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let \(\phi_{0}(t)=0\) and use the method of successive approximations to approximate the solution of the given initial value problem. (a) Calculate \(\phi_{1}(t), \ldots, \phi_{4}(t),\) or (if necessary) Taylor approximations to these iterates. Keep tems up to order six. (b) Plot the functions you found in part (a) and observe whether they appear to be converging. Let \(\phi_{n}(x)=x^{n}\) for \(0 \leq x \leq 1\) and show that $$ \lim _{n \rightarrow \infty} \phi_{n}(x)=\left\\{\begin{array}{ll}{0,} & {0 \leq x<1} \\ {1,} & {x=1}\end{array}\right. $$

Short Answer

Expert verified
Question: Calculate φ₁(t), φ₂(t), φ₃(t), and φ₄(t) using the method of successive approximations and confirm the given limit. Answer: All the approximations φ₁(t), φ₂(t), φ₃(t), and φ₄(t) are equal to 0. The limit $$\lim_{n \rightarrow \infty} \phi_{n}(x)=\left\\{\begin{array}{ll}{0,} & {0\leq x<1} \\\ {1,} & {x=1}\end{array}\right.$$ holds true for the given function.

Step by step solution

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01

Define the function φ₀(t) and find its derivative

Given, φ₀(t) = 0. To apply the method of successive approximation, first calculate the derivative of φ₀(t), which should be straightforward: $$\frac{d\phi_{0}(t)}{dt} = \frac{d(0)}{dt} = 0$$
02

Calculate φ₁(t), φ₂(t), φ₃(t), and φ₄(t)

We'll find subsequent approximations by using the relationship between successive iterations: $$φ_{n+1}(t) = \int_a^t \frac{d\phi_n}{dt}(s) ds$$ Now, calculate each of the required approximations using this relationship. 1. For φ₁(t), $$ \phi_{1}(t) = \int_0^t \frac{d\phi_{0}}{dt}(s) ds = \int_0^t 0 ds = 0 $$ 2. For φ₂(t), $$ \phi_{2}(t) = \int_0^t \frac{d\phi_{1}}{dt}(s) ds = \int_0^t 0 ds = 0 $$ 3. For φ₃(t), $$ \phi_{3}(t) = \int_0^t \frac{d\phi_{2}}{dt}(s) ds = \int_0^t 0 ds = 0 $$ 4. For φ₄(t), $$ \phi_{4}(t) = \int_0^t \frac{d\phi_{3}}{dt}(s) ds = \int_0^t 0 ds = 0 $$
03

Consider the Taylor approximations

Since the approximations φ₁(t), φ₂(t), φ₃(t), and φ₄(t) are all equal to 0, there's no need to apply the Taylor approximations, as there are no terms up to order six.
04

Plot the functions

Plot the functions φ₀(t), φ₁(t), φ₂(t), φ₃(t), and φ₄(t) to observe if they appear to be converging. In this case, all the functions are equivalent to 0, thus making them the same function. Therefore, it can be said that the approximations are converging.
05

Show the limit

To show that $$\lim_{n \rightarrow \infty} \phi_{n}(x)=\left\\{\begin{array}{ll}{0,} & {0\leq x<1} \\\ {1,} & {x=1}\end{array}\right,$$ it's necessary to consider that, in this case, $$\phi_n(x) = x^n$$. So, we need to evaluate the limit as n approaches infinity for the given function: $$ \lim_{n \rightarrow \infty} x^n=\left\\{\begin{array}{ll}{0,} & {0 \leq x < 1} \\\ {1,} & {x = 1}\end{array}\right. $$ This confirms the given limit holds true for the function, as shown above.

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding the Initial Value Problem
An initial value problem (IVP) in the realm of differential equations is a problem where you are given a differential equation along with the value of the unknown function at a specific point, referred to as the initial value. The goal is to find a function that satisfies both the differential equation and the initial condition. Many real-life phenomena, from physics to finance, can be modeled using IVPs, providing a starting point for predictions and analyses.

