Gauss Elimination Method

a₁₁x₁  +  a₁₂x₂ +  a₁₃ x₃      = b₁

a₂₁x₁  +  a₂₂x₂ + a₂₃x₃        =b₂

a₃₁x₁  + a₃₂x₂  + a₃₃x₃        =b₃

[a₁₁         a₁₂         a₁₃]   [x₁]        [b₁]

[a₂₁             a₂₂       a₂₃]   [x₂]    =  [b₂]

[a₃₁         a₃₂         a₃₃]   [x₃]        [b₃]

To  eliminate coefficient of x₁ from 2^nd and 3^rd Row Recursive formula :-

[ R_i – a_i1 × R_i                    for [ I = 2 & 3]

        a₁₁

[a₁₁        a₁₂          a₁₃]  [x₁]         [b₁]

[a₂₁        a₂₂          a₂₃]  [x₂]    =   [b₂]

[a₃₁        a₃₂          a₃₃]  [x₃]          [b₃]

To eliminate coefficients of x₂ from 3^rd row

[R_i – a_i2 × R₂]

       a₂₂

Ques:-

           2x₁  +  2x₂  +  2x₃  =  14

              x₁  +  6x₂   -    x₃  =  13

             2x₁  -   x₂   +  2x₃  =   5

Sol:-

To eliminate coefficient of x₁ from 2^nd and 3^rd row.

For |i=2|

            R₂ – a₂₁ ×  R₁

                   a₁₁

  a₂₁ = a₂₁ – 1 × a₁₁

                      2

         = 1 – 1 × 2

                 2

|a₂₁=0|

a₂₂= a₂₂ – a₂₁ × a₁₂

               a₁₁

   = 6 – 1 × 8

           2

|a₂₂ = 2|

a₂₃= a₂₃ × a₂₁  × a₁₃

               a₁₁

      = - 1 – 1 × 2

                2

|a₂₃=- 2|

[2      8        2] [x₁]        [14]

[0       2     -2]  [x₂]   =   [13]

[2      -1      2]  [x₃]        [5]

For |i = 3|

        R₃ – a₃₁ × R₁

              a₁₁

a₃₁ = a₃₁ –  a₃₁ × a₁₁

                 a₁₁

      = 2 – 2 × 2

              2

|a₃₁ = 0|

a₃₂ = a₃₂ – a₃₁ × a₁₂

                a₁₁

       = - 1 – (-1) × 8

                   2

|a₃₂= -9|

a₃₃ = a₃₃ –  a₃₁ × a₁₃

                 a₁₁

       = 2 – 2  × 2

               2

|a₃₃=0|

b₃ = b₃ – a₃₁ × b₁  => 5 – 2 ×14

             a_{11                  2} 

                     

     = 5 – 14

|b₃= -9|

[2       8        2]  [x₁]        [14]

[0       2       -2]  [x₂]   =   [ 6 ]

[0      -9        6]  [x₃]        [-9]

To eliminate x₂ from 3^rd row.

| R_i –  a_i2 × R₂ |

          a₂₂

for |i = 3|

       R₃ –  a₃₂ × R₂

              a₂₂

 a₃₂ = a₃₂ –  a₃₂ × a₂₂

                  a₂₂

        = - 9 – (-9) × 2

                   2

|a₃₂=0|

a₃₃ = a₃₃ –  a₃₂ × a₂₃

                 a₂₂

      = 0 – (-9) ×(-2)

                2

|a₃₃= - 9|

b₃ = b₃ – a₃₂  × b₂

             a₂₂

     = - 9 – (-9) × 6

                 2

     = -9 – (-27)

|b₃= 18|

[2       8        2]  [x₁]       [14]

[0       2       -2]  [x₂]  =   [ 6 ]

[0       0       -9]  [x₃]       [18]

-       9x₃ = 18

  x₃ = - 18

            9

|x₃= 2|

  2x₂ + (2x₃)  = 6

  2x₂ – 2x₃     = 6

    x₂ – x₃       = 3

           x₂      = 3 – 2

|x₂ = 1|

2x₁   +   8x₂   +   2x₃   =   14

2x₁   +  8(1)  +   2(-2)=   14

2x₁   =  14 – 8 + 4

   x₁  = 10

            2

|x₁ = 5|

[x₁ = 5   ,   x₂ = 1   ,   x₃ = -2 ]

Newton Rapshon Method

Formula:-

          x = x_n –  f(x_n)

                          f '(x_n')

Ques.1:-

              x² – 5x + 2

           f(x) = x² – 5x + 2

           f '(x) = 2x – 5

           x₀ = 0 – 0 + 2 = 2                       (+ ve)

           x₁ = 1 – 5 + 2 = -4 +2 = - 2        (- ve)

so a root of f(x)=0 , lies b/w 0 and 1.

So, the mid value in 0 and 1 is 0.5

  

          So,

                  x₀ = 0.5 

1^st        

      x₁ = x₀ – f(x₀)

                        f '(x₀)

             = 0.5 – (0.5)² – 5(0.5) + 2

                                2(0.5) – 5 

       [x₁ = 0.4375]

2^nd

        x₂ = x₁ – f(x₁)

                          f '(x₁)

              = 0.4375 – (0.4375)² – 5(0.4375) + 2

                                       2(0.4375) – 5

              =0.4375  –  0.191406 – 2.1875 + 2

                                          0.875 – 5

              =0.4375 – 0.0039

                               -4.125

  

              =0.437 + 0.00094

         [x₂= 0.4379]

Clearly, x₁ = x₂   . Hence the required root is 0.437

Secant Method

 Que:- x³ = 5x + 1

  x₂ = x₁ – [x₁ – x₀       ]  f(x₁)

