The arithmetic geometric mean (Agm) Tamar Ziegler PicturesbyDavidLehavi 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