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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 \(h=0.1 .]\) \(y^{\prime}=y+\sin t ; \quad y(t)=\frac{3 e^{t}-\cos t-\sin t}{2}\)

Short Answer

Expert verified
Question: Find the Taylor polynomial of order 4 for the given differential equation \(y'(t) = y(t) + \sin(t)\) with initial condition \(y(0) = 1\). Verify that the given function \(y(t) = \frac{3e^t - \cos(t) - \sin(t)}{2}\) is the solution to the initial value problem. Evaluate the exact solution and the Taylor polynomial at \(t=0.1\), and compute the error. Answer: The Taylor polynomial of order 4 for the given differential equation is \(P_4(t) = 1 + t + \frac{t^2}{2} + \frac{2t^3}{6} + \frac{t^4}{24}\). The given function is indeed the solution to the initial value problem. The exact solution at \(t=0.1\) is approximately 1.0280199, and the Taylor polynomial at \(t=0.1\) is approximately 1.0279167. The error between the exact solution and the Taylor polynomial is approximately 0.0001032.

Step by step solution

01

First given the differential equation \(y'=y+\sin t\), we can directly compute \(y'(0) = y(0) + \sin(0) = 1\) as \(y(0) = 1\). #Step 2: Compute \(y''(0)\)

To find the second derivative of \(y\), we differentiate the equation \(y'=y+\sin t\) with respect to \(t\) and get \(y'' = y' + \cos t\). From the previous step we can find \(y''(0) = y'(0) + \cos(0) = 1+1 = 2\). #Step 3: Compute \(y'''(0)\)
02

To find the third derivative of \(y\), we differentiate the equation \(y'' = y' + \cos t\) with respect to \(t\) and obtain \(y''' = y'' - \sin t\). We replace the values we already have and get \(y'''(0)=y''(0)-\sin(0)=2-0=2\). #Step 4: Compute \(y^{(4)}(0)\)

To find the fourth derivative of \(y\), we differentiate the equation \(y''' = y'' - \sin t\) with respect to \(t\) again, and we get \(y^{(4)} = y'''-\cos t\). Then, we substitute the values to obtain \(y^{(4)}(0) = y'''(0)-\cos(0)=2-1=1\). #Step 5: Form the Taylor polynomial
03

Using the values we found in previous steps, we can form the Taylor polynomial \(P_4(t)\): $$ P_{4}(t)=1 + 1t + \frac{2}{2!} t^{2} + \frac{2}{3!}t^{3} + \frac{1}{4!}t^{4} $$ #Step 6: Verify the solution

The given function that we would like to verify as the solution of the initial value problem is \(y(t)=\frac{3e^{t}-\cos t - \sin t}{2}\). We first need to check whether it satisfies the initial condition \(y(0)= 1\). Indeed, by evaluating \(y(t)\) at \(t=0\), we obtain \(y(0) = \frac{3-\cos 0 - \sin 0}{2} = 1\), so it holds true. To confirm that it also satisfies the differential equation, we can find its derivative and check if it obeys the given equation. Differentiating \(y(t)\) with respect to \(t\), we get: $$y'(t)=\frac{3e^{t}-\sin t+\cos t}{2}$$ We can rewrite the differential equation as follows to match the structure of our solution: $$y'(t) = y(t)+\sin t = \frac{3e^{t}-\cos t+\sin t+\sin^2 t}{2}$$ Comparing this with our computed derivative, we can see that it matches. Therefore, the given function is indeed the solution of the initial value problem. #Step 7: Evaluate the exact solution and Taylor polynomial at \(t=0.1\)
04

Using the provided function and the Taylor polynomial derived before, we can compute their values at \(t=0.1\): Exact solution: $$ y(0.1)=\frac{3e^{0.1}-\cos 0.1 - \sin 0.1}{2} \approx 1.0280199$$ Taylor polynomial: $$ P_{4}(0.1) \approx 1 + 0.1 + \frac{2}{2!}(0.1^2) + \frac{2}{3!}(0.1^3) + \frac{1}{4!}(0.1^4) \approx 1.0279167$$ #Step 8: Calculate the error

Now, we can compute the error between the exact solution and the polynomial representation: $$E(0.1) = y(0.1) - P_{4}(0.1) \approx 1.0280199-1.0279167 \approx 0.0001032$$ The error between the exact solution and the Taylor polynomial is approximately 0.0001032.

