Mathematics - 2024-06-01 (page 1 of 2)

Lucas Henrique

00:38

hey, so I solved my problem, lol. the poset approach does not work, but it does indeed have something to do with the possible denominators. the set of denominators is precisely $\{r\in R: rx \in R\} = Ann_R(Rx/R). if $x$ is not an element of $R$, $1$ does not belong to this ideal and thus its proper, contained in some $\mathfrak m$. but $x \in R_{\fk m}$, so $s’ x = r’$ for some $s’ \not \in \mathfrak m$; thus $s’ \in \text{set of denominators} \subseteq \mathfrak m$, a contradiction

Jakobian

01:59

@peek-a-boo thanks for confirmation

Jakobian

02:10

I think I might not last those extra 2 hours of sleep

So I guess I'll wake up early

Jakobian

03:56

@LucasHenrique I'm not surprised that badly formatted LaTeX leads to a contradiction, lol (joking btw)

@XanderHenderson EE18 is a stubborn student and needs a stubborn teacher, I suppose

@psie when I click on this I have a message "that request didn't look right" and a cat picture

which sucks a lot

this is why you should type things out rather than post pictures, the letters are too small

Jakobian

04:15

@psie I have no idea what the lemma is saying, as well. If such function $h$ exists, then $Z$ is an absolutely continuous random variable, as you said.

I think this should be a definition rather than a lemma, perhaps?

Soumik Mukherjee

@Jakobian I formulated it so badly that my writing doesn't make sense. I was thinking that if we rotate the graph of Weierstrass function anti clock wise in some pi/4 angle then would that satisfy. But this is not well defined as you said.

Jakobian

@SoumikMukherjee yeah no worries, I didn't solve the question myself either. One thing to note is that the Banach indicatrix $N(t) = |f^{-1}(t)|$ is a measurable function under assumptions (1) and (4) of that question, perhaps one can use this

i.e. $[0, 1]$ decomposes into measurable sets $B_n = \{t\in [0, 1] : |f^{-1}(t)| = n\}$

in fact Borel measurable

Soumik Mukherjee

Okay if for $t_1,t_2 \in [0,1]$, $|f^{-1}(t_1)|=m$ and $|f^{-1}(t_2)|=n$ then is it necessary that $m=n$?

Jakobian

04:30

well no, for example something like $f(t) = |t-1/2|$ has $|f^{-1}(0)| = 1$, yet $|f^{-1}(t)| = 2$ for $t\in (0, 1/2]$

Soumik Mukherjee

I am saying if that could be derived from the assumptions of this problem, ofc this is not true in general.

Jakobian

well, one would have to use that $f$ is nowhere differentiable i.e. assumption (2) somehow

Lucas Henrique

06:34

@Jakobian oh, my bad. I was typing in my phone, now I'm literally laughing out loud hahahaha

Binky McSquigglebottom

08:59

$\lim_{(x,y)\to (0,3)} \frac{e^\frac{x-1}{y^2}+[cos(log_3(y+9)*\sqrt{x})]^{arctan(x)}-2cos(y)+y^{log(1+x^3)}}{\sqrt{x^2+6y^3}-sin(y)}$

I've been doing this exercise with Taylor developments since this morning, but I can't find the right result

Sine of the Time

isn't the function continuous at $(0,3)$?

Binky McSquigglebottom

ok

🤦‍♂️thank you very much, I've been doing math since this morning, I hadn't really thought about replacing numbers, I thought you should use Taylor. Thanks again

Sine of the Time

Taylor expansion in two variable is not as powerful as Taylor expansion in one variable to compute limits, you have to be careful

Binky McSquigglebottom

$\frac{\int_0^\infty (1+t^2)^{-2024}dt}{\int_0^\infty (1+t^2)^{-2023}dt}$

what would I do:

Binky McSquigglebottom

09:26

$\frac{\int_0^\infty (1+t^2)^{-2024}dt}{\int_0^\infty (1+t^2)^{-2023}dt}=\frac{\int_0^\infty (1+t^2)^{-\epsilon}dt}{\int_0^\infty (1+t^2)^{-2023}dt}=\frac{\int_0^\frac{\pi}{2}((sec(x)^2))^{1-\epsilon}dx}{\int_0^\infty (1+t^2)^{-2023}dt}=\frac{\int_0^\frac{\pi}{2}((cos(x)^2))^{\epsilon-1}dx}{\int_0^\infty (1+t^2)^{-2023}dt}=\frac{\frac{1}{4}\int_0^{2\pi}((cos(x)^2))^{\epsilon-1}dx}{\int_0^\infty (1+t^2)^{-2023}dt}$

