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Q5.4 Consider what you already know about the Euler and midpoint's method. Are these two methods actually an improvement? You should consider the accuracy and efficiency of each method. You may simply state if each method is an improvement without reasoning

Q5.4 Consider what you already know about the Euler and midpoint's method. Are these two methods actually an improvement? You should consider the accuracy and efficiency of each method. You may simply state if each method is an improvement without reasoning

Database System Concepts
Database System Concepts
7th Edition
ISBN:9780078022159
Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Chapter1: Introduction
Section: Chapter Questions
Problem 1PE
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What is the order of each method?

Q5.4 Consider what you already know about the Euler and midpoint's method. Are these two methods actually an improvement? You should consider the accuracy and efficiency of each method.

You may simply state if each method is an improvement without reasoning.

For the example f(t, y) = 1 - y² take one step of the improved midpoint method to compute the solution at t = 0.5 starting from y(0) = 0. Give your answer to 3 decimal places. Q5.3 A Python imlpementation of each method has been run and have produced the following error tables. The headings are: dt: the time step y(1.0): the estimate of the solution at the final time dt 5.0000e-01 Error: the absolute error of the method against a known exact solution feval: the number of right-hand side evaluations taken. Improved Euler: 2.5000e-01 1.2500e-01 6.2500e-02 3.1250e-02 Improved midpoint: dt 5.0000e-01 2.5000e-01 k₁ = f(t(¹), y(i)) k₂= f(t) + dt/2, y(¹) + dtk₁/2) k3= f(t) + dt/2, y(¹) + dtk₂/2) k₁ = f(t) + dt, y) + dtk3) y(i+1)= y(i) + (dt/6) (k₁ + 2k2 + 2k3 + ks). 1.2500e-01 6.2500e-02 3.1250e-02 y(1.0) 0.823867 0.837084 0.840375 0.841197 0.841403 y(1.0) 0.841489 0.841472 0.841471 0.841471 0.841471 Error 1.760413e-02 4.387233e-03 1.095951e-03 2.739342e-04 6.848020e-05 Error 1.839786e-05 1.143445e-06 7.136556e-08 4.458792e-09 2.786504e-10 feval 4 8 16 32 64 feval 8 16 32 64 128
Transcribed Image Text:For the example f(t, y) = 1 - y² take one step of the improved midpoint method to compute the solution at t = 0.5 starting from y(0) = 0. Give your answer to 3 decimal places. Q5.3 A Python imlpementation of each method has been run and have produced the following error tables. The headings are: dt: the time step y(1.0): the estimate of the solution at the final time dt 5.0000e-01 Error: the absolute error of the method against a known exact solution feval: the number of right-hand side evaluations taken. Improved Euler: 2.5000e-01 1.2500e-01 6.2500e-02 3.1250e-02 Improved midpoint: dt 5.0000e-01 2.5000e-01 k₁ = f(t(¹), y(i)) k₂= f(t) + dt/2, y(¹) + dtk₁/2) k3= f(t) + dt/2, y(¹) + dtk₂/2) k₁ = f(t) + dt, y) + dtk3) y(i+1)= y(i) + (dt/6) (k₁ + 2k2 + 2k3 + ks). 1.2500e-01 6.2500e-02 3.1250e-02 y(1.0) 0.823867 0.837084 0.840375 0.841197 0.841403 y(1.0) 0.841489 0.841472 0.841471 0.841471 0.841471 Error 1.760413e-02 4.387233e-03 1.095951e-03 2.739342e-04 6.848020e-05 Error 1.839786e-05 1.143445e-06 7.136556e-08 4.458792e-09 2.786504e-10 feval 4 8 16 32 64 feval 8 16 32 64 128
Predictor corrector methods In the notes you are given a method to construct a second order method, the midpoint method, by leveraging Euler's method. This question presents an alternative way to achieve higher order methods using the "predictor-corrector" method. We consider a generic ordinary differential equation y'(t) = f(t, y), subject to y(0) = yo. Q5.1 In this approach, we take one step of the original method and then do some averaging to get the final result. For example for Euler's method at each time step we perform: k = y(i) + dtf (t(i),y(i)) y(i+¹) = y(i) + (dt/2) (f(t(i),y(i)) + f (t(i+¹), k)). For the example f(t, y) = 1 - y² take two steps of the improved Euler method to compute the solution at t = 0.5 starting from y(0) = 0. Give your answer to 3 decimal places. Q5.2 We consider a similar approach for the midpoint method. This results in the following scheme:
Transcribed Image Text:Predictor corrector methods In the notes you are given a method to construct a second order method, the midpoint method, by leveraging Euler's method. This question presents an alternative way to achieve higher order methods using the "predictor-corrector" method. We consider a generic ordinary differential equation y'(t) = f(t, y), subject to y(0) = yo. Q5.1 In this approach, we take one step of the original method and then do some averaging to get the final result. For example for Euler's method at each time step we perform: k = y(i) + dtf (t(i),y(i)) y(i+¹) = y(i) + (dt/2) (f(t(i),y(i)) + f (t(i+¹), k)). For the example f(t, y) = 1 - y² take two steps of the improved Euler method to compute the solution at t = 0.5 starting from y(0) = 0. Give your answer to 3 decimal places. Q5.2 We consider a similar approach for the midpoint method. This results in the following scheme:
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