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12:00 AM
I will be making it to NY/Philadelphia one of these days ... Too many friends I have to visit.
 
That could work @Ted. @Null, I have dinner, will be back soon
 
@TedShifrin did you have cancer, or just a check?
@KajHansen good meal
 
Bon appétit, @Kaj.
I had an operation 5 years ago, @Null. It was pretty bad. But so far so good ...
 
@TedShifrin care to say which organ? You don't have to.
 
Left arm, @Null. Very rare soft-tissue sarcoma (if that makes any sense in English).
 
12:02 AM
@TedShifrin I really just hope that I never get braincancer
 
He was back in lecture that Monday @Null
 
wow
 
(Ok, I'm off for real now)
 
No, no, @Kaj.
 
Oh?
 
12:03 AM
After the heart bypass, I was back 3 days later.
 
It was on a Friday though, right?
 
But I did make it back post-cancer for the beginning of spring semester in January.
I missed a bunch of classes with the cancer, @Kaj. Jason covered my classes and I could barely grade exams lying at home.
You're thinking of the heart bypass one ... it was a piece of cake :D
 
Ah
Yes I was
 
Happy dinner.
 
Yep, I'm off
 
12:05 AM
@TedShifrin is there any reason why the locus of intersections is a conic?
it's quite surprising :P
 
That's because the coefficients of that system of linear equations must satisfy determinant = 0, and each coefficient is effectively a linear function. So you get a quadratic. It's worth doing the proof and a different example :)
 
(or is that covered in the book)
 
Of course it's covered in the book.
 
also, is there any reason we're doing this in real projective space instead of $\mathbb{R}^2$?
 
Because all this projective transformation stuff is really projective. Otherwise you are missing points.
Parallel lines cause you problems in $\Bbb R^2$, but not in $\Bbb P^2$, etc.
Time to do exercises relating to Desargues, Pappus, etc. :)
 
12:15 AM
sounds like you're h saving fun
 
Who?
You're the one climbing glaciers!
 
a surjective function from A to B means $|B|\geq |A|$ ?
 
No, if A hits everything in B, do you need more in A or more in B?
 
ah ok
more in A
 
Right.
This is one of those situations where intuition is totally correct. :)
 
12:19 AM
@Ted We were disappointed with the glacier experience.
Tomorrow will hopefully be more satisfied.
man my brain no work
 
I hope so. I remember some serious ice and caves in the French Alps that were damn impressive.
 
and injective means then $|B|\geq |A|$, as no b in B can be hit twice
 
right, @Null: If you like, you can set $A$ into $B$.
 
@TedShifrin does a (non-degenerate) conic when placed in $\mathbb{R}^3$ just look like the standard conic, with a linear transformation applied?
 
Placed in $\Bbb R^3$?
 
12:27 AM
like
$\mathbb{P}^2$ have element which are sets of points in $\mathbb{R}^3$
 
But you have a conic's worth of lines in $\Bbb R^3$, so it's a conical surface.
 
i think the following belongs to philosophic SE: what if we are all gods trying to prove our points among gods?
 
Not helpful thinking about that, I don't think.
@meow: The point is that a nondegenerate conic in $\Bbb P^2$ is basically unique. But it can look quite different in $\Bbb R^2$: parabolas, hyperbolas, and ellipses look very different.
 
@TedShifrin what is wrong in defining "be hit by" as "the image of"? (just asking)
 
@TedShifrin so it depends on what basis vectors we choose?
 
12:34 AM
It's just not formal mathematical language, @Null.
What do you mean, @meow?
 
welll
i mean
hmm
it depends how we map $[s,t] \mapsto L_{[s,t]}$
 
Oh, I see what you mean.
 
that is, which projective transformation we use from $\lambda_P$ to $\lambda_Q$
also, pascal is kind of cool :)
 
So you're back to choosing the right basis vectors in order to make the projective transformation the identity.
 
how can a 14 year old make more advanced stuff than me. that rustles my jimmys
 
12:36 AM
@Null 13, and what makes anybody's mental capabilities less than someone else's because of their age?
 
