Newton-Raphson Method

Newton-Raphson Method

)(xf

)f(x - = xx

i

iii 1

f ( x )

f ( x i )

f ( x i - 1 )

x i + 2 x i + 1 x i X

ii xfx ,

Figure 1 Geometrical illustration of the Newton-Raphson method.2

Derivation

f(x)

f(xi)

xi+1 xi

X

B

C A

)(

)(1

i

iii xf

xfxx

1

)()('

ii

ii

xx

xfxf

AC

ABtan(

Figure 2 Derivation of the Newton-Raphson method.3

Algorithm for Newton-Raphson Method

4

Step 1

)(xf Evaluate symbolically.

5

Step 2

i

iii xf

xf - = xx

1

Use an initial guess of the root, , to estimate the new value of the root, , as

ix

1ix

6

Step 3

0101

1 x

- xx =

i

iia

Find the absolute relative approximate error asa

7

Step 4

Compare the absolute relative approximate error with the pre-specified relative error tolerance .

Also, check if the number of iterations has exceeded the maximum number of iterations allowed. If so, one needs to terminate the algorithm and notify the user.

s

Is ?

Yes

No

Go to Step 2 using new estimate of the

root.

Stop the algorithm

sa

8

Convergence

1 1'

1

2

( )Comparing x with x ( )of the iteration method,

( )

( )( )

'( )

( )( )

'( )

( ) "( )'( )

'( )

Since iteration method converges if '( ) 1

Newton's method con

n n n n

nn n n

n

f xx x

f x

f xx x x

f x

f xIn general x x

f x

f x f xx

f x

x

2verges if ( ) "( ) '( )

in the interval considered.

f x f x f x

The rate at which the iteration method converges if the initial approximation to the root is sufficiently close to the desired root is called the rate of convergence.

Example 1

Q1. Using Newton-Raphson method find the real root of the equation 3x=cosx + 1 correct to four decimal places.

11

Solution

A root of f(x)=0 lies between 0 and 1.

( ) 3 cos 1

(0) 2( )

(1) 1.4597( )

f x x x

f ve

f ve

Cont… Also f’(x)=3+sin x

New Iteration formula gives,

1

( )

'( )

3 cos 1...............(1)

3 sin

nn n

n

nn

n

f xx x

f x

x xx

x

Cont… Put n=0 the first approximation is given by

0 0 01

0

sin cos 1

3 sin

0.6sin 0.6 cos 0.6 1

3 sin 0.60.6071

x x xx

x

Cont.. Put n=1,Then second approximation is

1 1 12

1

sin cos 1

3 sin

.6071sin(.6071) cos(.6071) 1

3 sin(.6071)

0.6071

x x xx

x

Cont..Clearly X1 = X2 .

Hence desired root is 0.6071 correct to 4 decimal places.

Example 2Q 2: Find to four places of decimal, the

smallest root of the equation

sinxe x

Solution

The given equation is

So that,

Take x0 = 0.6, then

( ) sin 0xf x e x

1

sin

cos

n

n

xn

n n xn

e xx x

e x

1

2

0.58848

0.588559

x

x

Cont..Hence desired value of the root is a

0.5885

http://numericalmethods.eng.usf.edu19

20

2

2

Example 3

Evaluate 12 to four decimal places by Newton's iteration method.

Sol. Let 12 so that 12 0 .........(1)

( ) 12, ' ,

n

x x

Take f x x Newton s iteration formula gives

x

2

1

0

1 00

( ) 12 1 12........(2)

'( ) 2 2

Now since (3) 3 and (4) 4

The root of (1) lies between 3 and 4.

take 3.5, from eqn.(2),

1 12 1 123.5 3.4643

2 2 3.5

n nn n n

n n n

f x xx x x

f x x x

f f

x

x xx

21

2 11

3 22

2 3

1 12 1 123.4643 3.4641

2 2 3.4643

1 123.4641

2

, 4 .

, 12 3.4641.

x xx

x xx

Since x x upto decimal places

So

Newton Iterative formulaThe reciprocal or inverse of s number ‘a’ can

be considered as a root of the equation which can be solve

by Newton’s method.

10a

x

2

1 1( ) , '( )f x a f x

x x

Cont… Newton’s Formula gives,

1

2

1

1

nn n

n

ax

x x

x

1 (2 )n n nx x ax

Square Root The square root of ‘a’ can be

considered as a root of equation solved by Newton’s method.

2 0x a

2( ) , ( ) 2f x x a f x x 2

1

1

2

1

2

nn n

n

n nn

x ax x

x

ax x

x

Inverse Square Root Equation is

Iterative Formula is

2

10a

x

21

1(3 )

2n n nx x ax

General Formula for pth root

1 1

( 1) pn

n pn

p x ax

px

Newton’s Iterative Formulae

Inverse:

Square Root:

Inverse Square Root:

For pth root :

27

1 (2 )n n nx x ax

1

1

2n nn

ax x

x

21

13

2n n nx x ax

1 1

( 1) pn

n pn

p x ax

px

Example 1 Cont.

The equation that gives the depth x in meters to which the ball is submerged under water is given by

423 1099331650 -.+x.-xxf

Use the Newton’s method of finding roots of equations to find a)the depth ‘x’ to which the ball is submerged under water. Conduct three iterations to estimate the root of the above equation. b)The absolute relative approximate error at the end of each iteration, andc)The number of significant digits at least correct at the end of each iteration.

