Mathematics - 2018-05-12 (page 2 of 3)

Ted Shifrin

05:07

@HarryEvans: Well, sure.

Why wouldn't it be?

So, if $u,v\in C$, why is $su+(1-s)v\in C$ for any $0\le s\le 1$?

Harry Evans

So my idea is I let $a, b \in [0,1]$ and then define $u(at + (1-b)t)$, integrate, etc

Ted Shifrin

No, no. Wrong notion of convex.

You want $C$ to be convex. So take two points of $C$.

Harry Evans

OK, let me see if this is correct

Consider
$$\int_0^1 \left| su(t) \right|^2 dt \quad \text{and} \int_0^1 \left| (1-s) v(t) \right|^2 dt$$

Ted Shifrin

No.

Look what I wrote up there ^^^^ and use the definition of $C$.

Harry Evans

Yup, hear me out, see if it works

I will show the sum is less than 1

Ted Shifrin

05:14

But you're missing the cross-term.

Remember how to square a sum?

Harry Evans

But can't I just factor our $(1-s)^2$?

Ted Shifrin

$[su+(1-s)v]^2 = $ ?

Harry Evans

$s^2u^2 + 2s(1-s)uv + (1-s)^2v^2$

Ted Shifrin

OK ... Now integrate.

Harry Evans

$u < v$, so $uv \le v^2$, correct?

Ted Shifrin

05:19

No.

Where did $u<v$ come from? These are functions in $C$.

Harry Evans

for all $t\in [0,1]$, can we assume that $u(t) \le v(t)$

Ted Shifrin

Say what?

NO.

Harry Evans

The definition of convexity I know is this: if $a,b \in C$, then $[a,b] := \{ta + (1-t)b : t \in [0,1] \} \subseteq C.$

Ted Shifrin

Right, but I used the letter $s$ for a good reason.

Harry Evans

So from that definition, one of the functions must be less than the other

Ted Shifrin

05:22

This holds for any (even infinite-dimensional) vector space.

NOOOOO.

You are totally confused.

Harry Evans

If $u$ is $a$ and $v$ is $b$, right?

Ted Shifrin

In our function-space, yes. But the interval notation $[a,b]$ is just notation. It's not about real numbers and an interval.

Harry Evans

OK, prof @TedShifrin, help me understand so I can come out of my confusion :)

Ted Shifrin

Just use the definition without making unwarranted assumptions.

Harry Evans

I have always thought of the interval as a subset of the set $C$

Ted Shifrin

05:25

Yes, but we're in an infinite-dimensional space, so there's no $<$.

You do need to know a basic fact about $\int_0^1 |u(t)v(t)|dt$.

Harry Evans

Can we use the Holder inequality there?

Ted Shifrin

Aha.

Harry Evans

Actually Cauchy Schwarz?

Ted Shifrin

Sure.

Harry Evans

Thanks, Prof @TedShifrin, I think I'm good! :)

Ted Shifrin

05:29

I knew you'd be fine once you stopped making silly mistakes.

BTW, I'll be up around Stanford for a few days in August. :)

Harry Evans

Oh nice!

I will see if I can drop by and have you sign an autograph

Ted Shifrin

LOL ... you around in summer?

Harry Evans

Yes, I only live near my school

Ted Shifrin

oh cool.

Harry Evans

Thanks much, Prof!!!

Ted Shifrin

05:30

you're welcome.

Daminark

Rehi Ted!

BAYMAX

Hmm..thinking of how ?

$p_{k}$ be $k$th prime number

to show that there exists infinte number of $k $ sch that $p_{k+1}-p_{k} > 2$

?

Tug' Tegin

05:47

Hi! I'd like somebody for a hint to this word problem:
Several timbers each 2 m long were cut into small pieces each 50 sm long in 4.5 hours. How many hours would it have taken to cut if the pieces were 40 sm long each? (I think i have translated it correctly.)
The answer is 6 but i can't get to it, i get 5.5 and other numbers.

abhishek chaudhary

06:21

hi i am abhishek a new user

skull

welcome

abhishek chaudhary

can we prove e to be irrational using basic ideas

skull

Proof that e is irrational

The number e was introduced by Jacob Bernoulli in 1683. More than half a century later, Euler, who had been a student of Jacob's younger brother Johann, proved that e is irrational; that is, that it cannot be expressed as the quotient of two integers. == Euler's proof == Euler wrote the first proof of the fact that e is irrational in 1737 (but the text was only published seven years later). He computed the representation of e as a simple continued fraction, which is e = [ 2 ; 1 , 2 , 1 , ...

Leaky Nun

06:51

@BAYMAX twin prime conjecture?

Silent

I am followig Artin's Algebra, but having really hard time understanding eigenvalues-vectors, etc. Please someone recommend a book or online source that covers eigenvalues, eigenvectors, char. polynomial, maximal polynomial and Calyey-Hamilton theorem well.

Alessandro Codenotti

@BAYMAX even better, for all $n$ you can show that there are consecutive primes distant more than $n$

Secret

Leaky Nun

@Silent 3b1b?

