## Poll

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**23 members have voted**

Quote:ThatDonGuyI'm not entirely sure, as I am having a little problem simulating it to confirm the answer...

Let E(N) be the expected number of rolls needed if the N most recent are all different

Suppose the last 5 rolls are 1, 2, 3, 4, 5, in order.

If the next roll is 1, then you have 2, 3, 4, 5, 1; if it is 2, you have 3, 4, 5, 2; if it is 3, you have 4, 5, 3; if it is 4, you have 5, 4; if it is 5, you have 5; if it is 6, you are done.

E(5) = 1 + 1/6 E(1) + 1/6 E(2)+ 1/6 E(3) + 1/6 E(4) + 1/6 E(5) + 1/6 x 0

Similarly, if the last 4 are 1, 2, 3, 4, then a 5 or 6 results in a "run" of 5:

E(4) = 1 + 1/6 E(1) + 1/6 E(2)+ 1/6 E(3) + 1/6 E(4) + 1/3 E(5)

E(3) = 1 + 1/6 E(1) + 1/6 E(2)+ 1/6 E(3) + 1/2 E(4)

E(2) = 1 + 1/6 E(1) + 1/6 E(2)+ 2/3 E(3)

E(1) = 1 + 1/6 E(1) + 5/6 E(2)

E(0) = 1 + E(1)

The first five can be written as five equations in five unknowns:

1/6 E(1) + 1/6 E(2) + 1/6 E(3) + 1/6 E(4) - 5/6 E(5) = -1

1/6 E(1) + 1/6 E(2) + 1/6 E(3) - 5/6 E(4) + 1/3 E(5) = -1

1/6 E(1) + 1/6 E(2) - 5/6 E(3) + 1/2 E(4) = -1

1/6 E(1) - 5/6 E(2) + 2/3 E(3) = -1

-5/6 E(1) + 5/6 E(2) = -1

Solving results in E(1) = 52.8, so the expected number = E(0) = E(1) + 1 = 53.8

Also incorrect, but also hot on the trail.

Assuming this answer is correct (or close) I find it very interesting that the number of rolls is this high. Intuitively, I'm sure I would have guessed something much lower.

Quote:GialmereAlso incorrect, but also hot on the trail.

I think there's a bug in my equation solver - when I do it by hand, I get:

416/5 = 83.2

Found the problem in my equation solver: it panicked if, when trying to normalize row N, column N was already zero.

Quote:ThatDonGuyI think there's a bug in my equation solver - when I do it by hand, I get:

416/5 = 83.2

Found the problem in my equation solver: it panicked if, when trying to normalize row N, column N was already zero.

Correct!

It's interesting you had to solve it by hand. I guess that, even in this high tech age, John Henry will still occasionally beat the steam drill.

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So here's the question: Is there a value x=t so that if Gabriel's Horn goes from x=1 to x=t, then the volume and surface area of Gabriel's Horn are equal? If so, can you solve for this value?

Quote:teliotFor this question, watching this video is absolutely encouraged!

So here's the question: Is there a value x=t so that if Gabriel's Horn goes from x=1 to x=t, then the volume and surface area of Gabriel's Horn are equal? If so, can you solve for this value?

Cool video. Though it left my questioning why you can get away with using dx to integrate the volume instead of needing to use ds. Likewise they made a big deal of needing to use ds to integrate for surface area yet they could have ignored that complicating factor and used dx and still shown it was infinite. In fact that’s what they did at the end by showing integers of 1 to infinity of 2piR was infinite.

The paradox further messes with me. If pi units of paint fills the horn, then it must coat all the interior surfaces of the horn. Yet the interior surfaces of the horn have area of infinity. But we painted them!

As to your question, need coffee and paper and pencil before attempting. The video gives the necessary equations so shouldn’t be difficult to attempt to set them equal to one another.

