New Township Costs / Sizes

[agentjwall]

See? that's why I called -1 "coding jargon" because I knew that most people wouldn't appreciate the math...
-.-

 

[Kainzo]

Wait I heard someone mention HTTP parent/child regions, is this in?

Not yet.

 

[Predator2010]

OMG WE DO THAT AT MY SCHOOL TOO!

We do Mechanics/Robotics at College, its great fun
You can terrorise staff with them >

 

[deadandown]

Are the required mats and building still relevant for having a town?


@Kainzo

 

[gooscar]

Are the required mats and building still relevant for having a town?

I don't think the mats are, however we couldn't get regions until we built a town hall.

 

[Matt5327]

You are getting infinity larger, not canceling numbers out. The answer to the equation is infinity. (2-1) + (4-2) + (8-4) + (16-8) +..... [The correct usage of the communitive property in this instance].

Because you are adding the (2-1), you are in fact doing the exercise 2A - A + 2B - B + 2C - C + 2D - D ... when utilizing the generic form of the equation, it is obvious that the numbers do not cancel out, but instead you are adding A + B + C +D
[2-1] [ A+ B + C +D]
[1] [A + B +C +D...]
I really find it preposterous that people are actually thinking that adding up an infinite series of POSITIVE integers somehow produces a negative number after some hand waving. This would be equal to saying 2 + 2 = -4.

While you could argue that it is -1 due to both quantities approaching infinity [2 Quantities going to +infinity and I am actually trying to sound as if it is logical for them to somehow appear at -1], the rate at which they approach infinity is different, and because infinity is not actually a number it can not be held as an "ultimate equalizer." In fact, the item must be evaluated forever, which would show that it is in fact approaching infinity because by pairing the terms as they appear you would get [2-1] + [4 -2] + [8-4] considering that you evaluate the function at n intervals at an indefinite amount of intervals, this process would be repeater forever, but the pattern of positively adding the numbers would still hold true.

I'd actually like to clarify that at the best case the equation would = Infinity - Infinity, which is definitely not -1. Infinity is not a concrete number, it is a concept.

You are assuming that the two sets have to line up. They don't, because of the communitive property.
And as far as infinity being a concept, this is true only to a limit (see what I did there?). Calculus exists because it treats infinity as a number. Calculus is basically the application of algebra to the idea of counting to infinity and dividing by zero. Thus, limits.

 

[Hydroking77]

As being in AP Calculus, I would have to agree with Matt on this one.

 

[AlexDaParrot]

You are assuming that the two sets have to line up. They don't, because of the communitive property.
And as far as infinity being a concept, this is true only to a limit (see what I did there?). Calculus exists because it treats infinity as a number. Calculus is basically the application of algebra to the idea of counting to infinity and dividing by zero. Thus, limits.

I'm in second semester Calculus - I understand what it is.
Calculus treats infinity as an infinity number of repetitions as categorized by the utilization of Riemann sums. An integral is simply the limit of an infinite number of rectangles placed under a curve. [Summation of an infinite limit]
However, on to my main point, Calculus acknowledges (Infinity) - (Infinity) as an indeterminate form. However, that is excluding your obvious failure of logic that you are trying to apply a limit to an equation without a variable. A limit is defined as (x) as it approaches a number [Thus (limit(x)]-> Infinity). The limit of a constant is a constant [Limit of infinity is infinity]. Because you are dealing with limits in this case, you either result in the limit being equivalent to infinity or being an indeterminate form of (Infinity) - (Infinity) as I noted before.

All you accomplish by doing (2-1) incorrectly is that you form an indeterminate form that has no variable so it can not be broken utilizing l'hospital's rule.

As being in AP Calculus, I would have to agree with Matt on this one.

Cool.

 

[gooscar]

So, towns huh.

 

[Matt5327]

I'm in second semester Calculus - I understand what it is.
Calculus treats infinity as an infinity number of repetitions as categorized by the utilization of Riemann sums. An integral is simply the limit of an infinite number of rectangles placed under a curve. [Summation of an infinite limit]
However, on to my main point, Calculus acknowledges (Infinity) - (Infinity) as an indeterminate form. However, that is excluding your obvious failure of logic that you are trying to apply a limit to an equation without a variable. A limit is defined as (x) as it approaches a number [Thus (limit(x)]-> Infinity). The limit of a constant is a constant [Limit of infinity is infinity]. Because you are dealing with limits in this case, you either result in the limit being equivalent to infinity or being an indeterminate form of (Infinity) - (Infinity) as I noted before.

All you accomplish by doing (2-1) incorrectly is that you form an indeterminate form that has no variable so it can not be broken utilizing l'hospital's rule.




Cool.

Looks like we've reached an impass. Probably because I recognize Newtonian Calculus, and you recognize that other guy's version (a bit more populat, but…)

 

[Hydroking77]

Oh, what is that guy's name? It slipped my mind.

 

[GreekCrackShot]

I feel like a novice at math now

 

[Hydroking77]

S 4(upper limit) -4(lower limit) 4x^5-9x^3-x dx= ?
Really easy

 

[AlexDaParrot]

S 4(upper limit) -4(lower limit) 4x^5-9x^3-x dx= ?
Really easy

Polynomial integration - Boring :\.

Try 1/[x^4 +1]. That's a fun one we had in class on Wednesday . [Just the indefinite integral, definite integral would be a even more monotonous than the indefinite integral]

 

[Psychokhaos]

I'm gonna stick with my Advanced Algebra for now, thank you very much. Taking Pre-Cal next year, though.

(Hey, give me credit for being 2 years ahead in math.... please? )

 

[toadsworth29]

I'm gonna stick with my Advanced Algebra for now, thank you very much. Taking Pre-Cal next year, though.

(Hey, give me credit for being 2 years ahead in math.... please? )

There are some freshmen in my school that are taking calc next year.

EDIT: And there's a junior who's past Calc 3, the highest my school offers.

 

[Psychokhaos]

There are some freshmen in my school that are taking calc next year.

EDIT: And there's a junior who's past Calc 3, the highest my school offers.

My district doesn't care as much, though. You're in that smarty school, though, plus our town is only really known right now because of the Powell case. And not too widely. (R.I.P)

 

[Oxea]