Silent

@LeakyNun I have seen those videos. Any reading material?

Leaky Nun

07:00

@Silent 3b1b should give you a clear picture of matrices (as linear transformations) and eigenvectors (as vectors whose directions do not change)

Silent

ok

Leaky Nun

and the best proof of Cayley-Hamilton to me is: it holds for the diagonalizable matrices, which is dense in the Zariski topology, so it holds for every matrix.

(because Cayley-Hamilton is a closed condition)

Daminark

@Silent I'm not sure if it'll do these things in the order you want, it is quite long, but my favorite linear algebra book is Hoffman and Kunze

Silent

@Daminark Thank you!

Secret

It still not clear to me what Cayley hamilton geometrically means though, as well zeros of matrix polynomials in general

Daminark

07:07

Is there a geometric interpretation?

Leaky Nun

@Daminark yes

@Secret you see, the matrices of dimension n over a field form a topology isomorphic to F^(n^2)

isomorphism by transport of structure

the topology on the latter is the Zariski topology

we will try to establish these facts in order to prove the Cayley-Hamilton theorem

1. The diagonalizable matrices is dense.

2. Cayley-Hamilton is a closed condition.

3. Cayley-Hamilton holds for the diagonalizable matrices.

3 is trivial

1 is because its complement is the zero of some polynomial (specifically, the discriminant)

correction: replace "diagonalizable" with "matrices of distinct eigenvalues"

@Daminark do I make sense

The char-poly of $\begin{bmatrix}a&b\\c&d\end{bmatrix}$ is $\lambda^2 - (a+d)\lambda + (ad - bc)$, whose discriminant is $(a+d)^2 - 4(ad-bc) = a^2+d^2-2ad+4bc$

The discriminant is zero iff the eigenvalues are not distinct

The discriminant is non-zero iff the eigenvalues are distinct

2 is because it is the zero of some polynomial

for the 2x2 case, 2 is the zero of $\begin{bmatrix}a&b\\c&d\end{bmatrix}^2 - (a+d)\begin{bmatrix}a&b\\c&d\end{bmatrix} + (ad-bc)I = \begin{bmatrix}
0 & 0 \\ 0 & 0 \end{bmatrix}$, so it turns out it is the zero of the zero polynomial

what a surprise, since CH holds for F[a,b,c,d] as well

but if we didn't know CH holds, we can still say that it is the zero of a polynomial with degree at most n^3

in at most n variables

so it defines a closed set in F^(n^2)

abhishek chaudhary

07:22

is there a nice career in mathematics

Secret

I wish I knew the details of algebraic variety is, at this level, it is still mostly incomprehensible to me. I'll worry about polynomials later after my PhD

Daminark

Well, you can have non-distinct eigenvalues and still be diagonalizable

But you can take the minimal polynomial

That has 0 discriminant iff the matrix is diagonalizable

(That's a very clever way to think about it, I like it)

Leaky Nun

but you can't easily write the polynomial defining that

I don't think it exists, to be frank

Daminark

Yeah that's the problem

Leaky Nun

whereas the matrices with distinct eigenvalues is the complement of the zero of a polynomial that one can write

Daminark

07:32

The point is that there's more work to verify if you want to say that being diagonalizable is a closed condition, it seems

Leaky Nun

are we arguing for the same thing

Daminark

Err I meant to say open condition, sorry

But I guess you can take a proper subset which is open and you're good

Density still checks out

Leaky Nun

24

Q: A question about "Zariski dense" arguments

This question is a little basic, but I think it is consistent with the goals of MO. My question is about a certain type of argument in algebraic geometry which exploits the abundance of dense sets in the Zariski topology. A classical example is the Cayley-Hamilton theorem, and I will frame my ...

The set of diagonalizable matrices is Zariski dense in F^(n^2) because it contains the complement of the zero locus of the discriminant polynomial.

@Daminark I still don't know how to argue that the complement of the zero is dense though

I know some fields for which it cannot be dense, but how do I argue for the rest of the fields?

Daminark

Isn't there something about how for infinite fields, open sets are dense?

Evinda

Hello!!! I have a question

Leaky Nun

07:44

20

A: Zariski Open Sets are Dense?

As an exercise, let's reduce everything to statements about polynomials. Every open set contains a basic open set $U$, which is the complement of the zero set of some nonzero polynomial $f$, so it suffices to show that these are Zariski dense. The Zariski closure of a set is the intersection of t...

@Daminark probably this

lemma: if f(x) is nonzero, then f(a) is nonzero for infinitely many a

Evinda

We have that U is the interior of the curve C and C has the parametric representation $r(t)=2 \cos^3 t i +2 \sin^3 t j$, $0 \leq t \leq 2 \pi$. We cannot find U, can we?

