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I've thought it was paradoxical that infinitely long curves could have finite integrals ever since I first took calculus. For example, the integral of 1/x is |l

by jveld 11y ago

I've thought it was paradoxical that infinitely long curves could have finite integrals ever since I first took calculus. For example, the integral of 1/x is |ln x|. I wonder why it takes three dimensions before people start getting upset.

parallel 11y ago
Or even just a convergent infinite series. Infinitely many positive numbers that sum to a finite number.
- reagency 11y ago
  Or in general, any infinite size n-dimensional structure can be embedded in some finite size n+1 dimensional structure.
- wz1000 11y ago
  I don't think that's paradoxical. We deal with convergent infinite series all the time in everyday life. 10/3(3.333...) is 3 + 3/10 + 3/100/ + 3/1000 + 3/10000... π(3.1415...) is 3 + 1/10 + 4/100 + 1/1000 + 5/10000... Its quite easy to see that both of these series will be finite, and many numbers, lets say 4 or 3 + 5/10 will be greater than either of them.
- thaumasiotes 11y ago
  People don't tend to find that unintuitive. The Greeks had a nice geometric example in a square of side length 1: ┌┬┬─┬───┬───────┐ ├┘│3│ │ │ ├─┘-│ │ │ │ 64│ │ │ ├───┘ │ │ │ │ │ │ 3/16 │ │ │ │ │ ├───────┘ │ │ │ │ │ │ │ │ 3/4 │ │ │ │ │ │ │ └───────────────┘ 3/4 + 3/16 + 3/64 + 3/256 + ... is easy to visualize as successively filling in three quarters of an ever-smaller residual square. Intuitively, no matter how finely you detail it, you're never going to stop fitting inside the original area-1 square. edit: better text art
  - lmm 11y ago
    Now squeeze the top-left piece so it's half as wide but twice as tall (i.e. the same height as the full rectangle), and do this recursively. Same area, right? Then stack the L shapes on top of each other. Then you have one of these horns.
    - thaumasiotes 11y ago
      Huh? The horn is a three-dimensional object; the square exists in 2-space. You can't make the horn from pieces of the square even if you allow deformation.
- raldi 11y ago
  This sum's finiteness seems pretty intuitive: 9 + 0.9 + 0.09 + 0.09 + 0.009...
  - sesquipedalian 11y ago
    Good example. Minor nitpick, you listed 0.09 twice.
crimsonalucard 11y ago
The paradox is not the convergence of infinite long curves.. Read the paradox description. To fill the horn with paint you need a finite amount. To paint the outside of the horn you need an infinite amount. That is the paradox. The surface area of the inside is equal to the outside, yet one side requires a finite amount of paint while the other side requires an infinite amount. At least that's how I understand the paradox.
- om2 11y ago
  That's not really the paradox. The amount of paint required depends on the thickness of the coat, and the inside is forced to get a thinner and thinner coat (since the space available gets thinner) while the outside is assumed to be painted with an even thickness of paint. If you painted the outside with a layer of paint whose thickness is proportional to the thickness of the curve, you could do it with a finite amount. (This is all assuming idea "paint" that is continuous and arbitrarily subdivisible, real paint is made of molecules.) The intended "paradox" is that the surface area is infinite but the volume is finite.
- gohrt 11y ago
  It's the same thing. Imagine that you need one blob of paint for every tick-mark on the real axis. Infinite blobs of paint. (The resolution, in both scenarios, is that the size of a blob is not constant, and goes to infinitesimal)
- xerophyte12932 11y ago
  From the article: >Since the Horn has finite volume but infinite surface area, it seems that it could be filled with a finite quantity of paint, and yet that paint would not be sufficient to coat its INNER surface
UncombedCoconut 11y ago
I think you're misunderstanding the example. An anti-derivative of 1/x is ln |x| (very different from |ln x|) when the domain excludes x=0. By the fundamental theorem of calculus, you can use it to integrate over intervals that exclude x=0. (On positive-real intervals F(x)=ln(x) works, and on negative-real intervals F(x)=ln(-x) works. In either case, F(x)=ln|x| is an equivalent formula. If you try a domain that includes x=0, you risk not just a problem with technicalities but also the practical problem that F(x)=ln|x| and G(x)=ln|x|+sgn(x) look no different...)
nhaehnle 11y ago
I'm not sure how you intended the example, but it's probably a bad one: If you take the integral of 1/x, say from x=1 to infty, you don't get a finite amount (because ln x goes to infinity). On the other hand, if you take 1/x^2 as the integrand, you get the desired effect: its antiderivative is -1/x, so the integral of 1/x^2 from x=1 to infty is finite (actually, it's 1). By the way: For good fun with convergent series, take a look at the problem of escaping a lion in a circular arena at equal maximum speeds. Here's an example link to both question and answer: http://puzzling.stackexchange.com/questions/8140/escaping-a-hungry-lion-you-cant-outrun http://puzzling.stackexchange.com/questions/8140/escaping-a-... IIRC, there is an even more fascinating story associated to the history of this problem, as the first published solution was actually incorrect.
- jveld 11y ago
  Thanks for the correction! It's been a couple years since I've actually done any calculus. I couldn't remember the exact function, so I had to google - I guess I read too quickly.

