Regula - Falsi Method

Formula:-

            x₁  f(x₂) – x₂ f(x₁)

                  f(x₂) – f(x₁)

Que:- x³ – x² – 2 = 0

Sol:- 

   f(x)=x³ – x² – 2 

     f(0)= -2 

     f(1)= -2 

     f(2)= + 2

x₁ = 0 ,                      f(x₁)=-2

x₂ = 2 ,                      f(x₂) = 2

1^st approx :-

            1(2) – 2(-2)

                         2 –(-2)

                    2 + 4

                       4

                    6 = 1.5

                    4

      (1.5)³ – (1.5)² – 2 = 0

        3.37 – 2.25 – 2 =0

     f(x₁)= - 0.875

2^nd approx:-

x₁ = 1.5 ,                   f(x₁) = - 0 .875

x₂ = 2 ,                      f(x₂) = 2

          =       1.5(2) – 2(-0.875)

                         2 –(-0.875)

            =       4.75

                    2.875

         x₁= 1.65217

   (1.652)³ – (1.625)² – 2 = 0

    4.50847 – 2.729104 – 2 = 0

    f(x₁) = - 0.220634

3^rd approx:- 

x₁ = 1.652    ,             f(x₁) = - 0.220634

x₂ = 2          ,             f(x₂) = 2

                =  1.652(2) – 2(-0.2206)

                      2 –(-0.2206)

              = 3.7452

                 2.2206 

          x₁ = 1.68657

       = (1.6865)³ – (1.6865)² – 2 = 0

        =  4.79688 – 2.84428 – 2

     f(x₁)  = –  0.0474

4^th approx:- 

x₁ = 1.68657   ,          f(x₁) = - 0.0474

x₂ = 2              ,           f(x₂) = 2

           =  1.68657(2) – 2(-0.0474)

                2 –(-0.0474)

        = 3.4678

           2.0474

     x₁ = 1.693757

        

     = (1.693757)³ – (1.693757)² – 2 = 0

      = 4.85907 – 2.868812 – 2

f(x₁)= -0.0101

5^th approx:-

x₁ = 1.693757   ,          f(x₁) = -0.0101

x₂ = 2                ,           f(x₂) = 2

       =  1.69375(2) – 2(-0.0101)

                2 –(-0.0101)

       = 3.4077

          2.0101  

       = 1.6952

The real root is 1.69  

        

                     

x3 – 18 = 0 lying between 2.5 and 3 up-to two decimal places.?

Sol:-   x³ – 18 = 0

        f(x)=x³ – 18 

        f(2.5)= (2.5)³ – 18 

                 = 15.625-18

                 = -2.375 

       f(2.6)= (2.6)³ – 18 

                =   17.575-18

                = -0.424

      f(2.7)= (2.7)³ – 18

               = 19.683-18

               = 1.683

1^st:-  

       2.7 + 2.6     =  5.3  =  2.65

                 2                2 

      =(2.6)³ – 18 

      = 18.609625 - 18 

      = 0.60925 

2^nd:- 

         2.65 + 2.6     =  5.5  =  2.625

                    2                 2

        = (2.625)³ – 18 

        = 0.087890624

3^rd:- 

         2.625 + 2.6     =  2.6125

                  2 

        = (2.6125)³ – 18 

        = -0.169279296

4^th:- 

        2.6125 + 2.625     =  2.61875

                    2

        = (2.61875)³ – 18 

        = -0.04100122

5^th:- 

        2.61875 + 2.625     =  2.621875

                     2

        = (2.621875)³ – 18 

        = 0.023367889

6^th:- 

        2.621875 + 2.61875     =  2.6203125

                         2

     = (2.6203125)³ – 18 

     = -8.835857391 

Root is 2.62 .

  

Ques. x3 - 3x -5 =0

Sol:-

       f(x)= x³ – 3x – 5 

       f(0)= -5

       f(1)= -7

       f(2)= -3

       f(3)= 13

the root lies b/w 2&3.

1^st :- 2+3   =  2.5 = x

                 2

      f(2.5)= (2.5)³  – 3(2.5) – 5 

               = 3.125

the root lies b/w 2.5&2 .

   

2^nd  :- 2.5+2   = 2.25 = x

                      2

      f(2.25)= (2.25)³  – 3(2.25) – 5 

                 = -0.359375

the root lies b/w 2.25&2.5 .

