Mathematics - 2019-11-21

johnny09

12:48 AM

suppose we have the cubic polynomial function $y=f(x)=-0.06 x^3+0.9x^2+3x$ with $x\in[0,5]$ and $y\in[0,30]$

how can we find the inverse function?

I have tried the cardano method and the hyberbolic functions method

but both fail

as the coefficients violate the conditions (eg the discriminant is negative so no real root)

are there any other methods to find the inverse function?

Galois in the Field

2:18 AM

Imagine if this was modern game advertising:

2

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Vladimir Reshetnikov

3:11 AM

Leaky Nun

3:34 AM

$$\sum_{n \in \Bbb Z} \operatorname{sinc}(n) = \pi$$

Leaky Nun

5:15 AM

$\widehat{\delta_{x_0}}(s) = \exp(-2\pi i x_0 s)$

$\widehat{\delta_{x_0/2\pi}} + \widehat{\delta_{-x_0/2\pi}} = 2 \cos$

$\newcommand{\sgn}{\operatorname{sgn}}\newcommand{\sinc}{\operatorname{sinc}}$$\displaystyle \frac{4}{\pi} \sum_{n=0}^\infty (-1)^n \frac{\cos\left(\left(2n+1\right)x\right)}{2n+1} = \sgn(\sinc(x)\sinc(2x))$

$\sgn(\sinc(x)\sinc(2x)) = \displaystyle \mathcal F\left\{ \frac2\pi \sum_{n=0}^\infty (-1)^n \frac1{2n+1} \left( \delta_{(2n+1)/2\pi} + \delta_{-(2n+1)/2\pi} \right) \right\}$

$\displaystyle \sinc(x) = \dfrac{\sin(x)}{x} = \dfrac{\exp(ix) - \exp(-ix)}{2ix} = \pi \int_{-1/2\pi}^{1/2\pi} \exp(-2 \pi ixt) \ \mathrm dt = \mathcal F \left\{ \pi \chi_{\left[-\frac1{2\pi}, \frac1{2\pi}\right]} \right\}$

$\displaystyle \sinc(2x) = \frac12 2 \sinc(2x) = \frac12 \mathcal F \left\{ \pi \chi_{\left[-\frac1{2\pi}, \frac1{2\pi} \right]}\left( \frac12 t \right) \right\} = \frac12 \mathcal F \left\{ \pi \chi_{\left[-\frac1{\pi}, \frac1{\pi} \right]} \right\}$

Abhas Kumar Sinha

5:33 AM

@LeakyNun What is sinc? hyperbolic sin?

Leaky Nun

$\chi_{[-1/2\pi, 1/2\pi]} \ast \chi_{[-1/\pi, 1/\pi]} = \begin{cases} 1/\pi & |x| \le 1/2\pi \\ 3/2\pi - |x| & 1/2\pi \le |x| \le 3/2\pi \\ 0 & |x| \ge 3/2\pi \end{cases}$

Abhas Kumar Sinha

or circular sin?

Leaky Nun

@AbhasKumarSinha no, $\sinc(x) = \dfrac{\sin(x)}x$

Abhas Kumar Sinha

@LeakyNun why?

Leaky Nun

> The term sinc /ˈsɪŋk/ was introduced by Philip M. Woodward in his 1952 paper "Information theory and inverse probability in telecommunication", in which he said the function "occurs so often in Fourier analysis and its applications that it does seem to merit some notation of its own",[2] and his 1953 book Probability and Information Theory, with Applications to Radar.[3][4]

Abhas Kumar Sinha

5:35 AM

@LeakyNun oh okay... :)

@LeakyNun Do you know functional analysis?

Leaky Nun

not much

Abhas Kumar Sinha

@LeakyNun hermitian operators?

