Let's discuss Mathematics

[ParadigmShifter]

Yeah ok that looks right

I only did a bit of differential equations in further maths A-level, of the form

Af(x) + Bf '(x) + Cf ' '(x) = D

 

[Mise]

Well I gave up and did it on a spreadsheet.

FU, algebra.

 

[ParadigmShifter]

I did a bit of the easy case (2 trapeziums equal area) and it was pretty long winded, yeah

I did get a formula for the intermediate point x as a function of m and c though. It's not pretty. Obviously it involves the other endpoint x's too but those are constant.

 

[ParadigmShifter]

The full formula I got was (x₀, x₁, x₂ are the points on the x axis, solve for x₁)

2mx₁² + 4cx₁ = 2cx₀ + 2cx₂ - mx₀² + mx₂²

Plug into quadratic formula... that's when it gets messy

The only sanity checking I did was verify the m = 0 case, i.e. a constant function... reduces to

4cx₁ = 2cx₀ + 2cx₂

=> 2cx₁ = cx₀ + cx₂

=> 2x₁ = x₀ + x₂

=> x₁ = (x₀ + x₂)/2

as expected.

EDIT: Whoops mistyped 0 as 1 before

EDIT2: Did another sanity check... chose the y = x case between 0 and 1, comes up with 1/(sqrt(2)) as the answer, which is correct.

 

[wer]

So there's no way of working out the x's without going through each step? There's no function that will magically generate them for me? Oh well.

That's elementary measure theory.

The magical function is called quantile function and it is nothing but the inverse of the standardized distribution function generated by your function.

If you start with normalizing your function to probability density, you can use all the instruments of probability theory designed for your very problem. (I guess you are more likely to find sources on quantil and distribution functions in probability theory than the genaralized concepts in measure theory.)

 

[ParadigmShifter]

*irrelevant stuff deleted*

EDIT2: Ahh it's the inverse of the cumulative density function, gotcha.

 

[ParadigmShifter]

Sooo... if the are under the curve is normalised (total area = 1) and we scale the endpoints so they are 0, 1 we get the CDF is integral(0 to x) of (mt+c)dt

which is 0.5mx^2 + cx

and wolfram alpha gives me the inverse of that as

So solve that for 1/N, 2/N, 3/N for N trapeziums then presumably.

EDIT: Wolfram Alpha kicks donkey

EDIT2: That looks wrong though... If m = 2, c = 0 (so area between 0 and 1 is 1) then this calculation

http://www.wolframalpha.com/input/?i=solve+(c/m+sqrt(c^2+2.+m+x)/m)+=+0.5

gives x = 1/4 which can't be correct

EDIT: Looks like I got the input wrong somehow. Working it through I get the correct answer x = 1/sqrt(2)

Whoops, you end up with 2sqrt(x) = 1 so x = 1/4. That isn't correct...

 

[wer]

Ahh it's the inverse of the cumulative density function

Cumulative density is oxymoron.

gotcha.

Sooo... if the are under the curve is normalised (total area = 1) and we scale the endpoints (...)
Whoops, you end up with 2sqrt(x) = 1 so x = 1/4. That isn't correct...

No, that’s you who ends up with incorrect solution. Your inverse is wrong (check minuses), your equation is wrong (the inverse appears in the solution, not in the equation) and your scaling of the endpoints is redundant. This way:


Let µ be the measure induced by non-negative integrable intensity function f defined on the interval

, i.e.

for every measurable subset M of Ω.

We are looking for points

such as

for i=1,...,n.

Let F be any primitive function of f, then the distribution function

of the measure µ is defined as

Then for every i=0,...,n

i.e.

and finally

(In the probabilistic approach we work analogically with measure

, distribution function

and density

)



In our particular case we have f(x)=mx+c, hence F(X) could be for example

and

is the one of the two solution of the equation

which lies in Ω.

 

[ParadigmShifter]

Oh, everyone is a critic!

Just kidding

 

[Mise]

Wow, that was incredibly straightforward. Thanks wer!!

