Mathematics - 2017-05-06 (page 2 of 4)

gansub

09:01

Jij(t) is the Jacobian of the transformation and contains terms like dxi/dai

user400188

Is anyone in this room familiar with the Isabelle programing language?

BAYMAX

09:19

$X = \{(x_{n}) : x_{n} \in \{0,1\} n \in \mathbb{N} \}$

Is this set uncountable ?

mercio

yes

Mudassir Malik

can we write math equations in title of question

?

mercio

yes but don't be excessive and make the whole title a single equation

Mudassir Malik

how? any example

mercio

???

half the questions on the main page have equations in them

Mudassir Malik

09:26

give a link of question having math equation in the title.

mercio

2

Q: Solvability of $X^p-X-a$

In remark 3.28 of J. S. Milne's online note, Fields and Galois Theory, it claims that if $F$ is a field with characteristic $p$, then $X^p-X-a=0$ is not solvable by radicals even though it is separable with abelian Galois group. It is obvious that $X^p-X-a$ is separable, but I do not know how t...

galois-theory

Mudassir Malik

thanks bro

BAYMAX

yes but how I can prove that uncountability

like that there is no such injective function from $X$ to $N$

Like suppose there exists an injective function

mercio

no

it's easier to go the othre way

suppose you have a function from $N$ to $X$

then show that it's not surjective

by providing an example of an element of $X$ that's different from all the $f(k)$ for $k \in N$

BAYMAX

$f : N \rightarrow X$

mercio

09:34

yeah

Santosh Linkha

hello

mercio

hi

BAYMAX

like which element of $X$ ? it has only two elements $0$ and $1$ ?

mercio

no

BAYMAX

like a sequence of

mercio

09:36

$\{ 0 ; 1\}$ has two elements

BAYMAX

$0$ and $1$

yeah

mercio

$X$ is the set of sequences with values in $\{0; 1\}$

BAYMAX

yes

mercio

or the set of functions from $N$ to $\{0;1\}$ if you want to be really precise

$0$ is not in $X$, it's not a sequence

BAYMAX

$X$ is the collection of sequences ?

mercio

09:38

yes

for example, $(0,1,0,1,\ldots), (1,1,1,1,1,\ldots), (0,1,0,0,1,0,0,0,1,\ldots)$ are elements of $X$

BAYMAX

yes

mercio

that's three different elements of $X$

and there are many many more

now if you have a function $f$ from $N$ to $X$

you can build a sequence $(x_n)$ that's different form every $f(k)$

BAYMAX

yes now I have to produce an element of $X$ (basically a sequence) that is not in $f(k),k \in N$

mercio

by making so that $x_k$ is different from $f(k)_k$, for example

So you simply define the sequence $(x_n)$ as $x_n = 1-f(n)_n$

and this is an element of $X$ that can't be in the image of $f$

BAYMAX

like where thee is 1 we make that position of sequence 0 ?

and when 1 we make it 0

mercio

09:43

yeah

BAYMAX

nice one!

mercio

we make it so $(x_n)$ disagree with $f(k)$ at position $k$

BAYMAX

yes

mercio

Cantor came up with it

BAYMAX

similar to Cantor Diagonalisation argument!

mercio

09:44

well it IS the diagonalisation argument

BAYMAX

oh

yeah

thanks mercio

mercio

np

in fact $N$ can be any set

this shows that the set of functions from $N$ to $\{0;1\}$ is always "strictly larger" than $N$, for any set $N$

Alessandro Codenotti

Another easy way to show uncountability of that set is to show a bijection with $\mathcal{P}(\Bbb N)$

mercio

that's assuming you have shown $P(N)$ is uncountable with the diagonal argument

BAYMAX

gotta go, bye

mercio

09:48

bye

Balarka Sen

Hi chat

Alessandro Codenotti

Hi @Balarka

mercio

hi

Astyx

Hi chat

Alessandro Codenotti

10:05

Hi @Astyx

Astyx

What's up ?

Serge Seredenko

10:21

Is it possible to explicitly construct a Hamel basis for R^inf or L_2 or l_2? I mean can I see such a set that spans a big space with finite linear combinations, or is this one of those things that just uselessly "exist" under AC?

mercio

I'm pretty sure they only uselessly exist under AC

Astyx

10:36

So am I

Martin Sleziak

@MudassirMalik See also Guidelines for good use of $\LaTeX$ in question titles and Title and $\LaTeX$ on meta.

@SergeSeredenko This answer shows that you need (some form of) AC for any Banach space. I have collected a few more related posts from the main site in the set theory chat room.

James Bond

10:59

All LaTeX questions should be directed to tex.stackexchange.com instead.

