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2:02 PM
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 $\sin x + \sqrt 3 \cos x$
 
@adeshmishra Are you sure of the numerator?
 
yeah
 
$\cos x + \sqrt 3\sin x%$ in the numerator would make it easier, I think.
 
Can we write it that way?
 
@MartinSleziak You mean $\cos x - \sqrt{3} \sin x$ right?
 
2:13 PM
Well, Abhigyan Chattopadhyay already asked whether the integral is copied correctly.
 
Yes, I'd have to put minus there to get $\cos(x+\frac\pi3)$.
 
@MartinSleziak How did you get $\cos \left(x+\pi/3 \right)$
?
 
Set the denominator to $t^2$, and the numerator to $At'+Bt+C$
 
2:20 PM
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
 
Right, now set $(1+\sin x + \sqrt{3} \cos x)$ as $t$
And, assume the numerator as $At'+Bt+C$
It'll work out
 
It's not coming, $$ u = 1+ \sin x + \sqrt 3 \cos x \\ du/dx = \cos x - \sqrt 3 \sin x$$
 
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 $x\in(0,\pi/2)$.
 
Do you also make mistakes?
 
So whatever the result is, it should be increasing on that interval.
I should probably go, have a nice day!
 
2:33 PM
Have a nice day Sir
 
3:18 PM
Greetings
 
3:45 PM
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.
 
4:06 PM
Oh hey, can do it myself now. Thanks for the rep
 
4:50 PM
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?
 
5:13 PM
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?
 
5:24 PM
What do you mean with diagonal concatenation?
 
@COTO do you mean something like $\begin{pmatrix} A & 0 \\ 0 & B\end{pmatrix}$ ?
(block matrix)
 
Hi @ÍgjøgnumMeg
 
Hey @Alessandro
 
Exactly, Alessando.
 
woah
$A \oplus B$ is standard I think
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:...
 
5:26 PM
That's exactly the matrix representing $A\oplus B$ on $V\oplus W$ so looks good to me
 
Ah. It is there on Wiki.
 
OK. I'll have to convince my supervisor that I'm not just making crap up. :D
 
"@AlessandroCodenotti Hi allesandro, could you provide some assistance with a notation problem i am having_
 
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?)
 
5:30 PM
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 $a_o+a_1x+a_2x^2....$
 
Thanks all. Cheers.
 
@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
 
No you are right. I wrote it badly.
So what would then $f(X+1)$ mean?
 
$a_n(X+1)^n + \dots + a_1(X+1) + a_0 $
 
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
 
5:38 PM
Ohh!!
 
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
 
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 Thank you! now I can solve the question <3
 
Nooo worries
 
6:07 PM
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?
 
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 $
 
6:25 PM
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
 
Nvm, think I've an argument for it being compact. Just need to show it is not H-S now
 
It's compact because it is a limit of finite rank operators
To show that it is not HS compute its norm
 
@MyWrath what is $n$?
 
The index
He's identifying $\ell^2$ with sequences $(x_n)_{n\in\Bbb N}$
 
6:34 PM
@Thorgott n is an integer number
 
your examples don't make sense without specifying what $n$ is in these examples
 
It's a sequence @Thorgott. $M$ maps the sequence $(x_n)_{n\in\Bbb N}$ to the sequence $(n^{-1/5}x_n)_{n\in\Bbb N}$
 
I don't think you two are talking about the same problem
 
Yeah, I'm talking about Wraths question
 
Oh lol my bad
I thought you were also answering to @TheTerriblePuddle
Sorry
 
6:37 PM
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
 
I think there are two numbers divisible by 7 in the range from 7 to 14, namely 7 and 14
 
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
 
Thanks again @Thorgott
 
As for your question, the answer should be $\max\left(0,\lfloor\frac{b}{n}\rfloor-\lceil\frac{a}{n}\rceil+1\right)$
 
6:44 PM
Thanks @Thorgott but I'm trying to install the MathJax renderer you linked because your code is difficult to parse
 
@AlessandroCodenotti You mean compute its Hilbert-Schmidt norm right? And if it is not finite it is not H-S
 
@Thorgott I can't get MathJax to install. After you add it as a bookmark how do you get it to run?
 
Maybe I should use that $\|M\|\leq \|M\|_{HS}$?
 
You click on the bookmark
 
6:51 PM
just to comfirm it's only start ChatJax that I need to install right, not render MathJax?
 
Yes
 
@Thorgott I had an add on that was blocking JavaScript, had to unblock it. Its working great and the equations look nice now
 
@TheTerriblePuddle Just write down the series
 
Nice
 
Hi, demonic @Alessandro, @Thorgott, Puddle, Wrath.
 
6:56 PM
Hi @Ted
 
Hi @TedShifrin
 
Hi Ted
 
@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
 
@AlessandroCodenotti $\|M\|^2_{HS}=\sum_i |\langle Me_i,e_i\rangle|^2$ ?
 