Daminark

Okay so in that case, we say okay, the map taking a matrix and spitting out the discriminant of its char poly should be continuous

Then diagonalizable matrices contain the complement of the 0 set of that map, so they contain an open set, and open sets are dense

Leaky Nun

@Daminark it's not just continuous, it's a fkin polynomial

Daminark

This is true, the point still stands

Leaky Nun

we've been repeating the same argument for some reason

Evinda

07:48

Hello @LeakyNun
Did you see my question?

Leaky Nun

yes I did

Evinda

Do you have an idea?

Leaky Nun

probably [0,1) multiplied by r

Daminark

So for some reason, earlier I had no idea what you were saying, so I was like wha

Leaky Nun

but we're always saying the same thing

BAYMAX

07:50

Hmm@LeakyNun if we use twin prime conjecture then how can we show infinite number of consecutive primes differing by 2

Secret

(The question dies not exist. In the next downtime, a list showing all the FAQ people in my view will be listed)

Daminark

Well the thing I thought you were doing for a second there was trying to say that the diagonalizable matrices were open

Leaky Nun

ok here goes my corrected proof

1. Cayley-Hamilton is a closed condition

Daminark

So that's why I brought up the minimal polynomial

Leaky Nun

2. Cayley-Hamilton holds for the matrices with distinct eigenvalues

3. Matrices with distinct eigenvalues is dense in an infinite field

BAYMAX

07:52

I meant differing by greater than 2

Leaky Nun

@BAYMAX I may have erred in my thinking

BAYMAX

hmm

Leaky Nun

@Daminark maybe you didn't see my correction

39 mins ago, by Leaky Nun

correction: replace "diagonalizable" with "matrices of distinct eigenvalues"

BAYMAX

That looks like general, how can we do that any all $n$ @AlessandroCodenotti

Daminark

Ah

Leaky Nun

08:01

@Daminark so do we agree lol

I apologize for the confusion

Daminark

Yeah now I'm happy

Leaky Nun

yay

I still want to prove properly that it is dense

Sylent Nyte

08:58

@TedShifrin that cross product that i sent you in my example of a contradiction is correct.

Владислав Харламов

https://math.stackexchange.com/questions/2777665/the-hypothesis-of-the-unbounded-product-of-the-digits-of-a-prime-number
If someone can help, I will be very happy!

2

skull

are you able to prove there are infinitely prime numbers?

*infinitely-many

Владислав Харламов

It does not help in any way, read the task again, please

Abhas Kumar Sinha

09:22

@AkivaWeinberger What's the problem?

Secret

09:38

@ВладиславХарламов Why it would not be bounded. For any two prime numbers, there are only finitely many digits and the product of two finite numbers is always finite hence have finite digits by definition?

Daminark

@Secret the points in the set being finite is very different from the set itself being bounded

The real numbers are definitely not a bounded set

Despite the fact that every real number is finite

ÍgjøgnumMeg

@Secret you can still ask whether or not the product of the digits of a prime can be made arbitrarily large

Secret

so we are considering the set of all possible products of primes?

Daminark

Not quite. Take a prime number, write it in digits, then take the product of those digits

ÍgjøgnumMeg

no, given a prime, the question asks whether or not the product of the digits has an upper bound

Daminark

09:41

Do that for all primes, you get a set of numbers, is that set bounded?

If we could show that you could an infinite set of primes which have a bounded number of 1s as digits and don't have any zeroes, we're done

But it's not immediately obvious to me that this can be done

Secret

hmm... it seems very nontrivial, and it kinda reminds me of the colatz conjecture a bit. For example, if after iterating this process of taking product of the digits the resulting prime grows without bound, then the set will be unbounded, The set will also be unbounded if some function that counts the number of digits stay positive and the set does not eventually start reproducing the numbers already there

The only thing I knew is every natural number can be decomposed into product of primes by the first fundemental theorem of arithmetic, but I don't know what function can capture the digits in terms of expansions of product of primes

For example, $124 = 2^2 \cdot 31$ but $1\cdot 2\cdot 4 = 8 = 2^3$

is there exists some function $f$ that relates e.g.

$2^2 \cdot 3^0 \cdot 5^0 \cdot 7^0 \cdot 11^0 \cdot 13^0 \cdot 17^0 \cdot 19^0 \cdot 23^0 \cdot 29^0 \cdot 31^1$ to

$2^3 \cdot 3^0 \cdot 5^0 \cdot 7^0 \cdot 11^0 \cdot 13^0 \cdot 17^0 \cdot 19^0 \cdot 23^0 \cdot 29^0 \cdot 31^0$

Other thoughts:

$124 = 2^2 \cdot 31 = 100+20+4 = 2^2 \cdot 5^2 + 2^2 \cdot 5^1 + 2^2 = 2^2 (5^2 + 5^1 + 5^0)$

Secret

10:06

$124 = 1 \cdot (2\cdot 5)^2 + 2 \cdot (2 \cdot 5) + 2^2 (2 \cdot 5)^0$

$1\cdot 2 \cdot 4 = 1 \cdot 2 \cdot 2^2 = 2^3$

$2^4 = 16 = 1 \cdot (2 \cdot 5) + 2 \cdot 3 \cdot (2 \cdot 5)^0$

mercio

o..o'

