Pages: 12
Assuming paint molecules are infinitely small, how much paint do you need to fill the volume of Gabriel's Horn? What happens when you pour the paint out? Have you coated the entire inside surface area?

(Do not reference the Wikipedia explanation, which I found after asking this question)
Last edited on
Surface area of Gabriels horn is infinite so you wound need an infinite amount of paint. And paint molecules are not infinitely small.

Edit: Actually, IIRC the volume is finite, so I guess you could coat the inside SA.
Last edited on
Also, you haven't ever painted the outside. XP
Wow I had never heard of Gabriel's Horn before. That's pretty interesting.
Came up in calc2 for me.
We were reviewing stuff in my AP Calculus BC class at my high school (we're preparing for the AP Test).

@naraku: Theoretical paint molecules are infinitely small in this case.

You're right that you can coat the inside. With infinitely small paint molecules, an infinite number of molecules would fill the finite volume, and with infinite paint molecules you have enough molecules to coat the infinite inside surface area and even turn it inside out.

Wikipedia's solution is that the paint would have to get thinner infinitely, and it also says it can't all be with the same thickness. This is, of course, not with theoretical infinitely small paint molecules.

Oh, and after you filled it, you'd have to make sure it didn't create a black hole, being infinitely dense and all.
> being infinitely dense and all.
┬┐where did you get that part?
If a paint molecule is infinitely small, it would require an infinite number of molecules to reach any finite number.
You're right, that's my mistake. Just because they're infinitely small doesn't mean the spacing between them is too - I guess that complicates my explanation quite a fact I think my whole explanation is wrong because of that. I'll do some thinking.

Good catch ne555!
I just looked up Gabriel's horn. It looks like a 3d version of the Dirac delta function. The Dirac delta function is a basic principle in calculus which says that the integral (area) under 1 dimensional line is 1. We use that in communication theory quite a bit when representing a simple sinusoid in the frequency domain.
the function is 1/x, quite different

> The Dirac delta function is a basic principle in calculus
┬┐? it's a really weird function, f(x) = 0, \forall x not_eq 0 and the integral in all the domain is 1

> when representing a simple sinusoid in the frequency domain
I suppose that you use the discrete version
Hmm... Well, OP, it would appear that you answered your own question. If the paint is infinitely small, then the question does not apply. If the paint is not infinitely small, then it simply cannot fill the entirety of the horn.

So, since the question appears to have been answered, I feel it to be appropriate to pose a new question- what is the sum where n goes from 1 to infinity of 1/n2? Preferably without trying to look up the solution on, say, Wikipedia. That takes all of the mathematics out of the mathematics- knowing the solution beforehand defeats the point.
ne555, it doesn't have to be discrete. The theoretical or analogue stuff can be done continuous. It's only when you go digital that you need to take the discrete FTs.
what is the sum where n goes from 1 to infinity of 1/n2?
Damn! I wrote a paper on function used in calculation of this in university, so I already know the answer.

I was late for first question discussion and I already know the answer for second. Guess I am just unlucky.
Ispil wrote:
what is the sum where n goes from 1 to infinity of 1/n2?
I happen to remember this by chance, Pi squared over six. Or maybe that's when n starts at 0.
Last edited on
LB you are right.
Or maybe that's when n starts at 0.
1/(02) makes so much sense :)
Ah, heh, yeah. So, it was when n started at 1 - I was remembering it backwards. Still pretty good seeing as I didn't even do anything in my head, I just happened to have done it before and remembered the interesting result because of ho rare it is to see Pi squared.
Well, do you remember how to get to that result? There are several ways not using any calculus or other advanced concepts, but the method I find easiest to understand is the one using Fourier series.
This is a programming site.

Just go in a loop, when the difference in calculation between this iteration and last iteration is less than your tolerance, stop. NUMERICAL METHODS :)

double answer = 0, prev = 0;
for (int i = 1; true ; ++i)
    prev = answer;
    answer += 1.0 / ( (double)( i * i ) );
    if (abs(answer - prev) < tolerance ) 

Otherwise use calculus.
Last edited on
But you will only get a bunch of digits. How could you know what number exactly do they represent?
Pages: 12