Show that a recursive sequence converges

  • Thread starter Mr Davis 97
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In summary, the conversation discusses the proof that the sequence (a_n) is decreasing and has a limit of √x, given that x and y are positive numbers and a_0 = y. There is some confusion about the initial condition and whether or not it should be exempt from the sequence. The expert provides a proof for both conditions and concludes that the sequence is bounded below and decreasing, with a limit of √x. The discussion also brings up the convention for the first index of a sequence and suggests that defining a_0 as a member of a set but not a member of the sequence could resolve any potential issues.
  • #1
Mr Davis 97
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Homework Statement


Let ##x,y## be positive numbers. Let ##a_0 = y## and let ##a_n = \frac{(x/a_{n-1})+a_{n-1}}{2}##. Prove that ##(a_n)## is a decreasing sequence with limit ##\sqrt{x}##.

Homework Equations

The Attempt at a Solution


I'm confused about the initial condition being an arbitrary positive real number. This is because it says to prove that ##(a_n)## is a decreasing sequence, but what if ##0<y< \sqrt{x}##? Then ##(a_n)## can't be decreasing...
 
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  • #2
You are right.
It should be a "converging" sequence.
Or perhaps they should exempt ##a_0## from the sequence.
 
  • #3
What you can show is ##a_n \ge \sqrt x## for ##n\ge 1## regardless whether or not ##a_0 = y < \sqrt x##, and that ##a_{n+1}\le a_n## for ##n\ge 1##.
 
  • #4
LCKurtz said:
What you can show is ##a_n \ge \sqrt x## for ##n\ge 1## regardless whether or not ##a_0 = y < \sqrt x##, and that ##a_{n+1}\le a_n## for ##n\ge 1##.
Here is my argument:

(1) WTS: ##\forall n \ge 1##, ##a_n \ge \sqrt{x}##.

Note that it is true that ##(x - y^2)^2 \ge 0##. This implies that ##x^2-2xy^2+y^4 \ge 0 \implies x^2 + y^4 \ge 2y^2x \implies x+y^2 \ge 2y\sqrt{x} \implies \frac{\frac{x}{y} + y}{2} \ge \sqrt{x} \implies a_1 \ge \sqrt{x}##. So the base case is satisfied. Now, suppose that for some ##k \in \mathbb{N}## we have that ##a_k \ge \sqrt{x}##. Then ##a_{k+1} = \frac{\frac{x}{a_k} + a_k}{2} \ge \frac{\frac{x}{\sqrt{x}} + \sqrt{x}}{2} = \sqrt{x}##. This completes the induction.

(2) WTS: ##\forall n \ge 1##, ##a_n \ge a_{n+1}##. We see that ##a_{n+1} = \frac{\frac{x}{a_n}+a_n}{2} \le \frac{\frac{a^2_n}{a_n}+a_n}{2} = a_n##.

So we see that ##(a_n)## is bounded below and is decreasing, so by the MCT it has a limit, L. Then ##L = \frac{\frac{x}{L}+L}{2} \implies L = \sqrt{x}##.
 
  • #5
Mr Davis 97 said:
Let ##a_0 = y##
What is the convention used by your text materials for the first index of a sequence? Is it 1? - or 0? - can it be either?

A nice legality could be that ##a_0## is defined as a member of a set, but it is not a member of the sequence if the text materials insist the first index of a sequence must be 1.
 

Related to Show that a recursive sequence converges

1. What is a recursive sequence?

A recursive sequence is a sequence of numbers where each term is defined by a formula that uses one or more of the previous terms. This means that the value of a term depends on the value of the previous term(s).

2. How do you determine if a recursive sequence converges?

To determine if a recursive sequence converges, you must find the limit of the sequence as n approaches infinity. If the limit exists and is a finite number, then the sequence converges.

3. What is the difference between a convergent and a divergent recursive sequence?

A convergent recursive sequence has a limit as n approaches infinity, meaning that the sequence approaches a finite number. A divergent recursive sequence does not have a limit, meaning that the values of the sequence become increasingly larger or smaller.

4. Can a recursive sequence converge to more than one value?

No, a recursive sequence can only converge to one value. This is because the limit of a sequence is a unique value.

5. What is the importance of showing that a recursive sequence converges?

Showing that a recursive sequence converges is important because it helps us understand the behavior and patterns of the sequence. It also allows us to make predictions about the future values of the sequence, which can be useful in various mathematical and scientific applications.

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