Author |
Topic: transcendence of the agm (Read 373 times) |
|
JocK
Uberpuzzler
Gender: 
Posts: 877
|
 |
transcendence of the agm
« on: Jan 16th, 2005, 12:00pm » |
 Quote  Modify
|
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)
|
|
 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.
|
|
|
Barukh
Uberpuzzler
Gender:
Posts: 2276
|
|
Re: transcendence of the agm
« Reply #1 on: Jan 18th, 2005, 10:52am » |
Quote Modify
|
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.
|
|
IP Logged |
|
|
|
|
RIDDLES SITE
WRITE MATH!
Home 
Help 
Search 
Members 
Login 
Register
Reply
Notify of replies
Send Topic
Print