xkcd

MartianInvader

I'd say it IS still possible for McCain to win. If he takes Ohio, Pennsylvania, and Florida, then the election is his. The machines in Florida already have reports of switching votes from Obama to McCain. Officials claim they need to be recalibrated, which would leave them pretty vulnerable to tampe...

MartianInvader

There was an interesting blog post by Scott Adams on this subject recently that, summed up, said that Big War is a thing of the past after Iraq and Afghanistan as they have shown us the even "victory" is no longer economically viable. Countries quite simply can't afford to go to war while...

MartianInvader

Also, do not make the mistake of thinking that because someone is begging, they are not committing crimes, or that because someone is not begging they are committing crimes. It's not a binary system. You seem to be making the assumption that because they are begging, they are comminting crimes, whi...

MartianInvader

Ah yes, of course. I probably should have been able to think of that. Thanks a bunch.

MartianInvader

Some human rights organizations recently have attacked violent acts against women, standing against any type of violence... If a daughter or sister makes a mistake – especially a moral one – that negatively affects the entire family and its reputation, what’s the solution by such organizations? Wow...

MartianInvader

A question for all the people who like set-theory: I was wondering if there exists (under ZFC) a set S with the following property: \forall A \in S, \{f | f:[0,1] \rightarrow A\} \in S That is, if A is an element of S, then the set of paths in A (the set of functions from the interval [0,1] into A) ...

MartianInvader

Saying the answer is 50% is saying that you ALWAYS see 50% of each color. This is false, as looking dead-on at a blue section will cause more of that section to take up the circle. I believe the exact percentage will be something like \sqrt(2)/2. You can get this answer by noting that the radial lin...

MartianInvader

What really pisses me off about this kind of operation, regardless of whether it fits your definition of "terrorism", is that it has the opposite effect of what was intended. I mean, even if the strike was perfect and all of the dead were terrorists, how many civilians saw the US fly into ...

MartianInvader

Ok, it's been a long time and no-one has gotten it. PaulT, do you know a solution? If so, perhaps it's time for a hint? This puzzle has been driving me crazy.

MartianInvader

The intro I like to talk about is gluing together edges of a square in order to make a sphere (all glued into a single point), a torus (opposite sides glued), a klein bottle (one pair of opposite sides glued with a flip), and real projective space (both opposite sides glued with flips).

MartianInvader

How are they not able to pass anything they want at all with a simple majority (51%) in both houses of Congress, though? Everybody's mentioned filibusters, but let's also remember the good ol' presidential veto. Doesn't that require 2/3 majority to overturn? And Bush has often flat-out stated "...

MartianInvader

So, just to make sure I'm understanding: If I have a set of points in the (n-1)-dimensional hyperplane plane through the origin given by x, and apply the rotation you have defined to them, their images will all be in the (n-1)-dimensional hyperplane through the origin defined by y? Yeah. You can se...

MartianInvader

Okay, so are you claiming this is a solution? I mean, if you slice the cake into two pieces and only one contains the center, then they're NOT congruent; this is easy to prove. I don't think you're allowed to cut up into lots of different pieces (since you did slice a couple of your congruent pieces...

MartianInvader

To put it algorithmically, get an orthonormal basis for your vector space so that the first two basis vectors span the plane generated by your two normal vectors. Say your vectors are at angle \theta apart. Then write the 2-dimensional rotation-by-\theta matrix in the upper-left corner of your n-by-...

MartianInvader

Hmm... Okay. There's something bothering me here. There's different standards being applied for what constitutes "sexist"... maybe not by the same person, but I thought it should be pointed out. The whole "bad mother" thing... this was being argued as sexist because one would nev...

MartianInvader

Yep, you're right, phlip. How about this? Let S_m be the number of sequences of length m so that no initial segment is repeated. Now we can obtain such a sequence of length m+1 by adding a new letter to such a sequence of length m. Then we have the following recurrence relations: If m is even, S_{m+...

MartianInvader

http://en.wikipedia.org/wiki/Riemann_sphere I think you might enjoy reading that. The circles which pass through the "point at infinity" are exactly the lines in the standard complex plane. Mobius transformations also give a nice feel for the similarities between lines and "infinite ...

MartianInvader

Isn't it as simple as (1/n) ^ x, where x ~ number of characters typed so far? Well, that's the chance of doing it on a certain step, but then you gotta sum it up. So you'd be looking at something like summing (1/n)^m from 1 to infinity, which would be 1/(n-1). Bu...

MartianInvader

That's a really nice development from the intuitive definition to the mathematical one.

MartianInvader

That proof strategy will probably work, though you have a few errors. You don't necessarily have a straight line to the center (maybe the pieces are all spiral-like), so you'll have to use a path. If you assume that all the "outer" edges correspond to each other, then you can assert that t...

