7:05 AM
Thank you very much all for your great answers! With the answers I got, I do not need to post any question in MSE:-)!
7:34 AM
0

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

3 hours later…
10:45 AM
Hi!

2 hours later…
1:13 PM
what does "finding meaning in life" mean according to u?
@Pizza hi
1:30 PM
@RyderRude it means having a purpose you're striving for
@Jakobian so it's something one can choose for themselves
Yes, you always choose it
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
2:25 PM
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.
3:04 PM
@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?
@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 $F \to F^{n \times n}$?
3:24 PM
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
isn't division actually just modulo math tho
like fraction fields or quotient things are just modular math
In a way?

okay, but $(4 mod(6)) / 2$
@HoushouRattengod nothing because it exists
Good job, I suppose?
@Jakobian Really? Aw... because I was lead to believe that it wasn't possible
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
3:38 PM
$(4 \pmod{6}) / 2 = 2 \pmod{6}, 5 \pmod{6}$
@HoushouRattengod what does the comma mean?
It's both
That's... true in some sense, if one were to squint their eyes, but formally makes no sense
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
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 $n$
3:49 PM
Uh, yeah. That's the basic gist of what I did. Although, I went more of the:
$4 \pmod{6} = (6x + 4) / 2 = 3x + 2$
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
Yeah, right now. I've only gotten it down where $\pmod{n} / x$ is a whole integer.
What do you mean by $\pmod{n}/x$?
Some modular base, divided by a random integer equals a whole integer...?
So you mean $n/x$?
3:59 PM
$2 \pmod{10} / 5$
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.
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
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: $1 \pmod{2}$
?
There are no solutions
And there it is.

I told you, I was under the impression that "division in Modulo Maths was not possible"
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 $k\neq 0$
4:10 PM
Is there a specific reason for the use of the equals sign over the equivalent sign?
Its easier to write = than \equiv
I figured that was the answer, but wanted to check
And both are used in practice
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 $\pmod{n/x}$ ?
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
4:22 PM
In your example, that would be 1. Since both 5 and 2 are prime.
$9x = 27 \pmod{6}$ into $x = 3 \pmod{2}$
The only requirement is that I can divide both sides. What $n$ is is irrelevant
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.
4:44 PM
? $(2 \pmod{10}) / 2 = 1 \pmod{5} \equiv 1 \pmod{10} & 6 \pmod{10}$

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

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

$2/2 = 1 \equiv 1 \pmod{10}$
$12/2 = 6 \equiv 6 \pmod{10}$
$22/2 = 11 \equiv 1 \pmod{10}$
$32/2 = 16 \equiv 6 \pmod{10}$
...
5:06 PM
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 $R^n$ ?
5:42 PM
i write :
D = { 1 β€ $x^2 + y^2$ β€ 4, y β₯ $\sqrt{3}$}
$\iint_D x^2 + y^2 dxdy$
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
$1 β€ r β€ 2$ 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?
6:34 PM
no
@Thorgott What did I do wrong :(?
6:52 PM
Is this how I would mathematically write "round up to the next whole integer"?

$\lceil \frac{x}{y} \rceil$
7:17 PM
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)
@leslietownes By chance you can also take a look at my message above, pls
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?
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
regardless is it's $x + 0.001$
pizza: that's too many calculations for me to hold in my head right now :)
7:31 PM
@leslietownes Okay, don't worry!
pizza imagine having to teach multivariable calculus in the state of "too many calculations to hold in my head right now"
Can someone help me pls , im in very difficulty
@BinkyMcSquigglebottom How it got 28 views? there are not that much people currently in this room :0
@SoumikMukherjee i dont know
7:39 PM
Random bots.
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
The former
either $\forall a,b \in \mathbb{R}$ or $\forall (a,b) \in \mathbb{R}^2$
@BinkyMcSquigglebottom did you try to solve it with Taylor expansions?
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
7:49 PM
@SineoftheTime I can't use it
@Obliv You should add a quantifier
$\forall a,b \in R$ and $\forall c,d \in S$
ok
@Obliv $\forall a,b\in R$ and $\forall c,d\in S$
@BinkyMcSquigglebottom why?
@Obliv $\forall (a,b,c,d) \in R^2 \times S^2$
is it an indeterminate for to begin with? I can't read the last terms
7:53 PM
@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
so you don't have an indeterminate form right?
I have a 0^0
I see
why do you think you can't expand?
Because there are functions elevated to other functions
Hi! @SineoftheTime
How is it going ?
8:05 PM
@BinkyMcSquigglebottom what is $\lim\limits_{x\to0}x^3\log(x)$?
@robjohn 1
really?
Wait
Its 0
yes, and $\log\left(x^{\log\left(1+x^3\right)}\right)\to0$
since $\log\left(1+x^3\right)\sim x^3$ near $0$
8:16 PM
D = { 1 β€ $x^2 + y^2$ β€ 4, y β₯ $\sqrt{3}$}
$\iint_D x^2 + y^2 dxdy$
$x = r\cos(\theta) , y= r\sin(\theta), dxdy= r \ dr \ d\theta, x^2 + y^2 = r^2$
So I write the domain like this
$1 β€ r β€ 2$ 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 :(?
Nice πΊπ² π @robjohn
How do you determine which system belongs to each figure? (2 figures will be left)
@Pizza hii
I'm fine, how are you progressing with differential geometry?
8:30 PM
I mean, is it going well?
yes, I still have to study the last part
@BinkyMcSquigglebottom Do you know what those figure means? in terms of solutions of a system of equations
@SineoftheTime nice!
Do you happen to have time to see what I posted above?
@SoumikMukherjee yes
Ok, so figure 10 means that there is a unique solution, so is there any of the 3 systems that has unique solution?
8:36 PM
@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
@Pizza I'm a little busy right now, maybe I can take a look in a while
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
I think 8 will be the solution of 1
@SoumikMukherjee there is no this amount of people in the world
the more interesting interpretation of the problem would be to assume that the red coloring has no meaning
8:42 PM
@leslietownes Oh, true
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)
@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 $\sin(\theta) \geq \frac{\sqrt{(3)}}{r} \Leftrightarrow \arcsin(\sin(\theta)) \geq \arcsin(\frac{\sqrt{(3)}}{r})$
I think I'll stop contributing to this site
@Jakobian Why?
@leslietownes in the case of 9 or 11 the normals all lie on some plane, therefore they are linearly dependent????
@Jakobian why?
8:50 PM
@leslietownes Marx would like to have a chat with you
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
@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
@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
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...
9:07 PM
@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.
@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.
@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!
πͺ π€Ίπ₯π€π€
oh, i definitely saw that coming
π
9:21 PM
@leslietownes I assume you did, since you always know.
@leslietownes Thank you so much !!!
"I" solved
10:09 PM
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 β€ $x^2 + y^2$ β€ 4, y β₯ $\sqrt{3}$}
$\iint_D x^2 + y^2 dxdy$
$x = r\cos(\theta) , y= r\sin(\theta), dxdy= r \ dr \ d\theta, x^2 + y^2 = r^2$
So I write the domain like this
$1 β€ r β€ 2$ and $r \sin(\theta) \geq \sqrt{3} \Rightarrow \sin(\theta) \geq \frac{\sqrt{3}}{r}$
I think I corrected it?
10:34 PM
@XanderHenderson I'm at a point where apologies mean nothing to me