Mathematics - 2020-01-16 (page 2 of 2)

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adesh mishra

14:02

I have to solve this integral $$ \int \frac{\cos x + \sqrt 3}{1 + 4\sin \left(x+\pi/3\right) +4 \sin^2 \left(x+\pi/3\right)}$$ I could see that the denominator is can be written as $ \left(1+ 2\sin \left(x+\pi/3)\right \right) ^2$

I can see that the denominator can be written as $ \left(1+2\sin\left(x+\pi/3 \right)\right)^2$

writing $2\sin \left(x+\pi/3)\right)$ as $\sin x + \sqrt 3 \cos x$

Abhigyan Chattopadhyay

@adeshmishra Are you sure of the numerator?

adesh mishra

yeah

Martin Sleziak

$\cos x + \sqrt 3\sin x%$ in the numerator would make it easier, I think.

adesh mishra

Can we write it that way?

Abhigyan Chattopadhyay

@MartinSleziak You mean $\cos x - \sqrt{3} \sin x$ right?

Martin Sleziak

14:13

Well, Abhigyan Chattopadhyay already asked whether the integral is copied correctly.

adesh mishra

Yes

Martin Sleziak

Yes, I'd have to put minus there to get $\cos(x+\frac\pi3)$.

adesh mishra

user image

@MartinSleziak How did you get $\cos \left(x+\pi/3 \right)$

?

Abhigyan Chattopadhyay

Set the denominator to $t^2$, and the numerator to $At'+Bt+C$

Martin Sleziak

This is what WolframAlpha says

adesh mishra

14:20

Well, finally our integral reduces to $$ \int \frac{\cos x+ \sqrt 3}{\left( 1+\sin x+ \sqrt 3 \cos x \right)^2}$$

Thank you for pointing that out

Abhigyan Chattopadhyay

Right, now set $(1+\sin x + \sqrt{3} \cos x)$ as $t$

And, assume the numerator as $At'+Bt+C$

It'll work out

adesh mishra

It's not coming, $$ u = 1+ \sin x + \sqrt 3 \cos x \\ du/dx = \cos x - \sqrt 3 \sin x$$

Martin Sleziak

I am not really sure whether some of those four options is a correct result.

Anyway, we have $\int \frac{\cos x + \sqrt 3}{1 + 4\sin \left(x+\pi/3\right) +4 \sin^2 \left(x+\pi/3\right)}>0$ for $x\in(0,\pi/2)$.

adesh mishra

Do you also make mistakes?

Martin Sleziak

So whatever the result is, it should be increasing on that interval.

I should probably go, have a nice day!

adesh mishra

14:33

Have a nice day Sir

The Terrible Puddle

15:18

Greetings

The Terrible Puddle

15:45

Would someone be so kind as to ask how this guy has shown $f_N$ to be extreme points of $K$? https://math.stackexchange.com/questions/3505781/finding-extreme-points-of-closure-of-convex-hull
I don't yet have the reputation to make a comment.

The Terrible Puddle

16:06

Oh hey, can do it myself now. Thanks for the rep

MadSpaceMemer

16:50

Good evening! I keep forgetting what the notation of the polynom $f(X)$ should mean. Is it saying that $f(X) := 0.x+1.x+0.x^2.....$ or is it just a symbol for a random polynom and it is unknown at which $x$ the highest power exists?

COTO

17:13

Figured I'd ask this simple question in here rather than plugging up the board: Is there a standard infix operator notation for the diagonal concatenation of two square matrices? For example, $A \oplus B$? I've been told that $diag[A,B]$ is standard in some cases, but I'm hoping there's some standard that doesn't take up more space than A (+) B?

I've seen operators like [+] and [] in LaTeX. Any of them stand for diagonal concatenation?

Alessandro Codenotti

17:24

What do you mean with diagonal concatenation?

ÍgjøgnumMeg

@COTO do you mean something like $\begin{pmatrix} A & 0 \\ 0 & B\end{pmatrix}$ ?

(block matrix)

Alessandro Codenotti

Hi @ÍgjøgnumMeg

ÍgjøgnumMeg

Hey @Alessandro

COTO

Exactly, Alessando.