Secret

I have no idea if there exists a function that relates a number's prime expansion to its base b expansion

And the reason why it reminds of the Collatz conjecture is if only a single digit number is left behind the process, then the process terminates and the set will be bounded

O wait, using Daminark's advice, we need to seek all prime numbers with a binary expansion a bounded number of 1s and no zeros, whether zeros can be produced

but we have problem, given any prime number that has binary 11...1 its product of its digits will always be one and given any prime number which has at least one zero, then its product will always be zero

Multiplicative digital root

The multiplicative digital root of a positive integer n is found by multiplying the digits of n together, then repeating this operation until only a single digit remains. This single-digit number is called the multiplicative digital root of n. Multiplicative digital roots depend upon the base in which n is written. If the term is used without qualification, it is assumed that n is written in base 10. Multiplicative digital roots are the multiplicative equivalent of digital roots. == Example == 9876 would be reduced as 9876 -> 9×8×7×6 = 3024 -> 3×0×2×4 = 0. So the multiplicative digital root of...

Thus the problem reduces to one which to find a prime number that has no multiplicative digital root

Repunit

In recreational mathematics, a repunit is a number like 11, 111, or 1111 that contains only the digit 1 — a more specific type of repdigit. The term stands for repeated unit and was coined in 1966 by Albert H. Beiler in his book Recreations in the Theory of Numbers. A repunit prime is a repunit that is also a prime number. Primes that are repunits in base 2 are Mersenne primes. == Definition == The base-b repunits are defined as (this b can be either positive or negative) R n ( b ...

The process seemed to be base dependent. For example:

$32=100000_2$

but $3\cdot 2 = 6$ and $1\cdot 0^5 = 0$

O we can take: $99=1100011_2$

But $9^2 = 81$, $8\cdot 1 = 8$ and meanwhile $1^4 \cdot 0^3 = 0$

so the first set has elements $99,81,8$ but in binary it is $1100011_2, 0_2$

In fact in binary, given any natural $n$ the set is always bounded and has the form $\{n, \delta\}$ where $\delta = 0 \lor 1$

Leaky Nun

11:26

@Secret you know what a polynomial is right

I think I've translated the proof so that it should be easier to understand

Secret

I am familar with the discriminant of quadratics, but not so much on general polynomials. That said, let me reread your proof again, I might have overlooked something when I am reading about the line with the 2x2 matrix polynomial

Mary Star

11:54

Let $C[0,1]$ and $C^1[0,1]$ be the space of continuous and continuously differentiable (respectively) functions $x:[0,1]\rightarrow \mathbb{R}$ with the supremum norm $\displaystyle{\|x\|=\sup_{t\in [0,1]}|x(t)|}$ and $T_0, T_1, T_2: C^1[0,1]\rightarrow C[0,1]$ maps with \begin{equation*}T_0(x)(t)=x'(t), \ \ T_1(x)(t)=\int_0^tsx(s)\, ds \ \ \text{ und } \ \ T_2(x)(t)=\int_0^tsx^2(s)\, ds\end{equation*}
How could we check the continuity of $T_0, T_1, T_2$ ? Could you give me a hint?

Silent

Is it true that if $p$ is prime and $P$ is $p$-Sylow subgroup of some finite group $G$, then for every subgroup $H$ of $G$, $H\cap P$ is a $p$-Sylow subgroup of $H$?

Silent

12:18

I think it is not true, because, eg, $S_3$, we see that $<(1\,2\,3)>$ is a 3-sylow subgroup, but $<(1\,2)>\cap <(1\,2\,3)>$ is not 3-sylow. Am i correct? @LeakyNun

Silent

12:51

@Semiclassical, will you please confirm if i am correct or not, above?

Semiclassical

I don't know sylow groups, so I can't confirm

Leaky Nun

13:17

@Silent you're correct

Silent

13:38

thank u

philmcole

14:10

Hi, I want to show that the following function is differentiable but has no continuous partial derivatives

$$f(x,y)=\begin{cases}
(x^2+y^2)\sin\frac{1}{\sqrt{x^2+y^2}},&(x,y)\neq(0,0)\\
0,&(x,y)=(0,0).
\end{cases}$$

How do I show that $f$ is differentiable at the origin?

I can't take the right and left limit like in the 1D case but must show that there exists a linear map $Df$ such that the definition is satisfied...

How would you do it?

ahorn

14:35

Sorry, it's been a while... how do I make a [hyperlink](www.facebook.com) again?

ÍgjøgnumMeg

If $(X , d)$ is a metric space, is the "topology induced by $d$" just the topology where the open sets are open balls wrt $d$?

Alessandro Codenotti

yes, to be precise the balls are a basis for the topology actually

philmcole

aren't there also open sets which need not to be balls in a metric space?

ÍgjøgnumMeg

Ahhhh

idk

philmcole

like a union of two balls...

