Mathematics - 2024-06-25

Dimitris

07:05

Thank you very much all for your great answers! With the answers I got, I do not need to post any question in MSE:-)!

Thomas Finley

07:34

0

Q: Why this volume problem in Multivariate Calculus seems to have an anomaly?

Find the volume of the solid inside the cylinder $x^2+y^2-2ay=0$ and between the plane $z=0$ and the cone $x^2+y^2=z^2.$ I tried solving this problem as follows: Equation of the cylinder $x^2+(y-a)^2=a^2.$ Now, if we are able to find the volume $V$ between the plane $z=0$ and $z=\sqrt{x^2+y^2},$ ...

This is so strange.

I really don't have an answer to such an anomaly

@XanderHenderson Can you please help me with this?

Pizza

10:45

Hi!

Ryder Rude

13:13

what does "finding meaning in life" mean according to u?

@Pizza hi

Jakobian

13:30

@RyderRude it means having a purpose you're striving for

Ryder Rude

@Jakobian so it's something one can choose for themselves

Jakobian

Yes, you always choose it

Ryder Rude

if life is a movie, purpose is like trying to significantly shape the plot, either for urself or for others

but no one gets to watch the full plot

e.g. some people may try to reverse global warming but die before they see the result for themselves

Obliv

14:25

I don't get this argument (I also don't see why we even have to prove that there is no upper bound for the positive integers)

I get the argument I guess it just seemed weird that we proved midway the existence of $n$ larger than $b-1$ but yeah it makes sense.

EE18

15:04

@leslietownes It's actually a super messy definition now that I glance back at it

i.e. the definition is restricted to $K^{n \times n}$ it seems?

Ryder Rude

This Disease is Deadlier Than The Plague

►

EE18

@leslietownes This part I do not get though. "Matrices over $F[x]$" means maps from $n \times n \to F[x]$ while "matrix valued functions on F" would mean maps from $F \to F^{n \times n}$?

Houshou Rattengod

15:24

Question: What would you say if I "found" a way to do division in Modular Math? (Currently within a certain set of rules/restrictions)

I honestly don't know if this is something or not. Because I was under the impression that Division in Modulo Maths was not possible

Obliv

isn't division actually just modulo math tho

like fraction fields or quotient things are just modular math

Houshou Rattengod

In a way?

okay, but $(4 mod(6)) / 2$

Jakobian

@HoushouRattengod nothing because it exists

Good job, I suppose?

Houshou Rattengod

@Jakobian Really? Aw... because I was lead to believe that it wasn't possible

Jakobian

Well it depends on what you mean

It exists in the sense that $kx = ky \pmod{n}$ iff $x = y \pmod{n/\gcd(n, k)}$ for $k\neq 0$

This is well-known

Even if you are re-proving results thats not a bad thing

Good job

Houshou Rattengod

15:38

$(4 \pmod{6}) / 2 = 2 \pmod{6}, 5 \pmod{6}$

Jakobian

@HoushouRattengod what does the comma mean?

Houshou Rattengod

It's both

Jakobian

That's... true in some sense, if one were to squint their eyes, but formally makes no sense

Houshou Rattengod

when $4 \pmod{6}$ is divided by 2, it produces 2 answers whose equivalents are evenly divided between.

I'm sorry. I don't know how to formally write out my maths to make it understandable on a wider scale. But that is what I meant by Modular Math Division

Jakobian

Formally, you are looking for solutions to $4 = 2x \pmod{6}$, and those are $x = 2\pmod{3}$. Or, going back to mod $6$ we can write the solutions similarly to yours

The expression $x = y\pmod{n}$ means that $x$ and $y$ have the same remainder when divided by $n$

Houshou Rattengod

15:49

Uh, yeah. That's the basic gist of what I did. Although, I went more of the:
$4 \pmod{6} = (6x + 4) / 2 = 3x + 2 $

Jakobian

You should look into modular arithmetic to make your attempt at division in modular arithmetic formally correct

Should be in either a course into abstract algebra or introduction to number theory

Houshou Rattengod

Yeah, right now. I've only gotten it down where $\pmod{n} / x$ is a whole integer.

