Question

Consider the following sequences defined by a recurrence relation. Use a calculator, analytical methods, and/or graphing to make a conjecture about the limit of the sequence or state that the sequence diverges. $$a_{n+1}=2 a_{n}\left(1-a_{n}\right) ; a_{0}=0.3$$

Short answer

Answer: The limit of the sequence is 0.5.

Step-by-step solution

Understand the sequence definition

The sequence is defined by the recurrence relation $$a_{n+1}=2 a_{n}\left(1-a_{n}\right),$$ with the initial term $$a_0=0.3$$.

Calculate the first few terms of the sequence

Let's calculate the first few terms of the sequence using a calculator: 1. \(a_1 = 2(0.3)(1-0.3) = 0.42\) 2. \(a_2 = 2(0.42)(1-0.42) = 0.4856\) 3. \(a_3 = 2(0.4856)(1-0.4856) = 0.49946176 \) 4. \(a_4 = 2(0.49946176)(1-0.49946176) = 0.499999852273 \)

Analyze the trend and make a conjecture

From the calculated terms, we can observe that the sequence appears to be converging towards 0.5. To gain a better understanding, let's use some analytical methods: Let's consider the function \(f(x) = 2x(1-x)\). Our sequence can be re-written as \(a_{n+1} = f(a_n)\). To find the limit of the sequence, if it exists, we have to find the fixed points of the function \(f(x)\), which are the points where \(f(x) = x\).

Find the fixed points of the function

We will find the fixed points of the function by solving the equation \(f(x)=x\): $$2x(1-x) = x$$ Factoring x out, we get: $$ x(2-2x)=x$$ Now we can find the fixed points by solving for x: 1. \(x = 0\) 2. \(2-2x=1 \Rightarrow x = 0.5\) So our fixed points are \(x = 0\) and \(x = 0.5\).

Determine stability of the fixed points

To determine the stability of the fixed points, we need to look at the derivative of the function: $$f'(x) = \frac{d}{dx} (2x(1-x)) = 2(1-2x)$$ Now, let's evaluate the derivative at each fixed point: 1. \(f'(0) = 2(1-0) = 2\) 2. \(f'(0.5) = 2(1-1) = 0\) Since \(|f'(0)| > 1\), the fixed point at x = 0 is unstable. However, the fixed point at x = 0.5 is stable since \(|f'(0.5)| < 1\). This indicates that the sequence converges to 0.5. #Conjecture# The limit of the sequence is 0.5, as the sequence appears to be converging towards this value and it is a stable fixed point for the corresponding function.

Key concepts

Sequence Convergence

When dealing with sequences in mathematics, particularly those defined by recurrence relations, a common question is whether these sequences converge to a limit or diverge. Convergence of a sequence means that as we progress through the terms of the sequence, they get closer and closer to a single value known as the limit. This is akin to the terms of the sequence 'settling down' to a specific number as they increase in index.

In our exercise, the sequence is defined by the recurrence relation:
\[a_{n+1}=2a_{n}(1-a_{n})\].
The terms seem to be getting closer to 0.5, indicating that the sequence might converge to this limit. To ascertain whether a sequence converges, we calculate several terms and observe if there's a pattern where the distance between successive terms and a particular value gets smaller. If there is, it's likely that the sequence is converging to this value, though rigorous mathematical proofs are generally required for complete certainty.

Fixed Points

In the context of sequences and functions, a fixed point is a point at which the value of a function equals the point itself. In other words, it is where \(f(x)=x\).

Identifying fixed points is crucial when analyzing the behavior of sequences defined by recurrence relations. If a sequence converges, its limit is a fixed point of the function that defines the recurrence.

In our example, we seek the fixed points of the function \(f(x)=2x(1-x)\).
By setting \(f(x)=x\)
and solving for x, we found two fixed points: x = 0 and x = 0.5. These points are potential candidates for the limit of the sequence, where the sequence might settle.

Stability of Fixed Points

Once we've identified fixed points, the next crucial step is to determine their stability. A fixed point is considered stable if small deviations from it tend to return to the fixed point, whereas an unstable fixed point would see such deviations grow larger.

The stability can often be determined by examining the derivative of the function at the fixed points. A common rule of thumb is:

• If the absolute value of the derivative at the fixed point is less than 1, the point is stable.
• If the absolute value is greater than 1, the point is unstable.

In our exercise, the derivative of the function at x = 0.5 is 0, which is less than 1, indicating a stable fixed point. This does not just hint at where the sequence is headed but provides confidence that deviations from this fixed point will not cause the sequence to 'escape' to a different behavior.

Derivative of a Function

The derivative of a function is a fundamental concept in calculus, representing the rate at which a function's output changes with respect to changes in its input. Intuitively, you can think of the derivative as the slope of the function's graph at any given point.

In practice, finding the derivative, which is denoted as \(f'(x)\),
involves calculation according to rules of differential calculus. For our function \(f(x)=2x(1-x)\),
the derivative is \(f'(x)=2-4x\).
Calculating the value of this derivative at the fixed points helps us understand the behavior of the sequence at those points. If the sequence does indeed converge at a stable fixed point, the derivative here will tell us about the 'pull' the function exerts on the sequence terms to bring them back to the fixed point, hence reinforcing the notion of stability.