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pjs36
pjs36
12:00 AM
That is probably the best idea, @PaulPlummer. I wonder if there's an area 51 for that...
The answer is no, not at all... wow, Sage is like on the 2nd page, as far as "How far along" the sites are. That's pretty surprising, considering it doesn't have enough upvoted questions yet...
 
Mike Miller
Mike Miller
@PaulP: I wouldn't be surprised if Mathematica had enough users to support its own site, but that the others combine to something with slightly (but not much) more users.
 
Paul Plummer
Paul Plummer
Well given that it does have its own site, I tend to agree, but it seems to make more sense too have all computer-math sites put together, as they would not have to create their own site, just make a tag. They solve similar problems and ask similar questions, maybe in slightly different ways.
@MikeMiller
 
Mike Miller
Mike Miller
I agree
But w/e
 
Paul Plummer
Paul Plummer
Haha: Science Exchange
Proposed Q&A site for scientists of all domains to publish their work openly and truly peer reviewed.
the next viXra
 
pjs36
pjs36
12:16 AM
Oh man, now that... well, I wouldn't pay to see it, but I could definitely find enjoyment.
 
Mike Miller
Mike Miller
I would give five bucks to see that
 
Paul Plummer
Paul Plummer
Well you can follow it for free
35
Science Exchange

Proposed Q&A site for scientists of all domains to publish their work openly and truly peer reviewed.

Currently in definition.

 
pjs36
pjs36
That was misleading. It seems to be largely non-cranky. But, I'll follow anyway...
 
Paul Plummer
Paul Plummer
For now...
 
pjs36
pjs36
21% of those committed are self-proclaimed "experts" though, so you may be on to something
 
Karim Mansour
Karim Mansour
12:22 AM
@PaulPlummer or @pjs36 can you guys check this question
0
Q: $X_{2n}$ be group presentation as displayed below

Karim MansourHi so I am solving problems in dummit and foote, however this problem I am not able to do it Show that if $n = 3k$, then $X_{2n}$ has order 6, and it has same generators and relations as $D_6$ when x is replaced by r and y is replaced by s. where $X_{2n} = <x,y | x^n = y^2 = 1, xy = yx^2>$ So ...

abstract-algebra group-presentation
 
pjs36
pjs36
Karim, I briefly took a look. I don't have a lot of experience with presentations, but I find calculating the commutator $[x, y] = xyx^{-1}y^{-1}$ to be pretty helpful in this kind of situation
Because it lets you "move" the $x$'s and $y$'s across each other, since $[xy]yx = xy$
 
Paul Plummer
Paul Plummer
@KarimMansour You said that you could get derive the relations for $D_6$ and vice versa, that basically means you constructed a map $X_6 \to D_6 \to X_6$
and the composition is a bijection
 
pjs36
pjs36
But Paul's idea sounds better, quite frankly.
 
Paul Plummer
Paul Plummer
so they have the same order
 
Karim Mansour
Karim Mansour
oh I see makes sense
yeah so they have same cardnality yeah makes sense
ok I will write the proof down and modify answer my own question
 
robjohn
robjohn
12:46 AM
@anon Thanks for taking that on. I just got back from a very upsetting drive home from UCLA. One of my tires blew out.
 
Karim Mansour
Karim Mansour
oke done
writing it now
 
anon
anon
sorry to hear
 
Paul Plummer
Paul Plummer
Also to explain what anon said is that, due to universal property, if you show there is a group that satisfies those relations where $x$ is not trivial, then you have shown $x$ can not be trivial. So a standard trick is to map to a simpler group (although I am not sure if anon is on the right track with what he said)
@anon hey
 
anon
anon
hey
 
Paul Plummer
Paul Plummer
Just commented on that post, If Karims's work is right $x$ can not have order two, if it has order one or three
 
N3buchadnezzar
N3buchadnezzar
12:49 AM
heya
 
anon
anon
yes, I know that 2 does not equal 1 or 3, that does not require any group theory to understand...
 
N3buchadnezzar
N3buchadnezzar
Is therea simple way to find a number $k\in\math{N}$ such that $\log k > n$ for some integer $n$?
 
robjohn
robjohn
@anon perhaps "blew out" is not quite right. I heard what sounded like a gunshot hitting the left rear side of my car. I did not see any decrease in the pressure at that time, but about 10-15 minutes later, I noticed it was down over 15 psi. I took it off and replaced it with the under-sized spare, whose presence angers me. The tire dropped to 1 psi by the time I got to the tire store. They gave me a loaner tire, so that I did not need to drive on the small spare.
 
