Mathematics - 2013-05-14 (page 3 of 5)

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5:06 PM

Hi

@Lord_Farin hehe that reminds me Asaf's blog "I have no choice"

@somaye hi

@Charlie how are you?

@somaye fine, and you?

me too@Charlie

@somaye good, Sommy :)

Dominic Michaelis

5:21 PM

@charlie i posted my questions on the meta

answers*

if a,b,c are three odd natural numbers how could I prove that: $a^2+b^2+c^2$ is not a full square

Dominic Michaelis

BTW Helfgott uploaded a proof of the weak goldbach conjecture on arxiv

@DominicMichaelis good :)

:(

@somaye what happened?

5:33 PM

Hi all

@OldJohn Hi :)

Hi @pourjour @somaye @Charlie

@pourjour Have you worked out why it cannot be a perfect square?

@OldJohn no

Look at the values mod 4 :)

@OldJohn is it obligatory to be mod 4

5:36 PM

@pourjour It is the easiest way to get the result you want. There might be other ways, but mod 4 is so simple, when you see it

if $a$ is odd, then $a^2 \equiv 1 \pmod{4}$

and similarly for $b$ and $c$

so $a^2+b^2+c^2 \equiv 3 \pmod{4}$, but that is impossible

@OldJohn HI JOHN

Hi @Charlie

Hi@OldJohn :)

how are you?

Hi @somaye Fine here, thanks - and you?

@OldJohn so if a is a full square $\Longleftrightarrow a\equiv 0[4]$

5:39 PM

Spent most of the day working in the garden

@OldJohn i am fine too:)

@somaye Excellent

what is excellent ?@Charlie

@pourjour any square must be equiv to either 0 or 1 $\pmod{4}$

@OldJohn

5:40 PM

@somaye what?

@somaye It is excellent that you are fine :)

@OldJohn thak you

@somaye :)

@OldJohn ok thanks for the trick :D

hey @OldJohn look at my pic and tell me wow what a pretty girl ok?

:)

5:41 PM

@pourjour It is worth remembering that when considering squares, mod 4 is always worth looking at

Ok?

@OldJohn

@somaye Yes, indeed - very!

@OldJohn :) as angle?

Maybe an angel :)
not an angle :)

@OldJohn hey i am

5:43 PM

@somaye an angle or an angel?

@OldJohn hehe angel

@somaye hehe

Chris's wise sister

@Lord_Farin Here is another cute question I think of right now $$\int_0^{\pi/2}\log(\log(\cot (x/2))) \ dx=\frac{\pi}{2} \log\left(\frac{4 \pi^3}{\Gamma(1/4)^4}\right)$$
It's not that hard as it might seem as first sight. It's related to the previous question.

Dominic Michaelis

@oldjohn there is a proof of goldbachs weak conjecture on arxiv

@DominicMichaelis Yes, I saw that on MO - interesting, but I am not too excited by it, as such a result seems unlikely to have further consequences

Dominic Michaelis

5:51 PM

now this proof is possible mathoverflow.net/questions/42512/… :D

I understand it uses the Hardy circle method, but better than has been done previously - I find that results about adding primes are likely to be a bit peripheral to serious progress in number theory

@DominicMichaelis Cute! But as you say, possibly circular

@DominicMichaelis I have spent the last 3 evenings try to teach myself Python programming by solving Project Euler questions using it

Dominic Michaelis

Thats how i learend the bit of Mathematica I know

@DominicMichaelis But I have found that pari/gp has been the quickest method for some of the ones I have solved

Dominic Michaelis

with a score index of 100 It is quiet likely to be so :D

@DominicMichaelis score index??

Dominic Michaelis

6:01 PM

projecteuler.net/languages

But don't ask me what it stands for :D

@DominicMichaelis OK - just found it :)

Dominic Michaelis

79 upvotes and 23 downvotes for my nomination till now

compared to the upvotes i got pretty many downvotes

@DominicMichaelis oh

Dominic Michaelis

@Lord_Farin jupp they are still countable :D

@DominicMichaelis :P

Chris's wise sister

6:07 PM

@Lord_Farin I think you 'd have had good chances for a mod position.

@Chris'swisesister I continue to think that my relative inexperience would have been a limiting factor.

But thanks -- again. :)

Brian could be

Hi @will

@somaye Yesterday Jonas said:
Jonas Teuwen: @GustavoBandeira Girls are just underdeveloped boys.
and later you said:
somaye: your self are underdeveloped girl @JonasTeuwen
So, is Jonas also an angel?

