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3.3.5 Example Let a > 0; we will construct a sequence (s) of real numbers that converges to √a. 2 Let s₁ > 0 be arbitrary and define Sn+1 := (Sn+a/sn) for n = N. We now show that the sequence (Sn) converges to √✓a. (This process for calculating square roots was known in Mesopotamia before 1500 B.C.) - We first show that s² 2 ≥ a for n ≥ 2. Since s, satisfies the quadratic equation s² − 2Sn+1Sn + a = 0, this equation has a real root. Hence the discriminant 45% 24+ must be nonnegative; that is, s² (24+1 ≥ a for n ≥ 1. To see that (Sn) is ultimately decreasing, we note that for n ≥ 2 we have n+1 - 4a a 1 (s² - a) Sn Sn+1 = Sn Sn + ≥ 0. Sn 2 Sn Hence, Sn+1 ≤ sɲ for all n ≥ 2. The Monotone Convergence Theorem implies that s := lim(Sn) exists. Moreover, from Theorem 3.2.3, the limit s must satisfy the relation whence it follows (why?) that s = S= 2 (+9 a/s or s² = a. Thus s = √a. For the purposes of calculation, it is often important to have an estimate of how rapidly the sequence (Sn) converges to √a. As above, we have √√α ≤ sn for all n ≥ 2, whence it follows that a/s, ≤ √√α ≤ Sn. Thus we have 0 ≤ Sn – √√α ≤ sn – a/Sn = (s²/ − a)/Sn for n ≥2. - Using this inequality we can calculate ✓a to any desired degree of accuracy.

3.3.5 Example Let a > 0; we will construct a sequence (s) of real numbers that converges to √a. 2 Let s₁ > 0 be arbitrary and define Sn+1 := (Sn+a/sn) for n = N. We now show that the sequence (Sn) converges to √✓a. (This process for calculating square roots was known in Mesopotamia before 1500 B.C.) - We first show that s² 2 ≥ a for n ≥ 2. Since s, satisfies the quadratic equation s² − 2Sn+1Sn + a = 0, this equation has a real root. Hence the discriminant 45% 24+ must be nonnegative; that is, s² (24+1 ≥ a for n ≥ 1. To see that (Sn) is ultimately decreasing, we note that for n ≥ 2 we have n+1 - 4a a 1 (s² - a) Sn Sn+1 = Sn Sn + ≥ 0. Sn 2 Sn Hence, Sn+1 ≤ sɲ for all n ≥ 2. The Monotone Convergence Theorem implies that s := lim(Sn) exists. Moreover, from Theorem 3.2.3, the limit s must satisfy the relation whence it follows (why?) that s = S= 2 (+9 a/s or s² = a. Thus s = √a. For the purposes of calculation, it is often important to have an estimate of how rapidly the sequence (Sn) converges to √a. As above, we have √√α ≤ sn for all n ≥ 2, whence it follows that a/s, ≤ √√α ≤ Sn. Thus we have 0 ≤ Sn – √√α ≤ sn – a/Sn = (s²/ − a)/Sn for n ≥2. - Using this inequality we can calculate ✓a to any desired degree of accuracy.

Algebra & Trigonometry with Analytic Geometry
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter10: Sequences, Series, And Probability
Section10.2: Arithmetic Sequences
Problem 67E
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Use the method in Example 3.3.5 to calculate Sqrt(2), correct to within 4 decimals. 

3.3.5 Example Let a > 0; we will construct a sequence (s) of real numbers that converges to √a. 2 Let s₁ > 0 be arbitrary and define Sn+1 := (Sn+a/sn) for n = N. We now show that the sequence (Sn) converges to √✓a. (This process for calculating square roots was known in Mesopotamia before 1500 B.C.) - We first show that s² 2 ≥ a for n ≥ 2. Since s, satisfies the quadratic equation s² − 2Sn+1Sn + a = 0, this equation has a real root. Hence the discriminant 45% 24+ must be nonnegative; that is, s² (24+1 ≥ a for n ≥ 1. To see that (Sn) is ultimately decreasing, we note that for n ≥ 2 we have n+1 - 4a a 1 (s² - a) Sn Sn+1 = Sn Sn + ≥ 0. Sn 2 Sn Hence, Sn+1 ≤ sɲ for all n ≥ 2. The Monotone Convergence Theorem implies that s := lim(Sn) exists. Moreover, from Theorem 3.2.3, the limit s must satisfy the relation whence it follows (why?) that s = S= 2 (+9 a/s or s² = a. Thus s = √a. For the purposes of calculation, it is often important to have an estimate of how rapidly the sequence (Sn) converges to √a. As above, we have √√α ≤ sn for all n ≥ 2, whence it follows that a/s, ≤ √√α ≤ Sn. Thus we have 0 ≤ Sn – √√α ≤ sn – a/Sn = (s²/ − a)/Sn for n ≥2. - Using this inequality we can calculate ✓a to any desired degree of accuracy.
Transcribed Image Text:3.3.5 Example Let a > 0; we will construct a sequence (s) of real numbers that converges to √a. 2 Let s₁ > 0 be arbitrary and define Sn+1 := (Sn+a/sn) for n = N. We now show that the sequence (Sn) converges to √✓a. (This process for calculating square roots was known in Mesopotamia before 1500 B.C.) - We first show that s² 2 ≥ a for n ≥ 2. Since s, satisfies the quadratic equation s² − 2Sn+1Sn + a = 0, this equation has a real root. Hence the discriminant 45% 24+ must be nonnegative; that is, s² (24+1 ≥ a for n ≥ 1. To see that (Sn) is ultimately decreasing, we note that for n ≥ 2 we have n+1 - 4a a 1 (s² - a) Sn Sn+1 = Sn Sn + ≥ 0. Sn 2 Sn Hence, Sn+1 ≤ sɲ for all n ≥ 2. The Monotone Convergence Theorem implies that s := lim(Sn) exists. Moreover, from Theorem 3.2.3, the limit s must satisfy the relation whence it follows (why?) that s = S= 2 (+9 a/s or s² = a. Thus s = √a. For the purposes of calculation, it is often important to have an estimate of how rapidly the sequence (Sn) converges to √a. As above, we have √√α ≤ sn for all n ≥ 2, whence it follows that a/s, ≤ √√α ≤ Sn. Thus we have 0 ≤ Sn – √√α ≤ sn – a/Sn = (s²/ − a)/Sn for n ≥2. - Using this inequality we can calculate ✓a to any desired degree of accuracy.
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