I think what you meant was that the topology induced by the metric space are exactly the sets which are open in the metric space

ÍgjøgnumMeg

14:43

Fair, I'm an absolute beginner in topology so I'm just reading through definitions at the moment

philmcole

I find it confusing too when there are different notions of openness and one is more standard then the other ones

ahorn

Nevermind

philmcole

Your open sets in the metric space are defined by the open balls wrt. to $d$ so there you have your connection to the metric again.

ÍgjøgnumMeg

I see

user131753

I am studying Lee's Introduction to Smooth Manifolds. His proof of the result every smooth atlas $\mathscr{A}$ for $M$ (a manifold) is contained in a unique maximal smooth atlas is given below:

user131753

14:49

"Let $\mathscr{A}$ be a smooth atlas for $M$; and let $\bar{\mathscr{A}}$ denote the set of all charts that are smoothly compatible with every chart in $\mathscr{A}$. To show that $\bar{\mathscr{A}}$ is a smooth atlas, we need to show that any two charts of $\bar{\mathscr{A}}$ are smoothly compatible with each other, which is to say that for any $(U\varphi), (V,\psi)\in \bar{\mathscr{A}}$, the map $\psi\circ \varphi^{-1}: \varphi(U\cap V)\to\psi(U\cap V)$ is smooth.

Let $x=\varphi(p)\in\varphi(U \cap V)$ be arbitrary. Because the domains of the charts in $\mathscr{A}$ cover $M$; there is s…

user131753

My question is: What is the justification of the next-to-last two lines?

Praphulla Koushik

Can you just write which is not clear.

user131753

@PraphullaKoushik Was it addressed to me?

user131753

15:05

In any case, "Since $p∈U∩V∩W$ , it follows that $ψ∘φ^{−1}=(ψ∘θ^{−1})∘(θ∘φ^{−1})$ is smooth on a neighborhood of $x$. Thus, $ψ∘φ^{−1}$ is smooth in a neighborhood of each point in $U∩V$." - I don't understand how from the first line, the second line follows.

Praphulla Koushik

Yes yes, it was addressed to you. I thought it will show as it is addressed. I realise it is not showing like that....isn’t it just composition of smooth functions?? Am I missing something?

user131753

@PraphullaKoushik It is just the composition of smooth functions but locally. How does one move from local smoothness to global one?

Praphulla Koushik

Smoothness is seen at a point. Given a point he is saying function is smooth at that point. That is how you prove some map is smooth.

A map is smooth globally if it is smooth at each point.. isn’t it so.?

user131753

More specifically, @PraphullaKoushik, how can I prove the following proposition: "Let $U$ and $V$ be non-empty open subsets of $\mathbb{R}^n$ and $\mathbb{R}^m$ respectively . Let $f: U\to V$ be a map. If every point $p \in M$ has a neighborhood $U$ such that the restriction of $f$ to $U$ is smooth, then $f$ is smooth."?

Praphulla Koushik

You say f is smooth if it is smooth at each point . You say a function is smooth at a point if there exists some open set contain g that point where it is smooth. I do not know what more to say

user131753

15:26

@PraphullaKoushik I see. Thanks.

Praphulla Koushik

Does it really not answer your question.

user131753

@PraphullaKoushik Yes. It does. I got confused since I found no explicit definition of local smoothness in Lee (at least so far as I have read).

user131753

With that definition it is trivial.

Praphulla Koushik

There is no definition of local smoothness because there is no concept of local smoothness. There is a concept of smooth at a point and smooth map..

user131753

15:44

@PraphullaKoushik By "local smoothness" I meant "smooth at a point".

Praphulla Koushik

Ok ok. Hope it helped. Let me know if you have any more questions

user131753

@PraphullaKoushik Yes it did. I also have one more question.

Praphulla Koushik

Ask

user131753

15:59

Suppose $M$ is a smooth $n$-manifold, $k\in\mathbb{N}$, and $f: M\to \mathbb{R}^k$ is a smooth map. Let $(U,\varphi)$ be any smooth chart on $M$. Is $f\circ \varphi^{-1}:\varphi(U)\to\mathbb{R}^k$ smooth @PraphullaKoushik?

Leaky Nun

16:19

Doesn't smooth mean that derivatives of arbitrarily high order exist?

which isn't really a local condition?

Joe Shmo

well if the smooth embeddings are local, then the condition is local

user131753

Are you there @PraphullaKoushik?

philmcole

@Leaky Hey, how would you show that a certain function is differentiable although its partial derivatives are not continuous which means you can't use the theorem that says "all partials exists and are continous $\implies f$ is differentiable"?

Leaky Nun

no idea

philmcole

really? :D

2 hours ago, by philmcole

Hi, I want to show that the following function is differentiable but has no continuous partial derivatives

$$f(x,y)=\begin{cases}
(x^2+y^2)\sin\frac{1}{\sqrt{x^2+y^2}},&(x,y)\neq(0,0)\\
0,&(x,y)=(0,0).
\end{cases}$$

Leaky Nun

16:32

apply the definition of differentiable

philmcole

yeah but how?

the definition is with the linear map and the limit

My linear map would be a piecewise function too...

how do you take the limit then

iamacomputer

Hey there, I'm trying to find vocabulary to google a class of math problem. I'm sure it has a proper name, and if I find that name, there are probably 100s of papers. But I as of yet do not know the name.