ÍgjøgnumMeg

woah

$A \oplus B$ is standard I think

Matrix addition

In mathematics, matrix addition is the operation of adding two matrices by adding the corresponding entries together. However, there are other operations which could also be considered as a kind of addition for matrices, the direct sum and the Kronecker sum. == Entrywise sum == Two matrices must have an equal number of rows and columns to be added. The sum of two matrices A and B will be a matrix which has the same number of rows and columns as do A and B. The sum of A and B, denoted A + B, is computed by adding corresponding elements of A and B:...

Alessandro Codenotti

17:26

That's exactly the matrix representing $A\oplus B$ on $V\oplus W$ so looks good to me

COTO

Ah. It is there on Wiki.

ÍgjøgnumMeg

Aye

COTO

OK. I'll have to convince my supervisor that I'm not just making crap up. :D

MadSpaceMemer

"@AlessandroCodenotti Hi allesandro, could you provide some assistance with a notation problem i am having_

Alessandro Codenotti

I don't know

You should just ask the question and maybe me or someone else will be able to help

(also you don't have to answer if you don't want to share personal informations, but are you German?)

MadSpaceMemer

17:30

No but i do study in germany

@AlessandroCodenotti i have a function that is doing the following $g(f(X)) = f(X + 1)$ i am not sure what is happening here. I was googling what the notation $f(X)$ is (i know its a polynomial but i wasnt exactly sure how it looked like ) turns out that f(X) is just $a_o+a_1x+a_2x^2....$

COTO

Thanks all. Cheers.

ÍgjøgnumMeg

@MadSpaceMemer something important: a polynomial is a necessarily finite sum, and secondly, if your polynomial is called $f(X)$ then your variables should match, i.e. $f(X) = a_nX^n + \dots + a_1 X + a_0$

rather than $f(X) = a_nx^n + \dots + a_1 x + a_0 $

unless $X := X(x)$ but idk

MadSpaceMemer

No you are right. I wrote it badly.

So what would then $f(X+1)$ mean?

ÍgjøgnumMeg

$a_n(X+1)^n + \dots + a_1(X+1) + a_0 $

MyWrathAcademia

Who can answer this: Given a range of numbers in the range a to b all inclusive. Find the count of numbers divisible by n in this range? e.g. 1: 7 14 7 the answer is 1, e.g.2: 6 14 6 the answer is 2

MadSpaceMemer

17:38

Ohh!!

MyWrathAcademia

I know that if n > b then no number in that range is divisible

but I'm trying to obtain a general formula that works for all cases

Semiclassical

That itself is a polynomial of the form $\sum_{k=0}^n b_k X^k$, of course.

with some linear transformation between the $a_k$'s and the $b_k$'s

ÍgjøgnumMeg

Aye

MadSpaceMemer

@ÍgjøgnumMeg Thank you! now I can solve the question <3

ÍgjøgnumMeg

Nooo worries

The Terrible Puddle

18:07

Operator $M\in\mathcal{L}(l_2,l_2)=\{T:l_2\to l_2 \text{ continous linear functions}\}$ defined by $M(x)=(n^{-1/5}x_n)_{n\geq 1}$ for $x=(x_n)_{n\geq 1}\in l_2$. How would you justify $M$ being compact but not Hilbert-Schmidt?

Thorgott

This is very much pedantic, so feel free to ignore the following. If you have a ring $R$, you can construct the corresponding polynomial ring $R[X]$. Take a polynomial $P\in R[X]$; this can be written in the form $\sum_{k=0}^na_kX^k$ with $a_1,...,a_n\in R$.
The polynomial ring is characterized by a universal property, which allows you to evaluate a polynomial $P\in R[X]$ in an element $x\in S$ ($S$ being another ring) via a ring homomorphism $\varphi\colon R\rightarrow S$ (ommitting a small technical detail). This is given by applying $\varphi$ to the coefficients of $P$ and substituting $…

MyWrathAcademia

18:25

hey @Thorgott do you have any idea how to approach the following problem:

Given a range of numbers in the range a to b all inclusive. Find the count of numbers divisible by n in this range? e.g. 1: 7 14 7 the answer is 1, e.g.2: 6 14 6 the answer is 2

The Terrible Puddle

Nvm, think I've an argument for it being compact. Just need to show it is not H-S now

Alessandro Codenotti

It's compact because it is a limit of finite rank operators

To show that it is not HS compute its norm

The Terrible Puddle

alrighty

Thorgott

@MyWrath what is $n$?

