Mathematics - 2017-01-01 (page 4 of 4)

Studentmath

19:00

So I did get it right after all

Daminark

And since $f$ is only defined on $[0,3]$, $x = |x| \implies h_2(x) = f(x)$

Studentmath

Maybe we are talking about the same subject after all.

Daminark

I was talking about philosophy :/

Studentmath

Well, that changed quickly

I have a nice one

DHMO

my program reached 240! but i have to go now

might try implementing some algorithms later

Studentmath

19:11

Consider $E$ to be a field of characteristic $p\neq 0$.
Now we look at $F={a^p|a\in E}$. One can easily show that $F$ is a subfield of $E$.
Now say $E$ is a finite extension of $F$

OneRaynyDay

hey guys? Was wondering if there is a convergence value for $n*d^{n-1}$

where sum from n = 1 to infinity.

and $ d \lt 1 $

Studentmath

I want to show $[E:F]$ is a $p$. Now I can easily show that for $a\in E$, $a$ is algebraic over $F$ with degree at most $p$. But I can't figure out why it is precisely $p$.

@oneray I think $1/(d-1)^2$. Maybe.

arctic tern

@OneRaynyDay you are not asking if there is a "convergence value for nd^(n-1)," which is nonsensical. you are asking if sum nd^(n-1) exists.

and yes, just differentiate the geometric series term by term

@Studentmath are you sure it's precisely p? it seems if E=K(t,s), then F=K(t^p,s^p) has index p^2 inside E

also use \{ \} for curly braces in latex

OneRaynyDay

ah I see... so what do you mean by differentiate the geometric series? I know that for this it's true: $\sum_{n=1}^\infty x^n = 1/(1-x)$

Studentmath

Thanks! Been trying to figure out how to do that. I used to recall..

arctic tern

19:19

@OneRaynyDay take the derivative of both sides

OneRaynyDay

oh what the heck - gotcha

Studentmath

@arctic Sorry, you're right. $[E:F]$ is a power of $p$. I want to show that for every $a\in E$, it is either algebraic of degree 1 or $p$ over $F$ in order to prove that

Unless I am missing something.

arctic tern

suppose a is in E\F. then a is a root of (x-a)^p=x^p-a^p in F[x]. it must be irreducible over F, since (x-a)^r won't be in F[x] for any r<p.

Studentmath

The irreducible part is what I can't figure out - why won't it be?

arctic tern

any polynomial factor of (x-a)^p will be (x-a)^r for some r<p. but the second-to-leading coefficient will by ra (up to sign), which is not in F since r<p, hence (x-a)^r will never be in F[x] for r<p.

OneRaynyDay

19:29

@Studentmath It's actually $ln(1/1-d)$

but I see where you got that from :)

arctic tern

@OneRaynyDay no. I told you to differentiate, not integrate

(so it is actually me who sees where you got that from)

OneRaynyDay

But by differentiating on both sides, we don't get the correct answer, right?

arctic tern

@OneRaynyDay huh?

OneRaynyDay

so like, we get the $d(answer)/dx = \sum_{n=1}^\infty x^n$

so shouldn't we integrate the result?

arctic tern

differentiating $$\sum_{n=0}^\infty x^n=\frac{1}{1-x}$$ gives $$\sum_{n=1}^\infty nx^{n-1}=\frac{1}{(1-x)^2}$$

OneRaynyDay

19:31

just differentiating gives us the geometric series, where I got $1/(1-x)$

Oh holy derp - I see where I made the mistake

thank you embarrassed

Studentmath

@Arctic Oh I get it now. I was completely over-complicating it, thank you very much!

arctic tern

np

Daminark

= p

@arctictern I just had to take that opportunity

Mahmoud

$\cos(x+y)=\cos(x)\cos(y)-\sin(x)\sin(y)$, $\sin(x+y)=\sin(x)\cos(y)+\cos(x)\sin(y)$ Any intuition behind those ? I find them very hard to memorize properly.

@Daminark Thanks.

arctic tern

I can multiply complex numbers in my head, $(a+bi)(c+di)=(ac-bd)+(ad+bc)i$. Those identities are just the real and imaginary parts of $e^{i(x+y)}=e^{ix}e^{iy}$ (using the identity $e^{ix}=\cos(x)+\sin(x)i$). That's how I have them memorized.

GFauxPas

19:41

well yeah, polar form is a lot easier to multiply than rectangular form

multiply the magnitudes, add the arguments

Mahmoud

Trivial ! But very useful, I'm a fan !