The method of successive approximations, also known as Picard iteration, is a technique to tackle IVPs. It involves starting with an initial guess for the function, often dictated by the problem's initial conditions, and then refining this guess iteratively. Each step of iteration involves using the prior approximation to generate a new one that is closer to the true solution. In the provided exercise, we start with \( \phi_{0}(t)=0 \) and use this method to create subsequent approximations \( \phi_{1}(t), \phi_{2}(t), \ldots \) until they converge to a solution.
Taylor Approximations Simplified
Taylor approximations are a powerful mathematical tool used to estimate the values of functions. By expanding a function into a series at a particular point, Taylor approximations can give us a polynomial that represents the function's behavior near that point. It's like sketching a simple outline that captures the essence of a more complex shape.

In the context of differential equations and successive approximations, if our initial estimates for the solution are not zero—unlike in our exercise where each \( \phi_{n}(t) \) was zero—we would use Taylor approximations to refine these estimates. The idea is to take the known value of a function at a point and then add terms that incorporate the rates of change of the function at that point to better approximate the function nearby. Up to order six means we would include terms from the constant (the value at the point) up to the sixth derivative in our approximation.
Convergence of Functions: What Does it Mean?
In mathematics, the convergence of functions refers to the idea that a sequence of functions gets arbitrarily close to a given function as the sequence progresses. This is a crucial concept when discussing iterative processes like the method of successive approximations.

In our exercise, we calculate several iterations \( \phi_{1}(t), \phi_{2}(t), ... \) using the initial guess \( \phi_{0}(t) = 0 \) and observe whether the sequence is converging towards a solution to the initial value problem. This is equivalent to observing whether the functions in the sequence are becoming indistinguishable from each other as we proceed with iterations, which in our case they are, since all \( \phi_{n}(t) \) are zero. Convergence can be analyzed graphically, as requested in the exercise, or by using mathematical tests for convergence.
Differential Equations and Their Significance
Differential equations are equations that involve an unknown function and its derivatives. They are integral in modeling systems where change is continuous and can occur in various scientific fields such as physics, engineering, biology, and economics.

The initial value problem is a type of differential equation problem that includes conditions at a single point, embodying the principle that specifying the state of a system at one time can determine its future behavior. To solve a differential equation means to find all functions that satisfy the equation, with or without those initial conditions.

In the given exercise, the successive approximations method applied to solve the differential equation illustrates the power of iterative techniques in handling such problems. While this specific example resulted in a straightforward set of zero functions, more complex problems would yield a sequence of increasingly accurate approximations that approach the true solution.

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Most popular questions from this chapter

Involve equations of the form \(d y / d t=f(y)\). In each problem sketch the graph of \(f(y)\) versus \(y,\) determine the critical (equilibrium) points, and classify each one as asymptotically stable or unstable. $$ d y / d t=-2(\arctan y) /\left(1+y^{2}\right), \quad-\infty

Show that the equations are not exact, but become exact when multiplied by the given integrating factor. Then solve the equations. $$ (x+2) \sin y d x+x \cos y d y=0, \quad \mu(x, y)=x e^{x} $$

let \(\phi_{0}(t)=0\) and use the method of successive approximations to solve the given initial value problem. (a) Determine \(\phi_{n}(t)\) for an arbitrary value of \(n .\) (b) Plot \(\phi_{n}(t)\) for \(n=1, \ldots, 4\). Observe whether the iterates appear to be converging. $$y^{\prime}=t y+1, \quad y(0)=0$$

Show that if \(\left(N_{x}-M_{y}\right) / M=Q,\) where \(Q\) is a function of \(y\) only, then the differential equation $$ M+N y^{\prime}=0 $$ has an integrating factor of the form $$ \mu(y)=\exp \int Q(y) d y $$

Consider the initial value problem $$ y^{\prime}=3 t^{2} /\left(3 y^{2}-4\right), \quad y(1)=0 $$ (a) Use the Euler formula ( 6) with \(h=0.1\) to obtain approximate values of the solution at \(t=1.2,1.4,1.6,\) and 1.8 . (b) Repeat part (a) with \(h=0.05\). (c) Compare the results of parts (a) and (b). Note that they are reasonably close for \(t=1.2,\) \(1.4,\) and 1.6 but are quite different for \(t=1.8\). Also note (from the differential equation) that the line tangent to the solution is parallel to the \(y\) -axis when \(y=\pm 2 / \sqrt{3} \cong \pm 1.155 .\) Explain how this might cause such a difference in the calculated values.

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