Learning is not a competition, @Null.
 
^ that too
 
@meow-mix I think you live in a very good environment, I envy you for that. Don't waste your talents and keep doing what you do :)
 
But nor is it good to try to learn things for which one is not really prepared ...
 
the only thing wrong with my environment is my family doesn't support me
 
12:38 AM
They want you to be a bit more well-rounded, @meow?
 
@TedShifrin they want me to be like my brother
 
@meow-mix I really don't want to get into it, but consider the following: maybe the notsupporting of your family has made you the way you are ;)
 
Which means what?
A jock?
 
yeah, pretty much
 
But I've tried to tell you to slow down, ease up, and not push too hard, @meow. So maybe you should listen a little bit.
 
12:40 AM
i've been trying to
 
Kids are different, but you should try to have friends and do things other than math. Being a jock might be too far of a stretch ... although exercise is important, regardless.
 
@meow-mix Take this with a grain of salt. You reaaally don't want to be that one in class. ;)
 
@Null you don't know how much i don't give a shit in class
besides in classes i don't do well in, i just kind of zone out
which i guess isn't too good, but i don't really umm
 
Well, in all seriousness, if you want to go to a college/university that you will be happy at, you need all-round good grades.
 
care. and i can already see @Ted telling me "bad boy!"
@TedShifrin yeah, i get around 95s and stuff
 
12:42 AM
@meow-mix why are you not doing well in other classes that don't involve mathematics?
 
No, I'm not scolding you. Oh, hell, fine.
There are a lot of boring classes out there ...
 
@Null i'm doing well; its just that i have to actually try in those classes
 
I do hope you have some friends to do stuff with.
 
yeah
 
Then cool.
 
12:43 AM
we usually play football at recess, so i guess that also covers the whole exercise bit
 
wow, american football?
 
yeah
we're not good or anything, just for fun.
 
Well, some. I would be a lot better off if I had played tennis more throughout my life. That's the only sport I care about at all.
Just for fun is all that matters.
 
i don't watch football however; i don't find it very interesting
 
I don't either :P
 
12:44 AM
meh, i just think i wasted my life, so take everything from me with a big grain of salt as i'm biased
 
i'm also part of my school's band, so i made a few friends there
 
Good, meow. That's cool.
@Null: How old are you?
 
@TedShifrin 28
 
and joined the chess club recently
 
OK, you're older than usual. If you're motivated now, that's all that matters.
 
12:45 AM
@Null are you attending any educational institution?
 
Often older students want to be there and do better because of it.
 
@meow-mix you mean, becoming a teacher? (sorry i dont understand the question)
 
@Null: Do you go to university?
 
@Null i.e. university
 
yes
 
12:46 AM
ah. major?
 
first semester bachelor
 
European systems are rather different, meow.
 
and i can just nsay im overloaded
@TedShifrin we now have bachelor and master
 
@TedShifrin i wasn't aware. i do, however, have a swedish friend who's majoring in applied math
apparently, analysis is much more popular there.
 
But in France you have to pick science or humanities, etc. Totally different universities.
 
12:48 AM
mmh
 
@meow: In Europe calculus courses are like Spivak's book ... much more like analysis.
 
(than algebraic studies)
 
I really just want to get my doctor in math to say "fuck you" and go to a nicer place
 
That's a long road to travel yet, @Null.
 
@TedShifrin for me with some os more haha. loooong
 
12:49 AM
@TedShifrin i prefer rigorous calculus; teaching students that math is about plugging in and that you just use "formulae" for everything leaves students with a coarse understanding of the content
 
hi@TedShifrin
 
@meow-mix true
 
@meow: We've had arguments about this here before. Proofs are not necessary for everyone. Understanding should be in every class, but just doing proofs doesn't necessarily enhance understanding for most students.
 
i'm interested in obtaining a doctorate, but im about 2 decades away :P
 
hi @Kasmir
No, you're not 2 decades away, @meow.
 
12:51 AM
really?
when do people usually obtain their doctorates?
 