28

Figure 3 Floating ball problem.

Example 1 Cont.

423 1099331650 -.+x.-xxf

29

To aid in the understanding of how this method works to find the root of an equation, the graph of f(x) is shown to the right,

where

Solution

Figure 4 Graph of the function f(x)

Example 1 Cont.

30

x-xxf

.+x.-xxf -

33.03'

10993316502

423

Let us assume the initial guess of the root of is . This is a reasonable guess (discuss why and are not good choices) as the extreme values of the depth x would be 0 and the diameter (0.11 m) of the ball.

0xfm05.00 x

0x m11.0x

Solve for xf '

Example 1 Cont.

06242.0

01242.00.05 109

10118.10.05

05.033.005.03

10.993305.0165.005.005.0

'

3

4

2

423

0

001

xf

xfxx

31

Iteration 1The estimate of the root is

Example 1 Cont.

32

Figure 5 Estimate of the root for the first iteration.

Example 1 Cont.

%90.19

10006242.0

05.006242.0

1001

01

x

xxa

33

The absolute relative approximate error at the end of Iteration 1 is

a

The number of significant digits at least correct is 0, as you need an absolute relative approximate error of 5% or less for at least one significant digits to be correct in your result.

Example 1 Cont.

06238.0

104646.406242.0

1090973.8

1097781.306242.0

06242.033.006242.03

10.993306242.0165.006242.006242.0

'

5

3

7

2

423

1

112

xf

xfxx

34

Iteration 2The estimate of the root is

Example 1 Cont.

35

Figure 6 Estimate of the root for the Iteration 2.

Example 1 Cont.

%0716.0

10006238.0

06242.006238.0

1002

12

x

xxa

36

The absolute relative approximate error at the end of Iteration 2 is

a

The maximum value of m for which is 2.844. Hence, the number of significant digits at least correct in the answer is 2.

ma

2105.0

Example 1 Cont.

06238.0

109822.406238.0

1091171.8

1044.406238.0

06238.033.006238.03

10.993306238.0165.006238.006238.0

'

9

3

11

2

423

2

223

xf

xfxx

37

Iteration 3The estimate of the root is

Example 1 Cont.

38

Figure 7 Estimate of the root for the Iteration 3.

Example 1 Cont.

%0

10006238.0

06238.006238.0

1002

12

x

xxa

39

The absolute relative approximate error at the end of Iteration 3 is

a

The number of significant digits at least correct is 4, as only 4 significant digits are carried through all the calculations.

Advantages and Drawbacks of Newton

Raphson Method

40

Advantages

Converges fast (quadratic convergence), if it converges.

Requires only one guess

41

Drawbacks

42

1. Divergence at inflection pointsSelection of the initial guess or an iteration value of the root that is close to the inflection point of the function may start diverging away from the root in their Newton-Raphson method.

For example, to find the root of the equation .

The Newton-Raphson method reduces to .

Table 1 shows the iterated values of the root of the equation.

The root starts to diverge at Iteration 6 because the previous estimate of 0.92589 is close to the inflection point of .

Eventually after 12 more iterations the root converges to the exact value of

xf

0512.01 3 xxf

2

33

113

512.01

i

iii

x

xxx

1x

.2.0x

Drawbacks – Inflection Points

Iteration Number

xi

0 5.0000

1 3.6560

2 2.7465

3 2.1084

4 1.6000

5 0.92589

6 −30.119

7 −19.746

18 0.2000 0512.01 3 xxf

43

Figure 8 Divergence at inflection point for

Table 1 Divergence near inflection point.

2. Division by zeroFor the equation

the Newton-Raphson method reduces to

For , the denominator will equal zero.

Drawbacks – Division by Zero

0104.203.0 623 xxxf

44

ii

iiii xx

xxxx

06.03

104.203.02

623

1

02.0or 0 00 xx Figure 9 Pitfall of division by zero or near a zero number

Results obtained from the Newton-Raphson method may oscillate about the local maximum or minimum without converging on a root but converging on the local maximum or minimum.

Eventually, it may lead to division by a number close to zero and may diverge.

For example for the equation has no real roots.

Drawbacks – Oscillations near local maximum and minimum

02 2 xxf

45

3. Oscillations near local maximum and minimum

Drawbacks – Oscillations near local maximum and minimum

46

-1

0

1

2

3

4

5

6

-2 -1 0 1 2 3

f(x)

x

3

4

2

1

-1.75 -0.3040 0.5 3.142

Figure 10 Oscillations around local minima for .

2 2 xxf

Iteration Number

0123456789

–1.0000 0.5–1.75–0.30357 3.1423 1.2529–0.17166 5.7395 2.6955 0.97678

3.002.255.063 2.09211.8743.5702.02934.9429.2662.954

300.00128.571 476.47109.66150.80829.88102.99112.93175.96

Table 3 Oscillations near local maxima and mimima in Newton-Raphson method.

ix ixf %a

4. Root JumpingIn some cases where the function is oscillating and has a number of roots, one may choose an initial guess close to a root. However, the guesses may jump and converge to some other root.

For example

Choose

It will converge to instead of

-1.5

-1

-0.5

0

0.5

1

1.5

-2 0 2 4 6 8 10

x

f(x)

-0.06307 0.5499 4.461 7.539822

Drawbacks – Root Jumping

0 sin xxf

47

xf

539822.74.20 x

0x

2831853.62 x Figure 11 Root jumping from intended location of root for

. 0 sin xxf

THE END