Of course, the LaTeX on math.stackexchange.com is not a standalone LaTeX installation, so there are differences.

We should all just read 'More math into LaTeX' by Gratzer, fifth edition, lol.

Mats Granvik

11:46

@TedShifrin I have this integral from Raymond Manzonis zeta zero counting formula that when integrated gives the 126 first zeta zeros, and a few more. I don't see how to solve the integral symbolically. The approach I have been trying is power series expansions, and later I would try integration by parts applied to multiple products. You can find the counting functions here: https://math.stackexchange.com/q/2136668/8530

I entered the Manzoni variant in the OEIS. I should probably have given him credit for it. I don't expect to prove the Riemann hypothesis, but I hope that one could at leas…

oeis.org/A058303 <--The OEIS entry.

Related: mathoverflow.net/a/261734/25104

Simply Beautiful Art

@Secret @Benjamin Hope I didn't miss needing to help anyone with ordinal collapsing functions?

Mats Granvik

$$\small \gamma_n=\int _0^k\frac{1}{2} \left(1-\text{sgn}\left(\left\lfloor \frac{t \log \left(\frac{t}{2 e \pi }\right)}{2 \pi }+\frac{7}{8}\right\rfloor +\frac{1}{2} \left(-1+\text{sgn}\left(\Im\left(\zeta \left(\frac{1}{2} + i t\right)\right)\right)\right)-n+\frac{3}{2}\right)\right)dt
$$

$k>\gamma_n$

Secret

A bit busy at chemistry lately due to many weird results, which is why I have not touched any maths. I do managed to clarified why $\omega^{\epsilon_0+1}$ does not collapse, because 1 is not a omega exponential map fixed point

Simply Beautiful Art

Oh jesus, that is a lot

@Secret Nice then :-)

Mats Granvik

$\gamma_n$ is the n-th zeta zero.

Learning user

11:55

Hi

sorry for interuppting

Simply Beautiful Art

Hi

o/

Learning user

one small doubt

How to find the square root of a number for 224 using long division method,i just little forgot about these things

Simply Beautiful Art

What is 224 divisible by?

Hint: It's even

Learning user

yeah tried ,i have got confused

for 224 , we have to seperate the digits like 2 and 24

Simply Beautiful Art

224/2

??

Learning user

11:59

for 2 - 1 we have to quotient and divisible

Simply Beautiful Art

Wow, I think you're over complicating division

Learning user

to find the square root using long division method

Simply Beautiful Art

Just divide 224 by two

Learning user

by explaining to you, i got the answer

thank you so much, i have rectified my mistake

Astyx

112

Learning user

12:03

ok

Konformist Liberal

A question of mine has been closed and reopened and I suspect this prevent some regular users who follow the tag to skip mz question

Learning user

what 112

Konformist Liberal

my*

Learning user

?

Konformist Liberal

Any suggesstions?

Learning user

12:06

How to find the square of 132, is there any shortcut is there

by reducing the time

without doing normal Multiplication

Simply Beautiful Art

@Learninguser Wait, do you want the decimal expansion of √(224)?

Astyx

I'd write $132 = 100+32$

And use $(a+b)^2 = a^2 + b^2 + 2ab$

Learning user

then any other method

Astyx

here 10000 + 6400 + 960 + 64

Learning user

How ?

Simply Beautiful Art

12:08

@Astyx's method is the general way I square/cube/higher power my numbers by hand

Learning user

you done above steps

how? 10,000 + 3200 + 960 +64

?

Astyx

10000 = 100^2

Learning user

please explain

ok

Simply Beautiful Art

$132 = 100+32$

$132^2=(100+32)^2$

Astyx

I meant $6400 = 2 \times 32 \times 100$ for the second one

and $32^2 = 32 \times 30 + 32 \times 2$

Simply Beautiful Art

12:10

or $32=30+2$

and $32^2=(30+2)^2$

$=900+120+4$

Astyx

It's just practice really

Learning user

is anyother method, for doing fast manner

Astyx

Not that I know of

Doing calculations fast isn't really that important

Knowing how to do them is

Learning user

ok

6400 - 2 times of 32 and 32 times of 100

?

Simply Beautiful Art

Still interested in √224?

Learning user

12:12

I have got it

wait i will complete this one

and i will come

Astyx

$4\sqrt{14}$ ?

Learning user

6400 - 2 times of 32 and 32 times of 100?

please explain

one more time

Simply Beautiful Art

@Astyx numerical expansion

Note $15^2=225$

so binomial theorem is good

Astyx

Hehe I'll skip

$6400 = 2\times 32\times 100$

Learning user

I am not understand

for 2 ab

Astyx

12:15

What don't you understand ?

