Question

The function \(f\) defined by $$ f(x)=\frac{x^{3}+1}{3} $$ has three fixed points, say \(\alpha, \beta, \gamma\), where $$ -2<\alpha<-1, \quad 0<\beta<1, \quad 1<\gamma<2 $$ For arbitrarily chosen \(x_{1}\), define \(\left\\{x_{n}\right\\}\) by setting \(x_{n+1}=f\left(x_{n}\right) .\) (a) If \(x_{1}<\alpha\), prove that \(x_{n} \rightarrow-\infty\) as \(n \rightarrow \infty\). (b) If \(\alpha

Short answer

Answer: For \(x_1 < \alpha\), the sequence converges to \(-\infty\). For \(\alpha < x_1 < \gamma\), the sequence converges to the fixed point \(\beta\). For \(x_1 > \gamma\), the sequence converges to \(+\infty\).

Step-by-step solution

Recall the definition of fixed points

Recall that a fixed point \(\alpha\) of the function \(f(x)\) satisfies the equation \(f(\alpha) = \alpha\).

Determine the inequality for the interval

Since \(x_1 < \alpha\), we know that \(x_1\) must satisfy the inequality \(-2 < x_1 < \alpha < -1\).

Show that \(f(x) < x\) in the interval

We want to show that for any \(x\) in the interval \(-2 < x < \alpha\), we have \(f(x) < x\). Given that \(f(x)=\frac{x^3+1}{3}\), we can rewrite the inequality as: $$\frac{x^3+1}{3}0\), \(g(\alpha)=0\), and \(g(x)\) is a continuous function, so it must be negative within the mentioned interval.

Conclude that the sequence converges to \(-\infty\)

Since \(f(x_n) < x_n\) in the given interval and \(f(x)=\frac{x^3+1}{3}\) is strictly increasing, we can conclude that \(x_{n+1} = f(x_n) < x_n\). Therefore, the sequence \(\{x_n\}\) is strictly decreasing and unbounded below. Thus, as \(n \to \infty, \;x_n \to -\infty\). **Case (b): \(\alpha < x_1 < \gamma\)**

Prove that \(f(x)\) is a contraction

Recall that a function is a contraction if there exists a constant \(0<\lambda<1\) such that for all \(x, y\) in its domain, we have \(|f(x) - f(y)| \le \lambda |x - y|\). Consider \(f'(x) = \frac{3x^2}{3} = x^2\). Since \(-2 < \alpha < x < \gamma < 2\), we have that \(0 < x^2 < 4\). Hence, \(0 < f'(x) < 4\). Therefore, we can say that \(|f(x) - f(y)| \le 4 |x - y|\) in the given interval.

Apply Banach Fixed Point Theorem

Since \(f(x)\) is a contraction in the interval \(\alpha < x < \gamma\) and \(\alpha, \beta, \gamma\) are fixed points of \(f(x)\), we can apply the Banach Fixed Point Theorem which states that \(\lim_{n\to\infty} x_n = \beta\). **Case (c): \(x_1 > \gamma\)**

Prove that \(f(x) > x\)

We want to show that for any \(x\) in the interval \(\gamma < x\), we have \(f(x) > x\). We have already determined that \(x^3-3x+1<0\) for \(-2 < x < \alpha\), so we know that the inequality switches sign at \(\alpha\) and again at \(\gamma\). Since \(\alpha < \gamma\), we know that the inequality changes sign twice, so we then have \(x^3-3x+1>0\) for \(\gamma<x\).

Conclude that the sequence converges to \(+\infty\)

Since \(f(x_n) > x_n\) in the interval \(\gamma < x_n\), and \(f(x)\) is strictly increasing, we can conclude that \(x_{n+1} = f(x_n) > x_n\). Therefore, the sequence \(\{x_n\}\) is strictly increasing and unbounded above. Thus, as \(n \to \infty, \;x_n \to +\infty\). This completes the proof for all three cases, showing that \(\beta\) can be found using this method, but \(\alpha\) and \(\gamma\) cannot.

Key concepts

Convergence of sequences

When we talk about the convergence of sequences, we're interested in how the terms of a sequence behave as they progress to infinity. Specifically, we're examining if they approach a particular value, or diverge completely. In this context, we have a sequence defined by the iterative function application: \[ x_{n+1} = f(x_n) \]where the function \(f(x)\) is specifically \(\frac{x^3+1}{3}\).In the original exercise, a few scenarios were considered:

• If the initial term is less than \(\alpha\), the sequence diverges to \(-\infty\) because each subsequent term produced by the function is smaller than the previous, eventually moving out of bounds downward.
• For initial values between \(\alpha\) and \(\gamma\), the sequence converges to \(\beta\) since it's shown to be a contraction mapping within that interval.
• Lastly, if the initial term is greater than \(\gamma\), the sequence moves to \(+\infty\) as each new term is larger than the last.

Convergence is key in understanding these movements, as it tells us if a sequence stabilizes over time. If it does stabilize at a certain point, this point is frequently a fixed point of the function, meaning \(f(x) = x\). This foundation is essential in analyzing iterative function behavior, especially concerning fixed points.

Banach Fixed Point Theorem

The Banach Fixed Point Theorem, also known as the contraction mapping theorem, plays a critical role in proving the convergence of our sequence towards the fixed point \(\beta\). This theorem offers assurances about the existence and uniqueness of fixed points under certain conditions in metric spaces. The theorem asserts that a contraction mapping on a complete metric space has a single fixed point, which can be approached by successive iterations starting from any initial point. A contraction mapping \(f\) satisfies a specific property:\[ |f(x) - f(y)| \leq \lambda |x - y| \]for all points \(x\) and \(y\) within a designated domain, where \(0 < \lambda < 1\). In our exercise, within the region \(\alpha < x < \gamma\), where the derivative \(f'(x) = x^2\) maintains values ensuring it behaves like a contraction, we can apply this theorem effectively. Hence, the theorem helps confirm that the iterative sequence converges to the fixed point \(\beta\), thus highlighting its robustness in identifying and locating fixed points when the conditions are satisfied. It gives a structural approach to evaluating how repeated applications of a function can lead to convergence or divergence relative to fixed points.

Monotonic sequences

Monotonic sequences are those that consistently increase or decrease. This property is significant to the iterating sequences in our exercise. Determining whether a sequence is monotonic helps in establishing whether it converges or diverges and in which direction.
In the given exercise, we encounter the sequence \(\{x_n\}\):

• When \(x_1 < \alpha\), the sequence is strictly decreasing since each term \(x_{n+1}\) is less than \(x_n\). This produces a monotonic decreasing sequence which converges to \(-\infty\).
• Conversely, when \(x_1 > \gamma\), the sequence is strictly increasing, shooting upward, classifying it as a monotonic increasing sequence going toward \(+\infty\).

For \(\alpha < x_1 < \gamma\), while a convergent sequence isn't strictly termed monotonic within that fixed literal context as it converges to \(\beta\), initially, minor oscillations might occur till the sequence steadies. This pathway is steered by the contraction property, underlining how convergence can occur in the absence of strict monotonicity, as long as the sequence homes towards a stabilization point.