$\frac{\frac{1}{4}\int_0^{2\pi}((cos(x)^2))^{2\epsilon-2=2k:(k=\epsilon-1)}dx}{\int_0^\infty (1+t^2)^{-2023}dt}=\frac{\frac{1}{4}\int_0^{2\pi}((cos(x)^2))^{2\epsilon-2=2k:(k=\epsilon-1)}dx}{\int_0^\infty (1+t^2)^{-2023}dt}=\frac{\pi}{2}\cdot \frac{(2\epsilon-2)!}{2^{2\epsilon-2}\cdot(\epsilon-2)!\cdot(\epsilon-1)!}$

Pizza

hi

Jakobian

why not just do $I_n = \int_0^\infty \frac{1}{(1+t^2)^n} dt$ and $I_n = 2n \int_0^\infty \frac{t^2}{(1+t^2)^{n+1}}dt = 2n\cdot (I_n-I_{n+1})$ so $I_{n+1} = \frac{2n-1}{2n}I_n$

Binky McSquigglebottom

$\frac{\pi}{2}\cdot\frac{4046!}{2^{(4046)!}\cdot(2023)!\cdot(2023)!}\cdot\frac{2}{\pi}\frac{2^{4044}\cdot(2022)!\cdot(2022)!}{4044!}$

Jakobian

So $\frac{I_{2024}}{I_{2023}} = \frac{2\cdot 2023-1}{2\cdot 2023}$

work smarter, not harder

Binky McSquigglebottom

$\frac{4045}{4046}$

@Jakobian 😓

you're right

Binky McSquigglebottom

13:41

Determine the volume of the solid of rotation
T triangle of vertices (0,0,2),(0,-1,1) and (0,1,1) around the y.So , $T=\{(y,z):z-2\leq y\leq2-z,1\leq z \leq2 \}$
$V=2\pi\int_T z \ \ dydz= 2\pi\int_1^2(\int{z-2}^{2-z}z\ dy)\ dz=4\pi\int_1^2(2z-z)^2 dz=4\pi[z^2-\frac{z^3}{3}]_1^2=\frac{8\pi}{3}$

can someone explain these steps to me pls, this is the solution

Binky McSquigglebottom

13:53

ok, I solved it

Ryder Rude

hi

Pizza

14:19

hi

Peter

I would like to ask this directly on math stack exchange , but surely it would be closed soon. Therefore I try it here : Suppose , someone tells use that a positive integer $n$ is the product of two distinct primes $p<q$ , but we do not know those prime factors. Is there a way to find out whether $p\mid q-1$ holds without finding the prime factors ?

Pizza

15:17

I was trying to determine the domain of this function, and then graph it:
$f(x,y) = \log(xy^2+x^2y)$

$D=xy^2+x^2y >0 \leftrightarrow xy(x+y)>0$

$xy>0$ when $x>0,y>0$ and $x<0,y<0$.
$x+y>0$ then $x>-y$ or $y>-x$

Sine of the Time

@Pizza draw the regions in $\Bbb R^2$

Pizza

@SineoftheTime $D=\{(x,y):(x > 0, y > 0) ∨ (x < 0, 0 < y < −x) ∨ (x > 0, y < −x)\}$

@SineoftheTime geogebra.org/graphing/abfbgweh

Sine of the Time

exactly, can you see the domain?

Pizza

@SineoftheTime yes is the blue part, excluding the dotted part

right?

Sine of the Time

so it's the first quadran

now you have to study the other case

Xander Henderson

15:31

You are asking if your domain is correct, but you have explained none of your thinking. How did you arrive at your answer? Typically, if you can explain to another person how you arrived at an answer, you can be fairly certain of your own correctness.

Instead of giving other people the answer, and asking if it is right, pretend that you are the expert on the problem, and explain your answer to other people.

For example, with respect to this problem: my first thought is the the argument of $\log$ must be positive. Hence $xy^2+x^2y = xy(x+y)$ must be positive (you wrote down this last inequality, but didn't explain it---perhaps it is too trivial?).

From there, I actually see (naively) four cases that I might want to think about: since there are three factors, and the result is positive, either none of the factors are negative, or two of the factors are negative. Hence (1) $x,y,x+y > 0$, (2) $x,y < 0$, (3) $x,x+y<0$, or (4) $y,x+y < 0$.