Gave the same answer as my recursive spreadsheet

 

[Atticus]

It's almost Friday, so I'll get the party started with little exercise:
Show that there are Lebesgue measurable sets which aren't Borel sets.

 

[ParadigmShifter]

That's probably a tad too technical I did pure maths for 3 years at Uni and we never touched measure theory and I have no idea what a Borel set is. I'll wiki it of course.

 

[Atticus]

Hmmm, it seems so. Our department was focused on analysis, so I naturally assume that everyone with education in maths know these things too. I've noticed that it isn't true, but manage always to forget it.

Here's the answer:

Spoiler :

Let's call the non-measurable set E (and let's assume that we already know what it is). Because E isn't measurable, it can't be a Borel set, so neither is A:={0}XE in R^2. Now A's pre-measure is 0, so according to Caratheodory's criterion it's measurable.

Borel-sets are simply those that can be obtained as countable intersections or unions of open and closed sets. In practice that means every set that comes across. The question I posted was meant to be introduction to couple questions more, to which I don't know the answer:
1. Is it possible to construct a Borel set without the axiom of choice?
2. How do you prove that without the axiom of choice every set in R is measurable? (I've heard a rumor that this is the case).

Let's hope we have a measure theory guru here, who has just been busy with civ 5 lately, and comes to solve all our problems.

 

[pboily]

1. Yes. Consider the set {0}. It is a Borel set, being a finite set (hence closed). The axiom of choice wasn't used in it's construction. I think there's a typo in your question, unless I misread something.

2. I am no guru, but I do know that if you assume AC, then there are non-(Lebesgue) measureable subsets of R. I think the best (and equivalent) thing that could be said is that if every subset of R was Lebesgue measurable, then AC couldn't possibly hold. But that is a far cry from saying that if AC doesn't hold, then all sets are Lebesgue measurable.

EDIT: I seem to recall that if the continuum hypothesis holds, then it is for sure impossible to have a translation-invariant, interval-respecting, countably additive measure on the power set of R: some set has to be non-measurable. But that's clearly not what you had in mind.

If you replace countable additivity by finite additivity, then the resulting measure is such that every subset is measurable. And if you replace countable additivity with countable sub-additivity, then the resulting measure (the outer measure) is such that every subset is measurable. But in both instances, that is not the Lebesgue measure.

EDIT the second: The rumour you heard seems to have been true: Solovay's model "only" requires an inaccessible cardinal.

 

[Atticus]

Unfortunately I'm too sleep deprivated at the moment to understand that wikipage. And probably it's pretty hard anyhow, since I'm not that good with logic.

There indeed was a typo in the first question, I meant to ask, whether it is possible to construct non-Borel set without the use of AoC. I thought this question could be good way to attack the question #2.

 

[pboily]

Sleepy or not, I don't think anyone could understand the nitty-gritty of that page in less than a few days, at the very least. This is one of those instances (like Wiles' proof, say) where I will trust the work of the author and the referees.

 

[Atticus]

It seems so. At that moment even your post was difficult to read, so I didn't quite notice, how difficult that wikipage could be. Now on second inspection I noticed the word "forcing", and understood that there's no chance I could understand it without some studies.

 

[ParadigmShifter]

Mandelbrot has died

Tribute thread started in OT

http://forums.civfanatics.com/showthread.php?t=392160

 

[Spoonwood]

Mandelbrot has died

Tribute thread started in OT

http://forums.civfanatics.com/showthread.php?t=392160

Julia has been dead for years. Menger has been dead for years. Koch has been dead for over a century. Cantor... sorry I mean Henry John Stephen Smith... has been dead for years. Sierpinski is long gone also. Hilbert lives in Cantor's Paradise these days. Peano is dead.

Which fractal do you think hardest to invent or discover? Do you have a fractal which you like best, and if so, which one?

 

[ParadigmShifter]

I think the Mandelbrot set is my fave since it isn't entirely self similar.

Of the self similar ones my favourites are the Sierpinski Triangle

the Menger Sponge

and the Barnsley Fern

EDIT: You forgot Poincarre too