Mathematics - 2015-01-18 (page 2 of 6)

On twitter they write: We have enabled Cloudflare for stackoverflow.com while solving the routing issue. There are known SSL issues but it should help. I am not sure what that means, but at least I see that I can access the site.

Hippalectryon

@MartinSleziak It's a cached version

user112495

Let G be a group with $x, y \in G$ and $r, s, n \in \mathbb{N}$. If $|x|=r \cdot s$ and $|y| = n$, find the orders of $x^r$ and $y^r$.

I got that $|x^r| = s$.

For the second one, I got $|y^r| = \frac{n}{r}$, but I know this is not correct as if I take $D_4$ with $n=4$ and $r=3$ I get the wrong answer.

Martin Sleziak

@user112495 What is the smallest multiple of r which is divisible by n?

user112495

@MartinSleziak $lcm(n, r)$

Martin Sleziak

10:21 AM

Correct. Does this help with your problem?

user112495

@MartinSleziak Okay, so it looks like the answer should be $\frac{lcm(n,r)}{r}$ then. But I'm still not sure how I would go about properly proving this.

@MartinSleziak Oh, is it just because we need it to be an integer.

And so $\frac{lcm(n,r)}{r}$ is the smallest integer (as we are using lcm) such that ${y^r}^{{\frac{lcm(n,r)}{r}}} = 1_G$?

Martin Sleziak

I think that the basic fact is that for any element of the form $y^k$ there exists unique $j\in\{0,1,\dots,n-1\}$ such that $y^k=y^j$. Namely, $j$ is the remainder of $k$ when dividing by $n$.

After using this fact, the problem is change from group theoretic problem to question from elementary number theory.

@user112495 Indeed, you get $1_G$ in this way. When you write down the proof, you should also explain that $y^{rk}$ is not $1_G$ for smaller $k$'s.

But I think that the argument should be clear from what we said above.

David Zhang

Hey, does anybody remember where the following problem was posted on MSE?

Let f: R -> R be a C^\infty function with the following property:

For each real number x, there exists a natural n such that f^(n)(x) (that is, the nth derivative of f at x) vanishes.

Is f necessarily a polynomial?

I know I've seen it posted before, but I can't find it now.

Martin Sleziak

10:38 AM

@DavidZhang What about Baire: Show that $f\colon \mathbb{R}\to\mathbb{R}$ is a polynomial in an open bounded set?

It was among the first hits when I put derivative vanishes polynomial site:math.stackexchange.com in Google.

If you try some similar queries or have a look at the other results from that search, you might find a few more copies of that question.

David Zhang

Yeah, I think that's it. Looks like I remembered a few details wrong. Thanks!

Martin Sleziak

10:52 AM

The routing issues should be resolved now. Sorry for any inconvenience.

Tweeted by StackStatus on January 18, 2015 at 10:32 AM

pourjour

11:09 AM

Hi

$g(y)f^{''}(x)+f(x)g^{''}(y)+ (\alpha ^2 + \beta ^2)f(x)g(y)= 0 \implies f^{''}(x)+\alpha ^2f(x) = 0 , g^{''}(y)+ \beta ^2 g(y)=0$

can someone help me to prove this

Jasper Loy

@Alessandro You don't like Shoenfield? Take a look at Mendelson then. But it might be equally hard. But I think these two are the best.

Alessandro

@JasperLoy I wouldn't say that I don't like it, but it's a bit too advanced for my current level

Hippalectryon

11:30 AM

@pourjour If you don't give us the context, we can't help.. what are $f,g,x,y$ ?

pourjour

that's what is it all about

@Hippalectryon the electrostatic field is $E(x,y)= f(x).g(y)$

maybe it's about implicit function theory

Hippalectryon

@pourjour Is your property above verified for all $x,y$ ?

pourjour

@Hippalectryon yes

Hippalectryon

11:50 AM

Hmm so $g(f''+\alpha^2f)+f(g''+\beta^2g)=0$

Which is pretty close

@pourjour I can give an answer from Physics' point of view, but it might be trickier from Maths' point of view

anon

@Hippalectryon it's not a good idea to suppress the arguments if there are two different arguments at play

then a reader would be unable to recover the original equation from seeing your equation

