify ymin and ymax in a (2d) plot in mathematica?

Peter Tamaroff

2:38 AM

I getting $|f(x)|<M_N+\epsilon$ @robjonh

For any given $\epsilon$

and large $N$

cassandra0

@Argon, that's really the heart of it. For fibonacci numbers $F_{n} - F_{n-1} - F_{n-2} = 0$ we get $$(1 - x - x^2) G = (F_{0} - F_{-1} - F_{-2}) x^0 + (F_{1} - F_{0} - F_{-1}) x^1 + (F_{2} - F_{1} - F_{0}) x^2 + \ldots = 1$$

therefore $$G = \frac{1}{1 - x - x^2}$$

Argon

Cools

cassandra0

I'm wrong about this $F_1 - F_0 - F_{-1} = 0$

I corrected it now

Argon

$F_{-n} = 0$?

cassandra0

yes

@Argon, so that's how to turn recurrences into generating functions. Then performing long division computes fibonacci numbers (or powers of two or whatever, any linear recurrence)

Peter Tamaroff

2:41 AM

@robjohn ?

Argon

So shouldn't it equal $\frac{1}{1-x}$?

cassandra0

we can use $G$ to find closed form formulas too

$\frac{1}{1-x} = 1 + x + x^2 + x^3 + \ldots$

robjohn

@PeterTamaroff Sorry, I was afk. What's up?

cassandra0

this is the recurrence relation $a_n - a_{n-1} = 0$

Argon

@cassandra0 I mean, why does the sum equal 1?

cassandra0

2:42 AM

which sum?

love joe satriani

$$ (F_{0} - F_{-1} - F_{-2}) x^0 + (F_{1} - F_{0} - F_{-1}) x^1 + (F_{2} - F_{1} - F_{0}) x^2 + \ldots = 1$$

Isn't this just $1+x+x^2+\cdots$?

cassandra0

@Argon, $F_i - F_{i-1} - F_{i-2} = 0$ for all $i > 2$ because we're doing fibonicci numbers

Peter Tamaroff

@cassandra0 There was a show here! G3

robjohn

2:43 AM

@PeterTamaroff I haven't gotten into his music, but I haven't heard enough yet

cassandra0

that means there's just couple terms at the very beginning (the initial conditions of the recurrence) that aren't zero

Peter Tamaroff

@robjohn I really meant to tell you to "Just look up" and see what I had asked

Hehehehehe

Argon

@cassandra0 So why does it equal 1?

Peter Tamaroff

@Argon Each term is zero

cassandra0

@Argon, well the fibonacci sequence $F_{-2},F_{-1},F_0,F_1,F_2,$... is 0,0,1,1,2,...

Peter Tamaroff

2:45 AM

Except the first

Argon

@PeterTamaroff Ok, now I see

cassandra0

so $(F_{0} - F_{-1} - F_{-2}) x^0 + (F_{1} - F_{0} - F_{-1}) x^1 + (F_{2} - F_{1} - F_{0}) x^2 + \ldots$ = $(1 - 0 - 0) 1 + (1 - 1 - 0) x + (2 - 1 - 1)x^2 + (3 - 2 - 1)x^3 + (5 - 3 - 2)x^4 + ...)$

Peter Tamaroff

@robjohn

8 mins ago, by Peter Tamaroff

Suppose $\{f_n\}$ are bounded over $[a,b]$ and $f_n\to f$ uniformly. THen $f$ is bounded.

7 mins ago, by Peter Tamaroff

I getting $|f(x)|<M_N+\epsilon$ @robjonh

7 mins ago, by Peter Tamaroff

For any given $\epsilon$

7 mins ago, by Peter Tamaroff

and large $N$

robjohn

@PeterTamaroff for some $n$, $\|f_n-f\|_\infty<1$ put that together with $\|f_n\|_\infty=M$ and what do you get?

Peter Tamaroff

@robjohn Oh, man,

Well, I like incognito.

hhahaha

cassandra0

2:48 AM

still no one answered my question :(

Peter Tamaroff

@robjohn Why do you write $||_{\infty}$??

$\infty$ norm is $|x|$?

robjohn

@PeterTamaroff because that is what you use for uniform convergence, is it not?

Peter Tamaroff

@robjohn Well, UC is $|f-f_n|<\epsilon$ for all $x$, given any $\epsilon$

robjohn

@PeterTamaroff which is saying the same thing?

Peter Tamaroff

@robjohn I thought the $\infty$ norm was $\max$ =P

cassandra0

2:50 AM

@Argon, anything unclear left or am I just making it more confusing each time I say something more?