Paul Plummer
Paul Plummer
Just saying your comment implies that
 
anon
anon
@N3buchadnezzar assuming that's natural log, you can set k=3^n
if it's common log (base 10) then k=1+10^n will do
 
robjohn
robjohn
12:51 AM
@anon 18000 miles on what are supposedly 40000 mile tires. I was about to take them in anyway since they are getting very thin.
 
anon
anon
@PaulPlummer I am not sure what the point of your comment is
I should probably pay more attention to car stuff and be a responsible adult. ugh.
 
Paul Plummer
Paul Plummer
You are saying that the image of $x$ can have order 2, since it is not the identity in a group of order 2
 
robjohn
robjohn
@N3buchadnezzar $k=\left\lceil e^n+1\right\rceil$ or $k=\left\lceil10^n+1\right\rceil$, depending on your base of log
 
anon
anon
@PaulPlummer no I am not saying that
what group of order 2 are you talking about?
 
N3buchadnezzar
N3buchadnezzar
@anon Thanks =)
 
Paul Plummer
Paul Plummer
12:55 AM
Oh I just misunderstood what you wrote, a group that has two elements, I thought you meant a group that only has two elements
 
Karim Mansour
Karim Mansour
@PaulPlummer
I answered it
0
Q: $X_{2n}$ be group presentation as displayed below

Karim MansourHi so I am solving problems in dummit and foote, however this problem I am not able to do it Show that if $n = 3k$, then $X_{2n}$ has order 6, and it has same generators and relations as $D_6$ when x is replaced by r and y is replaced by s. where $X_{2n} = <x,y | x^n = y^2 = 1, xy = yx^2>$ So ...

abstract-algebra group-presentation
 
anon
anon
@PaulPlummer ha, no
 
Karim Mansour
Karim Mansour
@anon nice profile pic
 
anon
anon
thx
 
Paul Plummer
Paul Plummer
Did you switch because some other guy is using your profile pic?
 
Karim Mansour
Karim Mansour
12:58 AM
this one is actually much cooler
than his old one
 
anon
anon
this is my 12th. I change every once in awhile when the mood strikes.
 
Paul Plummer
Paul Plummer
@KarimMansour So you are trying to show that $X_{2\cdot 3k}$ is isomorphic to $D_6$ for all $k$or just $k=1$?
 
Karim Mansour
Karim Mansour
for all k
well 1 is easy
you could show it directly using the relations
 
Paul Plummer
Paul Plummer
So now I am Paul Hammer?
 
Karim Mansour
Karim Mansour
yeah :D
I liked discipline of barney too
how come did you decide to change it @PaulPlummer ?
 
Paul Plummer
Paul Plummer
1:10 AM
I don't know, just decided to
 
Karim Mansour
Karim Mansour
I see
 
N3buchadnezzar
N3buchadnezzar
Is there a simpler way to write
$W( e^{a} )$ where $a$ is some real number?
 
 
1 hour later…
user147690
2:24 AM
What does V with two underlines refer to in regards to complex analysis
 
user147690
$$\frac{\partial u}{\partial \underline{\underline{V}}} = \psi$$
 
user147690
Neumann's boundary condition^^
 
Mike Miller
Mike Miller
who knows
neumann boundary condition should be specifying what the normal derivative is on the boundary
 
user147690
$\frac{\partial u}{\partial \underline{\underline{V}}} = \psi$ on $\partial\Omega$
 
user147690
How would you have written that @MikeM?[E.g. what did you learn Neumann's BC to be written as]
 
Clarinetist
Clarinetist
2:29 AM
Is there a generally accepted shorthand notation to denote that $\mathscr{V}^{\prime}$ is a subspace of $\mathscr{V}$?
where $\mathscr{V}$ is a vector space
 
Ted Shifrin
Ted Shifrin
some people use underlines to denote a vector, @Alex.
 
Clarinetist
Clarinetist
Good evening @Ted
 
Ted Shifrin
Ted Shifrin
I would typically write this $\dfrac{\partial u}{\partial n}$ or perhaps $\dfrac{\partial u}{\partial\mathbf n}$.
hi @Clarinetist
 
user147690
@TedShifrin Where $\mathbf n$ is as Mike says, the normal derivative on the boundary?
 
Ted Shifrin
Ted Shifrin
Whew ... back from my 4-hour retirement shindig. Emotionally exhausted.
yes, @AlexC, precisely
 
Clarinetist
Clarinetist
2:34 AM
@TedShifrin Wait, what is this?
 
user147690
Thanks @Ted. How did the shindig go?
 
