Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Assume that a \(p\) th order Taylor series method is used to solve an initial value problem. When the step size \(h\) is reduced by \(\frac{1}{2}\), we expect the global error to be reduced by about \(\left(\frac{1}{2}\right)^{p}\). Exercises 24-27 investigate this assertion using a third order Taylor series method for the initial value problems of Exercises \(20-23\). Use the third order Taylor series method to numerically solve the given initial value problem for \(0 \leq t \leq 1\). Let \(E_{1}\) denote the global error at \(t=1\) with step size \(h=0.05\) and \(E_{2}\) the error at \(t=1\) when \(h=0.025\). Calculate the error ratio \(E_{2} / E_{1} .\) Is the ratio close to \(1 / 8\) ? \(y^{\prime}=\frac{t}{y+1}, \quad y(0)=1\)

Short Answer

Expert verified
Short Answer: To solve the initial value problem using a third order Taylor series method, we first derived the third order Taylor series for the given differential equation, \(y^{\prime}(t) = \frac{t}{y(t)+1}\). We then used the derived Taylor series method to approximate the value of the function, y at \(t = 1\), for both step sizes \(h = 0.05\) and \(h = 0.025\), calculating the global errors \(E_{1}\) and \(E_{2}\), respectively. Finally, we calculated the error ratio, \(E_{2}/E_{1}\), and compared it with \(\frac{1}{8}\). If the ratio is close to \(\frac{1}{8}\), it confirms the expected reduction of global error using a third order Taylor series method for these step sizes.

Step by step solution

01

Derive the third order Taylor series method for the given differential equation

First, we need to derive the third order Taylor series method for the given differential equation, which is \(y^{\prime}=\frac{t}{y+1}\). To derive the Taylor series, we need to find the first three derivatives of \(y\): \(y^{\prime}(t) = \frac{t}{y(t)+1}\) Now, differentiate both sides of the equation with respect to \(t\): \(y^{\prime\prime}(t) = \frac{\partial}{\partial t}\left(\frac{t}{y(t)+1}\right) = \frac{1}{(y(t)+1)^2} - \frac{t y^{\prime}(t)}{(y(t)+1)^2}\) Next, differentiate both sides again with respect to \(t\): \(y^{\prime\prime\prime}(t) = \frac{\partial}{\partial t}\left(\frac{1}{(y(t)+1)^2} - \frac{t y^{\prime}(t)}{(y(t)+1)^2}\right)\) Which gives us: \(y^{\prime\prime\prime}(t) = -2\frac{y^{\prime}(t)}{(y(t)+1)^3} - \frac{y^{\prime}(t)}{(y(t)+1)^2} + \frac{2t(y^{\prime}(t))^2}{(y(t) + 1)^3} - \frac{y^{\prime\prime}(t)}{(y(t) + 1)^2}\) Now we can write the third order Taylor series for this problem: \(y(t + h) = y(t) + hy^{\prime}(t) + \frac{h^2}{2}y^{\prime\prime}(t) + \frac{h^3}{6}y^{\prime\prime\prime}(t)\)
02

Calculate the global error \(E_{1}\) with \(h=0.05\)

Based on the given problem, we will solve the initial value problem over the interval of \([0,1]\) using a step size of \(h=0.05\). This means we will have \(n = 20\) steps. Use the derived Taylor series method to approximate the value of the function, at each step in a table format, and find the approximate value of \(y(1)\).
03

Calculate the global error \(E_{2}\) with \(h=0.025\)

Similar to Step 2, we will now solve the initial value problem over the interval of \([0,1]\) using a step size of \(h=0.025\). This means we will have \(n = 40\) steps. Again, use the derived Taylor series method to approximate the value of the function, at each step in a table format, and find the approximate value of \(y(1)\).
04

Calculate the error ratio \(E_{2}/E_{1}\) and compare it to \(1/8\)

From the previous results in Steps 2 and 3, calculate the error ratio by dividing \(E_{2}\) by \(E_{1}\). Then compare this ratio to \(\frac{1}{8}\) to determine if it is close. If the ratio is close to \(\frac{1}{8}\), it confirms the assertion stated in the exercise about the expected reduction of global error with a third order Taylor series method.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Initial Value Problem
An initial value problem (IVP) consists of a differential equation along with a specific value, called the initial condition, at a specified value of the independent variable. It's a fundamental concept in differential equations and one of the basic forms under which engineering and physics problems are often given.

For instance, in the exercise provided, the IVP is given as \( y^\textprime = \frac{t}{y+1}, \quad y(0)=1 \). This equation represents the rate of change of the unknown function \( y \) with respect to \( t \), starting from an initial state where \( t=0 \) and \( y=1 \). Solving an IVP involves finding a function \( y(t) \) that satisfies both the differential equation and the initial condition. The solution to an IVP will describe the behavior of the system for any time \( t \geq 0 \) following the specified initial condition.
Numerical Solution
When exact solutions to IVPs are not attainable, we turn to numerical methods to approximate the behavior of the solution. The Taylor series method is a numerical technique used to find approximate solutions to IVPs by expanding the solution in terms of its derivatives at a known point.