@Charlie, good morning

@WillJagy how are you? Have you been scratching lots of dogs lately?

I think Somaye left

6:13 PM

hi

Dominic Michaelis

@somaye you changed your picture

@Charlie, probably today. Sunny out but not too hot.

i come back

yeah

@somaye I put a message for you, just about 7 messages back on this page

yeah @janos is an angel

@WillJagy

6:15 PM

@somaye, good, we should tell him when he returns.

Hi @WillJagy

@WillJagy sure tell him

@WillJagy i like hangover movie

it is really funny movie

@WillJagy you has doubth that he is an nagel?

really?

I must go to eat - back later, folks

Dominic Michaelis

see you

@somaye @jonas :) not janos :D

Later @old

6:19 PM

@Charlie later :)

@Charlie ok!

@OldJohn, hello.
@somaye, I think I have seen the first two hangover movies.
Jonas is a complicated figure. I am reading a book about Leonardo da Vinci. They have the same difficulty finishing projects. If Jonas gets some good people to help him finish writing mathematics articles, things should work out. Right now, there is a young man in Finland writing up a joint paper with Jonas. This is very good. Have you read this http://en.wikipedia.org/wiki/Layla_and_Majnun

@WillJagy you have doubth that @janos is an angel?

@somaye, I just left a long answer to that, above, it took me a while to type.

?????

????????

6:23 PM

@WillJagy you mean that @janos is working on poem ?persian story?

Dominic Michaelis

@peter where is the problem it is a twin paper

Julian Kuelshammer

hi

@DominicMichaelis Ah?

any reference for the manner used by MathGems in this answer:

math.stackexchange.com/questions/296987/…

@WillJagy Ok no problem

@WillJagy i put a new pic in my profile see it and tell me wow what a pretty girl OK?

Dominic Michaelis

6:26 PM

as the first one only gives new bounds for the conjecture the second really proves it

@somaye, no, Jonas is working on mathematics, always. But he has trouble actually writing complete mathematics manuscripts, I think because he starts to think about something else before he finishes writing.
I thought of the Persian poem yesterday and mentioned it as something Old John could read to practice Persian poetry. I am not sure you will know this, there is a famous song that uses the same story by Eric Clapton...http://en.wikipedia.org/wiki/Layla

@somaye, wow, what a pretty girl. You in a nice chair, I think a dark blue dress and a necklace

@Charlie, anyway, it has begun as a nice day. I buy this Black Chinese tea that is fermented a little, it is usually written Pu-Erh. It is available as little compressed pellets or loose, en.wikipedia.org/wiki/Pu-erh_tea but I buy it as loose leaves. I have been trying to get the amount of leaves correct, I think I have figured it out.

@PeterTamaroff, look for the Second Edition of The Hardy-Littlewood Method by Robert C. Vaughan. If the library does not have it, get Mariano to order it. I'm in it.

@WillJagy "I'm in it?"

@pourjour It is pretty straightforward. if $d\mid7a+5$ and $d\mid4a+3$, then $\begin{bmatrix}7&5\\4&3\end{bmatrix}\begin{bmatrix}a\\1\end{bmatrix}\equiv0\pmod{‌d}$

@robjohn how about the column where (a and 1) I never saw such matix

This can only happen if $\det\begin{bmatrix}7&5\\4&3\end{bmatrix}\equiv0\pmod{d}$

6:36 PM

@WillJagy yeah really i do not hear that song by Eric

@PeterTamaroff, pages 127 and 146. My name is only on page 146.

@pourjour you've never seen a column vector before?

@WillJagy How nice.

@robjohn not yet

I just started learning algebraic structures

@pourjour have you read any Linear Algebra books/papers?

6:37 PM

@WillJagy So I guess you know what all this is about?

About the proof of the conjecture?

@robjohn number theory only :)

@pourjour you should learn linear algebra before you tackle number theory, it would seem to me.

@WillJagy interesting

Dominic Michaelis

@robjohn yeah linear algebra before weak goldbach conjecture seems reasonable for me :D

@PeterTamaroff, not in the first edition. We told Kevin Ford about the little result, he told us to contact Vaughan, and that was just in time for it to appear in the second edition.
It appears they are talking about odd numbers as the sum of three primes. Vinogradov showed this was true for sufficiently large numbers, but with no explicit bound for "large"

6:39 PM

@WillJagy Ah, so this is not something particularly "new"?