3^rd  :- 2.5 + 2.25    =  2.375 = x

                       2

       f(2.375)= (2.375)³  – 3(2.375) – 5 

                    = 1.2714844

4^th  :- 2.25 + 2.375   =  2.125 = x

                        2

        f(2.125)= (2.125)³  – 3(2.125) – 5 

                     = 0.42896508

the root lies b/w 2.315&2.25 .

5^th  :- 2.3125 + 2.25   =  2.28125 = x

                          2

         f(2.28125)= (2.28125)³  – 3(2.28125) – 5

                          = 0.02810

the root lies b/w 2.28125&2.25 .

6^th  :- 2.28125 + 2.25    =  2.265625 = x

                         2

  f(2.265625)= (2.265625)³  – 3(2.265625) – 5 

                     = -0.167293548

7^th  :- 2.265625 + 2.28125  = 2.2734 = x

                           2

     f(2.2734)= (2.2734)³  – 3(2.2734) – 5 

                    = -0.07047866 .

8^th  :- 2.2734 + 2.28125  = 2.277325

                            2

Hence, the root is 2.27 .

Bisection Method

Que:- 

         x³ – x – 1 = 0

Ans:- 

       f(x)=  x³  – x – 1 

       f(1)= 1³ – 1 – 1 

             = -1.

       f(2)= 2³ – 2 – 1 

             = +5

∴ f(1) is -(negative) and f(2) is +(positive). So, the root lies b/w 1&2 .

1^st approx.:- 

             1+2  = 1.5 = x. 

                          2

                   f(x)= (1.5)³ – 1.5  – 1

                         = 0.875

∴ f(1) is -(negative) and f(1.5) +(positive). So, the root lies b/w 1&1.5.

2^nd approx.:-

                1+1.5  = 1.5 = x

                                 2

                   f(x)= (1.25)³ – 1.25  – 1 

                         = -0.297

∴ f(1.25) is -(negative) and f(1.5) is +(positive). So, the root lies b/w 1.25&1.5 .

3^rd approx. :-

                    1.25+1.5  = 1.375 = x

                                    2

                     f(x)= (1.375)³ – 1.375  – 1 

                        = 0.2246

∴ f(1.375) is +(positive) and f(1.25) is -(negative). So, the root lies b/w 1.375&1.25 .

4^th approx.:-

                      1.375 + 1.25  = 1.313 = x

                                      2                   

                      f(x)=(1.313)³ – 1.313  – 1

                          = -0.494

∴ f(1.313) is -(negative) and f(1.375) is +(positive). So, the root lies b/w 1.375&1.313 . 

         

    

Pitfall of Floating Point Representation

1. Rearrange the formula so that you can avoid subtraction of two nearly equal numbers.

for Ex:- x² – y²  can be replaced by x+y.

             x – y

2.Avoid multiplication of large numbers that make left to overflow.

3.Use alternative arithmetic such as internal arithmetic if necessary.

4.Wherever possible use integer arithmetic to avoid conversion and round off errors.

5.Do-not test a floating point number for zero in your algorithm.

6.Wherever possible rearrange your formula so that you use the original data rather than derived data.

7.When finding the sum of set of members arrange the set so that they are in the ascending order of absolute value.

i.e. when | a | > | b | > | c |  , then (c - b)+a is better than (a - b)+c.

8.It necessary use double precision for floating point calculations this would include the accuracy considerably but would take more time on compute the memory space.

approx. value =0.333 and true value = 1/3, find ea ,er and ep..?

Solution:-

      x=0.333    ,    x'=1/3

         

            e_a= ₁ - _{0.333 × 3}

                      3     1 × 3

                =0.333 - 0.333

           e_a= 0

             

           e_r= 1/3  –  0.333

                       1/3

             =_0.00033

                   _1/3 

                 e_r =0.000999

          e_p=  0.000333   _{× 100}

                     _0.333

          e_p= 0.09999.

  

                               

        

      

       

             

                      

 

Errors and Floating Point Representation

Inherit Errors:-

                      Errors which are already present in the statement of a problem before a solution is called Inherit Errors.

Rounding Errors:-

                      The errors on the process of the rounding of the number during the computation.

Truncation Errors:-

                      They are caused by using approx. results or replacing a infinite process by a finite one.

Absolute Errors:-

           True Value  –  approx. value  

                      X=True Value

                      X'= approx. Value

                         |e_a =X-X'|

        e_r=true value  –  approx. value 

                         True value      

                  =      Ea

                     True value

       e_p=true value  –  approx. value   ×100

                        True value                
       

                  =    e_a/ū × 100      

                                    Here u = x .

