Continued Fractions Consider again continued fraction expansion of x, x = [a_0;

Question

Continued Fractions Consider again continued fraction expansion of x, x = [a_0; a_1, a_2, a_3, ...]. Compute arithmetic mid geometric mean of a_1, a_2, ctdot, a_n for x = squareroot 2, and x = e. Does the n rightarrow infinity limit of the arithmetic mean diverge in both of these cases? Which of these two numbers is typical with respect to the behaviour of the arithmetic mean. Does the n rightarrow infinity limit of the geometric mean converge to K_0 in both of these cases. Which of these two numbers is typical with respect to the behaviour of the geometric mean.

Explanation / Answer

Arithmetic-and-Geometric-mean-converge

anan+1bn+1bnanan+1bn+1bn with a1=a and b1=b

an+1= an+bn / 2

and

bn+1= squrt of anbn

To show that sequences an and bn converges and that an and b have the same limit.

monotonic convergence theorem to prove that both sequences converges we have the following proof:

{an} is monotonically decreasing while {bn} is monotonically increasing. Since {an} is bounded above by supremum a1below by its infimum b1, {an} according to the monotonic convergence theorem has to converge.

Similarly, we notice that {bn} is bounded below by infimum b and supremum a By monotonic convergence theorem {bn} must also converge as well.

show that {an} and {bn} have the same limit. In other words, if [anbn] as n tends to infinity must be 0. For this part, the case that one can prove it by just showing that an+1bn+1(1/2)(anbn)an+1bn+1(1/2)(anbn). show this by using the definition of the arithmetic mean, which is an+1bn+1an+1bn=(1/2)(anbn)an+1bn+1an+1bn=(1/2)(anbn).

show that {an} and {bn} have the same limit. In other words, if [anbn] as n tends to infinity must be 0. For this part, prove it by just showing that an+1bn+1(1/2)(anbn)an+1bn+1(1/2)(anbn). And I know you can just show this by using the definition of the arithmetic mean, which is an+1bn+1an+1bn=(1/2)(anbn)an+1bn+1an+1bn=(1/2)(anbn).

a1=ab1=b0a1=ab1=b0 and an+1bn+11/2(anbn)an+1bn+11/2(anbn) for all n1n1, we have:

0an+1bn+11/2n(ab),0an+1bn+11/2n(ab),

leading to

limn(anbn)=0.

For every xNxN an0,bn0using well known inequetion

aba+b/2,a0,b0

we get

an+1=an+bn/2 squrt anbn=bn+1

since bn+1=anbnb2nbn+1=anbnbn2=bn and an+1=an+bn/2an+1=an+bn2an and bnana1,anbnb1bnana1,anbnb1 so (an)(an) and (bn)(bn) are monoton and bounded so they have finit limits A and B.when nn

an+1=an+bn/2

we get

A=B

Related Questions

Navigate

• Browse (All)
• Browse C
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.