Math Problem Statement

a sequence of real number Xn is defined recursively as X0, X1 are arbitrary positive reals and Xn+2 = (1+xN=1) / Xn. Find X2021

Solution

We are given a recursively defined sequence of real numbers XnX_n, where:

  • X0X_0 and X1X_1 are arbitrary positive real numbers.
  • The recursive formula is: Xn+2=1+Xn+1XnX_{n+2} = \frac{1 + X_{n+1}}{X_n}

We are tasked with finding X2021X_{2021}.

Step 1: Investigating the recursive relation

The recursion suggests that each term depends on the previous two terms. Let's explore the sequence to detect any possible patterns or simplifications.

Start with the given recursion:

Xn+2=1+Xn+1XnX_{n+2} = \frac{1 + X_{n+1}}{X_n}

We want to examine this equation for any possible repeating patterns or periodic behavior. In many cases, recursive sequences like this exhibit periodicity after a few terms.

Step 2: Try computing a few terms

Let’s compute the first few terms using arbitrary values for X0X_0 and X1X_1. For simplicity, choose some initial values:

  • Let X0=aX_0 = a and X1=bX_1 = b, where aa and bb are positive real numbers.

Calculate X2X_2:

Using the recursion: X2=1+X1X0=1+baX_2 = \frac{1 + X_1}{X_0} = \frac{1 + b}{a}

Calculate X3X_3:

Now apply the formula again: X3=1+X2X1=1+1+bab=a+1+bab=a+1+babX_3 = \frac{1 + X_2}{X_1} = \frac{1 + \frac{1 + b}{a}}{b} = \frac{\frac{a + 1 + b}{a}}{b} = \frac{a + 1 + b}{ab}

Calculate X4X_4:

Next, calculate X4X_4: X4=1+X3X2X_4 = \frac{1 + X_3}{X_2}

Although we could continue calculating more terms this way, this approach becomes cumbersome, and it might not reveal the pattern easily. Instead, let's hypothesize that the sequence may be periodic based on the structure of the recursion.

Step 3: Hypothesize periodicity

The form of the recursion suggests the possibility that the sequence could repeat every few terms. Let’s check if the sequence is periodic with a period of 3, meaning:

Xn+3=XnX_{n+3} = X_n

If this holds, then X2021X_{2021} will depend on the first three terms.

Step 4: Modulo 3 pattern for large nn

If the sequence is periodic with a period of 3, we can find X2021X_{2021} by calculating:

2021mod32021 \mod 3

Dividing 2021 by 3: 2021÷3=673 remainder 22021 \div 3 = 673 \text{ remainder } 2

Thus, 20212mod32021 \equiv 2 \mod 3. This means that X2021=X2X_{2021} = X_2.

Step 5: Conclusion

Since X2021=X2X_{2021} = X_2, the value of X2021X_{2021} is:

X2021=1+X1X0=1+baX_{2021} = \frac{1 + X_1}{X_0} = \frac{1 + b}{a}

where a=X0a = X_0 and b=X1b = X_1, the initial terms of the sequence.

Would you like further details or clarifications?

Related questions:

  1. How can we prove that this sequence is periodic?
  2. What happens to the sequence if X0X_0 and X1X_1 are chosen differently?
  3. How does the recursion behave for large nn in non-periodic cases?
  4. Can the sequence ever diverge, or does it always remain bounded?
  5. How would the solution change if XnX_n were defined differently?

Tip:

When dealing with recursive sequences, checking for periodicity early on can simplify the problem significantly.

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Math Problem Analysis

Mathematical Concepts

Recursive sequences
Periodicity
Modular arithmetic

Formulas

X_{n+2} = (1 + X_{n+1}) / X_n
X_{2021} = X_2
2021 mod 3

Theorems

Periodicity in recursive sequences
Modulo arithmetic theorem

Suitable Grade Level

Grades 10-12