Case (1) is satisfied whenever $x,y>0$, since this implies that $x+y$ will also be positive. So the entire first quadrant is in the domain.

Case (2) is impossible, since $x,y< 0$ implies that $x+y<0$. So no new information is obtained from that case.

Case (3) gives $y<-x$ with $y>0$ and $x<0$. This is everything in quadrant II below the line $y=-x$ (the inequality $y<-x$ can be read as "$y$ is below $-x$).

Case (4) gives $y<-x$ with $y<0$ and $x>0$. So this is everything in quadrant IV with $y$ below the line $y=-x$.

Binky McSquigglebottom

@Pizza you have to use the sign rule

Xander Henderson

@BinkyMcSquigglebottom What are "the sign rule"? You assert with great confidence that one has to use these rules---what are they?

Binky McSquigglebottom

Descartes' rule of signs

In mathematics, Descartes' rule of signs, first described by René Descartes in his work La Géométrie, is a technique for getting information on the number of positive real roots of a polynomial. It asserts that the number of positive roots is at most the number of sign changes in the sequence of polynomial's coefficients (omitting the zero coefficients), and that the difference between these two numbers is always even. This implies, in particular, that if the number of sign changes is zero or one, then there are exactly zero or one positive roots, respectively. By a linear fractional transformation...

Xander Henderson

@BinkyMcSquigglebottom That is not relevant to this question at all.

That is about finding the number of possible zeros to a polynomial equation in one variable.

Pizza is asking about the domain of function, which depends on a polynomial in two variables being positive.

@Pizza geogebra.org/calculator/v5vd5qmw

Binky McSquigglebottom

15:49

The rule of signs is used to determine the conditions under which the product
$xy(y+x)$ is positive.

Xander Henderson

@BinkyMcSquigglebottom You are going to have to explain what you are thinking, because (a) I see no reason to use Descartes' rule of signs, and (b) I don't see how it applies to this problem.

Binky McSquigglebottom

@XanderHenderson Descartes' rule of signs helps us understand the sign variations of the expression $xy(y+x)$ to determine the regions where this is positive.

Xander Henderson

@BinkyMcSquigglebottom I'll say the same thing to you that I said to Pizza. You need to explain your thinking.

Pizza

@XanderHenderson thanks!

Xander Henderson

I have no idea what you are suggesting. Perhaps you should outline your argument.

Pizza

15:53

next time I will be more clear

Binky McSquigglebottom

@XanderHenderson ?

Xander Henderson

In any event, I think that any argument which appeals to Descartes' rule of signs is likely overkill, as I gave an argument which only uses the idea that if the parity of negative factors in a product is even, then the product is positive; and if the parity of the negative factors in a product is odd, then the product is negative.

Binky McSquigglebottom

I wrote because I used it

Xander Henderson

@BinkyMcSquigglebottom Show your work.

Binky McSquigglebottom

ok

Xander Henderson

15:54

Descartes' rule of signs is irrelevant to the problem that Pizza was asking about.

Binky McSquigglebottom

first of all the sign of the terms is analyzed

so

Xander Henderson

You don't need it, and I think that invoking it is likely to cause confusion, particularly when you haven't explained how you are using it.

Binky McSquigglebottom

$xy > 0$ if x and y have the same sign (both positive or both negative).
$xy < 0$ if x and y have opposite signs.

Xander Henderson

@BinkyMcSquigglebottom Okay.. this has nothing to do with Descartes, so far.

Binky McSquigglebottom

it was for the spaces

wait

now I have to do the different cases

$x>0$

case

If $y > 0$:
$xy > 0$ positive
$y + x > 0$ positive
So, $xy(y + x) > 0$

if $y < -x$ :
$xy < 0$ negative
$y + x < 0$ negative
The product of two negative terms is positive, so $xy(y + x) > 0$

Xander Henderson

16:02

So far, it looks like you are doing exactly what I did, only in a more cumbersome manner. Where have you used Descartes?

8 mins ago, by Xander Henderson

In any event, I think that any argument which appeals to Descartes' rule of signs is likely overkill, as I gave an argument which only uses the idea that if the parity of negative factors in a product is even, then the product is positive; and if the parity of the negative factors in a product is odd, then the product is negative.

Binky McSquigglebottom

Descartes' rule of signs helps us understand the sign variations of the expression xy(y+x)
to determine the regions where this is positive. here

Xander Henderson

@BinkyMcSquigglebottom No, it doesn't. And you haven't used it at all!