Clarinetist
Clarinetist
Ooh
I learned a new word
Lol
 
Ted Shifrin
Ted Shifrin
Saw some students there I hadn't seen in 20+ years. It was wonderful, very embarrassing, honestly, but like two dozen speeches ...
Vocabulary is important, @Clarinetist :P
 
Clarinetist
Clarinetist
I'm super, super excited to leave actuarial science and for my month-long vacation! Haven't really had a vacation since 2010 :)
 
user147690
@TedShifrin Embarrassing why? Embarrassing in the same way that it is embarrassing sitting there while people sing you happy birthday?
 
Ted Shifrin
Ted Shifrin
2:36 AM
Have a wonderful vacation, @Clarinetist. But don't spend it here.
 
Clarinetist
Clarinetist
Lol of course not @TedShifrin
Might consider going to San Francisco
 
Ted Shifrin
Ted Shifrin
No, @AlexC, lots of personal compliments and "life-changing" stories. Very touching.
Nice, @Clarinetist. I'll meet you there in a few months :P
 
Clarinetist
Clarinetist
Haha
 
Owatch
Owatch
Ok, where is immediate-assist night shift crew?
 
Ted Shifrin
Ted Shifrin
LOL, smacks @Owatch
isn't it like early morning where you are?
 
Clarinetist
Clarinetist
2:40 AM
Okay, so here's a stupid question. The book I have says (in my reworded way):

Let $X = \{x_1, \dots, x_r\} \subset {}_{\mathscr{F}}\mathscr{V}$. If there exist scalars $\{\alpha_i\}_{i \in \{1, 2, \dots, r\}} \subset \mathscr{F}$ not all equal to $0$ such that $\sum\limits_{i=1}^{r}\alpha_i x_i = \mathbf{0}$ (the additive identity), then $X$ is said to be linearly dependent. Otherwise, if such scalars do not exist, $X$ is said to be linearly independent.
 
Owatch
Owatch
@TedShifrin No.
This should be easy.
 
Ted Shifrin
Ted Shifrin
Like 5 or 6 AM, @Owatch? No?
 
Owatch
Owatch
22:40
 
Ted Shifrin
Ted Shifrin
Oh, I thought you were in Africa. Whom am I thinking of?
 
Clarinetist
Clarinetist
I'm used to the $\sum\limits_{i=1}^{r}\alpha_i x_i = \mathbf{0} \implies \alpha_i = 0$ for all $i$ definition for linear independence. How do I know that these definitions are equivalent?
 
Ted Shifrin
Ted Shifrin
2:41 AM
@Clarinetist: What does "if such scalars do not exist" mean?
So what's your question, @Owatch?
 
Owatch
Owatch
I was given $x = (sin\theta)^{3}, y = (cos\theta)^{3}$, and asked to find the equation for the tangent at point $\theta = \pi /6$. I have the slope at that point, but I can't figure out what + b is in y = mx + b.
 
Ted Shifrin
Ted Shifrin
Plug in the $x$ and $y$, @Owatch.
 
Owatch
Owatch
Oh.
 
Clarinetist
Clarinetist
@TedShifrin That's from my stats book. Directly from the book, it says "if such $\alpha_i$s do not exist, $X$ is linearly independent."
Maybe I should just go with Insel for reviewing
 
anon
anon
$A\iff B$ is equivalent to $\neg B\iff \neg A$
 
Clarinetist
Clarinetist
2:45 AM
So...
OH I SEE
 
Ted Shifrin
Ted Shifrin
No, no, @Clarinetist. I'm asking you to interpret that sentence.
 
Clarinetist
Clarinetist
Got it
 
Ted Shifrin
Ted Shifrin
hi @anon :)
 
anon
anon
hi
 
Clarinetist
Clarinetist
Thanks @TedShifrin @anon
 
Owatch
Owatch
2:46 AM
So, $(cos(\pi/6))^{3} = (-1.73)(sin(\pi/6))^{3}+b$
 
Ted Shifrin
Ted Shifrin
But, yes, you should not be learning linear algebra from your stats book, @Clarinetist :P
Right, @Owatch.
Or use point-slope formula from the beginning.
 
Owatch
Owatch
ChatJax won't stay on :<
 
Clarinetist
Clarinetist
Now I have to learn about Gran-Schmidt... which unfortunately, my course 3-4 years ago did not cover. My impression is that it's just a way to create an orthonormal basis
 
Ted Shifrin
Ted Shifrin
@Clarinetist: It's about the easiest thing in the whole course.
 