Peter Tamaroff

@robjohn But I now see it is the uniform norm too.

I mean, it has another meaning.

Just the same word.

Uniform norm

In mathematical analysis, the uniform norm (or sup norm) assigns to real- or complex-valued bounded functions f defined on a set S the non-negative number :\|f\|_\infty=\|f\|_{\infty,S}=\sup\left\{\,\left|f(x)\right|:x\in S\,\right\}. This norm is also called the supremum norm, the Chebyshev norm, or the infinity norm. The name "uniform norm" derives from the fact that a sequence of functions \{f_n\} converges to f under the metric derived from the uniform norm if and only if f_n converges to f uniformly. If we allow unbounded functions, this formula does not yield a norm or metric i...

robjohn

@PeterTamaroff I am not talkiing about a.e.

Argon

@cassandra0 Can we just review what was covered?

cassandra0

@Argon, I'd love to!

Peter Tamaroff

@robjohn Almost everywhere?

Argon

2:52 AM

$$F_n-F_{n-1}-F_{n-2}=0$$

$\forall n \ge 2$

Peter Tamaroff

@robjohn What you mean is that one can write $$\sup_{n \geq N}|f-f_n|<\epsilon$$ correct?

Argon

?

cassandra0

@Argon, that's the definition of fibonacci numbers, right?

Argon

@cassandra0 Yes

cassandra0

along with $F_0 = 1$, $F_1 = 1$

Peter Tamaroff

2:53 AM

@Argon $\forall n\in \Bbb Z$!

cassandra0

@Argon, so this is just a nice example but this theory works for any linear recurrence

Argon

@PeterTamaroff Not by our defintion - $F_{-1} = 0$

cassandra0

my use of negatives is not standard and maybe confusing.. it's just how I do it :

user19161

I am going to take a shower and nap now. See you guys in my dreams.

Peter Tamaroff

@cassandra0 It is really useful.

cassandra0

2:54 AM

I find it good

Argon

@WillHunting Bye

user19161

@Argon I will start work on the book cover! =)

cassandra0

Fundamental Theorem: $$x^k \sum_{i=0}^\infty c_{i} x^i = \sum_{i=0}^\infty c_{i-k} x^i$$

Argon

@WillHunting Hahaha!

@cassandra0 Ok, got that part

cassandra0

(There is also a nice $\frac{d}{dx} \sum_{i=0}^\infty c_{i} x^i = \sum_{i=0}^\infty i c_{i-1} x^i$)

Argon

2:56 AM

@cassandra0 Hahaha cool!

user19161

@amWhy That would be my secret for now...

amWhy

@WillHunting What would be what secret for now? ;-)

user19161

@amWhy You need to click on the left arrow to see which I am replying to. That's another tip for you. =)

Argon

@cassandra0 $$t_n - 2 t_{n-1} - (-1)^n = 0$$

cassandra0

@Argon, so this is not linear we can't apply the technique

Argon

2:59 AM

DARRRRRRRRNNN

cassandra0

wait!!

amWhy

@WillHunting Thanks, got it. Well, I hope all is Good Will Hunting, soon! I'm not so good myself, but that would be my secret...

cassandra0

there isa very beautiful theorem

Argon

Good @Will Hunting

cassandra0

Let a_n and b_n be linear recurrences

Argon

2:59 AM

Done :)

user19161

@amWhy OK, perhaps we might share some secrets in future...

cassandra0

then c_n defined by c_{2n} = a_n, c_{2n+1} = b_n is a linear recurrence

amWhy

@WillHunting Do take a nice "nap"...I think I'm heading there myself.

Argon

@cassandra0 ok

amWhy

@WillHunting Yes, let's do. When you're ready ;-)

Argon

3:00 AM

This is what you did before?

cassandra0

so let's just think about it a bit.....

Argon

@WillHunting en.wikipedia.org/wiki/File:Jasper.pebble.600pix.bkg.jpg

cassandra0

if $A(x) = a_0 + a_1 x + a_2 x^2 + a_3 x^3 + \ldots $ generates $(a_n)$

and $B(x) = b_0 + b_1 x + b_2 x^2 + b_3 x^3 + \ldots $ generates $(b_n)$

user19161

@Argon =)

cassandra0

then $C(x) = A(x^2) + x B(x^2)$ generates $(c_n)$

proved!

Argon

3:02 AM

Good night everybody!

I need to sleep now :)

cassandra0

ok

amWhy

@Argon g'night. Me too.