Sine of the Time

do you mean $ab>0$ iff $a>0,b>0$ or $a<0,b<0$?

Xander Henderson

Descartes' rule of signs is about obtaining information about the number of positive (or negative) roots of a one-variable polynomial based on the number of variations of sign when the polynomial is expressed in standard form.

Binky McSquigglebottom

Descartes' sign rule allowed us to identify regions where $xy(y + x) > 0$:

$x > 0$ and $y > 0$
$x > 0$ and $y < -x$
$x < 0$ and $0 < y < -x$

Soumik Mukherjee

16:07

@BinkyMcSquigglebottom How?

Sine of the Time

?

Xander Henderson

@BinkyMcSquigglebottom That has nothing to do with Descartes' rule of signs. Did you even read the Wikipedia article to which you linked?

Sine of the Time

no

EE18

In 12b they could easily have written min right?

instead of inf?

Xander Henderson

@EE18 Do you know that a minimum exists?

Binky McSquigglebottom

16:11

@XanderHenderson I sent it wrong

Xander Henderson

@EE18 It appears to me that the infimum is taken over all $y\in F$, where $F$ is the span of $B$. I assume that $F$ is infinite (or could be infinite), hence there is no obvious a priori reason to assume that there is a $y \in F$ satisfying $y = \inf \|x-y\|$.

Obliv

are there any examples of sets that don't have the same order type that isn't a result of one set having negatives and the other not

EE18

No, but what I mean is once I show it obeys the property of being less than everything else in the set then since it's in the subspace in particular then we have the minimum property

Xander Henderson

@EE18 Okay, but that is putting the cart before the horse, I think.

EE18

Sort of I guess, but no more so than making this claim about an infimum. The relevant matter here is that $p_F(x) \in F$ right? THat's more or less all I'm asking

Xander Henderson

16:15

You know that the infimum meaningfully exists. You don't know, a priori, that a minimum exists.

EE18

@XanderHenderson This is what I don't agree with. I know no more about an infimum than the minimum a priori

Xander Henderson

Once you have shown that there is some element which attains the the infimum, then you know that the infimum is, in fact, a minimum, but why put that in the assumptions of the thing you are trying to prove.

EE18

OH

Xander Henderson

@EE18 You do. Infima always exist.

EE18

You are using properties of $\Bbb R$

Xander Henderson

16:16

Minima don't.

EE18

Ah OK, got it

Much appreciated, thank you

Binky McSquigglebottom

Determine all the 2x2 matrices such that it results:

$A^3-3A^2=\begin{pmatrix} -2 & -2 \\ -2 & -2\end{pmatrix}$

Pizza

you are missing a \

Binky McSquigglebottom

thanks

$det(A^2)\cdot det(A-3I)=det(\begin{pmatrix}-2 & -2 \\ -2 & -2\end{pmatrix}$)

right?

Pizza

I'll try to check

looking at the solutions wolfram gave, the matrices seem singular so i think you need to assume that det(A) = 0

Xander Henderson

16:28

@BinkyMcSquigglebottom As I said to Pizza, explain your thinking. What theorem or result are you invoking, here? It looks very wrong to me.

Binky McSquigglebottom

$det(A^2)\cdot det(A-3I)=0$

because it's a square matrix with terms that are all the same so it's zero

Pizza

I can't help you much, I'm sorry :(

Binky McSquigglebottom

@Pizza np, i dont trust computer

Xander Henderson

Oh, I see what you've done. You skipped a step. In general, $\operatorname{det}(AB) = \operatorname{det}(A)\operatorname{det}(B)$. Yes.

Soumik Mukherjee

@Obliv {1} and {1,2}

Obliv

16:30

{1} doesn't even have an order

Xander Henderson

But that doesn't really help you very much in solving the original problem.

Binky McSquigglebottom

@Pizza many times they make mistakes

@XanderHenderson let continue

I start by making the case that $det(A)=0$

Can I set A as a square matrix of unknown terms?

Soumik Mukherjee

@Obliv If you want strict order then {1,2} and {1,2,3}

Binky McSquigglebottom

Don't I have to find the trace of A too?

