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CHAPTER 3 NUMERICAL METHOD
3.1  The Trapezium Rule ,[object Object],[object Object],[object Object],[object Object]
Introduction ,[object Object],[object Object],[object Object]
[object Object],[object Object]
The trapezium rule ,[object Object],[object Object],[object Object],[object Object]
 
Let y 0 ,  y 1  , y 2 , …., y n   be the values of the function f (x) . These correspond to the (n+1) ordinates x 0  , x 1 , x 2 , ……..,x n   respectively.
[object Object],Area of trapezium  =  ( width x sum of parallel sides ) The area of the first trapezium is A 1  =  h ( y 0  + y 1  ) The areas of the second and the third trapezia are then A 2  =  h ( y 1  + y  2  ) A 3  = h ( y  2  + y  3  )
[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]
[object Object],[object Object],[object Object]
[object Object],[object Object],[object Object],Solution The integration interval ( b- a ) = 6 –1 = 5 units So h =  =   = 1
The value of x at which y is calculated are : 1, 2, 3, 4, 5, 6 Tabulating the results as follows help the final calculation : 4.787 1.792 Totals 0.693 1.099 1.386 1.609 0 1.792 y 0 y 1 y 2 y 3 y 4 y 5 1 2 3 4 5 6 Remaining ordinates First and last ordinates y X
dx      ( y 0  + y 5 ) + 2 ( y 1  +… +y 4  )    =     1.792 + 2 ( 4.787 )   =  5.683
Example 2 ,[object Object],[object Object],Solution: For five ordinates, ( y o   , y 1  , y 2   , y 3  , y 4  ) evenly spaced in the range  0 ≤ x ≤ 0.8 , we need to divide this range into four equal parts The integration interval ( b- a ) = 0.8 - 0 =  0.8 units
[object Object],[object Object],[object Object],3.6476 2.8965 Totals 1.0408 1.1735 1.4333 1 1.8965 y 0 y 1 y 2 y 3 y 4 0 0.2 0.4 0.6 0.8 Remaining ordinates First and last ordinates Y x
[object Object],[object Object],[object Object],[object Object]
3.2  Solutions of non – linear equations   ,[object Object],[object Object]
Introduction ,[object Object],[object Object],[object Object],[object Object]
[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]
Example 1 ,[object Object],y = 4-x y 0  1  2  3   4  the intersection   is at x    2.9 x y y = ln x
[object Object],[object Object],[object Object],[object Object]
[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]
[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]
Example 2 ,[object Object],Solution Iteration method Write the given equation as x = - e x      x = g(x) g (x) = - e x g  (x) = - e x ,   g  (-0.6)    =   -e-0.6   = 0.5488  ( < 1 )
[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]
Example 3 ,[object Object],[object Object]
[object Object],[object Object],[object Object],[object Object],[object Object],=  - 0.96  ( negative )
   g(x)  =   g'(x)  =   =   =  g'(0.2) =   =  0.0245  ( <1 )  therefore, g(x) = iteration function
[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],x 3  and x 4  have same value when round up to three sig.figs. Hence we should stop the working.
3.3  Solutions of non-linear equations ,[object Object],[object Object],[object Object]
Newton-Raphson Method ,[object Object],[object Object],[object Object],[object Object]
Example 1 ,[object Object],[object Object],[object Object],[object Object]
[object Object],[object Object],[object Object],[object Object],[object Object]
[object Object],[object Object],[object Object],[object Object],for x < 2  to three decimal places.
2 1 Solution :   y y = e x   1   x y = 2 - x 0 0.4
[object Object],[object Object],[object Object],[object Object],[object Object]
0.4 - [ ]  = =  0.4043 =  0.4043 - [ ]   =  0.4433  =  0.4433 - [ ]   =  0.4428 =  0.4428 - [ ]  =  0.4428 The required solution is 0.443 ( three decimal places ).
EXERCISES ,[object Object],[object Object],[object Object]
[object Object],Show that the equations  2sin x – x = 0 has a root between  x = 1(radian) and x= 2 (radian). Find the root of the equation by using  i) iteration method ii) Newton Raphson method Giving your answer to two decimal places.

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Chapter 3

  • 2.
  • 3.
  • 4.
  • 5.
  • 6.  
  • 7. Let y 0 , y 1 , y 2 , …., y n be the values of the function f (x) . These correspond to the (n+1) ordinates x 0 , x 1 , x 2 , ……..,x n respectively.
  • 8.
  • 9.
  • 10.
  • 11.
  • 12. The value of x at which y is calculated are : 1, 2, 3, 4, 5, 6 Tabulating the results as follows help the final calculation : 4.787 1.792 Totals 0.693 1.099 1.386 1.609 0 1.792 y 0 y 1 y 2 y 3 y 4 y 5 1 2 3 4 5 6 Remaining ordinates First and last ordinates y X
  • 13. dx   ( y 0 + y 5 ) + 2 ( y 1 +… +y 4 )  =  1.792 + 2 ( 4.787 )  = 5.683
  • 14.
  • 15.
  • 16.
  • 17.
  • 18.
  • 19.
  • 20.
  • 21.
  • 22.
  • 23.
  • 24.
  • 25.
  • 26.
  • 27.
  • 28. g(x) = g'(x) = = = g'(0.2) = = 0.0245 ( <1 ) therefore, g(x) = iteration function
  • 29.
  • 30.
  • 31.
  • 32.
  • 33.
  • 34.
  • 35. 2 1 Solution : y y = e x 1 x y = 2 - x 0 0.4
  • 36.
  • 37. 0.4 - [ ] = = 0.4043 = 0.4043 - [ ] = 0.4433 = 0.4433 - [ ] = 0.4428 = 0.4428 - [ ] = 0.4428 The required solution is 0.443 ( three decimal places ).
  • 38.
  • 39.