@robjohn I'm just interested in this cramer's rule for now because I don't have enough time since the national exam is coming by the end of this month

If we upvote the answer of a suspended user, do those votes count when they become unsuspended?

@Charlie see if you can view this: youtube.com/watch?v=rTEMRo-IcGk

@WillJagy ok

@pourjour you want to learn Cramer's Rule without knowing about determinants and column vectors? That would be kind of hard, unless all you want is the owner's manual (how to use Cramer's Rule)

6:42 PM

The possible benefits of thinking outside the box: xkcd.com/936

@WillJagy "...this shows that every odd number $N ≥ 10^{30}$ can be written as the sum of three odd primes."

"Since the ternary Goldbach conjecture has already been checked for all $N ≤
10^{30}$ [HP]..."

@robjohn I learnt about determinants and matrix :)

@PeterTamaroff, the consideration is certainly not new. Success would be very different. There are many results where they can prove something for "sufficiently large" numbers, but there is no specific lower bound. Such results, or the implied constant in the result, is called "ineffective."
I see you put 10^30, and it has been checked. Yes, that would complete the job.

@pourjour column vectors are matrices, too. You add and multiply them just the same.

@WillJagy Aren't you excited about this?

@WillJagy Oh, I want to ask something concerning summing over divisors and usual sums.

6:45 PM

@WillJagy hehe it's hilarious

@robjohn so how did he use it in this example

I bumped into $$\sum_{a=1}^{[x]}\sum_{d\mid a} F(d)$$ a few times. This is $$\sum_{d=1}^{[x]} F(d)\left\lfloor \frac{x}{d}\right\rfloor$$ yes?

@PeterTamaroff, I guess I am not very generous in that way, I get excited about things I do.
@Charlie, every home should have one.

@WillJagy Well, I understand. Mathematics is egocentric.

Cramer's Rule: if $$\begin{bmatrix}a&b\\c&d\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix} =\begin{bmatrix}e\\f\end{bmatrix}$$ then $$x=\left.\det\begin{bmatrix}e&b\\ f&d\end{bmatrix} \middle/\det\begin{bmatrix}a&b\\c&d\end{bmatrix}\right.$$ you substitute the result vector for the column corresponding to the variable you want

6:47 PM

@WillJagy YES! My dog loves her tennis ball and wants to play all the time so.funny

@Charlie, found it, they sell these handles, you can throw the bal over and over without your arm getting tired, petsmart.com/product/index.jsp?productId=3207014

@robjohn hmm good I get it and how this related to "d"

@WillJagy hahaha

@pourjour Cramer's rule is one way of showing that a matrix is invertible if its determinant is invertible

@Charlie, arm fatigue is a real proble. Iused to take Abby to the park with a heavy rubber ball used for an American sport called Lacrosse. Then I would use a baseball bat to hit it, because Abby was so very, very fast. If she lost the ball, she would run around in circles to search, always the same, clockwise I think. And her tail kept rotating. She did not exactly wag her tail, it went around like an airplane propeller

A lacrosse ball is the solid rubber ball that is used, in conjunction with a lacrosse stick, to play the sport of lacrosse. It is typically white, but is also produced in a wide range of colors. The old NCAA specifications are: ;Weight: 140g - 147g ;Diameter: 62.7 mm - 64.7 mm ;Rebound: From 1,800 height 1,092 - 1292 mm (70% rebound from falling point) ;Rubber content: 65% The new NCAA year 2000 specs states: "Section 17. The ball shall be white, yellow or orange solid rubber between 7 3/4 and 8 inches in circumference, between 5 and 5 1/2 ounces in weight and when dropped from a hei...

@PeterTamaroff, it will not let me make a jpeg of your formulas, I will need to write it out. Elementary transform and summation formulas for number theory sums should be in Hardy and Wright, it is a matter of finding where

6:57 PM

@WillJagy Cannot you use chatJAX?

Hardy and Wright? I take it it is a good resource? I am using Landau's "Number Theory" now.

@WillJagy hahahahs yes,.when they are too happy! I read a reasearch that said that if their tail "goes" more to the right,.usually means they are happy, if it.is goes to left,.angry.

@PeterTamaroff, yes, but for some reason it is blocking the middle with a dark rectangle in the jpeg. Chatjax is working. Hardy and Wright is the first book I would give to anyone looking into number theory. There is a sixth edition with new references by Montgomery, I think.

