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Topic: transcendence of the agm (Read 365 times) |
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JocK
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transcendence of the agm
« on: Jan 16th, 2005, 12:00pm » |
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Is the arithmetic-geometric mean* of any two positive integers M and N other than M = N transcendental?
* The arithmetic-geometric mean agm(A,B) of two numbers A and B is defined as:
agm(A,B) = limn[to][infty] an
with:
a0 = A, an+1 = (an + bn)/2
b0 = B, bn+1 = [sqrt](an bn)
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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: transcendence of the agm
« Reply #1 on: Jan 18th, 2005, 10:52am » |
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My guess is that the answer is yes, but I don't have any idea how to prove this. It looks like a very tough problem...
The only approach I tried is to show that every iteration of the AGM sequence increases the degree of the polynomial that an, bn can be. But this is not true.
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