                        f(x₁) – f(x₀)

Sol:-

       f(0)= 0 - 0 +1 = 1

       f(1)= 1 - 5 + 1= -3

I^st

  x₀ = 0             ,            f(x₀) = 1

  x₁ = 1              ,            f(x₁) = - 3

x₂ = x₁ – [x₁ – x₀       ]  f(x₁)

                f(x₁) – f(x₀)

  

= 1 – [ 1 – 0    ] – 3                         

           -3 – 1

= 1 – [ -3 ]

          -4

x₂= 1 – 3  = 1 – 0.75 = 0.25

           4

|x₂= 0.25|

F(x₂) = (0.25)³ – 5(0.25) + 1

         = 0.015625 – 1.25 + 1

|f(x₂)= - 0.234375|

2^nd

x₃ = x₂ – [x₂ – x₁       ]  f(x₂)

               f(x₂) – f(x₁)

x₂ = 0.25      ,         f(x₂) = - 0.234

x₁ = 1           ,          f(x₁)= - 3

x₃ =  0.25 – [ 0.25 – 1 ] (-0.234)

                  -0.234 – (-3)

= 0.25 – [- 0.75] (-0.234)

               2.766

= 0.25 – [0.1755]

                2.766

x₃ = 0.25 – 0.06344=0.18656

     |x₃ = 0.18656|

f(x₃) = (0.18656)³ – 5(0.18656) + 1

         = 6.4900 – 0.93265 + 1

|f(x₃) = 6.55735|

3^rd

    x₄ = x₃ – [x₃ – x₂       ]  f(x₃)

                        f(x₃) – f(x₂)

x₃= 0.18656    , f(x₃)= 6.55735

x₂ = 0.25         , f(x₂) = -0.234

x₄ = 0.18656 – [0.18656 – 0.25] (6.55735)

                         6.55734 + 0.234

     = 0.18656 – [-0.06344](6.55735)

                            6.79135

   

     = 0.18656 – [-0.4159]

                            6.7913

     = 0.18656 + 0.06124

|x₄ = 0.2478|

f(x₄) = (0.024978)³ – 5(0.2478) + 1

         = 0.015216 – 1.239 + 1

|f(x₄) = -0.223784|

4^th

   x₅ = x₄ – [x₄ – x₃       ]  f(x₄)

                     f(x₄) – f(x₃)

x₄ = 0.2478      ,     f(x₄) = -0.223784

x₃ = 0.18656    ,     f(x₃) = 6.5573

x₅ = 0.2478 – [0.2478 – 0.18656] (-0.2237)

                        -0.2237 – 6.5573

     = 0.2478 – [- 0.01369]

                        -6.781

     = 0.2478 – 2.01887

|x₅ = -1.77107|

                                                       and so on …..

Program Of bisection Method

# include <stdio.h>

# include <conio.h>

#include  < math.h>

# include <process.h>

#include <string.h>

# define EPS  0.00000005

# define  f(x)     x³ – x – 1

   void  Bisect ()

   int count  = 1 , n

   float root = 1;

  void main ()

  {

     clrscr (),

   priftf(“\n solution by Bisection method\n”);

    printf(“Equation”);

    printf(“\n\t\t\t x³ –x – 1 =0 \n\n”)

    printf(“Enter the no. of iteration:”);

    scanf(“%d”,&n);

    Bisect ();

    getch ();

 }

     void Bisect ()

 {

      float x₀, x₁, x₂ ;

      float f₀, f₁, f₂ ;

      int I =0 ;

     for (x₂ = 1; x₂++)

{

     f₂  =  f(x₂);

     if (f₂ < 0);

     {

         Break ;

      }

}

     printf(“\n\t\t_ _ _ _ _ _ _ _ _ _ _ _ _”);

     printf(“\n\t\t   Root  = %  f ”,x₀);

     printf(“\n\t\t  ITERATION = %d\n”, count -1 );

      printf(“\t\t_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ ”);

      getch();

}

         Algorithm of Bisection Method

Step1:- Start of the program.

Step2:- I/P the various x₁, x₂, for the task.

Step3:- Check f(x₁) * f(x2) < 0

Step4:- If yes proceed.

Step5:- If no exit & print error message.

Step6:- Repeat 7 – 11 if condition not satisfied.

Step7:-  x₀= (x₁ + x₂)/2

Step8:- if f(x₀) * f(x₁) < 0

Step9:- x₂ = x₀

Step10:- ELSE

Step11:-   x₁ = x₀

Step12:-   Condition

Step13:- |(x₁ – x₂ ) /x₁ | < max. possible error or f(x₀) = 0.

Step14:-    printf o/p

Step15 :-   End of program.

Find the relative percentage error in an error representation of/ by 1.33

Sol:- 

e_p= True value – approx. value × 100

                     True value

True value = 4

                   3

Approx. value= 1.33

So,

True value – approx. value

     = 4 – 1.33

        3

     = 0.0033

So,

  e_p= 0.0033 × 100

         1.3333

      = 0.249975

2. add .4327 E 7 and .0048 E 8

Sol:- 

   =   .4327 E 7 and.0048 E 8

   = .04327 E 8 and .0048 E 8

   = 0.0048 E 8 

3. subtract .8756 E-3 from .4492 E -9  

Sol:- 

       = .8756 E -3 from .0044 E -3

                   = .8712 E -3 

4. Division .6259 E 12 ÷ .0018 E 6

Sol:- 

 = .6259 E^12-6

      .0018 E

   = .6259 E⁶

      .0018

      = 347.72 E 6

         = .34772 E 9

5. Multiplication .9999 E -99 and .1148 E 96

Sol:-

   =  .9999 × .1148 E^-94+96

        = 0.1147 E ⁺²