Showing a limit of the recurrence relation

In summary, the given problem involves finding the limit of a sequence defined by a recursive equation. It is shown that if the limit exists, it must be either 0 or 1/3, but it is also possible for the limit to not exist, in which case a third solution, +∞, can be considered in the extended real-number system. The correct value of the limit in this problem is +∞.
  • #1
Mr Davis 97
1,462
44

Homework Statement


Let ##x_1=1## and ##\displaystyle x_{n+1} = 3 x_n^2## for ##n \ge 1##.
a) Show if ##a = \lim x_n##, then ##a = 1/3## or ##a = 0##.
b) Does ##\lim x_n## actually exist?

Homework Equations

The Attempt at a Solution


I have proven before that, in general, ##\lim s_{n+1} = \lim s_n##. Hence to answer a), we simply take the limit of both sides and solve the quadratic equation. Easy. But clearly, since ##x_n## is not bounded, the limit doesn't actually exist. Why am I getting values ##1/3## or ##0## when the limit doesn't actually exist?
 
Physics news on Phys.org
  • #2
What happens if ##x_1=1/3##? What happens if ##x_1<1/3##?
 
  • #3
I found the ##a=0## case, but not the ##a=\frac{1}{3}##. Anyway, if it this editor here allowed me to write letters upside down, I will win in the lottery tomorrow. That's 100% certain and true. What changed between a) and b) and have you used all information given for a) same as for b)?
 
  • #4
I think I see... In part a), are we saying that if ##x_n## converges, one of those would be the resulting value. In part b), I guess the question of whether the sequence actually converges is dependent on the initial value, as if ##|x_1| < 1/3## it converges to 0, ##|x_1| = 1/3## it converges to 1/3, and if ##|x_1| > 1/3## it diverges.
 
  • #5
Mr Davis 97 said:
I think I see... In part a), are we saying that if ##x_n## converges, one of those would be the resulting value. In part b), I guess the question of whether the sequence actually converges is dependent on the initial value, as if ##|x_1| < 1/3## it converges to 0, ##|x_1| = 1/3## it converges to 1/3, and if ##|x_1| > 1/3## it diverges.
No, in b) you used ##x_1=1## to see that it does not converge. In a) you reasoned by ##\lim s_n = \lim s_{n+1}##. Where's the initial value here? It is the other way around. And in a) we have if then, but if is never true.
 
  • Like
Likes Mr Davis 97
  • #6
Mr Davis 97 said:

Homework Statement


Let ##x_1=1## and ##\displaystyle x_{n+1} = 3 x_n^2## for ##n \ge 1##.
a) Show if ##a = \lim x_n##, then ##a = 1/3## or ##a = 0##.
b) Does ##\lim x_n## actually exist?

Homework Equations

The Attempt at a Solution


I have proven before that, in general, ##\lim s_{n+1} = \lim s_n##. Hence to answer a), we simply take the limit of both sides and solve the quadratic equation. Easy. But clearly, since ##x_n## is not bounded, the limit doesn't actually exist. Why am I getting values ##1/3## or ##0## when the limit doesn't actually exist?

If the limit exists (call it ##a##) we have ##a = 3 a^2##, so ##a = 0## or ##a = 1/3##.

However, what if the limit does not exist? In that case, a third solution to ##a = 3 a^2## "exists" within the extended real-number system, namely: ##a = +\infty##.

What do you think is the correct value of ##a## in this problem?
 

Related to Showing a limit of the recurrence relation

1. What is a recurrence relation?

A recurrence relation is a mathematical formula that defines a sequence of numbers, where each subsequent number is dependent on the previous numbers in the sequence.

2. How do you show a limit of a recurrence relation?

To show a limit of a recurrence relation, one must first find the general formula for the sequence and then use mathematical techniques, such as induction, to prove that the sequence approaches a specific value as the number of terms in the sequence increases.

3. What is the importance of showing a limit of a recurrence relation?

Showing a limit of a recurrence relation is important because it helps us understand the behavior of a sequence as the number of terms increases. It also allows us to make predictions about the future terms of the sequence and determine the overall trend of the sequence.

4. Can a recurrence relation have multiple limits?

Yes, a recurrence relation can have multiple limits. This can occur when the sequence has a pattern that alternates between two or more values, or when there are multiple variables in the recurrence relation that can affect the limit.

5. Are there any real-world applications of recurrence relations?

Yes, recurrence relations have many real-world applications, such as in finance, physics, and computer science. For example, the Fibonacci sequence is a famous recurrence relation that is used to model population growth and financial markets.

Similar threads

  • Calculus and Beyond Homework Help
Replies
13
Views
2K
  • Calculus and Beyond Homework Help
Replies
11
Views
1K
  • Calculus and Beyond Homework Help
Replies
1
Views
415
  • Calculus and Beyond Homework Help
Replies
1
Views
1K
  • Calculus and Beyond Homework Help
Replies
3
Views
1K
  • Calculus and Beyond Homework Help
Replies
11
Views
1K
  • Calculus and Beyond Homework Help
Replies
9
Views
1K
  • Calculus and Beyond Homework Help
Replies
9
Views
966
  • Calculus and Beyond Homework Help
Replies
1
Views
1K
  • Calculus and Beyond Homework Help
Replies
2
Views
903
Back
Top