   Chapter 11.1, Problem 91E

Chapter
Section
Textbook Problem

Let a and b be positive numbers with a > b . Let a 1 be their arithmetic mean and b 1 their geometric mean: a 1 = a + b 2       b 1 = a b Repeat this process so that, in general, a n + 1 = a n + b n 2       b n + 1 = a n b n (a) Use mathematical induction to show that a n > a n + 1 > b n + 1 > b n (b) Deduce that both { a n } and { b n } are convergent.(c) Show that lim n → ∞ a n = lim n → ∞ b n Gauss called the common value of these limits the arithmetic-geometric mean of the numbers a and b.

To determine

Part (a)

To show:

an>an+1>bn+1>bn using mathematical induction

Explanation

1) Concept:

Use induction to show an>an+1>bn+1>bn

2) Given:

a1=a+b2,  b1=ab

3) Calculation:

For n=1, show that a>a1>b1>b

a1-b1=a+b2-ab

=12a-2ab+b

=12a-b2>0[Since a>ba-b>0a-b2]

a1>b1

Also a-a1=a-12a+b=12a-b>0

a>a1

and

b-b1=b-ab=b b-a<0

b<b1

b1>b

Therefore,

a>a1>b1>b

In the same way, we show that a1>a2>b2>b1, and so, the given assertion is true for n=1

To determine

Part (b)

To deduce:

an and  bn are convergent

To determine

Part (c)

To show:

limnan=limnbn

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