Argon

@cassandra0 Bye

cassandra0

but $C(x) = a_0 + b_1 x + a_2 x^2 + b_2 x^3 + \ldots$

Argon

@amWhy Good night!

cassandra0

3:03 AM

just so you see it

Peter Tamaroff

What about coffee people!? What about coffee!?

amWhy

@Argon Sleep well!

Argon

@amWhy You too!

user19161

Let's all meet in our dreams. Over and out!

cassandra0

my question

@robjohn, i was hoping ,if you saw it

> 15 years old, Gauss proposed that pi(n)∼n/(ln n).

15 years old..

> Gauss later refined his estimate to pi(n)∼Li(n)

anon

3:08 AM

I wrote up a quick spiel

Peter Tamaroff

@cassandra0 He had tables and tables and calculated and calculated.

Yes. One hell of a mathematician.

@cassandra0 Let it be recorded that I upvoted you first.

First

I think I'll take a look into this

anon

In conclusion, cass, if you want to understand the meaning behind elements of analytic number theory, it is necessary to understand the meaning and motivation behind asymptotic thinking.

Peter Tamaroff

@anon Well said.

anon

@PeterTamaroff Eh? I'm the only one that's upvoted the question...

Peter Tamaroff

@anon Your answer

anon

3:13 AM

But you linked to cass. :-)

robjohn

@cassandra0 I just got back from fixing dinner for my dog. I answered your question, but anon has a much larger post.

anon

I would have answered it sooner, too, if it weren't for meddling plot options in mathematica.

#nostalgia

Peter Tamaroff

This is interesting: Let $x_n$ be such that $|x_{n+1}-x_n|<c^n$ for some $0<c<1$ then $x_n$ is Cauchy.

(I proved it, but yet...)

anon

sure, geometric series bounding, yadda yadda

Peter Tamaroff

I was thinking about finite calc

finite differences

cassandra0

3:18 AM

is $x^2 \sim 2x^2$?

oh they don't have the same leading coeffcient

anon

@cassandra0 No, they are not $\sim$ to each other.

cassandra0

thanks so much

why would $\pi(n)$ actually be $\sim$ to anything, let alone such a simple function as $\frac{x}{\log(x)}$

anon

Because primes are musical and grow swimmingly on large scales.

Also I like the words musical and swimmingly.

cassandra0

you know what's really weird

$$\pi(x) \sim \frac{x}{\log(x)}$$

now let $\pi_{a,q}$ count the number of primes $\equiv a \pmod q$ for $(a,q) = 1$. We have $$\pi_{a,q}(x) \sim \frac{1}{\varphi(q)} \frac{x}{\log(x)}$$

so the prime number theorem isn't just a simple function like that it's ONE multiplied by a simple function

if there was some irrational number like $\gamma$ it would seem less weird

anon

well, in a sense there is a weird constant in the expression - the natural base $e$.

cassandra0

3:24 AM

oh that's a really good point, it must be the natural log, musn't it? Any other wouldn't be $\sim$.

I was wondering earlier if I could bound like $$c_1 2^n < 3.7^n < c_2 2^n $$

and you can't

anon

changing the base of your logarithms results in a scalar multiple change in the function. that is, $$\pi(x)\sim (\log_e b)\frac{x}{\log_b x}.$$

cassandra0

wow

I'm really amazed now

anon

this is due to the change-of-base formula $$\log_bc=\frac{\log_ac}{\log_ab}$$

cassandra0

this is truly a result how the primes behave in the large scale, not just a measure of their growth

anon

behavior, growth, poh-tay-to, poh-tah-to.

cassandra0

3:28 AM

well I mean Chebychev is just a growth result

asymptotic growth

thanks very much I am really glad to understand this now

I learned about big O in computing so I've a had a different view on these asymptotic notations

this $\sim$ is much more fine

viewing it as an equiv. class is very helpful too.

