I hate products that tries to jail users. Is there some policy about data liberation (something like with Google products here) with iPad? I have an iPad without 3G but I was considering a new iPad with 3G. But I have no interest to upgrade to a new iPad without proper data-liberation policies (o...
This is a common problem with many other so-called high-end products, even cyanogenmod where you need to parse all kind of obfuscated drivers if you really want to do something more serious with them. Everything is dxmn closed -- if you are perfectionist, you cannot do them the slightest better -- makes me always feel bad when I find bugs...
Are they any good?
My answer is about Nokias and Androids. I recommend you to wait with them until the problems fixed below.
Poor Keyboards with Nokias but not with Androids, at least G1. Poor usability in both camps however will hinder your productivity
The family, N8XX and N9XX, has ver...
Most of this marketing facade around Nokia, Android and Apple are just junk -- I wish there was something better :P ...I have stopped to waste time on them, perhaps some day.
@hhh The point with that floor function is that the $2\pi/3(3\theta/2\pi - \lfloor 3\theta/2\pi\rfloor)$ starts off the same as $\theta$, until $\theta$ gets to $2\pi/3$, at which point it jumps back to zero.
Ciguli Miguli is a 1952 Yugoslav political satire film directed by Branko Marjanović and written by Joža Horvat. It was meant to be the first satirical film of the post-World War II Yugoslav cinema, but its sharp criticism of bureaucracy was politically condemned by the authorities and the film was banned as "anti-socialist".
Plot
Ivan Ivanović, a party functionary, arrives in a provincial town as a temporary replacement for a cultural official. The newcomer is fanatically eager to reform the town's cultural life in accordance with socialist ideals. He abolishes all five music societies ...
Gigli () is a 2003 romantic comedy film written and directed by Martin Brest and starring Ben Affleck, Jennifer Lopez, Al Pacino, Christopher Walken, and Lainie Kazan.
After a protracted battle between studio and director, a radically revised version of the original film was released. There was significant media attention and popular interest prior to its release, primarily because Affleck and Lopez, the film's stars, were romantically involved at the time. However, critical reception was extremely poor, and in the years since its release Gigli has frequently been cited as among the wor...
@DavidWallace, in short he's been trolling the chatroom for the past few months and now he cries that even he is trying to contribute to the chat in a meaningful way - people get angry at him.
@Skullpatrol No. I talk about myself, and about the rather starred comments by several people that asked you to be less trolling. I don't know, but I can't be accounted for all those stars.
This chatroom is hardly ever serious. Not all trolling is enjoyable.
user19161
I thought mathematicians had a great sense of humour, maybe I was wrong.
If Skully's comments were actually offensive (racist, sexist or whatever), it would be a different story. But basically, if they're just irrelevant and slightly annoying, well who really cares?
I know that I do talk for myself, and I do talk for the several different people which starred the comment of Skullpatrol when he was "Oh lord, if several people star this comment I am leaving forever". I also know that in private conversations several people have expressed distraught about his presence at times.
@DavidWallace Ever had someone stand by you and say "David" in a monotonic voice for half an hour? That is not sexist, racist, or otherwise offensive. However if you don't want to punch that person in through the back of his skull then you must be deaf to begin with.
@DavidWallace See? I don't see why I should give up the entire room just because of one pesky little troll. I do see the reason, however, to defend my stand for disliking him.
Yeah, I don't see how I am being selective. If it were up to me you would have been removed from the chatroom altogether, not just parts of what you say.
@DavidWallace Seeing how I am the Owner of the chatroom, might be nearly meaningless - and as Skullpatrol would say I am completely power tripping here - but it still means that it is not I who should walk away.
@DavidWallace That's what I'd like to do but that little X up in the top right corner is my best friend. I'd throw something at the person but I never have done it. I have no choice there.
I do believe that when someone which supposes to maintain the room ignores a person then that person is better removed from the room altogether. Either way, now that I can effectively remove his posts from the main room I see no need to ignore him. If he crosses a line, his posts will be removed.
user19161
I see that @jonas has changed his avatar into the very good-looking one!
@DavidWallace Just ignore someone in the middle of a semi-war and see all @THEDAMNPERSON messages getting on your nerves. You ignored them and other users start to get attracted by that person at the same time. It's exactly how it works.
@DavidWallace Well that is totally different what floor -function usually mean! It usually mean that to get back to the nearest integer...very well there must be some flag to handle this. Actually, is Modulo what you are looking here instead of the floor -function? Start again from 0 when $\frac{c\pi}{b}$ -some-condition?
@Skullpatrol No, and that is the sad part. The only way you can stop being a troll is if you take a few years to grow up. That's just how life is in these parts of the internet. I should know.
hi folks. given the position of a particle in parameterized form (x(t), y(t)), the distance the particle travels from t=a to t=b is the integral of sqrt(x^2+y^2)dt? Or is it the integral of sqrt((x')2+(y')^2)dt? If the latter, why do you have to take the derivative first when you have position functions and you're calculating distance?
@robjohn ok. right. but then you take really small increments of those distances. so you're still not working with the derivative. you still have that the hypotenuse of the small distance is $\sqrt(x^2+y^2)$. How do you get to using the derivative?
@Jeff Travelling around the unit circle, you can travel from $(1,0)$ around the circle to $(1,0)$, travelling $2\pi$, but ending a distance of $0$ from where we started.
@DavidWallace Yes but now we have radians with that condition $\left(\frac{3\theta}{2\pi}-\left\lfloor\frac{3\theta}{2\pi}\right\rfloor\right)\in[0,1)$, $1 \text{ rad } <\frac{2\pi}{3}$ -- how does it work now?
OK, start at $\theta = 0$. Run around to $2\pi/3$, and the expression in brackets is just $\theta$. But as soon as you get to $2\pi/3$, it becomes $\theta - 2\pi/3$, and you start running along the second side of the triangle.
2
@Gigili Aha! I just found it.
I put it in a folder and forgot which one.
Umm, I when I said "expression in brackets", I meant with the factor of $2\pi/3$ included. Otherwise what I said makes no sense.
I ran into this teacher recently and mentioned this quiz problem. She said she thought my son had written "8" and didn't know that a sideways "8" means infinity. I don't even
@Jeff You need to partition the curve into pieces at points $\{(x_i,y_i)\}$ so that $|(x_{i+1}-x_i,y_{i+1}-y_i)|<\epsilon$ and so that for almost all $i$, $\left|\frac{(x_{i+1}-x_i,y_{i+1}-y_i)}{|(x_{i+1}-x_i,y_{i+1}-y_i)|}-\frac{(x_i-x_{i-1},y_i-y_{i-1})}{|(x_ix_{i-1},y_i-y_{i-1})|}\right|<\epsilon$
My son is in 2nd grade. His math teacher gave the class a quiz, and one question was this:
If a triangle has 3 sides, and a rectangle has 4 sides,
how many sides does a circle have?
My first reaction was "0" or "undefined". But my son wrote "$\infty$" which I think is a reasonable answer. ...
The only thing you need to know is that closed subgroups of $\mathbb{R}$ are cyclic if they aren't $\mathbb{R}$ or $0$. That's straightforward to prove: you have a smallest positive elements and that one generates.
@david i have. i know the how. i have a tutoring student who got this question wrong and asks me why i'm using the derivative when it seems like it s/b $\int{\sqrt{x^2+y^2}}dt$ :)
I guess that brute force computation was not what got the votes on this problem. Davide Giraudo posted after and got 2 votes, so it is not a matter of being too late.
However, I do like the Cayley-Hamilton answer, but that was already given. :-)