MartianInvader

I like to view things topologically, so I just use the word "closed interval" and forget about this "extended real line" nonsense.

MartianInvader

Yeah, Token didn't subtract enough. I think we want {n \choose p}p^m - {n \choose p-1}(p-1)^m - {n \choose p-2}(p-2)^m - ... - {n \choose 1}(1)^m This might have some nice expression... I'm not sure. In the given case, that's the same as mine. I'm wrong because I missed some double counting, which ...

MartianInvader

WOW, this thread has gotten real crazy real fast. I wanna join in! I can't believe nobody has responded to this little tidbit yet- The blame for high gas prices in Europe falls on the metric system. Our government adds maybe $1 in taxes to every gallon. Europe adds about $1 per litre. I think in Fra...

MartianInvader

Man, reading that article just makes the lawsuit seem full of AWESOME. All the posts put in this thread talking about how this is wasting taxpayer money etc. are exactly the questions he was trying to get people to ask! If, as a result of this lawsuit, stricter measures get placed on what types of l...

MartianInvader

Yeah, Token didn't subtract enough. I think we want

{n \choose p}p^m - {n \choose p-1}(p-1)^m - {n \choose p-2}(p-2)^m - ... - {n \choose 1}(1)^m

This might have some nice expression... I'm not sure.

MartianInvader

Every time the McCain campaign has tried something like that, it's blown up in their face. I think they've learned their lesson by now.

MartianInvader

If I recall correctly, diamonds are only expensive now because of hoarding from the diamond companies... artificial diamonds, while cool, aren't exactly necessary yet. Would artificial diamonds be cost effective if natural diamonds were priced normally? I think the point is the companies are drivin...

MartianInvader

Belial wrote:The press's fabled "liberal bias" or "conservative bias" (depending on whom you ask) is extremely, ridiculously secondary to its laziness and sensationalism biases.

Oh man, I am so sigging that.

MartianInvader

I don't think he's full of shit, but it's just not that surprising. I mean, if I was at a rally for a candidate I supported, and someone I knew in the crowd yelled something like that, I might not approve of it, but I certainly wouldn't give his name to the FBI, especially if I thought he wasn't ser...

MartianInvader

Thanks for your input, weosdfl34! And thanks for those e-mails with the links to penis enlargement product sites! But seriously, we all know what this hexagon is. It's clearly some sort of underground alien complex, constructed by Saturnian life forms vastly different, and vastly superior to us. Our...

MartianInvader

Aw c'mon, now you're splitting hairs. Your (22/7) question clearly was phrased to try to discredit his (GreaterSteven's) original point by making it sound nigh-impossible to accomplish; he responded to this point by citing an example where it had been done in less time with less resources. Now you'r...

MartianInvader

I seem to recall that linear algebra works perfectly fine over division rings (i.e., fields without the communativity axiom for multiplication). So yeah, some of the axiomatic stuff is just thrown in there for comfort. But it probably wouldn't be a bad idea to just mentally note what axiom you're us...

MartianInvader

Yup. That's also why I was careful to say "linear in the size of the input". The size of the input would be the number of bits used to store it.

MartianInvader

Yeah, implicit and explicit differentiation are the same operation. The words "implicit" and "explicit" refer to what you're differentiating , not to how you're doing it. If you have a function defined explicitly, it means something like f(x) = x^3 +sin(x). A function defined imp...

MartianInvader

But doesn't everything that uses arbitrarily large input take at least linear time in the size of the input? I mean, it takes that much time just to read in the data.

MartianInvader

The differentials dx and dy are smooth sections of the cotangent bundle of R 2 . In the case you have a function f: R -> R , the graph of f is a one-dimensional submanifold, and dx and dy restrict to smooth sections of the cotangent bundle of graph(f). They are (locally) related to each other by dy...

MartianInvader

I understand what the problem is intended to be (and I'm working on it), but you may want to alter its set-up a smidgen. As it is, there's no basis for saying "we'll never know them", since there's nothing keeping A from just telling B what the sum of the numbers are and B from telling A t...

MartianInvader

Even worse - The function which is zero for x \leq 0 and e^{-1/x} for x > 0 is continuous and differentiable, but isn't even equal to its Taylor expansion around zero (indeed, every derivative at zero is zero, so the Taylor expansion would be the zero polynomial). Also, plenty of functions are once-...

MartianInvader

Sure! Your Russian-ness is just the average of the Russian-ness of all your ancestors at some fixed level (it's the average of your parents Russian-ness, which is the same as the average of your grandparents' Russian-ness, which is the same as the average of your great-grandparents Russian-ness, etc...

MartianInvader

Indeed. I count that as "doing something tricky." I purposely used the word "typically" when I said you get a k+m-n sphere. How about I phrase it this way: The intersection is is a k+m-n sphere 100% of the time

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