In a nutshell, I have a skeleton which builds a body. Each joint-node has several "measurements" for possible coordinates. Each measurement has a confidence which is indicative but not absolute.

I need to collapse this graph of probabilities, to a "most probable" skeleton

Given the measurements of each joint and a set of likely bone lengths (between joints)

Any body have ideas for terms to google?

user131753

I think I have figured it out @PraphullaKoushik. Let $(V,\psi)$ be any smooth chart on $M$ and $x=\psi(p)\in \psi(V)$. Since $f$ is smooth, there is a smooth chart $(U,\varphi)$ such that $p\in U$. Hence $p\in U\cap V$. Now on $\psi(U\cap V)$ we have $$f\circ \psi^{-1}=(f\circ \varphi^{-1})\circ (\varphi\circ \psi^{-1})$$since both $f\circ \varphi^{-1}$ and $\varphi\circ \psi^{-1}$ are smooth, we have shown that $f\circ \psi^{-1}$ is smooth at $p$. Is it correct?

user131753

16:49

Sorry, smooth at $x$ (instead of $p$).

feynhat

hello

I came across the following statement in Munkres' Topology
> "To say that $p$ is a quotient map is equivalent to saying that $p$ is continuous and $p$ maps saturated open sets of $X$ to open sets of $Y$."
Are these conditions enough to ensure that $p$ is surjective?

How do I blockquote in chat?

mercio

what's a saturated open ?

it doesn't look like it implies that $p$ is surjective

Praphulla Koushik

17:05

Yes you are correct

feynhat

17:17

@mercio saturated open = saturated and open (at least that's what I think the author is implying)

Silent

18:24

True or False: For any polynomial $f(x)$ with real coefficients and of degree $2011$, there is a real number $b$ such that $f(b)=f'(b)$.

My attempt: $f(x)-f'(x)$ is a polynomial with odd degree and real coefficients, so it has a real root.

@XanderHenderson , will you please check this?

Poline Sandra

hello, need help on second order differential equation

B. Mehta

@Silent looks good to me!

Silent

@B.Mehta Thank you very much

Mary Star

19:05

Let $x:[0,1]\rightarrow \mathbb{R}$ with the supremum norm $\displaystyle{\|x\|=\sup_{t\in [0,1]}|x(t)|}$ and $T_0: C^1[0,1]\rightarrow C[0,1]$ with $T_0(x)(t)=x'(t)$. I want to find a counterexample, that shows that that map is not continuous. For that I have to find functions $x(t)$ and $y(t)$ such that $\|x-y\|=0\nRightarrow \|T(x)-T(y)\|=0$, or not?
What functions could we take?

Leaky Nun

@MaryStar I think you confused continuity with well-defined

Mary Star

Ah ok. How can we check the continuity of that map then? @LeakyNun

Leaky Nun

epsilon-delta.

Mary Star

Ok! We have that $\|T_0(x)-T_0(y)\|=\sup_{t\in [0,1]}|T_0(x)(t)-T_0(y)(t)|=\sup_{t\in [0,1]}|x'(t)-y'(t)|$, right? How can we relate this with $\|x-y\|$ ? Could you give me a hint? @LeakyNun

Leaky Nun

I don't want to think about this right now

watchme

19:13

Can you know everything in Math!?

philmcole

no

at least I believe you can't since there are constantly new things invented

watchme

Why not? I mean in Physics I would accept the answer "no", but in Math, I think you can know everything which is "knoweable" in this moment (so which has been found out by someone already).

(I must say I am not a professional math-guy as you can see)

philmcole

Don't think math is static. There is constantly new things invented.

watchme

I mean shure, there are many many many methods to get to one solution (some people have their own methods, so you cannot know them ALL). But you know what I mean...

philmcole

The thing is in math you don't just "find out" but you create

You have your axioms and then you can do whatever you want. It's like painting a picture. You have the colors and then you can draw what you want.

There is no bottom you know

user10478

19:21

Hello, I am having some conceptual difficulty with dependent and independent variables.

On one hand, their supposedly determined by the pure mathematical function line tests.

watchme

@philmcole what are axoims? (I am from Austria, sorry :P )

axioms*

user10478

On the other hand, *they're supposedly determined by the real world problem being modeled, where the independent variable is the one that can be directly controlled.

philmcole

How do you start from nothing? You need to have something you can work with. So you specify some rules which are true by definition on which you build everything else. @watchme

Those are called axioms.

user10478

So given an equation, especially in implicit form, there seem to be two separate ways to determine the dependent and independent variables, which could contradict each other.

philmcole

@user10478 Can you give an example?

watchme

19:28

@philmcole yeah thats why I love math... ok now I know what you mean :)

philmcole

Cool. Yeah it's a rather different type of science from other fields which often don't even know why their foundation is like it is. In math you know exactly what you assumed and what you derived. So you know on what soil you build your house.

konoa

hi chat

watchme

@konoa hi konoa

konoa

hi @watchme, nice to meet you

watchme

@philmcole You know, with all this formulars and high-math-things, sometimes the easy things are forgotten.

