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| Acmlm's Board - I3 Archive - - Posts by beneficii |
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beneficii |
 Broom Hatter                  Since: 11-18-05 Last post: 5911 days Last view: 5907 days
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Oh God, I can practically feel the testosterone pumping in this thread.
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Anyway, Vyper, it's not humans' responsibility to ensure that karma is brought about. We have a criminal justice system and we bring criminals to justice, but that's as punishment for a crime and our activities in that regard are and should be tightly regulated. If this was karma, then it wasn't the humans' responsibility in carrying it out. And that last statement ("Now that you know these cops will fire 100+ rounds into your bitch ass, are going going to kill one any time soon? Didn't think so.") seems like it was meant to be very menacing. Sheesh, that's why these debates are always so wearing. (restricted) |
beneficii |
Broom Hatter Since: 11-18-05 Last post: 5911 days Last view: 5907 days
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Hmm, another bumped thread. This person was working feverishly. |
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beneficii |
Broom Hatter Since: 11-18-05 Last post: 5911 days Last view: 5907 days
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Let us accept that I will be keeping a very close eye on this thread. |
beneficii |
Broom Hatter  Since: 11-18-05 Last post: 5911 days Last view: 5907 days
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Well, according to this:
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http://en.wikipedia.org/wiki/International_reactions_to_the_2006_North_Korean_nuclear_test Condemnation has been nearly universal among the international reactions. Even China, North Korea's ally, condemned it. (edited by beneficii on 10-10-06 01:30 AM)
beneficii |
Broom Hatter Since: 11-18-05 Last post: 5911 days Last view: 5907 days
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OK. This conversation has chugged on long enough... |
beneficii |
Broom Hatter  Since: 11-18-05 Last post: 5911 days Last view: 5907 days
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I don't know. It looks like the defendant never had a chance to defend herself. She didn't have any money and couldn't hire a lawyer. By the time she can hire a lawyer, it will be too late to appeal and so she's probably going to be paying a substantial amount of money out of each of her paychecks to pay these damages for the rest of her life. Seems pretty harsh for a fricking Internet post. |
beneficii |
Broom Hatter Since: 11-18-05 Last post: 5911 days Last view: 5907 days
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OK, no posts for a few days except for the last reply, with it being worthless; and Libertarian in the poll being spelled wrong--plus the starter being permabanned: I think it's safe to say that this can be closed. |
beneficii |
Broom Hatter Since: 11-18-05 Last post: 5911 days Last view: 5907 days
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God, that's just terrible. I mean, some people just get so obsessed it seems that they're willing to do anything to accomplish it.
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Well, he shall see what we prescribe for such actions.  (edited by beneficii on 10-13-06 01:51 AM) (restricted) |
beneficii |
Broom Hatter Since: 11-18-05 Last post: 5911 days Last view: 5907 days
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Hmm, we all know that 1/0 is undefined, but what about (1/0)^(-1)? In other words, the inverse of (1/0)? Well, it would seem if you divided 1 by 1/0 the answer should be zero because it seems that:
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1/(1/0) = 1 * (0/1) = 0 What do y'all think? I tried looking this up on Dr. Math, but I could not find anything that discussed this. (edited by beneficii on 10-28-06 11:25 PM) (edited by beneficii on 10-28-06 11:26 PM) (edited by beneficii on 10-29-06 12:00 AM)
beneficii |
Broom Hatter Since: 11-18-05 Last post: 5911 days Last view: 5907 days
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Er, made an error I apologize. The final result was to be zero, not one. (Edited it.)
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Uly, What I'm thinking is that, following from if you do a division with fractions like so: (a/b) / (c/d) then that is the same as: (a/b) * (d/c) where you take the reciprocal of the second number, dividing 1 by (1/0) like so: 1 / (1/0) should be 1 * (0/1) and then 0 because in the second number you do the reciprocal which comes out to being valid.
beneficii |
Broom Hatter Since: 11-18-05 Last post: 5911 days Last view: 5907 days
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Originally posted by rubixcuber But that doesn't prove it wrong. Your error was that you multiplied both sides by what is clearly an undefined number (1/0)^(1) which clearly equals just (1/0), which is clearly undefined. MON, Your answer in that last one probably makes the most sense. If you look at the problem itself, and you distribute the exponent into the parantheses, you will have: 1 ^ -1 / 0 ^ -1 and within you're going to have 0^-1 power, which is undefined. Thank you for helping.