Alessandro Codenotti

The index

He's identifying $\ell^2$ with sequences $(x_n)_{n\in\Bbb N}$

MyWrathAcademia

18:34

@Thorgott n is an integer number

Thorgott

your examples don't make sense without specifying what $n$ is in these examples

Alessandro Codenotti

It's a sequence @Thorgott. $M$ maps the sequence $(x_n)_{n\in\Bbb N}$ to the sequence $(n^{-1/5}x_n)_{n\in\Bbb N}$

The Terrible Puddle

I don't think you two are talking about the same problem

Thorgott

Yeah, I'm talking about Wraths question

Alessandro Codenotti

Oh lol my bad

I thought you were also answering to @TheTerriblePuddle

Sorry

MyWrathAcademia

18:37

Oh I see, apologies. Let me fix it: e.g. 1: 7 14 7 where a is 7, b is 14 and n is 7 the answer is 1. Likewise for e.g.2: 6 14 6 where a is 6, b is 14 and n is 6 the answer is 2

respectively

Thorgott

I think there are two numbers divisible by 7 in the range from 7 to 14, namely 7 and 14

MyWrathAcademia

Oops, your'e right. Thanks.

Oh and by the way I keep seeing dollar signs. I'm assuming code or mathematical equations are not being displayed correctly. How do I get math to display correctly here

Thorgott

See tinyurl.com/cfqcvpc

MyWrathAcademia

Thanks again @Thorgott

Thorgott

As for your question, the answer should be $\max\left(0,\lfloor\frac{b}{n}\rfloor-\lceil\frac{a}{n}\rceil+1\right)$

MyWrathAcademia

18:44

Thanks @Thorgott but I'm trying to install the MathJax renderer you linked because your code is difficult to parse

The Terrible Puddle

@AlessandroCodenotti You mean compute its Hilbert-Schmidt norm right? And if it is not finite it is not H-S

Alessandro Codenotti

yes

MyWrathAcademia

@Thorgott I can't get MathJax to install. After you add it as a bookmark how do you get it to run?

The Terrible Puddle

Maybe I should use that $\|M\|\leq \|M\|_{HS}$?

Thorgott

You click on the bookmark

MyWrathAcademia

18:51

just to comfirm it's only start ChatJax that I need to install right, not render MathJax?

Thorgott

Yes

MyWrathAcademia

@Thorgott I had an add on that was blocking JavaScript, had to unblock it. Its working great and the equations look nice now

Alessandro Codenotti

@TheTerriblePuddle Just write down the series

Thorgott

Nice

Ted Shifrin

Hi, demonic @Alessandro, @Thorgott, Puddle, Wrath.

Alessandro Codenotti

18:56

Hi @Ted

MyWrathAcademia

Hi @TedShifrin

Thorgott

Hi Ted

The Terrible Puddle

Hi Ted

MyWrathAcademia

@Thorgott Let me try to understand your solution by first saying that I attempted to solve this problem before asking here and got (b - a) / n but obviously this does not work for a = 7 b = 14 and n = 7 which should give 2 numbers in the range a to b that are divisible by 7

The Terrible Puddle

@AlessandroCodenotti $\|M\|^2_{HS}=\sum_i |\langle Me_i,e_i\rangle|^2$ ?

Thorgott

19:01

@MyWrath when is a number divisible by $n$?

MyWrathAcademia

@Thorgott unless you add 1 to get (a - b) / n + 1 but then this modified formula does not work for a = 7 b= 14 and n = 6

@Thorgott when n is a factor, i.e. when the remainder of dividing a number by n is 0

Alessandro Codenotti

@TheTerriblePuddle $\|M\|_{HS}=\sum_i\|Me_i\|^2$

Thorgott

Right, so a number is divisible by $n$ if it has the form $nk$ for some natural number $k$ and we want to find the numbers of this form amidst $a,a+1,...,b$. What is the smallest such number?