GFauxPas

@Mahmoud there are geometric proofs of them but they won't help you memorize the rule

Mahmoud

@GFauxPas Yes, they suck, I've seen two of them,

GFauxPas

well what you can do is write $\cos \theta = \operatorname{Re} \left({e^{i \theta}}\right)$

so $\cos \left({x + y}\right) = \operatorname{Re} \left({e^{i(x+y)}}\right)$

see if that helps you at all

expand out $e^{i(x+y)}$ and collect all real terms

Daminark

Lol after seeing the total clunkiness by which Spivak defined trigonometry I've decided to just say screw it, I'm sticking with defining it through $e^ix$

GFauxPas

19:47

real trig?

I'd rather Euler's formula be a theorem rather than a definition

if it's a definition it's less profound

Daminark

I mean I tend to prefer defining sine/cosine by their power series

GFauxPas

the approach I like best is

define it for triangles for $(0..\pi/2)$

then show how that's equivalent to the unit circle definition, with standard convention for signed distances

then prove $\dfrac {\sin x}{x} \to 1$ geometrically

that's enough to get you $D_\theta \sin \theta = \cos \theta$

and likelywise for cosine

then you make the power series

show error goes to zero

and then put $z$ in the power series and bam it's complex

Mahmoud

@GFauxPas Can we also do $i\sin(x+y)=\Im(e^{i(x+y)})$ ?

GFauxPas

why would you want to do that

Daminark

Yup

GFauxPas

19:51

I mena you can

Daminark

Or no

GFauxPas

but just multiply out all terms and pick the ones that are wholly real

Daminark

Not $i\sin(x+y)$, the imaginary part is just $\sin(x+y)$

GFauxPas

$e^{i(x+y)}=e^{ix}e^{iy}$

Mahmoud

@GFauxPas To get the formula for $\sin(x+y)$

GFauxPas

19:53

that's one way to do it, yup

Mahmoud

And you can do \Re and \Im

For $\Re$ and $\Im$

GFauxPas

that font looks ugly

for uppercase letters at least

fraktur

Mahmoud

Unless I'm mistaken and there are for something completely different

GFauxPas

they're synanomous, but I don't like how those letters look

Mahmoud

I find them ... ancient.

My English vocabulary failed to describe them.

arctic tern

19:57

heh

GFauxPas

"ugly"

Kaj Hansen

Cannot be reproduced by hand

GFauxPas

another reason why they're bad, yes

arctic tern

well $\frak I$ is just $J$ with a big top

Kaj Hansen

I usually see lowercase fraktur

Astyx

20:05

$\frak J$ is very similar to $\frak I$

Jessy Cat

20:23

1

Q: Show that Centralizer is a subgroup of Normalizer - Intermediate Step

I am in the process of proving that $C_{G}(H)$ is a normal subgroup of $N_{G}(H)$, where $C_{G}(H)$ is the centralizer - i.e., the set of elements $g \in G$ that commute with all $h \in H \leqslant G$ - and $N_{G}(H)$ is the normalizer - i.e., the set of all $g \in G$ such that $g^{-1}Hg = H$. I...

Group theory question this time. I know y'all know a lot about that.

GPhys

Hello

2017 is the maximum number of slices you can cut a pizza into with 63 straight cuts :D

Fargle

@JessyCat To show that $H \leq K \leq G$ it is enough to show that $H \leq G$ and $H \subset K$.

GPhys

imo the most interesting property is 2017 is the smallest positive integer whose third root begins with all ten decimal digits without repetition

Astyx

@GPhys Is that only with 2 dimensionnal cuts ?

GPhys

correct @Astyx

2017 is not the maximum for any number of slices for the 3D equivalent

(known as a "cake number")

Astyx

20:28

Is the other one a pizza number ?

GPhys

no

Astyx

Would have been fun :p

GPhys

I've seen it called a pancake number, but the wiki page is something else

let me find it

the 2D one: en.wikipedia.org/wiki/Lazy_caterer's_sequence

Astyx

Nice

arctic tern

@JessyCat if C(H) is a subset of N(H), and C(H) is a group under the group operation it shares with N(H) and G, then yes C(H) is a subgroup of N(H).

Fargle

20:35

$2017 = 2 \cdot 33^2 - 161$, and $161$ was my old bus number.

Am I a math yet?

4

Jessy Cat

@arctictern is the group operation the same? In C(H) it's multiplication

Also, what was that other guy saying about how if H and K are both subgroups of G. but K is contained in H, K is also a subgroup of H?

I like that, and now I'm thinking I want to prove it. Not sure I know exactly how, but I'm going to take a stab at it.