Unless you burn out, you should be done by 25 or 26.
 
@TedShifrin Did you make a series about analytic functions , complex derivative and line integral on complex curves?
 
@TedShifrin well, proofs should go hand in hand with implications, teachingwise. there's really no point in proving something if you don't know what it might implie
 
@TedShifrin oh, wow :P
 
LOL, @Kasmir, you mean lectures? Not really. Just the last lecture of the second semester course, I did a little on how that stuff ties in with Green's Theorem.
 
12:52 AM
@TedShifrin when students are taught to just plug-and-chug formulas, i believe questions like this come up: math.stackexchange.com/questions/2014635/…
 
Some proofs give no insight into why something is true and how to use it. Other proofs do give insight. For the average student, I don't see the point of including the first sort.
I think there should be different courses for different students, within reason.
 
@TedShifrin So our maxim should be to only write proofs which also explain well and at the very least show what could be done with them?
 
@TedShifrin yes that :D can you please give me some insight on how greens theorem ties with complex line integration , and abit about what is complex integrals and CR - criteria :D
 
I'm just giving my personal opinion. I'm not advocating anything I say as an absolute.
@Kasmir: I have to leave in about 5 minutes.
 
me neither, just asking if it is a good mantra^^
 
12:55 AM
If one wants to be a mathematician, as opposed to an engineer, then one should like different things about mathematics, @Null.
 
@TedShifrin thanks anyway, i hope i talk to you other day , you are the only proffesor that explains things clear to me :D
 
You need to find better professors at your school, @Kasmir !
 
@TedShifrin they are not even trying to make us understand , just gives the lecture and the most simple examples they can find and there you go
on exam they put the hardest
like on greens theorem , was not a single example that a curve must be closed
 
You can only apply Green's Theorem when the curve is closed.
ONLY.
 
yes yes :D
but what i meant to say
they made line integral examples without mention that criteria for green
but on test we should know about green
 
12:58 AM
Let $t\in \mathbb{R}$ and the vectors $$v_1=\begin{pmatrix}0\\ 1\\ -1\\ 1\end{pmatrix}, v_2=\begin{pmatrix}t\\ 2\\ 0\\ 1\end{pmatrix}, v_3=\begin{pmatrix}2\\ 2\\ 2\\ 0\end{pmatrix}$$ in $\mathbb{R}^4$.

I want to determine a maximal linearly independent subset of $\{v_1, v_2, v_3\}$ and to extend these to a basis of $\mathbb{R}^4$.

I have shown the following:

If $t\neq 1$ the $3$ vectors are linearly independent, so the linearly independent subset is the whole set.

If $t=1$ we get for example $v_3=-2v_1+2v_2$ and $v_1$ and $v_2$ are linearly independent, so a linearly independent subset
 
I dont want to see the full text on that one , just send link instead
 
@TedShifrin i was intrigued when i found out the contour integral of a complex entire function is always 0
maybe i shouldnt but
i thought it was kind of cool
 
@KasmirKhaan The curve being closed is stated in every definition of Green's Theorem I've seen... why would it need an example?
 
Oh, entire, yes. Well, it's all about Cauchy Riemann saying that $f$ is holomorphic is equivalent to saying the $1$-form $f(z)\,dz$ is closed. :P
 
@fluffy_muffin its first time we do that type of integral , it would have been cool to show an exemple where you have to close the curve in order to make it easiar
like parabola and add horizantal line
but all we did were simple parametrization of curves and did line integrals
where on test we wont get that kind of questions
 
1:01 AM
Well, I usually do examples where you use symmetry to see the double integral very easily. You'll find some of those in my lectures on Green's Thm, @Kasmir.
 
@meow-mix is that true ? all line integrals of comples function = 0 ?
 
if the function is holomorphic/analytic everywhere
 
@TedShifrin i will watch them now :)
 
@KasmirKhaan only when the functions are holomorphic everywhere
 
all line integrals around closed curves.
 