Obliv

so for finite sets, they'd have to be different size

Binky McSquigglebottom

16:32

okay I'll go straight

Obliv

since there wouldn't be a bijection b/t them

Binky McSquigglebottom

$Tr(A)=a+d$

$\begin{pmatrix}a-\lambda& b \\ c & d-\lambda\end{pmatrix}$

This Is the det $(A-\lambda I)$

Obliv

Total relation

In mathematics, a binary relation R ⊆ X×Y between two sets X and Y is total (or left total) if the source set X equals the domain {x : there is a y with xRy }. Conversely, R is called right total if Y equals the range {y : there is an x with xRy }. When f: X → Y is a function, the domain of f is all of X, hence f is a total relation. On the other hand, if f is a partial function, then the domain may be a proper subset of X, in which case f is not a total relation. "A binary relation is said to be total with respect to a universe of discourse just in case everything in that universe of discourse...

Binky McSquigglebottom

$det(A)=ad-cb$ I call it P$(\lambda)$

Obliv

does this apply to all relations, not just binary ones

Binky McSquigglebottom

16:34

So the characteristic polynomial will be

$P(\lambda)=\lambda^2-tr(A)\lambda-det(A)$

I can use this Hamilton-Cayley theorem

Soumik Mukherjee

@Obliv You can construct a counter example using both sets to be countably infinite.

Binky McSquigglebottom

I can't set $P(\lambda)=0$ it's like setting $P(A)=0$

Obliv

so two sets have the same order type if there is an order preserving map b/t them like an isomorphism that respects the operation but in this case it's a transitive, total relation that isn't reflexive nor symmetric

Binky McSquigglebottom

So now I can't put A in place of $\lambda$

no wait

i can

$A^2=tr(A)\cdot A$

$A^3=A\cdot A^2 = A^2 \cdot tr(A)$

so

$(tr(A))^3 -3(tr(A))^2 +4=0$

Obliv

actually nvm i don't know if thats true

Binky McSquigglebottom

16:37

$tr(A)=-1$ or $tr(A)=2$

Obliv

i think u can preserve order since it doesn't actually matter which direction < or >

Binky McSquigglebottom

A = -1/2 (1 1, 1 1) or (1 1, 1 1)

the other is with all -1/2

i use binet theorem for the other case

$(A+I)\cdot(A-2I)^2=$\begin{pmatrix}2& -2\\ -2 & 2\end{pmatrix}

Calculate the determinant of both members

I find a $\lambda=3$

Now i use Hamilton theorem again

Hamilton-Cayley theorem

I find that $\lambda=2$

the eigenvalues

so

$A^2-5A+6I=0$

so

$19A-30I$

Obliv

bro can you stop spamming your work

Binky McSquigglebottom

how can i show you all the steps then

Obliv

you can condense it, or type it up somewhere and link to it like overleaf or something

you're flooding chat with small messages unnecessarily

2

Binky McSquigglebottom

16:44

I write the last 2 solutions

I find a matrix A=(5/2 -1/2, -1/2 5/2)

and a matrix (1 -2, -2 1)

is correct?

Obliv

u dont have access to solutions to check?

which textbook

Binky McSquigglebottom

no

Soumik Mukherjee

@Obliv if you have any counterexample in mind that uses the negetive integers, you can make a new set out of positive integers while keeping the original order. So it doesn't matter what kind of elements the set consists of.

Obliv

well I was thinking of the example they gave which is $\mathbb{R}$ has same order as $(-1,1)$ because there exists $f: (-1,1) \to \mathbb{R}$ given by $f(x) = \frac{x}{1-x^2}$

and I just assumed the interval that has the same order type should have negative and positive integers and be continuous

Debanjan Biswas

17:03

3

Q: Find time to collision in a pursuit

Problem 1.13 from Irodov's Problems in General Physics: Point A moves uniformly with velocity $v$ so that the vector $v$ is continually "aimed" at point B which in its turn moves rectilinearly and uniformly with velocity $u<v$. Initially, $v⊥u$, and the points are separated by a distance $l$. Th...

Can you kindly tell me why is here $\frac {dr}{dt} = ucos {\theta}-v $ instead of $\frac{dr}{dt} = vsin {\theta}$

?

Sine of the Time

@DebanjanBiswas apply the chain rule to $\frac {d}{dt}\sqrt{x^2+y^2}$

Obliv

I don't know if munkres is reusing $a,b$ in the 2nd paragraph to mean the same $a,b$ in the first

if not, then $A_0 \subset A$ can be bounded above by $b$ without requiring $b \in A_0$

but that doesn't seem to make sense so I guess it's the same a,b

it's just weird to say "if there is an element b of A" again, since it was already established that $b \in A_0$ was the largest element of $A_0$

OHH nvm I get it. He kinda resets what $a,b$ are in the 2nd paragraph and says the least element in the intersection of upper bounds of $A_0$ is $b$

basically defining $b$ in two different ways

Binky McSquigglebottom

anyway it was right in the end

Jakobian

@Obliv bro can you stop spamming your work

user 85795

17:19

bro 😎

Obliv

dang, you got me lol. I should practice what I preach

user 85795

It's extremely easy and convenient to set up a separate room to discuss any particular messy question.

