/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 18 Local versus Global Error. In de... [FREE SOLUTION] | 91Ó°ÊÓ

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Local versus Global Error. In deriving formula (4) for Euler's method, a rectangle was used to approximate the area under a curve (see Figure 3.14). With \( g(t) :=f(t, \phi(t)) \), this approximation can be written as $$ \int_{x_{n}}^{x_{n+1}} g(t) d t \approx h g\left(x_{n}\right), \quad \text { where } \quad h=x_{n+1}-x_{n} $$ (a) Show that if g has a continuous derivative that is bounded in absolute value by B, then the rectangle approximation has error \( \mathrm{O}\left(h^{2}\right) \); that is, for some constant M, $$ \left|\int_{x_{n}}^{x_{n+1}} g(t) d t-h g\left(x_{n}\right)\right| \leq M h^{2} $$ This is called the local truncation error of the scheme. [Hint: Write $$ \int_{x_{n}}^{x_{n+1}} g(t) d t-h g\left(x_{n}\right)=\int_{x_{n}}^{x_{n}+1}\left[g(t)-g\left(x_{n}\right)\right] d t $$ Next, using the mean value theorem, show that \( \left|g(t)-g\left(x_{n}\right)\right| \leq B\left|t-x_{n}\right| \). Then integrate to obtain the error bound \( (B / 2) h^{2} \cdot ] \) (b) In applying Euler's method, local truncation errors occur in each step of the process and are propagated throughout the further computations. Show that the sum of the local truncation errors in part (a) that arise after \( n \) steps is \( O(h) \). This is the global error, which is the same as the convergence rate of Euler's method.

Short Answer

Expert verified
The local truncation error in Euler's method is \( Mh^{2} \) where M is a constant, and the global truncation error, which is the sum of the local truncation errors after n steps, is of order \( O(h) \). This indicates that as the step size decreases, the global error also decrease.

Step by step solution

01

Determining Local Truncation Error (Part a)

Rewrite the given integral: \[\int_{x_{n}}^{x_{n+1}} g(t) d t-h g\left(x_{n}\right)=\int_{x_{n}}^{x_{n}+1}\left[g(t)-g\left(x_{n}\right)\right] d t \] Next, using the mean value theorem, show that:\[\left|g(t)-g\left(x_{n}\right)\right| \leq B\left|t-x_{n}\right|\] Now, evaluating the integral:\[\int_{x_{n}}^{x_{n+\delta x}}\left[g(t)-g\left(x_{n}\right)\right] d t \leq \int_{x_{n}}^{x_{n+\delta x} B |t - x_{n}| dt = \left.\frac{B}{2}(t - x_{n})^{2}\right|_{x_{n}}^{x_{n+\delta x} = \frac{B}{2}h^{2}\] where \( h = x_{n+1}-x_{n} \). So the local truncation error is \( Mh^{2} \) for some constant M.
02

Summing the Local Truncation Errors (Part b)

Add up the local truncation errors that arise after n steps: \[\sum_{i=0}^{n-1}Mh^{2}=nMh^{2}=Mh(nh)=O(h)\] since nh is the total interval of integration and is constant for a given problem, and O(h) indicates the order of magnitude of the error. Hence, the global error is proportional to the step size, h, which is the same as the convergence rate of the Euler's method.

