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 …..