$(\sin(x)+5)x^2$ is bounded between $4x^2$ and $6x^2$ but isn't $\sim a x^2$

for any a

anon

right

cassandra0

3:46 AM

wikipedia says "
$ \pi(x)\sim\frac{x}{\ln x}.\!$

This notation (and the theorem) does not say anything about the limit of the difference of the two functions as x approaches infinity."

this is actually wrong, we have $|f(x)-g(x)| = o(g(x))$

whenever $f(x) \sim g(x)$

that's just proved by multiplying $f(x)/g(x) - 1 \to 0$ by $g(x)$ and taking absolute value

anon

right. the riemann hypothesis puts some bounds on the error term too.

cassandra0

in terms of chebychev I think we get some negative $-0.94 x/\log(x) < \pi(x) - x/\log(x) < 2^{342} x/ \log(x)$

the constants are made up but supposed to be realistic

I think I'd need absolute value I'm a bit confused

user19161

@anon What has potato got to do with what you are saying?

cassandra0

I found in apaper c1 = 0.92129, c2 = 1.10555

holds for $x > 96097$

user19161

@anon I like musical and swimmingly too!

cassandra0

3:55 AM

so yeah subtracting x/log(x) really just tells us that pi(x) - x/log(x) is between $-1/10$th and $1/10$th of $x/log(x)$

wait no that's wrong

anon

@WillHunting http://english.stackexchange.com/questions/53019/is-po-tah-to-an-acceptable-pronunciation-for-potato

"poh-tay-toe, poh-tah-toe" is a common phrase

cassandra0

so |pi(x) - x/log(x)| < 1/10 x/log(x)... but PNT gives |pi(x) - x/log(x)| = o(x/log(x))

i.e. the constant 1/10th gets smaller

anon

for any $\epsilon>0$, there is an $X$ such that $\left|\pi(x)-\frac{x}{\log(x)}\right|<\epsilon\frac{x}{\log x}$ for all $x>X$. this is a restatement of PNT.

cassandra0

omg!

You can get Bertrands Postulate from Chebyshevs bound!

mathoverflow.net/questions/70692/…

$$\sum_{p^k\leq x} \log p= x -\sum_{\rho:\ \zeta(\rho)=0} \frac{x^{\rho}}{\rho} +\frac{\zeta'(0)}{\zeta(0)}.$$

that's such a cool formula

PNT doesn't imply bertrands postulate though

or does it?

anon

4:20 AM

PNT is asymptotic, so no

if you remove that "for all" part on bertrands postulate and replace it with "eventually," you can not only use PNT but obtain stronger results with it.

cassandra0

hmm

so I get $$\pi(2n) - \pi(n) \ge \frac{2n}{\log(2) + \log(n)} - \frac{n}{\log n} + o(\frac{2n}{\log(2n)})$$

if I could show that last term $\ge 1$ (and I guess that holds if we take n large enough so that the implied constant is $< 1/2$) that would give bertrand for n large

anon

that's pretty simple. pick u in between 1 and 2. then log(2)+log(n)<u*log(n) eventually, so the RHS above is >= (2/u-1)n/log(n)+o(blah), which is eventually >1.

cassandra0

ahh very nice

I guess the thing about betrand is, as you said, that we really want to have it for all n though

leo

4:45 AM

have you guys seen parametrizations $\delta$, $\gamma$ of the unit circle in $\Bbb R^2$ (or part of it) so that $\delta\circ\gamma^{-1}$ is not of class $C^\infty$?

I'm looking for two inyective immersions from $\Bbb R$ into $\Bbb R^2$, $F$ $G$ so that $F\circ G^{-1}$ is not a diffeomorphism

anon

by parametrizations do you mean functions $\Bbb R/\Bbb Z\to S^1$? You may as well view $S^1$ as a subset of $\Bbb C$, set $\gamma$ to be the usual parametrization and $\delta$ something like the complex exponential of say the cantor function

leo

5:17 AM

@anon sorry for the delay. I mean functions $\Bbb R\to S^1$. So far I managed to find $\delta:(-1,1)\to S^1;\theta\mapsto (\cos\theta,\sin\theta)$ and $\gamma:(-1,1)\to S^1;t\mapsto \left(\dfrac{2t}{1+t^2},\dfrac{1-t^2}{1+t^2}\right)$, as shown here

anon

neither of those is surjective. are you okay with that?

leo

Now I have to find $\gamma^{-1}$

@anon yep. Indeed I want something from $\Bbb R$ or an open interval to a some part of $S^1$ so that the map is injective, as in those cases

the domain of $\delta$ is wrong

it must be $]0,\pi[$

if $(\cos\theta,\sin\theta)=\left(\dfrac{2t}{1+t^2},\dfrac{1-t^2}{1+t^2}\right)$, it is enough to show that the function from $(-1,1)\to(0,\pi)$ is not a diffeomorphism

@anon the important part is that they have the same image set

$\lt$

leo

5:48 AM

it seems that the function of $\theta$ in terms of $t$ is $$\theta=2\arctan\left( \dfrac{1-t}{1+t} \right)$$

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