Bahamut

Yeah upscaling can get very ugly. Playing old PS1 games on newer TVs also get just as ugly - I remember playing Breath of Fire IV in the summer on my HDTV at home and wow did it look horrible. That can be true, but it all depends on the device doing the upscaling. The XRGB2/3 for instance does a really good job all around, but it would cost me ~$300 to get, So I figured I'd wait for the PS3. Now I don't claim to know what hardware makes a good upscaler, but I think the PS3 could probably do a really good job deinterlacing/upscaling those older games. I can understand them saying "no enhancements like texture filtering/FSAA" but upscaling can be done to all games with no chance of there being any compatibility problems (Unlike that new Xploder HDTV accessory for the PS2) Well, Sony is a company that seems focused on having consumers excited about future releases and not nostalgia for old games, so I can see why they wouldn't go for upscaling old PS1 games - besides, it takes time from their R&D that could be better spent improving the PS3 hardware.

Is there any particular date & time suggestion (I'm guessing we'd meet in Grand Central Terminal again?) that'd work for everyone though?

What the hell is that?

Yeah upscaling can get very ugly. Playing old PS1 games on newer TVs also get just as ugly - I remember playing Breath of Fire IV in the summer on my HDTV at home and wow did it look horrible.

It does have to do with the distribution of prime numbers - in fact, the Riemann zeta function is used in several proofs of the Prime Number Theorem, including the first proof. The PNT says that pi(x), the number of prime numbers less than or equal to x, is asymptotic to x/log x, also written as pi(x) ~ x/log x (since being asymptotic is an equivalence relation). Two functions f(x) & g(x) are asymptotic if lim f(x)/g(x) = 1, where the limit is as x is approaching infinity. It turns out that 1/zeta(1) = 0 is equivalent to the Prime Number Theorem. The zeta function isn't discussed in Calc 2 because it is impractical for most students in Calc 2, such as engineers & pre-med students. Also, the uses of the zeta function are considered fairly advanced, since its definition involves complex numbers and so anything involving the zeta function requires complex analysis (i.e. concepts of analytic continuation and the theory of Dirchlet series).

The time is almost here! Date & Time: Dec. 28th noon @ Grand Central Terminal by the Information booth with the clock on top of it in the main floor Fun stuff to do: And also a stop by Multimedia 1.0 by 6th St.! zircon's cell: 215-531-0798 <- call if you are having problems! Definites Athair (with friend) Atma Weapon Axel B. Universe (with friend) Bahamut bustatunez Captain Huge! DarkeSword DrumUltimA epinephrin Geoffrey Taucer José the Bronx Rican Katsurugi (with a friend) Kroze (possibly with friend) Leon K. Pi_R_[]ed (with 1 - 3 friends) pixietricks Shnabubula VRemedyz zircon Tentatives Bren Staying at zircon's overnight zircon pixietricks GeoffreyTaucer Bahamut Athair (with friend) bustatunez José the Bronx Rican DrumUltimA VRemedyz

Let's not forget that racist billboard advertising the white PSP. Is that real? Yep and sure enough, a lot of people in the U. S. complained about them.

Another doublepost! But I figured out how to do this problem, so it is solvable. The offer is still out there I guess for the ambitious people, but I highly doubt anyone on the forums will figure out how to do it, and if you do, you might want to consider going into math.

Well Gamespy gave it a 2/5 and the current GameRankings average is 61%... I'd buy it if I weren't saving money... Yeah even 1UP gave it a 55% too.

Aren't Kirby games supposed to be easy in general?

Power series looks too ugly for this to work, since you'd have to multiply two power series, and that doesn't go well for solving differential equations. Edit: Alright so here's a math problem for you people, and if you're able to solve it by Sunday @ midnight, then I will paypal you $20. Let m be a fixed natural number. Find the infinite sum of 1/(m * n)^s, where s is just a complex exponent (it's not important), and n is all such numbers such that (m, n) = 1, or in otherwords, m & n are relatively prime/coprime. Edit #2: Expressions with the infinite sum of 1/k^s for k = 1, 2, .... = zeta(s) btw, so the infinite sum of 1/k^(2s) = zeta(2s) for example, where zeta(s) is the Riemann zeta function.