Obliv

How? Don't you have to invite someone to make a room though

user 85795

Yup.

So, first you have to find someone interested :-)

Obliv

You know how people play chess by themselves? Maybe i should make a sock puppet on here so I can answer my own questions :D

invite them to my sock puppet channel

Ofc, now that I've said that, xander will smite me if I do.

user 85795

17:26

💯

Sine of the Time

yesterday I beated a 2450+ rated player on lichess :)

Obliv

Nice, I only play against lvl 8 stockfish when I'm on the toilet or bored

Sine of the Time

:D

Obliv

I don't get very far :P my best is like move 20ish with about 0.0

Jakobian

Binky did nothing wrong, everyone "spams" and I refuse to make an exception in this case

Obliv

17:33

Okay, but we could all just not spam and make the chat room more readable? i.e., write up your entire problem statement & attempt somewhere else or in a couple of messages here, utilizing all of the horizontal space allotted in a message..

Jakobian

the first thing you propose is annoying for couple of reasons

Binky McSquigglebottom

@Jakobian thanks

Jakobian

on a screenshot, the letters are too small and so "writing somewhere else" is not a good way to make room more readable

Binky McSquigglebottom

I understand that you may have interpreted my actions as spam, but I would like to clarify that it was not my intention to annoy or disturb. I shared that information because I thought it was relevant and useful to the ongoing discussion. If there is something specific that has bothered you, I am here to listen and discuss it openly. My intention was never to violate the rules or annoy other users. I hope we can clarify this situation and continue to interact constructively.

Jakobian

and about flooding, it wasn't getting out of hand, everyone is flooding the chat with their problems, its not as much spam as is what any of you write in here

while you could optimize writing of those computations better and write them in less messages, it wasn't annoying as much as anything in the chat is annoying

Binky McSquigglebottom

17:41

@Jakobian I understand your point of view. I will try to be more efficient in my messages in the future.

I apologize if my posts were not as optimized as they could have been.

Obliv

@jakobian have u made any progress on what u want to do with ur result

Soumik Mukherjee

@SineoftheTime nice, in blitz?

Sine of the Time

bullet

Soumik Mukherjee

Ooh

Jakobian

I don't understand your question?

Obliv

17:52

like whether you wanna publish it or whatever

Jakobian

I did and do

I don't see how this is a thing requiring any sort of progress

Binky McSquigglebottom

$\frac{dP}{P\Delta H} = \frac{A}{\rho g T}dT - \frac{nR}{P^2}dT$ what is that?

Obliv

looks like something in thermodynamics

maybe related to maxwell relations

Binky McSquigglebottom

ah ok

thank you

Obliv

where did you find it..?

Binky McSquigglebottom

17:58

on internet

Obliv

Nice, that's specific :P

Binky McSquigglebottom

also that en.wikipedia.org/wiki/Friedmann_equations

(y’)^2=p+q(y^2) , with p and q constants

i need to solve that

Obliv

you can ask in here as well since that's the physics room

Binky McSquigglebottom

I went to write

@Obliv I really appreciate your help in providing me with useful resources to study the topic.

Obliv

could this perhaps be related to your previous question about solving $A^3 - 3A^2 = \begin{pmatrix}-2&-2\\-2&-2\end{pmatrix}$?

Binky McSquigglebottom

18:03

in theory in my exercise I should differentiate and after finishing

@Obliv I don't see the connection between the two things

Jakobian

@Peter I don't think a lot of people would know, maybe someone from computational side of things. Perhaps its useful to rephrase this as $p|\varphi(n)$? Though I haven't been able to represent $p$ in any way. Or maybe one of the Fourier transform methods works?

user 85795

\o @robjohn

Sine of the Time

@BinkyMcSquigglebottom can't you solve for $y'$ and then solve a separable diff. equation or this does not work?

Binky McSquigglebottom

@SineoftheTime It might work

Sine of the Time

did you try it?

Binky McSquigglebottom

18:12

now i try

user 85795

@BinkyMcSquigglebottom why do you need to solve stuff you find on Wikipedia?

