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Since the integral $$y(x) :=\int_{0}^{x} f(t) d t$$ with variable upper limit satisfies (for continuous f) the initial value problem $$y^{\prime}=f(x), \quad y(0)=0$$ any numerical scheme that is used to approximate the solution at $$x=1$$ will give an approximation to the definite integral $$\int_{0}^{1} f(t) d t$$ Derive a formula for this approximation of the integral using (a) Euler's method. (b) the trapezoid scheme. (c) the improved Euler's method.

Short Answer

Expert verified
The approximations using Euler's method, Trapezoid scheme and Improved Euler's Method can be expressed as \(\sum_{n=0}^{N-1} \frac{1}{N} f \left (\frac{n}{N} \right)\), \(\frac{1}{2N} \left[f(0) + 2\sum_{n=1}^{N-1}f \left (\frac{n}{N} \right) + f(1)\right]\), and \(\frac{1}{N}\left[f(0)+\sum_{n=1}^{N}\frac{1}{2}\left[f\left(\frac{n-1}{N}\right)+f\left(\frac{n}{N}\right)\right]\right]\) respectively.

Step by step solution

01

(a) Euler's method

Euler's method is one of the simplest methods to solve first-order, ordinary differential equations. Here, \(y^\prime =f(x)\) and \(y(0)=0\). One step of Euler's method from \(x_n\) to \(x_{n+1}\) is: \(y_{n+1}=y_n+h*f(x_n)\). If we choose \(h=1/N\) and \(x_n=n/N\) for \(n=0,...,N\), we get our approximate sum for the integral as \(\sum_{n=0}^{N-1} \frac{1}{N} f \left (\frac{n}{N} \right)\).
02

(b) Trapezoid scheme

The trapezoid rule gives a better approximation of a definite integral by summing the areas of the trapezoids connecting the points. The trapezoidal rule sums an average of the two end-point function values times the step width. This can be formulated as \(y_{n+1}=y_n+h*\frac{1}{2}*[f(x_n)+f(x_{n+1})]\). With the similar replacement \(h=1/N\) and \(x_n=n/N\), the integral approximation reads: \(\frac{1}{2N} \left[f(0) + 2\sum_{n=1}^{N-1}f \left (\frac{n}{N} \right) + f(1)\right]\).
03

(c) Improved Euler's Method

Improved Euler's method, or Heun's method, has an additional correction step which makes it more accurate than the standard Euler's method. It is an explicit method for solving initial value problems. The method can be formulated as follows: initial guess by euler's method \(\bar{y}_{n+1}=y_n+h*f(x_n)\), and then corrector as \(y_{n+1}=y_n+h*\frac{1}{2}*[f(x_n)+f(\bar{y}_{n+1})]\). Again, substituting \(h=1/N\) and \(x_n=n/N\), we have the integral approximation formula \(\frac{1}{N}\left[f(0)+\sum_{n=1}^{N}\frac{1}{2}\left[f\left(\frac{n-1}{N}\right)+f\left(\frac{n}{N}\right)\right]\right]\).

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Key Concepts

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

Euler's Method
Euler's method is a foundational numerical technique for approximating solutions to first-order differential equations. It's straightforward and serves as a great introduction to numerical integration. Here's how it works.

In Euler's method, you move step-by-step from an initial point to approximate the function's value at another point. You start with a known value, which in this case is the initial value of the integral at zero, given by the problem as \( y(0) = 0 \). With this method, you approximate the function increment by increment.

Here's the formula to follow:
  • From an initial point \( x_n \) to \( x_{n+1} \), use the equation: \( y_{n+1} = y_n + h \, f(x_n) \).
  • Choosing a step size \( h = 1/N \), and having the points \( x_n = n/N \) for \( n = 0, ..., N \).
This results in the approximation of the integral to be a sum:
\[ \sum_{n=0}^{N-1} \frac{1}{N} f \left (\frac{n}{N} \right) \]
which represents the Riemann sum for the integral \( \int_{0}^{1} f(t) \, dt \).