1:02 AM
holomorphic is what?
 
more generally, as long as the function is analytic/holomorphic everywhere inside the curve, you're cool.
holomorphic = complex differentiable
 
(or analytic; which you can say because in $\mathbb{C}$ holomorphic functions are equal to their laurent series expansions)
 
oh we call complex differentiable for analytic
 
@TedShifrin in $\mathbb{R}$ are all infinitely differentiable functions necessarily analytic?
 
OK, I need to get going
NOOOOO @meow ... very much not. Hardly any at all.
 
1:02 AM
@TedShifrin thanks Ted good bye :)
 
Bye, @Kasmir.
 
@TedShifrin are there any examples?
oh, he left :(
 
I am really sad John Nash died
his story makes me feel really cringy
 
Algebraic question. Consider the proof here: calpoly.edu/~brichert/teaching/oldclass/s2004482/Handouts/… for number 30. Why is a^m(a^(n-m)) acceptable here?
 
@Null what's his story?
 
1:10 AM
@meow-mix he has been diagnosed mental illness and got therapied with insolinshocks. That alone makes me sad. Furthermore he made one of the biggest results for gametheory.
 
@fluffy_muffin note that $a^{-m}a^{m} = 1$
 
So in esscence a great mind destroyed by society
 
@meow-mix Yes, but this is my question. We don't know that inverses exist?
 
thus, multiplying $a^n$ by that doesnt affect its value and we get $a^{n-m}a^m$
because... its a ring?
oh shit
 
Should have been clearer. Meant for a^m. It's a ring but that doesn't mean every element has a multiplicative inverse.
That would be a field, which we don't know that the ring is.
 
1:14 AM
@fluffy_muffin what is the statement?
 
If there is a fixed integer n > 1 such that x^n = x for all elements (x) in a ring and m is any positive integer such that a^m = 0 prove that a = 0.
The first case is trivial, but in the other two cases I don't see how it follows due to the previous.
 
@fluffy_muffin i believe $x^{n-2}$ is its inverse
 
Hi everyone! Try to wish his/her birthday with a downvote
http://math.stackexchange.com/questions/192205/where-does-feigenbaums-constant-4-6692-originate/2051937#2051937
 
you can conclude that $x^n = x^{n-1}x = x$
thus, by uniqueness of identity element
$x^{n-1}$ is the identity element
 
@Null @DHMO Are you familiar with linear algebra? Have you seen my question above?
http://chat.stackexchange.com/transcript/message/34024893#34024893
Do you have an idea?
 
1:18 AM
@fluffy_muffin Well, the addition law of exponents are still true in a ring
 
Here also, Mr. President Donald Trump Himself :

http://math.stackexchange.com/review/first-posts/728626
 
don't view $a^{n-m}$ as $a^n\cdot a^{-m}$
$a^{n-m}$ is just $a^{n-m}$. $n-m$ exists because ring under addition forms abelian group
 
@fluffy_muffin agree?
 
@MaryStar bear with me, what is "maximally" linearly independant?
 
@Null Isn't it the maximal set, i.e., the set with the most possible elements, so that they are linearly independent?
 
1:23 AM
@fluffy_muffin because that's the definition of ring
 
@MaryStar can it be translated to the minimal span?
 
@MaryStar i.e. a set of linearly independent vectors whose span is the vector space?
 
I think so. @Null @meow-mix
 
or, more simply known as a basis
@MaryStar since a basis is the minimal set of vectors whose span is the vector space; and since linearly dependent vectors don't affect the span, it would be the maximal linearly independent set, as you say
 
But this is not a basis of $\mathbb{R}^4$, since we have 3 vectors, right?
So, which is the maximal linearly independent subset? Does it depend on $t$ or do we say that since for $t\neq 1$, we get the whole set then that is the maximal linearly independent subset? @meow-mix
 
1:29 AM
@MaryStar simply pic any vector that is independent, it will be a basis. so its maximally independent. as I might add, this si confusing BS
(with it I mean the 4 vectors)
 
@MaryStar there are infinitely many
infinitely many bases, in fact
at least in infinite vector spaces
 
2:29 AM
Are rational exponents acceptable in groups / rings?
 