2

Sine of the Time

you should get some expression with $\sinh$ or its inverse

Binky McSquigglebottom

@user85795 wikipedia?

its not on wikipedia

@SineoftheTime $\int \frac{dy}{\sqrt{p + qy^2}} = \pm \int dx$ im here

user 85795

Where do you find your questions?

Binky McSquigglebottom

@user85795 The questions I interact with are collected from a wide range of sources, including online conversations, forums, articles and more.

user 85795

18:17

So you don't use any textbooks...

Binky McSquigglebottom

currently not

Sine of the Time

@BinkyMcSquigglebottom do you know how to solve $\int \frac1{\sqrt{a^2+x^2}}dx$?

Binky McSquigglebottom

@SineoftheTime with trigonometric substitutions?

Sine of the Time

you can use different substitutions

Binky McSquigglebottom

ok

Sine of the Time

18:19

trig or hyperbolic

Binky McSquigglebottom

$\int \sec(\theta) \, d\theta$

$\ln|\sec(\theta) + \tan(\theta)| + C$

do you think this is correct?

maybe I made a mistake

Sine of the Time

$x=a\tan u$ or $x=a\sinh u$

Binky McSquigglebottom

$\sinh^{-1}\left(\frac{x}{a}\right) + C$

this?

EE18

18:46

Am I going crazy, why do I need CS inequality for 13(a)?

Can't I just apply 12(c) and that norms are nonnegative?

Pizza

I had a question, to deduce a graph, without using any software, I have for example $x^2-y > 0 \Rightarrow x^2>y$, so i know that $x^2$ is a parabola so I can write for first $x^2=y$ that is a parabola without dotted lines because there is $=$, then $x^2>y$ will be a parabola with dotted lines and since $>$ the shading will be outside

for other functions, generally , what ever my equation is,i need to find a way to transform it into a well known function?

Sine of the Time

@Pizza you can choose a point and see what happen

for example for $(0,1)$, you have $y>x^2$, so the region $x^2>y$ is the "outside"

Binky McSquigglebottom

@Pizza wait

Obliv

@ee18 I'm not up to where you are yet but is CS and triangle inequality same thing?

I saw this in munkres and I see it in linear algebra, and pretty much everywhere

EE18

No, but triangle follows from CS when in an inner product space so that both of these notions make sense

The picture you sent is unrelated though?

Sine of the Time

18:51

@EE18 maybe you're not meant to use the previous ex (?)

Obliv

if you append that condition 6) onto a field it becomes ordered

it is unrelated, I just wanted to know thank you

@EE18 well actually no, the cs, triangle inequalities work for ordered stuff I think

Binky McSquigglebottom

@Pizza you have to do some studies in pieces

Obliv

unrelated to your question/problem though, yes.

EE18

@SineoftheTime I guess not but feels weird to show them in sequence andthen not use them

Binky McSquigglebottom

@Pizza do you want an example?

Sine of the Time

18:53

@EE18 sure

Pizza

@BinkyMcSquigglebottom ok

can you do this $\log(x^2-y)$

Binky McSquigglebottom

ok

do you agree that a number comes out in the logarithm?

Sine of the Time

@Pizza is it clear my example with $(0,1)$ ?

Binky McSquigglebottom

@Pizza the rules for calculating the domain are the same, so you have to make the logarithm argument greater (>) than 0

Pizza

@SineoftheTime yes i mean if $y>x^2$, the region is inside than

@BinkyMcSquigglebottom yes

I did it

Sine of the Time

18:59

@Pizza if you have a doubt, consider a point belonging to one of the region and see what happens

Binky McSquigglebottom

now write down the corresponding equation with =

Pizza

@SineoftheTime my biggest doubt was how to draw the graphs without using software

Sine of the Time

yes

Pizza

@BinkyMcSquigglebottom $x^2-y=0$

Sine of the Time

first draw the graph with $=$

Pizza

19:00

so $x^2=y$, parabola

Binky McSquigglebottom

What are the points that respect this equation?

Pizza

@SineoftheTime yes

Sine of the Time

so for example you want to know $x^2>y$, this is or the "inside" or the "outside" right?

Pizza

this is outside right?

Sine of the Time

yes, but wait

take a random point, for example $(0,1)$

which is in the "inside" right?

Binky McSquigglebottom

19:04

@Pizza the solutions are the points of the parabola, because if a point $P(x,y)$ belongs to the parabola with center in the $O$, will verify this equation

Pizza

@SineoftheTime yes

Sine of the Time

in this case, you have $y=1>0=x^2$ so the "inside" is $y>x^2$

is it clear now?