Hippalectryon

@anon But I'm lazy :P

$g(y)(f″(x)+\alpha^2f(x))+f(x)(g″(y)+\beta^2g(y))=0$

anon

you can rewrite $A(x)B(y)+C(x)D(y)=0$ in the form $F(x)=G(y)$; what do we all know that implies?

pourjour

@Hippalectryon thanks

Hippalectryon

@anon But here we don't exactly have that form

anon

11:58 AM

2 mins ago, by Hippalectryon

$g(y)(f″(x)+\alpha^2f(x))+f(x)(g″(y)+\beta^2g(y))=0$

look at it

Hippalectryon

Ooh lol

Of course

Ramanewbie

hi hippa !

anon

anyway I get that there exists a constant $k$ so that $f''(x)+(\alpha^2-k)f(x)=0$ and $g''(y)+(\beta^2+k)g(y)=0$; without boundary conditions I don't think there's anything to imply $k=0$ necessarily

Hippalectryon

@Ramanewbie >.>

Ramanewbie

? bananas ?

why did you tell @KajHansen you were 14 years old ???

Hippalectryon

12:01 PM

@Ramanewbie I didn't

Ramanewbie

That's what he thinks

"Yeah. I think it says on his MSE profile or something."

Hippalectryon

Hippalectryon, France

3k 2 15 43

The age section is empty

Ramanewbie

Where is it ?

Chris's sis

Greetings

Hippalectryon

@Chris'ssis Hello :DDDD

Chris's sis

12:08 PM

@Hippalectryon Hello :-)

Ramanewbie

whut

Hippalectryon

@Chris'ssis bbl

Chris's sis

@Hippalectryon OK :-)

evinda

Hallo @DanielFischer !!! :D

Jasper Loy

Hello @Chris'ssis. I think your book should be published this year.

2

Chris's sis

12:17 PM

@JasperLoy Hello! I agree with you! :-) How are you these days?

evinda

    @DanielFischer Could we modify the Binary Search also like that?
     int binary_search(int A[], int key, int low, int high, int K)
    {
      if (high < low) return 0;
      else
        {
          int mid=low+floor((high-low)/2);
          if (abs(A[mid]-key) > K) return binary_search(A, key, low, mid - 1);
          else if (abs(A[mid]-key)<=K) return mid;

        }
    }

Jasper Loy

@Chris'ssis Not good. I need to find out answers to 2 questions (1) Why I am still not well after so many years of trying (2) What I must do to get well. I am writing Jonas an email now. Do you remember him?

Chris's sis

@JasperLoy Yeah, a bit.

Daniel Fischer

1:07 PM

@evinda No, not like that, abs(A[mid]-key) > K doesn't tell you in which half of the array you should look further. You must distinguish between A[mid] > key + K and A[mid] < key - K.

Alessandro

1:18 PM

@JasperLoy from the preview available on google books Mendelson seems to be more accesible than Shoenfield. What do you think about Enderton's "introduction to mathematical logic" (and the following "elements of set theory)?

Jasper Loy

@Alessandro Mendelson is more modern than Shoenfield, obviously. Enderton's books are great too. All these are classics.

Lord_Farin

A simple reference request:

After about one-and-a-half year of hiatus, I'm starting to look into maths more seriously again. My favourite topics are logic, category theory, and topology.

I'm looking for new books to expand my library.

Jasper Loy

@Lord_Farin What kind of logic and what kind of topology?

Lord_Farin

@JasperLoy Model theory / formal logic, and mostly general topology.

Jasper Loy

@Lord_Farin Willard's General Topology is good.

Lord_Farin

1:25 PM

I also ventured in the direction of algebraic topology.

Jasper Loy

@Lord_Farin Bredon's Topology and Geometry is good.

Lord_Farin

@JasperLoy I already have Switzer for AT but I notice it's too advanced for me at this point, I lack topological background/experience.

Do you know Switzer's book?

Jasper Loy

@Lord_Farin Bredon's book is 3 in 1, point set, differential and algebraic. Try it.

@Lord_Farin Yes, it is very good but very advanced.

Lord_Farin

@Jasper Thanks for the recommendations, I'll definitely look into them! I had the impression that Switzer is good, but it was a bit of a let-down that I wasn't able to handle it :(.

I shouldn't have been surprised though, jumping in after essentially two years without much topology.

@Jasper Are you a topologist? You seem very knowledgeable.