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Key Concepts

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

Local Truncation Error
In numerical methods like Euler's Method, local truncation error is the error made in a single step of the method. Euler's Method uses a rectangle to approximate the area under a curve.
This approximation introduces a small error in each step, known as the local error.
To derive this, consider the integral \[\int_{x_{n}}^{x_{n+1}} g(t) dt - h g(x_{n}) = \int_{x_{n}}^{x_{n+1}} [g(t) - g(x_{n})] dt\]Using the mean value theorem, we can show\[|g(t) - g(x_{n})| \leq B |t - x_{n}|\]where \( B \) is a constant.
When we integrate this, we find the local truncation error is bounded by \( Mh^2 \), where \( M \) is a constant and \( h \) is the step size. This means the error gets smaller as the step size decreases.
Understanding this concept helps improve the accuracy of numerical solutions by highlighting the importance of choosing an appropriate step size.
Global Error
While local truncation error examines a single step, global error examines the accumulated error after multiple steps. In Euler's Method, local truncation errors happen at every step and their effects add up over all steps.
If the local error is \( Mh^2 \), then after \( n \) steps of size \( h \), the sum of these errors is:\[\sum_{i=0}^{n-1} Mh^2 = nMh^2 = Mh(nh)\]Here, \( nh \) is the total range over which the method is applied. This shows global error is \( O(h) \), which grows linearly with \( h \). Reducing \( h \) decreases global error, making calculations more accurate.
This is crucial for long calculations that involve many steps. Understanding global error aids in making decisions about step size to maintain precision across iterative methods.
Mean Value Theorem
The Mean Value Theorem is a core tool in minimizing errors in numerical methods. It states that for a differentiable function, there is at least one point where the instantaneous rate of change (derivative) of the function equals the average rate of change over an interval.In Euler's Method, the Mean Value Theorem helps express the difference \( g(t) - g(x_{n}) \) as \( B|t-x_{n}| \), with \( B \) being the bound on the derivative.
This helps us relate how changes in the function's behavior between two points affect the overall approximation error.
Understanding the role of the Mean Value Theorem allows students to comprehend how variations in functions contribute to approximation accuracy. It also underscores how derivatives help us manage these errors.
Convergence Rate
The convergence rate describes how quickly a numerical method approaches the exact solution as the step size decreases. In the case of Euler's Method, the convergence rate is directly linked to the global error.Since the global error is \( O(h) \), it shows that Euler's Method is first-order accurate.
This means as the step size \( h \) is halved, the error also halves. The convergence rate is considered linear because the error decreases linearly with the step size.
A higher convergence rate is desirable as it indicates faster and more efficient attainment of the true value for smaller step sizes.
Understanding these rates gives insight into the efficiency and effectiveness of various numerical methods when computing solutions.

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Most popular questions from this chapter

The solution to the initial value problem $$ \frac{d y}{d x}=(x+y+2)^{2}, \quad y(0)=-2 $$ crosses the x-axis at a point in the interval 30, 1.44 . By experimenting with the improved Euler's method subroutine, determine this point to two decimal places.

By experimenting with the fourth-order Runge-Kutta subroutine, find the maximum value over the interval 31, 24 of the solution to the initial value problem $$ y^{\prime}=\frac{1.8}{x^{4}}-y^{2}, \quad y(1)=-1 $$ Where does this maximum occur? Give your answers to two decimal places.

Use the fourth-order Runge-Kutta algorithm to approximate the solution to the initial value problem $$ y^{\prime}=y \cos x, \quad y(0)=1 $$ at \(x=\pi .\) For a tolerance of \(\varepsilon=0.01,\) use a stopping procedure based on the absolute error.

Use Euler's method (4) with \( h=0.1 \) to approximate the solution to the initial value problem $$ y^{\prime}=-20 y, \quad y(0)=1 $$ on the interval \( 0 \leq x \leq 1 \) 1 (that is, at \( x=0,0.1, \ldots, 1.0 \)). Compare your answers with the actual solution \( y=e^{-20 x} \). What went wrong? Next, try the step size \( h=0.025 \) and also \( h=0.2 \) What conclusions can you draw concerning the choice of step size?

Early Monday morning, the temperature in the lecture hall has fallen to $$40^{\circ} \mathrm{F}$$, the same as the temperature outside. At 7:00 a.m., the janitor turns on the furnace with the thermostat set at $$70^{\circ} \mathrm{F}$$. The time constant for the building is 1>K = 2 hr and that for the building along with its heating system is 1>K1 = 1>2 hr. Assuming that the outside temperature remains constant, what will be the temperature inside the lecture hall at 8:00 a.m.? When will the temperature inside the hall reach $$65^{\circ} \mathrm{F}$$?

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