Lecture 44Lecture 44

NumericalAnalysis

NumericalAnalysis

Solution of Solution of

Non-Linear Non-Linear EquationsEquations

Bisection MethodBisection MethodRegula-Falsi MethodRegula-Falsi MethodMethod of iterationMethod of iterationNewton - Raphson MethodNewton - Raphson MethodMuller’s MethodMuller’s MethodGraeffe’s Root Squaring Graeffe’s Root Squaring MethodMethod

Bisection MethodBisection MethodRegula-Falsi MethodRegula-Falsi MethodMethod of iterationMethod of iterationNewton - Raphson MethodNewton - Raphson MethodMuller’s MethodMuller’s MethodGraeffe’s Root Squaring Graeffe’s Root Squaring MethodMethod

Newton -Raphson Method

Newton -Raphson Method

An approximation to An approximation to the root is given bythe root is given by

01 0

0

( )

( )

f xx x

f x

Better and successive Better and successive approximations approximations xx22, , xx33, …, , …, xxnn

to the root are obtained from to the root are obtained from

1

( )

( )n

n nn

f xx x

f x

N-R Formula

Newton’s algorithmNewton’s algorithm

To find a solution to To find a solution to f(x)=0f(x)=0 given an initial given an initial approximation approximation pp00

INPUTINPUT initial approximation initial approximation pp00; tolerance TOL; maximum ; tolerance TOL; maximum

number of iterations Nnumber of iterations N00

OUTPUTOUTPUT approximate approximate solution p or message of failuresolution p or message of failure

Step 1Step 1Set Set I = 1I = 1

Step 2Step 2While i While i << N0 do Steps 3-6 N0 do Steps 3-6

Step 3Step 3 SetSet p = pp = p0 0 – f ( p– f ( p0 0 ) / f’ ( p) / f’ ( p0 0 ))

(compute p(compute pi i ).).

Step 4Step 4 If Abs (p – pIf Abs (p – p00) < TOL) < TOL

OUTPUTOUTPUT ( p ); ( p );(The procedure was successful.)(The procedure was successful.)

STOPSTOP

Step 5Step 5 SetSet i = i + 1i = i + 1

Step 6Step 6 Set pSet p00 = p = p (Update p(Update p0 0 ))

Step 7 Step 7 OUTPUTOUTPUT (The method failed after N(The method failed after N00 iterations, N iterations, N00

= ‘,N= ‘,N00 ) )

The procedure was unsuccessfulThe procedure was unsuccessful

STOPSTOP

ExampleExampleUsing Maple to solve a Using Maple to solve a non-linear equation.non-linear equation.

cos( ) 0x x

Solution Solution The Maple command will be as The Maple command will be as follows,follows,Fsolve ( cos (x) -x);Fsolve ( cos (x) -x);

alg023(); alg023(); This is Newton's MethodThis is Newton's MethodInput the function F(x) in terms of xInput the function F(x) in terms of xFor example:For example:> cos(x)-x> cos(x)-xInput initial approximationInput initial approximation> 0.7853981635> 0.7853981635Input toleranceInput tolerance> 0.00005> 0.00005

Input maximum number of Input maximum number of iterations - no decimal pointiterations - no decimal point> 25> 25Select output destinationSelect output destination1. Screen1. Screen2. Text file2. Text fileEnter 1 or 2Enter 1 or 2> 1> 1

Select amount of outputSelect amount of output1. Answer only1. Answer only2. All intermediate 2. All intermediate approximationsapproximationsEnter 1 or 2Enter 1 or 2> 2> 2

Newton's MethodNewton's Method I P F(P)I P F(P)

1 0.739536134 1 0.739536134 -7.5487470e-04-7.5487470e-04

2 0.739085178 2 0.739085178 -7.5100000e-08-7.5100000e-08

3 0.7390851333 0.739085133 0.0000000e-01 0.0000000e-01

Approximate solution = 0.73908513Approximate solution = 0.73908513

with F(P) = 0.0000000000with F(P) = 0.0000000000

Number of iterations = 3Number of iterations = 3

Tolerance = 5.0000000000e-05Tolerance = 5.0000000000e-05

Another ExampleAnother Example> > alg023(); alg023(); Input the function F(x) in Input the function F(x) in terms of x , terms of x , > sin(x)-1> sin(x)-1Input initial approximationInput initial approximation> 0.17853> 0.17853Input toleranceInput tolerance> 0.00005> 0.00005

