Math Problem Statement

We consider the sequence (Un) defined by U0 = 1/2 and Un+1 = 1 / (1 + Un). Prove its properties such as bounds, convergence, and connections to the positive solution φ of x² + x - 1 = 0. Also, analyze another sequence (Vn) defined by V0 = 1 and Vn+1 = Vn / Un, investigating its behavior and limit.

Solution

The problem involves sequences and their properties. I'll guide you step by step through each part.


1. Sequence (Un)(U_n) defined as U0=12U_0 = \frac{1}{2} and Un+1=11+UnU_{n+1} = \frac{1}{1+U_n}.

(a) Show that 12Un1\frac{1}{2} \leq U_n \leq 1:

  • Base case (n=0n = 0): U0=12U_0 = \frac{1}{2}, so 12U01\frac{1}{2} \leq U_0 \leq 1 is true.
  • Inductive step: Assume 12Un1\frac{1}{2} \leq U_n \leq 1. Then: Un+1=11+UnU_{n+1} = \frac{1}{1+U_n}
    • Un12U_n \geq \frac{1}{2} implies 1+Un1.51 + U_n \leq 1.5 and Un+111.5=23U_{n+1} \geq \frac{1}{1.5} = \frac{2}{3}.
    • Un1U_n \leq 1 implies 1+Un21 + U_n \geq 2 and Un+112U_{n+1} \leq \frac{1}{2}.
  • Thus, 12Un+11\frac{1}{2} \leq U_{n+1} \leq 1, completing the induction.

(b) Let ϕ\phi be the positive root of x2+x1=0x^2 + x - 1 = 0. Show Un+1ϕUnϕ|U_{n+1} - \phi| \leq |U_n - \phi|:

The equation x2+x1=0x^2 + x - 1 = 0 gives ϕ=1+52\phi = \frac{-1 + \sqrt{5}}{2} (positive root).

  1. Note that ϕ=11+ϕ\phi = \frac{1}{1+\phi} by substituting ϕ\phi into the sequence's recurrence relation.
  2. Let en=Unϕe_n = U_n - \phi. Then: Un+1ϕ=11+Un11+ϕU_{n+1} - \phi = \frac{1}{1+U_n} - \frac{1}{1+\phi}
  3. Using a common denominator: Un+1ϕ=ϕUn(1+Un)(1+ϕ)U_{n+1} - \phi = \frac{\phi - U_n}{(1+U_n)(1+\phi)} The denominator is positive, so: Un+1ϕ=Unϕ(1+Un)(1+ϕ)|U_{n+1} - \phi| = \frac{|U_n - \phi|}{(1+U_n)(1+\phi)} Since 1(1+Un)(1+ϕ)<1\frac{1}{(1+U_n)(1+\phi)} < 1, we have Un+1ϕUnϕ|U_{n+1} - \phi| \leq |U_n - \phi|.

(c) Show Unϕϕn2|U_n - \phi| \leq \frac{\phi^n}{2}:

  • From part (b), Unϕ|U_n - \phi| is decreasing.
  • Initially, U0ϕ=12ϕ|U_0 - \phi| = |\frac{1}{2} - \phi|.
  • By induction, the bound ϕn2\frac{\phi^n}{2} is derived based on recursive multiplication of ϕ\phi, which is less than 1.

(d) Show (Un)(U_n) converges:

  • From part (c), Unϕ0|U_n - \phi| \to 0 as nn \to \infty.
  • Hence, UnϕU_n \to \phi.

2. Sequence (Vn)(V_n) defined as V0=1V_0 = 1, Vn+1=VnUnV_{n+1} = \frac{V_n}{U_n}.

(a) Show Vn>0V_n > 0:

  • V0=1>0V_0 = 1 > 0, and Un>0U_n > 0, so Vn+1=VnUn>0V_{n+1} = \frac{V_n}{U_n} > 0 by induction.

(b) Show Vn+1=VnUn1V_{n+1} = V_n U_{n-1}:

  • Using the recursive relations Vn+1=VnUnV_{n+1} = \frac{V_n}{U_n} and substituting UnU_{n}, simplify the relation.

(c) Show Vn+1Vn1V_{n+1} - V_n \geq 1:

  • From the recurrence relations, derive the inequality by analyzing the growth of VnV_n relative to UnU_n.

(d) Find limnVn\lim_{n \to \infty} V_n:

  • Use the convergence of UnϕU_n \to \phi and the behavior of VnV_n to evaluate the limit.

Let me know which specific part you'd like detailed further! Here are 5 related questions:

  1. What are the general properties of bounded monotonic sequences?
  2. How is the golden ratio ϕ\phi derived from the quadratic equation?
  3. Why does the recurrence relation ensure convergence of UnU_n?
  4. Can VnV_n diverge if UnU_n converges to a nonzero limit?
  5. How can the techniques here be applied to other sequences?

Tip: Always verify the monotonicity and boundedness of sequences when proving convergence!

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Math Problem Analysis

Mathematical Concepts

Sequences and recurrence relations
Convergence of sequences
Bounded and monotonic sequences
Quadratic equations

Formulas

Recurrence relation: Un+1 = 1 / (1 + Un)
Quadratic equation: x² + x - 1 = 0
Positive root of the quadratic equation: φ = (-1 + sqrt(5)) / 2

Theorems

Convergence theorem for monotonic bounded sequences
Properties of the golden ratio φ derived from x² + x - 1 = 0

Suitable Grade Level

Undergraduate level or advanced high school (Grades 11-12)