Leaky Nun

there's the spectral theorem

Abhas Kumar Sinha

For some operator $\hat f$ $$ \left < \phi_1 , \hat f \phi_2 \right> = \left < \phi_2, \hat f \phi_1 \right> $$

1. Is it true for all $a, b$? $$ \left < a, b \right> = \left < b, a \right> $$

@LeakyNun

Leaky Nun

Let $\displaystyle g = \frac2\pi \sum_{n=0}^\infty (-1)^n \frac1{2n+1} \left( \delta_{(2n+1)/2\pi} + \delta_{-(2n+1)/2\pi} \right)$

@AbhasKumarSinha no

so I think you got the definition wrong

Abhas Kumar Sinha

5:40 AM

@LeakyNun no inner products are not commutative?

Leaky Nun

it should be $\langle \phi_1, \hat f \phi_2 \rangle = \langle \hat f \phi_1, \phi_2 \rangle$

they're anti-commutative

$\langle a,b \rangle = \overline{\langle b, a \rangle}$

Abhas Kumar Sinha

@LeakyNun But, $$ \int a b dx = \int b a dx $$?

Leaky Nun

so $|\sinc(x) \sinc(2x)| = \mathcal F \left\{ g \ast \chi_{[-1/2\pi, 1/2\pi]} \ast \chi_{[-1/\pi, 1/\pi]} \right\}$

@AbhasKumarSinha well are you real or complex?

Abhas Kumar Sinha

@LeakyNun complex

Leaky Nun

then the inner product is $\displaystyle \int a \overline{b}$

Abhas Kumar Sinha

5:42 AM

@LeakyNun ugh!

Leaky Nun

correction: let $g = \displaystyle \sum_{n=0}^\infty (-1)^n \frac1{2n+1} \left( \delta_{(2n+1)/2\pi} + \delta_{-(2n+1)/2\pi} \right)$ and $|\sinc(x) \sinc(2x)| = \pi \mathcal F \left\{ g \ast \chi_{[-1/2\pi, 1/2\pi]} \ast \chi_{[-1/\pi, 1/\pi]} \right\}$

so $\displaystyle \int_{-\infty}^\infty |\sinc(x) \sinc(2x)| \ \mathrm dx = \pi (g \ast \chi_{[-1/2\pi, 1/2\pi]} \ast \chi_{[-1/\pi, 1/\pi]})(0) = \pi \left( \frac1\pi + \frac1\pi \right) = 2$

so $\displaystyle \int_0^\infty |\sinc(x) \sinc(2x)| \ \mathrm dx = 1$

@VladimirReshetnikov there you go

phew

Abhas Kumar Sinha

@LeakyNun Bro, how are dot products defined in complex numbers?

Leaky Nun

so on $\Bbb C^n$?

Abhas Kumar Sinha

@LeakyNun yes

@LeakyNun ..?

Leaky Nun

@AbhasKumarSinha $\langle v, w \rangle = \displaystyle \sum_{i=1}^n v_i \overline{w_i}$

Abhas Kumar Sinha

5:52 AM

@LeakyNun proof?

Leaky Nun

it's a definition

Abhas Kumar Sinha

@LeakyNun does it works same with vectors?

Leaky Nun

what?

Abhas Kumar Sinha

@LeakyNun see $$ \left< a, b \right> = \int a.b^* \, dx $$?

Leaky Nun

that's the definition of inner product for functions

Abhas Kumar Sinha

5:54 AM

$a, b$ are funcs

@LeakyNun then prove that $$ \int a^* (\hat f b) \, dx = \int b^* (\hat f a) \, dx $$ is always a real valued eigenoperator $\hat f$ for $a(x), b(x)$ as complex valued functions.

Leaky Nun

your question doesn't make sense

Abhas Kumar Sinha

^The def of hermitian operators

@LeakyNun When $\hat f$ operator acts on functions $a, b$ it always gives real values

Leaky Nun

it doesn't

Abhas Kumar Sinha

@LeakyNun strange....

