lec 7 newton raphson

7/27/2019 Lec 7 Newton Raphson

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Roots of Nonlinear Equations

Topic: Newton-Raphson Method

Dr. Nasir M Mirza

Numerical Methods

Email: [email protected]


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Newton-Raphson Method

This method is one of the most widely used methods to solveequations also known as the Newton-Raphson.

The idea for the method comes from a Taylor series expansion,

where you know the function and its first derivative.

f(xk+1) = f(xk) + (xk+1 - xk)*f (xk) + ...

The goal of the calculations is to find af(x)=0, so setf(xk+1) =

0 and rearrange the equation. [f (xk) is the first derivative of

f(x). ]0 = f(xk) + (xk+1 - xk)*f (xk)

xk+1 = xk - [ f(xk) / f (xk) ]


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Newton-Raphson Method

f(x)

f(xi)

f(xi-1)

xi+2 xi+1 xix

ii xfx ,

)(xf

)f(x-= xx

i

iii

1

The method uses the slopeof the line to project to the

x axis and find the root.

The method converges on

the solution quickly.


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Derivation

f(x)

f(xi)

xi+1 xix

B

C

A

)('

)(1

i

iiixf

xfxx

1

)()('

ii

ii

xxxfxf

AC

ABtan(


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Algorithm for Newton-RaphsonMethod


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Step 1

Evaluate df/dx = f(x) symbolically: that is find an analyticalexpression for the first derivative of the function with respect to x.


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Newton-Raphson Method: Step 2

)f'(x

)f(x-= xx

i

iii 1

010x

1

1 x

- xx=

i

iia

Calculate the next estimate of the root

Find the absolute relative approximate error


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Step 3

Find if the absolute relative approximate error is greater thanthe pre-specified relative error tolerance.

If so, go back to step 2, else stop the algorithm.

Also check if the number of iterations has exceeded themaximum number of iterations.


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Newton-Raphson Method: Example You are working for DOWN THE TOILET COMPANY that makes

floats for ABC commodes. The ball has a specific gravity of 0.6

and has a radius of 5.5 cm. You are asked to find the distance to

which the ball will get submerged when floating in water.


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Newton-Raphson Method: Example 1

The equation that gives the depth x to which the ball is submergedunder water is given by

Use the Newtons method of finding

roots of equations to find the depth xto which the ball is submerged underwater. Conduct three iterations toestimate the root of the aboveequation.

x.-xxf

.+x.-xxf

-

3303

10x99331650

2

423

Solution:


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Graph of function f(x)

-0.02 0.00 0.02 0.04 0.06 0.08 0.10 0.12

-0.0004

-0.0003

-0.0002

-0.0001

0.0000

0.0001

0.0002

0.0003

0.0004

0.0005

f(x)

x

4

23

10x9933

1650

-.+

x.-xxf


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Iteration #1

%96.75

0.08320

10x4.5

10.413x302.0

'

02.0

3

4

1

0

0

01

0

a

x

xf

xfxx

x


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Iteration #2

%86.42

0.05824

10x689.610x670.108320.0

'

08320.0

3

4

2

1

112

1

a

x

xf

xfxx

x


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Iteration #3

%592.6

0.06235

10x043.9

10x717.305284.0

'

05824.0

a

3

5

2

223

2

xf

xfxx

x


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Newtons Method: computer algo

Do while |x2 - x1| >= tolerance value 1

or |f(x2)|>= tolerance value 2

or f(x1) ~= 0

Set x2 = x1 - f(x1)/f(x1)

Set x1 = x2;

END loop


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Example 2:

Let us solve the followingproblem:

x5 + x3 + 4x2 - 3x2 = 0

Then f(x) = x5 + x3 + 4x2 - 3x2

df/dx = 5x4 + 3x2 + 8x - 3

Let us first draw a graph andsee where are the roots;

The roots are between (-1.7,-

1.3), (-1,0), & (0.5,1.5)

Then let us write a computer

program in MATLAB thatwill compute the root using

Newton-Raphson method.

-2.0 -1.5 -1.0 -0.5 0.0 0.5 1.0 1.5 2.0

-20

-10

0

10

20

30

40

50

f(x)

x


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Computer Program in MATLAB

% A program that Uses Newton-Raphson method% to find the roots of x = x^3 +x^2 -3x -3

% Input: x0 = initial guess;

% n = number of iterations; default: n = 10;

% Output: x = estimate of the root;

n = 10; x0 = -2.0

x = x0; % Initial Guess

fprintf(' k f(x) dfdx x(k+1)\n');

for k=1:n

f = x^5 + x^3 + 4*x^2 - 3*x - 2;

dfdx = 5*x^4 + 3*x^2 + 8*x - 3;

x = x - f/dfdx;fprintf('%3d %12.3e %12.3e %18.15f\n',k-1,f,dfdx,x);

end


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Newtons Method

k f(x) dfdx x(k+1)

0 -2.000e+001 7.300e+001 -1.726027

1 -5.367e+000 3.651e+001 -1.579022

2 -1.043e+000 2.293e+001 -1.533545

3 -8.054e-002 1.944e+001 -1.5294014 -6.276e-004 1.914e+001 -1.529368

5 -3.911e-008 1.914e+001 -1.529368

6 -2.665e-015 1.914e+001 -1.529368

7 3.553e-015 1.914e+001 -1.529368

8 -2.665e-015 1.914e+001 -1.5293689 3.553e-015 1.914e+001 -1.529368

Result of the computer program: x0 = -2.0 and iterations n = 10.

