- QQ：99515681
- 郵箱：[email protected]
- 工作時間：8:00-23:00
- 微信：codehelp

David Siˇ ska School of Mathematics Sciences, University of Edinburgh ˇ

2018/19 Semester 2

Object Oriented Programming with Applications

Problem Sheet 4 - Wednesday 24th October 20181

Exercise 4.1. Get a working implementation of the CompositeIntegrator class.

Use the lecture material.

You will notice that the class, as presented in lectures is not robust. Consider and

modify your code according for the following situations (i.e. throw appropriate exceptions).

1. What if newtonCotesOrder in the is a number other than 1, 2, 3, 4?

2. What if N in the Integrate method is less than or equal to 0?

3. Does the code work if a > b in the Integrate method?

Exercise 4.2. The Newton’s method for approximating solutions x to f(x) = 0, where

f : R → R is assumed to be differentiable is an iterative method where, given an initial

guess for the solution x0 the next entry in the sequence of approximations is calculated

as:

xn := xn1

f(xn 1)

f

0(xn 1)

, n = 1, 2, . . . .

Do the following:

1. To see that you understand how this works calculate the first three approximations

to 0 = x

2 2 =: f(x) with x0 = 2 (this is a nice way of approximating √

2 if

you ever have to do this by hand).

2. Create a method with the following “signature”:

static double NewtonSolver(Func<double, double> f,

Func<double, double> fPrime,

double x0,

double maxError, int maxIter)

It should have the following properties: the return value should be the approximation

xn of solution to f(x) = 0 such that |f(x)| <maxError. If the number of

iterations has reached or exceeded maxIter then an exception should be thrown.

If fPrime != null then it should be used, otherwise the derivative should be

approximated using symmetric finite differences. That is, for a given δ > 0 we

say that

f0(x0) is approximately 1

2δ

(f(x0 + δ) f(x0 ? δ)).

3. Test it by approximating the solution to 0 = x

2 2 with x0 = 2.

1Last updated 9th October 2018

1

Exercise 4.3. Newton’s method generalises to higher dimensions. Indeed consider

a differentiable F : R

d → R

d

. Let JF (x) denote the Jacobian matrix of this function

evaluated at x. That is:

JF (x)

Given an initial guess x0 ∈ R

d we obtain successive approximations to x such that

F(x) = 0 by solving

JF (xn?1)(xn ? xn?1) = ?F(xn?1), n = 1, 2, . . . .

Your task is to complete the class below:

public class NewtonSolver

{

private const double delta = 1e-8; // for approximating partial derivatives

private double tol;

private int maxIt;

public NewtonSolver(double tolerance, int maximumIterations)

{

tol = tolerance;

maxIt = maximumIterations;

}

public Matrix<double> ApproximateJacobian(Func<Vector<double>,

Vector<double>> F, Vector<double> x)

{

/* ... write the code ...*/

}

public Vector<double> Solve(Func<Vector<double>, Vector<double>> F,

Func<Vector<double>, Matrix<double>> J_F, Vector<double> x_0)

{

/* ... write the code ...*/

}

}

The Newton Method should stop if either maxIt is reached or if the l

2 norm (i.e. the

usual Euclidean norm) of F(xn) is smaller than tol.

Now test it by approximating the solution to F(x, y) = 0 where

with x0 = (1, ?1)T

. Note that it is easy to see that at least one solution is x = 4, y = 3.

Hints.

To define something like F : R

d → R

d

in C# use:

Func<Vector<double>, Vector<double>> F = (x) =>

{

Vector<double> y = Vector<double>.Build.Dense(x.Count);

y[0] = x[0]*x[0] + x[1]*x[1] - 2*x[0]*x[1] - 1;

y[1] = x[0]*x[0] - x[1]*x[1] - 7;

return y;

};

To define something like the Jacobian Matrix use:

2

Func<Vector<double>,Matrix<double>> J_F = (x) => {

int d = x.Count;

Matrix<double> J_F_vals = Matrix<double>.Build.Dense (d, d);

J_F_vals[0,0] = 2*x[0] -2*x[1];

J_F_vals[0,1] = 2*x[1]-2*x[0];

J_F_vals[1,0] = 2*x[0];

J_F_vals[1,1] = -2*x[1];

return J_F_vals;

};

You can see above how to define an empty vector and matrix respectively.

