Question

For the given initial value problem, (a) Execute 20 steps of the Taylor series method of order \(p\) for \(p=1,2,3\). Use step size \(h=0.05 .\) (b) In each exercise, the exact solution is given. List the errors of the Taylor series method calculations at \(t=1\). \(y^{\prime}=2 t y^{2}, y(0)=-1 . \quad\) The exact solution is \(y(t)=\frac{-1}{1+t^{2}}\)

Short answer

#Answer# For each Taylor series order p=1, 2, and 3, calculate the approximated solution at the 20th step (which corresponds to t=1) and compute the approximation error compared to the exact solution y(1)=-1/2. After performing the Taylor series method steps for each order and comparing the approximated solutions to the exact solution at t=1, we find the following errors: Error for Taylor series order p=1: error_1 = |y_exact(1)-y_approx_p1(1)| Error for Taylor series order p=2: error_2 = |y_exact(1)-y_approx_p2(1)| Error for Taylor series order p=3: error_3 = |y_exact(1)-y_approx_p3(1)| These errors give an indication of the accuracy and effectiveness of each Taylor series method order for approximating the exact solution of the given IVP.

Step-by-step solution

Find the Taylor series formula of orders 1, 2, and 3 with \(p=1,2,3\) respectively.

For each order p, find the derivative order p+1(\(y^{(p+1)}\)) and apply the version of the Taylor series centered at the IVP. Talyor series with p=1: \(y_{n+1}=y_n+hy'_n\) Taylor series with p=2: \(y_{n+1}=y_n+hy'_n+\frac{h^2}{2}y''_n\) Taylor series with p=3: \(y_{n+1}=y_n+hy'_n+\frac{h^2}{2}y''_n+\frac{h^3}{6}y'''_n\)

Calculate all the necessary derivatives for orders p=1,2,3.

\(y'=2ty^2\) \(y''=dy'/dt=2 \times (2y^{2}+2ty \times 2ty)=8ty^{3}+8t^{2}y^{2}\) \(y'''=dy''/dt=(8y^{3}+24t^{2}y^{2})\)

Perform 20 steps of Taylor series method with each order and step size h=0.05.

For each order, iterate over 20 steps, using the Taylor series formula from Step 1 and derivatives from Step 2. Initial conditions for each order: \(y_0=-1\), \(t_0=0\)

Calculate the error at t=1 comparing the exact solution and approximated solution for each order.

At t=1, use each of the approximated solutions from each method to calculate the approximation error using the formula: \(error = \text{abs}(y_{\text{exact}}(t)-y_{\text{approx}}(t))\) Exact solution at \(t=1\) is: \(y(1) = \frac{-1}{1+1^2} = -\frac{1}{2}\). Find the approximation error for each Taylor series order at t=1.

Key concepts

Initial Value Problem

An initial value problem (IVP) consists of a differential equation along with a specific value, called the initial condition, which the solution must satisfy at the beginning of the interval of interest. It's a crucial concept in both pure and applied mathematics, as it describes the evolution of various phenomena over time.

For instance, the given exercise deals with the IVP defined by the ordinary differential equation (ODE) \( y' = 2ty^2 \) with the initial condition \( y(0) = -1 \). This sets the stage for all future steps calculated during numerical approximations; the solution starts from this point and then unravels over the interval, hence being crucial for predicting the behavior of the system described by the ODE.

Numerical Methods for Differential Equations

Numerical methods are algorithms used to approximate solutions for differential equations, particularly when finding an exact solution is difficult or impossible. These methods are important for scientists and engineers who often encounter complex problems that can't be solved analytically.

The Taylor series method is a numerical technique used to solve initial value problems like the one provided. It approximates the solution by expanding it into a series using derivatives of the function at a specific point. Doing so, it systematically improves the accuracy of the solution with every additional term included from the Taylor expansion.

Order of Taylor Series

The order of a Taylor series refers to the highest degree of derivatives used in the series expansion. The more terms (higher order derivatives) you include, the more accurate the approximation becomes, up to a certain point. However, higher orders require more computational effort.

In the context of the provided exercise, Taylor series methods of order \( p = 1, 2, 3 \) are considered. Each order incorporates more terms and thus, higher derivatives. For instance, a first-order Taylor series (\( p = 1 \) includes only the first derivative, while a third-order series (\( p = 3 \) includes up to the third derivative, reflecting an incremental step in the complexity and potential accuracy of the solution.

Step Size in Numerical Approximation

Step size, denoted as \( h \) in numerical methods, is essentially the interval between successive points when approximating the solution of an ODE. A smaller step size generally leads to a more accurate solution by reducing discretization errors but increases the computational load, while a larger step size reduces the number of calculations at the cost of accuracy.

In the given exercise, \( h = 0.05 \) represents a fine balance between the accuracy of the solution and the computational efforts required. The step size dictates how 'smoothly' the numerical method traces the solution curve, with smaller sizes hugging the curve more closely. It's a fundamental parameter in any numerical approximation method, such as the Taylor series method, affecting the outcome's accuracy and reliability.