Leaky Nun

I think I know what you mean to ask, but the way you say it makes no sense

so you misunderstood the statement / question

Abhas Kumar Sinha

5:58 AM

@LeakyNun perhaps...

$$ \langle a, b \rangle \stackrel{def}{=} \int a^*.b \, dx $$ Is this true?

@LeakyNun ?

Leaky Nun

yes

"def" means it is a definition

Abhas Kumar Sinha

@LeakyNun yes

Leaky Nun

definitions are definitions

they are not true or false

Abhas Kumar Sinha

@LeakyNun Then $$ \langle b, a \rangle = \int b^*.a \, dx$$?

Leaky Nun

sure

Abhas Kumar Sinha

6:04 AM

@LeakyNun For an operator $\hat f$ to be hermitian, $$ \langle a, \hat f b\rangle = \langle b, \hat f a \rangle $$?

Leaky Nun

that's an incomplete statement (missing quantifiers)

Abhas Kumar Sinha

@LeakyNun what? then please complete it

Leaky Nun

you need $\langle a \hat f b \rangle = \langle b, \hat f a \rangle$ for all $a$ and $b$

Abhas Kumar Sinha

@LeakyNun yes, $a, b$ are functions...

Leaky Nun

the point is "for all $a$ and $b$"

Abhas Kumar Sinha

6:06 AM

@LeakyNun yes

then

$$ \int a^*. \hat f b \, dx = \int b^*. \hat f a \, dx $$ for all complex valued $a, b$ for an operator $\hat f$?

Leaky Nun

no, $\hat f$ needs to be hermitian (that's the whole point)

Abhas Kumar Sinha

@LeakyNun Yes, sorry forget to mention that..

@LeakyNun $\hat f$ is hermitian...

Leaky Nun

then that's what you get by expanding the definition

Abhas Kumar Sinha

Then, $$ \hat f a = \alpha a $$ for some constant $\alpha$, prove it is real!

Leaky Nun

that's the right question

Abhas Kumar Sinha

6:10 AM

@LeakyNun yeaaaaaaaaah!! $\ddot \smile$

@LeakyNun proof? :P

Leaky Nun

ah you need $a \ne 0$ also

Abhas Kumar Sinha

@LeakyNun yes yes

Leaky Nun

$\langle a, \hat f a \rangle = \langle \hat f a, a \rangle$ by hermitian

Abhas Kumar Sinha

yes

Leaky Nun

on the other hand, $\langle a, \hat f a \rangle = \overline{\langle \hat f a, a \rangle}$ by property of inner product

Abhas Kumar Sinha

6:14 AM

okay!

Leaky Nun

so $\langle \hat f a, a \rangle$ is real

and $\langle \hat f a, a \rangle = \langle \alpha a, a \rangle = \alpha \langle a, a \rangle$

so $\alpha$ is real

Abhas Kumar Sinha

yes

@LeakyNun damn... I was integrating whole night for this!! Night Wasted

It's so easy!!?

Leaky Nun

rip

Abhas Kumar Sinha

hehhehe XD

@LeakyNun Can you help me to find some good references for the introduction of innerproduct of vectors and complex numbers on web?

Leaky Nun

Inner product space

In linear algebra, an inner product space is a vector space with an additional structure called an inner product. This additional structure associates each pair of vectors in the space with a scalar quantity known as the inner product of the vectors. Inner products allow the rigorous introduction of intuitive geometrical notions such as the length of a vector or the angle between two vectors. They also provide the means of defining orthogonality between vectors (zero inner product). Inner product spaces generalize Euclidean spaces (in which the inner product is the dot product, also known as the...

Leaky Nun

6:32 AM

it's not true

Abhas Kumar Sinha

@LeakyNun now?

Leaky Nun

now it doesn't even make sense

Abhas Kumar Sinha

@LeakyNun $\left < a, b \right> \in \mathbb{R}$

proof

Leaky Nun

that's not true

Abhas Kumar Sinha

@LeakyNun Inner products can be complex too?