The Terrible Puddle

@AlessandroCodenotti should I fix an $x\in \ell^2$ with $\|x\|=1$ then?

Alessandro Codenotti

You fix an orthonormal basis $e_i$ and take the sum over those

Ted Shifrin

19:12

@Alessandro: Didn't you forget a square root?

Alessandro Codenotti

Oh yeah you're right

Well we're just trying to decide whether the norm is finite so it's not too important

Ted Shifrin

Well, you're not usually sloppy :P

Alessandro Codenotti

Can't be (too) sloppy in logic/set theory!

Ted Shifrin

Are you implying that we can be sloppy in geometry?

Hrumph.

The Terrible Puddle

$\|Me_i\|^2=\sum^\infty_{n=1}|n^{-1/5}x_ne_i|^2$ right?

MyWrathAcademia

19:18

@Thorgott I think the smalllest such number is $a + n$

Ted Shifrin

What are $x_n$, @TheTerriblePuddle?

MyWrathAcademia

@Thorgott So the smallest number should satisfy $ a + n <= b $

The Terrible Puddle

@TedShifrin My sequence in $\ell^2$

Ted Shifrin

But Alessandro told you to compute $Me_i$. What does $e_i$ stand for?

Sorry, I shouldn't be butting in. shuts up

The Terrible Puddle

ONB

Ted Shifrin

19:21

Where does $x$ appear when you do $Me_i$?

Thorgott

@MyWrath no, consider the first example again

Alessandro Codenotti

@TedShifrin The Italian school of geometry was pretty famous for that!

The Terrible Puddle

@TedShifrin In how I defined the operator $M:\ell^2 \to \ell^2$

Ted Shifrin

Not being Bourbaki-rigorous doesn't necessarily mean sloppy, although some notions were indeed sloppy.

The Terrible Puddle

by $M(x)=(n^{-1/5}x_n)_{n\geq 1}$

Alessandro Codenotti

19:23

Ok, so what is $M(e_i)$? What is $e_i\in\ell^2$?

Ted Shifrin

Comment to Puddle: If I give you $f(x)=\sin(x)$ and I ask you for $f(2)$, you'd better not have an $x$ in the answer.

@Alessandro: It's true that a lot of the degenerating/specializing arguments in Italian algebraic geometry took a fair amount of work the last 50 years to make careful sense of. But, as is often the case (still the case with physicists, scarily), they mostly got the right answers.

Alessandro Codenotti

I see, I don't know any of the details to be fair

Ted Shifrin

It's called Enumerative Geometry.

Some of my research dealt with such things.

Alessandro Codenotti

Oh I see

There was a course here in enumerative geometry this semester

Ted Shifrin

Beautiful stuff. Lots of chern class computations with Grassmannians.

The Terrible Puddle

19:27

Okay, I don't know anymore

MyWrathAcademia

@Thorgott Is the smallest such number $nk >= a$?

And the largest $nk <= b$?

Thorgott

Yes, now for which $k$ do these happen?

Ted Shifrin

Puddle: Answer Alessandro's question. What is $e_i\in\ell^2$?

The Terrible Puddle

Some sequence

Ted Shifrin

You have to do better than that.

The Terrible Puddle

19:35

I suppose $\|e_i\|=1$

Alessandro Codenotti

Well which sequence?

@TheTerriblePuddle It better be the case since it's supposed to be an element of an orthonormal base

MyWrathAcademia

@Thorgott $k$ is smallest at $smallest number / n$ and largest at $largest number / n$? not sure how to represent this mathematically

Thorgott

that $k$ is not necessarily a natural number!

The Terrible Puddle

(0,0,0,0,...,1,0,0,0,...) for the i'th place

MyWrathAcademia

I see. In the examples I made $k$ is a natural number. Do you have an example when $k$ is not a natural number and is my above comment correct about when $k$ is smallest and biggest?

Ted Shifrin

19:46

Yippee, @Puddle, so what is $\|Me_i\|$?

Thorgott

no, I mean that what you said is not the proper $k$, since it's not necessarily a natural number

MyWrathAcademia

Wait so are you saying that what I wrote means that k may not be a natural number?