Input maximum number of Input maximum number of iterations – no decimal pointiterations – no decimal point> 25> 25Select output destinationSelect output destination1. Screen1. Screen2. Text file2. Text fileEnter 1 or 2Enter 1 or 2>2>2

Select amount of outputSelect amount of output1. Answer only1. Answer only2. All intermediate 2. All intermediate approximationsapproximationsEnter 1 or 2Enter 1 or 2> 2> 2

Newton's MethodNewton's Method I P F(P)I P F(P) 1 1.01422964e+00 -1.5092616e-011 1.01422964e+00 -1.5092616e-01 2 1.29992628e+00 -3.6461537e-022 1.29992628e+00 -3.6461537e-02 3 1.43619550e+00 -9.0450225e-033 1.43619550e+00 -9.0450225e-03 4 1.50359771e+00 -2.2569777e-034 1.50359771e+00 -2.2569777e-03 5 1.53720967e+00 -5.6397880e-045 1.53720967e+00 -5.6397880e-04 6 1.55400458e+00 -1.4097820e-046 1.55400458e+00 -1.4097820e-04 7 1.56240065e+00 -3.5243500e-057 1.56240065e+00 -3.5243500e-05

More…More…

8 1.56659852e+00 -8.8108000e-068 1.56659852e+00 -8.8108000e-06 9 1.56869743e+00 -2.2027000e-069 1.56869743e+00 -2.2027000e-06 10 1.56974688e+00 -5.5070000e-0710 1.56974688e+00 -5.5070000e-07 11 1.57027163e+00 -1.3770000e-0711 1.57027163e+00 -1.3770000e-07 12 1.57053407e+00 -3.4400000e-0812 1.57053407e+00 -3.4400000e-08 13 1.57066524e+00 -8.6000000e-0913 1.57066524e+00 -8.6000000e-09 14 1.57073085e+00 -2.1000000e-0914 1.57073085e+00 -2.1000000e-09 15 1.57076292e+00 -6.0000000e-1015 1.57076292e+00 -6.0000000e-10

Approximate solution Approximate solution = 1.57076292= 1.57076292with with F(P) =6.0000000000e-10F(P) =6.0000000000e-10Number of iterations = 15Number of iterations = 15Tolerance = Tolerance = 5.0000000000e-055.0000000000e-05

Bisection MethodBisection Method

> alg021();> alg021();This is the Bisection Method.This is the Bisection Method.Input the function F(x) in Input the function F(x) in terms of xterms of xFor example:For example:> x^3+4*x^2-10> x^3+4*x^2-10

Input endpoints A < B Input endpoints A < B separated by blankseparated by blank> 1 2> 1 2Input toleranceInput tolerance> 0.0005> 0.0005Input maximum number of Input maximum number of iterations - no decimal pointiterations - no decimal point> 25> 25

Select output destinationSelect output destination1.1.Screen , Screen , 2. Text file2. Text fileEnter 1 or 2Enter 1 or 2> 1> 1Select amount of outputSelect amount of output1. Answer only1. Answer only2. All intermediate approximations2. All intermediate approximationsEnter 1 or 2Enter 1 or 2> 2> 2

Bisection MethodBisection Method I P F(P)I P F(P) 1 1.50000000e+00 2.3750000e+00 1 1.50000000e+00 2.3750000e+00 2 1.25000000e+00 -1.7968750e+00 2 1.25000000e+00 -1.7968750e+00 3 1.37500000e+00 1.6210938e-013 1.37500000e+00 1.6210938e-01

4 1.31250000e+00 -8.4838867e-01 4 1.31250000e+00 -8.4838867e-01 5 1.34375000e+00 -3.5098267e-01 5 1.34375000e+00 -3.5098267e-01 6 1.35937500e+00 -9.6408842e-02 6 1.35937500e+00 -9.6408842e-02 7 1.36718750e+00 3.2355780e-02 7 1.36718750e+00 3.2355780e-02 8 1.36328125e+00 -3.2149969e-02 8 1.36328125e+00 -3.2149969e-02 9 1.36523438e+00 7.2030000e-05 9 1.36523438e+00 7.2030000e-05 10 1.36425781e+00 -1.6046697e-02 10 1.36425781e+00 -1.6046697e-02 11 1.36474609e+00 -7.9892590e-03 11 1.36474609e+00 -7.9892590e-03