Clarinetist
Clarinetist
Some of these definitions in the stats book are silly. For example:

If $\mathscr{N}$ is a subspace of $\mathscr{M}$ and if $\{x_1, \dots, x_r\}$ is a linearly independent spanning set for $\mathscr{N}$, then $\{x_1, \dots, x_r\}$ is called a basis for $\mathscr{N}$.
 
pjs36
pjs36
2:51 AM
You think so? I was never a big fan of Gram-Schmidt. I don't mind some computations, but that's just plain too many!
 
Ted Shifrin
Ted Shifrin
That's not silly, @Clarinetist. That's the definition.
But it's totally algorithmic, @pjs36, and what's going on is geometrically clear.
 
Clarinetist
Clarinetist
@TedShifrin Meh, I don't agree with using $\mathscr{M}$ as a vector space, and I personally think there's no need to mention the subspace (maybe I'm wrong?). Couldn't you just say if $\{x_1, \dots, x_r\}$ is a linearly independent spanning set of a vector space $\mathscr{V}$ that $\{x_1, \dots, x_r\}$ is a basis for $\mathscr{V}$?
 
Ted Shifrin
Ted Shifrin
Of course you're right. But pedagogically it's sometimes better to think of subspaces of a fixed vector space.
 
Clarinetist
Clarinetist
What I really don't get is why they use the letter $\mathscr{M}$ for vector spaces and not $\mathscr{V}$.
 
Ted Shifrin
Ted Shifrin
I have no clue.
 
pjs36
pjs36
2:56 AM
@TedShifrin You're right, I just got tired of computing after one or two bases :P
 
Paul Plummer
Paul Plummer
@AlexClark I think I filled the gap, so I am going to check over everything again, and then post it (it might take a little while, around 2700 words)
 
user147690
@PaulPlummer Nice work, how long did you spend on it, do you know?
 
Paul Plummer
Paul Plummer
No. A lot longer than I thought it was going to be though
 
Owatch
Owatch
Why must these problems be long and take up pages
It is discouraging.
 
Maximilian
Maximilian
Owatch!
 
Owatch
Owatch
3:06 AM
How's it Max?
 
Maximilian
Maximilian
Bad
 
Owatch
Owatch
Why is that?
 
Maximilian
Maximilian
For math and personal reasons
but to stick with math, i am confused with this log
we have to solve without a calculator
 
Owatch
Owatch
Post it
 
Maximilian
Maximilian
log base 3 of 9 = (x+1)
 
Owatch
Owatch
3:08 AM
Shouldn't be hard
THere's a rule for that
 
Maximilian
Maximilian
Really?
Know the name of it so i can look it up?
 
Owatch
Owatch
Hold on.
 
pjs36
pjs36
Try to simplify $\log_3(9)$, it's a nice round number. Remember, logarithms are like the answer to a question. In this case, the question is, "What power do I need to raise $3$ to, to get $9$?" The answer is $\log_3(9)$. Another acceptable answer would be $___$?
 
Owatch
Owatch
It'a easy
log base 3 to the 9
Means that 3 to X power = 9
 
Maximilian
Maximilian
I haven't taken Algebra 2 in like 2 years so... i got no idea about logs and its on the final:P which is monday
2
 
Owatch
Owatch
3:12 AM
Yes
So 2 = (x+1)
Because it is the equivalent of 2, which means that 2 is the equivalent of (x+1), which means x = 1.
 
Karim Mansour
Karim Mansour
hi @TedShifrin
@TedShifrin check this question I think its cool
2
Q: $X_{2n}$ be group presentation as displayed below proof verification

Karim MansourHi so I am solving problems in dummit and foote, however this problem I am not able to do it Show that if $n = 3k$, then $X_{2n}$ has order 6, and it has same generators and relations as $D_6$ when x is replaced by r and y is replaced by s. where $X_{2n} = <x,y | x^n = y^2 = 1, xy = yx^2>$ So ...

abstract-algebra group-presentation
 
Owatch
Owatch
Which is correct, right? Because if (x = 1), then log base 3 to the 9 = ((1)+1), which is 2. It proves itself correct.
 
Maximilian
Maximilian
The x+1 are in parenthesis in the equation, can i still subtract 1?
 
Owatch
Owatch
Yeah
There's nothing being multiplied, they don't really need to be there.
More of a habit of mine, since I factor a lot.
 