                

Operations of Normalized Floating Point

1.Addition

2.Subtraction

3.Multiplication

4.Division.

1.Addition:-

        (i). .4546 E 5 and .5433 E 5

                  = .9979 E 6

          (ii). .459 E 5 and .5433 E 7

                  = .5478 E 5

2.Subtraction:-

       (i). .9432 E -4 from .5452 E -3

           =.9432 E -4 from .5452 E-3 

                                              |
                                       greater no.

              .09432 from.5424 E -3

                           + 5424 

                            . 0993   

                 = .4509 E -3 

       (ii). .5424 E 3 from .5452 E 3

                 = . 5452

                    . 5424 
              . 0028

                 = .0028 E 3

                 = .28 E 1

    (iii). .5424 E -99 from .5452 E -99

                 = . 5424

                    . 5424       

                    .0028 E -99

                = .28 E -101 

                                               underflow

3.Multiplication:-

       (i). .5543 E 12 and .4111 E -15

             .5543 E 12  ×  .4111 E -15

                = 0.2278 E -3

      (ii). .1111 E 10 and .1234 E 15

             .1111 E 12  ×  .1234 E 15

                = 0.01370974 E 10 + E 15

                = 0.01370974 E 25 

                = 1.37094 E 23

      (iii). .1111 E 51 and .4444 E 50

                =.1111 E 51 × .4444 E 50

                =.4437284 E 100

                                               overflow

  (iv). .12134 E -49 and .1111 E -54

         = .12134 E-49  ×  .1111 E -54

                = .1370974 E -104

                                             underflow

4.Division:-

       (i). .9998 E 1  ÷  .1000 E -99

                = .9998 E 1  ÷  .1000 E -99

                = 9.998 E 1-(-99)

                = 9.998 E 100

                = .9998 E 101

                                             overflow

     (ii). .1000 E 5  ÷  .9999 E 3

                = .1000 E 5  ÷  .9999 E 3

                = .100010001 E 5-3

                = .100010001 E 2 .

                                     

Floating Point Representation

1. Integer Arithmetic 

2.Real Floating Point Arithmetic

                                   word
                | 1   | 2 | 3 | 4 | 5 |
        word is a memory location 

Real no. consist of 2 combinations:-
1.mantissa,
2.Exponent.
        

                  Mantissa

                 <1  . < .1

(1).If a no. is greater then one do left left shift (+).

(2).If a no. is less then one do right shift (-).

                     43.86×10⁶

                         43.86×E 6

(1)  837.149 E 4

                = . 83714 E 7

                   .| 8 | 3 | 7 | 1 | 4 | 9 | E | 7 |

(2) .0004378 E 8

                      =  .4378 E 5

                          .| 4 | 3 | 7 | 8 | E | 5 |

(3) 4296.632 E 2

          = .4296632 E 6

             .| 4 | 2 | 9 | 6 | 6 | 3 | 2 | E | 6|

(4) .00004318 E 9

          = .4318 E 5

             .| 4 | 3 | 1 | 8 | E | 5 |

(5) 1.69 E 2

          =.169 E 3

            .| 1 | 6 | 9 | E | 3 |

                

3.Numerical Differentiation and Integration

Introduction Numerical Differentiation , Numerical Integration , Trapazoidal rule, Differential Equations : Picard's Method , Euler's Method , Taylor's Method , Runge-Kutta methods.

2.Simultaneous Linear Equations

Solutions of system of linear equations , Gauss Elimination direct method and pivoting , III conditioned system of equations , Refinement of solution . Gauss  iterative method , Rate of convergence .Interpolation and approximation : Finite Differences , Difference tables . Polynomial Interpolation : Newton's forward and backward formula Central Difference Formula-e : Gauss forward and backward formula , stir-ling's Bass-else Everett's formula. Interpolation with unequal intervals: Language's Interpolation , Newton Divided difference formula .

Floating Point Arithmetic

Representation of Floating Point numbers , Operation , Normalization , Pitfalls of floating point representation , Errors in numerical computation . Iterative Methods Zeros of a single transcendental equation and zeros of polynomial using , Bisection Method . Iteration method , Regula-falsi method , Newton Raphson method , Secant method , Rate of convergence of iterative methods .

Meaning of C.B.N.S.T

C.B.N.S.T
Full Form:- Computer Based Numerical Techniques.
C.B.N.S.T is a subject when we doing some graduations like B.C.A.

In this subject we read three sub-subjects
1:- Floating Point Arithmetic.
2:- Simultaneous Linear Equations.
3:- Numerical Differentiation and Integration .