I sure hope you don't go into business. I was talking from a consumer's perspective. I don't care who innovates more than whom (and who copies whom), or who has bigger balls than whom, or who makes the best ads, or who cares the most about children, I just want to play good games. I'm well aware that marketing has a huge impact on sales, I'm just saying that for us, who are relatively well informed about the industry, it shouldn't. I don't really care if people from Sony say : "We want this system to be expensive and we want to fuck our customers up the ass" whereas people from Nintendo say : "We want a revolution in the gaming world and we care about the gamers, especially the hard core, since we work hard to make the best games for them so that they can have a thoroughly enjoyable experience". A company exists to make profits, and it's all just a bunch of guys working on projects. There are just different ways to pass it on the public. They don't wanna be friends with us, they want us to buy their games. In the end, all that matters to us are the game themselves, not what the company that makes the console looks like. Your claim about being for the consumers falls apart simply since Sony went after a company offering a useful service for consumers. Their action was anti-consumer, there is no arguing that point. An analogous situation is the RIAA going after P2P services for not conforming to their way - it is 100% anti-consumer. I have zero tolerance for any company/organization that goes out of their way to stifle consumer offerings with lies & abuse of the civil system. If Sony does not want problems with limited supplies of consoles going overseas, they should exercise other fine methods such as tighter control over the suppliers. Anyone who is interested in entrepreneurship or interested in keeping companies as honest as possible should be able to understand this simple concept. Otherwise, you assent that they should be allowed to fuck you up the ass. But again, we are going off-topic - any more responses should be delegated to the Lik-sang thread.

Think of differential equations. In this case, the equation would be f' = (u')^2. I can't think of any generalized solution to this, can you?

So Kamoh and I traded games.

I sure hope you don't go into business.

For those who want to oogle at the PS3, here's some pics of the contents of the package: http://media.ps3.ign.com/articles/741/741368/imgs_1.html That looks as shiny as the DS Lite.

Awwwwwww yeeeeaaaahhhh. Too bad its been delayed to February .

No, I addressed this in the pertinent thread though. Sony did essentially one of the lowest blows possible in the business world - corporate blackmail. The previous efforts by Nintendo, Microsoft, and Sony are of a different issue, and even then it is still legitimate (selling modchips). But again, this should be in the proper thread, not here. Sony literally has the whole company on the line with the PS3 - boycotting them can have a tremendous effect on their future.

This isn't a Nintendo thread. And there is absolutely no excuse for Sony's actions. But this is a PS3 thread, not a Lik-Sang thread.

But what if the sequence I chose was {1/4, 1/1, 1/6, 1/1930, 1/205, -4, 1/356!, ...}. I can't order it so that it is always decreasing, because of that -4. Eventually I'll have to use that -4, and then the next number will be greater than it. All that to say, the axiom of choice only lets me pick arbitrary numbers out of the sequence (I think?), and that -4 makes life difficult. Edit: Hmm, I'd have to think on it more. That -4 should be irrelevant though. Any subset of the real numbers has a limit point - each element is a limit point. That doesn't help you at all though. I'd think more on this, but I got a test tomorrow so maybe tomorrow night I can think on that more.

So Bahamut wins two vs Keegan. The first game I won in some good play, and the second one I was losing, but Keegan forgot about the time.

Oh wow that's dirtbag. Guess I'll hold off on that PS3 until the end of the generation then.

I already preordered it in store earlier today. Great deal, even though it's $40.

Lunar IS awesome. Lunar Dragon Song is fucking awful, though. Biggest disappointment of all time. I'm enjoying it though. Plus, I got it cheap. I get all my games cheap. Because I'm a tightfisted bastard, who only buys when it's on special and pre-owned. I cannot stress enough how much you need to play each of the other Lunar games. Lunar 2 is an exercise in an RPG's tediousness. I had very mixed feelings about the game - loved the OST, and the story was intriguing, but the combat...holy shit was it shitty.

Is H a subset of the real numbers? If so, then with a bounded set, you have the least upper bound and greatest lower bound. There must exist sequences in H that converge to the least upper bound and greatest lower bound. Then, there exist subsequences of each sequences such that the subsequences are monotone and converge to the least upper bound and greatest lower bound, respectively. Yeah, H is a subset of the reals. I don't follow your argument, though. Let's say H is the union of {1-1/n|n is a natural number} and {1+1/n|n is a natural number}. It's bounded by 0 and 2, but I need a sequence that is always increasing or always decreasing, rather, x_n>x_n+1 always or x_n<x_n+1 always. No such sequence converges to 0 or 2... the only way I can find such a sequence is if it converges to 1... I can't see how talking about the least upper bound or greatest lower bound helps me.. Yes - I'm just talking about the existence, guaranteed by theorems. If you don't have the particular theorems that I'm referring to, then just prove them - if you have an element that is not the least upper bound or greatest lower bound, you can find another element closer, and keep constructing it that way. Um. I need the sequence to be strictly increasing or strictly decreasing, not monotone. I think that's our delimma. If the least upper bound and the greatest lower bound isn't in your sets (i. e. if it's an open bounded set), then my method works in constructing a strictly increasing or decreasing sequence. That's true. But I need it in general. For instance, the example above. It's not open bounded, so that proof doesn't help me. Oh wait you have an infinite set. The real numbers have the well-ordering property, so you can take that set and find a sequence and order it so that it is always increasing or decreasing then. This argument is quite strong though...(uses the Axiom of Choice)