find the Ist term of the the series who's 2nd and subsiquent term are 8,3,0,-1,0

x              f(x)               ∆ f(x)             ∆²f(x)

2                8

                                       -5

3                3                                           2

                                       -3

4                0                                           2

                                       -1

5              -1                                            2

                                       -1

6                0

F(1) = E^-1  f(2)

        = (1 + ∆)^-1 f(2)

        = (1 - ∆ + ∆²)  f(2)

     = f(2) - ∆ f(2) + ∆² f(2)

      = 8 – (-5) + 2

      =15

F(2) = E^-1 f(3)

        = (1 +∆)^-1 f(3)

        = (1 - ∆ + ∆²) f(3)

       

        = f(3) - ∆f(3) + ∆² f(3)

        = 3 – (-3) + 2

        = 3 + 3 + 2

F(2) = 8

find the value of E2 x2 when the value of x vary by a constant increment of 2.?

E² x² = E.Ex²

         = E(x+2)²

         = (x + 2 + 2)²

         = (x + 4)²

         = x² + 16 + 8x

         = x² + 8x + 16

             [Difference Table]

x   |   f(x)    |  ∆ f(x)   |  ∆² f(x)

1   |      8     |              | 

     |             |        4    |

2   |      12   |              |      3

     |             |        7    |

3   |      19   |              |      3

     |             |        10  |

4   |       29  |              |      3

     |             |        13  |

5   |        42 |              | 

     |             |        16  |      3

6   |        58 |              |              

Step.1:- Constant the difference table for the   given value of the series.

Step.2:- Note that values of ∆² f(x) are all equal to 3.

Step.3:- Write the sixth row and put 3 in the Column of ∆² f(x). 

Step.4:- add 3 to 13 in column of ∆ f(x) and write 16 below 13.

Step.5:- add 16 to 42 in column of f(x) to get 58. Which is the required 6^th term.

Calculses of Finite Differences

E f(x)  =  f ( x+n )

Operator of E and ∆

Meaning of E

                        if y = f (x) is any function then operation of E an f (x) implies that gives an increment .

Given increment to the value of x in the function .

If this increment is of quantity H, then the operation for E means that put (x + h) in the function whenever there is x

i.e.,  |E f(x) = f(x+n)|

E f(x) does not mean that E is multiply to f(x) but this means that E is operator on f(x) and as such it is only a symbol and not an algebric number.

 E²    f(x) means  that operator E is applied twice only function f(x)

i.e.,         E²     f(x)  = E    E f(x)

                            = E    f(x+n)

                            = E     (x + n + n)     [x=x+n]

                            = E     (x + 2n)

Relatives   or    ∆= E - 1

        ∆ f(x) = f(x+n) - f(x) 

                      = E f(x)  - f(x)

                      = E f(x)  - [ E - 1]

  

           ∆ f(x)  =  f(x) [E -1]

              ∆      = E - 1 

               |E = 1+∆|

    

Question solved by Gauss Seidal Iterative Method

Que :-

          8x - 3y + 2z    =  20

              4x + 11y - z    =  33

              6x + 3y + 12z = 35

Sol:-

         x = 1 (20 + 3y - 2z)       ......(1)

                       8

         y = 1 (33 - 4x + z)                 ......(2)

              11

         z = 1 (35 - 6x - 3y)                ......(3)

              12

1^st approx.

                  put y = 0, z = 0 in Eq.(1)

                       x₁ = 1 ( 20 + 0 + 0 )

                              8

                     |x₁ = 2.5|

                     y₁ = 1 ( 33 - 4(2.5) + 0)

                           11

           

                    |y₁ = 2.091|

                    z₁    = 1 ( 3.5 - 6(2.5) - 3(2.091)

                            12

                   |z₁ = 1.144|

2^nd approx.

                 put y₁ = 2.091, z₁ = 1.144 in Eq.(1) 

                       x2 = 1 ( 20 + 3y₁ - 2z₁)

                               8

                      |x2 = 2.998|

                       y2 = 1 ( 33 - 4x2 + z₁ )

                              11

                      |y2 = 2.014|

                       z2 = 1 ( 35 - 6x2 - 3y2 )

                              12

                      |z2 = 0.914|

3^rd approx.

                 x₃ = 1 ( 20 + 3y2 - 2z2 )

                                        8

                 x₃ = 1 [ 20 + 3(2.014) - 2(0.914) ]

                        8

                |x₃ = 3.026|

                 y₃ = 1 ( 33 - 4x₃ + z2 )

                        11

                 y₃ = 1 [ 33 - 4(3.028) + 0.914 ]

                        11

                |y₃ = 1.982|

                 z₃ = 1 ( 35 - 6x₃ - 3y₃ )