Approximate solution P = 1.36474609 Approximate solution P = 1.36474609 with F(P) = -.00798926with F(P) = -.00798926Number of iterations = 11 Number of iterations = 11 Tolerance = 5.00000000e-04Tolerance = 5.00000000e-04

alg021(); Another example of the alg021(); Another example of the Bisection Method.Bisection Method.Input the function F(x) in terms Input the function F(x) in terms of x,of x,> cos(x)> cos(x)Input endpoints A < B separated Input endpoints A < B separated by blankby blank> 1 2> 1 2Input toleranceInput tolerance> 0.0005 > 0.0005

Input maximum number of iterations - Input maximum number of iterations - no decimal pointno decimal point> 25> 25Select output destinationSelect output destination1. Screen , 2. Text file1. Screen , 2. Text fileEnter 1 or 2Enter 1 or 2> 1> 1Select amount of outputSelect amount of output1. Answer only1. Answer only2. All intermediate approximations2. All intermediate approximationsEnter 1 or 2Enter 1 or 2> 2> 2

Bisection MethodBisection Method 1 P F(P)1 P F(P) 1 1.50000000e+00 7.0737202e-02 1 1.50000000e+00 7.0737202e-02 2 1.75000000e+00 -1.7824606e-01 2 1.75000000e+00 -1.7824606e-01 3 1.62500000e+00 -5.4177135e-02 3 1.62500000e+00 -5.4177135e-02 4 1.56250000e+00 8.2962316e-03 4 1.56250000e+00 8.2962316e-03 5 1.59375000e+00 -2.2951658e-02 5 1.59375000e+00 -2.2951658e-02

6 1.57812500e+00 -7.3286076e-03 6 1.57812500e+00 -7.3286076e-03 7 1.57031250e+00 4.8382678e-04 7 1.57031250e+00 4.8382678e-04 8 1.57421875e+00 -3.4224165e-03 8 1.57421875e+00 -3.4224165e-03 9 1.57226563e+00 -1.4692977e-03 9 1.57226563e+00 -1.4692977e-03 10 1.57128906e+00 -4.9273519e-04 10 1.57128906e+00 -4.9273519e-04 11 1.57080078e+00 -4.4542051e-06 11 1.57080078e+00 -4.4542051e-06

Approximate solution P = 1.57080078Approximate solution P = 1.57080078

with F(P) = -.00000445with F(P) = -.00000445

Number of iterations = 11Number of iterations = 11

Tolerance = 5.00000000e-04 Tolerance = 5.00000000e-04

alg025(); This is the Method of False alg025(); This is the Method of False

PositionPosition

Input the function F(x) in terms of xInput the function F(x) in terms of x

> cos(x)-x> cos(x)-x

Input endpoints Input endpoints P0 < P1P0 < P1 separated by a separated by a

blank spaceblank space0.5 0.78539816350.5 0.7853981635

Input toleranceInput tolerance

>0.0005>0.0005

Input maximum number of Input maximum number of iterations - no decimal pointiterations - no decimal point> 25> 25Select output destinationSelect output destination1. Screen1. Screen2. Text file2. Text fileEnter 1 or 2Enter 1 or 2>1>1

Select amount of outputSelect amount of output1. Answer only1. Answer only2. All intermediate 2. All intermediate approximationsapproximationsEnter 1 or 2Enter 1 or 2> 2> 2

METHOD OF FALSE POSITION METHOD OF FALSE POSITION I P F(P)I P F(P) 2 7.36384139e-01 4.51771860e-032 7.36384139e-01 4.51771860e-03 3 7.39058139e-01 4.51772000e-053 7.39058139e-01 4.51772000e-05 4 7.39084864e-01 4.50900000e-074 7.39084864e-01 4.50900000e-07

Approximate solution P = .73908486Approximate solution P = .73908486with F(P) = .00000045with F(P) = .00000045Number of iterations = 4 Number of iterations = 4 Tolerance = 5.00000000e-04Tolerance = 5.00000000e-04

System of System of Linear Linear

EquationsEquations

System of System of Linear Linear

EquationsEquations

Gaussian Elimination Gauss-Jordon EliminationCrout’s Reduction Jacobi’sGauss- Seidal Iteration RelaxationMatrix Inversion