Thorgott

yes

for example, 1/2 is not a natural number

MyWrathAcademia

I see. I didn't notice that $smallestnumber/n$ could give $1/2$

Thorgott

it can give any positive rational, we're not placing any restrictions on $a,b,n$ except being natural after all

The Terrible Puddle

19:53

$\|Me_i\|=|n^{-1/5}x_i|$

Ted Shifrin

shrieks

The Terrible Puddle

he

MyWrathAcademia

So then $k$ has to be natural number, correct?

Ted Shifrin

Puddle, you have two things on the right hand side that make no sense. You have $x_i$ and you have $n$.

The answer can depend ONLY on $i$.

Thorgott

Yes, that's the definition of one number dividing another

The Terrible Puddle

19:55

But what about how M was defined?

$|i^{-1/5}x_i|$?

Ted Shifrin

A half hour ago I said you could not have $x_i$ in your answer.

The Terrible Puddle

Then what do I do with M?

Ted Shifrin

You have to understand what symbols actually mean.

The Terrible Puddle

right

Ted Shifrin

When you have a formula for $f(x)$ and you put in a value for $x$, say $x=2$, what do you do?

The Terrible Puddle

20:00

put it in

Ted Shifrin

So?

What is $Me_1$?

Eran

Hey, I'm trying to understand the solution of this exercise:

user image

The Terrible Puddle

1^{-1/5}x_1

Ted Shifrin

You keep repeating the thing I've told you is wrong. This is the fifth time. You have to be self-critical here.

You do NOT understand what $e_1$ means.

Eran

user image

The Terrible Puddle

20:03

So I don't

Eran

Why is the integrand is entire?

Ted Shifrin

Removable singularity at $0$, @Eran.

Thorgott

The singularity at $0$ is removable

got sniped

Eran

This is not entire though

You have to redefine it in order to be entire

Ted Shifrin

Yes, that's completely understood when there's a removable singularity.

MyWrathAcademia

20:07

@Thorgott The smallest such $k$ for the smallest such number $nk>=a$ is $k >= a/n$ hence all values of $k$ greater than or equal to $a/n$ satisfy $$nk>=a$$ and for the largest such number $nk<=b$ the largest such $k$ is $k <= b/n$ hence all values of $k$ less than or equal to $b/n$ satisfies $nk <= b$

Thorgott

That is correct. The usual notation for the smallest natural $k\ge\frac{a}{n}$ is $\lceil\frac{a}{n}\rceil$ and the usual notation for the largest natural $k\le\frac{b}{n}$ is $\lfloor\frac{b}{n}\rfloor$. Based on this, can you count the number of numbers divisible by $n$ between $a$ and $n$?

MyWrathAcademia

Thanks a lot! I've not seen those kind of brackets before so that notation is new to me. I will try to count the divisible numbers in that range based on those conditions.

manooooh

20:42

Hello all!

How can we find $$L=\lim_{t\to+\infty}\ln(t)-t/2$$ without using L'Hoptial's Rule? Thank you!

I tried the following: $$\ln(t)-t/2=\frac{\ln(t)}{2}\left(2-\frac{t}{\ln(t)}\right).$$ Since $\lim_{t\to+\infty}\ln(t)/2=+\infty$ we need to show that $\lim_{t\to+\infty}(2-\frac{t}{\ln(t)})=-\infty$ to have $L=+\infty\cdot(-\infty)=-\infty$

But I am struggling not with $\lim_{t\to+\infty}2$ but $\lim_{t\to+\infty}(-t/\ln(t))$

MyWrathAcademia

@Thorgott But first I have a question about the condition for the smallest number. For the example $a = 7$ $b = 14$ $n = 6$ the smallest $k$ is $2$ because $6 * 2 = 12$ which is the smallest number divisible by $6$. So why does plugging in $a = 7$ and $n = 6$ into $k >= a /n$ give $k>=1$ but the smallest $k$ is 2 and also plugging in $k >=1$ into $nk>=a$ gives $6x1>=7$.

Or is the point of this to say that $k == 1$ is not the smallest $k$ for $a = 7$ and $b = 14$. Wait since $k$ increments then the count of numbers divisible by $n$ between $a$ and $b$ is when $k <= b$ is true - when $k >= a is true + 1 (i.e. $(k <= b$ - $k >= a) + 1). Is this correct @Thorgott?

