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[FearfulJesuit]

I love maths. Do you love maths?

I'd like to start with a conjecture that came upon me whilst doodling in class a few days ago. Since LaTeX isn't supported here, I'll just link to the post I made on the XKCD forums.

Anyway, is anybody else here some variety of maths nerd?

[EDIT] There is now a prize for solving my conjecture, just like for all sorts of other conjectures. However, instead of getting a nice new shiny Fields Medal or a million dollars, you get a chicken hat. Oh well.

[Max White]

I miss Vector too, but we have to move on...

[mainiac]

Take the case of a=0
f(x)=2x

I[f(x)]=1
I[I[f(x)]=1
...

Doesn't solve the general case but it gives me the suspicion that this problem isn't as complicated as it seems on the surface.

[ed boy]

I haven't read through the whole XKCD thread, so I might be repeating some point that appears there.

I would be inclined to disagree with you there, on the question of the existance of such a limit of the stack of integrals.

Let An and Bn be sequeces of real numbers such that
An is the integral between 0 and A(n-1) of f(x), with A0=1
Bn is the integral between 0 and B(n-1) of f(x), with B0=1/2

Now let f(x) be the cantor function multiplied by two (There are probably a lot simpler examples, but I happened to be thinking of the cantor function at the time).

A0=1, and the integral of the cantor function over [0,1] is 1/2, so A1=1. Similarly, An=1 for all n.
Now note than B0=1/2. We can see that B1 is going to be less than half of A1, and since the cantor function is positive, we get that Bn will be at most 1/2 for all n (In fact, I woudl conjecture that Bn tends to zero).

As you can see, when n goes to infinity you get an infinite stack of integrals in both cases, but they look very different.

(Also, maths undergrad here)

[Virex]

OK, take 2 on this.


First, let's introduce a notation that is actually readable without a Tex interpreter:


I[a,b->n]{f(x) dx} means a stack as you described, of a height of n integrals with a base of a and an upper value for the top integral of b (so I'm handling a more general case than the case you detailed, which is for b = 1)


We want to know if the following limit converges for any value of a and b:


Limit 1: lim (n->inf) I[a,b->n]{f(x) dx}




I will first attempt to find the properties of f(x) for which limit 1 will be bound, as that is easier than to prove it will actually converge. I also assume that a and b are non-complex.


1.) If for any combination of a and b, I[a,b->1]{f(x) dx} is undetermined, we cannot evaluate Limit 1 for that combination of a and b.
Therefor f(x) must be continuous and exist for all x in R.


2.) For limit 1 to converge, for every value of a, there must exist a value b' and b'' for which I[a,b->1]{f(x) dx} =< b >= b''-b for any b => b' and I[a,b->1]{f(x) dx} =>b <= b for any b <= b''. b and b'' may be the same value.


Any function that adheres to these requirements should cause limit 1 to be bound.

[MagmaMcFry]

Let g_a(b) = I[a,b]f(x)dx = F(b) - F(a).
Then your question is: Is there an f(x), so that for every real a, repeated application of g_a onto 1 will result in a convergent sequence?
The answer is yes: Let c be a real number in the range (-1,1). Then f(x) := c satisfies the conditions.
Proof: Let (k(n)) be this sequence, with k(0) := 1 and k(n) := g_a(k(n-1)). Then F(x) = cx, therefore g_a(b) = cb - ca, and k(n) = ck(n-1) - ca, therefore k(n) converges to ac/(c-1).

Q.E.D, bitches.

[Shook]

Maths? Ah luff me sum trigs. It's really useful in a huge number of ways, and i actually have a solid understanding of it (because it's so basic, herf durf). Sine, cosine and tangent are super good things that have served me in many ways during both modding and game developing.

[Virex]

Let g_a(b) = I[a,b]f(x)dx = F(b) - F(a).
Then your question is: Is there an f(x), so that for every real a, repeated application of g_a onto 1 will result in a convergent sequence?
The answer is yes: Let c be a real number in the range (-1,1). Then f(x) := c satisfies the conditions.
Proof: Let (k(n)) be this sequence, with k(0) := 1 and k(n) := g_a(k(n-1)). Then F(x) = cx, therefore g_a(b) = cb - ca, and k(n) = ck(n-1) - ca, therefore k(n) converges to ac/(c-1).

Q.E.D, bitches.

Wait, that can't be true. If we start the sequence from the top instead from the bottom, we get I[a,1]{f(x) dx} > 1 for a < 0 or a > 2, so the sequence diverges for some value of a.