What I'm trying to do is find the value of 1 / (-k)!, where k is an integer > 0. Basically, I'm working on the inverse of a negative factorial. The reason goes to the binomial coefficients (Pascal's triangle), where I was trying to find the value of C(0, 1)*. My best guess was to take C(1, 1) - C(0, 0), which is 1 - 1, and 0. I noticed if you simplified the problem, it would come out to 1 / (-1)!. I wondered if then the inverse of a factorial of a negative integer would be 0. *C(n, k) = n! / (k! (n - k)!)
beneficii |
Broom Hatter Since: 11-18-05 Last post: 5911 days Last view: 5907 days
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rubixcuber,
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Because, at the time (1/0) ^ (-1) was not clearly undefined, but you then multiplied both sides by (1/0), which we all know was undefined. It was the step should not have even been there. Still, it is clear now that (1/0) ^ (-1) is undefined.
Re EDIT: That's what I was worried about. Still, from what I understand, for combinations (which are also called binomial coefficients), 0 <= k <= n. But it causes us to wonder what's outside of those ones in Pascal's triangle. In that case also, what would be: SIGMA [j = 0 to k - 1] C(j, k) As you know: 
Referencing article: http://en.wikipedia.org/wiki/Binomial_coefficient But, if you started j from 0 instead, and went to k - 1, what should the sum be? For one particular problem I'm working on to work out (which I'll try to get uploaded here), it would seem that such sum from 0 to k - 1 should be 0. Which would mean, that each case to the right of the triangle should be 0. Question: What program would be good for generating images containing mathematical notation? EDIT: Added article link. (edited by beneficii on 10-30-06 02:35 PM)
beneficii |
Broom Hatter Since: 11-18-05 Last post: 5911 days Last view: 5907 days
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Hmm, the common belief is that, in the original Famicom version of Super Mario Bros. 2 (J), to get to World 9, you must simply abstain from using any warp zone up to beating World 8--effectively meaning you beat every level. In the All-Stars version, if you have World 9 already, and use a warp zone on a save and then save it, you will lose World 9. But in the original Famicom version, it seems that going into a warp zone won't necessarily make you forfeit World 9.
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In the original Famicom version of SMB2J, the criterion is not abstaining from the use of warp zones and just going into beat World 8, but rather that you must beat every castle from Worlds 1 through 8. When you beat a castle (or more precisely touch the axe and cause Koopa to fall into the lava pit) in a world, a bit is set on byte $07FA corresponding to the world you've beaten. Beating World 1 will set the lowest bit, while beating World 2 will set the next lowestt and so on. After the scene with the Princess at the end of World 8, the game performs a check of byte $07FA, seeing if it is #$FF or not (i.e. seeing that all the bits were set or not). If it is #$FF, then the game will proceed to load World 9, while if it is not, then it will proceed to reload the title screen. It seems that using a warp zone doesn't have an effect on byte $07FA, so if you warped forward and then warped back so you can still beat every castle, then theoretically it would seem that you could still get to World 9, in the original Famicom version of SMB2J. Perhaps someone can try this out, hmmm?
(edited by beneficii on 11-09-06 03:43 AM)
beneficii |
Broom Hatter Since: 11-18-05 Last post: 5911 days Last view: 5907 days
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Er, and with n3g-z3ro's and Tauwasser's posts I think this can be closed now. |
beneficii |
Broom Hatter Since: 11-18-05 Last post: 5911 days Last view: 5907 days
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Oh Lord. Not this discussion again. To prevent the same stuff from being recycled again and again, this'll be closed. |
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| Acmlm's Board - I3 Archive - - Posts by beneficii |