Jakobian

What do you mean by $\pmod{n}/x$?

Houshou Rattengod

Some modular base, divided by a random integer equals a whole integer...?

Jakobian

So you mean $n/x$?

Houshou Rattengod

15:59

$2 \pmod{10} / 5$
In this example, the Division number is 5, so there are 5 answers in $\pmod{10}$

The beginning part is $2/5$, which I have been rounding up to the next whole number.

Jakobian

As I wrote earlier, you divide what can be divided in a sense that you replace $n/x$ by $n/\gcd(n, x)$

Oh thats what you mean

$5x = 2 \pmod{10}$ has no solutions

There are 0 answers

Houshou Rattengod

And yet, I say that the answers are:

$1 \pmod{10}, 3 \pmod{10}, 5 \pmod{10}, 7 \pmod{10}, 9 \pmod{10}$

Because all of these are equivalent to: $1 \pmod{2}$

Jakobian

?

There are no solutions

Houshou Rattengod

And there it is.

I told you, I was under the impression that "division in Modulo Maths was not possible"

Jakobian

It is possible. Just not always

32 mins ago, by Jakobian

It exists in the sense that $kx = ky \pmod{n}$ iff $x = y \pmod{n/\gcd(n, k)}$ for $k\neq 0$

Houshou Rattengod

16:10

Is there a specific reason for the use of the equals sign over the equivalent sign?

Jakobian

Its easier to write = than \equiv

Houshou Rattengod

I figured that was the answer, but wanted to check

Jakobian

And both are used in practice

Houshou Rattengod

I see that, in a sense. I got the same rule... But mine is not dependent on the greatest common denominator between $n$ & $k$. Only that the $\pmod{n} /x$ results in a whole integer

Oh... should I be writing it as $\pmod{n/x}$ ?

Jakobian

My formula is more general

$5x = 5\pmod{2}$ can be simplified to $x = 1\pmod{2}$ for example

By the way gcd stands for greatest common divisor

Houshou Rattengod

16:22

In your example, that would be 1. Since both 5 and 2 are prime.

Jakobian

$9x = 27 \pmod{6}$ into $x = 3 \pmod{2}$

The only requirement is that I can divide both sides. What $n$ is is irrelevant

Houshou Rattengod

Okay. I see what you're saying...

And I'm talking about division only within the $y \pmod{n}$ side.

just like you would see $6 / 2 = ?$ in a grade school math test.

Solve: $y/x \pmod{n/x}$

Is what I have potentially found a way to do.

Houshou Rattengod

16:44

? $(2 \pmod{10}) / 2 = 1 \pmod{5} \equiv 1 \pmod{10} & 6 \pmod{10}$

${2, 12, 22, 32,...} \equiv 2 \pmod{10}$

But if we were to divide this by 2 the values at the front would be divided evenly between the previously mentioned answers.

$2/2 = 1 \equiv 1 \pmod{10}$
$12/2 = 6 \equiv 6 \pmod{10}$
$22/2 = 11 \equiv 1 \pmod{10}$
$32/2 = 16 \equiv 6 \pmod{10}$
...

nickbros123

17:06

under the definitions given in rudin, loosely: E is connected if its not covered by the union of two separated sets A and B (separated here is defined as $\text{closure}A \cap B=\text{closure}B \cap A= \Phi$, and one useful characterisation given in rudin is: if $E \subset R$ is connected $\iff$ if $x,y \in E$, $x<z<y \implies z \in E$. Is there a similar characterisation for $R^n$ ?

Pizza

17:42

Hi, I had a question

i write :

D = { 1 ≤ $x^2 + y^2$ ≤ 4, y ≥ $\sqrt{3}$}

$\iint_D x^2 + y^2 dxdy$

Can I do it this way ?(I write)

$x = r\cos(\theta) , y= r\sin(\theta), dxdy= r \ dr \ d\theta, x^2 + y^2 = r^2$

So I write the domain like this

$1 ≤ r ≤ 2$ and $ r \sin(\theta) \geq \sqrt{3} \Rightarrow \sin(\theta) \geq \frac{\sqrt{3}}{r} \Rightarrow \sin(\theta) \geq \frac{\sqrt{3}}{2} $

$\theta = \frac{\pi}{3} , \frac{2\pi}{3}$ so $ \frac{\pi}{3} \leq \theta \leq \frac{2\pi}{3}$

So the integral becomes $\int^{\frac{2\pi}{3}}_{\frac{\pi}{3}}\int^2_1 r^3 \ dr \ d\theta$

Did I do it right or did I make a mistake?