The root value is x = -1.529368 and it was obtained after 4 iterations.


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Advantages

Converges fast, if it converges

Requires only one guess


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Drawbacks

-20

-15

-10

-5

0

5

10

-2 -1 0 1 2

1

2 3

f(x)

x Inflection Point

01 3 xxf


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Drawbacks (continued)

-1.00E-05

-7.50E-06

-5.00E-06

-2.50E-06

0.00E+00

2.50E-06

5.00E-06

7.50E-06

1.00E-05

-0.03 -0.02 -0.01 0 0.01 0.02 0.03 0.04

x

f(x)

0.02

Division by zero

010x4.203.0 623 xxxf


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-1.5

-1

-0.5

0

0.5

1

1.5

-2 0 2 4 6 8 10

x

f(x)

-0.0630690.54990 4.462 7.53982

Root Jumping

0 xSinxf


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-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

Oscillations near Local Maxima or Minima

022 xxf


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Example 3:

Let us solve the following

problem:

x3 - 4x2 + x10 = 0

Then f(x) = x3 - 4x2 + x10

df/dx = 3x2 -8 x + 1

Let us first draw a graph andsee where are the roots;

The roots are between (3,6).


program in MATLAB that

will compute the root usingNewton-Raphson method.

-6 -4 -2 0 2 4 6

-400

-350

-300

-250

-200

-150

-100

-50

0

50

100

f(x)

x


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Example 3: Computer Program

% A program that Uses Newton-Raphson method% to find the roots of x = x^3 +x^2 -3x -3




n = 10; x0 = 3.0


fprintf(' k f(x) dfdx x(k+1)\n');

for k=1:n

f = x^3 - 4*x^2 + x 10.0;

dfdx = 3*x^2 - 8*x + 1.0;


end


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Example 3: results

k f(x) dfdx x(k+1)

0 -1.600e+001 4.000e+000 7.000000

1 1.440e+002 9.200e+001 5.434783

2 3.781e+001 4.613e+001 4.615102

3 7.716e+000 2.798e+001 4.339292

4 7.280e-001 2.277e+001 4.3073275 9.181e-003 2.220e+001 4.306913

6 1.526e-006 2.219e+001 4.306913

7 4.796e-014 2.219e+001 4.306913

8 3.553e-015 2.219e+001 4.306913

9 3.553e-015 2.219e+001 4.306913

>>


The root value is x = 4.306913and it was obtained after 5 iterations.


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Example 4:

Let us solve the following

problem:

exp(x)+x - 10 = 0

Then f(x) = exp(x) + x10.

df/dx = exp(x) + 1.

Let us first draw a graph and seewhere are the roots;

The roots are between (0, 4).


program in MATLAB that will

compute the root using Newton-Raphson method.

-4 -2 0 2 4

-20

-10

0

10

20

30

40

50

f(x)

x


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Example 4: Computer Program

% A program that Uses Newton-Raphson method

% to find the roots of x = x^3 +x^2 -3x -3




n = 10; x0 = 0.0


fprintf(' k f(x) dfdxx(k+1)\n');

for k=1:n

f = exp(x) + x 10.0;

dfdx = exp(x) + 1.0;


end


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Example 4: results

k f(x) dfdx x(k+1)

0 -9.000e+000 2.000e+000 4.500000

1 8.452e+001 9.102e+001 3.571415

2 2.914e+001 3.657e+001 2.774566

3 8.806e+000 1.703e+001 2.257515

4 1.817e+000 1.056e+001 2.085456

5 1.337e-001 9.048e+000 2.070678

6 8.746e-004 8.930e+000 2.070580

7 3.803e-008 8.929e+000 2.070580

8 0.000e+000 8.929e+000 2.070580

9 0.000e+000 8.929e+000 2.070580

>>


The root value is x = 2.070580


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Why convergence is so rapid?

Let us assume that exact root is a and error in nth iteration isen . Then

However,

nn eax

)()(

1n

n

nn xf

xf

eax

The Taylor Series for f(a - en) and its derivative is

af

eafeafeaf n

nn 2

2

afe

afeafeaf nnn2

2


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Why convergence is so rapid?

)(

)(1

n

nnn

xf

xfeax

Putting the Taylor Series for f(a - en) and its derivative in the formula

)(2

)(2

2

2

1

af

afea

af

afe

eeax

n

n

nnn

Hence error in (n+1)th iteration is proportional to (en)2 and that is why it

converges rapidly.

lec 7 newton raphson

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