@fluffy_muffin can you stop deleting everything you say?
@fluffy_muffin no it is not
 
I deleted because I realized it wasn't relevant since my question was simply the first part.
 
it's still annoying
 
Apologies. Why isn't it acceptable? Can it not be interpreted simply as the solution to (x^(m/n))^n = x^m?
 
well firstly you need to have square roots
and they are not guaranteed to exist in rings
 
2:40 AM
Hmm, makes sense but I don't see how you'd approach that problem if rational exponents are not allowable.
 
is there a rigid polyhedron whose faces are not all triangles?
 
why do you need them? @fluffy_muffin
@Sophie depends on your definition of "rigid"
 
ok
 
2:46 AM
@DHMO I'm sure I don't, but I don't see an alternative solution at the moment whereas mine does. Just failing to see an alternative approach.
 
@fluffy_muffin are you still talking about q30?
 
@Sophie Does the dodecahedron not qualify as such?
 
@DHMO q30? Not sure I understand.
 
@Fargle is it rigid?
@fluffy_muffin you know, question 30
 
Ah, no. This problem is slightly different. If there's a positive even integer n such that a^n = a then I'm trying to show that -a = a for all a in some ring.
 
2:48 AM
@DHMO no, wait. The same as used in that wikipedia page but I only require the length of the edges to be fixed, you can deform the faces. As if you were playing with that magnet toy
 
@Sophie if you can deform the faces, is there any non-rigid polyhedron?
 
tetrahedron, icosahedron, octahedron are all rigid. Do you mean a rigid polyhedron?
so for example, you could get two opposite faces of a cube and slide them relative to one another, while keeping all the edges the same length
 
@Sophie but it's still rigid in your definition
 
the cube?
 
yes
 
2:52 AM
why?
 
never mind
 
The only solution I could come up with was a^n = (a^2)^(n/2) and then from using properties of ring multiplication a^2 = -a^2 and so we'd get a^n = -a implying a = -a. But, this uses rational exponents which as you noted I can't do.
 
@fluffy_muffin n/2 is an integer
 
@DHMO right... because it's even so n = 2k and over n/2 = k which is an integer... Yeap definitely missed that.
 
@fluffy_muffin I don't think the theorem is correct
consider the integer ring by itself
let a=1 and n=4
 
3:00 AM
@Sophie My gut is still telling me that the dodecahedron is rigid by your definition, but I don't know how to check this.
Other than by making a dodecahedron out of straws and toying with it myself.
 
I have a proof by magnet that it isn't
 
Ah, you've done your due diligence then, haha.
The only other idea that comes to mind is the rhombic tricontahedron.
 
polyhedra have scary names
 
@DHMO But in the integer ring a^n \ne a for all a in the ring?
 
@fluffy_muffin I thought 1^4 = 1
 
3:06 AM
@Fargle the one with 30 rhombic faces?
 
@fluffy_muffin you never wrote "for all a in the ring" in your hypothesis for the problem, instead you waited to write it at the end of the problem where it asks you to show something.
words matter
It would be correct to say "If there's a positive even integer n such that a^n = a for all a in the ring, then -a=a for all a in the ring."
 
@arctictern Your right it was bad wording on my part. I stated it at the end rather than writing it twice, but I now see the problem you pointed out.
 
Hi chat
anyone here?
 
@fluffy_muffin anyway, try a=-1
yes people are here
 
:)
can you please help me understadn this
z= z not + delta z
i mean why is this true?
 
3:12 AM
what is z? what is z_0? what is delta z?
 
am working on the proof of derivative of complex numbers
z is complex
 
@arctictern Do you mean in the integers still implying that the theorem isn't true? Sorry, I got a bit lost in the discussion here and not sure I'm following. If so, then the integers don't apply to the theorem since a^n \ne a for all a in Z.
 