Maximilian
Maximilian
alright... cool.. is there a better way to show $log_3(9)$ without just putting 2? he says without calculator and to provide the steps to prove we didn't use one
 
pjs36
pjs36
3:16 AM
The typical "conversion" is that the statements $\log_b(n) = p$ is equivalent to $b^p = n$, and most people find that question much easier to figure out
 
Maximilian
Maximilian
also, the other problem I have is $ln\sqrt{3}=x$, i googled and did $ln(e^(1/2))=x$, then $(1/2)ln(e)=x$ which I got $1/2=x$ based on what the website had?
 
pjs36
pjs36
So for you, $\log_3(9) = p \iff 3^p = 9$, and then you can tell what $p$ has to be.
 
Maximilian
Maximilian
alright cool, thanks pjs36
 
Owatch
Owatch
@pjs36 Might look more confusing as (x+1) tho, but yes.
If you can see that (x+1) must be 2, then you're good.
 
Maximilian
Maximilian
he is very strict on showing work
 
Owatch
Owatch
3:19 AM
That should be suffecient.
 
Maximilian
Maximilian
I got -1/2 point when proving tan(a+b)
because I when i made the fractions, I didn't cross them out
 
Owatch
Owatch
Like, $9 = 3^{x+1}$, and $9 = 3^{2}$, so $2 = x + 1$
 
Maximilian
Maximilian
Im like, I assumed you could have figured that out
:O Owatch
That makes sense
 
Karim Mansour
Karim Mansour
you know @PaulPlummer even though we proved the problem I am still bothered about why we can't prove that ord(x) = 3 like one should be able to do it that way too
then ord(G) = lcm(|x|,|y|)
 
Maximilian
Maximilian
Thanks @Owatch and @pjs36
 
Owatch
Owatch
3:24 AM
DO another.
 
JMoravitz
JMoravitz
@Max "also the other problem I have is $\ln\sqrt{3}=x$..." I assume you were using an anecdote to see how one problem might relate to the other, but just to confirm, $\sqrt{3}\neq e$, so they are two separate problems. You can use the property that $\log_b x^y = y\cdot \log_b x$, so $\ln\sqrt{3}$ simplifies to $\frac{1}{2}\ln 3$
 
Owatch
Owatch
log base 4 (16) = x + 2
Solve that
 
Paul Plummer
Paul Plummer
@KarimMansour You might be able to prove it directly. The tricky thing about presentations, in general, is that in some sense, they are not concrete descriptions of a group, so to say things about the elements of a group given by presentation, you sometimes have to give it a concrete description by looking at homomorphisms to concrete groups.
So we basically end up look at a homomorphism from that group to the dihedral group, and see that $x$ did not get mapped to the identity, so it is not order one.
 
Maximilian
Maximilian
@JMoravitz Oops, i meant to type in $ln\sqrt{e}$
 
Paul Plummer
Paul Plummer
For example there is no algorithm that can determine, all groups, whether or not they are the trivial group, just by looking at the presentation.
 
Owatch
Owatch
3:26 AM
@Maximilian You might already know this, but ln is log (base e), it's just a special case.
I didn't realise that for way too long.
 
JMoravitz
JMoravitz
It is an unfortunate abuse of notation as well that $\log$ without a specified base is ambiguous depending on which field you are working in. math.stackexchange.com/questions/293783/…
 
Karim Mansour
Karim Mansour
oh I see
 
Owatch
Owatch
I'm used to it being base 10
If it's log.
 
Karim Mansour
Karim Mansour
presentation are nice though I wonder if there is other abstractions of group representation that capture many properties of group @PaulPlummer
 
Paul Plummer
Paul Plummer
So for example on of the "standard" ways of looking at some nice presentations, is that you can often say that all the elements can be made to look like $x^ay^b$, (for example dihedral groups are like that), but the problem you need to figure out is when the exponents on the elements are different then the elements are different, so one of the ways to determine that is look at the homomorphic image of the group to a "concrete" group
 
Owatch
Owatch
3:34 AM
Given that I am finding the tangent for a parametric curve, I was provided with a value of $t$ and asked to find $\frac{dy}{dx}$ at $t = somevalue$, how do I find this point if I am given the coordinates in terms of $y,$ and $x$, when they are both defined in terms of $t$?
I could set both equations equal to those coordinates, and find $t$, but wouldn't that give me two different values or should it give me the same>
 
Paul Plummer
Paul Plummer
@KarimMansour Sometimes if you can have a group that has a certain kinds of presentation, then you can say quite a bit about the group, and in some sense make the presentation "concrete". For example if you can show the presentation satisfies small cancellation conditions, or it describes amalgamated product of groups, or things like that. But that requires quite a bit of theory, and you still can not grapple with arbitrary prestations
 