Starfall

i don't see how there's anything to prove here

Jessy Cat

20:52

Yeah neither do I now that I just wrote it all out.

If there was an award for overthinking, I'd have won it many times over.

Oh somebody please star that - that's golden

Mahmoud

Hi.

Jessy Cat

Hi

Mahmoud

How can one write a formula for $24$ including the digits, $5$, $5$, $5$ and $1$ ?

Starfall

does [(5+5)/5] [5 - 1] count

in base 10, obviously

oops, that's one 5 too many

Krijn

5*5-1^5

Starfall

21:01

that sounds better

or maybe (5 - 1^(5+5))!

Mahmoud

Nice.

Astyx

$(55-51)!$

Mahmoud

@Astyx Most creative one.

Astyx

Do I get an award for this ? :D

Mahmoud

Ehm .. like what ?

Astyx

21:06

No clue

Maybe 2 more weeks of holidays ?

I wish

Mahmoud

I don't control that, sorry.

What about this link

Astyx

I'll take it

Mahmoud

@Astyx Did you try one of the videos ?

Astyx

I am right now

Mahmoud

21:25

Thoughts ? @Astyx

Astyx

The video I watched was nice

Mahmoud

Which one ?

Astyx

Riemann zeta function

ie the first one that showed up

Mahmoud

He is brilliant, fighting for ''intuitive understanding'', and tries to make things as simple as they can get. @Astyx

Starfall

how expensive is it to compute the minimal polynomial of a square matrix

Astyx

21:35

True

Minimal polynomial ?

Starfall

i think it can be done in O(n^5) time for a n x n matrix, can we improve on this

@Astyx yes

Astyx

Do you know for sure it can be done in $O(n^5)$ ? And could you give such an algorithm ?

Starfall

O(n^2)?

i said O(n^5)

Mahmoud

$\operatorname{O}()$ is an operator ?

Starfall

it's the order of complextiy

complexity*

Astyx

21:41

My bad, typo

Mahmoud

Of .. .

Astyx

the algorithm

Mahmoud

Order of complexity of the algorithm ?

Starfall

yes

Astyx

Basically if your matrix is $n\times n$, you expect a number of elementary operations that is asymptotically a $O(n^5)$

So what algorithm are you thinking about @Starfall ?

Starfall

21:48

@Astyx do you actually know a better algorithm

Astyx

No

But I'd be interrested to see yours

Starfall

compute the minimal polynomial in a bunch of cyclic subspaces, then take lcms

row reduction is O(n^3), and you row reduce O(n^2) times at worst, so O(n^5) is a good upper bound on the complexity

if you're lucky then you could get away with O(n) row reductions

Akiva Weinberger

Hello!

How can I show that no three points in the image of the curve $f(t)=(t,t^2,t^3)$ are colinear?

This problem came from Ted but I have no idea how to approach it

Astyx

Suppose they are

That's absurd

Akiva Weinberger

Maybe I take three points on it, $a$, $b$, and $c$, and show that $(a-b)\times(a-c)$ is nonzero? (Cross product)

Astyx

21:52

Thus they're not

Akiva Weinberger

@Astyx Thanks. -_-

:P

Starfall

@AkivaWeinberger the idea is to write down a square matrix whose rows would be linearly dependent

Astyx

That's actually nearly a Vandermonde matrix there

Starfall

and then use column rank = row rank to pass to "columns are linearly dependent"

Astyx

You can compute the determinant

See it's non zero

Oh wait I'm being silly

Akiva Weinberger

21:54

The determinant of what? $[\vec a\mid\vec b\mid\vec c]$? What if $\vec a=0$?

Astyx

No I was thinking of something else

Akiva Weinberger

Hm. $(a-b)\times(a-c)=a\times a-a\times(b+c)+b\times c$

So I want to show $a\times(b+c)\ne b\times c$ perhaps

I need paper…

Wait. Hold on.

Mahmoud

$\stackrel{y}{_x\triangle_{z}}$ Testing some commands.

Akiva Weinberger

The projection onto the $y$-coordinate is $f_y(t)=t^2$, yeah?

That's a parabola

@Mahmoud Cool. (Is this a reference to the 3Blue1Brown video?)

Mahmoud

$\log_x(z)=\stackrel{}{_x\triangle_{z}}$

Akiva Weinberger

21:58

No three points on a parabola are colinear, right?

Mahmoud

@AkivaWeinberger Probably, I wanted to test the $\LaTeX$

Akiva Weinberger

So no three points on $(t,t^2,t^3)$ can be colinear, right?