Learning user

then how? 960+ 64

formula is only for three digits, but for calculating it is coming four digits how?

Astyx

$32^2 = 32\times(30+2) = 32\times 30 + 32\times 2 = 960 + 64$

Learning user

here 10000 + 6400 + 960 + 64

Astyx

The formula applies for all numbers

Simply Beautiful Art

o/ @Daniel

Learning user

12:18

oh understand thank you @Astyx

Another doubt

Simply Beautiful Art

no problem :-)

(because I'm rude and I take congrats that aren't mine)

Learning user

can you please this sum

3

Q: If the average age decreased by 6 months, calculate the average age of new students

The average age of 12 students in exam hall is decreased by 6 months, when two of them aged 13 years and 15 years are replaced by two new students. The average age of new students is ? I am not able to solve this problem and I am facing little ambiguity for solving, how the age will be decre...

algebra-precalculus average word-problem

Astyx

@Simply concerning the question you linked to me earlier (about asymptotical equivalence), I think your idea is too restricitve. you did say $\forall x\gt x_0, g(x+a) \le f(x) \le g(x+b)$

Learning user

after seeing this sum, can you please remind me

I have so many doubts

Simply Beautiful Art

@Astyx Yes, I did. What is wrong with it?

Astyx

12:20

But that implies $g(x+a)\le g(x+b)$

You'd need increasing functions

And many are not

Simply Beautiful Art

@Astyx Yes, they should be increasing functions

Astyx

(otherwise it's not reflexive and you definitely don't want that)

mercio

you could say $f(x) = g(x+O(1))$

Simply Beautiful Art

Well, if you read my question, its built for functions that grow fast. Particularly, much faster than exponential functions. I assume these are strictly increasing.

mercio

if $g$ is bijective it kinda makes sense

Simply Beautiful Art

12:21

@mercio Ah, that's nice way to put it.

mercio

or $g^{-1} \circ f = x + O(1)$

Learning user

@Astyx after completing your disussion @SimplyBeautifulArt, please my sum also

Astyx

Right then I guess it is a well hidden Big Theta after all

Simply Beautiful Art

Yup

@mercio math.stackexchange.com/questions/2258932/…

Astyx

@Learninguser Are the five answers not enough ?

Should I give this as an answer ?

It isn't really one

Learning user

12:24

please I am not understand

I am a learner

Simply Beautiful Art

If you guys want to answer it it'd be great

Learning user

@Astyx

Simply Beautiful Art

if not, I'll post that answer in a few days

Astyx

@Learning let $a_1, \dots, a_{12}$ be the 12 ages of the original students

Learning user

wait

my queries

Astyx

12:26

You have ${1\over 12}\sum_{i=1}^{12} a_i = m$ where $m$ is the original mean age (in years)

Simply Beautiful Art

lol

Learning user

wait please

Simply Beautiful Art

did that guy come to leave? O.o

@mercio You don't mind if I wait a while?

mercio

not at all

Astyx

I'll answer then @Simply

Simply Beautiful Art

12:27

@Astyx He beat you to it

Astyx

Oh mercio already has

Cool :)

mercio

i almost have the silver badge for number theory

Astyx

It does feel like your answer is also very restrictive @mercio

Albert Einstein

@SimplyBeautifulArt, hello

Astyx

For instance take $f(x)=\log x$ and $g(x) = \lfloor\log x\rfloor$

Learning user

12:29

Hi @Astyx

let we start

my doubt

mercio

how so ? I think it's equivalent to what he's asking for assuming $g$ is increasing

Astyx

Or maybe you don't want these to be comparable (that seems weird to me)

Simply Beautiful Art

@Astyx Hi

Astyx

I'm listening @Learning

Learning user

ok

mercio

12:31

hmm

Learning user

How the average age of 12 students decreased by 12 months

sorry 6 months

mercio

but your $f$ isn't a bounded variation of $g$

Learning user

after replacing or before replacing

Question itself I am not able to understand

Astyx

Right

Say we have 12 students of which the average age is $m$

Learning user

ok, how it will decrease by six months

Astyx

12:32

2 of them go, we get another average age for the ten remaining students, say $m'$

Two other join, now the average is $m'' = m-6\text{ months}$

Learning user

average number decrease by each student

Simply Beautiful Art

@mercio I think $g(x)\le f(x)\le g(x+1)$

Astyx

That does not mean anything

Learning user

How it will decrease by six month

Astyx

@Simply No he's right, i's not bounded above

@Learning that is the question

Learning user

12:34

their age ah? weight reducing

Simply Beautiful Art

Oh, right

Learning user

for each student

Astyx

I have no idea what you mean Learning

mercio

your example has $g(x) \le f(x) \le g(x)+1$ and $g(x) \le f(x) \le g(ex)$

Learning user

ok Thank you @Astyx,please i am not able to understand

Simply Beautiful Art

12:37

Hm, what about this?

