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Chapter 11

Newtons

Method At the end of this chapter, students should be able to:

Derive Newtons method graphically and algebraically

Apply Newtons method

11.1 Introduction

In Chapter 10, the bisection method of solving non-linear equations has been

discussed. Based on the given examples it can be concluded that if theexistence of a root is guaranteed then the method will always converge to the

required root. Nevertheless one of the disadvantages is that the convergence

rate is slow.

Is there a method which has a better convergence rate? One of the favourite

numerical methods of solving non-linear equation is known as the Newton s

method or the Newton-Raphson. This method is attributed to Sir Isaac

Newton (1643-1727) and Joseph Raphson (1648-1715). If the bisection

method applies a vertical line to do its approximation, Newtons method on

the other hand will make the use of a tangent line at a point.

11.2 Newtons Method

Being one of the most widely used method of root finding, the procedure

attempts to find a solution of the equation 0)( x f where )( x f is a continuous

and differentiable function of one variable.

http://www-groups.dcs.st-and.ac.uk/~history/Mathematicians/Newton.htmlhttp://www-groups.dcs.st-and.ac.uk/~history/Mathematicians/Newton.htmlhttp://www-groups.dcs.st-and.ac.uk/~history/Mathematicians/Raphson.htmlhttp://www-groups.dcs.st-and.ac.uk/~history/Mathematicians/Raphson.htmlhttp://www-groups.dcs.st-and.ac.uk/~history/Mathematicians/Newton.htmlhttp://www-groups.dcs.st-and.ac.uk/~history/Mathematicians/Newton.html

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In the Newtons method, a pproximation is done by using tangential lines. The

solution process begins with choosing a value as the first estimate of the

solution (normally obtained from graphing). This initial value is sometimes

called the initial guess. The second estimate is obtained by taking the

tangent line to )( x f at the initial value. The third estimate is obtained by taking

the tangent line to )( x f at the second estimate. The process goes on and on

until desired accuracy is achieved.

Geometr ic Derivat ion

Let us consider a differentiable function )( x f and the problem of

approximating a zero r of )( x f which is known to lie between the

numbers a x and b x . Newtons method of approximating r calls for an

initial guess 0x to be made. The line 0L tangent to the graph of )( x f y at

))(,(x 00 x f is then constructed. By finding the x-intercept 1x of 0L , we obtain

a second approximation to r. Newtons famous observation is simply that for

many functions the second approximation 1x is better than the first 0x .

Figure 11.1

If the procedure is then repeated by finding the line 1L tangent to the graph

of f at ))(,(x 11 x f , the x -intercept of 1L (say 2x ) provides an even closer

approximation to r than 1x . The procedure is repeated again and again,

until an approximation of sufficiently high accuracy is obtained.

0L

0 x

1 x

2 x x

y

Actualroot r

)( x f y

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We can obtain a simple equation for obtaining the (n + 1) th approximation,

1nx , from the nth approximation, nx by considering Figure 11.2.

Figure 11.2

Since the slope of line nL tangent to the graph of )( x f y at ))(,(xn n x f is

given by the derivative )(' n x f , the equation for nL can be written as:

nnn x x x f x f y )(')( (1)

To find 1nx , the intercept of nL , we set y = 0 and solve for x in equation

(1). We obtain:

)(')(

xx nn

n x f x f

; 0 )(x'f n

Since this is the desired approximation 1nxx , we have the approximation

scheme for Newtons met hod:

)(')(

xx n1nn

n x f x f

, 0 )(x'f n

Algebraic Derivat ion

Another possibility is to derive the Newtons method is based upon the Taylor

polynomial. Suppose that the function )( x f is twice continuously

differentiable on the interval [a, b]. Let ],[xn ba be an approximation to r

such that 0 )(x'f n and r -xn is small. Consider the second -degree

Taylor polynomial for )( x f , expanded about nx

)("f

2

)x-(r )(x')f x-(r )f(xf(x)

2n

nnn (1)

where in between x and nx . Since 0)( r f , with x = r , gives

nL

n x

1n x x

y

Actualroot r

1nL

)( x f y

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)f(2

)x-(r )(x')f x-(r )f(x0

2n

nnn

Since r -xn was assumed to be small, r -xn is even smaller, and if this

quantity is assumed to be negligible, then)(x')f x-(r )f(x0 nnn

Solving for r in this equation gives:

)(')(

xr nn

n x f x f

which should be a better approximation to r than is nx . This sets the stage

for the Newton-Raphson or Newtons method

)(x'f

)f(x

- x x nn

n1n

Example 1

For the equation 0cosx-x , use the Newtons method with threeiterations to estimate the root given .4.10 x What can be said about the

accuracy of the estimated value (root)?

