Problem 8
Question
Suppose \(f: X \rightarrow X\) is a contraction for \(k<1 .\) Suppose you use the iteration procedure with \(x_{n+1}:=f\left(x_{n}\right)\) as in the proof of the fixed point theorem. Suppose \(x\) is the fixed point of \(f .\) a) Show that \(d\left(x, x_{n}\right) \leq k^{n} d\left(x_{1}, x_{0}\right) \frac{1}{1-k}\) for all \(n \in \mathbb{N}\). b) Suppose \(d\left(y_{1}, y_{2}\right) \leq 16\) for all \(y_{1}, y_{2} \in X,\) and \(k=1 / 2 .\) Find an \(N\) such that starting at any point \(x_{0} \in X,\) \(d\left(x, x_{n}\right) \leq 2^{-16}\) for all \(n \geq N\)
Step-by-Step Solution
Verified Answer
a) Use contraction to show the bound. b) Choose \( N = 16 \).
1Step 1: Understand the Contraction Mapping
In a contraction mapping, there’s a constant \( k < 1 \) such that for any two points \( x, y \in X \), the distance \( d(f(x), f(y)) \leq k \cdot d(x, y) \). This means the function \( f \) brings points closer together with each iteration.
2Step 2: Initial Setup for the Iteration
The sequence is defined as \( x_{n+1} = f(x_n) \). Our goal is to find a bound for \( d(x, x_n) \), where \( x \) is the fixed point such that \( f(x) = x \).
3Step 3: Use the Contraction Property to Bound Distances
We begin by showing \( d(x, x_{n+1}) \leq k \cdot d(x, x_n) \). This is due to the property \( d(f(x), f(x_n)) \leq k \cdot d(x, x_n) \) and since \( f(x) = x \), we have \( d(x, x_{n+1}) = d(f(x), f(x_n)) \leq k \cdot d(x, x_n) \).
4Step 4: Bound the Distance Using Geometric Series
By induction or iterating the contraction property: \( d(x, x_n) \leq k^n d(x_1, x_0) \). Recognizing it as a geometric series, the sum becomes \( d(x, x_0) \cdot \sum_{i=0}^{n-1} k^i \). The sum of an infinite geometric series \( \sum_{i=0}^{\infty} k^i = \frac{1}{1-k} \), so the distance can be bounded by \( d(x, x_n) \leq k^n d(x_1, x_0) \cdot \frac{1}{1-k} \).
5Step 5: Apply Given Conditions for Part b
For part b, we have \( d(y_1, y_2) \leq 16 \) and \( k = \frac{1}{2} \). We need \( d(x, x_n) \leq 2^{-16} \), set \( k^N d(x_1, x_0) \cdot \frac{1}{1-k} = 2^{-16} \). Solving for \( N \) gives \( N = \lceil 16 \rceil \), since \( k=\frac{1}{2} \) leads to \( \left(\frac{1}{2}\right)^N \). The series converges rapidly, and using these approximations, \( N = 16 \) should suffice for \( 2^{-16} \).
Key Concepts
Fixed Point TheoremDistance BoundingGeometric SeriesIteration Procedure
Fixed Point Theorem
The Fixed Point Theorem is an important concept in mathematics, particularly in the context of contraction mappings. A fixed point for a function \( f \) is a value \( x \) such that \( f(x) = x \). In simpler terms, this means that when \( f \) is applied to \( x \), it gives back \( x \) unchanged. This theorem assures us of the existence of such a point under certain conditions.
In the specific case of contraction mappings, if \( f: X \rightarrow X \) is a contraction, then it guarantees that there is exactly one fixed point for \( f \). This unique existence is crucial because it provides a predictable outcome in iterative processes.
In the specific case of contraction mappings, if \( f: X \rightarrow X \) is a contraction, then it guarantees that there is exactly one fixed point for \( f \). This unique existence is crucial because it provides a predictable outcome in iterative processes.
- Contraction means that distances between points decrease by a consistent factor \( k \)
- This shrinking effect results in the sequence generated by repeated application of \( f \) converging to the fixed point
- The condition \( k < 1 \) ensures that \( f \) brings any two points closer together
Distance Bounding
Distance Bounding helps us understand how close our iterative sequence is to the fixed point. In a contraction mapping \( f \), the distance between the fixed point \( x \) and the current position \( x_n \) can be effectively bounded. The bound is given by \( d(x, x_n) \leq k^n d(x_1, x_0) \frac{1}{1-k} \). This equation provides a way to anticipate how rapidly a sequence is converging to its fixed point.
In practical terms, you are given:
In practical terms, you are given:
- \( k^n \): The contraction factor raised to the power of the iteration number \( n \)
- \(d(x_1, x_0)\): The initial distance between the starting points of the sequence
- \(\frac{1}{1-k}\): The factor derived from the infinite geometric series
Geometric Series
Understanding geometric series is key in estimating distances in contraction mappings. A geometric series is a series of terms in which each term is a constant multiple, \( k\), of the preceding one. In our context, the sum \( \sum_{i=0}^{n-1} k^i \) represents the cumulative effect of these iterations when bounding the total distance to the fixed point.
The magic of geometric series is in its convergence properties. When the ratio \( k \) is less than 1, the series \( \sum_{i=0}^{\infty} k^i \) converges to \( \frac{1}{1-k} \). This allows us to replace an infinite sequence with a finite, manageable constant for iterations:
The magic of geometric series is in its convergence properties. When the ratio \( k \) is less than 1, the series \( \sum_{i=0}^{\infty} k^i \) converges to \( \frac{1}{1-k} \). This allows us to replace an infinite sequence with a finite, manageable constant for iterations:
- Each term \( k^i \) represents the diminishing distances through iterations
- The series sum \( \frac{1}{1-k} \) bounds total distance over infinite iterations
Iteration Procedure
In mathematics, the Iteration Procedure is a tool used to iteratively apply a function to approximate a solution. Consider a sequence defined by \( x_{n+1} = f(x_n) \). It incrementally gets closer to the fixed point \( x \) of \( f \). The power of this procedure lies in its simplicity and effectiveness at solving problems iteratively.
How it works:
How it works:
- The sequence starts from an initial guess \( x_0 \)
- Each subsequent \( x_{n+1} \) is obtained by applying the function \( f \) to the current point \( x_n \)
- Convergence is assured due to the contraction property, depending on the constant \( k \)
Other exercises in this chapter
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