Bounded by an iteration:
$g(x) \le f(x) \le g(g(x))$
With $g(x)$ and $f(x)$ increasing functions

Astyx

In any case, I don't think these are sufficiently reasonnable to become truly useful for comparing functions assymptotically

mercio

then you have to define some kind of $\log_g$ function lol

Simply Beautiful Art

@Astyx Then what do you recommend for functions that grow faster than tetration?

Astyx

Are $\sim$ and $\Theta$ not enough ?

Simply Beautiful Art

That is, functions that satisfy the following:
$f(x+1)\gg 2^{f(x)}$

Yes, $\sim$ and $\Theta$ are not enough for functions like these

For the most part, these functions are usually only defined over $\mathbb N$.

Astyx

12:43

Why are they not enough ?

Simply Beautiful Art

$f(x)$ and $g(x)$ grow too fast

Astyx

Too fast for what ?

Could you perhaps give a concrete example ?

Simply Beautiful Art

$f_0(n) = n+1$

$f_{k+1}(n) = f_k(\dots f_k(n)\dots)$ with $n$ amount of $f$'s

I want to show that $f_k(n)\approx 2\uparrow^{k-1}n$

Using Knuth's up-arrow notation

The above is the fast growing hierarchy for finite inputs

You will note that the relationship is not of any big-O notation nor is it $\sim$, so I must use something else

Astyx

Mmm I don't quite have the time to think about it right now but I will later

I have to go - bye !

Simply Beautiful Art

Bye!

Julius

12:56

Hi. Let $P$ be a $K^3\times K$ binary matrix whose columns sum to $K$, and $D$ be a $K^2\times 1$ nonzero real vector. Any ideas on how to check whether $P\otimes D'$ has any negative eigenvalues? $\otimes$ is the Kronecker product.

BAYMAX

13:21

(x,sin(1/x)) , 0<x$\leq$1 ?

is this closed?

happyEddie

BAYMAX, no

BAYMAX

Like reasoning?

@happyEddie

i have the graph in mind

happyEddie

for example the point (0,0) is in the closure but not in the set

take any open ball centered at (0,0), there are points of your set in that ball

Konformist Liberal

can anyone able to open nLab right now or is it just me?

happyEddie

BAYMAX, can you see this?

BAYMAX, let me know if you need more explanation...

BAYMAX

13:28

yeah I am trying to recollect the theorems

@happyEddie

yes it is the limit point

origin is the limit point

now a set is closed if it contains all its limit points

happyEddie

isn't is "iff" ?

BAYMAX

but now there exists a limit point (origin ) which is not in the set {(x,sin(1/x)) , 0<x$\leq$1}

hyes

now a set is closed iff it contains all its limit points

happyEddie

a set is closed iff it contains all its limit points, right

BAYMAX

so it is not closed

right?

happyEddie

yes, we found a limit point that is not in the set, conclude that the set is not closed

the direction we use is "if a set is closed then it contains all its limit points"

i have a question of my own, let R be a domain, c an element of R different from 0, let phi be the homomorphism R[X] -> R[1/c] sending X to 1/c, i'm trying to see that the kernel of phi is the ideal generated by the polynomial cX - 1, how to see that if phi(F) = 0, then F must be in this ideal?

BAYMAX

13:35

negation of which we get is "If there exist a limit point which is not in the set then the set is not closed."

Leaky Nun

@BAYMAX right

happyEddie

BAYMAX, i'm not sure that the correct term is negation, you mean contraposition?

BAYMAX

so a new question - is negation same as contrapositive statement ?

happyEddie

BAYMAX, but anyway the theorem that you state is correct and equivalent to "if a set is closed then it contains all its limit points"

BAYMAX

:)

Alessandro Codenotti

13:37

If 1/c is in R isn't cX-1 its minimal polynomial hence a generator of that ideal?