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Key Concepts

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

Taylor Series
The Taylor Series is a powerful tool in mathematics that allows us to express a function as an infinite sum of terms. These terms are calculated from the values of the function's derivatives at a single point. In this exercise, we focus on forming a Taylor polynomial, which is a finite approximation of a Taylor series. Specifically, we use derivatives evaluated at the initial point, in this case, zero, to construct the polynomial.
For a function \(f(t)\), the Taylor series expansion around a point \(t = a\) can be written as:
\[ f(t) = f(a) + f'(a)(t-a) + \frac{f''(a)}{2!}(t-a)^2 + \cdots \]
In our problem, we're constructing the fourth-degree Taylor polynomial \(P_4(t)\), using the derivatives calculated at \(t = 0\). This polynomial provides a close approximation of the original function for small values of \(t\).
Initial Value Problem
An Initial Value Problem (IVP) involves solving a differential equation given an initial condition. This condition specifies the value of the function at a certain point, often making the problem unique and solvable.
In this exercise, the differential equation is \(y' = y + \sin t\), with the initial condition \(y(0) = 1\). The task is to find a function \(y(t)\) that satisfies both the equation and the initial value.
This method is a common approach in mathematical modeling and physics, where the behavior of dynamic systems over time is determined from known starting conditions. The specific solution is verified by substituting it back into the differential equation and ensuring both the equation and initial condition are satisfied.
Local Truncation Error
Local truncation error measures how much error is introduced when approximating a function with a finite Taylor polynomial instead of using the full series. It indicates the difference between the exact solution and the polynomial approximation at a specific point.
In this context, the error \(E(0.1)\) at \(t = 0.1\) represents the local truncation error incurred when using the fourth-order Taylor series approximation. It's calculated as the difference between the exact value \(y(0.1)\) and the Taylor polynomial \(P_4(0.1)\):
\[ E(0.1) = y(0.1) - P_4(0.1) \approx 0.0001032 \]
This error helps us understand the accuracy of our approximation for small intervals and assists in estimating the potential error in predictive models.
Derivative Computation
Calculating derivatives is essential when working with Taylor series and differential equations. Derivatives give us information about the function's rate of change, enabling us to construct Taylor polynomials and solve differential equations.
In this exercise, the derivatives \(y'(0), y''(0), y'''(0), \text{and } y^{(4)}(0)\) are computed by differentiating the given differential equation \(y' = y + \sin t\) multiple times.
  • \(y'(0)\) is calculated directly using \(y(0) = 1\).
  • \(y''(0)\) uses the result from \(y'(0)\) and differentiating again.
  • Similarly, \(y'''(0)\) and \(y^{(4)}(0)\) are computed by continuing this differentiation process.
This step-by-step computation ensures the precise formation of the Taylor polynomial and aligns with the given initial condition.

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Most popular questions from this chapter

