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[MagmaMcFry]

Let s_n be the partial sums of a_n, and let s be the limit of s_n. Let c = 0.0000001.

Since the a_n have strictly decreasing absolute value, the s_n alternate between being higher than and lower than the final value s, growing strictly closer to s from above and from below (not necessary with the same speed). This means that if a_5 is less than c, then s_4 and s_5 must have distance less than c to s, since s lies between s_4 and s_5.

Now all you need to do is check that s_2 and s_3 must have distance greater than c to s, and you're done because no s_i can lie between s_2 and s_3 for i < 2.

To check this fact, you don't need to know s directly: Since s_2 < s_4 < s < s_5 < s_3, a sufficient condition for the fact would be s_2 + c < s_4 and s5 < s_3 - c. From there s_2 + c < s < s_3 - c immediately follows.

[Parsely]

How do I solve for the coefficient of kinetic friction given initial velocity (20 m/s) and time (5 seconds)?

friction = frictionCoefficient * NormalForce, N = mass * gravity
f = fCoef * m * 9.8

So I need to solve for mass and friction force. I could figure the friction force if I knew the horizontal force but I'm not sure how to get there? I also have no idea how I can solve for mass with the given information.

We have some equations that might get us to m:
force = mass * acceleration
weight = mass * gravity

We can solve for acceleration using:
finalVelocity = acceleration * time + initialVelocity
0 = 5a + 20
-5a = 20
a = -4 m/s/s

But that doesn't help because I still don't know what force or weight is to solve for mass!

We can also solve for horizontal distance using:
x = 1/2 * a * t^2 + v0 * t + x0

But is that even useful? I'm so stumped on this one.

[WealthyRadish]

0.5 * m * v^2 = F * d = mμgd

Edit:
This is assuming the object comes to a rest at the end.

[askovdk]

I guess the missing key is that the friction is the horizontal force (assuming that the motion is horizontal, - in general friction is parallel to the motion).

You already have the effective accelleration, so the effective force on the object must be the unknown mass times this, but luck (or rather the beauty of physics) will have it that the mass is part of the equation for the friction which is the only possible source for the the effective force. i.e.

m? * a = F_effective = F_friction = fCoef * m? * 9.8  m/s2
=>
fCoef = a  / 9.8 m/s2   =  -4 m/s2 / 9.3 m/s2 = (-)0.41

Independently of the mass.
I.e. the basic friction formula says that it doesn't matter if you have a 1g cube or a 20 tonnes sheet, - starting at the same velocity they will come to rest at the same position after the same time. 

[NB: This doesn't contradicts UrbanGiraffe's nice method which looks at the energy conversion, but that equation requires the travel distance.]

[Parsely]

Okay I had to really struggle to understand but now I get it, I didn't realize you can just sub in friction in the force equation (cuz friction is a force DUH)

This is how it eventually made sense to me:
force = ma
uN = ma
umg = ma
ug = a
u = a / g
u = -4 / 9.8
u = -0.41

Thanks y'all and also thanks Xanmyral who talked me through it on Discord too!!!

[Reelya]

Ricardo noticed that the bargaining power of laborers can never dip below the produce obtainable on the best available rent-free land, because whenever rent leaves them with less than they could get on that free land, they can simply move to the new location

Oh, to live in the 1800s when there was such a thing as "free land".

That said, Ricardian rent is a pretty cool theory.  Just make sure you adhere to its assumptions when applying it...

I think we're missing how generalizable it is. People have a house they live in, that is their land. It's reading it too literally to think that you need actual land for that to hold true. The assumptions of that were in the time when agriculture was still the dominant part of the economy. You needed land to grow things. But that isn't relevent to modern labourers.

The modern version is that the bargaining power of wages is constrained by what you could earn working for yourself. Sure, some people exceed wage-level working for themselves, but most would be unable to do so, so they choose to earn wages since that exceeds what they could make in self-employment. Visually, you can imagine that as an American refusing to take a job that would pay $1 an hour, even illegally, since it's literally "not worth their time" and they could in fact do some sort of labor or make something that exceeds $1 in value per hour.

[WealthyRadish]

I was playing minesweeper while rebooting my router, and came across an interesting question I hadn't considered before:

What is the probability of the tiles labelled A, B, C, and D containing a mine in this corner section of the board? (not knowing the total number of remaining mines or state of the rest of the board)

ABCD
2X3X

An 'X' indicates a known mine.
My first thought was to consider the number of unique arrangements, which would be:

X X
2X3X

X  X
2X3X

 X 
2X3X


-- and from this, using the assumption that each outcome is equally likely, conclude that the probabilities of containing a mine would be:
A: 2/3
B: 1/3
C: 1/3
D: 1/3

-- but I'm not certain if it's correct here to only consider unique combinations or if repeated permutations may matter (where the configurations with two mines would then have "double" weight), giving:

A: 4/5
B: 1/5
C: 2/5
D: 2/5

It seems to me that for this problem it would matter how the algorithm places the mines; if it's picking pseudo-random indices on the grid to place mines until it runs out (the most likely way to write minesweeper), then repeated permutations could matter, as they are at least distinct in the sequence in which the mines were placed. It seems absurd that A would be a mine 4/5 of the time empirically, however, and now I think that the underlying assumptions may have muddled the whole issue.