Leaky Nun

6:34 AM

yes

Abhas Kumar Sinha

@LeakyNun sorry, I forgot what to ask...

@LeakyNun What is the graphical meaning of inner products for 2D complex plane?

Leaky Nun

so $\Bbb C$

Abhas Kumar Sinha

@LeakyNun yes

Leaky Nun

so that's $\langle z, w \rangle = \overline{z} w$

so you reflect $z$ along the real axis

and then multiply by $w$

brb

Abhas Kumar Sinha

@LeakyNun ok

Leaky Nun

6:49 AM

so the magnitude is the product of the magnitudes (which agrees with the dot product of the two complex numbers viewed as elements of $\Bbb R^2$)

but now you have a complex number

the argument is the difference of the two arguments

@AbhasKumarSinha ^

Abhas Kumar Sinha

ok

maths student

7:43 AM

0

Q: equivalence relation over real number

$ xRy$ iff $x^2 -y^2 = x-y$ is relation we have defined over $\mathbb{R} $ I have shown this is equivalence relation. Now we have asked to find equivalence class of 3 which can be found out to be -2 and 3 . Also next we need to find equivalence class of general x which turn out out to be 1-x...

real-analysis number-theory elementary-number-theory equivalence-relations

RockPaperLizard

8:18 AM

My calculations indicate that 25.8db is about 5% quieter than 26db. Am I correct?

Jack M

9:06 AM

This question was posted a few days ago with a theorem I haven't seen before in finite group theory

has anyone come across this result anywhere?

Adam

11:54 AM

$${\{a_j}\}_{j=1..n} \subset \mathbb N$$
$$n\,\varphi\Bigl(\sum^n_{j=1}a_j\Bigr) \leq \gcd({\{a_j}\}_{j=1..n})$$

Has to correct the previous time I put this up to put emphasis that uniqueness for ${\{a_j}\}_{j=1..n}$ is necessary

user193319

12:54 PM

What exactly is $\Bbb{Z}[\frac{1}{2}]$? Is it the polynomial ring $\Bbb{Z}[x]$ with $x = \frac{1}{2}$?

Balarka Sen

$\Bbb Z[x]/(2x - 1)$, yes.

Ring of rational numbers with denominator some power of $2$

Alternatively $\Bbb Z$ localized at the multiplicative subset $\{1, 2, 2^2, \cdots\}$

user193319

1:20 PM

Thanks @BalarkaSen

Suppose I have two matrices $A, B \in SL_2(\Bbb{Z}[\frac{1}{2}])$ that satisfy the relationship $BAB^{-1} = A^2$. Can I say that they generate a group isomorphic to the abstract group $\langle a,b \mid bab^{-1} = a^2 \rangle$?

Balarka Sen

You can't say a-priori that they don't satisfy any other relations. You need something like the ping-pong lemma.

What are your matrices?

user193319

$A = \begin{pmatrix} 1 & 1 \\ 0 & 1 \\ \end{pmatrix}$ and $B = \begin{pmatrix} 2 & 1 \\ 0 & 1 \\ \end{pmatrix}$

So showing they satisfy the same relationship isn't sufficient. Okay. That was my gut feeling. How exactly does ping pong help. I thought that concerned free groups.

Balarka Sen

Hm, I am not sure how to go about this

Semiclassical

“Ping-pong lemma”

That’s a great name

Mike Miller

From the relation $BA = A^2B$ we see that any word in $A$ and $B$ is actually of the form $A^k B^j$ for some integers $k,j$. So it suffices to check that this is never the identity.

Balarka Sen

1:34 PM

Meh, nevermind, nonsense.

One should be able to ping-pong this

feynhat

Hello

$P \subset N \subset M$ are modules. If P is a direct summand in N, then is it also direct summand in M?

Mike Miller

Why should it be?

Take P = N

Then you're asking if every submodule is a summand

feynhat

Oh...

Balarka Sen

Hm. Let $G = \langle A, B \rangle$. We know $H = \langle A \rangle$ is an infinite cyclic normal subgroup of $G$.

