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Unit – IV: Numerical Techniques•Zeroes of transcendental and polynomial equations using •Bisection method, •Regula-falsi method•Newton-Raphson method,• Rate of convergence.• Interpolation:• Finite differences,•Newton’s forward and backward interpolation,• Lagrange’s and Newton’s divided difference formula forunequal intervals.

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A root, r, of function f occurs when f(r) = 0.For example:

f(x) = x2 – 2x – 3has two roots at r = -1 and r = 3.

f(-1) = 1 + 2 – 3 = 0f(3) = 9 – 6 – 3 = 0

We can also look at f in its factored form.

f(x) = x2 – 2x – 3 = (x + 1)(x – 3)

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Bisection MethodInitial two values of x taken: x1=a & x2=b such that if

y(a) is +ve then y(b) is –ve. Then new value c=(a+b)/2

a bc

f(a)>0

f(b)<0

f(c)>0

If c is +ve, then replace the value of a by c. If c is –ve, then replace the value of b by c.Then continue to find the next value of c (End of 1st iteration) with c=(a+b)/2Guaranteed to converge to a root if one exists within these initial two values.

4

Regula Falsi

a bc

( ) ( )( ) ( )

( ) ( )( ) 0 ( )

0 ( )( ) ( )

( )( ) ( )

f a f by x f b x ba bf a f by c f b c ba b

a bf b c bf a f bf b a b

c bf a f b

f(c)<0

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Newton-Raphson Method• Only one current guess of x .• Consider some point x0.– If we approximate f(x) as a line about x0,

then we can again solve for the root of the line.

0 0 0( ) ( )( ) ( )l x f x x x f x Solving, leads to the following

iteration:0

1 00

1

( ) 0( )( )( )( )i

i ii

l xf xx xf xf xx xf x

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Unit – V: Numerical Techniques –II

•Solution of system of linear equations, •Matrix Decomposition methods, •Jacobi method,• Gauss- Seidal method.• Numerical differentiation, •Numerical integration, •Trapezoidal rule, •Simpson’s one third and three-eight rules,• Solution of ordinary differential equations (first order,second order and simultaneous)•by Euler’s, Picard’s and• fourth-order Runge- Kutta methods.

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Systems of Non-linear EquationsConsider the set of equations:

1 1 2

2 1 2

1 2

, , , 0

, , , 0

, , , 0

n

n

n n

f x x x

f x x x

f x x x

11 1 12 2 1 1

21 1 22 2 2 2

1 1 2 2

The system of equations:

N N

N N

N N NN N N

a T a T a T Ca T a T a T C

a T a T a T C

A total of N algebraic equations for the N nodal points and the system can be expressed as a matrix formulation: [A][T]=[C]

11 12 1 1 1

21 22 2 2 2

1 2

= , ,

N

N

N N NN N N

a a a T Ca a a T C

where A T C

a a a T C

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Jacobi Iteration

nnnnnn

nn

nn

bxaxaxa

bxaxaxabxaxaxa

2211

22222121

11212111

0

02

01

0

nx

xx

x

)(1 01

02121

11

11 nnxaxaba

x

1

1 1

1 1 i

j

n

ij

kjij

kjiji

ii

ki xaxab

ax

)(1 011

022

011

1 nnnnnn

nnn xaxaxaba

x

)(1 02

0323

01212

22

12 nnxaxaxaba

x

9

Gauss-Siedel Method

33

23213131

22

32312122

11

31321211

axaxabx

axaxabx

axaxabx

3333232131

2323222121

1313212111

bxaxaxa

bxaxaxa

bxaxaxa

Now we can start the solution process by choosing guesses for the x’s. A simple way to obtain initial guesses is to assume that they are zero. These zeros can be substituted into x1 equation to calculate a new x1=b1/a11.

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Gauss-Seidel (GS) iteration

nnnnnn

nn

nn

bxaxaxa

bxaxaxabxaxaxa

2211

22222121

11212111

0

02

01

0

nx

xx

x

1

1 1

11 1 i

j

n

ij

kjij

kjiji

ii

ki xaxab

ax)(1 0

102121

11

11 nnxaxaba

x

)(1 111

122

111

1 nnnnnn

nnn xaxaxaba

x

)(1 02

0323

11212

22

12 nnxaxaxaba

x

Use the latestupdate

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Difference between the Gauss-Seidel and the Jacobi iterative methods

The Gauss-Seidel method

The Jacobi iteration method

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ExampleSolve the following system of equations using (a) the Jacobi methos, (b) the Gauss Seidel iteration method.

4 2 112 0 3

2 4 16

X Y ZX Y ZX Y Z

,* ,

(a) Jacobi method: use initial guess X0=Y0=Z0=1, stop when maxXk-Xk-1,Yk-Yk-1,Zk-Zk-1 0.1First iteration: X1= (11/4) - (1/2)Y0 - (1/4)Z0 = 2Y1= (3/2) + (1/2)X0 = 2Z1= 4 - (1/2) X0 - (1/4)Y0 = 13/4

Reorganize into new form:

X = 114

- 12

Y - 14

Z

Y = 32

+ 12

X + 0 * Z

Z = 4 - 12

X - 14

Y

4 2 1 111 2 0 3

2 1 4 16

XYZ

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Second iteration: use the iterated values X1=2, Y1=2, Z1=13/4X2 = (11/4) - (1/2)Y1 - (1/4)Z1 = 15/16Y2 = (3/2) + (1/2)X1 = 5/2Z2 = 4 - (1/2) X1 - (1/4)Y1 = 5/2

Final solution [1.014, 2.02, 2.996] Exact solution [1, 2, 3]

5 4 5 4 5 4

Converging Process:13 15 5 5 7 63 93 133 31 393[1,1,1], 2,2, , , , , , , , , ,4 16 2 2 8 32 32 128 16 128

519 517 767, , . Stop the iteration when 512 256 256

max , , 0.1X X Y Y Z Z

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Example (cont.)(b) Gauss-Seidel iteration: Substitute the iterated values into the iterative process immediately after they are computed.