So $z_0$ is some fixed complex number, $z$ is something near it, and $\Delta z$ is the change from $z_0$ to $z$? Then that equation is true by definition.
@fluffy_muffin No, I mean take the additive inverse of the ring's identity, which will be called -1, then write (-1)^n=-1 and go from there.
 
so z is the one "moving here" ?
 
yes
the subscript of 0 is conventional for "initial value," which means the same letter without the subscript will mean "final value"
common in many places
 
3:14 AM
what comfuses me is that , limz_0 = f (z_0 +$\Delta z$ ) - f((z_0 )
i dotn see any moving z here
 
What you wrote is nonsense.
 
forgot to divide by delta z sorry
z_0 approches 0
 
I don't see any $z$ in that equation. Does $z$ appear later or something? I am very distrustful of your ability to accurately describe your situation. Are you reading from a text you can cite or upload a scan of, or a document you can link to?
 
I know am typing bad, i will send a pic
sorry again
 
It is not just the formatting that is bad, even if you formatted everything correctly you are trickle-truthing me instead of giving me the whole picture.
 
3:18 AM
@arctictern Been staring at your comment for awhile, and I'm not seeing where I would go from there.
 
the main part i dont understand is , z =z_0+$\Delta z$ , that comfuses me because it should be the derivative at z_0
 
@fluffy_muffin do you know what (-1)^n is when n is even?
 
@arctictern It's just 1?
 
yes. so 1=-1, multiply by a, get a=-a.
 
Ah, so if we take a(-1)^n = a(-1) then we have a = -a
 
3:20 AM
staring is no substitute for thinking and doing :P
 
@arctictern i understand it now thanks man
am just way too tired it s 4 am here
 
@arctictern That is very true D: Although, as a note... Since this is any ring where a^n = a for all a (that need not have unity) I don't know that still holds?
 
your rings don't need to have unity?
that's annoying
well, then do stuff to (-a)^n = -a
 
is it correct to say that f^-1(x)'s domain is the same as the image set of f(x) and that the image set of f^-1(x) is equal to the domain of f(x)

or are there situations where this is not the case that i've yet to learn
 
@arctictern The problem statement doesn't say so... So, without unity would the approach be the one I stated earlier? Resposting: a^n = (a^2)^(n/2) and a^2 = (-a)^2 so we get a^n = -a implying a = -a since a^n = a by the problem statement? And, as noted earlier since n is even we have that n/2 is still an integer and avoid rational exponent problems.
* positive even integer.
 
3:25 AM
@WDUK usually f^-1 isn't taken to exist unless the codomain equals the range (image), in which case it f^-1 has the same domain and range as f but swapped
 
@fluffy_muffin (-a) = (-a)^n = ((-a)^2)^(n/2) = (a^2)^(n/2) = a^n = a, ok?
no unity needed
 
@fluffy_muffin yes. that's equivalent to what I just said: -a=(-a)^n=a^n=a
(this uses (-a)^even = a^even, which you might have only proved for even=2, in which case you should use your proof)
 
@arctictern Apologies, didn't see what you had posted until I had commented. Was just confused earlier about rational exponents in rings as @DHMO pointed out. Since I had missed that n/2 was still an integer. But, that makes sense now. I over complicated the rpoblem.
 
ah so its only true if f(x) is one to one
 
what's the problem with having a ring that $a^{odd}=(-a)^{odd}$, won't we get the same conclusion of $a=-a$?
(I knew that does not hold in $\mathbb{Z}$, but does it hold in general?)
 
3:33 AM
@Secret (-a) = (-a)^n = (-a)((-a)^2)^m = -(a(a^2)^m) = -a
you can't prove that -a = a
 
I have a question
Not to interupt
 
@Secret consider the ring 2Z/8Z. here, 2 does not equal -2, but x^3=(-x)^3 for all x in the ring.
 
So I came to the thought of this, how fast is the tip of the trimmer line traveling at. What would be the formula to figure that out? The RPM is unknown but can not measure for a while. But say it is spinning at 7000 RPM at 12 inches diameter. Thanks!
 
