#92 PART B ONLY **Text for Educational Website:**---### Arithmetic-Geometric Mean ProcessIn general, the process is defined as:\[a_{n+1} = \frac{a_n + b_n}{2} \quad \quad b_{n+1} = \sqrt{a_n b_n}\]#### Tasks:(a) **Use mathematical induction to show that:**\[ a_n > a_{n+1} > b_{n+1} > b_n \](b) Deduce that both sequences \(\{a_n\}\) and \(\{b_n\}\) are convergent.(c) Show that \(\lim_{n \to \infty} a_n = \lim_{n \to \infty} b_n\). Gauss referred to the common value of these limits as the **arithmetic-geometric mean** of the numbers \(a\) and \(b\).---### Continued Fraction Expansion92. **Problem:**(a) Show that if \(\lim_{n \to \infty} a_{2n} = L\) and \(\lim_{n \to \infty} a_{2n+1} = L\), then \(\{a_n\}\) is convergent and \(\lim_{n \to \infty} a_n = L\).(b) If \(a_1 = 1\) and \[ a_{n+1} = 1 + \frac{1}{1 + a_n} \]Find the first eight terms of the sequence \(\{a_n\}\). Then use part (a) to show that \(\lim_{n \to \infty} a_n = \sqrt{2}\). This gives the **continued fraction expansion**:\[\sqrt{2} = 1 + \frac{1}{2 + \frac{1}{2 + \cdots}}\]---### Fish Population Model93. **Problem:**The size of an undisturbed fish population has been modeled by the formula:\[ p_{n+1} = \frac{b p_n}{a + p_n} \]where \(p_n\) is the fish population after \(n\) years, and \(a\) and \(b\) are positive constants dependent on the species and environment.---

Name: #92 PART B ONLY **Text for Educational Website:**---### Arithmetic-Geo
Uploaded: 2024-03-20T06:46:03.000Z
Channel: Bartleby
Description: #92 PART B ONLY **Text for Educational Website:**---### Arithmetic-Geometric Mean ProcessIn general, the process is defined as:\[a_{n+1} = \frac{a_n + b_n}{2} \quad \quad b_{n+1} = \sqrt{a_n b_n}\]#### Tasks:(a) **Use mathematical induction to show t

(b) limn→∞ a2n=L & limn→∞ a2n+1=L Here a1 = 1, and an+1=1+11+an To find the first eight terms, a1 = 1 For n = 1, a2=1+11+1=1+12=32=1.5 For n = 2, a3=1+11+32=1+152=1+25=75=1.4 For n = 3, a4=1+11+75=1+1125=1+512=1712=1.4167 For n = 4, a5=1+11+1712=1+12912=1+1229=4129=1.4138 For n = 5, a6=1+11+4129=1+17029=1+2970=9970=1.4143 For n = 6, a7=1+11+9970=1+116970=1+70169=239169=1.4142 For n = 7, a8=1+11+239169=1+1408169=1+169408=577408=1.4142 Therefore, the first eight terms are, 1, 1.5, 1.4, 1.4167, 1.4138, 1.4143, 1.4142, 1.4142 Now, from part (a), limn→∞ a2n=L & limn→∞ a2n+1=L The sequence {an} is convergent and limn→∞an=L Since a2n & a2n+1 approaches to L a2n+1=1+11+a2n⇒L=1+11+L⇒L=2+L1+L⇒L2+L=2+L⇒L2=2⇒L=2 Therefore, limn→∞an=2