0 0 0

1 0 0

1 1

1 1 1

Use initial guess X 111 1 1 3 1 1 1, , 44 2 4 2 2 2 4

11 1 1First iteration: X = ( ) ( ) 24 2 43 1 3 1 5(2)2 2 2 2 2

1 1 1 1 5 194 4 (2)2 4 2 4 2 8

5 19Converging process: [1,1,1], 2, ,2 8

Y Z

X Y Z Y X Z X Y

Y Z

Y X

Z X Y

29 125 783 1033 4095 24541, , , , , ,32 64 256 1024 2048 8192

The iterated solution [1.009, 1.9995, 2.996] and it converges faster

Immediate substitution

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Numerical Differentiation: [Find dy/dx or f’(x)]

Forward DifferenceBackward Difference

hhxfhxfxf

2)()()(

Central Difference

hhxfhxfxf

2)()()(

hxfhxfxf )()()(

3-points Difference Formulas (More Accurate )

hxfhxfhxfxf

2)(3)(4)2()(

Forward Difference :

Backward Difference: h

xfhxfhxfxf2

)()(4)2(3)(

2-Points Difference Formulas

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Example:Find estimates for the derivative with the data set given to the right. We notice the x values are evenly spaced.Two-Point Estimates:

i xi yi

1 3 1

2 5 4

3 7 3

21

5743)5('

23

3514)3('

f

f

21

5743)7('

23

3514)5('

f

f

Forward Backward

21

42

3713)5('

f

Central

Three Point Estimates:

25

410

37)1(3)4(43)3('

f23

46

37)1(1)4(4)3(3)7('

f

Forward Backward

1441)4(23

21)5('' 2

f

Second Derivative (h=2)

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Numerical Integration

Simpson’s 1/3rd rule

nn

nnn

b

an

b

a

xaxaxaaxf

dxxfdxxfI

1

110)(

)()(

Trapezoidal rule

2

)()(4)(3 210

abhxfxfxfhI

1

10

12110

)(2)()(2

2)()(

2)()(

2)()(

n

iin

nn

xfxfxfnabI

xfxfhxfxfhxfxfhI

)()(3)(3)(83

3210 xfxfxfxfhI Simpson’s 3/8th rule 3abh

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Integration with Unequal Segments

Example 21.7

2)()(

2)()(

2)()( 121

210

1nn

nxfxfhxfxfhxfxfhI

x f(x) x f(x)0.0 0.2 0.44 2.8420.12 1.309 0.54 3.5070.22 1.305 0.64 3.1810.32 1.743 0.70 2.3630.36 2.074 0.80 0.2320.40 2.456

594.112975.01307.00905.0 2

363.2232.010.02

181.3363.006.02

309.1305.110.02

2.0309.112.0

I

Data for f(x)= 0.2+25x-200x2+675x3-

900x4+400x5

Using Trapezoidal Rule

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Example: xxexf Find the approximate value of

2f

1.9 12.703199

2.0 14.778112

2.1 17.148957

2.2 19.855030

x xf

with

10.h

Using the 1st Three-point formula:

032310.22 855030.19

148957.174778112.1432.0

1

)2.2()1.2(4)2(31.02

12

ffff

hxfhxfxfh

xf 24321

0000

Using the 2nd Three-point formula:

228790.22

703199.12148957.172.0

1

)9.1()1.2(1.02

12

fff

The exact value of 22.167168 :is 2f

hxfhxfh

xf 000 21

20

by dividing the interval into 20 subintervals.

Find π

sin0

(x)dx

2021020

20

20

....., , , , ,π

π

kkkhax

nabh

n

k

995886120

2040

22

19

1

1

10

.

πsinπsinsinπ

)()()sin(π

k

n

kk

k

bfxfafhdxx

000006220

124

20220

6010

1

9

10

.

)πsin(π)(sin

πsinsinπ)sin(π

k

k

k

kdxx

By Simpson’s Rule

By Trapizoidal Rule

Example-2

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Numerical Solutions for Partial Differential Equations by 4th order Runga-Kutta Method

hkkkkyy ii )22(61

43211

),(

)21,

21(

)21,

21(

),(

34

23

12

1

hkyhxfk

hkyhxfk

hkyhxfk

yxfkwhere

ii

ii

ii

ii

Consider Exact SolutionInitial condition :Step size :

2txdtdx

t222 ettx

10 x1.0h

104829.1226

109499.0104988.1,1.01.0,

104988.02/.1,05.01.021,

21

10475.005.1,05.01.021,

21

1.0011.01,01.0,

432101

34

223

12

21

kkkkhxx

fkxhtfhhk

kfkxhtfhhk

fkxhtfhhk

fxtfhhk