To fill in a whole jth column in a matrix you can use: M.SetColumn(j,v) assuming

M is a matrix and v is a vector of appropriate length.

To solve a linear system Ax = b (with A a d × d matrix and b a vector in R

d

, x

unknown) use: Vector<double> x = A.Solve(b);.

To find a determinant of a matrix A use double determinant = A.Determinat();

Note that A.Solve(b) might also fail when the determinant is very small rather

than exactly 0.

Exercise 4.4. In fact our interest in the Newton’s method is also for minimization

problems. A differentiable function f : R

d → R will have a local minimum (or maximum

or saddle point) at x if

is equal to zero. If f is convex then this will be the global minimum.

So we are approximating solutions to F(x) = ?f(x) = 0 using Newton’s method.

The Jacobian matrix of F is now

which is the Hessian matrix of f.

In some cases one would have to approximate the Hessian using finite differences. Let

us define Tδ,iy : R

d → R

d as

Tδ,iy = (y1, . . . , yi?1, yi + δ, yi+1, . . . , yd)T.

(f(Tj,hTi,hx) f(Tj,hTi,hx))

1

2h

(f(Tj,hTi,hx) f(Tj,hTi,hx))

.

Your task is to develop a Newton based minimizer by completing the class below

3

public class NewtonMnimizer

{

private const double delta = 1e-7; // for approximating grad

private NewtonSolver solver;

public NewtonMnimizer(double tol, int maxIt)

{

solver= new NewtonSolver(tol, maxIt);

}

private Vector<double> ApproximateGrad(Func<Vector<double>, double> f,

Vector<double> x)

{

/* ... write the code ... */

}

private Matrix<double> ApproximateHessian(Func<Vector<double>, double> f,

Vector<double>> grad_f,

Vector<double> x)

{

/* ... write the code, allow for grad_f == null ... */

}

public Vector<double> Minimize(Func<Vector<double>, double> f,

Func<Vector<double>, Vector<double>> grad_f,

Func<Vector<double>, Matrix<double>> hessian_f,

Vector<double> x_0)

{

/* ... write the code,

allow for grad_f == null and also hessian_f == null ... */

}

}

Test it by finding the global minimum of the function f given by f(x, y) = x

2 + y

2 by

running the following code:

NewtonMnimizer minimizer = new NewtonMnimizer (1e-5, 100);

Func<Vector<double>, double> f = (x) => x [0] * x [0] + x [1] * x [1];

Vector<double> startPt = Vector<double>.Build.Dense (2);

startPt[0] = 1; startPt[1] = -1;

Console.WriteLine ("With approximate grad of f and hessian of f using f.d.");

Console.WriteLine(minimizer.Minimize(f, null, null, startPt));

Console.WriteLine ("With exact grad_f and approximate hessian f using f.d.");

Func<Vector<double>, Vector<double>> grad_f = (x) => {

Vector<double> grad = Vector<double>.Build.Dense(x.Count);

grad[0] = 2*x[0]; grad[1] = 2*x[1];

return grad;

};

Console.WriteLine(minimizer.Minimize(f, grad_f, null, startPt));

Console.WriteLine ("With exact grad of f and hessian of f.");

Func<Vector<double>, Matrix<double>> hessian_f = (x) => {

Matrix<double> hessian = Matrix<double>.Build.Dense(x.Count,x.Count);

hessian[0,0] = 2; hessian[0,1] = 0;

hessian[1,0] = 0; hessian[1,1] = 2;

return hessian;

};

Console.WriteLine(minimizer.Minimize(f, grad_f, hessian_f, startPt));

Console.ReadKey ();

版權所有：編程輔導網 2018 All Rights Reserved 聯系方式：QQ:99515681 電子信箱：[email protected]

免責聲明：本站部分內容從網絡整理而來，只供參考！如有版權問題可聯系本站刪除。