Thorgott

18:34

no

Pizza

@Thorgott What did I do wrong :(?

Houshou Rattengod

18:52

Is this how I would mathematically write "round up to the next whole integer"?

$\lceil \frac{x}{y} \rceil$

leslie townes

19:17

ceiling doesn't perform what many people usually have in mind when they think of "rounding," e.g. ceiling(1.1) is 2, although it does capture the "up"

flloor(x + 1/2) will do the thing where anything between 1 and 1.5 (not including 1.5) goes to 1, while 1.5 and above go up to 2

if you look around at canned implementation of that kind of rounding there is sometimes variation in where the exact halves go (always up, always down, alternating between the two, who knows what else)

i guess my one comment is, aside from how you write it, take care in how you talk about it, because what you have in mind when you say 'rounding' might not be what someone else has in mind. i don't know if there is a simple verbal formula that removes the ambiguity for ceiling. (floor is pretty unambiguously described as 'discarding' the non-integer part instead of 'rounding' it down)

Pizza

@leslietownes By chance you can also take a look at my message above, pls

Houshou Rattengod

So, in a paper I would need to define that I am using this $\lceil x \rceil$ command as a means to indicate the next higher integer?

leslie townes

if you use the "floor" and "ceiling" symbols as they are defined on e.g. wikipedia, i think that would probably be understood as long as you said "these are the usual floor and ceiling operations," but providing an actual definition would be even better. i wouldn't think of these operations at the same level as, say, division, where you can just pop "3/5" into a paper without ever saying what / is

Houshou Rattengod

regardless is it's $x + 0.001$

leslie townes

pizza: that's too many calculations for me to hold in my head right now :)

Pizza

19:31

@leslietownes Okay, don't worry!

Binky McSquigglebottom

leslie townes

pizza imagine having to teach multivariable calculus in the state of "too many calculations to hold in my head right now"

Binky McSquigglebottom

Can someone help me pls , im in very difficulty

Soumik Mukherjee

@BinkyMcSquigglebottom How it got 28 views? there are not that much people currently in this room :0

Binky McSquigglebottom

@SoumikMukherjee i dont know

user 85795

19:39

Random bots.

Obliv

how do people write "for all a,b in R" in symbolic form usually? is it $\forall a,b \in R$ or do you put parentheses somewhere like $\forall(a,b)\in R$ etc

Soumik Mukherjee

The former

Binky McSquigglebottom

either $\forall a,b \in \mathbb{R}$ or $\forall (a,b) \in \mathbb{R}^2$

Sine of the Time

@BinkyMcSquigglebottom did you try to solve it with Taylor expansions?

Obliv

what about if you wanna say $\forall a,b \in R, c,d \in S$ or something do you stick to one quantifier for each set/thing ur quantifying over

Binky McSquigglebottom

19:49

@SineoftheTime I can't use it

@Obliv You should add a quantifier

$\forall a,b \in R$ and $\forall c,d \in S$

Obliv

ok

Soumik Mukherjee

@Obliv $\forall a,b\in R$ and $\forall c,d\in S$

Sine of the Time

@BinkyMcSquigglebottom why?

Binky McSquigglebottom

@Obliv $\forall (a,b,c,d) \in R^2 \times S^2$

Sine of the Time

is it an indeterminate for to begin with? I can't read the last terms

Binky McSquigglebottom

19:53

@SineoftheTime I don't think it can be used in this case

@SineoftheTime $x^\ln(1+x^3)$

This term is the problem

@SineoftheTime Is an indeterminate form

Sine of the Time

so you don't have an indeterminate form right?

Binky McSquigglebottom

I have a 0^0

Sine of the Time

I see

why do you think you can't expand?