$G/H$ is obviously $\langle B \rangle$, which includes into $G$, giving a section, right?

So $G$ is a semidirect product of $H$ and $G/H$.

Yeah, because $G/H$ is a free - there is always a section

This forces the group to be $\langle a, b | bab^{-1} = a^k \rangle$ for some $k$. You already know the relator $bab^{-1} = a^2$ is in the presentation.

Semiclassical

Fun pictures here: BS groups

Ffs, I hate when links break like that

Balarka Sen

1:47 PM

@MikeMiller's argument is much more down to earth though, but I think what I said works (right?)

Semiclassical

Nicely, the first pic there is for B(1,2) which is the group of interest here

feynhat

I am reading proof of a baby version of Universal Coefficient Theorem, in which $H_{n-1}$ is assumed to be free and we show that $H^n \approx Hom(H_n, \mathbb{Z})$. The isomorphism is given by the map $k([\phi])([c]) = \phi(c)$, where $\phi$ is a cocycle and $c$ a cycle.

Rajesh Dachiraju

0

Q: Solving a set of simultaneous equation for periodic boundary conditions.

Given $$g(x) = e^{-|x-x_0|/\alpha} + Ae^{-x/\alpha} + Be^{x/\alpha}$$ $\alpha >0$, $x_0 \in (0,1)$ Constraints are $$g(0) = g(1)$$ and $$g'(0) = g'(1)$$ Solve for $A$ and $B$. My solution : $$A = \frac{e^{(x_0-1)/\alpha}}{1-e^{(-1/\alpha)}}$$ and $$B = \frac{e^{-x_0/\alpha}}{e^{1/\alpha}-1}$$...

calculus algebra-precalculus

feynhat

$k$ takes a cohomology class $[\phi]$ and map it to the element in $Hom(H_n, \mathbb{Z})$, given by $[c] \mapsto \phi(c)$. It is well-defined, surjective.

I am trying to show that $k$ has kernel $0$.

Suppose $k([\phi]) = 0$. This means that $\phi(c)$ is zero for all cocycles $c$, ie. $\phi$ vanishes on kernel of $\partial: C^n \to Im \partial$. So, it factors through $Im \partial$.

What I mean is, it induces a map $\tilde \phi : Im \partial \to \mathbb{Z}$.

such that $\tilde\phi \circ \partial = \phi$.

Since $H_{n-1} = \ker\partial / Im \partial$ is free. $Im\partial$ is a summand in $\ker\partial$.

What I'd like to show is that $Im\partial$ is a summand in $C_{n-1}$.

Because then $\tilde\phi$ can be extended to $f : C_{n-1}\to \mathbb{Z}$. Then $\delta f(c) = f(\partial c) = \tilde\phi(\partial c) = \phi(c)$.

so $\phi = \delta f$

or $[\phi] = 0$.

6 mins ago, by feynhat

What I'd like to show is that $Im\partial$ is a summand in $C_{n-1}$.

@MikeMiller

feynhat

2:23 PM

$\ker\partial$ is a direct summand in $C_{n-1}$.

because, $C_{n-1}/\ker\partial = Im\partial$ which is free, being a submodule of a free-module.

$Im\partial$ is summand in $\ker\partial$ is summand in $C_{n-1}$.

So, $Im\partial$ is summand in $C_{n-1}$.

Is this good?

Obviously, $C_n$'s are singular chain modules with coefficients in PID, $R$. I don't know why I wrote $\mathbb{Z}$ all over. Please read it as $R$.

user193319

3:03 PM

What is a formula for $\begin{pmatrix} 2 & 1 \\ 0 & 1 \\ \end{pmatrix}^i$? I thought I was able to discern a pattern for $i=1,2,3,4$; but after that the (1,2) entries blows up.