[Reelya]

Yup, also don't fall into the logical trap that if there are e.g. 4 possible unique outcomes then each one must have a 1/4 chance of occurring, because it's not certain, especially when interchangeable things are being distributed. A counter-example here is the Monty Hall problem with some people arguing that because only two choices remain, and those choices were evenly distributed to start with, the final chance must be 50/50.

For example, if the choices are X_X, X__X and _X__ then you can be pretty certain that the first two choices are equally likely, however the third option is not guaranteed to be equally likely to the others.

What you in fact need to do is count the number of ways that bombs could have fallen in those four squares and work out how many ways match each outcome. Say there were 10 bombs scattered originally, and label them A to J. You'd have to account for patterns such as A_J_ and J_A_ occuring, which means some patterns need to be double-counted and the like. The total amount of bombs original scattered can change the relative probabilities of local patterns, because you need to account for those patterns which give the same result.

[bloop_bleep]

I was reminded of this thread and wanted to share something interesting that I learned recently.

Specifically that measurable sets in R with positive measure are "locally intervals". In particular if you have a measurable set A subset of R with a measure greater than 0, then "almost every" (meaning all but a measure-zero set) x in A gives you that the limit of

mu(A intersect (x - epsilon, x + epsilon))/(2 * epsilon)

goes to 1 as epsilon goes to 0 (where mu is the measure in question.) This is very weird. Basically it's saying that around almost any x, A behaves increasingly like an interval as epsilon decreases. You can use this to prove this other interesting thing:

If A subset of R is a measurable set with positive measure, then for some positive epsilon the set {x - y | x, y in A } contains the interval (-epsilon, epsilon). Which I find rather surprising.

[Vector]

That IS rather surprising and I thank you for bringing it to my attention!

[MaximumZero]

I do not understand any of that. Please make either it dumber or me smarter. Also, I could have sworn that everyone moved on to another math thread.

[MagmaMcFry]

Consider a measurable subset A of R. Measurable implies that you can assign a measure, a real number (call it length or area or volume) to the intersection of A and any interval in a sensible way.

At any point x in R, A may or may not have a density at x, which is the limit, as an interval I containing x becomes arbitrary small, of the measure of the intersection of A and I divided by the length of I.

The claim is that this density is 1 at almost all points in A, which means that the subset of points in A where the density does not exist or is less than 1 has measure zero.

This statement implies for example that there is no measurable subset of R that has density 1/2 everywhere.

[da_nang]

Sounds almost like, in the range [0,1], the amount of "length" you gain at x with a small perturbation is equal to that perturbation.

In other words, dist(x,x+eps) = eps for almost all x in [0,1].

A bit of handwavey obvious intuition that makes you go "well, duh!" At least until you realize you need to prove it for every measurable subset and measure.

[bloop_bleep]

Sounds almost like, in the range [0,1], the amount of "length" you gain at x with a small perturbation is equal to that perturbation.

In other words, dist(x,x+eps) = eps for almost all x in [0,1].

A bit of handwavey obvious intuition that makes you go "well, duh!" At least until you realize you need to prove it for every measurable subset and measure.

... not quite. Yes measure of (x - epsilon, x + epsilon) is 2*epsilon, that's given. What it's saying is if you have positive measure set A, even if the measure is 1/2 or 1/4 or something like that, if you consider the intersection of A with the interval (x - epsilon, x + epsilon), that *intersection* is similar in measure to the *whole interval* (x - epsilon, x + interval) for small enough epsilon. The limit goes to 1, not the measure of A or something like that. That's the interesting part.

EDIT: I might have been using two different measures here; A has positive density in R, you could say, or A subset of [0, 1] has positive measure on [0, 1]. The concept is the same. I confused myself for a moment there. It works for A in R or [0, 1] or [0, 2] with positive Lebesgue measure. So intervals, the thicc Cantor set or stuff like that. Doesn't need to span all of R. Also note x in the statement is in A.

[Vector]

I do not understand any of that. Please make either it dumber or me smarter. Also, I could have sworn that everyone moved on to another math thread.

Y'all should know that MZ doesn't know what "measure" is. Maybe just a one-liner explaining why it's interesting to a lay person would be a good thing.

(I would do it myself but I'm too tired).