Jasper Loy

@Lord_Farin No, after I finished my undergrad and worked as a teacher for a while, I stopped working and have been trying to recover from my mental illness the past many years. I don't know what will happen to me, the future is bleak. I hope that by the time I get well, grad schools will still accept me.

Lord_Farin

1:31 PM

@Jasper :( I'm sad to hear that.

Alessandro

@JasperLoy many thanks for the recommendations, I'll look more into Mendelson and Enderton, but considering the prices I might stick to Shoenfield and ask (a lot of) questions on math.stackexchange when I can't understand something :)

Lord_Farin

@JasperLoy What are the troubles you're dealing with? (Naturally, don't answer if you don't feel like it.)

Jasper Loy

@Lord_Farin OCD, PTSD mainly.

Lord_Farin

@JasperLoy Mmh, I can imagine that to be really bad because people tend not to be very understanding about these problems.

:(

evinda

@DanielFischer We want to have $-K+key \leq A[mid] \leq K+key$, right?
So, if $A[mid]>key+K$, we have to look at the interval $[low,mid-1]$, right?
If $A[mid]<key-K$ the desired property is satisfied.. Or am I wrong?

Daniel Fischer

1:39 PM

@evinda If A[mid] < key -K, then we have |A[mid] - key]| = key - A[mid] > K.

Lord_Farin

@Jasper I have to go now, thanks once more for your recommendations. I wish you strength in overcoming your illness.

Jasper Loy

@Lord_Farin Thank you.

skullpatrol

see you pal

evinda

1:58 PM

@DanielFischer So, when we have $A[mid]<key-K$ then it cannot be that $A[mid+i]-key<K$, right?

Jasper Loy

2:10 PM

Hi @user130018

Ramanewbie

2:26 PM

hi

is there anyone here ?

Hippalectryon

@Ramanewbie no

Ramanewbie

Hippa I've a question :

I'ld like to find a number which respects these conditions :

$1+4x$ must be a natural positive root

is it clear ?

natural ?

yes

you mean integer ?

2:35 PM

the number ensemble

like decimals, reals...

Hippalectryon

??

$\mathbb{N}$

Ramanewbie

yes integer sorry ^^

Hippalectryon

and what does "being a root" mean to you ?

Ramanewbie

being an integer root

I mean :

If you take the root of this number, you get an integer

Hippalectryon

Ah, so it's called a perfect square

Ramanewbie

2:38 PM

yes... I don't know all the vocabulary yet ^^

Hippalectryon

Well, take $9=4*2+1$

Ramanewbie

hum...

I would like to know how you make a proper equation with what I told

Hippalectryon

You asked for a number satisfying the property

Ramanewbie

That's not what I meant

I made a mistake

I wanted to make an equation

Hippalectryon

$4x+1=a^2,a\in\mathbb{N}$

Ramanewbie

2:41 PM

what ?

why does it fit with what I wanted ?

are you sure it works ?

Hippalectryon

That's exactly what you asked... $4x+1$ is a perfect square in this case...

Ramanewbie

you just added a∈N !

Hippalectryon

That was obvious

Ramanewbie

^^ for you

I remind you I'm a noob !

2

so $ x=\dfrac{a^2-1}{4}$

But there are two variables, how can find x ?

Gato

heyheyhey

Hippalectryon

2:48 PM

@Ramanewbie Your set of solutions is $\{\dfrac{a^2-1}4\mid a\in\mathbb{N}\text{ and }\dfrac{a^2-1}4\in\mathbb{N}\}$

Ramanewbie

ok now

what's the point of the pipe ?

Hippalectryon

For instance, $\dfrac{3^2-1}4=2$

@Ramanewbie pipe????

Ramanewbie

| don't you call it a pipe ? |

Hippalectryon

No, it's just a middle bar ...

@Huy o/

Huy

sup @Hippalectryon

Hippalectryon

2:51 PM

@Huy $\inf$

Ramanewbie

so what's a pipe ?

Hippalectryon

wordreference.com/enfr/pipe

Ramanewbie

I know that

but I thought it was also a keyboard key

isn't it ??

Hippalectryon

No idea

Ramanewbie

@everybody how can I change my avatar ?