by jveld 11y ago

parallel 11y ago

Or even just a convergent infinite series. Infinitely many positive numbers that sum to a finite number.

reagency 11y ago
Or in general, any infinite size n-dimensional structure can be embedded in some finite size n+1 dimensional structure.
wz1000 11y ago
I don't think that's paradoxical. We deal with convergent infinite series all the time in everyday life. 10/3(3.333...) is 3 + 3/10 + 3/100/ + 3/1000 + 3/10000... π(3.1415...) is 3 + 1/10 + 4/100 + 1/1000 + 5/10000... Its quite easy to see that both of these series will be finite, and many numbers, lets say 4 or 3 + 5/10 will be greater than either of them.
thaumasiotes 11y ago
People don't tend to find that unintuitive. The Greeks had a nice geometric example in a square of side length 1: ┌┬┬─┬───┬───────┐ ├┘│3│ │ │ ├─┘-│ │ │ │ 64│ │ │ ├───┘ │ │ │ │ │ │ 3/16 │ │ │ │ │ ├───────┘ │ │ │ │ │ │ │ │ 3/4 │ │ │ │ │ │ │ └───────────────┘ 3/4 + 3/16 + 3/64 + 3/256 + ... is easy to visualize as successively filling in three quarters of an ever-smaller residual square. Intuitively, no matter how finely you detail it, you're never going to stop fitting inside the original area-1 square. edit: better text art
- lmm 11y ago
  Now squeeze the top-left piece so it's half as wide but twice as tall (i.e. the same height as the full rectangle), and do this recursively. Same area, right? Then stack the L shapes on top of each other. Then you have one of these horns.
  - thaumasiotes 11y ago
    Huh? The horn is a three-dimensional object; the square exists in 2-space. You can't make the horn from pieces of the square even if you allow deformation.
raldi 11y ago
This sum's finiteness seems pretty intuitive: 9 + 0.9 + 0.09 + 0.09 + 0.009...
- sesquipedalian 11y ago
  Good example. Minor nitpick, you listed 0.09 twice.

crimsonalucard 11y ago

The paradox is not the convergence of infinite long curves.. Read the paradox description. To fill the horn with paint you need a finite amount. To paint the outside of the horn you need an infinite amount. That is the paradox. The surface area of the inside is equal to the outside, yet one side requires a finite amount of paint while the other side requires an infinite amount. At least that's how I understand the paradox.

om2 11y ago
That's not really the paradox. The amount of paint required depends on the thickness of the coat, and the inside is forced to get a thinner and thinner coat (since the space available gets thinner) while the outside is assumed to be painted with an even thickness of paint. If you painted the outside with a layer of paint whose thickness is proportional to the thickness of the curve, you could do it with a finite amount. (This is all assuming idea "paint" that is continuous and arbitrarily subdivisible, real paint is made of molecules.) The intended "paradox" is that the surface area is infinite but the volume is finite.
gohrt 11y ago
It's the same thing. Imagine that you need one blob of paint for every tick-mark on the real axis. Infinite blobs of paint. (The resolution, in both scenarios, is that the size of a blob is not constant, and goes to infinitesimal)
xerophyte12932 11y ago
From the article: >Since the Horn has finite volume but infinite surface area, it seems that it could be filled with a finite quantity of paint, and yet that paint would not be sufficient to coat its INNER surface

UncombedCoconut 11y ago

I think you're misunderstanding the example. An anti-derivative of 1/x is ln |x| (very different from |ln x|) when the domain excludes x=0. By the fundamental theorem of calculus, you can use it to integrate over intervals that exclude x=0. (On positive-real intervals F(x)=ln(x) works, and on negative-real intervals F(x)=ln(-x) works. In either case, F(x)=ln|x| is an equivalent formula. If you try a domain that includes x=0, you risk not just a problem with technicalities but also the practical problem that F(x)=ln|x| and G(x)=ln|x|+sgn(x) look no different...)

nhaehnle 11y ago

I'm not sure how you intended the example, but it's probably a bad one: If you take the integral of 1/x, say from x=1 to infty, you don't get a finite amount (because ln x goes to infinity). On the other hand, if you take 1/x^2 as the integrand, you get the desired effect: its antiderivative is -1/x, so the integral of 1/x^2 from x=1 to infty is finite (actually, it's 1). By the way: For good fun with convergent series, take a look at the problem of escaping a lion in a circular arena at equal maximum speeds. Here's an example link to both question and answer: http://puzzling.stackexchange.com/questions/8140/escaping-a-hungry-lion-you-cant-outrun http://puzzling.stackexchange.com/questions/8140/escaping-a-... IIRC, there is an even more fascinating story associated to the history of this problem, as the first published solution was actually incorrect.

jveld 11y ago
Thanks for the correction! It's been a couple years since I've actually done any calculus. I couldn't remember the exact function, so I had to google - I guess I read too quickly.