The arithmetic geometric mean (Agm)

85 Pages·2010·1.47 MB·English

by  Tamar Ziegler

The arithmetic geometric mean (Agm)
Tamar Ziegler
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TamarZiegler Agm
 ￿
a1=1 ￿ ￿
1 1
an+1=2 βa2n+β
β>0 (i
1
β nl→im∞an β>0 (ii
0<β≤1 (iii
β>1 (iv
n≥1 an+1=33++3aann ￿an￿∞n=1 ￿
nl→im∞an
￿
a1=1
an+1=sinan
This is part of exerci￿se 5 question 4 in myc>ca1lculus class Fall 2010:
a1=c
an+1=5an−2
nl→im∞an=∞
￿
a1=a b1=b
an+1=an+2bn bn+1=√anbn
a>b>0
(n+1)n−(n−1)n
lim
n→∞ nn+n2
￿ ￿(n+1)2
Here is the translation (and indexlimchann+g2e):n+2 ￿
n→∞ n+1
aan0=(=−1)an+sin￿n4bπ￿0 = b (cid:112)
a +b
a = n n b liminf=an,limasupban n = 0,1,...
n+1 n+1 n n
2
￿ ￿
n0∈N ε>0 an ∞n=1 ￿
For a > b > 0 show that|abn−oatmh|<sεequences cno,nmv>enr0ge and calculate
their limits.
TamarZiegler Agm
By the mean inequality:
√
a+b
≥ ab
2
Therefore
a ≥ b
1 1
and by induction
a ≥ b .
n n
Also
(cid:112)
a +b
a ≥ n n = a ≥ b = a b ≥ b
n n+1 n+1 n n n
2
Why do the limits exist ?
TamarZiegler Agm
Therefore
a ≥ b
1 1
and by induction
a ≥ b .
n n
Also
(cid:112)
a +b
a ≥ n n = a ≥ b = a b ≥ b
n n+1 n+1 n n n
2
Why do the limits exist ?
By the mean inequality:
√
a+b
≥ ab
2
TamarZiegler Agm
Also
(cid:112)
a +b
a ≥ n n = a ≥ b = a b ≥ b
n n+1 n+1 n n n
2
Why do the limits exist ?
By the mean inequality:
√
a+b
≥ ab
2
Therefore
a ≥ b
1 1
and by induction
a ≥ b .
n n
TamarZiegler Agm
Why do the limits exist ?
By the mean inequality:
√
a+b
≥ ab
2
Therefore
a ≥ b
1 1
and by induction
a ≥ b .
n n
Also
(cid:112)
a +b
a ≥ n n = a ≥ b = a b ≥ b
n n+1 n+1 n n n
2
TamarZiegler Agm
The sequence {a } is decreasing and bounded below by b, and the
n
sequence {b } is increasing and bounded above by a. So both
n
sequences converge !
Actually, they converge to a common limit:
a +b a −b
0 ≤ a −b ≤ a −b = n n −b = n n
n+1 n+1 n+1 n n
2 2
Inductively:
a−b
0 ≤ a −b ≤
n n 2n
We have
a ≥ a ≥ ... ≥ a ≥ a ≥ b ≥ b ≥ ... ≥ b ≥ b
1 n n+1 n+1 n 1
TamarZiegler Agm
Actually, they converge to a common limit:
a +b a −b
0 ≤ a −b ≤ a −b = n n −b = n n
n+1 n+1 n+1 n n
2 2
Inductively:
a−b
0 ≤ a −b ≤
n n 2n
We have
a ≥ a ≥ ... ≥ a ≥ a ≥ b ≥ b ≥ ... ≥ b ≥ b
1 n n+1 n+1 n 1
The sequence {a } is decreasing and bounded below by b, and the
n
sequence {b } is increasing and bounded above by a. So both
n
sequences converge !
TamarZiegler Agm
Inductively:
a−b
0 ≤ a −b ≤
n n 2n
We have
a ≥ a ≥ ... ≥ a ≥ a ≥ b ≥ b ≥ ... ≥ b ≥ b
1 n n+1 n+1 n 1
The sequence {a } is decreasing and bounded below by b, and the
n
sequence {b } is increasing and bounded above by a. So both
n
sequences converge !
Actually, they converge to a common limit:
a +b a −b
0 ≤ a −b ≤ a −b = n n −b = n n
n+1 n+1 n+1 n n
2 2
TamarZiegler Agm
We have
a ≥ a ≥ ... ≥ a ≥ a ≥ b ≥ b ≥ ... ≥ b ≥ b
1 n n+1 n+1 n 1
The sequence {a } is decreasing and bounded below by b, and the
n
sequence {b } is increasing and bounded above by a. So both
n
sequences converge !
Actually, they converge to a common limit:
a +b a −b
0 ≤ a −b ≤ a −b = n n −b = n n
n+1 n+1 n+1 n n
2 2
Inductively:
a−b
0 ≤ a −b ≤
n n 2n
TamarZiegler Agm