Astyx

Yeah but I think that's the point

Mahmoud

$x^y=\stackrel{y}{_x\triangle}$

Akiva Weinberger

So I don't need to do any cross product stuff, then. If three points on it were colinear, so would their projection onto the $xy$ plane, which is impossible because that's a parabola.

Also, small error above, I meant that projecting onto the $xy$-plane gives $f_{12}(t)=(t,t^2)$ which is a parabola.

Starfall

22:01

@AkivaWeinberger that works

Mahmoud

$\stackrel{y}{_x\triangle_z}$

$\stackrel{y}{_x\triangle_{z}}$

$\stackrel{y}{_x\triangle_{z}}\cdot \stackrel{b}{_a\triangle_{c}}$

Astyx

I'm pretty sure you can work with determinants too

Take three points in the curve

Mahmoud

I still have no idea how this \stackrel{y}{_x\triangle_{z}} works and why is it like this.

Astyx

If none of them are 0, they are colinear iff their determinant is non zero

Mahmoud

Like, what does \stackrel do ?

Astyx

22:05

And if one is zero, the other two are trivially equal if they are colinear

Starfall

@Astyx you don't need determinants, although vandermonde does give the result

ya

Astyx

Sure, you don't need them

Starfall

you find that one of them are zero

and that implies that the other two are equal

Astyx

Just my proposition

Starfall

that's the idea i had in mind originally

Sophie

22:05

does $\int_1^\infty\frac{1}{\lfloor t \rfloor^x} - \frac{1}{t^x} dt$ always diverge when $x<0$?

Alessandro Codenotti

I think that problem can be approached also by projectivising (is that even a word?) the curve and doing some Bezout's theorem magic

Astyx

@Sophie you might better write it as $\int_{1}^{+\infty} {\lfloor t\rfloor}^x - t^x$ for $x\ge 0$

Mahmoud

Test $\mathop{\sum \sum}_{i,j=1}^{N} a_i a_j$

Akiva Weinberger

@Astyx Not true. Take $(1,0,0)$, $(0,1,0)$, and $(1,1,0)$

Astyx

Oh yeah

Starfall

22:09

@AkivaWeinberger it's true for your curve

Astyx

My brain isn't fully functionnal tonight it seems

Mahmoud

$\bar{z}$

Starfall

oh, wait, @Astyx wasn't making any sense

Alessandro Codenotti

You can take $3$ points $x,y,z$, write a $2\times \text{something}$ matrix with $x-y$ and $y-z$ on the rows and see if the rank of this matrix is $1$ (collinear) or $2$ (not collinear)

Astyx

Let's just say the determinant being 0 if they are colinear is sufficient

We do not have equivalence

Akiva Weinberger

22:10

@Mahmoud Test: $\stackrel ab$

Astyx

But is they are colinear, the determinant is 0, then that means two points are equal

Akiva Weinberger

$\stackrel{\rm one~thing}{\rm another}$

Astyx

Hope I'm not being more and more wrong

Akiva Weinberger

@Mahmoud OK. \stackrel puts one thing on top of another.

Mahmoud

$\stackrel{Mathematics}{Physics}$

Assuming no physicist is here.

Right ?

Akiva Weinberger

22:12

Compare \over: $Mathematics\over Physics$

Mahmoud

$A\over B$

Akiva Weinberger

(\over is essentially \frac)

Alessandro Codenotti

$\mathrel{Mathematics}{Physics}$

Mahmoud

It works better than \frac

Astyx

$$\underbrace{Mathematics}_{Physics}$$

Mahmoud

22:13

$\underbrace{A}$

Sophie

$\stackrel{\stackrel{\stackrel{s}{h}}{\stackrel{i}{t}}}{\stackrel{\stackrel{p}{o‌}}{\stackrel{s}{t}}}$

Astyx

Gotta go

Bye

Akiva Weinberger

@Sophie I see.

Mahmoud

$\underbrace{A}_b$

$a_n=\underbrace{0.111\cdots 111}_{n \; \text{times}}$

That's an interesting sequence !