[MagmaMcFry]

Erm... what? Start the sequence from the top? Could you be a little more formal? Also, g_a(1) > 1 doesn't imply that the sequence diverges.

[darkrider2]

Maths? Ah luff me sum trigs. It's really useful in a huge number of ways, and i actually have a solid understanding of it (because it's so basic, herf durf). Sine, cosine and tangent are super good things that have served me in many ways during both modding and game developing.

I once played a programming game (they exist) called mindrover, and I was finally able to pinpoint x,y coords for the other robots after I understood trig. Super useful maths.

[Virex]

Erm... what? Start the sequence from the top? Could you be a little more formal? Also, g_a(1) > 1 doesn't imply that the sequence diverges.

Well, if we define the sequence as I[a,b]{f(x) dx}, I[a,I[a,b]{f(x) dx}]{f(x) dx}, I[a,I[a,I[a,b]{f(x) dx}]{f(x) dx}]{f(x) dx}...
Then this can be evaluated by treating the inner integrals first and working our way outwards. Since the sequence is the same, this should yield the same result. However, if we do that for f(x) dx, we see that for a value of a < 1-1/c, I[a,1]{f(x) dx} > 1, so I[a,I[a,b]{f(x) dx}]{f(x) dx} >  I[a,1]{f(x) dx} and therefore  I[a,I[a,I[a,b]{f(x) dx}]{f(x) dx}]{f(x) dx} > I[a,I[a,b]{f(x) dx}]{f(x) dx}. From this it becomes obvious that the sequence cannot converge.

Edit: a < 1-1/c of course, not a < 0.

[ed boy]

It might not be a convergent sequence, but it'll be bounded and thus it'll have a convergent subsequence, which is better than nothing.

[MagmaMcFry]

Well, if we define the sequence as I[a,b]{f(x) dx}, I[a,I[a,b]{f(x) dx}]{f(x) dx}, I[a,I[a,I[a,b]{f(x) dx}]{f(x) dx}]{f(x) dx}...
Then this can be evaluated by treating the inner integrals first and working our way outwards. Since the sequence is the same, this should yield the same result. However, if we do that for f(x) dx, we see that for a value of a < 1-1/c, I[a,1]{f(x) dx} > 1, so I[a,I[a,b]{f(x) dx}]{f(x) dx} >  I[a,1]{f(x) dx} and therefore  I[a,I[a,I[a,b]{f(x) dx}]{f(x) dx}]{f(x) dx} > I[a,I[a,b]{f(x) dx}]{f(x) dx}.

That's all nice and correct if you limit yourself to positive c.

From this it becomes obvious that the sequence cannot converge.

Ah, here's where you've totally lost it.

For example, take the sequence (2-2^-n).
The sequence looks like this: 1, 1.5, 1.75, 1.875, 1.9375, 1.96875, ...
See a pattern? The sequence is increasing and all members are >= 1, but it still converges to 2.

Note: if you choose a=-2 and c=1/2, you will get exactly that sequence.

[Virex]

Well, if we define the sequence as I[a,b]{f(x) dx}, I[a,I[a,b]{f(x) dx}]{f(x) dx}, I[a,I[a,I[a,b]{f(x) dx}]{f(x) dx}]{f(x) dx}...
Then this can be evaluated by treating the inner integrals first and working our way outwards. Since the sequence is the same, this should yield the same result. However, if we do that for f(x) dx, we see that for a value of a < 1-1/c, I[a,1]{f(x) dx} > 1, so I[a,I[a,b]{f(x) dx}]{f(x) dx} >  I[a,1]{f(x) dx} and therefore  I[a,I[a,I[a,b]{f(x) dx}]{f(x) dx}]{f(x) dx} > I[a,I[a,b]{f(x) dx}]{f(x) dx}.

That's all nice and correct if you limit yourself to positive c.

From this it becomes obvious that the sequence cannot converge.

Ah, here's where you've totally lost it.

For example, take the sequence (2-2^-n).
The sequence looks like this: 1, 1.5, 1.75, 1.875, 1.9375, 1.96875, ...
See a pattern? The sequence is increasing and all members are >= 1, but it still converges to 2.

Erm, yes. you're right. I need to go sit in this corner for a while...

Note: if you choose a=-2 and c=1/2, you will get exactly that sequence.

You can't chose a. The challenge is to find an f(x) that converges for any a.

[MagmaMcFry]

Note: if you choose a=-2 and c=1/2, you will get exactly that sequence.

You can't chose a. The challenge is to find an f(x) that converges for any a.

Yes, and f(x) := 1/2 will converge for any a, including a = -2. It just happens that the resulting sequence (k(n)) is exactly (2-2^-n).