Binky McSquigglebottom

Because there are functions elevated to other functions

Pizza

Hi! @SineoftheTime

How is it going ?

robjohn

20:05

@BinkyMcSquigglebottom what is $\lim\limits_{x\to0}x^3\log(x)$?

Binky McSquigglebottom

@robjohn 1

robjohn

really?

Binky McSquigglebottom

Wait

Its 0

robjohn

yes, and $\log\left(x^{\log\left(1+x^3\right)}\right)\to0$

since $\log\left(1+x^3\right)\sim x^3$ near $0$

Binky McSquigglebottom

Yes

Pizza

20:16

D = { 1 ≤ $x^2 + y^2$ ≤ 4, y ≥ $\sqrt{3}$}
$\iint_D x^2 + y^2 dxdy$
$x = r\cos(\theta) , y= r\sin(\theta), dxdy= r \ dr \ d\theta, x^2 + y^2 = r^2$
So I write the domain like this
$1 ≤ r ≤ 2$ and $ r \sin(\theta) \geq \sqrt{3} \Rightarrow \sin(\theta) \geq \frac{\sqrt{3}}{r} \Rightarrow \sin(\theta) \geq \frac{\sqrt{3}}{2} $
$\theta = \frac{\pi}{3} , \frac{2\pi}{3}$ so $ \frac{\pi}{3} \leq \theta \leq \frac{2\pi}{3}$
So the integral becomes $\int^{\frac{2\pi}{3}}_{\frac{\pi}{3}}\int^2_1 r^3 \ dr \ d\theta$

What did I do wrong :(?

user 85795

Nice 🇺🇲 👍 @robjohn

Binky McSquigglebottom

How do you determine which system belongs to each figure? (2 figures will be left)

Sine of the Time

@Pizza hii

I'm good, what about you?

Pizza

I'm fine, how are you progressing with differential geometry?

Sine of the Time

yes

Pizza

20:30

I mean, is it going well?

Sine of the Time

yes, I still have to study the last part

Soumik Mukherjee

@BinkyMcSquigglebottom Do you know what those figure means? in terms of solutions of a system of equations

Pizza

@SineoftheTime nice!

Do you happen to have time to see what I posted above?

Binky McSquigglebottom

@SoumikMukherjee yes

Soumik Mukherjee

Ok, so figure 10 means that there is a unique solution, so is there any of the 3 systems that has unique solution?

leslie townes

20:36

@BinkyMcSquigglebottom a solution to a simultaneous system of linear equations, by definition, has to satisfy every one of the equations. graphically, a point will be in the solution set if it lies on every one of the planes represented by the individual equations. the red sets in figures 8 and 9 do not depict points that lie on every one of the planes so they are not going to be the answers

as you seem to be starting to point out, the other figures depict situations where there are no, exactly one, or infinitely solutions, respectively, so you can calculate with those systems and see which of those possibilities you get

Sine of the Time

@Pizza I'm a little busy right now, maybe I can take a look in a while

leslie townes

i guess i misspoke, the 'no solutions' case could "look like" either figure 7 or figure 8 or figure 9 graphically, depending on whether pairs of the equations have a solution or not

but the use of the red coloring in figures 8 and 9 just really puts me off, in suggesting that the red points are the solution set

you might have to be more careful depending on whether the red coloring in the figure is just there, or is intended to represent points in the solution set

Soumik Mukherjee

I think 8 will be the solution of 1

Jakobian

@SoumikMukherjee there is no this amount of people in the world

leslie townes

the more interesting interpretation of the problem would be to assume that the red coloring has no meaning

Binky McSquigglebottom

20:42

@leslietownes Oh, true

leslie townes

binky: in the more interesting case, you would look at the system with no solutions and see if every choice of pairs of equations has solutions (figure 9), only some pairs do (figure 8), or none of them do (figure 7)

Pizza

@SineoftheTime Okay! Maybe I understood where I went wrong, I'll update if necessary

I think : Its correct until $\sin(\theta) \geq \frac{\sqrt{(3)}}{r}$. Then i cant just plug in the value $r=2$.

