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Topic: pi in various metrics (Read 1882 times) |
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JocK
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pi in various metrics
« on: Dec 8th, 2005, 2:30pm » |
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The metric on a plane is given by a general vector norm.
What norm leads to pi = 4?
What norm was once used by the State of Indiana legislature? 
What is the minimum and maximum values of pi that can be reached for arbitrary norms?
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| « Last Edit: Dec 8th, 2005, 2:53pm by JocK » |
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solving abstract problems is like sex: it may occasionally have some practical use, but that is not why we do it.
xy - y = x5 - y4 - y3 = 20; x>0, y>0.
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Barukh
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Re: pi in various metrics
« Reply #1 on: Dec 9th, 2005, 5:22am » |
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Applying the definition of a unit circle as a set of points at a unit distance from a given point, and the definition of pi being half a perimeter of the unit circle, the answer to the first question seems to be Norm(x, y) = max(x, y).
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Barukh
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Re: pi in various metrics
« Reply #2 on: Dec 10th, 2005, 7:51am » |
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By the way, the so called taxicab norm T(x, y) = |x| + |y| gives the same value for pi. It's not a coincidence: in both cases unit circles are squares.
Very interesting topic! 
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JocK
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Re: pi in various metrics
« Reply #3 on: Dec 10th, 2005, 9:13am » |
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on Dec 10th, 2005, 7:51am, Barukh wrote:By the way, the so called taxicab norm T(x, y) = |x| + |y| gives the same value for pi. It's not a coincidence: in both cases unit circles are squares.
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You're on the right track... 
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IP Logged |
solving abstract problems is like sex: it may occasionally have some practical use, but that is not why we do it.
xy - y = x5 - y4 - y3 = 20; x>0, y>0.
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Barukh
Uberpuzzler
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Posts: 2276
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Re: pi in various metrics
« Reply #5 on: Dec 11th, 2005, 11:27pm » |
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on Dec 10th, 2005, 9:13am, JocK wrote:| You're on the right track... |
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I wish I was on a right track! All the facts I’ve got from the survey "On the Perimeter and Area of Unit Disc”. The question posted is related to the 4–th Hilbert problem and Minkowsky geometries.
One of the results stated there (and proven more than 70 years ago) answers JocK's last question: pi may be any number in the interval [3, 4].
I’ve found the following amazing: For any curve satisfying the following three conditions:
1) It’s closed.
2) It’s symmetric w.r.t. 0.
3) It’s convex.
there exists a norm for which this curve is a unit circle in corresponding Minkowsky geometry. As such, every regular 2n-gon is a unit circle in some geometry. Two examples of squares were given earlier.
Now, here’s a question: can you give an explicit formula for the norm in geometry with unit circle being regular hexagon (for which pi is always 3)?
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Eigenray
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For the p-norm |(x,y)| = (|x|p + |y|p|)1/p, we find
Pi(p) = [int]-11 [ 1 + (|x|-p-1)1-p ]1/p dx.
Pi(p) is apparently minimized when p=2. Is there an easy way to show this?
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| « Last Edit: Dec 13th, 2005, 12:28pm by Eigenray » |
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JocK
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Re: pi in various metrics
« Reply #7 on: Dec 13th, 2005, 3:27pm » |
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A p-norm and a q-norm metric are each others duals if 1/p + 1/q = 1. Dual metrics have a common "pi-value". So, when you plot your pi-curve as function of x = 1/p - 1/2, the curve will be symmetric.
What remains to be proven is that for p>2 the pi-function is monotonous. 
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solving abstract problems is like sex: it may occasionally have some practical use, but that is not why we do it.
xy - y = x5 - y4 - y3 = 20; x>0, y>0.
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