Mike Miller

3:28 PM

@BalarkaSen Seems fine to me

My argument takes longer

I'd find a formula for the left column of B^k and then a formula for the top-left term of A^j B^k --- if A^j B^k = I, what is k? Then simply find a formula for A^j for all j.

Whence we get that none are the identity except j = k = 0

@feynhat Sorry, this will take too long for me to think about right now

though actually now that I look I do think we have explicitly $B^k = \begin{pmatrix}2^k & 2^k - 1 \\ 0 & 1\end{pmatrix}$ for all $k$

Adam

4:40 PM

Well if I digress enough to reveal my personality I get banned so ill keep booking this up since I get told to only discuss math $${\{a_j}\}_{j=1..n} \subset \mathbb N$$
$$n\,\varphi\Bigl(\sum^n_{j=1}a_j\Bigr) \leq \gcd({\{a_j}\}_{j=1..n})$$

just a counter example and ill leave everyone alone

and that can be spun into a multidimensional argument no doubt which is popular with folk hmmm? …. hmmmm?

and then Iran hikes their petrol prices by 300 percent all the while plastic <---> petroleum

Rithaniel

6:07 PM

So, the powerset of a set equipped with the union operation qualifies as a monoid. However, there don't exist any rings with this monoid as their multiplicative structure, correct?

Poline Sandra

Hello, please I need help in this simple question, il must use the definition to prove that $\lim_{x\to 1} (x^3+2x^2-2} =1$ I started by

$|x^3-1+2x^2-2|\leq |x^3-1|+2|x^2-1|=|x^3-1|+2 |(x-1)| |(x+1)|=|x-1|(|x^2+x+1|+2|x+1|)$

how to find $\delta>0$ such that $|x-1|<\delta$

Thorgott

@Rithaniel trivial case: the null ring and the powerset of the empty set

Rithaniel

Ah, yeah, fair enough. I still forget to rule out trivial cases

Thorgott

6:22 PM

$\mathbb{Z}/2\mathbb{Z}$ works too

Poline Sandra

someone help me on analysis ?

Thorgott

generally, such a ring would only have one invertible element (corresponding to the empty set)

so it must be of characteristic $2$

Rithaniel

I suppose the set of a single element corresponds to the multiplicative structure on $\mathbb{Z}/2\mathbb{Z}$?

Thorgott

@PolineSandra I don't think a $\delta$ such that $|x-1|<\delta$ is what you want. You want a $\delta>0$ such that, if $|x-1|<\delta$, you can bound the RHS with a given $\varepsilon>0$.

Rithaniel

So, any set union itself is itself, so every element in this ring is idempotent, so this is a boolean ring

Thorgott

6:28 PM

That checks out, cause $0^2=0$ and $1^2=1$

I'm not sure about examples with more elements

Leaky Nun

so you have stone's representation theorem

Stone's representation theorem for Boolean algebras

In mathematics, Stone's representation theorem for Boolean algebras states that every Boolean algebra is isomorphic to a certain field of sets. The theorem is fundamental to the deeper understanding of Boolean algebra that emerged in the first half of the 20th century. The theorem was first proved by Marshall H. Stone (1936). Stone was led to it by his study of the spectral theory of operators on a Hilbert space. == Stone spaces == Each Boolean algebra B has an associated topological space, denoted here S(B), called its Stone space. The points in S(B) are the ultrafilters on B, or equivalently...

given any compact Hausdorff space, its clopen sets form a Boolean algebra

and every Boolean algebra arises in this way

Thorgott

I was just reading up on that, though I'm not familiar with the result

But I think that gives you a contradiction with more than two elements

Rithaniel

Yeah, if $A\bigcup B=S$ then $A\bigcup(A+B)=A\bigcup A+A\bigcup B=A+S=A$ because $S$ would have to be the "zero" of the ring, so $A+B=A$ and, by a symmetric argument, $A+B=B$

Leaky Nun

so given a set X, the power set of X can be equipped with a ring structure that makes multiplication the union

by having symmetric difference as addition

and the whole set the zero

the empty set the one

hi @Semiclassical

Semiclassical

hi

Poline Sandra

6:39 PM

@Thorgott yes that is it

how to do please

to prove using the definition that $\lim_{x\to 1} (x^3+2x^2-1)=2$

Leaky Nun

@Semiclassical what's the eigenvalue of a symbol?