For instance:

Alessandro Codenotti

19:36

Hi @konoa

konoa

@AlessandroCodenotti you should be drinking

user10478

Okay, so given the equation a + b^2 = 4, one might be taught in school to immediately determine which variable is a function of the other using function line tests. However, this same equation could model a number of unrelated real world scenarios. In some, a would be the variable under the researcher's control, while it would be b in others (perhaps both or none in others), so the dependent and independent variables should be determined by the real world scenario.

philmcole

@user10478 Of course this equation could mean a lot of different things in the real world. But I don't think about it that much. If you want to "solve" an equation you look at what you want (here maybe a, or b or both) and what you have and then express what you want in terms of what you have.

user10478

So, for example, it's not clear to me when implicit differentiation VS partial differentiation should be used, which depends on the types of variables and how they relate as functions of each other.

philmcole

I guess this depends on context.

watchme

19:52

@philmcole You are able to explain to a child why multiplications exists: 4*a tells us, that you have to add up "a" four times (a+a+a+a)! Things with decimals (rationals) get a little bit weird and you have to tell the kid (if it knows what a decimal number is): 3.2*a means, that you have to add up "a" three times, and add to those three apples 1/5 of an apple!

Thats easy.

Whats also easy is to explain exponents: a^2 means that you have to multiply a once by itself.
But whats not easy to explain are rational exponents. What does a^(1/2) mean? We KNOW what it means, because we learned it. …

user10478

If I understand correctly, the line tests are considered to be not only a necessary condition of a relation being a function, but also a sufficient condition. This is bizarre to me; it would seem much more sensible for, i.e., a vertical line test to provide a necessary condition for y being a function of x, but for the sufficient condition to be provided only in light of the relation's real world application. But as I understand things, this isn't the case.

B. Mehta

@watchme You could introduce areas, and say $a^{\frac{1}{2}}$ is the side length needed to make a square of area $a$

that's what i'd do at least

@user10478 that's a fair concern, but not all functions come from a real world application - not obviously at least

and so when writing down a mathematical definition of a function, you'd like it to be separate from any real world concerns like that

then the application of the maths can worry about dependencies

in my opinion, at least

philmcole

@user10478 The vertical line test just allows you to see if some curve is the graph of a function. It has nothing to do with relations. A graph is just a visualization technique which allows us to see what a function does.

B. Mehta

@philmcole I don't think that's fair - you can have functions which aren't graphs. Also, a graph is ultimately describing a relation, so the characterisation of a function as a relation satisfying the vertical line property is entirely fair

to further this point, a typical definition of a function is as a relation which satisfies (the algebraic version of) the vertical line test

philmcole

Ok say for simple and nice functions

watchme

19:59

@B.Mehta very good example, thanks indeed, yes! But what I have experienced is, that when a kid learns "a^2 = a*a", it wants connect "a^(1/2)" to that what it has learned, as it is also an exponent. But it can't! Where is the pattern when going down from "a^(m/n)" in "a^(1/n)" steps, and why is this the case...

I know I am asking to much here... but I haven't been able to let go of knowing math from ground on, to not learn anything by hard... (That I finally can explain math with a full picture to some kid, to answer EVERY (why this why that)-question), if you know what I mean...

Is late again in Austria, I shouldn't write texts at that time :D

Its currently 22:00

10:00 pm^^

Do you think you know math from ground on?:P

philmcole

That's an abituous goal @watchme but I think you can get quite far and answer most questions if you study long enough :P If I know math from the ground up? No, certainly not.

user10478

@B.Mehta That's how I would like to view it, there are no dependent or independent variables, no functions, inputs, or outputs, until the application "layer," but throughout many levels of Calculus one is routinely, perhaps even more often than not, committed to treating equations differently in light of these concepts. Does this force us to say that half or more of a college level vector calc or diff eq course doesn't really work as pure math?

B. Mehta

@user10478 It's hard to say - I haven't taken math courses in the US so I can't really comment on how they're structured

user10478

okie

watchme

I know @philmcole....
But thats when I can play around with math, when I can connect by myself, when I don't have to ask "is this even possible", because I KNOW it must be, as it fits in with all the rules I have learned from ground up. Do you study math!?

philmcole

20:15

No, I do physics. That's why I'm not so interested in the fundamentation of math as I want to apply it rather to real world scenarios. But it's certainly interesting to see what concepts one can come up with if one isn't restricted by the real world. But I think even in math you're still biased when inventing new things to keep them useful and connected to the real world. Just theoretically you could do almost anything.

Ted Shifrin

@SylentNyte Sylent, my apologies. I misread your vector. Your cross product is in fact correct. However, note that both lines are contained in the plane $x=y$. So unless the lines are parallel, they must intersect. In fact, they do intersect, at the point $(1,1,1)$ !!!

