Newton-Raphson Method Nonlinear...

Newton-Raphson MethodMajor: All Engineering Majors

Authors: Autar Kaw, Jai Paul

http://numericalmethods.eng.usf.eduTransforming Numerical Methods Education for STEM

Undergraduates

1/10/2010 1http://numericalmethods.eng.usf.edu

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Newton-Raphson Method

http://numericalmethods.eng.usf.edu


Newton-Raphson Method

)(xf)f(x - = xx

i

iii ′+1

f(x)

f(xi)

f(xi-1)

xi+2 xi+1 xi X θ

( )[ ]ii xfx ,

Figure 1 Geometrical illustration of the Newton-Raphson method.http://numericalmethods.eng.usf.edu3

Derivation

f(x)

f(xi)

xi+1 xi X

B

C A α

)()(

1i

iii xf

xfxx′

−=+

1

)()('+−

=ii

iixxxfxf

ACAB

=)αtan(

Figure 2 Derivation of the Newton-Raphson method.4 http://numericalmethods.eng.usf.edu

Algorithm for Newton-Raphson Method

5 http://numericalmethods.eng.usf.edu

Step 1

)(xf ′Evaluate symbolically.

http://numericalmethods.eng.usf.edu6

Step 2

( )( )i

iii xf

xf - = xx′+1

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

ix1+ix


Step 3

0101

1 x

- xx = i

iia ×∈

+

+

Find the absolute relative approximate error asa∈


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 >∈∈


Example 1

You are working for ‘DOWN THE TOILET COMPANY’ that makes floats for ABC commodes. The floating ball has a specific gravity of 0.6 and has a radius of 5.5 cm. You are asked to find the depth to which the ball is submerged when floating in water.

Figure 3 Floating ball problem.http://numericalmethods.eng.usf.edu10

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.


Figure 3 Floating ball problem.

Example 1 Cont.

( ) 423 1099331650 -.+x.-xxf ×=


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.


( )( ) x-xxf

.+x.-xxf -

33.03'1099331650

2

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.

( ) 0=xfm05.00 =x

0=x m11.0=x

Solve for ( )xf '

Example 1 Cont.

( )( )

( ) ( )( ) ( )

( )06242.0

01242.00.05 10910118.10.05

05.033.005.0310.993305.0165.005.005.0

'

3

4

2

4230

001

=−−=

×−×

−=

−×+−

−=

−=

−

−

−

xfxfxx


Iteration 1The estimate of the root is

Example 1 Cont.


Figure 5 Estimate of the root for the first iteration.

Example 1 Cont.

%90.19

10006242.0

05.006242.0

1001

01

=

×−

=

×−

=∈x

xxa


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 1090973.81097781.306242.0

06242.033.006242.0310.993306242.0165.006242.006242.0

'

5

3

7

2

4231

112

=×−=

×−×−

−=

−×+−

−=

−=

−

−

−

−

xfxfxx



Example 1 Cont.


Figure 6 Estimate of the root for the Iteration 2.

Example 1 Cont.

%0716.0

10006238.0

06242.006238.0

1002

12

=

×−

=

×−

=∈x

xxa



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

06238.033.006238.0310.993306238.0165.006238.006238.0

'

9

3

11

2

4232

223

=×−−=×−

×−=

−×+−

−=

−=

−

−

−

−

xfxfxx



Example 1 Cont.


Figure 7 Estimate of the root for the Iteration 3.

Example 1 Cont.

%0

10006238.0

06238.006238.0

1002

12

=

×−

=

×−

=∈x

xxa



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



Advantages

Converges fast (quadratic convergence), if it converges.

Requires only one guess


Drawbacks


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 ther 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

1 13512.01

−+−

−=+i

iii x

xxx

1=x

.2.0=x

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 =+−= xxf26 http://numericalmethods.eng.usf.edu

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


ii

iiii xx

xxxx06.03

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


3. Oscillations near local maximum and minimum

Drawbacks – Oscillations near local maximum and minimum


-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

–1.00000.5

–1.75–0.303573.14231.2529–0.171665.73952.69550.97678

3.002.255.0632.09211.8743.5702.02934.9429.2662.954

300.00128.571476.47109.66150.80829.88102.99112.93175.96

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


( )xf

539822.74.20 == πx

0=x

2831853.62 == πx Figure 11 Root jumping from intended location of root for

.( ) 0 sin == xxf

Additional ResourcesFor all resources on this topic such as digital audiovisual lectures, primers, textbook chapters, multiple-choice tests, worksheets in MATLAB, MATHEMATICA, MathCad and MAPLE, blogs, related physical problems, please visit

http://numericalmethods.eng.usf.edu/topics/newton_raphson.html

http://numericalmethods.eng.usf.edu/topics/newton_raphson.html�

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