Euler's method is simple but may not always be the most accurate, especially for functions with high variability within the interval. However, it can be useful for providing a quick, rough estimate.
Trapezoid Rule
The trapezoid rule is a more powerful method for numerical integration than Euler's method. It leverages the fact that a function can be approximated as a series of linear trapezoids, rather than rectangles as in Euler's method.

Here's what makes this method special:
  • Unlike Euler's method, which only considers the function's value at the start of each step, the trapezoid rule factors in the function's value at both the start and end of each interval.
  • This results in a more accurate approximation of the area under the curve, because it accounts for the slope between these two points.
Mathematically, for each step:
\[ y_{n+1} = y_n + h \cdot \frac{1}{2} \cdot [f(x_n) + f(x_{n+1})] \]
With \( h = 1/N \) and \( x_n = n/N \) as before, the approximation becomes:
\[ \frac{1}{2N} \left[f(0) + 2\sum_{n=1}^{N-1}f \left (\frac{n}{N} \right) + f(1)\right] \]
This formula provides a much finer estimate of the integral \( \int_{0}^{1} f(t) \, dt \) because the trapezoid rule effectively smooths out small variations in the function's value across the intervals. This makes it particularly useful for functions that change quickly.
Improved Euler's Method
Also known as Heun's method, the improved Euler's method introduces a correction step that enhances the accuracy of the standard Euler's method. It's an excellent balance between simplicity and precision.

The method involves two steps for each interval:
  • First, make an initial estimate of the next value using Euler's method: \( \bar{y}_{n+1} = y_n + h \cdot f(x_n) \).
  • Second, apply a correction to this estimate using the trapezoid concept: \[ y_{n+1} = y_n + h \cdot \frac{1}{2} \cdot [f(x_n) + f(\bar{y}_{n+1})] \].
Incorporating \( h = 1/N \) and points \( x_n = n/N \), this gives us:
\[ \frac{1}{N}\left[f(0) + \sum_{n=1}^{N}\frac{1}{2}\left[f\left(\frac{n-1}{N}\right) + f\left(\frac{n}{N}\right)\right]\right] \]

This approach refines the initial guess and corrects it by considering the average slope of the tangent between the old and predicted new points, leading to a more accurate approximation of the integral. The improved Euler's method is great for obtaining higher precision without adding much complexity to the calculations, making it very efficient for various practical scenarios.

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Most popular questions from this chapter

A sailboat has been running (on a straight course) under a light wind at 1 m/sec. Suddenly the wind picks up, blowing hard enough to apply a constant force of 600 N to the sailboat. The only other force acting on the boat is water resistance that is proportional to the velocity of the boat. If the proportionality constant for water resis- tance is $$b=100 \mathrm{N}-\sec / \mathrm{m}$$ and the mass of the sailboat is 50 kg, find the equation of motion of the sailboat. What is the limiting velocity of the sailboat under this wind?

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.

A $$400-1 \mathrm{b}$$ object is released from rest 500 ft above the ground and allowed to fall under the influence of gravity. Assuming that the force in pounds due to air resistance is -10v, where v is the velocity of the object in ft/sec,determine the equation of motion of the object. When will the object hit the ground?

The solution to the initial value problem $$ \frac{d y}{d x}+\frac{y}{x}=x^{3} y^{2}, \quad y(1)=3 $$ has a vertical asymptote ("blows up") at some point in the interval 31, 24 . By experimenting with the improved Euler's method subroutine, determine this point to two decimal places.

An object of mass $$5 \mathrm{kg}$$ is given an initial downward velocity of $$50 \mathrm{m} / \mathrm{sec}$$ and then allowed to fall under the influence of gravity. Assume that the force in newtons due to air resistance is -10v, where v is the velocity of the object in m/sec. Determine the equation of motion of the object. If the object is initially 500 m above the ground, determine when the object will strike the ground.

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