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

In most applications of numerical methods, as in Exercises 1619 , an exact solution is unavailable to use as a benchmark. Therefore, it is natural to ask, "How accurate is our numerical solution?" For example, how accurate are the solutions obtained in Exercises 16-19 using the step size \(h=0.05\) ? This exercise provides some insight. Suppose we apply Heun's method or the modified Euler's method to the initial value problem \(y^{\prime}=f(t, y), y\left(t_{0}\right)=y_{0}\) and we use a step size \(h\). It can be shown, for most initial value problems and for \(h\) sufficiently small, that the error at a fixed point \(t=t^{*}\) is proportional to \(h^{2}\). That is, let \(n\) be a positive integer, let \(h=\left(t^{*}-t_{0}\right) / n\), and let \(y_{n}\) denote the method's approximation to \(y\left(t^{*}\right)\) using step size \(h\). Then $$ \lim _{h \rightarrow 0 \atop t^{*} \text { fixed }} \frac{y\left(t^{*}\right)-y_{n}}{h^{2}}=C, \quad C \neq 0 $$ As a consequence of this limit, reducing a sufficiently small step size by \(\frac{1}{2}\) will reduce the error by approximately \(\frac{1}{4}\). In particular, let \(\hat{y}_{2 n}\) denote the method's approximation to \(y\left(t^{*}\right)\) using step size \(h / 2\). Then, for most initial value problems, we expect that \(y\left(t^{*}\right)-\hat{y}_{2 n} \approx\left[y\left(t^{*}\right)-y_{n}\right] / 4 .\) Rework Example 1 , using Heun's method and step sizes of \(h=0.05, h=0.025\), and \(h=0.0125 .\) (a) Compare the three numerical solutions at \(t=0.05,0.10,0.15, \ldots, 0.95 .\) Are the errors reduced by about \(\frac{1}{4}\) when the step size is reduced by \(\frac{1}{2}\) ? (Since the solution becomes unbounded as \(t\) approaches 1 from the left, the expected error reduction may not materialize near \(t=1\).) (b) Suppose the exact solution is not available. How can the Heun's method solutions obtained using different step sizes be used to estimate the error? [Hint: Assuming that $$ y\left(t^{*}\right)-\hat{y}_{2 n} \approx \frac{\left[y\left(t^{*}\right)-y_{n}\right]}{4} $$ derive an expression for \(y\left(t^{*}\right)-\hat{y}_{2 n}\) that involves only \(\hat{y}_{2 n}\) and \(\left.y_{n} .\right]\) (c) Test the error monitor derived in part (b) on the initial value problem in Example \(1 .\)

For the given initial value problem, an exact solution in terms of familiar functions is not available for comparison. If necessary, rewrite the problem as an initial value problem for a first order system. Implement one step of the fourth order Runge-Kutta method (14), using a step size \(h=0.1\), to obtain a numerical approximation of the exact solution at \(t=0.1\). \(y^{\prime \prime}+z+t y=0\) \(z^{\prime}-y=t, \quad y(0)=1, \quad y^{\prime}(0)=2, \quad z(0)=0\)

Let \(h\) be a fixed positive step size, and let \(\lambda\) be a nonzero constant. Suppose we apply Heun's method or the modified Euler's method to the initial value problem \(y^{\prime}=\lambda y, y\left(t_{0}\right)=y_{0}\), using this step size \(h\). Show, in either case, that \(y_{k}=\left(1+h \lambda+\frac{(h \lambda)^{2}}{2 !}\right) y_{k-1}\) and hence \(y_{k}=\left(1+h \lambda+\frac{(h \lambda)^{2}}{2 !}\right)^{k} y_{0}, \quad k=1,2, \ldots\)

In each exercise, (a) Solve the initial value problem analytically, using an appropriate solution technique. (b) For the given initial value problem, write the Heun's method algorithm, $$ y_{n+1}=y_{n}+\frac{h}{2}\left[f\left(t_{n}, y_{n}\right)+f\left(t_{n+1}, y_{n}+h f\left(t_{n}, y_{n}\right)\right)\right] . $$ (c) For the given initial value problem, write the modified Euler's method algorithm, $$ y_{n+1}=y_{n}+h f\left(t_{n}+\frac{h}{2}, y_{n}+\frac{h}{2} f\left(t_{n}, y_{n}\right)\right) . $$ (d) Use a step size \(h=0.1\). Compute the first three approximations, \(y_{1}, y_{2}, y_{3}\), using the method in part (b). (e) Use a step size \(h=0.1\). Compute the first three approximations, \(y_{1}, y_{2}, y_{3}\), using the method in part (c). (f) For comparison, calculate and list the exact solution values, \(y\left(t_{1}\right), y\left(t_{2}\right), y\left(t_{3}\right)\). \(y^{\prime}=-y, \quad y(0)=1\)

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