Gaussian Elimination Gauss-Jordon EliminationCrout’s Reduction Jacobi’sGauss- Seidal Iteration RelaxationMatrix Inversion

> alg061();> alg061();This is Gaussian Elimination to solve a This is Gaussian Elimination to solve a linear system.linear system.The array will be input from a text file in The array will be input from a text file in the order:the order:A(1,1), A(1,2), ..., A(1,N+1), A(2,1), A(1,1), A(1,2), ..., A(1,N+1), A(2,1), A(2,2), ..., A(2,N+1),..., A(N,1), A(N,2), ..., A(2,2), ..., A(2,N+1),..., A(N,1), A(N,2), ..., A(N,N+1)A(N,N+1)

Place as many entries as desired on each Place as many entries as desired on each line, but separate entries withline, but separate entries withat least one blank.at least one blank.

Has the input file been created? - enter Y or N.Has the input file been created? - enter Y or N.> y> yInput the file name in the form - drive:\Input the file name in the form - drive:\name.extname.extfor example: A:\DATA.DTAfor example: A:\DATA.DTA> d:\maple00\dta\alg061.dta> d:\maple00\dta\alg061.dtaInput the number of equations - an integer.Input the number of equations - an integer.> 4> 4Choice of output method:Choice of output method:1. Output to screen1. Output to screen 2. Output to text file2. Output to text filePlease enter 1 or 2.Please enter 1 or 2.> 1> 1

GAUSSIAN ELIMINATIONGAUSSIAN ELIMINATION

The reduced system - output by rows:The reduced system - output by rows: 1.00000000 -1.00000000 2.00000000 -1.00000000 -8.000000001.00000000 -1.00000000 2.00000000 -1.00000000 -8.00000000

0.00000000 2.00000000 -1.00000000 1.00000000 6.000000000.00000000 2.00000000 -1.00000000 1.00000000 6.00000000

0.00000000 0.00000000 -1.00000000 -1.00000000 -4.000000000.00000000 0.00000000 -1.00000000 -1.00000000 -4.00000000

0.00000000 0.00000000 0.00000000 2.00000000 4.000000000.00000000 0.00000000 0.00000000 2.00000000 4.00000000

Has solution vector:Has solution vector: -7.00000000 3.00000000 2.00000000 2.00000000-7.00000000 3.00000000 2.00000000 2.00000000

with 1 row interchange (s)with 1 row interchange (s)

> alg071();> alg071();This is the Jacobi Method for Linear This is the Jacobi Method for Linear Systems.Systems.The array will be input from a text file The array will be input from a text file in the orderin the orderA(1,1), A(1,2), ..., A(1,n+1), A(2,1), A(1,1), A(1,2), ..., A(1,n+1), A(2,1), A(2,2), ..., A(2,2), ..., A(2,n+1),..., A(n,1), A(n,2), ..., A(n,n+1)A(2,n+1),..., A(n,1), A(n,2), ..., A(n,n+1)Place as many entries as desired on Place as many entries as desired on each line, but separateeach line, but separateentries with at least one blank.entries with at least one blank.

The initial approximation should follow The initial approximation should follow in the same format has the input file in the same format has the input file been created? - enter Y or N.been created? - enter Y or N.> y> yInput the file name in the form - drive:\Input the file name in the form - drive:\name.extname.extfor example: A:\DATA.DTAfor example: A:\DATA.DTA> d:\maple00\alg071.dta> d:\maple00\alg071.dtaInput the number of equations - an Input the number of equations - an integer.integer.> 4> 4

Input the tolerance.Input the tolerance.

> 0.001> 0.001

Input maximum number of iterations.Input maximum number of iterations.

> 15> 15

Choice of output method:Choice of output method:

1. Output to screen1. Output to screen

2. Output to text file2. Output to text file

Please enter 1 or 2.Please enter 1 or 2.

> 1> 1

JACOBI ITERATIVE METHOD JACOBI ITERATIVE METHOD FOR LINEAR SYSTEMSFOR LINEAR SYSTEMS

The solution vector is :The solution vector is :1.00011860 1.99976795 1.00011860 1.99976795 -.99982814 0.99978598-.99982814 0.99978598using 10 iterationsusing 10 iterationswith Tolerance 1.0000000000e-03with Tolerance 1.0000000000e-03

Lecture 44Lecture 44

NumericalAnalysis

NumericalAnalysis