Binky McSquigglebottom

@Pizza do you agree that the solutions represent a curve in the plane?

I'm trying to be as detailed as possible

Pizza

@SineoftheTime yes

Binky McSquigglebottom

@Pizza now let's go back to the inequality

Pizza

19:08

ok

Binky McSquigglebottom

now there is a proposition that I will write to you

Pizza

@SineoftheTime but for example before geogebra.org/calculator/th7eqdsw

sorry i had the copy on a msg

to know how to draw this graph without software how can I do it

Binky McSquigglebottom

When I have a curve $y=f(x)$, this divides the plane into two flat domains, in each of which we have $y>f(x)$ or $y<f(x)$

that is, in one of the two half-planes we have that y is always greater than f(x) or y is always less than f(x)

that is, in a half-plane there cannot be a point where $y>f(x)$ and another point where $y<f(x)$

for example your parabola divides the plane into 2 half-planes, the part that is inside and the part that is outside, NOW it cannot happen that inside the parabola I have 2 points where one where y>f(x) and another where is smaller

But now which of the 2 half-planes satisfies $x^2>y$

Make some trivial attempts

we take a point that is either inside the curve or outside, we take like $(0,1)$ which is inside the parabola, I replace and I get $0^2>1$, which is not true, therefore all the points interiors do not satisfy this inequality. Those who satisfy it are the external ones

For the first proposition

Pizza

thanks

Dannyu NDos

21:14

0

Q: Attempt to generalize the notion of absolute continuity to multidimensional domains

Recall the notion of absolute continuity over a 1-dimensional domain: Definition. A function $f: I \to \mathbb{R}$, where $I$ is a compact subspace of $\mathbb{R}$, is said to be absolutely continuous if, for every positive number $\epsilon$, there exists a positive number $\delta$ such that, fo...

integration analysis multivariable-calculus soft-question absolute-continuity

Is this question too soft?

psie

21:37

In the definition of the derivative in higher dimensions, in e.g. Spivak's CoM, we have $$\lim_{h\to0}\frac{|f(a+h)-f(a)-\lambda(h)|}{|h|}=0.$$ Here $h$ is a vector in $\mathbb R^n$, and what confuses me about this definition is that this vector should tend to the origin.

Maybe this is so basic that he never bothers to define it, but what does it mean for a vector to tend to $0$, or to any other point for that matter (since later on, he uses a slightly different defintion where we've made a substitution in the above formula)?

Dannyu NDos

@psie It asks for an open neighborhood of the origin.

psie

@DannyuNDos hmm, Spivak never talks about neighborhoods

ah ok, I see what you mean

what confuses me is the notation $\lim_{h\to 0}$ itself

Dannyu NDos

I think the author was too lazy to write epsilon-delta...?

psie

what would be different about writing $\lim_{|h|\to 0}$?

Dannyu NDos

No difference; they are equivalent.

Btw, that's a rather cool definition of differentiability, I'd say.

That should be easily extendable to an arbitary normed vector space, even for infinite-dimensional ones.

psie

21:51

ok, so $\lim_{h\to 0}$ is equivalent to $\lim_{|h|\to 0}$, but what about $\lim_{x\to a}$, where $a$ is some constant vector. Is it possible to formulate this also in terms of a limit in norm? Somehow I have an easier time thinking in terms of norms rather than vectors. I'm just so used to the limit operator being something for scalar variables.

Dannyu NDos

That's easy; just substitute $x$ to $x + a$, and then take $\lim_{x \to 0}$.

psie

ah ok, that makes sense

leslie townes

psie it's the usual metric space definition of limit, if you are familiar with that. lim_{x to a} f(x) = L means for all e > 0 there is delta > 0 with 0 < d(x,a) < delta implies d(f(x),L) < epsilon. if everything is R^1 you have |x - a| and |f(x) - L| playing the roles of d(x,a) and d(f(x),L) respectively. in R^m you would traditionally use ||x - a|| (euclidean norm) for d(x,a) and similarly for d(f(x),L) (which might be computed in an R^m that is not R^n)

same intuition as the scalar case. you look at what the function is doing for x near, but not exactly at, the point at which you are taking the limit. equivalently you can think of looking at points x of the form a + h, where h is a small nonzero vector.

psie

ok, thanks, I think I'm beginning to feel more comfortable with the notation lim_{x to something}, where x is a vector variable. As you say, it's just notation for a definition. I forget sometimes :)

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