# Showing a limit of the recurrence relation

• Mr Davis 97
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 +∞.
Mr Davis 97

## 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?

## 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?

What happens if ##x_1=1/3##? What happens if ##x_1<1/3##?

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)?

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.

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.

Mr Davis 97
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?

## 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?

## 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.

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