I completely agree with this observation. That part of the video was lame for the infinite horn. On the other hand, maybe you need that full formula for the finite case? Come to think of it, maybe the whole problem is much tougher than I originally thought ... maybe showing that such an x=t exists is all you can do.Quote:unJonCool video. Though it left my questioning why you can get away with using dx to integrate the volume instead of needing to use ds.

And yes, filling the volume with paint should include the paint touching the surface, ergo painting it by default. That's what makes this a paradox.

Quote:teliotSo here's the question: Is there a value x=t so that if Gabriel's Horn goes from x=1 to x=t, then the volume and surface area of Gabriel's Horn are equal? If so, can you solve for this value?

The only value where they are equal is t = 1 (i.e. both volume and surface area are zero), and even this assumes that the disc of radius one at t = 1 is not included in the area.

The surface area = INTEGRAL[1, t] (2 PI / t) dt = 2 PI INTEGRAL[1, t] (1/t) dt = 2 PI (ln t - ln 1) = 2 PI ln t.

The volume = INTEGRAL[1, t] (PI / t^2) dt = PI INTEGRAL[1, t] (1/t^2) dt = PI (-1/t + 1/1) = PI (1 - 1/t).

These are equal when 2 ln t = 1 - 1/t.

2 ln 1 = 1 - 1/1, so they are equal when t = 1.

If t > 1, then they are equal when 2 ln t + 1/t - 1 = 0.

Let f(t) = 2 ln t + 1/t - 1

df/dt = 2/t + 1/t^2; since this is positive for all t > 1, f(t) is strictly increasing for all t > 1, which means f(t) > f(1).

If the disc is included, then the surface area = 2 ln t PI (2 ln t + 1), and they are equal when ln t = -1/t, but for t > 1, ln t > 0 and -1/t < 0.

Another thing that makes it paradoxical: if you have another horn that is y = 2/x rotated around the x-axis, the volume between the two is finite, so you can "paint the outside" of the original horn by filling the gap between the two.

Your solution was my original thought on this. But now, I disagree about the surface area calculation you've given ... thinking out loud, if you replace your formula with "The Surface Area > ..." and the rest of your argument works.Quote:ThatDonGuy

The only value where they are equal is t = 1 (i.e. both volume and surface area are zero), and even this assumes that the disc of radius one at t = 1 is not included in the area.

The surface area = INTEGRAL[1, t] (2 PI / t) dt = 2 PI INTEGRAL[1, t] (1/t) dt = 2 PI (ln t - ln 1) = 2 PI ln t.

The volume = INTEGRAL[1, t] (PI / t^2) dt = PI INTEGRAL[1, t] (1/t^2) dt = PI (-1/t + 1/1) = PI (1 - 1/t).

These are equal when 2 ln t = 1 - 1/t.

2 ln 1 = 1 - 1/1, so they are equal when t = 1.

If t > 1, then they are equal when 2 ln t + 1/t - 1 = 0.

Let f(t) = 2 ln t + 1/t - 1

df/dt = 2/t + 1/t^2; since this is positive for all t > 1, f(t) is strictly increasing for all t > 1, which means f(t) > f(1).

If the disc is included, then the surface area = 2 ln t, and they are equal when ln t = -1/t, but for t > 1, ln t > 0 and -1/t < 0.

I guess I have always disagreed with this as a "paradox". The problem I always had was with equating surface area to volume. Gabriel's Horn hides this in a calculation where we concentrate on one integral being finite but another is infinite. But consider the x-y plane. The surface area of that is infinite, but the volume of paint required to "paint" it is 0. In the case of Gabriel's Horn, you go out and buy your can of paint with volume, pi "units", you paint the inside (or outside) using up 0 "units", and you still have pi "units" of paint left to fill the horn with.

As to Teliot's original question, I agree with the calculations Teliot and ThatDonGuy discussed...but if I go back to comparing surface area to volume, does equating them have some physical or geometric meaning I am missing?