Identify the equation and determine )( x f

0cosx-x cosx-x)( x f

Find the derivative )(' x f

sinx2

1)(' x

x f

Solut ion

Steps : Newtons method

Identify the equation and determine )( x f Find the derivative )(' x f Identify initial value 0 x Evaluate )( x f and )(' x f at 0 x Apply Newtons method

)(')(

1n

nnn x f

x f x x

Repeat until desired iterations/accuracy

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Identify initial value 0 x

4.10 x

Evaluate )( x f and )(' x f at 0 x

013.1)4.1( f 408.1)4.1(' f

Apply Newtons method

)(')(

0

001 x f

x f x x

681.0408.1013.1

4.11 x

Repeat until desired iterations/accuracy

i x

cosx-x)( x f

sinx2

1)('

x x f

Newton

0 1.4 1.013 1.408 0.681

1 0.681 0.048 1.235 0.642

2 0.642 0.000 1.223 0.642

Hence, after 3 iterations, the estimated root is 0.642 (accurate to 3D).

Example 2

The solution of the equation x e x 2 is known to exists between 0 and

1. Estimate the root to 2 decimal places using Newtons method.

Identify the equation and determine )( x f

x e x 2

02 x e x 2)( x e x f x

Find the derivative )(' x f

1)(' x e x f

Identify initial value; 0x

5.00 x

Evaluate )( x f and )(' x f at 0x

Solut ion

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149.0)5.0( f 649.2)5.0(' f

Apply Newtons method

)(' )( 0001 x f x f x x

444.0649.2149.0

5.0

Repeat until desired iterations/accuracy

i x 2)( x e x f x 1)(' x e x f Newton0 0.5 0.149 2.649 0.444

1 0.4439 0.003 2.559 0.443

Hence, the estimated root correct to 2 decimal places is 0.44.

Example 3

Apply Newtons method to approximate the root of 1)( 3 x x x h to 2

significant digits.

Given

1)( 3 x x x h 13)(' 2 x x h

Let 10 x , then

5.12

11

)(')(

0

001 x f

x f x x

x )( x f )(' x f Newton-1 1 2 -1.5

-1.5 -0.875 5.750 -1.348

-1.348 -0.101 4.451 -1.325

-1.325 -0.001 4.267 -1.325

Hence, the estimated root correct to 2 significant digits is -1.3.

Solut ion

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Example 4

Determine the formula for Newtons method to approximate 3 a .

Identify the equation and determine )( x f

The problem is to determine the root 3 a x , which can be rewritten

as

03

3

a x

a x then a x x f 3)(

Find the derivative )(' x f

23)(' x x f

Identify initial value

n x

Evaluate )( x f and )(' x f at n x

a x x f nn 3)(

23)(' nn x x f

Apply Newtons method

2

2

3

1

33

3

n

nn

n

nnn

x

a x x

x

a x x x

Let 05-2x-x3 where 3 r 2 .

(i) Identify the equation and determine )( x f (ii) Find the derivative )(' x f (iii) Identify initial value 0 x (iv) Evaluate )( x f and )(' x f at 0 x (v) Apply Newtons method to determine

)(')(

0

001 x f

x f x x

(vi) Determine the third iteration.

Solut ion

Warm up exercise

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Exercise 11

1. Apply Newton s method to the equation 01663 x x with 00 x and try

to determine the root correct to three decimal places.

2. Determine the first four approximations of the actual root of the equation

0213 x in the interval [0, 1], using the Newton s method.

3. Use Newtons method to find the point in the right half -plane where the graphs

of 32)( x x f and 15)( x x g intersects.

4. Apply Newton s method to the equation 075.0sin x .5. By Newton s method, determine the first three approximations of the

following equations in the given interval.(a) 01sin 3 x x ; [-2, 0]

(b) 03ln x x ; [4, 5]

(c) x x x f 3coscos)( ;2

,2

(d) x x tan2 ; [0, 3]

6. Show that 023 x x has a root between 1 and 2. Use Newton method, to

approximate the root accurate to two decimal places.

7. Using Newton s method, solve 3ln x x , given that the root is close to 2.Obtain the root correct to two decimal places.

8. Use Newton s method to approximate the zero of 21)( x x x f . If

4.00 x give your answers accurate to 2 significant digits.

9. Solve x e x cos2 accurate to three decimal places using Newton s method.

10. By Newton's method, estimate the solution to 0772 x . Give your answer

correct to 3 decimal places.

11. Approximate by using Newton's method to find a solution of 0sin x ,starting with 30 x . How many iterations are needed until the solution is

accurate to 3 decimal places?

12. Approximate the root(s) of 242)( 3 x x x h . How many real roots does h

have?

13. Find a solution to the equation 13 x x that is near 5.10 x using threeiterations.