Can't edit previous post so here's a fix to the part that wasn't formatted correctly: Since $k$ increments then the count of numbers divisible by $n$ between $a$ and $b$ is when $k<=b$ is true - when $k >=a$ is true $+ 1$ (i.e. $(k <= b−k >= a) + 1)$. @Thorgott is this correct?

Ted Shifrin

20:58

@manooooh: Do you know anything about exponential/logarithmic rate of growth?

Forget about L'Hôpital's rule.

Thorgott

@MyWrath you have to look at the smallest $k$ that is greater than $\frac{a}{n}$. in your example, the smallest $k$ greater than $7/6$ is...?

manooooh

@TedShifrin not much Ted. What kind of rate of growth will you propose?

Ted Shifrin

Although you can use it to answer your final question, there. I just have different ways of doing it.

manooooh

@TedShifrin alternative solution?

Ted Shifrin

Exponentials grow faster than any polynomial, and polynomials grow faster than any power of $\log$.

What's your actual definition of logs and exponentials in your course?

manooooh

21:00

@TedShifrin okay

@TedShifrin I am not in a course anymore. I am helping to a student which said he can't use L'Hopital's Rule

MyWrathAcademia

In the example $a = 7$ $b = 14$ $n = 6$ the smallest $k$ greater than $7/6$ is $2$. I understand it I made a mistake in the last few comments. Here's what I meant to write: Since $k$ increments then the count of numbers divisible by $n$ between $a$ and $b$ is when $k<=b/n$ is true - when $k>=a/n$ is true $+ 1$ (i.e. ($k<=b − k >=a) + 1)$. @Thorgott is that now correct?

Ted Shifrin

Well, then the student should know something about these functions. People usually prove what I said with L'Hôpital, but I hate doing that.

If you define $\log x = \int_1^x dt/t$, note immediately that $\log x < \int_1^x dt/\sqrt t = 2(\sqrt x-1)$, so $(\log x)/x\to 0$ as $x\to\infty$.

It just depends what you and said student actually know.

Thorgott

@MyWrath what you wrote after i.e. is not the same as what you wrote before, but the earlier statement is indeed true: the number of natural numbers between a and b that are divisible by n is the number of natural numbers k between a/n and b/n

They could use Taylor, but if you want something elementary, it does indeed depend on how log has been defined

Ted Shifrin

@Thorgott: Yeah, Taylor for the exponential is sort of the flip-side of my approach.

manooooh

@TedShifrin that's a very good approach I think. But we have $x/\ln(x)$ not $\ln(x)/x$

Ted Shifrin

21:07

@manooooh: Duh. Stop and think.

manooooh

Ok. Stop. Hardest part: Think

Ted Shifrin

Today has been high frustration for me in this chatroom.

manooooh

We are on holidays probably?

At least in Latinamerica

Ted Shifrin

I don't think too many are on holidays.

manooooh

and a part of Europe

Thorgott

21:09

It's the first week of uni for me

manooooh

The teachers (I assist to a teacher, so technically I am not a teacher) are on holidays. We return to class on Feb 13

MyWrathAcademia

Yes I understand that the number of natural numbers between $a$ and $b$ that are divisible by $n$ is the number of natural numbers $k$ between $a/n$ and $b/n$ or in other words mathematically written as $(k <= b/n − k >= a/n) + 1$. @Thorgott is this equation a correct mathematical representation of the statement I quoted in the first sentence of this comment?

Thorgott

That statement is nonsense to me

Ted Shifrin

I guess this is your summer, so I'm surprised the schedule isn't flipped compared to the northern hemisphere.

MyWrathAcademia

@Thorgott what I mean is the largest $k$ that satisfies $k <= b/n$ minus the smallest $k$ that satisfies $k >= a/n$ plus $1$. Is this clearer?