@Ljk2000 if the diameter is 12 then the circumference is 12pi and a point on the outside of the circle travels 7000 times 12pi inches every minute. (not sure why you'd say "the rpm is unknown" and then tell us what the rpm is, that's kinda confusing)
 
hmm, makes sense
 
3:39 AM
So 12XPi = 37.68. And then 37.68X7000 = 263,760 In per min.
I say 7000 RPM because I simply do not know how fast my trimmer is spinning or how many RPM it is rated for. Did my math make sence?
 
you should probably wait to round until after you multiply by 7000
 
?
 
12pi does not equal 37.68 exactly, you rounded that. then you multiplied by 7000. see what happens if you wait to round until after you multiply by 7000.
 
oh, hold on
That got 263,893.78290
well after converting that means the tip of the line is traveling about 250 miles per hour
I think just in case anybody does end up wanting to know I will use this equation and after figuring the RPM out on both my trimmer to put how many MPH the line is traveling at (the tip). I have a 18v ryobi and a craftman gas trimmer.
And thanks that was helpful!
 
hello
 
3:49 AM
hi...
 
find out how many RPM needed to reach 767 mph and then you'll know how fast it has to spin to hit the sound barrier
 
sure, at what diamiter
 
the one you have
 
I spelled diameter wrong. hold on
this should be interesting
 
thats the fun side of maths :P
 
3:53 AM
[Division by zero] Caption:
1. The mere existence of zero divisors in associative algebras places heavy restrictions on the zero terms. When these were given inverses, further restrictions were imposed by associativity.
2. There are a total of 7 distinct cases for each sided multiplicative identity (and the two sided identity cases are given by taking the intersection of the suitable pairs of the total of 14 case). Of the 7 base cases, if any product xy equals a simple cube e.g. zzz, then xy=yx for the given x and y involved. One consequence is that at least one zero term commutes if a nilp
 
Well actually based off the fact that it takes 7000 RPM at 12in diamiter to reach 250 MPH simple math would be that you triple that. So there would be 21000 RPM at this point. Leaving out only 16 MPH. That would really take down weeds and really hurt if it hit your skin. If the line does not brake :P
 
fairly sure you would go deaf too :P
 
Also you would have to run at 54volts. And very true.
My gas trimmer runs 18 in diamiter
that little difference going from 12 to 18in in diamiter would make it spin at 1100+ MPH
*18in diameter
 
Does someone of you have an idea about my question: math.stackexchange.com/questions/2051986/… ?
 
i'll look
That is way to complicated for me. Only have basic knowledge. Sorry! I am off to sleep peeps.
 
3:59 AM
Ok, no problem.
 
Hi folks. :)
 
4:23 AM
.http://math.stackexchange.com/questions/2052075/constructing-the-eigensystem-of-a-matrix/2052085#2052085

^Some people :rolleyes:
 
@MaryStar I have no idea what your question is asking. What does $\Theta$ mean for example?
 
Hi guys
My question hasn't been prove for days.
For a day.
 
I'm still the only upvote. I'm sorry to see that @kayak. Best of luck
 
Oh thanks again Kaj.
 
5:08 AM
.https://www.youtube.com/watch?v=qWiVAeJhRpc
^dope trax
 
5:39 AM
@kayak I am not sure whether it improves your chances too much - since that room is mostly inactive - but you can also post the question to the Calculus and Analysis chat room
Your question is relatively new, but I guess that (at least some of) the advice given here about how to get more attention to a question applies to new questions as well: How to grab users' attention on an old question?
And if you look at unanswered tab you can see that one day is not that long in the context on getting answers. There were questions which have been unanswered for years.
Also on meta: Distribution of time to first answer in MathStackExchange? At the time of that post, the record time between posting the question and the first answer was 38 months.
Also s seemed to be one of the areas which are underrepresented on this site.
It has been my observation that MSE has relatively many experts on number theory and algebra, and relatively few on PDEs and differential geometry. I'm not sure if this is representative of the demographics within mathematics as a whole, or something particular to this site. — user7530 Jun 19 '14 at 22:02
 

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