@WillJagy Nice! Will check it out.

any one bye for now

@WillJagy

@Charlie

AND .................

OTHER FRIENDS

@somaye good night

7:03 PM

@somaye bye bye

@WillJagy Could you not explain why such sums are equal? I think I know why, but maybe you have a better or clearer explanation.

The number of multiples of $k$ that are $\leq x$ are $$\lfloor x /k\rfloor$$

I'll go do something useful instead.

That is, $\sum_{d\mid a\;,\; a\leq x} 1=\lfloor x/a\rfloor$

@Peter still looking. Why don't you continue explaining, maybe I will learn something. I have never paid any attention to summation techniques.

Oh, is your F(a) restricted, multiplicative maybe?

@WillJagy Just an arithmetic function. Well, the author has used this with $\mu(n)$ which is multiplicative.

He wrote

$$\sum\limits_{a = 1}^{\left\lfloor x \right\rfloor } {\sum\limits_{d\mid a} {\mu \left( d \right)} } = \sum\limits_{d = 1}^{\left\lfloor x \right\rfloor } {\mu \left( d \right)\left\lfloor {\frac{x}{d}} \right\rfloor } $$

7:11 PM

@PeterTamaroff, Much better. just a minute

start with tthis en.wikipedia.org/wiki/M%C3%B6bius_inversion_formula

@WillJagy Yes, I know about that.

But the author proves that without MIF.

He argues about the amount of $d\leq x$ such that $d\mid a$

(Landau is the guy)

I know that $U(n)=\lfloor n^{-1}\rfloor$ is such that $a\star U=U\star a= a$.

@PeterTamaroff, tell you what, give me some wider context. I have later books but not Landau

Yes.

@JonasTeuwen, note that Somaye has included you in the category of angel/pretty girl. Go back through the posts a bit.

Yes.

I am such a lady boy.

2

7:22 PM

I may be slightly off, it seems she asked me whether I doubted your beauty. Anyway, today's task is to look at the new profile picture she has put on Main and give her your comments. It takes a day or more for such pictures to appear in Chat, so you need to open a window in Main and look at her profile there.

@WillJagy Will you be around at night? I am busy now, sadlyt.

Say around 10 pm.

It is 4:20 pm here now.

@PeterTamaroff,, just look up my profile on Main and send me email at the gmail address from the CML, or ask Mariano

@WillJagy Heh, OK.

Thanks, anyways.

Chris's wise sister

@PeterTamaroff Have you seen this? $$\int_0^{\pi/2}[\log(\log(\cot (x/2)))]^2 \ dx$$

@PeterTamaroff, I meant ask Mariano for my email address. I really do prefer email for this kind of thing. For most things, in fact.

7:27 PM

@WillJagy OK.

CML? @WillJagy

@PeterTamaroff, if possible, make pdfs of the few pages of Landau before the troubling formula.

@WillJagy OK.

@Peter ams.org/cml

@WillJagy I think I can stick for a while longer.

Let me send you that e-mail.

@Peter, I'm just trying x=1,2,3, it appears there should be a pure induction proof.

Dominic Michaelis

7:40 PM

i jsut realised i posted kind of homework on MO and got upvoted for it :D

@PeterTamaroff, yes, proof by induction works perfectly. Just write it out for $x=1,2,3,$ both sides, then see what happens to both sides if $x$ is increased to $x+1$ Similar to that theorem about the prime factors of $n!$

@WillJagy Purrfect.

I just realised my left leg does not listen to me.

@JonasTeuwen What?

@WillJagy So any arithmetic function will do yes? We do not want $F$ multiplicative

@PeterTamaroff, any function. Let me write the lemma the way I like it...

7:53 PM

@JonasTeuwen Are you going to go to the medic?

Why?

@JonasTeuwen Dunno, just asking.

Will send me back home; all is known.

@PeterTamaroff, If $c |(x+1),$ then $$ \left\lfloor \frac{x+1}{c} \right\rfloor = 1 + \left\lfloor \frac{x}{c} \right\rfloor. $$
If $c$ does not divide $(x+1),$ then
$$ \left\lfloor \frac{x+1}{c} \right\rfloor = \left\lfloor \frac{x}{c} \right\rfloor. $$

@WillJagy OK.

7:58 PM

$35u-96v=1$

is there any way to find this (u,v) of this Diophantine equation

@PeterTamaroff youtube.com/watch?v=YrLk4vdY28Q

@pourjour Yes, very carefully.

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