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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
- 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 .
- 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.
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 )
- 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 )
- 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 )
- =1.019 ( 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.
- 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
- 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:
- (a)Plot ( or sketch ) the graph of y = f(x).
- The real roots are the points where the curvecuts the x axis.
- (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).
- 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.
- 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.
- 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(