Karim Mansour
Karim Mansour
oh I see
that makes sense @PaulPlummer
nice stuff I can't wait to get more into this stuff
 
Paul Plummer
Paul Plummer
Of course there is representation theory, which is all about studying representations of groups (classically as matrix groups, but also permutation groups)
 
Karim Mansour
Karim Mansour
oh I see yeah there is a section here on representation theory in DF
 
Paul Plummer
Paul Plummer
I don't know much about that stuff though, sadly
 
Karim Mansour
Karim Mansour
3:38 AM
oh :S
 
pjs36
pjs36
That was the pinnacle of my time in school, doing a bit of character theory (representation theory for lazy people), so I know a bit about how parts of it works
You definitely need a lot of algebra in order to talk about that stuff though, beyond just group theory. Modules in particular come to mind
 
Karim Mansour
Karim Mansour
yeah representation theory in DF come at the end of the book
 
pjs36
pjs36
The basic idea isn't too bad - homomorphisms from your group into the group of invertible matrices of some dimension over some field - but to prove the theorems that let you say much involves a lot of other ideas
 
Owatch
Owatch
I can't think straight. .
 
Karim Mansour
Karim Mansour
I see
 
Owatch
Owatch
3:47 AM
I don't see how to solve for the tangent if I don't have a value of $t$ to plug in.
I've only got an (x,y) coordinate pair. But this is parametric. .
Both are defined in terms of T.
 
Rammus
Rammus
@Owatch are you using $\frac{dy}{dx} = \frac{dy}{dt}\frac{dt}{dx}$?
 
Owatch
Owatch
I can plug (x,y) into their respective equations which will spit out a different value for $t$ each, but how can I use them to solve?
 
Paul Plummer
Paul Plummer
Maybe you are thinking straight, you just happen to be in a space that is curvy
 
Owatch
Owatch
@Rammus Yes.
 
pjs36
pjs36
Unless your curve $(x(t), y(t))$ crosses itself at $(x_0, y_0)$, you should be able to solve the system $\begin{cases} x(t) = x_0 \\ y(t) = y_0\end{cases}$ and get a unique solution $t_0$, provided you know $(x_0, y_0)$ is indeed a point on your curve.
 
Owatch
Owatch
3:50 AM
@Rammus $\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}$
Same thing
NVM.
@pjs36 So, plug whatever I get for $t$ into their derivatives for the numerator and denomenator?
 
pjs36
pjs36
Yeah, then it just involves finding $\frac{dy}{dx}$ at $t = t_0$, however you do that :)
 
Owatch
Owatch
So if y = t + 1, for example, and I was given point (-1,2), I'd have t =1. Then put that in $\frac{dy}{dt}$, and the same thing would be done for $\frac{dx}{dt}$ with whatever t is when solved for x = -1
Kind of crudely articulated.
Annnd I get undefined when attempting to solve for T.
Screw this.
 
pjs36
pjs36
You could parameterize $y = x^2$ by $(x(t), y(t)) = (t^3, t^6)$ and make sure what you're getting with this method agrees with what you'd normally get, differentiating $y = x^2$. Or the upper semi-circle $y = \sqrt{1 - x^2}$ parameterized by $(\cos t, \sin t),\ \ t \in [0, \pi]$.
 
Owatch
Owatch
wat
 
pjs36
pjs36
Just to have concrete stuff to play around with, if you haven't given up :P
 
Owatch
Owatch
4:01 AM
I just have absolutely no idea how to get T given two different equations for it that I can't solve.
And the examples don't have this case.
 
Rammus
Rammus
@Owatch Post the parametric equations?
I'm intrigued
 
Owatch
Owatch
$x = cost+cos2t$, $y = sint+sin2t$
(-1,1) is the point.
$-1 = cost+cos2t$
$-1 = cost + (cost)^{2}-(sint)^{2}$
$1 = -cost - (cost)^{2} + (sint)^{2}$
$1 = -cost - 1$
$2 = -cost$
undef
Unless I did something wrong with my identities there.
 
Rammus
Rammus
The line after $1=-\cos t - \cos^2 t + \sin^2 t$ is wrong
 
Owatch
Owatch
I thought that was sketchy.
I guess the - sign doesn't allow me to make that simplification?
 