                        12

                |z₃ = 0.908|

4^th approx.

                   x₄ = 1 ( 20 + 3_y3  - 2z3  )

                                           8

                  |x₄ = 3.016|

                   y₄ = 1 ( 33 + 6x₄  - z₃ )

                          11

                   y4 = 1 ( 33 + 6(3.016) - 0.908 )

                          11

                  |y4 = 1.986|

                   z4 = 1 ( 35 - 6x4 - 3y4 )

                          12

                   z4  = 1 [ 35 - 6(3.016) - 3(1.986) ]

                           12

                  |z4 = 0.912|

5^th approx.

                   x₅ = 1 ( 20 + 3y4 - 2z4 )

                                            8

                   x₅ = 1 [ 20 + 3(1.986) - 2(0.1986)]

                           8

                  |x₅ = 3.016|

                   y₅ = 1 ( 33 - 4x₅ + z4 )

                          11

                  |y₅ = 1.986|

                   z₅ = 1 [ 35 - 6(2.988) - 3(2.014) ]

                          12

                 |z₅ = 0.914|

Thus, in last 2 approx. value of x, y and z are matched. 

  

                                                         

Gauss Seidal Iterative Method

a₁x + b₁y + c₁z = d₁               ……(1)

a₂x + b₂y + c₂z = d₂              ……..(2)

a₃x + b₃y + c₃z = d₃             ……….(3)

from eq(1)

                  x =  1  ( d₁ – b₁y – c₁z)

                                         a₁

from eq(2)

                 

                         y = 1 ( d₂ – a₂x – c₂z )

                               b₂

from eq(3)

                z = 1 ( d₃ – a₃x – b₃y )

                                    c₃

                      

One more question based on Gauss Elimination Method

Que:-

           x + 2y + z    =3

           2x + 3y +3z = 10

            3x – y + 2z = 13

 Sol:-

[1        2        1]  [x]     [3]

[2        3        3]  [y]  = [10]

[3      -1         2]  [z]     [13]

[a₁₁     a₁₂       a₁₃]  [x]    [b₁]

[0       a₂₂        a₂₃]  [y] = [b₂]

[0       0           a₃₃]  [z]     [b₃]

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

R_i  -  a_i1 × R₁              for [ I = 2^nd & 3^rd ]

        a₁₁

For| i = 2|

               R₂ – a₂₁ × R_i

                      a₁₁

               R₂ – 2  ×  R₁

                      1

               R₂ – 2R₁

               a₂₁ = a₂₁ – a₁₁

                     = 2 – 2(1)

              |a₂₁= 0|

a₂₂ = a₂₂ – 2a₁₂

     

       = 3 – 2(2)

       = 3 – 4

|a₂₂= -1|

a₂₃ = a₂₃ – 2a₁₃

       = 3 – 2(1)

|a₂₃ = 1|

          

               b₂ = b₂ – 2b₁

                    = 10 – 2(3)

              |b₂= 4|

| i = 3 |

               R₃ – a₃₁ ×  R₁

                      a₁₁

                       R₃ – 3 × R₁

                          1

                R₃ – 3R₁

a₃₁ =  a₃₁ – 3a₁₁

       = 3 – 3(1)

|a₃₁= 0|

a₃₂= a₃₂ – 3a₁₂

      = -1 – 3(2)

          |a₃ = -7|

a₃₃= a₃₃ – 3a₁₃

      = 2 – 3(1)

    |a₃₁ = -1|

b₃ = b₃ – 3b₁

     = 13 – 3(3)

|b₃= 4|

[1        2        1]     [x]     [3]

[0       -1        1]    [y] =  [4]

[0       -7       -1]    [z]     [4]

To eliminate coefficient of y from 3^rd row.

| i = 3|

          R_i -  a_i2 × R₂

                 a₂₂

         R₃ – a₃₂ × R₂

                a₂₂

         R₃ – (-7) × R₂

                (-1)

         R₃ – 7R₂

a₃₂ = a₃₂ – 7a₂₂

      = -7 – 7(-1)

|a₃₂ = 0|

a₃₃ = a₃₃ – 7a₂₃

      = -1 – 7(1)

|a₃₃ = -8|

b₃ = b₃ – 7b₂

     = 4 – 7(4)

|b₃ = -24|

[1        2        1]  [x]      [3]  --------(i)

[0       -1        1]  [y] =  [4]  ---------(ii)

[0        0       -8]  [z]      [-24] -------(iii)

-8z = -24 

    z = 24

           8

       |z = 3|

-y + z = 4

-y = 4 – 3

  

 |y = -3|

x + 2y + z =3

x + 2(-1) + 3 = 3

         x = 3 + 2 – 3

|x = 2|

[x = 2     ,     y = -1     ,     z = 3]

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 ]