Thorgott

21:13

Yes, but it's not the same as the first sentence

It's nearly the same

MyWrathAcademia

Which part of the first sentence? Is it this part $k<=b/n−k>=a/n)+1$? I'm trying to be better at explaining math

manooooh

@TedShifrin let me use this property: $a<b\implies1/b<1/a$

Ted Shifrin

Assuming $0<a$?

manooooh

So if $\ln(x)=\int_1^x1/t\,dt$ then $\int_1^x1/t\,dt<\int_1^x1/\sqrt{t}\,dt$ so: $$\frac{1}{\int_{1}^{x}\frac{1}{\sqrt{t}}dt}<\frac{1}{\int_{1}^{x}\frac{1}{t}dt}$$

so $$\frac{x}{\int_{1}^{x}\frac{1}{\sqrt{t}}dt}<\frac{x}{\int_{1}^{x}\frac{1}{t}dt}$$

Ted Shifrin

First. Don't use $x$ as a limit of integration and as a dummy variable.

What are you doing here?

If my inequality shows that $(\log x)/x\to 0$, then what does $x/(\log x)$ do?

manooooh

21:19

@TedShifrin done, thank you

Thorgott

@MyWrath sorry, I was trying to be cautious of an error that could not happen. Your last formulation was correct.

Ted Shifrin

No need to be writing all these integrals over and over.

manooooh

@TedShifrin I don't know. I mean numerator and denominator are flipped

But I think in the limit it can be anything; not just $\pm\infty$

Ted Shifrin

So, if $f(x)\to 0^+$ as $x\to\infty$ (or indeed $x\to a$), what does $1/f(x)$ do?

manooooh

@TedShifrin as $x\to\infty$ and $f(x)\to0^+$ then $1/f(x)\to\infty$

MyWrathAcademia

21:23

@Thorgott Thanks a million @Thorgott. No need to apologize. I'm glad to know that the formula $k <= b/n − k >= a/n) + 1$ is correct. Unfortunately it's really late so I have to go but I'll be back tomorrow.

Ted Shifrin

Yes, of course. And you can do as careful an $\epsilon$-$N$ argument as you want.

Thorgott

No no, that formula is not correct

It doesn't make sense

The sentence you wrote is correct

Semiclassical

hi chat

manooooh

hello

anakhro

HEY EVERYONE.

MyWrathAcademia

21:24

Please link to which sentence you mean?

anakhro

BYE EVERYONE.

Ted Shifrin

hi/bye @anakhro

manooooh

@TedShifrin ok. Thank you!

Thorgott

@MyWrathAcademia this is correct

(click on that arrow to get to the message)

MyWrathAcademia

Oh you mean basically this "Yes I understand that the number of natural numbers between a and b that are divisible by n is the number of natural numbers k between a/n and b/n"? How do I represent the sentence you linked mathematically so that it makes sense?

Thorgott

21:30

The number is $\lfloor\frac{b}{n}\rfloor-\lceil\frac{a}{n}\rceil+1$

MyWrathAcademia

That's the notation you showed me earlier. Please bear with me and I know the following is wrong because as you say it doesn't make sense (mathematically) but isn't that the same as this $k <= b/n − k >= a/n ) + 1$ which is to say the value of the left operand - the value of the right operand + 1

Thorgott

You mean to express the same, but since your expression doesn't make sense, it's not the same. $k\ge a/n$ is an inequality, not a number.

MyWrathAcademia

Oh right it evaluates to a boolean (truth value), thanks for pointing that out and the binary operator $-$ only works on numbers

We'll continue this chat later @Thorgott , hopefully tomorrow. Thank you.

Abwatts

21:52

I am not really sure if I solved this question correctly: If $1<x<2$ find a bound for $|\frac{x^3+x^2-1}{x-6}|$. The answer I got is $\frac{11}{7}$. Is that right?

Thorgott

upper bound or lower bound

Abwatts

upper bound

Semiclassical

maybe try a few values between 1 and 2

that doesn't guarantee you're right, but it gives a good cheap way to invalidate it if it's wrong

Abwatts

Thank you! I didn't think about it this way.

Does my answer seem correct though?

Thorgott

maybe plugging $1$ and $2$ themselves in will be even more instructive to start with

Abwatts

21:59

I see. Could we also use the triangle inequality? Because when I plug x = 2, I get $11/4$, but using the triangle inequality I got $11/7$

Semiclassical

the thing to consider is to what extent the absolute value signs actually matter

for instance, where is the numerator positive vs. negative? what about the denominator?

