Math Problem Statement

Soit a ∈ [0; 1]. On définit la suite (u_n) par u_0 = a et, pour tout n ∈ N, u_(n+1) = 3u_n(1 - u_n). Montrer que, pour tout x ∈ [0 ; 1], x(1 - x) ≤ 1/4. Démontrer que, pour tout n ∈ N, u_n ∈ [0; 1].

Solution

Let's go through the problem presented in the image step by step.

Given Information:

  • Let a[0;1]a \in [0; 1].
  • We define the sequence (un)(u_n) by u0=au_0 = a and for all nN, un+1=3un(1un)n \in \mathbb{N}, \ u_{n+1} = 3u_n(1 - u_n).

Tasks:

  1. Prove that for all x[0;1]x \in [0; 1], we have: x(1x)14.x(1 - x) \leq \frac{1}{4}.

  2. Prove that for all nNn \in \mathbb{N}, we have: un[0;1].u_n \in [0; 1].

Solution Outline:

1. Proving x(1x)14x(1 - x) \leq \frac{1}{4} for x[0;1]x \in [0; 1]:

Let f(x)=x(1x)f(x) = x(1 - x) for x[0,1]x \in [0, 1].

  • First, we calculate the derivative of f(x)f(x): f(x)=12x.f'(x) = 1 - 2x.
  • Setting f(x)=0f'(x) = 0, we find the critical point: 12x=0    x=12.1 - 2x = 0 \implies x = \frac{1}{2}.
  • Now, evaluate f(x)f(x) at x=0x = 0, x=1x = 1, and x=12x = \frac{1}{2}: f(0)=0,f(1)=0,f(12)=14.f(0) = 0, \quad f(1) = 0, \quad f\left( \frac{1}{2} \right) = \frac{1}{4}.
  • Since f(x)f(x) is a quadratic function opening downwards and attains its maximum value at x=12x = \frac{1}{2}, the maximum value of f(x)=x(1x)f(x) = x(1 - x) on the interval [0,1][0, 1] is 14\frac{1}{4}.

Therefore, for all x[0,1]x \in [0, 1], x(1x)14.x(1 - x) \leq \frac{1}{4}.

2. Proving un[0;1]u_n \in [0; 1] for all nNn \in \mathbb{N}:

  • Base case (for n=0n = 0): We are given that u0=a[0;1]u_0 = a \in [0; 1], so the base case holds.

  • Inductive step: Assume un[0;1]u_n \in [0; 1] for some n0n \geq 0. We need to prove that un+1[0;1]u_{n+1} \in [0; 1].

    From the recurrence relation: un+1=3un(1un).u_{n+1} = 3u_n(1 - u_n).

    • If un=0u_n = 0, then un+1=0u_{n+1} = 0.
    • If un=1u_n = 1, then un+1=0u_{n+1} = 0.

    Now, consider un(0,1)u_n \in (0, 1). We know from part 1 that for un[0,1]u_n \in [0, 1], un(1un)14.u_n(1 - u_n) \leq \frac{1}{4}.

    Therefore, un+1=3un(1un)3×14=34.u_{n+1} = 3u_n(1 - u_n) \leq 3 \times \frac{1}{4} = \frac{3}{4}.

    Since un+10u_{n+1} \geq 0 and un+134u_{n+1} \leq \frac{3}{4}, we have un+1[0,1]u_{n+1} \in [0, 1].

By induction, un[0,1]u_n \in [0, 1] for all nNn \in \mathbb{N}.


Would you like a more detailed explanation of any part? Here are some related questions that expand on this topic:

  1. How can you find the global maximum of any quadratic function?
  2. Can you generalize the recurrence relation for different initial conditions?
  3. What is the long-term behavior of the sequence unu_n as nn \to \infty?
  4. Could we use other methods (e.g., graphical approaches) to verify the inequality x(1x)14x(1 - x) \leq \frac{1}{4}?
  5. How does the recurrence relation change if we modify the coefficient from 3 to another constant?

Tip: When solving recurrence relations, always try to check boundary and critical points to ensure the correct behavior over an interval.

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Math Problem Analysis

Mathematical Concepts

Sequences
Quadratic Functions
Induction

Formulas

u_(n+1) = 3u_n(1 - u_n)
x(1 - x) ≤ 1/4

Theorems

Maximum of quadratic functions
Mathematical induction

Suitable Grade Level

Grades 11-12