lecture 31

MA/CS 375Fall 2002Lecture 31

MA/CS375 Spring 2002

Root FindingGiven a real valued function f of one variable (say x), the idea is to find an x such that:

f(x) = 0


Newtons Method cont.Repeat the formula to create an algorithm:

If at each step the linear model is a good approximation to f then xn should get closer to a root of f as n increases.


Newton in Matlab


Newtons Method Without Knowing the DerivativeRecall: we can approximate the derivative to a function with:


Modification


Team Exercise10 minutesModify your script to use the approximate derivative (note you will require an extra parameter delta)Test it with some function you do not know the derivative of.


Convergence Rate For Newtons MethodTheorem 8 (van Loan p 285)Suppose f(x) and f(x) are defined on an interval where and positive constants rho and delta exist such that

If xc is in I, then is in I and

That is x+ is at least half the distance to x* that xc was. Also, the rate of convergence is quadratic.


Convergence Rate of Newtons Method contThe proof of this theorem works by using the fundamental theorem of calculus. All of the restrictions are important and can be fairly easily broken by a general functionThe restrictions amount to:f does not change sign in a neighbourhood of the root x*f is not too non-linear (Lipschitz condition)the Newtons iteration starts close enough to the root x* then convergence is guaranteed and the convergence rate is quadratic.


Finding A Root Of A Two-dimensional Function of Two VariablesSuppose:

Specifically:


Finding A Root Of A Two-dimensional Function of Two VariablesSuppose:

Specifically:

We can construct a Taylor series:


Constructing A Newton SchemeCreate a sequence by using this linear approximation to update from (xn,yn) to (xn+1,yn+1)


Constructing A Newton SchemeWe figure that if the linear approximation is good then the f(xn+1,yn+1) and g(xn+1,yn+1) should be small


Two-dimensional Newton Method


Jacobian Matrix


Team Exercise (Part 1)Code up the two-dimensional Newton solver. Use the approximation for small delta:

Use it to find x,y such that:


Team Exercise (Part 2)Generalize your Newton Solver to solve:

Construct a 3-vector function and find its roots.

Plot the position of (xn,yn,zn) using sphere


Problems with Multi-D NewtonThe method relies on inverting the Jacobian matrix.

Recall that matrix inversion is strongly dependent on the condition number of the matrix.

Clearly, if the root is near a region where the gradient is small then we will run into slow convergence when the search nears the root.


Team Exercise:Dodgy ConvergenceTry to find the (0,0) root of

Compare the rate of convergence for the (0,0) root of: For each iteration plot the condition number of the Jacobian matrix. Try several different starting positions.


Roots of a PolynomialSuppose we wish to find all the roots of a polynomial of order P

Then there are going to be at most P roots!.

We can use a variant of Newtons method.


Roots of a Polynomial cont.Suppose we have an initial guess for one of the roots of the polynomial function f Then we can use Newtons method, starting at this guess to solve for f(x)=0 Once we have found the first root x1 we apply polynomial deflation to remove this root and then repeat the process to find the next root.


Algorithm1) We are seeking the roots x1,x2,..xP of a polynomial f 2) We find x1 using Newtons method.3) We then use Newtons method to find the next root of f(x)/(x-x1)4) Then we find x3 as a root of f(x)/((x-x1)(x-x2))5) Repeat until all roots found


Details of Newtons AlgorithmAt the kth step, we need to find a root of For Newtons we need:


Product rule fordifferentiation


Newton Scheme For Multiple Root Finding


Multiple Root Finder(applied to find roots of Legendre polynomials)


Recall: recurrence relation of Legendre polynomials


Recall: Recurrence Relation for Gradient of Legendre Polynomials


Roots of the 10th Order Legendre PolynomialNotice how they cluster at the end points


Numerical QuadratureA numerical quadrature is a set of two vectors. The first vector is a list of x-coordinates for nodes where a function is to be evaluated. The second vector is a set of integration weights, used to calculate the integral of a function which is given at the nodes


Example of QuadratureSay we wish to calculate an approximation to the integral of f over [-1,1] : Suppose we know the value of f at a set of N points then we would like to find a set of weights w1,w2,..,wN so that:


Newton-Cotes FormulaThe first approach we are going to use is the well known Newton-Cotes quadrature.

Suppose we are given a set of points x1,x2,..,xN. Then we require that the constant is exactly integrated:


Now we require that 1,x,x2,..,xN-1 are integrated exactly


In Matrix Notation: Notice anything familiar?


Its the transpose of the Vandermonde matrix


Integration by InterpolationIn essence this approach uses the unique (N-1)th order interpolating polynomial If and integrates the area under the If instead of the area under f Clearly, we can estimate the approximation error using the estimates for the error in the interpolation we used before.


Newton-Cotes Weights


Using Newton-Cotes Weights


Using Newton-Cotes Weights(Interpretation)i.e. we calculate the coefficients of the interpolating polynomial expansion using the Vandermonde, then since we know the integral of each term we can sum up the integral of each term to get the total.


Matlab Function for Calculating Newton-Cotes Weights


Demo: Matlab Function for Calculating Newton-Cotes Weightsset N=5 pointsbuild equispaced nodescalculate NC weights

evaluate F=X^3 at nodesevaluate integral

F is anti-symmetric on [-1,1] so its integral is 0Answer correct


Team ExerciseGet the directory Lecture19m from the cd-rom make sure your matlab path points to the copy of this directory using a script figure out what order polynomial the weights can exactly integrate for a given set of N points (say N=6).


Gauss QuadratureThe construction of the Newton-Cotes weights does not utilize the ability to choose the distribution of nodes for greater accuracy. We can in fact choose the set of nodes to increase the order of polynomial that can be integrated exactly using just N points.


Suppose:


Suppose:Remainder term, which must have p roots located at the interpolating nodes


At this point we can choose the nodes {xi}. If we choose them so that they are the p+1 roots of the (p+1)th order Legendre function then s(x) is in fact the N=(p+1)th order Legendre function itself!.


But we also know that if r is a lower order polynomial than (p+1)th order, it can be expressed as a linear combination of Legendre polynomials and is in fact orthogonal to Lp+1


i.e. the quadrature is exact for all polynomials of order up to p=(2N-1)Hence:


Summary of Gauss QuadratureWe can use the multiple root finder to locate the roots of the Nth order Legendre polynomial. We can then use the Newton-Cotes formula with the roots of the Nth order Legendre polynomial to calculate a set of N weights.

We now have a quadrature !!! which will integrate polynomials of order 2N-1 with N points


Team ExerciseUse the root finder and Newton-Cotes routines to build a quadrature for N points (N arbitrary).

Test it on some functions you know the integral of (sin(x) or tan(x) or exp(x) or ) over the interval [-1,1]


SummaryWe have looked at two ways to find the root of a single valued, single parameter function We considered a robust, but slow bisection method and then a faster but less robust Newtons method which we generalized to vector functions of multiple arguments

We discussed the theory of convergence for Newtons method.


lecture 31

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