$\stackrel{\cdot \cdot}{\smile}$

Pissed off layman

(removed)

Mahmoud

22:28

$\text{(removed)}$

${\tt a}$

Pissed off layman

that trophy is a nice fit on your avatar

Mahmoud

Thank you ${\tt :)}$

$\displaystyle \sum$

$\sum$

Akiva Weinberger

$\sum\Sigma$

Mahmoud

$\huge\color{blue}{\displaystyle \aleph_0}$

Akiva Weinberger

$\color{Red}r\color{Orange}o\color{Yellow}y\color{Green}g\color{Blue}b\color{ind‌igo}i\color{violet}v$

Mahmoud

22:37

$\displaystyle{\partial \over \partial{x}}$

$\displaystyle{\zeta(1)=\infty}$

$a/_b$

$\text{$N/_{Kg}$}$

Mike Miller

@Balarka

Mahmoud

$\rm{E}$

$E$

$\text{E}$

$\int_0^\infty \mathrm{e}^{-x}\,\mathrm{d}x$

$\int_0^\infty \rm{e}^{-x}\, \rm{d}x$

$\iiint$

$\int\dots \int$

$\idotsint$

Akiva Weinberger

$sk\dot\imath\ddot\imath ng$

Mahmoud

$\big($

Akiva Weinberger

skiïng

Mahmoud

22:51

$\bigg($

$( \big( \Big( \bigg( \Bigg($

$\circ \big( \Big( \bigg( \Bigg($

$\cdots \Bigg) \bigg) \Big) \big) ) \; \infty \; ( \big( \Big( \bigg( \Bigg( \cdots$

@AkivaWeinberger What do you think ?

Akiva Weinberger

I think that's enough LaTeX for one day

Mahmoud

It's so complex, a lot of commands

$\mathfrak{A}$

$\mathfrak{M}$

$\mathfrak{I}$

$\mathfrak{J}$

$\emptyset$ vs $\varnothing$ ?

What is $\wp$ for ?

Akiva Weinberger

Weierstrass p function, I think it's called

If I recall correctly, it's a complex function with two periods, so that $f(z+a)=f(z+b)=f(z)$ for some distinct complex numbers $a$ and $b$

so its values on a parallelogram in $\Bbb C$ determine it everywhere

It's complex-differentiable, and it has a pole somewhere in that parallelogram, I think

GFauxPas

23:07

guys what the heck is going on here, complex functions are so surprising

Mahmoud

@AkivaWeinberger Lot more to learn about Complex Analysis.

GFauxPas

I lost 3 lines???

it could be my code is wrong

Akiva Weinberger

Is this Lambert W?

GFauxPas

yes

curves and the images of the curves under W

let me check the data

Akiva Weinberger

I didn't realize you could do $W_0$ on the complex plane. I guess you do a branch cut somewhere?

GFauxPas

23:10

no they're overlapping, they're not hidden

they're piling on top of each other

yes you do a branch cut $(-\infty..-\exp(-1))$

yeah the problem is they're merging on top of one another

Akiva Weinberger

So I needed to find a function $\Bbb R^2\to\Bbb R$ that's continuous everywhere except for the origin, such that it's continuous at $\bf0$ on every line through the origin, and somehow unbounded in every neighborhood of the origin.

GFauxPas

whoa

Akiva Weinberger

I think this works:

$\displaystyle f(x,y)=\frac{2x^4y}{x^8+y^4}$

and $f(0,0)=0$.

On every line through the origin, the limit is continuous… but on the parabola $y=x^2$, it takes the value $1/y$ and so approaches infinity at the origin.

(And the same at $y=-x^2$, so it approaches minus infinity from there.)

Mahmoud

23:33

Very interesting : $\prod_{i=1}^{n} (r_i, \theta_i)=(r_1\dots r_n, \theta_1+\dots +\theta_n)$

Adeek

hi everyone

Mahmoud

Hi @Adeek

Adeek

hi @Mahmoud

Benjamin R

23:51

Hi Beautiful Math Geniuses.

Happy New Year to y'all!

Quick question, probably not worthy of posting:

In *Introduction to Abstract Algebra* (2014) by Fine, Gaglione, Rosenberger, p.459, we have a very clear definition of the properties of the **norm** on a vector space such that *"Any inner product space $V$ is a normed linear space with |v| = $\sqrt{<v,v>}\ $ for* [all] *v $\in V$."*

Where *A* **norm** *on a vector space is a function $|\ |\ :\ V \rightarrow \mathbb{R}\ \ $satisfying:*

$(1)\ \ |v| \geq 0 \ \ for\ all\ v \in V$

$(2)\ \ |v| = 0 \ \ if\ and\ only\ if\ \ v = 0_{v}$

$(3)\ \ |tv| = |t||v|\ \ for\ \text{[all]}\ v\ \in V\ and\ t\ \in \mathbb{R}$

(why is the formatting syntax different in chat from SE posts? that's dumb)

Alessandro Codenotti

$\sqrt{t^2}=|t|$, you want the positive one (if we had $|tv|=t\cdot|v|$ then it'd be hard to respect $(1)$ )

Transcript for

$Mathematics$ Mathematics