Astyx

yes it is @Alessandro

At least in $\Bbb R[X]$

happyEddie

1/c is not necessarily in R, rather in the field of fractions of R, i should have stated that

Astyx

Oh it wasn't actually a question you were asking ... my bad :p

happyEddie

and R is not the real numbers, just a general domain

clearly phi sends cX - 1 to zero, but how to see that if an element is sent to zero, it must be a multiple of cX - 1

Alessandro Codenotti

cX-1 is the minimal polynomial

Simply Beautiful Art

13:42

Hello sir @Ben

happyEddie

Alessandro, do we need the coefficient ring to be a field to be able to use the theorems dealing with the minimal polynomial, no?

it's just that i've used to dealing with the minimal polynomials in the setting of fields and field extensions, perhaps it works in this more general setting too, i have to see

for example R[X] is not necessarily a principal ideal domain now

Alessandro Codenotti

You can generalize it to more general settings, but I don't remember the details

happyEddie

Alessandro, ok thanks, I'll take a look

Alessandro Codenotti

You probably need the base ring to be a field, but the extension not necessarily

(You lose some very nice properties, minimal polynomials don't even need to be irreducible at this point, think about $M_2(\Bbb Q)$ as a ring extension and consider the minimal polynomial of $\begin{pmatrix} 0 &1\\0& 0\end{pmatrix}$

Actually I'm quite sure now that you mention it that this won't work for a generic domain because you want R[X] to be a PID

happyEddie

Alessandro, right, that was my worry too

Alessandro Codenotti

13:54

Maybe you can still do something here thinking about it as an "extension" of the field of fractions? I'm not sure, I'm afraid you need someone who actually knows abstract algebra

happyEddie

if F = sum{i=0...n} a_i X^i, then phi(F) = sum{i=0...n} a_i (1/c^i) = 0 implies sum{i=0...n} a_{n-i} c^i = 0 in R: c is a root of the polynomial a_0 X^n + a_1 X^{n-1} + ... + a_n

BAYMAX

Ok now if I add the point (0,0) , then will it be closed ?

I think yes as now it contains all the limit points!

Simply Beautiful Art

Essence of calculus, chapter 1

►

Leaky Nun

@BAYMAX no it doesn't

any point of the form (0,y) where -1<=y<=1 is a limit point

BAYMAX

oh like why?

Mehdi Slimani

14:08

Hi ! Q : Was the definition of complex of differentiable made in the complex plane without "room" for different partial derivatives because the complex plane is 1 dimensional ?

Leaky Nun

@MehdiSlimani what do you mean by "dimension"? It is 1D over $\Bbb C$ but 2D over $\Bbb R$.

BAYMAX

oh

nice1@LeakyNun

Leaky Nun

@BAYMAX It's well-known. It's the topologist's sine-curve.

BAYMAX

yes,never thought about it deeply!

Mehdi Slimani

n dimensional iff n elements are sufficient to span

Leaky Nun

14:11

@MehdiSlimani yes, but over which field?

Mehdi Slimani

C

how was it again to read the math stuff

one sec

Leaky Nun

look at the room description

I don't understand what you mean by 'without "room" for different partial derivatives'

BAYMAX

Wow a Wormhole @MarcusS

Blackhole1

Mehdi Slimani

before starting the subject i expected something very similar to the structure of R2 but when we arrived to definition of differentiable i was kinda surprised

room for partial different partial derivatives means same definition of differentiable as R2

my q better formulated is why the definition of differentiablein C wasnt chosen to be the same as in R2

Leaky Nun

because the functions in C are C->C while the functions in R2 are R2->R

Mehdi Slimani

14:19

what about a function R2 R2 ? it is still differentiable if all the numbers in its jacobian are different

i think maybe my question is pointless

ok i think i found a better formulation : is the definition of differentiable in a space a consequence of the dimension of the space ? (where dimension means cardinality of a basis)

Konformist Liberal

ncatlab.org is it only down for me?

Mehdi Slimani

i can access it

BAYMAX

like you write notes in that ?

Mehdi Slimani

thanks for trying @LeakyNun, maybe my question really doesnt make sense. have a nice afternoon :) !

Santosh Linkha

15:27

hello anyone here?

if $\alpha$ is algebraic over $K \subset L$, then how to show $K[\alpha]=K(\alpha)$?

Eric Stucky

Santo: take a minimal polynomial for $\alpha$, and note that it has a nonzero constant term. Can you take it from here?

Santosh Linkha

hold on a sec ... lemme try @EricS

Alessandro Codenotti

15:55

Take an nonzero element of $K[\alpha]$, that is the image of a polynomial $g(\alpha)$. Let $m$ be the minimal polynomial of $\alpha$, what's $\gcd(g,m)$? What does Bezout's identity tell you? Can you conclude from there?

Santosh Linkha

hmm ...

m(alpha) = 0

is $gcd(g,m) = 1$ ?? since $m$ is irreducible in $K[\alpha]$?

Alessandro Codenotti

m is an element of $K[x]$, not of $K[\alpha]$ (and so is g)

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$Mathematics$ Mathematics