7:01 PM
@MyWrath when is a number divisible by $n$?
 
@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
 
@TheTerriblePuddle $\|M\|_{HS}=\sum_i\|Me_i\|^2$
 
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?
 
@AlessandroCodenotti should I fix an $x\in \ell^2$ with $\|x\|=1$ then?
 
You fix an orthonormal basis $e_i$ and take the sum over those
 
7:12 PM
@Alessandro: Didn't you forget a square root?
 
Oh yeah you're right
Well we're just trying to decide whether the norm is finite so it's not too important
 
Well, you're not usually sloppy :P
 
Can't be (too) sloppy in logic/set theory!
 
Are you implying that we can be sloppy in geometry?
Hrumph.
 
$\|Me_i\|^2=\sum^\infty_{n=1}|n^{-1/5}x_ne_i|^2$ right?
 
7:18 PM
@Thorgott I think the smalllest such number is $a + n$
 
What are $x_n$, @TheTerriblePuddle?
 
@Thorgott So the smallest number should satisfy $ a + n <= b $
 
@TedShifrin My sequence in $\ell^2$
 
But Alessandro told you to compute $Me_i$. What does $e_i$ stand for?
Sorry, I shouldn't be butting in. shuts up
 
7:21 PM
Where does $x$ appear when you do $Me_i$?
 
@MyWrath no, consider the first example again
 
@TedShifrin The Italian school of geometry was pretty famous for that!
 
@TedShifrin In how I defined the operator $M:\ell^2 \to \ell^2$
 
Not being Bourbaki-rigorous doesn't necessarily mean sloppy, although some notions were indeed sloppy.
 
by $M(x)=(n^{-1/5}x_n)_{n\geq 1}$
 
7:23 PM
Ok, so what is $M(e_i)$? What is $e_i\in\ell^2$?
 
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.
 
I see, I don't know any of the details to be fair
 
It's called Enumerative Geometry.
Some of my research dealt with such things.
 
Oh I see
There was a course here in enumerative geometry this semester
 
Beautiful stuff. Lots of chern class computations with Grassmannians.
 
7:27 PM
Okay, I don't know anymore
 
@Thorgott Is the smallest such number $nk >= a$?
And the largest $nk <= b$?
 
Yes, now for which $k$ do these happen?
 
Puddle: Answer Alessandro's question. What is $e_i\in\ell^2$?
 
Some sequence
 
You have to do better than that.
 
7:35 PM
I suppose $\|e_i\|=1$
 
Well which sequence?
@TheTerriblePuddle It better be the case since it's supposed to be an element of an orthonormal base
 
@Thorgott $k$ is smallest at $smallest number / n$ and largest at $largest number / n$? not sure how to represent this mathematically
 
that $k$ is not necessarily a natural number!
 
(0,0,0,0,...,1,0,0,0,...) for the i'th place
 
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?
 
7:46 PM
Yippee, @Puddle, so what is $\|Me_i\|$?
 
no, I mean that what you said is not the proper $k$, since it's not necessarily a natural number
 
Wait so are you saying that what I wrote means that k may not be a natural number?
 
yes
for example, 1/2 is not a natural number
 
I see. I didn't notice that $smallestnumber/n$ could give $1/2$
 
it can give any positive rational, we're not placing any restrictions on $a,b,n$ except being natural after all
 
7:53 PM
$\|Me_i\|=|n^{-1/5}x_i|$
 
shrieks
 
So then $k$ has to be natural number, correct?
 
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 $i$.
 
Yes, that's the definition of one number dividing another
 
7:55 PM
But what about how M was defined?
$|i^{-1/5}x_i|$?
 
A half hour ago I said you could not have $x_i$ in your answer.
 
Then what do I do with M?
 
You have to understand what symbols actually mean.
 
When you have a formula for $f(x)$ and you put in a value for $x$, say $x=2$, what do you do?
 
8:00 PM
put it in
 
So?
What is $Me_1$?
 
Hey, I'm trying to understand the solution of this exercise:
 
1^{-1/5}x_1
 
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.
 
 
8:03 PM
So I don't
 
Why is the integrand is entire?
 
Removable singularity at $0$, @Eran.
 
The singularity at $0$ is removable
got sniped
 
This is not entire though
You have to redefine it in order to be entire
 
Yes, that's completely understood when there's a removable singularity.
 
8:07 PM
@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 $nk <= b$
 
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 $n$?
 
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.
 
8:42 PM
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 $\lim_{t\to+\infty}(-t/\ln(t))$
 
@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?
 
8:58 PM
@manooooh: Do you know anything about exponential/logarithmic rate of growth?
Forget about L'Hôpital's rule.
 
@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...?
 
@TedShifrin not much Ted. What kind of rate of growth will you propose?
 
Although you can use it to answer your final question, there. I just have different ways of doing it.
 
@TedShifrin alternative solution?
 
