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## Newton Method and Loops

Newton’s Method and Loops::Solving equations numericallyFor the subsequently few lectures we will focus on the problem of solving an equation:

f(x) = 0.

As you erudite in calculus, the final step in several optimization problems is to solve an equation ofthis form where f is the derivative of a function F that you desire to maximize or minimize. In realengineering problems the function you wish to optimize able to come from a large variety of sources,including formulas, solutions of differential equations, experiments, or simulations.

:Newton iterationsWe will signify an actual solution of equation by x∗. There are three methods which you may perhapshave discussed in Calculus: the bisection method the secant method with Newton’s method. Allthree depend on commencement close (in some sense) to an actual solution x∗.

Recall Newton’s method youmust know that the basis for Newton’s method is approximation ofa function by it linearization at a point that is:

f(x) ≈ f(x_{0}) + f′(x_{0})(x − x_{0}).Since we wish to find x therefore that f(x) = 0 set the left hand side (f(x)) of this approximation equalto 0 along with solve for x to obtain

x≈ x0−f(x_{0})/f′(x_{0})We begin the method with the initial guess x

_{0}, which we hope is fairly close to x∗. Then we definea sequence of points {x_{0}, x_{1}, x_{2}, x_{3}, . . .}from the formula:x_{i+1}= x_{i}−f(x_{i})/f′(x_{i})If f(x) is reasonably well-behaved near x∗as well asx0 is close enough to x∗subsequently it is a fact that the sequence will converge to x∗and will do it very quickly.

:The loop for ... endIn order to do Newton’s method we requirerepeating the calculation in a number of times. Thisis accomplished in a program using a loop which signifies a section of a program which is repeated.The simplest manner to accomplish this is to count the number of times through. In Mat labafor ... end statement makes a loop as in the subsequent simple function program:

function S = mysum(n)% gives the sum of the first n integers

S = 0; % start at zero

% The loop:

for i = 1:n % do n times

S = S + i; % add the current integer

end % end of the loop

Call this function in the command window as:

>mysum(100)

The result will be the total of the first 100 integers. All for ... end loops have the similar format itbegins with for followed by an index (i) and a range of numbers (1:n). Afterward come the commandsthat are to be repeated. Last come the end command.

Loops are one of the major ways that computers are made to do calculations that humans cannot.

Any computation that involves a repeated process is easily done by a loop.

At the moment let’s do a program that does n steps (iterations) of Newton’s method. We will requires to inputthe initial guess, the function, its derivative, and the number of steps. The output will be the absolutevalue of xthat isxn. If we are merely interested in the final approximation not the intermediate stepswhich is usually the case in the real world then we can use a single variable x in the program andchange it at each step:

function x = mynewton(f,f1,x0,n)% Solves f(x) = 0 by doing n steps of Newton’s method starting at x0.

% Inputs: f -- the function, input as an inline

% f1 -- it’s derivative, input as an inline

% x0 -- starting guess, a number

% n -- the number of steps to do

% Output: x -- the approximate solution

format long % prints more digits

format compact % makes the output more compact

x = x0; % set x equal to the initial guess x0

for i = 1:n % Do n times

x = x - f(x)/f1(x) % Newton’s formula, prints x too

end

In the command window define an inline function: f(x) = x

^{3}− 5 i.e.> f = inline(’x^3 - 5’)

and define f1 to be its derivative, i.e.

> f1 = inline(’3*x^2’).

Then run my Newton on this function. By trial as well as error what is the lowest value of n for which theprogram converges (stops changing). Through simple algebra the true root of this function is 3√ 5. Howaccurate is the program’s answer to the true value?

:ConvergenceNewton’s method converges rapidly when f′(x∗) is nonzero and finite, and x0 is close enough to x∗that the linear approximation is valid. Let us take a look at what is able to go wrong.

For f(x) = x

^{1/3}we have x∗= 0 but f′(x*) = ∞. If you try> f = inline(’x^(1/3)’)

> f1 = inline(’(1/3)*x^(-2/3)’)

> x = mynewton(f,f1,0.1,10)

then x explodes.

For f(x) = x

^{2}we have x= 0 but f′(x) = 0. If you try> f = inline(’x^2’)

> f1 = inline(’2*x’)

> x = mynewton(f,f1,1,10)

Then x does converge to 0 however not that rapidly.

If x

_{0}isn’t close enough to xthat the linear approximation is valid, then the iteration gives some x_{1}that may or mayn’t be any better than x_{0}. If we keep iterating afterwards either• x

_{n}will ultimately get close to xand the method will then converge (rapidly) or• The iterations will not approach x.

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