Problem 7
Question
Consider \(y^{\prime}=y^{2}, y(0)=1 .\) Use the iteration scheme from the proof
of the contraction mapping principle. Start with \(f_{0}(x)=1 .\) Find a few
iterates (at least up to \(f_{2}\) ). Prove that the pointwise limit of \(f_{n}\)
is \(\frac{1}{1-x},\) that is for every \(x\) with \(|x|
Step-by-Step Solution
Verified Answer
The limit \(f_n(x)\) is \(\frac{1}{1-x}\) for \(|x| < 1\).
1Step 1: Understand the Problem
We are given a differential equation \(y' = y^2\) with an initial condition \(y(0) = 1\). We need to apply an iterative scheme to approximate the solution, starting with \(f_0(x) = 1\). The goal is to establish that the limit of these approximations is \(\frac{1}{1-x}\) for \(|x|
2Step 2: Initialize Iteration Sequence
Our initial function according to the problem is \(f_0(x) = 1\). This serves as the starting point for the iterative process.
3Step 3: Define the Iteration Scheme
Using the contraction mapping principle, our iteration scheme involves updating \(f_n(x)\) to \(f_{n+1}(x)\) by integrating the differential equation. The general transformation is given as \(f_{n+1}(x) = 1 + \int_0^x f_n(t)^2 \, dt\).
4Step 4: Compute First Iteration
Calculate \(f_1(x)\). Start with \(f_0(x) = 1\), so \(f_1(x) = 1 + \int_0^x 1^2 \, dt = 1 + x\).
5Step 5: Compute Second Iteration
Now, compute \(f_2(x)\) using \(f_1(x) = 1 + x\): \(f_2(x) = 1 + \int_0^x (1+t)^2 \, dt\). Calculate this integral to find that \(f_2(x) = 1 + x + \frac{x^2}{2} + \frac{x^3}{3}\).
6Step 6: Identify General Pattern
Observe that each iteration adds terms of the binomial expansion of \(\frac{1}{1-x}\), suggesting that \(f_n(x)\) accumulates terms of \(1 + x + x^2 + \cdots\) up to the polynomial corresponding to \(x^n\).
7Step 7: Prove Convergence to Limit
Use the property of geometric series: \(\frac{1}{1-x} = 1 + x + x^2 + x^3 + \cdots\). Show that as \(n\) increases, \(f_n(x)\) approaches the series expansion of \(\frac{1}{1-x}\) termwise because it approximates successive terms \(x^n/n\). As \(n\to\infty\), \(f_n(x)\to\frac{1}{1-x}\) for \(|x| < 1\).
8Step 8: Establish Radius of Convergence
Confirm that the series converges for \(|x| < 1\), ensuring \(h = 1\). Thus, \(\lim_{n \to \infty} f_n(x) = \frac{1}{1-x}\) holds within this interval.
Key Concepts
Iterative MethodDifferential EquationSeries ExpansionRadius of Convergence
Iterative Method
The iterative method is a process used to approach the solution of a problem through successive approximations. In this context, it involves starting with an initial guess, denoted as \( f_0(x) \), and iteratively refining this guess to get closer to the actual solution of a differential equation. In our exercise, the iterative method is applied starting with \( f_0(x) = 1 \).
Through each iteration, a new function \( f_{n+1}(x) \) is generated by computing an integral involving the square of the previous function \( f_n(x) \).
Through each iteration, a new function \( f_{n+1}(x) \) is generated by computing an integral involving the square of the previous function \( f_n(x) \).
- Start with an initial function: \( f_0(x) = 1 \).
- Update this function iteratively: \( f_{n+1}(x) = 1 + \int_0^x f_n(t)^2 \, dt \).
Differential Equation
A differential equation involves functions and their derivatives. Here, we are dealing with an ordinary differential equation (ODE) written as \( y' = y^2 \), with an initial condition \( y(0) = 1 \).
This specific equation requires finding a function \( y \) such that its derivative with respect to \(x\), \( y' \), equals the square of \( y \).
This problem is tackled using our iterative process to approximate the function that satisfies both the differential equation and its initial condition.
This specific equation requires finding a function \( y \) such that its derivative with respect to \(x\), \( y' \), equals the square of \( y \).
This problem is tackled using our iterative process to approximate the function that satisfies both the differential equation and its initial condition.
- The derivative \( y' \) is given as \( y^2 \).
- An initial condition is stated: \( y(0) = 1 \).
- The solution must satisfy these conditions through successive approximations.
Series Expansion
Series expansions allow functions to be expressed as infinite sums, enabling easier manipulation and understanding. Here, the function \( \frac{1}{1-x} \) is key, which can be expanded into a geometric series:
- The series expansion is \( \frac{1}{1-x} = 1 + x + x^2 + x^3 + \cdots \).
- Each term corresponds to powers of \( x \), providing a polynomial approximation when summed up to \( n \) terms.
Radius of Convergence
The radius of convergence is a concept that determines where a series converges to a finite value. For the series \( \frac{1}{1-x} = 1 + x + x^2 + x^3 + \cdots \), it converges when \(|x| < 1\).
- This radius, here \( h = 1 \), sets the boundary within which we can safely say \( f_n(x) \to \frac{1}{1-x} \) as \( n \to \infty \).
- Convergence within this interval ensures the validity of our iterative method over a range \(|x| < 1\).
Other exercises in this chapter
Problem 6
Let \(\left(X, d_{X}\right)\) and \(\left(Y, d_{Y}\right)\) be metric spaces. a) Show that \((X \times Y, d)\) with \(d\left(\left(x_{1}, y_{1}\right),\left(x_{
View solution Problem 7
In any metric space, prove: a) \(E\) is closed if and only if \(\partial E \subset E\). b) \(U\) is open if and only if \(\partial U \cap U=\emptyset\).
View solution Problem 7
Let \(C([a, b], \mathbb{R})\) be the metric space as in Example 7.1.8. Show that \(C([a, b], \mathbb{R})\) is a complete metric space.
View solution Problem 7
A set \(S \subset X\) is said to be dense in \(X\) if \(X \subset \bar{S}\) or in other words if for every \(x \in X,\) there exists a sequence \(\left\\{x_{n}\
View solution