Alessandro

2:58 PM

can you show us which key or symbol you meant with "pipe"? So that we can tell you the right name

Ramanewbie

alt-gr + 6

|

on French keyboards ^^

on a US one, I've no idea

kb.mailchimp.com/merge-tags/using/where-to-find-the-pipe-key

Alessandro

As far as I know that's usually called a vertical bar

Ramanewbie

maybe...

but I heard lots of people call it a pipe

On YouTube for instance

Alessandro

anyway when describing a set it means "such that"

Ramanewbie

ok

how can I change my avatar ?

Alessandro

3:02 PM

$\{x|\sqrt{x}\in\Bbb{N}\}$ is the of all the $x$ such that their square roots is an integer (i.e. the set of all the perfect squares)

Ramanewbie

@Alessandro do you know how I can change my avatar ?? I searched in my profile but cannot find it !!

Alessandro

in the main site (math.stackexchange.com) click on "user" in the upper bar, then click on edit and here you can change nickname, personal information, avatar and such

Ramanewbie

tks

@Hippalectryon does that mean I can choose any integer to remplace a ?

Hippalectryon

Yes

Ramanewbie

hum...

doesn't work

Hippalectryon

3:15 PM

It does, if you follow the two conditions

Ramanewbie

what ones ? I know only one condition

Hippalectryon

$\{\dfrac{a^2-1}4\mid a\in\mathbb{N}\text{ and }\dfrac{a^2-1}4\in\mathbb{N}\}$

Ramanewbie

so the condition is that $a\in\mathbb{N}$

right ?

Hippalectryon

and that $\dfrac{a^2-1}4\in\mathbb{N}$...

Ramanewbie

that changes everything

I understand now

I changed my avatar, why is it still the same on here ?

Hippalectryon

3:20 PM

The database is cached up to some point, hopefully...

Ramanewbie

what must I do ? clear the cache ?

Hippalectryon

No. Just wait.

Ramanewbie

pff

my computer is making weird noises

Hippalectryon

This is supposed to me a maths chat q_q

Ramanewbie

haha

supposed to, yeah...

grr... the text on my avatar isn't visible !

Hippalectryon

3:33 PM

It's plagiarism in English

And plagiat in French

Ramanewbie

really?

I'm fixing that... ^^

Hippalectryon

You could as well have written "illiterate"

Ramanewbie

why ?

what's the link ?

Hippalectryon

You can't spell -__-

No

Ramanewbie

pff

I just forgot a "t"

It won't stop the spin of the Earth !

evinda

3:35 PM

How could we show that for $n \geq 3$, it holds that $\lg^{\star}{\lg n}+1 \leq \lg^{\star}n$ ?

Hippalectryon

@evinda $lg$ ?

evinda

@Hippalectryon $\lgn=\log_2n$

Hippalectryon

Oh ok

What about $\lg^\star$ ? @evinda

Alessandro

@evinda Hi! preparing for your exam?

evinda

@Hippalectryon Neither I got taught it, but I found this on wikipedia: en.wikipedia.org/wiki/Iterated_logarithm

@Alessandro Hi!!! Yes, I am looking at some older exam questions..

Ramanewbie

3:42 PM

@Hippalectryon time ^^ don't kill me

how to determine if 3 points (or more) belong to a same straight line, from their coordonates ?

Huy

@Ramanewbie: Write down a parameter form of the line running through two of them, then check if the third lies on that line.

(PS: This was an exercise at my high schoolers' exam too)

Alessandro

@Ramanewbie find the equation of the line passing through 2 of them and see if the third one also satisfies this equation

evinda

@Hippalectryon $$\lg^{\star} n =\min \{ k \geq 0 | \lg^{(k)} n \leq 1\}$$

Alessandro

@Huy woops you ninja!

Ramanewbie

@Huy what's a parameter form ??

Huy

3:50 PM

@Ramanewbie: If you're only in 2 dimensions, use the equation for lines, $y = ax + b$, in higher dimensions you will need the parameter form.

Ramanewbie

ok, well I'm in 2 dimensions then

Axoren

There was an elegant method I found for n-dimensional vectors. To find if they were colinear.

Let me see if I can find it.

Ramanewbie

@Alessandro there's a problem then

let's choose A(0;−1), B(−2;−4) and C(23;0)

I calculate a for A and B :

$a=\dfrac{-4-0}{-2+1}$

$a=4$

and A(0;−1), so $b=-1$

so f(x)$=4x-1$

@Alessandro right ?

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$Mathematics$ Mathematics