@B.Mehta: I asked my expert on probability and matrices and he didn't know the answer to your question.

Mary Star

Hello @TedShifrin !!

Let $x:[0,1]\rightarrow \mathbb{R}$ with the supremum norm $\displaystyle{\|x\|=\sup_{t\in [0,1]}|x(t)|}$ and $T_0: C^1[0,1]\rightarrow C[0,1]$ with $T_0(x)(t)=x'(t)$.

Could you give me a hint how we could check the continuity of $T_0$ ?

We have that $\|T_0(x)-T_0(y)\|=\sup_{t\in [0,1]}|T_0(x)(t)-T_0(y)(t)|=\sup_{t\in [0,1]}|x'(t)-y'(t)|$, right? Do we have to relate this with $\|x-y\|$ ?

B. Mehta

20:31

@TedShifrin Hi, thanks very much for asking anyway!

Alessandro Codenotti

$T_0$ is linear, showing that it is continuous in $0$ is enough

watchme

MathJax isnt displayed correctly (in this chatroom), does anybody know what I can do?

Mary Star

@AlessandroCodenotti Why is this enough?

MatheinBoulomenos

@MaryStar That's not continuous. Consider the sequence $f_n(x)=(\frac{x}{2})^n$

skull

here @watchme

MatheinBoulomenos

20:34

it's continuous if you put the $C^1$-norm on $C^1[0,1]$, though

skull

8

A: How can I enable MathJax in chat?

Update 2017-05-01 The MathJax CDN retired and the javascript-URL idea is not so easy any more, because of browser security. (Chrome stips away any leading javascript: when pasting into the URL line. SE modified the javascript: link so that it does not work.) So here is my take. I modified the ...

MatheinBoulomenos

$f_n$ converges pointwise on $[0,1]$ to $0$. Since the convergence is montonous, the convergence is uniform by Dini's theorem

Alessandro Codenotti

See here, thm 1.17 people.math.ethz.ch/~salamon/PREPRINTS/funcana.pdf

MatheinBoulomenos

but the sequence of derivatives doesn't converge

watchme

thanks! @skull

skull

20:39

np

Kasmir Khaan

@MatheinBoulomenos mathein :D got time ?

Mary Star

Do you mean to consider $x(t) = (\frac{x}{2})^n$ for $y(t)$ the limit of $(\frac{x}{2})^n$ as $n$ goes to infinity?
Or what do we have to do? I got stuck right now.

MatheinBoulomenos

@KasmirKhaan a bit

@MaryStar continuous maps send convergent sequences to convergent sequences

I gave an example that shows that this map is not continuous

Kasmir Khaan

@MatheinBoulomenos okay >< I got few problems I wish you could help me with :D I got exam very soon and I want to see your solutions if that is fine with ya :D

MatheinBoulomenos

oh, sorry for the confusion, I meant $x_n(t)=(t/2)^n$

that's a sequence of elements in $C^1[0,1]$

you can check that it converges pointwise on $[0,1]$ to $0$. It even converges uniformly (i.e in the $\sup$-norm) by Dini's theorem

@KasmirKhaan I can't promise that I can do them all, but go ahead

Kasmir Khaan

20:47

@MatheinBoulomenos No need for all >< I ll just send 2-3 on email :D

thanks mathein you are the best _

Alek Murt

Hey guys, if i have $A \subset B \subset X$ and $A$ dense in $X$ then $B$ is dense in $X$

B. Mehta

@AlekMurt $X = \overline{A} \subset \overline{B} \subset X$

MatheinBoulomenos

@MaryStar okay that sequence may not work. Try $x_n(t)=\frac{\sin(nx)}{n}$ instead

Alek Murt

@B.Mehta how do i have $\overline{B} \subset X$?

Kasmir Khaan

@MatheinBoulomenos Sent! :)

B. Mehta

20:56

I assumed this is inside the topological space $X$

Mary Star

Ah ok! So, we have that $$\|T_0(x_n)-T_0(x)\|=\sup_{t\in [0,1]}|T_0(x_n)(t)-T_0(x)(t)|=\sup_{t\in [0,1]}|x_n'(t)-x'(t)|=\sup_{t\in [0,1]}|\cos (nt)-0|=1$.
We also have that $\|x_n-x\|=\sup_{t\in [0,1]}\left |\frac{\sin (nt)}{n}-0\right |=\sup_{t\in [0,1]}\left |\frac{\sin (nt)}{n}\right |$, right? This must be equal to $0$, or not? But why is this $0$ and not $1$ ? @MatheinBoulomenos

B. Mehta

if it's not, you can say $\overline{B} \subset \overline{X}$

and if $X$ is the whole space, $\overline{X} = X$ clearly

MatheinBoulomenos

the map is not continuous, that's the reason

Daminark

Aw man some nerds

Alessandro Codenotti

@Daminark And the nerdest of them all is here too, now that you're here we only miss Balarka

How is it going?

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$Mathematics$ Mathematics