William Sun

Leaky Nun

atiyah macdonald!

William Sun

In part ii) shouldn't it be under the assumptions that \mathfrak q is primary?

Yes it is!

Leaky Nun

"Let $\mathfrak q$ be a $\mathfrak p$-primary ideal"

William Sun

6:43 PM

Yes but if q is not primary then there are easy conterexamples in the ring of integers

Leaky Nun

"Let $\mathfrak q$ be a $\mathfrak p$-primary ideal" means that $\mathfrak q$ is primary

William Sun

Oh I didn't see that assumption in the book so just to confirm. Thank you;)

Leaky Nun

it's literally the first sentence

Semiclassical

@LeakyNun what

Leaky Nun

@Semiclassical of an parabolic PDE

Semiclassical

6:45 PM

ah. dunno

Thorgott

@PolineSandra try factorizing $x^3+2x^2-3$

Sayan Chattopadhyay

7:02 PM

How does one integrate a function in a complicated open set. For instance, I have been given $f(x,y) = \frac{1}{(y+1)^2}$ and have been asked to integrate this in the region $A(x, y ) = \{ (x,y) | x >0 , x^2 < y < 2x^2\}$.

Theoretically I need to find a nested sequence of open sets sets $\{U_n\}$ such that $\int_{U_{n}} f$ is bounded. But how do I find such sets here

Also I need to use fubini's theorem but again I need to give proper limits. And I dont get how to do that for a relatively complicated open set that needs to be taken given my set $A$

Thorgott

7:21 PM

Since $f(x,y)$ doesn't depend on $x$, it seems reasonable to just cut off large values of $x$ to achieve such a sequence

Leaky Nun

how is $\int_{U_n} f$ defined? this seems a little circular to me

Thorgott

$\int_{\mathbb{R}^2}1_{U_n}f$, id say

Sayan Chattopadhyay

@LeakyNun you define the integral in terms of compact rectifiable sets.

@Thorgott Could you elaborate on this?

Leaky Nun

compact sets can't be open

Thorgott

take $A_n=\{(x,y)|0<x<n,x^2<y<2x^2\}$, then $A_n\uparrow A$

Sayan Chattopadhyay

7:28 PM

@LeakyNun Yes but using the compact set definition you can arrive at this one.

Thorgott

@LeakyNun what about the empty set

Leaky Nun

@Thorgott don't be that guy

:P

Sayan Chattopadhyay

@Thorgott I see. Cool. That's an obvious choice

Thorgott

i had to

Sayan Chattopadhyay

@Thorgott though if I take the limits to be $0$ and $n$ for $x$ in the integral, you arrive at having to evaluate $1/x$ at $0$ which is a problem.

Ted Shifrin

7:35 PM

Huh? @Sayan

Thorgott

where does that term come from?

Sayan Chattopadhyay

Oh never mind. I am making calculation errors

Ted Shifrin

There's a $1+$ in there.

Sayan Chattopadhyay

Yeah, I didn't notice that. I should sleep lol

Ted Shifrin

Mathing without sleep isn't good.

Sayan Chattopadhyay

7:39 PM

Yeah, especially if its analysis

Sayan Chattopadhyay

7:57 PM

This is weird. If I take the same $A_n$ but instead of $x^2 < y <2x^2$ I take $x < y <2x$ (i.e change the region to $\{ x>0 , x<y<2x\}$) the integral $\int_{A_{n}} f$ is bounded for all $n$. Specifically it is $\log(\frac{n+1}{2n+1})$. But the text says that $f$ is not integrable in the changed region.