Rammus
Rammus
$-\cos^2 t +\sin^2 t \neq -1$
 
pjs36
pjs36
4:06 AM
Yeah, $\pm(\sin^2(t) + \cos^2(t)) = \pm 1$, your signs didn't agree
 
Owatch
Owatch
Well then.
Not sure how to simplify.
 
pjs36
pjs36
Since you can anticipate solving a quadratic equation in $\cos t$, use the identity $\cos(2t) = 2\cos^2(t) - 1$.
Alternatively, from what you had, $\cos^2(t) - \sin^2(t) = \cos^2(t) - (1 - \cos^2(t)) = 2\cos^2(t) - 1$, if you didn't remember the "shortcut"
 
Maximilian
Maximilian
@Owatch I don't know anything:P I haven't done logs in awhile
 
Owatch
Owatch
@pjs36 That helped, I got some numbers now.
Well, I got two numbers for t
1.57, and 2.09
 
pjs36
pjs36
Nice! So if you've got some $t$-values from $x(t)$, you'll get some from $y(t)$ as well, and your $t$-value should be the $t$-value the solutions have in common.
 
Owatch
Owatch
4:11 AM
Which ought I to pick of the two X's to use?
 
pjs36
pjs36
One's certainly $\pi/2$, I don't know the other. It might be helpful to keep it exact, if possible
I'd solve your equation $y(t) = 1$, and get two more $t$-values. If the two sets of two $t$-values have anything in common, that's the $t$-value you want to use
 
Owatch
Owatch
I'm not good at keeping it exact.
@pjs36 I'll do that.
 
pjs36
pjs36
Fair enough, hopefully it'll work out!
 
Owatch
Owatch
Thanks.
I'll be back tomorrow. Thanks for the assistance.
@pjs36 That is some nested logic.
 
Paul Plummer
Paul Plummer
4:37 AM
@AlexClark Well it is up. Apparently the version that is hosted on the server of the software is older than the one that I am using, so it did not like some of my latex settings, and it didn't like it in a very inconsistent way, sometime it thought it was okay, and sometimes it didn't (thank god for search and replace). So if you see any typographical errors let me know
 
Karim Mansour
Karim Mansour
I should get the online version of DF coz that book is very heavy to carry
gonna go pirate it
:D
just kidding
 
Paul Plummer
Paul Plummer
May as well, especially since you already got a hard copy
 
Karim Mansour
Karim Mansour
yeah
 
Australia
Australia
4:54 AM
 
Karim Mansour
Karim Mansour
hey @PaulPlummer I am proving the following let $\sigma$ be m cycle show that $\sigma^{i}$ is also an m-cycle iff (i,m) = 1. So my idea first for the --> implication is that ord(G) = m | im now suppose gcd(m,i) = d. this means d | m and d |i so d | im. so we have m | d by definition of order so d = m however this can't happen unless d = 1 since d | i < m so gcd(m,i) = 1
what do you think ?
for this implication
 
Paul Plummer
Paul Plummer
I am not following why $m|d$ by definition of order
@KarimMansour
 
Karim Mansour
Karim Mansour
5:14 AM
the reason m | d is that d | im and and $\sigma^{im} = 1$ so we must have m | d
since m | im
 
Paul Plummer
Paul Plummer
Maybe I am missing something, but d | m and $\sigma^m =1$, so how does that mean m|d
gcd(16,4)=4 and 4 | 4 and 4 | 16, so how do you get that 16|4
 
Karim Mansour
Karim Mansour
hm
 
anon
anon
@Australia no
 
Australia
Australia
why?
 
Karim Mansour
Karim Mansour
d | m and d | im right ?
 
anon
anon
5:21 AM
@Australia because the $a_i$s and $b_i$s can be different numbers
 
Australia
Australia
can you give a counter-example
 
anon
anon
have you tried making one yourself?
 
Australia
Australia
sorry,I can't
 
anon
anon
what happens if $a_n>b_n$ for every $n$?
 
Karim Mansour
Karim Mansour
yeah I see now what your saying @PaulPlummer yeah that doesn't guarantee
 
Australia
Australia
5:22 AM
But this condition $a_{n}\approx b_{n}$
 
Karim Mansour
Karim Mansour
what I said
 
anon
anon
@Australia what's you're point? $a_n/b_n\to 1$ does not say that $a_n$s and $b_n$s are equal. they can still be different values.
 
Paul Plummer
Paul Plummer
@iwriteonbananas Hello. Do you graffiti on bananas too?
 
Karim Mansour
Karim Mansour
but I am thinking I am going in the right direction in this proof right @PaulPlummer?
 
