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/sci/ - Science & Math

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3993432 No.3993432 [Reply] [Original]

I expect you guys to respond better to this than /b/

>> No.3993440 [DELETED] 
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3993440

>at a finite stage, i can tell you which finite ball number was thrown out
not gonna fly OP, i'm high as fuck, and even i'm not falling for this...

>> No.3993444

are people dense enough to be confused by this?

>> No.3993452

>implying that throwing ball 543 out at stage 543 means that all of the balls are thrown out.

>> No.3993456 [DELETED] 
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3993456

>>3993444
>mfw not on /sci/

>> No.3993463

Supertasks are impossible.

>> No.3993473

>stage N

N can only approach infinity, never reach it

p.s. no matter how much you approach infinity, you never reach infinity

>> No.3993483

>>3993452

I should clarify. Your logic is,
"At stage 545 there are 545 balls in the room"
"But ball 545 was thrown out at stage 545"
"Then there are 543 balls"
"But ball 543 was thrown out the stage previous to that"
etc...

No. At stage n, there are n balls in the room, and the nth ball is thrown out. The balls remaining in the room are numbers (n+1) through 2n. So yes, while at stage 543 you throw out ball 543, there are balls 544 through 1086 in the room.

>> No.3993490

The first limit is the limit of the number of balls in the room. This is f(n) = 2n - n. This limit is clearly infinite.

The second limit you took was the amount of balls that have been thrown out. f(n) = n. This limit is also infinite.

The fact that these are both infinite does not mean the functions are the same, nor does it mean you can commit a fallacy of decomposition.

>> No.3993501

lim(x-> inf) 3x - x = inf

>> No.3993506

In case anybody was wondering, the correct answer to this is that after a truely infinite amount of time, there are no balls left in the room.

>> No.3993515

>>3993506
I'd like to see how you came to this one.

>> No.3993523

<span class="math">\infty - \infty = 0[/spoiler] because <span class="math">\infty + 0 = \infty[/spoiler]
<span class="math">\infty - \infty = \infty[/spoiler] because <span class="math">\infty + \infty = \infty[/spoiler]
<span class="math">\require{AMSsymbols} \therefore 0 = \infty[/spoiler]

>> No.3993531

>>3993506
Nope, the correct answer is that supertasks are impossible.

>> No.3993533

>>3993506
I Lold heartily.

If the limit tends to zero prove it using delta/elison

>> No.3993537

Division by Three:
www.math.dartmouth.edu/~doyle/docs/three/three.pdf

covers infinite sets, enjoy

>> No.3993539

>>3993523
I'm being trolled.

1 + inf = inf
1 = 0

QED

>> No.3993576

INFINITY
IS
SHIT

IT DOESN'T MATTER, IT'S NOT REAL, AND YOU SUCK

>> No.3993585

Lol. Is it that difficult?

At each stage, two new balls are thrown in. Even if a ball gets thrown out each stage, it still gets replaced by another one. There is a net increase with each stage.

>> No.3993606 [DELETED] 

>>3993506
no, as the amount of 'rounds' tends to infinity, the amount of balls in the room tends to infinity

>> No.3993615

Infinity being used for a numerical value, or anything for that matter is retarded

stop using it, its a lazy button

>> No.3993640

I climb out a window and let go. My kinetic energy increases at the rate F*v where F is my weight and v is my speed. I can calculate my speed from my kinetic energy: <span class="math">v = \sqrt{2T/m}[/spoiler] where T and m are my kinetic energy and mass. So the rate at which my kinetic energy increases depends on how much kinetic energy I currently have:
<div class="math">{dT \over dt} = F\sqrt{2T/m}</div>
At first, my kinetic energy is zero[*]. That makes the rate of increase of my kinetic energy zero. Therefore I never start moving, and am able to hover in midair.

Problem, gravity?

[*] Yes, exactly zero, because there is an inertial frame in which I am at rest.

>> No.3994012

>>3993576

it is not clear what you mean by 'real'. Aleph Null is the Cardinality of the counting numbers. It seems mathematically perverse to say 'The Cardinality of the counting numbers' doesn't exist.