Well it's not exactly that but the integral is bounded for all $n$

Thorgott

It should be $\log\left(\frac{n+1}{\sqrt{2n+1}}\right)$

Sayan Chattopadhyay

Yeah

Thorgott

That's not bounded

Sayan Chattopadhyay

Oh I am messing up big time. Shit. I have to sleep

k170

8:13 PM

6

Q: How to evaluate $\lim_{x \to 0}\left(\frac{1}{x} - \frac{1}{x^2}\right)$

I have to find the limit of following $$\lim_{x \to 0}\left(\frac{1}{x} - \frac{1}{x^2}\right)$$ I have no idea how to start this one off. How would I do it? Do I just substitute the $0$? It doesn't look that easy and simple. The answer says it's negative infinity. Please show me a solution w...

limits

Do you guys see anything wrong with my answer here?

topologicalmagician

8:56 PM

Let $A \in \mathbb{Z}+\cup$ $\{$ $0$ $\}$ and $B \in \mathbb{Z}^+$ then there exists $C\in \mathbb{Z}$^+ $\cup$ $\{$ 0$\}$ with $C\leq A$ and $P\in \mathbb{Q} \cup [0,1)$ so that $A=BC+BP$

Is this statement true?

Semiclassical

In simpler terms: Suppose $A,B$ are nonnegative integers with $B\neq 0$. Then there exists a nonnegative integer $C\leq A$ such $C\leq A/B<C+1$.

topologicalmagician

@Semiclassical

How'd you prove it though?

Semiclassical

actually. should your initial statement have had $\mathbb{Q}\cup [0,1)$ or $\mathbb{Q}\cap[0,1)$?

topologicalmagician

$\cap$

Rithaniel

Break it up into cases. If $A<B$ and if $B<A$

Semiclassical

9:03 PM

okay. I want to sat that's just the archimedian principle?

Rithaniel

(and if $A=B$)

$A=B$ should be very easy, though

Well, I shouldn't say that. All three are easy, if you already know the answer

Semiclassical

The version I wrote can be summed up as this: For every nonnegative rational number, there is some smallest integer which is bigger than it.

Ted Shifrin

9:34 PM

@Semiclassic: Why are you saying $C\le A$?

Semiclassical

45 mins ago, by topologicalmagician

Let $A \in \mathbb{Z}+\cup$ $\{$ $0$ $\}$ and $B \in \mathbb{Z}^+$ then there exists $C\in \mathbb{Z}$^+ $\cup$ $\{$ 0$\}$ with $C\leq A$ and $P\in \mathbb{Q} \cup [0,1)$ so that $A=BC+BP$

Is this statement true?

@TedShifrin because it was in there

shrug

Ted Shifrin

oh ...

@k170 I write such proofs very differently. I would also try to make it all about positive numbers, instead of negative ones, because inequalities are a pain with negatives. In particular, since $|x-1|>3/4$, I think your $5/4$ should be $3/4$ in your final stuff.

k170

10:10 PM

@TedShifrin Thanks for the feedback. I agree that these edits would make it smoother to read.

Ted Shifrin

But I think you did mess up your inequalities, getting the $5/4$.

I didn't take the time to check line by line.

k170

Are you referring to this part?

$$\left|x\right|\lt\frac14$$
$$-\frac54\lt x-1\lt -\frac34$$

Ted Shifrin

No, that part is fine. It's when you're bounding $(x-1)/x^2$.

Note that $3/4<|x-1|<5/4$. So $|x-1|>3/4$.

We're doing an > argument, not an < argument.

k170

Ahhh Yes that makes sense

Thanks again. I'll edit my post

Ted Shifrin

Adam

10:25 PM

why is the content of the chat room considerably less impressive and or helpful compared to responses for questions I post? Like how come only a tiny selection of users use the chat feature to commutate almost entirely exclusively with one another?

amanuel2

11:34 PM

Can anyone explain how answer key got the arctan value here? Im super confused

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$Mathematics$ Mathematics