Maybe $\sin(\theta) \geq \frac{\sqrt{(3)}}{r} \Leftrightarrow \arcsin(\sin(\theta)) \geq \arcsin(\frac{\sqrt{(3)}}{r})$

Jakobian

I think I'll stop contributing to this site

Soumik Mukherjee

@Jakobian Why?

Binky McSquigglebottom

@leslietownes in the case of 9 or 11 the normals all lie on some plane, therefore they are linearly dependent????

@Jakobian why?

Soumik Mukherjee

20:50

@leslietownes Marx would like to have a chat with you

Jakobian

Because of people and their unreasonable, contradicting each other policies

It has an identity crisis and you'll never know when someone will attack you over something petty

leslie townes

@BinkyMcSquigglebottom in figuring out the picture for the 'no solutions' case (which is one of 7-9) looking at the normals is one way to do it. 7 is all the normals are multiples of one another, 8 is a pair of normals are multiples of one another, 9 is the normals are not multiples of one another, but all lie in a plane. 11 is like 9 in this last respect, but those are distinguishable based on the solution set (nonempty in 11 vs. empty in 9)

@Jakobian yes i do. i always know when someone will attack me over something petty. who are you to suggest otherwise

Binky McSquigglebottom

@leslietownes ok,yes

So for 9, all normals lie in a plane

How do we deduce which system that is

Or atleast the possibilities @leslietownes

psie

Anyone keen on hearing my $\LaTeX$ problem? I've been stuck on this for some time now, but I feel like I really need to solve this issue...

The code is fairly simple, but the solution is maybe harder, I'm not sure.

I'll just tell it. I have two different enumerating lists, one with 1a), 1b), etc. and the other with 2a), 2b), etc. When I \label an item in the list and use \ref, I only get b) in the output, so you can't tell from which list it comes from...

Soumik Mukherjee

21:07

@Jakobian I am not sure what to write but I will mention that your contributions are helpful to many people(including me). And about people who are attacking over petty reasons, maybe just ignoring them is the best way to deal with it.

Xander Henderson

@Jakobian If you are referring to the discussion in CURED earlier today, I certainly do not believe that I was "attacking" you, and if I made you feel attacked, I apologize.

From my perspective, a user (you) did something which skirted the rules of the site, and I stepped in, in my capacity as a moderator, to explain the rule. Nothing personal was meant by it, and I certainly was not intending to "attack" you.

robjohn

@leslietownes Your avatar is fuzzy. It's hard to tell what it is. I throw stones at it!

Now my screen is all scratched and I blame you!

user 85795

💪 🤺🥊🤛🤜

leslie townes

oh, i definitely saw that coming

user 85795

👀

robjohn

21:21

@leslietownes I assume you did, since you always know.

Binky McSquigglebottom

@leslietownes Thank you so much !!!

"I" solved

Pizza

22:09

I think I found the errors

$\arcsin(\frac{\sqrt(3)}{r}) \leq \theta \leq \pi - \arcsin(\frac{\sqrt (3)}{r})$

arcsin is only defined between -1 and 1. so i have to consider $\frac{\sqrt{3}}{r} \leq 1 \Leftrightarrow r \geq \sqrt{3}$

This means my bounds for r are $\sqrt{3}$ and 2.

So the integral Is $\int_{\sqrt{3}}^2 \int_{\arcsin(\frac{\sqrt(3)}{r})}^{\pi-\arcsin(\frac{\sqrt(3)}{r})} r^3 d\theta dr$

I was referring to here:D = { 1 ≤ $x^2 + y^2$ ≤ 4, y ≥ $\sqrt{3}$}
$\iint_D x^2 + y^2 dxdy$
$x = r\cos(\theta) , y= r\sin(\theta), dxdy= r \ dr \ d\theta, x^2 + y^2 = r^2$
So I write the domain like this
$1 ≤ r ≤ 2$ and $ r \sin(\theta) \geq \sqrt{3} \Rightarrow \sin(\theta) \geq \frac{\sqrt{3}}{r}$

I think I corrected it?

Jakobian

22:34

@XanderHenderson I'm at a point where apologies mean nothing to me

Transcript for

$Mathematics$ Mathematics