On the contrary, mine does not round the wrong way, as it does not round at all! It truncates!
One might define the “diameter” of a circle lying on a sphere as the length of an arc segment passing through its center and also lying on the sphere, which I suppose one might be able to manipulate in order to get the ratio to come out to exactly three. But there’s no indication in the passage you cite that the authors meant one should measure the distance from one side of a concave vessel to the other by measuring along the closed bottom surface of said vessel rather than using the intuitive method of measuring across its open top. I’ll stick with the above until a different conclusion can be demonstrated.
A circle in spherical geometry is not the same as a circle on the surface of a sphere. Rather, it resembles two circles of equal diameter running parallel to a great circle (I think–this is a semi-educated guess), and its radius is the distance from any point on either circle to the closest pole. I’m not gonna work out the area or circumference right now, but I suspect that Chronos is correct.
In that case it seems to me we now have two questions:
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Might a circle in spherical geometry as described above have a diameter exactly three times its radius?
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Could there be a physical object such as a vessel for holding liquid in three-dimensional, non-spherical Euclidian space that could have a diameter exactly three times its radius?
Even if the answer to 1. is, “yes,” I’m sticking to my guns that the answer to 2. would still be, “not unless we redefine some terms.”
I presume that you meant “circumference three times its diameter” there, Tom. Diameter is always twice radius, by definition.
For a circle in spherical geometry, the ratio of circumference to diameter can be anything greater than or equal to 2 and less than the Euclidian value of [symbol]p[/symbol], depending on the size of the circle, so 3 is a possibility.
As for you second question in normal, 3-d Euclidean space, it depends on how one defines “the distance across” something. If you find the distance across a bowl by stretching a string across, then you’re in flat geometry, and [symbol]p[/symbol] = 3.14159265358979323846… . If you do it by rolling a measuring wheel along the bottom of the bowl, then it’s less.
By the way, I was not trying to imply that this was the True Explanation for the passage in Kings: Roundoff error is a far more plausible explanation. I was just saying that it could be an answer.
Yeah, whatever. BTW, whatever happened to:
`` . . . it’s actually possible that those measurements are exactly and precisely correct. The object in question is bowl-shaped, so if measurements are restricted to the surface of the bowl, then we’re in spherical geometry, and pi is less than the Euclidian value.’’
Pi refers to exactly one number, the ratio of the circumference of a circle to its diameter in Euclidean space.
Mmmm… pi.
Incidentally, the appproximations for [symbol]p[/symbol] come from its continued fraction expansion
[symbol]p[/symbol] = [3,7,15,1,292,…].
The 1st convergent is 3 + 1/7 = 22/7. This is the best possible approximation to [symbol]p[/symbol] by a rational number with denominator no greater than 56.
The 2nd is 3 + 1/(7 + 1/15) = 333/106
The 3rd is 3 + 1/(7 + 1/(15 + 1/1)) = 355/113. This approximation differs from [symbol]p[/symbol] by less than 5*10[sup]-7[/sup] and was discovered by the Chinese mathematician Chao Jung-Tze c. AD 500.
The 4th is 3 + 1/(7 + 1/(15 + 1/(1 + 1/292))) = 103993/33102.
After that it starts to get silly.
I just remember e^(pi i) + 1 = 0 and solve for whichever term I need.
[symbol]p[/symbol] doesn’t have to be represented by a continued fraction. There are books filled with various expansions for [symbol]p[/symbol], some of them extremely clever and some impossible (for me) to understand how anyone could come up with them.
Here’s a simple one:
[symbol]p[/symbol] = 4(1 – 1/3 + 1/5 – 1/7 + …)
And it doesn’t even have to be a summation. You can find [symbol]p[/symbol] through multiplication (and square roots), as Euler did:
[symbol]p[/symbol] [sup]2[/sup] = 6[2[sup]2[/sup] / (2[sup]2[/sup] – 1) x 3[sup]2[/sup] / (3[sup]2[/sup] – 1) x 5[sup]2[/sup] / (5[sup]2[/sup] – 1) x 7[sup]2[/sup] / (7[sup]2[/sup] – 1) x …]
Usually, you don’t start by saying “I want to know what the value of [symbol]p[/symbol] is”. You start with the infinite series, or the infinite product, or a definite integral, or whatever it is, and you want to find the value of it. Eventually, your number crunching leads you to some trignometric function, and that function gives you an answer expressed in terms of [symbol]p[/symbol]. If you want to publish a book, you then say “Aha! That means that I can also express [symbol]p[/symbol] as (long complicated expression that was difficult to solve in the first place)”.
I memorized pi to 200 digits when I was in high school. Do I get a cookie?
No, you get a piece of pi.
:smack: