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Transcript of Chapter 3

  • 1. CHAPTER 3 NUMERICAL METHOD

2. 3.1The Trapezium Rule

  • OBJECTIVES :The objectives of thislesson are to enable students to
  • a) derive the trapezium rule by dividing the area represented bydx into n ( n7 )
  • trapezium each with width h.
  • b)use the trapezium rule to approximate

3. Introduction

  • What happens when a mathematical function cannot be integrated ? In these cases a numerical method can be used to find an approximate value for the integral
  • As the definite integralis a numberwhich represents the area betweeny = f (x), thex-axis and the lines
  • x = aand x =b .

4.

  • Therefore , evenifcannot be found, approximate value forcan be found by evaluating the appropriate area using another method.The two common methods are the trapezium rule and Simpsons rule.
  • However, here we will discuss the trapezium ruleas requested by our syllabus .

5. The trapezium rule

  • This method divides the area A under the curve
  • y = f(x) is into n vertical strips.The width of each strip is h.
  • The area under the curvesum of areas of n trapezia.
  • Note : approximately equal to

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.

  • Using

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+ y2) A 3= h ( y2+ y3) 9.

  • Verify that if this process is continued then
  • An-1=h ( yn-2+ yn-1)
  • An =h ( yn-1 + yn)
  • Now add all of these separate areas together. The approximate value of A is given by
  • AA 1+ A 2 + A 3 + +An-1+ An
  • h ( y0 + y 1) +h ( y 1 + y2) +h ( y 2+ y3) + +h ( yn-2+ yn-1) +h ( yn-1+ yn)
  • h ( y0 + 2y 1+2y 2+2y 3 + +2y n-1+ y n )

10.

  • This method of approximating the area under a curve by n trapezia of equal width h is called the trapezium rule, and the result can be summarized as :
  • dx hy0 + yn + 2 ( y 1 + y 2 + + yn-1 )
  • where h =and y r= f( x r )

11.

  • Example 1
  • Evaluatedxusing 5 strips by the trapezium rule.
  • Notation: Working must have 4 or more decimal places.

Solution The integration interval ( b- a ) = 6 1 = 5 units So h == = 1 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. Example 2

  • Use the trapezium rule, with five ordinates, to evaluateex dx.
  • Correct your answer to three decimal places.

Solution: For five ordinates, ( y o , y 1, y 2 , y 3, y 4) evenly spaced in the range0 x 0.8 , we need to divide this range into four equal parts The integration interval ( b- a ) = 0.8 - 0 =0.8 units 15.

  • So,h === 0.2
  • The value of x at which y is calculated are :
  • 0 , 0.2, 0.4, 0.6 , 0.8

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 16.

  • e x dx( y0 + y4 ) + 2 ( y1+ +y 3)
  • =2.8965 + 2 ( 3.6476 )
  • =1.0192
  • =1.019 ( three decimal places )

17. 3.2Solutions of non linear equations

  • OBJECTIVE :The objective of thislesson is to enable students to understand the method of finding roots using the iteration method by writing:
  • f(x) = 0 in the form of x = g(x). The iteration scheme is x n+1= g(x n ) , n = 1,2,3.. The method fails when g (x)> 1 in the neighborhood of the roots of the equation.

18. Introduction

  • Many equations cannot be solved exactly, but various methods of finding approximate numerical solutions exist.
  • The most commonly used methods have two main parts:
  • (a) finding an initial approximate value
  • (b) improving this value by an iterative process

19.

  • Initial Values:
  • Theinitial value of the roots of f(x) = 0 can be located approximately by either a graphical or an algebraic method.
  • Graphical Method:
  • Either
  • (a)Plot ( or sketch ) the graph of y = f(x).
  • The real roots are the points where the curvecuts the x axis.
  • Or
  • (b)Rewrite f(x) = 0 in the form F(x) = G(x).
  • Plot ( or sketch ) y =F(x) and y = G(x).The real roots are at the pointswhere these graphs intersect.

20. Example 1

  • Find the approximate value of the equation ln x + x - 4 = 0 by using the graphical method.

y = 4-x y 0123 4the intersection is at x2.9 x y y = ln x 21.

  • Algebraic Method
  • Find two values a and b such that f(a) and f(b) have different signs.
  • At least one root must lie between a and b if f(x) is continuous.
  • If more than one root is suspected between a and b, sketch a graph of y = f(x).

22.

  • Iterative Method
  • All iterative methods follow the same basic pattern. A sequence of approximations
  • x 1 , x 2, x 3, x 4.. is found , each one closer to the rootof f(x)=0.
  • Each approximation is found from the one before it using a specified method.
  • The process is continued until the required accuracy is reached.
  • Two methods we have to discuss are Iteration Method and Newton-Raphson Method.

23.

  • Iteration Method
  • Rewrite the equation f(x) = 0 in the form x = g(x).
  • If the initial approximation is x 1, then calculate
  • x 2= g(x 1 )
  • x 3= g(x 2 )
  • x 4= g(x 3 )
  • and so on ..
  • This method fails if g (x)> 1 near the root.
  • So, the value of g (x1)should be< 1.

24. Example 2

  • Using the iteration method, find the solution of f(x) = x + e x near x = -0.6 to three decimal places.

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 ) 25.

  • x 1 = -0.6
  • x 2 = -e-0.6 = -0.5488
  • x 3 = -e-0.5488 = -0.5453
  • x 4 = -e-0.5453 = -0.5797
  • x 5 = -e-0.5797 = -0.5601
  • x 6 = -e-0.5601 = -0.5712
  • x 7 = -e-0.5712 = -0.5648
  • x 8 = -e-0.5648 = -0.5684
  • x 9 = -e-0.5684 = -0.5664
  • x 10= -e-0.5664= -0.5676
  • x 11= -e-0.5676 = -0.5668
  • x 12= -e-0.5668 = -0.5673 the required solution is x = - 0.567 ( three decimal places )#

26. Example 3

  • Show that the equationhas a root between0.2 and0.3.
  • Taking 0.2 as first approximation find the root of the equation ,giving your answer to three significant figuresby using the iteration method.

27.

  • Solution
  • f(x) = x 2-+ 4
  • f(0.2) =
  • f(0.3) ==0.756 ( positive )
  • therefore f(x) has a root betweenx = 0.2 and x = 0.3.

=- 0.96( negative ) 28. g(x)= g'(x)= = =g'(0.2) = =0.0245(