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?
 
9:00 PM
@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
 
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?
 
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.
 
@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
 
@Thorgott: Yeah, Taylor for the exponential is sort of the flip-side of my approach.
 
@TedShifrin that's a very good approach I think. But we have $x/\ln(x)$ not $\ln(x)/x$
 
9:07 PM
@manooooh: Duh. Stop and think.
 
Ok. Stop. Hardest part: Think
 
Today has been high frustration for me in this chatroom.
 
We are on holidays probably?
At least in Latinamerica
 
I don't think too many are on holidays.
 
and a part of Europe
 
9:09 PM
It's the first week of uni for me
 
The teachers (I assist to a teacher, so technically I am not a teacher) are on holidays. We return to class on Feb 13
 
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?
 
That statement is nonsense to me
 
I guess this is your summer, so I'm surprised the schedule isn't flipped compared to the northern hemisphere.
 
@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?
 
9:13 PM
Yes, but it's not the same as the first sentence
It's nearly the same
 
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
 
@TedShifrin let me use this property: $a<b\implies1/b<1/a$
 
Assuming $0<a$?
 
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 $$\frac{x}{\int_{1}^{x}\frac{1}{\sqrt{t}}dt}<\frac{x}{\int_{1}^{x}\frac{1}{t}dt}$$
 
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?
 
9:19 PM
@TedShifrin done, thank you
 
@MyWrath sorry, I was trying to be cautious of an error that could not happen. Your last formulation was correct.
 
No need to be writing all these integrals over and over.
 
@TedShifrin I don't know. I mean numerator and denominator are flipped
But I think in the limit it can be anything; not just $\pm\infty$
 
So, if $f(x)\to 0^+$ as $x\to\infty$ (or indeed $x\to a$), what does $1/f(x)$ do?
 
@TedShifrin as $x\to\infty$ and $f(x)\to0^+$ then $1/f(x)\to\infty$
 
9:23 PM
@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.
 
Yes, of course. And you can do as careful an $\epsilon$-$N$ argument as you want.
 
No no, that formula is not correct
It doesn't make sense
The sentence you wrote is correct
 
hi chat
 
hello
 
HEY EVERYONE.
 
9:24 PM
Please link to which sentence you mean?
 
BYE EVERYONE.
 
hi/bye @anakhro
 
@TedShifrin ok. Thank you!
 
@MyWrathAcademia this is correct
(click on that arrow to get to the message)
 
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?
 
9:30 PM
The number is $\lfloor\frac{b}{n}\rfloor-\lceil\frac{a}{n}\rceil+1$
 
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
 
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.
 
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.
 
9:52 PM
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?
 
upper bound or lower bound
 
upper bound
 
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
 
Thank you! I didn't think about it this way.
Does my answer seem correct though?
 
maybe plugging $1$ and $2$ themselves in will be even more instructive to start with
 
9:59 PM
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 $11/7$
 
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
 
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.
 
and the bound you purport to get is actually too strong
 
That makes sense! So, 11/7 would actually be considered an upper bound, just not the smallest one?\
 
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
 
10:04 PM
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.
 
well, how were you trying to use the triangle inequality?
 
so, to begin with I tried to find the upper bound for $|x^3+x^2-1|$ like so: $|x^3+x^2-1| \leq |x|^3 + |x|^2 - 1 \leq 8 + 4 - 1 = 11$
 
that's an immediate red flag, unfortunately
the triangle inequality states that |x+y| <= |x|+|y|
not |x-y|<=|x|-|y|
 
i'm sorry i meant to write
$|x^3+x^2-1| \leq |x|^3 + |x|^2 + |-1| \leq 8 + 4 + 1 = 13$
 
13, but okay
 
10:10 PM
so, now that we have 13
 
yeah, that's legitimate.
 
can't we find the bound on $|\frac{1}{x-6}|$?
and multiply their bounds?
 
Absolute value, of course.
 
sure. this will give a legitimate bound, though not the best possible
 
I see, thanks a lot!
 
10:13 PM
Best possible involves dividing .. yuck.
 
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
 
I understand. I just wanted to apply it to practice it and see how it can be used :]
 
mmkay
 
Thanks!
 
what do you get as an upper bound on |1/(x-6)|, just to check?
 
10:19 PM
notes silence
 
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.
 
they didn't? what
i mean, that's the most basic thing that a linear map does
 
There are millions of great uses the triangle inequality has, but this isn't one of them
 
10:27 PM
Well, one can certainly use it, but one gets an ineffective bound. But meh.
 
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.
 
How is that $\det(A)$?
 
isn't that just the Leibniz formula for the determinant?
missing a factor of 1/6, though
 
What is the Leibniz formula?
 
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...
 
10:39 PM
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 $\Bbb R^3$?
 
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.
 
later
i think their formula leads to a 6-fold overcounting because of the two Levi-Civitas
 
Bob
11:39 PM
good evening
 
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