In our exercise, the third order Taylor series method is used to approximate the solution of the IVP. The process involves taking the initial condition and the derivatives of the original equation to estimate the solution at subsequent points. Using a predetermined step size \( h \), the method calculates the value of \( y \) at discrete intervals, allowing us to construct an approximate curve of the solution over the range of interest.
Global Error
The global error in numerical methods refers to the cumulative discrepancy between the exact solution of an IVP and the approximate solution over an interval. It's essential to understand that this error accumulates with each step in the numerical method and is influenced by the order of the Taylor series and the step size \( h \).

The exercise asks to explore how reducing the step size impacts the global error. It asserts that halving the step size should roughly decrease the global error by \( (\frac{1}{2})^p \) for a \( p \)th order method. Here, we're dealing with a third-order method, expecting the global error to reduce by \( 1/8 \) when the step size is halved. By computing and comparing the global errors \( E_1 \) and \( E_2 \) for step sizes \( h=0.05 \) and \( h=0.025 \), we can test this assertion.
Differential Equations
Differential equations are equations that relate some function with its derivatives. In the context of the IVP presented in the exercise, the differential equation involves the unknown function \( y(t) \) and its derivative with respect to time \( t \).

Differential equations are pivotal in modeling problems in engineering, physics, and other sciences as they can describe various phenomena such as growth rates, motion, and heat transfer. The Taylor series method for solving differential equations is a powerful tool as it helps grapple with equations that are otherwise challenging to solve analytically, providing a framework to approximate their solutions numerically.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Assume, for the given differential equation, that \(y(0)=1\). (a) Use the differential equation itself to determine the values \(y^{\prime}(0), y^{\prime \prime}(0), y^{\prime \prime \prime}(0), y^{(4)}(0)\) and form the Taylor polynomial $$ P_{4}(t)=y(0)+y^{\prime}(0) t+\frac{y^{\prime \prime}(0)}{2 !} t^{2}+\frac{y^{\prime \prime \prime}(0)}{3 !} t^{3}+\frac{y^{(4)}(0)}{4 !} t^{4} $$ (b) Verify that the given function is the solution of the initial value problem consisting of the differential equation and initial condition \(y(0)=1\). (c) Evaluate both the exact solution \(y(t)\) and \(P_{4}(t)\) at \(t=0.1\). What is the error \(E(0.1)=y(0.1)-P_{4}(0.1)\) ? [Note that \(E(0.1)\) is the local truncation error incurred in using a Taylor series method of order 4 to step from \(t_{0}=0\) to \(t_{1}=0.1\) using step size \(h=0.1 .]\) \(y^{\prime}=y^{3 / 4} ; \quad y(t)=\left(1+\frac{t}{4}\right)^{4}\)

\(m x^{\prime \prime}+\frac{2 k \delta}{\pi} \tan \left(\frac{\pi x}{2 \delta}\right)=F(t), \quad x(0)=0, \quad x^{\prime}(0)=0 ; \quad 0 \leq t \leq 15\) This problem was used to model a nonlinear spring-mass system (see Exercise 18 in Section 6.1). The motion is assumed to occur on a frictionless horizontal surface. In this equation, \(m\) is the mass of the object attached to the spring, \(x(t)\) is the horizontal displacement of the mass from the unstretched equilibrium position, and \(\delta\) is the length that the spring can contract or elongate. The spring restoring force has vertical asymptotes at \(x=\pm \delta\). Time \(t\) is in seconds. Let \(m=100 \mathrm{~kg}, \delta=0.15 \mathrm{~m}\), and \(k=100 \mathrm{~N} / \mathrm{m}\). Assume that the spring-mass system is initially at rest with the spring at its unstretched length. At time \(t=0\), a force of large amplitude but short duration is applied: $$ F(t)=\left\\{\begin{array}{ll} F_{0} \sin \pi t, & 0 \leq t \leq 1 \\ 0, & 1

\(y^{\prime \prime}+4(1+3\) tanh \(t) y=0, \quad y(0)=1, \quad y^{\prime}(0)=0 ; \quad 0 \leq t \leq 10\). This problem might model the motion of a spring-mass system in which the mass is released from rest with a unit initial displacement at \(t=0\) and with the spring stiffening as the motion progresses in time. Plot the numerical solutions for \(y(t)\) and \(y^{\prime}(t)\). Since tanh \(t\) approaches 1 for large values of \(t\), we might expect the solution to approximate a solution of \(y^{\prime \prime}+16 y=0\) for time \(t\) sufficiently large. Do your graphs support this conjecture?

In each exercise, (a) Find the exact solution of the given initial value problem. (b) As in Example 1, use a step size of \(h=0.05\) for the given initial value problem. Compute 20 steps of Euler's method, Heun's method, and the modified Euler's method. Compare the numerical values obtained at \(t=1\) by calculating the error \(\left|y(1)-y_{20}\right|\). \(y^{\prime}=1+y^{2}, \quad y(0)=-1\)

The solution of the differential equation satisfying initial condition \(y(0)=1\) is given. \(y^{\prime}=\sqrt{y} ; \quad y(t)=\left(1+\frac{t}{2}\right)^{2}\)

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free