@Abwatts 11/7 < 11/4, so if that's what the triangle inequality gives you then it's not giving the right bound

Thorgott

You could use the triangle inequality, but in this case, it's easily possible to not only get an upper bound, but the smallest possible upper bound through elementary methods. Using the triangle inequality makes your bound worse.

Semiclassical

and the bound you purport to get is actually too strong

Abwatts

That makes sense! So, 11/7 would actually be considered an upper bound, just not the smallest one?\

Semiclassical

no

it's not a valid upper bound in the first place, because as x->2 you get values which are larger than 11/7

so your function is not bounded above by 11/7

Abwatts

22:04

Oh, I see..

Could you perhaps show me how to find the upper bound using the triangle inequality? I just recently got exposed to the triangle inequality and wanted to genuinely understand how to apply it.

Semiclassical

well, how were you trying to use the triangle inequality?

Abwatts

so, to begin with I tried to find the upper bound for $|x^3+x^2-1|$ like so: $|x^3+x^2-1| \leq |x|^3 + |x|^2 - 1 \leq 8 + 4 - 1 = 11$

Semiclassical

that's an immediate red flag, unfortunately

the triangle inequality states that |x+y| <= |x|+|y|

not |x-y|<=|x|-|y|

Abwatts

i'm sorry i meant to write

$|x^3+x^2-1| \leq |x|^3 + |x|^2 + |-1| \leq 8 + 4 + 1 = 13$

Semiclassical

13, but okay

Abwatts

22:10

so, now that we have 13

Semiclassical

yeah, that's legitimate.

Abwatts

can't we find the bound on $|\frac{1}{x-6}|$?

and multiply their bounds?

Ted Shifrin

Absolute value, of course.

Semiclassical

sure. this will give a legitimate bound, though not the best possible

Abwatts

I see, thanks a lot!

Ted Shifrin

22:13

Best possible involves dividing .. yuck.

Semiclassical

that said, I think the triangle inequality is not a good idea here

it's better to consider whether either the numerator or denominator can possibly be positive/negative over the interval 1<x<2

Abwatts

I understand. I just wanted to apply it to practice it and see how it can be used :]

Semiclassical

mmkay

Abwatts

Thanks!

Semiclassical

what do you get as an upper bound on |1/(x-6)|, just to check?

Ted Shifrin

22:19

notes silence

Semiclassical

yup

Ted Shifrin

Of course, I was a mean a**hole earlier, expecting someone doing functional analysis to know what it means to evaluate a linear map on a basis vector. Sigh.

Semiclassical

they didn't? what

i mean, that's the most basic thing that a linear map does

Thorgott

There are millions of great uses the triangle inequality has, but this isn't one of them

Semiclassical

yep

Ted Shifrin

22:27

Well, one can certainly use it, but one gets an ineffective bound. But meh.

Semiclassical

this question frustrates me: math.stackexchange.com/q/3511732/137524

on the one hand, i think the OP is being quite lazy

but I am rather curious about the answer.

Ted Shifrin

How is that $\det(A)$?

Semiclassical

isn't that just the Leibniz formula for the determinant?

missing a factor of 1/6, though

Ted Shifrin

What is the Leibniz formula?

Semiclassical

Leibniz formula for determinants

In algebra, the Leibniz formula, named in honor of Gottfried Leibniz, expresses the determinant of a square matrix in terms of permutations of the matrix elements. If A is an n×n matrix, where ai,j is the entry in the ith row and jth column of A, the formula is det ( A ) = ∑ τ ∈ S n sgn ⁡ ( τ ) ∏ i...

Ted Shifrin

22:39

Oh, the usual definition. I sure don't recognize what the OP has written down as anything like that. So we're saying it's a $2$ tensor on $\Bbb R^3$?

Semiclassical

yeah

Ted Shifrin

And I don't get the product of $\epsilon$s being the right thing. But I have to go get my real ID done, so I'm gone.

Semiclassical

later

i think their formula leads to a 6-fold overcounting because of the two Levi-Civitas

Bob

23:39

good evening

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