Australia
Australia
for example? @anon,Thank you
 
anon
anon
5:24 AM
@Australia make you're own example
 
Australia
Australia
...
 
anon
anon
Oct 21 '13 at 5:03, by anon
do not wait for math to penetrate you. you must go out of your way to penetrate it.
 
Australia
Australia
oh ,I have 1/(n^2) and 1/(n^2+1)?
 
anon
anon
sure
 
Australia
Australia
oh, Thank you
 
anon
anon
5:25 AM
then 1/n^2 > 1/(n^2+1) so the product prod(1+1/n^2) will be bigger than prod(1+1/(n^2+1))
 
Paul Plummer
Paul Plummer
@KarimMansour I am not seeing where it is going, but I have not really thought about it that much. For example you have not even brought up $\sigma^i$ is an $m$ cycle
 
anon
anon
@KarimMansour if $\sigma$ is an $m$-cycle then it is $(k~\sigma(k)~\sigma^2( k)~\cdots~\sigma^{m-1}(k))$ for some $k$. if $(i,m)\ne1$ then that list of numbers (with $\sigma$ replaced with $\sigma^i$) repeats so can't define an $m$-cycle
 
Karim Mansour
Karim Mansour
oh
 
anon
anon
noting that cycles are in bijection with orbits, one can rephrase this as a question about Z/mZ if you want
 
Karim Mansour
Karim Mansour
I see
 
anon
anon
5:30 AM
now you have to say that if $i$ is coprime to $m$ then that list of numbers is actually the same list but rearranged, and so it does define and $m$-cycle
 
Paul Plummer
Paul Plummer
Your picture keeps changing
 
Karim Mansour
Karim Mansour
yes
 
Paul Plummer
Paul Plummer
Did you guys know Zorns Lemma is a title of a film?
 
anon
anon
did not
doubtful the movie has anything to do with it
 
Paul Plummer
Paul Plummer
Yah, looks like an avant guarde art film "about" the alphabet
 
Karim Mansour
Karim Mansour
6:32 AM
lol
Paul you should watch big hero 6
 
Samuel Yusim
Samuel Yusim
if you guys want a laugh you should look up this short film called The Secret Number
 
Karim Mansour
Karim Mansour
very nice movie
or the movie PI @SamuelYusim
 
Samuel Yusim
Samuel Yusim
it's about math but made by people who have no idea about math
 
Karim Mansour
Karim Mansour
same as PI
Pi (film)
Pi, also titled π, is a 1998 American surrealist psychological thriller film written and directed by Darren Aronofsky in his directorial debut. The film earned Aronofsky the Directing Award at the 1998 Sundance Film Festival, the Independent Spirit Award for Best First Screenplay and the Gotham Open Palm Award. The title refers to the mathematical constant pi. Like most of Aronofsky's films, Pi centers on a protagonist whose obsessive pursuit of ideals leads to severely self-destructive behavior. == Plot == Max Cohen is the story's protagonist and unreliable narrator. Unemployed, and living in...
 
iwriteonbananas
iwriteonbananas
6:48 AM
@PaulPlummer yes, absolutely. bananas can be used in many different ways that most people are not aware of
 
Karim Mansour
Karim Mansour
6:59 AM
lol
I should probably go to sleep I am asking super stupid questions now
fighting sleepiness and being tired is bad idea when you do math xD
 
anon
anon
your question was hard to read
was my interpretation correct?
 
Karim Mansour
Karim Mansour
yeah
yeah that is really stupid sorry
 
anon
anon
and your reference to DF was about $S_n\not\cong S_m$ if $n\ne m$ I'm guessing?
or no, it's about ${\rm Perm}(X)\cong{\rm Perm}(Y)$ if $X\cong Y$ have the same size as sets, right?
 
Karim Mansour
Karim Mansour
yeah the second statement @anon
Perm(X) $\cong$ Perm(Y) if X $\cong$ y have same size as sets
 
anon
anon
one never refers to a group as bijective, one says a pair of groups are in bijection
 
Karim Mansour
Karim Mansour
7:04 AM
I see
but yeah I should delete that question
 
anon
anon
if you feel you understand what's going on, then that'd be good
 
Karim Mansour
Karim Mansour
yeah
 
anon
anon
there are typically numerous abelian groups of a given order too
 
Karim Mansour
Karim Mansour
yeah
 
Chris's sis
Chris's sis
7:56 AM
Greetings
@robjohn last night I created some advanced integrals that can be done by real analysis although I bet that 99% of all mathematicians (or even more) would try some complex analysis - otherwise no hope.
@robjohn see above
 
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